Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Authors: Hongyi Wan
Shiyuan Ren
Wei Huang
Miao Zhang
Xiang Deng
Yixin Bao
Liqiang Nie
Venue: ICML 2025 poster (2025)
Keywords: Continual learning
Feature learning theory
Replay method
Primary area: theory->learning_theory

Abstract: Continual learning (CL) is crucial for advancing human-level intelligence, but its theoretical understanding, especially regarding factors influencing forgetting, is still relatively limited. This work aims to build a unified theoretical framework for understanding CL using feature learning theory. Different from most existing studies that analyze forgetting under linear regression model or lazy training, we focus on a more practical two-layer convolutional neural network (CNN) with polynomial ReLU activation for sequential tasks within a signal-noise data model. Specifically, we theoretically reveal how the angle between task signal vectors influences forgetting that: *acute or small obtuse angles lead to benign forgetting, whereas larger obtuse angles result in harmful forgetting*. Furthermore, we demonstrate that the replay method alleviates forgetting by expanding the range of angles corresponding to benign forgetting. Our theoretical results suggest that mid-angle sampling, which selects examples with moderate angles to the prototype, can enhance the replay method's ability to mitigate forgetting. Experiments on synthetic and real-world datasets confirm our theoretical results and highlight the effectiveness of our mid-angle sampling strategy.

Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Hongyi Wan I Shiyuan Ren 1
Wei Huang?
Miao Zhang'
Xiang Deng'
Yixin Bao
Liqiang Nie
Abstract
challenge in CL is the catastrophic forgetiing phenomenon (McCloskey & Cohen, 1989), where learning from new distributions often causes a significant decline in the model's ability to retain knowledge from previous tasks.
Continual learning (CL) is crucial for advanc- ing human-level intelligence, but its theoretical understanding, especially regarding factors influ- encing forgetting, is still relatively limited. This work aims to build a unified theoretical frame- work for understanding CL using feature learning theory. Different from most existing studies that analyze forgetting under linear regression model or lazy training, focus on more practical two- layer convolutional neural network (CNN) with polynomial ReLU activation for sequential tasks within a signal- data model. Specifically, we theoretically reveal how the angle between task signal vectors influences forgetting that: acute small obtuse angles lead to benign forgetting, whereas larger obtuse angles result in harmful forgetling. Furthermore, we demonstrate that the replay method alleviates forgetting by expand- ing the range of angles corresponding to benign forgetting. Our theoretical results suggest that mid-angle sampling, which selects examples with moderate angles to the prototype, can enhance the replay method's ability mitigate forgetting. Experiments on synthetic and real-world datasets confirm our theoretical results and highlight the effectiveness of our mid-angle sampling strategy.
Catastrophic forgetting in CL has prompted various empir- ical methods, which can be broadly categorized into three types: regularization-base methods, replay-based meth- ods, and architecture-based methods. Regularization-based methods (Kirkpatrick et al., 2017; Chaudhry et al., 201 8a; Benzing, 2022) selectively regularize parameter changes to balance the learning of new and old tasks. Replay-based methods (Rebuffi et al., 2017: Lopez-Pazz Ranzato, 2017; Chaudhry al., 2018 8b; Rolnick al., 2019; Buzzega et al., 2020; Boschini al., 2022) approximate and restore old data distributions by storing training samples gradient in- formation from previous tasks. Architecture sbased methods (Mallya et al., 2018; Ostapenko et al., 2019; Qin et al., 2021; Kumar et al., 2021) allocate specific network parameters to each task to counteract forgetting.
Despite the success of experimental studies in addressing forgetting, theoretical understanding of forgetting in CL remains limited. Most of the recent works concentrate on linear models (Evron et al., 2022: 2023; Lin et al., 2023; Ding et al., 2024; Zhao et al., 2024). Analyses beyond linear models are primarily confined to the teacher-student setup (Lee et al., 2021) and the neural tangent kernel (Bennani et al., 2020; Doan et al., 2021). However, these theories often rely on strong assumptions, such as infinitely wide net- works linearizing assumptions in the weight space, which may not fully capture the generalization and forgetting dy- namics in CL. As a result, a unified theoretical framework for CL with practical neural networks remains absent.
1. Introduction
Continual learning (CL) involves learning series of tasks in sequence (Wang al., 2024), where the model adapts if all tasks are learned simultaneously. Unlike traditional machine learning models which deal with static data, CL focuses on dynamic and evolving distributions. A major
In this paper, we fill the theoretical gap in CL with neural networks through the application of feature learning theory (Allen-Zhu & Li, 2020; Cao et al., 2022; Huang al., 2023), establishing a unified theoretical framework to analyze the convergence, generalization and forgetting in CL. We pro- vide precise characterization to how the angle between task signal vectors influences forgetting in polynomial ReLU neural networks during continual learning. Moreover, we reveal the mechanisms by which replay-based methods help alleviate forgetting. Our theoretical findings inspire mid-
'Equal contribution 'Harbin Institute of Technology (Shen- zhen). Shenzhen, China 'Independent Researcher. Correspon- dence to: Miao Zhang Shhangmia@@hi.ccu..coo Wei Huang <weihuang, uts gmail.com>>
Proceedings of the nnnd International Conference Machine Learning. Vancouver, Canada. PMLR 267, 2025. Copyright 2025 by the author(s).

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
angle sampling, which further enhances the effectiveness of replay-based methods in mitigating forgetring.
as cos01.2 (11, (0);)2)/// ([l || We prove the following Theorem 1.2, which characterizes the training loss and test loss on the current task 72, well as the true eтror on the previous task Ti of the two-layer polynomial ReLU CNN. Figure 1 visually illustrates the relationship between forgetting and cos 0122 from Theorem 1.2.
1.1. Problem Setup
Our focus is primarily on two binary classification tasks in CL, using data distribution similar to that in Cao et al. (2022). The input data consists of two components: sig- nals which correspond to image labels, and noises which represent background information unrelated to the labels.
CL:
Harmful Forgetting
Benign Forgetting
Let μa E IRd be a fixed vector representing the signal contained in each data point of the k-th task Ti, and let U = orth( {1+})) E gddk represent the collection of signal vectors, where orth(.) denotes Gran-Schmidi or- thogonalization and normalization. Then each data point (xK, yks) of task Tr with X2 = =[x
Definition 1.1.
001 9gg
-2 -1
Replay:
Benign Forgetring
(1) x TTT E Rdd and Uk E [- is generated the distribution Dy:
1. Label yj is generated as a Rademacher random variable.
Figure I. Illustration of the relationship between forgetting and cosine similarity in CL wo/w replay. The blue region represents the cosine setting with benign forgetting, while the yellow region corresponds the setting with harmful forgetting. The gray region indicates the setting where forgetting is not well characterized.
2. A noise vector Ex is generated from the Gaussian distribu- tion o0, -I--UU A--UUTUT U.)-U") where U E Rd
3. One x signal, the other is given by &k. which represents noises.
(1) x
given as ykc pt, which represents the
Theorem 1.2 (Informal). For any 6 > O, under certain regularity condition, if SNR. = = for k E [2], with probability at least 1 1-8, there exists Lend such that the train- ing loss of task T2 converges to e, i.e., Ls( L (T(Tteeal)))) S e. Furthermore, the model obtained through CL achieves small test loss on the current task: LDz "Dg(W'r (W(Tz))od)) < Ge + exp(-nī)- Regarding forgetting on the previous task in CL, we present the following statements to illustrate its relationship with cosine similarity.
Definition 1.1 adapts the data distribution from Cao et al. (2022) for the CL. task setting, with the noise patch following U-U U---U ensuring orthogonality to the signal vector f4x of each task Tg. We train two-layer CNN with polynomial ReLU activation (ReLU9, q 2) by minimizing empirical cross-entropy loss:
N NO0,0 o/2
Ls, (W))
1 TI
mi bln f(W(T)) X..).. (1.1)
1. When -C₁ S cos coso1,2 S 1, the forgetting on task Ti: T, (W(T2)lend)) S exp(-CC 22 /q2))
where ((2) = log(1 + exp(-2)). Sx = (x yh Ma is the training data set for task TR. and S(W(Te), x is the two-layer CNN defined in Section 2. We use gra-
2. When -1 -1 cos 1,2 < -C2, the forgetting on task Ty: 1%,11 (W(Tastona)) 2 -expp-C- 2 lq2).
dient descent to minimize (1.1). Additionally, we define the true loss (test loss) LD. (W(T+))= == B(xx.Bx)>, E f(W(Th), xx)) and the true error (test error):
3. With replay, when - C2+1 S cos 01.2 2 < 0. the forget- 2
ting on task Ti is bounded by: 1b (W[(Tddd)))) < exp(-C CC m22 (00-2224/42
18.' (W.(()) : [yn+ # sign (f(W(T)), x.
Then the forgetting is naturally measured by the true error on the first task:
Here, C1 and C2 are positive constants with C1 < C2 <
The significance of Theorem 1.2 is summarized as follows:
L0.1 (W)()) = ((x (y) + sign (f(W(Ta),xi) xi))]-
Under the condition on the signal-to -o-uil- ratio that na SNR; = Q(1). the learned CNN can achieve both small training and test losses on the current task.
1.2. Main Contributions
For ease of discussion, we define the signal-to-noise ratio for task Tę as SNR, = (1 vd), and the cosine similarity between the signal vectors of task Tj and T2
When the angle 0,,2 between two tasks satisfies -C₁ S cos 01 2 A 1, the CNN can achieve benign forgetting,
2

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
where performance on the previous task slightly deterio- rates to accommodate the new task, but test error remains small. This contrasts with harmful forgetting, which oc- curs when -1 I cos f1.2 < -C2 leading to significant test emrrr on the old task. Notably, Jeon et al. (2022) also introduce the benign forgetting, but they refer to the for- getting of misguidance learned from biased datasets rather than the balance between old and new tasks.
FWI(W Xx) are defined as:
1 m
F (W) xa)
M
L ()) o( (T)) {
where m is the network width, and w j.r represents the r-th neuron in the first layer during task Tg. The entire parameter set of the model is denoted by W(2)),
(Tx)
Training Algorithm. We use gradient descent to optimize the objective defined in (1.1). For the first task, the weights are initialized using Gaussian initialization, where each en- try of (T..)) and W independently drawn from a Gaussian distribution N (0, oz) In the context of CL, the ini- tial weight of the current task naturally equal the final weight of the previous task, i.e. w jrr
With replay, benign forgetting occurs when the angle be- tween two tasks satisfies C2+1 _C,+1 2 I cos 01,2 & 1, indi- cating an expanded angular range compared to the case without replay (-C1 S cos 2 - 1).
(f2.0) = w (Te (Te-)) (k 2 2). j.r
To the best of our knowledge, we are the first to apply feature learning theory to analyze the convergence, gen- eralization, and forgetting of the polynomial ReLUU CNN in CL. This analysis extends beyond the previous works that rely on the linearization assumption in lazy training (Bennani et al., 2020; Doan et al., 2021), where network parameters remain close to their initial values throughout the training process. Our theoretical results demonstrate that replay-based methods mitigate forgetting by expanding the angular range of benign forgetting and inspire mid-angle sampling to enhance replay strategies. Furthermore, we conduct comprehensive experiments on both synthetic and real-world datasets to validate our theoretical findings and the effectiveness of mid-angle sampling.
Note that (Tx: represents the final iteration of task Tke where the corresponding tend can vary for different tasks. Then the update of the filters in the CNN during learning task Ta can be written
(Tr (T)) ju
(Tv.))
T] V Vw Lsv L...Lss..... (W(TA.)
w 6r
(Tv.)) w Ar
T
(Tv.t) jio (low(wwll (Te.t)
((w) En.i))
7ig
Se, + a (ow(WW/
(Te.t) is sBss
(2.1)
for j E (+1) and r E [m], where the derivative of the loss is defined as &,
((TTe.t) = l' [un ((W(le.)) xx,i)]-
3. Main Results
2. Preliminaries
In this section, we present our main theoretical results, which are established based on the following conditions. Condition 3.1. Suppose that
In this section, we present the notation, the neural network model, and the training algorithm based on gradient descent.
Notation. We represent scalars with lowercase letters, vec- tors with lowercase boldface letters, and matrices with up- percase boldface letters. The le norm of a vector is de- noted as ||2/|12 2-. For a matrix A, we denote its spectral norm by || and its Frobenius norm by ||]| To compare two sequences, we use standard asymptotic notations such as o()). O()), 2(-) and ((-) characterize their limiting behavior. Additionally, we utilize Ö(-), (()) and 6(.) to suppress logarithmic factors in these notations. Moreover, we write In = poly if x. Ovv") for some positive constant D. Lastly, sequences of integers are represented as [n] 2,, (1.2,..., n2). and V denotes max(a, b).
1. The dimension d is sufficiently large: d 211/424/))2 max ( 20122-2)/0))))))
2. The neural network width m and training sample size my of task Ta satisfy m, ng +D(olyyog(d)) for k E [2].
3. The learning rate Tik of task Tk satisfies mh S Ö(min()) [amiin||lxxxl"""" 2xll=2 ooddd")) for k E [2]. The standard deviation of Gaussian initialization To is chosen such that do S 30mm-2//e (22/(g-2)]v) max -1/o-222-2))) min( vd) -1 |1
Neural Network Model. We consider a two-layer CNN model with ReLU activation function, defined as (2)) = (ma 2. When learning task Tk, the neural network's output for the input data X3 is expressed as (WW())), xx) =
The first two conditions regarding d, ny are designed to ensure sufficiently over-parameterized learning setup, as well as certain statistical properties of the training data and weight initialization based on concentration inequalities. Similar conditions have also been employed in Chatterji & Long (2021): Frei et al. (2022); Cao et al. (2022); Kou et al.
F41 F+1 (T.) xxk) (W(() xx). where the sign cor- responds to the neuron fixed as either +1/m or - -1/m (Tx)
in the second layer. The terms F (W((T) xx) and
3

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
(2023). Condition on ai and oo is imposed guarantee that gradient descent can effectively minimize the training loss.
3. The CNN achieves small test error on task Tí:
T (W((aeea)))) S exp(=CC -0.---m.-- m22-2229/4
The definition of cos 1,, follows Subsection 1.2, represent- ing the cosine similarity between task I1 and T2. Based on the aforementioned conditions and definitions, we present the main result in the following theorems.
The first two points of Theorem 3.3 demonstrate that the CNN trained with replay in CL can achieve small training and test losses on the current task, similar to Theorem 3.2. The final point states that if the replay buffer size for task T exceeds certain threshold, the replay method can mitigate forgetting by expanding the range of angles corresponding to benign forgetting.
Theorem 3.2 (standard CL), For any e > O. under Con-
dition 3.1, if -NNR = -(1), then with probability at least 5 there exists tend Omm15 to (1-2)) ||2ll + m² 1/2 E 2|2122 such that the training loss on task T2 converges to c, i.e., Ls( Ls(WWTaaeen)) I c. Moreover, the trained CNN will generalize with small test loss on task I 6E + exp(-nā) The connection between forgetting and cosine similarity can be summarized in the following three points:
4. Proof Roadmap
Ty: LD,(W(Ta,taw
In this section, we provide a proof roadmap for Theorem 3.2 and Theorem 3.3. Based on the gradient descent update rule in (2.1), it can be observed that the weights w (Tx.() for task
jir
Ty are a linear combination of the initialization w (Tx.0) the j.r
1. When cos 1, 2 O, the trained CNN achieves a small test loss on task Ty: LDD (W(aalen))) 18c + exp( nn))-
signal vectors μAs, and the noise vectors Éx., for i E [me] and E [2]. Therefore, for r E [m], the weight vector can be decomposed as such:
2. When -C1 S cos 0,2 < O, the trained CNN achieves a small test error on task Tr: LT, (W(Tateol) S
W ja((n) (h.4 As
(TL.0)
(TL.)) Ens
exp(-C 1--mmmm n2 /g2).
mmlll
3. When - cos 01,2 < -C2, the test error on task Tı:
Further denote p(Ex) (Tv.t) == p(Ex) (pp(E))),r'" (Tre(t)
T.2.1 (W((Tstee)))) 2 - exp(-C m22 22 (q²).
0), p(Ex) )... == p(Ek) (Tx,t) ((p(Ex)
((p(E))))T))) (iTe))) S 0). Then
Here, C1 and C2 are positive constants, with C < C2 < 1.
(T.) (..0) j- 2(1a)ff 10. l|mella² 2 ph
> p (EN) + plta)) 1. 1Exlll - Eks (4.1)
Theorem 3.2 shows that the learned CNN can achieve small training and test losses on the current task if the signal-to- noise ratio is large. As the training loss converges to e, the angle f1.2 plays a crucial role in determining the nature of forgetting: acute or small obtuse angles lead to benign forgetting (small test error the previous task), while larger obtuse angles cause harmful forgetting (large test error on the previous task). Note that small test loss implies a small test error, as the loss function in (1.1) penalizes incorrect classifications more heavily. Based on Theorem 3.2, we further examine the relationship between forgetting and cosine similarity in CL with replay.
where 714) (Te.0) ( (Ts.)) p(Ex) represent the
(Ta.t)
relevant coefficients. We refer to Equation (4.1) as the signal-noise decomposition of (T..t) where || and [ 2 serve as normalization factors. The key aspect of the signal-noise decomposition model is to simplify the dynamic process of CNN learning into the dynamic update of the coefficients, which forms the basis of our analysis.
4.1. Proof Sketch for Continual Learning
Theorem 3.3 (Replay-base based CL). For any c > O, un- der the same conditions as Theorem 3.2, when 1+Ca
Based on the decomposition in (4.1) and gradient descent update (2.1), I), the iteration of coefficients are given:
2 <
cos L,2 < and replay buffer size ni 2 n2
Lemma 4.1. The coefficients ((W)) (T~-)) Pará p(Tu.t) in
(TT..)
θ cos 01.2) then with probability at least
1 6, there exists tend 60%5'mo 2-9 |2 +
decomposition (4.1) satisfy the following equations:
m 1722 k such that:
2
(T.0) (m)) 3r Jorce -(En)
((E))
(T,.0)
(T..0)
0,
(4.2)
1. The training loss on task T2: Ls, (W(Talend)) S C,, where C = O(1) is positive constant.
(m (T.11) = ((mn)
(Th
Σ
mmm
d ((w (Ta..) ,yk. pa))) 1laxx
(4.3)
2. The trained CNN achieves a small test loss on task To: L.,, (W(Tatena)) S Gc + exp(-n?)
P(ex) (T,tt)) = p(Ex) (Te.))
((T2)))
mmm
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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
-d'( (v.))) E (yn. = 5), (4.4)
studying a subset of neurons for single task in Cao et al. (2022). Specifically, Cao et al. (2022) focus on neurons Wir for E 1111 = {r E [m]] (w]r j.r
Jir
((En))Ta(tt)) = p(EN) + T7s ((T..))
(21:0) .jmm) > 03. These
mum
-'((W por Ex,))) | (y = -)) (4.5)
neurons induce a negative effect (w (1.0)) -jp2) during the initial phase of training task T2. As training progresses, (2b.1) (w Ar > --jut)) gradually decays to O, and the correspond- ing ((2)) (Tzifon)) that characterizes signal learning for task
Lemma 4.1 describes the iteration rule for the coefficients under gradient descent update. When cos f1.2 0 (right angle), the absence of interaction between signal vectors en- sures small test loss on both tasks following the approach of Cao et al. (2022). We then analyze the convergence, general- ization, and forgetting in CL under two cases: cos 01,2 > (acute angle) and cos 01,2 < 0 (obtuse angle).
T/ is bounded by ->(11) (. (feme)))) | l2 cos /1.2/ f1.2/11/ull2 In contrast, the neurons wir for r E 1,22 = {r E [m]] (W/ .jm))) - 03 {r [m] : (W) (f.-0) jju)) > 0}.
where μ M2 l|llll2 cos 81.2 /1/ll actually
Acute angle. When cos f1. > O, apply two-stage technique to analyze the convergence and generalization the current task T2 similarly to Cao et al. (2022). The key difference is that since task Ti has already been learned, a component ((1)) ) (fae)))))) || | ||2 cos 01.2/ (| ||2 is present in the signal vector μ2 during the initial phase of training task T2. If this component is large, the training process can skip the first stage and proceed directly to the second stage. As a result, small training and test losses on task T2 in Theorem 3.2 naturally follow. The complete proof can be found in Appendix E.2.
learn the feature f4z and exhibit increasing inner products (w (w/FF (T2.)) , The learning of neurons Wj.r for E I32
is key to achieving small training and test losses on task T- in Theorem 3.2. Note that while the two-stage signal learning for the tasks is similar, the neurons responsible for learning signals μr and μ2 are distinct. The complete proof and detailed explanation can be found in Appendix F.
By the definition of forgetting, (WF(((aa)e)))))
IP ()),, ly # sign()(W(Taaen x1)) in Subsec- tion I.I, we know that forgetting is closely related to the sign of yi (WW())eee)))) x1). Under the condition that the signal-to-noise ratio ny SNR; = Ω(1), the noise memory (o(E)))T will be bounded by oo0 Vd, which allows us to approximate (T(((a)a x1) (Ta.tend)
After learning task T2, forgetting can be analyzed by exam- ining the true loss on task Ti, i.e., LD, (W(Taffena)) ). Under the same conditions as Theorem 3.2, Cao et al. (2022) show that learning a single task can result in a small true loss, i.e., LD, (f(a))) I 6c + cxp((-n)). Consequently, we can bound LD,(WWGss..).. (W(G))tnn)) by analyzing its relationship with LD, (2,(Wi (fer)) When cos 01.2 > O, due to the posi- tive interaction between the signal vectors of the two tasks, the inequalities (w.) (Ta.fend)
Em=rlrr( (orllo (W (1pp))) - a((w -n -m (Taten)) w1))
Through a comprehensive analysis of neuron learn- ing, Em Er= y.r (Taifena) .1111)) can be characterized by O(m7(1)) )n (1 - cos201,2)9) where 7(1))) = m EErE,,, (r()) (f deddan for j = 1/1 represents the average value of M(2)) (Titeema) corresponding to the neurons that learn the signal 141 Sim- ilarly, Σm Ex-ro((w (( (Trifena) ,1111)) can be character- "yir ized by ((m)((2)) in [1 cos 21//a) l2), where ((22)-n = 1 m AErer,a(uaa) j.r (Taten)) for j "1 represents the average value of (Tataa)) corresponding to the neurons that learn the signal H2. Based on the above analysis, the study of forgetting can be transformed into the study of the antagonism between Em= (f..f. y1 and Σx 1r (a(((w (( (Taifend) ,V1H1)), which is directly related to cos 01,2. -yi,r Specifically, the relationship between Em (o (Tzitena) (1)) - a((w"((w "w ((W[(T..een)) (Tz.fand) ,11 and cos f1,2 can be characterized by the following lemma.
111111) 2 (w dend) v1111) (Ti.tena) and (w (w-g1. (Ta,tond) 111 I (w -w.r..r (1) can be estab- lished, which naturally lead to the following lemma.
Lemma 4.2. Under the same conditions as Theorem 3.2, when cos 0,2 > O, the following bounds hold for lend!
y1 f( (WW(TT (,)) - fi yn (WW((sssaaa))) xxi) < K. (4.6)
where K= = (1) is positive constant.
Note that (4. essentially characterizes the connec- tion between LD, (T(aafend)) and LD, (W(T))fen)) by the definition I.D,( 1.D,(W (W(D)))oa)) = (x fly
((. xx]] in Subsection 1.1. Then with the prop- erty of cross-entropy loss that {(2) S exp(-2) for all 2, the test loss on the previous task LD, (W(T).ffom) < 18c + exp(-nf) in Theorem 3.2 naturally follows. The
complete proof's is provided in Appendix E.3.
Lemma 4.3. Under the same conditions as Theorem 3.2. when - C & cos 0, < 0, it holds that
Obtuse angle. Due to the complex interaction between the signal vectors when cos f1, < 0, a comprehensive analysis of neuron behavior is required to characterize the conver- gence and generalization of the current task T2, rather than
(Iztendd) ,21M1)) (ail)))o((())) > To ((w 01.7
(Taifans) -915 , .w1 2 Ca.
5

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
where C < 1. and C3 are positive constants.
inner products is dominated by the largest inner prod- uct. This allows to simplify the derivative as follows: p(T2,.0) 1/ exp( m (( (75.0) (jm)))][ + (1)]. When 1. task Ti trained until the training loss reaches 6, we can estimate (w (. (i..t .ipe)) by (m log(1/e))//4 Con- sequently, the derivative of task Tj becomes very small. Thus, when switching to task training T2, neither the coeffi- cients 7(11) }|, (T2.t) nor p( (Ta.t) )/: of task Ti will change significantly. During the initial phase of task T2 train- ing, the maximum inner product j. j will ini- tially decrease and then increase again, a phenomenon we prove in the Appendix F. Additionally, it shown that the changes in (w),r (T2,)) .juz)) are essentially similar to those in (Wrr the higher order infinity of the other inner products. The (T2.t)
In contrast, under the condition that the cosine similar- ity cos cos01,2 is relatively small, the relationship between "or=|or((w' (Taifena), (1)) (( (Tadtena) v1)) and ((
j.r
cos 0,, is described by the following lemma.
Lemma 4.4. Under the same conditions as Theorem 3.2, when- -1 < cos 01.2 S - -C2, it holds that
m [lo((w (Tadend)) M1.
(aa))))o((())) (1111))) wowwo(WwWW (Wytee) D1
[1 N C3+
where C1 C2 < 1 and C3 are positive constants.
(T..)) .jmi),, meaning that the largest (w (T2.0) .ju2)) is
Using the antagonistic relationship in Lemma 4.3, along with the Gaussian concentration of Lipschitz function, we can directly derive an upper bound for benign forgetting (the second part of Theorem 3.2), following similar idea to the proof of Theorem 1 in Kou et al. (2023). Similarly, by Lemma 4.4, the lower bound for harmful forgetting (the third part of Theorem 3.2) naturally follows. The complete proof can be found in Appendix F3.
overall analysis of (w .juu)) is also based on the two- stage approach.
(Ta.() Stage 1. When (w)r"" (fu2) is large enough, similar to (TT..) we can give rough estimate of (((22))
((Tait) 1/expt (aap(Haatt (( (Ta.t) .juu))}. Thus, when
(w)r (Ta)) .jp2) S <m,, we can assert that - -K(TTa O())
4.2. Proof Sketch for Replay Methods
The following lemma gives our specification of stage 1:
We also analyze the replay method using signal-noise de- composition and a two-stage approach. After applying signal-to-noise decomposition, we derive the inner prod-
Lemma 4.6. Under the same condition as Theorem 3.3. there exists time
(Ta.t) and signal vector p1: uct of network weights w jir
C(ni + m2)22 log /1 - (ros 01.2 )2 oo 1eal22
T
(w (a.t) .um)
(Tz.0) = (w 3.r
jpzz) + (41)) (T2.6)
m29m2 (cos 22 1211
+ cos 01,2
(T2.)) |2|||
| y (142)
such that
From this formula, it is evident that the inner product (T2.1)
(1 max, 7 (122) 2 1c(oos for j (+1).
m
(w jir jjuz) is influenced by two dynamic coefficients, (())) (Ta.t) and y(22) (T2.4) These coefficients evolve dur- ing neural network training, and analyzing the changes in the cross-entropy loss derivatives of task 7i and task T2 is complex. To simplify the derivative form, we first introduce the following lemma.
max (p(p)))(
(Ta.t) I po Vd all e (=1).
r Te [m]] in E [nx] and 0 S ST T
Lemma 4.6 gives a significant difference between
T(1)) j.r and p( (Tait) within time T,t, during which
Lemma 4.5. Under Condition 3.1, during task Ti training, when t (L/ee,, we have
m 7(22) (22 reaches (-(cos0) while (p(p))) (T2.0) is up- per bounded by Juopi vd. After this, as 7(22) (Tz,1) j.r grows,
(T1.) .jmi) = o((w) (Ti.) J.r ,jmm))
the derivative -((T7.4) cannot be maintained at a constant level so that the training for task T2 enters stage 2.
(w fir
jor E {+1} and r o tr" r"
Stage 2. At this stage, 7(12) (T2,1) reaches a higher value, but p(HN) jari is still very different from ((H2) (Ta.1) Specifically, 7(2) jm
r* represents the maximum value of all (w (Ti.)) jiz)), where r E [m]. The proof of this lemma relies on the relationship between infinite series and infinite prod-
(Ta.t" reaches a large and reasonable
A(mm log( Mmmlog( value
ucts, which is detailed in the Appendix G. By this (Tist)
where k is positive constant, and 1-(00s0) p(px)) j.rci, can be bounded by 2000p2 v The following lemma gives some specifications.
(T.() lemma, we can show that: Em ir ..m))) = a(w (71.9) (1p)))(1 + o(1),, indicating that the sum of
6

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Lemma 4.7. Under Condition 3.1, let Ts T 2 1 )²). Then we have maxjar (p(pus)j.rs p000p vd for all T S S Tg. And we can also find a time t" that ((2z) (Tadt) N
to benign forgetting. These results are attributed to a com- prehensive analysis of neuron behavior. The second plot of Figure 2 verifies that (w (fi.)) .jmm) "jjm))
(Ti.1) = ol j.r jin)) holds for j E {+1] and r * r" as t approaches ((1/c) in Lemma 4.5 during the training of task Ti, where r" (Ti,t)
k(m log( 2 where I-(cos01 23
T t s TTg.
arg |a (w) jor juz). The neuron behavior during task T2 in the case of obtuse angle can be verified by the third plot of Figure 2. The descending curve corresponds to neurons jar for r E I1,1, which learn the signal μ4 of task Ti. These neurons initially induce negative effect (w(77,00 - which gradually decays to O. The ascend- jr ing curve corresponds to neurons Wj,r for E 1,,2, which learn the signal #2 of task T2 and exhibit an increasing (w(T),)) (w j.r .juu).. The sets I,,1 and 122 are defined as in Sub- section 4.1. The relationship between forgetting and the angle 0,, arises from the antagonism between these two types of neurons.
In Lemma 4.7, k is a parameter that measures feature learn- ing on task T2. (p(ux) jr.iy can be bounded by 20 vd. This suggests that the model is far inferior to feature Iearn- ing in terms of noise memorization. The time to stop train- (Ta.(") reaches ing for task T2 is the moment when 7(12) j,r
kfm ( T- (_(co88.2)) for the first time, which we denote by lend- At the moment fend, loss of training for task T2 will be reduced to a minimal level: Ls,(Wi (W(3s,dene) S ////2 Further, we will prove in the Appendix H.5 that when -3111))) -
L+C2 S cos 01 S 0, Er= [a (Tattnna) 2 (T2.feme), 3111 2 C3. ((w -yn."
Real-world experiments We validate our conclusions on the MNIST dataset (LeCun, 1998) using a two-layer ReLU CNN for binary classification with labels +1 and -1 (See Section 1.1). We denote the average flattened image of all positive labeled examples (label +1) in task Ti as the signal vector μ1, Similarly, we obtain μ22 for task T2. Then the angle 01,2 between the two signal vectors naturally follows. We conduct the following four sets of experiments:
It should be clear that our proof for the replay method is
based on the fact that (w (Tb.1) E1,i) and (w (22.1) : E2,1) can
be bounded. The following lemma will give this fact.
Lemma 4.8. Under Condition 3.1, when O S S T
Ö(12 1 moj (m00-11222)) "1 + m/ -t---k k-"k),,, we have
(w/rr") (T2.0 Ei,x), (w,))(,,))) j.r
(T2.1) .2,,)= S Ce: (w.r.r
0g=mmm where Cg = Õ(m - 12/12 2]]v1 max(mx) } 11/(0 ) for
1. Experiment 1: Task Ti: classification of the original digit 5 (label +1) and its inverted version (label 1). Task Ty: classification of the original digit 8 (label +1) and its inverted version (label -1). Then the angle between two tasks 0, = 30.199.
all j E 1=1] and r E [m].
In this case, we complete the proof of the replay method.
5. Experiments
2. Experiment 2: Task Ti: classification of the original digit 0 (label +1) and its inverted version (label -1). Task Ty: classification of the original digit 1 (label 1) and its inverted version (label +1). Then the angle between two tasks 0,, 102.97
Synthetic experiments We validate theoretical findings with synthetic data as per Definition 1.1, setting training data size to n = 100 for e [2] and dimension d = 100. Signal vector μ4i for task Ti is || [1, 0, ..... 0] with [| ll2 8, and #2 for T2 is obtained by rotating μn by angle 2, Noise variance is pp. = 1 for l E [2].
3. Experiment 3: Task Ti: classification of the original digit 2 (label +1) and its inverted version (label -1). Task Ty: classification of the original digit 6 (label 1) and its inverted version (label +1). Then the angle between two tasks 01,2 = 140.26..
We sequentially train a two-layer ReLU CNN defined in Section 2 on the datasets of the two tasks described above, where width m = 10 and q = 3. We initialize the model with a Gaussian distribution with op = 0.02 and train it using full-batch gradient descent with a learning rate of 0.001 for 3000 iterations on each task.
4. Experiment 4: Task Ti: classification of the original digit 4 (label +1) and its inverted version (label -1). Task T2: classification of the original digit 9 (label 1) and its inverted version (label +1). Then the angle between two tasks 0,,2 = 153.90
After training two tasks, we estimate forgetting (test er- ror on the previous task Ti) using 1000 test data points. In Figure 2, we see that acute or small obtuse angles are associ- ated with benign forgetting, while large obtuse angles lead to harmful forgetting. Moreover, the replay method alleviates forgetting by expanding the range of angles corresponding
Figure 3 illustrates the relationship between test accuracy on task Ti and the replay buffer size across four settings. For tasks with acute angles (30.199) or smaller obtuse angles
7

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
331
Repley
The for beger ( aun
The taxt laguut 24 (vgg)
The secual hages (1127) 3001
(012) (PD))
500
largua 15 Aec)
lerge (7-0
0.4
boatt (8/2
The tht lurget (1
The zith langut 14/5.
0.2
Figure 2. validation of the relationship between forgetting and the angle 01,2 between the signal vectors of two tasks under both CL and CL with replay in Section 3, and we provide experimental support for the key analysis of neuron behavior discussed in Section 4.
(102.979), the CNN exhibits benign forgetting even with- out replay, supporting Theorem 3.2. In contrast, for larger obtuse angles (140.26 and 153.90%), harmful forgetting occurs unless sufficient replay data is provided. When the replay buffer size reaches threshold, forgetting signifi- cantly alleviated, aligning with Theorem 3.3. These results highlight the angle-dependent effectiveness of replay in mit- igating forgetting.
Table I. Accuracy comparison on previously trained classes us- ing different replay buffer sampling methods within the iCaRL. framework the CIFAR 100 dataset (percentage omitted).
Sampling
CIFARI00-10
CIFARI00-5
Random
47.17 + 0.45
56.08 1 0.12
Small-angle
45.63 + 0.12
54.36 + 0.35
Mid-angle
48.02 - 0.27
56.51 = 0.06
Sig-lngle
45.34 0.76
54.77 : 0.29
Arcuracy on Task
Herding
47.40
56.12 0.20
1.0
0.8
validate the universality of mid-angle sampling, we conduct experiments on the CIFAR100 dataset (Krizhevsky et al., 2009) following the iCaRL replay framework from Rebuffi et al. (2017). The data split into T tasks, denoted as dataset-T. For example, CIFARI00-AA represents 10 tasks, each containing 10 distinct classes. Specifically, model f consisting of feature extractor yp and a classifi- cation layer e. The prototype of the j-th class is defined as μi T3.T E. (( where Sj represents the dataset of class j. Based on the prototype, mid-angle sampling, hig- angle sampling and small-angle sampling for class j can be performed according to sim(sp(Z),,). 14j), where sim(.) denotes the cosine similarity function. Further details can be found in Appendix B.2. Table 1 shows the test accuracy on previously trained classes for the three above sampling methods, random sampling, and herding. In addition, we in- clude in Table 2 the average forgetting metric used by Wang et al. (2024) on previously learned tasks to further validate the effectiveness of mid-angle sampling. As shown, mid- angle sampling outperforms the other sampling methods, aligning with our theoretical predictions.
Acmrory
0.6
0.2
97'
26
train
$50.90
0.0
0
1000
2000
3000
4000
5000
Roplay Buffer Size
Figure 3. Experimental results of CL the MNIST dataset with varying angles between task signal vectors: reported are accuracies task Yi for different replay buffer sizes.
Mid-angle sampling: Inspired by our theoretical results, leverage the angle 01,2 to balance the stability and plasticity. Instead of random sampling old task examples, we select examples that have moderate cosine similarity (mid-angle) with the prototype for each class, which we refer as mid-angle sampling. Examples with high or low cosine similarity typically correspond to higher or lower levels of forgetting during new task training respectively. In contrast, mid-angle sampling selects examples with moderate degree of forgetting, which proves to be more cost-effective approach.
6. Conclusion and Future Work
This paper establishes unified theoretical framework for understan understanding CL through feature learning theory. By ana- lyzing neuron behavior, we reveal the relationship between forgetting and the angle between task signal vectors. Fur- thermore, we demonstrate that the replay method alleviates
We conduct an initial validation of the effectiveness of mid-angle sampling using a two-layer CNN on the MNIST dataset, which can be found in Appendix B.1. To further
8

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
the extended der-verse. IEEE transactions on pattern analysis and machine intelligence, 45(5):5477-5512, 2022.
Table Average forgetting comparison on previously learned tasks using different replay buffer sampling methods within the iCaRL framework on CIFARI00 dataset (percentage omitted).
Buzzega, P., Boschini, M.. Porrello, A., Abati, D., and
Sampling
CIFAR100-10
CIFAR100-5
Calderara, S. Dark experience for general continual learn- ing: a strong, simple baseline. Advances in neural infor- mation processing systems, 33:15920-15990 2020.
Random
15.72 + 0.31
11.15 + 0.46
Small-angle
19.47 + 0.39
14.10 0.26
Mid-angle
14.84 + 0.26
10.15 0.28
Big-angle
18.04 1 0.49
12.50 0.72
Cao, Y., Chen, Z., Belkin, M., and Gu, Q. Benign overfitting in two-layer convolutional neural networks. Advances in neural information processing systems, 35:25237-25250, 2022.
Herding 15.51 1: 0.08
10.64 + 0.13
forgetting by expanding the range of angles associated with benign forgetting. Our theoretical results also inspire a mid- angle sampling strategy, which can help the replay method better mitigate forgetting. As a pioneering study in feature learning within the context of CL, our theoretical frame- work is limited to binary classification under a single-head setting. Moreover, the contribution of mid-angle sampling is relatively modest. Future work can extend framework to more complex task settings and explore more effective replay methods guided by theoretical insights into how the angle between task signal vectors influences forgetting.
Chatterji, N. S. and Long, M. sinte-ssampee analysis of interpolating linear classifiers in the overparameterized regime. Journal of Machine Learning Research, 22(129) (12)): 1-30, 2021.
Chaudhry, A., Dokania, P. K., Ajanthan, T., and Torr, P. H. Riemannian walk for incremental learning: Understand- ing forgetting and intransigence. In Proceedings of the European conference on computer vision (ECCV), pp. 532-547, 2018a.
Chaudhry, A., Ranzato, M., Rohrbach, M., and Elhoseiny, M. Efficient lifelong learning with a-gem, arXiv preprint arXiv: 08/2.00220 2018b.
Acknowledgements
Miao Zhang was partially sponsored by the National Nat- ural Science Foundation of China under Grant 62306084 and U and Shenzhen College Stability Support Plan under Grant GXWD20231 12881224112222 and Grant ZDSYS2022306660910000000..
Darst, B. F., Malecki, K. C., and Engelman. C. D. Using recursive feature elimination in random forest account for correlated variables in high dimensional data. BMC genetics, 19:1 2018.
Ding, M., Ji, K., Wang, D., and Xu, Understanding forgetting in continual learning with linear regression. arXiv preprint arXiv:2405. 17583, 2024.
Impact Statement
This paper presents work whose goal advance the field of Machine Learning. There are many potential societal consequences of our work, none which we feel must be specifically highlighted here.
Doan, T., Bennani, M. A., Mazoure, B., Rabusseau, G., and Alquier, P. A theoretical analysis of catastrophic forgetting through the ntk overlap matrix. In International Conference on Artificial Intelligence and Statistics, pp. 1072-1080. PMLR, 2021.
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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
A. Related Work
A.1. Feature Learning Theory
Unlike some of the more traditional feature extraction methods, such as recursive feature elimination (Darst et al., 2018), which typically require the knowledge and experience of domain experts to guide the selection and construction of features, feature learning theory (Cao et al., 2022) reveals an efficient method to automatically extract useful features from raw data. It aims find feature representations that better represent the data by mining the intrinsic structure of the data. Feature learning theory has wide range of applications. Cao et al. (2022) and Kou et al. (2023) address the phenomenon of benign overfitting that may occur in modern deep learning models, and study benign overfitting in two-layer ReLU convolutional neural networks through a feature learning theoretical approach to reveal changes under different conditions; Huang et al. (2023) and Pan et al. (2024) establish a unified theoretical foundation for understanding federated learning through feature learning theory, revealing that federated averaging improves the signal-to-noise ratio to achieve near-zero test error and weighted federated averaging to solve the problem of data heterogeneity in feature learning; Yang & Hu (2021) point out that the standard and NTK parameterisations of neural networks cannot reach the infinite-width limit at which feature learning can take place, and propose a simple modification of the standard parameterisation to achieve feature learning at the limit, which is found to outperform the neural tangent kernel baseline and finite-width networks, and the finite width networks converge to the infinite-width feature learning performance as the width increases; Zou et al. (2023) analyse Mixup theoretically from the perspective of feature learning, and construct a feature-noise data model to carry out the study, which reveals that Mixup training can effectively learn rare features from a mixture of rare and common features, while standard training cannot do so, which in turn leads to the problem of poor generalisation performance: Jelassi et al. (2022); Jiang et al. (2024) conduct in-depth research on Vision Transformers (ViTs). Given that ViTs can achieve comparable or even better performance than Convolutional Neural Networks (CNNs) in computer vision without incorporating the visual inductive bias of spatial locality and can learn spatially localized patterns, the authors aim to provide theoretical explanations for this phenomenon. In contrast, our research employs feature learning theory to investigate how the angle between task signal vectors relates to benign and harmful forgetting in continual learning (CL), introducing additional techniques designed for CL. Furthermore, it reveals the theoretical mechanism underlying replay- methods, which mitigate forgetting by expanding the angular range associated with benign forgetting.
A.2. Theoretical Analysis of CL
Since McCloskey & Cohen (1989) first put forward the phenomenon of catastrophic forgetting in continual learning in 1989, researchers have been committed to finding effective methods to balance the learning plasticity of models on new tasks and the catastrophic forgetting of old tasks. In recent years, researchers have started from the theory of continual learning to conduct theoretical interpretations of continual learning and explore new methods. Bennani et al. (2020) proposed that in wver-parameterised neural networks, the OGD algorithm can effectively solve the catastrophic forgetting problem in continuous learning, and also revealed the need to take into account the effect of NTK variations on the algorithm's 's performance in practical applications. Doan et al. (2021) proposes a task similarity metric based on the neural tangent kernel (NTK) overlap matrix, analyses how common projected gradient algorithms can mitigate catastrophic forgetting, and proposes a variant of Orthogonal Gradient Descent (OGD) using Principal Component Analysis (PCA) structured data, with experimental results supporting the theoretical findings and demonstrating the approach's forgetting potential. Evron al. (2022) studys the problem of catastrophic forgetting when fitting parrparraeterrieed linear models to cope with sequences of tasks with different input distributions, analysing the extent of forgetting, establishing relevant domain links and demonstrating situation-specifc upper bound on forgetting and pointing out how it varies when the tasks are differently ordered. Lin et al. (2023) conducts a theoretical analysis based on an over-parameterised linear model, give explicit forms of the expected forgetting and generalisation errors under a general CL setting, analyse the effects of relevant factors on CL forgetting and generalisation errors, and show that some of the insights can be used in practice and can contribute to the design of CL algorithms through experiments with deep neural networks on real datasets. Ding et al. (2024) provides generalised theoretical analysis of forgetting in linear regression models via stochastic gradient descent (SGD), revealing interesting insights into the relationship between task sequences and algorithm parameters, showing the impact of task alignment and the choice of an appropriate step size on forgetting in large data volumes. Zhao et al. (2024) presents statistical analysis of continuous learning in a sequence of regularisation-base linear regression tasks, compares the minimum-pe paradiigg estimator with the lower bound of continuous ridge regression to reflect its non-optimality, and concludes with an experimental validation of its theory.
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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
B. Supplements for Experiments
B.1. Mid-angle Sampling with a Two-layer CNN on MNIST
We conduct experiments to validate the effectiieneess of mid-angle sampling using a two-layer ReLU CNN with width m 256 and q on the MNIST dataset (LeCun, 1998). We define two tasks, where task T₃ involves classifying digits 4 and 5, and task 72 involves classifying digits and 9. We denote the flattened image of the i-th example in the j-th class as Mj... and the prototype of the j-th class is represented pij 1sp]" SE M3 where Sj is the dataset of class j. We perform sampling based on the cosine similarity sim( M2), where sim(-) denotes the cosine similarity function. Examples with moderate similarity sim(p',,s Mj) are selected for mid-angle sampling, while examples with higher or lower similarity correspond to small- anglle sampling and big-angle sampling respectively. We compare the three sampling methods above with random sampling in terms of the test accuracy on the previous task Ti in Figure 4. As shown, mid-angle sampling outperforms the other three methods, which is consistent with our analysis.
Task Accuracy
85
80
Accuracy (%)
55
Small-angle
50
Mid-angle
Big-angle
random
1000
1100
1200
1300
Replay Buffer Size
Figure 4. Experimental results the MNIST dataset using different replay buffer sampling methods: reported accuracies on task T} for various replay buffer sizes.
B.2. Experimental Details of Mid-angle Sampling on CIFAR-100
We evaluate five replay buffer sampling methods within the iCaRL framework (Rebuffi et al., 2017): small-angle, mid-angle, big-angle, random sampling and herding. We rely on PyTorch (Paszke et al., 2019) and train an 18-layers CBAM-ResNN (He al., 2016; Woo al., 2018) as the feature extractor yp, allowing iCaRL to store up K 5000 exemplars. Each task is trained for 100 epochs, with the learning rate starting at 2.0 and divided by 5 after 48, 62 and 80 epochs. The network is trained via standard backpropagation with minibatches of size 128 and a weight decay of 0.00001.
C. Preliminary Lemmas
Lemma C.1 (Cao al. (2022)). Suppose that 8 and = ((og((nxx/))) Then with probability at least 1
o d/2 < E 112 < 30 d/2,
K Ev < 20 pw v Volog(4n2/0)
fori E [mx].i' E [mw) and k. k' E [2]. Lemma C.2 (Cao et al. (2022)). Suppose that d 2 20lg(mmms/8) 22(oo(( (1)). Then with probability at least
1 1-8, 8,
(Ti.0) (w j.r -px) S 22log(8m/5) a 0allMx112
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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
(TL.0) (w)) J.r Ex,I) S 2/1oo(8mmx//)) 0p0p, vd
for all E [m], j E {=1} and E [Nu]. Moreover,
0o |4 122 2
max rE(m)
(w (1i.0) pux) V2log(8m/0) colluu|12
pn vd/4 < max j (w (Ti.0) Ev,a) S 2/ 22/lo(8mma//0 po Vd rE(m]
for all j {=1} and i [nx]-
D. Signal-noise Decomposition for Task T1
Based on the analysis of training task Ti by Cao et al. (2022) we know that
w (7200) (erdend) w/ (T).0) + ((())(()))))))))) ( dend).
11
Σ
11
(
(Tnde
f1, E
Σ
11
p(s)
jort
f1,
1
Jan
(D.1)
And if ni- SNR, 9 = (211,, the learned neural network can achieve small training and test losses. So at the beginning of training task T2, we have following property holds:
max, (H1)) (Trdlndd)
Vie {+1}.
ma (p(E)) (,1 Mend)) S Jo0pp Vd.
We also know that
| cos 01,2
(w (T2.0)
u2) =(w (w/ .22) + 73)7)(1))))))) ((u))) (Tritend)
(Ti.0)
|
(E1,1, E2.1)
(Tz.0) (w fr E2,1) (w (Ti.0) E2,1) +
p(E1) (Tirenn))
j.r
jirs
|1E,, 112
By Condition 3.1, Lemma C.1 and C.2, we further have
max (w (T2.0)
(22) =(mllo((1))))))) Vj (+1)
(D.2)
max (w (T2.0) 2 4/ la//lo((Smmnn (800028m22/00) 000p2 vd
(D.3)
E. Learning of Task T2 (Standard CL with Acute Angle)
In this section, we consider continual learning where the signal vectors between task Ti and task T3 form an acute angle, i.., (M1, (2) 0. Denote cos 01,2 (Mz ((p)((l [|2z)|2).. Then have that cos 01,2 0. We remind the readers that the proofs in this section are based on the results in Section D, which can achieve small training loss and test loss during learning task Ti.
E. Signal-noise Decomposition Analysis
Lemma E.1 (Restatement of Lemma 4.1). The coefficients 7112)
(Ta.) in decomposition (4.1) (Ta.1) (((2) (((a))
(T2.0)
satisfy the following iterative equations:
((p)) .P((2) (T.a) p((2)
(Tz.0)
(T2.0)
(T2.0) j.nt
= 0.
((12) (22,+1) = ((22) (T.00)
(T2.t)
122 Σ
(T2.t) d'(ww (Ta))
jrr U2 (z)) 11
2,
n2m
14

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
((E2) 1ir
(T2,4+1) = p((2))
(T.a)
1/2
(T22))
o d'(w
(Ta.0 ((w (2,))) 112,, 1 = j),
2,
Jar
T2
_ jrrd ((E2) (T2,+1)
r2
(T22) of ((w)r (2,))) 1122.112 (12, -j) 2,
(Tai)
+
ngm
for all r E [m], j {+1] and [n].
Proof of Lemma E.I. First, we iterate the gradient descent update rule (2.1) t times and get
w (22.0)
(12.0)
72
((22.) 2
(T2,6) o' d'((W ((w (2,1)) J22, jr ,2,,),
12
8=0
==1
m2
(T2.9)
((W .322;122) j12..
(T2.6)
2,4
12mm
According to the decomposition (4.1);
w (22.0) J.r
w (25.0) J,T
+
(12))
(T2.)
122l22² p2 +
Σ
(lE2)
((E2)
(Ta.4) jad
1162,111²² E2.
i=1
Note that the vectors are linearly independent with probability 1, under which condition have the unique representation t-1
7 (12)
jrr
(T2.t)
172
(T2.s) 2,
o ((w()) (W (222112)) ||
(T2.s)
(E.1)
12
==0
((Ea)
plEz) (1o.))
72
(Ta.)) 2,1
of o'((w) [w , 1| ju2.i.
(T2..)
m2
8=0
Now with the notation ((22)
(T2.1)
= p(E2)
(T2,t)
(pp(t))
(T2d))
2 0), p(E2) (T2,)) = p(Ea) (T2,t) I(p(Ee) (T 2 and the
(T2.))
jir
jan
fact (T22s)) O, we get
2,
( (mni)
(T2.t)
722
(T2,*)
of d((w (T2.s) Ea,))) 1|E2. 1 = =),
(E.2)
2,1
122m7
==0
e(S2) j.r
(T2.0)
(T2,)) 2,3
o' ((w
(Ta.s) E2,)) ||62,113 Il (12,1 = -j).
(E.3)
mm
Writing out the iterative versions of (E.1), (E.2) and (E.3) completes the proof.
We will prove the Proposition E.2 following Cao et al. (2022), which shows that the coefficients in the signal-noise decomposition (4.1) stay reasonable scale for long time of training. Consider the learning period 0 l < T", where T poly( 77 -e_,, || || || d
ddlooo "07?" d 0 on ,n,,7 722, m, d) is the maximum admissible iterations. Note that we can consider any polynomial training time T Denote a = 4 log((T) (*)). Here we list the exact conditions on 72, oo, for the proofs, which are part of Condition 3.1:
d), 172 O mmin(1gm/((55 d), ngm/(g20+2 d.. nym n) 2 1 12)1)
(E.4)
σo < (82glog(8mn min p1 || | (ap, V -1)
(E.5)
d 2 1024 1224log(4gg/4 0
(E.6)
(1(w ( on mmxx(lll 12, op Vd]. Then, by (E.5) and (E.6), it straightforward verify the following inequality:
Denote 30 = 2 maxi
j.r u22) ([ (T..0)
(T)
E2,))]- By Lemma C.2, we can upper bound 3n by 4 /log(mmn2/0
4 max { Bo. 8m2
log (gg(4ng/))
1
(E.7)
15

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Denote 31 = 2 max (I(w) (7200) ,12))), = 2 max (I(w)) 4." (E.5), have that
(2.00) -2,,)l)Thhen, by (D.2), (D.3) and conditions listed in
31 = 0 (og(me)))(//))//
(E.8)
32 -
(E.9)
Proposition E.2 (Acute-angle Case). Under Condition 3.1, and when cos 01. for 0 S t T", we have that
(T2.1) 0 S (12) jrr .P(E2) S a,
(Ta.6)
(E.10)
log (444////
(Tz.t) 0 2 p(aa) (,r5) - -16m2 V
2 -0.
(E.11)
d
for all r [m], j {=1] and [m2]
We prove Proposition E.2 by induction, first introducing several technical lemmas for the proof.
Lemma E.3 (Cao al. (2022)). Fort 2 0, holds that (w]rr
(T2.t)
(T2.0)
jrr .22) = j 7(22) ((2)))
(T2.0) for E [m], j E (+1).
j.r
Lemma E.4 (Cao al. (2022)). Under Condition 3.1, suppose (E.10) and (E.11) hold iteration t. Then
log log(4n2/0)
log(4n3/ a, j + y2
g(Ez) (Tz.t) 8m2
d a S (w (w(Tait) Wi E2,) S g(E2)
(Tz.1) (Tg.0)
(Tz.t) + 8mz
d
log(4n2; a S (w!121)) w jr
llo 1og(4n2/0)
p(Ez) fri 8n2
(T2.0) Eax)) S p\E2) jard + 8mZ
a, j = 12,1
d
d
for all r E [m]) { and i E [12] Lemma E.S. Under Condition 3.1, suppose (E.10) and (E.11) hold at iteration t. Then
(w (Ta.t) j.r
(2.2422 I (w) (Ti.0)
(2,))
j.r
log 1og(4433/
(w. (Ta.1) E2,1) S (T2.0) E2,i) ,2,1) + 8m2 V
d
E3 FW x2,1) 51
for all r E [m] and + V2
Proof of Lemma E.5. For j + 42 we have that
(w
(T.T2.)) (T2.t) ji (222112) (w!r jir (2. V2 i- N(m Drr
(Ti.0)
(Tidana).
| 2 cos 01.2
+ V2 J (22)) (T2.1)
11 l2
(w 2222112),
(T.
(E.12)
(T2.0) 2 0. In addition, we have where the inequality is by (11) (Ti,tend) 2 0 and 7(22)
(0g(4)3/0)
log( (og(422/0)
(w (12.1) E2,1) (w (22.0)
E2,) + ((.)) Snzy
S (w E2,1) + 8m2
(12.0)
d az (E.13)
d
where the first inequality is by Lemma F.4 and the second inequality is due to ((E2)
(T2.t)
< 0. Then we can get that
F,(W] (T..t)
x2,)
o ((
(T..4) -j (22)) + o ((w)) E2,)))
(T..))
)at
+
(T. u2)1 kw (T2.0) E2.)). 8nn
max don
d
1,
where the first inequality by (2. (3. and the second inequality by (E.7), (F.9).
16

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Now are ready to prove Proposition E.2.
Proof of Proposition E.2. Our proof proceeds by induction. The claim is trivially true at t 0 since all coefficients are initially zero. Suppose that there exists T < T such that the results in Proposition E.2 hold for all time 0 < I 1. We aim to prove that they also hold for t = T.
We first prove that (E.11) holds for t T, i.e. g(E2)),r3
(Tz,6)
- 1
log 10g(4n2/0) for T,r r E [m], j E {=1}
d
andt 0 for j 12 it suffices to consider the case where j # V2,2- When p p(Ez) (T: (T2.F-1) -0.5,2 / log(4n2/0) a, by Lemma E.4 we have that
and i [n2]. Since g(Ez)
12
E2,1) S p(E2)
p(E2)
112
+ (w (W (T2.0) E2,1) + 8m2
log (2(44n//)) a S 0,
j.n
d
and thus ((:))
(12.3) = el
5,r,
log(An3/00 a by induction hypothesis. When g(<2) (( (2.T-1) (2. 2 32 16nz /
-0.533 8n2 1oR(4n3/6) Ct, we have that d
(Iz.T)
(T2
1/2
-12
.2,1)) 1122, 12) o (ww ((w
11.
_(E2)
j.r.5
jrr
n2m
log( 4 /8)
12 d
log (og(4n2/0) a
2 -0.533 8m2
D2
0'
0.5,92 + 8m2
d
122m
d
log 4n2//8) α
d 1290 pz
log 1og(4nz/0)
-0.33 812
0.5,2 + +822/
d
12m
d
log (226/8) a2
2 - 16
d
where we use
(T2, T 1)
< 1 and ||Ea;
O d) in the first inequality, the second inequality is by 0.5,2 +
8n2 log(In3/0 /6)
< 1, and the last inequality is by 7/2 ( ( in (F.4).
Next prove (E. holds for t T. By Lemma E.5, we have
(T2..)
2,5
=
+ exp(U2. [F+1 W (Ta.t)
+1 x2.) - F-1 F-I(W(T.))) x2,))) x2,))
S expp y2,1 [F+1 (T( x2,1)
(Ta.t)
-F x2,))
+1
expl-F
- FF,,, (W(T2.1) x22;) + +1..
(E.14)
Following Cao al. (2022), let (j,r, denote the last time < T that p p(E2) (T2,4) 0.5a. By the update rule of ((E)) we have that
(TaTT) Pira
(Tastaina)
112
(Tataaa))) d' ((W
(Tadsira)) E2.)) (yy2, = 12
2
12m
T2
(Ta.)) o' d(w(w ((w
(T2.1) (2.,)) 1 = 12.
(E.15)
2,
12m
We first bound / as follows,
log(4n2/0)
9-1
M1 < 2qn mmm''mmmmmp n' (P(E))
(22tsira)) + 0..33 + S
a)
1000mmmmm999-00 of d S q2nn 1 1122a-1102 1-00- < 0.25a,
d
17

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
where the first inequality is by Lemmas E.4 and C.I, the second inequality by 32 S 0.1a and / log(An2/0) the last inequality is by 1/2 S /gmm (ggn///22200--2 d).
d a S 0.1a,
For r, <t < T and y2 = j. applying Lemma E.4 and a technique from Cao et al. (2022), (w) (72.6) (w!??:)) .2,,) is bounded by
0.25a S (w) (T2.4) 22.a) S 2a. Plugging these bounds into 12 yields
122 S
Σ
122
exp(-o((w))
(T2.3) jr E2,i)) + 1) of d((w ((w,rr .2,,)) (12, - E2.1lé (T2,t)
122
eq2912T
exp( exp(-a"/49)a-
R2y
< 0.251 exp(-ad /4")aaa
< 0.257 exp( (T*)(
< 0.25a,
where the first inequality is by (E.14), the second inequality is by Lemma C.I, the third inequality is by 722 (12m/(g29** 2 20-20² 2 d)) in (E.4), the fourth inequality is by a = 4lg((T)) and the last inequality is by the fact that log(T") j 2 l.. Plugging the bound of I1, /2 into 15) completes the proof for p. Similarly, using 31 S 0.1a in (E.8) and 7/2 = O (1gmm/ (u (1ma11)))) (E.4), we obtain ((22) (T3..) S a. Therefore Proposition E.2 holds for t= T, which completes the induction.
E.2. Signal Learning
p2 (vd))). We remind the readers that the proofs in this section are based on the results in Section and D, which hold with high probability. For the ease of discussion, we decompose μ4 into components along #22 and the direction orthogonal to μ1z. Then by (2.1) and (4.1), we have that
In this section, we consider the signal learning case under the condition that nzll 12 2 Ma
(Tait)
(T..0) w jrr
+ j. (((11)
(oa))) j.r
|122ll2 cos 01.2
(12))
142
+
(11)) (Tad)
Aelll2 I sin 0122
|11112
|
1
| 12
Σ ((En
ten))
S1.1
R1
p(E))
(Titend)
122
(T.)) jont
(
(T2.0)
E2,
p(Ez)
(12.4)
$2.4
jint
|
junt
Jirsh
E2,
where 2nz E (span } and p1
is orthogonal to P2, with || ll2 = || 1u2ll2- Denote 7(122 jrr
(T2id) = ((1)) (T (Ten))
| ll cos 01.2/11u1l1: + ((12) j.r
(T2.0)
By (4.3), then
7(12) (T2++11 (12)
(T2it)
122
(2
(U2,
(T..)) (w j.r (2) + v2. . -7())) (T2.))) ||
2,
12m
We know that if (12) Jrr (12)) 7(1) (T) Jend),
|| cos 0122/1/41/22 < 2, the learning process enters the first stage;
otherwise, it directly proceeds to the second stage.
E. FIRST STAGE
Lemma E.6 (Cao et al. (2022)). Under the same conditions as Theorem 3.2, in particular if we choose
n2 SNR2 2 C C'log66/og /
(E.16)
where C. = (1) is a positive constant, there exists time
Cloo(6//0) | [22l12)27+1 m
t =
1o0g
2
|2
such that
max, ((12)
(T2.11) 2 for E (+1).
1p (T2:0) S G00pp vd for all j E (+1),r E [m], i E [nz] and S 1 S t.
18

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
E.2.2. SECOND STAGE
By the results in the first stage, the following property holds the beginning of the second stage:
max, (142) (T2,ts)
2 2, e
max 10 1
I 3 where 3 000 vd/2.
Lemma 4.1 ensures that the learned feature (12)) (.2.0) will not deteriorate, i.e., (22) j.r
(T2,0+1)
{12))
(T2.0)
for 41
Thus, max, 7(22) (rr0)) Now we choose W follows:
W
(T..0) j.r
2gm log(2a/c) j.
+
J- (1 ) (3) (and)
|| sin 01,2
|
1122
|
((
tend))
E
+ (T).
n1
Ei1 l|E1.:l12
Based on the above definition of W we have the following lemmas.
[
-w'
Lemma E.7 (Cao et al. (2022)), Under the same conditions as Theorem 3.2, let tt = t +
2
t + Om - 1 11221122
Then we have maxj,n, (T,t)) s <23 000p2 for
for all t S t S tt. Besides,
| W* 12
Ls, (W. < (2q (t
6
+
29 -
s=t
for all t Ststt, tt, and we can find an iterate tend with training loss smaller than e within t iterations.
Lemma E.8 (Cao et al. (2022)), Let tt and tend be defined in Lemma E.7 respectively. Under the same conditions as Theorem 3.2, for any
t+ with Ls,(W (W)(d.))) S holds that LD. (W(Taa)) S 6. Ls,(Wi,,,,), (W(Ts.t) + Furthermore, since Ls, (W(I)))ne)) S e, it follows that LD, (W(T).) < 6e exp( exp(-n2).
E.3. Forgetting
Consider a new data point (X1, /1) drawn from the distribution of task Ti defined in Definition 1.1. Without loss of generality, we suppose that the first patch is the signal patch and the second patch is the noise patch, i.e., X [y Ei}. Moreover, based on the analysis of signal learning we know that
(Ta)fmma)
(Ti.0) j.r + j. ((11)
(Tist
1&1
+ - (112) (Ta (Taifoma)
f12
jrr
||
p(E:)
(T.))
(Tidtna))
E1.1
Σ p(E2)
n2
(Tzieen))
jind
E
(E.17)
E2,
1=1
And at the end of training task Tz, we have following property holds:
(Tidenn)) max (p(E:) j.r.
S Toop, Vd.
(Ta.tnn)) ma (2(E2)
I
000p2 Vd.
Lemma E.9. Under Condition 3.1, we have
a (w
(T2land)) -uiit ;11 (1)) S
Proof of Lemma E.9, We have that
|1 2|2 cos 01.2
(w (Izdend)
-W-r V1 m1) (w (T..0)
"y1.r 31 m1) ((11) (T (Tnda)d)
(122)
(F2ilad)) (2
-
12
19

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
S (w (TI.0)
31rr V1 f11);
(E.18)
where the inequality by (11) -91- 91.7
(Ti (ena)
0 and 7(12) (aatand)
(0155 2 0. Then we can get that
m
o ((w ((w (Tastand) -B.r 111 (1)) S max (w (T,.0) -1 D1 m1))
m
1
where the first inequality is by (E.18) and the second inequality is by (E.5) and Lemma C.2.
Lemma E.10 (Cao et al. (2022)). Under the same conditions as Theorem 3.2. with probability least 4m
exp(-C2 do pm? ") we have that maxj.r (w]] (ww.Tiden (ieeed)) .E1il S 1/2, where C2 = Õ(1).
J.r
Lemma E.11. Under the same conditions as Theorem 3.2, with probability at least 1 4m -xp(pppp Cz 0 have that maxj,r (w (WT..l (Tz.land) &1)1 1/2, where C2 = (1).
20000 2p' 2d we
Proof of Lemma E. Let W (T2ifand)
(Tzitena) -j j. (111) (Te W f.r
|| - 7122)
(Tzstan)
142 1
then we have
jir
that (w) (Tadennd) {1) (w (1rsaand)
(1) and || (antend)
j.r
2|2 Ö(ooVd mir0 + n200) = Ō(ooVd),
(E.19)
where the equality is due to d 2 50(2² max(mi; nz]) by Condition 3.1. Then similar to Cao et al. (2022), applying the properties of Gaussian distribution completes the proof.
Lemma E.12. Let tend be defined in Lemma E.7. Under the same conditions as Theorem 3.2, it holds that LD, (W(Tafeeal) < 18c +
Proof of Lemma E.8. By Lemma D.8 in Cao et al. (2022), we know that Lp, (W(TLdena)) I 6c + exp(-nf)- We intend to determine an upper bound for LD, (W(T)))) by leveraging the relationship between Lp, (W(I2,lend)) and ToD, (W(Tiffed)) According to the definition of true loss in Two-layer CNNs, we have
LD. (W(Ts (x.. flyn ((W(Ti)ten)) x1)]
Ld, (W (W(Tsifena)) = (xI Cly1 (altnff(WWWaaaa ((W(en)))))) x1)]-
When y1 ((a(ena)en))) xx1) 2 V1 f(W(Ti x1). by {(2) log(1 exp(-2)), we have
L,, (W(Tedtnne) < TDD, ( (W(li) < 6e + e
When y1 - (Usiion))))) x1) < y1 ((i(ti)))))) x1), let event be the event that Lemma E.10 and E.11 holds. Then can divide LD, (W(faffen)) into parts:
B[e(yn (W
((L(E)(()) (T x1))]
x1)]]
(E.20)
12
In the following, we bound I, and /2 respectively. Bounding T, By the signal-noise decomposition (D.1) and (E.17), we have
(w (Tid)))) 91.t
3/ 11) (T..).) U1 11) + + (Ti,fend)
yi.
(w (Tittnnd)
D11 (A)) (w (21.0)
(w-ys.r 31 m1) ( -y1.r
(2i.tand)
(W-y1.r
1/1 11) (w(T).0) V1 m1) + (( (Tr.fand) + (H2z) on
(T2dend) (Pse))
| 22 cos 01.2
[2 122l2
(Tadtan)) v1.T
M1 (1) (w(2) (T:.0)
(w-nirr 31 11) (())) (iitend)) ((22)
( (T2naan))
|1 2|2 cos 01.2
122la
20

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
(T2stend) Since (122), (,
(Ta.fend) 2 and cos 01,2 2 0, we know that (Wgi.r U1 p1) 2 (Wii.r 31 p1) and (w (T2tona) yn (1).. Then if event holds, we have that
(Tr.fand)
un. 91
11) S (w (Tiifand)
m
-91 (W x11= y (W x1) +
To (o(Ww(),eend)) 21."
1/2 (1)) o ((w!l2
yr [1l)
o (w (T2tena) v1.
111 -11111)) - o
(Tiifen)) yi.r
91 m1)))))]]]] +
o
W (Tnlenn)) E1))
1 m m
m m
(w
(22land) yi.r
Eii)
((w
(Ti)) yi.r
E1i)
((w (Totendd) E1)
62.r {1)
171
yi (WW((( xx1) + m
a (/w.TT
(Ti.band)
E1))
-wi. a
((w
(Taitene) y2.
11.0
S (Wi((W(Th.eem)) x1) +
1 (11/2) +
(1/2)
x1) + where the second inequality is by maxj.r Kw (Ti.fend) Lemma 1.11. Therefore, we have that
S (Wi
(E.21)
E1>1 S 1/2 in Lemma E.10 and maxj,r ( (w .ii) S 1/2 in
(Tafena)
1 S
yn (W(tatetn))))))) x1)]]
$ (x) (W(( )]]
18e,
where the first inequality is by the property of cross-entropy loss that {(2) S exp(=2) for all z, the second inequality is
by (E.21), and the third inequality is by ((1 exp(-xi (o ((W(Ti)fen))) x1)]] 6e in the proof of Lemma D.8 in Cao et al. (2022).
Bounding I2: Next we bound the second term 1z. We choose an arbitrary training data (xr, such that ya = y. Then we have
(WW x1)) S log(1 + exp(F=p) (Widen) x1)))
< 1 + F v1
(W(Tedend) x1))
= -1
((w
(Tzifand)
(11111) + - > ( -1.r
1
(Tzifan) (i)) 21.""
m71 p=
m
< 2 +
(w
Ei))
m
< 2 + (((00 11111
(E.22)
where the first inequality is due to Fy, (Tatet)))e) x1)) 2 O, the second inequality is by the property of cross-entropy loss, m -
i.., log(1 + exp(=)) < 1 + 2 for all 2 2 0, the third inequality is by 1 the last inequality is due to (W (Tarfena)) .Ei = (w jor
(T2itema))
- -n _31/41) S 1 in Lemma E,9, and
(Tastem))
E1) I
| Ar
(Tastea))
|| I ((olV))\ in (E.19). Then
we further have that
I/ < v [[(]
E (WWi)((((Tatend xx))
s /P(9))
V 44 (4+0(ooV()))) B [129]
< eexpl-Ω(oo
-2 07 -d- (d))
< exp(-nf),
21

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
where the first inequality is by Cauchy-Schwarde inequality, the second inequality is by (E.22), the third inequality is by -1 condition do v 00m=2210-2) (Op, v vd)= dd) in Condition 3.1. Plugging the bounds of 71 12 into (F.20) completes the proof.
the fact that / 4 ((ovv [ = O(poly(d)) Lemma E.10 and Lemma E. 11, and the last incquality is by our
F. Learning of Task T2 (Standard CL with Obtuse Angle)
In this section, we consider continual learning where the signal vector of task Ti forms an obtuse angle with the signal vector of task T2. When the angle between the two task signal vectors 0 2 cos 81,2 2 - -o2log(8m/// Folluxlla//00 log log({/)))///), through Condition 3.1, we know that - log(Sm/o) collmilla/(/ log(1/e)) - and 01.2 - 7/2. For continual learning where the signal vectors of the two tasks are orthogonal, since the components in the signal directions ((1) and (122) do not interfere with each other, the analysis of training task T2 is similar to the analysis of task Ti by Cao et al. (2022), resulting in small amount of forgetting. Next, we consider a less extreme scenario where -1 - cos 1,2 I - log (8) ao ool| /l2/ ([m (0mm//)))//))))) (9) We intend to investigate the mechanisms of signal learning in task T3 and forgetting on task Ti by conducting a detailed analysis of the behavior of various types of neurons during the training process of task Tg.
F.. Behavior of Different Types of Neurons
By the decomposition 14, we have that
N2
n22
j.r w Jir + ((22) (Ta.t), ||u2l122 p2 + _
(T.t1) (T2,0)
((E2)
11,2,1122 E2, + > plEz) Joi I|E2,ell2? E2.--
(Ta,t)
Next, we intend to categorize the neurons w (Ta)
j.r into three groups based on the following conditions: r E fr E [m]
(w (Ti.0)
Jrr jomz) > 0}, r E fr E [m] (w)
J.r (Pm) I 0) fr E [m] (w/rr ,jm') > 0} and r E fr E [m]
(T..0)
(w j.r j s fr e [m] (w (T..0)
(T1.0)
fr (P S 0}. Subsequently, we will analyze their behavior during the training
process of task T2.
jum) S 0}, we have that Lemma F.1 Under the same conditions as Theorem 3.2, when r fr E [m] (w (71.0)
(Titena) (w jrr jp1)..
(Ti.0)
j.r .jmz) for S fend based on induction. The result is obvious at = O as all the coefficients are zero. Suppose that there exists [ S lend such that the result holds for all time 0 <tt t < - 1. We aim to prove that it also holds for t = F. By the update formula for the signal vector of task Ti similar to that in Equation (4.3), we have that
Proof of Lemma F.I. We first prove that 7(1)) (T).)) S - (T."))
( (T,
(Ti,)) = -(m1) (T
T-)
M
R2
(Ty
0' d'((w -1, pi) + +i. j ((41) (11- ||mm
(Ti.0)
1,
mmm
=>(m)) (2 (2)
m
I-1,
I-1 o' d'(-(w (fi .j p2) ((1)) [2)) |1
mim
=I-==j
(T).0) Denote (2 -7(11) 9/3000 - (w .jm)), then have
((1r) (T(.0) = -7(11) (T1,-11) 1 Σ
(Ti
o (5(()))) (5((41) (T, -1144
1
mim
:1.==-
2 (7( ( (1),
mm mm11
(Tr (3)
(7(1)) 3.ر
m
= -5(m1) (21,-11)
9th 1
(8(1))
(Ti 10
jrr
22

---

Lemma F22 Under the same conditions as Theorem 3.2, when r E fr E [m]] : (1.0).) (T2.0) Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
9Th 11 12
(14)
frr [
(Ti
(Ti.0) (-(w)) 3rr (o2
20,
(Ti (Ti--1) where we use 0 > 1, 2 -1 and (11) 3,r F-) 2 in the first inequality, the second inequality is by 7(11)
I)
jrr
(21)) or (TI.0) = - j.r
(T.0)
.)) and the last inequality is by 7n 5 m/ mm/ga9=11| ll2) in Condition 3.1. Now we know that
HH1)) (TT.)) is monotonically decreasing and bounded below, thus {(1)) (T) must converge to a limit, which we denote by
C. Then we have that
R1 Σ M1,
C= C+
T1 7
G
o d'(C) 11m
By (,2 ((Tittond) < 0 and C' 2 O, we know that C' must be equal to O. It follows that ((m)) (7 converges to 0, implying that (T) ((1)) 1 (Tistend) - (w jor jipi), which completes the proof.
jrr .mm))) > 0}, we have that
(w jr .122) S and (nd2)((1)))))))) ((122) jo (Izdend) = -w jrr .jpz) for E {+1}.
(Ta.))
proof of Lemma F.2. By (D.1), we have that (w (T2.0) -jm2) = (w jr
(T.0) .jpz) + 7(pr) (Tr.ena)) || 2 cos 01.2
jrr
|1 12
S (wirr .jzz) / 122lo(88m/6) collme2 l2
())
S 0,
where the first inequality is by cos 01,2 / logoo88m 1log(88m0)) oo 1ollma1 0oll/u1lae ([m log( and the second inequality
is by ((w]) ((w(T (Ti.0) p2Yl S logo 8m/0) 0)- Goll l2 in Lemma C.2. Then, we can further prove that ((22) (Tatend)
- -(w), (T) (uz) by 7/2 S m//gaa || in Condition 3.1, which is analogous to the proof of Lemma F.I.
(Ti.0) (im)) 03 fr E [m] Lemma F3. Under the same conditions as Theorem 3.2, when {r E [m]] : (w (w(T1.0)
(w (T),0) .jut) > 0}. we have that (w jrr .ju2) (w)r
(Tz.0)
(T,.0) .jut), ((w[7.-)) u2)) I v (2log(Sm))) (8/08m/m ao vollut ll2 and
max+e|m] (w (T2.0) juzz) 2 oollut for j E {+1}, where 1t μ2 - || cos 01.2 |Milaa is orthogonal to the signal vector pi.
proof of Lemma F33 By (w (Ty.0) jr .ppz)
, S and Lemma F.1, we know that ((1)) ) (Tnn)) -(w
(Ti.0) || la cos 01,2
(w(Th.0) frr .jp1). Then
(w (T2.0)
jr ,juz)) (w (Ti.0) .ju2) + 7 (7 (Tede
| 1|
||2 12 cos 61,2
(Ti.0) .jpz) - (w (Ti.0) .jm
|Mill 1|2
j. (1) jr
.12 - || cos 01.2
11
|/ 1
(w jor
(T.0).
.jnt)
O.
By Lemma C.2, we further have that (( (w (T.00) oollut ll2/2, which completes the proof.
ا2)ي < 22 solleo! ll and max/e[m) max/elm|(w][? .juz) 2
(w/?3:))
jor
Lemma F44 Under the same conditions as Theorem 3.2, when r E fr E [m] : (w (w (T..0) jjm) s 0) fr E [m]
(w (T..0) j.r jut)) S 0}, we have that (w (T2.0)
.j12) (w[Th.0) (T.0)) .iut) S 0 and (12)) j.r
(Tatend) = -(w (T,00)
j.r .jut) for
j.r
j E (+1}, where ut = f12 2 cos cos0 f1
Ta orthogonal to the signal vector p1.
23

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
proof of Lemma F.4. By Lemma F.3, we have that (w) (W.rr .Jpz) = (w (Ti.0) .jPT) S 0. Then, we can further prove that (T1.0)
(T2.0)
(T2.0) T(U2)) (Tasend) - -w jr jp2) = - (w jr fftt,, which similar to the proof of Lemma F.1.
Proposition F,5 (Obtuse-angle Case). Under Condition 3.1, and when cos 01.2 < for 0 S t S T*, we have that
(T2.0) 0 S (122) jrr .T(22)
(T2.4)
jrr S a,
log (423/5)
0 z (lE)) j.r,t - 16m2 V
(T2.))
2- -0.
d
for all r E [m], j {=1} and i [m2]]
Proof of Proposition E5. It is known that the noise vector orthogonal to the signal vector, which implies that the proof
for 0 S p(5))),
(T2.4) I and 0 2 2(E2) 2 B2 16nz V log( (2ngg/2))
(72.t)
follows directly from the proof
provided for the Acute Angle Case in Proposition E.2. Therefore, it remains to prove S (12)) I a. When (Ti.0)
(72.1)
r E (r E [m] (w .jm) > 0}. by the proof of Lemma F.2, we have that
(Tridend) | cos 01,2
((12) (T2.2) < -w(w (Ti.0) .jp2) - 7(1) (
|| 22
S a.
where the last inequality is by (( (T)
U221 I / (llog))m/0) u2lle in Lemma C.2 and (T (Tri)ene) S a in (7).0)
Proposition 5.3 by Cao et al. (2022). When E {r [m] (w.rr jmm) s n [m] (w (T.,0) proof of Lemma F.4, we have that
..m) 0}. by the
(T2.t) S -
(T..0) ju+tt S a,
where the last inequality is by (w (T.0) Mt < 2 lolog88m/ oo |all|tt ll2 in Lemma C.2. When E fr E [m] (w] (Ti.0) .jmi) - 03 fr [m] (w jr ,juut 0}. by Lemma F.3, we have that
(T)
(T2.0) kw j.r
|22 S /2 1lg((8m/ collut ll2
(Tz.0) rE[m]
max (w)r .juz) 2 oo | 112/2.
(T2.1) Then we can use induction analogous to Proposition 5.3 in Cao et al. (2022), to prove 0 S 7(22) (Tat)) S a for fr
[m]:
(w (Ti.0) jr jpm) s 0) [m] (w (T..0)
ja .jpt) > 0}, which completes the proof.
F.2. Signal Learning
In this section, we consider the signal learning case under the condition that n n|| M(03 (vd)9). Based on the analysis of neural behavior in Subsection F.1, we have that
kw (T2.0) j.r
(2)1 222lo(88m/0) Gollet12
max (w (T2.0) rE(m] jor .juz) on polluttlaa 112/2,
for (r E [m] (w.3..) .jm) 03 n (r [m] (wff..)) jjrtt 0} and j (+1), where nt n2
|| 2 cos 01,2 | is orthogonal to the signal vector μ41 Note that the component of these neurons along the signal direction exhibits an evolution similar to that of the growing signal component in task Ti, as analyzed by Cao al. (2022) Furthermore, we know that the noise vector is orthogonal to the signal vector. Thus, similarly to the acute angle case, (T2.0)
we have maXj.r. (w j.r E2,4) S (og(8m22/8) duop2 vd in D.3. Then we utilize a two-stage technique similar to Lemma 5.5 and Lemma 5.6 in Cao et al. (2022) to analyze the learning process.
24

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
1.2.1. FIRST STAGE
Lemma F.6 (Cao et al. (2022)), Under the same conditions Theorem 3.2, in particular if we choose
log( (mm 1)/2 | (z vdy 2 C Coloo(6/oo) ut 112)2
(F.I)
where C = ( is a positive constant, there exists time
C'log(6/ (ogg(6/oollze oool ( 12))2+)
4
1200
|1
such that
max, 7 (12,6s) 2 2 j E {+1}.
1iina S p2 /d/2foo for all E {+1),r [m], i e [m2] and 0 S ti.
F.2.2. SECOND STAGE
By the results we get the first stage, the following property holds:
max, (122) (T2.6)
2 E
(T2.11) ma jp.n
S 3 where = Jo vd/2.
Now we choose W as follows:
w (Ti,0) + 2qm log(22/e) j- A42
|
Based on the above definition of W we have the following lemmas.
w 11
Lemma F.7 (Cao et al. (2022)). Under the same conditions as Theorem 3.2, let tt = 1 +
27p
+ Om )1 Then we have maxj,ra 10 s 23 000.2 vd for all 4 St +t. Besides,
(Ta,t)
(2q 1)me(t [ W
Σ Ls, (W smtj
c
<
+1
2q-1
for all 1 51 S (t, and we can find an iterate lend with training loss smaller than within (+ iterations.
Lemma F88 (Cao et al. (2022)), Let it and lend be defined in Lemma F.7 respectively. Under the same conditions as Theorem 3.2, for any 0 < t < Et with Ls, (T),..t) < 1, it holds that LD, (W(T2.4) S 6 Ls, (W(T2.1) + exp(-n?). Furthermore, since Lg, (T(Teeee))) I c, it follows that LDz
(D (T & 6c + exp(-n3).
F.3. Forgetting
Similar to the acute-angle case, consider new data point (x1, V1) with x1 = [y1 E1] following the distribution of task Ti as defined Definition 1.1. Furthermore, based on the analysis signal learning, it follows that
(Ta)fand) jrr
(Ti.0) w j.r + 7141)) tend)
M1
+ (112) (Ta (T2))))a)
112
|
12 2221
+
((E)) j.nk
(Trstend).
E1.
+
p
j.rd
(Tzifena)
11,,,
(F22
im1
maxj.ra (e(E)) ( (Tridend) < po Vd.
maxj.ra (p(E2) (T
(Tz.teme)) S 00 Va.
25

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Lemma E9. Under the same conditions as Theorem 3.2, when E {r E [m] (War jir (im) 0}, we have that
(T1.0)
(w) (lidnnd) jrr (w) (Tidend) -رH1] S S, 1r
.j1m1) = (w (1.00) j.r .jm) + (11)) (2i.)a)))
for j {+1}.
Proof of Lemma F.9. By the signal-noise decomposition (D.1), we have that (2i.0)
(and)) jr
_jm1] = (wrrr .jm)) + 7(p1)) dona)
(w
Then we further have that
dendd
-jH11 (w (1i.0)
-jpp) - -7(p22) (1 (Fi
jo
S 00
(T,,0) where the inequality by (w jr .jm)) 2 and y(
f/r (Tnd)) 2 which completes the proof.
Lemma F.10. Under the same conditions as Theorem 3.2, when r E fr E [m] (w (Ti.0) jrr .jpi) s 0}, we have that
(w (Tidend) .jpp)
(w (Tidtend) -jphn
for j {+1).
Proof of Lemma F.10. By the signal-noise decomposition (D.1) and Lemma F.l, we have that (w (T1.)
(Ti.fend) .jui)
(w jr .jui) + ((1)) (
(Ti dend)) 0 and (w,rr
(Ti.tend) -j -- j.r
(Tr,fena) .jpi) O, which completes the proof.
Lemma F.11. Under the same conditions as Theorem 3.2, when r E {r E [m] (W (2i.0) "2i.,,,je .jm) > 0}, we have that
(andend)
jM1) 2 (111) j.r
(Fi,aan)) (1- cos² (o_cos² 01,2) -
(Th dendd)
jjm1) S 2 lo2g//(m/0) oo 0ollmel12
jz2) 0
j,r
-ju12) 0, (w) jor
(T2)ten))
for j E {+1], where n = max(mn n2) and μ = f1 ||l11l/ 2|2 cos 01,2
44 Tpalla
is orthogonal to the signal vector μ22
Proof of Lemma F.11. By Equation (F.2) and Lemma F.2, have that
(w (2n))nd) -1Mz) (w (T(T1.0) (Ti.0) (TT.))....) .jmz) + ((2))) (T (Tadenn)) + *122) (14)) (Taitene) || ll 01.2
|22l2
-w ji
-1-cos² 01,2) (Ti.0) jumtt + ((11))11))))))) (T (Tistnn))
2 (())) jrr
dema) (1 - cos² 01,2) - -0(m-14 'nn nn),,
where the inequality is by ((w j. in Condition 3.1. Similarly,
(Ti .0
/2
-01|/1 |l2 similar to Lemma C.2 and Tol| ll s
(w (aa)end)
-jp11) = (w (T (Ti .0
j.r -j ((1)) jrr
(Tisfena) (1 cos2 01,2)
S / log (8m/8//mmmp /5) aoll
26

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Understanding the Forgetting of (Replay-basedd Continual Learning via Feature Learning: Angle Matters
(2loog(m/O) oollu
where the first inequality is by (w,,, (2).0)
|42 1 I / logg(m//)) a Goll ll2 similar to Lemma C.2, the second inequality
is by the definition of pzz Then by Equation (4.1) and Lemma F.2, we have that (w), (TaAtema) .juz) (w) (7..0) (Tasfend)
j.r .ipe) +
(Tadena) = 0, which completes the proof. (( ( = 0 and (w)), (w)rr (Ta.fend) -jp2) = j.r -jm2) (
(72.0)
7
Lemma F.12. Under the same conditions as Theorem 3.2, when r E (r E [m] (w(0).0) (im) & 0f n fr E [m]
(w (21.0) Jar
) > 0}, we have that
(w (Trdena) fr
.1p1) 50
|| 2l2 cos 01.2
(w) (Toitona) 3.r
jjHz) = (12 (
(Taitene),
12l12
(Tz,tand)
.jpu2) ((12) ((42))77 (Ta,tand)
(w) jrr
-jpb)) S 0, (w jr
(To,tand)
for j {+1}, where pi 42 |2ll2 cos 01.2
Mi is orthogonal the signal vector p1. Tellla
Proof of Lemma F.12. By Equation (F.2) and Lemma F.1, have that
(w) (T2)) .(m1) = (w (Ti.0) 3,r
..p1) + ((11) ) (Trtteme) + 7 ((42) ((22))
(Tzitene) | cos
jrr
|42 122
(Taifand) ( fr
| cos 01.
142122
S 0,
where the inequality is by cos 11, O. Similarly,
(Tattnnd) (w jrr
= -jpnm) -ju11) (
( (Tdlena) | l|2 cos 01.
112 12
Then by Equation (F.2) and Lemma F.3, we have that
(Tatfon))
.jpz) = (ww) (Ti.0) )ittt + 7 (142) (12) (Tgteen)
(w jr
2 ((uz))(T (Tatten))
where the inequality is by (w (Ti.0)
j.r .jp > 0. Then we further have that (w (Talend) jrr
-jH2) = -(w (2.dand)
j.r .j42) S 0.
which completes the proof.
Lemma F.13. Under the same conditions as Theorem 3.2, when r E {r E [m] (w (T..)) jjmm) I 5 [m] (w (Ti.)) (Ti.0) jrr jutt S 0}, we have that
( (Tatona) fr ..p1) 50
(w (Tzslond)
-jpp) s (2log(Sm/8) collmal12
jir
(W (Tzstand) jrr
.j122) =0 (T2,land)
-jpp)) 0, (w jrr
for j {+1}, where pi 422 || cos 01,2
11 is orthogonal the signal vector pi
Mell2
27

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters Proof of Lemma F.13. Similar the proof of Lemma F.12, have that
(w (Tzifona) 3.r
,jui) - ((122) (T.da)d)
p 2|2 cos 01.2
< 0,
1
|e 1/2 cos 01.2
(w) (Tasend)
-jjut) (112) (Tastana)
|
| 2|2 cos 01.2
((w]7 .j+t)
12212
S /2 lomm/mmm// 0ollu1211
where the third equality is by Lemma F.4, the inequality is by kw j.r
(Ti.0) .iTi S (2log(Bm/8) oollui ll2 similar
to Lemma C.2 and the definition of pi: Then by Equation (F.2) and Lemma F.4, we have that (w ( .izz) =
(Tz.lens)
(T5,0) (w) Jir . +
((p2)) (1 (Iz.Aen)) = 0 and (w) jir
(22:len)) (T24ena) -ju2) = (w jit
.jpz) = 0, which completes the proof. O
for Lemma E.1. Under Condition 3.1, we have that kw (T2itend) Ei)l 6 log (8mn/0)000 vd for all r [m] and j E [+1}, where max(n1, n2)-
Proof of Lemma F.14. By (F.2), we have that
122
-1) = (w) (71.0) (Tasena) (w) jr
E1)
+
||51,1112
(EI.1, E1) +
(Todend) Pira
||62,r1122 .E2,, E1)
jr
log(4n+/0)
log An
2 log (0)00/ Vd + (4n1)
+ 1)000p, Vd + (4m2))
+ 1)000p. Vd
d
d
2/log log (8mn1/6)000 Vd + 2 log (8mn1/0)000 vd + 2 2/ (8mng/6)a00pp, Vd
< 66 (8mn/8)0u0p, Vd.
(Tr.fand) S po v and maxj,r, (p(E2) where the first incquality is by maxj,r, (p(E)) fmn inequality is due to d 2 Q(m"² max m²}) by Condition 3.1, and the last inequality is by = max (11,
jand (p(E2) (Ta.fund) S Tn Vd, the second
Using a proof technique similar to that in Lemma F.14, we can further prove the following lemma.
Lemma F.15. Under Condition 3.1, we have that kw
jrr (Tidenn) ,1,'), I 6 (8mm/0)0o0p. Vd and (odenna) E2,2r) S 6y/og (8mn/8) vd for all r E [m)) j E {+I), i' [m]] and i" [m2)). where
(w jor
max(ni, n2).
For ease of discussion, let 4,, = (r E [m] (w (T..0)
jrr .jmi > 0}. 15,1 = (r [m] (w (w (Ti.0)
j.r .jmi) S 0}. 1,2 = {r "..p).pp S 03 fr E [m]]
[m]: (w!7h (T2.0)
jar .jmi S 0} n {rr E [m]: (wfr (Ti.0) .jmt) > 0} and 43.3 = {r E [m] (w jr .jpi)
(Ti.0)
(Th.0)
jar ,5t) < 0} for j E [=1], then the following lemma holds.
Lemma E16 (Cao et al. (2022)). Suppose that 8 > 0 and n 2 32lg///))) Then with probability at least 8.
11 | 2 m/4 and 11,2, | 2 m/8for j E {+1).
Lemma F.17. Under the same conditions as Theorem 3.2, denote (pr); 1 1.E Ere,, (711) (T (T.eon)) Then we have
log( 1
c - -m(m 1mmmmmn--)) < 7(11) S log( +
Co ) + (mm-1n11) 1I1,)
for j E {+1), where Co >6>1 1 is a constant and n = max max(m1, (n1 n22
Proof of Lemma F.17. Since Lis, (W(Ti))eaa) < e, there must exist one (x1,1, and constant Co 2 0 such that e/Co < ( (I x))+)) S which implies that
1
(Ti dema), x1.)) I log( Co
S M1.i log( + o(E)
(TW
+ o(e)
28

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
By the definition of ((W(erlfed)) x1,5), we have that
V1, f( f(W(TLiend) x1,5) F a
(Wid))f x1,i) F- F-vi, (W
(Fisend) -y1,
x1,s)
[o ((w (Trndan)) M1.0
U1,5 (1)) (aw (endead)
mait E1,1)
( (( 21,r
(Tistena)
M1. x1)) + o(( (Frdtnna) ye.e
E1.a))
m
m 1 m
((w)))
Vi. (1)) + Mi.eif
(Tidend)
[o((w (Tistend) Di,i
E1,1) a(ww (( (aid)
,L.,))]s
m
where the last equality by (w (afded))) //oog log (8mn/8)oug pup vd in Lemma F.15 and puTp, Vd s Ō(m-'n=) in Condition 3.1, we have that
-Mi,er Mi, p1) S 0 Lemma F.9 and Lemma F.10. Then by ([ (Tiieen)) jir
E1,) s
1 Σ m
Co + 0(m (annnnn1)
log( E + o(t))
Õ(m "non"nnnn n
(Ti.fena) mi.e 1/1 m1)) S log( + o(t)
By Lemma F.10, we have that
m
(w (and)))) D1,s.
31.4 M1)) = m
Σ
o((w (Tidee))) Bisst M1 (11)
relvi
a(w((h..) mar Vi.5 pi) + r(p ( Hidena))
m vE
Then by 1(w (T.0) (T1.0) dur
pIll & (2log(8m/0) Golle ll in Lemma C.2 and o Pollmil12 s Ö(m "onn-nn11 'n=)) in Condition 3.1, we
have that
log( c 016)) 1mmmmmm-1
Σ
(7 ( (iifonal log( waar
Cu ct(()) Om--'n
o(c)
m rElpI,
which completes the proof.
Using a method analogous to that Lemma F.17, we can prove the following lemma.
Lemma F.18. Under the same conditions Theorem 3.2, denote ((u2); Er ( (122) (12)) (22eeee)) Then we have
log( + o(e) (
- 0mmmmmn {(122 S log( + o(e)
Co + Ō(m (mmm1nn1)
for j E {+11), where Co > is a constant and n. max max(n1, n22). Lemma F.19. Under the same conditions as Theorem 3.2, when - C1 < cos cos 01, < O, we have that
[lo((w (Tadaad) y1f1)) - o(( (Taifana) oi.r
3111111]] 2 Cs,
where < Ci < and C3 are positive constants.
Proof of Lemma F.19. By Lemma E.12 and Lemma F.13, we have that
a((w (Tateend) 81.t
(111i)
Σ
a((w ((( (endena))
11
rel
29

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
2 m
7(11)
-1- -1s- 01.2) Ö(n
4
where the inequality is by Lemma F.1l and 1yr.al 2 m/4 m/44n in Lemma F.16. Then by Lemma F.11 and Lemma F.13, we have that
Σ
(Tzifand) o -yii
(1111))
Σ
(Ta.tana) o((W-ri. -vi.
(1111)) Ö(n=l)
r v1
M1 12 (- cos 01,2)
m (12)
Ö(n
| 12
where the first inequality is by colluil1a O0 in 2) in Condition 3.1, the second inequality is by Lemma F.12. Then we have that
((w
(T2 yirt
(11111)) o -wir -1111))]]
(T2.tena)
m
(11)g) (1 (1-cone cos² 01,2) (H(2))
p1 12
cos cos$1.2) Ö(.
4
122
By Lemma F.17 and Lemma F.18, we have that
log( (
1n=)
1 <
(122)
log(Co) + log( (((()))))))) + Ömm 11
log(Co) + log( -o(()) O(m-In
7(11)
log( cto(e)) O(m--mmm) in
1/9 is a constant, denoted as C4. Now we choose C1 as follows:
Then we know that (0000vi//(11)) (11))
CC4s C3C + -11603 ((n- (7()))))))))))))) 1/0)
C
where C3 = 4/9, //mll///l/// ((r) Erel,, (((1)))) (iptonnd 2 llloo( (og(()))))))) -mmm'nn11) by Lemma F.17.
Then when
(1/9) ||12112 5-cos0 Ci,
1/9 :1/11112
41/9 - (122) -v1
it is easy to verify that
q/4 (((22) 1/9 ||
(- cos 01,2)19-1
||112122
7m1)m 1/9 (1 - cos? 01,2) 4/14 (122)
1/9
|| l12 H2 l2
4[Cs Ö(n 1)
cos 01,2) 2
m
where 7(42)
Erer ban,,,,((4 (T((12) (Tzifund) (T Then we have that
m
m
(Hi)y -11- cos² 01.2) m ((12)
(-
-os 2
4
||
2 m
9/42/1 . ,7(z)) -vi 1/4 ||11|/2 (- (-cos (2.2) -(-cos
4
Hell
(Hr))n 1/9 (1 (1- cos2 01,2) - 41/9 7742))
1/4
(- cos 01.2)
12l12
2 C3s Ö(n-)),
where the first inequality is by the convexity of ReLU9. Denote h(x)) as
m h(x)) 4 (11)y (1 (1- x²)9 (
| x
Ö(n x E (0.
M2
30

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
We know that h()) is monotonically decreasing. So when
(1/q) F. ||2||
co00012 S
1/0
42/9 (142)
111201122
we have that
778 4
(111) (11 cos 01,2) m m. (12)
1 (- cos 012)
2 C3 Ö(n 1)
12
Then when - cos012 S C₁, we have that
m
Σ
[o((w (o((w(Tattna)) (Taifnna) (1H1)) - o((w War
(Tadand) yi.r
(1111111 2 Co.
which completes the proof.
Lemma E200 Under the same conditions as Theorem 3.2, when -1 < cos 01, S -O2, we have that
Elo((w (Tzdand) -nn Mi.
(111)) - o((w (Ta.eaa)))) ,1141)1] 2 C3.
where 0 < C2 < 1 and C3 are positive constants.
Proof of Lemma F.20. By (z) 2 O, we have that
Σ o (w ())n)))
(T2ifand) -i.r
Σ o(w (( (Tzifand) -vir (11
rEl-v1.2
2 m 8 7(12)
|1 11 cos 01.02
|
where the second inequality is by Lemma F.12 and we have that
| 2 m/8 in Lemma F.16. Then by Lemma F.12 and Lemma F.13,
o (w (Tz.ona)) V1.r
-111)))
Σ a(( ((W (111))
r
$ m 7(1 ) (1 cos² 01,2) Ö(n-1),
where the inequality by Lemma F.11 and 00l/21112 S Õ(m -' -n1 1) Condition 3.1. Then we have that
m
(Taifena) -1ir
(1111) ((ww(!!)
(Tz.fand) (1111))) v1.
m
((2))
|mil (-cos cos 21.))
m ((1) -11-oss" -1--co=80,))9 - 1t.
8
|22
Now we choose C2 as follows:
C2 -C4C + C3C2 + +4+ (320s +(n--))/(8W/mgg (00m0////m/m0p)))))) 1/0)
and 7(11) = in Erery ( ( (Tiaame) 2 where C₁ = (7(2) ( ( 1/9 Cs = | 18//// |||2||2)
log( ( (m -inn-nn11 11) by Lemma F.17. Then when C2 - cos 01,2 < 1. using proof technique similar to that in Lemma F.19, we have that
m (12)
| 112 (- cos 01
m mm (mn cos² cos2012)) 2 C3 Ō(n -1.
8
[2
31

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Then when C2 cos 112 1, we have that
Σ
[lo((w (Fa.ten)) -yir (Vmli) o ((Ww!l (Zaleaa)) m)mmi)]] 2 C2.
which completes the proof.
Lemma F.21. Let C1 be defined in Lemma F.19. Under the same conditions as Theorem 3.2, when -O1 S cos 01,2 < 0, we have that P ((xLn))DD (y) + ( x1))) I exp( 22202222 22 /²",, where C. = (1))
Proof of Lemma F.21. For the sake of convenience, we use (x1, V1) ~ D1 to denote the following: data point (x1, v1) follows distribution D1 of Task T₁ defined in Definition 1.1. We can write out the test error as
P P(LLLDDD (v1 *
x1))) = ((),n)) (Unf(W(Zaiean) x1) S 50).
It therefore suffices to provide upper bound for I (()))))) (y1 (W x1) S 50). To achieve this, we write
X (M1 E1), and get
f1f(WW (Taten))))) x1) = Fyi (Tuien))
x1) - F-2. (2.teen x1)
-M1
m
To( (2ndean)) (1)) ol ((w((
(WWTr (Ta.een)) [1))
m1.
m
(T2ifena) [o - ((1m1))) + a(( (W [1)
(T2tend)
m
Then
P (x (W x11 50
$ PM
Σ
[o(((ro
(Tattene)
[Eii]] 2 >
[((((W!l vi.
(1111)) - o((w (Ta.aan))
(F.3)
Mir
Let W (T2.1) jr
w jrr
(Ta.t)
-
(Trtena)
[A1
15
(T2.1)
Teal
then we have that (w (Ta.)) E1) = (w (Ta.t)
jrr
Ei) and
|
(T2
1l2 Õ(0ovo Vd + max n2) o0) Ö(ooVd),
(F.4)
jrr
where the equality is due to d 2 A(m² max(ni: n23²) by Condition 3.1. By (F.4), let maxj,- 1 (T
ll2 S Cra Vd.
where C7 = Ō(1). Denote g(E1) as E, a ((W (Ta)tana) holds that
Ei)). According to Theorem 5.2.2 in, we know that for any x 2 0t
v1."
P(g(E)) Rg((i) 2 x) exp (-
(F.5)
pr.
11
Lip
where e a constant. To caleulate the Lipschitz norm, we have
m
m
Ig(E) g(E')
Σ
a (Ta.fema) vi.r
4)) Σ
(T's (oma)))) vr.
E2))
|o (and)
{) ((W (22.lana)
E1))
-viit
I Tog (8mn/5)ouc vaya-1
W
(Tzitent) 21.""
: --')
I < (8mn/6)ou0y, Vaja-1
w
(Tztona) vi.r
[| 116 E'112,
32

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
where the first inequality is by triangle inequality; the second inequality is by the convexity of ReLU9; the last inequality is by Cauchy-Schwartz inequality. Therefore, we have
m
1,я| I ((6/log (8mn/6)000,, vdje 9-1
W
(F.6)
2
and since (w (Tsteaa) vi.t
(Txtene) E1) ~ (0, |w
110 p1 ). we can get
v1.r
E1) 2 T(
((
w
(T2))) -iir
120pz)
Eg(E1) = Σ Bo (ow
(Taifend) =1.r
2
Cs.
W
(antend)
11
2t
2
U1.
) is a constant, with [(2) = ft e- dt denoting the Gamma function. When -C1 1 cos 112 < 0, by (F.3) and (w (22.1)
where Cs = 2 I( 2
E1) = (w) (2211) E1), we have that
P(x),11)211 (WI Wulf(W x1) 00)
< (
[o(( (T2daand) 21.5
_Ei)]] 2 [or (( o(( -wir -b/pr))]
(Tztend)
%
= -P
g(E)) - Eg(Ei) N
Σ
(1111)) - o(
(Tziland) oirr
[1111)]]]] - Css 22
w
T22 (and)))) y/.
m [o((w (((W211)) (( (Tzstana) (1p2)))
.81141)] a(WWTTitt (T2itone) (Cs/v22 Em= oi
./
(T3.
11
2
exp
q2 (Volo (8mm/ (ooum//5000p:1 vd) c
20-2
Er= pp || -Mi.r
(12tieor))))) l12) (11111 o ((w ((w(3 (32.tand)
2
Em [o{ yo.
(22dand)
27
S exxp
-81.
Cs/ Cs/V2TT)
42 ( log (Smm//5oooo vil
24-2"
(22ddad)) w m Pp
==
l|/
m -r=1
[o (w
(Tzifend)
21 (1111)
-01." : []111))]]] 2
(Tzitena)
yir
exp
92 (6/log (8mm/6)000 vdl
20-2
2x
2
(I2dene)) pp Er
ll2
v
where the second inequality is by (F.5) and (F.6), the third inequality is by op w (extana)) 1l2 S Ö(aoppi Vd) according to (F.5) and vd & Ömm=2n=1 in Condition 3.1, the last inequality is due to the fact that (s- 2 8²/2 12, Vs, t 2 0. Then by Lemma 1.19, we further have that
P P(x)m)> (2if(W (T2tem)) x1) 0))
1
(C2
C3
< <xpp
0.5c
9² (6 (8mn// 6 oo vd)
27-2
2T
0pi E
(Tz.tena)
l12
[C3
cC3
Sexp
q² ( log
vd)
20-2
2n
2m 2 C9 (aua p. vdjz
C
Sexp
9² 6oolo (8mn//)o vd)
29-2
4m
Am22C(aaop Vayz
[-C . 2n
< exp
q²
where the second inequality is by maxj.r Condition 3.1. Then we have that
| (T2.t) jr
1|2 S CranVd, the third inequality is by 0oom Vd < Ō(m - 'n '1= in
(( (y1 A Sigu(f(W(Ulittee x1))) S exp((C 24 -2 2, /92)
which completes the proof.
33

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Lemma F.22. Let C2 be defined in Lemma F.20. Under the same conditions as Theorem 3.2, when -1 cos 0080 S -C2. have that IP (x) (y1 # (ogn(((WWaaaeendd x1))) 2 exp(-C m22--2222 //²), where C O(1).
Proof of Lemma F.22. We can write out the test as
P PoLLDDD (y1 sign( x1))) P P(x ( (Wr(feon)))))) x1) S 50)
=1 =1 -1-P((x (XL.Ii/D (11 (W (W x1) 2 0).
It therefore suffices to provide upper bound for (())))))D (v1 (W (T x1) 0).
To achieve this, we write
X1 = (11 Ei), and get
y1 f( (Znna)))
x1) = Fei (W(Ta,ten))
x1) - F- (W x1)
1
m
[o (w (2.dona) vi.r
(1111)) + o
(an))) E1))
91.
m
[a (Taitend) -21
.y1111)) o c((w ([: (Tasfene) _1i)]--
Then
P (( (V1 (Wn((((esten))) x11 xx12 20
< ((L
Σ
[(i)]] 2 [(o((w [o Mir
(bidend)
(Tzifena) (aaeaa)) 91."
-11 - o o( ((
(Ti,tene) m1."
When -1 < cos 1,2 $ - C2. using a proof technique similar to that in Lemma F.21, by Em [o((w
(Tastens) ,11
Bi.r
a(w (Ta.fand) 11111))] 2 Cg in Lemma F.20, we have that
P P(XLM)>DD (yn (Wr((W(a))n)))) x1) 0) S <exp(-C -opp-Cmm=4-2p" 2229/2))9/
Then we further have that
Pix (y1 F sign(] sign()(((Zddenn) x1))) 2 1 exp(- C 11,00-22 20 ///))
which completes the proof.
G. Insights into Task T1 Learning
In this section, our intention is to present the specific details associated with the learning of task Ti. Specifically, the focus is on the maximum value of the inner product between the convolutional filters within F41 and μı, which is denoted as (Ty.e) m| (w +1,r p1). If we suppose that r" is the value for which the inner product (w (w+1, p1) attains its maximum, then it follows that (w (Ti.))
(Ti.))
+iir M1). Moreover, r** is the value that results in the second largest +1,r .1). In such a situation, we can state that (w (Ty.)) +1,r""; p1) max, +1, f1)- Similarly, we
+1.r* p41) ma (w (Ty.e)
inner product (w (w (T...)
(Ti.6)
(T..) can also obtain the values of (w[Ti,e), -1) and (w -w.r". -p1).
Lemma G.1. Under the previous conditions,
get that when tends to infinity
(Tu,)) (w jrn .jmz) + 400
for j E {+1}.
proof of Lemma G.I. Based on the analysis of training task Tj before, we know the update rule for w (Ti.).
as follows:
(T.++1) w (Ti.)
jrr 171 Vw Ls, (W(T..0)
34

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Understanding the Forgetting of (Replay-based Continual Learning via Feature Learning: Angle Matters
B1
(Ti.0)
7
(T..0 ((T))) o' d'(( jrr E.a)) sun
Jir
mm
71
(T:.
o' d'w (), pr
(Tr.6)
mim
Perform the inner product operation on the vector μ4 on both sides of the equation, we can get
nh Il 12
(w) (11))) .jpz) = (w)rr (T.4)) (mm) +
((Ti))))
o' d'((w (Ti.)) jr s1.HH))).
Jrr
1$
nmm
Then we have
.jm1) (w (T.)) jum) + m11l
Σ
((Ti)))) 0' d'(w (T. j(m))..
(G.1)
nim
01,
(
(1)++1)
(Ti.t) .jpz) = jorr jmmz +
mh
11112112
Σ
((T)::)) o' ((w (Ti,t)
(G.2)
j,
nim
Dividing the two equations (G.1) and (G.2), we can get (Ti.0)
(w (1,,++1)
.im))
(w (w jrr .JM1) +
2 Al
1,1 Em (-1,,
((i..t) o' ((w j,r .111))) (2.
mamm
(G.3)
(T)++1)
.fp)) (W j.mi + milles 112
Em, = 11,5
((Ti.t) o' d'((w) ((w) (Ti11) jpm)))
we can judge when & tends to infinity, (w j.r jppz) tends to infinity. According to (G.1), we know that (T..f (1) an increasing sequence, so its limit is either a constant or positive infinity. Assume it converges to a positive constant M, take the limit of both sides of the equation (G.1) simultancously, we can get that
(T,,))
Th m1122 Σ B1,
(Tr,)) d'((W) jr .jm))
M = M + lim
+++00 nmm
By the fact that ((..)) O, get that
M
Σ 01
((Ti.)) (- (M))
7mm
then we will come to a wrong conclusion: 0
m
E Eg.:= (Ti,)) Jo' (M), so (w (Ti,)) jpm) tends to infinity.
mg
For our further work, need to review the following mathematical lemma:
Lemma G.2. Let On 0, then the infinite product IIt (1 + an) converges if and only if the series > n== On converges.
The proof of this lemma can be found in many mathematics books. With this tool, we will prove the following lemma. Lemma G.3. We can get that when tends to infinity
(Ti.)) (w J jpm) = o((w/TTh))
(j (jm)))
for j E {+1}.
proof of Lemma G.3. If, as approaches infinity, the inner product (w (i.f) proof of this lemma becomes straightforward. Subsequently, we will investigate the case where (w ,.uz) approaches
jjm)) converges to a positive constant, then the
infinity. To simplify the expression, we let aj = (w (T.,4) .ميz) and be = (w (Ti.)) jpz)). According to equation (G.3), we can deduce that:
35

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
(111.) a 17:llAu113 1 p ( 1, )a?
1
Oc-1 b(+1
Gt
nam
be
1
9n 1
Ey, )( (- 12,6 yog-
Let Ct 8t, b,
then we have
1
90n p
Σ :n
((22.1)
1, )az 9-2
C++1 = C
1 guallet, 1
2m.
-1y (T,,
1,
)b² 19-2
2 Ct:
where the inequality is by 0e 2 by, so it is obvious that c is increasing sequence and o 2 for all / 2 0. Combining equation (G.1) and the definition of Ct, we can obtain
9Th Ml 11
c44bb++1 = cebr +
Σ
((Titt) (-(((i))))
19-1 1,2
11
V1
Dividing this equation by (G.2) gives
C1++1
b,
qun
| 12
Σ (- (Ti)))
9 D
Ct
71m b
bl+1
11,
by + on |
E Em
(T q- 20
num
b + 9uh Ipa ll
Σ yy
C1.t
((T
ngm
1 + 9m 12
En
(TT3.
1,
200-2
Em,=,(
1,
(Ti.1) 1b8
Next, we analyze the convergence and divergence of the infinite series:
+00
Mi
((T).)) Ey,,= (-
1)68
1
9Th ||A
Enn 1,5
t=
m1m
+00
9 11
(T:.))
M
71
1,2
$1.=
9-2 2
bet1
-1) 1).
M
by
where the inequality is by e is increasing sequence, and Em. decaying tail. Since the infinite product I1: diverges: by
(
19-2 1,
M-1 because El has exponentially
hil
4000
b++1
b box to0
b:
bn
t=0
according to lemma G.2, the infinite series
4:00 Σ
b- -1) 1) diverges. Thus, we can conclude that the infi-
t==
1 en1
Σ Em,,=, (T Xo 2 1)69
1+ 0n1 19iler
13
Σ
vi. =y (- (( ((T 208-2
nite series Σ D
mim
diverges. Notice that
mim
1+ amilial
Σn, En,,=,( 2)
14 goaling
12
Σ
v1 = ( 1171 1)
mmm
mjm
mym
391 1 1
E
Σn En,,= ( (-477) 11--0111122
MT7 (2) ((9 2 1)b8
1, so the infinite product
t00 G++1 de
11
toc
291 Mil mym
9%, Tell
Σ
1
Σ Emn 1
11 = ( --
(- 2, (1100)80-2
27 (1) 11 200
36

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
diverges. Therefore,
C++1
C+00
CD
t00
1=0 Ct
so lim be lim, lim + too
O, that is to say (w (wi)
jum) = ol (w (Tr.t) jon ((H1))
(T1.) In this way we get the relationship between the maximum inner product (w jrr jpp) and other inner products, that is, (Ti.1)
(w ji .jp)) = ((w j. ,,n)),, + r". Before the next proof begins, we need to first prove the following Lemma: (Ti.))
Lemma G.4. For j E {=1}, it holds with probability at least å that
711
R1
2 log
n1.J
2
8
where mij = ((, j, [m]]H.
Proof of Lemma G.4. Note that nıj E T(y i] where yı takes label +1 -1 with equal probability 1.
according to Hoeffding's inequality, have
n2
Σ [yi, = 3 21) S 2exp (- n
21
1y1, = 5
E
i (+1),
and it follows that
2t
IP
11.
21 1)
2exp
. E {+1},
11
leading to
2
M1
<
log
2
with probability at least - 8.
Then we have
n1.j
2
2
1
log
n2/2
n1
11
leading to
11
M1.
n1.
1
2
n1/2
R1
1
1 θ
2
1I [1 + o(1)],
2
where the last equation follows by n1 = ((polylog( (d). Similarly, under Condition 3.1, we have
n2 n2. + o(1) [[1)]
(G.4)
for j (+1). Next, will prove that the maximum value of the inner product of the positive and negative convolutional filters is approximately equal, that is,
(T.O (11)
Lemma G.5. When i tends to infinity, lim
1.
(w
1,y= pi)
37

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
proof of lemma G.5. Using the lemma G.3, We can simplify the equation (G.1):
d d'(w (T2,))
Th T1 l2
1r" -11))
(w) 1,r
(1,1+11 _111 = (w) (T1.4) 1, (41 +
Σ
mm
+ expl o (( 1,r" (1))(1 + o()]
(li,))
01,==1
p4n + 7h M = (w (wl.r (T1.0)
113
n1
[1 + o(1)]
o' d'(ww(, i, (1,))
mm
2
+ exp( m o ((W], (Ti.1)
mn))( + +()
(Wi.r p1)) + I 112
o' (( (Ti.1) (22)
(T).))
(11)
(G.5)
2m
1 exp( o ((w (Ti,t) pli)
(wi."""
we can also simplify the equation (G.2)
(w (7).4+1)
-1,r" -11) = (w (T).d)
Mi
Σ
-1,r" - (11)
o' (1.0) (2..0)
-1.r -11)
71777
+ expl(+) -1,- -I, -11
- o()]]
(Ti,))
-1.
d d((wTTi.t)) -1, -11)
(Ti.t)
(w (Ti.t)
11, -11) +
mi
n1
[1 + o(1)]
7m
2
1 + +xpl (1 In ( (Ti.0)
-1, -11 -11 + +()]]
mil mllaali
d'( -1.", (Ti.t)
-1))
(w -1r" p1)
(Ti.t)
o())
(G.6)
2m
1 1 exp( m (( (Irr -11)
-1,r"
We have the function f(x) 1+exp (,, 24) further we get derivative of this function f'(z) =
---1
(1+epp ( 29)22 (Td +(
x x²--2
1) exp 1 x)). When x 2 (2 max/m, q})= à we have that m
1
1
1
1)
exp
- I9)
399
9
9
1)(
x9)
17
9
9
1
1 x
x
171
m
4
1
2
11
171
9
9
29
2m
< 0,
where the first inequality due to the fact that et > x. for x: > 0, and the last inequality follows by the condition that t 2 (2 max(m, /.) So when x 2 (2 max(m, q} 9})=, f'(x) < O, f(x decreases monotonically. By Lemma G.I, we (Ti.)) (w. .jpz) will reach (2 max(m, g))², Suppose that at some point,
know that at some point in the iteration max, j.r
(w (T--0)
(W (T, + C, where C = o(1) is positive constant, then we subtract the two equations (G.5) and (G.6), we
have that
(w (T.e+11 M1) - (w
(T.,++1)
-11 = (wf.r- (Ti.t)
11) - (w (Ti.)) -1,-- (wr. -11
-1.-"
|| 13
d'((w ((W (Ti,1) 111))
17 A1lIH1113
o' (((w -1,1o -11)) + o()
((
2m
1 + exp( 1 ( ((w (1,
(wir"" (1)) (T)
2m
+ expl 1 o ((w (w-,,re, -m1))
(Ti,))
< (w i,rr" m1)
(Ti.))
(w (w-1,r-1 -11)
C,
where the equation is due to the fact that f(x)
T ||l
(m)
is monotonically decreasing function when
2m
T+exp( - (x))
38

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
2 (2 max g}) So when tends to infinity, (w(fp="") 141) (w-I,rs;-p -111 S C, we have that
(Tr.))
(wirr-) ( pe) lim (w (w-,, -41) lim
(Ti.6)
(wi,r- Hb1 - (w (25.0)
-,r -11)
lim 1-++00
1,
(w (W-1,r" p1)
(Ti.)
1-+000
(w (7i.))
(w-1,r"> -11)
1-+000
(w -1," -M11
(71.))
which completes the proof.
H. Proof of Replay-based Methods
The replay method here refers to combining the partial data of task 7 and all data of task T3 for model training. The
(0) initial weight of the model is the weight value retained at the end of the previous task: wj.r W.r (fk--s)enmd) Let consider the case of two tasks, task Ti and task T2. The signal vectors for the two tasks are μ4 and μ2, and the noise vector are f1, and E2.s. The empirical replay method is mainly to solve the forgetting situation where the angle between two tasks is obtuse, so we might as well make the angle between the two signal vectors obtuse 0. To simplify the mathematical form, we let represent the - cos 1, expressed in terms of the formula À = cos 01,2. So the inner product of μe and μ2z is
(1i, 12) = =1 la U2ll2: A (0.
Then w jr
W) (Tidens) j.r at this time, the loss function is defined as
n2 Σ
Ls.us, (W)
Cba f(W, +
eluz, f(W, x2,)))
ni + n22
Moreover,
we
define a number k,
which is expressed by the formula k l| ll
min
I-Ca I+C2 8(C++))
(3+02))+022)
(3+C2)(1-C2)
where the C1 and C2 the same as that in Lemma F.19.
(s((1aa))) " (1+02)
The number l is used to describe the growth of w) two below inequality holds:
j,r on the vector μ2z Then we can verify that when 0 < A < 1+C2 the
(3.4)
1l 2 2)8
(1 C2 jays
+
C2
(1
23)5 (k
N (1
y==
2δ
|
2
|
2
1 C2
||
ll2
1
+
C2
|
ll2
1 C2 y--1
2 (1
(
(k
2
|
la
2
|2
2
O(),
(H.1)
where the first inequality is by the decreasing nature of the function A(A) = -1- (k: ( A)9, the second inequality due to the Lagrange Median Theorem, and the last inequality follows by the definition kt and we denote the positive constant by 2C, and we can obtain
Č 2 4k.
(H.2)
where the inequality by the definition of k.
Lemma H.l. The gradient of loss function Ls.us, (W(Tz.)) with respect to weight parameters (Ta.9) w is
V (2 Ls, w Li,ss (W(Ta.0)
(TTtt)) o' o'((w ((w:T2.)) EEI,)) j1,,
(T2,t)
(nī + m2)m
n2
(Tzzt)) o ((w E2,)) E2, 2,
(Tb,t)
(nf + 12g)m
i=1
n²
+ (nf + n2 )
(T2,t))
1,5
o'((w'))))))
j.r (1,, jpm
(T211)
i=1
39

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Understanding the Forgetting of (Replay-base) based) Continual Learning via Feature Learning: Angle Matters
11
(TI.))) o ((WW) (T2.)) (225l1a)) jp2,
(nf + m2
2,3
for j =1,1 I r 2 m, where f' x1,))) (f22.f(Wit) .2,))) 1 exp [f f (W(1), x2,))l) denoted by 2,5
1/((1 exp (/1.. (W())x))))) is denoted by C and
proof of Lemma H.I. When j == r m, w eo /WW(200) x1,1) (vi,s (WV(2.)) x1,)))) Vi V w. x1.1)
y1 (j of ((w jr (n.,1)))i + o' d' (22.0) 1,1
2(22.00
((w (1,)))
12 f(W(Ta)), x2,1) (12, (WV(a.)) x2,1 y2. V (2.0)) f( (T2.0) x2.1)
(Ta.6) 2, y2.5 (j of ((w) jr (2. 122 + (( j.r (Ta.1) E2.))
(172.1)
Note that Vw, Ww uss (W)
ni+nng wy, Ayr. f(W. x1,0 + Eie V Vwy,. Cluz. - f(W x2,)), and bringing
the results of our caleulations into this equation, we can get that
V (T Lius (W(Tait)
(TTz)t) 0' d'((w], j,r .I.))) S1.
(T2.0)
(nī + m2)
1,
1=1
((T2.))) . o' o'((w ((w(T (Ta.)) E2,i)) j22.
+
(ni + 122)
==1
1,1 (Tz.))) ((w (T2.t) ,)1,)11)) jmm
(nf + (22)
n2
((T2210) of (lw (( (Tai)) (22 jpz.
2,5
(H.3)
(nf + m2)m
which completes the proof.
When the model is trained by gradient descent, the update rule can be formulated by
(Ta,+1)
(T2.t) w j.r
122 Vws., Lsjus, (W(Ta:)),
(H.4)
for j = +1 and r E [m]] In the empirical replay method, we place equal emphasis on the laws of change in the inner (Ta.1) .jmz) and (w).r (Ta.4) .jmz) represent feature learning, while (w),r (Ty.4) E1,s) and (w (Ta.t)
products. Specifically, (w
E2,)
pertain to noise memorization. And then we have following lemma for the inner product update rule.
Lemma H.2. The performance of gradient descent with respect to feature learning and noise memorization can be formulated by
ni
(w (22.)) jor .jmi) (w jrr
(T2.0)
. +
172
((T2.)) o ((W
(22.)) -),
(n] + m2) ==0
A |lle|lllelll (ni + n2)m
t-1 1
(T22.))
0 o'(ww!!f (( -222(12))
(T210)
(H.5)
2,
Aup2 le 1|21|1222211
2|
na
(T24) (T.t). (umz) = (w)) (w]72:00 for
(T2.0) .ipz)
((T21t))
o'(WW v...11))
(T2.0)
(nj + n2)m
1.8
40

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
1/2 1221 +
1-1
(T22))) 2,
o'((W),, d((w]7? (22.H12))
(T2:s)
(H.6)
(ni + n2 m
== ==1
(w (2.1)
(T2.0) E1,)) (Wrr
E1,2
2
Σ
((T22))) 0' d'((w,,
((W]T (T2.0) ,T jyn,
1,8
m2 Jm
a 2m
1-1 m2
1/2
2,
d'((w)
(T2,s) ((w
E2,,)) j32, E2,);
(H.7)
(nj + n2 n2)m
3rr
t- 1 n2j
(w (Tait) Jin
(2.1) = (w jrr E2,x) + (nī + 12 m
(T2.0)
172
((220))) o (( (T2.3) (i,))) i
am 0
+ 1/2 E2,
(T2)))) of ((
(22is)
.2,x)) jU2.i.
(H.8)
(nj + n2 m
2,5
==0 i=1
proof of Lemma H.2. From equation (H.4) and lemma H.1, we can get that
(T.)+1)
(Tg.t)
(2 LS (W(T2.))
W
jir
1/2
ni
(22.t)
722
(T2,t)) o' o'((w) (T2.)) E1,))) 211
jr
(ni + n2 m
1.5
fgg
1/2
(T2.))
a'((w (Ta.t) E2,i)) f2,
(n] + 02
2,1
172
(Tz.t) - (((W (1. jul 1,1
(22.0)
(ni + (12 mm
7/2
(T22)) 2,
- o'(WW]r
(T2.t) j.r v2. ju2.
(H.9)
(ni + ri2 mm
1=1
Adding up the left side of the equation (H.9) gives us the following equation:
(22.6)
(T2.0) w j.r
722
1, ((l))))) α d'l(w] j.r .I.i)) ,..i)) $1,
(T2,0)
jr
(nj + n2) m2)m m
n2
172
(T2.*) E2.i)) f22 ((T21))) α d'((W].r 2,1
[n] + m2)
0=0 im1
1/2
((T2,))) 0' o'((w (Tx0)) -.1.44a)) jun 1,4
(nf + n2g m
I-1
n2
772
((22.)) 0 d'((w]
(T2.2) jrr (22,H22)) juz
(H.
(nj + 122 mm
2,
By doing a vector inner product on each side of the equation, we can get the Lemma H.2. This completes the proof.
We define the coefficients ((11) (12) p(pr) p(12)) jurd as follows, which satisfy the following iterative equations:
(T2.))
(T2.e)
(T2.0)
(T2,0)
(111) , (142) (T22 (((()) (T21) _((12) (Ta.0)
(T2.0)
(T2.0)
41

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
{(1))/r (22.141) ->(11)
(22.0)
+
(nī + m
72
((12.1)) (22.1)
1,
(' " (),a))))) |1
n2
((12))/ (x(22 )!4."))
1/2
((22)0)
2, ) a' ((w,, (T2.1)
+
(ni + n2 m
((n))tr((tt)) p(ue)) (Tree))
712
(T22)) ((w)) jo
(T2.)) (..))
(ni + n2 )
1/2
(T2
)o' ((
(T2.4) le..
2,1
(nj + n2 m
Lemma H.3. By the definition of coefficients ((1))/r (T2.1) the equation (H.10)
(22.0) ,7 (12)
p (2.00) p(pz)
(ra as above, We can simplify
jrr
(2.t)
(T2.0)
+ j ((1)) (T2.t) 111m 2 pm + i (122) (72.1) || 122
jr
W jr
na
p(11)
(T2.1)
|1,all E1,c + Σ pluz)
(T2.1) 1
(H.11)
proof of Lemma H.3. According to the equation, we can easily caleulate that
t-1
ni
((11)/220) = =((11) (12.0)
n22
(
((22))) . o' o'(w or
(( (1. ||
(Ta,0)
(nī +12) m
t- 1
{(12) 7(122) ((20)) (22.0)
1/2
(T2) 0' o'(lw 2,
(2b,2) (22,paa) ||
(ni + 122 )
sm0 0
i=1
p(mi))res) (T2.2) p(mx)) (1rrs 177210)
T22
((T2)+) 1,
Yo' ((w (w/!7-.)) (T210)
+
[ni n2 12
p(u2))TTat) (T2.))
1/2
(T22:)) 2,
(o')))''''non (w!.Tx..) (T2.>) E2,,)) 1B.v2...
(nj + m2)
Bringing these equations into equation (0. completes the proof. Through the action of these four coefficients, we have succeeded in simplifying the equation 10). Further, the inner product update rule can also be simplified, they can be rewritten as
1 ll
(T2.0) (w jrr .jpz) = -w . + (11) (T2.0)
(T2.0)
(1)))
(22))
((22)
(T2.1)
(H.12)
122ll2
122ll2
(w j.r
(T..). .jpz)) = (w jrr
(T2.0)
.juu) + (1122 (T3.0)
((11) (12r))/07) (T2,1)
(H.13)
|1ll2
(T.)) E1,x) = (w/TT..) (2.00) E1,) + p( (T2.))
Σ p((11) )
(Tap)) 11
1 + Σ ((12)) (Ta.1) |62, 2 E1.))
E2.1) = (w (Tz.0) E2x) + p(pz) (T2.) (w jr
(Tait)
Σ p p(
(T2,
1|.. (E,, E2.) + Σ p(1))
Saa | 11 2 (E2,
When performing training for task Tı, let's assume that we plan to stop training when the training loss reaches e. Then at the end of the task Ti training, we have that
Ls, (W(Tidene)
1 Σ E(vi. (W (W(Ti dame)) 21,))
=
n1
42

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
1
log(1 + o(w (eddd) "i. (1.01))12
n1
c.
(H.14)
Co
(Tid))))) where Co is a positive constant and Co 1. Suppose the maximum value (.rr jom) has reached Amaz, then get
log(1 + - (
E
Co
Amaz Im log(
1 e
1
+ [1)) loxx
{
o()],
(H.15)
where the last equation is due to the fact that e" - = e when converges to 0.
Next we will give a lemma, which will be used in the proof that follows.
Lemma H.4. Under Condition 3.1, if (())) (T.t) > م
Tell ((2)
(T2.t) when tends infinity, for r E I5,2. we have
max, ?(u2))? (T2.1) + for E {+1].
((u)) o(?(p2) where argmax (122) (Tg,t) r + r and j {+1).
( (T20)) (122) -1,0 + o(1)]
(T2.0)
Proof of Lemma H.4. Recall the update rule of 7(12) jir we can get
(T2.t)
((22) (T2,t+1)
(T2.6) + 7(22) (T2.))
1/2
Σ
((2.)) o' d'(ww)
(T2.0) j.r .im2) + (1 (2)( ([2.))) 1|221
2,
(nj + 12)m
#2,==
Similar to the analysis of the max(w (T2.6) when tends to infinity
frr jum) in the previous section G, using the same proof method, we can prove that
max, (1x +00 for (+1).
(1z) (T2.0) (12) where r" argmax (12)) (T2,t) r r" and {=1}.
((22)(7 (T2.t) (12)) (t) 1 [1 + o(1).
Here the proof completes.
Proposition H.5. Under Condition f.I. 0 < t Tž, we have that
(41) (122)
(T2.t) (T2,)) S 02
(H.16)
for all j {+1} and E [m]) and
0 < (p(1)) ( 17
(Ta.9) (e( S a2
(H.17)
for all e {=1}, E [m], i [nt] and d' E [nz].
We will use induction to prove Proposition H.5. We will elaborate on the following theorems, which will be used to prove Proposition H.5.
43

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Lemma H.6. For any 2 we have that
log( /0)
log (annn///))
(w)r (221)) (wrrr " E1,) + AniV
E1,z) E1,a)
(T200)
(22
4nz
pn
022
d
ps
log( (og(Ang/0)
log (g(AAng/0 /6)
(w)/ E2,2) (w/rr (T2.0) E2,2) + 4nj
p2
C2 4m21
02
pp1
for any r E [m] and y1.3, V2,1 = j.
Proof of Lemma H.6. Recall the update rule for (w (T2.t)
Ei,a)=
jir
12
(w/TT.)) .E,,)) (E1,)) =
(Ta.0) E1,) E..)) + (Taa)
Σ
p(()))
(or 1 (E,,
Er,)) + > p( (Ta
(Ta.))
2 (E2,)
so we have
(w (T2.)) Jr
E1,1) S (w Jir ,1,s) + L p(H)) (T2.t)
(T2.0)
| 11 E1,) + Σ
p
(T2.t)
11,2,;112"(E2,: E1.
8'zs
log (og(4ng/0) /5)
log(4n5/0)
n2
.1,s) (w/fr" (T2.0) (T..0),,11.)) + +4V
p(11) (T2it)
pm
p(
(T2.)) j.r.
pp
S ((w! jr S1,) + 4n²y
log( (n4//5)
a2 + 4nz
pp
1og(4n3/6)
0.
d
0p
where the first inequality is by that p( j.r. last inequality is due to our induction hypothesis. Similarly, we can prove that
()
< 0 when V/i,i = -j, the second inequality follows by Lemma C. and the
(w (Ta.t) Jar
E2.1) S (w
(Tx.0) fr E2,s) + 4n D2
10g(4n3/8)
log log(4n3/0 (4n3/8) a2 + 4n2y
Q2
pp1
d
d
which completes the proof.
Lemma H.7. Under Condition 3.1, 0 E tt3 T2, have
(1 /2llla a)",
(W (Ta.t) x2,) S 12+1(2
H1ll2
for j -/2, where the constant defined in Cao et al. (2022)
Proof of Lemma H.7. Firstly, we know that ((1)) jr
(end) S a, which is proved in Cao et al. (2022) So when /2,1 = -j.
we calculate (w (T2.0)
jr -ju2) as follows:
(2.0) jr -juz) (w (T.0), (T.0))
jjpz) + ||12|122 (Tatena)) 2X
|| l1
(11))
|12||2
|1112 a,
(H.18)
where the inequality is due to Lemma C.2 that (w (T.0)
juzz) / log(8mm0) ao 0ol|Mel12 S A l|aall2 a. We also have
pilla
(w)) (T2.0) fr
E2.1) = -(w!? (T.0) E2i) + Σ Pand
mm E1,,ill" (E1. E2,1)
Mog(4nf/0)
I 221oog(8mm Toop Vd + 4n
T
pp
d
44

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
21
|alla a,
(H.19)
where the second inequality is by Lemma C.I and Lemma C.2, the third inequality is due to Condition 3.1 that do (mm-(B/to2))vV 11//o-20lV1mmmx/mm max(n1, m2) 11111(1-2111)))) - min (o vd)= (id)=",llee ,1lm l2 (Opp vd)= |2l2) and d 2
max(ni, (11, n2) )2) (((((22)2((1))))) Then we can get
m
(W x2,)
m Σ
(a(
((
(2.1) 1.r
-jpu)) + ((((()) (2.4) E2,))
|og 1og(4n8/0)
1og(4n3/0)
S 2
max Jon
(( jr
(ww - kk (T2.0)
E2,)),, 4n
An
0pi V
02 4m2 V
a2
d
122l2 |1ll2
a)";
where the first inequality is due to (H.18) and (H.19), and the last inequality is by Condition 3.1, which completes the proof.
Then we will prove Proposition H.5.
Proof of Proposition H.S. When U2 j, denote fo,, the last time that p(112) (T2.8) 0.502, and recall the update rule for (T2.) ((1
have that
p((22)
(12,T)
S p((22) j,r3
(12d2
+
(nī + m2)m
172
(- (23.e) ) -o' d'wW,, (2,,,1
2,7
122
(- (-172100) o' d'((W for
(T2,t)
(H.20)
(n] + n2 n2)m
for any T s T2. Then we will give I an upper bound:
72l E2.i 11 d (I(w (22.0) E2,1)l + p((22) )(1e)) (Taitana)) + An Tp2
log (gg(4n7/0)
log 4n2/5) o2))
I 5
o2 4n2y
(ni + 112) m22mm
0m
o p2 9-1
(ni + 122mm
< 0.25az
(Tz.0) where the second inequality is due to that (( j.r E2.1) S 0.102, p(u2) ( (Tadsins))
log(An? (6) 02 +
< 0.5az and 4n
dp
d
4 /0)(425/8)
02 S 0.10g, and the last inequality follows by Condition 3.1 that n72 S
(m2+m 8qm do?
When the time t
reaches 1.,. we can lower bound (w!2 (T2.0)
E2,1) as follows,
log (og(4n]/0)
log (og(4//0))
(w (T2.t) jrr E2,) 2 p(112) j.r.
Kw
(22.0) E2,13l -4n
T2 pp d
4m2
02
0.5ag 0.10g 0.102
0.25oz:
where the second inequality is due to p(uz) (Tads.ra)) jes
> 0.50g, (w (T2.0)
E2,1)l < 0. and 4n
Ooa mm
log(An3 (8)
02
+
4m2y 1og(An2/8) d az S 0.102. Then we will upper bound (w (w jrr E2,) in the same way,
(Taat)
log (og(4n//)) + 4n2
log log(4n3/0)
(w) is
(22.)) E2,1) S (1
(T2.t) + (w/T2.0) .E2s) + 4n
(Tz.0)
r
d
d
45

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
02 0.1org + 0.1a2
202,
where the second incquality is by the induction hypothesis (.s S 02 and (W)r (w)rr.)) .2,x), 4nj S2a
(Tz.0)
log( (n(4
pp
d
a2
+
4m2y V
1og(4n2//)) (8)
d
02
I 0.1.2. Then we can plug the lower bound and upper bound of (w), Jrr E2,2) into 12:
(T.1t)
12 S
Σ
722
expl- a(w 3rr E2,i))
(T2.t)
22
(A
| la a))} of (2nx)
(ni +
12)m
|
24 29 d
M22 a)9
T2 (nj + n2 m 2olla///4 + 22
|1ll l2
< 0.257 exp(-1i /2 Jaz
< 0.25T² expl (log T2 J002
< 0.250z,
where the second inequality is by Lemma C.I, the third inequality follows by the definition of a2 = max | + 1)a, 6 log(TT) } and Condition 3.1 on T2, the fourth inequality due 02 2 6log(T)) and the last inequality follows by the fact that (log TV ya > log T2. When 42 = =j. we can use the similar method in Cao et al. (2022) to prove that
||
p(pu)) (T2.t) 2 agg, so as to ((11),,) Proposition H.5.
(T2.)) ((22) (T2.t) and ( jind (7( for all j {+1}. Here we complete the proof of
H.1. Noise Analysis
(T."), .e..). At the end of In this subsection, we will give an upper bound for the noise inner product (w/ E1,x) and (w)) the task Ti training, we get an upper bound for the max,,r, 1p (Tx
max (0(ux) (T2.0) (1an) 3.322
S apam Vd,
for all j E (=1),r im [nx] and ls (1,2). We have similar conclusion when training the experience replay method.
Lemma H.8. There exists a time T 22 2 max bd such that
30 log (8m(11) //5)/
(T.1) (w jr (1.) (Tat) E2,iil S 6 (8m(n1 n2)/8) ouop Va
for any j E +l,r E [m], i e [mi] and S TT
proof of Lemma H.8. This lemma requires mathematical induction prove. When the task Ti training ends, we have
E1,ill < (( E1.i + max (Tidema) 1( jr
(12.0)
(2..0)
Pija
+
Σ
P
dend))
-11 (E1.11 E1.>
J.r,8
log 4mf/8)
(( (1.0)
E1,i
max jor.
dend)))) pp
+ 4nny
max dand))
d
5.r.8
1og(4n2/5)
< 2 log (8mm1/5)- ooop Vd + ooop, vd 4m1y
000 Vd
d
< log (8mm1/5) ooop, vd.
(H.21)
where the first inequality is by the absolute value inequality, the second inequality is due to Lemma C.I and Lemma C.2, the third inequality follows by maxari 1p).r: (fi.fendd) I 000p, vd proved in Cao et al. (2022), and the last inequality is by the Condition 3.1 that 2 1024 log(4 max /8)(8 max (ni nz) max "e 2
46

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Recall the rules for updating the inner product of the noise vector for the empirical replay method
R2
(w) (a.)) jir E1,4) = (w]r jo
(2.00) E1,1)
p(pr)
(Ta.)) 1eapPI1S., E1. 21, (S1,0, E1.) +
p(pe)
1112,, 2 E1.)) jand
From the absolute inequality we can get
(( (Tat)) S kw
(Ta.t) jr E1,a
(T2.0) E1,i
+ lo(ua)) (
( Th 11 E1,1) + Σ (((p))))r (Ta.t) 1112,1112 (H.22)
Then we can give (((1))))))) (72,t)
an upper bound. Firstly we have that
(0(11)
11)
1 (
(T2.t)
7/2
(- (0
jirss
j.r.
[n] + 12)m
I ( (Ta+t)
729
((w
(T.t),
(n) + 122)m
Jer
312q0 2 d
s ( (43.)))))
pt
(1(w E1,e)> + |p(1)) (
(22.0)
2(nj + n2)
(p(p))
(22.()
||111.111 ( E1
l
|1..2.112 " (12.. (22,4) jar
jast
37290 d
s $ip(H1) (T2.1)
[4 /log (8m max(ni, 0o0p. Vd
+ 2(ni + m22m
+ p00 Vd + 4m1
log (n) d 31240² d p1
poopi Vd + 4nz
dpi
log 1og(4n3/0) 0))
000 vdr-
d
pz
< (p(()) (43
(1
2(n + m2)m [5 log (8m(n) m2)/0) OoOp, Vd
where the third inequality is by (H.22) and Lemma C. and the fourth inequality follows by the induction hypothesis that (Ta,)) S pa vdl and Lemma C.1 and Lemma C.2, and the last inequality is due to Condition 3.1 on d, specifically
d 2 1024 m4x mf. 22)/8)a3 max(n];ne n²} max o
When Tit we have
372402 d p.
(112)) s (
(T2.0-1)
[5 [5 (8m(n1+ n2)/6) oop,, Vd!-
2(n + n2
31240 d
S (e(w))
jard j.rc
(T2.0)
[5 log (8m(n) + n2)/8) ooop vd1-1
2(n + n2)
31240 p1 d
< 5t+t.
[5 log (8m(n1 + n2)/8) p vap-1
2(n + m2)m
3n2q0 p1 d
< T
[ log (8m(n1 + n2)/8) 0o0p, vd
2(n + n2)
< T p. Vd,
where the last inequality is by that t < T,t. and the last inequality is due to the definition of 7
Based on the above analysis, we can give gor E1.a) an upper bound
(w (2.4) E1,1 S 5 (8m(n1 + n2)/8) Jpap Vd.
(H.23)
47

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
In the same way, we can also give (w], (w]r (T2.4)
-2,x) an upper bound. We have
(22.0) (w j.r E2,aa1 S ((
(2) 0)
E2,1)1+ +
Pjan P
dond))
|- 11E1,:11 (S1,
jor
log (og(An9/8)
(7).0) (w jr
E2,1) +
4n
P2
max a0n
|P
(T's (end)))))
pp
d
log (o44n]//)
2/ log (8mm2/6)
p0 Vd + 4n1
000p, Vd
0p1
d
s 4Vlog (8m max(ni, / J00 Vd.
where the first inequality is by the absolute value inequality, the second inequality is due to Lemma C.I and Lemma C.2, the third inequality follows by max lpg.r: ( S pp0p1 va proved in Cao et al. (2022), and the last inequality is by the Condition 3.1 that d 2 1024 log(4 max nz) /6) max n²) max
Te 2... Further we have p2 m1
jir (2.1 + 1 M {2,1) S kw (Tz.0) kw (Ta.t) fr
ni
I0(pn) 1f 2 E2,i1++ + Ele(pe) (Ta.))
(Tr 11162, E2,)),
(H.24)
which can be proved by the update rule of (w]fr (Ta.t)
E2.1)
Then we can get
(0( (T2,+1)
s (p( (T2.t)
722
(- 1-12,
((T2.))) ((w (w/TT.
(T2.0)
(2.:))
(nī + m2)m
W (0(p2)) (To.t)
129
((w)
(T.t)
(Ta.t)
(mi + m2)m
2 d 3129072
s (p((2)) (Ta
pz
(I(w/T
(Tb.0)
E2,x>l + (o(
2(n + m2)
|
(T2.t)
||| 11 11,, E2.1 + Σ
(p(
(T2it) j.rt
|11Ea, (2.)))-))))))-
31290 d +
S 1p( (T2.t)
p2
[4 Vlog (8m max(ni, ne) 180) 00 Vd
2(n + n2 m2)
log 1og(4n2/0)
log ( Jo0p, Vdr-
+ p00 Vd + An
P2
poo Vd + 4nl
pp
d
d
3n2go
$ <Ip(pr) (12
/lo (8m(n1 + n2)/5) Tu Vd--1,
2(n + n2mm
where the third inequality is by (H.24) and Lemma C.I, and the fourth inequality follows by the induction hypothesis that p(p)) (Ta.) S Soopp vid and Lemma C.I and Lemma C.2, and the last inequality is due to Condition 3.1 on d. Lastly, we have
31240 2
l0(mz) (T.0)) < (p(p2))
j.r. (Ta
(T2.-11)
+
P2
[5 [5 (8m(n1 + n2)/5) p2 Vdla-1
2(n n2 m
3n290
2
< (0(w)) (73 (T..0)
p2
[5 log (8m(n) + n2)/8) 00 vdrai
2(n + 122m
3m2q0
pa d
50+1
55llg (8m(n1 + n2)/6) ou0 vd
2(n + mm)
311240 p-
2 d
< T 2(n + m2)m
[5 log (8m(ni + n2)/8) 000 vd
48

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
po Vd,
where the last inequality is by that & I Tit, and the last inequality is due to the definition of 7/. Then we can get that
(72.4) (( jr E2,,i) S 5yllog (8m(n1 + n2)/6) puT Vd.
(H.25)
Here completes the proof.
We have obtained the upper bounds of (w), (72.0) E1,z) and (w (T2.0) E2,1), which we can mark Cg. The formula to express this meaning is
Cg = log (8m(n1 + n2)/6) do : Jv.
Recall the definition in the subsection F.I, we categorize the neurons w (Ti.0) into three sets: J = fr E [m] jir
(w (T1.0)
jrr jum) > 0). 11 = fr E [m]] (w (T..0)
jr .im) S 0}, J22 = fr E [m] : (w (Ti.0)
jo jim) S 03 n E [m]]
(w (21.0) j > 0} and f,,3 = fr e [m] (w (21.0) .jm) < 50 n (r E [m] (w (1.00 jjrtt x 0} for j e (+1). Before jir
jr
doing the next analysis of the changes the following three neurons, we first want to show that at within a period time
from the start of training for task T2, the value of (Taii)) is very small. We have that
((2.))
(724 (..))))) 1/((1 + (a((W (2. jm jin)) (.,))))) 2o E. (a( (aa)) sim) ta (W-
< 1/((1 + e E. a(( (72 jun E, a(w (2ng jmm)) 202)
$ 1/(1 + 1 [11-A9pe | AA mu. 203)
< 1/(1 + (1 ( A) log log(1/e)
(H.26)
for j = the equation is the specific formula of - (2..)) the first inequality is by the upper bound of (w (Ta.4)
jur E1,i)
and (w (T2.1) E2.i) proved before, the third inequality follows by the definition of Amaz, and the last inequality is by H. The proof of upper bounds for derivatives is based primarily on induction and will be addressed in subsequent proofs for correctness and specific details.
H.2. The First Type of Neuron
The first class of neurons is those belonging to the set 4,,1, which is expressed by the formula as follows:
(w .jpz) > O.
(T..))
By the previous analysis, we can get that
(w) (Tz.0) .jmz) > 0, (w j.Hz) < O.
(T2.0)
Then, we will first study the maximum value of the inner product max, (Wi,r (Tidend) jimi) of this set, which is denoted as dona)) jim).. Then we can get the value of the inner product (w j.rr"" (w jis
(Tidend)
.jM2).
Lemma H.9.
(w (Tiitona) jrr
.juz) -XAmax [1+ o(1)],
jor E {+1).
proof of Lemma H.9, We know that for the maximum inner product,
(w (end)))) jor
.jpmi) = max (w (Tidend) jor jpm) = Amaz > 0,
(H.27)
49

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters (Ti.9)
And as for task T1, recall that (w)
(w)r jir :jjm1) = (W)r (T1.0) (jm) + 7(1)). (Ti.0)
so we have
>(11) ((41) (end)))) = mmxx(w).r (and))
j111) max (T..0) .im)) jr
J.r
mmx((w (T (ond)))) jum)[l max, (w (T.0)
.141)
6.r
max, (w (Ths (end))))
.jm1
Amarl [ + o(1)]
(T:,0) where the last equation is by that max, (w),, jir jjp is definite constant.
(Tn)) Then, we can do a calculation of the value of the inner product (w jor
.ju2):
(Tid) (W) jrr
_112) max max(w)) (7)
( (Ti.fend) .1122
max((w) j,r .مmz) +1(11)
(Ti (Tistond) |al1 Fei, u2)
3r
jom)) || 2 (Tistene)
m 5.r
(T.0) (7)
||12||2 (112) jor
=- || 1l ((12))
max, (w jrr .jzz)
(Ti.0)
(Ti.dana) [1 orr
|| 1
l 12 2 ) dend)
M la
|| 12
Amarl [1 + o(1)]
| ||
this completes the proof.
Lemma H.10. There exists time
T m (nj + n2 0/2
(H.28)
34 p12 112
such that
(w/TT (Tai)) .jm) .
(1 )²) + +1)]
(w] (Tit)) .juz) = =(().
for j {+1} and 0 Sts Tz¹
proof of Lemma H.10. In this case, the update method of the (41) (330)) (122) is follows
(221)
((1) (H1) (T2++1)
7(41) (Ta.))
1/2
Σ
(-1(T2)0)) o' a'Amez + ((1)) (Ta.tt)
M1 l|2
(122) (T2.t)) 11
(ni + 722m
[1zl12
V1,==j
(122) (12,411 = 7(42) ) (t)
T22 (nj m22
Σ
(-
((22))) a'(2 H2lle
Amaz (1 + o(1)] - (122)) (2.1) + X H2lle ( (11) ((3.))) 11
2,
|1ll22
|1112
Firstly, we will use induction to show that when S To. (11)) (Tz.6)
has a very little change
(141)
() m 2
50

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
By the definition, we know that (( (T.00) 0. According to the inductive hypothesis (H.26), we can get an upper bound for 1, (Ta.t) Thus, plugging the upper bound of -4,, ((aa.)) into r(p1) jr
(T2,++1) update rule gives
(T,.+11) ((11) 1r
= -(11)),r (22.4)
1/2 14 112
((Ta))) jo' ((w)
(T:.0)
jpm) + ((22)(()))))) (T2,1)
> | ll (1ez) (112)) (
[ni + 12 m yー=j
1,8
|2222
3mp uu [gni C (Amax + ((1))) ((11) (0)
s ((11) ((111) ((2.11)
4(n] + 122 722)m
3m221/m 11//11119m0 l:gmi Ag-1 max
S $7(11) (T7..)
2(n + 1222 m22m
3rp u1 Ban
S { (7a.))
(nī + m/ log(1/e)1-1/9,
(H.29)
where the first inequality is by the upper bound of - 1, (Ta.t) and the last inequality is by the induction. For any 0 s / s TT.
taking a telescoping sum over = 0, ...,, T 1 then gives
(T2 I {7(11) j.r {11)
(T2.0)
1= (nj + n2 m2)m (og(1/e)**
312llp [2 e
W ((11) jrr + T 3//2 | 111m e log(1/e)*-1/6
(nj (2)mm//
3/21114 18gm 6/2
I T2 (nj n2) 1/4
S m 20,
where the third inequality follows by the T S T in our induction hypothesis and 6/2 log(1/c))-1/ s for 2 exp(Ce(), and the last inequality is due to the definition of Tx*
Secondly, we are going to analyze update changes of 7 (12 (T2.t)
Lemma H.11. Under Condition 3.1. we suppose 7 ( (Ta,t)
// I
Then there exists time
1-x
A(ni + n2 m log(A X T
| mAmaz)
q2o722 C% 12
(Tzdaad) ju2z for all I11 such that (w) [
Proof of Lemma H.ll. When T we have
(Tg
(7b.1) 1/(1 + m (a(l wa ( - # E.la((w!T?!)) (.."((( "0.""-v,,.mmm +ooem))) -pir (c,))))) -3,"
2,
i (( s.r" الاد) , (1
2 1/(1 + e m
1/(1 m 1 (1 | + [1))
1
C22y
2(1 +)
where the second inequality is by Lemma H.4, and we denote 2(+e) by C²2- Within the Txt, (11) (7.0))
(22.0) is no more than
m. 20, so we have
(12)) ((12))a2
(T2,4+1)
((12) (ni
(T2.0)
me || 12
Σ
(-e(Tai
d'(1 |elll
(Amar + ((21)))12000) (12)) (oa0))
m2
|1ll
=
51

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
= [
(T2.4)
T2
Σ
(-
o'(2
[22l2
Amaz [1 + o(1) - ((ue)) (?2.0) || 12222
[n] n2) m2m
2,8
11l12
22,
From this equation, we can see that ((22) (Tat)) is an increasing sequence. In order to simplify the form, order lll
I la Amaz [1 + o[1)] (u2)) (Ta.)) ae, so that
1/2
E
(-12) o' d(ro)) 1
a
[nj + n2 m
12
Giving the specific conditions for Theorem 3.3 ni 2 n2 {2(- cos 21, 1 12112111111 "2mmmm 2/9 llo (2) -1]. the
inequality (mj fm
M la 2 2(A
I la Amaz)" ) holds, we have that
n29l 22
Σ
p(0) d'te)) 11
a++1 a2 [nf 72 m
:2,1=-
IV
a: [ni + n2 m2)
122
Σ
(as))
1
-2,==1
n |2 112 (at)
IV IV
(nf + m2)m
0,
where the last inequality is due to that a, S 0g
> ||لaعla Amar I + o(1).. 1f0 S az S || Aans we can have that |paill
|
0 S ae+1 S X || Ama for that ag is a decreasing series. Since ag = > Imill
|ezlla
Amarl + o11), according to the iterative
method, we can know that ae 2 0 for t 2 0. Denoting by T the last time in [0, I that satisfying 2 1 then for t S T, we have
m
Σ
((0) 21
o'(e)) 11
a++1 az
[nī (2 =2
9man2Cy Ch 12 (ar)
< 0j
4(n; + mm2m
T22 C2llu213
S sa
4(n] m2 )m
where the first inequality is by (-111) 2 C22 and the last inequality is due to the assumption that a, 2 1 Taking telescoping multiplication over t = 0, 1.... - 1 for all < T, we obtain
27 12212
a $ ao(1 4(nī + n2)
> ||22|22
Amau (1-
|1///2
4(ni n2 )m
log(> Mezlla m.mmz))
where the last inequality is by a. S > [22lla
Amaz. So thent=
we have that a, S m We can
malla
1og(11(1- ung of l02
verify that
4(n) + n2)m log(> [| la mmmma) l|
log(A m m.maa) ||ril|
||a22lla
Ts
m22 C% 12
1log(1((1
guang || 4in + 1m
where the inequality is by log( -
I for x > O. By the definition of T. we have T 5 Tst. Now let us sort
log (8mm(n) + +20)/89)9--1)/ minfo3, -0%,
Hlmelly 20m 2/2 I1 1330% SNN308
out the time relationship. B <
we have T 2 T By do S
52

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Understanding the Forgetting of (Replay-based. Continual Learning via Feature Learning: Angle Matters
0
mggmn max (op o7 p. Jd v/ 1/1l1
we have 77 2 25.
Here we would like to make it clear that these
2 log (8m(n)) m2]/6] log(A lealz mndmos)
two inequalities imply when A Iaaall
Amaz(1 [1 + o()] - ((22) (Tz,t) down to 1 m. maxgra (( (7:00] Σ Ooopp vd and < m ze still hold. Here we complete the proof.
||
(um) (f
(T.10) ( and (w[[?.) , Then we calculate (wrr
> |ulll,,
(T.0) (w) j.r .ozz = (w (T.00)
jir .Hm)) + 7(1) )/7 (Ta.0)
H2212 ((12) (Taffend)
jire
(Ti.0) } + Amaz [1 + o[1] ^² Amazl [1 (())
[W jrr
= (1 (1-X²)A >22) [1 + o(1)];
(w) (Tai)) j.r .j12) = (w) (T).0)
jor ..122) + (122)
j.r" (Ta.t)
> |
|1l1 (ML) (Taifond)
j,r
O(m #)
m
= o())
Here the proof of Lemma 10 completed.
When 7(11) (T2.)) S m 24 no longer holds, we still have
(w (Tait) jir .(m)) = (w) jor ..m)) + (11)) (T2.0) - | (Taifene)
(T..0)
| (12))
j,r
2 (1 >²)) (1 (11 (11) (Tart)
[1 + o(1)]
2 (1 - (2)=2)Amazlll o()]
j.r ((m) will enter the re-ascending phase again. And we still have (w) (Tattoua) ..pz] = o11). As for the other neurons inside the set I,1, using the similar method in our previous section G, we can get (Tadand)
From here, we can see that (T
(Trdaad)) (Wer jjm) = o((((( j,r" jum)) and (w jr .مز2) o(1) for *r.
(Ta))ena)
H.3. The Second Type of Neuron
The second class of neurons is those belonging to the set I2,2, which is expressed by the formula follows:
(w (T..) .jm) < 0, (w jo .jntt > 0.
(T.0)
By the previous analysis, we can get that
(T2.0) (w jrr
.1m1) = -C1 < 0, (w .jm = C2 > 0.
(Tz.0)
In this case, the update method of the (m jir
(T2.1)
(112))
(T2.1) is as follows
((1)) (T =1(1) (T2.))
172
(- (TTa)))) o'(Ci -2(1) (T21)) + |11ll2 > (12) (T.0)) 11|41113
(ni + n2 m
H2ll2
O'(C2 + ( ) (22.) > ||12||2
((12) (T2.+1) = ((12)(4 (1z.1)
7a
Σ
(- ((T2)))) 2,
Mel1 (H4 (41) ((2)))) -11
(nī n2 m
=
The following lemma shows that with a certain time, 7 (11)
(T2,t) (T2.6)
changes little for r 4,.2- First of all, similar to the proof
of Lemma H. we have (11) (T2.t) X
[ualla (Mz) )rr + (w (w(T..) .jpz), plugging this into (1u)) (T2.)) update rule,
(Tz00)
53

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
can obtain
(122) (1
(112)) (7000)
(22.0)
1/2 11
Σ V2,1
(Tx 2,
o' d'(( ((w]7:00 ,0(22 + X |2lll (1)) + (1 2²) (22)())) (?200
(7i.0)
(nj + 122 -22)m
21ll
= =>(122) (22.0)
72
((t d' + (1- - X² (12) 12
(nī + n2 m
2,
12.
Then we can estimate (w (22.4) jipz) by (1122) (22.1) we have
(T.tt) (w) jr .jpz) = (w.""" (T:.0) .j12) + (1222) ).r
(T2.))
|1eell2, 7
((11) (T2,4)
=(1->2)((22) (/10:)) o(1)]
where the equation is due to Lemma H.4 that (122) (Tr to0 and (w (T2,0) jjuz), ((11) (T2.)) < 01).. Then by Lemma H.4 we can get
(Ta,)) 2,
11((1 + e (o(o (T210) (2. (1 (a,)))- #ob (o "o (7..0) (e.i))))))
=1/(1 + th (ww (T.0 (un o(1)
11(( + e ( (11)))))r))))))) _. ) + (1))
Lemma H.12. With probability at least 1 - 6,
122l22
V1 1200111221
max (w (Ti.0) rE(m]
,i(2 + >
11)
|1ll2 1|2
for all j E {+1}.
Proof of Lemma H.12. Similar to the previous proof of Lemma C.2, we have for each E [m]. (w (T.0) : |leella 11))
a Gaussian random variable with mean zero and variance (1 (1 (
1213 And,
> (w (T,.0)
5rr
llka (z)))) is an absolute constant, and therefore by the condition on m, we have M
P(V1 - XPool|Hu222 < max (w) (T)0)
rE(m] J.r .jpp |
||11112 11))))
> max (w (73.0) rElm] jir , + > U2el2 )))
= 1 - P(v 1 111---1--oollll 1000l122212
|1l1
= 1 - P(V 1 - 100--3oolleea ABoullHel122 > (w jjrr 1 + X |
(T),))
|1l12 11)))
2 1- 5,
here we completes the proof.
Lemma H.13. Under Condition 3.1, if choose
52SNR? 2 C 23 min 0z 0g1 log /1 - X20ol Hzll2
2m 1
[lo((S((1 + 4 1)/2
(H.30)
(1 (1-13) 22 0/
where C. (() is a positive constant, there exists time
T C(ni + n2)22 log
vI x
(1 22)0/20
pgg
such that
54

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
max, ( 2
(T2.1)
1- for j E {+1].
m
(T2,)) max ( j.rik
S pu vd all E [=1], r E [m], is E [mx] and 0 stsT7
Proof of Lemma H.13. Denote by T3 the last time satisfying that ((1z) (Ta,))
-
m [1-
Thus, we can have an upper
bound for -22,
(T 1/(1+ + e=oo(1-XX910n))) (2(1a) 2 [1 o(1)]
1
2
2(1 + C22y
2(1 + e ahe o (111))))())))))"
where the second inequality is by our hypothesis, and we denote 2(1+e) as C2. Thus we get C22 as the upper bound for (T224) Then we compute the growth of (122) as below: 2,5
(12)) (T2.++1) = =/ (22,0)
122 22l1
Σ
(T2. 2,
)o Jo (C2 (C2 (T21t) - ||lalll ((11)))
(mj + 12)m
||11112
yj
12 1211 1
= =((m)) (12.1)
2, 1o (C2 + (1 X2 (222)) (120))
(1b.1)
(1 +1++00 o(1))
(ni + m2)m n==i
C2 μ2) 1222
2 2(122) (1200)
(22.0)
(C2 + (1 - X )2) (122 (12
8(n1 + n2) m
where the inequality is by (G.4) and
2,1 2 C¹2 Let A max, (C2 + (1 (uz)) (T2.1)) then we have that
(T2,1))
A + (1 X²) C/g p2 B (A/0))-1
A (
8(n] + 1/2 )m
[1 + 17211 X²) C²)Ca p2 1B:12 (--))-2
2 1
8(0 + 122) m
2 /4/011 [1+ 12(11 X² >2)Cklll |2 B22 (4 (222
8(n + n2 )
[1+ + Ma(1 - X2 p2 1132
2 4
(V1 - X2 12001112112))
8(n + n2) )
where the third inequality is due to that 4 is an increasing sequence, and the last inequality follows by Lemma H. and A!0) (w (T),0)
", -j(12 + > | (1))) )) Then we have
[1+ + m2 X3) 12 12
A/00 2 4 -
X 1011 12)12)
8(nj + n2)
2 4 [1+ 12(1 x²) p2 11:12
- Xo00 11 |2222 (|) 2- 27
8(n] + mm2m
2 4 exp { m(1 x²)C22 (4 112n2
X²
16(n; + 122) m
2 exp { 11(1 X2 C3)C/9l| 12
1
x2o0loll))2 --))) 21 v 1- 1 X2o0llu222,
16(n) + n2)m
where the second inequality is due to the cumulative product of the series A the third inequality is by 1 + x 2 éă for x < 2 and our condition of 1/2 and oo listed in Condition 1. and the last inequality follows by Lemma H. Therefore
16(2 +2)) log (1 x Halls
will reach 2m within T
iterations. Therefore, we can get 7112)
(12.8) A,-C 2
12 (015mma11130//90000 X3
p
T-
55

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
m i| -9 log vT x lingll
2m C 2
m I-x2 By (H.30), we can obtain that 7 S TT. By 00 S 0
we have
(1-22j0/2 log (A
lize mAmux)
T 2 T3, so Lemma H.11 hypothesis holds and max ( (1:s0l
S Jopps Pa vd obviously holds. Here the proof
completes.
6(nj =aaja aalaaaaaaapa aap) m2l p Biggee
Lemma H.14. Under Condition 3.1, let T55t = Txx= +
= =T + 6mm c k-)). Then we have
max ( j.r. S 2000 vd for all T S t S T5.
And we can also find a time t" that (122) (T2.)) z k.maz
where T S S TT.
Proof of Lemma H.14. When (12)) (Ta.0)
reaches
m 1- we have
1/2 1211
Jo'((2 + ( (Tx.0) -x ||lealla ((1 (/r 197200)
( (T2,+1)
7(12)) (/x01)
(T2,t)
(T2.)) 2,
(ni n22) n22 yi=j
M1llz
M2ll 142 1B
((1 X²) 12) 10-
= =>(1>) (3),1)
[1 + o(1))
2(ni + m
= ((1 - A2) X2م 9 y
where the last equation is by (G.4). Let az = (1 -1-18 (²)) verify that
( (22.t) and function {(x) log(1 + exp(- x))) then we can
m2llu12 122 p (at). 2(n + n22)
01+1 = at
Firstly, we will give an upper bound for (e()))
(((=))²= (
mi x
2
S
m
2 29-
?(x) S (((())),
(H.31)
1 + expl 2
1 + exp( =
where the second inequality is due to Itz S log(1 + 1 for x > 0, and the last inequality is by 1+expl < too for
20
3 ạtr { m log
1/4
0. Denote max
22hAmaz by a", then we have
(a" - az)² - (a" a(41)² (a141 az) a2a at aa41))
-(a"-
1221 1221 /222Bl122111 ((a)) 220 -2at + 12 m221|12 |23n2 p (at))
2(nj + 122)
2(nz + +n2))
12 m2ll 1, l' (((a)) -aa--aaa-a11a + o[1]] -
1/2 |2z 11ān22 p' (ax) + +o(1))
(ni + n2)
2(n + n22))
3722 p2 11 i(a.)
37211/2211 11:02 ((aa)
1/ 122l|eel1 12 1B:22
2
((((a)))
4(n] + +n2)
4(ng + -22)
[ni + m2)
3721l/22 18 ((a"), 4(nī + +22)
~ Elar)
2(n + +n2)
where the first inequality follows by the convexity of {(x) and (H.31), and the last inequality is by Condition 3.1 that 1/2 S Ö[min(]luall ;o Taking a summation over T T + 1,.., T5. we have
+ 2(n + (12)
i(a.) s 2(T -T + +1)((a))
(a" a )2
nyl 72l1u1 |22 12l 1
+ 2(n + 122) (a* aT62+ )2
s 2 - T + 1)ē(a*)
n2l |2 E:n
56

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
S <2( - IT + +)((aa) + 2(nī + m2) ((m²). 1/2 1211 1m
where the last inequality is by (a* - aT,7 = 6(m²). Let = =
6(n 6(mi+ -ppt
= 25 +6
[nj (my)m
na malluea Engekt
72 7all Biggek
then dividing T3 -T + 1 on both side of the last inequality gives us
To
2(n + 122) 22 (T² - -Tt + 1)
(ar) S 20(a")
+
((m²)
T5
12l1
t=
1
_ n'
+
3
s 49
where the second inequality is by log(1 + x) S x for x > 0 and the definition of a". Since the mean is less than ch. there
must exist a at (TT S t" s 155) such that
((r.)) S SC
and we can obtain that ((2)) (12)1)
AAmge By the definition of T55 we can obtain
T
Σ ((a+) 2 - T + +1)((a) + 2(n + m2) ((m)) 66 [ni (((mi + 112 )m
-
(H.32)
m/2 |22 Bmz
1/2 |2z 1B
Letting ((1) = m ( g obviously we have v (M1) S 000p, vd S 2000p. Vd. Now suppose that there
(1)
exists T E ( TTg1] such that v (()) I 2000p, Vd for all t E [T T - 1]. Then for t E [T . 1], we have
(H1) s (())((())
+ max
1/2
(-2,,
((T2..)) )o'( (T2.1) E2.11)
(ni + 122 m
< ((1))(())
+ 2(n m2
37290 pt
maxx
{
(- (- ((
2,
((w E1,'' + (p(()))))))))) 9020))
n12 Elx))+
+ Σ (p(p))
(22.0) ||| (EE, 11
Σ
|1 12 ( E1,03 (1<2)
(22.0)
3n2q03 de
s {((())
[4 /log (8m n23x/1, 0o0m Vd
2(n n2)
log 4n5/8)
log Anz/6) L (11) (0)00-
(1)) 4nf
T )()) + 4ng
Tp
pm
where the the second inequality is by Lemma C.I, and the last inequality follows by (H.26) and Lemma C.I. Taking a telescoping sum over L = T T + T 1, we have
--1
nage
(1) < (((()) (T)
(ni + n2)m
T
n2ge
s (TT))
(T² O1))) Vajo-1
(ni + 12)m
S wooos Vd δ
n1 2/4 10 1SRR 2 (000 Vajs-2 poom vd
m
P2
S Sooop. Vd 0
0
pı m
2/4-11
2/q 1
000p, vd (aoop, vajo-2 v -000)
02
57

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
< p1 vd.
where the first inequality follows by the assumption that d 2 1024 log(4 n2) max(n?, nž} max _n 0 }.
30 21
the second inequality is by T < Tigt. the third inequality is due to Lemma H.8, the fourth inequality is due to n2 !(1) and the last inequality is by Condition 3.1 on do. Similarly we will use the same way to obtain that 4 (p2) < 200 Vd
for all Ti I f T5a. Obviously we have (TT) I ouTp2 Vd I 2000pa Vd. Now suppose that there exists T E 25"] such that (pa))() - 22000p vd for all t E [T T - 1]. Then for t E 1ra T - 1]. we have
((2))(1-1) S ((22)
+ max f.n
(nf + 122 )
1/2
(-
(ol Jo (( ((w.
2,1
(25.0) .62.)))((..1
31220 2 d
I ((z2)(())
+ max sin
p2
(- 2, ((T2)0))
[<w/570000 .E2.)) + ++((()2)) (2
2(n + m2)mm
"i
j.rd" 11|.2.111 3 9-1
| 11 I1| 112 "E,, E2,121 +
Σ
10(12)
31240 d
2 (12)
p2
maxt m (173:001 4 log (8m max Vd
2(11 + m2)m
p0 vd + Anj
p2 log (447//0)
To (() + \
log( 4n3/0)
V ((p
9-1
o
d
d
where the the second inequality and the last inequality follows by (H.26) and Lemma C.I. Taking telescoping sum over t T7. T + I,..., T 1, we have
312q0 d 0((000p
(( s ((2))(
max { ((22.t) Σ
2,0
2(n + n2)m
Tt
3n2a0 d
(nj + n2 m2)
S ((22)
Ō((00pp vā p2
2(n + n22m
1/2 12 1212
S 000m Vd Ō 2/24--10 n SSRR, 2 (0 vajo-2
1 Vd
5000p vd δ
m n2
2/ 2/4 (000p)
t vājo-2
on0 Vi
200 Va.
where the first inequality is by the assumption that d 2 1024 log(4 (n) max nz) max of o,
02
_ ). the
o2
second inequality is due to (H.32), the third inequality is due to Lemma H.8, the fourth inequality is due to n2 = 2(1)
and the last inequality is by Condition 3.1 on oo. Therefore, ((()))()) S 0p0ps vd for E [1,2] and T < t S T which completes the proof.
Moreover, we can bound (w
(T.T2.)) (T.()) j.r E1,i) and (w (w.4))
jr E2,)) by Ce. When T S S TT', we have
kw (Tatt) Jir
(Tstt)
E1,,Dl S kw (T2.0) EI.i) + (p(m))) jnn
n2
Σ (p( (T2.0) I1 3 +
Σ
((
(Ta.) j.r.
|1E2, 11 KK.,, E1.0)
< 4 4vlog (8m max m2 ooop, p. Vd
(og(4n]/0) (8)
1og(4n3/
- 2000p Vi + 4njy
2000 p. Vd + 4nz
Tp
200 pa Vd
d
p2
d
58

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
<6ylog (8m max n2) p1 Vd.
where the second inequality is by Lemma C.I, and the last inequality follows by d > 1024 log(4 max (nq, /0)02 max(n?; nB) max (Ta.0)
dh P2 on
Thus, when T S s T5. can bound the (w)r .L.,) by Cg. Using the same method, we can also bound the (War (w)r) E2,) by Ce.
H.4. The Third Type of Neuron
The first class of neurons is those belonging to the set I3,, which is expressed by the formula follows:
(TI.0) (w jrr .imi < 0, (w)r (Ti.0)
jr .imi) ) < 0
By the previous analysis, we can get that
(T2.0) (w jr jim) = -C== < 0, (w (22.0) _jmu) = -C2 = -0(1 < 0.
Lemma H.15. For any 1. 2 O, we have the following bounds
||12||20 + +|l1mll 1|2C2
0 V (14)) (T2.t)
(1 > ||2
|| 11222 + 211222122C
0 S ( (T2 (T2,t)
(1 (1- x2)|| ||2
We will prove this lemma by mathematical induction.
proof of Lemma H.15. In this case, the update method of the ((1)) j.r ( (Tx1)) is as follows
(T2.))
(T2.4)
((11) ((p=)) (To.++1) = = (To.))
72
Σ
(- (Tx(T)))) o'(O1 - (1(1)
jrr + > |1
(Ta.t)
U2llz 7 (12)) (2) (T.tt) 11111112 -1/0)))11111)
(nī + m2)m
7(12) (2, ((122) (4r
(Ta.t)
172
Σ
(- 2,3
((T of (C2 - (112) (2.00 +
|22lle
|1112 7 ( ((3)))))))))) ((2t))) || 1
[ni + n2 m
=2
When t = 0, the inequality clearly holds. Suppose that the inequality holds true when 0 s t S T - 1, then we can get that
1/2 1
o'O + || + > 111
(12, ((1) jrr
< ((())
jrr
Σ
(ni + m22)m
(1->
S (11) (72.1)
1/2 1 11
(Taxt)) o'( | + Allmmllle (111)!T21))
(nj + 122 m
1,
(1- |2212
=1
5>(m))
(T2.0)
qaqni |1
|ella + Allm1llaC2 (411) (720))0--1
(H.33)
(ni + mm
X2] p2 12
where the first inequality follows by the induction hypothesis H.10, the second inequality is by the fact that o'xx) is a monotonically increasing function, and the last inequality is by the fact that Epr,=-; S ni and (T2.)) < After the same derivation, we can get that
(T2.0 + T29 1 12 | 12212 + Allleell20 ((2)) (1 (12) (2,.411) S <7(22)
(ni + m2)m
13 l2
In order to prove the lemma H. we need more mathematical transformations. Let's make a. (p (t) then according to the inequality (H.33), we can have
||02 |20 Nal
(1 (1-X") Halla
M2gni 11 (at)
a++1 2
(H.34)
(ni + n22 m2)
59

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
Then, we will use the inductive method to prove a₁ 2 0. This known from the ((41) (Taw)s iterative equation, 2(11) (T2.0) an increasing sequence, so de 2 d++1- When S de S
(n. (n7+n2)m
14-1, can be seen from the inequality H.34, we have
yagn M l
that d++1 2 0. And because 0 s dg - (1-22) |
[n +in mm 1-11 which can be proved by Condition 3.1 on 72.
Then by induction, we can know that ac+1 2 0, for any 2 O. So is obvious that 0 S "(H1)) In the same way, we can get 0 S 7122) (22.0) (12)
(T2.))
< l| JaC +
(1-XP)
< Il + Al
(1-X") palla
Taking it a step further, we can bound the inner products (w (T2.4) Lemma H.16. for any 2 O,
jrr jumi) and (w (w ..u2) as follows.
jjm) 0(1) Kw (T2,0)
kw (Ta.0) Jir
.5p2) - O(1). Jir
proof of Lemma H. Combining Lemma H.16 and the two equations (H. and (H.13), we are able to derive that
| ll
(w (To.t) jrr
.ju)) = (w (w. (Ta.0) .مm) + ((1))
(T2.4)
|2212 (12)
(T2,1) 3rr
- + ((1) ((2200)
S C1 + || X1|11112202
(1- |2ll
= 0(1),
(H.35)
where the first inequality by the fact that (2z)) (T700) and the second is by known result that (u1) Call:Ch Further, get that
(T2,4)
(1-X)]lmal2
|u1ll2
(w (Tb.4) Jr -jm) = (w (w!?2.0) 122.0)
.jmm)) + 7 (22.t)
(11)
(112) (12))
(22.0)
H2212
|1 112
2 -
((122) jr"
(T2.t)
|
[1 l|2
||
|,nlla02 All
2 -
|42||2
(1 - X2) {
=0((1:
(H.36)
where the first inequality by the fact that ((1)) ),rr Ie.ll:Ca+N/leellaC;; ,he proof for (w (Tz.t) (1-X2)][m)
(Tz.t) and the second is by known result that (1z)) (T2,t)
((22))(2
j.r .izz) is similar, we can prove that -0(1) S (w (Ta.t)
j.r jizz) and (w (T2.t)
jrr ..42) S
O(1), which completes the proof.
We denote the time to stop training for task T2 by fand, which represents the first time (11)) (T2.0) j.r reaches kAmazs and we
know I < fend < TT5
For 1,,1, we can calculate the following summation the neuron output:
Σ
a((w! jrr
(Tz.tand) CllW-Tadane (aadamm)) .jM1)) o (W .jur)) Σ
o((W (((wfffrdann (oatean)) .jun)) [(1 A²)Amaz] o(1))
r
rzpt
and
Σ a(lw!l ((wW(Tailnn)) jor
(Tzilane)
(122)
Σ
(o((1) = o(m).
relga
ret,
60

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
We also have
Σ r
o((w a(( -jHu)) o(m),
(Taitens)
and
Σ o((w),
((w (Tziland) -juz))
E
o(o(1) o(m).
r
retga
Now we obtain the neurons in the set f,, output values task 71 and T2.
For I5,2, we can calculate the following summation of the neuron output:
Σ
o(( (T2itona) jr
j(m)), Σ
((w
(Tzilend) -
Σ
a(( (Taifand) -jjuz)) = o(m),
a((w)
rElsa
rEIga2
rElya
and
Σ a(w a(WW[Taiten) j(uz)) = (( (antene) j22)) + Σ o o((
(Tailend)
(Tzitend) .(z))) [kAmax] 1 o(1)]
rET,a
rept"
Now we obtain the neurons in the set f,2 output values in task Ti and T2. For 15,3, we can calculate the following summation the neuron output:
Σ
o((w!!'. ,jmi)) Σ a0(1)) = O(m),
regys
reI,,3
and
Σ rET,33
a(ww! (T2itend) o(WW??atent .(1z)) = Σ
-(((1))=00 =
rETz,3
We also have
Σ a(w (Taitens) -jpn)))
Σ
o(O(1)) 0mm),
rEI,.
and
Σ
a(w ji
a((W(Tatennd) (Tastena) -jpp))
Σ a(O())) (m))
retga
retga
Now we obtain the neurons in the set 133 output values in task Ti and T2.
H.S. Loss Analysis
In the previous section, we divided the neurons into three sets: 15,1 1Ú,2 and 15.3. For the set 1,,1, conclude that
Σ a((w ((WWTTstemm (T2stend) .jpi)) = [(1 - A²)) + [
Σ o((w (Trifand)
((W((Tiitene)) jun))) = o(m),
rE4,1
rEl,a
al(w ((W[Fadden)) reh,
Σ
(Tzdlnnd) -jpn)) o(m), reh,
Σ
a((w (((Wfllilend) jor
-j o(m).
61

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
for j =1.
For the set 12, we conclude that
Σ
a(w (Irdend) or
_jmi)) = o(m),
Σ
((w (aldend) ,(1z)) = [kA [kAmaz]] + o(1)}
((
rEl,i2
Σ re4ya
a((w))
jor
(aahene) jjm()) o(m), r
Σ
a((w
(T2itend)
- -جu o(m),
for j E =1.
For the set 1,,3, we conclude that
Σ a(w ((ll)lalmm)) ((m1) 0mm), rEtg,9
Σ
o((w))
((WWlattomd) .juz)) 0(m),
Σ rE
o(ww Jir
((WW)?2ifom)) -j = 0(m), rElga
Σ
((w 3o
(2ntens) - = 0(m),
for j E =1.
Then we can calculate the Er=1 o(w ((w!Ta ,j (1)), E ((W .j Er= ((w (Tademe) ,5 p2)) and
E o ((W((Ta (T.ten)) . p2))):
f.r
m
Σ
( ((w
(Tzitand) v1))) oor
Σ a (w (T2itend) Jr
. (11)))++++ Σ ol(w (Tadend) .j (-m)))++
E o( (Tatend) .j p1))
rEl,,2
rEl;a
= [[1 - X²) Amas ( + +((m) + 0(m)
[[1 - 22) Amaz [[1+o(1)]]
Σ o(WW-
(T2stend) .f -11))
Σ a((w" (Tzitena) rEI- -j.r
.f (1))) +
Σ a((w (Tasennd) -j.r .f m1)) +
Σ
a(w (Teifend)
-j.r ... (11)
rEI-
rEI
o(m) + +(m) (())
0(m),
J. (12))) a(lw ((W
(22tene) j.r
Σ Ca((w(1) jr
(a(W(Taadend)) (artand) .j (121)))+++ + r
Co((w (Tadand) .j (222)) + E
(Tadend) .j 122)
r
rEI,a
= =(m) + [kAmaz]] 1 +(m)
= [kAmazl [1 + o(1)],
Σ o (Iddndd) :f 222)) jjr
Σ
a((w (((W!Tadend) (Tdennd) .j. 122) + Σ
a((w! a(wwlZadendd (Tllndd) .f (22)))
+ Σ
a((w (((W(lilled)) .j p2))
rel_ja
reI_;,2
reI
62

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Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
o(m) ((m) ((m)
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63

---

Understanding the Forgetting of (Replay-based) Continual Learning via Feature Learning: Angle Matters
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64


Reviews:
[{"review_id": "sUnK4kbn5c", "reviewer": "Reviewer_2zRv", "summary": "The paper provides a mathematical framework of forgetting in continual learning, for the specific case of a two-layer convolutional neural network with polynomial ReLU activation. The authors show that replay has the effect of increasing the range of settings under which forgetting is limited. Based on their analysis, they also propose a scheme for sampling mid-angle examples for the buffer, which has a slightly positive effect in an experiment on Cifar 10 and Cifar 100.", "questions": "", "limitations": "", "rating": 2, "confidence": null, "soundness": null, "presentation": null, "contribution": null, "strengths": "", "weaknesses": "", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "Thanks for your constructive feedback! We address your questions as follows.\n> **Q1. How the mid-angle sampling strategy relates to the theoretical theorems**\n\nWe sincerely appreciate the opportunity to clarify the connection between our theoretical findings and mid-angle sampling. Our theoretical results show that smaller angles between task signal vectors lead to benign forgetting, while larger angles cause harmful forgetting. In practice, we treat each class prototype—the mean feature of its examples—as the signal vector.\n\nWe focus on the case where the angle between task prototypes is obtuse, as harmful forgetting arises only in this setting. If a sample’s feature forms a larger angle with its own prototype, it tends to form a smaller angle with the second task’s prototype, likely falling within the benign forgetting range in standard CL (i.e., it can be remembered without replay). Conversely, if the angle with its own prototype is smaller, the angle with the second task’s prototype may be larger—possibly beyond the benign range under CL with replay (i.e., it will be forgotten despite replay).\n\nIn contrast, samples with mid-range angles are more likely to fall outside the benign range of standard CL but within that of replay-based CL—making them the most effective candidates for replay. Thus, mid-angle sampling offers a more efficient and targeted replay strategy.\n\n> **Q2. Discuss the choice for a single head setup.**\n\nWe adopt the single-head setting primarily to facilitate theoretical analysis. However, our theoretical framework can extend to the multi-head setting. In fact, our final forgetting results are derived by analyzing the behavior of individual neurons, through which we characterize the angle-dependent antagonism between tasks. While the multi-head setup may increase the number of neurons involved in learning the feature of the second task, the core antagonism between tasks (driven by the angle) still persists.\n\nMoreover, We emphasize that slight differences between theoretical settings and practical setups are common to ensure analytical robustness. As Reviewer #2 (qzcp) noted, \"For a primarily theoretical analysis, using a simplified architecture that captures core nonlinear effects (rather than a purely linear or kernelized model) is a rational choice.\" In this regard, our framework goes beyond linear and NTK-based models, making it more aligned with practical scenarios.\n\n> **Q3. The noise that is orthogonal to all signal vectors seems restrictive.**\n\nWe adopt the orthogonality assumption primarily to simplify the proof and reduce the length of the manuscript. In fact, by techniques similar to those used by Kou et al. (2023) [1], we can extend our theoreical results to non-orthogonality case.\n\n> **The signal and noise vector are processed separately by the network.**\n\nThis setting, commonly used in analyses of feature learning theory [1,2], mainly serves to simplify the proof. In practice, we can extend to multi-patch model using techniques similar to those of Allen-Zhu et al. (2020) [3].\n\n> **The definition of Forgetting is weird.**\n\nAs you noted, the approximation using $L_{D_1}^{0-1} (W^{(T_2)})$ has limited impact since $L_{D_1}^{0-1} (W^{(T_1)})$ approaches zero. We commit to clarifying this to avoid any misunderstanding. \n\n> **Experiment 3 lacks details with mid-angle sampling a slightly positive effect**\n\nThe experiment follows the classical iCaRL framework for CL based on CNNs [4], which consists of a feature extractor and a classification layer. Since iCaRL allows only a fixed number of replay samples to be stored, our Mid-angle sampling strategy selects the most intermediate examples by first sorting all examples within each class based on their cosine similarity to the class prototype, and then selecting those closest to the median. As shown in Table 1 in our response to Reviewer #4 (z4gT), Comment Q2, our mid-angle sampling outperforms herding—a nontrivial result given that herding is a widely used sampling method in CL.\n\n> **Condescending to refer to methods such as EWC as \"empirical methods\".**\n\nHere, we follow the description from Ding et al. (2024) [5]. We will remove the misleading term \"empirical methods\" and revise the description for clarity.\n\n> **The structure of the paper should be revised.**\n\nThanks for your suggestion regarding the structure of our manuscript. We will move Section 2 before 1.2 to improve readability.\n\n**References**\n\n[1] Kou, Y., et al. Benign overfitting in two-layer ReLU convolutional neural networks. In ICML 2023\n\n[2] Cao, Y., et al. Benign overfitting in two-layer convolutional neural networks. In NeurIPS 2022\n\n[3] Allen-Zhu, Z., et al. Towards understanding ensemble, knowledge distillation and self-distillation in deep learning. arXiv, 2020\n\n[4] Rebuffi S A, et al. icarl: Incremental classifier and representation learning. In CVPR 2017\n\n[5] Ding, M., et al. Understanding forgetting in continual learning with linear regression. arXiv, 2024", "claims_and_evidence": "The main claims of the paper seem accurate. \n\nMy main concern is related to the relevance of the studied setting, which is quite different from a practical neural network setup. This applies not so much to the fact that it's only two layers, as most theoretical works focus on simplified architectures, even just one layer or linear models. More importantly, they seem to work with a shared head, whereas separate heads for each tasks is mostly used in practice, and they assume the noise is orthogonal to the signal vectors from all tasks (definition 1.1). I realize the latter setting is adopted from earlier work (Cao et al., 2022), but still I find it a strong assumption that is not well motivated. Finally, the signal and noise vector are processed separately by the network, which again is not how it works in practice.", "methods_and_evaluation_criteria": "1. The definition of Forgetting (end of section 1.1) is weird. Instead of measuring the true error on the first task, it should measure the increase in test error on the first task, i.e. L_D1^{0-1}(W^(T2)) - L_D_1^{0-1}(W^(T1)). In practice, in the simple setting and with all the assumptions made, the test error after training task 1 (second term) is close to zero, so it doesn't impact too much, but still...\n\n2. Initially, I was charmed by the idea of a theoretical paper that, at the same time, included a practical algorithm evaluated on more common CL setups using CIFAR and some larger network (not specified though). However, he proposed method applying mid-angle sampling is only very weakly related to the theory given earlier in the paper. I would appreciate if the authors could elaborate how the proposed strategy follows directly from the earlier theorems.", "theoretical_claims": "I did not check all proofs in the appendix (50 pages!). As far as I checked, the math is correct under the defined setting / assumptions.", "experimental_designs_or_analyses": "Experiments are limited, as this is mostly a theoretical paper. Experiment 1 illustrates the theoretical setup.  Experiment 2, classifying MNIST digits vs. their inverse, is quite extreme, yet by design still very close to the theory. A multi-head setup would have made it more realistic. Experiment 3 lacks details (in main paper): what network is used ? what range is actually sampled for \"moderate cosine similarity\"  ? The differences between the different sampling methods are very small, making one wonder if they are significant at all.", "relation_to_broader_scientific_literature": "Overall, the relation to the broader scientific literature is well described.\n\nI just found it a bit condescending to refer to methods such as EWC as \"empirical methods\".", "essential_references_not_discussed": "NA", "other_strengths_and_weaknesses": "The structure of the paper should be revised. \nSection 1.2 ('Main Contributions') is impossible to follow, as many of the symbols used are only introduced later in Section 2.", "other_comments_or_suggestions": "## update after rebuttal ##\nI stick to my original score, as I'm still not convinced what the added value of this paper is. \n\nFirst, I don't think the theoretical analysis really brings us more insights. That angles matter is, in fact, rather intuitive. It's equivalent to saying something like 'distance to the decision boundary matters', but then on a unit sphere. I'm not impressed.\n\nSecond, I emphasize the impact of the 'single head' setting. It's not just a \"slight difference between theoretical settings and practical setup\", as stated by the authors. It's a completely different setup that influences the analysis drastically, and it wasn't even discussed in the paper as being a deviation from the practical setting. At the very least authors should be transparant about all the simplifications they make, rather than hiding them and hope readers overlook. Without such transparancy, papers like this bring a false impression of theoretical foundation. Using a single head only makes sense to me for domain incremental settings, where the angles are typically small anyway. No one with some common sense in continual learning would try a domain incremental setting with extreme domain changes. If that were the case, probably a task incremental setting would be selected instead (i.e., first identifying the task).", "questions_for_authors": "1. How does the proposed 'mid-angle' sampling strategy relate to the theoretical theorems given earlier ? \n2. Please discuss the choice for a single head setup.\n3. The choice for noise that is orthogonal to all signal vectors (from all tasks) seems a very restrictive one. Please discuss.", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {"ethical_review_concerns": "/"}}, {"review_id": "GERSEsy6sJ", "reviewer": "Reviewer_qzcp", "summary": "The paper develops a theoretical framework for understanding continual learning (CL) and catastrophic forgetting using a two-layer polynomial ReLU CNN. It focuses on how the angle between two tasks’ “signal vectors” (representing core features for each task) influences forgetting: if the angle is acute or only mildly obtuse, forgetting from the first task remains “benign,” but if the angle is large (i.e., vectors are nearly opposite), the model experiences “harmful” forgetting. The authors prove these claims by characterizing network training through a signal-versus-noise decomposition and analyzing neuron behavior under gradient descent. They also show that replay-based methods expand the range of angles for which forgetting stays benign and introduce a “mid-angle sampling” strategy that selects replay samples with moderate angles to their class prototypes, demonstrating an improvement over standard sampling techniques in empirical tests on synthetic data, MNIST, and CIFAR100. ## update after rebuttal", "questions": "", "limitations": "", "rating": 4, "confidence": null, "soundness": null, "presentation": null, "contribution": null, "strengths": "", "weaknesses": "", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "Thanks for your constructive feedback! We address your questions and concerns as follows.\n> **Q1. Your theory focuses primarily on two binary tasks. Could you outline how you would expect the angle-based framework and replay analysis to extend if there were multiple sequential tasks or multi-class tasks for each session?**\n\nWe validate the relationship between forgetting and angle in multi-task settings using the synthetic dataset. We conduct two sets of experiments, each consisting of three tasks. In Experiment E1, Task 3 forms angles $\\theta_{1,3} = 150°$ and $\\theta_{2,3} = 100°$ with Tasks 1 and 2, respectively. In Experiment E2, these angles are $\\theta_{1,3} = 170°$ and $\\theta_{2,3} = 80°$. Let $Acc_{1}$ and $Acc_{2}$ denote the accuracy on Tasks 1 and 2, respectively, after learning all three tasks. As shown in Table 1, the conclusion consistently holds.\n\n||$(\\theta\\_{1,3},Acc\\_{1})$|$(\\theta\\_{2,3},Acc\\_{2})$|\n|-|-|-|\n|$E\\_1$|$(150°,2.5\\\\%)$|$(100°,99.7\\\\%)$|\n| $E\\_2$|$(170°,0.3\\\\%)$|$(80°,100\\\\%)$|\n\n*Table 1: Experimental Results in multi-task settings on the synthetic dataset.*\n\nWhile our current analysis focuses on two binary tasks for clarity and tractability, the angle-based framework naturally extends to multi-class and multi-task settings. Specifically, complex configurations can be decomposed into pairwise class-level interactions across tasks, with angular relationships capturing the core learning dynamics. This forms the basis for our analysis. We plan to extend our theory by studying how the accumulation of such pairwise interactions drives forgetting and how subproblems influence one another in more general settings.\n\n> **Q2. Do you see a straightforward way to relax the “signal + noise” assumption to capture more varied real-world data distributions? Or do you view the current model as primarily a stepping-stone for further exploration?**\n\nThe signal-noise data model takes inspiration from image data, where the inputs are composed of various patches, and only certain patches are relevant to the class label of the image. This model has been widely adopted in recent theoretical studies, including our references [1,2]. Furthermore, the two-patch setting can be extended to a multi-patch model by techniques similar to those in Allen-Zhu et al. (2020) [3].\n\nJiang et al. recently employed a feature learning theory based on signal-noise decomposition to study benign overfitting in Vision Transformers [4]. Our exploration of continual learning in CNNs through this lens may serve as a foundation for extending such analysis to more complex models like Transformers.\n\n> **Q3. How significant is the computational cost of measuring and comparing angles (or proxies for them) in mid-angle sampling, particularly for large networks or large-scale datasets?**\n\nWe primarily implement the mid-angle sampling strategy by computing cosine similarity, whose computational cost should be comparable to the original herding strategy adopted in the iCaRL framework (which relies on Euclidean distance) [5].\n\n> **Typos and Minor Clarifications**\n\nWe appreciate the thoroughness of the review and the opportunity to improve the accuracy and professionalism of our paper. We commit to fixing all minor writing issues and ensuring consistent notation to further improve clarity.\n\n**References**\n\n[1] Cao, Y., et al. Benign overfitting in two-layer convolutional neural networks. In NeurIPS 2022\n\n[2] Kou, Y., et al. Benign overfitting in two-layer ReLU convolutional neural networks. In ICML 2023\n\n[3] Allen-Zhu, Z., et al. Towards understanding ensemble, knowledge distillation and self-distillation in deep learning. arXiv, 2020\n\n[4] Jiang J, et al. Unveil benign overfitting for transformer in vision: Training dynamics, convergence, and generalization. In NeurIPS 2024\n\n[5] Rebuffi S A, et al. icarl: Incremental classifier and representation learning. In CVPR 2017", "claims_and_evidence": "The paper’s main claims revolve around (1) the theoretical relationship between the angle of two tasks’ signal vectors and the severity of forgetting, (2) the ability of replay methods to expand the range of angles over which forgetting remains benign, and (3) the benefit of “mid-angle sampling.” Below is how the paper substantiates these claims:\n\n1.Angle and Forgetting: The authors provide a formal derivation under a polynomial ReLU two-layer CNN. They track the gradient-based evolution of weights with a “signal-noise decomposition,” proving that the inner-product behaviors align with angle-based predictions. They construct a controlled “signal + noise” dataset with varying angles. The observed forgetting closely matches the theoretically predicted angular thresholds.\n\nComments: The proofs are detailed and logically consistent with prior feature-learning analyses. The empirical curves on synthetic data indeed show changes in forgetting severity at about the angles predicted.  \n\n2.Replay Expands the Benign-Forgotting Range: The authors augment their theoretical framework to account for stored samples of the previous task and show that, under certain buffer size conditions, harmful forgetting is avoided even when the angle is moderately large. They run experiments (both synthetic and on MNIST, CIFAR100) contrasting the “no replay” condition against “with replay,” then measure performance on the old task. \n\nComments: The paper does not fully generalize this to many tasks, but the data for the two-task scenario supports the claim well.\n3.Mid-Angle Sampling: On MNIST and CIFAR100, they compare mid-angle sampling with random sampling and herding. Results show small but consistent accuracy improvements on older tasks.\n\nComments: margin of improvement is not enormous. The mechanism for why it works is rooted in their angle-based theoretical analysis, which is coherent within their two-task scope.\n\nOverall, the authors present a coherent chain of reasoning plus experiments that affirm their principal angle-based explanations for forgetting, along with replay’s benefits and the utility of mid-angle sampling. However, there are several weaknesses:\n●Simplicity of the Data Model: The paper’s theoretical and synthetic experiments heavily rely on a fairly stylized “signal + noise” data model. Real-world data can be more varied, so it may be difficult to guarantee the same clean angle-based properties in practice.\n●Limited Improvements in Experiments: While the mid-angle sampling approach does show some gains over standard replay sampling, the performance boost is not markedly large. The experiments, though suggestive, do not represent a major breakthrough in empirical performance.", "methods_and_evaluation_criteria": "In the context of a theoretical study on continual learning, the paper’s methods and chosen evaluation criteria are generally reasonable for the goals it aims to achieve, but there are also some limitations:\n\nSignal+Noise Synthetic Model: The authors use a carefully controlled synthetic dataset to verify their angle-based theoretical predictions. Because the paper is largely focused on proving formal guarantees, having a simple generative setup where signal vectors, noise, and angles can be precisely controlled is sensible. It helps isolate and confirm the paper’s core theory on how angles influence forgetting.\n\nTwo-Layer Polynomial ReLU CNN: This is a restricted but analytically tractable network architecture. For a primarily theoretical analysis, using a simplified architecture that captures core nonlinear effects (rather than a purely linear or kernelized model) is a rational choice. It allows them to go beyond lazy training assumptions and linear analyses.\n\nNevertheless, the choices of methods and datasets reflect a clear effort to validate both theoretical and practical aspects of the approach. By anchoring the proofs in a specifically designed synthetic model, the authors can rigorously pinpoint the conditions under which angle-based insights hold. At the same time, employing MNIST and CIFAR100—despite being relatively standard benchmarks—demonstrates that the proposed ideas and replay strategy are not confined to purely toy examples.", "theoretical_claims": "The paper’s most notable highlight is how it thoroughly compares two methods—standard continual learning versus a replay-enhanced version—and demonstrates, both theoretically and empirically, how replay expands the range of angles for benign forgetting. Additionally, the technical derivation that underpins these findings is quite detailed, showcasing a clear, step-by-step structure. The logical flow—spanning from the setup of the signal-noise decomposition, to the rigorous lemmas about inner products and gradient dynamics, and ultimately to the theorems on angle-based forgetting—reflects a methodical and well-organized presentation. Together, these aspects make the core results not only transparent but also easy to follow.", "experimental_designs_or_analyses": "Strengths\n●Clear Connection to Theory: The synthetic setup precisely matches the assumptions in the paper, making it easy to see the influence of angles on forgetting.\n●Use of Standard Benchmarks: Validating on MNIST and CIFAR100 shows that the angle-based findings and replay strategy improvements hold in relatively common experimental contexts, beyond purely synthetic data.\n●Systematic Comparisons: They compare no replay vs. replay, as well as different sampling approaches (including mid-angle sampling), which cleanly highlights each method’s impact on forgetting.\nWeaknesses\n●Limited Scope: The experiments focus on binary classification in a two-task scenario. It’s unclear how the angle-based conclusions might extend to more tasks or multi-class settings.\n●Incremental Gains: While mid-angle sampling does yield improvements, the empirical boost over standard sampling strategies (like random or herding) is not very large.\n●Data Model Simplifications: The synthetic data strictly follows a “signal + noise” model, which may not capture all complexities of real-world datasets.", "relation_to_broader_scientific_literature": "The paper’s central focus—analyzing catastrophic forgetting through the lens of feature learning and the geometry between tasks—connects directly to several threads in the existing continual learning literature.", "essential_references_not_discussed": "It appears that the paper adequately addresses the relevant prior work for its main theoretical results and replay-based methods.", "other_strengths_and_weaknesses": "Other Strengths\n1.Originality of Angle-Based Analysis: Although researchers have long recognized that task similarity can affect forgetting, framing this in terms of a precise angle between “signal vectors” provides a fresh, more mathematically rigorous viewpoint.\n2.Balanced Theoretical and Empirical Components: By combining rigorous proofs with both synthetic and real-data experiments, the paper goes beyond many purely theoretical treatments and offers a more complete picture.\nOther Weaknesses： Focus on Two-Task Setting: While the theoretical insights may be extended, the paper’s primary focus is a two-task scenario, which leaves open how well these angle-based insights hold for longer task sequences.", "other_comments_or_suggestions": "Additional Suggestions\n●Typos and Minor Clarifications: A quick proofreading pass could help catch minor linguistic issues, particularly in the theorem statements and figure captions. Ensuring complete alignment of notation between the main text and supplementary would also enhance clarity.\n●Further Exploration: For readers seeking deeper insight into how angles evolve across multiple tasks, the paper could briefly outline potential extensions beyond the two-task, two-layer setup—even if only at a conceptual level.", "questions_for_authors": "Question: Your theory focuses primarily on two binary tasks. Could you outline how you would expect the angle-based framework and replay analysis to extend if there were multiple sequential tasks or multi-class tasks for each session?\nQuestion: Do you see a straightforward way to relax the “signal + noise” assumption to capture more varied real-world data distributions? Or do you view the current model as primarily a stepping-stone for further exploration?\nQuestion: How significant is the computational cost of measuring and comparing angles (or proxies for them) in mid-angle sampling, particularly for large networks or large-scale datasets?", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "4JUCm8AU3R", "reviewer": "Reviewer_N92G", "summary": "The authors propose a theoretical analysis of catastrophic forgetting in the two class setup for two layer convolutional neural networks, with polynomial RELU activations. They prove that for rehearsal free CL, forgetting is significant when the angle is between the new task and previous task is small enough. They also prove that if this angle is large enough, the forgetting can be upper bounded. \nFor CL methods with rehearsal, the authors prove that the benign forgetting range is larger therefore incurring less forgetting for more dissimilar tasks. \\\nBased on the findings above, the authors present a rehearsal method with mid angle rehearsal, to mitigate forgetting more effectively, compared to random rehearsal. This method is compared experimentally against other baselines on CIFAR-100 CL benchmarks.", "questions": "", "limitations": "", "rating": 4, "confidence": null, "soundness": null, "presentation": null, "contribution": null, "strengths": "", "weaknesses": "", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "Thanks for your constructive feedback! We address your questions and concerns as follows.\n> **Is the grey area a significant forgetting region ?**\n\nThe grey area is a region for uncertainty, either harmful or benign forgetting, ensuring that our claims remain rigorous. The yellow area is for harmful forgetting.\n\n> **Reporting the std and Average Forgetting in Table 1.**\n\nDue to space limitations, the answer is provided in our response to Reviewer #4 (z4gT), Comment Q2.\n\n> **Experimentally validating Theorem 3.2 and 3.3.**\n\nAs Reviewer #2 (qzcp) noted, \"the authors present a coherent chain of reasoning plus experiments that affirm their principal angle-based explanations for forgetting\". Figure 2 shows that benign forgetting leads to near-zero error, while harmful forgetting approaches one, validating our Theorem 3.2 and 3.3. Figure 3 confirms both on MNIST. The second plot in Figure 2 shows that our over-parameterized model avoids the lazy training regime, with significant weight increase in the signal direction.\n\n> **I may have misunderstood the rightmost plot in Figure 2.**\n\nBlue indicates the maximum value, and green indicates the second-largest value, rather than a single continuous line. The intersection of the blue and green lines shows where the maximum value shifts between the two variables.\n\n> **Why is x_k subdivided into two vectors ?**\n\nDue to space limitations, the answer is provided in our response to Reviewer #2 (qzcp), Comment Q2.\n\n> **What are the assumptions about the distribution D_k ?**\n\nWe assume $\\xi_k \\sim N\\left(0, \\sigma_{p_k}^2 \\cdot \\left(I - U (U^T U)^{-1} U^T\\right)\\right)$. The label $y_k$ is a Rademacher random variable. One of $x^{(1)}_k$ and $x^{(2)}_k$ is $y_k\\cdot\\mu_k$, the other $\\xi_k$, with $(x_k, y_k)\\sim D_k$.\n\n> **Explain the choice of the logistic loss. Is it restrictive ?**\n\nLogistic loss is a special case of cross-entropy loss for binary classification. We use $L_{CE}=-\\frac{1}{n}\\sum_{i = 1}^{n}[y_i\\log(\\hat{y}_i)+(1- y_i)\\log(1-\\hat{y}_i)]$, where $\\hat{y}_i$ is softmax-normalized. After softmax calculations in $L\\_{CE}$, we get the logistic loss used in our analysis.\n\n> **Theorem 3.1 : is the upper bound on cos theta a typo, shouldn't it be 1 ?**\n\nTheorem 3.1 does not appear in the paper. If the intended reference is Theorem 3.2 or 3.3, the range should be $1 \\geq \\cos\\theta_{1,2} \\geq 0$ for Theorem 3.2 and $-\\frac{1+C_2}{2} \\leq \\cos\\theta_{1,2} \\leq 1$ for 3.3.\n\n> **Does the network fall in the lazy regime ?**\n\nOur approach avoids the lazy training regime by using smaller initialization, in contrast to the NTK setting. This allows the weights to move significantly and learn the signal ($\\gamma_{j,r}^{(t)}$), rather than staying near initialization. Large initialization typically induces NTK-like behavior with minimal updates, while small initialization enables meaningful parameter growth along the signal direction, thus avoiding lazy regime.\n\n> **Why is the $Relu^q$ activation chosen ? Is it restrictive or does it translate to commonly used activations ?**\n\nPolynomial RELU can speed up both signal learning and noise memorization, to further boost the gap between them. Our theoretical framework can be extensible to RELU activation by techniques similar to those by Kou et al. (2023) [1]. \n\n> **Does the analysis apply to non convolutional models ?**\n\nThe answer is provided in our response to Reviewer #2 (qzcp), Comment Q2.\n\n> **Why the network is a CNN ? What happens if the convolutions are not split between the signal and noise ?**\n\nOur network structure, a common choice for theoretical analysis (including our references [1,2]), retains the core features of a CNN. We use a dual-channel input model $\\mathbf{x}=[\\mathbf{x}^{(1)},\\mathbf{x}^{(2)}]$ with shared weights across channels. The output is:\n$ f=\\frac{1}{m}\\sum_{r = 1}^m\\left[\\sigma(\\mathbf{w}\\_{+1,r}^\\top\\mathbf{x}^{(1)})+\\sigma(\\mathbf{w}\\_{+1,r}^\\top\\mathbf{x}^{(2)})\\right]-\\frac{1}{m}\\sum_{r = 1}^m\\left[\\sigma(\\mathbf{w}\\_{-1,r}^\\top\\mathbf{x}^{(1)})+\\sigma(\\mathbf{w}\\_{-1,r}^\\top\\mathbf{x}^{(2)})\\right] $.\nHere, $m$ is the number of filters, $\\sigma(z)=(\\max\\{0,z\\})^q$ ($q > 2$), and $\\mathbf{w}\\_{j,r}$ are filter weights. The structure approximates $\\sigma(\\mathbf{w}\\_{j,r}^\\top(\\mathbf{x}^{(1)}+\\mathbf{x}^{(2)}))$, which is commomly used in feature learning theory [1,2]. Furthermore, We can extend the two-patch data to multi-patch data model similar to Allen-Zhu et al. (2020) [3]. \n\n> **In the multitask setup, is the final accuracy impacted by the angle ?**\n\nThe answer is provided in our response to Reviewer #2 (qzcp), Comment Q1.\n\n**References**\n\n[1] Kou, Y., et al. Benign overfitting in two-layer ReLU convolutional neural networks. In ICML 2023\n\n[2] Cao, Y., et al. Benign overfitting in two-layer convolutional neural networks. In NeurIPS 2022\n\n[3] Allen-Zhu, Z., et al. Towards understanding ensemble, knowledge distillation and self-distillation in deep learning. arXiv, 2020", "claims_and_evidence": "The main claims stated in the contributions sections are supported with proofs.\n\nIn terms of experimental evidence : \n\nConvincing evidence\n- the claim about the forgetting regions with and without replay (Fig 1) is verified experimentally in Figure 2. However, while the experiments show that the replay setup has a larger range of non forgetting, it looks like the region that was identified as a grey area in the analysis is partially a significant forgetting region.\n\nMissing evidence : \n- It would be very informative to validate experimentally the tightness of the forgetting bounds in Theorems 3.2 and 3.3\n- To validate the proposed rehearsal scheme, the std is missing from Table 1. The std is necessary to conclude on the significance of the mean improvement. More so given that the metrics are very close across the baseline. Also, I think that it's important to report the Average Forgetting as well, because conclusions cannot be made only based on the Average Accuracy.", "methods_and_evaluation_criteria": "The proposed evaluation criteria is sensible overall.\n\nThe missing critical elements are the following : \n- Reporting the std in Table 1\n- Reporting the Average Forgetting in Table 1\n\nNice to have evaluations : \n- Experimentally validating the tightness of the forgetting bounds in Theorems 3.2 and 3.3\n- Experimentally validating the over-parameterisation lower bound for the Theorems", "theoretical_claims": "- Definition 1.1 : Could you clarify the design choice of splitting x into two vectors ?\n- Sec 2. Could you clarify why the proposed neural network definition is a convolutional neural network ? Also isn’t it too simplistic to define one convolution wrt the signal and the second convolution wrt the noise ? Could you discuss the assumption and its possible limitations ?\n- Could you discuss the loss assumption, and to which extent it is limiting to generalise the takeaways to more commonly used losses such as the cross entropy loss ?\n- Is the analysis extensible to the multitask learning setup, where the data mixture has large obstute angles ? \n\nI haven’t checked the proof in the Appendix.", "experimental_designs_or_analyses": "I checked all the experiments and shared some related comments in the \"Methods And Evaluation Criteria\" section.\n\nSome additional comments : \n- I may have misunderstood the rightmost plot in Figure 2, but isn't there a color error, shouldn't the blue curve decay to zero and the green increase monotonically ?\n- Could you clarify how the experiment in Figure 3 translates to or validates the analysis ?\n- In Table 1, do you consider the full CIFAR dataset or only two tasks ? Could you clarify it ?\n- Optional : Could you run the synthetic experiments in the multitask setup, to see the impact of the angle on the final accuracy ? Do you think it would be sensible ?", "relation_to_broader_scientific_literature": "This work relates to the theoretical Continual Learning literature. Several works quantify the impact of task similarity on CF under different task, model and data assumptions : [2], [3], [4], [5], [6].\n\nThe analysis is also based on Feature Learning Theory. I am not familiar with this research area, however the analysis is significantly inspired by [1].\n\n\n- [1] Cao, Yuan et al. “Benign Overfitting in Two-layer Convolutional Neural Networks.” ArXiv abs/2202.06526 (2022): n. Pag.\n- [2] Bennani, Mehdi et al. “Generalisation Guarantees for Continual Learning with Orthogonal Gradient Descent.” ArXiv abs/2006.11942 (2020): n. Pag.\n- [3] Doan, Thang Van et al. “A Theoretical Analysis of Catastrophic Forgetting through the NTK Overlap Matrix.” International Conference on Artificial Intelligence and Statistics (2020).\n- [4] Lee, Sebastian et al. “Continual Learning in the Teacher-Student Setup: Impact of Task Similarity.” International Conference on Machine Learning (2021).\n- [5] Evron, Itay et al. “How catastrophic can catastrophic forgetting be in linear regression?” ArXiv abs/2205.09588 (2022): n. Pag.\n- [6] Evron, Itay et al. “The Joint Effect of Task Similarity and Overparameterization on Catastrophic Forgetting - An Analytical Model.” ArXiv abs/2401.12617 (2024): n. pag.\n- [7] Hiratani, N. (2024). Disentangling and Mitigating the Impact of Task Similarity for Continual Learning. ArXiv, abs/2405.20236.", "essential_references_not_discussed": "I am not aware of any missing references.", "other_strengths_and_weaknesses": "- Strengths : \n   - Theoretical analysis of forgetting for a practical architecture (CNN), and interesting derivation of bounds on forgetting regimes depending on the angle between the tasks\n   - Experiments in the same setup as the theory to validate some analytical observations\n   - Deriving a practical application from the analysis - though the significance of the improvement is still unclear for now\n   - Overall clear presentation of the intuition behind the theorems even though the notation is heavy\n- Weaknesses : \n   - Very restrictive assumptions (overparameterisation and architecture), it's unclear to which extend they could apply to more complex and widely used architectures.\n   - Missing std in Table 1, therefore no conclusion is possible yet about this experiment\n   - Unclear tightness of the bounds", "other_comments_or_suggestions": "- Definition 1.1 : I would suggest clarifying that the intuition behind the covariance matrix is the orthogonality wrt the U\n- Definition 1.1 : I suggest clarifying the intuition behind mu, it only became clear to me in the experiments section", "questions_for_authors": "In addition to the questions in the other sections, I wanted to ask the following questions : \n\n- Definition 1.1 : why is x_k subdivided into two vectors ? Is it the definition of a CNN in this analysis ?\n- Definition 1.1 : what are the assumptions about the distribution D_k ?\n- L88 : Could you explain the choice of the loss function and to which extend it is restrictive compared to the cross entropy loss ?\n- Theorem 3.1 : is the upper bound on cos theta a typo, shouldn't it be 1 ? if not why ?\n\nGeneral questions : \n- A large enough over-parametrisation is one of the assumptions of the theorems, is the lower bound not too large enough to fall in the lazy regime ?\n- Why the choice of polynomial RELU activations, to which extent is it restrictive or does it translate to commonly used activations ?\n- Does the analysis apply to non convolutional models ?\n- What happens if the convolutions are not split between the signal and noise ?\n- Optional :  In the multitask setup, under the same assumptions, is the final accuracy impacted by the angle ?", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "Cr8X4xiyvm", "reviewer": "Reviewer_z4gT", "summary": "The paper develops a unified theoretical framework for understanding catastrophic forgetting in continual learning through the lens of feature learning. The authors focus on a two-layer convolutional neural network with a polynomial ReLU activation function that is trained sequentially on binary classification tasks. Their key findings are:\n\n- The extent of forgetting on previously learned tasks is critically influenced by the cosine similarity (i.e., the angle) between the task signal vectors. Specifically, when the angle is acute or only slightly obtuse, the network experiences benign forgetting with only minor performance degradation on old tasks. In contrast, larger obtuse angles lead to harmful forgetting with significant performance loss.\n- Replay-based continual learning methods are shown to mitigate forgetting by effectively expanding the angular range that corresponds to benign forgetting. This insight leads to the proposal of a “mid-angle sampling” strategy, where examples are selected based on having a moderate cosine similarity to the class prototype. This strategy aims to balance stability and plasticity, further enhancing the effectiveness of replay methods.\n- Theoretical results are rigorously supported by a detailed analysis of neuron behavior during training (via a signal-noise decomposition) and by characterizing how the network’s weight updates interact with task signal angles.\n- Experimental validations on both synthetic datasets and real-world benchmarks (such as MNIST and CIFAR100) confirm the theoretical predictions, illustrating the relationship between task angles, forgetting, and the beneficial impact of replay and mid-angle sampling.\n\nIn summary, the paper contributes a novel theoretical perspective that links the geometric relationship between tasks to the phenomenon of forgetting in continual learning, and it introduces practical replay strategies inspired by this analysis.\n\n## update after rebuttal\nThe authors address most of my concerns. So I keep my positive score.", "questions": "", "limitations": "", "rating": 4, "confidence": null, "soundness": null, "presentation": null, "contribution": null, "strengths": "", "weaknesses": "", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "Thanks for your constructive feedback! We address your questions and concerns as follows.\n> **Q1. The current analysis focuses on binary classification with two-layer CNNs. Could you elaborate on how the framework might extend to multi-class classification or deeper architectures? A clear discussion on this could enhance the practical impact and generalizability of your work.**\n\nWhile our current analysis focuses on two binary tasks for clarity and tractability, the angle-based framework naturally extends to multi-class and multi-task settings. Specifically, complex configurations can be decomposed into pairwise class-level interactions across tasks, with angular relationships capturing the core learning dynamics. This forms the basis for our analysis. We plan to extend our theory by studying how the accumulation of such pairwise interactions drives forgetting.\n\nAdditionally, Jiang et al. recently employed a feature learning theory based on signal-noise decomposition to study benign overfitting in Vision Transformers [1]. Our exploration of continual learning in CNNs through this lens may serve as a foundation for extending such analysis to more complex models like Transformers.\n\n> **Q2. The mid-angle sampling strategy appears promising. Can you provide additional insights on how sensitive the method is to the choice of cosine similarity thresholds? Is there any theoretical or empirical guidance on selecting these thresholds optimally?**\n\nIn fact, our mid-angle sampling strategy does not require threshold selection. The experiment follows the classical iCaRL framework for CL based on CNNs [2], which consists of a feature extractor and a classification layer. Since iCaRL allows only a fixed number of replay samples to be stored, our Mid-angle sampling strategy selects the most intermediate examples by first sorting all examples within each class based on their cosine similarity to the class prototype, and then selecting those closest to the median. We conduct experiments on the CIFAR100-5 (5 tasks with 20 classes each) and CIFAR100-10 benchmarks. As shown in Table 1, our mid-angle sampling outperforms herding—a nontrivial result given that herding is a commonly used sampling method in CL with replay.\n\n|Sampling|Random|Small-angle|Mid-angle|Big-angle|Herding|\n|-|-|-|-|-|-|\n|ave-accuracy ↑ (CIFAR100-10)|47.17 ± 0.45|45.63 ± 0.12|**48.02 ± 0.27**|45.34 ± 0.76|47.40 ± 0.17|\n|ave-forgetting ↓ (CIFAR100-10)|15.72 ± 0.31|19.47 ± 0.39|**14.84 ± 0.26**|18.04 ± 0.49|15.51 ± 0.08|\n|ave-accuracy ↑ (CIFAR100-5)|56.08 ± 0.12|54.36 ± 0.35|**56.51 ± 0.06**|54.77 ± 0.29| 56.12 ± 0.20|\n|ave-forgetting ↓ (CIFAR100-5)|11.15 ± 0.46 |14.10 ± 0.26| **10.15 ± 0.28**|12.50 ± 0.72|10.64 ± 0.13|\n\n*Table 1: Experimental Results with std and Average Forgetting on CIFAR100.*\n\n> **Q3. Your proofs rely on a signal-noise decomposition of network weights. Could you offer more intuition or illustrative examples (e.g., visualizations) that clarify how this decomposition relates to neuron behavior and forgetting in continual learning?**\n\nWe first provide an intuitive explanation of the relationship between signal-noise decomposition and neuron behavior. In our setting, the signal vector $\\mu$ is orthogonal to the noise $\\xi$, forming a basis for a plane. After training, the CNN weights evolve as $\\mathbf{w}_{j,r} = j\\gamma\\mu + \\rho\\xi$, where $\\gamma \\gg \\rho$, indicating that the weights grow predominantly along the signal direction. This suggests that the network has effectively learned the signal. In multi-task scenarios, the weights adjust according to the signal vectors of tasks.\n\nRegarding forgetting in CL, we show that harmful forgetting mainly arises in the obtuse-angle case. As stated in Lemma 4.3 (page 5), if $\\sum\\_{r = 1}^{m}[\\sigma(\\langle\\mathbf{w}\\_{y\\_{1},r}^{(T\\_{2},t\\_{end})}, y\\_{1}\\mu_{1}\\rangle)-\\sum\\_{r = 1}^{m}\\sigma(\\langle\\mathbf{w}\\_{-y\\_{1},r}^{(T\\_{2},t\\_{end})}, y\\_{1}\\mu\\_{1}\\rangle)] \\geq C\\_{3}$, then the CNN will achieve benign forgetting. We further derive that $\\sum\\_{r = 1}^{m}\\sigma(\\langle\\mathbf{w}\\_{y\\_{1},r}^{(T\\_{2},t_{end})}, y\\_{1}\\mu\\_{1}\\rangle)$ and $\\sum\\_{r = 1}^{m}\\sigma(\\langle\\mathbf{w}\\_{-y\\_{1},r}^{(T\\_{2},t\\_{end})}, y\\_{1}\\mu\\_{1}\\rangle)$ can be characterized by $\\Theta(m\\overline{\\gamma(\\mu\\_{1})\\_{y\\_{1}}}(1 - \\cos^{2}\\theta\\_{1,2})^{q})$ and $\\Theta(m\\overline{\\gamma(\\mu\\_{2})\\_{-y\\_{1}}}\\frac{||\\mu\\_{1}||\\_{2}^{q}(-\\cos\\theta\\_{1,2})^{q}}{||\\mu\n\\_{2}||\\_{2}^{q}})$ respectively. Then we obtain the angle range corresponding to benign forgetting; the analysis for harmful forgetting follows similarly.\n\n**References**\n\n[1] Jiang J, et al. Unveil benign overfitting for transformer in vision: Training dynamics, convergence, and generalization. In NeurIPS 2024\n\n[2] Rebuffi S A, et al. icarl: Incremental classifier and representation learning. In CVPR 2017", "claims_and_evidence": "The paper’s claims are largely supported by a combination of rigorous theoretical analysis and comprehensive experimental validations. In particular:\n\n- The core claim—that the cosine similarity between task signal vectors critically influences the degree of forgetting—is backed by detailed theoretical derivations (e.g., Theorem 3.2 and Theorem 3.3) and is further substantiated through experimental evidence on both synthetic and real-world datasets.\n- The analysis of neuron behavior via signal-noise decomposition provides a convincing mechanism for how different angles lead to either benign or harmful forgetting.\n- The effectiveness of replay-based methods and the proposed mid-angle sampling strategy is supported by experiments that demonstrate improved performance under various task settings.\n\nOne potential concern is that the theoretical results rely on strong assumptions (such as over-parameterized two-layer CNNs and binary classification settings), which may limit the direct generalizability to more complex or different learning scenarios.", "methods_and_evaluation_criteria": "The methods and evaluation criteria appear well-tailored to the problem. The paper employs a rigorous theoretical framework—developed for a two-layer CNN with polynomial ReLU activation—and supports its findings with experiments on both synthetic data and established benchmarks like MNIST and CIFAR100. These datasets are standard in continual learning research, providing a reasonable basis for evaluating both forgetting and the effectiveness of replay-based methods. Additionally, the evaluation metrics (training loss, test loss, and test error on old tasks) align closely with the objectives of mitigating catastrophic forgetting. While the setting is somewhat restricted (e.g., binary classification and overparameterized networks), within this scope, the methods and criteria make sense and are appropriate for the application at hand.", "theoretical_claims": "I reviewed the proof sketches provided for the main theoretical results—specifically Theorem 3.2 (for standard continual learning) and Theorem 3.3 (for replay-based continual learning)—along with the supporting lemmas (e.g., Lemma 4.1 through Lemma 4.8). Within the framework of their stated assumptions (such as the over-parameterization of a two-layer polynomial ReLU CNN, binary classification settings, and certain conditions on the signal-to-noise ratio and network initialization), the proofs are logically consistent and appear to be correctly derived.", "experimental_designs_or_analyses": "The experimental designs appear sound and well-motivated for validating the theoretical claims. Here are some key points:\n\n- Synthetic experiments were designed following a controlled data distribution (as described in Definition 1.1) with parameters (e.g., training sample size, dimension, noise variance) that allow a clear examination of the relationship between task angles and forgetting. This controlled setup helps in isolating the effects predicted by the theory.\n\n- Real-world experiments on benchmark datasets such as MNIST and CIFAR100 are standard in continual learning research. They not only test the basic hypothesis regarding the cosine similarity between task signals but also validate the effectiveness of replay-based methods and the proposed mid-angle sampling strategy.\n\nOne potential limitation is that the experiments focus on binary classification and over-parameterized two-layer networks. While this aligns with the theoretical framework, it may limit the direct applicability of the findings to more complex settings (e.g., multi-class scenarios or deeper architectures).", "relation_to_broader_scientific_literature": "The paper’s contributions are deeply connected with several strands of prior work in continual learning and theoretical deep learning. In particular:\n\n- It extends earlier theoretical studies on catastrophic forgetting, which often rely on linear models or lazy training regimes (e.g., Evron et al., 2022; Doan et al., 2021), by analyzing a two-layer CNN with a polynomial ReLU activation in a feature learning setting. This move addresses limitations in capturing the dynamics of practical neural networks.\n- The work builds on feature learning theory advances (e.g., Allen-Zhu & Li, 2020; Cao et al., 2022; Huang et al., 2023), adapting these ideas to the continual learning scenario. This integration allows for a unified framework that links the geometry (via cosine similarity between task signals) to the degree of forgetting.", "essential_references_not_discussed": "While the paper cites a broad range of works on catastrophic forgetting, continual learning, and feature learning theory, there are a few related lines of research that could further contextualize its key contributions:\n\n- There is a growing body of work examining the geometric properties of learned representations and their impact on transfer or interference between tasks. For instance, studies on Neural Collapse (e.g., Papyan et al., 2020) reveal that deep networks tend to organize their features in a highly symmetric and clustered manner, which could be directly related to how task signal vectors interact. These works might provide additional insights into the role of angular relationships in feature representations.\n- In the realm of deep metric learning, methods such as CosFace or ArcFace explicitly incorporate angular margins to improve discrimination between classes. Although these works focus on face recognition, the idea that angular separation can enhance class separability is relevant to understanding why certain angles between task signals might lead to benign versus harmful forgetting.", "other_strengths_and_weaknesses": "Strengths:\n- Novel Theoretical Perspective: The paper presents a unique theoretical framework that links the geometric relationship (cosine similarity) between task signal vectors to catastrophic forgetting. This approach provides a fresh angle on understanding when forgetting is benign versus harmful.\n - Integration of Theory and Practice: By combining rigorous proofs with experimental validation on both synthetic and benchmark datasets (MNIST, CIFAR100), the work offers a comprehensive study that spans theory and practical implementation.\n\nWeaknesses:\n- Restrictive Assumptions: The theoretical analysis relies on assumptions such as over-parameterized two-layer CNNs, binary classification tasks, and specific conditions on the signal-to-noise ratio. These may limit the direct applicability of the results to more complex architectures or real-world scenarios that involve multi-class problems.\n- Complexity of Theoretical Arguments: While the proofs are thorough, the level of mathematical complexity and the reliance on heavy technical machinery may make it challenging for practitioners who are less familiar with the theoretical underpinnings of deep learning.", "other_comments_or_suggestions": "Overall, the paper is solid with clear theoretical contributions and extensive experimental validations. Here are a few additional suggestions and minor comments:\n- Consider adding a brief discussion that highlights potential avenues for extending the analysis beyond binary classification and two-layer networks, which could help contextualize the work for a broader audience.\n- A few sections of the proofs are mathematically dense; including a high-level summary or intuition behind the most critical steps could further aid readers who are less familiar with the technical details.", "questions_for_authors": "- The current analysis focuses on binary classification with two-layer CNNs. Could you elaborate on how the framework might extend to multi-class classification or deeper architectures? A clear discussion on this could enhance the practical impact and generalizability of your work.\n\n- The mid-angle sampling strategy appears promising. Can you provide additional insights on how sensitive the method is to the choice of cosine similarity thresholds? Is there any theoretical or empirical guidance on selecting these thresholds optimally?\n\n- Your proofs rely on a signal-noise decomposition of network weights. Could you offer more intuition or illustrative examples (e.g., visualizations) that clarify how this decomposition relates to neuron behavior and forgetting in continual learning?", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}]

Author rebuttal:

Meta-review: The paper develops a unified theoretical framework for continual learning (CL) based on feature learning theory, focusing on a two-layer CNN with polynomial ReLU activation. It shows that the angle between task signal vectors significantly affects forgetting. The study also explains how replay methods reduce forgetting by expanding the range of benign angles. A new mid-angle sampling strategy is proposed to further improve replay performance. Experimental results on both synthetic and real-world data validate the theoretical findings.

While reviewers acknowledge the value of the proposed framework,  they share some common concerns, including the strong assumptions and lack of generalization to multiple sequential tasks. The authors' rebuttal has addressed these important concerns to certain extent  and most reviewers support the acceptance of the paper in the end. 

Overall, the theoretical results are not entirely surprising as pointed some reviewers. Meanwhile, the use of a single-head setup also deviates from practical continual learning settings, particularly in class-incremental scenarios. The paper does not clearly state its simplifying assumptions, which may undermine its theoretical contribution and create a less strict sense of generality.

The authors are suggested to carefully incorporate these comments and further clarify the scope and significance of their contributions to strengthen the presentation.
Decision: Accept (poster)