nyuuzyou/wikis huggingface dataset:
Sufficient Condition for Weak Convergence of Bounded Sequence in Hilbert Space in terms of Everywhere Dense Set

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $\HH$.
Let $D$ be an everywhere dense set in $\HH$ such that: 
:$\innerprod {x_n} v \to \innerprod x v$
for all $v \in D$, for some $x \in \HH$.
Then: 
:$x_n \weakconv x$
where $\weakconv$ denotes weak convergence.
== Proof ==
We show the theorem in the case $x = 0$ first for neatness.
From Weak Convergence in Hilbert Space, we want to show that: 
:$\innerprod {x_n} z \to \innerprod x z$ 
for all $z \in \HH$.
Let $\epsilon > 0$. 
Since $D$ is everywhere dense in $\HH$, for each $n \in \N$ we can pick $z_n \in D$ such that: 
:$\norm {z - z_n}_\HH < \dfrac 1 n$
We now have, for each $n \in \N$ and $m \in \N$:
:$\cmod {\innerprod {x_n} z} = \cmod {\innerprod {x_n} {z - z_m} - \innerprod {x_n} {z_m} }$
due to the conjugate linearity of the inner product in its second argument.
Pick $M > 0$ such that: 
:$\norm {x_n}_\HH \le M$ 
for each $n \in \N$. 
We can then bound:
Fix $m \in \N$ such that: 
:$m > \dfrac {2 M} \epsilon$
Since we have: 
:$\innerprod {x_n} {z_m}_\HH \to 0$ 
for this fixed $m$, we can pick $N$ such that: 
:$\cmod {\innerprod {x_n} {z_m}_\HH} < \dfrac \epsilon 2$
for $n \ge N$.
Then for $n \ge N$ we have: 
:$\cmod {\innerprod {x_n} z_\HH} < \epsilon$
So:
:$\innerprod {x_n} z_\HH \to 0 = \innerprod 0 z$
giving:
:$x_n \weakconv 0 = x$ 
Now we show the theorem for general $x$. 
Suppose that:
:$\innerprod {x_n} v \to \innerprod x v$
for all $v \in D$.
Then, using the linearity of the inner product in the first argument, we have: 
:$\innerprod {x_n - x} v \to 0$ 
Then by our previous calculations: 
:$x_n - x \weakconv 0$
so:
:$x_n \weakconv x$
from Linear Combination of Weakly Convergent Sequences is Weakly Convergent.
Category:Hilbert Spaces
Category:Weak Convergence (Normed Vector Spaces)
Sufficient Conditions for Basis of Finite Dimensional Vector Space

== Theorem ==
Let $K$ be a division ring.
Let $n \ge 0$ be a natural number.
Let $E$ be an $n$-dimensional vector space over $K$.
Let $B \subseteq E$ be a subset such that $\card B = n$.
:$(1): \quad$ $B$ is a basis of $E$.
:$(2): \quad$ $B$ is linearly independent.
:$(3): \quad$ $B$ is a generator for $E$.
== Proof ==
=== 1 implies 2 and 3 ===
Let $B$ be a basis of $E$.
Then conditions $(2)$ and $(3)$ follow directly by the definition of basis.
=== 2 implies 1 ===
Let $B$ be linearly independent.
Suppose $B$ does not generate $E$.
Then, because $\card B = n$, by Linearly Independent Subset also Independent in Generated Subspace there would be a linearly independent subset of $n + 1$ vectors of $E$.
But this would contradict Size of Linearly Independent Subset is at Most Size of Finite Generator.
Thus condition $(2)$ implies $(1)$.
=== 3 implies 1 ===
Let $B$ be a generator for $E$.
By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, $B$ contains a basis $B'$ of $E$.
But $B'$ has $n$ elements and hence $B' = B$ by Bases of Finitely Generated Vector Space have Equal Cardinality.
Thus condition $(3)$ implies $(1)$.
== Examples ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.12$
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.7$
* 
Category:Sufficient Conditions for Basis of Finite Dimensional Vector Space
Category:Generators of Vector Spaces
Category:Bases of Vector Spaces
Category:Finite Dimensional Vector Spaces
Sufficient Conditions for Basis of Finite Dimensional Vector Space/Examples/2 Dimensions

== Examples of Use of Sufficient Conditions for Basis of Finite Dimensional Vector Space ==
Let $V$ be a vector space of $2$ dimensions.
Let $\mathbf u, \mathbf v \in V$ be vectors
Let $\mathbf u$ and $\mathbf v$ be such that neither $\mathbf u$ nor $\mathbf v$ is a scalar multiple of the other..
Let $\mathbf w \in V$.
Then there exist scalars such that:
:$\mathbf w = a \mathbf u + b \mathbf v$
== Proof ==
== Sources ==
* : $\text{A}.2$: Linear algebra and determinants: Corollary $\text{A}.8$
Category:Sufficient Conditions for Basis of Finite Dimensional Vector Space
Sufficient Conditions for Uncountability

== Theorem ==
Let $X$ be a set.
The following are equivalent:
:$(1): \quad X$ contains an uncountable subset
:$(2): \quad X$ is uncountable
:$(3): \quad $ Every sequence of distinct points $\sequence {x_n}_{n \mathop \in \N}$ in $X$ omits at least one $x \in X$
:$(4): \quad $ There is no surjection $\N \twoheadrightarrow X$
:$(5): \quad X$ is infinite and there is no bijection $X \leftrightarrow \N$
Assuming the Continuum Hypothesis holds, we also have the equivalent uncountability condition:
:$(6): \quad $There exist extended real numbers $a < b$ and a surjection $X \to \closedint a b$
== Proof ==
Recall that $X$ is uncountable if there is no injection $X \hookrightarrow \N$.
=== $(1)$ implies $(2)$ ===
Suppose there exists an injection $f: X \hookrightarrow \N$.
Let $Y \subseteq X$ be uncountable.
Then $\hat f: Y \to \N$ defined as:
:$\hat f = \set {\map f x: x \in Y}$
is an injection $Y \hookrightarrow N$, which is a contradiction.
=== $(2)$ implies $(3)$ ===
Suppose $\sequence {x_n}_{n \mathop \in \N}$ is a sequence of distinct points such that every $x \in X$ equals $x_n$ for some $n$.
Since the $x_n$ are distinct, this $n$ is unique.
Let $f: X \to \N$ be defined by $\map f x = n$, where $n$ is the unique natural number such that $x_n = x$.
Then $f$ is injective because $n = m$ implies that $x_n = x_m$.
So there is an injection $X \hookrightarrow \N$, a contradiction.
=== $(3)$ implies $(4)$ ===
Suppose there exists a surjection $f: \N \to X$.
Therefore every $x \in X$ is $\map f n$ for some $n \in \N$.
Define $g: X \to \N$ by $\map g x = \inf \set {n \in \N : \map f n = x}$.
Then if $\map g {x_1} = \map g {x_2} = n$, we have $\map f n = x_1$ and $\map f n = x_2$.
So by the definition of a mapping, $x_1 = x_2$.
Therefore $g$ is an injection $X \to \N$, a contradiction.
=== $(4)$ implies $(5)$ ===
Since there exists no surjection $\N \to X$, and a bijection $\N \to X$ must be surjective, there is no such mapping.
=== $(5)$ implies $(1)$ ===
Suppose $f: X \hookrightarrow \N$ is an injection.
We essentially relabel the image of $f$ to obtain a bijection, and thus a contradiction.
Construct a map $g: X \to \N$ as follows.
Let $S_1 = \Img f$ and:
:$x_1 = \map {f^{-1} } {\inf \set {n \in \N : n \in S_1} }$
and define $\map f x = 1$.
Note that $x$ is a singleton because $f$ is injective.
Given $S_n \subseteq \Img f$, let:
:$x_n = \map {f^{-1} } {\inf \set {n \in \N : n \in S_n} }$
and set $S_{n + 1} = S_n \setminus x_n$.
The mapping:
:$n \stackrel {f^{-1} } \longrightarrow x_k \stackrel g \longrightarrow k$
defines a bijection from $\Img f$ to $\N$, so $g$ is injective because $f$ is.
Furthermore $g$ assigns each $n \in \N$ to some $x \in X$ (because $X$ is infinite) so $g$ is surjective.
Thus $g$ is a bijection, a contradiction.
=== $(6)$ implies $(1)$ ===
Category:Uncountable Sets
Sufficient Conditions for Weak Extremum

== Theorem ==
Let $J$ be a functional such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
:$\map y a = A$
:$\map y b = B$
Let $y = \map y x$ be an extremum.
Let the strengthened Legendre's Condition hold.
Let the strengthened Jacobi's Necessary Condition hold.
Then the functional $J$ has a weak minimum for $y = \map y x$.
== Proof ==
By the continuity of function $\map P x$ and the solution of Jacobi's equation:
:$\exists \epsilon > 0: \paren {\forall x \in \closedint a {b + \epsilon}:\map P x > 0} \land \paren {\tilde a \notin \closedint a {b + \epsilon} }$
Consider the quadratic functional:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x - \alpha^2 \int_a^b h'^2 \rd x$
together with Euler's equation:
:$-\dfrac \rd {\rd x} \paren{\paren {P - \alpha^2} h'} + Q h = 0$
The Euler's equation is continuous $\alpha$.
Thus the solution of the Euler's equation is continuous $\alpha $.
Since:
:$\forall x \in \closedint a {b + \epsilon}: \map P x > 0$ 
$\map P x$ has a positive lower bound in $\closedint a {b + \epsilon}$.
Consider the solution with $\map h a = 0$, $\map {h'} 0 = 1$.
Then
:$\exists \alpha \in \R: \forall x \in \closedint a b: \map P x - \alpha^2 > 0$
Also:
:$\forall x \in \hointl a b: \map h x \ne 0$
By Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:
:$\ds \int_a^b \paren {\paren {P - \alpha^2} h'^2 + Q h^2} \rd x > 0$
In other words, if $c = \alpha^2$, then:
:$(1): \quad \exists c > 0: \ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > c \int_a^b h'^2 \rd x$ 
Let $y = \map y x$ be an extremal.
Let $y = \map y x + \map h x$ be a curve, sufficiently close to $y = \map y x$.
By expansion of $\Delta J \sqbrk {y; h}$ from lemma $1$ of Legendre's Condition:
:$\ds J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$
where:
:$\ds \forall x \in \closedint a b: \lim_{\size h_1 \mathop \to 0} \set {\xi,\eta} = \set {0, 0}$
and the limit is uniform.
By Schwarz inequality:
Notice that the integral on the right does not depend on $x$.
Integrate the inequality $x$:
Let $\epsilon \in \R_{>0}$ be a constant such that:
:$\size \xi \le \epsilon$, $\size \eta \le \epsilon$
Then:
Thus, by $(1)$:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$ 
while by $(2)$:
:$\ds \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$ 
can be made arbitrarily small.
Thus, for all sufficiently small $\size h_1$, which implies sufficiently small $\size \xi$ and $\size \eta$, and, consequently, sufficiently small $\epsilon$:
:$J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x > 0$
Therefore, in some small neighbourhood $y = \map y x$ there exists a weak minimum of the functional.
== Sources ==
: $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points
Category:Calculus of Variations
Sufficient Conditions for Weak Stationarity of Order 2

== Theorem ==
Let $S$ be a stochastic process giving rise to a time series $T$.
Let the mean of $S$ be fixed.
Let the autocovariance matrix of $S$ be of the form:
:$\boldsymbol \Gamma_n = \begin {pmatrix}
\gamma_0 & \gamma_1 & \gamma_2 & \cdots & \gamma_{n - 1} \\
\gamma_1 & \gamma_0 & \gamma_1 & \cdots & \gamma_{n - 2} \\
\gamma_2 & \gamma_1 & \gamma_0 & \cdots & \gamma_{n - 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\gamma_{n - 1} & \gamma_{n - 2} & \gamma_{n - 3} & \cdots & \gamma_0
\end {pmatrix} = \sigma_z^2 \mathbf P_n = \begin {pmatrix}
1 & \rho_1 & \rho_2 & \cdots & \rho_{n - 1} \\
\rho_1 & 1 & \rho_1 & \cdots & \rho_{n - 2} \\
\rho_2 & \rho_1 & 1 & \cdots & \rho_{n - 3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\rho_{n - 1} & \rho_{n - 2} & \rho_{n - 3} & \cdots & 1
\end {pmatrix}$
Then $S$ is weakly stationary of order $2$.
== Proof ==
Follows from the definition of weak stationarity.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Weak stationarity
Category:Stationary Stochastic Processes
Sufficient Statistic/Examples/Mean of Normal Distribution

== Example of Sufficient Statistic ==
Let $\mu$ be the expectation of a normal distribution $N$.
Let $S$ be a sample from $N$.
Then the sample mean $\overline x$ is a sufficient statistic for estimation of $\mu$.
This is because knowledge of the individual sample values provides no further information about $\mu$.
This in turn is because the distribution of $S$, conditional upon $\overline x$, is independent of the population mean.
== Sources ==
* 
* 
Category:Examples of Sufficient Statistics
Suffix/Examples/Arbitrary Example 1

== Examples of Suffixes ==
Let $abc$ be a string.
Then the suffixes of $abc$ are:
:$\epsilon$, $c$, $bc$ and $abc$
where $\epsilon$ is the null string.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Suffixes
Suffix of String is Substring

== Theorem ==
Let $S$ be a string.
Let $T$ be a suffix of $S$.
Then $T$ is a substring of $S$.
== Proof ==
By definition of substring, there exists a string $T'$ such that:
:$S = T'T$
Hence $S$ is the concatenation of the null string, $T'$, and $T$.
Thus by definition of substring, $T$ is a substring of $S$.
Category:Collations
Sum Formulas for Sine and Cosine

== Theorem ==
=== Sine of x plus Cosine of x: Sine Form ===
=== Sine of x plus Cosine of x: Cosine Form ===
=== Sine of x minus Cosine of x: Sine Form ===
=== Sine of x minus Cosine of x: Cosine Form ===
=== Cosine of x minus Sine of x: Sine Form ===
=== Cosine of x minus Sine of x: Cosine Form ===
== Multiple of Sine plus Multiple of Cosine ==
Category:Trigonometric Identities
Category:Hyperbolic Functions
Sum Less Maximum is Minimum

== Theorem ==
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \max \set {a, b} = \min \set {a, b}$.
== Proof ==
From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \max \set {a, b} = \min \set {a, b}$ follows by subtracting $\max \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets.
However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
This is because the fact that $a + b \ge \max \set {a, b}$ follows immediately again from $a + b = \max \set {a, b} + \min \set {a, b}$.
Category:Algebra
Sum Less Minimum is Maximum

== Theorem ==
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b - \min \set {a, b} = \max \set {a, b}$
== Proof ==
From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.
Thus $a + b - \min \set {a, b} = \max \set {a, b}$ follows by subtracting $\min \set {a, b}$ from both sides.
It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets.
However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
This is because the fact that $a + b \ge \min \set {a, b}$ follows immediately again from $a + b = \max \set {a, b} + \min \set {a, b}$.
Category:Algebra
Sum Over Divisors Equals Sum Over Quotients

== Theorem ==
Let $n$ be a positive integer.
Let $f: \Z_{>0} \to \Z_{>0}$ be a mapping on the positive integers.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum of $\map f d$ over the divisors of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \map f d = \sum_{d \mathop \divides n} \map f {\frac n d}$.
== Proof ==
If $d$ is a divisor of $n$ then $d \times \dfrac n d = n$ and so $\dfrac n d$ is also a divisor of $n$.
Therefore if $d_1, d_2, \ldots, d_r$ are all the divisors of $n$, then so are $\dfrac n {d_1}, \dfrac n {d_2}, \ldots, \dfrac n {d_r}$, except in a different order.
Hence:
Category:Number Theory
Category:Divisors
Sum Over Divisors of Multiplicative Function

== Theorem ==
Let $f: \Z_{>0} \to \Z_{>0}$ be a multiplicative function.
Let $n \in \Z_{>0}$.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum over the divisors of $n$.
Then $\ds \map F n = \sum_{d \mathop \divides n} \map f d$ is also a multiplicative function.
== Proof ==
Let $\ds \map F n = \sum_{d \mathop \divides n} \map f d$.
Let $m, n \in \Z_{>0}: m \perp n$.
Then by definition:
:$\ds \map F {m n} = \sum_{d \mathop \divides m n} \map f d$
The divisors of $m n$ are of the form $d = r s$ where $r$ and $s$ are divisors of $m$ and $n$ respectively, from Divisors of Product of Coprime Integers.
It is noted that $r \perp s$, otherwise any common divisor of $r$ and $s$ would be a common divisor of $m$ and $n$.
Therefore:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f {r s}$
So, as $f$ is multiplicative:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f r \map f s$
But at the same time:
:$\ds \map F m \map F n = \paren {\sum_{r \mathop \divides m} \map f r} \paren {\sum_{s \mathop \divides n} \map f s}$
Multiplying out the product on the , $\map F {m n}$ and $\map F m \map F n$ are seen to be the same.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $31$
Category:Multiplicative Functions
Sum Over Divisors of von Mangoldt is Logarithm

== Theorem ==
Let $\Lambda$ be von Mangoldt's function.
Then for $n \ge 1$:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$
== Proof ==
Let $n \ge 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct primes and $e_1, \ldots, e_k > 0$.
Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.
By the definition of $\Lambda$, for such $d$ we have $\map \Lambda d \ne 0$ there is exactly one $i \in \set {1, \ldots, k}$ such that $f_i > 0$.
If this is the case, let $d = p_i^{f_i}$.
Then:
:$\map \Lambda d = \ln p_i$
Therefore:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \sum_{i \mathop = 1}^k e_i \ln p_i$
Also, we have:
Thus we indeed have:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$
Category:Analytic Number Theory
Category:Von Mangoldt Function
Sum Rule for Convergent Generalized Sums

== Theorem ==
Let $\struct {G, +}$ be a commutative topological semigroup.
Let $\family {g_i}_{i \mathop \in I}$ and $\family {h_i}_{i \mathop \in I}$ be an indexed family of elements in $G$.
Let the generalized sums $\ds \sum_{i \mathop \in I} g_i$ and $\ds \sum_{i \mathop \in I} h_i$ be convergent to the following limits:
:$\ds \sum_{i \mathop \in I} g_i = a$
:$\ds \sum_{i \mathop \in I} h_i = b$
Then:
:the generalized sum $\ds \sum_{i \mathop \in I} \paren{g_i + h_i}$ converges to the limit:
::$\ds \sum_{i \mathop \in I} \paren{g_i + h_i} = a + b$
== Proof ==
Consider the set $\FF$ of finite subsets of $I$.
By definition of Definition:Generalized Sum:
:$\ds \sum_{i \mathop \in I} g_i$ is the net $\ds \family{\sum_{i \mathop \in F} g_i}_{F \mathop \in \FF}$
and
:$\ds \sum_{i \mathop \in I} h_i$ is the net $\ds \family{\sum_{i \mathop \in F} h_i}_{F \mathop \in \FF}$
By definition of convergence:
:$\ds \lim_{F \mathop \in \FF} \paren{\sum_{i \mathop \in F} g_i} = a$
and:
:$\ds \lim_{F \mathop \in \FF} \paren{\sum_{i \mathop \in F} h_i} = b$
From Sum Rule for Convergent Nets:
:$\ds \lim_{F \mathop \in \FF} \paren{\sum_{i \mathop \in F} g_i + \sum_{i \mathop \in F} h_i} = a + b$
We have:
:$\forall F \in \FF : \ds \paren{\sum_{i \mathop \in F} g_i} + \paren{\sum_{i \mathop \in F} h_i} = \sum_{i \mathop \in F} \paren{g_i + h_i}$
Hence:
:$\ds \lim_{F \mathop \in \FF} \sum_{i \mathop \in F} \paren{g_i + h_i} = a + b$
By definition of generalized sum, $\ds \sum_{i \mathop \in I} \paren{g_i + h_i}$ converges to the limit:
:$\ds \sum_{i \mathop \in I} \paren{g_i + h_i} = a + b$ 
Category:Generalized Sums
Sum Rule for Convergent Nets

== Theorem ==
Let $\struct {G, +}$ be a commutative topological semigroup.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {g_\lambda}_{\lambda \mathop \in \Lambda}$ and $\family {h_\lambda}_{\lambda \mathop \in \Lambda}$ be an indexed family of elements in $G$.
Let $\family {g_\lambda}_{\lambda \mathop \in \Lambda}$ and $\family {h_\lambda}_{\lambda \mathop \in \Lambda}$ be convergent to the following limits:
:$\ds \lim_{\lambda \mathop \in \Lambda} g_\lambda = a$
:$\ds \lim_{\lambda \mathop \in \Lambda} h_\lambda = b$
Then:
:the indexed family $\family {g_\lambda + h_\lambda}_{\lambda \mathop \in \Lambda}$ converges to the limit:
::$\ds \lim_{\lambda \mathop \in \Lambda} g_\lambda + h_\lambda = a + b$
== Proof ==
Let $U$ be an open neighborhood of $a + b$.
By definition of topological semigroup:
:the binary operation $+ : G \times G \to G$ is continuous.
By definition of continuous mapping:
:$\exists W, V$ open neighborhoods of $a$ and $b$ respectively:
::$+ \sqbrk {W \times V} \subseteq U$
By definition of convergence:
:$\exists \lambda_1 \in \Lambda : \forall \mu \in \Lambda : \lambda_1 \preceq \mu \implies g_\mu \in W$
and:
:$\exists \lambda_2 \in \Lambda : \forall \mu \in \Lambda : \lambda_2 \preceq \mu \implies h_\mu \in V$
By definition of directed set:
:$\exists \lambda \in \Lambda : \lambda_1, \lambda_2 \preceq \lambda$
By definition of directed set:
:$\forall \mu \in \Lambda : \lambda \preceq \mu \implies \lambda_1, \lambda_2 \preceq \mu$
Hence $\forall \mu \in \Lambda$ such that $\lambda \preceq \mu$:
:$g_\mu + h_\mu \in U$
Since $U$ was arbitrary, for all open neighborhoods $U$ of $a + b$:
:$\exists \lambda \in \Lambda : \forall \mu \in \Lambda : \lambda \preceq \mu \leadsto g_\mu + h_\mu \in U$
By definition of limit of net:
:$\ds \lim_{\lambda \in \Lambda} g_\lambda + h_\lambda = a + b$
Category:Nets (Set Theory)
Sum Rule for Derivatives

== Theorem ==
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x + \map k x$.
Then $f$ is differentiable at $\xi$ and:
:$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
:$\forall x \in I: \map {f'} x = \map {j'} x + \map {k'} x$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.2$
* : $5$. Differential Calculus: Appendix: Differentiation Rules: $2.$ Sum (algebraic) rule
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
Category:Differential Calculus
Category:Sum Rule for Derivatives
Sum Rule for Derivatives/General Result

== Theorem ==
Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions all differentiable.
Then for all $n \in \N_{>0}$:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$
$\map P 1$ is true, as this just says:
:$\map {D_x} {\map {f_1} x} = \map {D_x} {\map {f_1} x}$
which is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\ds \map {D_x} {\map {f_1} x + \map {f_2} x} = \map {D_x} {\map {f_1} x} + \map {D_x} {\map {f_2} x}$
which has been proved in Sum Rule for Derivatives.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^k \map {f_i} x} = \sum_{i \mathop = 1}^k \map {D_x} {\map {f_i} x}$
from which it is to be shown that:
:$\ds \map {D_x} {\sum_{i \mathop = 1}^{k + 1} \map {f_i} x} = \sum_{i \mathop = 1}^{k + 1} \map {D_x} {\map {f_i} x}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$
== Sources ==
* : $\S 13$: General Rules of Differentiation: $13.5$
Category:Differential Calculus
Category:Sum Rule for Derivatives
Sum Rule for Sequence in Normed Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences such that: 
:$x_n \to x$
and:
:$y_n \to y$
Then: 
:$x_n + y_n \to x + y$ 
== Proof ==
For each $n \in \N$, we have: 
So from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$x_n + y_n \to x + y$
Category:Convergent Sequences (Normed Vector Spaces)