nyuuzyou/wikis huggingface dataset:
Combination Theorem for Bounded Real-Valued Functions

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $\lambda \in \R$.
Then the following results hold.
=== Sum Rule ===
=== Negation Rule ===
=== Difference Rule ===
=== Product Rule ===
=== Multiple Rule ===
=== Absolute Value Rule ===
=== Maximum Rule ===
=== Minimum Rule ===
Category:Combination Theorem for Bounded Real-Valued Functions 
Category:Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Absolute Value Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be a bounded real-valued function.
Let $\size f : S \to \R$ denote the absolute value of $f$, that is, $\size f$ denotes the mapping defined by:
:$\forall s \in S : \map {\size f} s = \size{\map f s}$
 
Then:
:$\size f$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M$
We have:
It follows that $\size f$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Difference Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f - g : S \to \R$ be the pointwise addition of $f$ and $g$, that is, $f - g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f - g} } s = \map f s - \map g s$
Then:
:$f - g$ is a bounded real-valued function
== Proof ==
From Pointwise Difference is Pointwise Addition with Negation:
:$f - g = f + \paren{-g}$
where:
:$-g$ denotes the pointwise negation of $g$
:$f + \paren{-g}$ denotes the pointwise addition of $f$ and $-g$
From Negation Rule for Bounded Real-Valued Function:
:$-g$ is a bounded real-valued function
From Sum Rule for Bounded Real-Valued Functions
:$f + \paren{-g}$ is a bounded real-valued function
The result follows.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Maximum Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f \vee g$ denote the pointwise maximum of $f$ and $g$, that is, $f \vee g$ is the mapping defined by:
:$\forall s \in S : \map {\paren{f \vee g} } s = \max \set{\map f s, \map g s}$
Then:
:$f \vee g$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
and
:$\exists M_g \in \R_{\ge 0} : \forall s \in S : \size{\map g s} \le M_g$
From Negative of Absolute Value:
:$\forall s \in S : \map f s \le \size{\map f s}$
and
:$\forall s \in S : \map g s \le \size{\map g s}$
Let $M = \max \set{M_f, M_g}$.
We have:
It follows that $f \vee g$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Minimum Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f \wedge g$ denote the pointwise minimum of $f$ and $g$, that is, $f \wedge g$ is the mapping defined by:
:$\forall s \in S : \map {\paren{f \wedge g} } s = \min \set{\map f s, \map g s}$
Then:
:$f \wedge g$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
and
:$\exists M_g \in \R_{\ge 0} : \forall s \in S : \size{\map g s} \le M_g$
From Negative of Absolute Value:
:$\forall s \in S : \map f s \le \size{\map f s}$
and
:$\forall s \in S : \map g s \le \size{\map g s}$
Let $M = \min \set{M_f, M_g}$.
We have:
It follows that $f \wedge g$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Multiple Rule

== Theorem ==
Let $\struct{S, \tau_S}$ be a topological space.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be bounded real-valued function.
Let $\lambda \in \R$.
Let $\lambda f : S \to \R$ be the pointwise scalar multiplication of $f$ by $\lambda$, that is, $\lambda f$ is the mappping defined by:
:$\forall s \in S : \map {\paren{\lambda f} } s = \lambda \map f s$
Then:
:$\lambda f$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
Let $M = \size{\lambda} M_f$.
We have:
It follows that $\lambda f$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Negation Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be bounded real-valued function.
Let $-f : S \to \R$ denote the pointwise negation of $f$, that is, $-f$ denotes the mapping defined by:
:$\forall s \in S : \map {\paren{-f} } s = - \map f s$
Then:
:$-f$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M$
We have:
It follows that $-f$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Product Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f g : S \to \R$ be the pointwise product of $f$ and $g$, that is, $f g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f g} } s = \map f s \map g s$
Then:
:$f g$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
and
:$\exists M_g \in \R_{\ge 0} : \forall s \in S : \size{\map g s} \le M_g$
Let $M = M_f M_g$.
We have:
It follows that $f g$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Bounded Real-Valued Functions/Sum Rule

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f + g : S \to \R$ be the pointwise addition of $f$ and $g$, that is, $f + g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f + g} } s = \map f s + \map g s$
Then:
:$f + g$ is a bounded real-valued function
== Proof ==
By definition of bounded real-valued function
:$\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
and
:$\exists M_g \in \R_{\ge 0} : \forall s \in S : \size{\map g s} \le M_g$
Let $M = M_f + M_g$.
We have:
It follows that $f + g$ is a bounded real-valued function by definition.
Category:Combination Theorem for Bounded Real-Valued Functions
Combination Theorem for Cauchy Sequences

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n}$ be Cauchy sequences in $R$.
Let $a, b \in R$.
The following results hold:
=== Constant Rule ===
=== Sum Rule ===
=== Difference Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Inverse Rule ===
=== Quotient Rule ===
== Sources ==
* : $\S 3.2$: Completions
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Cauchy Sequences in Normed Division Rings
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Combined Sum Rule

== Theorem == 
Let $\struct {R, \norm{\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n}$ be Cauchy sequences in $R$.
Let $a, b \in R$.
Then:
:$\sequence {a x_n + b y_n }$ is a Cauchy sequence.
== Proof ==
From the Multiple Rule for Normed Division Ring Sequences:
:$\sequence {a x_n}$ is a Cauchy sequence
:$\sequence {b y_n}$ is a Cauchy sequence.
The result now follows directly from the Sum Rule for Normed Division Ring Sequences:
:$\sequence {a x_n + b y_n}$ is a Cauchy sequence.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Constant Rule

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
The constant sequence $\tuple {a, a, a, \dots}$ is a Cauchy sequence.
== Proof ==
Let $\sequence {x_n}$ be the constant sequence: 
:$\forall n, x_n = a$
Given $\epsilon > 0$:
:$\forall n, m \ge 1: \norm {x_n - x_m} = \norm {a - a} = \norm {0} = 0 < \epsilon$
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Difference Rule

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n}$ be Cauchy sequences in $R$.
Then:
:$\sequence {x_n - y_n}$ is a Cauchy sequence.
== Proof ==
From Multiple Rule for Normed Division Ring Sequences:
:$\sequence {-y_n} = \sequence {\paren {-1} y_n}$ is a Cauchy sequence. 
From Sum Rule for Normed Division Ring Sequences:
:$\sequence {x_n - y_n} = \sequence {x_n + \paren {-y_n} }$ is a Cauchy sequence.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Inverse Rule

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ be a Cauchy sequences in $R$.
Suppose $\sequence {x_n}$ does not converge to $0$.
Then:
:$\exists K \in \N: \forall n > K : x_n \ne 0$
and the sequence:
:$\sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is well-defined and a Cauchy sequence.
== Proof ==
Since $\sequence {x_n}$ does not converge to $0$, by Cauchy Sequence Is Eventually Bounded Away From Non-Limit then:
:$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C < \norm {x_n}$
or equivalently:
:$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: 1 < \dfrac {\norm {x_n} } C$
By :
:$\forall n > K : x_n \ne 0$
Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ defined as:
:$y_n = x_{K + n}$
By Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
:$\sequence {y_n}$ is a Cauchy sequence.
So $\sequence { {y_n}^{-1} }$ is well-defined and $\sequence { {y_n}^{-1} } = \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$.
Let $\epsilon > 0$ be given.
Let $\epsilon' = \epsilon C^2$, then $ \epsilon' > 0$.
Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N$ such that:
:$\forall n, m > N_2: \norm {y_n - y_m} < \epsilon'$
Thus $\forall n, m > N$:
:$(1): \quad 1 < \dfrac {\norm {y_n} } C, \dfrac {\norm {y_m} } C$
:$(2): \quad \norm {y_n - y_m} < \epsilon'$
Hence:
So:
:$\sequence { { {y_n}^{-1} } }$ is a Cauchy sequence in $R$ by definition.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Multiple Rule

== Theorem == 
Let $\struct {R, \norm{\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Let $a \in R$.
Then:
:$\sequence {a x_n}$ is a Cauchy sequence.
== Proof ==
Follows directly from Product Rule for Normed Division Ring Sequences, setting
:$\sequence {y_n} := \sequence {x_n}$
and:
:$\sequence {x_n} := \tuple {a, a, a, \ldots}$
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Product Rule

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be Cauchy sequences in $R$.
Then:
:$\sequence {x_n y_n}$ is a Cauchy sequence.
== Proof ==
Because $\sequence {x_n} $ is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.
Suppose $\norm {x_n} \le K_1$ for $n = 1, 2, 3, \ldots$.
Because $\sequence {y_n} $ is a is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.
Suppose $\norm {y_n} \le K_2$ for $n = 1, 2, 3, \ldots$.
Let $K = \max \set {K_1, K_2}$. 
Then both sequences are bounded by $K$.
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2K} > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
:$\forall n, m > N_1: \norm {x_n - x_m} < \dfrac \epsilon {2 K}$
Similarly, $\sequence {y_n} $ is a Cauchy sequence, we can find $N_2$ such that:
:$\forall n, m > N_2: \norm {y_n - y_m} < \dfrac \epsilon {2 K}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n, m > N$, both the above inequalities will be true.
Thus $\forall n, m > N$:
Hence:
:$\sequence {x_n y_n}$ is a Cauchy sequence in $R$.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Quotient Rule

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Suppose $\sequence {y_n}$ does not converge to $0$.
Then:
:$\exists K \in \N: \forall n > K : y_n \ne 0$
and the sequences:
:$\sequence { {x_{K + n} } \paren {y_{K + n} }^{-1} }_{n \mathop \in \N}$ and $\sequence {\paren {y_{K + n} }^{-1} {x_{K + n} } }_{n \mathop \in \N}$ are well-defined and Cauchy sequences.
== Proof ==
By the Inverse Rule for Normed Division Ring:
:$\exists K \in \N : \forall n > K : y_n \ne 0$.
and the sequence:
:$\sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is well-defined and a Cauchy sequence.
By Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_{K + n} }_{n \mathop \in \N}$ is a Cauchy sequence.
By Product Rule for Normed Division Ring Sequences:
:the sequences $\sequence { {x_{K + n} } \paren {y_{K + n} }^{-1} }_{n \mathop \in \N}$ and $\sequence {\paren {y_{K + n} }^{-1} {x_{K + n} } }_{n \mathop \in \N}$ are Cauchy sequences.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Cauchy Sequences/Sum Rule

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Then:
:$\sequence {x_n + y_n}$ is a Cauchy sequence.
== Proof ==
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
:$\forall n, m > N_1: \norm{x_n - x_m} < \dfrac \epsilon 2$
Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N_2$ such that:
: $\forall n, m > N_2: \norm{y_n - y_m} < \dfrac \epsilon 2$
Now let $N = \max \set {N_1, N_2}$.
Then if $n, m > N$, both the above inequalities will be true.
Thus $\forall n, m > N$:
Hence:
:$\sequence {x_n + y_n}$ is a Cauchy sequences in $R$.
== Sources ==
* : $\S 3.2$: Completions
Category:Combination Theorem for Cauchy Sequences
Combination Theorem for Complex Derivatives

== Theorem ==
Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $z \in D$.
Let $w, c \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Quotient Rule ===
== Also see ==
* Sum Rule for Derivatives
* Derivative of Constant Multiple
* Product Rule for Derivatives
* Quotient Rule for Derivatives
These theorems show that identical results hold for derivatives of real functions.
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Category:Complex Differential Calculus
Combination Theorem for Complex Derivatives/Combined Sum Rule

== Theorem ==
Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$.
Let $c, w \in \C$.
Let $\map {\dfrac \d {\d z} } {c f + w g}$ denote the derivative of $c f + w g$.
Then:
:$\map {\map {\dfrac \d {\d z} } {c f + w g} } z = c \dfrac \d {\d z} \map f z + w \dfrac \d {\d z} \map g z$
for all $z \in D$.
== Proof ==
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Multiple Rule

== Theorem ==
Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex-differentiable function on $D$.
Let $w \in \C$.
Then $w f$ is complex-differentiable in $D$, and its derivative $\paren {w f}'$ is defined by:
:$\map {\paren {w f}'} z = w \map {f'} z$
for all $z \in D$.
== Proof ==
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map \epsilon h}$
where $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is a complex function such that $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
Then:
:$w \map f {z + h} = w \map f z + h \paren {w \map {f'} z + w \map \epsilon h}$
From Multiple Rule for Limits of Complex Functions, it follows that:
:$\ds \lim_{h \mathop \to 0} w \map \epsilon h = \paren {\lim_{h \mathop \to 0} w} \paren {\lim_{h \mathop \to 0} \map \epsilon h} = 0$
Then the Epsilon-Function Complex Differentiability Condition shows that:
:$\map {\paren {w f}'} z = w \map {f'} z$
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Product Rule

== Theorem ==
Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $f g$ denote the pointwise product of the functions $f$ and $g$.
Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
:$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$
for all $z \in D$.
== Proof 1==
== Proof 2==
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Product Rule/Proof 1

== Theorem ==
Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $f g$ denote the pointwise product of the functions $f$ and $g$.
Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
:$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$
for all $z \in D$.
== Proof ==
Define $k: D \to \C$ as the pointwise product of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Product Rule/Proof 2

== Theorem ==
Let $D$ be an open subset of the set of complex numbers.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $f g$ denote the pointwise product of the functions $f$ and $g$.
Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
:$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$
for all $z \in D$.
== Proof ==
Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + \map h {\map {f'} z + \map {\epsilon_f} h}$
:$\map g {z + h} = \map g z + \map h {\map {g'} z + \map {\epsilon_g} h}$
where $\epsilon_f, \epsilon_g: \map {B_r} 0 \setminus \set 0 \to \C$ are complex functions that converge to $0$ as $h$ tends to $0$.
Then:
Define $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ by:
:$\map \epsilon h = h \paren {\map {f'} z + \map {\epsilon_f} h} \paren {\map {g'} z + \map {\epsilon_g} h}$
From Product Rule for Limits of Complex Functions and Combined Sum Rule for Limits of Complex Functions:
:$\ds \lim_{h \mathop \to 0} \map \epsilon h = \paren {\lim_{h \mathop \to 0} h} \paren {\lim_{h \mathop \to 0} \paren {\map {f'} z + \map {\epsilon_f} h} } \paren {\lim_{h \mathop \to 0} \paren {\map {g'} z + \map {\epsilon_g} h} } = 0$
Then the Epsilon-Function Complex Differentiability Condition shows that:
:$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Quotient Rule

== Theorem ==
Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions.
Let $\dfrac f g$ denote the pointwise quotient of the functions $f$ and $g$.
Then $\dfrac f g$ is complex-differentiable in $D \setminus \set {x \in D: \map g z = 0}$.
For all $z \in D$ with $\map g z \ne 0$:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$
where $\paren {\dfrac f g}'$ denotes the derivative of $\dfrac f g$.
== Proof ==
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D \setminus \set {x \in D: \map g z = 0}$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in \map {B_{r_0} } 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$
:$\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$
where $\epsilon_f, \epsilon_g: \map {B_{r_0} } 0 \setminus \set 0 \to \C$ are complex functions that converge to $0$ as $h$ tends to $0$.
From Complex-Differentiable Function is Continuous, it follows that $g$ is continuous at $z$.
Then there exists $r_1 \in \R_{>0}$ such that for all $h \in \map {B_{r_1} } 0$, we have $\size {\map g {z + h} - \map g z} < \size {\map g z}$.
Then by Backwards Form of Triangle Inequality:
:$\map g {z + h} \ne 0$ for all $h \in \map {B_{r_1} } 0$
Put $r = \min \set {r_0, r_1}$.
Then for all $h \in B_r \setminus \set 0$:
where $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is defined by:
:$\map \epsilon h = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}$
From Combination Theorem for Limits of Complex Functions, it follows that:
Then the Epsilon-Function Complex Differentiability Condition shows that:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Sum Rule

== Theorem ==
Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Then $f + g$ is complex-differentiable in $D$, and its derivative $\paren {f + g}'$ is defined by:
:$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$
for all $z \in D$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives
Combination Theorem for Complex Derivatives/Sum Rule/Proof 2

== Theorem ==
== Proof ==
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$
:$\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$
where $\epsilon_f, \epsilon_g: \map {B_r} 0 \setminus \set 0 \to \C$ are complex functions that converge to $0$ as $h$ tends to $0$.
Then:
From Sum Rule for Limits of Complex Functions, it follows that $\ds \lim_{h \mathop \to 0} \map {\paren {\epsilon_f + \epsilon_g} } h = 0$.
By the Epsilon-Function Complex Differentiability Condition, it follows that:
:$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$
== Sources ==
* $\S 1.1$
Category:Combination Theorem for Complex Derivatives