nyuuzyou/wikis huggingface dataset:
Exponent Combination Laws/Power of Power/Proof 2

== Theorem ==
Let $a \in \R_{>0}$ be a (strictly) positive real number.
== Proof ==
We will show that:
:$\forall \epsilon \in \R_{>0}: \size {a^{x y} - \paren {a^x}^y} < \epsilon$
, suppose that $x < y$.
Consider $I := \closedint x y$.
Let $I_\Q = I \cap \Q$.
Let $M = \max \set {\size x, \size y}$
Fix $\epsilon \in \R_{>0}$.
From Real Polynomial Function is Continuous:
:$\exists \delta' \in \R_{>0}: \size {a^x - a^{x'} } < \delta' \leadsto \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } < \dfrac \epsilon 4$
From Power Function on Strictly Positive Base is Continuous:
Further:
and:
Let $\delta = \max \set {\dfrac {\delta_1} {\size x}, \dfrac {\delta_2} M, \delta_3, \delta_4}$.
From Closure of Rational Interval is Closed Real Interval:
:$\exists r, s \in I_\Q: \size {x - r} < \delta \land \size {y - s} < \delta$
Thus:
Hence the result, by Real Plus Epsilon.
Category:Exponent Combination Laws
Exponent Combination Laws/Power of Product

== Theorem ==
Let $a, b \in \R_{\ge 0}$ be positive real numbers.
Let $x \in \R$ be a real number.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$\paren {a b}^x = a^x b^x$
== Proof ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.6$
* : $\S 14.7 \ (1) \ \text{(ii)}$
* : $\S 13$: Laws of Exponents: $13.6.$
* 
Category:Exponent Combination Laws
Exponent Combination Laws/Power of Quotient

== Theorem ==
Let $a, b \in \R_{>0}$ be (strictly) positive real numbers.
Let $x \in \R$ be a real number.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$\paren {\dfrac a b}^x = \dfrac {a^x} {b^x}$
== Proof ==
Category:Exponent Combination Laws
Exponent Combination Laws/Product of Powers

== Theorem ==
Let $a \in \R_{> 0}$ be a positive real number.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$a^x a^y = a^{x + y}$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.1$
* : Chapter $1$: Scientific Notation
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* 
* 
* : $\S 13$: Laws of Exponents: $13.1.$
* 
Category:Exponent Combination Laws
Exponent Combination Laws/Product of Powers/Proof 1

== Theorem ==
Let $a \in \R_{> 0}$ be a positive real number.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$a^x a^y = a^{x + y}$
== Proof ==
== Sources ==
* : $\S 14.7 \ (1) \ \text{(i)}$
Category:Exponent Combination Laws
Exponent Combination Laws/Product of Powers/Proof 2

== Theorem ==
Let $a \in \R_{> 0}$ be a positive real number.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$a^x a^y = a^{x + y}$
== Proof ==
Let $x, y \in \R$.
From Rational Sequence Decreasing to Real Number, there exist rational sequences $\sequence {x_n}$ and $\sequence {y_n}$ converging to $x$ and $y$, respectively.
Then, since Power Function on Strictly Positive Base is Continuous: Real Power:
Category:Exponent Combination Laws
Exponent Combination Laws/Product of Powers/Proof 2/Lemma

== Theorem ==
Let $x_1, x_2, y_1, y_2 \in \R_{>0}$ be strictly positive real numbers.
Let $\epsilon \in \openint 0 {\min \set {y_1, y_2, 1} }$.
Then:
:$\size {x_1 - y_1} < \epsilon \land \size {x_2 - y_2} < \epsilon \implies \size {x_1 x_2 - y_1 y_2} < \epsilon \paren {y_1 + y_2 + 1}$
== Proof ==
First:
The same logic, mutatis mutandis, shows that $0 < y_2 - \epsilon$.
From Negative of Absolute Value: Corollary 3:
Hence:
Subtracting $y_1 y_2$ from all sections of the inequality:
:$-\epsilon \paren {y_1 + y_2} - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \paren {y_1 + y_2} + \epsilon^2$
If follows that:
Hence the result.
Category:Exponent Combination Laws
Exponent Combination Laws/Quotient of Powers

== Theorem ==
Let $a \in \R_{>0}$ be a positive real number.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$\dfrac{a^x} {a^y} = a^{x - y}$
== Proof ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.2$
* 
* 
* : $\S 13$: Laws of Exponents: $13.2.$
* 
Category:Exponential Function
Exponent Not Equal to Zero

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x \ne 0$.
Then:
:$x^y \ne 0$
== Proof ==
The proof shall proceed by Transfinite Induction on $y$.
=== Basis for the Induction ===
:$x^0 = 1$ by the definition of ordinal exponentiation.
Therefore, $x^0 \ne 0$.
This proves the basis for the induction.
=== Induction Step ===
The inductive hypothesis supposes that $x^y \ne 0$.
This proves the induction step.
=== Limit Case ===
The inductive hypothesis says that:
:$\forall z \in y: x^z \ne 0$
This proves the limit case.
== Sources ==
* : $\S 8.32$
Category:Ordinal Arithmetic
Exponent of Convergence is Less Than Order

== Theorem ==
Let $f: \C \to \C$ be an entire function.
Let $\omega$ be its order.
Let $\tau$ be its exponent of convergence.
Then $\tau \le \omega$.
== Proof ==
We may assume $\map f 0 \ne 0$.
Let $f$ have finitely many zeroes.
Then:
:$\tau = 0 \le \omega$
Let $f$ have infinitely many zeroes.
Let $\sequence {a_n}$ be the sequence of nonzero zeroes of $f$, repeated according to multiplicity and ordered by increasing modulus.
Let $r_n = \size {a_n}$ and $R_n = 2 \size {a_n}$.
By Jensen's Inequality:
:$n \le \dfrac {\map \log {\max_{\size z \le R_n} \size f} - \log \size {\map f 0} } {\log 2}$
for all $n \in \N$.
Let $\epsilon > 0$.
Because $f$ has order $\omega$:
:$n \ll_\epsilon \size {a_n}^{\omega + \epsilon}$
Thus:
:$\size {a_n}^{-\paren {\omega + \epsilon} } \ll_\epsilon n^{-1}$
By Convergence of P-Series:
:$\omega + \epsilon \ge \tau$
Because $\epsilon$ is arbitrary:
:$\omega \ge \tau$
== Also see ==
* Order is Maximum of Exponent of Convergence and Degree
== Sources ==
* : Chapter III: Further Theory of $\zeta(s)$. Applications: $\S 7$: Integral functions: Theorem $\text F 2$
Category:Entire Functions
Exponential Distribution in terms of Beta Distribution

== Theorem ==
Let $\sequence {X_n}$ be a sequence of independent random variables with:
:$X_n \sim \BetaDist 1 n$
for each natural number $n$, where $\BetaDist 1 n$ denotes the beta distribution with parameters $1$ and $n$. 
Then: 
:$n X_n \xrightarrow d X$
with:
:$X \sim \Exponential 1$
where: 
:$\Exponential 1$ denotes the exponential distribution with parameter $1$,
:$\xrightarrow d$ denotes convergence in distribution.
== Proof ==
We aim to show that for each real $x > 0$, we have: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {X_n \le x} = \map \Pr {X \le x}$
From the definition of the exponential distribution, we have: 
:$\map \Pr {X \le x} = 1 - e^{-x}$
Note that, from the definition of the beta distribution:
:$0 \le X_n \le 1$
So, if $n \le x$, we have: 
:$\map \Pr {X_n \le x} = 1$
Take $n \ge x$. 
Then, we have: 
From the definition of the exponential function as a sequence, we have:
:$\ds \lim_{n \mathop \to \infty} \paren {1 - \frac x n}^n = e^{-x}$
so:
:$\ds \lim_{n \mathop \to \infty} \paren {1 - \paren {1 - \frac x n}^n} = 1 - e^{-x} = \map \Pr {X \le x}$
That is: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {X_n \le x} = \map \Pr {X \le x}$
as required.
Category:Exponential Distribution
Category:Beta Distribution
Exponential Distribution in terms of Continuous Uniform Distribution

== Theorem ==
Let $X \sim \mathrm U \hointl 0 1$ where $\mathrm U \hointl 0 1$ is the continuous uniform distribution on $\hointl 0 1$.
Let $\beta$ be a positive real number.
Then: 
:$-\beta \ln X \sim \Exponential \lambda$
where $\Exponential \cdot$ is the exponential distribution.
== Proof ==
Let $Y \sim \Exponential \lambda$.
We aim to show that:
:$\map \Pr {Y < -\beta \ln x} = \map \Pr {X > x}$
for all $x \in \hointl 0 1$.
We have:
Category:Exponential Distribution
Category:Continuous Uniform Distribution
Exponential Dominates Polynomial

== Theorem ==
Let $\exp$ denote the real exponential function.
Let $k \in \N$.
Let $\alpha \in \R_{>0}$.
Then: 
:$\exists N \in \N: \forall x \in \R_{>N}: \map \exp {\alpha x} > x^k$
== Proof ==
Choose any $N > \dfrac {\paren {k + 1}!} {\alpha^{k + 1} }$, where $!$ denotes the factorial.
By Taylor Series Expansion for Exponential Function we have for $x \in \R_{\ge 0}$:
:$\ds \map \exp {\alpha x} = \sum_{m \mathop \ge 0} \frac {\paren {\alpha x}^m}{m!} > \frac {\paren {\alpha x}^{k + 1} } {\paren {k + 1}!}$
Therefore, for $x > N$:
Category:Exponential Function
Category:Polynomial Theory
Exponential Form of Complex Conjugate

== Theorem ==
Let $z := r e^{i \theta} \in \C$ be a complex number expressed in exponential form.
Then:
:$\overline z = r e^{-i \theta}$
where $\overline z$ denotes the complex conjugate of $z$.
== Proof ==
== Sources ==
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation
Category:Complex Conjugates
Exponential Function as Limit of Gaussian Hypergeometric Function

== Theorem ==
:$\ds e^x = \lim_{p \mathop \to \infty} \map F {1, p; 1; \dfrac x p}$
where:
:$x$ is a real number
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Exponential Function in terms of Hypergeometric Function
== Sources ==
* : $\S 31$: Hypergeometric Functions: Special Cases: $31.5$
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Exponential Function in terms of Hypergeometric Function

== Theorem ==
:$e^x = \map { {}_1 \operatorname F_1} {p; p; x}$
where:
:$x$ is a real number with $x \in \R$
:$e^x$ denotes the exponential function
:$\map { {}_1 \operatorname F_1} {a; b; x}$ is the generalized hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} } { b^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Exponential Function as Limit of Gaussian Hypergeometric Function
* Sine Function in terms of Hypergeometric Function
* Cosine Function in terms of Hypergeometric Function
* Hyperbolic Sine Function in terms of Hypergeometric Function
* Hyperbolic Cosine Function in terms of Hypergeometric Function
Category:Hypergeometric Functions
Category:Exponential Function
Exponential Function is Continuous/Complex

== Theorem ==
The complex exponential function is continuous.
That is:
:$\forall z_0 \in \C: \ds \lim_{z \mathop \to z_0} \exp z = \exp z_0$
== Proof ==
This proof depends on the differential equation definition of the exponential function.
The result follows from Complex-Differentiable Function is Continuous.
Category:Exponential Function is Continuous
Exponential Function is Continuous/Real Numbers

== Theorem ==
The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Exponential Function is Continuous
Category:Continuous Real Functions
Exponential Function is Continuous/Real Numbers/Proof 1

== Theorem ==
== Proof ==
This proof depends on the limit definition of the exponential function.
Let:
:$\ds \exp x = \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n$
Fix $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
From Closed Bounded Subset of Real Numbers is Compact, $I$ is compact.
From Exponential Sequence is Uniformly Convergent on Compact Sets:
:$\paren {1 + \dfrac x n}^n$ is uniformly convergent on $I$.
By the Uniform Limit Theorem:
:$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \exp x$
is continuous on $I$.
In particular, $\exp x$ is continuous at $x_0$.
Category:Exponential Function is Continuous
Exponential Function is Continuous/Real Numbers/Proof 2

== Theorem ==
== Proof ==
This proof depends on the definition of the exponential function as the function inverse of the natural logarithm.
From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$.
From Real Natural Logarithm Function is Continuous, $\ln$ is continuous on $\R_{>0}$
Thus, from the Continuous Inverse Theorem, $\exp := \ln^{-1}$ is continuous.
Category:Exponential Function is Continuous
Exponential Function is Continuous/Real Numbers/Proof 5

== Theorem ==
== Proof ==
This proof depends on the series definition of $\exp$.
That is, let:
:$\ds \exp x = \sum_{k \mathop = 0}^ \infty \frac {x^k} {k!}$
From Series of Power over Factorial Converges, the radius of convergence of $\exp$ is $\infty$.
Thus, from Power Series Converges to Continuous Function, $\exp$ is continuous on $\R$.
Category:Exponential Function is Continuous
Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation

== Theorem ==
Let:
:$(1): \quad y' ' + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Then:
:$y = e^{m_1 x}$ is a solution to $(1)$
:$m_1$ is a root of the auxiliary equation $m^2 + p m + q = 0$
== Proof ==
Consider the equation:
:$y = e^{m_1 x}$
Differentiating $x$:
=== Sufficient Condition ===
Let $y = e^{m_1 x}$ be a solution to $(1)$.
Substituting for $y$ and its derivatives in $(1)$:
That is, by definition, $m_1$ is a root of $m^2 + p m + q = 0$.
=== Necessary Condition ===
Let $m_1$ be a root of $m^2 + p m + q = 0$.
Thus:
Thus $y = e^{m_1 x}$ satisfies $(1)$.
== Sources ==
* : $\S 3.17$: The Homogeneous Equation with Constant Coefficients
Category:Constant Coefficient Homogeneous LSOODEs
Exponential Function is Superfunction

== Theorem ==
The function $f : \C \to \C$, defined as:
:$\map f z = c^z$
is a superfunction for any complex number $c$.
== Proof ==
Define $h: \C \to \C$ by $\map h z = z \times c$.
Then:
Thus $\map f z = c^z$ is a superfunction and $\map h z = z \times c$ is the corresponding transfer function.
Category:Superfunctions
Exponential Function is Well-Defined/Real

== Theorem ==
Let $x \in \R$ be a real number.
Let $\exp x$ be the exponential of $x$.
Then $\exp x$ is well-defined.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Exponential Function
Category:Exponential Function is Well-Defined
Exponential Function is Well-Defined/Real/Proof 1

== Theorem ==
== Proof ==
This proof assumes the power series definition of $\exp$.
From Series of Power over Factorial Converges:
:$\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges 
Hence the result, from Convergent Real Sequence has Unique Limit.
Category:Exponential Function is Well-Defined
Exponential Function is Well-Defined/Real/Proof 2

== Theorem ==
== Proof ==
This proof assumes the sequence definition of $\exp$.
Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R \to \R$ defined as:
:$\map {f_n} x = \paren {1 + \dfrac x n}^n$
Fix $x \in \R$.
Then:
Thus, $\sequence {\map {f_n} x}$ is bounded above.
From Exponential Sequence is Eventually Increasing:
:$\exists N \in \N: \sequence {\map {f_{N + n} } x}$ is increasing
From Monotone Convergence Theorem (Real Analysis), $\sequence {\map {f_{N + n} } x}$ converges to some $z \in \R$.
From Tail of Convergent Sequence, $\sequence {\map {f_n} x}$ converges to $z$.
Hence the result, from Limit of Real Function is Unique.
Category:Exponential Function is Well-Defined
Exponential Function is Well-Defined/Real/Proof 3

== Theorem ==
== Proof ==
This proof assumes the continuous extension definition of $\exp$.
Let $e$ denote Euler's number.
Let $f: \Q \to \R$ be the real-valued function defined as:
:$f \left({r}\right) = e^r$
From Euler's Number to Rational Power permits Unique Continuous Extension, there exists a unique continuous extension of $f$ to $\R$.
Hence the result.
Category:Exponential Function is Well-Defined
Exponential Function is Well-Defined/Real/Proof 4

== Theorem ==
== Proof ==
This proof assumes the definition of the exponential as the inverse of the logarithm.
From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$.
From Inverse of Strictly Monotone Function, $f$ permits an inverse mapping.
Hence the result, from Inverse Mapping is Unique.
Category:Exponential Function is Well-Defined
Exponential Function is Well-Defined/Real/Proof 5

== Theorem ==
== Proof ==
This proof assumes the definition of $\exp$ as the solution to an initial value problem.
That is, suppose $\exp$ solves:
:$(1): \quad \dfrac \d {\d x} y = \map f {x, y}$
:$(2): \quad \map \exp 0 = 1$
on $\R$, where $\map f {x, y} = y$.
From Derivative of Exponential Function: Proof 4, the function $\phi : \R \to \R$ defined as:
:$\ds \map \phi x = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$
satisfies $\map {\phi'} x = \map \phi x$.
So $\phi$ satisfies $(1)$.
From Exponential of Zero: Proof 3:
:$\map \phi 0 = 1$
So $\phi$ satisfies $(2)$.
Thus, $\phi$ is a solution to the initial value problem given.
From Exponential Function is Continuous: Proof 5 and $(1)$:
:$\phi$ is continuously differentiable on $\R$.
Because $\map f {x, \phi} = \phi$:
:$f$ is continuously differentiable on $\R^2$.
Thus, from Uniqueness of Continuously Differentiable Solution to Initial Value Problem, this solution is unique.
Category:Exponential Function is Well-Defined
Exponential Generating Function for Boubaker Polynomials

== Theorem ==
The Boubaker polynomials, defined as: 
:$\map {B_n} x = \begin {cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end {cases}$
have as an exponential generating function:
:$\ds \map {f_{B_n, \operatorname {EXP} } } {x, t} = \sum_{n \mathop = 0}^\infty \map {B_n} x \frac {t^n} {n!} = 4 x e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} } - 3$
== Proof ==
 
From the definition of the Boubaker polynomials, we have: 
:$\begin {cases} \map {B_0} x = 1 \\ \map {B_1} x = x \end {cases}$
We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind: 
:$\begin {cases} \map {T_0} x = 1 \\ \map {T_1} x = x \end {cases}$
and:
:$\begin {cases} \map {U_{-1} } x = 0 \\ \map {U_0} x = 1 \\ \map {U_1} x = 2 x \end {cases}$
Then, from the definition of the exponential generating functions of the Chebyshev polynomials of the first kind:
:$\ds \map {f_{T_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} }$
and the Chebyshev polynomials of the second kind: 
:$\ds \map {f_{U_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} }$
 
and from the relation:
:$\map {B_n} x = 4 x \map {U_{n - 1} } {\dfrac x 2} - 2 \map {T_n} {\dfrac x 2}$
which is valid only for $n > 1$, we have:
:$\ds \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} + \sum_{n \mathop = 2}^\infty \map {B_n} x \dfrac {t^n} {n!} = 4 x \sum_{n \mathop = 2}^\infty \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 2}^\infty \map {T_n} x \dfrac {t^n} {n!} + \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} - 4 x \sum_{n \mathop = 0}^1 \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 0}^1 \map {T_n} x \dfrac {t^n} {n!}$ 
Finally, by simplifying the two sides and considering the exponential generating functions and the values of the first terms of $T_n$ and $U_n$, we obtain: 
:$\ds \sum_{n \mathop = 0}^\infty \map {B_n} x \frac{t^n} {n!} = 4 x e^{t \frac x 2} \dfrac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\dfrac x 2}^2} } - 3$
Hence the result. 
 
Category:Boubaker Polynomials
Category:Examples of Exponential Generating Functions
Exponential Growth Equation/Special Case

== Theorem ==
All solutions of the differential equation $y' = y$ take the form $y = C e^x$.
== Proof ==
Let $\map f x = C e^x$.
Then by Derivative of Exponential Function:
:$\map {f'} x = \map f x$
From Exponential of Zero:
:$\map f 0 = C$
Hence $C e^x$ is a solution of $y' = y$.
Now suppose that a function $f$ satisfies $\map {f'} x = \map f x$. 
Consider $\map h x = \map f x e^{-x}$.
By the Product Rule for Derivatives:
From Zero Derivative implies Constant Function, $h$ must be a constant function.
Therefore, $\map h x = \map h 0 = \map f 0$.
Recalling the definition of $h$, it follows that:
:$\map f x = \map f 0 e^x$
== Sources ==
* : $\S 8.1$
Category:First Order ODEs
Exponential Sequence Converges Compactly

== Theorem ==
For each $n \in \N$, let $f_n : \C \to \C$ be defined as:
:$\map {f_n} z = \paren {1 + \dfrac z n}^n$
Then, $\sequence {f_n}$ converges compactly to the complex exponential function.
== Proof ==
Let $D$ be a compact subset of $\C$.
By definition, $D$ is bounded, so there is some $M \in \R$ such that:
:$\forall z \in D: \size z \le M$
Let $\epsilon > 0$ be arbitrary.
By definition of the exponential function, there is some $N \in \N$ such that:
:$\forall n \ge N: \map \exp M - \paren {1 + \dfrac M n}^n < \epsilon$
Now, for any $z \in D$ and $n \ge N$:
As $\epsilon > 0$ was arbitrary, $\sequence {f_n}$ converges uniformly to $\exp$ on $D$.
But as $D$ was also arbitrary, $\sequence {f_n}$ converges compactly.
Category:Compact Convergence
Category:Exponential Function
Exponential Sequence is Eventually Increasing

== Theorem ==
Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for sufficiently large $n \in \N$, $\sequence {\map {E_n} x}$ is increasing $n$.
That is:
:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x \le \map {E_{n + 1} } x$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
Fix $x \in \R$.
Then:
So we may apply the AM-GM inequality, with:
:$x_1 := 1$
:$x_j := 1 + \dfrac x n$ for each $j \in \set {2, 3, \dotsc, n + 1}$
to obtain that:
:$\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$
After simplification:
:$1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$
From Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
:$\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$
where we have raised both sides to the power of $n + 1$.
Hence the result.
Category:Exponential Function
Exponential Sequence is Eventually Strictly Positive

== Theorem ==
Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for each $x \in \R$ and for sufficiently large $n \in \N$, $\map {E_n} x$ is positive.
That is:
:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x > 0$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
Fix $x \in \R$.
Then:
Category:Exponential Function