Equivalence of Definitions of Real Exponential Function/Power Series Expansion equivalent to Differential Equation

From ProofWiki
Jump to navigation Jump to search

Theorem

The following definitions of the concept of real exponential function are equivalent:

As a Power Series Expansion

The exponential function can be defined as a power series:

$\exp x := \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

As the Solution of a Differential Equation

The exponential function can be defined as the unique solution $y = \map f x$ to the first order ODE:

$\dfrac {\d y} {\d x} = y$

satisfying the initial condition $\map f 0 = 1$.


Proof

Power Series Expansion implies Solution of Differential Equation

Let $\exp x$ be the real function defined as the sum of the power series:

$\exp x := \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$


Let $y = \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.


Then:

\(\ds \dfrac {\d y} {\d x}\) \(=\) \(\ds \dfrac \d {\d x} \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)
\(\ds \) \(=\) \(\ds \map {\dfrac \d {\d x} } {\frac {x^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!} }\) extracting the zeroth term
\(\ds \) \(=\) \(\ds 0 + \sum_{n \mathop = 1}^\infty \frac {n x^{n - 1} } {n!}\) Power Rule for Derivatives and Derivative of Constant
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {\paren {n - 1}!}\) simplifying
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds y\)


Setting $x = 0$ we find:

\(\ds \map y 0\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {0^n} {n!}\)
\(\ds \) \(=\) \(\ds \frac {0^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {0^n} {n!}\)
\(\ds \) \(=\) \(\ds \frac {0^0} {0!}\) as $0^n = 0$ for all $n > 0$
\(\ds \) \(=\) \(\ds 1\) Definition of $0^0$

$\blacksquare$


This article needs to be linked to other articles.
In particular: uniform convergence
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.


That is:

$\exp x$ is the particular solution of the differential equation:

$\dfrac {\d y} {\d x} = y$

satisfying the initial condition $\map f 0 = 1$.

$\Box$


Solution of Differential Equation implies Power Series Expansion

Let $\exp x$ be the real function defined as the particular solution of the differential equation:

$\dfrac {\d y} {\d x} = y$

satisfying the initial condition $\map f 0 = 1$.


We have Taylor Series Expansion for Exponential Function:


From Higher Derivatives of Exponential Function, we have:

$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$


Since $\exp 0 = 1$, the Taylor series expansion for $\exp x$ about $0$ is given by:

$\ds \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$


From Radius of Convergence of Power Series over Factorial, we know that this power series expansion converges for all $x \in \R$.

From Taylor's Theorem, we know that

$\ds \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$

where $0 \le \eta \le x$.

Hence:

\(\ds \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }\) \(=\) \(\ds \size {\frac {x^n} {n!} \map \exp \eta}\)
\(\ds \) \(\le\) \(\ds \frac {\size {x^n} } {n!} \map \exp {\size x}\) Exponential is Strictly Increasing
\(\ds \) \(\to\) \(\ds 0\) \(\ds \text { as } n \to \infty\) Series of Power over Factorial Converges


So the partial sums of the power series converge to $\exp x$.

The result follows.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Real_Exponential_Function/Power_Series_Expansion_equivalent_to_Differential_Equation&oldid=508876"