Equivalence of Definitions of Real Exponential Function/Limit of Sequence implies Power Series Expansion

From ProofWiki
Jump to navigation Jump to search

Theorem

The following definition of the concept of the real exponential function:

As the Limit of a Sequence

The exponential function can be defined as the following limit of a sequence:

$\exp x := \ds \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$


implies the following definition:

As a Power Series Expansion

The exponential function can be defined as a power series:

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


Proof

Let $\exp x$ be the real function defined as the limit of the sequence:

$\exp x := \ds \lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n$

From the General Binomial Theorem:

\(\ds \paren {1 + \frac x n}^n\) \(=\) \(\ds 1 + x + \frac {n \paren {n - 1} x^2} {2! \ n^2} + \frac {n \paren {n - 1} \paren {n - 2} x^3} {3! \ n^3} + \cdots\)
\(\ds \) \(=\) \(\ds \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds \paren {1 + \frac x n}^n - \paren {\frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots}\)

From Power over Factorial, this converges to:

$\exp x - \paren {\dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \dfrac {x^3} {3!} + \cdots} = 0$

as $n \to +\infty$.

\(\ds \leadsto \ \ \) \(\ds \exp x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)

$\blacksquare$