ProofWiki:Sandbox

Theorem
The following definitions of the exponential function are equivalent.

1 implies 2
Fix $x \in \R$.

Let $\left \langle{a_n}\right \rangle$ and $\left \langle{b_n}\right \rangle$ be real sequences defined by:
 * $\displaystyle a_n = \sum_{k = 0}^{n} \frac{x^k}{k!}$

and:
 * $\displaystyle b_n = \lim_{n \to \infty} \left({ 1 + \frac{x}{n} }\right) ^n$

Then:

Also, as $n$ runs through $\N$, the difference of the $k$th summands goes to:

Now, fix $\epsilon \in \R^{ > 0 }$.

From Tail of Convergent Series, it follows that we can find $M \in \N$ such that for all $m \ge M$:
 * $\displaystyle \sum_{k = m}^{n} \left\vert{ \frac{x^k}{k!} }\right\vert < \frac{\epsilon}{2}$

From the second string of equations above, for all $k \in \left\{ {0, 1, \ldots, M-1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$:
 * $\displaystyle \left\vert{ \frac{x^k}{k!} - \frac{x^k}{k!} \prod_{j = 1}^{k-1} \left({ 1 - \frac{j}{n} }\right) }\right\vert < \frac{ \epsilon }{ 2 \left({ M-1 }\right) }$

So for $n \ge \max \left({ N_0, N_1, \ldots, N_{M-1} }\right)$,

Thus, for all $n \ge \max \left({N_0, N_1, \ldots, N_{M-1}, M }\right)$, we have:

Whence: $\lim_{n \to \infty} \left({ a_n - b_n }\right) = 0$

Further, $\left \langle{a_n}\right \rangle$ converges by Series of Power over Factorial Converges.

So:

Therefore $\left \langle{a_n}\right \rangle$ converges, in particular to $ \displaystyle\ lim_{n \to \infty} b_n$.

Hence the result.

2 implies 3
Since $\exp$ is continuous, it is a  continuous extension of $\exp \restriction_\Q$.

inverse implies ode
This proves that $y$ is a solution.

It remains to be proven that $y$ fulfils the initial condition:

ode implies inverse
To solve for $C$, put $\left({x_0, y_0}\right) = \left({0, 1}\right)$:

sequence implies series
From the Binomial Theorem:

From Power of Number less than One, this converges to:


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

as $n \to +\infty$:

Compare Series of Power over Factorial Converges.

sequence implies ode
The application of Derivative of Exponential at Zero is not circular as the referenced proof does not depend on $D_x \exp x = \exp x$.

Also see

 * Equivalence of Definitions of Euler's Number
 * Equivalence of Definitions of Natural Logarithm