# exponential function defined as limit of powers

Two basic results which are needed to make this development possible are the following:

###### Theorem 1.

Let $x$ be a real number and let $n$ be an integer such that $n>0$ and $n+x>0$. Then

 $\left({n+x\over n}\right)^{n}<\left({n+1+x\over n+1}\right)^{n+1}.$
###### Theorem 2.

Suppose that $\{s_{n}\}_{n=1}^{\infty}$ is a sequence  such that $\lim_{n\to\infty}ns_{n}=0$. Then $\lim_{n\to\infty}(1+s_{n})^{n}=1$,

For proofs, see the attachments. From them, we first conclude that a sequence converges.

###### Theorem 3.

Let $x$ be any real number. Then the sequence

 $\left\{\left({n+x\over n}\right)^{n}\right\}_{n=1}^{\infty}$

The foregoing results show that the limit in the following definition converges, and hence defines a bona fide function.

###### Definition 1.

Let $x$ be a real number. Then we define

 $\exp(x)=\lim_{n\to\infty}\left({n+x\over n}\right)^{n}.$
###### Theorem 4.

For any two real numbers $x$ and $y$, we have $\exp(x+y)=\exp(x)\exp(y)$.

###### Proof.

Since

 ${n(n+x+y)\over(n+x)(n+y)}=1-{xy\over(n+x)(n+y)}$

and

 $\lim_{n\to\infty}{n\over(n+x)(n+y)}=0,$

theorem 2 above implies that

 $\lim_{n\to\infty}\left({n(n+x+y)\over(n+x)(n+y)}\right)^{n}=1.$

Since it permissible to multiply convergent sequences termwise, we have

 $\displaystyle\exp(x)\exp(y)$ $\displaystyle=\lim_{n\to\infty}\left({n+x\over n}\right)^{n}\left({n+y\over n}% \right)^{n}\left({n(n+x+y)\over(n+x)(n+y)}\right)^{n}$ $\displaystyle=\lim_{n\to\infty}\left({n+x+y\over n}\right)^{n}=\exp(x+y)$

###### Theorem 5.

The function $\exp$ is strictly increasing.

###### Proof.

Suppose that $x$ is a strictly positive real number. By theorem 1 and the definition of the exponential   as a limit, we have $1+x<\exp(x)$, so we conclude that $0 implies $1<\exp(x)$.

Now, suppose that $x$ and $y$ are two real numbers with $x>y$. Since $x-y>0$, we have $\exp(x-y)>1$. Using theorem 4, we have $\exp(x)=\exp(x-y)\exp(y)>\exp(y)$, so the function is strictly increasing. ∎

###### Proof.

Suppose that $0. By theorem 1 and the definition of the exponential as a limit, we have $1-x<\exp(-x)$ and $1+x<\exp(x)$. By theorem 4, $\exp(x)\exp(-x)=\exp(0)=0$. Hence, we have the bounds $1+x<\exp(x)<1/(1-x)$ and $1-x<\exp(-x)<1/(1+x)$. From the former bound, we conclude that $\lim_{x\to 0-}\exp(x)=1$ and, from the latter, that $\lim_{x\to 0+}\exp(x)=1$, so $\lim_{x\to 0}\exp(x)=1$.

Suppose that $y$ is any real number. By theorem 4, $\exp(x+y)=\exp(x)\exp(y)$. Hence, $\lim_{x\to 0}\exp(x+y)=\exp(y)\lim_{x\to 0}\exp(x)=\exp(y)$. In other words, for all real $y$, we have $\lim_{x\to y}\exp(x)=\exp(y)$, so the exponential function is continuous. ∎

###### Theorem 7.

The function $\exp$ is one-to-one and maps onto the positive real axis.

###### Proof.

The one-to-one property follows readily from monotonicity — if $\exp(x)=\exp(y)$, then we must have $x=y$, because otherwise, either $x or $x>y$, which would imply $\exp(x)<\exp(y)$ or $\exp(x)>\exp(y)$, respectively. Next, suppose that $x$ is a real number greater than $1$. By theorem 1 and the definition of the exponential as a limit, we have $1+x<\exp(x)$. Thus, $1; since $\exp$ is continuous, the intermediate value theorem asserts that there must exist a real number $y$ between $0$ and $x$ such that $\exp(y)=x$. If, instead, $0, then $1/x>1$ so we have a real number $y$ such that $\exp(y)=1/x$. By theorem 4, we then have $\exp(-y)=x$. So, given any real number $x>0$, there exists a real number $y$ such that $\exp(y)=x$, hence the function maps onto the positive real axis. ∎

###### Theorem 8.

The function $\exp$ is convex.

###### Proof.

Since the function is already known to be continuous, it suffices to show that $\exp((x+y)/2)\leq(\exp(x)+\exp(y))/2$ for all real numbers $x$ and $y$. Changing variables, this is equivalent     to showing that $2\exp(a+b)\leq\exp(a)+\exp(a+2b)$ for all real numbers $a$ and $b$. By theorem 4, we have

 $\displaystyle\exp(a+b)$ $\displaystyle=\exp(a)\exp(b)$ (1) $\displaystyle\exp(a+2b)$ $\displaystyle=\exp(a)\exp(b)^{2}.$ (2)

Using the inequality  $2x\leq 1+x^{2}$ with $x=\exp(b)$ and multiplying by $\exp(a)$, we conclude that $2\exp(a+b)\leq\exp(a)+\exp(a+2b)$, hence the exponential function is convex. ∎

Defining the constant $e$ as $\exp(1)$, we find that the exponential function gives powers of this number.

###### Theorem 9.

For every real number $x$, we have $\exp(x)=e^{x}$.

###### Proof.

Applying an induction  argument   to theorem 4, it can be shown that $\exp(nx)=\exp(x)^{n}$ for every real number $x$ and every integer $n$. Hence, given a rational number $m/n$, we have $\exp(m/n)^{n}=\exp(m)=\exp(1)^{m}=e^{m}$. Thus, $exp(m/n)=e^{m/n}$ so we see that $\exp(x)=e^{x}$ when $x$ is a rational number. By continuity, it follows that $\exp(x)=e^{x}$ for every real number $x$. ∎

Title exponential function defined as limit of powers ExponentialFunctionDefinedAsLimitOfPowers 2013-03-22 17:01:35 2013-03-22 17:01:35 rspuzio (6075) rspuzio (6075) 27 rspuzio (6075) Definition msc 32A05 ExponentialFunction ComplexExponentialFunction ExponentialFunctionNeverVanishes