exponential function defined as limit of powers

It is possible to define the exponential function and the natural logarithmin terms of a limit of powers. In this entry, we shall present thesedefinitions after some background information and demonstrate the basicproperties of these functions from these definitions.

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

Theorem 1.

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

Theorem 2.

Suppose that ${\mathrm{\{}{s}_n\mathrm{\}}}_{n\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}$ is a sequence such that${\lim }_{n\mathrm{\to }\mathrm{\infty }}\mathit{}n\mathit{}{s}_n\mathrm{=}\mathrm{0}$. Then ${\lim }_{n\mathrm{\to }\mathrm{\infty }}\mathit{}{\mathrm{(}\mathrm{1}\mathrm{+}{s}_n\mathrm{)}}^n\mathrm{=}\mathrm{1}$,

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

Theorem 3.

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

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

is convergent.

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

Definition 1.

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

$$\exp (x)=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+x}{n}\right)}^n.$$

We may now derive some of the chief properties of this function.starting with the addition formula.

Theorem 4.

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

Proof.

Since

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

and

$$\underset{n\to \mathrm{\infty }}{lim}\frac{n}{(n+x)(n+y)}=0,$$

theorem 2 above implies that

$$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n(n+x+y)}{(n+x)(n+y)}\right)}^n=1.$$

Since it permissible to multiply convergent sequences termwise, we have

$\exp (x)\exp (y)$$=\underset{n\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{n+x}{n}}\right)}^n{\left({\displaystyle \frac{n+y}{n}}\right)}^n{\left({\displaystyle \frac{n(n+x+y)}{(n+x)(n+y)}}\right)}^n$

$=\underset{n\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{n+x+y}{n}}\right)}^n=\exp (x+y)$

∎

Theorem 5.

The function $\exp$ is strictly increasing.

Proof.

Suppose that $x$ is a strictly positive real number. By theorem 1 andthe definition of the exponential as a limit, we have ,so we conclude that implies .

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.∎

Theorem 6.

The function $\exp$ is continuous.

Proof.

Suppose that . By theorem 1 and the definition of the exponentialas a limit, we have and . By theorem 4,$\exp (x)\exp (-x)=\exp (0)=0$. Hence, we have the bounds and . From the former bound, we concludethat ${lim}_{x\to 0-}\exp (x)=1$ 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 or$x>y$, which would imply 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. Thus, ; since $\exp$ is continuous,the intermediate value theorem asserts that there must exist a realnumber $y$ between $0$ and $x$ such that $\exp (y)=x$. If, instead,, 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 realnumber $x>0$, 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 showthat $\exp ((x+y)/2)\le (\exp (x)+\exp (y))/2$ for all real numbers $x$and $y$. Changing variables, this is equivalent to showing that $2\exp (a+b)\le \exp (a)+\exp (a+2b)$ for all real numbers $a$ and $b$.By theorem 4, we have

$\exp (a+b)$$=\exp (a)\exp (b)$(1)

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

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

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

Theorem 9.

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

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 arational 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 arational number. By continuity, it follows that $\exp (x)=e^x$ for everyreal number $x$.∎