elementary proof of growth of exponential function

Proposition 1.

If $x$ is a non-negative real number and $n$ is a non-negative integer, then ${\mathrm{(}\mathrm{1}\mathrm{+}x\mathrm{)}}^n\mathrm{\ge }\mathrm{1}\mathrm{+}n\mathit{}x$.

Proof.

When $n=0$, we have ${(1+x)}^0=1\ge 1+0\cdot 0$. If, for some natural number $n$, it is the case that ${(1+x)}^n\ge 1+nx$ then, multiplying both sides of the inequality by $(1+x)$, we have

$${(1+x)}^{n+1}\ge (1+x)(1+nx)=1+(n+1)x+n{x}^2\ge 1+(n+1)x.$$

By induction, ${(1+x)}^n\ge 1+nx$ for every natural number $n$. ∎

Proposition 2.

If $b$ is a real number such that $b\mathrm{>}\mathrm{1}$ and $n$ and $k$ are non-negative integers, we have $b^n\mathrm{>}{\mathrm{(}\frac{b\mathrm{-}\mathrm{1}}{b\mathit{}k}\mathrm{)}}^k\mathit{}{n}^k$.

Proof.

Let $x=b-1$. Write $n=mk-r$ where $m$ and $r$ are non-negative integers and .

By the preceding proposition, ${(1+x)}^m>mx$. Raising both sides of this inequality to the $k^{\text{th}}$ power, we have ${(1+x)}^{mk}>{(mx)}^k$. Since , we also have ${(1+x)}^{-r}>{(1+x)}^{-k}$; multiplying both sides by this inequality and collecting terms,

$${(1+x)}^{mk-r}>{(\frac{x}{1+x})}^k{m}^k.$$

Multiplying the right-hand side by $k^k/k^k$ and rearranging,

$${(\frac{x}{1+x})}^k{m}^k={(\frac{x}{(1+x)k})}^k{(mk)}^k.$$

Since $mk\ge mk-r$, we also have

$${(\frac{x}{(1+x)k})}^k{(mk)}^k\ge {(\frac{x}{(1+x)k})}^k{(mk-r)}^k.$$

Recalling that $mk-r=n$ and $1+x=b$, we conclude that

$$b^n>{(\frac{b-1}{bk})}^k{n}^k.$$

∎

Proposition 3.

If $a$, $b$, and $x$ are real numbers such that $a\mathrm{\ge }\mathrm{0}$, $b\mathrm{>}\mathrm{1}$ and $x\mathrm{>}\mathrm{0}$, then

$$b^x>(\frac{{(b-1)}^a}{b^{a+1}{(a+1)}^a})x^a.$$

Proof.

Let $k$ and $n$ be integers such that and $x\le n\le x+1$. Since $x+1>n$, we have $b^{x+1}>b^n$. By the preceeding proposition, we have

$$b^n>{(\frac{b-1}{bk})}^k{n}^k.$$

Since , we have $1/k^k>1/{(a+1)}^k$, so

$${(\frac{b-1}{bk})}^k>{(\frac{b-1}{b(a+1)})}^k.$$

Since $k\ge a\ge 0$, we have

$${(\frac{b-1}{b(a+1)})}^k{n}^k\ge {(\frac{b-1}{b(a+1)})}^a{n}^a.$$

Summarrizing our progress so far,

$$b^{x+1}>{(\frac{b-1}{b(a+1)})}^a{n}^a.$$

Dividing both sides by $b$ and simplifying,

$$b^x>(\frac{{(b-1)}^a}{b^{a+1}{(a+1)}^a})x^a.$$

∎

Proposition 4.

If $a$ and $b$ are real numbers and $b\mathrm{>}\mathrm{1}$, then

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^a}{b^x}=0.$$

Proof.

Substituting $a+1$ for $a$

$$b^x>(\frac{{(b-1)}^{a+1}}{b^{a+2}{(a+2)}^{a+1}})x^{a+1}.$$

Dividing by $x$ and rearranging,

Since ${lim}_{x\to \mathrm{\infty }}0=0$ and ${lim}_{x\to \mathrm{\infty }}\frac{1}{x}=0$, we also have ${lim}_{x\to \mathrm{\infty }}\frac{x^a}{b^x}=0$ by the squeeze rule.

∎

Title	elementary proof of growth of exponential function
Canonical name	ElementaryProofOfGrowthOfExponentialFunction
Date of creation	2014-03-10 17:57:26
Last modified on	2014-03-10 17:57:26
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	28
Author	rspuzio (6075)
Entry type	Definition
Classification	msc 26A12
Classification	msc 26A06