# Real Analysis/Exponential Function

 Real Analysis Exponential Function

Our aim in this chapter is to formally define the very interesting exponential and logarithmic functions for all reals. Several authors approach this by defining the logarithm as an integral and the exponent as its inverse. We will, however, follow the reverse approach by constructing the exponential function of reals from that of rationals. (Unfortunately, in this chapter, the reader will have to bear with some tedious computations!)

We begin with construction of rational powers of arbitrary positive reals.

## Construction of Rational Powers

We can now use what we know about continuity to construct rational powers of positive real numbers. We've already defined the integer powers; let's show they're continuous first.

### Continuity of x^n

• $f(x)=x^{0}=1$ is continuous.

Given $\epsilon >0$, $|f(x)-f(c)|=|1-1|=0<\epsilon$. So, $\forall \delta >0:|x-c|<\delta \implies |f(x)-f(c)|<\epsilon$.

• $f(x)=x$ is continuous.

Given $\epsilon >0$, let $\delta =\epsilon$. Then $|x-c|<\delta \implies |x-c|<\epsilon \implies |f(x)-f(c)|<\epsilon$ .

• $f(x)=x^{n}$ is continuous for all $n\in {\mathbb {N}}$ and all $x\in {\mathbb {R}}$.

We proceed by induction. We have already seen that $f(x)=x^{{1}}$ is continuous. Assuming $f(x)=x^{{n-1}}$ is continuous, we use the fact that continuity is preserved under algebraic operations to see that $xf(x)=x^{n}$ is continuous.

• $f(x)=x^{{-n}}$ is continuous for all $n\in {\mathbb {N}}$ and all $x\in {\mathbb {R}}\setminus {0}$.

Since $x^{n}$ is continuous and nonzero on the set in question, ${\frac {1}{x^{n}}}=x^{{-n}}$ is continuous since continuity is preserved under division by a nonzero function.

We can now use the continuity of $x^{n}$ together with the intermediate value theorem to construct positive nth roots. As promised, this is much nicer than the construction of square roots in the first chapter:

### Construction of nth roots

Given $c>0$, consider the function $f(x)=x^{n}-c$(it is clear that 0 has a unique nth root, so we do not consider this case). $f(0)=-c<0$ and since $1+c>1$, $f(1+c)=(1+c)^{n}-c>(1+c)^{n}-(1+c)>0$. By the Intermediate Value Theorem, $\exists x\in (0,1+c):f(x)=x^{n}-c=0$. Thus c has a positive nth root.

To prove uniqueness, let x and y be two nth roots of c. If $x>y>0$, then $x^{n}>y^{n}>0$. But then it would follow that $c>c$, a contradiction. Similarly we cannot have $x, so it follows that $x=y$.

### Definition and Properties of Rational Powers

Given $n\in {\mathbb {N}}$ we define $x^{{{\frac {1}{n}}}}={\sqrt[ {n}]{x}}$ to be the unique nonnegative nth root of x. We then define all rational powers as follows:

If $r={\frac {p}{q}}$ is in lowest terms(i.e. p and q have no common factors and $q>0$), we define $x^{r}=({\sqrt[ {q}]{x}})^{p}$.

Our definition would work just as well if ${\frac {p}{q}}$ were not in lowest terms, as we'll see in a minute. First we must prove some basic facts:

• ${\sqrt[ {m}]{{\sqrt[ {n}]{x}}}}={\sqrt[ {mn}]{x}}$

Note that $({\sqrt[ {m}]{{\sqrt[ {n}]{x}}}})^{{mn}}=(({\sqrt[ {m}]{{\sqrt[ {n}]{x}}}})^{m})^{n}={\sqrt[ {n}]{x}}^{n}=x$. Thus ${\sqrt[ {m}]{{\sqrt[ {n}]{x}}}}$ is an mn-th root of x. The result follows immediately from uniqueness of positive roots.

• $x^{{{\frac {p}{q}}}}={\sqrt[ {q}]{x^{p}}}={\sqrt[ {q}]{x}}^{p}$

Using what we know about integer powers we see that $(x^{{{\frac {1}{q}}}})^{p})={\sqrt[ {q}]{({\sqrt[ {q}]{x}}^{p})^{q})}}={\sqrt[ {q}]{({\sqrt[ {q}]{x}}^{q})^{p})}}={\sqrt[ {q}]{x^{p}}}$

As promised, our definition does not depend on the fraction representing r:

• If ${\frac {p}{q}}={\frac {m}{n}}$, then $x^{{{\frac {p}{q}}}}={\sqrt[ {n}]{x^{m}}}$.

If ${\frac {p}{q}}={\frac {m}{n}}$, then $m=cp$ and $n=cq$ for some $c\in {\mathbb {Z}}$. Thus ${\sqrt[ {n}]{x^{m}}}=({\sqrt[ {c}]{{\sqrt[ {q}]{x}}}}^{c})^{p}={\sqrt[ {q}]{x^{p}}}=x^{{{\frac {p}{q}}}}$.

Now we'll prove the standard algebraic facts about rational powers:

• If $r,s\in {\mathbb {Q}}$ and $x>0$, then $x^{r}x^{s}=x^{{r+s}}$ and $(x^{r})^{s}=x^{{rs}}$

Proof: Let $r={\frac {a}{b}}$ and $s={\frac {c}{d}}$. Then $x^{r}x^{s}=x^{{{\frac {a}{b}}}}x^{{{\frac {c}{d}}}}=x^{{{\frac {ad}{bd}}}}x^{{{\frac {bc}{bd}}}}=(x^{{{\frac {1}{bd}}}})^{{ad}}(x^{{{\frac {1}{bd}}}})^{{bd}}=(x^{{{\frac {1}{bd}}}})^{{ad+bc}}=x^{{{\frac {ad+bc}{bd}}}}=x^{{(r+s)}}$

Also, $x^{{rs}}$ $=x^{{{\frac {a}{b}}{\frac {c}{d}}}}$ $=x^{{{\frac {ac}{bd}}}}$ $=(x^{{ac}})^{{{\frac {1}{bd}}}}$ = $(((x^{{a}})^{{c}})^{{{\frac {1}{b}}}})^{{{\frac {1}{d}}}}$ $=(((x^{a})^{{{\frac {1}{b}}}})^{c})^{{{\frac {1}{d}}}}$$=(x^{{{\frac {a}{b}}}})^{{{\frac {c}{d}}}}$

• If $r={\frac {a}{b}}>0\in {\mathbb {Q}}$ and $y>x>0$, then $y^{r}>x^{r}>0$.

Proof: If $y^{{{\frac {1}{b}}}}\leq x^{{{\frac {1}{b}}}}$, then $(y^{{{\frac {1}{b}}}})^{b}\leq (x^{{{\frac {1}{b}}}})^{b}$, and $y\leq x$, contradicting the assumption $y>x>0$. So $y^{{{\frac {1}{b}}}}>x^{{{\frac {1}{b}}}}>0$. Since a > 0, $y^{{{\frac {a}{b}}}}>x^{{{\frac {a}{b}}}}>0$. Thus $y^{r}>x^{r}>0$

### Continuity of rational powers

Now we'll use the preceding algebraic properties to prove continuity of all rational powers:

• $f(x)=x^{{{\frac {1}{n}}}}$ is continuous for all $n\in {\mathbb {N}}$and $x\geq 0$.

Proof: Given $\epsilon >0$, let $\delta =|c^{{{\frac {n-1}{n}}}}|\epsilon$. Then $|x-c|<\delta \implies$

$|x^{{{\frac {1}{n}}}}-c^{{{\frac {1}{n}}}}|={\frac {|x^{{{\frac {1}{n}}}}-c^{{{\frac {1}{n}}}}||x^{{{\frac {n-1}{n}}}}+x^{{{\frac {n-2}{n}}}}c^{{{\frac {1}{n}}}}+\cdots +x^{{{\frac {1}{n}}}}c^{{{\frac {n-2}{n}}}}+c^{{{\frac {n-1}{n}}}}|}{|x^{{{\frac {n-1}{n}}}}+x^{{{\frac {n-2}{n}}}}c^{{{\frac {1}{n}}}}+\cdots +x^{{{\frac {1}{n}}}}c^{{{\frac {n-2}{n}}}}+c^{{{\frac {n-1}{n}}}}|}}$$={\frac {|x-c|}{|x^{{{\frac {n-1}{n}}}}+x^{{{\frac {n-2}{n}}}}c^{{{\frac {1}{n}}}}+\cdots +x^{{{\frac {1}{n}}}}c^{{{\frac {n-2}{n}}}}+c^{{{\frac {n-1}{n}}}}|}}$$\leq {\frac {|x-c|}{|c^{{{\frac {n-1}{n}}}}|}}<{\frac {|c^{{{\frac {n-1}{n}}}}|\epsilon }{|c^{{{\frac {n-1}{n}}}}|}}=\epsilon$.

The preceding argument works for $c\not =0$. If $c=0$, then let $\delta =\epsilon ^{{n}}$. Then:

$|x-0|<\delta \implies |x|<\epsilon ^{n}\implies |x|^{{{\frac {1}{n}}}}<\epsilon \implies |x^{{{\frac {1}{n}}}}-0|<\epsilon$

So, $x^{{{\frac {1}{n}}}}$ is continuous for all $x\geq 0$.

• $x^{q}$ is continuous for all $q\in {\mathbb {Q}}$

Proof: If $q={\frac {a}{b}}$, where a and b are integers and $b>0$, then $x^{q}=(x^{a})^{{{\frac {1}{b}}}}$. Thus $x^{q}$ is the composition of continuous functions, and therefore is continuous itself.

## Real Powers

We will define arbitrary real exponents as the supremum of the exponents of rational members of the "Cut" that corresponds to the given real number. But first, we need to establish that this operation indeed produces a unique real number.

### Theorem

Let $a,x\in {\mathbb {R}}$ and let $a>0$

Let $A=\{a^{q}|q\in {\mathbb {Q}};q\leq x\}$
Let $B=\{a^{q}|q\in {\mathbb {Q}};q\geq x\}$

Let $\alpha =\sup A$ and $\beta =\inf B$

Then, $\alpha =\beta$