# How to construct the definition of $a^x,a\in\mathbb{R}^+,x\in\mathbb{R}$ with the exponential function?

I have some Questions marked by $$a,b,c,d$$

I know that the exponential function $$\displaystyle\exp(x):=\sum_{n=0}^{\infty}\frac{1}{n!}x^n$$ is a strictly rising and continious function with $$\exp:\mathbb{C}\rightarrow \mathbb{C}$$, but for me only the constraint $$\exp:\mathbb{R}\rightarrow\mathbb{R}$$ is relevant.

Euler's number $$e$$ is defined as $$\exp(1)$$.

With the functional equation $$(*)$$ one can show that $$\exp(nx)=(\exp(x))^n,n\in \mathbb{Z}$$

(My) Proof by induction:

Inductionbase: $$\exp(0)=1$$, because for $$n>0,\frac{1}{n!}x^n=0$$

Induction step (1): $$n\Rightarrow n+1$$

$$exp((n+1)x)=exp(nx + x)\stackrel{*}{\Rightarrow}exp(nx)exp(x)\stackrel{IH}{\Rightarrow}exp(x)^nexp(x)=exp(x)^{n+1}$$

Induction step (2): $$n\Rightarrow n-1$$

$$\exp(nx)=\exp(x)^n\iff \exp((n-1)+1)x)=\exp(x)^n= \exp((n-1)x+x)$$

$$\stackrel{*}{\Rightarrow}\exp(x)^n=\exp((n-1)x)\exp(x)\Rightarrow \exp(x)^{n-1}=\exp(n-1)x$$ $$\blacksquare$$

From this we should conclude that $$(exp(x))^{1/m}=exp(\frac{x}{m})$$ (a)

From (a) we should conclude that $$exp(qx)=((exp(x))^q,q\in\mathbb{Q}$$ (b)

How can I do that?

In the lecture note it says :

In particular $$exp(q)=e^q$$. Consistent with that we define $$e^x:=exp(x)$$ (c)

I don't understand the 'In particular' and the 'Consistent with that' in the Statement. It implies that (c) is a result of (a) and (b). Why is that so? And also Consistent with what exactly?

Applying theorems about continuity and $$(*)$$ we deduce that $$exp(x)$$ has an inverse function $$ln(x)$$, whit the functional equation:

$$\ln(yy')=\ln y +\ln y',\forall y,y'>0$$

The general exponentiation with the exponent $$r\in\mathbb{R}$$ can then be defined by:

$$\mathbb{R}^{+}\rightarrow \mathbb{R},x\mapsto x^r:=e^{r\ln(x)}$$ (d)

Why is the domain only $$\mathbb{R}^{+}$$ and what if $$r\ln(x)\in\mathbb{R}\backslash\mathbb{Q}$$, i.e. why is $$exp(a*b)=exp(a)^b\forall a,b\in\mathbb{R}$$?

Thank you for reading I hope you can help me to clarify (a) to (d)

• Just as a comment, speaking of an increasing map $f : \mathbb C \to \mathbb C$ doesn’t make sense. – mathcounterexamples.net Jan 13 at 10:31
• It has to be something with roots – RM777 Jan 13 at 10:44
• The title refers to $a^x$ with $a$ real, not only positive real, but the body nowhere refers to this case (which is fortunate, since this case is absurd). Modify the title? – Did Jan 13 at 10:54
• What should I change in the title? $x\in \mathbb{R}^{+}$ or $x\in \mathbb{R}$ – RM777 Jan 13 at 11:02
• $a\in\mathbb R_+$. – Did Jan 13 at 11:05

You should begin your journey by showing that $$\exp(z+w) =\exp(z) \exp(w)\,\forall z,w\in\mathbb {C} \tag{1}$$ if $$\exp(z)$$ is defined via the series $$\exp(z) =1+z+\frac{z^2}{2!}+\dots=\sum_{n=0}^{\infty} \frac{z^n} {n!}, \, z\in\mathbb {C} \tag{2}$$ The functional equation $$(1)$$ is easily established by multiplying the series for $$\exp(z)$$ and $$\exp(w)$$.

As mentioned in question let's restrict our attention only to functions of a real variable. Then we show that $$\exp (x) >0$$ for all $$x\in\mathbb {R}$$. First note that $$\exp(x) \exp(-x) =\exp(0)=1$$ and hence $$\exp(x) \neq 0$$ for all $$x\in\mathbb {R}$$ and then we have $$\exp(x) =\exp(x/2)\exp(x/2)>0\,\forall x\in\mathbb {R} \tag{3}$$ Using the series definition $$(2)$$ we can see that if $$|x|<1$$ then \begin{align*} \left|\frac{\exp(x) - 1}{x}-1\right|&=\left|\frac{x}{2!}+\frac{x^2}{3!}+\dots\right|\\ &\leq\frac{|x|}{2!}+\frac {|x|^2}{3!}+\dots\\ &\leq\frac{|x|}{2}+\frac{|x|^2}{2^2}+\dots \\ &=\frac{|x|} {2-|x|}\\ &<|x| \end{align*} and thus by Squeeze theorem we get $$\lim_{x\to 0}\frac{\exp(x)-1}{x}=1\tag{4}$$ Using the above limit and functional equation $$(1)$$ we can easily show that $$\frac{d} {dx} \exp(x) =\exp(x),\, x\in\mathbb {R} \tag{5}$$ Further $$\exp(x) >0$$ it follows from the above equation that $$\exp$$ is strictly increasing on whole of $$\mathbb {R}$$.

After this initial groundwork is complete we can deal with your individual questions. Your proof by induction for $$\exp(nx) =\{\exp(x) \} ^n,\, n\in\mathbb {Z} \tag{6}$$ is fine and it can be easily extended to the case when $$n\in\mathbb {Q}$$. Let $$n=r/s$$ where $$r\in\mathbb {Z}, s\in\mathbb {Z} ^{+}$$. Replacing $$x$$ in $$(6)$$ by $$rx/s$$ and $$n$$ by $$s$$ we get $$\{\exp(rx/s) \} ^s=\exp(rx) =\{\exp(x) \} ^r$$ and note that both $$\exp(rx/s),\{ \exp(x) \} ^r$$ are positive and hence $$\exp(rx/s) =\sqrt[s] {\{\exp\} ^r} =\{\exp(x) \} ^{r/s}$$ so that we have $$\exp(nx) =\{\exp(x) \} ^n, \, n\in\mathbb {Q}, x\in\mathbb {R} \tag{7}$$ Your questions $$(a)$$ and $$(b)$$ are handled together.

Next we define the number $$e$$ as $$\exp(1)$$ and note that the equation $$(7)$$ above implies $$\exp(x) =e^x$$ for all $$x\in\mathbb {Q}$$. Your question $$(c)$$ is about "consistency" with this equation which is desired to hold not just for rational $$x$$ but for all $$x\in\mathbb {R}$$. In order to do that we must be able to define irrational exponents. To proceed in that direction we need to introduce another function called logarithm.

Since $$\exp(x)$$ is strictly increasing and positive on $$\mathbb{R}$$ we have the existence of its inverse function denoted by $$\log$$ and the function $$\log:\mathbb{R} ^{+} \to\mathbb{R}$$ is defined by $$\log x=y\iff x=\exp(y) \tag{8}$$ It is important to understand that the domain of $$\log$$ is the set of positive reals because the range of $$\exp$$ is the same. The restriction can not be removed if we deal with real variables only.

We next define a general exponent $$a^b$$ with the restriction $$a>0,b\in\mathbb {R}$$ as $$a^b=\exp(b\log a) \tag{9}$$ This definition makes sense because $$a>0$$ implies that $$\log a$$ exists. Further this definition is consistent with equation $$(7)$$ if $$b\in\mathbb {Q}$$ (check by putting $$x=\log a, n=b$$ in $$(7)$$). One can also see that the base $$a$$ in $$(9)$$ must be positive because $$\log a$$ is defined only when $$a$$ is positive. It can now be proved that $$(7)$$ holds not only when $$n\in\mathbb{Q}$$ but also when $$n\in\mathbb {R}$$. To change symbols as per question let $$a, b\in\mathbb{R}$$ and then $$\exp(a) >0$$ and we have by definition $$(9)$$ $$\{\exp(a) \} ^b=\exp(b\log\exp(a))=\exp(ab)$$

Considering the definition $$\displaystyle \exp(x):=\sum_{n=0}^\infty\frac{x^n}{n!},$$ you want to show that $$\exp(qx)=[\exp(x)]^q\tag{0}$$ You showed that following lemma

Lemma. For $$x\in{\mathbb R}$$ and $$n\in\mathbb{Z}$$, $$\exp(nx)=[\exp(x)]^n\tag{1}$$

But this lemma implies that $$\exp(y)=\exp(n\cdot\frac{y}{n})=[\exp(\frac{y}{n})]^{n}$$ and thus $$\exp(n\cdot\frac{y}{n})= [\exp(y)]^{1/n},\quad n\in\mathbb{Z}\tag{2}$$ Combining (1) and (2), one has $$\exp (\frac{m}{n} x)=[\exp(\frac{x}{n})]^m=\bigg[[\exp(x)]^{1/n}\bigg]^m=[\exp(x)]^{m/n}.$$ Writing $$q=m/n$$ gives you the disired identity.

"In particular", (0) implies by setting $$x=1$$ that $$\exp(q)=[\exp(1)]^q.$$ But the constant $$e$$ is defined as $$\exp(1)$$. So you have $$\exp(q)=e^q\tag{3}$$

"Consistent with" means the identity (3), which is proved using the definition $$e=\exp(1)$$ is consistent with the definition $$e^x:=\exp(x)$$. Note this "definition" means one takes $$e^x$$ symbolically as $$\exp(x)$$.

Why is the domain (of $$x\mapsto\ln x$$) $$\mathbb{R}_+$$?

Because the range of the exponential function $$\exp(x)$$ is $$\mathbb{R}_+$$ (Exercise! Hint: show that $$\exp(x)>0$$ for $$x\ge 0$$ and consider $$\exp(x)\cdot \exp(-x)=1$$ for $$x<0$$.) and the logarithm is its inverse.

Why is $$\exp(a\cdot b)=[\exp(a)]^b$$?

This is an instructive exercise: basically, you need to establish continuity of the function $$\exp(x)$$ and use the fact that the set of rational numbers is dense in the real line.

Some hints:

a) both sides to the $$m$$-th power equal $$exp(x)$$, then uniqueness of positive real roots.

b) Take $$q=n/m$$ and apply the two previous results one at a time.

c) In particular: Take $$x=1$$ in (b). Consistent with: Since $$exp$$ is continuous and agrees with $$e^x$$ when $$x$$ is rational, it makes sense to define $$e^x=exp(x)$$ for all real $$x$$ (and it's the only possible way to get $$e^x$$ to be continuous since the rationals are dense in the reals). It's just a notation, but a very useful one.

d) How would you define $$(-1)^x$$, where $$x$$ is 1/2 or 1/4 or 2/6 or any irrational or ...? Use continuity of $$exp$$ to argue that other properties that hold at rationals in fact work for all reals.

Well, To proove (a) you can see that : $$\exp(x) = \exp(x/m)^m$$

To proove (b) you should take $$q = \frac{n}{m}$$

From (b) you conclude that : $$\forall q \in \mathbb{Q} \quad \exp(xq) = \exp(x)^q$$, and we know that $$\mathbb{R} - \mathbb{Q}$$ is dense in $$\mathbb{Q}$$

That mean : $$\forall x \in \mathbb{R} - \mathbb{Q} \, : \, x = \displaystyle \lim_{n \rightarrow +\infty} q_n$$, with $$(q_n)_{n \in \mathbb{N}}$$ a sequence in $$\mathbb{Q}$$

$$x \rightarrow \exp(x)$$ is continue in $$\mathbb{R}$$, then you get the result $$\forall q \in \mathbb{R} \quad \exp(xq) = \exp(x)^q$$

Then we have $$\exp(x) = \exp(1)^x = e^x > 0$$

We have $$x \rightarrow \exp(x)$$ continue and strictly increasing for $$x \in \mathbb{R}$$, then accept an inverse function $$x \rightarrow \ln(x)$$ for $$x \in \mathbb{R}^{*}_+$$

You can also define the function $$x \rightarrow -\exp(x)$$ continue and strictly decreasing for $$x \in \mathbb{R}$$, then accept an inverse function $$x \rightarrow \ln(-x)$$ for $$x \in \mathbb{R}^{*}_-$$