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)

  • 2
    $\begingroup$ Just as a comment, speaking of an increasing map $f : \mathbb C \to \mathbb C$ doesn’t make sense. $\endgroup$ – mathcounterexamples.net Jan 13 at 10:31
  • $\begingroup$ It has to be something with roots $\endgroup$ – RM777 Jan 13 at 10:44
  • 1
    $\begingroup$ 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? $\endgroup$ – Did Jan 13 at 10:54
  • $\begingroup$ What should I change in the title? $x\in \mathbb{R}^{+}$ or $x\in \mathbb{R}$ $\endgroup$ – RM777 Jan 13 at 11:02
  • $\begingroup$ $a\in\mathbb R_+$. $\endgroup$ – 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 $(9)$ $$\{\exp(a) \} ^b=\exp(b\log\exp(a))=\exp(ab)$$


Many questions in your post!

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 $x \in \mathbb{R}^{*}_-$


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