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

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

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

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

(My) Proof by induction:

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

Induction step (1): $nRightarrow n+1$

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

Induction step (2): $nRightarrow n-1$

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

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

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

From (a) we should conclude that $exp(qx)=((exp(x))^q,qinmathbb{Q}$ (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 $rinmathbb{R}$ can then be defined by:

$mathbb{R}^{+}rightarrow mathbb{R},xmapsto x^r:=e^{rln(x)}$ (d)

Why is the domain only $mathbb{R}^{+}$ and what if $rln(x)inmathbb{R}backslashmathbb{Q}$, i.e. why is $exp(a*b)=exp(a)^bforall a,binmathbb{R}$?

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

You should begin your journey by showing that $$exp(z+w) =exp(z) exp(w),forall z,winmathbb {C} tag{1}$$ if $exp(z) $ is defined via the series $$exp(z) =1+z+frac{z^2}{2!}+dots=sum_{n=0}^{infty} frac{z^n} {n!}, , zinmathbb {C} tag{2}$$ 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 $xinmathbb {R} $. First note that $$exp(x) exp(-x) =exp(0)=1$$ and hence $exp(x) neq 0$ for all $xinmathbb {R} $ and then we have $$exp(x) =exp(x/2)exp(x/2)>0,forall xinmathbb {R} tag{3}$$ Using the series definition $(2)$ we can see that if $|x|<1$ then
begin{align*}
left|frac{exp(x) - 1}{x}-1right|&=left|frac{x}{2!}+frac{x^2}{3!}+dotsright|\
&leqfrac{|x|}{2!}+frac {|x|^2}{3!}+dots\
&leqfrac{|x|}{2}+frac{|x|^2}{2^2}+dots \
&=frac{|x|} {2-|x|}\
&<|x|
end{align*}
and thus by Squeeze theorem we get $$lim_{xto 0}frac{exp(x)-1}{x}=1tag{4}$$ Using the above limit and functional equation $(1)$ we can easily show that $$frac{d} {dx} exp(x) =exp(x),, xinmathbb {R} tag{5}$$ 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 $$exp(nx) ={exp(x) } ^n,, ninmathbb {Z} tag{6}$$ is fine and it can be easily extended to the case when $ninmathbb {Q} $. Let $n=r/s$ where $rinmathbb {Z}, sinmathbb {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 $$exp(nx) ={exp(x) } ^n, , ninmathbb {Q}, xinmathbb {R} tag{7}$$ 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 $xinmathbb {Q} $. Your question $(c) $ is about "consistency" with this equation which is desired to hold not just for rational $x$ but for all $xinmathbb {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} ^{+} tomathbb{R} $ is defined by $$log x=yiff x=exp(y) tag{8}$$ 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,binmathbb {R} $ as $$a^b=exp(blog a) tag{9}$$ This definition makes sense because $a>0$ implies that $log a$ exists. Further this definition is consistent with equation $(7)$ if $binmathbb {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 $ninmathbb{Q}$ but also when $ninmathbb {R} $. To change symbols as per question let $a, binmathbb{R} $ and then $exp(a) >0$ and we have by definition $(9)$ $${exp(a) } ^b=exp(blogexp(a))=exp(ab)$$

Many questions in your post!

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

Lemma. For $xin{mathbb R}$ and $ninmathbb{Z}$,
$$
exp(nx)=[exp(x)]^ntag{1}
$$

But this lemma implies that
$$
exp(y)=exp(ncdotfrac{y}{n})=[exp(frac{y}{n})]^{n}
$$
and thus
$$
exp(ncdotfrac{y}{n})= [exp(y)]^{1/n},quad ninmathbb{Z}tag{2}
$$
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^qtag{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 $xmapstoln x$) $mathbb{R}_+$?

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

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

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