Exponent of Sum property

Question

I was reading the exponent Combination law in proofwiki.org and got confused in one part of the proof. The proof is as follows:

Let $a \in R{> 0}$

Let $x, y \in R$

Let $a^x$ be defined as $a$ to the power of $x$

Then:

$a^x a^y = a^{x + y}$

Proof:

$a^{x+y}= $ $\displaystyle \exp \left({\left({x + y}\right) \ln a}\right)$ Definition of Power to Real Number

$= \displaystyle \exp \left({x \ln a + y \ln a}\right)$

$=\displaystyle \exp \left({x \ln a}\right) \exp \left({y \ln a}\right)$ Exponent of Sum

$= \displaystyle a^x a^y$

My question is with this part: $\displaystyle \exp \left({x \ln a}\right) \exp \left({y \ln a}\right)$. I read that $\exp \left({x + y}\right) = \left({\exp x}\right) \left({\exp y}\right)$

How can I proof that because I can't find any explanation or proof for that. Also which book would be good to start reading regarding this type of math. I don't know if this is part of abstract algebra or what math background I need to understand it. Just to let know I am just a person who has developed a great interest for math in my 30's and I am learning by myself and doing what I can with my limitations.

Answer 1

in Analysis you define the exponential function mostly over the exponential series: $$\exp(z) := \sum_{k=0}^{\infty}{\frac{z^k}{k!}}$$ you can proove that this is equivalent to $$\exp(x)=\lim_{k\rightarrow\infty}{\left(1+\frac{x}{k}\right)^k}$$ (and yes this is if you set x = 1 you get one definition for e) $exp(z)*exp(w) = exp(z+w)$ can then be 'simply' proved with the Cauchy Product for series (because they are absolute convergent): therefor we define: $a_k := \frac{z^k}{k!}$ and $b_k:=\frac{w^k}{k!}$

Now it's just a lot of calculation: $$\exp(z)*\exp(w)=\left(\sum_{k=0}^{\infty}{\frac{z}{k!}}\right)*\left(\sum_{k=0}^{\infty}{\frac{w}{k!}}\right) = \left(\sum_{k=0}^{\infty}{a_k}\right)* \left(\sum_{k=0}^{\infty}{b_k}\right) \overset{Cauchy}{=} \sum_{k=0}^{\infty}{\sum_{j=0}^{k}{a_{k-j}b_j}} = \sum_{k=0}^{\infty}{\sum_{j=0}^{k}{\frac{z^{k-j}}{(k-j)!}*\frac{w^j}{j!}}} = \sum_{k=0}^{\infty}{\frac{1}{k!}\sum_{j=0}^{k}{\begin{pmatrix}k\\j\end{pmatrix}z^{k-j}w^j}} \underset{formula}{\overset{binomial}{=}}\sum_{k=0}^{\infty}{\frac{(z+w)}{k!}} = \exp(z+w)$$

.... there you can see, why you haven't found a proof yet.

Answer 2

Well if you use that definition (how you prove it depends on the route you take when you define things) then you have that

$(1+{x\over n})^n(1+{y\over n})^n = (1+{x+y\over n} + {xy\over n^2})^n = (1+{x+n\over n})^n(1+{xy \over n^2 (1+(x+y)/n)})^n$

Now if we know that $lim_{n\to\infty}(1+{xy \over n^2 (1+(x+y)/n)})^n = 1$ it would be easy to see that $\exp(x)\exp(y)=\lim_{n\to\infty}(1+x/n)^n\lim_{n\to\infty}(1+y/n)^n = \lim_{n\to\infty}(1+(x+y)/n)^n = \exp(x+y)$.

But let's look at the fraction inside the last factor ${xy \over n^2 (1+(x+y)/n)}$ it can be limited by noticing that $1\preceq1+(x+y)/n\preceq1+(x+y)/N$ (where $\preceq$ is $\le$ or $\ge$ depending on the sign of $x+y$). That is for the last factor we have

$(1+K/n^2)^n\le(1+{xy \over n^2 (1+(x+y)/n)})^n\le(1+L/n^2)^n$

For some (nonzero) constants $K$ and $L$. By using binomial expansion of we can get a estimate that shows that this must go to 1.

Normally I'd go from defining $a^b$ by starting with the integer definition and prove the equation there by induction and then extend it to dense subset of $\mathbb Q$ (ie for all rational numbers $r$ and $q$ such that $r^q$ could be considered rational) and then let completion to $\mathbb R$ finish it off.