0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What definition of $exp(x)$ are you used to? $\endgroup$ – A.S. Oct 2 '15 at 11:58
  • $\begingroup$ To be honest it was the first time I saw it, actually I thought it was $e$ but then I read this definition:$\exp x := \displaystyle \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$ which I think I understand because I see a relation with $e$ and its limit definition. $\endgroup$ – kprincipe Oct 2 '15 at 12:44
  • $\begingroup$ exp(x) is just another way to write $e^x$. $\endgroup$ – skyking Oct 2 '15 at 12:47
  • $\begingroup$ It would be useful if you provide a link to the side you refer to. Expecting people to find the link you have in mind is going to turn helpful people off. $\endgroup$ – skyking Oct 2 '15 at 12:48
0
$\begingroup$

in Analysis you define the exponential function mostly over the exponential series: $$\exp(z) := \sum_{k=0}^{\infty}{\frac{z}{k!}}$$ you can proof that this is equivalent to $$\exp(x)=\lim_{k\rightarrow\infty}{\left(1+\frac{x}{k}\right)^k}$$ (and yes this if you set x = 1 you get one definition for e) $exp(z)*exp(w) = exp(z+w)$ can then be 'simply' proofed with the Cauchy Product for series (because they are absulute 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 havn't found a proof yet ;)

PS: that was a lot of code... so I appologize for any mistakes I've propably made ..

$\endgroup$
  • $\begingroup$ I really didn't understand the explanation because is the first time I see something like that and I don't have to much knowledge on series. I wanted to ask you, can you recommend me a book I can start reading to start learning whats needed for this kind of proof. I bought one for introduction in mathematical analysis, but don't know if is a correct one. Thanks in advance $\endgroup$ – kprincipe Oct 7 '15 at 19:55
  • $\begingroup$ I don't know what you already know but I think If the book is the book from springer with the Authors: Kriz, Igor, Pultr, Ales then it is much to high for you (I'm sorry) I would stick to a little easier one.... I would recommend you this book for a start link.springer.com/book/10.1007/978-1-4939-2651-0 because it is 'simple' and correct. But even though I have to warn you... You won't be working through the book in one week,... You won't even make it in one year...students who study math nead at least one semester and then study all the time..... But in the end it's worth it :) $\endgroup$ – Börge Oct 8 '15 at 7:51
  • $\begingroup$ And if you have any questions, don't hesitate to ask me or someone else here... Good luck and have fun $\endgroup$ – Börge Oct 8 '15 at 7:51
  • $\begingroup$ Ok, thank you very much, I will follow the advise and is as you say, learning never ends, more if it is something so complex as math $\endgroup$ – kprincipe Oct 8 '15 at 9:59
  • $\begingroup$ I know this question was made some time ago but I have a question, how did you get from the part after Cauchy that I see 1/k!, how that 1/k! got factored out and what does the k and j on the parenthesis means $\endgroup$ – kprincipe Jan 12 '16 at 11:58
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.