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In order to show $e^{x+y}=e^{x}e^{y}$ by using Exponential Series, I got the following:

$$e^{x}e^{y}=\Big(\sum_{n=0}^{\infty}{x^n \over n!}\Big)\cdot \Big(\sum_{n=0}^{\infty}{y^n \over n!}\Big)=\sum_{n=0}^{\infty}\sum_{k=0}^n{x^ky^n \over {k!n!}}$$

But, where should I go next to get $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.

Thanks in advance.

Martin Sleziak
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Scorpio19891119
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3 Answers3

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You should have gotten $$ \sum_{n=0}^\infty \sum_{k=0}^n \frac{x^k}{k!}\cdot\frac{y^{n-k}}{(n-k)!}. $$ After that, you can write $$ \sum_{n=0}^\infty \sum_{k=0}^n \frac{1}{n!} \cdot \frac{n!}{k!(n-k)!} x^k y^{n-k}. $$ Since the factor $\dfrac{1}{n!}$ does not depend on $k$, you can pull it out: $$ \sum_{n=0}^\infty \left(\frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!} x^k y^{n-k}\right). $$ Then you have $$ \sum_{n=0}^\infty \frac{1}{n!} (x+y)^n. $$

How does $\displaystyle\left(\sum_{n=0}^\infty b_n\right) \left(\sum_{m=0}^\infty c_m\right)$ become $\displaystyle\sum_{n=0}^\infty \sum_{m=0}^\infty (b_n c_m)$?

And how does $\displaystyle\sum_{n=0}^\infty \sum_{m=0}^\infty a_{n,m}$ become $\displaystyle\sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}$?

In the first sum above, notice that $\sum_{m=0}^\infty c_m$ does not depend on $n$ so it can be pushed inside the other sum and become $$ \sum_{n=0}^\infty \left( b_n \sum_{m=0}^\infty c_m \right). $$ Then the factor $b_n$ does not depend on $m$, so that expression becomes $$ \sum_{n=0}^\infty \sum_{m=0}^\infty (b_n c_m). $$ That answers the first bolded question above.

Next consider the array $$ \begin{array}{ccccccccc} a_{0,0} & a_{0,1} & a_{0,2} & a_{0,3} & \cdots \\ a_{1,0} & a_{1,1} & a_{1,2} & a_{1,3} & \cdots \\ a_{2,0} & a_{2,1} & a_{2,2} & a_{2,3} & \cdots \\ a_{3,0} & a_{3,1} & a_{3,2} & a_{3,3} & \cdots \\ \vdots & \vdots & \vdots & \vdots \end{array} $$ The sum $\displaystyle\sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}$ runs down diagonals: $$ \begin{array}{ccccccccc} & & & & & & & & n=3 \\ & & & & & & & \swarrow \\ a_{0,0} & & a_{0,1} & & a_{0,2} & & a_{0,3} & & \cdots \\ \\ & & & & & \swarrow \\ a_{1,0} & & a_{1,1} & & a_{1,2} & & a_{1,3} & & \cdots \\ \\ & & & \swarrow \\ a_{2,0} & & a_{2,1} & & a_{2,2} & & a_{2,3} & & \cdots \\ & \swarrow \\ a_{3,0} & & a_{3,1} & & a_{3,2} & & a_{3,3} & & \cdots \\ \vdots & & \vdots & & \vdots & & \vdots \end{array} $$

Michael Hardy
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Try to expand $\displaystyle\frac{(x+y)^n}{n!}$.

zji
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Let $f(t)=e^t$, as defined by the power series. Then by differentiating term by term, we find that $f'(t)=e^t$. Now let $y$ be fixed, and let $$g_y(x)=\frac{f(x+y)}{f(x)}.$$ Differentiate. We get $g_y'(x)=0$, so $g_y$ is a constant function. Now let $x=0$. We find that $g_y(x)=e^y$, and the result follows.

André Nicolas
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  • First, in the case that $t\in\mathbb{R}$, shouldn't $f'(t)=f(t)=e^{t}$ (using the power series definition). We can differentiate term-by-term as long the given series is uniformly convergent (of differentiable functions) as well as the derivative series is also uniformly convergent. The complex series ${\displaystyle{\sum\limits_{n=0}^{+\infty}\dfrac{t^{n}}{n!}}}$ uniformly converges on bounded subsets of $\mathbb{C}$, and should $t$ lie in an unbounded subset in the complex plane, then we can not differentiate term-by-term – Procore Jan 17 '16 at 22:52
  • Thank you for pointing out the typo. Amazing that it survived so long. The argument is for the reals. – André Nicolas Jan 17 '16 at 22:57
  • No problem, I liked your approach, but I recently proved the complex exponential uniformly converges on bounded subsets in $\mathbb{C}$, and I was going to try to use your result in another problem, but I caught what I mentioned in my original comment. Maybe I'm wrong? – Procore Jan 17 '16 at 23:00
  • For most questions in complex analysis, I cannot answer quickly. Have taught real analysis many times, and even (somewhat accidentally) have written a couple of papers in the subject, so am comfortable there. But what I wrote is not even at the level of analysis, it is already in the (slightly unusual) books that introduce the exponential and trig functions via series. – André Nicolas Jan 17 '16 at 23:04
  • I'm a beginner at Complex Analysis as well...I'm currently taking an undergraduate course in Fourier Analysis, and we are working a lot in the complex plane. Overall, I wasn't sure if the complex case should be considered or not, but I figured I'd point out what I noticed above; I like the method by its elegance, and that's why I figured I mentioned something. – Procore Jan 17 '16 at 23:10
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    The same idea is used early in calculus in showing that if $f'(x)=f(x)$ then $f(x)$ is a constant times $e^x$. For let $g(x)=\frac{f(x)}{e^x}$. Then we find that $g'(x)=0$, so $g$ is a constant, and the result follows. – André Nicolas Jan 17 '16 at 23:15
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    if the relation holds on every bounded subset, it holds everywhere so that's enough. Also if two analytic functions agree on the reals they agree everywhere. Finally, if you have the proven the result for reals then you have proven the results for the coefficients which is enough. – Mark Joshi Jan 17 '16 at 23:15
  • Thank you Andre and Mark for your comments! I voted up accordingly, and that you for clearing up query. – Procore Jan 17 '16 at 23:55