Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series

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 $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.

• en.wikipedia.org/wiki/Cauchy_product – xavierm02 Jun 7 '13 at 18:15
• There's a simple mistake in your displayed sum. An answer that doesn't explain that already has three up-votes. You migth take a look at the other answer. – Michael Hardy Jun 7 '13 at 18:59

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: $$\begin{array}{ccccccccc} & & & & & & & & n=3 \\ & & & & & & & \swarrow \\ a_{0,0} & & a_{0,1} & & a_{0,2} & & a_{0,3} & & \cdots \\ \\ & & & & & \swarrow \\ a_{1,0} & & a_{1,1} & & a_{1,2} & & a_{1,3} & & \cdots \\ \\ & & & \swarrow \\ a_{2,0} & & a_{2,1} & & a_{2,2} & & a_{2,3} & & \cdots \\ & \swarrow \\ a_{3,0} & & a_{3,1} & & a_{3,2} & & a_{3,3} & & \cdots \\ \vdots & & \vdots & & \vdots & & \vdots \end{array}$$

• I've posted something on a related theoretical issue here: math.stackexchange.com/questions/414130/… – Michael Hardy Jun 7 '13 at 22:42
• @MinimusHeximus : Fixed: $\displaystyle\sum_{k=0}^n$ ${{}\qquad{}}$ – Michael Hardy Jun 8 '13 at 15:25
• Nice $\TeX$ing! – Pedro Tamaroff Jun 8 '13 at 15:35
• 'Then the factor $b_n$ does not depend on $n$', I think you made a typo here and it should be 'm' instead. – dreamer Jun 8 '13 at 15:36
• +1 for an extended, understandable answer that addresses even some points that where not explicitly asked but are necessary for a complete understanding for those not acquainted with it. – Gyro Gearloose Jan 17 '16 at 23:07

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

• 8 upvotes for this !! – Ash Apr 15 '17 at 15:09

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.

• 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
• 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
• 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