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Assuming $$e^{x}=\sum_{i=0}^{\infty}\frac{x^i}{i!},$$

show that $e^{x}e^{y}$ = $e^{x+y}$.

I have expanded both but am not sure if I need to multiply the two summations and show that product or if there is a simpler way of showing this.

Mee Seong Im
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Megan
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  • Can you show what you mean by expanded sums? Also use $ signs around your equations to make the math formatting work. :) – futurebird Jul 02 '17 at 04:25
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    $e^x$ = 1 + $x^2/2$ + $x^3/3!$ + ......... and $e^y$ = 1 + $y^2/2$ + $y^3/3!$ + .... – Megan Jul 02 '17 at 04:31
  • Thanks the question is much more clear now! – futurebird Jul 02 '17 at 04:33
  • There is a shorter notation if you are familiarized with ODEs: for every $y\in \mathbb{R}$ both are solutions to the IVP $\frac{d\phi}{dx}=\phi$, $\phi(0)=e^y$, which has locally unique solution. This is essentially the same. – Koto Jul 02 '17 at 04:37
  • Use the Cauchy product: https://en.wikipedia.org/wiki/Cauchy_product (this exact example is given on that page) – User8128 Jul 02 '17 at 04:42
  • @User8128 is the Cauchy Product the same as a multiplication? It's defined as a convolution. That may not mean the same as multiplying two sums of real numbers, correct? – user1952500 Jul 02 '17 at 04:43
  • You can prove that $e^x e^y = e^{x+y}$ by multiplying the power series for $e^x$ and $e^y$ using the Cauchy product. – User8128 Jul 02 '17 at 04:45
  • Thank you! Cauchy Product was exactly what I needed to see to get it. – Megan Jul 02 '17 at 04:47
  • You can also refer to "Real and Complex Analysis" by Walter Rudin. He proves this result in the prologue itself using the properties of absolutely convergent series. – jgsmath Jul 02 '17 at 04:50
  • @User8128 I got the answer to my question. The fact that the Cauchy Product (which is actually a convolution) is equal to the product is Merten's theorem which holds under some particular convergence rules. – user1952500 Jul 05 '17 at 09:40

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