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As an alternative to normal rules for powers giving $e^xe^y=e^{(x+y)}$ I am multiplying the long polynomial of the taylor series of $e^x$ and $e^y$. I only take the first three terms: $$ \left(1+x+\frac{x^2}{2!}+\cdots\right)\left(1+y+\frac{y^2}{2!}+\cdots\right).$$ With this I try to reach $$1+(x+y)+\frac{(x+y)^2}{2!}+\cdots$$ I collect a lot of terms and take the terms I know to be equal to $(x+y)^2=x^2+2xy+y^2$, but I'm then left with terms like $yx^2/2!+xy^2/2!$ These last terms I cant solve away. Am I doing something wrong or can I just say that I'll keep those for the $+\cdots$ bit?

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4 Answers 4

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Those terms belong to the $(x+y)^3$ term. The way to see that is the their total power is 3 a la the binomial theorem. The proper way to do this problem is with Cauchy products.

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  • $\begingroup$ thank you, googling cauchy products does look useful $\endgroup$
    – Leo
    Commented Jun 1, 2013 at 20:29
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You are doing right. If you expand $e^{(x+y)}$ you get $1+(x+y)+\frac 12(x+y)^2+\frac 16(x+y)^3 \dots$ as you say. Expanding the $\frac 16(x+y)^3$ term gives $\frac 16x^3+\frac 12 x^2y+\frac 12xy^2+\frac 16y^3$. The middle two of these are the ones you are worried about.

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$$(1+x+\frac{x^{2}}{2}+...)(1+y+\frac{y^{2}}{2}+...)=1+x+y+\frac{x^{2}}{2}+\frac{y^{2}}{2}+xy+...=$$ $$=1+(x+y)+\frac{(x+y)^{2}}{2}+...$$ The other terms, such as $yx^2/2!+xy^2/2!$ will apear as the parts of the expression $\frac{(x+y)^{3}}{3!}$.

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  • $\begingroup$ Thank you, i was wondering how to formalise that $\endgroup$
    – Leo
    Commented Jun 1, 2013 at 20:32
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$$\frac{y^3}{3!}+\frac{y^2 x}{2!} +\frac{y x^2}{2!}+\frac{x^3}{3!}=\frac{y^3+3y^2x+3yx^2+x^3}{3!}=\frac {(y+x)^3}{3!}$$

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