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It's been a while since I did any of this. I have the following product: $\exp(-j2 \pi u|k|x) \cdot \exp(-j2 \pi v |k|x)$. This seems like it is something that can be simplified, but how? Note, that is not convolution, it is simple multiplication. Thanks!

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Just recall that $\exp(x)=e^x$. It immediately comes natural to manipulate $\exp(x)$. – 000 Mar 26 '12 at 21:51
Please: To use an asterisk for ordinary multiplication within $\TeX$ is uncouth and vulgar. It amounts to eating mashed potatoes with your fingers when silverware is available. Or to putting your face into the plate and eating like a horse from a a trough. The asterisk is a workaround for occasions when you're restricted to the symbols on the keyboard and can't use a lower-case "x" because that's being otherwise used. In $\TeX$ you can write $a\cdot b$ or $a\times b$ or $a\otimes b$, etc. etc. – Michael Hardy Mar 27 '12 at 4:25

2 Answers

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Since $e^x\cdot e^y=e^{x+y}$, we get $$ \exp(-j2 \pi u|k|x) \cdot \exp(-j2 \pi v |k|x)=\exp(-j2 \pi (u+v)|k|x) $$

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up vote 4 down vote

$$\exp(-j2 \pi u|k|x) \cdot \exp(-j2 \pi v |k|x)=\exp\big(-j2\pi u|k|x-j2\pi v|k|x\big) $$

$$=\exp(-j2\pi (u+v)|k|x).$$

Exponentials obey $a^ba^c=a^{b+c}$ (though there can be branching issues for complex $a$).

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