Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

2 Answers

up vote 4 down vote accepted

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

share|improve this answer
add comment

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.