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.

I have encountered a problem where I get the correct outcome, but I am uncertain as to whether or not my steps are logically justified. I would really appreciate some input regarding this! The problem is as follows:

Let $f$ and $g$ be $2 \pi$-periodic, piecewise smooth functions having Fourier series $f(x) = \sum_n \alpha_{n}e^{inx}$ and $g(x) = \sum_n \beta_{n}e^{inx}$, and define the convolution of $f$ and $g$ to be $f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f * g$ is

$$f * g(x) = \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}$$

SOLUTION:

Since $f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t)g(x-t)dt$, we have:

$$f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \sum_n \alpha_{n}e^{int} \sum_n \beta_{n}e^{in(x-t)}dt$$

$$ = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{int} e^{in(x-t)}dt$$

$$= \frac{1}{2 \pi}\int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{in(t + x - t)}dt$$

$$= \frac{1}{2 \pi}\int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}dt$$

$$= \frac{1}{2 \pi} \cdot \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx} \int_{- \pi}^{\pi} dt$$

$$= \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}$$

What I am unsure about here is the step where I merge the two sums into one (step 1 - 2 in the solution above). This doesn't intuitively quite make sense to me based on the definition of the Cauchy product, but it gets me to the correct answer. If anyone can explain whether or not this proceude is valid or not, and how this "merging" of sums can be justified, I would be extremely grateful!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

You should use a different summation index for each sum: $f(x) = \sum_n \alpha_{n}e^{inx}$ and $g(x) = \sum_m \beta_{m}e^{imx}$; you need to consider these as separate sums.

After some rearranging you'll eventually get the following integral: $\int_{-\pi}^{\pi}e^{i(n-m)t}dt$. This integral is the key here to "collapsing" the two sums to just one:

$$\int_{-\pi}^{\pi}e^{i(n-m)t}dt=2\pi \delta_{mn}$$

Here's a simple calculation of what the Kronecker delta does in sums:

$$\displaystyle \sum \limits_{i=0}^n a_i \delta_{ij}=a_0 \delta_{0j}+a_1 \delta_{1j}+a_2 \delta_{2j}+ \ldots +a_j \delta_{jj}+ \ldots +a_n \delta_{nj} =a_j$$

share|improve this answer
    
Thanks a lot! I will try to work more on the problem following your hints! –  Kristian Oct 26 '12 at 7:26
    
Here's a follow up (manipulating indices, or index gymnastics, was always a pain for me). –  BobaFret Oct 26 '12 at 17:44

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.