# College Math Teaching

## September 2, 2014

### Using convolutions and Fourier Transforms to prove the Central Limit Theorem

Filed under: probability — Tags: , , — collegemathteaching @ 5:40 pm

I’ve used the presentation in the our Probability and Statistics text; it is appropriate given that many of our students haven’t seen the Fourier Transform. But this presentation is excellent.

Upshot: use the convolution to derive the density function for $S_n = X_1 + X_2 + ....X_n$ (independent, identically distributed random variables of finite variance), assume mean is zero, variance is 1 and divide $S_n$ by $\sqrt{n}$ to obtain the variance of the sum to be 1. Then use the Fourier transform on the whole thing (the normalized version) to turn convolution into products, use the definition of Fourier transform and use the Taylor series for the $e^{i 2 \pi x \frac{s}{\sqrt{n}}}$ terms, discard the high order terms, take the limit as $n$ goes to infinity and obtain a Gaussian, which, of course, inverse Fourier transforms to another Gaussian.

## November 12, 2013

### Why I teach multiple methods for the inverse Laplace Transform.

I’ll demonstrate with a couple of examples:

$y''+4y = sin(2t), y(0) = y'(0) = 0$

If we use the Laplace transform, we obtain: $(s^2+4)Y = \frac{2}{s^2+4}$ which leads to $Y = \frac{2}{(s^2+4)^2}$. Now we’ve covered how to do this without convolutions. But the convolution integral is much easier: write $Y = \frac{2}{(s^2+4)^2} = \frac{1}{2} \frac{2}{s^2+4}\frac{2}{s^2+4}$ which means that $y = \frac{1}{2}(sin(2t)*sin(2t)) = \frac{1}{2}\int^t_0 sin(2u)sin(2t-2u)du = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$.

Note: if the integral went too fast for you and you don’t want to use a calculator, use $sin(2t-2u) = sin(2t)cos(2u) - cos(2t)sin(2u)$ and the integral becomes $\frac{1}{2}\int^t_0 sin(2t)cos(2u)sin(2u) -cos(2t)sin^2(2u)du =$

$\frac{1}{2} (sin(2t))\frac{1}{4}sin^2(2u)|^t_0 - cos(2t)(\frac{1}{4})( t - \frac{1}{4}sin(4u)|^t_0 =$

$\frac{1}{8}sin^3(2t) - \frac{1}{4}tcos(2t) +\frac{1}{16}sin(4t)cos(2t) =$

$\frac{1}{8}(sin^3(2t) +sin(2t)cos^2(2t))-\frac{1}{4}tcos(2t)$

$= \frac{1}{8}sin(2t)(sin^2(2t) + cos^2(2t))-\frac{1}{4}tcos(2t) = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$

Now if we had instead: $y''+4y = sin(t), y(0)=0, y'(0) = 0$

The Laplace transform of the equation becomes $(s^2+4)Y = \frac{1}{s^2+1}$ and hence $Y = \frac{1}{(s^2+1)(s^2+4)}$. One could use the convolution method but partial fractions works easily: one can use the calculator (“algebra” plus “expand”) or:

$\frac{A+Bs}{s^2+4} + \frac{C + Ds}{s^2+1} =\frac{1}{(s^2+4)(s^2+1)}$. Get a common denominator and match numerators:

$(A+Bs)(s^2+1) + (C+Ds)(s^2+4) = 1$. One can use several methods to resolve this: here we will use $s = i$ to see $(C + Di)(3) = 1$ which means that $D = 0$ and $C = \frac{1}{3}$. Now use $s = 2i$ so obtain $(A + 2iB)(-3) = 1$ which means that $B = 0, A = -\frac{1}{3}$ so $Y = \frac{1}{3} (\frac{1}{s^2+1} - \frac{1}{s^2+4}$ so $y = \frac{1}{3} (sin(t) - \frac{1}{2} sin(2t)) = \frac{1}{3}sin(t) -\frac{1}{6}sin(2t)$

So, sometimes the convolution leads us to the answer quicker than other techniques and sometimes other techniques are easier.