# College Math Teaching

## November 25, 2013

Filed under: differential equations, Laplace transform — Tags: — collegemathteaching @ 10:33 pm

Consider: $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}......$

Now take the Laplace transform of the right hand side: $\frac{1}{s^2} - \frac{3!}{s^4 3!} + \frac{5!}{s^6 5!} .... = \frac{1}{s^2} (1 -\frac{1}{s^2} + \frac{1}{s^4} ...$.

This is equal to: $\frac{1}{s^2} (\frac{1}{1 + \frac{1}{s^2}})$ for $s > 1$ which is, of course, $\frac{1}{1 + s^2}$ which is exactly what you would expect.

This technique works for $e^{x}$ but gives nonsense for $e^{x^2}$.

Update: note that we can get a power series for $e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + ....$ which, on a term by term basis, transforms to $\frac{1}{s} + \frac{2!}{s^3} + \frac{4!}{s^5 2!} + \frac{6!}{s^7 3!} + ... = \frac{1}{s} \sum_{k=0} (\frac{1}{s^2})^k\frac{(2k)!}{k!})$ which only converges at $s = \infty$.