Question

hyperbolic sin/cos question

Answer 1

http://imgur.com/a/cZi3Z

Answer 2

I'm having some trouble understanding how they got from the top row to the bottom row. where do the n^2 and the n in the denominator go?

Answer 3

Looks like they just did some simple algebra:

So for the first term

\([- \sinh x \cos nx \cdot \frac{1}{n}]_{- \pi}^{\pi}\)

\( = (- \sinh \pi \cos n \pi \cdot \frac{1}{n}) - (- \sinh (-\pi) \cos n (-\pi) \cdot \frac{1}{n})\)

\( = (- \sinh \pi \cos n \pi \cdot \frac{1}{n}) - ( \sinh \pi \cos n \pi \cdot \frac{1}{n})\)

\(= \dots\)

What is weird is that both functions are odd and so the whole thing should just cancel out due to symmetry.

Certainly, for the second term, you have \(\sin n \pi = 0\) but you also have the indeterminate form at the origin itself so maybe this is work around.

NB: the Latex is not appearing in my browser as I type ..... so this could look like a complete car-crash once Post button is pressed....

[**Presses "Post"**]

:)

Answer 4

holy fook.....it looks OK !!!!!

good luck :)

Answer 5

erratum

the indeterminate thingy is not a Origin but at n = 0

but you're building a Fourier Series so you are rolling with \(n = 1 \to \infty\) i guess, so forget that bit