Infinite Series for Sine and Cosine
By Finn on Wednesday, October 03, 2001 - 07:25 pm:

Sin and cos can be expressed as a sum of an infinite series. So can Pi. Why does this work? What is the proof for it?

By Tristan Marshall on Friday, October 05, 2001 - 02:37 pm:

Not quite sure whether you mean 'How can an infinite series have a finite sum' or 'how do we know that the series has sum Pi (say)'. If you could just clarify this I'll try to explain

By Fionn on Friday, October 05, 2001 - 07:05 pm:

Sorry I meant did people find the series, tot it up and go ooh that's Pi , Sin, Cos etc. Or did they look at angles, circles, equations, differentials and stuff and think that logically e.g. Pi must be expressed as this number because of this or that property. Is it possible to prove for example that Pi can be expressed as the series?
Fionn

By James Myatt on Thursday, October 11, 2001 - 02:19 pm:
You'll probably have to ask them how they came up with it, but it certainly wasn't a case of toting it up, as it's an infinite series, so they'll have used something more sophisticated if they wanted to do it that way round.

More likely is that they wanted to know what the power series for functions like sin-1 x, and then used the fact that sin-1 1/2 = p/6.

It is certainly possible to prove that certain series sums to p, as otherwise you'd not have been told it was the case; that's how maths works. Stuff isn't true until it's proved rigorously. The method of proof varies from case to case.

James

By Tristan Marshall on Tuesday, October 16, 2001 - 01:36 pm:

'Proving' the power series for sin and cos is a little awkward, since that's actually how sin and cos are defined (once you get beyond A-level). However, it's fairly easy to justify:

If we assume for a moment that sin(x) has a power series expansion, ie:

sin(x) = (a0) + (a1)x + (a2)x^2 + ...

Where this series goes on forever, and (ai) are coefficients.

Now note that if we set x = 0, all terms but the first vanish!

Hence (a0) = sin(0) = 0

Now if we differentiate both sides (we'll assume it's ok to differentiate an infinite series term by term for now) and set x = 0, we get:

(a1) = cos(0) = 1

By repeating this process, we can recover all the coefficients of the series, hence:

sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - ...

(the same method works for cos)

The only thing that might spoil the party is if it's not ok to differentiate an infinite series term by term (in which case our derivation would be wrong). It turns out that we're allowed to do this whenever the series converges (ie has a finite sum). This is a little more tricky to check, but still fairly straightforward. It turns out that this series (and the one for cos) converge for all x, so everything's ok after all.



The series that sum to pi are a little more tricky. You almost certainly won't manage to prove these directly (eg by trying to sum to N terms and letting this expression -> 0. These are almost always proved indirectly, eg by finding some function with a series expansion which can be manipulated to give this result as a special case. This sort of thing often involves complex variables as well.