Converting Trigonometric Series to Closed Form: Strategies and Techniques

[moo5003]

Homework Statement

Prove that the series:
(K from 0 to n) cos(2kx) = [cos(nx)sin((n+1)x)]/sin(x)

The Attempt at a Solution

So, on my first attempt I simply wrote out the series as 1+cos2x+cos4x... and looked up the double angle formula's for someinsight on how to simplify the sums into the wanted answer, after not finding any obvious way to continue with this line of thought I stopped.

On my second try I tried to convey the series and an exponential expression:
(k from 0 to n) e^(2kix), the point being is that I would find the solution to that series and then take the real part of both sides, though again I'm not sure how to find its equivalence.

Basically I'm having a hard time showing what the series evaluates to in a closed form with any certain method, any help would be appreciated. Namely what are some good ways of finding closed forms, the exponential series seemed like it would be an easier series to find a closed form for though Its been awhile since complex analysis.

 

[Dick]

You've already found the good way. cos(2kx)=Re(exp(i2kx)). The exp form is a geometric series. Look up how to sum a geometric series.

 

[bel]

Why not try mathematical induction?

 

[moo5003]

You've already found the good way. cos(2kx)=Re(exp(i2kx)). The exp form is a geometric series. Look up how to sum a geometric series.

exp(2kix) = Series (n from 0 to infinity) [(2ki)^n/n!]*x^n

Let a_n = (2ki)^n/n!
a_(n+1)/a_n = 2ki/(n+1)

Doesnt a_(n+1)/a_n need to be constant for this to be a geometric series?

As for math induction:

The base case seems pretty straightforward though I would be unsure how to proceed to the induction step. Basically you can say the series from 0 to n+1 equals:

cos(nx)sin((n+1)x)/sin(x) + cos(2(n+1)x)

I have no idea how I would reformulate that into:

cos((n+1)x)sin((n+2)x)/sin(x).

 

[genneth]

\sum_{k=0}^{n} \cos \left(2kx\right) = \Re \left\{ \sum_{k=0}^{n} e^{2ikx} \right\} = \Re \left\{ \sum_{k=0}^{n} \left(e^{2ix}\right)^k \right\}

I leave you to go nuts with the geometric series.

 

[NateTG]

Telescoping sum.

A generic method for showing that a particular series \sum_n a_n has a closed form s(n), is show that:
s(n)-s(n-1)=a_n

If you're given the closed form this should be relatively easy.
In this case, you should be able to simplify:
\frac{\cos(nx)\sin((n+1)x)}{sin(x)}-\frac{\cos((n-1)x)\sin(nx)}{sin(x)}
and end up with:
\cos(2nx)

 

[moo5003]

Well.. I believe I have the right form for the geometric series but I'm having trouble simplifying it to the form of the answer. Just so I don't keep trying on something incorrec to begin with. The geometric series should go to something of the nature of:

Real Part of: (1-e^(2i(n+1)x))/(1-e^(2ix))

I'll keep working on simplification.

 

[genneth]

\frac{1-e^{2i(n+1)x}}{1-e^{2ix}} = \frac{e^{-ix} - e^{i(2n+1)x}}{e^{-ix} - e^{ix}} = \frac{e^{i(2n+1)x} - e^{-ix}}{2i \sin x}

The real part of which is

\frac{1}{\sin x} \frac{\sin\left((2n+1)x\right) - \sin (-x)}{2}

Now apply some trig identities!

 

[moo5003]

Sorry for the incredibly late response, I just did the problem using both methods thanks to your help. Overall I would say NateTG's method is a little bit easier and more intuitive since you don't have to go through the process of finding the real portion.

Basically from where Nate left off:
(Disregarding Denominator for now)

= 1/2[ (sin((2n+1)x)+sin(x)) - (sin((2n-1)x)+sin(x))]

= 1/2[sin((2n+1)x)-sin((2n-1)x)]

= cos(2nx)sin(x)

Sin(x) cancels due to denominator

= cos(2nx).

Note: The problems we are doing are practice for the Putnam exam near the end of the semester, I'm wondering how wise it would be to memorize the trig identities since it was obviously needed for this problem.

 

[Dick]

It's less important to memorize the trig identities than to know which ones exist. If you are practicing for the putnam you should be able to derive the one you want pretty fast. A recent thread showed that it's really handy to be able to remember sqrt(1+cos(t)) can be simplified with a half-angle formula. Once you know that its easy to derive. If you don't know such a thing exists, it can be hard.