# Thread: Fourier analysis; trigonometric sum.

1. ## Fourier analysis; trigonometric sum.

Sn = Sum from k=1 to n of Ck*cos (kt), where Ck = 1/2^k.

Prove Sn >= .7 for 0 <= t <= .1.

This question is killing me.

Here is what I've got so far:

I estimated cos kt > .95 for 0 <= kt <= .3 which hits the first three terms... now I need to find a lower bound for all the terms, and I think (from previous examples) I'm going to have to use the difference form of the triangle inequality...? Any guidance would be greatly appreciated. Thanks.

2. Hey again,

So, I have a solution, but something is seriously off. If anyone gets a chance please correct any logical errors you may come across.

Thank you.

Ok, so I estimated cos u > .995 for 0<= t <= .1.
Then, I set up the difference form of the triangle inequality:
|1/2*cos (t) + 1/4*cos(2t)+..+1/2^k*cos(kt)|

>= |1/2||cos(t)| - |1/4||cos(2t)|-..-|1/2^k||cos(kt)|.

>=1/2*.995 - 1/4*.995-...-1/2^k*.995 (for 0 <= t <= .1)

>=.7 ????--> (this is supposed to be my conclusion, but clearly this step is not true, or am I not seeing something).

Thanks again.