Question

WORLD'S HARDEST MATH PROBLEM
(maybe)
What is the solution to this in terms of n?

1 + cos (4x) + cos(8x) + cos(12x)....+ (cos nx).
It hurt my brain.

Answer 1

Don't worry if you can't solve it.. I have a Russian math teacher going nuts on us about calc.. Apparently this is one of the easier problems..

Answer 2

it's basically an expanded summation... except reducing it was interesting with half-angle and double angle formulas.. and all that fun stuff..

Answer 3

It's 1 + the summation of everything after that over sin (2x). Just can't get the top part now.

Answer 4

Teacher said to plug in for different values of n and expand it.. Fun.

Answer 5

I can solve it. I will write much. Do you really want to know the answer?

Answer 6

YES
lol

Answer 7

Nobody got the answer lol

Answer 8

\(1+\cos4x+\cos8x+\ldots+\cos4(n-1)x=\Re(1+(\cos4x+i\sin4x)+\ldots+\)
\((\cos4(n-1)+i\sin4(n-1)x)=\Re(1+e^{4ix}+e^{8ix}+\ldots+e^{4(n-1)ix})\)
This is a sum of a geometrical progression with a ratio \(e^{4ix}\). Now sum it. A formula will give an answer:
\[\Re(1+e^{4ix}+e^{8ix}+\ldots+e^{4(n-1)ix})=\Re(\frac{e^{4nix}-1}{e^{4ix}-1})\]Now try to give a fraction in another form:
\(\frac{e^{4nix}-1}{e^{4ix}-1}=\frac{e^{4(n-1)ix}-e^{4nix}-e^{-4ix}+1}{2(1-\cos4x)}\)
An now it is very easy to take a real part of the new fraction. So, the answer will be:
\[1+\cos4x+\cos8x+\ldots+\cos4(n-1)x=\frac{\cos4nx-\cos4(n-1)x+\cos4x-1}{2(\cos4x-1)}\]

Answer 9

there is a formula that you can use for cosines of angles in an arithmetic progression which is given by:\[\sum_{k=0}^{n-1}\cos(a+kd)=\frac{\sin(nd/2)}{\sin(d/2)}\cos\left(a+\frac{(n-1)d}{2}\right)\]
a proof of this can be found here: http://evergreen.loyola.edu/mpknapp/www/papers/knapp-sv.pdf

in your case we have \(a=4x\), \(d=4x\), leading to:\[\sum_{k=0}^{n-1}\cos(4x+k4x)=\frac{\sin(2nx)}{\sin(2x)}\cos\left(2(n+1)x\right)\]therefore:\[1+\cos(4x)+\cos(8x)+...+\cos(4nx)=1+\frac{\sin(2nx)}{\sin(2x)}\cos\left(2(n+1)x\right)\]