a) prove that 1+z+z^2+...+z^n = 1-z^(n+1) / (1-z) for z $\ne$ 1 b) Using the result of (a), prove that $1+cos\theta + cos2 \theta + ... + cosn \theta = \frac{1}{2} + \frac{sin(n+1/2)\theta}{2sin(\theta/2)}$

Question

<Prob. + Stat.>
a) prove that 1+z+z^2+...+z^n = 1-z^(n+1) / (1-z)  for z $\ne$ 1
b) Using the result of (a), prove that $1+cos\theta + cos2 \theta + ... + cosn \theta =  \frac{1}{2} + \frac{sin(n+1/2)\theta}{2sin(\theta/2)}$

anonymous · Answer

I've done part (a), but I don't know how to do part (b). Any ideas?

loser66 · Answer

how do you prove part a) ? by using generating function?

loser66 · Answer

or sum definition?

anonymous · Answer

^ sum of finite geometric series

loser66 · Answer

the problem ask us use part a to prove part b, it means they must relate to each other, right?

anonymous · Answer

GS for part (a)

anonymous · Answer

@RolyPoly: Is the LHS of part b $$1 + \cos \theta  + {\cos ^2}\theta  + ... + {\cos ^n}\theta $$?

loser66 · Answer

agree with drawar, unclear at that point . verify ,please

anonymous · Answer

@drawar No, cos(nθ)
It is Lagrange trigo's identity.

loser66 · Answer

take a look at http://math.stackexchange.com/questions/183859/how-to-prove-lagrange-trigonometric-identity

anonymous · Answer

So far, I got
1+z+z^2 + ... + z^n = 1-z^(n+1) / (1-z)
$$\frac{1-cos(n+1)\theta- i sin(n+1)\theta}{1-(cos+isin\theta)} = 1+cos \theta + cos2\theta + ... + cosn\theta + i(sin\theta + sin2\theta+... + sinn\theta)$$

anonymous · Answer

That's beyond my knowledge, I betta leave. :)

loser66 · Answer

@drawar do you see my link? exactly this problem

anonymous · Answer

Yes, but that's has something to do with complex analysis, which I haven't known of...

loser66 · Answer

ok

anonymous · Answer

$$LS = \frac{1-cos(n+1)\theta- i sin(n+1)\theta}{1-(cos+isin\theta)} $$$$= \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}} $$

anonymous · Answer

I didn't know it's complex analysis...

anonymous · Answer

Anyway, thanks~

loser66 · Answer

you mean you don't understand the way people come up with the problem at that site?

loser66 · Answer

Sorry for that, I just see the problem looks like yours and post the link for you to see. I didn't read it yet. Let see what can I do for you. give me time, I read and IF I can translate in simple way, I will show you mine. is it ok?

anonymous · Answer

You don't have to say sorry!! Thanks for giving me the link, even though it looks complicated.

loser66 · Answer

hand off, it sounds like only one way to come up with the problem as what people in that link said.

anonymous · Answer

Maybe... :(

anonymous · Answer

Thanks again!