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Hahaha, yeah I've started to play worm mostly now, I'll get better at it one day Mr. Snake pro.

Hehehe, did you see my attempts at snake?

Well I tried to make my integral definite, the limits being the same as yours -> theta and pi However I ended up on the RHS with one term being: \tan \frac{\pi}{2} Which I can't do. So I decided to put in the only other one that leave P on the LHS by itself and that was theta and 0. So I...

Oo, I see thank you. EDIT: Is there a way of doing something similar for my integral?

Thank you for taking the time to latex it all up, I appreciate it :) But, in your explanation to why the second term tends to zero, what do you mean it is bounded above by...?

Oh I completely stuffed up that interpretation.............. I took them as complex numbers... ಠ_ಠ

Was there any additional information missing in the question? At the moment the question doesn't make any sense. Because you got PA^2 = AP^2 And we can only derive that from some additional information..

Integrate the load of cosine functions basically Or if there is another method to acquiring this sum I would like to hear it.

Integration from when I have acquired: \sum_{k=1}^{n} \cos k\theta

If theta is zero, then z = 1 And we cannot have z = 1 due to using the geometric series formula, and the denominator being 1-z Yes, but how would I use HSC techniques to complete the problem (this is the point of the thread)

Not sure what you mean by that, but I only used complex numbers to get to the initial sum of cos identity to then integrate. I need help evaluating the limit/or another approach using integration.

Re: HSC 2013 4U Marathon Wow, very good. This is just a 'filler' question while I find/make something else to ask, but I want to stick with Complex Numbers/Polynomials so others can contribute. Also, this question is from another STEP paper but I forget which one.

Help - Real Analysis So I read somewhere that: \sum_{k=1}^{\infty} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2} However they derived it some weird way that I know nothing about (this is STEP Advanced Problems #43) Now I set out to derive this, I first though of...

Re: HSC 2013 4U Marathon $The nth Fermat number $ \ \ F_n \ \ $is defined by$ F_n= 2^{(2^n)} + 1 $Prove by Mathematical Induction or otherwise$ \\ \\ F_0 F_1 F_2 ... F_{k-1} \equiv \prod_{m=0}^{k-1}F_{k-1} = F_{k} - 2 $Hence deduce that no two Fermat numbers have a common factor greater...

Re: HSC 2013 4U Marathon Awww man :/ Heh, I guess I'll try make another one

Re: HSC 2013 4U Marathon I made this question just now: $If $ \ \ \ z= \cos \theta + i \sin \theta $Consider $ \ \ \ (z+\frac{1}{z})^{2n} $Show that$ \int_{0}^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n+1}(n!)^2}

Nah its not you, I really don't have any urge to go.

No thanks

Re: HSC 2013 4U Marathon Nice work! Your approach to the first and second one was different to mine, I actually ripped this question off of STEP 2008 III Q2

Re: HSC 2013 4U Marathon $Denote$ \ \ \ \ S_k(n) \equiv \sum_{r=0}^n r^k $You may assume$ \ \ \ S_1(n)=\frac{n}{2}(n+1) \ \ \ \ \ S_2 (n) = \frac{n}{6}(n+1)(2n+1) $i) By considering $ \sum_{r=0}^n (r+1)^k - r^k $Show that$ \\ \\ kS_{k-1}(n) = (n+1)^k - (n+1) -...