Recent content by Zetison

Thanks for the replies! The method hapefish proposed worked very vell. The sum diverges by the comparison test with the harmonic series. I also solved the problem using stirlings formula, thanks a lot guys! :)

Homework Statement Show that the folowing holds: \lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \inftyHomework Equations It can be shown that \frac{(n!)^2 4^n}{(2n)!} = \prod_{k=1}^n \frac{4k}{k+n}The Attempt at a Solution If I can proove that \lim_{n\to\infty} \ln\left[\frac{(n!)^2...

Yes. My english sucks, I know :P

Yes, I know. I'm a bit sceptical my self. But he is a really smart guy and has also study math like me. The problem is that he is no so old that he can't remember anything :P But if there is no solution without trig or i-s, then we shuld be able to prove it?

Well, the original problem is to solve for x here: http://folk.ntnu.no/jonvegar/images/flaggstangoppgaven.jpg Which lead to: [PLAIN][PLAIN]http://folk.ntnu.no/jonvegar/images/math.gif This will lead to a 3.degree-equation. And the result is a answer with either trigonometrical...

Well, it's just for fun actually. My grandfather is claim that he has solved it, but he can't recall how. So I only got he's word for it. But I really want to solve the problem. He claimed that he solved it when he was at my age (21), so I got a pressure to so too ;)

Yes ;)

Because I have tried for several years now :) But if you think it can be solved, be my guest ;)

I don't think it is possible to solve this problem, all tho my grandfather say he has done it. (he can't remember how, but I guess he cheated...)

But that doesn't give me an answer without trigonometrical components and imaginary numbers... And btw: [PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary9.jpg [SIZE="1"]Thank you for LaTeX tip :smile:

Why are we not interested in the negative root?

Homework Statement Find exact values for z without any trigonometrical components or imaginary numbers x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right) where z = \frac{473419}{6121\sqrt{6121}} Homework Equations cos\left(\frac{1}{3}arccos(z)\right) =...

The problem of this thread is actually solved. Maybe I shall make a new thread about my main goal...

Yes, and this is what I get. I guess that's the far as I am going to get: cos(\frac{1}{3}arccos(x)) = \frac{(x + \sqrt{x^2-1})^{1/3}}{2} + \frac 1 {2(x+\sqrt{x^2-1})^{1/3}} But again, my x is defined as x = \frac{473419}{6121\sqrt{6121}} approximate x = 0.9885806704. So my final goal here...

2y=eix+e-ix z=eix 2y=z+z-1 2yz=z2+1 z2-2yz+1=0 z=(y+(y2-1)0,5) or z=(y-(y2-1)0,5) eix=(y+(y2-1)0,5) or eix=(y-(y2-1)0,5) ix=ln(y+(y2-1)0,5) or ix=ln(y-(y2-1)0,5) x=i-1ln(y+(y2-1)0,5) or x=i-1ln(y-(y2-1)0,5)