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)!} = \infty
Homework 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, 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 [Broken]
Which lead to:
[PLAIN][PLAIN]http://folk.ntnu.no/jonvegar/images/math.gif [Broken]
This will lead to a 3.degree-equation. And the result is a answer with either...
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 ;)
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 [Broken]
Thank you for LaTeX tip :smile:
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) =...
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)
Well, I can make a new thread where you have to get rid of the i-s...
This is not any schoolwork so I have all the time in the world :smile:
I have tried what's in post #26 with no result
How do you get an expression for x, when you got both eix and e-ix? And even if I get an expression for...
How do you get an expression for x, when you got both eix and e-ix? And even if I get an expression for arccos without i-s, I still have to get rid of cos...
Ok, here is what I have done:
http://folk.ntnu.no/jonvegar/images/imaginary3.png [Broken]
http://folk.ntnu.no/jonvegar/images/imaginary2.png [Broken]
[PLAIN][PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary4.png [Broken]
But when I'm testing my result in maple I get "false"...
Belive me, I have tried. But I don't understand the answer i get if I use the 2. method:
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary.jpg [Broken]
I want a answer without any imaginary parts. My x is defined only for [0,9], but when I cancel the imaginary part I do not get the...
Hmmm, it shall be possible to write the equation without any logarithms.
If it is so easy, can you just write down the formula for me? :smile: I need it actually just to solv an other problem.
As Riachard Dawkins self put it: "Evolution is a fact"
http://richarddawkins.net/thegreatestshowonearth" [Broken]
The flying spaghettimonster is a argument atheist shall use.
http://www.youtube.com/watch?v=vk8EANdpAj0&playnext_from=TL&videos=9xrUGG_-QLU"
Invent your own monster :P
Oh, I'm sorry then. It's just my way to talk. I'm from Norway, so I suppose that our expressions can't be directly translated. And my english sucks. :smile:
I'm never rude :smile: