Let x 1 , x 2 , x 3 , ⋯ denote all the positive solutions of the equation tan x = x .
Find n = 1 ∑ ∞ cos 2 x n
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
This is nice!!!
Are you the same Sangchul Lee from MSE? If yes, I am a great fan of your integration skills. Your answers there are quite amazing. As far as this solution goes, I could just say one thing- Perfection at its best!!!!
Log in to reply
Thank you! I never expected anyone to recognize me, so I'm glad to see someone who knows me :)
Log in to reply
Genius always outshines the others... I know you from a quite long time. You have even answered some of my favourite questions very easily on MSE..... So how could I forget you....
The function f ( z ) = z cos z − sin z has a triple zero at z = 0 , and simple zeros at z = ± x n for n ∈ N . Moreover f ′ ( x ) = − z sin z . Thus the function g ( z ) = f ( z ) z sin z is an odd function with a simple pole at z = 0 and simple poles at z = ± x n for n ∈ N . Moreover R e s z = 0 g ( z ) = − 3 R e s z = ± x n g ( z ) = − 1 If we let C N be the square with vertices N π + i N π , N π − i N π , − N π − i N π , − N π + i N π , then we can show that there exists K > 0 such that ∣ g ( z ) ∣ ≤ K for all z ∈ C N for all positive integers N , and this implies that N → ∞ lim ∫ C N z 2 + 1 g ( z ) d z = 0 Now R e s z = i z 2 + 1 g ( z ) = 2 i g ( i ) = 4 1 ( e 2 − 1 ) R e s z = − i z 2 + 1 g ( z ) = − 2 i g ( − i ) = 4 1 ( e 2 − 1 ) and so 2 π i 1 ∫ C N z 2 + 1 g ( z ) d z = R e s z = 0 z 2 + 1 g ( z ) + R e s z = i z 2 + 1 g ( z ) + R e s z = − i z 2 + 1 g ( z ) + 2 n = 1 ∑ N R e s z = x n z 2 + 1 g ( z ) = − 3 + 4 1 ( e 2 − 1 ) + 4 1 ( e 2 − 1 ) − 2 n = 1 ∑ N x n 2 + 1 1 = 2 1 ( e 2 − 7 ) − 2 n = 1 ∑ N cos 2 x n Letting N → ∞ tells us that n = 1 ∑ ∞ cos 2 x n = 4 1 ( e 2 − 7 )
Problem Loading...
Note Loading...
Set Loading...
Here is an Euler-themed solution. First of all, notice that cos 2 x k = 1 + tan 2 x k 1 = 1 + x k 2 1 , and so, k = 1 ∑ ∞ cos 2 ( x k ) = k = 1 ∑ ∞ 1 + x k 2 1 . Next, introduce the function f ( z ) = z sin z − cos z . This is an entire function of order 1/2 . Moreover, a careful inspection shows that the only zeros of f are either 0 or x k 2 for some k ∈ N , all of which are simple. So by the Hadamard factorization theorem , f ( z − 1 ) = A ( z − 1 ) k = 1 ∏ ∞ ( 1 − x k 2 + 1 z ) . (Notice that the zeros of f are either 1 or x k 2 + 1 for some k ∈ N .) Plugging z = 0 determines the value of A as 1 / e , and so, k = 1 ∏ ∞ ( 1 − x k 2 + 1 z ) = e z − 1 f ( z − 1 ) = 1 − 4 e 2 − 7 z + O ( z 2 ) . Now comparing the coefficient of z in both sides, we obtain k = 1 ∑ ∞ 1 + x k 2 1 = 4 e 2 − 7 .