Question

Find \[\large{\sum_{n=1}^{\infty}\frac{1}{n^6}}\] using parsevals identity

Answer 1

hey funny bunny

Answer 2

lol hi

Answer 3

how is your day?

Answer 4

ok i don't really know this, but i am going to make a guess

Answer 5

ok go ahead:D

Answer 6

my guess is to use \(f(x)=x^3\)

Answer 7

and before i do this i guess we (or i) should go back and look to see how parseval uses \(f(x)=x^2\) to find
\[\sum \frac{1}{n^4}=\frac{\pi^4}{90}\]

Answer 8

i can find sum 1/n^2 =pi^2/3 using parsevals,,, but i was wondering do i use that with cubing it for example and do the same step or im not allowed to do that

Answer 9

well that was my guess, but i am still looking over the proof of
\[\sum \frac{1}{n^2}=\frac{\pi^2}{6}\] and
\[sum\frac{1}{n^4}=\frac{\pi^4}{90}\] to see if i understand exactly what is going on

Answer 10

i can show u the proof if u want

Answer 11

looks like we find \[c_n=\int_{-\pi}^{\pi}x^3e^{-inx}dx\] yes?

Answer 12

will require integration by parts i guess

Answer 13

ok
let f(x)=x^2
after doing all the steps to find the fourier series for f(x)
\[=\frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\]applying parsevals identity
\[\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)^2dx=2a_0^2+\sum_{n=1}^{\infty}(a_n^2+b_n^2)\]\[\frac{1}{\pi} \int\limits_{-\pi}^{\pi}x^4dx=2(\frac{\pi^2}{3})^2+\sum_{n=1}^{\infty}(\frac{16}{n^4)}\]now solve for the summation

Answer 14

\[-\pi<x<\pi\]

Answer 15

right, but now the real question boils down to how to get the coefficients to look like \(\frac{1}{n^6}\) and that is not so clear. doesn't look like \(x^3\)does that

Answer 16

i tried with x^3 but didnt work

Answer 17

yeah it seems like it does not. let me look around and see what i can find

Answer 18

wow this is epic stuff...beyond the limits of human intellect

Answer 19

ok thanku sat

Answer 20

when can i start to comprehend such logic? :P

Answer 21

got a lot of followers here
any ideas?
@Zarkon ?

Answer 22

it is not coming to me yet...been a long time.

Answer 23

well i am busy googling and cannot find anything similar,but it must be out there somewhere. trick is only to come up with appropriate function i think

Answer 24

what is \[\zeta(6)?\]

Answer 25

1

Answer 26

medal please?

Answer 27

http://math.stackexchange.com/questions/115981/computing-zeta6-sum-limits-k-1-infty-frac1k6-with-fourier-series

this is what i found but i dont know whats the zeta function

Answer 28

\[\Pi^6\div945\]

Answer 29

\[\zeta(6)=\frac{\pi^4}{945}\] which i think is your answer

Answer 30

http://www.math.ucsd.edu/~p1tong/MAT3093/Assign%205/assignment5.pdf

Answer 31

problem 8 b

Answer 32

isn't it pi^6 satellite?

Answer 33

nice @Zarkon thats exactly what i want
but is there any method or trick i can use to figure out functions for different sums for example if iw as asked for higher powers of n ...

Answer 34

I might have known some tricks a long time ago...but they have all faded away.

Answer 35

ok if i want to learn it ... what should i search for lol,,

Answer 36

not sure :(

Answer 37

yeah i saw the
\(\frac{1}{n^4}\) verson a few places

Answer 38

hehe thanks anyways youve done more than enough :D @Zarkon

And @satellite73 Thanku for spending this much time on it

you are both awesome:D

Answer 39

I have a nice little book on this stuff...I might look though it tonight and if I figure out anything I'll post it.

Answer 40

i'd appreciate that:)

Answer 41

page 5 has a formula for \(\zeta(2n)\)
http://numbers.computation.free.fr/Constants/Miscellaneous/zetageneralities.pdf