Question

show that
\( 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...=\frac{1}{6} \pi^2 \)

Answer 1

this is famous basel problem solved by euler i believe

Answer 2

aha then ?

Answer 3

there are many proofs online, wikipedia

Answer 4

mmm

Answer 5

http://en.wikipedia.org/wiki/Basel_problem

Answer 6

The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be π2/6 and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, and it was not until 1741 that he was able to produce a truly rigorous proof.

Answer 7

im given this Qn , and it asked to make usufull of fourier to show this

Answer 8

what Qn are you given, sorry i dont understand your english

Answer 9

you must write fourier series of \(x^2\) for some symmetric interval

Answer 10

use the problem
"
find the fourier series of the function f(x) with period p=2l
f(x)=x^2 (-1<x<1) p=2"

Answer 11

ohk cool @mukushla can u help me with that ?

Answer 12

It's a well known problem, yes, let's work it together if you want

Answer 13

yeah sure ^_^

Answer 14

i tried on x^2
ill type the final i got hope its right

Answer 15

\(\large \frac{\pi^2}{3}+\sum_{n=1}^{\infty} a_n \cos nx =\frac{\pi^2}{3} +a_1\cos x+a_2\cos2 x ++a_3\cos3 x\)

=\(\large \frac{\pi^2}{3} -4\cos x+\cos2 x -\frac{4}{9}\cos 3x+\frac{1}{4}\cos 4x+...\)

Answer 16

Can you provide general form of \(a_n\) ? :-)

Answer 17

yess
a_0=\(\pi^2/3\)
ill type a_n

Answer 18

\(4\frac{(-1)^n}{n^2}\)?

Answer 19

yes :D

Answer 20

well, now let \(x=0\), see what happens

Answer 21

for the series ?

Answer 22

yes

Answer 23

\(\large \frac{\pi^2}{3} -4 +1 -\frac{4}{9} +\frac{1}{4}-\frac{4}{25} +... \)

Answer 24

so\[0=\frac{\pi^2}{3}+\sum_{n=1}^{\infty} a_n \]\[\sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2}=-\frac{\pi^2}{3}\]\[\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}\]so we need to plug in another number for \(x\)

Answer 25

aha ic..

Answer 26

what do you think? what makes that \((-1)^n\) disappear?

Answer 27

this is fourier series, sorry i have not learned this :o

Answer 28

mmm pi ?

Answer 29

its ok perl :D

Answer 30

\(\pi\) is not on our interval

Answer 31

the interval is given for pre Qn

Answer 32

I think there is a little mistake in your final answer, that must be\[x^2=\frac{\pi^2}{3}+\sum_{n=0}^{\infty} \frac{4(-1)^n}{n^2} \cos n\color\red{\pi} x\]

Answer 33

mmm why ?

Answer 34

cuz of the interval ?

Answer 35

i solved this for -pi<x<pi

Answer 36

took period of 2 pi

Answer 37

aha i thought u used \((-1,1)\)

Answer 38

no ;)

Answer 39

so you are quite right :-) you know why? \[x^2=\frac{\pi^2}{3}+\sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cos n x\]let \(x=\pi\) \[\pi^2=\frac{\pi^2}{3}+\sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} (-1)^{n}\]\[\frac{\pi^2}{6}=\sum_{n=1}^{\infty} \frac{1}{n^2}\]

Answer 40

do u think i should make the interval -1<x<1
to have n pi x ?

Answer 41

ohh lol
i forget totake x^2 xD
ok i got it nww
tyyyyyy <3

Answer 42

very welcome <3

Answer 43

is there an introduction to fourier series online?

Answer 44

sure, just google it ;-)

Answer 45

that deserves a medal

Answer 46

:D