Question

\(\Large \dfrac{π^4}{90}=\dfrac{1}{1^4} +\dfrac{1}{2^4} +\dfrac{1}{3^4} +\dfrac{1}{4^4} +⋯\)

Answer 1

pretty

Answer 2

I have proved that
\(((π-x )/2)^2=π^2/12+[1/1^2 cos 1πx+1/2^2 cos 2πx+⋯]\)


from which i could deduce that
\([1/1^2 +1/2^2 +1/3^2 …]=(-π^2)/6 \)

now the 2nd part of deduction asks me to prove that.

Answer 3

thats supposed to be
\(\Large +\pi^2/6 \)

Answer 4

what could i do for the 2nd deduction ?

Answer 5

Are you familiar with \[ \sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2} \right), z\in \mathbb{C}\]

Answer 6

yes,
but i need to deduce that by using series for (pi - x)^2/2
its the 2nd part of that question.

i cannot use series for sin pi z

Answer 7

u can use parseval's indentity

Answer 8

question :
find fourier expansion for [(pi-x)/2 ]^2
and deduce those 2

Answer 9

there's a formula
\[\int\limits_{-c}^{c}[f(x)^2]dx=c[\frac{ a_0^2 }{2 }+\sum_{n=1}^{\infty}(a_n^2+b_n^2)]\]

Answer 10

yeah, i saw that, let me try that and get back to you

Answer 11

x --0 to (pi) ??

Answer 12

0 to 2pi

Answer 13

\(\Large ((\pi-x)/2 )^2 = \pi^2/12 +\sum \limits _{n=1}^\infty (1/n^2) \cos n\pi x \)

then what would be next ? how would i apply parseval here ?

Answer 14

bn = 0
an = 1/n^2

Answer 15

\(\Large \int\limits_{0}^{2\pi }[(\dfrac{(\pi-x)^2}{2})^2]dx=2\pi [\frac{ (\pi^2 /6)^2 }{2 }+\sum_{n=1}^{\infty}(1/n^4 )]\)

like this ?

Answer 16

and i need to solve the integral on left side ?

Answer 17

too much work for a couple of marks

Answer 18

yes
and it will be a dangerous one :(

Answer 19

hey what is a_0=pi^2/6 ?

Answer 20

yes

Answer 21

whats the exact formula for parsevals identity which i can use here...i would work it out, just wanna know the formula that i can use

Answer 22

the formula i wrote that's the exact form

Answer 23

ok, got my mistake
left side = pi^5/40

Answer 24

still i get -pi^4/720 ...

Answer 25

yes left side pi^5/40

Answer 26

then i get i get -pi^4/720

Answer 27

yes i'm also getting the same thing:(

Answer 28

that formula is for -c to c
is there some different formula for 0 to a
?

Answer 29

there could be three cases i think
1) if 0<x<2c
\[\int\limits_{0}^{2c}[f(x)]^2dx=c[\frac{ a_0^2 }{ 2}+\sum_{}^{}a_n^2+b_n^2]\]
2)if 0<x<c
\[\int\limits_{0}^{c}[f(x)]^2dx=c/2[\frac{ a_0^2 }{ 2 }+\sum_{}^{}a_n^2]\]
this is like halfrange cosine series
3)0<x<c (half range sine)
\[\int\limits_{0}^{c}[f(x)]^2dx=c/2[\frac{ a_0^2 }{ 2 }+\sum_{}^{}b_n^2]\]

Answer 30

so i just try using pi instead of 2pi on right...?
bn = 0 so 3) case =0

Answer 31

yep, just using pi gave me exact pi^4/90

Answer 32

yeah atlast :)

Answer 33

seriously, too much work for a couple of marks

Answer 34

yeah :)
are u preparing for ies or something

Answer 35

no no... i am writing past paper solutions ..

Answer 36

past papers of?

Answer 37

sem 3 maths, mumbai university

Answer 38

awesome :3