Prove-that-i-2-pi-2-6-ii-4-pi-4-90-

Question Number 2796 by Rasheed Soomro last updated on 27/Nov/15

P r o v e t h a t

(i) ζ (2) = \frac{π^{2}}{6} (i i) ζ (4) = \frac{π^{4}}{90}

Answered by prakash jain last updated on 28/Nov/15

(i) ζ (2) = \frac{π^{2}}{6}

There are several proofs available for ζ (2) . I

am giving a proof which utilizes some of

the question already asked Q 2735, Q 2751

and Q 2806 .

f (x) = \underset{n = 1}{\sum^{\infty}} \frac{\cos n x}{n^{2}} is uniformly converged on R . Q 2806

g (x) = \underset{n = 1}{\sum^{\infty}} \frac{\sin n x}{n} is convergent on (0, 2 π) Q 2735

\underset{n = 1}{\sum^{\infty}} \frac{\sin n x}{n} = \frac{π - x}{2} Q 2751

f^{'} (x) = - g (x) = \frac{x - π}{2}

\int_{0}^{π} (- g (x)) d x = f (π) - f (0) ∵ f (x) = \int - g (x) d x

\int_{0}^{π} (- g (x)) d x = \int_{0}^{π} \frac{x - π}{2} d x = - \frac{π^{2}}{4}

f (0) = ζ (2)

f (π) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = - η (2) = - \frac{1}{2} ζ (2)

η (2) is eta function

- \frac{1}{2} ζ (2) - ζ (2) = - \frac{π^{2}}{4}

ζ (2) = \frac{π^{2}}{6}

Answered by 123456 last updated on 27/Nov/15

(i i) ζ (4)

{(\underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{s}})}^{2} = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{2 s}} + 2 \underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{s}}

{[ζ (s)]}^{2} = ζ (2 s) + 2 \underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{s}}

ζ (4) = {[ζ (2)]}^{2} - 2 \underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{2}}

= {[\frac{π^{2}}{6}]}^{2} - 2 \cdot \frac{π^{4}}{120}

= \frac{π^{4}}{36} - \frac{π^{4}}{60}

= \frac{(60 - 36) π^{4}}{36 \times 60}

= \frac{24 π^{4}}{36 \times 60}

= \frac{6 π^{4}}{18 \times 30}

= \frac{π^{4}}{9 \times 10}

= \frac{π^{4}}{90}

- - - - - detail in comment - - - - -

Commented by 123456 last updated on 27/Nov/15

writing the power series of \sin x at x_{0} = 0

\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdot \cdot \cdot

writing it as

\sin x = x (1 + \frac{x}{π}) (1 - \frac{x}{π}) (1 + \frac{x}{2 π}) (1 - \frac{x}{2 π}) \cdot \cdot \cdot

\sin x = x (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) \cdot \cdot \cdot

evaluating x^{5} coeficient we gets

a_{5} = \frac{1}{π^{4}} \underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{2}}

so equalating the coeficients of x^{5}

\frac{1}{π^{4}} \underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{2}} = \frac{1}{5!}

\underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{1}{{(n m)}^{2}} = \frac{π^{4}}{120}

not a nice proof, but work (or i think

it work)

Commented by Rasheed Soomro last updated on 30/Nov/15

D e t e r m i n i n g c o e f f i c i e n t o f x^{5} f r o m

s i n x = x (1 + \frac{x}{π}) (1 - \frac{x}{π}) (1 + \frac{x}{2 π}) (1 - \frac{x}{2 π}) \dots

i s t o o c o m p l i c a t e d!

Commented by 123456 last updated on 01/Dec/15

there no need for this, use the fact

(a + b) (a - b) = a^{2} - b^{2}

wich turn more easy the determination

Commented by Rasheed Soomro last updated on 01/Dec/15

W i l l o n l y f i r s t f i v e f a c t o r s m a k e t h e t e r m c o n t a i n i n g

x^{5} ? T h e r e m a i n i n g f a c t o r s h a v e n o e f f e c t r e g a r d i n g

t h i s t e r m ?

Commented by 123456 last updated on 01/Dec/15

all terms have importance

note that 5 = 1 + 2 + 2

so we have (only loking for the potence)

1, (0, 2), (0, 2), (0, 2) \dots .

we have to make 5, and have to use

all elements above chosing the correct

from the parentys, the first choose is

1 2 2 0 0 \dots . .

1 2 0 2 0 \dots . .

1 2 0 0 2 \dots . .

⋮

1 0 2 2 0 \dots . .

1 0 2 0 2 \dots . .

⋮

1 0 0 2 2 \dots . .

as you can see, we gets

x \times - \frac{x^{2}}{{(n π)}^{2}} \times - \frac{x^{2}}{{(m π)}^{2}} m > n, (m, n) \in N

sums all these contribuitions and them

you gets

\underset{n = 1}{\sum^{+ \infty}} \underset{m = n + 1}{\sum^{+ \infty}} \frac{x^{5}}{{(n m)}^{2} π^{4}} wich is you factor of x^{5}

similiar think hold for x, x^{3}, x^{7}, \dots

not sure if that helped

ps : making it tl x^{3} is similiar to euler proof to basileia prolem

the only diference is that he used

\frac{\sin x}{x} to make the manipulation (and them

compared x^{2} and not x^{3})

Commented by Rasheed Soomro last updated on 01/Dec/15

T h i s {d o n}^{'} t r e q u i r e o n l y (a - b) (a + b) = a^{2} - b^{2}

T h i s i s s o m e w h a t d i f f i c u l t .

A n y w a y {THANK}^{S} f o r e x p l a n a t i o n i n d e t a i l .

I w i l l t r y m y b e s t t o u n d e r s t a n d i t .

Commented by 123456 last updated on 02/Dec/15

we can found x^{5} from

x (1 - \frac{x}{π}) (1 + \frac{x}{π}) \cdot \cdot \cdot

however its require a most hard work

than making same thing from

x (1 - \frac{x^{2}}{π^{2}}) \cdot \cdot \cdot

Commented by Rasheed Soomro last updated on 02/Dec/15

O f c o u r s e s i r!