let-put-x-n-1-1-n-x-with-x-gt-1-and-x-n-1-1-n-n-x-find-a-relation-between-x-and-x-

Question Number 26223 by abdo imad last updated on 22/Dec/17

l e t p u t ξ (x) = \sum_{n = 1}^{\propto} \frac{1}{n^{x}} w i t h x > 1

a n d δ (x) = \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n}}{n^{x}} f i n d a r e l a t i o n

b e t w e e n ξ (x) a n d δ (x) .

Commented by abdo imad last updated on 23/Dec/17

δ (x) = \sum_{p = 1}^{\propto} \frac{1}{{(2 p)}^{x}} - \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{x}} = 2^{- x} ξ (x) - \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{x}}

b u t ξ (x) = \sum_{p = 1}^{\propto} \frac{1}{{(2 p)}^{x}} + \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{x}}

\Rightarrow \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{x}} = ξ (x) - 2^{- x} ξ (x) = (1 - 2^{- x}) ξ (x)

\Rightarrow δ (x) = 2^{- x} - (1 - 2^{- x}) ξ (x) = (2^{1 - x} - 1) ξ (x)

Answered by prakash jain last updated on 22/Dec/17

δ (x) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{x}}

= - (\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{x}} - 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{x}})

- 1 + \frac{1}{2^{x}} - \frac{1}{3^{x}} + . . = - (1 + \frac{1}{2^{x}} + \frac{1}{3^{x}}) . . + 2 [\frac{1}{2^{x}} + \frac{1}{4^{x}} + . .]

= - (\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{x}} - 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{x}})

δ (x) = - ζ (x) + \frac{2}{2^{x}} ζ (x)

δ (x) = (2^{x - 1} - 1) ζ (x)

Commented by prakash jain last updated on 22/Dec/17

δ (x) = - η (x) (eta function)