developp-at-integr-serie-f-x-1-x-1-x-2-

Question Number 94337 by mathmax by abdo last updated on 18/May/20

d e v e l o p p a t i n t e g r s e r i e f (x) = \frac{1}{(x - 1) (x - 2)}

Answered by Rio Michael last updated on 18/May/20

if you mean develope a series then :

f (x) = \frac{1}{(x - 1) (x - 2)} \equiv \frac{1}{x - 2} - \frac{1}{x - 1}

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

= x^{- 1} {(1 - \frac{2}{x})}^{- 1} - x^{- 1} {(1 - \frac{1}{x})}^{- 1}

= x^{- 1} [1 + \frac{2}{x} + \frac{4}{x^{2}} + \dots] - x^{- 1} [1 + \frac{1}{x} + \frac{1}{x^{2}} + \dots]

= \frac{1}{x} + \frac{2}{x^{2}} + \dots - \frac{1}{x} - \frac{1}{x^{2}} - \dots = \frac{1}{x^{2}} + \dots ∣ \frac{2}{x} ∣ < 1

Commented by mathmax by abdo last updated on 18/May/20

sir rio this is not developpement at integr serie \dots

Answered by mathmax by abdo last updated on 18/May/20

we have f (x) = \sum_{n = 0}^{\infty} \frac{f^{n} (0)}{n!} x^{n} let determine f^{n} (0)

we have f (x) = \frac{1}{x - 2} - \frac{1}{x - 1} \Rightarrow f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x - 2)}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - 1)}^{n + 1}}

\Rightarrow f^{(n)} (0) = {(- 1)}^{n} n! {\frac{1}{{(- 2)}^{n + 1}} - \frac{1}{{(- 1)}^{n + 1}}}

= \frac{{(- 1)}^{n} n!}{{(- 1)}^{n + 1}} {\frac{1}{2^{n + 1}} - 1} = - n! {\frac{1}{2^{n + 1}} - 1} \Rightarrow

f (x) = - \sum_{n = 0}^{\infty} {\frac{1}{2^{n + 1}} - 1} x^{n} = - \sum_{n = 0}^{\infty} \frac{x^{n}}{2^{n + 1}} + \sum_{n = 0}^{\infty} x^{n} with

∣ x ∣< 1 .

Answered by mathmax by abdo last updated on 18/May/20

another way we have f (x) = \frac{1}{x - 2} - \frac{1}{x - 1}

= \frac{1}{1 - x} - \frac{1}{2 - x} = \frac{1}{1 - x} - \frac{1}{2 (1 - \frac{x}{2})} so for ∣ x ∣< 1 we get

f (x) = \sum_{n = 0}^{\infty} x^{n} - \frac{1}{2} \sum_{n = 0}^{\infty} {(\frac{x}{2})}^{n} = \sum_{n = 0}^{\infty} x^{n} - \sum_{n = 0}^{\infty} \frac{x^{n}}{2^{n + 1}}

= \sum_{n = 0}^{\infty} (1 - \frac{1}{2^{n + 1}}) x^{n}