let-f-x-e-2x-ln-1-x-2-developp-f-at-integr-serie-

Question Number 68239 by mathmax by abdo last updated on 07/Sep/19

l e t f (x) = e^{- 2 x} l n (1 + x^{2})

d e v e l o p p f a t i n t e g r s e r i e .

Commented by mathmax by abdo last updated on 09/Sep/19

f i r s t l e t d e t e r m i n e f^{(n)} (x) w e h a v e

f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(l n (1 + x^{2}))}^{(k)} {(e^{- 2 x})}^{(n - k)}

= l n (1 + x^{2}) {(- 2)}^{n} e^{- 2 x} + \sum_{k = 1}^{n} C_{n}^{k} {(l n (1 + x^{2}))}^{(k)} {(- 2)}^{n - k} e^{- 2 x}

l e t f i n d {(l n (1 + x^{2}))}^{(k)} f o r k ⩾ 1 l e t w (x) = l n (1 + x^{2}) w e h a v e

w^{^{'}} (x) = \frac{2 x}{1 + x^{2}} = \frac{2 x}{(x - i) (x + i)} = \frac{1}{x + i} + \frac{1}{x - i} \Rightarrow

w^{(k)} (x) = {(\frac{1}{x + i})}^{(k - 1)} + {(\frac{1}{x - i})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{k}} + \frac{{(- 1)}^{k - 1} (k - 1)}{{(x - i)}^{k}}

= {(- 1)}^{k - 1} (k - 1)! {\frac{1}{{(x + i)}^{k}} + \frac{1}{{(x - i)}^{k}}}

= {(- 1)}^{k - 1} (k - 1)! {\frac{{(x + i)}^{k} + {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}} = \frac{{(- 1)}^{k - 1} (k - 1)! 2 R e {(x + i)}^{k}}{{(x^{2} + 1)}^{k}}

w e h a v e x + i = \sqrt{1 + x^{2}} e^{i a r c t a n (\frac{1}{x})} \Rightarrow {(x + i)}^{k} = {(1 + x^{2})}^{\frac{k}{2}} e^{i k a r c t a n (\frac{1}{x})}

w^{(k)} (x) = 2 \frac{{(- 1)}^{k - 1} (k - 1)! {(1 + x^{2})}^{\frac{k}{2}} c o s (k a r c t a n (\frac{1}{x}))}{{(x^{2} + 1)}^{k}} \Rightarrow

f^{(n)} (x) = {(- 2)}^{n} e^{- 2 x} l n (1 + x^{2})

+ \sum_{k = 1}^{n} C_{n}^{k} \frac{2 {(- 1)}^{k - 1} (k - 1)! {(1 + x^{2})}^{\frac{k}{2}} c o s (k a r c t a n (\frac{1}{x}))}{{(x^{2} + 1)}^{k}} {(- 2)}^{n - k} e^{- 2 x}

Commented by mathmax by abdo last updated on 09/Sep/19

f^{(n)} (0) = 2 \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k - 1} (k - 1)! c o s (\frac{k π}{2}) {(- 2)}^{n - k}

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

= \sum_{n = 1}^{\infty} \frac{2}{n!} (\sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{n - k} c o s (\frac{k π}{2})) x^{n}