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Question Number 37838 by math khazana by abdo last updated on 18/Jun/18

l e t f (x) = e^{- 2 x} a r c t a n (x^{2})

1) c a l c u l a t e f^{(n)} (x)

2) f i n d f^{(n)} (0)

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

Commented by math khazana by abdo last updated on 20/Jun/18

L e i n i z f o r m u l a g i v e

f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n (x^{2}))}^{(k)} {(e^{- 2 x})}^{(n - k)}

w e h a v e {(e^{- 2 x})}^{(1)} = - 2 e^{- 2 x}

{(e^{- 2 x})}^{(2)} = {(- 2)}^{2} e^{- 2 x} \Rightarrow {(e^{- 2 x})}^{(p)} = {(- 2)}^{p} e^{- 2 x}

a l s o w e h a v e

{(a r c t a n (x^{2}))}^{(1)} = \frac{2 x}{1 + x^{4}} \Rightarrow

{(a r c t a n (x^{2}))}^{(n)} = {(\frac{2 x}{1 + x^{4}})}^{(n - 1)}

l e t w (x) = \frac{x}{1 + x^{4}}

w (x) = \frac{x}{(x^{2} - i) (x^{2} + i)} = \frac{x}{(x - \sqrt{i}) (x + \sqrt{i}) (x - \sqrt{- i}) (x + \sqrt{- i})}

= \frac{x}{(x - e^{\frac{i π}{4}}) (x + e^{\frac{i π}{4}}) (x - e^{- \frac{i π}{4}}) (x + e^{- \frac{i π}{4}})}

= \frac{a}{x - e^{\frac{i π}{4}}} + \frac{b}{x + e^{\frac{i π}{4}}} + \frac{c}{x - e^{- \frac{i π}{4}}} + \frac{d}{x + e^{- \frac{i π}{4}}}

λ_{i} = \frac{z_{i}}{4 z_{i}^{3}} = \frac{1}{4 z_{i}^{2}} \Rightarrow a = \frac{1}{4 {(e^{\frac{i π}{4}})}^{2}} = \frac{1}{4 i} = \frac{- i}{4}

b = \frac{1}{4 i} = \frac{- i}{4}, c = - \frac{1}{4 i} = \frac{i}{4}, d = - \frac{1}{4 i} = \frac{i}{4} \Rightarrow

w^{(n - 1)} (x) = \frac{- i}{4} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - e^{\frac{i π}{4}})}^{n}}

- \frac{i}{4} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + e^{\frac{i π}{4}})}^{n}} + \frac{i}{4} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - e^{- \frac{i π}{4}})}^{n}}

+ \frac{i}{4} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + e^{- \frac{i π}{4}})}^{n}} \Rightarrow

{(a r c t a n (x^{2}))}^{(n)}

= \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! {\frac{- 1}{{(x - e^{\frac{i π}{4}})}^{n}} + \frac{- 1}{{(x + e^{\frac{i π}{4}})}^{n}}

+ \frac{1}{{(x - e^{- \frac{i π}{4}})}^{n}} + \frac{1}{{(x + e^{- \frac{i π}{4}})}^{n}}} \Rightarrow

f^{(n)} (x)

= \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! \sum_{k = 0}^{n} {(- 2)}^{n - k} C_{n}^{k} {\frac{- 1}{{(x - e^{\frac{i π}{4}})}^{k}}

+ \frac{- 1}{{(x + e^{\frac{i π}{4}})}^{k}} + \frac{1}{{(x - e^{- \frac{i π}{4}})}^{k}} + \frac{1}{{(x + e^{- \frac{i π}{4}})}^{k}}} e^{- 2 x} .

Commented by math khazana by abdo last updated on 20/Jun/18

e r r o r o f c a l c u l u s a t t h e f i n a l l i n e

f^{(n)} (x) = \frac{i}{2} \sum_{k = 0}^{n} {(- 2)}^{n - k} {(- 1)}^{k - 1} (k - 1)! C_{n}^{k} {

\frac{- 1}{{(x - e^{\frac{i π}{4}})}^{k}} + \frac{- 1}{{(x + e^{\frac{i π}{4}})}^{k}} + \frac{1}{{(x - e^{- \frac{i π}{4}})}^{k}} + \frac{1}{{(x + e^{- \frac{i π}{4}})}^{k}}} e^{- 2 x}

Commented by math khazana by abdo last updated on 20/Jun/18

f^{(n)} (0) = \frac{i}{2} \sum_{k = 1}^{n} {(- 2)}^{n - k} {(- 1)}^{k - 1} (k - 1)! {

- {(- e^{\frac{i π}{4}})}^{- k} - {(e^{\frac{i π}{4}})}^{- k} + {(- e^{- \frac{i π}{4}})}^{- k} + {(e^{- \frac{i π}{4}})}^{- k}}

= \frac{i}{2} \sum_{k = 1}^{n} {(- 2)}^{n - k} {(- 1)}^{k - 1} (k - 1)! {- {(- 1)}^{k} e^{- \frac{i k π}{4}}

- e^{- \frac{i k π}{4}} + {(- 1)}^{k} e^{\frac{i k π}{4}} + e^{\frac{i k π}{4}}}

= \frac{i}{2} \sum_{k = 1}^{n} {(- 2)}^{n - k} {(- 1)}^{k - 1)} (k - 1)! {({(- 1)}^{k} + 1) (e^{\frac{i k π}{4}} - e^{- \frac{i k π}{4}})

= \frac{i}{2} \sum_{k = 1}^{n} {(- 1)}^{k - 1} (k - 1)! {(- 2)}^{n - k} {1 + {(- 1)}^{k}} 2 i s i n (\frac{k π}{4})

= \sum_{k = 1}^{n} {(- 1)}^{k} (k - 1)! {(- 2)}^{n - k} (1 + {(- 1)}^{k}) s i n (\frac{k π}{4})

Commented by math khazana by abdo last updated on 20/Jun/18

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

= 2 \sum_{n = 1}^{\infty} \frac{1}{n!} {\sum_{p = 1}^{[\frac{n}{2}]} (2 p - 1)! {(- 2)}^{n - 2 p} s i n (\frac{p π}{2})} x^{n}

Commented by math khazana by abdo last updated on 20/Jun/18

f^{(n)} (0) = \sum_{p = 1}^{[\frac{n}{2}]} (2 p - 1)! {(- 2)}^{n - 2 p} s i n (\frac{p π}{2}) .