let-f-x-x-2-1-x-4-1-calculate-f-n-x-2-find-f-n-0-3-developp-f-at-integr-serie-4-calculate-0-1-f-x-dx-

Question Number 37820 by prof Abdo imad last updated on 17/Jun/18

l e t f (x) = \frac{x^{2}}{1 + x^{4}}

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

4) c a l c u l a t e \int_{0}^{1} f (x) d x .

Commented by prof Abdo imad last updated on 18/Jun/18

2) f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{4} e^{- \frac{i π}{4}} ({(- 1)}^{n + 1} - 1) e^{- i \frac{n + 1}{4} π}

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

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

= \frac{{(- 1)}^{n} n!}{2} {{(- 1)}^{n + 1} - 1) c o s (\frac{(n + 2) π}{4})

\Rightarrow f^{(2 n)} (0) = \frac{(2 n)!}{2} (- 2) c o s \frac{(2 n + 2) π}{4}

= - (2 n)! c o s (\frac{n π}{2} + \frac{π}{2}) = (2 n)! s i n (\frac{n π}{2})

a n d f^{(2 n + 1)} (0) = 0

3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

= \sum_{n = 0}^{\infty} \frac{f^{(2 n)} (0)}{(2 n)!} x^{2 n}

= \sum_{n = 0}^{\infty} s i n (\frac{n π}{2}) x^{2 n} . b u t s i n (\frac{n π}{2}) = 0 i f n = 2 p

a n d s i n (\frac{n π}{2}) = s i n (\frac{(2 p + 1) π}{2}) = s i n (p π + \frac{π}{2})

= {(- 1)}^{p} i f n = 2 p + 1 \Rightarrow

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

= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n + 2} . t h e r a d i u s o f c o n v e r g e n c e

i s R = 1 .

Commented by prof Abdo imad last updated on 18/Jun/18

1) w e h a v e f (x) = \frac{x^{2}}{{(x^{2})}^{2} - i^{2}}

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

l e t p (x) = 1 + x^{4} w e h a v e

λ_{i} = \frac{z_{i}^{2}}{4 z_{i}^{3}} = \frac{1}{4 z_{i}} \Rightarrow

a = \frac{1}{4} e^{- \frac{i π}{4}}

b = - \frac{1}{4} e^{- \frac{i π}{4}}

c = \frac{1}{4} e^{\frac{i π}{4}}

d = - \frac{1}{4} e^{\frac{i π}{4}} \Rightarrow

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

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

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

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

Answered by behi83417@gmail.com last updated on 18/Jun/18

f (x) = \frac{1}{2} . \frac{2 x^{2}}{(x^{2} + i) (x^{2} - i)} = \frac{1}{2} . \frac{(x^{2} + i) + (x^{2} - i)}{(x^{2} + i) (x^{2} - i)} =

= \frac{1}{2} [\frac{1}{x^{2} + i} + \frac{1}{x^{2} - i}] = \frac{1}{2} [\frac{1}{(x + i \sqrt{i}) (x - i \sqrt{i})} +

+ \frac{1}{(x - \sqrt{i}) (x + \sqrt{i})}] = \frac{1}{2} [\frac{1}{2 i \sqrt{i}} [\frac{(x + i \sqrt{i}) - (x - i \sqrt{i})}{(x + i \sqrt{i}) (x - i \sqrt{i})}] +

+ [\frac{1}{2 \sqrt{i}} [\frac{(x + \sqrt{i}) - (x - \sqrt{i})}{(x + \sqrt{i}) (x - \sqrt{i})}]] \overset{\sqrt{i} = a}{=}

= - \frac{a}{4} [\frac{1}{x - a^{3}} - \frac{1}{x + a^{3}} + \frac{i}{x - a} - \frac{i}{x + a}]

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

f^{n} (x) = \frac{n! a {(- 1)}^{n}}{4} [\frac{1}{{(x - a^{3})}^{n + 1}} - \frac{1}{{(x - a^{3})}^{n + 1}} +

+ \frac{i}{{(x - a)}^{n + 1}} - \frac{i}{{(x - a)}^{n + 1}}] .

2) f^{n} (0) = 0

4) \int \frac{x^{2} d x}{1 + x^{4}} = - \frac{a}{4} [l n \frac{x + a^{3}}{x - a^{3}} + i . l n \frac{x + a}{x - a}]

F (1) = - \frac{a}{4} [l n \frac{1 + a^{3}}{1 - a^{3}} + i . l n \frac{1 + a}{1 - a}] =

- \frac{\sqrt{i}}{4} [i \sqrt{i} + \sqrt{i} - i + i (i \sqrt{i} + \sqrt{i} + i)] =

= - \frac{\sqrt{i}}{4} [i \sqrt{i} + \sqrt{i} - i - \sqrt{i} + i \sqrt{i} - 1] =

= - \frac{\sqrt{i}}{4} [2 i \sqrt{i} - i - 1] = - \frac{1}{4} (- 2 - i \sqrt{i} - \sqrt{i})

F (0) = - \frac{a}{4} [l n (- 1) + i . l n (- 1)] =

= - \frac{a}{4} (\frac{i π}{2} + i . \frac{i π}{2}) = - \frac{π \sqrt{i}}{8} (i - 1) = \frac{π \sqrt{i}}{8} (1 - i) .

I = F (1) - F (0) = \frac{1}{8} [4 + (i + 1) \sqrt{i} + π \sqrt{i} (i - 1)] .

i = c o s \frac{π}{2} + i s i n \frac{π}{2} = e^{\frac{i π}{2}} \Rightarrow \sqrt{i} = e^{\frac{i π}{4}}

\frac{1 + a^{3}}{1 - a^{3}} = \frac{{(1 + a^{3})}^{2}}{1 - a^{6}} = \frac{(1 + 2 i \sqrt{i} - i)}{1 + i} . \frac{1 - i}{1 - i} =

= \frac{1}{2} (1 - i + 2 i \sqrt{i} + 2 \sqrt{i} - i - 1) = i \sqrt{i} + \sqrt{i} - i

\frac{1 + a}{1 - a} = \frac{{(1 + a)}^{2}}{1 - a^{2}} = \frac{1 + 2 \sqrt{i} + i}{1 - i} . \frac{1 + i}{1 + i} =

= \frac{1}{2} (1 + i + 2 \sqrt{i} + 2 i \sqrt{i} + i - 1) = i \sqrt{i} + \sqrt{i} + i

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

s i r B e h i i f f (x) = \frac{1}{x + a} a f r o m C w e h a v e

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x + a)}^{n + 1}}!

Commented by behi83417@gmail.com last updated on 18/Jun/18

c o r r e c t e d! t h a n k s .

c o r r e c t e d! t h a n k s .