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Question Number 72260 by mathmax by abdo last updated on 26/Oct/19

l e t f (x) = \frac{2 x + 3}{x^{2} + 1}

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

2) f i n d f^{(10)} (x) a n d f^{(15)} (x)

3) c a l c u l a t e f^{(10)} (0) a n d f^{(15)} (0)

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

5) l e t g (x) = \int_{0}^{x} f (t) d t d e v e l o p p g a t i n t e g r s e r i e .

Commented by mathmax by abdo last updated on 27/Oct/19

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

a = \frac{2 i + 3}{2 i} = 1 + \frac{3}{2 i} = 1 - \frac{3 i}{2} = \frac{2 - 3 i}{2}

b = \frac{- 2 i + 3}{(- 2 i)} = \frac{2 i - 3}{2 i} = 1 - \frac{3}{2 i} = 1 + \frac{3 i}{2} = \frac{2 + 3 i}{2} \Rightarrow

f (x) = \frac{2 - 3 i}{2 (x - i)} + \frac{2 + 3 i}{2 (x + i)} \Rightarrow f^{(n)} (x) = \frac{2 - 3 i}{2} {(\frac{1}{x - i})}^{(n)} + \frac{2 + 3 i}{2} {(\frac{1}{x + i})}^{(n)}

= \frac{2 - 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x - i)}^{n + 1}} + \frac{2 + 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x + i)}^{n + 1}}

= \frac{{(- 1)}^{n} n!}{2} {\frac{2 + 3 i}{{(x + i)}^{n + 1}} + \frac{2 - 3 i}{{(x - i)}^{n + 1}}}

= \frac{{(- 1)}^{n} n!}{2} {\frac{(2 + 3 i) {(x - i)}^{n + 1} + (2 - 3 i) {(x + i)}^{n + 1}}{{(x^{2} + 1)}^{n + 1}}}

= \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} R e {(2 + 3 i) {(x - i)}^{n + 1}} b u t

2 + 3 i = \sqrt{13} e^{i a r c t a n (\frac{3}{2})} a n d x - i = \sqrt{x^{2} + 1} e^{- i a r c t a n (\frac{1}{x})} \Rightarrow

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

(2 + 3 i) {(x - i)}^{n + 1} = \sqrt{13} {(x^{2} + 1)}^{\frac{n + 1}{2}} e^{i (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} \sqrt{13} {(x^{2} + 1)}^{\frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))

f^{(n)} (x) = \sqrt{13} {(- 1)}^{n} n! {(x^{2} + 1)}^{- \frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))

Commented by mathmax by abdo last updated on 27/Oct/19

2) f^{(10)} (x) = \sqrt{13} (10)! {(x^{2} + 1)}^{- \frac{11}{2}} c o s (a r c t a n (\frac{3}{2}) - 11 a r c t a n (\frac{1}{x})

f^{(15)} (x) = - \sqrt{13} (15)! {(x^{2} + 1)}^{- 8} c o s (a r c t a n (\frac{3}{2}) - 16 a r c t a n (\frac{1}{x}))

Commented by mathmax by abdo last updated on 27/Oct/19

4) w e h a v e f^{(n)} (x) =

\sqrt{13} {(- 1)}^{n} n! {(x^{2} + 1)}^{- \frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x})) \Rightarrow

f^{(n)} (0) = \sqrt{13} {(- 1)}^{n} n! c o s (a r c t a n (\frac{3}{2}) \bar{+} (n + 1) \frac{π}{2})

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (x)}{n!} = \sum_{n = 0}^{\infty} \sqrt{13} {(- 1)}^{n} c o s (a r c t a n (\frac{3}{2}) \bar{+} (n + 1) \frac{π}{2}) x^{n}

Answered by mind is power last updated on 26/Oct/19

\frac{2 x + 3}{x^{2} + 1} = \frac{a}{x + i} + \frac{c}{x - i}

a + c = 2, i (c - a) = 3 \Rightarrow c = \frac{2 - 3 i}{2}, a = \frac{2 + 3 i}{2}

f (x) = \frac{2 + 3 i}{2 (x + i)} + \frac{2 - 3 i}{2 (x - i)}

f^{n} (x) = \frac{2 + 3 i}{2} \frac{d^{n} x}{{dx}^{n}} \frac{1}{x + i} + \frac{2 - 3 i}{2} \frac{d^{n} x}{{dx}^{n}} \frac{1}{x - i}

= \frac{2 + 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x + i)}^{n + 1}} + \frac{2 - 3 i}{2} . \frac{{(- 1)}^{n} n!}{{(x - i)}^{n + 1}}

= \frac{{(- 1)}^{n} n!}{2} (\frac{(2 + 3 i) {(x - i)}^{n + 1} + (2 - 3 i) {(x + i)}^{n + 1}}{{(x^{2} + 1)}^{n + 1}})

= \frac{{(- 1)}^{n} n! (2 . Re {(2 + 3 i) {(x + i)}^{n + 1}})}{{(x^{2} + 1)}^{n + 1}}

Re (2 + 3 i) {(x + i)}^{n + 1}

{(x + i)}^{n + 1} = Σ C_{n + 1}^{k} . i^{k} . {(x)}^{n + 1 - k}

= \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k} i . X^{n - 2 k}

Re (2 + 3 i) . {(x + i)}^{n + 1} =

2 \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + 3 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k + 1} . X^{n - 2 k}

f^{n} (x) = \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} . 2 (2 \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + 3 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k + 1} . X^{n - 2 k})

f^{(10)} (x) = 2 . \frac{10!}{{(x^{2} + 1)}^{11}} . (2 \underset{k = 0}{\sum^{5}} C_{11}^{2 k} . {(- 1)}^{k} X^{11 - 2 k} + 3 \underset{k = 0}{\sum^{5}} C_{11}^{2 k + 1} {(- 1)}^{k + 1} X^{10 - 2 k})

f^{15} (x) same idee

3)

f^{10} (x) = 2.10! . (3 .) = 6.10!

f^{15} (0) = 2 {(- 1)}^{15} . 15! . (2 .) = - 4.15!

4) f^{n} (0) if n = 2 s

f^{n} (0) = {(- 1)}^{2 s} . (2 s)! . 2 . (3 . {(- 1)}^{s + 1}) = \frac{{(- 1)}^{s + 1} . 6 . (2 s)!}{}

n = 2 s + 1 \Rightarrow f^{n} (0) = \frac{{(- 1)}^{2 s + 1} . (2 s + 1)! . 2 (2 . {(- 1)}^{s + 1})}{} = {(- 1)}^{s} . 4 (2 s + 1)!

f (x) = \sum_{n ⩾ 0} \frac{f^{n} (0) . x^{n}}{n!} = Σ \frac{f^{2 s} (0) x^{2 s}}{(2 s)!} + Σ \frac{f^{2 s + 1} (0) x^{2 s + 1}}{(2 s + 1)!}

= \underset{s = 0}{\sum^{+ \infty}} (\frac{f^{2 s} (0) x^{2 s}}{(2 s)!} + \frac{f^{2 s} (0) x^{2 s + 1}}{(2 s + 1)!}) = \underset{s = 0}{\sum^{+ \infty}} (6 {(- 1)}^{s + 1} x^{2 s} + 4 {(- 1)}^{s} x^{2 s + 1})

= \sum_{s ⩾ 0} {(- 1)}^{s} x^{2 s} (- 6 + 4 x)

5) gx = \int_{0}^{x} f

g (x) = \sum_{s ⩾ 0} (\int_{0}^{x} 6 {(- 1)}^{s + 1} t^{2 s} + 4 {(- 1)}^{s} t^{2 s + 1})

= \sum_{s ⩾ 0} . (6 {(- 1)}^{s + 1} . \frac{x^{2 s + 1}}{2 s + 1} + \frac{2 {(- 1)}^{s} x^{2 s + 2}}{s + 1})

Commented by mathmax by abdo last updated on 27/Oct/19

t h a n k y o u s i r .

Commented by mind is power last updated on 27/Oct/19

y^{'} re welcom

Commented by mind is power last updated on 27/Oct/19

y^{'} re welcom