let-f-x-arctan-3-x-1-calculste-f-n-x-and-f-n-1-2-developp-f-at-integr-seri-at-point-x-0-1-

Question Number 98722 by mathmax by abdo last updated on 15/Jun/20

let f (x) = \arctan (\frac{3}{x})

1) calculste f^{(n)} (x) and f^{(n)} (1)

2) developp f at integr seri at point x_{0} = 1

Answered by MWSuSon last updated on 16/Jun/20

y = \arctan (\frac{3}{x})

y^{'} = \frac{- 3}{(x^{2} + 9)} = \frac{1}{2 i} [\frac{1}{x + 3 i} - \frac{1}{x - 3 i}]

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

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

can be further simplified \dots

let x = rcos θ and 3 = rsin θ

y^{n} = \frac{1}{2 i} (n - 1)! {(- 1)}^{n - 1} [{(rcos θ + irsin θ)}^{- n} - {(rcos θ - irsin θ)}^{- n}]

y^{n} = \frac{1}{2 i} (n - 1)! {(- 1)}^{n - 1} r^{- n} [cosn θ - isinn θ - cosn θ - isinn θ]

y^{n} = - (n - 1)! {(- 1)}^{n - 1} r^{- n} \times sinn θ

y^{n} = (n - 1)! {(- 1)}^{n} r^{- n} sinn θ

[r = \sqrt{x^{2} + 9}]

[θ = \arctan (\frac{3}{x})]

2) y^{n} (1) = (n - 1)! {(- 1)}^{n} {(\sqrt{10})}^{- n} \times sinn (\arctan (3))

Answered by MWSuSon last updated on 15/Jun/20

i {don}^{'} t understand what your number 2 mean,

sir please retype .

Commented by mathmax by abdo last updated on 15/Jun/20

taylor serie at x_{0} = 1

Commented by MWSuSon last updated on 15/Jun/20

Taylor at x_{o} = 1 = \underset{n = 0}{\sum^{\infty}} f^{n} (1) {(x - 1)}^{n}

y = \arctan (\frac{3}{x})

y (x = 1) = \arctan (3)

y^{^{'}} = - \frac{3}{x^{2} + 9}

y^{^{'}} (x = 1) = - \frac{3}{10}

y^{″} = \frac{6 x}{{(x^{2} + 9)}^{2}}

y^{^{″}} = \frac{6}{100} = \frac{3}{50}

\dots .

\dots .

\arctan {(\frac{3}{x})}_{x = 1} = \arctan (3) - \frac{3 (x - 1)}{10} + \frac{3 {(x - 1)}^{2}}{100} + \dots

Answered by mathmax by abdo last updated on 15/Jun/20

1) f (x) = \arctan (\frac{3}{x}) \Rightarrow f^{^{'}} (x) = \frac{- 3}{x^{2} (1 + \frac{9}{x^{2}})} = \frac{- 3}{x^{2} + 9}

= \frac{- 3}{(x - 3 i) (x + 3 i)} = - 3 (\frac{1}{x - 3 i} - \frac{1}{x + 3 i}) \times \frac{1}{6 i} = - \frac{1}{2 i} (\frac{1}{x - 3 i} - \frac{1}{x + 3 i}) \Rightarrow

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

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

= \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! {\frac{{(x + 3 i)}^{n} - {(x - 3 i)}^{n}}{{(x^{2} + 9)}^{n}}}

= \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! \times \frac{2 i Im {(x + 3 i)}^{n}}{{(x^{2} + 9)}^{n}}

= {(- 1)}^{n} (n - 1)! \times \frac{Im {(x + 3 i)}^{n}}{{(x^{2} + 9)}^{n}} also we have x + 3 i = \sqrt{x^{2} + 9} e^{iarctan (\frac{3}{x})} \Rightarrow

{(x + 3 i)}^{n} = {(x^{2} + 9)}^{\frac{n}{2}} e^{inarctan (\frac{3}{x})} \Rightarrow Im {(x + 3 i)}^{n} = {(x^{2} + 9)}^{\frac{n}{2}} \sin (narctan (\frac{3}{x})) \Rightarrow

f^{(n)} (x) = {(- 1)}^{n} (n - 1)! \times \frac{\sin (narctan (\frac{3}{x}))}{{(x^{2} + 9)}^{\frac{n}{2}}}

Commented by mathmax by abdo last updated on 15/Jun/20

f^{(n)} (1) = {(- 1)}^{n} (n - 1)! \times \frac{\sin (n \arctan 3)}{10^{\frac{n}{2}}}

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

= \arctan (3) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \times \frac{\sin (narctan (3))}{10^{\frac{n}{2}}} {(x - 1)}^{n}