Question-135342

Question Number 135342 by Algoritm last updated on 12/Mar/21

Commented by Algoritm last updated on 12/Mar/21

\frac{d^{n} (y)}{{dx}^{n}} = ?

Answered by Ñï= last updated on 12/Mar/21

y = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k} x^{2 k}

\frac{d^{n} y}{{d x}^{n}} = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k} A_{2 k}^{n} x^{2 k - n}

Answered by mr W last updated on 12/Mar/21

y = \ln (1 + x^{2}) = \ln (x + i) + \ln (x - i)

y^{'} = {(x + i)}^{- 1} + {(x - i)}^{- 1}

y^{″} = (- 1) {(x + i)}^{- 2} + (- 1) {(x - i)}^{- 2}

y^{‴} = (- 1) (- 2) {(x + i)}^{- 3} + (- 1) (- 2) {(x - i)}^{- 3}

\dots

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

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

\Rightarrow y^{(n)} = 2 {(- 1)}^{n - 1} (n - 1)! \frac{\cos (n \tan^{- 1} \frac{1}{x})}{{(1 + x^{2})}^{\frac{n}{2}}}

Commented by mathmax by abdo last updated on 03/Apr/21

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

y^{(n)} = {(\frac{1}{x - i})}^{(n - 1)} + {(\frac{1}{x + i})}^{(n - 1)} (n ⩾ 1)

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

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

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

but {(x - i)}^{n} + {(x + i)}^{n} = 2 Res {(x + i)}^{n}

x + i = \sqrt{1 + x^{2}} e^{iarctan (\frac{1}{x})} \Rightarrow

{(x + i)}^{n} = {(1 + x^{2})}^{\frac{n}{2}} e^{inarctan (\frac{1}{x})} \Rightarrow Re {(x + i)}^{n} = {(1 + x^{2})}^{\frac{n}{2}} \cos (narctan (\frac{1}{x})) \Rightarrow

y^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{{(1 + x^{2})}^{\frac{n}{2}}} \cos (narctan (\frac{1}{x}))

Commented by mathmax by abdo last updated on 03/Apr/21

sorry y^{(n)} (x) = \frac{2 {(- 1)}^{n - 1} (n - 1)!}{{(1 + x^{2})}^{\frac{n}{2}}} \cos (narctan (\frac{1}{x}))