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

Commented by Algoritm last updated on 12/Mar/21

((d^n (y))/dx^n )=?

$$\frac{\mathrm{d}^{\mathrm{n}} \left(\mathrm{y}\right)}{\mathrm{dx}^{\mathrm{n}} }=? \\ $$

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

y=Σ_(k=1) ^∞ (((−1)^(k−1) )/k)x^(2k) (d^n y/dx^n )=Σ_(k=1) ^∞ (((−1)^(k−1) )/k)A_(2k) ^n x^(2k−n)

$${y}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}{x}^{\mathrm{2}{k}} \\ $$$$\frac{{d}^{{n}} {y}}{{dx}^{{n}} }=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}{A}_{\mathrm{2}{k}} ^{{n}} {x}^{\mathrm{2}{k}−{n}} \\ $$

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) ... y^((n)) =(−1)^(n−1) (n−1)![(x+i)^(−n) +(x−i)^(−n) ] y^((n)) =(−1)^(n−1) (n−1)!(1+x^2 )^(−(n/2)) [e^(intan^(−1) (1/x)) +e^(−intan^(−1) (1/x)) ] ⇒y^((n)) =2(−1)^(n−1) (n−1)!((cos (ntan^(−1) (1/x)))/((1+x^2 )^(n/2) ))

$${y}=\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\mathrm{ln}\:\left({x}+{i}\right)+\mathrm{ln}\:\left({x}−{i}\right) \\ $$$${y}'=\left({x}+{i}\right)^{−\mathrm{1}} +\left({x}−{i}\right)^{−\mathrm{1}} \\ $$$${y}''=\left(−\mathrm{1}\right)\left({x}+{i}\right)^{−\mathrm{2}} +\left(−\mathrm{1}\right)\left({x}−{i}\right)^{−\mathrm{2}} \\ $$$${y}'''=\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left({x}+{i}\right)^{−\mathrm{3}} +\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left({x}−{i}\right)^{−\mathrm{3}} \\ $$$$... \\ $$$${y}^{\left({n}\right)} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left[\left({x}+{i}\right)^{−{n}} +\left({x}−{i}\right)^{−{n}} \right] \\ $$$${y}^{\left({n}\right)} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{{n}}{\mathrm{2}}} \left[{e}^{{in}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}} +{e}^{−{in}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}} \right] \\ $$$$\Rightarrow{y}^{\left({n}\right)} =\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\frac{\mathrm{cos}\:\left({n}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} } \\ $$

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

y=log(1+x^2 ) ⇒y^′ =((2x)/(x^2 +1)) =(1/(x−i))+(1/(x+i)) ⇒ y^((n)) =((1/(x−i)))^((n−1)) +((1/(x+i)))^((n−1)) (n≥1) =(((−1)^(n−1) (n−1)!)/((x−i)^n ))+(((−1)^(n−1) (n−1)!)/((x+i)^n )) =(−1)^(n−1) (n−1)!{(1/((x−i)^n ))+(1/((x+i)^n ))} =(−1)^(n−1) (n−1)!×(((x−i)^n +(x+i)^n )/((x^2 +1)^n )) but (x−i)^n +(x+i)^n =2Res(x+i)^n x+i =(√(1+x^2 )) e^(iarctan((1/x))) ⇒ (x+i)^n =(1+x^2 )^(n/2) e^(inarctan((1/x))) ⇒Re(x+i)^n =(1+x^2 )^(n/2) cos(narctan((1/x))) ⇒ y^((n)) (x) =(((−1)^(n−1) (n−1)!)/((1+x^2 )^(n/2) )) cos(narctan((1/x)))

$$\mathrm{y}=\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow\mathrm{y}^{'} \:=\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}}+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}}\:\Rightarrow \\ $$$$\mathrm{y}^{\left(\mathrm{n}\right)} \:=\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}}\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \:+\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}}\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \:\:\:\:\left(\mathrm{n}\geqslant\mathrm{1}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}} }+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} } \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}} }+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} }\right\} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!×\frac{\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}} +\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{n}} } \\ $$$$\mathrm{but}\:\:\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}} \:+\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} \:=\mathrm{2Res}\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} \\ $$$$\mathrm{x}+\mathrm{i}\:=\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{e}^{\mathrm{iarctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \:\Rightarrow \\ $$$$\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} \:=\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{n}}{\mathrm{2}}} \:\mathrm{e}^{\mathrm{inarctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \:\Rightarrow\mathrm{Re}\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}} \:=\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{n}}{\mathrm{2}}} \:\mathrm{cos}\left(\mathrm{narctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right)\:\Rightarrow \\ $$$$\mathrm{y}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{n}}{\mathrm{2}}} }\:\mathrm{cos}\left(\mathrm{narctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right) \\ $$

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

sorry y^((n)) (x)=((2(−1)^(n−1) (n−1)!)/((1+x^2 )^(n/2) )) cos(narctan((1/x)))

$$\mathrm{sorry}\:\mathrm{y}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{n}}{\mathrm{2}}} }\:\mathrm{cos}\left(\mathrm{narctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right) \\ $$

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