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let-f-x-e-nx-ln-1-x-2-1-determine-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-n-integr-natural-

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Question Number 74502 by mathmax by abdo last updated on 25/Nov/19
let f(x)=e^(−nx) ln(1+x^2 ) 1)determine f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie (n integr natural)
$${let}\:{f}\left({x}\right)={e}^{−{nx}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right){determine}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:\:\:\left({n}\:{integr}\:{natural}\right) \\ $$
Commented by mathmax by abdo last updated on 25/Nov/19
1) f^((p)) (x)=Σ_(k=0) ^p C_p ^k (ln(1+x^2 ))^((k)) (e^(−nx) )^((p−k)) =ln(1+x^2 )(−n)^p e^(−nx) +Σ_(k=1) ^p C_p ^k (ln(x^2 +1))^((k)) (−n)^(p−k) e^(−nx) we have ln^′ (x^2 +1) =((2x)/(x^2 +1)) =(1/(x+i))+(1/(x−i)) ⇒ (ln(x^2 +1))^((k)) =((1/(x+i))+(1/(x−i)))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+i)^k )) +(((−1)^(k−1) (k−1)!)/((x−i)^k )) =(−1)^(k−1) (k−1)!((1/((x+i)^k ))+(1/((x−i)^k )))=(−1)^(k−1) (k−1)!×(((x+i)^k +(x−i)^k )/((x^2 +1)^k )) f^((p)) (x)=(−n)^p e^(−nx) ln(x^2 +1) +Σ_(k=1) ^p C_p ^k (−1)^(k−1) (k−1)!×(((x+i)^k +(x−i)^k )/((x^2 +1)^k ))(−n)^(p−k) e^(−nx)
$$\left.\mathrm{1}\right)\:{f}^{\left({p}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}} ^{{k}} \left({ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)^{\left({k}\right)} \:\left({e}^{−{nx}} \right)^{\left({p}−{k}\right)} \\ $$$$={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−{n}\right)^{{p}} \:{e}^{−{nx}} \:+\sum_{{k}=\mathrm{1}} ^{{p}} \:{C}_{{p}} ^{{k}} \left({ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\right)^{\left({k}\right)} \left(−{n}\right)^{{p}−{k}} \:{e}^{−{nx}} \\ $$$${we}\:{have}\:{ln}^{'} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\:=\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{1}}{{x}+{i}}+\frac{\mathrm{1}}{{x}−{i}}\:\Rightarrow \\ $$$$\left({ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right)^{\left({k}\right)} =\left(\frac{\mathrm{1}}{{x}+{i}}+\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({k}−\mathrm{1}\right)} \:=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+{i}\right)^{{k}} }\:+\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−{i}\right)^{{k}} } \\ $$$$=\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(\frac{\mathrm{1}}{\left({x}+{i}\right)^{{k}} }+\frac{\mathrm{1}}{\left({x}−{i}\right)^{{k}} }\right)=\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\frac{\left({x}+{i}\right)^{{k}} +\left({x}−{i}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{k}} } \\ $$$${f}^{\left({p}\right)} \left({x}\right)=\left(−{n}\right)^{{p}} \:{e}^{−{nx}} {ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{p}} \:{C}_{{p}} ^{{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\frac{\left({x}+{i}\right)^{{k}} \:+\left({x}−{i}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{k}} }\left(−{n}\right)^{{p}−{k}} \:{e}^{−{nx}} \\ $$
Commented by mathmax by abdo last updated on 25/Nov/19
f^((n)) (x)=(−n)^n e^(−nx) ln(x^2 +1) +Σ_(k=1) ^n C_n ^k (−1)^(k−1) (k−1)!×(((x+i)^k +(x−i)^k )/((x^2 +1)^k ))(−n)^(n−k) e^(−nx) and f^((n)) (0) =Σ_(k=1) ^n C_n ^k (−1)^(k−1) (k−1)!×(i^k +(−i)^k )(−n)^(n−k) e^(−nx) i^k +(−i)^k =2Re(i^k ) =2cos(((kπ)/2)) ⇒ f^((n)) (0) =2(−1)^(n−1) n^n e^(−nx) Σ_(k=1) ^n (k−1)! C_n ^k cos(((kπ)/2))
$${f}^{\left({n}\right)} \left({x}\right)=\left(−{n}\right)^{{n}} \:{e}^{−{nx}} {ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\frac{\left({x}+{i}\right)^{{k}} \:+\left({x}−{i}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{k}} }\left(−{n}\right)^{{n}−{k}} \:{e}^{−{nx}} \\ $$$${and}\: \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\left({i}^{{k}} \:+\left(−{i}\right)^{{k}} \right)\left(−{n}\right)^{{n}−{k}} \:{e}^{−{nx}} \\ $$$${i}^{{k}} \:+\left(−{i}\right)^{{k}} \:=\mathrm{2}{Re}\left({i}^{{k}} \right)\:=\mathrm{2}{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{{n}} \:{e}^{−{nx}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left({k}−\mathrm{1}\right)!\:{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right) \\ $$
Commented by mathmax by abdo last updated on 25/Nov/19
2)f(x)=Σ_(p=0) ^∞ ((f^((p)) (0))/(p!)) x^p we have f^((p)) (0) =Σ_(k=1) ^p C_p ^k (−1)^(p−1) (k−1)!((2cos(((kπ)/2)))/1)(n)^(p−k) =2Σ_(k=1) ^p ((p!)/(k!(p−k)!))(−1)^(p−1) (k−1)! cos(((kπ)/2)) n^(p−k) =2(−1)^(p−1) n^p Σ_(k=1) ^p ((p!)/(k(p−k)!n^k )) cos(((kπ)/2)) ⇒ f(x)=2Σ_(p=0) ^∞ (−1)^(p−1) n^p { Σ_(k=1) ^p ((cos(((kπ)/2)))/(k n^k (p−k)!))}x^p
$$\left.\mathrm{2}\right){f}\left({x}\right)=\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({p}\right)} \left(\mathrm{0}\right)}{{p}!}\:{x}^{{p}} \:\:{we}\:{have}\: \\ $$$${f}^{\left({p}\right)} \left(\mathrm{0}\right)\:=\sum_{{k}=\mathrm{1}} ^{{p}} \:{C}_{{p}} ^{{k}} \left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\frac{\mathrm{2}{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{1}}\left({n}\right)^{{p}−{k}} \: \\ $$$$=\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{p}} \:\frac{{p}!}{{k}!\left({p}−{k}\right)!}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:{n}^{{p}−{k}} \\ $$$$=\mathrm{2}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \:{n}^{{p}} \:\:\sum_{{k}=\mathrm{1}} ^{{p}} \:\:\:\frac{{p}!}{{k}\left({p}−{k}\right)!{n}^{{k}} }\:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\mathrm{2}\sum_{{p}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{p}−\mathrm{1}} {n}^{{p}} \left\{\:\sum_{{k}=\mathrm{1}} ^{{p}} \:\:\frac{{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{{k}\:{n}^{{k}} \left({p}−{k}\right)!}\right\}{x}^{{p}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 25/Nov/19
f^((n)) (0) =2(−1)^(n−1) n^n Σ_(k=1) ^n (k−1)!C_n ^k cos(((kπ)/2)).
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{n}^{{n}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \left({k}−\mathrm{1}\right)!{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right). \\ $$
Answered by mind is power last updated on 25/Nov/19
ln^((k)) (1+x^2 )=ln(1+x^2 ),k=0 =((1/(x+i))+(1/(x−i)))^((k−1)) ,k≥1 =((((−1)^(k−1) (k−1)!)/((x+i)^((k)) ))+(((−1)^(k−1) (k−1)!)/((x−i)^((k)) ))) =(−1)^(k−1) (k−1)!.((((x−i)^((k−1)) +(x+i)^((k−1)) )/((x^2 +1)^k ))) =(((−1)^(k−1) (k−1)!.2Re((x+i)^((k−1)) ))/((x^2 +1)^k )) x+i= { (((√(x^2 +1)).e^(iarctan((1/x))) ,x>0)),(((√(x^2 +1))e^(i((π/2)+arctan((1/x)))) ,x<0)) :} x≥0 (d/dx^k ){ln(1+x^2 )}=(((−1)^(k−1) (k−1)!(x^2 +1)^((k−1)/2) 2Re{e^(i(k−1)arctan((1/x))) })/((x^2 +1)^k )),k≥1 f(x)=e^(−nx) ln(1+x^2 ) f^((n)) (x)=Σ_(k=0) ^n C_n ^k .(e^(−nx) )^((n−k)) .(ln^((k)) (1+x^2 )) =Σ_(k=1) ^n C_n ^k (−n)^((n−k)) .e^(−nx) .(((−1)^(k−1) (k−1)!.cos((k−1)(arctan((1/x))))/((x^2 +1)^((k+1)/2) ))+(−n)^n e^(−nx) ln(1+x^2 ) f^((n)) (0) =Σ_(k=1) ^n C_n ^k (−n)^(n−k) .(−1)^(k−1) .(k−1)!.cos((k−1)(π/2)) 2) f(x)=Σ_(j=0) ^(+∞) ((f^j (0)x^j )/(j!)) f(x)=Σ_(j≥1) ((f^j (0)x^j )/(j!))=Σ_(j≥1) .(Σ_(k=1) ^j C_j ^k .(−n)^(j−k) (−1)^(k−1) cos(((k−1)/2)π))x^j .(1/(j!))
$${ln}^{\left({k}\right)} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right),{k}=\mathrm{0} \\ $$$$=\left(\frac{\mathrm{1}}{{x}+{i}}+\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({k}−\mathrm{1}\right)} ,{k}\geqslant\mathrm{1} \\ $$$$=\left(\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+{i}\right)^{\left({k}\right)} }+\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−{i}\right)^{\left({k}\right)} }\right)\: \\ $$$$=\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!.\left(\frac{\left({x}−{i}\right)^{\left({k}−\mathrm{1}\right)} +\left({x}+{i}\right)^{\left({k}−\mathrm{1}\right)} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{k}} }\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!.\mathrm{2}{Re}\left(\left({x}+{i}\right)^{\left({k}−\mathrm{1}\right)} \right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{k}} } \\ $$$${x}+{i}=\begin{cases}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}.{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} ,{x}>\mathrm{0}}\\{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{e}^{{i}\left(\frac{\pi}{\mathrm{2}}+{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)} ,{x}<\mathrm{0}}\end{cases} \\ $$$${x}\geqslant\mathrm{0} \\ $$$$\frac{{d}}{{dx}^{{k}} }\left\{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right\}=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{{k}−\mathrm{1}}{\mathrm{2}}} \mathrm{2}{Re}\left\{{e}^{{i}\left({k}−\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right\}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{k}} },{k}\geqslant\mathrm{1} \\ $$$${f}\left({x}\right)={e}^{−{nx}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} .\left({e}^{−{nx}} \right)^{\left({n}−{k}\right)} .\left({ln}^{\left({k}\right)} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−{n}\right)^{\left({n}−{k}\right)} .{e}^{−{nx}} .\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!.{cos}\left(\left({k}−\mathrm{1}\right)\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\right.}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{{k}+\mathrm{1}}{\mathrm{2}}} }+\left(−{n}\right)^{{n}} {e}^{−{nx}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−{n}\right)^{{n}−{k}} .\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} .\left({k}−\mathrm{1}\right)!.{cos}\left(\left({k}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right) \\ $$$${f}\left({x}\right)=\underset{{j}=\mathrm{0}} {\overset{+\infty} {\sum}}\:\frac{{f}^{{j}} \left(\mathrm{0}\right){x}^{{j}} }{{j}!} \\ $$$${f}\left({x}\right)=\underset{{j}\geqslant\mathrm{1}} {\sum}\frac{{f}^{{j}} \left(\mathrm{0}\right){x}^{{j}} }{{j}!}=\underset{{j}\geqslant\mathrm{1}} {\sum}.\left(\underset{{k}=\mathrm{1}} {\overset{{j}} {\sum}}{C}_{{j}} ^{{k}} .\left(−{n}\right)^{{j}−{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {cos}\left(\frac{{k}−\mathrm{1}}{\mathrm{2}}\pi\right)\right){x}^{{j}} .\frac{\mathrm{1}}{{j}!} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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