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Question Number 143381 by Mathspace last updated on 13/Jun/21
let f(x)=arctan((√2)x^2 ) 1) calculate f^((n)) (x)and f^((n)) (0) 2)if f(x)=Σa_n x^n find the sequence a_n
$${let}\:{f}\left({x}\right)={arctan}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right){and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){if}\:{f}\left({x}\right)=\Sigma{a}_{{n}} {x}^{{n}} \:\:{find}\:{the}\: \\ $$$${sequence}\:{a}_{{n}} \\ $$
Answered by TheHoneyCat last updated on 13/Jun/21
not in the right order, to make it more simple ∀x∈]−1,1[ (1/(1−x)) = Σ_(n=0) ^∞ x^n thus: (1/(1+x^2 ))=Σ_(n=0) ^∞ (−1)^n x^(2n) knowing that arctan′ = x→(1/(1+x^2 )) and that arctan(0)=0 arctan(x)=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) and therefor : arctan((√2)x^2 )=2(√2)Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(4n+2) so ∀n∈N if n≡2[4] a_n =(((−1)^((n−2)/4) )/(((n−2)/2)+1)) otherwise a_n =0 and ∀n∈N f^((n)) (0)=a_n as for f^((n)) (x) I am quite not courageous enought f^((0)) (x) =((2(√2)x)/(1+2x^4 )) from that point on, one must use Leibnietz′s formula for the derivation of a product (fg)^((n)) =Σ_(k=0) ^n ((n),(k) )f^((n−k)) g^((k)) with f: x→2(√2)x and g:x→(1/(1+2x^2 )) it obviously simplifies due to f′′=x→0 but like I said, I am not courrageous enougth maybe there is a faster method (for instance, if you guess a solution, you only have to verify it with a recurence) good luck to you with that last bit.
$$ \\ $$$$\mathrm{not}\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}\:\mathrm{order},\:\mathrm{to}\:\mathrm{make}\:\mathrm{it}\:\mathrm{more}\:\mathrm{simple} \\ $$$$\left.\forall{x}\in\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \\ $$$${thus}:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} \\ $$$${knowing}\:\:{that}\:\mathrm{arctan}'\:=\:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{and}\:{that}\:\mathrm{arctan}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{arctan}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${and}\:{therefor}\:: \\ $$$$\mathrm{arctan}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right)=\mathrm{2}\sqrt{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{4}{n}+\mathrm{2}} \: \\ $$$$ \\ $$$${so}\:\forall{n}\in\mathbb{N}\:\mathrm{if}\:{n}\equiv\mathrm{2}\left[\mathrm{4}\right]\:\:{a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{\frac{{n}−\mathrm{2}}{\mathrm{4}}} }{\frac{{n}−\mathrm{2}}{\mathrm{2}}+\mathrm{1}} \\ $$$$\mathrm{otherwise}\:{a}_{{n}} =\mathrm{0} \\ $$$$ \\ $$$$\mathrm{and}\:\forall{n}\in\mathbb{N}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)={a}_{{n}} \\ $$$$ \\ $$$${as}\:{for}\:{f}^{\left({n}\right)} \left({x}\right)\:{I}\:{am}\:{quite}\:{not}\:{courageous}\:{enought} \\ $$$${f}^{\left(\mathrm{0}\right)} \left({x}\right)\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}{x}}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} } \\ $$$${from}\:{that}\:{point}\:{on},\:{one}\:{must}\:{use}\:{Leibnietz}'{s}\:{formula}\:{for}\:{the}\:{derivation}\:{of}\:{a}\:{product} \\ $$$$\left({fg}\right)^{\left({n}\right)} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{f}^{\left({n}−{k}\right)} {g}^{\left({k}\right)} \\ $$$${with}\:{f}:\:{x}\rightarrow\mathrm{2}\sqrt{\mathrm{2}}{x}\:{and}\:{g}:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{it}\:\mathrm{obviously}\:\mathrm{simplifies}\:\mathrm{due}\:\mathrm{to}\:\mathrm{f}''={x}\rightarrow\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{like}\:\mathrm{I}\:\mathrm{said},\:\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{courrageous}\:\mathrm{enougth} \\ $$$$ \\ $$$$\mathrm{maybe}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{faster}\:\mathrm{method} \\ $$$$\left(\mathrm{for}\:\mathrm{instance},\:\:\mathrm{if}\:\mathrm{you}\:\mathrm{guess}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{you}\:\mathrm{only}\:\mathrm{have}\:\mathrm{to}\:\mathrm{verify}\:\mathrm{it}\:\mathrm{with}\:\mathrm{a}\:\mathrm{recurence}\right) \\ $$$$\mathrm{good}\:\mathrm{luck}\:\mathrm{to}\:\mathrm{you}\:\mathrm{with}\:\mathrm{that}\:\mathrm{last}\:\mathrm{bit}. \\ $$
Answered by mathmax by abdo last updated on 14/Jun/21
1)f(x)=arctan((√2)x^2 )⇒f^′ (x)=((2(√2)x)/(1+2x^4 ))=((2(√2)x)/(((√2)x^2 −i)((√2)x^2 +i))) =((2(√2)x)/(2(x^2 −(i/( (√2))))(x^2 +(i/( (√2))))))=(((√2)x)/((x−(1/( (√(√2))))e^((iπ)/4) )(x+(1/( (√(√2))))e^((iπ)/4) )(x−(1/( (√(√2))))e^(−((iπ)/4)) )(x+(1/( (√(√2))))e^(−((iπ)/4)) ))) (λ=(1/( (√(√2)))))=(a/(x−λe^((iπ)/4) ))+(b/(x+λe^((iπ)/4) ))+(c/(x−λe^(−((iπ)/4)) )) +(d/(x+λ e^(−((iπ)/4)) )) a=((λ(√2)e^((iπ)/4) )/(2λe^((iπ)/4) (λ^2 i+(i/( (√2))))))=(1/( (√2)(((2i)/( (√2))))))=(1/(2i)) its eazy to find other coefficient ⇒f^((n)) (x)=((a(−1)^(n−1) (n−1)!)/((x−λe^((iπ)/4) )^n ))+((b(−1)^(n−1) (n−1)!)/((x+λe^((iπ)/4) ))) +((c(−1)^(n−1) (n−1)!)/((x−λe^(−((iπ)/4)) )^n )) +((d(−1)^(n−1) (n−1)!)/((x+λe^(−((iπ)/4)) ))) =(−1)^(n−1) (n−1)!{(a/((x−λe^((iπ)/4) )^n ))+(b/((x+λe^((iπ)/4) )^n ))+(c/((x−λe^(−((iπ)/4)) )^n )) +(d/((x+λe^(−((iπ)/4)) )^n ))} ⇒ f^((n)) (0)=(−1)^(n−1) (n−1)!{((ae^(−((inπ)/4)) )/((−λ)^n ))+((be^(−((inπ)/4)) )/λ^n ) +((ce^((inπ)/4) )/((−λ)^n ))+(de^((inπ)/4) /λ^n )} (λ=(1/((^4 (√2))))) 2)if f(x)=Σa_n x^n ⇒a_n =((f^((n)) (0))/(n!))
$$\left.\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{x}}{\mathrm{1}+\mathrm{2x}^{\mathrm{4}} }=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{x}}{\left(\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{2}} −\mathrm{i}\right)\left(\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\mathrm{i}\right)} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{x}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)}=\frac{\sqrt{\mathrm{2}}\mathrm{x}}{\left(\mathrm{x}−\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}−\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\left(\lambda=\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\right)=\frac{\mathrm{a}}{\mathrm{x}−\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }+\frac{\mathrm{b}}{\mathrm{x}+\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }+\frac{\mathrm{c}}{\mathrm{x}−\lambda\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\:+\frac{\mathrm{d}}{\mathrm{x}+\lambda\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } \\ $$$$\mathrm{a}=\frac{\lambda\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{2}\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\lambda^{\mathrm{2}} \mathrm{i}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\frac{\mathrm{2i}}{\:\sqrt{\mathrm{2}}}\right)}=\frac{\mathrm{1}}{\mathrm{2i}}\:\:\mathrm{its}\:\mathrm{eazy}\:\mathrm{to}\:\mathrm{find}\:\mathrm{other}\:\mathrm{coefficient} \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{a}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }+\frac{\mathrm{b}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}+\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$+\frac{\mathrm{c}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\lambda\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }\:+\frac{\mathrm{d}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}+\lambda\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!\left\{\frac{\mathrm{a}}{\left(\mathrm{x}−\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }+\frac{\mathrm{b}}{\left(\mathrm{x}+\lambda\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }+\frac{\mathrm{c}}{\left(\mathrm{x}−\lambda\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }\right. \\ $$$$\left.+\frac{\mathrm{d}}{\left(\mathrm{x}+\lambda\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)=\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!\left\{\frac{\mathrm{ae}^{−\frac{\mathrm{in}\pi}{\mathrm{4}}} }{\left(−\lambda\right)^{\mathrm{n}} }+\frac{\mathrm{be}^{−\frac{\mathrm{in}\pi}{\mathrm{4}}} }{\lambda^{\mathrm{n}} }\:+\frac{\mathrm{ce}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} }{\left(−\lambda\right)^{\mathrm{n}} }+\frac{\mathrm{de}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} }{\lambda^{\mathrm{n}} }\right\}\:\:\left(\lambda=\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)=\Sigma\mathrm{a}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \:\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)}{\mathrm{n}!} \\ $$

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