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Question Number 94336 by mathmax by abdo last updated on 19/May/20
let f(x) =arctan(2x) e^(−3x)   1) determine f^((n)) (x) and f^((n)) (0)  2)developp f at integr serie
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{arctan}\left(\mathrm{2x}\right)\:\mathrm{e}^{−\mathrm{3x}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{determine}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$
Answered by prakash jain last updated on 18/May/20
e^x =Σ_(k=0) ^∞ (x^k /(k!))  We need (x^(4n) /((4n)!)) which is every 4^(th)   power.   Noting that i^4 =1 we start with   e^(ix) =1+ix−(x^2 /(2!))−i(x^3 /(3!))+(x^4 /4)+i(x^5 /(5!))+..  add e^(−ix)  to remove all odd power  e^(ix) +e^(−ix) =2(1−(x^2 /(2!))+(x^4 /(4!))−(x^6 /(6!))+(x^8 /(8!))..)  To get required terms we need to  add all even power of x  e^x +e^(−x) =2(1+(x^2 /(2!))+(x^4 /(4!))+−...)  sum  e^(ix) +e^(−ix) +e^x +e^(−x) =4(1+(x^4 /(4!))+(x^8 /(8!))...)  Σ_(k=0) ^∞ (x^(4n) /((4n)!))=(1/2)(cosx+cosh x)
$${e}^{{x}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}!} \\ $$$$\mathrm{We}\:\mathrm{need}\:\frac{{x}^{\mathrm{4}{n}} }{\left(\mathrm{4}{n}\right)!}\:\mathrm{which}\:\mathrm{is}\:\mathrm{every}\:\mathrm{4}^{\mathrm{th}} \\ $$$$\mathrm{power}.\: \\ $$$$\mathrm{Noting}\:\mathrm{that}\:{i}^{\mathrm{4}} =\mathrm{1}\:\mathrm{we}\:\mathrm{start}\:\mathrm{with}\: \\ $$$${e}^{{ix}} =\mathrm{1}+{ix}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−{i}\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+{i}\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+.. \\ $$$$\mathrm{add}\:{e}^{−{ix}} \:\mathrm{to}\:\mathrm{remove}\:\mathrm{all}\:\mathrm{odd}\:\mathrm{power} \\ $$$${e}^{{ix}} +{e}^{−{ix}} =\mathrm{2}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+\frac{{x}^{\mathrm{8}} }{\mathrm{8}!}..\right) \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{required}\:\mathrm{terms}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{add}\:\mathrm{all}\:\mathrm{even}\:\mathrm{power}\:\mathrm{of}\:{x} \\ $$$${e}^{{x}} +{e}^{−{x}} =\mathrm{2}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+−…\right) \\ $$$$\mathrm{sum} \\ $$$${e}^{{ix}} +{e}^{−{ix}} +{e}^{{x}} +{e}^{−{x}} =\mathrm{4}\left(\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{8}} }{\mathrm{8}!}…\right) \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{4}{n}} }{\left(\mathrm{4}{n}\right)!}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}{x}+\mathrm{cosh}\:{x}\right) \\ $$
Commented by abdomathmax last updated on 18/May/20
thank you sir prakach
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{prakach} \\ $$
Commented by mathmax by abdo last updated on 20/May/20
the question here is calculate Σ_(n=0) ^∞  (x^(4n) /((4n)!))
$$\mathrm{the}\:\mathrm{question}\:\mathrm{here}\:\mathrm{is}\:\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{4n}} }{\left(\mathrm{4n}\right)!} \\ $$
Commented by prakash jain last updated on 20/May/20
May be I answered agaimst a wrong  question.I was answering  Σ_(n=0) ^∞ (x^(4n) /((4m)!))
$$\mathrm{May}\:\mathrm{be}\:\mathrm{I}\:\mathrm{answered}\:\mathrm{agaimst}\:\mathrm{a}\:\mathrm{wrong} \\ $$$$\mathrm{question}.\mathrm{I}\:\mathrm{was}\:\mathrm{answering} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{4}{n}} }{\left(\mathrm{4}{m}\right)!} \\ $$
Answered by mathmax by abdo last updated on 20/May/20
1) we have f(x) =e^(−3x)  arctan(2x) ⇒  f^((n)) (x) =Σ_(k=0) ^n  C_n ^k  (arctan(2x))^((k))  (e^(−3x) )^((n−k))   =(−3)^n  e^(−3x)  arctan(2x) +Σ_(k=1) ^n  C_n ^k  (arctan(2x))^((k))  (−3)^(n−k)  e^(−3x)   we have (arctan(2x))^((1))  =(2/(1+4x^2 )) =(2/(4((1/4)+x^2 ))) =(1/(2(x−(i/2))(x+(i/2))))  =(1/(2i))((1/(x−(i/2)))−(1/(x+(i/2)))) ⇒(arctan(2x))^((k)) =(1/(2i)){((1/(x−(i/2))))^((k−1)) −((1/(x+(i/2))))^((k−1)) }  =(1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−(i/2))^k ))−(((−1)^(k−1) (k−1)!)/((x+(i/2))^k ))}  =(((−1)^(k−1) (k−1)!)/(2i))( (((x+(i/2))^k −(x−(i/2))^k )/((x^2  +(1/4))^k )))  =(−1)^(k−1) (k−1)!×((Im(  (x+(i/2))^k ))/((x^2  +(1/4))^k )) ⇒  f^((n)) (x) =(−3)^n  e^(−3x)  arctan(2x)  +Σ_(k=1) ^n  C_n ^k    (((−1)^(k−1) (k−1)! .Im((x+(i/2))^k ))/((x^2  +(1/4))^k ))×(−3)^(n−k)  e^(−3x)   ⇒ f^((n)) (0) =Σ_(k=1) ^n  C_n ^k  4^k (−1)^(k−1) (k−1)! ×(1/2^k )sin(((kπ)/2)) (−3)^(n−k)   ⇒f^((n)) (0) =Σ_(k=1) ^n  2^k  (−1)^(k−1) (k−1)! ((n!)/(k!(n−k)!)) sin(((kπ)/2))(−3)^(n−k)   =Σ_(k=1) ^n  (−1)^(k−1) 2^k  ((n!)/(k(n−k)!))(−3)^(n−k)   2) f(x) =Σ_(n=0) ^∞  ((f^((n)) (0))/(n!))x^n   ⇒  f(x) =Σ_(n=1) ^∞ (Σ_(k=1) ^n  (−1)^(k−1)  2^k  ×(1/(k(n−k)!))(−3)^(n−k) )x^n   =Σ_(n=1) ^∞  ( Σ_(k=1) ^n  (((−1)^(k−1)  2^k (−3)^(n−k) )/(k .(n−k)!))) x^n
$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\mathrm{3x}} \:\mathrm{arctan}\left(\mathrm{2x}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{k}\right)} \:\left(\mathrm{e}^{−\mathrm{3x}} \right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\left(−\mathrm{3}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{3x}} \:\mathrm{arctan}\left(\mathrm{2x}\right)\:+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{k}\right)} \:\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{e}^{−\mathrm{3x}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{1}\right)} \:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{x}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}\right)\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\left(\frac{\mathrm{1}}{\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}}\right)\:\Rightarrow\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{k}\right)} =\frac{\mathrm{1}}{\mathrm{2i}}\left\{\left(\frac{\mathrm{1}}{\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}}\right)^{\left(\mathrm{k}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}}\right)^{\left(\mathrm{k}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} }−\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!}{\mathrm{2i}}\left(\:\frac{\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} −\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{k}} }\right) \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!×\frac{\mathrm{Im}\left(\:\:\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} \right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{k}} }\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\left(−\mathrm{3}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{3x}} \:\mathrm{arctan}\left(\mathrm{2x}\right) \\ $$$$+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!\:.\mathrm{Im}\left(\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{k}} \right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{k}} }×\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \:\mathrm{e}^{−\mathrm{3x}} \\ $$$$\Rightarrow\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{4}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!\:×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{2}}\right)\:\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \\ $$$$\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{2}^{\mathrm{k}} \:\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \left(\mathrm{k}−\mathrm{1}\right)!\:\frac{\mathrm{n}!}{\mathrm{k}!\left(\mathrm{n}−\mathrm{k}\right)!}\:\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{2}}\right)\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \mathrm{2}^{\mathrm{k}} \:\frac{\mathrm{n}!}{\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)!}\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)}{\mathrm{n}!}\mathrm{x}^{\mathrm{n}} \:\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \:\mathrm{2}^{\mathrm{k}} \:×\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)!}\left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} \right)\mathrm{x}^{\mathrm{n}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \:\mathrm{2}^{\mathrm{k}} \left(−\mathrm{3}\right)^{\mathrm{n}−\mathrm{k}} }{\mathrm{k}\:.\left(\mathrm{n}−\mathrm{k}\right)!}\right)\:\mathrm{x}^{\mathrm{n}} \\ $$

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