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Question Number 33126 by prof Abdo imad last updated on 10/Apr/18
let give f(x)= (1/(2x^2 −3x+1)) 1) find f^((n)) (x) 2) find f^((n)) (0) 3) if f(x)=Σ a_n x^n calculate the sequence a_n
$${let}\:{give}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{3}{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{if}\:\:\:\:{f}\left({x}\right)=\Sigma\:{a}_{{n}} \:{x}^{{n}} \:\:{calculate}\:{the}\:{sequence}\:{a}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 12/Apr/18
1)poles of f? 2x^2 −3x+1 =0 Δ =9 −4 ×2 =1 ⇒x_1 =((3 +1)/4) =1 x_2 = ((3−1)/4) =(1/2) ⇒ f(x)= (1/(2(x−1)(x−(1/2)))) = (a/(x−1)) +(b/(x−(1/2))) a= (1/(2.(1/2))) =1 ,b= (1/(2.(−(1/2)))) =−1 ⇏ f(x)= (1/(x−1)) −(1/(x−(1/2))) ⇒ f^((n)) (x)= (((−1)^n n!)/((x−1)^(n+1) )) − (((−1)^n n!)/((x−(1/2))^(n+1) )) ⇒ 2)f^((n)) (0) = (((−1)^n n!)/((−1)^(n+1) )) −(((−1)^n n!)/((−(1/2))^(n+1) )) =−(n!) − (((−1)^n n!)/((−1)^(n+1) )) 2^(n+1) =n! 2^(n+1) −n! =n!( 2^(n+1) −1) 3) we have f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x)= Σ_(n=0) ^∞ (2^(n+1) −1)x^n .
$$\left.\mathrm{1}\right){poles}\:{of}\:{f}?\:\:\:\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{3}{x}+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta\:=\mathrm{9}\:−\mathrm{4}\:×\mathrm{2}\:=\mathrm{1}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{3}\:+\mathrm{1}}{\mathrm{4}}\:=\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\:\frac{\mathrm{3}−\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{f}\left({x}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\:\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${a}=\:\frac{\mathrm{1}}{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{1}\:\:\:,{b}=\:\frac{\mathrm{1}}{\mathrm{2}.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=−\mathrm{1}\:\nRightarrow \\ $$$${f}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:\:−\frac{\mathrm{1}}{{x}−\frac{\mathrm{1}}{\mathrm{2}}}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:−\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} }\:\Rightarrow \\ $$$$\left.\mathrm{2}\right){f}^{\left({n}\right)} \left(\mathrm{0}\right)\:\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} } \\ $$$$=−\left({n}!\right)\:\:−\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\mathrm{2}^{{n}+\mathrm{1}} \:\:={n}!\:\mathrm{2}^{{n}+\mathrm{1}} \:−{n}! \\ $$$$={n}!\left(\:\mathrm{2}^{{n}+\mathrm{1}} \:−\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\mathrm{2}^{{n}+\mathrm{1}} \:−\mathrm{1}\right){x}^{{n}} \:\:. \\ $$

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