Menu Close

let-f-x-x-2-x-3-4x-3-1-calculate-f-n-x-2-developp-f-at-integr-serie-

Tinku Tara 0 Comments
Pin




Question Number 35988 by abdo mathsup 649 cc last updated on 26/May/18
let f(x) = ((x+2)/(x^3 −4x +3)) 1) calculate f^((n)) (x) 2) developp f at integr serie.
$${let}\:{f}\left({x}\right)\:=\:\frac{{x}+\mathrm{2}}{{x}^{\mathrm{3}} −\mathrm{4}{x}\:+\mathrm{3}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by abdo mathsup 649 cc last updated on 27/May/18
1) we have x^3 −4x +3 =x^3 −x −3x +3 =x(x^2 −1) −3(x−1) = (x−1)( x^2 +x −3) ⇒ f(x) = ((x+2)/((x−1)(x^2 +x−3))) = (a/(x−1)) +((bx +c)/(x^2 +x−3)) a=lim_(x→1) (x−1)f(x) = (3/(−1)) =−3 lim_(x→+∞) x f(x)=0 = a +b ⇒b=−a = 3 ⇒ f(x) = ((−3)/(x−1)) + ((3x+c)/(x^(2 ) +x −3)) f(0) = (2/3) = 3 −(c/3) ⇒2= 9 −c ⇒ c =7 ⇒ f(x) = ((−3)/(x−1)) +((3x+7)/(x^2 +x−3)) let find the roots of x^2 +x−3 , Δ =1−4(−3) =13 x_(1 ) = ((−1+(√(13)))/2) , x_2 = ((−1−(√(13)))/2) so ((3x+7)/(x^2 +x−3)) = ((3x+7)/((x−x_1 )(x−x_2 ))) =((3(x−x_1 ) +3x_1 +7)/((x−x_1 )(x−x_2 ))) = (3/(x−x_2 )) + ((3x_(1 ) +7)/((x−x_1 )(x−x_2 ))) (3/(x−x_2 )) + ((3x_1 +7)/(x_1 −x_2 )) { (1/(x−x_1 )) −(1/(x−x_2 ))} ⇒ f(x) = −(3/(x−1)) +(3/(x−x_2 )) + (λ/(x−x_1 )) −(λ/(x−x_2 )) with λ = ((3x_(1 ) +7)/(x_1 −x_2 )) ⇒ f^((n)) (x) = −3 (((−1)^n n!)/((x−1)^(n+1) )) + (3−λ)(((−1)^n n!)/((x−x_2 )^(n+1) )) +λ (((−1)^n n!)/((x−x_1 )^(n+1) ))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{x}^{\mathrm{3}} −\mathrm{4}{x}\:+\mathrm{3}\:={x}^{\mathrm{3}} \:−{x}\:−\mathrm{3}{x}\:+\mathrm{3} \\ $$$$={x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:−\mathrm{3}\left({x}−\mathrm{1}\right)\:=\:\left({x}−\mathrm{1}\right)\left(\:{x}^{\mathrm{2}} \:+{x}\:−\mathrm{3}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{{x}+\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}−\mathrm{3}\right)}\:=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{bx}\:+{c}}{{x}^{\mathrm{2}} \:+{x}−\mathrm{3}} \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){f}\left({x}\right)\:=\:\frac{\mathrm{3}}{−\mathrm{1}}\:=−\mathrm{3} \\ $$$${lim}_{{x}\rightarrow+\infty} {x}\:{f}\left({x}\right)=\mathrm{0}\:=\:{a}\:+{b}\:\Rightarrow{b}=−{a}\:=\:\mathrm{3}\:\Rightarrow \\ $$$${f}\left({x}\right)\:\:=\:\:\frac{−\mathrm{3}}{{x}−\mathrm{1}}\:+\:\frac{\mathrm{3}{x}+{c}}{{x}^{\mathrm{2}\:} \:+{x}\:−\mathrm{3}} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:\mathrm{3}\:−\frac{{c}}{\mathrm{3}}\:\Rightarrow\mathrm{2}=\:\mathrm{9}\:−{c}\:\Rightarrow\:{c}\:=\mathrm{7}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{−\mathrm{3}}{{x}−\mathrm{1}}\:\:+\frac{\mathrm{3}{x}+\mathrm{7}}{{x}^{\mathrm{2}} \:+{x}−\mathrm{3}}\:{let}\:{find}\:{the}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} \:+{x}−\mathrm{3}\:\:\:,\:\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{3}\right)\:=\mathrm{13} \\ $$$${x}_{\mathrm{1}\:} =\:\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:\:,\:\:{x}_{\mathrm{2}} \:=\:\frac{−\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{2}}\:{so} \\ $$$$\frac{\mathrm{3}{x}+\mathrm{7}}{{x}^{\mathrm{2}} \:+{x}−\mathrm{3}}\:=\:\frac{\mathrm{3}{x}+\mathrm{7}}{\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{3}\left({x}−{x}_{\mathrm{1}} \right)\:+\mathrm{3}{x}_{\mathrm{1}} \:+\mathrm{7}}{\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)}\:=\:\frac{\mathrm{3}}{{x}−{x}_{\mathrm{2}} }\:+\:\frac{\mathrm{3}{x}_{\mathrm{1}\:} \:+\mathrm{7}}{\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)} \\ $$$$\:\frac{\mathrm{3}}{{x}−{x}_{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{3}{x}_{\mathrm{1}} +\mathrm{7}}{{x}_{\mathrm{1}} \:−{x}_{\mathrm{2}} }\:\left\{\:\:\frac{\mathrm{1}}{{x}−{x}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{x}−{x}_{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:−\frac{\mathrm{3}}{{x}−\mathrm{1}}\:\:+\frac{\mathrm{3}}{{x}−{x}_{\mathrm{2}} }\:\:+\:\frac{\lambda}{{x}−{x}_{\mathrm{1}} }\:−\frac{\lambda}{{x}−{x}_{\mathrm{2}} }\:{with} \\ $$$$\lambda\:=\:\frac{\mathrm{3}{x}_{\mathrm{1}\:} \:+\mathrm{7}}{{x}_{\mathrm{1}} \:−{x}_{\mathrm{2}} }\:\Rightarrow\: \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:−\mathrm{3}\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:+\:\left(\mathrm{3}−\lambda\right)\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−{x}_{\mathrm{2}} \right)^{{n}+\mathrm{1}} } \\ $$$$+\lambda\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{x}_{\mathrm{1}} \right)^{{n}+\mathrm{1}} } \\ $$
Commented by abdo mathsup 649 cc last updated on 27/May/18
2) we have f^((n)) (0) = 3n! +(3−λ) ((n!)/(−x_2 ^(n+1) )) +λ ((n!)/(−x_1 ^(n+1) )) f^((n)) (0) = 3(n!) +(λ −3) ((n!)/x_2 ^(n+1) ) −λ ((n!)/x_1 ^(n+1) ) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = Σ_(n=0) ^∞ (3 +((λ−3)/x_2 ^(n+1) ) −(λ/x_1 ^(n+1) )) x^n .
$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\mathrm{3}{n}!\:+\left(\mathrm{3}−\lambda\right)\:\frac{{n}!}{−{x}_{\mathrm{2}} ^{{n}+\mathrm{1}} } \\ $$$$+\lambda\:\:\frac{{n}!}{−{x}_{\mathrm{1}} ^{{n}+\mathrm{1}} } \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\mathrm{3}\left({n}!\right)\:\:+\left(\lambda\:−\mathrm{3}\right)\:\frac{{n}!}{{x}_{\mathrm{2}} ^{{n}+\mathrm{1}} }\:\:−\lambda\:\:\frac{{n}!}{{x}_{\mathrm{1}} ^{{n}+\mathrm{1}} } \\ $$$${f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(\mathrm{3}\:\:+\frac{\lambda−\mathrm{3}}{{x}_{\mathrm{2}} ^{{n}+\mathrm{1}} }\:\:−\frac{\lambda}{{x}_{\mathrm{1}} ^{{n}+\mathrm{1}} }\right)\:{x}^{{n}} \:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18
Dr=x^3 −4x+4−1 =x^3 −1−4(x−1) =(x−1)(x^2 +x+1)−4(x−1) (x−1)(x^2 +x+1−4) =(x−1)(x^2 +x−3) ((x+2)/((x−1)(x^2 +x−3)))=(A/(x−1))+((Bx+C)/((x^2 +x−3))) x+2=A(x^2 +x−3)+(x−1)(Bx+C) put x=1 3=−A so A=−3 x+2=−3x^2 −3x+9+Bx^2 +Cx−Bx−C x+2=x^2 (−3+B)+x(−3−B+C)+9−C −3+B=0 B=3 −3−B+C=1 −3−3+C=1 C=7 f(x)=((−3)/(x−1))+((3x+7)/(x^2 +x−3)) contf f(x)=u+v (u/(−3))=(x−1)^(−1) (u_1 /(−3))=(−1)(x−1)^(−2) (u_2 /(−3))=(−1)(−2)(x−1)^(−3) (u_3 /(−3))=(−1)(−2)(−3)(x−1)^(−4) ..... ..... (u_n /(−3))=(−1)(−2)(−3)...(−n)(x−1)^(−n−1) (u_n /(−3))=(((−)^n ×n!)/((x−1)^(n+1) )) u_n =(−3)×(((−1)^n ×n!)/((x−1)^(n+1) )) pls check contd
$${Dr}={x}^{\mathrm{3}} −\mathrm{4}{x}+\mathrm{4}−\mathrm{1} \\ $$$$={x}^{\mathrm{3}} −\mathrm{1}−\mathrm{4}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{4}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}−\mathrm{4}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{3}\right) \\ $$$$\frac{{x}+\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{3}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{Bx}+{C}}{\left({x}^{\mathrm{2}} +{x}−\mathrm{3}\right)} \\ $$$${x}+\mathrm{2}={A}\left({x}^{\mathrm{2}} +{x}−\mathrm{3}\right)+\left({x}−\mathrm{1}\right)\left({Bx}+{C}\right) \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\mathrm{3}=−{A}\:\:{so}\:{A}=−\mathrm{3} \\ $$$${x}+\mathrm{2}=−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{9}+{Bx}^{\mathrm{2}} +{Cx}−{Bx}−{C} \\ $$$${x}+\mathrm{2}={x}^{\mathrm{2}} \left(−\mathrm{3}+{B}\right)+{x}\left(−\mathrm{3}−{B}+{C}\right)+\mathrm{9}−{C} \\ $$$$−\mathrm{3}+{B}=\mathrm{0}\:\:{B}=\mathrm{3} \\ $$$$−\mathrm{3}−{B}+{C}=\mathrm{1} \\ $$$$−\mathrm{3}−\mathrm{3}+{C}=\mathrm{1}\:\:\:{C}=\mathrm{7} \\ $$$${f}\left({x}\right)=\frac{−\mathrm{3}}{{x}−\mathrm{1}}+\frac{\mathrm{3}{x}+\mathrm{7}}{{x}^{\mathrm{2}} +{x}−\mathrm{3}} \\ $$$${contf} \\ $$$$ \\ $$$${f}\left({x}\right)={u}+{v} \\ $$$$\frac{{u}}{−\mathrm{3}}=\left({x}−\mathrm{1}\right)^{−\mathrm{1}} \\ $$$$\frac{{u}_{\mathrm{1}} }{−\mathrm{3}}=\left(−\mathrm{1}\right)\left({x}−\mathrm{1}\right)^{−\mathrm{2}} \\ $$$$\frac{{u}_{\mathrm{2}} }{−\mathrm{3}}=\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{−\mathrm{3}} \\ $$$$\frac{{u}_{\mathrm{3}} }{−\mathrm{3}}=\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left({x}−\mathrm{1}\right)^{−\mathrm{4}} \\ $$$$….. \\ $$$$….. \\ $$$$\frac{{u}_{{n}} }{−\mathrm{3}}=\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)…\left(−{n}\right)\left({x}−\mathrm{1}\right)^{−{n}−\mathrm{1}} \\ $$$$\frac{{u}_{{n}} }{−\mathrm{3}}=\frac{\left(−\right)^{{n}} ×{n}!}{\left({x}−\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${u}_{{n}} =\left(−\mathrm{3}\right)×\frac{\left(−\mathrm{1}\right)^{{n}} ×{n}!}{\left({x}−\mathrm{1}\right)^{{n}+\mathrm{1}} }\: \\ $$$${pls}\:{check}\:{contd} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 26/May/18
dear mr tanmay and mr Rasheed! a comment have posted to Q.35844 please check it.thanks.
$${dear}\:{mr}\:{tanmay}\:{and}\:{mr}\:{Rasheed}! \\ $$$${a}\:{comment}\:{have}\:{posted}\:{to}\:{Q}.\mathrm{35844} \\ $$$${please}\:{check}\:{it}.{thanks}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *