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Question Number 58769 by Mr X pcx last updated on 29/Apr/19
decompose the fractions inside C(x) 1) (1/((x^2 +1)^3 )) 2) (1/((x^2 +1)^5 ))
$${decompose}\:{the}\:{fractions}\:{inside}\:{C}\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{5}} } \\ $$
Commented by maxmathsup by imad last updated on 01/May/19
1) F(x) =(1/((x^2 +1)^3 )) =(1/((x−i)^3 (x+i)^3 )) =Σ_(k=1) ^3 (a_k /((x−i)^k )) +Σ_(k=1) ^3 (b_k /((x+i)^3 )) =(a_1 /(x−i)) +(a_1 /((x−i)^2 )) +(a_3 /((x−i)^3 )) +(b_1 /(x+i)) +(b_2 /((x+i)^2 )) +(b_3 /((x+i)^3 )) we have conj(F(x))=F(x) ⇒b_k =a_k ^− let determine a_k changement x−i =t give F(x) =G(t) =(1/(t^3 (t+2i)^3 )) let find D_2 (0) for w(t) =(1/((t+2i)^3 )) ⇒ w(t) =w(0) +(t/(1!)) w^((1)) (0) +(t^2 /2) w^((2)) (0)+(t^3 /(3!)) ξ(t) but w(t) =(t+2i)^(−3) ⇒ w(0) =(2i)^(−3) w^′ (t) =−3(t+2i)^(−4) ⇒w^((1)) (0) =−3(2i)^(−4) w^((2)) (t) =12 (t+2i)^(−5) ⇒w^((2)) (0) =12(2i)^(−5) ⇒ w(t) =(2i)^(−3) −3(2i)^(−4) t +6 (2i)^(−5) t^2 +(t^3 /(3!))ξ(t) ⇒((w(t))/t^3 ) =(((2i)^(−3) )/t^3 ) −3(2i)^(−4) (1/t^2 ) +6(2i)^(−5) (1/t) +(1/(3!))ξ(t) ⇒ F(x) =((6(2i)^(−5) )/(x−i)) −((3(2i)^(−4) )/((x−i)^2 )) +(((2i)^(−3) )/((x−i)^3 )) +(1/(3!))ξ(x−i) ⇒ a_1 =(6/((2i)^5 )) =(6/(2.16 i)) =(3/(16i)) =((−3i)/(16)) a_2 =((−3)/((2i)^4 )) =((−3)/(16)) and a_3 =(1/((2i)^3 )) =(1/(−8i)) =(i/8) also b_1 =((3i)/(16)) b_2 =((−3)/(16)) and b_3 =−(i/8) ⇒ F(x) =((−3i)/(16(x−i))) −(3/(16(x−i)^2 )) +(i/(8(x−i)^3 )) +((3i)/(16(x+i))) −(3/(16(x+i)^2 )) −(i/(8(x+i)^3 )) .
$$\left.\mathrm{1}\right)\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\left({x}−{i}\right)^{\mathrm{3}} \left({x}+{i}\right)^{\mathrm{3}} }\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} \:\:\frac{{a}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{\mathrm{3}} \:\frac{{b}_{{k}} }{\left({x}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{{a}_{\mathrm{1}} }{{x}−{i}}\:+\frac{{a}_{\mathrm{1}} }{\left({x}−{i}\right)^{\mathrm{2}} }\:+\frac{{a}_{\mathrm{3}} }{\left({x}−{i}\right)^{\mathrm{3}} }\:+\frac{{b}_{\mathrm{1}} }{{x}+{i}}\:+\frac{{b}_{\mathrm{2}} }{\left({x}+{i}\right)^{\mathrm{2}} }\:+\frac{{b}_{\mathrm{3}} }{\left({x}+{i}\right)^{\mathrm{3}} } \\ $$$${we}\:{have}\:{conj}\left({F}\left({x}\right)\right)={F}\left({x}\right)\:\Rightarrow{b}_{{k}} =\overset{−} {{a}}_{{k}} \:\:\:\:{let}\:{determine}\:{a}_{{k}} \\ $$$${changement}\:{x}−{i}\:={t}\:{give}\:{F}\left({x}\right)\:={G}\left({t}\right)\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} \left({t}+\mathrm{2}{i}\right)^{\mathrm{3}} }\:\:{let}\:{find}\:{D}_{\mathrm{2}} \left(\mathrm{0}\right)\:{for} \\ $$$${w}\left({t}\right)\:=\frac{\mathrm{1}}{\left({t}+\mathrm{2}{i}\right)^{\mathrm{3}} }\:\Rightarrow\:{w}\left({t}\right)\:={w}\left(\mathrm{0}\right)\:+\frac{{t}}{\mathrm{1}!}\:{w}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)\:+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:{w}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}\:\xi\left({t}\right)\:{but} \\ $$$${w}\left({t}\right)\:=\left({t}+\mathrm{2}{i}\right)^{−\mathrm{3}} \:\Rightarrow\:{w}\left(\mathrm{0}\right)\:=\left(\mathrm{2}{i}\right)^{−\mathrm{3}} \\ $$$${w}^{'} \left({t}\right)\:=−\mathrm{3}\left({t}+\mathrm{2}{i}\right)^{−\mathrm{4}} \:\Rightarrow{w}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)\:=−\mathrm{3}\left(\mathrm{2}{i}\right)^{−\mathrm{4}} \\ $$$${w}^{\left(\mathrm{2}\right)} \left({t}\right)\:=\mathrm{12}\:\left({t}+\mathrm{2}{i}\right)^{−\mathrm{5}} \:\Rightarrow{w}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{12}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:\Rightarrow \\ $$$${w}\left({t}\right)\:=\left(\mathrm{2}{i}\right)^{−\mathrm{3}} \:−\mathrm{3}\left(\mathrm{2}{i}\right)^{−\mathrm{4}} {t}\:\:\:+\mathrm{6}\:\left(\mathrm{2}{i}\right)^{−\mathrm{5}} {t}^{\mathrm{2}} \:+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}\xi\left({t}\right)\:\Rightarrow\frac{{w}\left({t}\right)}{{t}^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2}{i}\right)^{−\mathrm{3}} }{{t}^{\mathrm{3}} }\:−\mathrm{3}\left(\mathrm{2}{i}\right)^{−\mathrm{4}} \:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:+\mathrm{6}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:\frac{\mathrm{1}}{{t}}\:+\frac{\mathrm{1}}{\mathrm{3}!}\xi\left({t}\right)\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{6}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} }{{x}−{i}}\:−\frac{\mathrm{3}\left(\mathrm{2}{i}\right)^{−\mathrm{4}} }{\left({x}−{i}\right)^{\mathrm{2}} }\:\:+\frac{\left(\mathrm{2}{i}\right)^{−\mathrm{3}} }{\left({x}−{i}\right)^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{3}!}\xi\left({x}−{i}\right)\:\Rightarrow \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{6}}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}.\mathrm{16}\:{i}}\:=\frac{\mathrm{3}}{\mathrm{16}{i}}\:=\frac{−\mathrm{3}{i}}{\mathrm{16}} \\ $$$${a}_{\mathrm{2}} =\frac{−\mathrm{3}}{\left(\mathrm{2}{i}\right)^{\mathrm{4}} }\:\:=\frac{−\mathrm{3}}{\mathrm{16}}\:\:\:\:{and}\:\:{a}_{\mathrm{3}} \:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{−\mathrm{8}{i}}\:=\frac{{i}}{\mathrm{8}}\:\:\:{also}\:\:{b}_{\mathrm{1}} =\frac{\mathrm{3}{i}}{\mathrm{16}} \\ $$$${b}_{\mathrm{2}} =\frac{−\mathrm{3}}{\mathrm{16}}\:\:\:\:{and}\:\:{b}_{\mathrm{3}} =−\frac{{i}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{−\mathrm{3}{i}}{\mathrm{16}\left({x}−{i}\right)}\:−\frac{\mathrm{3}}{\mathrm{16}\left({x}−{i}\right)^{\mathrm{2}} }\:+\frac{{i}}{\mathrm{8}\left({x}−{i}\right)^{\mathrm{3}} }\:\:+\frac{\mathrm{3}{i}}{\mathrm{16}\left({x}+{i}\right)}\:−\frac{\mathrm{3}}{\mathrm{16}\left({x}+{i}\right)^{\mathrm{2}} }\:−\frac{{i}}{\mathrm{8}\left({x}+{i}\right)^{\mathrm{3}} }\:. \\ $$$$ \\ $$

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