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Question Number 59580 by maxmathsup by imad last updated on 12/May/19

decompose at simple element the fraction F(x) =(1/(x^7 (x^2 −1)^2 ))

$${decompose}\:{at}\:{simple}\:{element}\:{the}\:{fraction}\: \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{7}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by malwaan last updated on 12/May/19

I need some time !

$${I}\:{need}\:{some}\:{time}\:! \\ $$

Answered by MJS last updated on 13/May/19

F(x)=(a/((x−1)^2 ))+(b/((x+1)^2 ))+(c/(x−1))+(d/(x+1))+(e/x)+(f/x^2 )+(g/x^3 )+(h/x^4 )+(i/x^5 )+(j/x^6 )+(k/x^7 ) the numerator of this fraction =N N−1=0 ⇒ the constant factors on the left handed side are zero: x^(10) : c+d+e=0 x^9 : a+b+c−d+f=0 x^8 : 2(a−b−e)−c−d+g=0 x^7 : a+b−c+d−2f+h=0 x^6 : e−2g+i=0 x^5 : f−2h+j=0 x^4 : g−2i+k=0 x^3 : h−2j=0 x^2 : i−2k=0 x^1 : j=0 x^0 : k−1=0 ⇒ a=(1/4); b=−(1/4); c=−2; d=−2; e=4; f=0; g=3; h=0; i=2; j=0; k=1 ⇒ F(x)=(1/(4(x−1)^2 ))−(1/(4(x+1)^2 ))−(2/(x−1))−(2/(x+1))+(4/x)+(3/x^3 )+(2/x^5 )+(1/x^7 )

$${F}\left({x}\right)=\frac{{a}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{c}}{{x}−\mathrm{1}}+\frac{{d}}{{x}+\mathrm{1}}+\frac{{e}}{{x}}+\frac{{f}}{{x}^{\mathrm{2}} }+\frac{{g}}{{x}^{\mathrm{3}} }+\frac{{h}}{{x}^{\mathrm{4}} }+\frac{{i}}{{x}^{\mathrm{5}} }+\frac{{j}}{{x}^{\mathrm{6}} }+\frac{{k}}{{x}^{\mathrm{7}} } \\ $$$$\mathrm{the}\:\mathrm{numerator}\:\mathrm{of}\:\mathrm{this}\:\mathrm{fraction}\:={N} \\ $$$${N}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factors}\:\mathrm{on}\:\mathrm{the}\:\mathrm{left}\:\mathrm{handed}\:\mathrm{side} \\ $$$$\mathrm{are}\:\mathrm{zero}: \\ $$$${x}^{\mathrm{10}} :\:\:{c}+{d}+{e}=\mathrm{0} \\ $$$${x}^{\mathrm{9}} :\:\:\:\:{a}+{b}+{c}−{d}+{f}=\mathrm{0} \\ $$$${x}^{\mathrm{8}} :\:\:\:\:\mathrm{2}\left({a}−{b}−{e}\right)−{c}−{d}+{g}=\mathrm{0} \\ $$$${x}^{\mathrm{7}} :\:\:\:\:{a}+{b}−{c}+{d}−\mathrm{2}{f}+{h}=\mathrm{0} \\ $$$${x}^{\mathrm{6}} :\:\:\:\:{e}−\mathrm{2}{g}+{i}=\mathrm{0} \\ $$$${x}^{\mathrm{5}} :\:\:\:\:{f}−\mathrm{2}{h}+{j}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} :\:\:\:\:{g}−\mathrm{2}{i}+{k}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} :\:\:\:\:{h}−\mathrm{2}{j}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} :\:\:\:\:{i}−\mathrm{2}{k}=\mathrm{0} \\ $$$${x}^{\mathrm{1}} :\:\:\:\:{j}=\mathrm{0} \\ $$$${x}^{\mathrm{0}} :\:\:\:\:{k}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{4}};\:{b}=−\frac{\mathrm{1}}{\mathrm{4}};\:{c}=−\mathrm{2};\:{d}=−\mathrm{2};\:{e}=\mathrm{4}; \\ $$$$\:\:\:\:\:\:{f}=\mathrm{0};\:{g}=\mathrm{3};\:{h}=\mathrm{0};\:{i}=\mathrm{2};\:{j}=\mathrm{0};\:{k}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{{x}−\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}+\frac{\mathrm{4}}{{x}}+\frac{\mathrm{3}}{{x}^{\mathrm{3}} }+\frac{\mathrm{2}}{{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{{x}^{\mathrm{7}} } \\ $$

Commented by ajfour last updated on 12/May/19

FABULOUS Sir!

$$\mathcal{FABULOUS}\:{Sir}! \\ $$

Commented by malwaan last updated on 15/May/19

thanks sir !

$${thanks}\:{sir}\:! \\ $$

Commented by MJS last updated on 13/May/19

yes, just a typo. I corrected. thank you!

$$\mathrm{yes},\:\mathrm{just}\:\mathrm{a}\:\mathrm{typo}.\:\mathrm{I}\:\mathrm{corrected}. \\ $$$$\mathrm{thank}\:\mathrm{you}! \\ $$

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