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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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