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


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Question Number 59580 by maxmathsup by imad last updated on 12/May/19
$decompose at simple element the fraction F(x) =(1/(x^7 (x^2 −1)^2 ))$
$d e c o m p o s e a t s i m p l e e l e m e n t t h e f r a c t i o n$ $F (x) = \frac{1}{x^{7} {(x^{2} - 1)}^{2}}$
Commented by malwaan last updated on 12/May/19

$I n e e d s o m e t i m e!$
	Answered by MJS last updated on 13/May/19
	$F(x)=(a/((x−1)^2 ))+(b/((x+1)^2 ))+(c/(x−1))+(d/(x+1))+(e/x)+(f/x^2 )+(g/x^3 )+(h/x^4 )+(i/x^5 )+(j/x^6 )+(k/x^7 ) the numerator of this fraction =N N−1=0 ⇒ the constant factors on the left handed side are zero: x^(10) : c+d+e=0 x^9 : a+b+c−d+f=0 x^8 : 2(a−b−e)−c−d+g=0 x^7 : a+b−c+d−2f+h=0 x^6 : e−2g+i=0 x^5 : f−2h+j=0 x^4 : g−2i+k=0 x^3 : h−2j=0 x^2 : i−2k=0 x^1 : j=0 x^0 : k−1=0 ⇒ a=(1/4); b=−(1/4); c=−2; d=−2; e=4; f=0; g=3; h=0; i=2; j=0; k=1 ⇒ F(x)=(1/(4(x−1)^2 ))−(1/(4(x+1)^2 ))−(2/(x−1))−(2/(x+1))+(4/x)+(3/x^3 )+(2/x^5 )+(1/x^7 )$
	$F (x) = \frac{a}{{(x - 1)}^{2}} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{x - 1} + \frac{d}{x + 1} + \frac{e}{x} + \frac{f}{x^{2}} + \frac{g}{x^{3}} + \frac{h}{x^{4}} + \frac{i}{x^{5}} + \frac{j}{x^{6}} + \frac{k}{x^{7}}$ $the numerator of this fraction = N$ $N - 1 = 0$ $\Rightarrow the constant factors on the left handed side$ $are zero :$ $x^{10} : c + d + e = 0$ $x^{9} : a + b + c - d + f = 0$ $x^{8} : 2 (a - b - e) - c - d + g = 0$ $x^{7} : a + b - c + d - 2 f + h = 0$ $x^{6} : e - 2 g + i = 0$ $x^{5} : f - 2 h + j = 0$ $x^{4} : g - 2 i + k = 0$ $x^{3} : h - 2 j = 0$ $x^{2} : i - 2 k = 0$ $x^{1} : j = 0$ $x^{0} : k - 1 = 0$ $\Rightarrow a = \frac{1}{4}; b = - \frac{1}{4}; c = - 2; d = - 2; e = 4;$ $f = 0; g = 3; h = 0; i = 2; j = 0; k = 1$ $\Rightarrow$ $F (x) = \frac{1}{4 {(x - 1)}^{2}} - \frac{1}{4 {(x + 1)}^{2}} - \frac{2}{x - 1} - \frac{2}{x + 1} + \frac{4}{x} + \frac{3}{x^{3}} + \frac{2}{x^{5}} + \frac{1}{x^{7}}$
	Commented by ajfour last updated on 12/May/19

	$FABULOUS S i r!$
	Commented by malwaan last updated on 15/May/19

	$t h a n k s s i r!$
	Commented by MJS last updated on 13/May/19

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