Sunday, January 19, 2014

`x/(16x^4 - 1)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`x/(16x^4-1)`


Let's factorize the denominator,


`16x^4-1=(4x^2)^2-1`


`=(4x^2+1)(4x^2-1)`


`=(4x^2+1)(2x+1)(2x-1)` 


Let `x/(16x^4-1)=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)`


`x/(16x^4-1)=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(A(8x^3+2x-4x^2-1)+B(8x^3+2x+4x^2+1)+(Cx+D)(4x^2-1))/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(A(8x^3-4x^2+2x-1)+B(8x^3+4x^2+2x+1)+4Cx^3-Cx+4Dx^2-D)/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D)/((2x+1)(2x-1)(4x^2+1))`


`:.x=x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D`


equating the coefficients of the like terms,


`8A+8B+4C=0`        ----- equation 1


`-4A+4B+4D=0`     ----- equation 2


`2A+2B-C=1`            ----- equation 3


`-A+B-D=0`             ------ equation 4


Now we have to solve the above four equations to find the solutions of A,B,C and D.


From equation 1,


`4(2A+2B+C)=0`


`2A+2B+C=0`


Subtract equation...

`x/(16x^4-1)`


Let's factorize the denominator,


`16x^4-1=(4x^2)^2-1`


`=(4x^2+1)(4x^2-1)`


`=(4x^2+1)(2x+1)(2x-1)` 


Let `x/(16x^4-1)=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)`


`x/(16x^4-1)=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(A(8x^3+2x-4x^2-1)+B(8x^3+2x+4x^2+1)+(Cx+D)(4x^2-1))/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(A(8x^3-4x^2+2x-1)+B(8x^3+4x^2+2x+1)+4Cx^3-Cx+4Dx^2-D)/((2x+1)(2x-1)(4x^2+1))`


`x/(16x^4-1)=(x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D)/((2x+1)(2x-1)(4x^2+1))`


`:.x=x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D`


equating the coefficients of the like terms,


`8A+8B+4C=0`        ----- equation 1


`-4A+4B+4D=0`     ----- equation 2


`2A+2B-C=1`            ----- equation 3


`-A+B-D=0`             ------ equation 4


Now we have to solve the above four equations to find the solutions of A,B,C and D.


From equation 1,


`4(2A+2B+C)=0`


`2A+2B+C=0`


Subtract equation 3 from the above equation,


`(2A+2B+C)-(2A+2B-C)=0-1`


`2C=-1`


`C=-1/2`


From equation 2,


`4(-A+B+D)=0`


`-A+B+D=0`


Now subtract equation 4 from the above equation,


`(-A+B+D)-(-A+B-D)=0`


`2D=0`


`D=0`


Now plug in the values of C in the equation 3,


`2A+2B-(-1/2)=1`


`2A+2B+1/2=1`


`2A+2B=1-1/2`


`2(A+B)=1/2`


`A+B=1/4`       ----- equation 5


Plug in the value of D in the equation 4,


`-A+B-0=0`


`-A+B=0`      ---- equation 6


Now add the equations 5 and 6,


`2B=1/4`


`B=1/8`


Plug in the value of B in the equation 6,


`-A+1/8-0`


`A=1/8`


`:.x/(16x^4-1)=(1/8)/(2x+1)+(1/8)/(2x-1)+((-1/2)x)/(4x^2+1)`


`x/(16x^4-1)=1/(8(2x+1))+1/(8(2x-1))-x/(2(4x^2+1))`


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