How do you express 1/(x^4 - 16)1x4−16 in partial fractions?
2 Answers
Derive a system of linear equations and solve to find:
1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))1x4−16=132(x−2)−132(x+2)−18(x2+4)
Explanation:
Assuming that we want partial fractions with Real coefficients, start by finding the Real factors of
Use the difference of squares identity, which can be written:
a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
twice, as follows:
x^4-16 = (x^2)^2-4^2x4−16=(x2)2−42
= (x^2-4)(x^2+4)=(x2−4)(x2+4)
= (x^2-2^2)(x^2+4)=(x2−22)(x2+4)
= (x-2)(x+2)(x^2+4)=(x−2)(x+2)(x2+4)
Since these factors are distinct, we are looking for a partial fraction decomposition of the form:
1/(x^4-16) = A/(x-2)+B/(x+2)+(Cx+D)/(x^2+4)1x4−16=Ax−2+Bx+2+Cx+Dx2+4
=(A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2))/(x^4-16)=A(x+2)(x2+4)+B(x−2)(x2+4)+(Cx+D)(x−2)(x+2)x4−16
=(A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x^2-4))/(x^4-16)=A(x+2)(x2+4)+B(x−2)(x2+4)+(Cx+D)(x2−4)x4−16
=((A+B+C)x^3+(2A-2B+D)x^2+(4A+4B-4C)x+(8A+8B-4D))/(x^4-16)=(A+B+C)x3+(2A−2B+D)x2+(4A+4B−4C)x+(8A+8B−4D)x4−16
Equating coefficients, we get the following system of linear equations:
{ (A+B+C=0), (2A-2B+D=0), (4A+4B-4C=0), (8A+8B-4D=1) :}
We can write this system of equations as an augmented matrix, then solve it by performing a sequence of row operations to make the left hand side into a
Apologies for the length, but Gaussian elimination is like that...
((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (4, 4, -4, 0, |, 0), (8, -8, 0, -4, |, 1))
Subtract
((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, -8, 0, |, 0), (8, -8, 0, -4, |, 1))
Divide row 3 by
((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (8, -8, 0, -4, |, 1))
Subtract row 3 from row 1 to get:
((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (8, -8, 0, -4, |, 1))
Subtract
((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, -8, |, 1))
Divide row 4 by
((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))
Subtract
((1, 1, 0, 0, |, 0), (0, -4, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))
Divide row 2 by
((1, 1, 0, 0, |, 0), (0, 1, 0, -1/4, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))
Add
((1, 1, 0, 0, |, 0), (0, 1, 0, 0, |, -1/32), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))
Subtract row 2 from row 1 to get:
((1, 0, 0, 0, |, 1/32), (0, 1, 0, 0, |, -1/32), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))
So we find:
((A),(B),(C),(D)) = ((1/32),(-1/32),(0),(-1/8))
Hence:
1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))
1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))
Explanation:
Use the difference of squares identity:
a^2-b^2=(a-b)(a+b)
to factor the denominator.
To make things easier, break down into partial fractions in two stages:
1/(x^4-16) = 1/((x^2-4)(x^2+4))
We find:
1/(x^2-4) - 1/(x^2+4) = ((x^2+4)-(x^2-4))/(x^4-16) = 8/(x^4-16)
So:
1/(x^4-16) = 1/(8(x^2-4))-1/(8(x^2+4))
Then note that:
1/(x-2)-1/(x+2) = ((x+2)-(x-2))/(x^2-4) = 4/(x^2-4)
So:
1/(x^2-4) = 1/(4(x-2))-1/(4(x+2))
And:
1/(8(x^2-4)) = 1/(32(x-2))-1/(32(x+2))
Putting it all together:
1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))
