Thread: Partial fractions

I have already found that both A and B are equal to 1/4 by multiplying both sides by z⁴-1 and setting z equal to -1 and 1. From here I am unsure how to find C and D.

Originally Posted by bikerboy2442

I have already found that both A and B are equal to 1/4 by multiplying both sides by z⁴-1 and setting z equal to -1 and 1. From here I am unsure how to find C and D.

Good- multiplying both sides by z⁴- 1 gives z= A(z-1)(z²+ 1)+ B(z+1)(z²+ 1)+ (Cz+ D)(z-1)(z+1)

Yes, setting z= 1 gives 1= A(0)(2)+ B(2)(2)+ (C+ D)(0)(2) or 1= 4B so B= 1/4. Setting z= -1 gives -1= A(-2)(2)+ B(0)(2)+ (-C+D)(0) or -1= -4A so A= 1/4.

There is no z that will make z²+ 1 equal to 0 but just choose 2 other numbers to get two more equatons. If we take z= 0 the equation becomes 0= A(-1)(1)+ B(1)(1)+ D(-1)(1), 0= -A+ B- D. Since A= B= 1/4, this is 0= -1/4+ 1/4- D= -D so that -D= 0. If we take z= 2, the equation becomes 2= A(1)(5)+ B(3)(5)+ (2C+D)(1)(3) or 2= 5A+15B+ 6C+ 3D. Since A= B= 1/4 and D= 0, this is 2= 5/4+ 15/4+ 6C= 20/4+ 6C= 5+ 6C. From 6C+ 5= 2, we have 6C= -3 and C= -1/2.
"0" and "2" were chosen simply because they are simple numbers. Choosing any two new numbers will give two new equations so that we have four equations in the four unknown values, A, B, C, and D.

An entirely different way to do this to multiply everything out and equate "corresponding coefficients".

z= A(z-1)(z²+ 1)+ B(z+1)(z²+ 1)+ (Cz+ D)(z-1)(z+1)
z= A(z³- z²+ z- 1)+ B(z³+ z²+ z+ 1)+ Cz³+ Dz²- Cz+ D
z= (A+ B+ C)z³+ (-A+ B+ D)z²+ (A+ B- C)z+ (-A+ B+ D)

For this to be true for all z, corresponding coefficents must be equal.
0= A+ B+ C, 0= -A+ B+ D, 1= A+ B- C, 0= -A+ B+ D, four equations to solve for A, B, C, D.