f
(
x
)
=
1
x
2
+
2
x
−
3
{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}}
Here, the denominator splits into two distinct linear factors:
q
(
x
)
=
x
2
+
2
x
−
3
=
(
x
+
3
)
(
x
−
1
)
{\displaystyle q(x)=x^{2}+2x-3=(x+3)(x-1)}
so we have the partial fraction decomposition
f
(
x
)
=
1
x
2
+
2
x
−
3
=
A
x
+
3
+
B
x
−
1
{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {A}{x+3}}+{\frac {B}{x-1}}}
Multiplying through by x 2 + 2x − 3, we have the polynomial identity
1
=
A
(
x
−
1
)
+
B
(
x
+
3
)
{\displaystyle 1=A(x-1)+B(x+3)}
Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that
f
(
x
)
=
1
x
2
+
2
x
−
3
=
1
4
(
−
1
x
+
3
+
1
x
−
1
)
{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {1}{4}}\left({\frac {-1}{x+3}}+{\frac {1}{x-1}}\right)}
f
(
x
)
=
x
3
+
16
x
3
−
4
x
2
+
8
x
{\displaystyle f(x)={\frac {x^{3}+16}{x^{3}-4x^{2}+8x}}}
After long-division , we have
f
(
x
)
=
1
+
4
x
2
−
8
x
+
16
x
3
−
4
x
2
+
8
x
=
1
+
4
x
2
−
8
x
+
16
x
(
x
2
−
4
x
+
8
)
{\displaystyle f(x)=1+{\frac {4x^{2}-8x+16}{x^{3}-4x^{2}+8x}}=1+{\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}}
Since (−4)2 − 4×8 = −16 < 0, the factor x 2 − 4x + 8 is irreducible, and the partial fraction decomposition over the reals has the shape
4
x
2
−
8
x
+
16
x
(
x
2
−
4
x
+
8
)
=
A
x
+
B
x
+
C
x
2
−
4
x
+
8
{\displaystyle {\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}-4x+8}}}
Multiplying through by x 3 − 4x 2 + 8x , we have the polynomial identity
4
x
2
−
8
x
+
16
=
A
(
x
2
−
4
x
+
8
)
+
(
B
x
+
C
)
x
{\displaystyle 4x^{2}-8x+16=A(x^{2}-4x+8)+(Bx+C)x}
Taking x = 0, we see that 16 = 8A , so A = 2. Comparing the x 2 coefficients, we see that 4 = A + B = 2 + B , so B = 2. Comparing linear coefficients, we see that −8 = −4A + C = −8 + C , so C = 0. Altogether,
f
(
x
)
=
1
+
2
(
1
x
+
x
x
2
−
4
x
+
8
)
{\displaystyle f(x)=1+2\left({\frac {1}{x}}+{\frac {x}{x^{2}-4x+8}}\right)}
The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system .
f
(
x
)
=
x
9
−
2
x
6
+
2
x
5
−
7
x
4
+
13
x
3
−
11
x
2
+
12
x
−
4
x
7
−
3
x
6
+
5
x
5
−
7
x
4
+
7
x
3
−
5
x
2
+
3
x
−
1
{\displaystyle f(x)={\frac {x^{9}-2x^{6}+2x^{5}-7x^{4}+13x^{3}-11x^{2}+12x-4}{x^{7}-3x^{6}+5x^{5}-7x^{4}+7x^{3}-5x^{2}+3x-1}}}
After long-division and factoring the denominator, we have
f
(
x
)
=
x
2
+
3
x
+
4
+
2
x
6
−
4
x
5
+
5
x
4
−
3
x
3
+
x
2
+
3
x
(
x
−
1
)
3
(
x
2
+
1
)
2
{\displaystyle f(x)=x^{2}+3x+4+{\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}}
The partial fraction decomposition takes the form
2
x
6
−
4
x
5
+
5
x
4
−
3
x
3
+
x
2
+
3
x
(
x
−
1
)
3
(
x
2
+
1
)
2
=
A
x
−
1
+
B
(
x
−
1
)
2
+
C
(
x
−
1
)
3
+
D
x
+
E
x
2
+
1
+
F
x
+
G
(
x
2
+
1
)
2
{\displaystyle {\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}={\frac {A}{x-1}}+{\frac {B}{(x-1)^{2}}}+{\frac {C}{(x-1)^{3}}}+{\frac {Dx+E}{x^{2}+1}}+{\frac {Fx+G}{(x^{2}+1)^{2}}}}
Multiplying through by (x − 1)3 (x 2 + 1)2 we have the polynomial identity
2
x
6
−
4
x
5
+
5
x
4
−
3
x
3
+
x
2
+
3
x
=
A
(
x
−
1
)
2
(
x
2
+
1
)
2
+
B
(
x
−
1
)
(
x
2
+
1
)
2
+
C
(
x
2
+
1
)
2
+
(
D
x
+
E
)
(
x
−
1
)
3
(
x
2
+
1
)
+
(
F
x
+
G
)
(
x
−
1
)
3
{\displaystyle {\begin{aligned}&{}\quad 2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+C(x^{2}+1)^{2}+(Dx+E)(x-1)^{3}(x^{2}+1)+(Fx+G)(x-1)^{3}\end{aligned}}}
Taking x = 1 gives 4 = 4C , so C = 1. Similarly, taking x = i i = (Fi + G )(2 + 2i ), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. With C = G = 1 and F = 0, taking x = 0 we get A − B + 1 − E − 1 = 0, thus E = A − B .
We now have the identity
2
x
6
−
4
x
5
+
5
x
4
−
3
x
3
+
x
2
+
3
x
=
A
(
x
−
1
)
2
(
x
2
+
1
)
2
+
B
(
x
−
1
)
(
x
2
+
1
)
2
+
(
x
2
+
1
)
2
+
(
D
x
+
(
A
−
B
)
)
(
x
−
1
)
3
(
x
2
+
1
)
+
(
x
−
1
)
3
=
A
(
(
x
−
1
)
2
(
x
2
+
1
)
2
+
(
x
−
1
)
3
(
x
2
+
1
)
)
+
B
(
(
x
−
1
)
(
x
2
+
1
)
−
(
x
−
1
)
3
(
x
2
+
1
)
)
+
(
x
2
+
1
)
2
+
D
x
(
x
−
1
)
3
(
x
2
+
1
)
+
(
x
−
1
)
3
{\displaystyle {\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+(x^{2}+1)^{2}+(Dx+(A-B))(x-1)^{3}(x^{2}+1)+(x-1)^{3}\\&=A((x-1)^{2}(x^{2}+1)^{2}+(x-1)^{3}(x^{2}+1))+B((x-1)(x^{2}+1)-(x-1)^{3}(x^{2}+1))+(x^{2}+1)^{2}+Dx(x-1)^{3}(x^{2}+1)+(x-1)^{3}\end{aligned}}}
Expanding and sorting by exponents of x we get
2
x
6
−
4
x
5
+
5
x
4
−
3
x
3
+
x
2
+
3
x
=
(
A
+
D
)
x
6
+
(
−
A
−
3
D
)
x
5
+
(
2
B
+
4
D
+
1
)
x
4
+
(
−
2
B
−
4
D
+
1
)
x
3
+
(
−
A
+
2
B
+
3
D
−
1
)
x
2
+
(
A
−
2
B
−
D
+
3
)
x
{\displaystyle {\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=(A+D)x^{6}+(-A-3D)x^{5}+(2B+4D+1)x^{4}+(-2B-4D+1)x^{3}+(-A+2B+3D-1)x^{2}+(A-2B-D+3)x\end{aligned}}}
We can now compare the coefficients and see that
A
+
D
=
2
−
A
−
3
D
=
−
4
2
B
+
4
D
+
1
=
5
−
2
B
−
4
D
+
1
=
−
3
−
A
+
2
B
+
3
D
−
1
=
1
A
−
2
B
−
D
+
3
=
3
,
{\displaystyle {\begin{aligned}A+D&=&2\\-A-3D&=&-4\\2B+4D+1&=&5\\-2B-4D+1&=&-3\\-A+2B+3D-1&=&1\\A-2B-D+3&=&3,\end{aligned}}}
with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = A − B = 1, F = 0 and G = 1.
The partial fraction decomposition of ƒ (x ) is thus
f
(
x
)
=
x
2
+
3
x
+
4
+
1
(
x
−
1
)
+
1
(
x
−
1
)
3
+
x
+
1
x
2
+
1
+
1
(
x
2
+
1
)
2
.
{\displaystyle f(x)=x^{2}+3x+4+{\frac {1}{(x-1)}}+{\frac {1}{(x-1)^{3}}}+{\frac {x+1}{x^{2}+1}}+{\frac {1}{(x^{2}+1)^{2}}}.}
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at x=1 and at x=i in the above polynomial identity. (To this end, recall that the derivative at x=a of (x−a)m p(x) vanishes if m > 1 and it is just p(a) if m=1 .)
Thus, for instance the first derivative at x=1 gives
2
⋅
6
−
4
⋅
5
+
5
⋅
4
−
3
⋅
3
+
2
+
3
=
A
⋅
(
0
+
0
)
+
B
⋅
(
4
+
0
)
+
8
+
D
⋅
0
{\displaystyle 2\cdot 6-4\cdot 5+5\cdot 4-3\cdot 3+2+3=A\cdot (0+0)+B\cdot (4+0)+8+D\cdot 0}
that is 8 = 4B + 8 so B=0 .
f
(
z
)
=
z
2
−
5
(
z
2
−
1
)
(
z
2
+
1
)
=
z
2
−
5
(
z
+
1
)
(
z
−
1
)
(
z
+
i
)
(
z
−
i
)
{\displaystyle f(z)={\frac {z^{2}-5}{(z^{2}-1)(z^{2}+1)}}={\frac {z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}}}
Thus, f (z ) can be decomposed into rational functions whose denominators are z +1, z −1, z +i, z −i. Since each term is of power one, −1, 1, −i and i are simple poles.
Hence, the residues associated with each pole, given by
P
(
z
i
)
Q
′
(
z
i
)
=
z
i
2
−
5
4
z
i
3
{\displaystyle {\frac {P(z_{i})}{Q'(z_{i})}}={\frac {z_{i}^{2}-5}{4z_{i}^{3}}}}
are
1
,
−
1
,
3
i
2
,
−
3
i
2
{\displaystyle 1,-1,{\tfrac {3i}{2}},-{\tfrac {3i}{2}}}
respectively, and
f
(
z
)
=
1
z
+
1
−
1
z
−
1
+
3
i
2
1
z
+
i
−
3
i
2
1
z
−
i
{\displaystyle f(z)={\frac {1}{z+1}}-{\frac {1}{z-1}}+{\frac {3i}{2}}{\frac {1}{z+i}}-{\frac {3i}{2}}{\frac {1}{z-i}}}
Limits can be used to find a partial fraction decomposition.[ 1]
f
(
x
)
=
1
x
3
−
1
{\displaystyle f(x)={\frac {1}{x^{3}-1}}}
First, factor the denominator:
f
(
x
)
=
1
(
x
−
1
)
(
x
2
+
x
+
1
)
{\displaystyle f(x)={\frac {1}{(x-1)(x^{2}+x+1)}}}
The decomposition takes the form of
1
(
x
−
1
)
(
x
2
+
x
+
1
)
=
A
x
−
1
+
B
x
+
C
x
2
+
x
+
1
{\displaystyle {\frac {1}{(x-1)(x^{2}+x+1)}}={\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}}
As
x
→
1
{\displaystyle x\to 1}
A term dominates, so the right-hand side approaches
A
x
−
1
{\displaystyle {\frac {A}{x-1}}}
1
(
x
−
1
)
(
x
2
+
x
+
1
)
=
A
x
−
1
{\displaystyle {\frac {1}{(x-1)(x^{2}+x+1)}}={\frac {A}{x-1}}}
A
=
lim
x
→
1
1
x
2
+
x
+
1
=
1
3
{\displaystyle A=\lim _{x\to 1}{\frac {1}{x^{2}+x+1}}={\frac {1}{3}}}
As
x
→
∞
{\displaystyle x\to \infty }
lim
x
→
∞
A
x
−
1
+
B
x
+
C
x
2
+
x
+
1
=
A
x
+
B
x
x
2
=
A
+
B
x
.
{\displaystyle \lim _{x\to \infty }{{\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}}={\frac {A}{x}}+{\frac {Bx}{x^{2}}}={\frac {A+B}{x}}.}
A
+
B
x
=
lim
x
→
∞
1
x
3
−
1
=
0
{\displaystyle {\frac {A+B}{x}}=\lim _{x\to \infty }{\frac {1}{x^{3}-1}}=0}
Thus,
B
=
−
1
3
{\displaystyle B=-{\frac {1}{3}}}
At
x
=
0
{\displaystyle x=0}
−
1
=
−
A
+
C
{\displaystyle -1=-A+C}
C
=
−
2
3
{\displaystyle C=-{\frac {2}{3}}}
The decomposition is thus
1
3
x
−
1
+
−
1
3
x
−
2
3
x
2
+
x
+
1
{\displaystyle {\frac {\frac {1}{3}}{x-1}}+{\frac {-{\frac {1}{3}}x-{\frac {2}{3}}}{x^{2}+x+1}}}
![{\displaystyle R(s)={\frac {P(s)}{[(x-a)^{2}+b^{2}]^{n}}}={\frac {A1}{[(x-a)^{2}+b^{2}]}}+{\frac {A2}{[(x-a)^{2}+b^{2}]^{2}}}+...+{\frac {An}{[(x-a)^{2}+b^{2}]^{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1087c669b3429c08bc3c7a2046872f08d0b981ab)
