Partial fraction decomposition

In summary, to decompose the function V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2} into simple fractions, we can use the formula V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}. The residues for the single pole at 3 and the double pole at 1 can be calculated as 1 and -2 respectively. To find the residue at the single pole at 1, we can use the formula and set it equal to the function in 0, giving us a value of -1 for C.
  • #1
lucad93
4
0
Hello everybody! I have to decompose to simple fractions the following function: \(\displaystyle V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2}\). I know I can see the function as: \(\displaystyle V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}\), and that the terms A, B, C can be calculated respectively as the residues in 3 (single pole), 1 (double pole), and 1 (single pole). So i calculate: \(\displaystyle A: Res(f, 3) = 1\), that's correct; \(\displaystyle B=Res(f,1) = \lim_{{z}\to{1}}\left[\d{}{x}z^2-4z+4\right]=-2\), that's correct. Finally I have to calculate the \(\displaystyle Res(f,1)\), single pole. I tried in some ways: by using the single pole formula and then using the "de l'hopital"'s theorem, by calculating the decomposed function in 0 and then equaling to the function in 0, but nothing was correct. Can you tell me a correct way to proceed? Thankyou
 
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  • #2
lucad93 said:
Hello everybody! I have to decompose to simple fractions the following function: \(\displaystyle V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2}\). I know I can see the function as: \(\displaystyle V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}\), and that the terms A, B, C can be calculated respectively as the residues in 3 (single pole), 1 (double pole), and 1 (single pole). So i calculate: \(\displaystyle A: Res(f, 3) = 1\), that's correct; \(\displaystyle B=Res(f,1) = \lim_{{z}\to{1}}\left[\d{}{x}z^2-4z+4\right]=-2\), that's correct. Finally I have to calculate the \(\displaystyle Res(f,1)\), single pole. I tried in some ways: by using the single pole formula and then using the "de l'hopital"'s theorem, by calculating the decomposed function in 0 and then equaling to the function in 0, but nothing was correct. Can you tell me a correct way to proceed? Thankyou

Put the partial fraction back over a common denominator:
$$
V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}=\frac{A(z-1)^2+[B+C(z-1)](z-3)}{(z-3)(z-1)^2}
$$

The coefficient of $z^2$ in the numerator is $A+C=1 $

.
 
  • #3
(z^2-4z+4)/(z-3)(z-1)^2 = 1/(z-3) - 2/(z-1)^2 + C/(z-1)

Now,plugging any number into z,say 0;

-4/3 = -1/3 -2 -C .Then,

C=-1
 

What is partial fraction decomposition?

Partial fraction decomposition is a mathematical process used to break down a rational function into smaller, simpler fractions. It involves rewriting a fraction as a sum of smaller fractions with different denominators.

Why is partial fraction decomposition useful?

Partial fraction decomposition is useful because it allows us to solve complex integrals and equations involving rational functions. It also helps us to better understand the behavior of these functions and identify any horizontal or vertical asymptotes.

What are the steps involved in partial fraction decomposition?

The first step is to factor the denominator of the rational function. Then, we determine the constants that will be in the numerator of each fraction. Next, we use the partial fractions to rewrite the original rational function. Finally, we solve for the constants using algebraic methods.

When is partial fraction decomposition not possible?

Partial fraction decomposition is not possible when the denominator of the rational function cannot be factored into linear or quadratic factors. In this case, the function is already in its simplest form and cannot be broken down into smaller fractions.

How is partial fraction decomposition related to complex numbers?

Partial fraction decomposition can be used to decompose rational functions with complex numbers in the denominator. This allows us to work with these functions in a more simplified form and make calculations easier. Additionally, the constants in the numerator can also be complex numbers.

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