How to solve this partial fraction decomposition?

Question

Please help me to solve the following partial fraction decomposition: $$\frac{1-v^2}{v+v^3} = \frac{A}{v}+\frac{Bv+C}{1+v^2}$$

Answer 1

The usual method would be to multiply both sides of this equation by $v+v^3$:

$$\begin{align}\frac{1-v^2}{v+v^3} &= \frac{A}{v}+\frac{Bv+C}{1+v^2}\\ 1-v^2 &= A(1+v^2) + (Bv+C)(v)\\ 1-v^2 &= Av^2 + A + Bv^2 + Cv\\ -v^2 + 0v + 1 &= (A+B)v^2 + Cv + A \end{align}$$

At this point, you can equate coefficients on the two sides. Since the $v^2$ coefficients must match, we have $-1=A+B$. Since the $v$ coefficients must match, we have $0=C$, and since the constant terms must match, we have $1=A$.

Do you see how that's working?

Answer 2

Multiply by $v $ to get

$$\frac {1-v^2}{1+v^2}=A+\frac{Bv^2+Cv}{1+v^2} $$

and then replace $v $ by $0$, to get

$$1=A $$

then multiply by $1+v^2$ to obtain

$$\frac {1-v^2}{v}=\frac {A(1+v^2)}{v}+Bv+C $$ and replace $v $ by the complex number $i $, to find

$$\frac {2}{i}=Bi+C=-2i $$ thus

$$B=-2 \;\;,\;C=0$$

Answer 3

You want to find values for $A$, $B$, and $C$ so that $$\frac{A}{v}+\frac{Bv+C}{1+v^2}=\frac{1-v^2}{v+v^3}$$ or $$A(1+v^2)+Bv^2+Cv=1-v^2$$ First, start by setting $v=0$. Then we see that $$A(1+0)+B(0)+C(0)=1-(0)$$ $$A=1$$ Now we have $$1(1+v^2)+Bv^2+Cv=1-v^2$$ $$Bv^2+Cv=-2v^2$$ $$Bv+C=-2v$$ Again, when we let $v=0$, we see that $$B(0)+C=-2(0)$$ $$C=0$$ And then, of course, we see that $B=-2$ since $C=0$. Your final answer should be $$\frac{1}{v}-\frac{2v}{1+v^2}=\frac{1-v^2}{v+v^3}$$

Answer 4

If you play around up- and downstairs you find:

\begin{align*} \frac{1-v^2}{v + v^3} & = \frac{1-v^2 + v^2 - v^2}{v \, (1+v^2)} \\ & = \frac{1+v^2}{v(1+v^2)} + \frac{-2v^2}{v (1+v^2)} \\ & = \frac{1}{v} - \frac{2 v}{1+v^2} \end{align*}

Answer 5

$\begin{cases} \displaystyle f(v)=\frac{1-v^2}{v+v^3}\\ \displaystyle g(v)=\frac{A}{v}+\frac{Bv+C}{1+v^2}\end{cases}$

With many simple partial fractions where there are not too much coefficients, it is generally possible to evaluate these coefficients by plugging some well chosen values for $v$ including equivalents in the poles and/or at infinity because they give precious information.

• In neighbourhood of $0$ we have $f(v)\sim \frac 1v$ and $g(v)\sim \frac Av$ thus $A=1$.
• at infinity we have $f(v)\sim -\frac 1v$ and $g(v)\sim \frac Av+\frac Bv$ thus $B=-2$.
• Finally $f(1)=0$ and $g(1)=A+\frac{B+C}2=1-1+\frac C2$ thus $C=0$