partial fraction decomposition

The general formula would be: \[\frac{p(x)}{q(x)}=\frac{A_{1}}{(a_{1}x+b_{1})}+\frac{A_{2}}{(a_{2}x+b_{2})}+\cdots+\frac{A_{n}}{(a_{n}x+b_{n})}\].
An example of such a case would be \[\frac{3x-1}{(x+2)(x-1)}\]. Using the general formula, we decompose it into \[\frac{A}{x+2}+\frac{B}{x-1}\].

\begin{align*} \frac{3x-1}{(x+2)(x-1)}&=\frac{A}{x+2}+\frac{B}{x-1} \\ 3x-1&=A(x-1)+B(x+2) \\ &=Ax-A+Bx+2B \\ &=(A+B)x-A+2B \\ \end{align*}

Therefore we get \[A+B=3\] and \[-A+2B=-1\] and we get \[A=\frac{7}{3}\] and \[B=\frac{2}{3}\]. Thus the partial fraction decomposition would be \[\frac{3x-1}{(x+2)(x-1)}=\frac{7}{3(x+2)}+\frac{2}{3(x-1)}\].

An example of such a case would be \[\frac{-x^{2}+2x+4}{x(x-2)^{2}}\]. Using the general formula, we can decompose it into \[\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}\].

\begin{align*} \frac{-x^{2}+2x+4}{x(x-2)^{2}}&=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}} \\ -x^{2}+2x+4&=A(x-2)^{2}+Bx(x-2)+Cx \\ &=Ax^{2}-4Ax+4A+Bx^{2}-2Bx+Cx \\ &=(A+B)x^{2}+(-4A-2B+C)x+4A \\ \end{align*}

Therefore we get \[A+B=-1\], \[-4A-2B+C=2\], \[4A=4\] and we will get \[A=1\], \[B=-2\] and \[C=2\]. Thus the partial fraction decomposition would be \[\frac{-x^{2}+2x+4}{x(x-2)^{2}}=\frac{1}{x}-\frac{2}{x-2}+\frac{2}{(x-2)^{2}}\].

It is to note that \[p(x)\] must be one degree less than \[q(x)\], e.g. \[\frac{1}{x^{2}+1}=\frac{Ax+B}{x^{2}+1}\] or \[\frac{1}{x^{3}+1}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+x-1}\], to account for all polynomial terms.

Additionally, the reason why \[\frac{Ax^{2}+Bx+C}{x^{3}}\] is that it is the sum of \[\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}\].

The general formula for polynomial division is:

\begin{align*} &\overbrace{\left( ax^{k+n}+f \right)}^{\rm dividend}-\frac{a}{b}x^{k}\overbrace{\left( bx^{n}+g \right)}^{\rm divisor}=f-\frac{a}{b}x^{k}g\\ \implies& \frac{ax^{k+n}+f}{bx^{n}+g}=\frac{a}{b}x^{k}+\underbrace{\frac{f-\frac{a}{b}x^{k}g}{bx^{n}+g}}_{\text{recurse on this}}\\ \end{align*}

Referencing to the above, for the first iteration, we would first identify the dividend, \[f(x)=3x^{5}+7x^{4}+8x^{2}+3\] and the divisor, \[d(x)=x^{2}-7\]. Then we separate the dividend into \[(3x^{5})+(7x^{4}+8x^{2}+3)\], thus \[ax^{k+n}=3x^{5}\] and \[f=7x^{4}+8x^{2}+3\]. From here, we know \[a=3\] and \[k+n=5\]. Next, we separate our divisor into \[(x^{2})+(-7)\], which tells us that \[bx^{n}=x^{2}\] and \[g=-7\]. By comparison, \[b=1\] and \[n=2\]. Now that we have \[n\], we can calculate \[k=3\]. To use the formula, we just need to calculate our scaling factor \[\frac{a}{b}x^{k}\], which is \[\frac{3}{1}x^{3}\] or \[3x^{3}\]. This matches our value obtained by long division above. To recursively find the next value, we just take:

\begin{align*} \frac{f-\frac{a}{b}x^{k}g}{bx^{n}+g}=\frac{(7x^{4}+8x^{2}+3)-\frac{3}{1}x^{3}(-7)}{1x^{2}+(-7)}=\frac{7x^{4}+21x^{3}+8x^{2}+3}{x^{2}-7} \end{align*}

and repeat this entire process until the degree of the dividend is smaller than the divisor.

• techniques of integration
  
• summation
  
• asymptote