High School Mathematics Extensions/Supplementary/Partial Fractions

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Method of Partial Fractions[edit]

All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision.

Introduction[edit]

Before we begin, consider the following: \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}......+\frac{1}{99\times100}

How do we calculate this sum? At first glance it may seem difficult, but if you use variables instead of numbers each term in the sum above would take the form:

\frac{1}{n\times(n + 1)}

which you can rewrite as

\frac{(n + 1) - n}{n\times(n + 1)} = \frac{(n+1)}{n\times(n + 1)}-\frac{n}{n\times(n + 1)} = \frac{1}{n}-\frac{1}{n+1}

Thus we can rewrite the original problem as follows:

(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+......+(\frac{1}{99}-\frac{1}{100})

We can regroup this sum as:

\frac{1}{1}+(-\frac{1}{2} + \frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+....+(-\frac{1}{99}+\frac{1}{99})-\frac{1}{100})

So all terms except the first and the last cancel out giving us:

\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}......+\frac{1}{99\times100}=1-\frac{1}{100}=\frac{99}{100}

In fact, you've just done partial fractions!

Partial fractions is a method of breaking down complex fractions that involve products into sums of simpler fractions.

Method[edit]

So, how do we do partial fractions? Look at the example below:
\frac{4z-5}{z^2-3z+2}

Factorize the denominator.
\frac{4z-5}{(z-1)(z-2)}

Then we suppose we can break it down into the fractions with denominator (z-1) and (z-2) respectively. We let their numerators be a and b.
\frac{4z-5}{(z-1)(z-2)} \equiv \frac{a}{z-1}+\frac{b}{z-2}

\frac{4z-5}{(z-1)(z-2)} \equiv \frac{a(z-2)}{(z-1)(z-2)} + \frac{b(z-1)}{(z-1)(z-2)}

\frac{4z-5}{(z-1)(z-2)} \equiv \frac{az-2a+bz-b}{(z-1)(z-2)}

\frac{4z-5}{(z-1)(z-2)} \equiv \frac{(a+b)z-(2a+b)}{(z-1)(z-2)}

4z-5 \equiv (a+b)z-(2a+b)

Therefore by matching coefficients of like power of z, we have:


\begin{cases}
a+b=4 & ...(1) \\
2a+b=5 & ...(2)
\end{cases}

(2)-(1):a=1

Substitute a=1 into (1):b=3

Therefore
\frac{4z-5}{z^2-3z+2}=\frac{1}{z-1}+\frac{3}{z-2}

(Need Exercises!)

More on partial fraction[edit]

Repeated factors[edit]

On the last section we have talked about factorizing the denominator, and have each factor as the denominators of each term. But what happens when there are repeating factors? Can we apply the same method? See the example below:

\frac{4x-1}{(x+2)^2(x-1)}

\equiv \frac{A}{x+2} + \frac{B}{x+2} + \frac{C}{x-1}

\equiv \frac{A+B}{x+2} + \frac{C}{x-1}

\equiv \frac{(A+B)(x-1)}{(x+2)(x-1)} + \frac{C(x+2)}{(x+2)(x-1)}

\equiv \frac{(A+B)(x-1)+C(x+2)}{(x+2)(x-1)}

\equiv \frac{(A+B+C)x+(2C-A-B)}{(x+2)(x-1)}

Indeed, a factor is missing! Can we multiply both the denominator and the numerator by that factor? No! Because the numerator is of degree 1, multiplying with a linear factor will make it become degree 2! (You may think:can't we set A+B+C=0? Yes, but by substituting A+B=-C, you will find out that this is impossible)

From the above failed example, we see that the old method of partial fraction seems not to be working. You may ask, can we actually break it down? Yes, but before we finally attack this problem, let's look at the denominators at more detail.

Consider the following example:
\frac{1}{2^{3}7^2} + \frac{1}{2^{5}7} =\frac{2^2}{2^{5}7^2} + \frac{7}{2^{5}7^2} =\frac{2^2 + 7}{2^{5}7^2}
We can see that the power of a prime factor in the product denominator is the maximum power of that prime factor in all term's denominator.

Similarly, let there be factor P_1,P_2,...,P_n, then we may have in general case:
\frac{A}{P_1^{\alpha_1}P_2^{\alpha_2}...P_n^{\alpha_n}} +
\frac{B}{P_1^{\beta_1}P_2^{\beta_2}...P_n^{\beta_n}} + ...
\frac{Z}{P_1^{\zeta_1}P_2^{\zeta_2}...P_n^{\zeta_n}}
If we turn it into one big fraction, the denominator will be:
P_1^{max(\alpha_1,\beta_1,...,\zeta_1)}
P_2^{max(\alpha_2,\beta_2,...,\zeta_2)}...
P_n^{max(\alpha_n,\beta_n,...,\zeta_n)}

Back to our example, since the factor (x+2) has a power of 2, at least one of the term has (x+2)^2 as the denominator's factor. You may then try as follows:

\frac{4x-1}{(x+2)^2(x-1)}

\equiv \frac{A}{(x+2)^2} + \frac{B}{x-1}

\equiv \frac{A(x-1)}{(x+2)^2(x-1)} + \frac{B(x+2)^2}{(x+2)^2(x-1)}

\equiv \frac{A(x-1) + B(x+2)^2}{(x+2)^2(x-1)}

\equiv \frac{Ax - A + Bx^2 + 4Bx + 4B}{(x+2)^2(x-1)}

\equiv \frac{Bx^2 + (A+4B)x + (4B-A)}{(x+2)^2(x-1)}

But again, we can't set B=0, since that would means the latter term is 0! What is missing? To handle it properly, let's use a table to show all possible combinations of the denominator:

Possible combinations of denominator
Power of (x+2) Power of (x-1) Result Used?
0 0 1 Not useful
1 0 (x+2) Not used
2 0 (x+2)^2 Used
0 1 (x-1) Used
1 1 (x+2)(x-1) Not useful
2 1 (x+2)^2(x-1) Not useful

So, we now know that X/(x+2) is missing, we can finally happily get the answer:

\frac{4x-1}{(x+2)^2(x-1)}

\equiv \frac{A}{(x+2)^2} + \frac{B}{x+2} + \frac{C}{x-1}

\equiv \frac{A(x-1)}{(x+2)^2(x-1)} + \frac{B(x+2)(x-1)}{(x+2)^2(x-1)} + \frac{C(x+2)^2}{(x+2)^2(x-1)}

\equiv \frac{A(x-1) + B(x^2+x-2) + C(x^2+4x+4)}{(x+2)^2(x-1)}

\equiv \frac{(B+C)x^2+(A+B+4C)x-(A+2B-4C)}{(x+2)^2(x-1)}

Therefore by matching coefficient of like power of x, we have


As a conclusion, for a repeated factor of power n, we will have n terms with their denominator being X^n, X^(n-1), ...,X^2, X

Works continuing, don't distrub :)

Alternate method for repeated factors[edit]

Other than the method suggested above, we would like to use another approach to handle the problem. We first leave out some factor to make it into non-repeated form, do partial fraction on it, then multiply the factor back, then apply partial fraction on the 2 fractions.

\frac{4x-1}{(x+2)^2(x-1)}

\equiv \frac{1}{x+2} \times \frac{4x-1}{(x+2)(x-1)}

Then we do partial fraction on the latter part:

\frac{4x-1}{(x+2)(x-1)} \equiv \frac{A}{x+2} + \frac{B}{x-1}

\frac{4x-1}{(x+2)(x-1)} \equiv \frac{A(x-1)}{(x+2)(x-1)} + \frac{B(x+2)}{(x+2)(x-1)}

\frac{4x-1}{(x+2)(x-1)} \equiv \frac{A(x-1)+B(x+2)}{(x+2)(x-1)}

\frac{4x-1}{(x+2)(x-1)} \equiv \frac{(A+B)x+(2B-A)}{(x+2)(x-1)}

4x-1 \equiv (A+B)x + (2B-A)

By matching coefficients of like powers of x, we have


\begin{cases}
A+B = 4 & ...(1) \\
2B-A = -1 & ...(2)
\end{cases}

Substitute A=4-B into (2),

2B-(4-B) = -1

Hence B = 1 and A = 3.

We carry on:

\equiv \frac{1}{x+2} \times \left ( \frac{3}{x+2} + \frac{1}{x-1} \right )

\equiv \frac{3}{(x+2)^2} + \frac{1}{(x+2)(x-1)}

Now we do partial fraction once more:

\frac{1}{(x+2)(x-1)} \equiv \frac{A}{x+2} + \frac{B}{x-1}

\frac{1}{(x+2)(x-1)} \equiv \frac{A(x-1)}{(x+2)(x-1)} + \frac{B(x+2)}{(x+2)(x-1)}

\frac{1}{(x+2)(x-1)} \equiv \frac{A(x-1)+B(x+2)}{(x+2)(x-1)}

\frac{1}{(x+2)(x-1)} \equiv \frac{(A+B)x+(2B-A)}{(x+2)(x-1)}

0x + 1 \equiv (A+B)x + (2B-A)

By matching coefficients of like powers of x , we have:


\begin{cases}
A+B = 0 & ...(1) \\
2B-A = 1 & ...(2)
\end{cases}

Substitute A=-B into (2), we have:

2B-(-B) = 1

Hence B=1/3 and A=-1/3

So finally,

\frac{4x-1}{(x+2)^2(x-1)} \equiv \frac{3}{(x+2)^2} - \frac{1}{3(x+2)} + \frac{1}{3(x-1)}