High School Mathematics Extensions/Supplementary/Partial Fractions

High School Mathematics Extensions

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Supplementary Chapters — Primes and Modular Arithmetic — Logic

Mathematical Proofs — Set Theory and Infinite Processes — Counting and Generating Functions — Discrete Probability

Matrices — Further Modular Arithmetic — Mathematical Programming — Markov Chains

Supplementary Chapters
Content
Basic Counting
Polynomial Division
Partial Fractions
Summation Sign
Complex Numbers
Differentiation
Problems & Projects
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Solutions
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Problem Set Solutions

Introduction

Before we begin, consider the following:

${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}......+{\frac {1}{99\times 100}}$

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:

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

Regrouping:

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

So all terms except the first and the last cancel out, yielding:

${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}......+{\frac {1}{99\times 100}}=1-{\frac {1}{100}}={\frac {99}{100}}$

Believe it or not, we've just done partial fractions!

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

General method

So, how do we do partial fractions? Look at the example below:

${\frac {4z-5}{z^{2}-3z+2}}$

Factorizing the denominator:

${\frac {4z-5}{(z-1)(z-2)}}$

Then we suppose we can break it down into two fractions with denominator (z - 1) and (z - 2) respectively, let their numerators be a and b:

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

Multiplying by (z - 1)(z - 2):

$4z-5=a(z-2)+b(z-1)$

$4z-5=az-2a+bz-b$

$4z-5=(a+b)z-(2a+b)$

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

${\begin{cases}a+b=4&({\text{A}})\\2a+b=5&({\text{B}})\end{cases}}$

Furthermore:

$({\text{B}})-({\text{A}}):a=1\to b=3$

Therefore

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

Generally speaking, this method only works with proper fractions. Numerators with greater exponents than their denominators need to be divided first.

1. Can you find an equivalent expression to $y={\frac {3p-21}{(p-5)p-14}}$ that is defined for $y={\frac {1}{3}}$ ?
TIP: Here, "defined" means having some value p for which the equation yields 1/3.

Power factors

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

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

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.

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

Partial fraction on the latter part:

${\frac {4x-1}{(x+2)(x-1)}}={\frac {a}{x+2}}+{\frac {b}{x-1}}$

Multiplying by (x + 2)(x - 1):

$4x-1=a(x-1)+b(x+2)$

$4x-1=a(x-1)+b(x+2)$

$4x-1\equiv (a+b)x+(2b-a)$

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

${\begin{cases}a+b=4&({\text{A}})\\2b-a=-1&({\text{B}})\end{cases}}$

Substitute a = 4 - b into (B),

$2b-(4-b)=-1$

Hence b = 1 and a = 3.

We carry on:

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

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

Now we do partial fraction once more:

${\frac {1}{(x+2)(x-1)}}={\frac {a}{x+2}}+{\frac {b}{x-1}}$

Multiplying by (x + 2)(x - 1):

$1=a(x-1)+b(x+2)$

$1=(a+b)x+(2b-a)$

$0x+1=(a+b)x+(2b-a)$

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

${\begin{cases}a+b=0&({\text{A}})\\2b-a=1&({\text{B}})\end{cases}}$

Substitute a = -b into (B), we have:

$2b-(-b)=1$

Hence b = 1/3 and a = -1/3.

So finally,

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

2. What about ${\frac {3q+2}{({\frac {3}{2}}q^{2}+{\frac {3}{2}}q-6)q-6}}$ ?

Quadratic factors

We should always try to factor the denominator, for the sake of simplicity. There are some cases, though, in which factoring a polynomial leads to complex coefficients. Since these do anything but simplify our task, we shall leave the polynomial as it is. That is, as irreducible quadratic factors:

${\frac {7x^{2}+22x+25}{x^{3}+3x^{2}+2x+6}}={\frac {7x^{2}+22x+25}{(x+3)(x^{2}+2)}}$

When dealing with the quadratic factor, we should use the following partial fraction:

${\frac {bx+c}{x^{2}+2}}$

Leading to:

${\frac {7x^{2}+22x+25}{(x+3)(x^{2}+2)}}={\frac {a}{x+3}}+{\frac {bx+c}{x^{2}+2}}$

Multiplying by (x + 3)(x^2 + 2):

$7x^{2}+22x+25=a(x^{2}+2)+(bx+c)(x+3)$

$7x^{2}+22x+25=a(x^{2}+2)+(bx+c)(x+3)$

$7x^{2}+22x+25=(a+b)x^{2}+(3b+c)x+(2a+3c)$

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

${\begin{cases}a+b=7&({\text{A}})\\3b+c=22&({\text{B}})\\2a+3c=25&({\text{C}})\end{cases}}$

Solving:

$({\text{C}})-2({\text{A}})-3({\text{B}})=({\text{C}})-2({\text{A}})-3({\text{B}})$

$(2a+3c)-2(a+b)-3(3b+c)=25-2(7)-3(22)$

$-11b=-55$

Therefore b = 5, a = 2 and c = 7.

Finally:

${\frac {7x^{2}+22x+25}{(x+3)(x^{2}+2)}}={\frac {2}{x+3}}+{\frac {5x+7}{x^{2}+2}}$

3. Try breaking down ${\frac {z^{2}+3z}{z^{3}+1}}$ .