Partial fractions
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Partial fractions
Sometimes it is useful to express a single fraction such as the sum of 2 (or more in other cases) separate fractions.
This is called decomposing a function, f(x), in partial fractions.
Consider this example:
This equation can be split into the sum of two single fractions.
Therefore
To find the value of the constants A and B depends on the factors in the denominator. Follow the examples below.
Using the example above:
Adding the two fractions on the right hand side gives:
As the denominators are now the same the numerators must match as well. Therefore:
If we choose to substitute x = -1, this will eliminate the constant B. This gives:
-2 − 1 = A (-3 + 2) |
-3 = -A |
A = 3 |
If we choose to substitute x = -2/3 to this will eliminate the constant A. This gives:
-2/3 × 2 − 1 = B (-2/3 + 1) |
-4/3 − 1 = B (1/3) |
-7/3 = 1/3 B |
-7 = B |
B = -7 |
Therefore:
Lets take the example:
As this has a quadratic factor in the denominator we need to write the partial equations in the following format:
Once again we need to add the two fractions on the right hand side. This gives:
Once again the denominators are now the same, so the numerators must be the same as well. Therefore:
By substituting x = 0, we eliminate B and C. This gives:
0^{2} + 3 = A (0^{2} + 2) |
3 = 2A |
A = 3/2 |
We can substitute two other values for x and solve B and C by simultaneous equations.
Consider this example:
The denominator contains (x − 2)^{2}. Since, (x − 2)^{2} = (x − 2 )(x − 2) it is called a repeated factor. When there is a repeated factor in the denominator we need to write the partial equations in the following format:
Once again we add the fractions on the right hand side together and thus the denominators are the same, so the numerators are the same, giving:
We again proceed by substitution to obtain the values of the constants.