4.8 Partial fractions
When we add fractions (here 77 7 Even though they’re in the set of real numbers, they don’t have to be written down directly as numbers (like or ) - they could be algebraic expressions (such as ), so long as when the value of those algebraic expressions is computed, the results are real numbers.), something like this happens
Sometimes we have fractions which are the "added together" form, which we’d like to turn into the "not added together" form. For example in the case of this fraction
We can break it apart. We can use Equation 4.71 to predict (if we set and , and ) that "splitting up" (aka partial fraction) will look like
To solve this, we take advantage of the fact that because the values of and are constants88 8 i.e. their value is always the same. This means that they must be the same for every value of , so if we plug in specific values of , and then find what and are given those values of we know that those values of and will be true for all values of .
We can also check the answer by adding the fractions back together again:
|Which is what we started out with!|
There’s another case, however, which we haven’t looked at yet; what if there’s a repeated root 99 9 Remember, from the factor theorem (see above if not) that roots, i.e. when a polynomial is zero, are also factors (e.g. is a factor of , and is a root.).
In this case we have something like
if we try to apply the previous technique, we do something like:
This is impossible to solve (because clearly values of and which satisfy one value of are not going to satisfy all values of ) - oops! The trick here is to do some clever algebra. We can rewrite in terms of (because ), and from there it becomes much easier.
The final thing is what happens when some other terms are mixed in, as in
We could try something like
however, this fails pretty quickly if we try to add the right-hand side back together
We know that when we add the fractions together, we should get as the numerator. Therefore the numerator of the partial fractions added together should be the same as the numerator of the original fraction.
We can "compare coefficients"1010 10 Note: this is explored further in the (not yet written) ”algebra techniques” section. and obtain that
Which simplifies to
If we substitute into the other equations, we get that
Which means that the equations are inconsistent 1111 11 i.e. are no solutions to the equations - more on inconsistent/consistent equations in the ”linear algebra” section (though that part of the ”linear algebra” section is yet to be written), and that there are no solutions satisfying all the equations at the same time.
Instead, we need to try to express this in a form similar to Equation 4.72, i.e. something like .
|If we set , then , and thus the whole left-hand part will be zero|
|Now find and|
There’s a caveat to all this, unfortunately! If the degree 1212 12 i.e. the highest power of the polynomial - e.g. for the degree would be and for it would be of the numerator is equal to or greater than the numerator, we first have to divide (using polynomial long division1313 13 Also in the yet-to-be-written algebra section) the numerator by the denominator to get an expression which isn’t a fraction, plus a remainder (where the degree of the numerator is less than the denominator).