In general, the original expression may have a polynomial on both the numerator and the denominator. If the degree of the numerator is equal or greater than ...
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Partial Fractions We can write an expression such as
in a simplified form
.
Working, backwards, we can easily show that the expressions are equivalent:
(Note the indentically equals sign ( is used because the expressions are the same for all values of , not simply particular values of .) In carrying out this process, the original expression partial fractions:
is said to be expressed as
.We need a method for obtaining the partial fractions, given the
original expression and this will be covered in this document. The first step is to express the denominator in its most simple factorised form, if it is not in this form already. For example, given the expression it in the form
the first step would be to write
and then determine the partial fractions.
In general, the original expression may have a polynomial on both the numerator and the denominator. If the degree of the numerator is equal or greater than that of the denominator then it will b possible to divide the numerator by the denominator, leaving a polynomial expression summed with an expression consisting of a polynomial numerator of lower degree to the polynomial denominator. For example, given the expression
, the degree of
the numerator is equal to the degree of the denominator. For example =
.
For the purposes of this document, and without loss of generality, it will be assumed that the numerator has a lower degree than the denominator. Note that there are three classes of partial fraction problems, that will be considered separately: 1. Linear, Unrepeated Factors, 2. Linear Repeated Factors, 3. Quadratic Factors. In each case there are two basic techniques that can be applied: the cover up rule and equating coefficients, sometimes both techniques are required for the most efficient means of solution. 1. Linear, Unrepeated Factors The example above
consists of two linear, unrepeated factors in the denominator.
In this most straightforward case we first write the partial fraction in the required form
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with
and
being constants that we need to determine.
With a common denominator (always the same as the original denominator) we can write the partial fractions expression in the following form
Hence we must have:
Since the denominators are equal then the numerators must be, so
The values of and can be determined by either the cover up rule or by equating coefficients, or a combination of the two. For illustration, let us apply both methods separately. Cover-up Rule This involves choosing convenient values of that reduces the number of unknowns ( eg A,B etc.) so that their values can be more easily determined. For example if we choose then the expression above becomes ‘B’ is removed from the equation and the value of A is determined: Similarly, if we choose then the expression above becomes and hence
.
Equating Coefficients This involves equating the coefficients of A,B(etc.). In the example above
Hence
(etc.)and constants to determine values for
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Equating coefficients in x: Equating coefficients in constants: In solving the two simultaneous equations it follows that
and
.
We have found A and B but A and B have only been introduced as a means to find the partial fractions. The question is to express
as partial fractions and a suitable answer
would be:
2. Linear, Repeated Factors If we start off with an expression of the form
then we write as partial fractiond in
the following way
where the squared factor occurs in one fraction and then it occurs squared in another. Note that if there were higher powers then we would step through from the linear factor, up through all the powers in the same way. We now need to determine the values of A,B and C to completely determine the partial fraction expansion. We proceed as before, recreating the original denominator from the partial fractions:
It then follows that the numerators of this and the original expression must be the same:
Using the cover-up rule, Letting : Letting : In order to determine A we could either choose any value of x, or we could use equating coefficients. Doing the former of these
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Letting
:
, hence
The solution is
3. Quadratic Factors If the denominator contains a quadratic or higher order factor, that cannot be factorised further then the partial fraction is written as follows:
It follows that
and that
Using the cover-up rule, putting
:
or
.
In order to find A and B, we could either choose other values of x or equate coefficients. For example setting x=0 gives Equating coefficients of
gives
. Since . Since
Hence, the complete partial fraction expansions is
then then
. .
