Out of the blue I wrote down a rather confusing mass of indices and summations on the board a few days ago. Writing this down at the last minute was perhaps a bad idea, but I wanted to show what the general form for expanding a fraction into partial fractions was. Here I’m just motivating it a little more. It’s not something that you will need to use, but it’s often good to write things down in as general a form as you can.

Let’s say that we have an expression of the form:




Where P(x) is some polynomial of degree less than 3 (because the denominator is degree 3). We can write this as:




To find A,  B and C, you cross-multiply, and then match coefficients of powers of x with those in P(x).  If you have an irreducible quadratic in the denominator you will have terms of the form:




in your partial fraction, and of course if it’s an irreducible quadratic to an integer power greater than one, you will have multiple terms, just as you do for the (x-2) expression in the example above. Thus, you will have terms of the form:




and terms of the form:




where A, B and C are found by cross-multiplying and matching with the original expression, and the constants a, b, c and l are fixed by the original expression. Because we will have a number of different terms in our partial fraction expression, rather than writing new letters each time, we can simply put an index on each letter to say that it’s coming from a new expression. So, the most general form of our partial fraction is going to be:


\sum_{i=1}^m \frac{A_i}{(x+a_i)^{k_i}}+\sum_{k=1}^n\frac{B_ix+C_i}{(x^2+b_ix+c_i)^{l_i}}


where the a_i,b_i,c_i all come directly from the original expression. The k_i, l_i are integers found simply by having multiple terms for each higher power of linear and irreducible quadratic terms in the original expression (a term cubed will lead to a cubic, square and first order term in the partial fraction) and A_i, B_i, C_i are found again by cross-multiplying and matching coefficients in the numerators. The limits of the sum m and n will depend on the powers in the original expression.


The expression I wrote down on the board was simply this last expression above which gives the most general form for a partial fraction.


How clear is this post?