Editing Rational function (section)

==Taylor series==
The coefficients of a [[Taylor series]] of any rational function satisfy a [[Recurrence relation|linear recurrence relation]], which can be found by equating the rational function to a Taylor series with indeterminate coefficients, and collecting [[like terms]] after clearing the denominator.

For example,

:<math>\frac{1}{x^2 - x + 2} = \sum_{k=0}^{\infty} a_k x^k.</math>

Multiplying through by the denominator and distributing,

:<math>1 = (x^2 - x + 2) \sum_{k=0}^{\infty} a_k x^k</math>

:<math>1 = \sum_{k=0}^{\infty} a_k x^{k+2} - \sum_{k=0}^{\infty} a_k x^{k+1} + 2\sum_{k=0}^{\infty} a_k x^k.</math>

After adjusting the indices of the sums to get the same powers of ''x'', we get

:<math>1 = \sum_{k=2}^{\infty} a_{k-2} x^k - \sum_{k=1}^{\infty} a_{k-1} x^k + 2\sum_{k=0}^{\infty} a_k x^k.</math>

Combining like terms gives

:<math>1 = 2a_0 + (2a_1 - a_0)x + \sum_{k=2}^{\infty} (a_{k-2} - a_{k-1} + 2a_k) x^k.</math>

Since this holds true for all ''x'' in the [[radius of convergence]] of the original Taylor series, we can compute as follows.  Since the [[constant term]] on the left must equal the constant term on the right it follows that

:<math>a_0 = \frac{1}{2}.</math>

Then, since there are no powers of ''x'' on the left, all of the [[coefficient]]s on the right must be zero, from which it follows that

:<math>a_1 = \frac{1}{4}</math>

:<math>a_k = \frac{1}{2} (a_{k-1} - a_{k-2})\quad \text{for}\ k \ge 2.</math>

Conversely, any sequence that satisfies a linear recurrence determines a rational function when used as the coefficients of a Taylor series. This is useful in solving such recurrences, since by using [[partial fraction|partial fraction decomposition]] we can write any proper rational function as a sum of factors of the form {{nowrap|1 / (''ax'' + ''b'')}} and expand these as [[geometric series]], giving an explicit formula for the Taylor coefficients; this is the method of [[generating functions]].