Editing Taylor series (section)

==Examples==
The Taylor series of any [[polynomial]] is the polynomial itself.

The Maclaurin series of {{math|{{sfrac|1|1 − ''x''}}}} is the [[geometric series]]

<math display="block">1 + x + x^2 + x^3 + \cdots.</math>

So, by substituting {{mvar|x}} for {{math|1 − ''x''}}, the Taylor series of {{math|{{sfrac|1|''x''}}}} at {{math|''a'' {{=}} 1}} is

<math display="block">1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots.</math>

By integrating the above Maclaurin series, we find the Maclaurin series of {{math|ln(1  − ''x'')}}, where {{math|ln}} denotes the [[natural logarithm]]:

<math display="block">-x - \tfrac{1}{2}x^2 - \tfrac{1}{3}x^3 - \tfrac{1}{4}x^4 - \cdots.</math>

The corresponding Taylor series of {{math|ln ''x''}} at {{math|''a'' {{=}} 1}} is

<math display="block">(x-1) - \tfrac{1}{2}(x-1)^2 + \tfrac{1}{3}(x-1)^3 - \tfrac{1}{4}(x-1)^4 + \cdots,</math>

and more generally, the corresponding Taylor series of {{math|ln ''x''}} at an arbitrary nonzero point {{mvar|a}} is:

<math display="block">\ln a + \frac{1}{a} (x - a) - \frac{1}{a^2}\frac{\left(x - a\right)^2}{2} + \cdots.</math>

The Maclaurin series of the [[exponential function]] {{math|''e''<sup>''x''</sup>}} is

<math display="block">\begin{align}
   \sum_{n=0}^\infty \frac{x^n}{n!} &= \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}+ \cdots \\
   &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots.
 \end{align}</math>

The above expansion holds because the derivative of {{math|''e''<sup>''x''</sup>}} with respect to {{mvar|x}} is also {{math|''e''<sup>''x''</sup>}}, and {{math|''e''<sup>0</sup>}} equals&nbsp;1. This leaves the terms {{math|(''x'' − 0)<sup>''n''</sup>}} in the numerator and {{math|''n''!}} in the denominator of each term in the infinite sum.