Power series

The exponential function (in blue), and its improving approximation by the sum of the first n + 1 terms of its Maclaurin power series (in red). So
n=0 gives ${\displaystyle f(x)=1}$,
n=1 ${\displaystyle f(x)=1+x}$,
n=2 ${\displaystyle f(x)=1+x+x^{2}/2}$,
n=3 ${\displaystyle f(x)=1+x+x^{2}/2+x^{3}/6}$etcetera.

Any polynomial can be easily expressed as a power series around any center c, although all but finitely many of the coefficients will be zero since a power series has infinitely many terms by definition. For instance, the polynomial ${\textstyle f(x)=x^{2}+2x+3}$ can be written as a power series around the center ${\textstyle c=0}$ as

$${\displaystyle f(x)=3+2x+1x^{2}+0x^{3}+0x^{4}+\cdots }$$

or around the center ${\textstyle c=1}$ as

$${\displaystyle f(x)=6+4(x-1)+1(x-1)^{2}+0(x-1)^{3}+0(x-1)^{4}+\cdots }$$

This is because of the Taylor series expansion of f(x) around ${\textstyle x=1}$ is

$${\displaystyle f(x)=f(1)+{\frac {f'(1)}{1!}}(x-1)+{\frac {f''(1)}{2!}}(x-1)^{2}+{\frac {f'''(1)}{3!}}(x-1)^{3}+\cdots ,}$$

as ${\textstyle f(x=1)=1+2+3=6}$ and the non-zero derivatives are ${\textstyle f'(x)=2x+2}$, so ${\textstyle f'(1)=4}$ and ${\textstyle f''(x)=2}$, a constant.

Or indeed the expansion is possible around any other center c.[1] One can view power series as being like "polynomials of infinite degree," although power series are not polynomials.

The geometric series formula

$${\displaystyle {\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+x^{3}+\cdots ,}$$

which is valid for ${\textstyle |x|<1}$, is one of the most important examples of a power series, as are the exponential function formula

$${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots ,}$$

and the sine formula

$${\displaystyle \sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots ,}$$

valid for all real x.

These power series are also examples of Taylor series.

Negative powers are not permitted in a power series; for instance, ${\textstyle 1+x^{-1}+x^{-2}+\cdots }$ is not considered a power series (although it is a Laurent series). Similarly, fractional powers such as ${\textstyle x^{\frac {1}{2}}}$ are not permitted (but see Puiseux series). The coefficients ${\textstyle a_{n}}$ are not allowed to depend on ${\textstyle x}$, thus for instance:

$${\displaystyle \sin(x)x+\sin(2x)x^{2}+\sin(3x)x^{3}+\cdots }$$

is not a power series.

When two functions f and g are decomposed into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if

$${\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}(x-c)^{n}}$$

and

$${\displaystyle g(x)=\sum _{n=0}^{\infty }b_{n}(x-c)^{n}}$$

then

$${\displaystyle f(x)\pm g(x)=\sum _{n=0}^{\infty }(a_{n}\pm b_{n})(x-c)^{n}.}$$

It is not true that if two power series ${\textstyle \sum _{n=0}^{\infty }a_{n}x^{n}}$ and ${\textstyle \sum _{n=0}^{\infty }b_{n}x^{n}}$ have the same radius of convergence, then ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}}$ also has this radius of convergence. If ${\textstyle a_{n}=(-1)^{n}}$ and ${\textstyle b_{n}=(-1)^{n+1}\left(1-{\frac {1}{3^{n}}}\right)}$, then both series have the same radius of convergence of 1, but the series ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}}}x^{n}}$ has a radius of convergence of 3.

The sum of two power series will have, at minimum, a radius of convergence of the smaller of the two radii of convergence of the two series (and it may be higher than either, as seen in the example above).[2]

With the same definitions for ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$, the power series of the product and quotient of the functions can be obtained as follows:

$${\displaystyle {\begin{aligned}f(x)g(x)&=\left(\sum _{n=0}^{\infty }a_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\\&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }a_{i}b_{j}(x-c)^{i+j}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}a_{i}b_{n-i}\right)(x-c)^{n}.\end{aligned}}}$$

The sequence ${\textstyle m_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}}$ is known as the convolution of the sequences ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$.

For division, if one defines the sequence ${\displaystyle d_{n}}$ by

$${\displaystyle {\frac {f(x)}{g(x)}}={\frac {\sum _{n=0}^{\infty }a_{n}(x-c)^{n}}{\sum _{n=0}^{\infty }b_{n}(x-c)^{n}}}=\sum _{n=0}^{\infty }d_{n}(x-c)^{n}}$$

then

$${\displaystyle f(x)=\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }d_{n}(x-c)^{n}\right)}$$

and one can solve recursively for the terms ${\displaystyle d_{n}}$ by comparing coefficients.

Solving the corresponding equations yields the formulae based on determinants of certain matrices of the coefficients of ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$

$${\displaystyle d_{0}={\frac {a_{0}}{b_{0}}}}$$

$${\displaystyle d_{n}={\frac {1}{b_{0}^{n+1}}}{\begin{vmatrix}a_{n}&b_{1}&b_{2}&\cdots &b_{n}\\a_{n-1}&b_{0}&b_{1}&\cdots &b_{n-1}\\a_{n-2}&0&b_{0}&\cdots &b_{n-2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{0}&0&0&\cdots &b_{0}\end{vmatrix}}}$$

Once a function ${\displaystyle f(x)}$ is given as a power series as above, it is differentiable on the interior of the domain of convergence. It can be differentiated and integrated quite easily, by treating every term separately:

$${\displaystyle {\begin{aligned}f'(x)&=\sum _{n=1}^{\infty }a_{n}n(x-c)^{n-1}=\sum _{n=0}^{\infty }a_{n+1}(n+1)(x-c)^{n},\\\int f(x)\,dx&=\sum _{n=0}^{\infty }{\frac {a_{n}(x-c)^{n+1}}{n+1}}+k=\sum _{n=1}^{\infty }{\frac {a_{n-1}(x-c)^{n}}{n}}+k.\end{aligned}}}$$

Both of these series have the same radius of convergence as the original one.

The sum of a power series with a positive radius of convergence is an analytic function at every point in the interior of the disc of convergence. However, different behavior can occur at points on the boundary of that disc. For example:

1. Divergence while the sum extends to an analytic function: ${\textstyle \sum _{n=0}^{\infty }z^{n}}$ has radius of convergence equal to ${\displaystyle 1}$ and diverges at every point of ${\displaystyle |z|=1}$. Nevertheless, the sum in ${\displaystyle |z|<1}$ is ${\textstyle {\frac {1}{1-z}}}$, which is analytic at every point of the plane except for ${\displaystyle z=1}$.
2. Convergent at some points divergent at others: ${\textstyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n}}}$ has radius of convergence ${\displaystyle 1}$. It converges for ${\displaystyle z=-1}$, while it diverges for ${\displaystyle z=1}$.
3. Absolute convergence at every point of the boundary: ${\textstyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n^{2}}}}$ has radius of convergence ${\displaystyle 1}$, while it converges absolutely, and uniformly, at every point of ${\displaystyle |z|=1}$ due to Weierstrass M-test applied with the hyper-harmonic convergent series ${\textstyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$.
4. Convergent on the closure of the disc of convergence but not continuous sum: Sierpiński gave an example[3] of a power series with radius of convergence ${\displaystyle 1}$, convergent at all points with ${\displaystyle |z|=1}$, but the sum is an unbounded function and, in particular, discontinuous. A sufficient condition for one-sided continuity at a boundary point is given by Abel's theorem.