Power series

This article deals with power series used to describe real or complex functions. For formal power series see there.

In analysis, a power series is an infinite series of the form ${\displaystyle P(x)}$

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

With

• any sequence of real or complex numbers ${\displaystyle (a_{n})_{n\in \mathbb {N} _{0}}}$
• the development point of the power series. ${\displaystyle x_{0}}$

Power series play an important role in function theory and often allow a meaningful continuation of real functions in the complex number plane. In particular, the question arises for which real or complex numbers a power series converges. This question leads to the concept of the radius of convergence .

Radius of convergence

→ Main article : radius of convergence

As the radius of convergence of a power series about the point of development the largest number is defined for which the power series for all to converge . The open sphere with a radius um is called the convergence circle. The radius of convergence is the radius of the circle of convergence. If the series converges for all , the radius of convergence is said to be infinite. If it only converges for , then the radius of convergence is 0, and the series is then sometimes not called convergent anywhere . ${\displaystyle x_{0}}$ ${\displaystyle r}$ ${\displaystyle x}$ ${\displaystyle |x-x_{0}|<r}$ ${\displaystyle U_{r}(x_{0})}$ ${\displaystyle r}$ ${\displaystyle x_{0}}$ ${\displaystyle x}$ ${\displaystyle x_{0}}$

When power series is the radius of convergence can with the formula of Cauchy-Hadamard calculated. The following applies: ${\displaystyle r}$

${\displaystyle r={\frac {1}{\limsup \limits _{n\rightarrow \infty }\ {\sqrt[{n}]{|a_{n}|}}}}}$

In this context one defines and ${\displaystyle {\tfrac {1}{0}}:=+\infty }$ ${\displaystyle {\tfrac {1}{\infty }}:=0}$

In many cases the radius of convergence can also be calculated more easily for power series with non-vanishing coefficients. It is true

${\displaystyle r=\lim _{n\to \infty }\left|{\frac {a_{n}}{a_{n+1}}}\right|,}$

if this limit value exists.

Examples

Every polynomial function can be understood as a power series in which almost all coefficients are equal to 0. Other important examples are the Taylor series and Maclaurin series . Functions that can be represented by a power series are also called analytic functions . Here is an example of the power series representation of some known functions: ${\displaystyle a_{n}}$

• Exponential function : for all , i.e. i.e., the radius of convergence is infinite. ${\displaystyle e^{x}=\exp(x)=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\dotsb }$ ${\displaystyle x\in \mathbb {R} }$

• Sine : ${\displaystyle \sin(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{(2n+1)!}}={\frac {x}{1!}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}\mp \dotsb }$

• Cosine : ${\displaystyle \cos(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}={\frac {x^{0}}{0!}}-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}\mp \dotsb }$

The radius of convergence is infinite for both the sine and the cosine. The power series representation results directly from the exponential function using Euler's formula .

• Logarithm function : ${\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {x^{k}}{k}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\dotsb }$

for , d. h .: The radius of convergence is 1, for is the series convergent, for divergent. ${\displaystyle -1<x\leq 1}$ ${\displaystyle x=1}$ ${\displaystyle x=-1}$

• Root function : for , d. that is, the radius of convergence is 1 and the series converges both for and for . ${\displaystyle {\sqrt {1+x}}=1+{\frac {1}{2}}x-{\frac {1}{2\cdot 4}}x^{2}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}x^{3}\mp \dotsb }$ ${\displaystyle -1\leq x\leq 1}$ ${\displaystyle x=1}$ ${\displaystyle x=-1}$

properties

Power series are normally convergent within their convergence circle . From this it follows directly that every function defined by a power series is continuous. It also follows that there is uniform convergence on compact subsets of the convergence circle. This justifies the differentiation and integration of a power series in terms of terms and shows that power series are infinitely differentiable.

There is absolute convergence within the convergence circle . No general statement can be made about the behavior of a power series on the edge of the convergence circle, but in some cases the Abelian limit theorem allows a statement to be made.

The power series representation of a function around a development point is clearly determined ( identity theorem for power series ). In particular, the Taylor expansion is the only possible power series expansion for a given expansion point.

Operations with power series

Addition and scalar multiplication

Are and by two power series ${\displaystyle f}$ ${\displaystyle g}$

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

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

represented with the radius of convergence and is a fixed complex number, then and are at least expandable in power series with radius of convergence and the following applies: ${\displaystyle r}$ ${\displaystyle c}$ ${\displaystyle f+g}$ ${\displaystyle cf}$ ${\displaystyle r}$

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

${\displaystyle cf(x)=\sum _{n=0}^{\infty }(ca_{n})(x-x_{0})^{n}}$

multiplication

The product of two power series with the radius of convergence is a power series with a radius of convergence that is at least . Since there is absolute convergence inside the convergence circle, the Cauchy product formula applies : ${\displaystyle r}$ ${\displaystyle r}$

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

The sequence defined by is referred to as the convolution or convolution of the two sequences and . ${\displaystyle \textstyle c_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}}$ ${\displaystyle (c_{n})}$ ${\displaystyle (a_{n})}$ ${\displaystyle (b_{n})}$

Concatenation

Admit it and two power series ${\displaystyle f}$ ${\displaystyle g}$

${\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}(x-x_{1})^{n}}$

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

with positive radii of convergence and the property

${\displaystyle b_{0}=g(x_{0})=x_{1}}$.

Then the concatenation of both functions is locally again an analytic function and can therefore be developed into a power series: ${\displaystyle f\circ g}$ ${\displaystyle x_{0}}$

${\displaystyle (f\circ g)(x)=\sum _{n=0}^{\infty }c_{n}(x-x_{0})^{n}}$

According to Taylor's theorem:

${\displaystyle c_{n}={\frac {(f\circ g)^{(n)}(x_{0})}{n!}}}$

With Faà di Bruno's formula , this expression can now be given in a closed formula depending on the given series coefficients, since:

${\displaystyle {\begin{aligned}f^{(n)}(g(x_{0}))&=f^{(n)}(x_{1})\\&=n!\cdot a_{n}\\g^{(m)}(x_{0})&=m!\cdot b_{m}\end{aligned}}}$

With multi-index notation you get :

${\displaystyle {\begin{aligned}c_{n}&={\frac {(f\circ g)^{(n)}(x_{0})}{n!}}\\&=\sum _{{\boldsymbol {k}}\in T_{n}}{\frac {f^{(|{\boldsymbol {k}}|)}(g(x_{0}))}{{\boldsymbol {k}}!}}\prod _{m=1 \atop k_{m}\geq 1}^{n}\left({\frac {g^{(m)}(x_{0})}{m!}}\right)^{k_{m}}\\&=\sum _{{\boldsymbol {k}}\in T_{n}}{\frac {|{\boldsymbol {k}}|!\cdot a_{|{\boldsymbol {k}}|}}{{\boldsymbol {k}}!}}\prod _{m=1 \atop k_{m}\geq 1}^{n}b_{m}^{k_{m}}\\&=\sum _{{\boldsymbol {k}}\in T_{n}}{{|{\boldsymbol {k}}|} \choose {\boldsymbol {k}}}\,a_{|{\boldsymbol {k}}|}\prod _{m=1 \atop k_{m}\geq 1}^{n}b_{m}^{k_{m}}\end{aligned}}}$

It is the multinomial to and is the set of all partitions (see Partition function ). ${\displaystyle {{|{\boldsymbol {k}}|} \choose {\boldsymbol {k}}}}$ ${\displaystyle {\boldsymbol {k}}}$ ${\displaystyle T_{n}=\left\{{\boldsymbol {k}}\in \mathbb {N} _{0}^{n}\,{\Big |}\,\sum _{j=1}^{n}j\cdot k_{j}=n\right\}}$ ${\displaystyle n}$

Differentiation and Integration

A power series is differentiable inside its convergence circle and the derivation results from differentiation in terms of terms:

${\displaystyle f^{\prime }(x)=\sum _{n=1}^{\infty }a_{n}n\left(x-x_{0}\right)^{n-1}=\sum _{n=0}^{\infty }a_{n+1}\left(n+1\right)\left(x-x_{0}\right)^{n}}$

It can be differentiated as often as required and the following applies: ${\displaystyle f}$

${\displaystyle f^{(k)}(x)=\sum _{n=k}^{\infty }{\frac {n!}{(n-k)!}}a_{n}(x-x_{0})^{n-k}=\sum _{n=0}^{\infty }{\frac {(n+k)!}{n!}}a_{n+k}(x-x_{0})^{n}}$

Similarly, an antiderivative is obtained by integrating a power series in terms of terms:

${\displaystyle \int f(x)\,{\text{d}}x=\sum _{n=0}^{\infty }{\frac {a_{n}\left(x-x_{0}\right)^{n+1}}{n+1}}+C=\sum _{n=1}^{\infty }{\frac {a_{n-1}\left(x-x_{0}\right)^{n}}{n}}+C}$

In both cases the radius of convergence is the same as that of the original series.

Representation of functions as power series

Often one is interested in a power series representation for a given function - especially to answer the question whether the function is analytical . There are several strategies for finding a power series representation, the most general using the Taylor series . Here, however, the problem often arises that a closed representation is required for the derivatives, which is often difficult to determine. However , there are some easier strategies for broken rational functions . The function

${\displaystyle f(z)={\frac {z^{2}}{z^{2}-4z+3}}}$

to be viewed as.

By means of the geometric series

By factoring the denominator and then using the formula for the sum of a geometric series , you get a representation of the function as the product of infinite series:

${\displaystyle {f(z)={\frac {z^{2}}{(1-z)(3-z)}}={\frac {z^{2}}{3}}\cdot {\frac {1}{1-z}}\cdot {\frac {1}{1-{\frac {z}{3}}}}={\frac {z^{2}}{3}}\cdot \left(\sum _{n=0}^{\infty }z^{n}\right)\cdot \left(\sum _{n=0}^{\infty }\left({\frac {z}{3}}\right)^{n}\right)={\frac {1}{3}}\left(\sum _{n=2}^{\infty }z^{n}\right)\left(\sum _{n=0}^{\infty }\left({\frac {z}{3}}\right)^{n}\right)}}$

Both series are power series around the point of expansion and can therefore be multiplied in the manner mentioned above. The Cauchy product formula delivers the same result ${\displaystyle z_{0}=0}$

${\displaystyle f(z)=\sum _{n=0}^{\infty }z^{n}\textstyle \sum _{k=0}^{n}a_{k}b_{n-k}}$

With

${\displaystyle a_{k}={\begin{cases}0&{\text{ für }}k\in \{0,1\}\\1&{\text{ sonst}}\end{cases}}}$

and

${\displaystyle b_{k}={\frac {1}{3^{k}}}.}$

From this it follows by applying the formula for the partial sum of a geometric series

${\displaystyle \sum _{k=0}^{n}a_{k}b_{n-k}=\sum _{k=2}^{n}\left({\frac {1}{3}}\right)^{n-k}={\frac {1}{3^{n-2}}}\sum _{k=0}^{n-2}3^{k}=-{\frac {1-3^{n-1}}{2\cdot 3^{n-2}}}}$

as a closed representation for the coefficient sequence of the power series. The power series representation of the function around the development point 0 is thus given by

${\displaystyle f(z)=\sum _{n=2}^{\infty }{\frac {1}{2}}\cdot \left(1-{\frac {1}{3^{n-1}}}\right)\cdot z^{n}}$.

By comparison of coefficients

Often the way over the geometric series is cumbersome and error-prone. Therefore the following approach is appropriate: It is assumed that a power series representation

${\displaystyle f(z)={\frac {z^{2}}{z^{2}-4z+3}}=\sum _{n=0}^{\infty }b_{n}z^{n}}$

the function with an unknown coefficient sequence exists. After multiplying the denominator and shifting the index, the identity results: ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$

${\displaystyle {\begin{aligned}z^{2}&=(z^{2}-4z+3)\sum _{n=0}^{\infty }b_{n}z^{n}\\&=\sum _{n=2}^{\infty }b_{n-2}z^{n}-\sum _{n=1}^{\infty }4b_{n-1}z^{n}+\sum _{n=0}^{\infty }3b_{n}z^{n}\\&=3b_{0}+z(3b_{1}-4b_{0})+\sum _{n=2}^{\infty }(b_{n-2}-4b_{n-1}+3b_{n})z^{n}\end{aligned}}}$

However, since two power series are exactly the same when their coefficient sequences match, this results from coefficient comparison

${\displaystyle b_{0}=0,\ b_{1}=0,\ b_{2}={\frac {1}{3}}}$

and the recursion equation

${\displaystyle b_{n}={\frac {4b_{n-1}-b_{n-2}}{3}}}$,

from which the above closed representation follows by means of complete induction.

The procedure by means of coefficient comparison also has the advantage that other development points are possible than are possible. As an example, consider the development point . First, the fractional rational function has to be represented as a polynomial in : ${\displaystyle z_{0}=0}$ ${\displaystyle z_{1}=-1}$ ${\displaystyle (z-z_{1})=(z+1)}$

${\displaystyle f(z)={\frac {z^{2}}{z^{2}-4z+3}}={\frac {(z+1)^{2}-2(z+1)+1}{(z+1)^{2}-6(z+1)+8}}}$

Analogous to the above one now assumes that a formal power series exists around the expansion point with an unknown coefficient sequence and multiplied with the denominator by:

${\displaystyle {\begin{aligned}(z+1)^{2}-2(z+1)+1&=((z+1)^{2}-6(z+1)+8)\sum _{n=0}^{\infty }b_{n}(z+1)^{n}\\&=8b_{0}+(z+1)(8b_{1}-6b_{0})+\sum _{n=2}^{\infty }(b_{n-2}-6b_{n-1}+8b_{n})(z+1)^{n}\end{aligned}}}$

Again this results from a coefficient comparison

${\displaystyle b_{0}={\frac {1}{8}},\ b_{1}=-{\frac {5}{32}},\ b_{2}=-{\frac {1}{128}}}$

and as a recursion equation for the coefficients:

${\displaystyle b_{n}={\frac {-b_{n-2}+6b_{n-1}}{8}}}$

By partial fraction decomposition

If you first apply polynomial division and then partial fraction decomposition to the given function , you get the representation

${\displaystyle f(z)={\frac {z^{2}}{z^{2}-4z+3}}=1+{\frac {4z-3}{(z-1)(z-3)}}=1+{\frac {1}{2}}\cdot {\frac {1}{1-z}}-{\frac {3}{2}}\cdot {\frac {1}{1-{\frac {z}{3}}}}}$.

Inserting the geometric series results in:

${\displaystyle {f(z)=1+{\frac {1}{2}}\cdot \sum _{n=0}^{\infty }z^{n}-{\frac {3}{2}}\cdot \sum _{n=0}^{\infty }{\frac {1}{3^{n}}}z^{n}=1+\sum _{n=0}^{\infty }{\frac {1}{2}}\cdot \left(1-{\frac {1}{3^{n-1}}}\right)z^{n}}}$

The first three elements of the sequence of coefficients are all zero and thus the representation given here agrees with the above.

Generalizations

Power series can not only be defined for, but can also be generalized. So are z. B. the matrix exponential and the matrix logarithm generalizations of power series on the space of the square matrices . ${\displaystyle x\in \mathbb {R} }$

If powers with negative integer exponents also appear in a series, this is called a Laurent series . If you allow the exponent to take on fractional values, it is a Puiseux series .

Formal power series are used, for example, as generating functions in combinatorics and probability theory (e.g. as probability-generating functions ). In algebra , formal power series over general commutative rings are examined.

literature

• Kurt Endl, Wolfgang Luh: Analysis II. Aula-Verlag 1973, 7th edition 1989, ISBN 3-89104-455-0 , pp. 85-89, 99.
• ED Solomentsev: Power series. In: Encyclopaedia of Mathematics .