Power Series and Analytic Functions

Power Series

http://mathonline.wikidot.com/power-series

Operations on power series

Addition and subtraction

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

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

then

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

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.

Multiplication and division

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

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

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

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

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

then

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

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

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

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

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

Differentiation and integration

Once a function $f(x)$ 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:

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

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

Analytic Functions

Template:Main A function f defined on some open subset U of R or C is called analytic if it is locally given by a convergent power series. This means that every a ∈ U has an open neighborhood V ⊆ U, such that there exists a power series with center a that converges to f(x) for every x ∈ V.

Every power series with a positive radius of convergence is analytic on the interior of its region of convergence. All holomorphic functions are complex-analytic. Sums and products of analytic functions are analytic, as are quotients as long as the denominator is non-zero.

If a function is analytic, then it is infinitely differentiable, but in the real case the converse is not generally true. For an analytic function, the coefficients a_n can be computed as

$a_{n}={\frac {f^{\left(n\right)}\left(c\right)}{n!}}$

where $f^{(n)}(c)$ denotes the nth derivative of f at c, and $f^{(0)}(c)=f(c)$. This means that every analytic function is locally represented by its Taylor series.

The global form of an analytic function is completely determined by its local behavior in the following sense: if f and g are two analytic functions defined on the same connected open set U, and if there exists an element c∈U such that fTemplate:I sup(c) = gTemplate:I sup(c) for all n ≥ 0, then f(x) = g(x) for all x ∈ U.

If a power series with radius of convergence r is given, one can consider analytic continuations of the series, i.e. analytic functions f which are defined on larger sets than { x : |x − c| < r } and agree with the given power series on this set. The number r is maximal in the following sense: there always exists a complex number x with |x − c| = r such that no analytic continuation of the series can be defined at x.

The power series expansion of the inverse function of an analytic function can be determined using the Lagrange inversion theorem.

Behavior near the boundary

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 $1$ and diverges at every point of $|z|=1$. Nevertheless, the sum in $|z|<1$ is ${\textstyle {\frac {1}{1-z}}}$, which is analytic at every point of the plane except for $z=1$.
2. Convergent at some points divergent at others: ${\textstyle \sum _{n=1}^{\infty }(-1)^{n+1}{\frac {z^{n}}{n}}}$ has radius of convergence $1$. It converges for $z=1$, while it diverges for $z=-1$
3. Absolute convergence at every point of the boundary: ${\textstyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n^{2}}}}$ has radius of convergence $1$, while it converges absolutely, and uniformly, at every point of $|z|=1$ 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 of a power series with radius of convergence $1$, convergent at all points with $|z|=1$, 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.

Licensing

Content obtained and/or adapted from:

• Power series, Wikipedia under a CC BY-SA license
• Power series, mathonline.wikidot.com under a CC BY-SA license

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