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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Zheng-Chao Han

Section 3.2 Properties of a Convergent Power Series

Proof.

Set \(c_{n}= a_{n}(y-x_{0})^{n}\) and consider the power series \(\sum_{n=0}^{\infty} c_{n}t^{n}\) in \(t\text{.}\) Then it equals the given power series at \(x\) with \(x-x_{0}=t(y-x_{0})\) and converges at \(t=1\text{.}\) It suffices to prove that \(\sum_{n=0}^{\infty} c_{n}t^{n}\) converges uniformly over \(0\le t \le 1\text{.}\)
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Set \(s_{N}= \sum_{n=0}^{N}c_{n}\) for \(n\ge 0\) and \(s_{-1}=0\text{.}\) Then under our assumption \(s_{N} \to s:= \sum_{n=0}^{\infty} c_{n}= \sum_{n=0}^{\infty} a_{n}(y-x_{0})^{n}\) as \(N\to\infty\text{.}\) For \(0\le t\lt 1\text{,}\) we have
\begin{equation*} \sum_{n=0}^{N}c_{n} t^{n}= \sum_{n=0}^{N}\left( s_{n}-s_{n-1}\right) t^{n}=(1-t)\sum_{n=0}^{N-1} s_{n} t^{n}+s_{N}t^{N}. \end{equation*}
Sending \(N\to \infty\) and using \((1-t)\sum_{n=0}^{\infty} t^{n}=1\) gives us
\begin{equation*} \sum_{n=0}^{\infty} c_{n}t^{n}= (1-t)\sum_{n=0}^{\infty} s_{n} t^{n}= (1-t)\sum_{n=0}^{\infty} \left(s_{n}-s\right) t^{n}+s. \end{equation*}
For any \(\epsilon \gt 0\text{,}\) let \(N\) be such that for \(n\gt N\text{,}\) we have \(\vert s_{n}-s \vert \lt \epsilon\text{.}\) Then
\begin{align*} \vert \sum_{n=0}^{\infty} c_{n}t^{n} -s \vert \amp \le |1-t| \sum_{n=0}^{\infty} \vert s_{n}-s \vert |t|^{n} \\ \amp = |1-t|\left( \sum_{n=0}^{N} \vert s_{n}-s \vert |t|^{n} + \sum_{n=N+1}^{\infty} \vert s_{n}-s \vert |t|^{n}\right)\\ \amp \le |1-t|\left( \sum_{n=0}^{N} \vert s_{n}-s \vert |t|^{n} + \epsilon \sum_{n=N+1}^{\infty} |t|^{n}\right)\\ \amp \le |1-t|\left( \sum_{n=0}^{N} \vert s_{n}-s \vert |t|^{n} \right) + \epsilon \frac{ |t|^{N+1} |1-t|}{1-|t|}. \end{align*}
If we take real \(t\) such that \(0\le t \lt 1\text{,}\) then \(\frac{ |t|^{N+1} |1-t|}{1-|t|} \le 1\text{,}\) and we can find some \(\delta \gt 0\) such that when \(|1-t|\lt \delta\text{,}\) we have \(|1-t|\left( \sum_{n=0}^{N} \vert s_{n}-s \vert |t|^{n} \right) \lt \epsilon\text{.}\) This shows that \(\sum_{n=0}^{\infty} c_{n}t^{n} \to s\) as \(t\to 1-\text{.}\) Note that the above argument works even for complex \(t\) as long as we restrict \(t\) to satisfy \(\frac{ |1-t|}{1-|t|} \le C\) for some \(C\gt 0\text{.}\) This is the case as long as \(t\to 1\) from within the unit disc in a non-tangential way.
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Remark 3.2.2.

Abel’s Theorem is a form of interchange of limits. Suppose that the series \(\sum_{n=0}^{\infty} x^{n}\) converges at one end, say at \(x=R\text{,}\) then the Theorem implies that for \(0\le x \le R\)
\begin{equation*} \sum_{n=1}^{\infty} a_n x^n =\lim_{N\to \infty} s_N(x) \end{equation*}
is a continuous function of \(x\in [0,R]\text{.}\) As a consequence,
\begin{equation*} \sum_{n=1}^{\infty} a_n R^n= \lim_{x\to R-} \sum_{n=1}^{\infty} a_n x^n\text{.} \end{equation*}
This can also be written as
\begin{equation*} \lim_{N\to \infty} s_N(R)= \lim_{N\to \infty} \lim_{x\to R-} s_N(x) = \lim_{x\to R-} \lim_{N\to \infty} s_N(x) = \lim_{x\to R-} \sum_{n=1}^{\infty} a_n x^n. \end{equation*}
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Proof.

This follows simply using the knowledge that the power series converges uniformly for \(x\) such that \(x_{0} \le x \le x_{0}+t\text{.}\)
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Example 3.2.4. The series \(\sum_{n=0}^{\infty} (-x^2)^n \).

It arises from the geometric series replacing \(x\) by \(-x^{2}\) when \(|x|\lt 1\text{:}\)
\begin{equation*} \frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}= 1-x^2+x^4-x^6+\cdots \end{equation*}
Its radius of convergence is \(1\text{.}\) Thus for any \(t, |t| < 1\text{,}\) we have
\begin{equation*} \int_{0}^{t} \frac{1}{1+x^2}\, dx = \sum_{n=0}^{\infty} \int_{0}^{t} (-1)^n x^{2n}\, dx = \sum_{n=0}^{\infty}\frac{(-1)^n t^{2n+1}}{2n+1}. \end{equation*}
Using calculus knowledge that \(\int_{0}^{t} \frac{1}{1+x^2}\, dx=\arctan t\text{,}\) and the series on the right converges at \(t=1\text{,}\) Theorem 3.2.1 implies that
\begin{equation*} \frac{\pi}{4}=\lim_{t\to 1-0} \arctan t= \sum_{n=0}^{\infty}\lim_{t\to 1-0} \frac{(-1)^n t^{2n+1}}{2n+1} =\sum_{n=0}^{\infty} \frac{(-1)^n }{2n+1}. \end{equation*}
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Proof.

One simply notes that the series \(\sum_{n=1}^{\infty} n a_{n}(x-x_{0})^{n-1}\) converges iff the series \(\sum_{n=1}^{\infty} n a_{n}(x-x_{0})^{n}\) converges, so these two series have the same radii of convergence. But the radius of convergence of the latter is given by
\begin{equation*} \limsup_{n\to\infty} \vert n a_{n} \vert^{1/n}= \limsup_{n\to\infty} n^{1/n} \vert a_{n}\vert^{1/n}= \limsup_{n\to\infty} \vert a_{n}\vert^{1/n} \end{equation*}
using \(\lim_{n\to \infty} n^{1/n}=1\text{.}\) As a result, for any \(0\lt r \lt R\text{,}\) the series \(\sum_{n=1}^{\infty} n a_{n}(x-x_{0})^{n-1}\) converges uniformly over \(x\) such that \(\vert x-x_{0}\vert \le r\text{,}\) and we can can apply the theorem on a sequence of functions whose derivative sequence converges uniformly to conclude that (3.2.1) holds for \(\vert x-x_{0}\vert \le r\text{.}\) Since this holds for any \(0\lt r \lt R\text{,}\) we conclude that (3.2.1) holds for \(\vert x-x_{0}\vert \lt R\text{.}\) The rest cases for \(k\gt 1\) follow by repeating this procedure.
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Remark 3.2.6.

The previous Theorem reveals that a convergent power series in \(x-x_{0}\) is the Taylor series of the power series at \(x_{0}\text{.}\) This raises the question: whether any infinitely differentiable function can be represented near any point \(x_{0}\) in its domain as a convergent power series in \(x-x_{0}\text{?}\)
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If this holds true for such a function \(f(x)\text{,}\) the power series must be the Taylor series of \(f\) at \(x_{0}\text{.}\) This raises a related question: does the Taylor series at any \(x_{0}\) in its domain of an infinitely differentiable function always converge? It turns out that, even if this Taylor series converges, the convergent power series may not be equal to \(f\)
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If \(f(x)\) is infinitely differentiable near \(x_{0}\text{,}\) then the Taylor expansion with remainder term gives us, for any \(N\text{,}\)
\begin{equation*} f(x)=\sum_{n=0}^{N} \frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n} +\frac{f^{(N+1)}(c)}{(n+1)!}(x-x_{0})^{N+1} \end{equation*}
for some \(c\) between \(x_{0}\) and \(x\text{.}\) So our questions above boil down to whether \(\frac{f^{(N+1)}(c)}{(n+1)!}(x-x_{0})^{N+1}\to 0\) as \(N\to \infty\text{.}\) This clearly requires some control on \(\vert f^{(N+1)}(c)\vert\text{.}\)
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Exercise 3.2.7. An infinitely differentiable function which does not equal its Taylor series at some point.

Consider
\begin{equation*} f(x)=\begin{cases} e^{-\frac{1}{x^{2}}} \amp\quad\text{if $x\ne 0$};\\ 0\amp\quad\text{if $x=0$}. \end{cases} \end{equation*}
Verify that this \(f(x)\) is infinitely differentiable and \(f^{(k)}(0)=0\) for all \(k\text{.}\) Note that its Taylor series at \(0\) is identically \(0\) so can’t equal \(f(x)\) for \(x\ne 0\text{.}\) We remark that if we choose any \(x_{0}\ne 0\text{,}\) then it turns out that \(f(x)\) equals its Taylor series at \(x_{0}\) in a neighborhood of \(x_{0}\text{.}\)
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Proof.

We use the binomial expansion
\begin{equation*} (x-x_{0})^{n}=(x-x_{1}+x_{1}-x_{0})^{n}=\sum_{k=0}^{n}\binom{n}{k} (x-x_{1})^{k} (x_{1}-x_{0})^{n-k} \end{equation*}
to write the power series \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) as a doubly indexed series. We show that this doubly indexed series converges absolutely for any \(x\) such that \(|x-x_{1}| \lt r=R- |x_{1}-x_{0}| \text{.}\) As a result, we can exchange the order of summation to obtain
\begin{align*} \amp \sum_{n=0}^{\infty} \sum_{k=0}^{n}a_{n} \binom{n}{k} (x-x_{1})^{k} (x_{1}-x_{0})^{n-k}\\ = \amp \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} a_{n} \binom{n}{k} (x_{1}-x_{0})^{n-k} (x-x_{1})^{k}, \end{align*}
which is the desired \(\sum_{k=0}^{\infty} b_{k}(x-x_{1})^{k}\text{.}\)
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The absolute convergence of the doubly indexed series is seen by noting that
\begin{equation*} \sum_{n=0}^{\infty} \sum_{k=0}^{n}|a_{n}| \binom{n}{k} |x-x_{1}|^{k} |x_{1}-x_{0}|^{n-k} =\sum_{n=0}^{\infty}|a_{n}| (|x-x_{1}| + |x_{1}-x_{0}|)^{n} \end{equation*}
and that \(|x-x_{1}| + |x_{1}-x_{0}| \lt R\) under our assumption \(|x-x_{1}| \lt r=R- |x_{1}-x_{0}| \text{,}\) and the knowledge that the series \(\sum_{n=0}^{\infty}|a_{n}| s^{n}\) converges for any \(s\lt R\text{.}\)
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Remark 3.2.9.

It is possible for the radius of convergence of the power series \(\sum_{k=0}^{\infty} b_{k}(x-x_{1})^{k}\) to be greater than \(R-|x_{1}-x_{0}| \text{,}\) as seen in the case of \(\sum_{n=0}^{\infty} x^{n}\) and \(x_{1}=-\frac 12\text{,}\) where, instead of computing the Taylor series of \((1-x)^{-1}\) centered at \(x_{1}=-\frac 12\text{,}\) we can use a geometric series to expand \(\sum_{n=0}^{\infty} x^{n}=(1-x)^{-1}\) as
\begin{equation*} (1-x)^{-1}=\frac{1}{\frac 32 -(x+\frac{1}{2})}= \frac 23 \frac{1}{1- \frac{x+\frac{1}{2}}{\frac{3}{2}}} =\frac 23 \left(\sum_{n=0}^{\infty}\left(\frac{x+\frac{1}{2}}{\frac{3}{2}} \right)^{n}\right), \end{equation*}
which converges as long as \(|x+\frac{1}{2}| \lt \frac{3}{2}\text{.}\)
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Proof.

We first represent the series as a power series in \(x-x_{1}\) for \(|x-x_{1}|\lt R-|x_{1}-x_{0}|\) by \(\sum_{k=0}^{\infty} b_{k}(x-x_{1})^{k}\text{.}\) We argue by contradiction. Suppose that the series is not identically \(0\text{,}\) then there is a smallest integer \(k=m\) such that \(b_{k}\ne 0\text{.}\) It follows that
\begin{equation*} \sum_{k=0}^{\infty} b_{k}(x-x_{1})^{k}=\sum_{k=m}^{\infty} b_{k}(x-x_{1})^{k} = (x-x_{1})^{m} \sum_{k=m}^{\infty} b_{k}(x-x_{1})^{k-m}. \end{equation*}
The series \(\sum_{k=m}^{\infty}b_{k}(x-x_{1})^{k-m}\) has the same non-zero radius of convergence, \(R-|x_{1}-x_{0}|\text{,}\) as that for \(\sum_{k=0}^{\infty} b_{k}(x-x_{1})^{k}\text{,}\) so defines a continuous function \(g(x)\) in a neighborhood of \(x_{1}\text{.}\) Since \(g(x_{1})=b_{m}\ne 0\text{,}\) we conclude that \(g(x)\ne 0\) in a neighborhood of \(x_{1}\text{.}\) But this would mean that \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}=(x-x_{1})^{m}g(x)\) has no zero beside \(x_{1}\) in this neighborhood, contradicting our assumption that a sequence of zeroes of \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) approach \(x_{1}\text{.}\) This shows that under our assumption the power series must have all its \(b_{k}=0\text{,}\) and that it equals \(0\) for all \(x\) such that \(|x-x_{1}|\lt R-|x_{1}-x_{0}|\text{.}\)
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The above argument applies to any \(z\) with \(|z-x_{0}|\lt R\) when \(z\) is a limit point of the set of zeroes of the power series. Let \(Z\) be the set of such points in \(|z-x_{0}|\lt R\text{.}\) Then \(Z\) is a non-empty closed subset of the disc \(|x-x_{0}|\lt R\text{.}\) But the above argument shows that \(Z\) is also open. This then implies that \(Z\) must consists of all points of the disc \(|x-x_{0}|\lt R\text{,}\) therefore, the power series equals \(0\) in this entire disc.
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Remark 3.2.11.

In the above theorem, it is important that the limit point referred to is in the disc of convergence of the power series. The function \(f(x)=(x+1)\sin\frac{1}{x+1}\) has a convergent power series expansion at \(x=0\) with radius of convergence equal to \(1\text{,}\) and \(f(x_{n})=0\) for \(x_{n}=-1+\frac{1}{n\pi}\to -1\text{,}\) yet \(f(x)\not\equiv 0\text{.}\)
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Proof.

For any \(x\) with \(|x|\lt R\text{,}\) we know that both \(f(x)=\sum_{n=1}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=1}^{\infty} b_{n} x^{n}\) converge absolutely, so \(f(x)g(x)\) equals the Cauchy product given as
\begin{equation*} \sum_{n=0}^{\infty} \left( \sum_{m=0}^{n} a_{m}b_{n-m}\right) x^{n}. \end{equation*}
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Combining this Theorem and Abel’s Theorem gives us
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Proof.

Set \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\text{.}\) Then the two power series converge at \(x=1\text{.}\) Thus \(f(x)g(x)= \sum_{n=0}^{\infty} c_{n} x^{n}\) for all \(x\) with \(|x|\lt 1\text{.}\) Abel Theorem is applicable to all three power series here for \(x=1\text{,}\) so
\begin{align*} \left(\sum_{n=0}^{\infty} a_{n} \right)\left( \sum_{n=0}^{\infty} b_{n} \right) \amp =\lim_{x\to 1-} \left(\sum_{n=0}^{\infty} a_{n} x^{n}\right)\left( \sum_{n=0}^{\infty} b_{n} x^{n}\right)\\ \amp =\lim_{x\to 1-} \sum_{n=0}^{\infty} c_{n} x^{n}= \sum_{n=0}^{\infty} c_{n}. \end{align*}
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Remark 3.2.14.

The condition that all three series converge can’t be dropped. For the case that \(a_{n}=b_{n}=\frac{(-1)^{n}}{n+1}\text{,}\) it turns out that the series of the Cauchy product \(\sum_{n=0}^{\infty} c_{n}\) does not converge.
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Remark 3.2.15.

It can be proved that the quotient of two convergent power series centered at some \(x_{0}\) has a convergent power series expansion centered at \(x_{0}\) in some neighborhood of \(x_{0}\text{,}\) provided that the denominator does not vanish at \(x_{0}\text{.}\) Similarly it can be proved that if \(f(x)\) has a convergent power series expansion centered at \(x_{0}\) with \(y_{0}=f(x_{0})\text{,}\) and \(g(y)\) has a convergent power series expansion centered at \(y_{0}\text{,}\) then the composition \(g\circ f (x)\) has a convergent power series expansion centered at \(x_{0}\) .
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