\(\blacksquare \) if we have a function \(f\left ( x\right ) \) represented as series (say power series or Fourier series), then we
say the series converges to \(f\left ( x\right ) \) uniformly in region \(D\), if given \(\varepsilon >0\), we can number \(N\) which depends
only on \(\varepsilon \), such that \(\left \vert f\left ( x\right ) -S_{N}\left ( x\right ) \right \vert <\varepsilon \).
Where here \(S_{N}\left ( x\right ) \) is the partial sum of the series using \(N\) terms. The difference between uniform
convergence and non-uniform convergence, is that with uniform the number \(N\) only depends
on \(\varepsilon \) and not on which \(x\) we are trying to approximate \(f\left ( x\right ) \) at. In uniform convergence, the
number \(N\) depends on both \(\varepsilon \) and \(x\). So this means at some locations in \(D\) we need much
larger \(N\) than in other locations to convergence to \(f\left ( x\right ) \) with same accuracy. Uniform
convergence is better. It depends on the basis functions used to approximate \(f\left ( x\right ) \) in the
series.
If the function \(f\left ( x\right ) \) is discontinuous at some point, then it is not possible to find uniform
convergence there. As we get closer and closer to the discontinuity, more and more terms
are needed to obtained same approximation away from the discontinuity, hence
not uniform convergence. For example, Fourier series approximation of a step
function can not be uniformly convergent due to the discontinuity in the step
function.
This work for positive and negative \(n\), rational or not. The sum converges when only for \(\left \vert x\right \vert <1\).
From this, we can derive the above sums also for the geometric series. For example, for \(n=-1\)
the above becomes
And now since \(\left \vert \frac {1}{x}\right \vert <1\), we can use binomial expansion to expand the term \(\left ( 1+\frac {1}{x}\right ) ^{n}\) in the above and
obtain a convergent series, since now \(\left \vert \frac {1}{x}\right \vert <1\,.\) This will give the following expansion
So everything is the same, we just change \(x\) with \(\frac {1}{x}\) and remember to multiply the whole
expansion with \(x^{n}\). For example, for \(n=-1\)
Where \(R_{n}\) is remainder \(R_{n}=\frac {\left ( x-a\right ) ^{n+1}}{\left ( n+1\right ) !}f^{\left ( n+1\right ) }\left ( x_{0}\right ) \) where \(x_{0}\) is some point between \(x\)
and \(a\).
\(\blacksquare \) \(\ \)Maclaurin series: Is just Taylor expanded around zero. i.e. \(a=0\)
\(\blacksquare \) \(\ \)This diagram shows the
different convergence of series and the relation between them
The above shows that an absolutely convergent series (\(B\)) is also convergent. Also a
uniformly convergent series (\(D\)) is also convergent. But the series \(B\) is absolutely convergent
and not uniform convergent. While \(D\) is uniform convergent and not absolutely
convergent.
The series \(C\) is both absolutely and uniformly convergent. And finally the series \(A\) is
convergent, but not absolutely (called conditionally convergent). Examples of \(B\) (converges
absolutely but not uniformly) is
For uniform
convergence, we really need to have an \(x\) in the series and not just numbers, since
the idea behind uniform convergence is if the series convergence to within an
error tolerance \(\varepsilon \) using the same number of terms independent of the point \(x\) in the
region.
\(\blacksquare \) The sequence \(\sum _{n=1}^{\infty }\frac {1}{n^{a}}\) converges for \(a>1\) and diverges for \(a\leq 1\). So \(a=1\) is the flip value. For example
Diverges, since \(a=1\), also \(1+\frac {1}{\sqrt {2}}+\frac {1}{\sqrt {3}}+\frac {1}{\sqrt {4}}+\cdots \) diverges, since \(a=\frac {1}{2}\leq 1\). But \(1+\frac {1}{4}+\frac {1}{9}+\frac {1}{16}+\cdots \) converges, where \(a=2\) here and the sum is
\(\frac {\pi ^{2}}{6}\).
\(\blacksquare \) Using partial sums. Let \(\sum _{n=0}^{\infty }a_{n}\) be some sequence. The partial sum is \(S_{N}=\sum _{n=0}^{N}a_{n}\). Then
If \(\lim _{N\rightarrow \infty }S_{n}\) exist and finite,
then we can say that \(\sum _{n=0}^{\infty }a_{n}\) converges. So here we use set up a sequence who terms are partial
sum, and them look at what happens in the limit to such a term as \(N\rightarrow \theta \). Need to find an
example where this method is easier to use to test for convergence than the other method
below.
\(\blacksquare \) Given a series, we are allowed to rearrange order of terms only when the series is
absolutely convergent. Therefore for the alternating series \(1-\frac {1}{2}+\frac {1}{3}-\frac {1}{4}+\cdots \), do not rearrange terms since
this is not absolutely convergent. This means the series sum is independent of the order in
which terms are added only when the series is absolutely convergent.
\(\blacksquare \) In an infinite series of complex numbers, the series converges, if the real part
of the series and also the complex part of the series, each converges on their
own.
\(\blacksquare \) Power series: \(f\left ( z\right ) =\sum _{n=0}^{\infty }a_{n}\left ( z-z_{0}\right ) ^{n}\). This series is centered at \(z_{0}\). Or expanded around \(z_{0}\). This has radius of
convergence \(R\) is the series converges for \(\left \vert z-z_{0}\right \vert <R\) and diverges for \(\left \vert z-z_{0}\right \vert >R\).
\(\blacksquare \) Tests for convergence.
Always start with preliminary test. If \(\lim _{n\rightarrow \infty }a_{n}\) does not go to zero, then no need to do
anything else. The series \(\sum _{n=0}^{\infty }a_{n}\) does not converge. It diverges. But if \(\lim _{n\rightarrow \infty }a_{n}=0\), it still can
diverge. So this is a necessary but not sufficient condition for convergence. An
example is \(\sum \frac {1}{n}\). Here \(a_{n}\rightarrow 0\) in the limit, but we know that this series does not converge.
For Uniform convergence, there is a test called the weierstrass M test, which
can be used to check if the series is uniformly convergent. But if this test fails,
this does not necessarily mean the series is not uniform convergent. It still can
be uniform convergent. (need an example).
To test for absolute convergence, use the ratio test. If \(L=\lim _{n\rightarrow \infty }\left \vert \frac {a_{n+1}}{a_{n}}\right \vert <1\) then absolutely
convergent. If \(L=1\) then inconclusive. Try the integral test. If \(L>1\) then not absolutely
convergent. There is also the root test. \(L=\lim _{n\rightarrow \infty }\sqrt [n]{\left \vert a_{n}\right \vert }=\lim _{n\rightarrow \infty }\left \vert a_{n}\right \vert ^{\frac {1}{n}}\).
The integral test, use when ratio test is inconclusive. \(L=\lim _{n\rightarrow \infty }\int ^{n}f\left ( x\right ) dx\) where \(a\left ( n\right ) \) becomes \(f\left ( x\right ) \).
Remember to use this only of the terms of the sequence are monotonically
decreasing and are all positive. For example, \(\sum _{n=1}^{\infty }\ln \left ( 1+\frac {1}{n}\right ) \), then use \(L=\lim _{N\rightarrow \infty }\int ^{N}\ln \left ( 1+\frac {1}{x}\right ) dx=\left ( \left ( 1+x\right ) \ln \left ( 1+x\right ) -x\ln \left ( x\right ) -1\right ) ^{N}\). Notice, we only use
the upper limit in the integral. This becomes (after simplifications) \(\lim _{N\rightarrow \infty }\frac {N}{N+1}=1\). Hence the
limit \(L\) is finite, then the series converges.
Radius of convergence is called \(R=\frac {1}{L}\) where \(L\) is from (3) above.
Comparison test. Compare the series with one we happen to already know it
converges. Let \(\sum b_{n}\) be a series which we know is convergent (for example \(\sum \frac {1}{n^{2}}\)), and
we want to find if \(\sum a_{n}\) converges. If all terms of both series are positive and if \(a_{n}\leq b_{n}\) for
each \(n\), then we conclude that \(\sum a_{n}\) converges also.
\(\blacksquare \) For Laurent series, lets say singularity is at \(z=0\) and \(z=1\). To expand about \(z=0\), get \(f\left ( z\right ) \) to look like \(\frac {1}{1-z}\) and
use geometric series for \(\left \vert z\right \vert <1\). To expand about \(z=1\), there are two choices, to the inside and to
the outside. For the outside, i.e. \(\left \vert z\right \vert >1\), get \(f\left ( z\right ) \) to have \(\frac {1}{1-\frac {1}{z}}\) form, since this now valid for
\(\left \vert z\right \vert >1\).
\(\blacksquare \) Can only use power series \(\sum a_{n}\left ( z-z_{0}\right ) ^{n}\) to expand \(f\left ( z\right ) \) around \(z_{0}\) only if \(f\left ( z\right ) \) is analytic at \(z_{0}\). If \(f\left ( z\right ) \) is not analytic at
\(z_{0}\) need to use Laurent series. Think of Laurent series as an extension of power series to
handle singularities.
Let us find the Laurent series for \(f\left ( z\right ) =\frac {5z-2}{z\left ( z-1\right ) }\). There is a singularity of order \(1\) at \(z=0\) and \(z=1\).
This makes \(g\left ( z\right ) \) analytic around \(z\), since \(g\left ( z\right ) \) do not have a pole at \(z=0\), then it is analytic around \(z=0\) and
therefore it has a power series expansion around \(z=0\) given by
The residue is \(2\). The above expansion is valid around \(z=0\) up and not including the next
singularity, which is at \(z=1\). Now we find the expansion of \(f\left ( z\right ) \) around \(z=1\). Let
This makes \(g\left ( z\right ) \) analytic around \(z=1\), since \(g\left ( z\right ) \) do not have a pole at \(z=1\). Therefore it has a power series
expansion about \(z=1\) given by
The residue is \(3\). The above expansion is valid around \(z=1\) up and not including the next
singularity, which is at \(z=0\) inside a circle of radius \(1\).
Putting the above two regions together, then we see there is a series expansion of \(f\left ( z\right ) \) that is
shared between the two regions, in the shaded region below.
Let check same series in the shared region give same values. Using the series expansion
about \(f\left ( 0\right ) \) to find \(f\left ( z\right ) \) at point \(z=\frac {1}{2}\), gives \(-2\) when using \(10\) terms in the series. Using series expansion
around \(z=1\) to find \(f\left ( \frac {1}{2}\right ) \) using \(10\) terms also gives \(-2\). So both series are valid produce same
result.
18.2.2 Method Two
This method is simpler than the above, but it results in different regions. It is based on
converting the expression in order to use geometric series expansion on it.
The above is valid for \(0<\left \vert z\right \vert <1\) which agrees with result of method 1.
Now, to find expansion for \(\left \vert z\right \vert >1\), we need a term that looks like \(\left ( \frac {1}{1-\frac {1}{z}}\right ) \). Since now it can be
expanded for \(\left \vert \frac {1}{z}\right \vert <1\) or \(\left \vert z\right \vert >1\) which is what we want. Therefore, writing \(f\left ( z\right ) \) as
With residue \(5\). The above is valid for \(\left \vert z\right \vert >1\). The following diagram illustrates the result
obtained from method 2.
18.2.3 Method Three
For expansion about \(z=0\), this uses same method as above, giving same series valid for \(\left \vert z\right \vert <1\,.\) This
method is a little different for those points other than zero. The idea is to replace \(z\) by \(z-z_{0}\)
where \(z_{0}\) is the point we want to expand about and do this replacement in \(f\left ( z\right ) \) itself. So for \(z=1\)
using this example, we let \(\xi =z-1\) hence \(z=\xi +1\). Then \(f\left ( z\right ) \) becomes
The above is valid for \(\left \vert \xi \right \vert <1\) or \(\left \vert z-1\,\right \vert <1\) or \(\,-1<\left ( z-1\right ) <1\) or \(0<z<2\). This gives
same series and for same region as in method one. But this is little faster as it uses
Binomial series short cut to find the expansion instead of calculating derivatives as in
method one.
18.2.4 Conclusion
Method one and method three give same series and for same regions. Method three uses
binomial expansion as short cut and requires one to convert \(f\left ( z\right ) \) to form to allow using
Binomial expansion. Method one does not use binomial expansion but requires doing
many derivatives to evaluate the terms of the power series. It is more direct
method.
Method two also uses binomial expansion, but gives different regions that method one and
three.
If one is good in differentiation, method one seems the most direct. Otherwise, the
choice is between method two or three as they both use Binomial expansion.
Method two seems a little more direct than method three. It also depends what
the problem is asking form. If the problem asks to expand around \(z_{0}\) vs. if it is
asking to find expansion in \(\left \vert z\right \vert >1\) for example, then this decides which method to
use.