27 - Power Series
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welcome to the second lecture in the
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last week of our course analysis of a
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complex kind today we'll learn about a
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special type of series namely the power
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series a power series also often call
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the Taylor series centered at a point Z
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zero in the complex plane it's a serious
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off the form sum K from 0 to infinity a
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K times Z minus e0 to the case is e 0 as
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this fixed number and Z's something
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arbitrary we can plug in the other so we
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can therefore vary this Z right here
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we've already seen a few examples of
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power series for example last class we
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looked at this series Z to the K and we
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found out that it converges as long as
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we plug in a 0 zap solut value is less
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than 1 so now our power series notation
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there's just a 1 here for the
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coefficient a K so the aks are all equal
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to 1 and the Z 0 is simply not there so
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Z 0 is equal to 0 here's another example
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the series minus 1 to the K over 2 to
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the K times Z to the 2k so this series
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only has even powers of Z so all the odd
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coefficients are equal to 0 K is equal
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to 0 if K is odd and the even
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coefficients of the form a 2 K are equal
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to minus 1 to the K divided by 2 to the
01:31
K and we have Z 0 is equal to 0 again
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how can we find out where the series
01:38
converges we can actually rewrite the
01:41
series of it by pulling out the exponent
01:44
K out of all of these individual terms
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there by finding the series equals minus
01:51
Z squared over 2 quantity to the K power
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because if I brought the K back in that
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would be back to the left hand side now
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if I call this entire term in
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parentheses W I get a series of W to the
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K which I completely understand already
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I know that series of the w to the case
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converges when
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who is less than one an absolute value
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and in divert just when W is greater
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than or equal to one what does that mean
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in terms of Z well the absolute value of
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W is the absolute value of minus e
02:25
squared divided by two
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when is that less than one well that is
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less than one when the absolute value of
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Z squared is less than two the negative
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sign goes away in the absolute value and
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then multiplied through by 2 when the
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see absolute value of Z squared less
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than 2 well that's the case when the
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absolute value of C is less than square
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root of 2 and so we know that the series
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this original series converges when the
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absolute value of Z is less than 2 and
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then diverges from the absolute value of
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C is greater than or equal to 2 so how
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do I find it in general for a given
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power series for what values of Z it
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converges there's a following theorem
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suppose I'm given the power series then
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there exists a number R in this arch
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between 0 and infinity infinity it's
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allowable as a number R and 0s as well
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such that this series converges
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absolutely as long as Z minus c0 is less
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than R an absolute value and it diverges
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when Z minus e0 is bigger than R in
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absolute value remember what absolute
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convergence meant that means if I put
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absolute values around all the terms of
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the series then the series still
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converts we noticed that absolute
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convergence implies regular convergence
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but the reverse is not necessarily true
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furthermore the convergence is uniform
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in the absolute value of Z minus is 0
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less than or equal to R for each R
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that's less than uppercase R uniform
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convergence means that if I plug in
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different values of Z's the convergence
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is kind of equally fast we won't go into
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details you so here's a little picture
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that hopefully helps an understanding
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this theorem this theorem predicts the
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existence of this radius R such that in
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the disc of radius R centered at the
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point Z 0 it's a little bit hard to read
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so this is
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right here in that disk I have
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convergence so all the blue shaded
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regions I get convergence outside of the
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disk
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I get divergence many plugins ease from
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out here the series of diverges the
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theorem says nothing about Z values on
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this boundary it gives us no information
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we've seen an example where the series
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diverges everywhere on the boundary but
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all kinds of things can happen on the
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boundary and lastly if I choose a radius
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little R that's strictly less than big R
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then if I look at the circle of radius
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little R which is contained in a disk of
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radius big R in that little disk I get
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uniform convergence we call this
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uppercase R the radius of the big disc
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the radius of convergence of the power
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series and you've probably heard of the
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radius of convergence before when you
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study the series on the real line and
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the word radius made very little sense
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because you got an interval of
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convergence and not a disk
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now the word radius makes sense because
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it's the radius of a disk we found that
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a power series converges for Z values
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from a disk and diverges outside of the
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disk this is called the disk of
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convergence and its radius is the radius
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of convergence
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let's look at some examples we already
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looked at the power series of the Z to
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the case and we saw it converges from
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the absolute value of Z is less than 1
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and diverges when the absolute value of
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Z is bigger than 1 therefore the radius
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of convergence is equal to 1 we also
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looked at this power series minus 1 the
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K over 2 to the K times Z to the 2k and
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we found that converges when the
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absolute value of Z is less than root 2
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and diverges when it's greater than
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which you and therefore a root 2 is the
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radius of convergence of this power
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series next let's look at the power
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series K from 0 to infinity K to the K Z
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to the K so these K to the case those
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are my a case if I wanted to write out a
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few terms of this series when K is equal
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to 0 0 to the zeroth power by definition
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is 1 so I get 1 times e to the 0 which
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is 1 when K is equal to 1 I get 1 to the
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first power times Z that's the when K is
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equal to 2 I get 2 to the second power
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that's 4 times Z squared when K is equal
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to 3 I get 3 cubed which is 27 Z cubed
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and so forth so the terms the
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coefficients in front of the powers of Z
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equal quite rapidly as K gets larger and
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larger so for what values of Z does this
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series converge well obviously when Z is
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equal to 0 all the terms are 0 so that's
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quite obviously a point of convergence
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let's not pick an arbitrary Z that's not
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0 and we observed the following if I
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look at the absolute value of K to the K
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times e to the K I can pull this
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exponent K outside and I get its K times
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the absolute value of Z to the power K
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but this K times the absolute value of Z
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notice that Z is fixed I can pick a
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large enough K for example greater than
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2 over the absolute value of Z and this
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term is greater than or equal to 2 to
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the K which means the individual terms
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of my power series they don't
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zero but a series whose terms don't go
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to zero cannot converge the terms going
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to zero wasn't necessary not a
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sufficient but a necessary condition for
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convergence
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since my term son go to zero the series
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does not converge and so can converge
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for any Z that's nonzero therefore the
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radius of convergence of this power
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series is 0 there is no disk around the
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origin in which the series converges so
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this power series has radius of
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convergence zero how about if I look at
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Z to the K divided by K to the K well
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again let's pick an arbitrary Z and I'm
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surf that the absolute value of Z to the
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K divided by K to the K I can do the
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same trick and pull these kids out it's
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the absolute value of Z divided by K the
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K power never when I pick a K large
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enough for example bigger than 2 times
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the absolute value of Z then this term Z
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over K to the K is founded about by 1/2
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to the K so most of the terms of the
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power series except for the first few
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are less than 1/2 to the K in absolute
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value but this series of 1/2 to the K is
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a geometric series and we know that
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converges some 1/2 to the K K from 0 to
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infinity converges and we actually know
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what it is it's sum is 1 over 1 minus
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1/2 which is actually 2 so since my
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absolute values of my series are bounded
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above by terms that belong to a series
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that converges the known theorem that
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therefore the original series converges
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as well even an absolute value so no
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matter what value of Z I choose
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eventually for large enough K this is
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going to be true and since finally many
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terms of a power series do not influence
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its convergence we find that this series
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converges no matter what Z I plug in
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there therefore it converges for all Z
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in the complex plane and that is an
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example of where the radius of
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convergence is infinity I can choose as
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large a discus I want to and the series
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will converge inside it
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here is the big theorem that makes power
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series so powerful
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suppose I'm given the power series and
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it's radius of convergence ins R and I'm
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gonna choose that R is greater than zero
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because when the radius of convergence
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is zero I only have this one point of
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convergence which is a bit boring and
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now I look at this function f of Z is
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given by the sum a KZ minus e0 to the K
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where Z varies and I let Z bury inside
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the disk of convergence and the claim is
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that the outcome is an analytic function
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in that disk of convergence and more is
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true an analytic function has a
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derivative and I can find that
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derivative by differentiating this
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series term by term so I can pull the
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derivative inside of this infinite sum
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we know that we can pull derivatives
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inside of finite sums but for infinite
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sum that's a whole different matter so
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that's something that needs to be proved
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but it can be proved that for this
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infinite sum which converges inside the
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disk of radius R I can differentiate
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inside the infinite sum and therefore
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find that F prime of Z is a K times the
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derivative of Z minus e0 to the K which
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is K times C minus e0 to the K minus
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first power notice I started this series
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had one here because when K is equal to
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0 the derivative of that term goes 0 and
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goes away the second derivative can be
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found similarly I just take another
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derivative inside so find a K times K
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times K minus 1 times Z minus a 0 to the
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K minus 2 and now it suffice us to start
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the summit to in particular if I look at
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the end derivative of that that's Z so I
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keep taking derivative the nth
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derivative it's going to be something
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like the sum K from n to infinity and
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then here I have a K times K times K
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minus 1 and so forth through K minus and
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plus 1
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z minus c0 to the K minus n when I plug
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in Z 0 for Z right there that makes this
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Z of Z 0 so as soon as this power K
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minus n is at least 1 the term vanishes
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because I have 0 to some positive power
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which is 0 only when this exponent is
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equal to 0 do I get a 1 so when the
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exponent is equal to 0 that's the case
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when K is equal to n so the first term
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of my powers here the end of the
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derivative of F at C 0 is the first term
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of the power series and that's gonna be
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a n times n times n minus 1 and so forth
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all the way through n minus n which is 0
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and so I just left with a 1 that's the
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factorial of n so that's a n times n
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factorial I wrote this out back here
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again the case derivative so I put it in
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terms of K and set up in terms of n the
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case the derivative of F at Z 0 is a K
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times K factorial and I can solve that
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for a k a k is therefore the case
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derivative of F at C 0 divided by K
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factorial for all case so that is
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amazing what this says that not only do
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I get an analytic function inside the
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disk of convergence but in fact these a
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case are strongly related to the
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derivatives of that analytic function
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namely the Keith's derivative of this
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analytic function at the points is they
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are the center of the disk divided by K
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factorial is equal to that a K let's
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look at some examples so remember this
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series e to the K has radius of
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convergence 1 therefore by the theorem
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this function f of C given by the sum K
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equals 0 to infinity Z to the K is an
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analytic function we can do that
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we found what the sum is right we know f
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of Z is 1 over 1 minus Z when your
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explicit formula and we know that it's
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analytic in fact we know this is
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actually an analytic function in the
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whole complex plane minus the value 1 we
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know this function agrees with the power
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series in the disk of radius 1 and the
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theorem says that's function is analytic
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but it also says I can differentiate
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term by term so let's take the
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derivative and differentiate term by
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term inside the sum as the theorem tells
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us we find that F prime of Z can be
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found as the sum K equals 1 to infinity
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K times e to the K minus 1 often it is
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desirable to express a power series
15:23
where the Z's are powers of K and that K
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minus 1 I therefore can do a shift of my
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summation index because the powers of Z
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when K is equal to 1 I get a power of 0
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in case you go to I get a power of 1 so
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I have all the powers of Z from 0 to
15:41
infinity here and therefore I can write
15:43
this sum in terms of powers of Z and I
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notice the number in front of Z is one
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bigger than the exponent so that must be
15:51
K plus 1 this is called a shift of the
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index if you wanted to do this in a
15:57
formal way you could briefly say well
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it's Albion new index and let L be K
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minus 1 so if I replace K with my L K is
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l plus 1 and therefore I get the sum and
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whenever I see ok I'm going to write L
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plus 1 so this K becomes L plus 1 times
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Z to the and the K minus 1 was my L now
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since Alice K minus 1 if K is equal to 1
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L is 0 so L must run from 0 to infinity
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so I first rewrite this first series in
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that way and then I rename my I'll back
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to K so that the series looks like like
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the one on the right hand side so now
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remember we also know that F of C is 1
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over 1 minus Yi and so we know the
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derivative of 1 over 1 minus C which is
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1 over 1 minus e quantity squared so I
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found two ways to express the derivative
16:58
of the function f and since both are the
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derivative of the seen function these
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two thinks must agree with each other
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and I therefore find that the sum K plus
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1 is either the K this 1 over 1 minus C
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quantity squared so I was actually able
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to evaluate a new series and find its
17:15
value using the fact that the derivative
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of my power series can be taken term by
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term
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note power series can similarly be
17:29
integrated turn-by-turn how does that go
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suppose you're given the power series of
17:36
radius of convergence R then for any W
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inside this disk of convergence oh
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here's the disc of convergence of
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centers you zero for any W in that disk
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we have that the integral from zero to W
17:54
off this whole series is equal to
17:58
pulling the integral inside so the
18:02
series a K of the integral Z minus e0 to
18:06
the K where the integral is taken over
18:08
any path from zero to W it doesn't
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matter which path I choose the integral
18:14
will always give you the same value
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because we're talking about an analytic
18:18
function in a disk I can actually
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evaluate this I have an antiderivative
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for Z minus e0 to the K namely 1 over K
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plus 1 times Z minus e0 to the k plus 1
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and then you need to evaluate that from
18:35
0 to W when they do that when they plug
18:40
in Z 0 it's nothing when I plug in W I
18:42
get 1 over k plus 1 times W minus Z 0 to
18:44
the k plus 1 let's apply this theorem
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again to our power series Z to the K
18:51
which we know has radius of convergence
18:53
R equals 1 so for any W whose absolute
18:57
value is less than 1 we can therefore
18:58
integrate this power series from 0 to W
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or again by that integral we mean the
19:04
integral over any path from 0 to W
19:09
within the disk and we find by this
19:14
previous fact that the interval from 0
19:16
to W of the serious e to the K can be
19:19
found by pulling the integral to the
19:21
inside therefore it's the series of the
19:25
integral from 0 to W of Z to the K D Z
19:27
but Z to the K has an antiderivative in
19:31
the disk where this path is located and
19:33
the antiderivative is 1 over k plus 1 z
19:38
to the k plus
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if we evaluate the antiderivative at the
19:43
two bounds we find one over k plus one
19:45
times w to the k plus one minus the
19:48
anti-derivative at zero which is zero
19:50
and we therefore find that the original
19:53
integral of the power series is equal to
19:55
the series of 1 over k plus 1 w to the k
19:58
plus 1 we can now do a shift of the
20:03
index of summation again as we did on
20:06
the previous page we notice we're
20:09
summing from k equals 0 to infinity but
20:12
if we look at these exponents here of W
20:15
they started 1 then 2 then 3 all the way
20:18
through infinity and that 1 over k plus
20:21
1 is the exact same number we could
20:22
therefore simply sum from 1 to infinity
20:25
and call the exponent K instead of k
20:27
plus 1 and we're then be dividing by K
20:31
furthermore we also know that the sum of
20:33
the Z to the case is 1 over 1 minus Z as
20:37
long as the absolute value of Z is less
20:39
than 1 of course and therefore instead
20:43
of integrating the whole series as we
20:45
did above we could just as well
20:47
integrate the function 1 over 1 minus C
20:49
well an antiderivative of 1 over 1 minus
20:53
e in the disk of radius 1 centered at
20:55
the origin is minus log of 1 minus C
20:58
because the derivative of minus log of 1
21:01
minus e is minus 1 over 1 minus C times
21:05
the derivative of the inside function
21:06
which gets rid of that minus n so the
21:09
derivative of minus log of 1 minus e is
21:11
indeed 1 over 1 minus C and if I again
21:14
plug in the W I get minus log of 1 minus
21:17
W and at the lower bound the primitive
21:20
is equal to 0 we used here that the
21:23
logarithm of 1 minus Z is analytic in
21:26
this disk let me remind you why that is
21:28
the case the logarithm function itself
21:32
most analytic in the complex plane minus
21:34
the negative real axis so that's more
21:38
log Z is analytic now we're looking at
21:43
logarithm of 1 minus C which shifts this
21:48
region to the complex plane minus the
21:51
portion of the
21:52
it relaxes starting at 1 so that's or
21:55
log of 1 minus these analytic or also -
22:01
a lot of one line you see that doesn't
22:02
change that - that and therefore in
22:05
particular its analytic in this disk of
22:07
radius 1 centered at the origin that's
22:10
where we're interested and that's where
22:11
our path is located that's why mine is
22:13
log of 1 minus e from set primitive of 1
22:16
over 1 minus e in that disk so we have
22:19
shown that the integral from 0 to W of
22:22
the power seriously the decay is equal
22:25
to the power series K from 1 to infinity
22:28
of W the K over K but we've also shown
22:32
that the same integral is equal to minus
22:35
log of 1 minus W since both of these
22:39
functions are equal to the integral of
22:41
the original power series the two
22:42
right-hand sides must agree so it must
22:45
be the case that the sum K from 1 to
22:47
infinity W to the K over K is equal to
22:50
minus log of 1 minus W we've therefore
22:53
found a power series expansion for the
22:56
logarithm function now log of 1 minus W
22:59
that sounds a little awkward so let's
23:01
set Z equal to 1 minus W in other words
23:06
replace this and call this actually to C
23:08
and then we're gonna bring this negative
23:11
sign to the other side and find a nicer
23:14
way of writing this for me that is log
23:16
of Z is equal to minus that's the minus
23:19
sign that came over some K from 1 to
23:21
infinity and of Z is equal to 1 minus W
23:23
then W is equal to 1 minus C so we find
23:27
1 minus e to the power K divided by K
23:30
and typically if you write a power
23:33
series you want it in the form Z minus Z
23:36
0 and at Z 0 minus C so our Z 0 is equal
23:39
to 1 but we want to interchange these
23:40
two terms here if I interchange these
23:43
two terms I have to pull out a negative
23:45
sign so I get a minus 1 to the K and I
23:48
also pulled this negative sign inside of
23:51
the Sun so I find minus 1 to the k plus
23:53
1 divided by K times Z minus 1 to the
23:56
power K so this is a power series for a
23:59
log of Z
24:05
in the disk of radius one centered at
24:08
one so the disk of radius one centered
24:11
at one is this disc right there and in
24:16
this disk I have that this is true I
24:19
have a power series that is equal to the
24:21
logarithm function there let's write
24:24
down the first few terms when K is equal
24:26
to one I find minus 1 squared that's 1
24:29
divided by 1 times Z minus 1 to the
24:33
first power so the first term is the
24:35
minus 1 the next term is minus 1 cubed
24:41
divided by 2 so minus Z minus 1 squared
24:48
over 2 next I get 2 plus Z minus 1 cubed
24:54
over 3 and so forth so it seems quite
25:00
important to find out where a power
25:03
series converges and we look at some
25:05
examples and found in our examples the
25:08
radius of convergence of the power
25:09
series how would you find it in general
25:12
though if I'm giving you an arbitrary
25:13
power series how would you determine its
25:16
radius of convergence that's a question
25:18
we're going to deal with in the next
25:20
lecture