
Title: Power Series Are Holomorphic  Proof

Series: Complex Analysis

YouTubeTitle: Complex Analysis  Part 11  Power Series Are Holomorphic  Proof

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Timestamps
00:00 Intro 00:21 Recap: results for power series 1:52 Proof of the first result (convergence of a kpower series) 6:50 Proof of the second result (convergence of a k1power series) 8:46 Proof of the third result (complex differentiability) 
Subtitle in English
1 00:00:00,486 –> 00:00:03,992 Hello and welcome back to complex analysis
2 00:00:04,814 –> 00:00:10,939 and of course, first I want to thank all the nice people that support this channel on Steady, via Paypal or by other means.
3 00:00:11,914 –> 00:00:19,730 Now, today in part 11 we will prove the important result about the uniform convergence and power series from the last video.
4 00:00:20,171 –> 00:00:23,586 In order to do this let’s quickly look at the statement again.
5 00:00:24,429 –> 00:00:30,312 So we have a power series named f, defined on the open disc of convergence.
6 00:00:30,512 –> 00:00:33,904 Then we are able to show 3 important statements here.
7 00:00:34,971 –> 00:00:41,742 The first one is that inside the open disc, for every closed ball, we have the uniform convergence.
8 00:00:42,500 –> 00:00:48,680 Next, the second statement tells us that this property also holds for this formal derivative
9 00:00:49,586 –> 00:00:55,904 and then in the last step we can show that this is indeed the derivative of our power series f
10 00:00:56,329 –> 00:01:02,628 or in other words, f is complex differentiable for any z and the derivative is as we expected.
11 00:01:03,729 –> 00:01:11,507 Now, I already told you, this is an important result, because it tells us that every power series is a holomorphic function
12 00:01:12,200 –> 00:01:18,091 and for this reason I think it’s very helpful that you can see, that we can actually prove it.
13 00:01:18,414 –> 00:01:23,680 The proof is not so complicated and indeed only part 3 will take some time.
14 00:01:23,880 –> 00:01:32,673 However also for the other 2 parts, to make our life a little bit easier let’s assume that the expansion point z_0 is equal to 0.
15 00:01:33,457 –> 00:01:38,230 This is not really a restriction, because the general case would work exactly the same.
16 00:01:39,000 –> 00:01:44,559 In the end this is just one translation in the complex plane to get any other expansion point.
17 00:01:45,500 –> 00:01:52,067 Still for us here everything is tidier and clearer when z_0 vanishes from all the equations.
18 00:01:53,057 –> 00:01:56,908 Ok, knowing this let’s start with the proof of part 1.
19 00:01:57,614 –> 00:02:05,142 There we need to show a uniform convergence and therefore we consider the difference (f  f_n) in the supremum norm.
20 00:02:06,114 –> 00:02:10,809 Here of course f_n is just the polynomial given by the partial sum.
21 00:02:11,543 –> 00:02:17,701 Also please note z_0 is set to 0 here and this one is our domain.
22 00:02:18,157 –> 00:02:22,389 Which means we have a closed ball inside the domain of convergence.
23 00:02:23,114 –> 00:02:29,537 Of course this is how you should understand the supremum norm here. It’s a supremum norm with respect to this domain.
24 00:02:30,200 –> 00:02:36,532 Hence we can immediately write this down as a supremum where z goes through all the points of this closed ball
25 00:02:37,343 –> 00:02:42,001 and it’s applied to the absolute value (f(z)  f_n(z)).
26 00:02:43,014 –> 00:02:47,592 Which is of course simply our power series that now starts with n+1.
27 00:02:48,329 –> 00:02:51,802 Simply because we subtracted the terms from 0 to n.
28 00:02:52,486 –> 00:02:58,038 Now, this is something we can work with, because for the absolute value we have the triangle inequality.
29 00:02:58,943 –> 00:03:04,689 Now, in order to apply it we first have to pull this limit here out of the absolute value.
30 00:03:05,214 –> 00:03:09,550 This is possible, because the absolute value is a continuous function.
31 00:03:10,614 –> 00:03:16,924 Ok, then the triangle inequality just means that now we have the absolute value inside the sum.
32 00:03:17,400 –> 00:03:23,371 So we simply have the absolute value of a_k times the absolute value of z to the power k.
33 00:03:24,114 –> 00:03:30,879 So what we can use now, is that the point z lies inside the closed ball with radius c.
34 00:03:31,343 –> 00:03:36,728 Hence the length of this point, the absolute value of z, is not greater than c.
35 00:03:37,600 –> 00:03:42,646 So we know we have the nice bound here. This is less or equal than c.
36 00:03:43,514 –> 00:03:48,529 Okay, so this is a nice result because now we can omit the supremum here,
37 00:03:48,700 –> 00:03:51,025 because there is no z involved anymore.
38 00:03:51,729 –> 00:04:00,983 Hence what we get is a power series that starts with n + 1 of the absolute value of a_k, times c to the power k
39 00:04:01,586 –> 00:04:10,933 and now you might already see, this looks close to a geometric series and indeed we can use a geometric series as a majorant here
40 00:04:11,857 –> 00:04:17,386 and please note here, because there is the absolute value involved, we are only working with real numbers now.
41 00:04:18,229 –> 00:04:22,624 For this reason you might recognize the whole argument here from real analysis.
42 00:04:23,243 –> 00:04:26,914 Still, I really think it’s helpful that we discuss it here.
43 00:04:27,686 –> 00:04:31,640 Okay, now the last step here I want to put as a remark on this side.
44 00:04:32,800 –> 00:04:41,181 By assumption we know that this power series is convergent, because the real number r tilde here lies in the region of convergence.
45 00:04:42,271 –> 00:04:47,003 So the point here is, we can choose a number that lies between c and r.
46 00:04:48,000 –> 00:04:53,875
So what we know is this series is convergent. Hence the sequence inside has to be bounded
47 00:04:54,743 –> 00:05:03,835 Hence we find a bound we can call B, such that the absolute value a_k r tilde to the power k is bounded by B.
48 00:05:04,771 –> 00:05:12,067 Now the lefthand side we can simplify, because r tilde is a positive real number which does not need the absolute value.
49 00:05:12,929 –> 00:05:16,357 Therefore in the next step we can write
this as :
50 00:05:17,143 –> 00:05:23,975 B is greater or equal than a_k in the absolute value, times r tilde to the
power k.
51 00:05:25,014 –> 00:05:30,341 Okay and now you see we have to bring c
to the power k into the game here .
52 00:05:31,057 –> 00:05:36,122 Of course we can just do this in this
way, but then we have to divide by c again
53 00:05:36,986 –> 00:05:41,707 Hence what we have here is the factor " a” tilde divided by c to the power k
54 00:05:42,186 –> 00:05:48,202
and here you see this number r tilde divided by c is a number that is greater than 1
55 00:05:49,200 –> 00:05:53,360 Therefore we can call this number q to
the power 1.
56 00:05:54,257 –> 00:05:58,665 Hence then q is a number strictly less
than 1.
57 00:05:59,714 –> 00:06:05,835 Okay and now you see this whole thing
here we can use for an estimate of this part there.
58 00:06:06,700 –> 00:06:14,424 This is now simply less or equal than B times
the series of q to the power k.
59 00:06:15,086 –> 00:06:24,503 and there you should see, this is just a
very nice geometric series. Which is of
course convergent, because q is less than 1.
60 00:06:25,143 –> 00:06:31,529 This means when we send the lower index
n to infinity this whole thing goes to 0
61 00:06:32,029 –> 00:06:38,095 and exactly this is what we wanted , because it means all the terms here are 0 in the limit.
62 00:06:39,057 –> 00:06:43,848 Hence the supremum norm goes to 0 and we
have the uniform convergence.
63 00:06:44,714 –> 00:06:48,031 Okay, there we have it. The first part is
proven.
64 00:06:49,100 –> 00:06:58,100 Now, fortunately the 2nd part can be proven with exactly the same steps
and in order to see this let’s go back
to the statement.
65 00:06:59,029 –> 00:07:04,926 So you see it looks exactly like part
 The only difference is that we have
another power series
66 00:07:05,714 –> 00:07:11,184 or in other words, we just have other coefficients, but these don’t change the
argument here.
67 00:07:12,014 –> 00:07:19,481 Therefore the only thing we actually
have to check is that this power series
has the same radius of convergence like
the first one
68 00:07:20,057 –> 00:07:23,894 and this is what we can do using the
Cauchy hardaman theorem.
69 00:07:25,086 –> 00:07:29,654 So please remind yourself, this is just
one lim sup we need to calculate.
70 00:07:30,557 –> 00:07:36,653 For this please note that our new
coefficients for this power series are given by a_k times k
71 00:07:37,714 –> 00:07:42,482 or more precisely we can give it a name
and call it b_(k1).
72 00:07:43,371 –> 00:07:48,468 The index is k1, because the power
here is also written as k1.
73 00:07:49,343 –> 00:07:50,769 Okay and now you know
74 00:07:50,969 –> 00:08:00,155 for the radius of
convergence we need to calculate this lim sup for k to infinity of the kth root of the coefficient in the absolute
value b_k.
75 00:08:00,957 –> 00:08:05,685 Hence what we have inside here is (a_k

 times (k + 1).
76 00:08:06,786 –> 00:08:13,527 Okay when we have this, I don’t have to
go into the details, because you see we can split it up into two parts.
77 00:08:14,386 –> 00:08:20,319 The first part where the coefficient a_k
is involved gives us the original radius
of convergence r
78 00:08:20,414 –> 00:08:24,714 and the 2nd part
just goes to 1 in the limit, k to
infinity.
79 00:08:25,829 –> 00:08:32,895 Therefore the radius of convergence of this new power series is simply the same as we had for the
original one
80 00:08:33,514 –> 00:08:39,512 and this means we can simply redo all
the steps, but now instead of a_k we would
write b_k.
81 00:08:40,486 –> 00:08:45,566 So I would say that’s enough for part 2. The interesting proof is
indeed part 3.
82 00:08:46,657 –> 00:08:52,065 For this reason let’s invest the next minutes to calculate the derivative of our function f.
83 00:08:53,086 –> 00:08:57,732 Of course what we want to get out, is
that the derivative is given by this
power series.
84 00:08:58,571 –> 00:09:03,179 Hence we need a good name for this function. Let’s call it f tilde.
85 00:09:04,171 –> 00:09:09,014 So in the end we want to get out that f
tilde is actually f’.
86 00:09:09,557 –> 00:09:13,566 However at the moment we don’t know if f ' exists.
87 00:09:14,543 –> 00:09:19,254 Therefore we are not able to assume it . We have to show it in this proof.
88 00:09:20,357 –> 00:09:23,903 However, of course we can look at the
difference quotient.
89 00:09:24,900 –> 00:09:31,027 this can be written as f(z + h)
 f(z) divided by h.
90 00:09:31,714 –> 00:09:38,846 Obviously here h is also a complex
number and in the limit h to 0 we get f ‘(z).
91 00:09:39,500 –> 00:09:44,360 For this reason a good idea would be to
subtract f tilde of z.
92 00:09:45,457 –> 00:09:51,666 Then the thing we want to show, is that
this term here goes to 0, when h goes to 0
93 00:09:52,757 –> 00:09:57,809 and of course in order to do this we
also will use a triangle inequality,
94 00:09:58,500 –> 00:10:03,504 but before we do this let’s split our
power series f into two parts.
95 00:10:04,343 –> 00:10:10,266 So you see here, instead of f, I write it
as the sum of p_N + q_N.
96 00:10:11,114 –> 00:10:18,122 Hence the idea here is to take the
infinite sum and to divide it at the
index N into two parts.
97 00:10:18,957 –> 00:10:24,809 More precisely this means p_N(z) as
given by the polynomial that ends with the index N.
98 00:10:25,771 –> 00:10:31,795 Accordingly q_N is simply defined as the whole rest that starts with N+1.
99 00:10:32,743 –> 00:10:37,329 Now, the reason why this splitting here
can be helpful, we have seen above.
100 00:10:38,029 –> 00:10:41,078 The derivative of the polynomial we can immediately calculate
101 00:10:42,343 –> 00:10:49,038 and exactly this derivative of the polynomial we want to include here and
then use the triangle inequality.
102 00:10:50,043 –> 00:10:55,324 However before we do that let’s first
put p_N at the front and q_N at the back.
103 00:10:56,014 –> 00:11:01,352 So here’s the difference quotient with p _ N and there is a difference quotient
with q_N.
104 00:11:01,814 –> 00:11:08,374 Of course we still subtract f_n, but here
we want to add and subtract the derivative of p_N now.
105 00:11:09,314 –> 00:11:12,905 Hence you could say we have simply added
a 0 here.
106 00:11:13,857 –> 00:11:19,808 Okay, now you should see that this really
helps, because now we have 3 different parts here.
107 00:11:20,800 –> 00:11:30,444 Also you surely remember our goal is to
apply the triangle inequality, so what we
have to do is to take the absolute value on both sides.
108 00:11:31,857 –> 00:11:39,814 Ok, then you should see this really
helps us, because now we can separate the 3 parts in the absolute value and get the inequality.
109 00:11:41,129 –> 00:11:46,456 Now, this is very nice, because we can
give the parts names and talk about them
separately.
110 00:11:47,386 –> 00:11:55,277 Of course we don’t need fancy names. So maybe we call the first one simply “A”, the
second one B and the last one C.
111 00:11:56,786 –> 00:11:59,709
So then let’s see which one of them
makes problems.
112 00:12:00,800 –> 00:12:07,674 For the first one, “A” we already know, when
we send h to 0 this one goes also to 0.
113 00:12:08,843 –> 00:12:12,378 Hence no problems there . What about the second one?
114 00:12:13,343 –> 00:12:22,080 Indeed by the property 2 from above, we
already know this thing here converges
to 0, when capital N goes to infinity.
115 00:12:23,129 –> 00:12:28,328 This is simply because the uniform
convergence implies the pointwise
convergence we have here.
116 00:12:29,629 –> 00:12:36,475 Of course in this case the question
remains. What happens with the last part C here, when N increases?
117 00:12:37,457 –> 00:12:44,055
If this one went bigger and bigger, it
would not help us that the first two
ones get smaller and smaller.
118 00:12:45,114 –> 00:12:48,435 Therefore we have to analyze part C in
detail.
119 00:12:49,543 –> 00:12:55,715 First i would say, inside the absolute
value, let’s substitute q_N again with the
power series.
120 00:12:56,943 –> 00:13:01,008 Hence we have one part with z + h
and one with just z.
121 00:13:01,900 –> 00:13:04,906
Then of course we should put both terms
together.
122 00:13:06,000 –> 00:13:15,276 So we simply have the series
with a_k
times (z + h) to the power k  z to
the power k, divided by h.
123 00:13:16,457 –> 00:13:24,821 Now, this looks much simpler than before
and indeed we can simplify this even
more, when we use the geometric sum
formula.
124 00:13:25,800 –> 00:13:31,125 Indeed this one is applicable, because we
have the same power and the difference involved.
125 00:13:32,100 –> 00:13:42,246 To refresh your memory, the geometric sum
formula holds for any number q and it
looks like: (1q) to the power k divided by (1q).
126 00:13:43,386 –> 00:13:47,770
So you see there’s one exception for q. q
is not allowed to be 1.
127 00:13:48,614 –> 00:13:55,814 However in all other cases this is indeed the finite sum starting with 0 and ending with k1
128 00:13:57,029 –> 00:14:00,271
and we simply sum up all the powers of q.
129 00:14:01,371 –> 00:14:06,675
Now, in order to apply this formula here , the trick is to use a suitable q.
130 00:14:08,000 –> 00:14:13,100 Maybe not so hard to see is that we need
both z+h and z in q.
131 00:14:13,900 –> 00:14:19,543 Hence what will work is, when we choose q
as z divided by (z+h).
132 00:14:20,829 –> 00:14:27,746 Then we can just multiply on both sides with the correct numbers, such that the
lefthand side here looks like this.
133 00:14:28,643 –> 00:14:34,312 Additionally and I simply tell you this,
now we find that the righthand side has
a very nice form.
134 00:14:35,271 –> 00:14:41,016
We start with the highest power of z +h which is the power k1.
135 00:14:41,657 –> 00:14:46,374 Then the next term in the sum is z+ h to the power k2.
136 00:14:47,471 –> 00:14:51,525
However, then in addition we also find z
to the power 1.
137 00:14:52,514 –> 00:14:56,685 Hence we can say here in the first term we had z to the power 0.
138 00:14:57,657 –> 00:15:07,393 So you see, this whole sum continues until
we reach the last term , z+h to the power 0 times z to the power k1.
139 00:15:08,329 –> 00:15:14,308 Now, please don’t forget ,this whole sum
here is only the part inside the series after a_k.
140 00:15:15,500 –> 00:15:20,299
This is important, because you know we apply the absolute value on the whole
term.
141 00:15:21,471 –> 00:15:24,795
Of course then we can use the triangle
inequality again.
142 00:15:26,114 –> 00:15:31,057 First we use it here to push the absolute value inside the series
143 00:15:32,029 –> 00:15:37,386 and then you see, for this whole sum here
we can use the triangular inequality again.
144 00:15:38,329 –> 00:15:45,353 Indeed what we can use there is the fact
that both z and z+h lie in our ball.
145 00:15:46,329 –> 00:15:52,094 Hence we know, the number r is an upper bound for the absolute value of both numbers.
146 00:15:53,214 –> 00:15:59,471 So you see for the absolute value we have r to the power k1 here, here and so on
147 00:16:00,000 –> 00:16:05,241 and because the sum has exactly k terms, we know we have this k times.
148 00:16:06,300 –> 00:16:09,327 In conclusion, we get a very nice inequality now.
149 00:16:10,500 –> 00:16:18,931 Of course we still have the series and the absolute value of a_k, but then comes r to the power (k1) times k
150 00:16:19,929 –> 00:16:22,647 and that’s a very nice series we get here.
151 00:16:23,243 –> 00:16:28,210 So you see this is related to the derivative, we already considered in part 2
152 00:16:29,229 –> 00:16:36,338 and then you should see, I need to correct myself here, because we need to lie inside the domain of convergence.
153 00:16:37,400 –> 00:16:45,547 Hence in order to do this correctly we need to choose a number r tilde which is less than the radius of convergence r.
154 00:16:46,486 –> 00:16:52,808 This is no problem at all, because z and z+h lie inside the domain of convergence.
155 00:16:54,000 –> 00:17:00,238 Hence instead of the number r we can choose a smaller number r tilde as the bound for all these numbers.
156 00:17:01,371 –> 00:17:07,518 Then we get this series here and the most crucial fact is: this is a convergent series
157 00:17:08,343 –> 00:17:15,268 and therefore this tail of the series has to go to 0 when capital N goes to infinity.
158 00:17:16,400 –> 00:17:22,305 Ok and with this you see, we have complemented the last missing part of our estimate above.
159 00:17:23,443 –> 00:17:28,933 To make it clearer, I would say for the end of the video let’s go through the argument again.
160 00:17:29,643 –> 00:17:38,336 So we start with a fixed point z and a small h. Then I just give you an arbitrary epsilon greater than 0
161 00:17:39,143 –> 00:17:46,038 and then you simply choose N so large, that this part and this part are both smaller than epsilon.
162 00:17:47,100 –> 00:17:52,919 After this, in the last part we simply take the limit h to 0 on both sides.
163 00:17:54,014 –> 00:18:00,301 Then A will vanish and we have the fact that this limit here is smaller than our epsilon.
164 00:18:01,229 –> 00:18:07,297 To put it in other words, the limit is arbitrarily small and therefore the limit has to be 0
165 00:18:08,400 –> 00:18:11,210 and there you see , this concludes the whole proof.
166 00:18:12,257 –> 00:18:18,168 Well, I still hope after this technical proof you are still interested in complex analysis.
167 00:18:19,343 –> 00:18:28,171 The next video will be easier, when we just look at examples for power series . So I see you there and have a nice day . Bye!