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00:26:750Paolo Guiotto: Okay.
00:29:840Paolo Guiotto: Good morning.
00:36:610Paolo Guiotto: The scope of this class is to extend
00:41:630Paolo Guiotto: The examples of functions of complex variables.
00:48:200Paolo Guiotto: We will, start, first with introducing
00:53:310Paolo Guiotto: A general class, which is the class of power series.
01:01:990Paolo Guiotto: And then, we introduce the most important elementary functions.
01:08:210Paolo Guiotto: Elementary…
01:12:780Paolo Guiotto: Functions… exactly as, for reals.
01:19:570Paolo Guiotto: For functions of real variables, we have a battery of important functions, like exponential.
01:29:530Paolo Guiotto: trigonometric functions, hyperbolic functions, powers with any exponent, logarithms, these are the most important functions. We have exactly the same type of functions, so exponentials.
01:44:980Paolo Guiotto: Sine and cosine… Cosine hyperbolic sine…
01:52:630Paolo Guiotto: And hyperbolic cosine. We have also powers.
01:59:250Paolo Guiotto: With any exponent, even complex, if you…
02:03:570Paolo Guiotto: want. And, log algorithms, so…
02:11:210Paolo Guiotto: So, this is the program of today. And then we… we see…
02:16:550Paolo Guiotto: We will see to take confidence with these functions.
02:19:990Paolo Guiotto: How to, solve some simple equations with this.
02:24:940Paolo Guiotto: with this function. So let's start talking about a power series. What is a power series?
02:31:630Paolo Guiotto: Well, first of all.
02:33:780Paolo Guiotto: You may imagine, as this part says, it will be a series. A series is something you studied in first-year calculus as an infinite sum, okay? Now, the idea is simply the following. We have seen polynomials, a polynomial.
02:55:370Paolo Guiotto: polynomial is a function, say, P.
02:58:450Paolo Guiotto: and X, to emphasize the degree of the polynomial, is a function of this type sum for k going from 0 to N, CKZ to… I'm sorry, we are using the complex variable, so we'll use letter Z.
03:14:100Paolo Guiotto: Z to K, where these, numbers, CK, are complex numbers, K from 0 to…
03:24:780Paolo Guiotto: And this is a well-defined function, there is no problem, because it's a combination is a sum of monomials. A monomial is a constant times a power, and the power is number multiplied by itself a certain number of times, so there is no problem with this.
03:43:50Paolo Guiotto: Now, what if we send this n the degree of the polynomial 2 plus infinity? So we want to define
03:52:300Paolo Guiotto: the polynomial of infinite degrees, so something like P of Z equals sum for k going from 0 to infinity, CKZ to K.
04:03:920Paolo Guiotto: Now, the reason of doing this, I'm anticipating the second part, is the following.
04:11:130Paolo Guiotto: So, why… We… interested.
04:22:190Paolo Guiotto: to… these… functions.
04:30:610Paolo Guiotto: Of course, there is a mathematical interest, but I'm…
04:34:290Paolo Guiotto: Probably it'd be so interesting. In general, it makes sense to study this type of things. But the idea is the following. We know that, and this is just one of the examples, and the same story will repeat for these functions here. So we know that
04:54:950Paolo Guiotto: Well, I start from what you… you know for sure. We know that,
05:02:760Paolo Guiotto: For example, for the exponential, we have the Maclaurin expansion that says that we can write E to X as sum for k going from 0 to n, and the expansion is x to k divided at k factorial plus a little correction
05:20:410Paolo Guiotto: Little O of X to power n.
05:23:930Paolo Guiotto: Now, I don't know if you have seen in first-year calculus, but this formula can be extended, letting n to plus infinity and getting an exact identity. So it can be proved.
05:43:210Paolo Guiotto: that if formally, this is not correct, what I'm saying, but the conclusion is correct. So if formally we send n to infinity, we would get this, e tox equal the sum for k going from 0 to infinity, x to k divided k factorial, and what happens to the little o? Now, the idea is that this little o is,
06:07:500Paolo Guiotto: is smaller when high N increases. So, the bigger is n, the smaller is O of X to N. So, in principle, we may expect that this disappears when n goes to infinity. That's not true, okay? So, do not think this is correct, but
06:26:480Paolo Guiotto: If this happened, we would have this formula that turns out to be true, okay? And this is an exact representation of the exponential through a formula that you can see, it's a type of function of that type. So if you write this as sum for k going from 0 to infinity, 1 over k factorial.
06:48:820Paolo Guiotto: X to K, you see exactly this type of structure.
06:54:150Paolo Guiotto: Where the coefficients CK are 1 over k factorial, so this is…
06:58:660Paolo Guiotto: the CK, and here we are considering X just real.
07:05:50Paolo Guiotto: But this is… now, we know that this is true, okay? Let's say, like, we accept this is a real, analysis story, so it's a first-year KICU story. This would suggest that perhaps we could define E to Z as
07:24:310Paolo Guiotto: The right-hand side, sum for k going from 0 to infinity, 1 over K factorial. Z to K, provided this infinite sum makes sense.
07:34:360Paolo Guiotto: So, you see, if I ask you, what does it mean to do E to I, none of you have a real meaning of what is E raised to complex number I, no, as a power.
07:48:660Paolo Guiotto: But if I say, this is what is written at right.
07:52:290Paolo Guiotto: Assuming that this makes sense, now you have a new interpretation. This is just a sum for k going from 0 to infinity.
08:02:30Paolo Guiotto: 1 over k factorial i to power k, which is something
08:06:580Paolo Guiotto: that we can, in principle, write i2K is i times iK times. We even know what… what is it, because E to 0 is 1, e to 1…
08:18:540Paolo Guiotto: I to 1 is i, i square is minus 1, i cubed will be i square times i, so minus i. i to power 4 goes back to 1.
08:31:830Paolo Guiotto: And so on. I to power 5 will be i again, etc. So we know these quantities. So, let's say that it seems that that series
08:42:460Paolo Guiotto: it's… it makes sense better than the left-hand side, so we can use the series to define the left-hand side. So this is exactly what we are going to do. So this is the way we define the exponential on complex numbers, so the exponential function as a function of a complex number Z.
09:02:290Paolo Guiotto: And to do this, we need to understand what does it mean to have a power series, so an infinite sum of this type.
09:11:80Paolo Guiotto: Okay, now, before we can define a power series, so to define power series.
09:26:640Paolo Guiotto: some…
09:27:470Paolo Guiotto: for K going from 0 to infinity, CKZ to K. We have to do one step back before, because this is an infinite sum of complex numbers. So, we have to say, what does it mean, the infinite sum of complex numbers? So…
09:47:700Paolo Guiotto: We need to… First at… to define… in… and find it.
10:03:590Paolo Guiotto: some… Or, in mathematics, we use better the term serious, off… complex.
10:18:770Paolo Guiotto: Numbers.
10:23:370Paolo Guiotto: Now, the definition is exactly the same you have seen in one variable analysis. So, we have a sequence of numbers. Let's call these numbers,
10:36:980Paolo Guiotto: This is the number I want to sum now, so let's call W these numbers. So, let WK be a sequence of complex numbers.
10:50:900Paolo Guiotto: So, we will be interested only in convergence here, so I will limit the definition only to convergence. I do not introduce divergence or indeterminacy. We just say that… we say…
11:08:650Paolo Guiotto: That… the… serious… sum for k going from 0 to infinity of WK, is convergent.
11:28:320Paolo Guiotto: So this is, the keyword…
11:33:530Paolo Guiotto: If, exactly as we do for real series, what we do is we start computing finite sums, and that can be done because we have a sum, we have an algebra.
11:45:640Paolo Guiotto: So, if, computing the sum for k going from 0 to certain N of these numbers, WK,
11:54:910Paolo Guiotto: Now, this makes sense, there is no problem. What we can do is, we do the limit when n goes to infinity.
12:03:740Paolo Guiotto: Of this sequence.
12:06:490Paolo Guiotto: This is a sequence of complex numbers. If this limit exists.
12:12:100Paolo Guiotto: And it is a complex number.
12:15:120Paolo Guiotto: We say that the series is convergent, and the value of the limit will be, by definition, the infinite sum of the
12:24:300Paolo Guiotto: Serious.
12:27:40Paolo Guiotto: Well, the… maybe the sum, let's just… let's eliminate that word infinitely. It would be the sum of the series.
12:34:300Paolo Guiotto: If this, is not verified, we just say that the series is not convergent.
12:42:230Paolo Guiotto: Yeah.
12:53:130Paolo Guiotto: So, we have, I think… No, real, maybe.
13:03:400Paolo Guiotto: Yeah, reals are considered as part of complex numbers. Complex numbers with imaginary part equals zero.
13:10:700Paolo Guiotto: Okay. So, if there exists not a limit.
13:17:250Paolo Guiotto: When n goes to infinity of these partial sums, WK, we say… Or…
13:30:560Paolo Guiotto: Well, let's say, R.
13:35:310Paolo Guiotto: Because the limit might exist and be infinite in the complex plane, or… If, limit.
13:44:590Paolo Guiotto: n goes… going to infinity of sum k from 0 to n of WK is, infinite.
13:53:180Paolo Guiotto: So now this is a sequence of numbers in complex plane, so there is not a plus-minus. Infinity, exactly as you have a sequence of vectors in the Cartesian plane, there is not a plus infinity minus infinity, so we say the infinity of the complex plane.
14:08:460Paolo Guiotto: So if that limit is not finite, so it means it does not exist, or that it is infinite, we say…
14:17:120Paolo Guiotto: that the series of the WK is not.
14:23:680Paolo Guiotto: convergent.
14:25:440Paolo Guiotto: So we will be mainly, clearly, interested in the case when the series is convergent. So, let's do an example, just to illustrate that this concept work
14:37:430Paolo Guiotto: works exactly as it works for real serious. And this is the traditional example, because it is one of the few examples for which we can show this definition in practice, doing all the calculations. It is the example of the geometric
14:59:70Paolo Guiotto: Serious.
15:01:180Paolo Guiotto: It is the series you already know, but now we will write in the form sum k going from 0 to infinity of Z to K, where Z is a complex number. By the way, this is an example of particular power series, because you see, it is that serious when all the coefficients CK are equal to 1.
15:23:530Paolo Guiotto: So, it is an example of… Power.
15:29:600Paolo Guiotto: Serious.
15:33:70Paolo Guiotto: Now, the argument that shows that this series is convergent… actually, we can prove that the series, the conclusion will be the serious
15:46:10Paolo Guiotto: sum for k going from 0 to infinity, W to K, Sorry, Z to K.
15:55:450Paolo Guiotto: is convergent if, and only if, the modulus of the complex number Z is strictly less than 1.
16:04:590Paolo Guiotto: And in that case, we also have the formula for the sum of the series.
16:09:700Paolo Guiotto: And the… For… Such.
16:15:850Paolo Guiotto: Z. We have that the sum for k going from 0 to infinity of Z to K, it's exactly the same formula you have for years, so 1 over 1 minus Z.
16:30:700Paolo Guiotto: This is the formula for the sum.
16:33:300Paolo Guiotto: Let's check this.
16:35:450Paolo Guiotto: So…
16:38:390Paolo Guiotto: So, to apply the definition, I have to start computing a finite sum, okay? Here, the WKRZ to KE.
16:46:940Paolo Guiotto: So if you want, here… WKs are Z to power K.
16:54:400Paolo Guiotto: The sum for k going from 0 to n of the WK is the sum for k going from 0 to n of z to power K.
17:06:920Paolo Guiotto: Now, this is, the calculation is the same, so I don't know if you remind of this example, but however, let's repeat. So this is 1 plus Z plus Z squared plus Z cubed, etc, until we arrive to Z to power n.
17:23:960Paolo Guiotto: Now, there is a trick to compute exactly this formula.
17:28:640Paolo Guiotto: well, actually, we may say that this is a formula for the sum. It's not wrong. But the problem is that since the next step I have to do is to take this limiter, so to send N to infinity.
17:42:500Paolo Guiotto: what happens here? If I use this formula and I send N to infinity, it's like an elastic. It means that when you send N to infinity, you just increase the number of elements of the sum, and you don't understand what happens, really, because
17:57:840Paolo Guiotto: It's like, it's not the limit of a sum, it's the sum of the limit, just to make clear, because that is the limit of the sum of 2 or 3 or 4, but the number is fixed.
18:08:680Paolo Guiotto: This is the limit of a sum where the number is variable, the number of elements in the sum is variable, so you cannot use that trick. So we need to rewrite this, and this is the major problem with Sirius.
18:21:610Paolo Guiotto: You can always write the partial sum, but you can almost never compute the limit, because you cannot use the definition of partial sum to do directly the limit. You need to represent the sum in some
18:35:650Paolo Guiotto: better way to do the limit. And this is the trick. So, if we call this sum
18:42:310Paolo Guiotto: Let's call it SN.
18:44:860Paolo Guiotto: We may notice that there are a couple of ways to do, which are perfectly equivalent, so we may notice
18:54:930Paolo Guiotto: That, this is,
18:59:620Paolo Guiotto: 1 plus Z, well, let's write like that.
19:03:520Paolo Guiotto: This, let's say there is also power n minus 1 here, so this part here is exactly the sum of the first n minus 1 powers up to the exponent z west, n minus 1, so this is the SN minus 1.
19:18:190Paolo Guiotto: So we have that this is SN minus 1 plus Z2N.
19:27:530Paolo Guiotto: Okay, so we have this first relation, but also another remark could be that we could rewrite this as 1 plus
19:34:880Paolo Guiotto: Factorizing a Z, we get 1 plus Z plus… you see, this becomes 1, this becomes Z, because I'm factorizing the Z, this becomes Z squared, and so on. The last one will become Z2N minus 1.
19:50:610Paolo Guiotto: So this second formula says that SN is also equal to 1 plus Z times SN minus 1.
19:59:900Paolo Guiotto: If you now impose the equality of the two, you have a sort of equation for this value that you find in the two lines. So this equation says that SN minus 1 plus Z to power n
20:14:980Paolo Guiotto: is equal to 1 plus ZSN minus 1. So, if we carry SN minus 1 on the same side, for example, the left side, you have SN minus 1 times 1 minus Z,
20:29:170Paolo Guiotto: equal 1, and then now I carry Z2N on the other side, 1 minus Z2N. And this is the identity. Now, I want to divide by 1 minus Z to extract as SN minus 1,
20:41:570Paolo Guiotto: I can do if Z is different from 1. So, let's note that if Z were 1, the value of the sum would be easy, because you were doing SN would be sum
20:54:660Paolo Guiotto: for K going… well, let's say it's SN minus 1,
20:58:730Paolo Guiotto: sum for k going from 0 to n minus 1 of 1 to power k. But 1 to power k is 1, so this is 1 plus 1 plus 1 plus 1, etc. How many times in that sum there are n elements, because you start from 0 to n minus 1, so they are n.
21:19:690Paolo Guiotto: So the sum is just N. So we have that. For Z equal 1, the value of Sn minus 1 is easy, it is equal to N.
21:29:870Paolo Guiotto: for Z different from 1,
21:33:220Paolo Guiotto: We can continue from this formula and say that SN minus 1 is equal 1 minus Z to power n divided 1 minus Z. So we have this formula.
21:47:460Paolo Guiotto: So, let's write for SN now, so let's shift the index of one unit. So this means that SN is equal to… when Z is 1, if SN minus 1 is N, SN will be N plus 1.
22:04:970Paolo Guiotto: And when Z is different from 1, it is 1 minus Z. Now, since here I have n minus 1 and here n, when I put N at left, I have n plus 1 there. So, 1 minus Z to N plus 1 divided 1 minus Z.
22:21:550Paolo Guiotto: So this is the formula for SN. And now this is better to compute the limit of this SN, which is the second step, no? So the first step is compute the partial sum, and then you want to send the number of terms of these sums to plus infinity.
22:41:130Paolo Guiotto: to see what happens. Now… If we do the limit.
22:47:680Paolo Guiotto: for n going to plus infinity of Sn, which is the limit when n goes to plus infinity of the sum for k going from 0 to N.
23:00:180Paolo Guiotto: of Z to K, so this limit is what we are computing. We have two possibilities, depending on what is the value of Z. For Z equal 1, I have the limit for n going to plus infinity.
23:14:150Paolo Guiotto: We say that SN is n plus 1, so clearly this limit is infinite.
23:20:530Paolo Guiotto: Since here we are in a complex plane, I do not write plus infinity, but just infinity of the complex plane.
23:27:620Paolo Guiotto: If Z is different from 1, we have to do the limit when n goes to infinity.
23:34:760Paolo Guiotto: of 1 minus Z to N plus 1 over 1 minus Z.
23:40:850Paolo Guiotto: Which, at the end, so you have 1 minus Z, the denominator is just a constant, it's a factor, so I could say this is equal 1 over 1 minus Z, the limit is in N, not in Z. Limit in n of 1 minus Z ton plus 1.
23:59:680Paolo Guiotto: But 1 is a constant, so at the end, this is 1 over 1 minus Z times 1 minus the limit in n of z to n plus 1. And now the problem is, what is this limit?
24:16:160Paolo Guiotto: Well, it happens similarly to what happens with reels now.
24:20:980Paolo Guiotto: the limit… for n going to infinity, or Z to N plus 1 is So…
24:30:860Paolo Guiotto: You know, these are Z times Z times Z, N plus 1 times.
24:35:630Paolo Guiotto: So, basically, what happens? If the number is smaller than 1,
24:41:150Paolo Guiotto: Well, actually, more precisely, if the modulus of Z is less than 1,
24:46:910Paolo Guiotto: that product will go to 0 when n goes to infinity, so the limit is 0 in this case.
24:52:790Paolo Guiotto: When a modulus of Z is equal to 1,
24:57:670Paolo Guiotto: You can understand that probably that limit won't exist.
25:02:450Paolo Guiotto: Because, imagine, for example, Z is, for some… for some… well, actually, for some Z, like Z equal 1, the limit exists and it is equal to 1. But here, we are considering Z different from 1, okay? So this story says for Z different from 1.
25:22:650Paolo Guiotto: Now, what happens? Imagine a Z different from 1. Well, there is a case you already know, which is Z equal minus 1.
25:31:660Paolo Guiotto: This was the case you have seen, probably, for reals. When you do minus 1 to n plus 1, this does not have any limit.
25:40:460Paolo Guiotto: Now, let's see what happens with the complex number, like Z equal i. It's a number with modulus 1. I'm choosing I because this is a modulus 1 number. And what happens here? I have that this is I to N plus 1.
25:56:840Paolo Guiotto: But we have seen above that this sequence, powers of I,
26:02:290Paolo Guiotto: is, particular, because I20 is 1.
26:06:520Paolo Guiotto: I to 1 is i, i square is minus 1, i cubed is minus i, and then we return back from I. So this is… the sequence is repeated, so I can see this sequence as, equal to 1…
26:22:590Paolo Guiotto: I minus 1, minus i, and then 1I minus, 1 minus I, and so on, forever.
26:31:390Paolo Guiotto: So this sequence in the complex plane is doing something like this. I start with, sorry, with 1,
26:39:50Paolo Guiotto: Then I flip to I, then I go to minus 1, then I go to minus I, and then I go back to here. So it's just jumping between these four points, so you may imagine that there won't be any limit here.
26:55:60Paolo Guiotto: Okay? And this can be proved in general, because if you take modulus Z equal 1,
27:02:40Paolo Guiotto: You know that, yes, there is the algebraic representation of complex numbers, but there is also the trigonometric representation. So, in this case, it is convenient, because this number is the number of this type cosine theta plus i sine theta.
27:19:920Paolo Guiotto: Because the point, cosine sine, belongs to the unitary circle. So now, when we do powers, Z, like Z ton plus 1,
27:31:30Paolo Guiotto: Well, you know that doing a power of a number written in algebraic form, it's complicated, but in trigonometric form, it is easy because of the addition formula for sine and cosine, and this turns out to be exactly cosine of n plus 1
27:47:170Paolo Guiotto: theta plus i sine N plus 1 theta.
27:53:820Paolo Guiotto: So, when you have to do the limit, you have to understand what happens to cosine of n plus 1 theta when n goes to infinity, and sine n plus 1 theta when n goes to infinity.
28:05:370Paolo Guiotto: Apart for theta equals 0, where this is 0 and the other one is 1, for all other theta, there is no limit for this, okay? So, you can easily see that 4 theta between 0 and 2 pi, so excluding the point exactly where Z is 1,
28:24:120Paolo Guiotto: This quantity has not a limit as well as this quantity. So, in particular, Z2N plus 1 won't have a limit, okay? Z2N plus 1
28:34:540Paolo Guiotto: won't have a limit. So the conclusion is that for module Z equal 1, since there is no limit for that term, there is no limit for the sequences n, so there is no limit.
28:49:760Paolo Guiotto: And finally, for modulus of Z greater than 1, it is easier, because when modulus of Z is greater than 1, so Z1 plus 1, if you look at this in absolute value, you see that it happens that this is the absolute value of Z to power n plus 1, because the modulus of the product is the product of the modulus, so in this case.
29:13:520Paolo Guiotto: We can flip models and power, and since this goes to plus infinity, you deduce that Z to N plus 1 will go to the infinity of the complex plane.
29:26:540Paolo Guiotto: And therefore, going back to our limit, this part goes to infinity, no? And therefore, all the limit is going to infinity. So in this case, the limit is infinite. So we may say that the unique case when the limit is finite is exactly this one.
29:46:840Paolo Guiotto: So when modulus of Z is 1.
29:50:70Paolo Guiotto: And therefore, what happens to the limit of SN in that case? So… the limit.
29:58:230Paolo Guiotto: in n of the Sn, which is the limit in N, as we said. Well, actually, it is equal to 1 over 1 minus Z times 1 minus the limit
30:10:480Paolo Guiotto: of Z ton plus 1. I'm just copying what we said, so we have that. For modulus of Z less than 1, this quantity, z to n plus 1, goes to 0, so the limit exists, and it is equal 1 over 1 minus Z, which is finite, is a complex number.
30:31:270Paolo Guiotto: Four modulus of Z equal 1, different from 1.
30:37:140Paolo Guiotto: The limit does not exist. If you want for Z equal 1, the limit is plus infinity, so we just write infinity of the complex plane. And for modulus of Z strictly greater than 1, again, the limit is infinite in the complex plane.
30:54:450Paolo Guiotto: So we can conclude that the limit conclusion
31:03:300Paolo Guiotto: the limit…
31:04:760Paolo Guiotto: of the sequence Sn, which is the limit of the partial sums, sum for k going from 0 to n minus… to N, sorry, of Z to K,
31:17:970Paolo Guiotto: exists.
31:20:00Paolo Guiotto: and it is finite, it's a complex number, if, and only if, modulus of Z is strictly less than 1, and in this case.
31:33:680Paolo Guiotto: the limit of these partial sums, which is what we call the sum of the series, so we write the sum as k going from 0 to infinity of Z to K is equal to 1 over 1 minus Z. So for every Z complex.
31:50:430Paolo Guiotto: with the modulus of Z strictly less than 1.
31:55:810Paolo Guiotto: This is, by the way, a remarkable formula, because it is one of the few formulas we have exactly for an infinite salmon.
32:04:380Paolo Guiotto: Of a power series.
32:06:530Paolo Guiotto: Okay, this closes this example that shows… Basically, as the definition of…
32:14:700Paolo Guiotto: Infinite sum of complex numbers works.
32:20:90Paolo Guiotto: Okay, so as the example shows, this example is based on the fact that we can literally apply the definition. So, step one, we compute the partial sum, the finite sum, the sum for k going from 0 to n of the numbers.
32:36:810Paolo Guiotto: And step two, we can do the limit of this sum when n goes to infinity.
32:41:730Paolo Guiotto: But as you have seen, this is very particular, tricky, so it's not a general situation.
32:48:460Paolo Guiotto: Okay? And that's exactly as it happens for real serious. Now, what do you do when you have a real serious? You know that there is a number of,
32:58:550Paolo Guiotto: test, afford convergence, okay? I don't want now to be detailed here, but a limit to one important test.
33:09:480Paolo Guiotto: Which is the following.
33:11:270Paolo Guiotto: Well, actually, let me refresh you something on series. You may remind that for numerical series, we have a number of tests for constant sine series, so series with positive terms or negative terms. For this, we have a good number of tests, but when the series has a generic term with variable sine.
33:34:310Paolo Guiotto: Well, yes, there is a particular test, like the liability test, when the sign is alternating.
33:40:130Paolo Guiotto: But, in general, we do not have a general test, apart from one very important test, which is the following. So let me refresh this. Recall…
33:53:730Paolo Guiotto: that… 4… Real… serious.
34:02:940Paolo Guiotto: So let's use this notation. Sum for k going from 0 to infinity of AK, so where now AK is a sequence of real numbers, we have…
34:17:520Paolo Guiotto: But let me refresh everything, so it's a sort of crash course, you know, in 10 minutes, in 5 minutes to make a series. Now, an accessory
34:33:670Paolo Guiotto: But not sufficient.
34:44:20Paolo Guiotto: Test… for… Convergence.
34:49:429Paolo Guiotto: So, what is this condition? The condition is the general term must go to zero. So, this says AK must go to 0 when K goes to plus infinity.
35:03:130Paolo Guiotto: This is required, but it is not sufficient. For example, the harmonic series sum of 1 over K, no, it's, as general term, going to 0, but it is not convergent. So, it's necessary, but not sufficient.
35:19:10Paolo Guiotto: Then we have, Sufficient…
35:27:990Paolo Guiotto: Conditions… for… Positive, let's say, positive… Terms serious.
35:43:910Paolo Guiotto: So… let me refresh this. So, if… we have that, DAK… All positive, in large sense.
35:58:580Paolo Guiotto: Then, a test is, the root test.
36:08:750Paolo Guiotto: This says that if you take the k root of the k term, so the k root of this, or if you want the AK to exponent 1 over K,
36:22:890Paolo Guiotto: And this goes to a number, say, a limit, L,
36:28:910Paolo Guiotto: This test tells something if L is less than 1 or greater than 1. If L is less than 1, it says that the series of the AK is convergent.
36:42:590Paolo Guiotto: If L is greater than 1, then it says that the series of the AIK is divergent.
36:51:170Paolo Guiotto: So, in particular, not convergent.
36:56:790Paolo Guiotto: If L is equal to 1, the test Fails.
37:03:250Paolo Guiotto: So it does not provide any information. A similar test, a twin test, in fact, is the ratio test.
37:13:380Paolo Guiotto: That says that you do something similar, but here you compute the limit of ratios between two consecutive terms of the series.
37:22:620Paolo Guiotto: Since you are having as denominator the AK, here you need that they are strictly positive. So, while the first, it's a little bit weaker, it works even if they are zero, some of them, this requires a stronger condition, slightly stronger.
37:39:860Paolo Guiotto: So, if you compute AK plus 1 over AK, and this goes to L, same conclusion as above.
37:47:350Paolo Guiotto: Same conclusion.
37:50:490Paolo Guiotto: So if L is less than 1, the series will be convergent. If L is greater than 1, the series will be divergent. If L is 1, the test fails. Then there are other tests, but since we are going to use these two in particular, I stop here, okay?
38:05:180Paolo Guiotto: for, positive term serious, okay?
38:10:650Paolo Guiotto: Now, this is, these are sufficient conditions, so this means if they work, fine. If they do not work, you don't know anything, okay?
38:20:790Paolo Guiotto: So the series… for example, the case when the test fails, the case L equal 1, does not provide any indication. You could have a test series which are convergent and L is 1, serious which are divergent and L is 1. So the test is not able to distinguish.
38:37:800Paolo Guiotto: Between convergent and divergent series.
38:40:450Paolo Guiotto: The test works when L is less than 1 or L is greater than 1, okay? In that case, it is able to classify convergence or not.
38:50:680Paolo Guiotto: Now, for generic,
38:54:340Paolo Guiotto: real-term series, I skipped the Leibniz test, because we don't, particularly need it here. Maybe if we need, I will return on this. So, for,
39:06:750Paolo Guiotto: 4.
39:08:90Paolo Guiotto: Let's say this is actually the unique sufficient condition, Oh, wait, 4… Real… serious.
39:23:420Paolo Guiotto: So where the AK are now generic, real, valued numbers.
39:30:500Paolo Guiotto: So, this is a sufficient condition that says if something is verified, the series is convergent, and the something is if the series of the absolute values
39:44:120Paolo Guiotto: K going from 0 to infinity is convergent. Then, also the series of the numbers.
39:54:560Paolo Guiotto: is convergent.
39:57:500Paolo Guiotto: Now, we call this condition, which is, because of this implication, stronger than the simple convergence of the series, we call this absolute
40:11:00Paolo Guiotto: convergence.
40:15:570Paolo Guiotto: And normally, when we do this, we also modify this. This would be convergence, but to distinguish from that one, we just call simple.
40:27:610Paolo Guiotto: convergence.
40:29:710Paolo Guiotto: So there is only the convergence of the series and nothing more.
40:33:960Paolo Guiotto: Now, this is interesting, why?
40:36:90Paolo Guiotto: Because that series is a series with positive terms, so I could use, to test convergence, the tests I mentioned above. So that's why it's important, because I don't know how to attack this one.
40:51:540Paolo Guiotto: Okay, but I can say, if I can show that that one is convergent, and that one is now a constant sine terms series, so I could use one of these tests to check convergence.
41:04:160Paolo Guiotto: And as you know, if you remind of this, as you know, these are relatively easy tests, because they
41:11:690Paolo Guiotto: means you have to compute a limit of the IK. You don't have to compute partial sums as here, as it happens in the definition. You don't have to do this
41:23:680Paolo Guiotto: Two-step work, and this sum is… this finite sum is generally difficult to handle in a usable way to compute the limit.
41:32:990Paolo Guiotto: So this is direct because you work directly on the terms of the series. So you compute some limit, the limit of the roots or the ratios, and this is the convergence of the series of the modulus, then you have automatically the convergence of the series of your numbers.
41:51:20Paolo Guiotto: Well, this is important because, now, if we move to a series with general terms, which are complex numbers.
41:59:820Paolo Guiotto: So, moving… to… See… serious.
42:08:700Paolo Guiotto: So I mean series of types, sum k going from 0 to infinity of WK, where WK is now a sequence of complex numbers.
42:18:250Paolo Guiotto: It is more or less like a sequence of reals, so where there is not a sign, because these have not even a sign in complex numbers, no? I is positive or negative. Sine is because there is an order, you can see there is not an order, so it looks like if it is an extension of this type of series.
42:36:330Paolo Guiotto: Now, the nice thing is that this test holds
42:39:730Paolo Guiotto: also for the C-valued series. We have this… I will not prove this, but we have this important fact.
42:49:650Paolo Guiotto: So, sufficient… test… for convergence.
42:58:830Paolo Guiotto: So if we still have that the series of the absolute values, and now the advantage is that this is no more a C-valued series, but it is a series of real numbers and even positive numbers.
43:12:350Paolo Guiotto: So it's a series to which you could, to discuss conversion, you could use root and ratio test to that series, okay? So if…
43:22:530Paolo Guiotto: The series of absolute values is convergent.
43:26:20Paolo Guiotto: Then, also the series of, the numbers, Plain numbers is convergent.
43:33:730Paolo Guiotto: And again, we use the name of absolute.
43:39:370Paolo Guiotto: Convergence for this.
43:45:110Paolo Guiotto: And for this, simple… convergence.
43:53:50Paolo Guiotto: Okay, so this is, this is, why is, important this, because,
44:00:940Paolo Guiotto: to study that convergence. Now, that will be a series with positive numbers, so I can treat as a series of this type. Of course, not with WK, no, but with modulus of WK.
44:16:790Paolo Guiotto: Okay, so now we can finally return to power series. We can… Apply.
44:28:560Paolo Guiotto: All this… to… power.
44:35:520Paolo Guiotto: Serious.
44:38:240Paolo Guiotto: So, sum k going from 0 to infinity, CK, Z to K.
44:50:140Paolo Guiotto: Well… Let's say that, actually
44:59:820Paolo Guiotto: Well, let's say later, because there is a slight generalization of this series.
45:06:650Paolo Guiotto: that change Z to K into Z minus something to K, but let's skip for this, for a moment, this, this formula. Now, since now the goal is to, to show, to illustrate how this, all these facts I mentioned here, applies on this type of series.
45:25:680Paolo Guiotto: And this will yield some interesting fact about the convergence of this.
45:31:290Paolo Guiotto: If you want, we can take 5 minutes back, and then we continue. Just 5 minutes.
45:45:50Paolo Guiotto: Okay, so…
45:46:570Paolo Guiotto: Let's apply these few general facts to this type of series. So, how can we discuss convergence for this series? We apply just this test.
45:58:210Paolo Guiotto: So… Applying.
46:06:920Paolo Guiotto: Absolute.
46:09:50Paolo Guiotto: convergence.
46:10:990Paolo Guiotto: Pastor.
46:13:430Paolo Guiotto: So, we discuss the convergence for the series, we discuss…
46:22:300Paolo Guiotto: the convergence for the series of the absolute values, so modulus CKZ to power K, which is equal to series K from 0 to infinity, modulus CK times modulus to Z to power K, that we can switch with the modulus and write in this way.
46:42:920Paolo Guiotto: Okay? Now, let's call this AK…
46:47:450Paolo Guiotto: And, an idea could be now to apply the root or ratio test to this series. What happens if we apply the root test?
46:56:700Paolo Guiotto: Now, by… D… Root test.
47:05:370Paolo Guiotto: If… let's just rewrite the root test. The root test says that if L, which is the limit
47:13:940Paolo Guiotto: in K of the AK to 1 over K, so we are applying the root test to the series of the AK, is this one exists, then
47:25:920Paolo Guiotto: We have L, less than 1, serious AK… converges.
47:32:500Paolo Guiotto: L greater than 1 series of AK diverges.
47:38:00Paolo Guiotto: Well, let's add one important fact that actually the root and ratio test says. This, in this second case, which is important for our discussion.
47:49:330Paolo Guiotto: Not only the series of the AK diverges, but in this second case, we also have this, and the general term AK goes to plus infinity.
47:59:660Paolo Guiotto: Okay?
48:02:350Paolo Guiotto: So, if we do now the limit for our case.
48:11:780Paolo Guiotto: the limit of… the AK to 1 over K,
48:18:170Paolo Guiotto: The AK is this number, modulus CK modulus z power k, so I have to do the limit.
48:24:900Paolo Guiotto: in K of modulus CK times modulus Z to power K. This is our AK.
48:34:880Paolo Guiotto: These two exponents, 1 over K, so the kth root of this.
48:40:00Paolo Guiotto: is the limit I need to compute.
48:42:290Paolo Guiotto: Now, you notice that if we do the root, we get the limit of, on K of, the root of the product is the product of the root, so this is modulus of CK to exponent 1 over K, and then I have modulus of Z to K to 1 over K.
48:59:110Paolo Guiotto: So if you want power of power, you multiply the exponents.
49:03:40Paolo Guiotto: K times 1 over K is 1, so Z to K times 1 over K, but that's 1.
49:10:180Paolo Guiotto: So at the end, we have that this is just
49:13:390Paolo Guiotto: modulus of Z, which is independent of K, so my limit, in fact, becomes modulus of Z times the limit in K of modulus CK to 1 over K. So this is the value of the limit of a K to 1 over K. So this is the number L.
49:34:210Paolo Guiotto: And this says that if this cell is less than 1, the series is convergent, So…
49:41:480Paolo Guiotto: If L, which is modulus of Z times the limit
49:46:580Paolo Guiotto: on K of CK to power, 1 over K. This is less than 1.
49:53:410Paolo Guiotto: We have that, the series, which series? We have two series here. The original power series, CKZ2K, and the series of the AK, which is the series of the models, is of these numbers.
50:06:920Paolo Guiotto: Now, the test is applied to the series of the AK, so the conclusion will be on the series of AK. So, series of AK…
50:16:650Paolo Guiotto: converges.
50:18:800Paolo Guiotto: But this means that the series of… the series of the AKAs, the series of the absolute values.
50:25:640Paolo Guiotto: of the CK times Z to K, so this is convergent.
50:32:760Paolo Guiotto: But, so this means that the original series is absolutely convergent, and because of this convergence test, whenever you have that a series is absolutely convergent, it is also convergent, we get that our initial series, so the power series.
50:51:840Paolo Guiotto: converges. CK, Z to K converges.
50:56:500Paolo Guiotto: Okay?
50:59:930Paolo Guiotto: Then, if L, which is still modulo Z limit in K of modulus CK to 1 over K, is now greater than 1,
51:13:80Paolo Guiotto: So we are in the second case. The limit of AK to 1 over K is greater than 1, and it says the series of AK diverges, and AK goes to plus infinity.
51:24:860Paolo Guiotto: So we have that. This series of AK divergences.
51:31:60Paolo Guiotto: and also AK goes to plus infinity. Now, let's translate back this into our original series. So this says that the series of the absolute values CKZ to K, diverges.
51:47:500Paolo Guiotto: Well, this alone is not sufficient to deduce anything, because we know that the two conditions, convergence and absolute convergence, are not equivalent. The vice versa is false.
52:00:950Paolo Guiotto: Okay? So you cannot say that one converges if and only if the other. So if the other does not converge, this one is not converging. This is not correct. You see? You can say, if that converges, this converges.
52:15:840Paolo Guiotto: But you cannot say that if that does not convergence, particular is divergent, this one is not convergent. This is false.
52:24:160Paolo Guiotto: Okay? Because this would be true if this implication would be an if-and-only if.
52:30:320Paolo Guiotto: One is convergent if and only if the other is convergent. So, if the other is not convergent, this one won't be convergent. But this is false. So, this is not new, it's not a novelty, it's also for numerical series, real series is the same music, no?
52:47:280Paolo Guiotto: I see that you are perplexed, but think about this example for this. You have that this series, the famous logarithm series, this one.
52:57:230Paolo Guiotto: minus n plus 1 over n, the alternating series, 1 plus, 1 minus 1 half plus 1 third minus 1 fourth, and so on. This is convergent.
53:08:930Paolo Guiotto: But the series of the absolute values is the harmonic series 1 over n, which is divergent.
53:15:830Paolo Guiotto: So it is non-true that you can have that a series is convergent, but the series of the absolute values is divergent. So, you cannot draw anything, in fact, from this information.
53:28:270Paolo Guiotto: But it is the second one which is important, because the second one says the modulus CK z to K goes to plus infinity.
53:37:620Paolo Guiotto: And why this is important? Because you remind that there is a necessary condition that we almost never use.
53:45:200Paolo Guiotto: Which is this one. If you pretend to have convergence, you must have that the general term goes to zero.
53:53:560Paolo Guiotto: So this, in particular, says that this does not go to zero, and this implies that the series of the CKZ2K is not convergent.
54:09:100Paolo Guiotto: Okay? So, let's do the summary of this.
54:12:870Paolo Guiotto: So, that is…
54:19:90Paolo Guiotto: So we have this. If L is less than 1, convergence. If L is greater than 1, non-convergence, okay? So.
54:28:820Paolo Guiotto: If L, which is, I repeat, modulus of Z, the limit in K of modulus CK to 1 over K is less than 1, then the series… let's write the final conclusion. The power series CKZ to K is convergent. Well, let's put in right this.
54:49:790Paolo Guiotto: convergent.
54:52:340Paolo Guiotto: While if L is greater than 1, then the power seated is not convergent.
55:03:870Paolo Guiotto: Now, how do we look at this condition? This is a condition for Z, in fact, because it says that… so when you have a power series, these numbers are fixed, and you look at this as a function of Z. So, it's defined for different values of Z, and this says that, equivalently.
55:24:200Paolo Guiotto: When the modulus of Z is less than 1 over that number, the limit in k of modulus CK to 1 over K, then this says that the series CKZ to K is convergent.
55:42:790Paolo Guiotto: When modulus of Z is larger than that number, 1 over the limiting K of modulus CK to 1 over K, then the series of CKZ to K is not convergent.
56:00:910Paolo Guiotto: So, in other words, that quantity that you have at right says, if you have less than that quantity, modulus of Z is less than that quantity, series is convergent.
56:11:320Paolo Guiotto: If the model of Z is greater, series is not convergent. So, in some sense, this decides where the series converges or not. So you can give an idea with a figure of this situation. So, this is now the complex plane.
56:27:830Paolo Guiotto: Numbers with modulus of Z less than something.
56:31:690Paolo Guiotto: R numbers that, are in a disk, in a ball. Here, we can say disc because we are in a plane. So, in a disk, centering the origin, and with number, that number as radius. So, if we call this, we baptize with the letter R,
56:49:650Paolo Guiotto: It's the same here.
56:51:560Paolo Guiotto: We see that there is a disk centered in the origin with the radius r, that number, for which we have the following alternative.
57:00:130Paolo Guiotto: So this is the disk. I dash points on the edge of the disk, because for those points, modulus of Z would be equal to R, and here you don't see any conclusion, but that's explainable easily, because that's the case when the limit alley is 1.
57:18:150Paolo Guiotto: And the test is not able to decide if the series is convergent or not in that case, so it's not a big problem.
57:24:860Paolo Guiotto: It says that if you are here with Z, in this case, modulus of Z here, modulus of Z is less than that number R, which is the radius of this disk.
57:37:420Paolo Guiotto: So it means that for all points of that disk inside, you have convergence. So here you have convergence.
57:46:150Paolo Guiotto: While, if you are here, for example, with a Z, in this case, the distance to the origin, so the modulus of Z here is larger than R. So this is the region where modulus of Z is greater than R. And for this region, it says not convergence.
58:05:10Paolo Guiotto: So… Not convergent.
58:09:680Paolo Guiotto: So, in other words, the set of Z for which a power series makes sense is a disk centered in the origin, for this case, with radius r, where R is computed by doing that limit.
58:23:640Paolo Guiotto: one over the limit of absolute value of the CK, the CK are the coefficients of the power series, to root the decayed root of this.
58:34:840Paolo Guiotto: Okay, so we got this proposition.
58:42:300Paolo Guiotto: So, let… sum CK, Z to K, B.
58:49:890Paolo Guiotto: power series.
58:56:80Paolo Guiotto: with the numbers, CK, the coefficients, complex numbers.
59:01:880Paolo Guiotto: Let.
59:03:290Paolo Guiotto: R?
59:04:300Paolo Guiotto: introduce this R as 1 over the limit in K of the modulus of CK to 1 over K.
59:14:390Paolo Guiotto: Now, we have to say something here, because there are two degenerate cases. The case is when that limit is infinite, and the case when that limit is zero. In general, downstairs, you have a limit of positive quantities, so this limit will be between 0 and plus infinity, but it can be 0 or plus infinity. Why not?
59:33:670Paolo Guiotto: So we have to say what this means, when this quantity is zero, or when this quantity is plus infinity, but the interpretation is natural. When the quantity is plus infinity, you have 1 over plus infinity, that number will be 0, okay? So with the agreement that…
59:50:510Paolo Guiotto: 1 over plus infinity makes 0.
59:54:650Paolo Guiotto: And when you have 1 over 0, since this is a 0 positive, it's like having 1 over 0 plus, this will be plus infinity.
00:05:250Paolo Guiotto: So, let R this… be this number.
00:09:260Paolo Guiotto: That is, in general, between 0 included and plus infinity included. It comes plus infinity if that denominator is 0.
00:20:180Paolo Guiotto: Then… We can say that.
00:25:130Paolo Guiotto: The, the power series
00:33:80Paolo Guiotto: sum of a K of CK, Z to K is… convergent.
00:46:800Paolo Guiotto: if modulus of Z is less than that number R.
00:52:160Paolo Guiotto: not convergent.
00:55:870Paolo Guiotto: if modulus of Z is greater than that number R.
01:01:70Paolo Guiotto: So this number, R, is called…
01:07:590Paolo Guiotto: radius… off… convergence.
01:18:70Paolo Guiotto: Because, you see, it's a radius. In fact, it says inside the disk of centrade in the origin, a radius R, you have convergence. Outside, you have not convergence. So it distinguishes, it separates where is convergent from where is not convergent, okay?
01:36:650Paolo Guiotto: And, the disk… the, say, the ball centered in the origin radius R,
01:44:650Paolo Guiotto: without the edges, so this is the set of Z, such that modulus of Z is less than R, strictly less, is called…
01:56:790Paolo Guiotto: disk… off.
02:00:440Paolo Guiotto: convergence.
02:07:90Paolo Guiotto: Okay.
02:08:260Paolo Guiotto: So, this is a first important fact. Basically, it says that this type of functions, these power series, make sense in a suitable disk, which disk… the disk with radius R determined by that formula.
02:23:610Paolo Guiotto: There is an alternative formula, let me show the two. Similarly.
02:34:490Paolo Guiotto: We could… Apply it.
02:41:360Paolo Guiotto: the, ratio test… To the series of the absolute values.
02:51:600Paolo Guiotto: modulus CKZ2K, which is the same as above, modulus CK
02:56:940Paolo Guiotto: times modules Z to power k, so if we call AK this. Now, we have to assume that this number is strictly positive to apply the ratio test, otherwise we cannot do. So, if AK is strictly positive.
03:14:140Paolo Guiotto: So this definitely is not the case when Z is 0, for example, no? But for Z equals 0, the discussion is trivia, because if you put Z equals 0 into that formula, all powers except Z to 0, which is assumed to be equal 1, but all the other powers are 0. So this infinite sum is a fake sum. There is only one term, which is non-zero, all the others are different from zero.
03:38:950Paolo Guiotto: You know, so… who cares? So in particular, in particular, modulus of Z.
03:47:910Paolo Guiotto: Greater than zero, except if… or if you prefer, Z non-zero.
03:55:600Paolo Guiotto: Otherwise, that term is, zero.
03:59:950Paolo Guiotto: Well, now the limit to be computed is the limit of…
04:05:520Paolo Guiotto: AK plus 1 over AK, according to the root test. But this limit…
04:13:70Paolo Guiotto: is the limit in k… if we plug AK plus 1, we get modulus of CK plus 1 times modulus of Z to exponent k plus 1, divided by the AK is modulus CK times modulus z to power k.
04:29:100Paolo Guiotto: So simplifying a bit here, we can cancel this with this. Basically, it remains modulus of Z. So we have modulus of Z, which is independent of K, so we can carry outside of the limit.
04:42:670Paolo Guiotto: limit modulus CK plus 1 divided the modulus CK. Now, this is the L.
04:49:660Paolo Guiotto: For which we have that if this is less than 1,
04:54:780Paolo Guiotto: Then we have that series of AK is convergent.
05:00:420Paolo Guiotto: And since AK is the series of the absolute value of these guys, so we have, by the same test we mentioned above, this one. From absolute convergence, it follows convergence.
05:20:520Paolo Guiotto: So we have that the series, the original series, CKZ2K, is converged.
05:26:720Paolo Guiotto: If this is greater than 1, again, we apply the root test, the ratio test says that the series of AK is divergent.
05:35:990Paolo Guiotto: But this is not the important information here, because this says that the series of… the power series is absolutely divergent, but it does not say what happens to the series, because we can have that it is convergent. And the fact that cuts the edge is this one.
05:53:890Paolo Guiotto: Also, the general term goes to plus infinity, no? So this is another information that comes from the ratio test. So, in particular, this means that CKZ to K in absolute value goes to plus infinity, and therefore, it cannot go to 0.
06:12:630Paolo Guiotto: So, the necessary condition for convergence is violated, and this means that the serious CKZ2K naught.
06:22:170Paolo Guiotto: convergence.
06:24:230Paolo Guiotto: Okay? So, the conclusion is similar.
06:27:910Paolo Guiotto: That is… So let's rewrite here.
06:35:920Paolo Guiotto: If modulus of Z times the limit in K of modulus CK plus 1 over modulus CK, this is the L. It's less than 1, then series CKZ2K
06:52:940Paolo Guiotto: convergent.
06:54:850Paolo Guiotto: If it is greater than 1, then the series is not convergent.
07:02:120Paolo Guiotto: And this yields to the same conclusion we have seen above, the same type of conclusion, not exactly the same, because it says, even though if… so, modulus of Z less than
07:14:770Paolo Guiotto: So I should say 1 over that limit, but since that is a limit of fractions, I can just rewrite as limit of when I flip numerator and denominator, so modulus CK…
07:26:770Paolo Guiotto: divided modulus CK plus 1.
07:31:320Paolo Guiotto: So if modus of Z is less than this number, then the series is convergent.
07:40:750Paolo Guiotto: And if it is greater than this number, then the series is not convergent.
07:46:600Paolo Guiotto: So this number here that we still baptize as are…
07:53:50Paolo Guiotto: does the same thing as for the application of the root test we have seen here. So I have a completely similar statement, proposition.
08:06:890Paolo Guiotto: So, so, let… CK be a sequence of complex coefficients, and define… R… as the limit
08:26:50Paolo Guiotto: in K of this modulus CK divided modulus CK plus 1.
08:33:840Paolo Guiotto: Well, actually, you see that to define this thing, you need that the CK are non-zero. So, let's add here this condition, such that CK…
08:45:550Paolo Guiotto: is… well, CK is different from 0 for every K. So it's a bit more restrictive than the previous test. For the previous test, we do not have such condition, but for this one, we are for every K.
09:00:290Paolo Guiotto: Now, define this number, which is a number that can be between 0 and plus infinity, and points included, because it's a limit of positive numbers, so the limit will be positive, it can be 0, it can be plus infinity, why not?
09:17:620Paolo Guiotto: Then… We have the following, that…
09:22:29Paolo Guiotto: If modulus of Z is less than R, the series CK, Z to K is convergent.
09:33:220Paolo Guiotto: If modulus of Z is greater than that R, then the series is not convergent.
09:41:479Paolo Guiotto: So the same thing. So this number is also called the radius of convergence.
09:49:290Paolo Guiotto: Ease.
09:50:529Paolo Guiotto: Called up.
09:54:970Paolo Guiotto: radius.
09:59:480Paolo Guiotto: of convergence.
10:03:870Paolo Guiotto: Now, you may… I don't know if you may wonder to this question, but apparently there are two different formulas for the same quantity, so maybe they can be different. No, it can be proved that if they both exist, they must be the same, okay? There is no possibility that one is different from the other.
10:22:770Paolo Guiotto: So let's see an example, a couple of examples.
10:27:80Paolo Guiotto: The example one is we review a moment the case of the geometric series, but let's see with these tests what we gain from these tests. Okay, so for the geometric series.
10:49:360Paolo Guiotto: So this is sum for k going from 0 to infinity of Z to K. So as you can see here, the coefficients are constantly equal to 1, so particularly simple.
11:01:470Paolo Guiotto: Here we have that the CK are identically equal to 1. So, let's see what comes out as radius of convergence with the two formulas. You will see that it comes the same number. So…
11:15:450Paolo Guiotto: We have…
11:18:970Paolo Guiotto: are equal. The first formula says 1 over the limit in k of modulus CK to 1 over K, right?
11:27:770Paolo Guiotto: But that CK is 1, so the modulus of CK is constantly equal to 1, to 1 over K is 1. So we are doing the limit of 1, we get 1 over 1, 1. So the radius of convergence is 1. So this says that the series Z to K converges
11:46:750Paolo Guiotto: for module Z less than 1 is not convergent
11:52:00Paolo Guiotto: for module Z greater than 1.
11:55:510Paolo Guiotto: And that's it. It does not tell anything about the value of the sum.
12:00:400Paolo Guiotto: Okay? So if you want the value, you have to sum the series, there is no other way, okay? But it tells where it is convergent much faster than what we did, as you can see, no?
12:12:00Paolo Guiotto: Of course, this confirms what we already obtained directly, because this was the example of the
12:19:750Paolo Guiotto: geometric series, we have seen that the series is convergent if and only if model Z is less than 1. In this case, we have also the formula for the sum that we do not have, we cannot have from these tests, because they just say where the series is convergent.
12:34:770Paolo Guiotto: If you repeat the calculation with the other formula, the formula provided by the application of the root, the ratio tester.
12:42:810Paolo Guiotto: Weed the… ratio test.
12:51:510Paolo Guiotto: the formula for R. Be careful, because the ratio test does the limit of a fraction, like, next element of a previous element, okay?
13:03:850Paolo Guiotto: So next, over the present element, but this is the contrary.
13:11:200Paolo Guiotto: Okay, and that's why, because you see, this was the original formula we inverted to get the condition for Z, so be careful, because you could get a wrong conclusion if you write, for example, limit in K. Maybe you think you are applying the… somehow the ratio test, and you do this. This is wrong. It's not the right formula. The right formula… for this case, nothing changed.
13:36:150Paolo Guiotto: Because it's one, but for other cases, it could change.
13:41:290Paolo Guiotto: So this is the formula we should apply. Now, they are both identically equal to 1, so clearly we are doing limit of 1s, and we get 1. As you can see, the same value, same conclusion.
13:57:140Paolo Guiotto: Okay, now, for example, it's time to introduce… to use this to introduce a definition of the exponential function. For a moment, I will use a different notation, because it is not yet clear what is it. So we call XPZ
14:17:40Paolo Guiotto: by definition, this infinites, the sum of this series, so sum for k going from 0 to infinity of Z to K divided by K factorial.
14:28:150Paolo Guiotto: Okay?
14:29:600Paolo Guiotto: Now, this is called the exponential… function.
14:37:110Paolo Guiotto: that soon will be denoted by E to Z. There is a reason for that. For a moment, let's define as XZ. It turns out that proposition
14:48:810Paolo Guiotto: Well, this function, exp, is XZ.
14:54:600Paolo Guiotto: is defined… is defined…
15:01:860Paolo Guiotto: for every Z complex. So it is the function defined in the intercom complex plane. So, in fact, this means that that series is a power series whose radius of convergence is equal to plus infinity. That's what we proved.
15:17:830Paolo Guiotto: Moreover.
15:23:230Paolo Guiotto: the… Following… Properties…
15:34:20Paolo Guiotto: Old?
15:36:930Paolo Guiotto: So, number one, EXPER.
15:40:780Paolo Guiotto: of 0,
15:42:650Paolo Guiotto: is 1, like, E to 0 is 1, okay? Number 2, X, if we restrict to reals, so X, when X is real, coincide with E to X.
15:59:540Paolo Guiotto: Number 3…
16:02:80Paolo Guiotto: We have what is called the group property, so if we do X of Z plus W, this is… so, like, E to Z plus W, the exponent… the…
16:13:680Paolo Guiotto: The power, when you do the exponent, the sum… the exponent is the sum of two exponents splits into the product, so this will become XZ times XW.
16:25:890Paolo Guiotto: for every Z and W.
16:29:740Paolo Guiotto: Complexer.
16:31:730Paolo Guiotto: Another formula is that the conjugate
16:36:500Paolo Guiotto: of X per, the conjugate number, is X of the conjugate, for every Z, complex.
16:47:970Paolo Guiotto: Let me see… We have any other property?
16:55:370Paolo Guiotto: No, it seems that these are the properties.
16:58:220Paolo Guiotto: Okay, let's see the proof of this.
17:04:380Paolo Guiotto: So, let's say that we check first, which is the application of what we have seen, we check, we start…
17:14:220Paolo Guiotto: Checking… that,
17:20:260Paolo Guiotto: this function, sum over k going from 0 to this series, 1 over k factorial Z to K, which is the X series, is convergent.
17:35:700Paolo Guiotto: for every Z complex.
17:38:910Paolo Guiotto: Now, this is a power series sum for k going from 0 to infinity. CKZ to K, where CK is exactly equal to 1 over k factorial.
17:52:600Paolo Guiotto: Now, we have to compute the radius of convergence, because the radius of convergence says that
18:00:20Paolo Guiotto: When modulus of Z is less than the radius, series is convergent. If modulus Z is greater than the radius, serious is not convergent. If we prove, as we will prove, that R is equal to plus infinity, you see that this becomes modulus Z less than plus infinity, and this is every Z.
18:17:610Paolo Guiotto: then you have convergence. A model Z greater than plus infinity is never.
18:21:880Paolo Guiotto: So the series is always convergent. So that's why we computed the radius of convergence.
18:27:710Paolo Guiotto: For this, we have two formulas for the radius. It is better to use this one because of the factorials, okay? It's much easier than the other one, so we do that. So, the…
18:43:580Paolo Guiotto: radius… of convergence, is R equal limit in K modulus CK divided modulus CK plus 1.
18:59:280Paolo Guiotto: So here we have limiting K of… seek is 1 over k factorial, so this is modulus of 1 over k factorial divided modulus of 1 over k plus 1 factorial.
19:14:610Paolo Guiotto: Now, of course, these are all positives, so we can cancel the modulus, and we can rewrite this as a limit in K,
19:23:380Paolo Guiotto: of… we can carry this K plus 1 at numerator divided by k factorial here. So now we can easily simplify this too, because k plus 1 factorial can be seen as k plus 1 times the product of all naturals between K and 1, so this is the factorial of K.
19:41:370Paolo Guiotto: That cancels with this one, and so we get the limiting K of K plus 1.
19:47:180Paolo Guiotto: And that's equal to plus infinity.
19:50:820Paolo Guiotto: So the radius is plus infinity. You see that in this case, if you write a wrong formula, so the formula with this red cross.
20:00:430Paolo Guiotto: it will get zero. So, it's an entirely different story, because this… when you have a radius of convergence zero, the series is… well, actually, you have to be careful, because at Z equals zero, they are always convergent, so you can… you have to interpret this when R is 0. When R is zero, this says that for module Z greater than zero, because model Z cannot be less than zero, it is at least zero.
20:25:220Paolo Guiotto: So only the second condition applies. This says, for model Z positive, so N is different from 0, the series is not convergent.
20:34:430Paolo Guiotto: So this basically says this series is converged at one single point, Z equals 0, and no other point. So it's an entirely different story. That's why I said be careful when you apply the formula, because
20:45:910Paolo Guiotto: The wrong formula leads to the wrong… the wrong result. So this says that the series 1 over k factorial Z to K converges for every Z,
21:00:120Paolo Guiotto: less than plus infinity, module Z less than plus infinity, that is for every Z complex. So, the function, exponential function, is well defined for every Z. Now, let's see the properties.
21:15:330Paolo Guiotto: Exposito…
21:17:360Paolo Guiotto: Well, that's easy because this is the sum as k goes from 0 to infinity of 0 to power k divided k factorial.
21:27:440Paolo Guiotto: Now, here, there is an agreement to remind that Z to 0 for us means
21:35:400Paolo Guiotto: identically 1, okay? So, here, it's needed, this, this, because 0 to 0 wouldn't make any sense in mathematics, okay?
21:46:70Paolo Guiotto: So, what is 0 to 0? Well, remind that the zero power is a simplification of 1, otherwise I should split the formula for k equals 0, say something, and k different from 0, something else, but it's better if we keep a unique formula. So, what happens here? For k equals 0, I get z to 0 is 1, so 1 over 0 factorial. Also, 0 factorial is in agreement, is
22:11:820Paolo Guiotto: And then, all the other terms, from 1 to plus infinity for k, 0 to k is just 0, no? Because it's 0 times 0 k times, so it's a sum of zeros over k factorial, and all this is equal to 0.
22:26:550Paolo Guiotto: So, it remains only the first term, which is one as expected.
22:31:150Paolo Guiotto: Number two, we say that exponential…
22:35:980Paolo Guiotto: on reals, well, this is the story I told at the beginning of this class. The exponential function, according to the definition of exponential function, when x is real, is just X to K divided k factorial, some of this.
22:52:980Paolo Guiotto: We accept that this is E to X for every X real, because this is something that comes from, let's say, first-year calculus.
23:03:120Paolo Guiotto: And that's it. Third, we have the, the, this, formula that,
23:11:800Paolo Guiotto: says this is basically working as an exponential. Now, because only exponential function does this. Exponential of the sum, F of the sum is products of F. So, this is… this requires a little bit of calculations, because
23:27:930Paolo Guiotto: We write this as the sum of a K of Z plus W to power K divided k factorial.
23:36:00Paolo Guiotto: Now, how do we handle this? Of course, what we do is we use the binomial expansion of Z plus W to power k. So the binomial formula, binomial…
23:51:700Paolo Guiotto: expansion. So, it is a sum of the binomial coefficients. Now, remind that this is… the exponent is k, so the binomial coefficient is something like k over J.
24:03:850Paolo Guiotto: Then we have Z to power J, and W to power k minus J. So we have this sum for J going from 0 to K.
24:13:890Paolo Guiotto: This is the binomial formula.
24:16:90Paolo Guiotto: Do you remind of this?
24:18:350Paolo Guiotto: I've never seen this, die. Newton formula.
24:22:230Paolo Guiotto: It's impossible, you have nothing.
24:25:570Paolo Guiotto: So, this is sum over K, sum over J going from 0 to K, of k over the binomial coefficient. Let's write explicitly, this is k factorial divided j factorial k minus j factorial. This is
24:41:830Paolo Guiotto: Z to J, W, K minus J.
24:47:220Paolo Guiotto: Okay.
24:50:40Paolo Guiotto: Now…
24:54:970Paolo Guiotto: What do we do here? We want to reconstruct the two exponentials. So let's put together this J factorial with this Z to J,
25:08:510Paolo Guiotto: That is the generic term of the… this… the exponential series for Z, and this seems to be…
25:18:170Paolo Guiotto: together with this, the generic term of the exponential series for, for W. So let's, for a second, write this. Sum over K, sum for J going from 0 to K. There is this K factor that we have to decide where to put.
25:33:650Paolo Guiotto: Then we have Z to J over J factorial, W to K minus J divided the k minus j factorial.
25:44:330Paolo Guiotto: Now, to recognize that this is really the product of the two exponentials, let's write the product here, XZXW.
25:53:410Paolo Guiotto: Now, this is the product of sum, of, so, let's say K
26:00:740Paolo Guiotto: or better. Let's use the letters used here. So for the Z, we use the Z to J divided J factorial times, for the other one, I don't use K minus J. For the second, let's use the letter H, W to H divided H factorial.
26:18:350Paolo Guiotto: Now, you may imagine that when you do this product, it's like the ordinary algebra. This is 1 plus Z plus Z squared over 2 factorial, and so on.
26:30:290Paolo Guiotto: This is an infinite sum times another infinite sum, 1 plus W plus W squared over 2 factorial, etc.
26:38:740Paolo Guiotto: So you may think that you have to multiply each term of the left parenthesis to each term of the right parenthesis.
26:44:770Paolo Guiotto: But you may regroup these terms in this way. Let's regroup terms whose the sum of the exponent is fixed. For example, the sum of the exponent of Z to the exponent of W is a number
27:00:890Paolo Guiotto: K, okay? So what happens? For example, you have to put this with the W to K over K factorial.
27:09:260Paolo Guiotto: So these two together. Then I have Z times that term with W to K-1. So these terms are of type. Z to J over J factorial times W to K minus J, because the sum of the two exponents must be equal to K times over k minus j factorial.
27:30:590Paolo Guiotto: Now, I have these terms, huh?
27:33:670Paolo Guiotto: And I have to do all possible combinations, so you see that, basically, you have, exactly k factorial of this, so that's why…
27:47:240Paolo Guiotto: we get that formula. So, this is the sum.
27:51:280Paolo Guiotto: For J, going from, 0 to, P.E.
27:57:610Paolo Guiotto: And then we sum up over all K. K going from 0 to infinity. And that's exactly the formula you have above, is this one.
28:07:190Paolo Guiotto: So, you see that XP, finally, XZ plus W equal XZ.
28:14:820Paolo Guiotto: times X… W.
28:18:170Paolo Guiotto: Okay, it remains the last one, you can check by yourself, do the conjugate of the series, you will see that it comes, the Z conjugate. Okay, so check number 4.
28:32:200Paolo Guiotto: Okay, let's stop here.
28:35:660Paolo Guiotto: And, I leave you to do the exercise on power series at the end of the chapter.
28:41:330Paolo Guiotto: So the first exercise, which is the number…
28:49:640Paolo Guiotto: It's an exercise on computing the radius of convergence.
28:54:190Paolo Guiotto: So that's 7-11-1.
28:58:160Paolo Guiotto: Do… 7, 11, 1.
29:02:820Paolo Guiotto: Okay, have a nice day.