Complex Power Series and their Derivatives

Key Takeaways

in this video we'll be going over power series for complex numbers and we've seen that throughout this series we've been doing a lot of counterparts for example we did uh limits for complex numbers just like those limits for real numbers we did uh functions for complex numbers just like those functions for real numbers so here we're gonna do power series for complex numbers just like there are power series for real numbers and the reason we're doing these over again after the real numbers case why we're doing it for complex numbers is that it's not necessarily always true if something is true for real numbers it's not necessarily always true for complex numbers there is a majority of cases where they are they do agree with each other but there are important cases where they don't so it's important that we prove things over again in complex numbers and don't just take them for granted so let's go ahead and start we'll start with the definition of power series so a power series is an infinite series of what form so the form is like this sum as n goes from zero to infinity of a sub n which is the coefficient z minus z naught to the power of n so let's just label some things so at 7 as i mentioned is the coefficients of this power series and these are important things note these are complex numbers so these can involve eyes and anything and then z naught is called the center of the power series okay so now i want to go straight into a theorem so this is going to be called theorem a and we'll go over the proof which i've written out after i write down the theorem so theorem a says that suppose there is a z one not equal to z naught such that the sum of a sub n uh z one minus z naught to the n converges so that's a supposition so suppose there is some z1 not equal z naught such that this converges and of course this is going to be n equals 0 to infinity and now what what the conclusion is is that then for all z so this upper uh this upside down a is a for all for all z with z minus z naught is less than z one minus z naught and this is the modulus so the modulus of z minus z naught is less than the modulus of z one minus z naught uh the sum a sub n z minus z naught power of n converges not just converges converges absolutely so this seems kind of messy right now maybe we're not really sure what's going on but in the proof it'll make more sense so i've written it again on this proof paper so let's just go through the outline of the proof so let's look at the statement again suppose we're going to look at this picture and the statement at the same time so we'll get a feeling for what's going on here so suppose there's a z1 not equal to z naught such that the sum from zero to infinity and sometimes i just write zero here but really you should think of it as n equals zero from zero to infinity of uh a sub n times z one minus z naught to the power of n converges okay so what does that mean so i've drawn a little picture here which i've enlarged the circle so here's z naught z naught remember is some point on the complex plane z one is some other point on the complex plane so if we take their difference and we raise it to the power of when n equals zero and then next we go n equals one n equals two and then we multiply that by the coefficient each time this will converge okay uh if that converges then for all z in which the modulus of z minus z naught so that means for all z in which the distance from that z to z naught uh is less than the distance from z one to z naught then that new series here concerning that new z is absolutely convergent so what does that all mean so here is uh i've drawn the circle and the size of the circle the radius is the distance from z 1 to xenon also known as modulus z 1 minus z naught right because modulus are distance and everything all the little purple lines in here is representing the inside of the circle only not the x not the not the exterior which is why it's dotted inside the circle every point in that inside of the circle every z in there satisfies this condition because the distance from any purple point to z naught is less than the distance from z 1 to z naught and we just see that geometrically and then if we take that distance and then we put it through the series that will be absolutely convergent so now we kind of see why this might be true because now we're taking a smaller distance and we're adding it over and over again so it seems like that should be convergent because that's smaller than the sum here so this sum converges that sum should converge but nonetheless we need to prove it so let's go right into the proof it's not that bad of a proof so the proof starts here suppose that z minus z naught the modulus is less than or equal to r and that r is less than the modulus z one minus z naught so to illustrate what that means i've enlarged this picture and i've added some extra details so again here's the z knot and on the orange circle on the outside here's z1 okay and now i've drawn some z any z that is inside this purple area so i've drawn it here for the sake of simplicity and now we have this new pink circle whose radius is some some r r doesn't consider us we don't need we don't need to care about exactly what value it is what we do need to care about is that r is bigger than the distance from z naught to z but it's smaller than the distance from z naught to z1 which is why i've drawn it intermediate and the important thing is we can always form a circle like that and the reason for that is because this purple disc right here is an open set so we can get infinitely close to z1 z can get infinitely close to z1 but there will always be enough distance in between so that a circle of some radius can fall between them so um try to try to wrap that around your head it just has to do with how this is an open disc and uh when we have numbers there's new there's no such thing as two numbers that are next to each other like there's no consecutive numbers because numbers have infinitely many gaps between them okay so next uh next step in our proof is to look at this statement since we know this is convergent why do we know this is convergent because that was uh suppose it's converging so we know this is convergent what do we know that means the limit as n approaches infinity of this inside approaches zero and why do we know that well that has to go back we go back to a single variable calculus and look at how if a series is convergent then the the ending the terms have to go to zero because if they don't go to zero then the series can be convergent because we're just going to keep adding things that are bigger than zero over and over again so that means that this sum the terms in the sum have to go to zero and that so that means this is true so since this is convergent the limit equals zero and furthermore there exists some constant m okay such that and now we're going to use absolute values we're going to have modulus a sub n times modulus z1 minus z naught to the power of n is less than or equal to m so now you're thinking why is what does this mean well think of it this way if you have a if you have a series like this there's always going to be a biggest term in the series right so you can't have the biggest term be infinity because this is a finite series so even the biggest term is huge it's still not infinity and since that's the biggest term all other terms are smaller than that so if we say that that means there exists that constant m and m will be that biggest term such that the absolute value of the terms in the sequence are always less than that biggest term and why would why we have to put absolute value is uh because first some some sequences let's say in real numbers can have an alternating negative positives so we can have like negative 5 4 negative 3 2 negative 1 and so on so the uh so if we take the absolute values of these it's going to be 5 4 3 2 1 and the biggest value will be 5 assuming that these just keep getting smaller and absolute value nothing weird happens down here so the same thing with complex numbers we'll take the absolute value and the absolute value will always have a maximum and the maximum will be m and this is true for all n so all terms in the series as we discussed so now we'll finish this proof off so now now i've written here uh absolute value a sub n times modulus z minus z naught to the power of n equals this and this seems kind of messy but if you just look at this look at this term right here z 1 minus z 0 to the power of n that's the same term as z 1 minus z naught to the power of n right here so these can technically cancel and you'll just be left with a sub n times z minus z naught modulus to the power of n n which is exactly what we have right here so i've just written this in expanded form and this right here is always less than or equal to m times p to the power of n i'll explain what p is in the second but uh what is m so we said that uh absolute value a sub n times z1 minus z naught power of n is always less than or equal to m so remember so we look at this and this is the same thing as this whole term right here so that is always less than or equal to m now what about this right here so p is defined as r remember what r is uh in fact i've written it here so r is that intermediate disk um and we we've uh remember that z minus z naught modulus is less than or equal to r which is what we have right here so z minus z naught is always less than or equal to r so that means that z minus z naught to the power of n is going to be less than r to the power of n which is exactly what we see when we have p to the power of n because that includes r to the power of n so just to recap we have modulus z minus z naught over modulus z 1 minus z naught all to the power of n and since r is greater than or equal to z minus z naught modulus p to the power of n has to be greater than z minus z naught modulus over c one minus z not modulus okay so this holds true so that means that this right here is less than or equal to this and now to finish this off since this sum right here so sum n equals 0 to infinity m times p to the power of n converges and why does that converge well let's look at it this this summer here m is a constant so we can factor it out it doesn't really affect the convergence or divergence of this series so what about p to the power of n so well we have this p and this p is a fixed number right because r is a fixed number z one minus z naught is a fixed number and furthermore we know that z one minus z naught modulus is greater than r which makes p less than 1 which means that this is a geometric series so go back to your algebra or geometry calculus to remember what geometric series is but the fact that this is a geometric series right here means that it will converge because the absolute value of p this ratio factor is always less than one which is what we said here okay so now the final the final uh statement that we have to say is that since we established here that each this is always less than or equal to this that means each term in this series so each term in this series right here is less than its corresponding term in this series and since this series converges and each of the terms in this series is less than this one this also converges and that means that our series right here that we're trying to prove is absolutely convergent remember what absolute convergent means it means that if we take the absolute value of the series which we did here that converges and since this converges that means that this converges because if a series is absolutely convergent it is convergent so i know that was a lot of that was a big mouthful if you if you want to understand this proof thoroughly go through these steps again it is important to understand it thoroughly for the sake of future proofs and things like that but the basic idea let me just go through it super fast is that we established an intermediate r and the reason we did that is because we want to use it for a geometric series test later we established that the limit has to go to zero and also that there's a maximum number and these are ideas from calculus and then we said that we equated we we said that this series is less than this series which is like a comparison test we said that this is less than this so we compared these two series and we said that this one converges therefore this one converges and since this one converges it is absolutely convergent which means the original series without the absolute values is also convergent so all that all that to prove that this is absolutely convergent so now let's get back to uh the new information in this video so we've gone over theorem a we proved it true the next topic to talk about is called radius of convergence so you might have remembered this from calculus the series part of calculus but we'll go over it again so radius convergence there are three different cases case one is that this series sum from uh zero to infinity a sub n z minus z naught power of n converges only when what well only when z equals z naught in which case uh the radius of convergence which we'll label r is gonna be zero okay and that's because uh if we draw a little graph and we have our z naught we can't go anywhere because no other points work so we just have the point z naught and the radius of a point is kind of you can say zero the next case is the this the same sum right here converges for all z and that is the extreme case the like the opposite case of this so that would be if you have z naught you can have any point of the complex plane you want it'll converge for that and then we kind of say here the radius convergence is infinity because you have an infinitely big radius because it's the whole plane okay in the last case which is the one that will be of interest to us mostly is that the same series converges four only some z that are not equal to z naught so the important thing that is that there's at least uh there's at least one z separate from z naught for which this converges so we can put at least one z different from xenon here such that this has a converging value okay and in that case we say the radius convergence will be some positive number for example this is just an example radius convergence equals five that means that the distance from z naught is five so everything within a five unit radius of z naught will be convergent inside that disk so we'll see more of that right here so now we're going to analyze case three a little bit further so what does radius of convergence r actually imply so i've written what it implies here so that means if this modulus z minus z naught is less than r so let me draw a picture because pictures usually help so here's our z naught and let's say this is our r so this is the distance r and so if we choose everything inside the circle not on the boundary so i should have drawn the boundary um open disk but not in the boundary so everything inside here these dashed lines then this then this series right here will what it will converge but if we choose something outside the boundary so if we choose anything out here if we use anything outside the disk then this series will do what it will diverge there will be no value that it is limiting to or converging upon now there is a third case of course so what if we have it equals r so what if we choose any point on the circle's boundary well that is the most that's the ambiguous case we do not know it could converge it could diverge we have to use some we have to use some different methods to figure out uh which which one it is so remember less than the radius convergence converges greater than radius convergence diverges equal to radius convergence ambiguous we don't know so now we're going to go over to familiar tests the ratio test and the root test root test is probably less famous but the ratio test probably you've heard of it so let's go over the ratio test so we're just going to do these via example so we want to find out whether this sum right here converges and we want to find out furthermore what is the radius convergence so the ratio test says that you take the limit as n approaches infinity of absolute value or modulus of the term n plus 1 over the term n so we're going to do exactly that we're going to do limit n approaches infinity so what is term n plus 1 that involves just putting n plus 1 everywhere i see an n so it's going to be negative 1 to the n plus 1 z to the n plus 1 over n plus 1 factorial and i'm going to multiply by the reciprocal so i'm going to multiply by the reciprocal of the nth term so that's going to be n factorial over z power of n negative 1 to the n now important to note since this is an absolute value or modulus everywhere you see a negative one to the power of n negative one to the power n plus one that just cancels out because it's in it's just gonna be a positive or absolute value number so next thing to do is rewrite this a little bit so we have z to the power of n plus one can be broken up as z to the n times z and now we have n factorial over n plus one factorial can be written as n plus one times n factorial and we have z to the power of n so lots of canceling going on here and we're left with z over n plus one and the limit as n approaches infinity so as n approaches infinity the denominator gets infinitely big and z over something infinitely big will be zero because z is going to be some some complex number and the absolute value or modulus of 0 will be 0. so this is going to be 0. and now what does the ratio test say the ratio test says that once we do this if we get something less than one then it's convergent if we get something greater than one then it's divergent if we get equal to one questionable we don't know we have to use some other test but in this case we got zero which is less than one so this is convergent and furthermore what is the radius convergence it's going to be infinity and y will be infinity that's because 0 is always less than 1. there is no condition so we'll see if you do the ratio test on some other examples you might get something some function of z for example let's say you got z minus one is has to be less than one that means or absolute value so that means that uh centered at one the disk centered at one with radius one everything inside there will be convergent but in this case since zero is always less than one it's going to be infinite because we can put any z we possibly want and zero will still be less than one as it always is and it'll be convergent so convergent radius of convergence is uh infinity now the root test is a little bit less known and you'll probably be using it less but it's nonetheless important we can uh we can go over it really quickly so root test says that you take the limit as n approaches infinity of the nth root of and we do absolute value of modulus again the nth term okay so let's try to do it for this example in the space we have right here so we're going to have limit as n approaches infinity and instead of putting nth root i'm going to put to the power of 1 over n to make this a little bit faster so we're going to have uh so t sub n will be 5 to the power of n z minus 1 to the power of n to the power of one nth and now since phi to the power of n is positive no matter what we can take it out of this absolute value all together uh and then we have 5 to the power vm 5 to the power of n times uh to the power of 1 to the power of n so this n and 1 over n cancel out to make simply 5 and then we have z minus 1 and that's gonna be i'm going to shift these absolute value of modulus inside of it and the same thing will happen n and 1 to the power of n so it's going to be to the power of one and we'll leave it like this and that has to be less than one so the conditions of the root test are the same as ratio test if this limit is less than one convergent greater than one divergent equal to one uh ambiguous so we have to have it less than one now we see here here's the case where the limit involves z so this is going to be z minus 1 is less than one-fifth so the radius convergence in this case will be one-fifth because we have this z centered at one so if we draw a little tiny picture centered at one and the how big can the circle be it can be one-fifth that's the radius so the radius convergence is one fifth so again ratio test more important root test it is useful in some cases but we could have just easily used the ratio test here actually so not not too big of a deal so now we're going to go into the main new topic of this video which is the derivative of a power series so i've written the gist of the what the information is in green here so it says that if f of z equals the sum from zero to infinity of a sub n times z minus z naught the power of n and let's just take this moment to talk about how how this is written so since this power series is it's really just a function of z correct because this a sub n these coefficients are fixed um and this z naught is fixed all you're really all you're really changing here is z because this is an infinite function it goes on forever but it's still a function you can plug in z and so it can be a function of z so it is f of z equals this series if this series or function has a positive or infinite radius of convergence okay so not zero uh then inside this disk modulus z minus z naught minus r so disc centered at z naught with radius r inside that disc f of z is what it is infinitely differentiable and what does infinite infinitely differential mean it uh it simply means that you can take the derivative once you can take it twice you can take it three times you can take it four times you get to get infinitely many times and you won't encounter any problems whatsoever it'll just keep being able to differentiate eventually maybe you might get to a point where the function some derivative is just zero well that's not a problem you can differentiate zero you just get zero and just keep doing that over and over again no problems so basically it means you run into no snags no division by zero errors nothing like that and furthermore each derivative is given by another power series so if you take the derivative of a power series with the infinite or positive radius of convergence you get another power series and then you take the derivative of that get a new power series and what is each derivative given by so this notation right here is the kth derivative of f evaluated at z so it'll be for example if k was 1 it'll be the first derivative of f and you put in whatever z you want so that is given by this new sum so the sum from should be at n equals k so the sum from n equals k to infinity of n times n minus one all the way up to n minus k plus one times a sub n times z minus z naught the power of n minus k so you're probably thinking right now that's very messy um so i agree it's very messy so we'll just do it we'll look at it by example and not worry too much about the exact definition so just go right into this example um so let me remind you again of course uh i put the zero here it should be at n equals k so you're starting at k you're not always starting zero okay okay so our example is here we have f of z equals uh the sum from n equals zero to infinity of n times z to the power of n so i've written a few terms out here so the first term would be zero times z to the power of zero so it would be zero next term would be one times e to the power of one so it's z then it's two z squared three z cubed and so on so that the exponent and the coefficient basically they match so what is the first derivative and we're not going to do it using this complicated formula we're just going to use it do it during using inspection right so we'll do a term by term what's the derivative of zero with respect to z well it's zero what's the derivative of z it's going to be one what's the derivative of 2z squared it's going to be 4z what's the derivative of 3z cubed so it's going to be 9z squared and so on now does that match what we want to see is does that match our definition right here if it matches then we see this definition is uh it's it's true okay so let's put this through the definition so we want the first derivative right so in this in this case we want k to be one so we're going to have k equals one and so n equals k is going to be n equals k equals one and then we're going to have n minus k is really n minus one and we're gonna have n minus k plus one is really n minus one plus one so it's just n so that means that all we're really dealing with is this this n none of these other terms so if we rewrite this it's going to be sum from n equals 1 to infinity of n times a sub n uh z minus z naught to the power of n minus one okay that should be the derivative so that should once we expand it out give us the same answer as zero plus one plus four z plus nine z squared and so on so now what is z naught in this case z naught is zero because there's no it's really z minus zero right here so this is nothing this is zero uh what is a sub n a sub n was n so this is going to be n so we'll just finish this up on a new paper and so this is the series we're left with after everything cancels out we have 2n so we have n squared and we have z to the power of n minus 1. and now we're going to put in 1 in here what do we get we get z 1 minus 1 z to the power of 1 minus 1 is e to the power of 0 so it's going to be 1 and we put in here we get 1. so we get one plus now when we put in n equals two we get two squared is four z to the power of two minus one is z to the power of one so just z now we put in z equals uh when we put z equals three we get 3 squared is 9 and then we have z squared so we see that it's matching up exactly with what we did using just observation using just term by term derivatives so the last instruction here we'll be doing is that we'll be seeing what happens if we plug in z equals z naught into this thing so i've written out the formula for the kth derivative again and we're going to see what happens if we want the kth derivative of z naught remember z naught is the center of the series so we're going to have let's see how this formula transforms we have n equals k infinity n n minus 1 goes all the way down a sub n but now here's the big difference since we have z minus z naught and z is z naught this is going to be zero so we're gonna have zero to the power of n minus k so you're probably thinking this is going to have to be 0 right because there's this big fat 0 here multiplied by anything well there is a slight nuance and let's find out right now so let's go ahead and just start writing down the terms of the series since we started n equals k let's plug in k so if we plug in um if we plug in k here we get n times n minus 1 all the way to if n is k then it's k minus k which is 0 plus 1 so this is really just k factorial so it's going to be k times k minus 1 all the way up to 1. so it's going to be equal to k factorial times 0 to the power of what so since n is k k minus k is going to be 0. so it's going to be 0 to the power of 0 plus and then what's the next term it's going to be k plus 1 so we're going to have k plus 1 and then k plus 1 minus 1 which is just k so it's going to be k plus 1 times k plus all the way up to and these are all times k plus 1 minus k plus 1 which is just 2. and then it's going to be 0 to the power of k plus 1 minus k which is 1. and 0 to the power of 1 is definitely 0. so it doesn't even matter where all this is this is going to be 0. and for that matter any further terms are just going to be 0 because the next thing the next term will have 0 to the power of 2 0 power 3 and they're all going to be 0. but let's look at this first term really fast what is 0 to the power of 0 if you use l'hopital's rule so if you evaluate this limit limit as n approaches 0 of n to the power of n so that's going to go to 0 to the power of 0 if you evaluate this using l'hopital's rule you're going to find that this is actually 1. so that means that this is actually k factorial and i think i left something out so this should have a a to the power of k here so it should be multiplied by a to the power of k and this should have been multiplied by a to the power of k plus 1 but that canceled anyway so it's going to be k factorial times a to the power of k so in the end what do we have we have the kth derivative at z naught equals a to the power of a sub k times k factorial and let's write this in a form that we're going to see in future videos so i'll write it up here the formula we're going to see in future videos is a sub k equals f kth derivative evaluated at z naught all over k factorial so this thing you see in black up here uh it's very important we're going to use later on for different series and uh evaluating new kinds of things so so again it's a sub k so the kth coefficient in this series is given by the kth derivative evaluated at z naught divided by k factorial okay so keep that in mind the last thing we'll be doing is a involved example which kind of puts together everything we've learned in this video so here we have a series we have sum from n equals zero to infinity of z to the power of n minus n factorial and we know this is convergent with radius of convergence infinity because if you go back to the part of the video that was about ratio test this is the this is the series we did except it had a negative 1 to the power of n in front of it but that went away anyway so they're both uh convergent with radius convergence infinity and here for emphasis i've written that the coefficient a sub n is given by 1 over n factorial so that's what this is and z naught the center of the series is going to be 0 because we see that it's just a z to the power of n so it's really z minus 0 to the power of n okay so now what do we want to do with this we want to we would like to find the derivative of this so let's use the formula right here that we have we want to find the first derivative only so we're going to go ahead and write f i'll put a 1 here instead of prime just to emphasize that it's the first derivative f prime of z well it's going to be what it's going to be first i put a sum and then n equals k and n equals 1 in this case to infinity uh times n times all this stuff but really um if we remember the first derivative problem we just did it's just going to be n and then what do we have we have a sub n so that i've written here is one over n factorial so one over n factorial times z minus z naught z naught is zero so just z to the power of n minus k so it's z to the power of n minus k which is one so now we have this is going to be and this n over n factorial will do what well we have n over n factorial which is n over n times n minus 1 factorial so that will cancel and we'll be left with 1 over n minus 1 factorial so this all after we're after this dust clears is sum n equals 1 to infinity of 1 over n minus 1 factorial z to the power of n minus 1. and since we're starting we're starting the series at one but let's just start it at zero so that we can take this n minus one this n minus one and put it n instead so n equals zero infinity z to the power of n over n factorial so uh so uh convince yourself that this right here and this right here are equal i just started this one at one i started this one at zero but if you write out the terms so the same exact series and furthermore here's the very very interesting thing to know this right here uh sum from zero to infinity z to n over n factorial is the exact same thing we started with which means that f if we call this f of z then this equals f prime f equals f prime it is its own derivative and what function do we know that is its own derivative in calculus and things like that well it's the exponential function so let's do as the last part of the video let's really fast prove that f is the exponential function that the exponential function is given by this power series by doing this so what we're going to do is we're going to look at the function e to the negative z times f of z and we're going to take its derivative so we'll do a power series really fast so it's going to be this times the derivative of this times this times the derivative of that so it's going to be minus e to the negative z f of z and since f prime of z is equal to f of z we could just put a prime right here so what we have is e to the negative z times f prime minus e to the power of negative z times f prime so this is just zero and what does that mean that means the derivative of this is zero and what what kind of things have derivatives of zero constants which means that this function inside here is a constant so e to the minus z times f of z is equal to some constant k and the last thing we want to do is try to figure out that constant so we're going to write down e to the power of negative z times f of z equals that constant you write that and so it doesn't matter what number we plug in here for z it's gonna be a constant so let's plug in zero so we get e to the power of negative zero times f of zero equals k and e to the power of zero is one times f of zero equals k what is f of zero well let's write down what f was again f of z we said was this sum uh we said it was sum zero to infinity z to the power n over n factorial right so if you plug in zero in here then the zero becomes a zero and we start with zero to the power of zero which is one we said using l'hopital's rule and so it equals one and now if we have if we put n is bigger than zero so if n is one two three four all these other terms are gonna be zero so that means f of zero is one so we have one times one equals k so k is itself 1. so that means that e to the minus z times f of z equals 1. so that's using this step right here and plugging in what k is what we found k to be so now if we just move this e to the negative z to the other side we get f of z equals e to the power of z and we're done we've proved that that uh that this infinite series right here represents e to the power of z so remember we did that for real numbers but as i noted at the beginning of the video we can't always assume things for complex numbers that are true for real numbers but now we have proved that this case in particular is true f of z does equal e to the z so in the next video we'll be going over a new topic calcium's theorem