An interesting infinite sum, featuring Fourier series.

in this video we're gonna use Fourier series to derive a nice closed form for this sum so we have the sum N equals 1 to infinity of cosine in over N to the fourth and here's the basic idea of a Fourier series which we won't prove but there's a bunch of videos on YouTube where you can see the derivation of this kind of formula and if f of X has period 2pi its Fourier expansion can be given by the following so we have 1/2 a 0 plus the sum N equals 1 to infinity of a n cosine NX + b + sine in X where a n including a 0 is 1 over pi the integral from 0 to 2pi of f of X cosine NX DX so notice if n is equal to 0 here we just get cosine of 0 which is 1 and then B in is 1 over pi the integral from 0 to 2pi of f of X sine and xdx great and we're actually going to apply this kind of expansion to the following function so this function pi minus X all over 2 is going to be equal to the sum of sine in x over N that's what we'll see and so what we'll do is we'll think about this as some sort of repeated function so let's get to that so let's go ahead and set G of X equal to this function so we'll say it's PI minus x over 2 on the interval X between 0 and 2pi and then repeat so in other words we have this kind of sawtooth functions so here's 2 pi so notice 0 this thing is PI over 2 and at 2 pi this thing is minus PI over 2 so that means it has this kind of picture right here great and then if we go to 4 pi and make this thing repeat it's gonna have that kind of picture right there and then so on and so forth so that is the shape of this function so now we can write a closed piecewise form of this function if we want to it's not too bad so we'll let this thing equal to PI times 2m plus 1 mine sex all over to where m is an integer and then X is going to be between two M PI and two times M plus 1 pi ok great so now notice if M equals zero then where PI minus X over 2 and that would be like this picture right here and that puts us between 0 & 2 pi and so that's what causes this thing to repeat ok so now what we want to do is calculate the Fourier expansion of this function so let's go ahead and get to that first we'll calculate this a zero term so notice that a zero is going to be 1 over pi and then the integral from zero to 2 pi of our function which is going to be PI minus x over 2 because we can just use this version of the function whenever we're inside that interval so that's nice and now we have this as times cosine of zero but that's just 1 so we have that so now notice that this is 1 over pi and then if we take the antiderivative of this we get pi minus x squared over 4 we have to evaluate that from 0 to 2 pi ok great let's see what we get so this is going to be 1 over 4 pi then if we evaluate this at 2 pi and then square it notice we're going to get PI squared then if we evaluate this at 0 and square we also get pi squared so we have PI squared minus PI squared so we have 0 we're going to need a little bit more room to calculate the other coefficients so I'll clean up the board and we'll do that so now let's calculate the rest of our a end coefficients so notice here we're gonna have 1 over pi then the integral from 0 to 2 pi PI minus x over 2 times cosine and X DX okay great I'm gonna go ahead and bring this two out maybe that'll simplify a little bit so 1 over 2 pi now we have the integral from 0 to 2 pi of Pi minus x times cosine of NX DX so to integrate this we want to use integration by parts and we know that because we have a polynomial function times a trans and dental function and those you usually tackle with integration of the parts furthermore we know that you needs to be the polynomial part because when you take the derivative it becomes simpler and that's a nice trick in order to figure out what U and DV is so let's go ahead and do that so here we have u equals pi minus x that makes d u equal minus DX and then we'll let DV equal Coates in X DX that's going to let V be sine and x over N okay good so now we have our whole integration by parts set up calculated and now we can use the formula so let's recall the formula is this should be u times V minus the integral of VDU so that's what we'll do so this is going to be 1 over 2 pi and now we have u times V so that's going to be PI minus x times sine and x over N we need to evaluate that from 0 to 2 pi great and now we have we have minus the integral of VDU but notice the D U has a minus sign into it so that'll cancel and give us a plus sign so it's really going to be Plus this times DX so here we have plus the integral from 0 to 2 pi of sine in x over n DX so we have something like that going on now notice if we plug 2 pi in for sine we get 0 we plug 0 into sine we get 0 so this whole thing cancels out and then we're left with this is 1 over 2 pi now take the antiderivative of this the antiderivative of sine is negative cosine so we have negative cosine and x over N evaluated from 0 to 2 pi but cosine has the same value on 2 n pi and 0 because it has a period of 2 pi so then those are going to cancel 2-0 so what that tells us is our a values are zero so recall that we had a zero with zero and now we just got a in is zero okay so now let's clean this up and we'll do the same calculation with B in all right we're doing the Fourier expansion of this function which is extended to repeat at a period of 2pi and we just found out that all of the a coefficients were zero by using this calculation over here now we're going to see what happens for the B coefficients so let's notice that BN is going to be one over pi the integral from zero to 2 pi of pi minus x over 2 times sine of NX DX ok so I'll do the same thing that I did in the last step and I'll take that 2 out so I have 1 over 2 pi then the integral from 0 to 2 pi of pi minus X sine NX DX and again we're going to use integration by parts because I have a polynomial function types of transcendental function and I'm going to do a very similar substitution method for my integration by parts so I'll let u equal X minus PI minus X that's going to make D u equal minus DX and then next I'm gonna let DV equal sign in X DX that's going to make V equal to minus cosine X over N using the fact that the antiderivative of sine is negative cosine great now we're set again with our integration by parts set up now I'll do the same thing again so u Z evaluated from 0 to PI minus the integral of VDU so let's see what we get for that so we have a 1 over 2 pi out in front of the whole thing and now u times V so I'm going to take that minus sign and let it flip the order here so that's gonna give me X minus PI x coast in X all over n evaluated from 0 to 2 pi and now I have minus V D u so notice I've got a minus sign an integral by parts formula a - they are - there so overall I have a minus sign so that's going to be minus the integral from 0 to 2 pi of coast in x over n DX great so now let's see what we get here so we've got a 1 over 2 pi in front of the whole thing now if we plug 2 Pi in here I've got 2 pi minus Pi which is just PI over N and then when I plug 2 Pi in there I get 1 so that's good now when I plug 0 in here I'm gonna get minus PI over N when I plug 0 into cosine I'm gonna get 1 so I have minus a negative PI over N great so just to reiterate that is what's happening with this guy right here now this is going to be minus so we need to take the antiderivative of that that's going to give me sine and x over N squared evaluated from 0 to 2 pi but evaluating that at 0 and 2 pi is going to give you 0 on both accounts and now notice what we get here this is going to give us 2 pi over 2 pi times n because notice this adds up to 2 pi over N but this is just 1 over N so just to reiterate what we did when we found our coefficients for our Fourier series we found that a 0 was equal to 0 a n was equal to 0 for n bigger than or equal to 1 and BN was equal to 1 over N for n bigger than or equal to 1 but now looking at the function that we started out with we see that we have derived this formula down here so the sum N equals 1 to infinity of sine and x over N is equal to pi minus x over 2 so we've got this okay now we're ready to move on to our main result so instead of tackling this directly we're actually going to insert an X here and find something slightly more general and then set X equal to 1 so let's see how that goes so we're going to look at the sum N equals 1 to infinity of host index over into the fourth but now I'm going to apply a trick that I'm gonna apply several times throughout the solution I'm gonna add 0 so I want to add 0 by adding 1 over into the 4th and subtracting 1 over N to the 4th so let's see what we get if we do that so N equals 1 to infinity of Coast and X over into the 4th now let's subtract 1 over N to the 4th but I'm gonna subtract it in this form this is cosine of n times 0 over in to the fourth plus 1 over n to the 4th great so notice I've added and subtracted 1 over N to the 4th just in slightly different forms ok great now I'm gonna rewrite this as the sum N equals 1 to infinity of cosine in Y over into the 4th evaluated from y equals 0 to y equals x so that takes care of these first terms great and then I'm gonna break the second part off into another sum which we can do because we know this thing absolutely converges you can check by a p-series test that this absolutely converges pretty up it's kind of pretty obvious great and so here we have this is 1 over N to the fourth now I'm gonna recognize this as the riemann zeta function evaluated at 4 so let's just work we're call real quick with the riemann zeta function is so that is the sum N equals 1 to infinity of 1 over N to the S so that's obviously the zeta function evaluated at 4 so that's cool but now this kind of evaluation gives us a hint that really underlying this is some sort of integral and notice we can rewrite this thing as an integral so this is the sum N equals 1 to infinity of the integral from 0 to X of minus sine NY over n cubed D Y so notice the antiderivative of negative sine is cosine but then when since we have this N and from the chain rule that into the cube becomes into the fourth good and then we've got this left over zeta evaluated at for now we're just gonna keep kind of doing this trick over and over and over again until we get down to being able to use this formula so let's see what we get for the next step so this is going to be N equals 1 to infinity and then the integral from 0 to X of minus sign in Y over N cubed plus sine n times 0 over N cubed dy plus the zeta function evaluated at 4 okay good but notice that this thing can be written in an integral similarly to the last step so this is the sum N equals 1 to infinity of the integral from 0 to Y and then the integral from 0 to X of minus cosine in Z over N squared D Z dy and then finally we've got this plus this zeta function evaluated a 4 at the outside so notice here I didn't have to add and subtract the same thing because sine of 0 is equal to 0 so now we have like one more step of doing something like this until we're back at this so let's go ahead and bring this up and then we'll move on on the last board we built up the following identity so we have the sum N equals 1 to infinity of cosine NX over N to the fourth was equal to the sum N equals 1 to infinity the integral from 0 to X then another integral from 0 to Y of the function minus cosine NZ over N squared then the Z integral is on the inside the Y integrals on the outside and then outside of the whole thing is the zeta function evaluated at 4 ok now we're going to do this kind of idea one more time to get us down to being able to use this formula which we derived earlier so notice that this thing is equal to minus I'll go ahead and take this minus sign out and now we have the sum N equals 1 to infinity of the sum of the integral from 0 to X the integral from 0 to Y of cosine in Z over N squared minus cosine N times 0 over N squared plus 1 over N squared and this is all within the DZ dy integral and then I have my plus zeta function evaluated a 4 on the outside and just to reiterate all of this was within the sum but the zeta function is not within the sum now just like we did with this 1 over N to the fourth term which became zeta evaluated a 4 and split that off from the sum we're also going to split this 1 over N squared term off for this zone so let's see what we get when we do that so this is going to be minus the sum N equals 1 to infinity of the integral from 0 to X the integral from 0 to Y I'm going to rewrite this as cosine in W over N squared evaluated from W equals 0 up to W equals E and this is still inside a DZ dy integral but then this is going to be minus the sum N equals 1 to infinity of the integral from 0 to X the integral from 0 to Y of 1 over N squared D Z D y plus the zeta function evaluated for ok so again this minus sign goes in on this 1 over N squared term which gives us that guy over there so the next thing that we can do is rewrite this as another integral so this is going to give us minus the sum N equals 1 to infinity of the integral from 0 to X the integral from 0 to Y the integral from 0 to Z of so we need to take the antiderivative of cosine or really we're taking the antiderivative of negative cosine because I can go ahead and bring that guy back in the antiderivative of negative cosine is sine so that'll give us sine in W over in after taking that antiderivative and this is going to be DW and then notice here we have - I'm gonna change the order of summation here I'm gonna change the order of the sum and the integral here and I get the integral from zero to X the integral from 0 to Y of the sum N equals 1 to infinity of 1 over N squared and then DZ dy and then Plus this zeta function evaluated at 4 now the next thing to notice is that this guy right here is actually the zeta function evaluated at 2 and so we'll use that in the next step as we continue to simplify ok so let's go ahead and bring this up and then we're almost home free we've done a bunch of work to get this thing down to the following form so we have the sum N equals 1 to infinity these three integrals in a row of sine NW over n DW and then DZ and then D Y but notice that's exactly this guy right here once we change the order of summation and integration which we're going to do and then we have this integral from 0 to X this integral from 0 to Y of the zeta function evaluated to DZ dy plus the zeta function evaluated at 4 ok so like I said let's change the order of summation and integration here that'll give us the integral from 0 to X 0 to y 0 to Z of the sum N equals 1 to infinity of sine in W over N and then we have DW DZ dy then minus so now let's get to integrating this out so we'll have the integral from 0 to X of zeta 2 times Z evaluated from 0 to y dy plus the zeta function evaluated at 4 so notice that's just a constant so there's nothing to do ok so now the next thing that we want to do is replace our sign in W over N with this thing that we derived on the beginning of the video so let's go ahead and do that ok that's good and now we can get to taking the integral of this so here we have the integral from 0 to X the integral from 0 to Y and so now we can do this as pi minus W squared over 4 evaluated from W equals 0 to W equals Z and then we have DZ dy great and then this is going to be minus the integral from 0 to X of the zeta function evaluated at 2 times y dy and then plus the zeta function evaluated at 4 okay so now let's evaluate Z in here and see what we get so this is going to be the integral from 0 to X the integral from 0 to Y of let's go ahead and take this quarter out so we've got 1/4 outside and now we have pi minus z quantity squared minus pi squared that's what we get from evaluating this thing at W equals 0 and then this is all inside a DZ dy integral good but now notice that this is going to be the zeta function evaluated at 2 times y squared over 2 evaluated from 0 to X and then plus this zeta function evaluated at 4 so now let's maybe do one more thing before we move everything to the top of the board so notice that this part can simplify and what can it simplify to well notice it can simplify to Z squared minus 2 pi Z and why is that well because notice we get a PI squared term here which is going to cancel with a pi squared term there but that's and then that's what we're left with and then notice that this thing right here is going to simplify to the zeta function evaluated 2 over 2 times x squared so let's bring those simplifications to the top and then we're almost done we've got two more integrals left to do and then we can start getting an idea of what our final goal will be so here we have this is going to be 1/4 the integral from 0 to X naught take the antiderivative of this with respect to Z we'll get one third Z cubed minus PI Z squared we need to evaluate that from 0 to Y and then that's still inside a y integral good and then minus so this is the zeta function evaluated 2 over 2 x squared plus 2 zeta function evaluated at 4 ok now we can go ahead and plug Y in there so here we have the integral from 0 to X of one-third Y cubed minus PI y squared dy and then minus these things that are that keep hanging on now this is going to give us 1/4 and then this is going to give us 1 over 12 Y to the fourth minus PI over 3 y cubed evaluate that from 0 to X and then we have still this minus zeta function evaluated at 2 over 2 x squared plus zeta function evaluated at 4 ok so now we can plug X into this and we finally have it so we're gonna have 1 over 48 X to the 4th minus PI over 12 X cubed minus the zeta function evaluated at 2 over 2 x squared plus the zeta function evaluated at 4 so that's our final form for this series so let's bring that up and then we'll evaluate at that at the appropriate place to get our goal on the last board we finally derived this formula if actually fix the sign error so look maybe in the last few minutes to see where I made a sign error but we have this sum N equals 1 to infinity of cosine NX over N to the fourth that is minus 1 over 48 X to the fourth plus PI over 12 X cubed minus the zeta function evaluated at 2 over 2 x squared plus the zeta function evaluated at 4 now the next thing that we're going to use is that the zeta function evaluated at 2 is PI squared over 6 so that's a famous some of oiler and less common but still pretty famous some of Euler is that the zeta function evaluated a for is PI to the fourth over 90 and then also will go ahead and set X equal to one and that's going to give us a closed form for our goal and I'm going to reorder the terms a little bit but notice the closed form for our goal is PI to the fourth over 90 minus PI squared over 12 that's what we get from there plus PI over 12 that's what we get from there minus one over 48 and now maybe a companion problem that you could try is this same exact thing with the sign there so maybe you guys want to try that using similar methods from this video and post in the comments what you get so that's a good place to stop

Channel: Michael Penn

Views: 17,959

Rating: 4.9556961 out of 5

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn, fourier series, sequences and series, infinite series, convergence

Id: LKCxNtBiIO8

Length: 25min 7sec (1507 seconds)

Published: Tue Apr 28 2020