An interesting approach to the Basel problem!

here we're going to take what i think is a pretty interesting and unique approach to the basil problem and that is to find the infinite sum of the reciprocals of the squares so it's pretty easy to show that this kind of series converges but it's fairly tricky to find out the sum there's a bunch of different ways to do it there's the famous way that euler did it and then there are some other ways as well using fourier series and so on and so forth here we're going to do something a little bit different but before we get started we need two following tools the first tool says that the sum of the reciprocal of the squares is equal to four thirds times the sum of the reciprocals of the squares of just the odd and positive integers and then secondly we're going to use the fact that 1 over m plus 1 squared equals the integral from 1 to 0 as of x to the m natural log of x dx and i should say here that i put 1 in the lower bound and 0 in the upper bound to keep a minus sign out of this if i put 0 to 1 i would have needed to include a minus sign and this statement is true for all m not equal to negative 1 although we'll only have um natural numbers to worry about here okay so let's go ahead and get into this first goal so here we have the sum as n goes from one to infinity of one over n squared so now what i want to do this is break this into even terms and odd terms so i can break it into even terms and it looks like this the sum as n goes from 1 to infinity of 1 over 2 n squared so those are going to be all the even terms plus the sum as n goes from 1 to infinity of 1 over 2 n minus 1 quantity squared so that's going to be all of the odd terms but now it's pretty easy to see that we can square this 2 and bring it out and that's going to give us one quarter the sum n equals 1 to infinity of 1 over n squared plus the sum n equals 1 to infinity of 1 over 2 n minus 1 squared and now we've got a simple algebraic equation that we can use to solve our sum of reciprocal of squares in terms of the sum of reciprocals of odd squares so in other words we have three quarters the sum n equals 1 to infinity of 1 over n squared equals the sum n equals 1 to infinity of one over two n minus one quantity squared then multiplying both sides by four thirds will get us to our solution okay so now we've proved this first tool and we're ready to move on to the second so i'll start with the right hand side and work towards the left hand side so i'm starting with the integral from 1 to 0 of x to the m natural log of x dx now i'm going to use integration by parts there's two big hints that we need to use integration by parts maybe the most glaring one is we are integrating something that looks like an inverse function so the natural log of x you want to think about the inverse of the exponential function and whenever you're integrating inverse functions integration by parts is a good choice and so we'll take u to be natural log of x and we do that because that's the inverse function and also because if we take the derivative of natural log we get something simpler so in this case we have d u is 1 over x dx and then next we'll take dv to be the rest so if u is natural log of x dv is x to the m times dx but then since we're assuming that m is not equal to negative one i haven't written that out but we'll just verbally confirm that we can see that v is equal to one over m plus one x to the m plus one and now we have all of our parts to do our integration by parts of this integral so let's see what we get so we're gonna have u times v so i'll go ahead and write that as 1 over m plus 1 x to the m plus 1 times natural log of x we need to evaluate that from 1 to 0 and then i have minus the integral of v d u so v d u i'll take this one over n plus one and bring it out and let this one over x multiply this x to the m plus one down to x to the m so in other words i have minus 1 over m plus 1 the integral from 1 to 0 of x to the m dx great now let's talk through what we get from evaluating this first bit so if we plug in zero we have a problem with natural log because that limit is going to be negative infinity but it's easy to check with l'hopital's rule or maybe something else that because we have this x to the m plus one here and now we'll assume that m is a positive integer this is gonna cancel out so we'll get zero plugging in x equals zero or x approaching zero then we know that the natural log of one is equal to zero so we get 0 for that term as well so in other words all of this cancels down to 0. next we can take the antiderivative of x to the m and that's going to give us minus 1 over m plus 1 squared we get an extra m plus one term in the denominator and now we have x to the m plus one we need to evaluate that from one to zero now i'll go ahead and take this minus sign get rid of it into a plus sign by changing the order of evaluation here but then evaluating that at 1 gives us 1 over m plus 1 squared which is exactly what we wanted to get okay so now we've got our second and last tool and now we're ready to move on to our main result so the first thing that we'll do is use the first tool to rewrite this sum of reciprocal of squares as a sum of reciprocal of odd squares so i'm going to go ahead and write this as 4 3 times the sum n equals 1 to infinity of 1 over 2 n minus 1 quantity squared now in order to make our calculation a little clearer i'm going to re-index this sum a little bit and i'm going to re-index this so i start at n equals zero but i can do that very easily by just putting an n two and plus one here here i'm replacing n with n plus one and that's what we get in this case okay great and now what i can do is use the second tool to rewrite each of these one over two n plus one quantity squared as this type of integral so now we have four thirds times the sum n equals zero to infinity of the integral from one to zero of x to the two n times natural log of x dx so again using the second tool i can replace this integral as exactly this one over two n plus one quantity squared okay fantastic now what i want to do is bring this sum inside the integral i can do that via something called the dominated convergence theorem so that's going to give me four thirds and now i have the integral from one to zero of the natural log of x times the sum n equals 0 to infinity of x to the 2n and then dx but now that is exactly a geometric series and in fact it's a kind of nice geometric series we can go ahead and add this up and we'll get one over one minus x squared so here we have four thirds the integral from one to zero now we have natural log of x over one minus x squared dx and just to reiterate what happened we took the sum of this geometric series notice the starting term is 1 and the common ratio is x squared and we sum that to give us this 1 over 1 minus x squared which showed up in the integral okay fantastic now the real trick is to rewrite this natural log of x as a difference of two things and we're going to do this in a couple of stages so here we have this is four thirds and now the integral from one to zero of one over one minus x squared i'm gonna bring that out and now i'm gonna rewrite natural log of x as the natural log of x minus the natural log of one notice the natural log of one is zero so that's easy to do and now we can use some logarithm rules so we'll use a logarithm rule to add a square right here at the cost of putting a one-half out front so notice if we bring this one-half inside the logarithm it'll cancel that square so i did not change anything here next we're going to use a pretty tricky fact from calculus 1 like when you're learning about limits of rational functions in order to rewrite this natural log of x squared in a certain way so we'll notice that the natural log of x squared can be rewritten as the limit as y goes to infinity of the natural log of 1 plus x squared y squared over 1 plus y squared so let's box that maybe so notice as y is approaching infinity the argument of this natural log is approaching x squared because there we've got a rational function in y so notice the leading coefficient in the numerator is x squared the leading coefficient and the denominator is 1 which tells you that this interior is approaching x squared over 1 which means the whole thing is approaching the natural log of x squared okay so that's one thing that we want to notice and another thing that we want to notice is that this natural log of 1 can be rewritten as this same term where we have evaluated y at zero so in other words we can rewrite all of this so let's take this one half out of this maybe in the following way so we've got four thirds and then we've got the integral from one to zero of 1 over 1 minus x squared and then we've got a half and then we've got the natural log of 1 plus x squared y squared over 1 plus y squared evaluated from y equals 0 up to y equals infinity okay good now i want to point out here that these two are the same if y is approaching infinity we talked our way through why we get natural log of x squared here and then if y is equal to zero we get natural log of one which that was zero and that's why that doesn't show up up here now i'll go ahead and bring this up and we'll move on to the next step so on the last board we arrived at our goal sum in this following integral form so i've actually done one step off of the board here we had natural log of this term over this term but i've used a log rule in order to split those up now the next thing that i want to do is turn this thing which looks like the evaluation of a single integral into a single integral and i can do that by taking the partial derivative of this with respect to y okay so let's go ahead and do that so we have four thirds then we have the integral from one to zero of one over one minus x squared that's kind of coming down for free great now like i said we're going to take the partial derivative of all of this with respect to y so let's put a partial with respect to y here and in order to do that we need to add a single integral in and that single integral will be with respect to y and it goes from 0 to infinity okay so let's see what we get so if we take the partial of this with respect to y we're going to send this 1 plus x squared y squared to the denominator and then we'll have uh the derivative of the inside with respect to y which will be 2 x squared y great now we're going to do the same kind of thing over here that's going to be minus now we'll have two y over one plus y squared and now we have d y d x okay great and now what i want to notice is that i can take this 2 and cancel it with this 2 and this 2 just by distributing that half through and then the next thing that i want to do is combine these two things via finding a common denominator so my common denominator will be clearly the product of these two so let's see what we get for that so we've got four thirds and now we have the integral from one to zero and the integral from zero to infinity of so let's write this out we've got x squared y times 1 plus y squared minus y times 1 plus x squared y squared all over 1 plus x squared y squared times 1 plus y squared and now we have d y dx so what i did is i took this numerator multiplied by this denominator this numerator multiplied by this denominator and that gives me the new numerator after subtracting them like that and then my new denominator is just that product of the old denominators okay so now let's see what we get from here so now we will have oh i just realized that i left out a 1 over 1 minus x squared from this guy right here okay so now multiplying this out i'll have x squared y plus x squared y cubed minus y minus x squared y cubed good but now some stuff cancels notice this x squared y cubed term cancels and then furthermore we can factor something out of this so i'm going to go ahead and factor a negative y out of this so if i take these two remaining terms and factor a negative y out i'm left with 1 minus x squared good so now this 1 minus x squared will cancel with this one minus x squared and then next this minus sign can be changed to a plus if we change this back ordering to zero to one instead of one to zero now let's see what we have so we have four thirds and now the integral is going to be zero to one and zero to infinity now after all of this carnage of canceling things out our numerator is just y and our denominator is one plus x squared y squared times 1 plus y squared and then i'm doing a y integral first and an x integral second now the next thing that i want to do is apply fubini's theorem which tells me i can change the order of integration to change this to an x integral and then a y integral but i'll do that as we move up to the next board okay so now we're picking up with the integral that we had on the last board i have changed the order of integration and i factored this function of y out of the interior integral which is possible because it is with respect to x now the antiderivative of this inside function is actually pretty standard so i'll let you guys look that up if you need to but suffice it to say we're going to get four thirds and then the integral from zero to infinity we've got y over one plus y squared and this is going to be one over y times arc tan of x times y we need to evaluate that from 0 to 1 and then we've got a y integral on the outside so notice if we take the derivative of this inverse tangent we're going to get 1 over x squared times y squared times the derivative of this with respect to x which will be y that'll cancel this 1 over y so that's why that works so notice that this y will cancel this y down to just the number one and then furthermore if we plug in x equals one we'll get arc tan of y if we plug in x equals zero we'll get arc tan of zero which is zero so in the end we'll have four thirds and then the integral from zero to infinity of arc tan of y over one plus y squared d y and now all that remains is a simple u substitution so we want to notice that the inverse tangent function is inside the integral and its derivative is also inside the integral so let's go ahead and set u equal to arc tan of y that's going to make d u equal to 1 over 1 plus y squared good and so that means that here this d y over 1 plus y squared is my d u term and this is my u term furthermore when y is equal to 0 that implies that u is also equal to 0 because arc tan of zero is zero like we discussed before and then as y approaches infinity u is approaching pi over two just by the asymptotic nature of the inverse tangent function so that means we can completely change this y integral to a u integral it'll be 4 3 times the integral from 0 to pi halves of u d u and now we're pretty much home free so this is going to be four thirds times u squared over two evaluated from zero to pi halves notice i can take this four thirds cancel the one sorry the two in the denominator down to a one and then i have two thirds so in the end i get that this is equal to two-thirds times pi squared over two squared in other words it's equal to pi squared over six which is the expected value of this sum and that's a good place to stop

Channel: Michael Penn

Views: 88,783

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: -Yy_Jsw0djM

Length: 19min 26sec (1166 seconds)

Published: Mon Aug 17 2020