Euler's infinite pi formula generator

crazy busy and very deadly times but be that as it may welcome to yet another Mathologer video. today I revisit the topic that I touched upon a couple of years ago. today we look at a mathematical leap of faith performed by the 18th century mathematician the amazing Leonhard Euler. it's Euler again. Euler's crazy leap of faith resulted in a magic key that opens the way to the most famous and most beautiful formulas for pi in particular what I'd like to do today is to again quickly perform that leap of faith together with you and then show you some breathtaking animated algebra that will get us once more the Leibniz-Madhava formula for pi and Wallis infinite product formula and William Brouncker's spectacular infinite fraction formula and Leonhard Euler's amazing formula solving the Basel problem which made the 28 year old Euler a mathematical superstar plenty of magic for today right but wait there's more I'd also show you how the infinitely many higher power counterparts of the Basel formula can be obtained that one there and that one and on and on as far as you want to go one Basel type identity for each even power of Pi amazing stuff huh really really really amazing stuff this video was inspired by a one-page article by Paul Levrie in the mathematical intelligencer in 2012 in this article Levrie sketches some lightning quick derivations of some of these results they really are a must animate I think anyway I'll put a link in the description okay here's a question for you let's say I tell you that my pet cubic has zeros at minus 1 0 & 1 is this information enough to pin down my cubic what do you think well the answer is almost here's one cubic that has these three zeros so the cubic is a product with one factor for each zero then you can get all the other possible cubics with those zeros by stretching in the vertical direction by scaling with nonzero constants easy stuff right and most of you will know that the same is true for polynomials in general if you know the n zeros of an nth degree polynomial then you can capture your polynomial as a product except for scaling now Euler and his buddies were experts at thinking of functions such as the exponential function and the trig functions as infinite polynomials there that's the famous Maclaurin series for the sine function one of these infinite polynomials now Euler wondered whether because functions like sine had representations as infinite polynomials they could also be written as products based on their zeros and as crazy as this idea may sound at first EulerI managed to get this to work for sine here's what he did ok are you ready for some seriously seriously mad flight of mathematical fantasy sure all right then madly we go forth let's ignore the Maclaurin series and hunt again for the polynomial for sine of X just as we did for the cubic so begin with the zeros sine has infinitely many of them and so our mad formula for sine should be an infinite product with one factor for each zero let's assemble them into a product from the inside out like this if all goes well sine should somehow be this product times some nonzero constant it looks pretty insane right but let's just tell our brains to shut up and let's keep following Euler on his flight of imagination to determine the constant Euler just goes for it and divides by X the new function on the left looks like this peters out right like that okay let's evaluate everything inside at zero there we go okay there on the right we've got oh well whatever that is it should be infinity hmm and on the Left we're dividing by zero double hmm okay left side sine X divided by X is definitely not defined at zero however although sine X divided by X is not defined at zero its limit at zero is one close our eyes close our brains and keep going divide by C flip both sides voila that's our C whatever the hell it means but don't worry mad and brainless we go on it ends soon I promise let's plug this crazy C into our original formula we'll do it carefully so the pi denominators and the two pi denominators and so forth line up with the other factors like that now little canceling there and there and there there there now of course the reasoning that let us see is completely crazy but believe it or not this final formula really makes sense and really is true I haven't given you anything like a proof of the formula but I can give you a sense of how the formula works to sneak up on that infinite product let's start multiplying out those factors and let's plot the partial product along the way then those partial products get closer and closer to the sine function so the first partial product is just X again it's very important to realize that the crazy flight of fantasy that led to this formula is very very very very far from a proof but it is also important to realize that mathematicians don't just work by trying to prove things the Nike method just do it and see what happens can result in great insights and a Nike method in the hands of a genius like Euler can result in miracles of course any discovery made the Nike Way still has to be confirmed by a rigorous proof there is no way around that in mathematics anyway first mission accomplished we have Euler's amazing product formula to play with our second mission today is to push and pull and prod Euler's formula to derive pretty much all the most famous infinite expressions for pi okay we're talking infinite sum product and fraction identities for pi what do you think is easiest to do well the right side of Euler's identity is a product and so a product should be easiest right and it is all right so let's chase down that product formula for pi actually really straightforward the most obvious thing to try is plugging in PI for X but that doesn't work we just get 0 for sine X and also on the other side 0 so 0 is equal to 0 no not good but the next most obvious thing to try works plug in PI over 2 for X and see what happens fantastic well as this product formula is really just Euler's product evaluated at PI over 2 pretty sure that Wallis would have enjoyed this what's next what sounds easier to do an infinite sum on infinite continued fraction the infinite sum right okay we are after a sum and how do you turn a product into a sum you know the answer to that or at least once upon a time in school you did logarithms remember log of a times B is equal to log a plus log B well we'll do exactly the same thing just with a slightly bigger product and we'll use the natural logarithm well that's great our product is now a sum and we can think about plugging in or whatnot but what about those logs how do we get rid of them well that's easy you simply differentiate here we go and now we can plug in what value should we choose for X well zero and PI actually explode the right side so they definitely won't work what about PI over two that makes the left side zero which feels like it might be good but we also end up with the right terms canceling in pairs so no dice probably the next most obvious values is to try as X is equal to PI over 4 and yep that works here we go Tada so you get the Leibniz Madhava formula from Euler's product formula by simply applying a logarithm differentiating and evaluating at PI over four simply magic next I'd like to show you an ingenious way to derive this amazing infinite fraction from our likeness Madhava infinite sum formula that one there so let's first recast this infinite fraction formula in terms of the infinite sum in the limits Madhava formula first off that one right at the beginning of the infinite fraction is a bit of an oddball right since there are 2s everywhere else we could add 1 on both sides of the equation to get all 2s but after some experimenting it turns out to be better to subtract the one from both sides like that now the Leibniz Madhava formula features PI over 4 and not 4 over pi that's easily fixed and now make the left side into one big fraction like that ok not so simple but at least it's PI over 4 --ish-- that means we can now plug in our infinite sum for PI over 4 so what we want to show is that that scary fraction on the left can be magically manipulated into the infinite fraction on the right ready not too scared well then let's do it ok a good easy step to begin with let's get rid of the brackets now I'm going to show you a clever mathematical combo move we'll be repeating this combo move over and over so pay careful attention the combo move begins by manipulating the denominator 2/3 hmm we want to end up with ones in all the numerators just as we started so to get there as the next part of the combo move we employ a sneaky trick adding and subtracting a clever expression right the pink and the green cancel so we haven't changed the denominator but how does this do nothing step help well let's see our do-nothing step arranged for us to have the same pink number at the top and at the bottom and our final step in the combo is to divide through by that pink number and that almost finishes our first magic combo movement remember we started with this fraction of infinite sums now just focusing on the green and pink fractions the new one on the right results from the one we started with on the left by discarding the leading terms and multiplying through with minus 1 and as I said what we'll do is to repeat this magical combo move over and over which will make the green and pink fraction disappear and what comes out is this beautiful infinite fraction but we're not quite ready we need one more mini step to control things down to infinity spot the difference not hard on the Left we've got a 3 and in the final fraction where after we've got a 3 squared to get the square happening our last mini step is very similar to our add and subtract trick here we multiply and divide by the same number have a look Neat that's the end result of our magical combo move and now we just start it up again just watch and Wonder brilliant or what does this actually show again well remember the oddball 1 that we moved to the left what we've shown is this and kicking the oddball one back to the right we get Lord Broucker's fantastic fraction but there's more let's do a reciprocals on both sides one over on both sides of course that means the infinite fraction and the infinite sum are equal right but not only that on close inspection it turns out that even all corresponding partial fractions and sums are equal there the first partial fraction and the first partial sum are both equal to 1 the second partial fraction and sum are both equal to 2/3 and so on amazing right just in case you were wondering whether this is obvious nope but it's also not too difficult to prove so we have our infinite fraction what's next our final goals Euler's Basel formula and it's infinitely many cousins these were really what Euler was after when he conjured up his product formative for sine the idea here is to expand the infinite product and collect together the different powers of X now before the monster expansion we first do an easy simplification of the product notice that each successive pair is a difference of two squares so we can simplify like this okay enough dilly-dallying time to expand to do this we take one term from each factor and multiply them and we do that in all possible ways let's go first let's take the X from the first factor no choice there right and choose the one and all other factors multiplying the X and all those ones together gives us yep you guessed it X next what if we change our choice in the first bracket to minus x squared divided by PI squared but keep all the other ones well then the resulting product is x times that x squared term which results in minus X cubed divided by PI squared got the idea great each sequence of choices gives us one term of our final sum so let's think about the terms that we'll get out starting with the lowest powers obviously this expansion only contains the one X term we've already obtained that one there what about x squared terms nope no way we can get those and the same for any even powers of X on to the X cubeds well there's the one we've already found and here's another one and another one and so on this means that the expansion of Euler's product starts out like this can you see what's coming Taylor series fans can take a shortcut here but let's take this slow and easy road to make the x-cubeds disappear will differentiate so differentiate once differentiate again and one last time done now set X equal to zero that wipes out everything on the right except for the first constant term and of course cos of zero on the left is 1 time to relax and go into easy autopilot and there you have it Euler's celebrated Basel formula and that's how Euler originally derived it simply magnificent isn't it but why stop there in for a penny in for infinitely many pounds Euler also noticed that we can get a whole lot more if we just keep going okay so where were we we differentiated three times and at this stage we plugged in X is equal to zero but instead of plugging in let's differentiate two more times that gets rid of the x squared we see here can you see it well differentiate once and one more time now we're ready plug in X is equal to zero as we did before impressed no don't worry you soon will be so what's the mystery stuff in the bracket well these are all the coefficients of the X to the power of five terms in the original expansion so let's go back to remind ourselves what happened there to get the X cubed terms we pick the X in front and one of the x squared terms and all the rest 1 so how do we get X to the power of five terms easy right we pick the X in front and two x squared terms and then all the rest ones like this or like this or like this in general these X to the power of five products look like this with I being less than J and so the bit in the brackets we're after is just the sum of all these coefficients there those ones and switch to algebra autopilot we get this that's great but that's not the sum we're after we get the sum we are after if we replace the less than sign by an equal sign like this there that's what we're really after the sum of the reciprocals of the fourth powers of the integers okay but how does the green less than sum help with that well the thing is we already know something else namely this sum there that third sum ranges over all possible pairs I and J without any restrictions hmm but how do we know that sum can you see it that third sum is just the square of the Euler Basel sum got it now that means the third sum is the square of PI squared over 6 can you see where I'm heading with all this if not don't worry from here it's easy peasy apart from the blue equal sum and the green less than sum there's also a red greater than sum yet another sum but this one we know right off because of the symmetry just swapping I and J the red greater than sum equals the green less than sum and now everything's falling into place right of course the blue plus the green plus the red is equal to the orange and now just plug in all the values of the sums we know and we get this and from this we get the value of the blue sum that we've been chasing very very pretty also don't you think and we can play the whole game once again if you differentiate our expanded product formula twice more and manipulate the result in the same way we just did here we get Euler's PI to the power of 6 formula keep going we get PI to the power of 8 formula and so on this gets more and more involved but in principle the type of manipulation needed to be done is always the same come to think of it I have not assigned any homework yet today terrible so for those of you desperate for some challenges why don't you try to derive the PI to the power of six formula up there also apart from all the craziness in the Nike derivation of the product formula for sine there are quite a few details that I glossed over in the derivation of the PI formulas that really need to be nutted out to make these derivations into rigorous proofs so here's another challenge then what are those details how many can you spot for example wasn't there a detail in the expansion of Euler's product for sine that needed some extra attention anyway returning to these Basel formulas Euler also managed to find a general super formula that encapsulates all these infinite sums most of you Mathologerers would have seen this super formula previously it's connected to the famous Riemann zeta-function and popped up in my video about the Bernoulli numbers and Euler Maclaurin formula check it out oh and one last note for experts among you all as infinite product formula for sine is just one incarnation of the so-called Weierstrass factorisation theorem this theorem says that every so called entire function can be represented as a product involving its zeros and that's it for today I trust you enjoyed this wonderful mathematical journey see you next time hopefully with the world in a safer and happier place

Channel: Mathologer

Views: 133,166

Rating: 4.9676285 out of 5

Keywords: Weierstrass factorization theorem, pi, Lord Brouncker, John Wallis, Euler, Leibniz formula, Madhava, product formula, Basel formula, Brouncker, infinite fraction, inifite series, Wallis formula

Id: WL_Yzbo1ha4

Length: 28min 57sec (1737 seconds)

Published: Sat May 02 2020

Reddit Comments

[u/cfoefs]

Gorgeous video.

[u/_selfishPersonReborn]

The continued fraction derivation is miraculous.

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[u/1Demerion1]

Now that they are uploading again, how has the argument between Burkhard and Giuseppe turned out?