The Gamma Function, its Properties, and Application to Bessel Functions

after a hiatus of a few months I'm finally back I'm sure you guys miss my charming personality dashing good looks which I haven't revealed yet and the inspiring intellectual stimulation that my lecture is provided anyway in today's video we're going to resume our mini lecture series on vessels equation now I would have stopped talking about vessel functions with the previous video unlinked out in the description but if you if you have requested me to continue giving lectures on vessels equation so I'm going to oblige and make you all happy however I won't directly talk about vessels equation and vessel functions except for the very end of this video because in this lecture we're going to take a bit of a detour and discuss the gamma function we'll start off by defining the gamma function then we'll show a few of its properties and then we'll apply the gamma function to the solutions of vessels equation we found in the last video which I linked in the description so let's start by defining the gamma function the gamma function evaluated at a complex numbers knee recall that complex numbers also include real numbers the gamma function evaluated at a complex number Z is just the integral from 0 to infinity of e to the negative T T to the power Z minus 1 DT where T is just a placeholder variable now you can think of the gamma function as a generalised factorial whereas the factorial applies only to non-negative integers like 0 1 2 so on a gamma function applies to any complex number except for the non positive integers so it doesn't apply to negative 1 negative 2 negative 3 0 so on specifically if Z is a positive or 0 integer then the factorial is just Z times Z minus 1 times Z minus 2 all the way to 1 so using the factorial function I can find 4 factorial which is just 4 times 3 times 2 times 1 which is just 24 but I cannot use the factorial function for a number like 4.7 I also can't use it for negative 4 points nor can I use the factorial for negative four since these last three are not non-negative integers on the other hand the gamma function can be applied to z equals four it can be applied to z equals four point seven and even to z equals negative four point seven however the gamma function is not defined at z equals negative four still the gamma function is defined for a much bigger selection of numbers than the factorial is so you can definitely say that it's more easily applied since it works for a larger more general set of numbers however this still doesn't prove that the gamma function of the generalized factorial because for the gamma function to be a generalized factorial it has to behave in a manner identical to the factorial function at least for values where the factorial function does apply the set of non negative integers this little dilemma leads us nicely to the second part of the video where we describe a few properties of the gamma function the first property is that the gamma function evaluated at Z plus 1 is just Z times the gamma function evaluated at Z so one unit back of Z plus one to prove this property we're going to use the gamma function formula defined above and perform the integration for gamma of Z plus 1 by the formula for the gamma function gamma of Z plus 1 is just the integral from 0 to infinity of e to the negative T times T to the power Z DT if we integrate by parts by picking our first function to be T to the power Z and our second function to be e to the power negative T we'll have t to the power Z times the integral from 0 to infinity of e to the negative T minus the integral from 0 to infinity of the integral of e to the negative T times the derivative of T to the power Z since the integral of e to the negative T is just negative e to the negative T we can simplify this expression to become the following now to prove this relation for the gamma function I'm going to evaluate these integrals for when Z is positive here as T approaches infinity T to the power Z also appear which is infinity but e to the power negative T approaches zero in this first term this negative e to the negative T times T to the power Z that first term so there's two opposing forces when T approaches infinity but if you use l'hopital's rule then you'll see that the exponential grows faster than T to the power Z for any positive value of Z therefore as T approaches infinity this negative e to the negative T times T to the power Z that term approaches zero now when T equals zero the answer is pretty obvious in this whole first term just becomes zero because of the T multiplying it there thus we can get rid of the first portion of the integral when Z is positive because it remains zero regardless of whether it's evaluated at T equals zero or as T gets prohibitively large this only leaves the second term which is the integral from zero to infinity of e to the negative T times V times T to the power Z minus one DT now go back to the original formula for the gamma function and you'll see that if you take the Z out of the second integral term you'll just be left with an integral that is in fact equal to the gamma function evaluated at Z therefore for positive C gamma of Z plus 1 is just Z times the gamma of Z but what about non integer negative values of Z like negative four point three negative four point seven well that's going to lead me to my second property which is the same formula but now for non integer negative see the proof however is a bit different in fact it's not really a proof because we can't really prove anything for negative values of C the reason is that if you try to integrate by parts using the integral formula we wrote down when defining the gamma function you'll find that this first portion becomes undefined at the limit of T equals zero however even for non-integer negative values we're going to simply suppose that the same property that we proved for positive z also applies this supposing that we did might seem a bit hand weighting and something conjured out of thin air that's probably true mainly because universities don't fund their math departments very much which leads to a collection of for depressed and lazy mathematicians that use a lot of hand wavy arguments nonetheless the supposing that we did to extend property one to negative non integers is still useful in fact it even has a name it's called analytic continuation what that means is just extending the domain of the function to apply to a larger set of values this analytic continuation for property two can actually be used to evaluate the gamma function at negative non integer values of Z for example if you want to evaluate gamma of negative zero point one you can just set up this equation from property two and rearrange M of negative zero point one in terms of gamma of 0.9 gamma of 0.9 can easily be arranged in terms of gamma one point nine by the same logic so ultimately you can find gamma of negative zero point one in terms of the gamma of some number greater that may be evaluated using the integral definition so while the integral definition of the gamma function may not work for negative non integers you can still extend the gamma function with this recursive relation to be able to find the gamma of negative values of Z the third property of gamma functions is that the gamma function at V equals 1 is just 1 this is fairly easy to prove just go back to the integral definition apply integration by parts notice that Z minus 1 is just 0 because Z equals 1 and then evaluate the first term at its limits property 4 is basically proving that the gamma function is a generalization of the factorial because what it says is that the gamma function evaluated at a non negative integer n plus 1 is just n factorial we can prove this using a combination of property 1 in property 3 first we'll show the extreme case of when any zero on the left-hand side we have gamma one which is just one from property 3 and on the right hand side we have zero factorial which is also one and since both sides are indeed equal to each other zero factorial is 1 we're consistent so we've proven this extreme case what about larger integers though well for larger integers we've shown in property 1 that the gamma of n plus 1 is n times the gamma n but what about gamma of n well the same rule applies so now we have the gamma of n plus 1 equals n times n minus 1 times the gamma and minus 1 we can keep going and since we're already added integer n we're going to eventually stop at 1 because gamma 1 is the end point since gamma of 1 is 1 from property 3 the gamma of n plus 1 finishes off at that value so this expression just becomes n times n minus 1 times n minus 2 all the way to gamma 1 which is just 1 you know that this whole expression is just n factorial so we've shown that gamma of n plus 1 where n is a positive integer is just n factorial property 5 is that the gamma function evaluated at Z equals 1 over 2 is the square root of pi now this property seems a bit specific but it's actually going to come in handy in the next video so I'll prove it here the procedure once again uses the definition of the gamma function however instead of integrating by parts I'm going to do something different I'm going to make a change of variables and set this dummy variable T equal to u squared my limits remain the same but my integral now becomes the integral from 0 to infinity of e to the negative u squared times u inverse DT we can change the DT to add EU just by taking the derivative of T with respect to U and what I'll get after substitution is that the gamma of 1 half is 2 times the integral from 0 to infinity of e to the negative u squared EU let's square both sides now and if we do that we'll get the following let's now take the second integral and just change variable from u to V it's a dummy variable so it doesn't really matter what we name it we can then combine the two integrals to get a double integral and here's the fun part if we suppose that U and V are Cartesian coordinates then we can convert them to polar coordinates using the substitutions u equals R cosine theta and V equals R sine theta if we do that then the entire expression becomes the following now because of this extra R we can easily evaluate this integral using the substitution method once again with Q set to be R squared which makes DQ equal to 2r times Zr note that we're only integrating over the first quadrant since it's just 0 to infinity on both integrals that's why we only have 0 to PI over 2 at the limits on R theta after evaluating the first integral you'll probably just get one and after evaluating the second integral you'll get PI over 2 times 2 which is just PI thus the value of gamma at 1 over 2 is just the square root of pi so we proven the fifth property and that should do it for the properties now we're not done yet because we still have to apply the gamma function notation of the solution of vessel's equation that we found last time recall that in my previous vessel equation video the first solution to vessels equation is given by this y1 now I mentioned in passing in the last video once again the links in the description I mentioned briefly that for a particular value of a not you get the Bessel function of the first kind JP that's why I put a tilde here instead of an equal sign because y1 is only proportional to JP for a general value a naught it only becomes identical for a particular value a naught so what is this particular value a naught I'm sure you're waiting with baited breath to know the answer well here it is a naught equals 1 over 2 to the power P times gamma of P plus 1 for this particular value of a naught y 1 becomes JP the Bessel function of the first kind this means that we can finally write down the exact expression for J sub P now if we combine the Ford of the K with the two to the P in the denominator we'll end up with two to the power two K plus P since 4 is just 2 squared we can also simplify the rest of the denominator using property number one of the gamma function that we showed above according to which gamma of P plus 1 times P plus 1 is just gamma of P plus 2 and gamma P plus 2 times P plus 2 is just gamma P plus 3 and so on we can keep applying property 1 until we get to P plus K so our final expression for J sub P the Bessel function of the first kind of order P is j p equals x to the power p times the sum from k equals 0 to infinity of negative 1 to the K over 2 to the power 2 K plus P times 1 over K factorial times the gamma of K plus P plus 1 times X to the power 2 K we can apply a similar procedure to get J negative P which is the second solution to Bessel's equation in the event that p and negative p do not differ by an integer to get J negative P all we have to do is replace the P and JP by negative P hopefully that shouldn't be too confusing if it is you can always ask in the comments and that should do it for this video I hope you all found it useful and thanks for watching

Channel: Faculty of Khan

Views: 68,792

Keywords: Gamma, Function, Bessel Function, Bessel Equation, Bessel, Factorial, Mathematics, Math, Science, Tutorial, Applied Mathematics, Series Solution, Frobenius Method, Equation, ODE, Differential Equation, Gamma Function, Gamma Function Introduction, Gamma Function Properties, Gamma Function Bessel, Generalized Factorial, (1/2)!, 1/2!, 1/2 factorial, Theorem, Proof

Id: PwCl7vVyXwY

Length: 14min 18sec (858 seconds)

Published: Sat Apr 01 2017