Can you actually recall that the Bessel function of the first order in Mu, that the J_Mu(x) is equal to sum of the k equal to 0 to infinity, minus 1^k, over k factorial Gamma of k plus Mu plus 1, and the x over 2^2k plus Mu? In general, this chain of x is an infinity series. But when the V is the half-integer. For V on half-integer. By which I mean, V is equal to 1/2, or 3/2, or 5/2, and so on. I claim the following things. Then J_plus or minus V(x). You can write it simply the plug in that value Mu is equal to 1/2, or 3/2, or 5/2 into that formula, then you get an infinite series. But we can recognize it as a trigonometric function. In other words, the sine or cosine times the sum powers of x. For those in my claim is produced the for V and half-integer then. It contains, I can say in the following way, can be written, can be expressed in a closed form. Now the infinite series close form involving trigonometric functions and the suitable powers of x. I will just see the couple of examples on here. For example, let's consider that J_1/2(x), then when V is equal to 1/2, 1/2(x) from this it will be a summation of k is equal to 0 to infinity of minus 1^k, over k factorial, and the Gamma of k plus 1/2 plus 1, and x over 2^2k plus 1/2. What is crucial to us is, compute Gamma of k plus 1/2 plus 1. Can you remind the following the videos for formula for the Gamma functions? All they recorded here. Gamma of Alpha times Alpha, this is equal to Gamma (Alpha plus 1), for Alpha positive number. Apply this relation to this one. Gamma over something plus 1 is equal to, something times the Gamma of something. That you can write it as k plus 1/2 of Gamma of k plus 1/2. You can go a little further. This is k plus 1/2 times of, express this one as k minus 1 plus 1/2. No, I'd rather to write it as this is k plus 1/2 is k minus 1. Let me see, k minus 1/2 plus 1, k minus 1/2 and plus 1. So that k over this one, that will be equal to, k minus 1/2 times of Gamma times the k minus 1/2. We can keep going until finally we get Gamma(1/2) so that doing the same process again and again, you will get, k plus 1/2 times k minus 1/2. Finally, you will end up by Gamma(1/2), then in front you need just 1/2. We simply have this formula. Let's try to figure out Simplify the product way, then this you can write k. That is equal to k plus 1 over 2 and 2k plus 1 over 2. This is the first one. Next one is 2k minus 1 over 2 and so on. Finally, we'll get 1.5 and times Gamma over 1.5. What is the Gamma over 1.5? On the other hand, we get Gamma over 1.5 is equal to integral from 0 to infinity, or by definition, t^1.5 minus 1 times the e to the minus t_dt way. Then you can compute using some calculus. I just acclaim that this is equal to square root of Pi. It's a good exercise in calculus, so prove it. Prove it by yourself. Pi which is Gamma over 1.5, I just a claim that is equal to square root of Pi. There is a Gamma over k plus 1.5 plus 1. Then this is equal to 1. In the top you have 2k plus 1 and the 2k minus 1 and so on. All the odd integers, so you'll have a 3 times 1 and we have many twos' in the bottom. How many twos' do we have down there? The easy way to compute it is, write this expression as 0 plus 1.5 times Gamma over 1.5. How many such products do we have from this one to that one? First one is a 0 plus 1.5. Next one is a 1 plus 1.5 and the so on. Finally, we have k plus 1.5. So you have totally, k plus 1, the products down there. That means that we have a 2 to the k plus 1 way. Go one step further, then that is equal to, it looks like a factorial but you are missing all the even part this way, so let's feed it up, 2k plus 1 times a 2k and the 2k minus 1 times and the so on, 3 times and the 2 times 1 and over, and 2 to the k plus 1. But because we multiply the top by 2k times 2 times over k minus 1 and so on. You must divide it through by 2k, two times over k minus 1 and so on and 2, and finally you get square root of Pi. Then this one, you can write it as, in the top you have a 2k plus 1 factorial, and in the bottom you have, there are how many twos' down there? We have exactly k two's, so they make a 2 to the k times 2 to the k plus 1. That is 2 to the 2k plus 1. Then remaining part is a k times a k minus 1 times, and finally 1, that is k factorial and the square root of Pi. We get this. I will plug this result to the infinite series for g over 1.5 of x. I think we should have to the k down there, and this is 2k plus Nu. 2k plus 1.5 of because there is 1.5. Now this is a summation. K is equal to 0 to infinity, and minus 1 to the k and the k factorial. Instead of 1 over Gamma k plus 1.5 plus 1, we can use this expression. We take the reciprocal of this, because if we have a one over Gamma something, so there we have 2 to the 2k plus 1 times k factorial over the 2k plus 1 factorial, and 1 over square root of Pi, then you have x over 2 and 2k plus 1.5. Test double what do we have now? This k factorial and that k factorial canceled out from the top and bottom. Then what? Let's do the final trick. That is summation k equal to 0 to infinity of minus 1^k and over 2k plus 1 factorial times the square root of Pi. I did not consider this term here. Instead of this, I write it as x over 2 and 2k plus 1. Then what? You should divide it by square root of x over 2. Because I multiply this by square root of x over 2 so that you should divide it by the square root of 2 over x and you still have a 2^2k plus 1 down there. They are equal. Then what? This is equal to square root of 2 over Pi times x, from this and that. Then the remaining is the summation from 0 to infinity of minus 1^k and 2k plus 1 factorial in the bottom and finally, 2^2k plus 1 times x over 2^2k plus 1 will be just this and the 2 to the that canceled out. You only have x^2k plus 1. Now we're at the final stage. Can you recognize this famous power series? This is a Taylor series expansion at point zero of sine of x. Finally, we get, so we have j over 1 1/2(x). This is equal to 2 over Pi(x) and square root of it times this. What is it? This is the sine of x so that we can confirm the hour declaim. Bessel function of first kind of four even integer known for the half-integer can be expressed as close to form containing determinative functions and the suitable powers of x. This is precisely such an example. Do the same trick for J of negative minus 1/2(x). You can do it by the exactly same way. Then I claim that. I claim the similarly you will get, for example, J over minus 1/2(x). This is equal to square root 2 over Pi(x) and then times the cosine of x. We can represent it, g of negative 1/2(x) in a closed-form. Similarly, we can expect a similar result for the Bessel function of the second kind. For example, think about y over 1/2(x). What is it? This is sine of 1/2 of Pi and cosine over 1/2of Pi times J over 1/2(x) and minus J of negative 1/2(x). By definition for the Bessel function of the second kind, how much is sine Pi over 2? This is one. How much is a cosine 1/2of Pi that is equal to 0? Immediately we'll get this is equal to minus J over negative 1/2(x). Using this one, you will get this is minus square root 2 over Pi of x and times the cosine of x. Similarly, you can find the Y over negative 1/2(x) will be equal to square root of 2 over Pi of x times sine of x. As the new is equal to three-halves or five-halves and the zone already you can expect that the such a close to form representation, even though the representation is a little bit longer. It may have a couple of several terms involving the sin and the cosine and the suitable powers of x.
