K.S. Narain , ICTP

This lesson continues with the study of the **hypergeometric differential equation. **First it is shown that dF(a,b,c,z)/dz proportional to F(a+1,b+1,c+1,z). First it is shown that dF(a,b,c,z)/dz proportional to F(a+1,b+1,c+1,z) and in general we have that dF^{n}(a,b,c,z)/d^{n}z is proportional to F(a+n,b+n,c+n,z).

Then the **hypergeometric function** F(a,b,c,z) was calculated for the case b=c and a=-m were m is a positive integer. This case is important because they **Jacobi polynomials . **In fact** **if in the **Jacobi differential equation **the change of variable z’=(1-z)/2 is done we obtain an **hypergeometric differential equation **for z’ with solution for the **Jacobi polynomials** ** via the hypergeometric function **is F(-?,?+?+?+1,?+1,(1-z)/2). The **Legendre polynomials** are obtained for ?=?=0.

In the second part of the lesson the **Confluent hypergeometric functions** were introduced it starts by the general definition of

**hypergeometric differential equation**given in lesson 30 and select the

**regular singular points**z

_{1}=0, z

_{2}and z

_{3 }then b is selected to be equal to b=z

_{2 }and b=2z

_{3 }and then take the limit b goes to infinity, the resultin equation is the

**confluent hypergeometric differential equation.**This equation will have only two parameters a and c and two singularities at z=0 and z=infinity. The singularity at z=0 is a

**regular singular**

**point**but z=infinity is an

**irregular singular**

**point.**

The solution for the case of z=0 can be found using the method explained in the previous lessons. Here is used the method of finding a solution is expressed in terms of an integral like in the case of lesson number 32. Here the solution is guest to be U_{1}(z)=Integral_{C} [Exp[zt]f(t)dt], this solution is substituted in the **confluent hypergeometric differential equation** and then the expression for f(t) is found to be f(t)=t^{a-1}(t-1)^{c-a-1}. So one solution is obtained when Re[c]>Re[a]>0 for this case the contour C is chosen to be from 0 to 1. This solution that is the analytic solution near z=0 is called **Confluent hypergeometric function **?(a,c,z). It was demonstrated that this is just the limit when b goes to infinity of the **hypergeometric function** F(a,b,c,z/b). The guest for the second solution is suposed to be of the form U_{2}(z)=(1-z)^{1-c}V(z) and then V(z) is found to be V(z)=?(a+1-c,2-c,z).

Finally the behavior near the **irregular singular point** z=infinity was studied for this case the same integral solution is proposed ?(z)=K Integral_{C} [Exp[zt]f(t)dt] but now the contour C is taken from minus infinity to zero and K is a constant and to the integrand be zero in the boundary we have to impose Re[a]>0 and Re[z]>0. Then the behavior of U_{1}(z), U_{2}(z) and ?(z) at z=infinity was studied and it was found that U_{1}(z) and U_{2}(z) grows exponentially while ?(z) does not, so as ?(z) can be written as linear combination of U_{1}(z) and U_{2}(z) this must be in such a way that the exponential in z cancels.

Mathematics for Physicists (Text Book from Google e-books preview)Written by<span class="addmd"> Philippe Dennery,Andre Krzywicki, Chapter 1, 2, 3 and 4</span>