No other cases can arise. Analytic varieties versus integral varieties of Lie algebras of vector fields with G. Characterizing singularities of varieties and of mappings with T. Theorem 3 Let the function f z be analytic everywhere inside the domain D except at a finite number of poles, continuous on the boundary C of this domain and not vanish on C ; moreover, let f ' z be continuous on C. Risler, B. Correspondingly, we must join across the shores of the cuts of the sheets D 0 and D 1 : 4 to 1 1 and 3 to 2 1. Theorem 1 Let there be given two simply connected regions D 0 and D 1 without common points such that their boundaries have one common section g and in these regions analytic functions f 0 z and f 1 z , respectively.
Don't consult these answers unless you are totally desperate. From finding the answer yourself you learn about all you don't know. From copying a given answer you learn nothing. Note 4: The lecture notes contain at the end of each chapter a section with physical applications.
This will allow us to check that we get the correct solution. We will be looking for a solution in the form,.
The next step is to combine everything into a single series. Notice that in the process of the shift we also got both series starting at the same place. We can now add up the two series. This gives,.
This gives us the following,. We wanted to find a series solution to the differential equation.
We are almost to the point where we can do that. Doing this gives,.
As you will see, we actually need these to be in the problem to get the correct solution. The next step is to collect all the terms with the same coefficient in them and then factor out that coefficient.
roots of Po(y) = 0. NOTE ON SINGULARITIES OF POWER SERIES. BY H. F. BOHNENBLUST. DEPARTMENT OF MATHEMATICS, PRINCETON UNIVERSITY. Full text. Full text is available as a scanned copy of the original print version. Get a printable copy (PDF file) of the complete article (K), or click on a page.
In the last step we also used the fact that we knew what the general formula was to write both portions as a power series. This is also our solution. We are done. First, we started out by saying that we wanted a series solution of the form,. We got a solution that contained two different power series. Also, each of the solutions had an unknown constant in them.
This is not a problem. From our work with second order constant coefficient differential equations we know that the solution to the differential equation in the last example is,. Solutions to second order differential equations consist of two separate functions each with an unknown constant in front of them that are found by applying any initial conditions.
Share full text access. Please review our Terms and Conditions of Use and check box below to share full-text version of article. Citing Literature. Volume s , Issue 4 October Pages Related Information. Close Figure Viewer. Browse All Figures Return to Figure. Previous Figure Next Figure.