The exponents differ by an integer so for a second

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ficients equal to zero, we find that , and , That is, for . , . Therefore one solution of the Laguerre equation is Note that if , a positive integer, then solution is a polynomial for . In that case, the ________________________________________________________________________ page 230 —————————————————————————— CHAPTER 5. —— Section 5.7 2. only for It follows that . Furthermore, and lim lim and therefore is a regular singular point. The indicial equation is given by , that is, , with roots 4. The coefficients no singular points. , 5. only for follows that and , and . are analytic for all . Furthermore, . Hence there are and It lim lim and therefore is a regular singular point. The indicial equation is given by , that is, 6. , with roots for and . and . We note that For the singularity at , , and lim lim and therefore is a regular singular point. The indicial equation is given by , that is, , with roots and . For the singularity at , ________________________________________________________________________ page 231 —————————————————————————— CHAPTER 5. —— lim lim lim and therefore lim is a regular singular point. The indicial equation is given by , that is, , with roots 7. only for follows that and . . Furthermore, and It lim lim and therefore is a regular singular point. The indicial equation is given by , that is, , with complex conjugate roots 8. Note that only for . It follows that . We find that lim , and lim lim and therefore . lim is a regular singular point. The indicial equation is given by , that is, 10. and , with roots for and lim which is undefined. Therefore at , and . We note that For the singularity at lim . , , , is an irregular singular point. For the singularity ________________________________________________________________________ page 232 —————————————————————————— CHAPTER 5. —— lim lim lim and therefore lim is a regular singular point. The indicial equation is given by , that is, 11. , with roots for and . and . We note that For the singularity at , lim lim lim and therefore , and lim is a regular singular point. The indicial equation is given by , that is, , with roots and lim . For the singularity at lim lim and therefore , lim is a regular singular point. The indicial equation is given by , that is, 12. , with roots for and . and . We note that For the singularity at , lim and therefore lim lim , and lim is a regular singular point. The indicial equation is given by ________________________________________________________________________ page 233 —————————————————————————— CHAPTER 5. —— , that is, , with roots and lim , lim lim and therefore . For the singularity at lim is a regular singular point. The indicial equation is given by , that is, 13 , with roots . Note the and and It follows that . . Furthermore, and lim lim and therefore...
View Full Document

This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue University.

Ask a homework question - tutors are online