Unformatted text preview: ind that
. In order to solve the second equation, set
remaining equations results in
,
,
Hence a second solution is . Solution of the
,. 16 . After multiplying both sides of the ODE by , we find that
. Both of these functions are analytic at
, hence
singular point.
. Furthermore,
and
. . So the indicial equation is . In order to find the solution corresponding to and
is a regular
, with roots , set Upon substitution into the ODE, we have
.
That is,
.
Setting the coefficients equal to zero, we find that for , .
It follows that ________________________________________________________________________
page 239 —————————————————————————— CHAPTER 5. —— .
Hence one solution is
.
The exponents differ by an integer. So for a second solution, set
.
Substituting into the ODE, we obtain
.
Since , it follows that
. Now Substituting for , the right hand side of the ODE is
. Equating the coefficients, we obtain the system of equations Evidently,
that . In order to solve the second equation, set
. We then find
,
,
, . Therefore a second solution is ________________________________________________________________________
page 240 —————————————————————————— CHAPTER 5. —— 19 . After dividing by the leading coefficient, we find that
lim lim lim
Hence
with roots
For lim is a regular singular point. The indicial equation is
and
.
,
lim lim lim
Hence , lim is a regular singular point. The indicial equation is
, with roots
. Given that and . is not a positive integer, we can set Substitution into the ODE results in That is, Combining the series, we obtain
,
in which ________________________________________________________________________
page 241 —————————————————————————— CHAPTER 5. ——
.
Note that
equal to zero, we have for . Setting the coefficients
, and . Hence one solution is Since the nearest other singularity is at
be at least
.
. Given that , the radius of convergence of is not a positive integer, we can set will Then Substitution into the ODE results in That is, After adjusting the indices, Combining the series, we obtain
,
in which
________________________________________________________________________
page 242 —————————————————————————— CHAPTER 5. ——
.
Note that
it follows that for . Setting , , Therefore a second solution is . Under the transformation , the ODE becomes
. That is,
.
Therefore is a singular point. Note that
and . It follows that
lim lim lim
Hence , lim is a regular singular point. The indicial equation is
, or
21 Evidently, the roots are and . . Note that
and . ________________________________________________________________________
page 243 —————————————————————————— CHAPTER 5. ——
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 Spring '08
 Staff
 Taylor Series, lim, Complex number

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