Review of Power Series

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Review of Power Series Real Analytic
functions Ordinary points of a 2nd Order
Homogeneous L.D.E.
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In this lecture we discuss the Power series
method of solving a second order homogeneous
linear differential equation First we spend some
time in reviewing the results about Power Series
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Definition An infinite series of the form
is called a Power Series centred at x0.
We usually deal with power series centered at the
origin
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Convergence of Power Series
The P.S.
certainly converges at the point x x0 to a0.

• One of the following happens
• It converges only at the point x x0
• It converges absolutely at all points x

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• There exists a R such that the P.S. converges
  absolutely for all x s.t. x- x0 lt R and
  diverges for all x s.t. x- x0 gt R

R is called the radius of convergence and the
interval (x0 R , x0 R ) is called the
interval of absolute convergence.
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Examples
1. The P.S.
converges only at x 0 (to 1).
2. The P.S.
converges absolutely for all x (to ex ).
3. The P.S.
converges absolutely only for x lt 1 to
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Formula for the Radius of convergence
Consider the P.S.
Suppose
Then the P.S. converges absolutely for all x s.t.
x- x0 lt 1/L if L gt 0. If L 0 it converges
absolutely for all x.
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Hence we can get R, the radius of convergence
from the formula
For the P.S.
Hence R 1.
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Properties of Power Series.
Fact 1. A P.S can be differentiated term by term
within its interval of convergence.
For example
Hence
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Fact 2. A P.S can be integrated term by term
within its interval of convergence.
For example
Hence
or
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Fact 3. If
and each of series converges for xltR
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then
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Real Analytic functions
A real-valued function f (x) is called to be
(real) analytic at the point x x0 if it can be
expanded in a Taylor series of powers of x -
x0 valid at least for all x near x0.
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Examples
sin x, cos x, ex are analytic for all x.
All polynomials are analytic for all x.
A rational function (which is the ratio of two
polynomials) are analytic for all x except where
the denominator is zero.
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Ordinary points of a second order homogenous
l.d.e.
Consider the second order homogeneous l.d.e. in
the standard form
A point x0 is said to be an ordinary point of the
above d.e. if both the coefficient functions
are real analytic at x0.
Otherwise x0 is said to be a singular point.
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Thus x0 will be a singular point if at least one
of the coefficient functions
is NOT analytic at x0.
Examples.
1. For the d.e.
every point is an ordinary point.
2. For the Bessels d.e.
every point ? 0, is an ordinary point.
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3. For the d.e.
x 0 is an ordinary point as
And hence is analytic at x 0 .
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We will state a theorem that asserts the
existence of Power series solutions in terms of
powers of (x x0) , whenever x0 is an ordinary
point. But we look at an example of finding Power
Series Solutions.
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Problem 1 Page 182
Find the solution of the d.e.
in terms of powers of x.
Since
are both analytic at all
points x, all points are ordinary points.
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We assume a solution as
Note that
Substituting for y, y?, y? in the given
differential equation, we get
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i.e.
In the first
we replace n with n2.
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We get
We now separate the terms corresponding to n 0
and n 1 and group the terms for n 2. Thus we
get
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i.e.
Equating the coefficients of 1, x, x2, on both
sides, we get
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.()
() is known as the recurrence relation for an.
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Since a3 0, the recurrence relation () gives
a5 0, a7 0,.
In other words, an 0 for all odd n gt 1.
Also,
(from () with n2)
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Thus the solution is
a0, a1 arbitrary constants.
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Theorem (Existence of Power Series Solutions)
Let x0 be an ordinary point of the d.e.
Let a0, a1 be two real numbers. Then there exists
a Power series solution (in powers of (x
x0)) of the above equation satisfying the initial
conditions When x x0, y a0 , y? a1.
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Further if the Power Series expansions of
converge for all x such that
then the power series solution
of the d.e. also converges for all x such that
and is valid in that interval.
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Example 2 Solve the d.e.
by Power Series method.
Solution Since P(x) x2 and Q(x) 2x are both
analytic at x 0, origin is an ordinary point of
the given d.e. We note that the Power Series
expansion of P(x) x2 and Q(x) 2x converge for
all x.
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Thus we assume a power series solution of the
form
Note that
Substituting for y, y?, y? in the given
differential equation, we get
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i.e.
we replace n by n2.
In the first
In the second
we replace
and third
n by n-1. We get
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We now separate the terms corresponding to n 0
and n 1 and group the terms for n 2. Thus we
get
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Equating the coefficients of 1, x, x2, on both
sides, we get
()
..()
i.e.
using ()
--- Recurrence Relation for an
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Since a2 0, the recurrence relation () gives
Putting n 4, 7, in () we get
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More generally
Putting n 2, 5, in () we get
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More generally
Thus the solution is
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a0, a1 arbitrary constants
Note that the above solution is valid for all x.
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We saw in the previous problems that the
coefficients an satisfied a two-term recurrence
relation. But being a second order equation,
normally we may expect the an s will satisfy a
three-term recurrence relation. The following
example problem illustrates this.
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Example 3. Solve the following d.e. by Power
Series method
Solution Here P(x) x 1, Q(x) 2x 3 are
polynomials and so are analytic at all points.
Hence all points (in particular x 0 ) are
ordinary points of the above d.e.
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Let us assume a solution as
Hence
Substituting for y, y? , y? in the given d.e. we
get
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i.e.
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we replace n by n 2
In the first
In the third
we replace n by n 1
In the fourth
we replace n by n -1. We get
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Isolating the terms corresponding to n 0 and
grouping the terms corresponding to n 1,
we get
Equating the coefficients of 1, x, x2, we get

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Hence
Note that a0, a1 are arbitrary and the an, n 2
can be expressed in terms of them.
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Problem 3(a), page 182 Verify that the equation
has a three term recursion formula and find
its series solution y1(x) and y2(x) such that
and
Solution
Since P(x) 1 and Q(x) -x are analytic every
where (in particular x 0)
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Thus we assume a power series solution of the form
Substituting for y, y? , y? in the given d.e. we
get
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we replace n by n 2
In the first
we replace n by n 1
In the third
In the last
we replace n by n -1. We get
Isolating the terms corresponding to n 0 and
grouping the terms corresponding to n 1,
we get
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or
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Legendres differential equation is the d.e.
where p is a constant.
Here
both of which are analytic at all points
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are ordinary points of
Thus all points
Legendrs equation. In particular, x 0 is an
ordinary point and we note the Power Series
expansion of
in powers of x converge absolutely for all
x lt 1. Hence power series solutions in powers
of x exist which converge absolutely
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for all x lt 1. We shall find them.
Let us assume a solution as
Hence
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Substituting for y, y? , y? in the given d.e. we
get
i.e.
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or
Isolating the terms corresponding to n 0, and n
1 and grouping the terms corresponding to n
2, we get
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Note
or
Equating the coefficients of 1, x, x2, we get

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Hence
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And so on.
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Hence the solution is
a0, a1 are arbitrary constants
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Note (i) If p is an even positive integer, then
y1(x) becomes a polynomial
(solution) of degree p containing only even
powers of x.
(ii) If p is an odd positive integer,
then y2(x) becomes a polynomial
(solution) of degree p containing only odd powers
of x.
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If p is a positive integer n , a suitable scalar
multiple of the polynomial solution obtained is
known as Legendre Polynomial of degree n and is
denoted by Pn(x). We multiply by a scalar so
that Pn(1) 1.
For example p 2 gives
Hence
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The first few Legendre Polynomials are
And so on
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Problem 6 Page 183
Chebyshevs d.e. is
where p is a non-negative constant.
Here
both of which are analytic at all points
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are ordinary points of
Thus all points
Chebyshevs equation. In particular, x 0 is an
ordinary point and we note the Power Series
expansion of
in powers of x converge absolutely for all
x lt 1. Hence power series solutions in powers
of x exist which converge absolutely
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for all x lt 1. We shall find them.
Let us assume a solution as
Hence
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Substituting for y, y? , y? in the given d.e. we
get
i.e.
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or
Isolating the terms corresponding to n 0, and n
1 and grouping the terms corresponding to n
2, we get
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or
Equating the coefficients of 1, x, x2, we get

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Hence
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And so on.
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Hence the solution is
a0, a1 are arbitrary constants
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Note (i) If p is an even positive integer, then
y1(x) becomes a polynomial
(solution) of degree p containing only even
powers of x.
(ii) If p is an odd positive integer,
then y2(x) becomes a polynomial
(solution) of degree p containing only odd powers
of x.
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If p is a positive integer n , a suitable scalar
multiple of the polynomial solution obtained is
known as Chebyshev Polynomial of degree n and is
denoted by Tn(x). We multiply by a scalar so
that Tn(1) 1.
For example p 2 gives
Hence
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The first few Chebyshev Polynomials are
And so on
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FROBENIUS SERIES SOLUTION OF A SECOND ORDER
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION
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An example.
Consider the second order homogeneous l.d.e.
Putting
the equation becomes
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whose solution is
or
c1, c2 arbitrary constants.
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nor
Now neither
can be expressed as a Power Series in
(non-negative integral powers of ) x. Thus the
given l.d.e. does not have a Power Seroes
Solution. We note that x 0 is NOT an ordinary
point of the given d.e. as
is NOT analytic at x 0.
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However we note that
and
are both analytic at x 0.
And the given d.e. has a Frobenius series
solution (i.e an infinite series solution where
the powers of x need not be non-negative
integers).
Such points are called regular singular points.
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Regular Singular points
A singular point x0 of the d.e.
is called a regular singular point if both

and
are analytic at x x0 .
Otherwise it is called an irregular singular
point.
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Problem 1(a) Page 191
Locate and classify the singularities of the
following d.e.
Here
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P(x) is analytic at all points except x 0.
Q(x) is analytic at all points except x 0,1.
Thus all points are ordinary points except x0, 1.
is NOT analytic at x 0.
Hence x 0 is an irregular singular point.
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But
are both analytic at x 1. Hence x 1 is a
regular singular point.
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For the Bessels d.e. of order p,
x 0 is the only singular point which is a
regular singular point, as you can easily verify.
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Look at the d.e.
Here
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P(x) and Q(x) are analytic at all points x except
x 0, 1, -1 where the denominator vanishes.
Hence the only singular points are
1,
x
0,
and
-1.
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Since x P(x) and x2 Q(x) are analytic at x 0,
x 0 is a regular singular point.
Similarly (x 1)P(x) and (x 1)2 Q(x) are
analytic at x 1. Or x 1 is a regular
singular point.
is NOT
analytic at x -1. So x -1 is an irregular
singular point.
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For the d.e.,
every point is an ordinary point.
For the d.e.,
x 0 is the only singular point which is a
regular singular point .
For the d.e.,
x 0 is the only singular point which is an
irregular singular point.
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We shall shortly state a theorem which assures us
that if x 0 is a regular singular point of the
d.e.
then there exists at least one Frobenius series
solution of the form
a0 ? 0, m a real number.
First we look at an example.
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Consider the d.e.
Clearly x 0 is a regular singualr point.
We assume a solution to the above equation as
a0 ? 0, m a real number.
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i.e.
Hence
Note As m need not be an integer, both
for y?, y? start from n 0.
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Substituting for y, y?, y? in the given d.e.,
we get
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i.e.
We divide throughout by xm. We get
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In the last
we replace n by n 1.
We thus get
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Isolating the terms corresponding to n0 and
grouping the terms for n ? 1, we get
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We shall now equate the coefficients of like
powers of x on both sides. We get
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as a0 ? 0
or
Indicial Equation
i.e.
m 1 and m -1/3 are called exponents of the
given d.e. at the regular singular point x 0.
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Also
( n ? 1)
Recurrence relation for an
Case (i)
m 1
The recurrence relation becomes
or
( n ? 1)
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( n ? 1)
n1 gives
n2 gives
n3 gives
And so on
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Hence one solution is
( a0 an arbitrary constant)
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Case (ii)
The recurrence relation becomes
i.e.
or
( n ? 1)
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( n ? 1)
n1 gives
n2 gives
n3 gives
And so on
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Hence a second solution is
( a0 an arbitrary constant)
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Note that y1(x) is bounded near x 0, but y2(x)
is unbounded near x 0.
Hence y1(x) and y2(x) are two LI solutions of
the given differential equation valid over the
interval (0, ?).
Hence the general solution of the given
differential equation valid over the interval (0,
?) is
( c1, c2 arbitrary constants)
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Here the roots of the indicial equation were not
equal and did not differ by an integer. Hence we
got two LI solutions. In general, a theorem
assures us that there always exists one solution
corresponding to the larger root of the indicial
equation.