Download Radius of Convergence

Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients
Mar. 07, 2012
•
Many examples here are taken from the textbook. The first number in () refers to the problem number
in the UA Custom edition, the second number in () refers to the problem number in the 8th edition.
0. Theoretical Issues
•
Radius of convergence.
Theorem 1. (Radius of convergence) For any power series
ρ ∈ [0, ∞] (meaning: ρ > 0 and can be infinity) such that
◦
◦
the power series converges for all x such that |x − x0| < ρ;
P
n
an (x − x0) , there is a number
the power series diverges for all x such that |x − x0| > ρ.
This particular number ρ is called the radius of convergence.
Remark 2. In other words, for any |x − x0| < ρ, the infinite sum of numbers
X
an (x − x0)n
(1)
indeed equals a number, while for any |x − x0| > ρ the infinite sum is either infinity or does not have
a limit at all. Consequently, inside |x − x0| < ρ, the power series
X
an (x − x0)n
(2)
indeed represent a well-defined function.
•
How to calculate radius of convergence. We have two ways. The first is for an arbitrary power series,
the 2nd deals with the situation when the power series is the result of solving a linear differential
equation.
◦
Method 1: Ratio test.
Theorem 3. (Ratio test) If, for n larger than a certain N, the coefficients an satisfy
an+1 =L
lim (0 6 L 6 ∞)
(3)
an n
∞
P∞
n
then the radius of convergence of the power series
n=0 an (x − x0) is ρ = 1/L, with ρ = ∞
if L = 0 and ρ = 0 if L = ∞. That is, the power series converges 1 for |x − x0| < ρ and diverges
for |x − x0| > ρ.
Example 4. (8.2 2; 8.2 2) Determine the convergence set of
∞
X
3n n
x .
n!
(4)
n=0
3n
Solution. We have an = n! . The ratio
an+1 3n+1/(n + 1)! 3 =
=
n + 1
an 3n/n!
So we have L = 0 and the radius of convergence is ∞.
1. In fact, absolutely converges.
1
0 as n
∞.
(5)
Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients
2
◦
Method 2: Checking singular points.
−
Only applies when the power series is the solution to a linear differential
equation.
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Analytic functions.
Definition 5. A function is analytic at x0 if, in an open interval about x0 this function
equals a power series about x0.
Theorem 6. If f (x) is analytic at x0 , then the power series it equals is exactly the
Taylor expansion of f at x0.
−
Checking analyticity. Given f (x).
•
Its analyticity at a point x0 cannot be checked by computing the Taylor
expansion of f and then study the radius of convergence of the latter.
•
Use the following rules to check analyticity.
1. ex, sin x, cos x, polynomials are analytic everywhere.
R
2. If f (x) is analytic at x0, so are f ′(x) and
f (x).
3. If f (x), g(x) are analytic at x0, so are f ± g,f g,f ◦ g.
4. If f , g are analytic at x0, then f /g is analytic at x0 if g(x0) 0.
−
Ordinary and singular points of a linear differential equation. For a linear homogeneous
differential equation,
an(x) y (n) + an−1(x) y (n−1) + + a1(x) y ′ + a0(x) y = 0.
(6)
We can write it into standard form:
y (n) + Pn−1(x) y (n−1) + + P0(x) y = 0.
(7)
Then a point x is singular if at least one of Pn−1, , P0 is not analytic at x. Otherwise
x is ordinary.
Example 7. Find the singular points of the equation
x2 y ′′ + 3 y ′ − x y = 0.
(8)
Solution. First write it into the standard form
y ′′ +
We have
p(x) =
3 ′ 1
y − y = 0.
x2
x
3
,
x2
1
q(x) = − .
x
(9)
(10)
As 3, 1, x2 and x are all analytic everywhere, their ratios are analytic at all points except
those making the denominator vanish.
Therefore the only singular point is x = 0.
−
Criterion for radius of convergence.
Theorem 8. (Existence of analytic solutions) Suppose x0 is an ordinary point for
equation
y ′′(x) + p(x) y ′(x) + q(x) y(x) = 0,
(11)
then it has two linearly independent analytic solutions of the form
y(x) =
∞
X
n=0
n
an (x − x0) .
(12)
Mar. 07, 2012
3
Moreover, the radius of convergence of any power series solution of the form given above
is at least as large as the distance from x0 to the nearest singular point (real or complexvalued).
Remark 9. It is important, when determining the radius of convergence, to remember
count in complex singular points.
Remark 10. The distance between any two complex numbers a + b i, c + d i can be
computed as
p
(a − c)2 + (b − d)2 .
(13)
Example 11. Find a minimum value for the radius of convergence of a power series
solution about x0.
(x + 1) y ′′ − 3 x y ′ + 2 y = 0,
x0 = 1.
(14)
Solution. Write the equation to standard form
y ′′ −
3x ′
2
y +
y = 0.
x+1
x+1
(15)
The only singular point is x = −1. Thus the minimum radius of convergence is the
distance between x0 = 1 and −1, which is 2.
Example 12. Find a minimum value for the radius of convergence of a power series
solution about x0.
(1 + x + x2) y ′′ − 3 y = 0;
x0 = 1.
(16)
Solution. Write the equation to standard form
y ′′ −
3
y = 0.
1 + x + x2
The singular points are roots of 1 + x + x2, which are
√
−1 ± 3 i
.
x1,2 =
2
(17)
(18)
To find out the closest singular point to x0, we compute
√ √
3 √
3
|x1 − x0| = − +
i = 3 ;
|x2 − x0| = 3 .
(19)
2
2 √
So√both are 3 away from x0. As a consequence, the minimum radius of convergence
is 3 .
1. Basic Information
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The Equation
We consider the following type of equations:
an(x) y (n) + + a1(x) y ′ + a0(x) y = b(x)
(20)
with an , , a0 and b all polynomials.
•
How to get general solution
1. Set ansatz
y=
Substitute into the equation.
2. Figure out the recurrence relation.
∞
X
n=0
an (x − x0)n.
(21)
Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients
4
3. Solve the recurrence relation.
4. Simplify.
•
How to solve initial value problem (IVP)
◦
Keep in mind that if the initial conditions are
then necessarily
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y ′(x0) = A1, y(x0) = A0,
(22)
a0 = A0, a1 = A1, (23)
How to check solutions
2. More Examples
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General formula required.
Example 13. (8.3 21; 8.3 21) Find a power series expansion about x = 0 for a general solution to
the given differential equation. Your answer should include a general formula for the coefficients.
y ′′ − x y ′ + 4 y = 0.
Solution. We write
y(x) =
∞
X
(24)
an xn.
(25)
n=0
Substituting into the equation, we have
0 = y ′′ − x y ′ + 4 y
∞
∞
∞
X
X
X
=
an n (n − 1) xn−2 − x
n an xn−1 +
4 an xn
=
n=2
∞
X
n=0
n=1
an+2 (n + 2) (n + 1) xn −
= (2 a2 + 4 a0) +
∞
X
n=1
This gives
∞
X
n=0
∞
X
n an x n +
n=1
4 an x n
n=0
[(n + 2) (n + 1) an+2 − (n − 4) an] xn.
(26)
2 a2 + 4 a0 = 0
(n + 2) (n + 1) an+2 − (n − 4) an = 0.
(27)
(28)
a2 = −2 a0,
n−4
an.
an+2 =
(n + 2) (n + 1)
(29)
Therefore
(30)
It is clear that we should discuss n = 2 k and n = 2 k − 1 separately.
For even n, we have
1
1
a2 = −2 a0, a4 = − a2 = a0, a6 = 0, a8 = 0, a10 = 0, 6
3
For odd n, we have
a2k+1 =
(2 k − 5) (2 k − 7)
(2 k − 5) (−3)
2k −5
a2k −1 =
a2k−3 = =
a1.
(2 k + 1) (2 k − 2)
(2 k + 1)!
(2 k + 1) (2 k)
(31)
(32)
Summarizing, we have
y(x) = a0
"
#
∞
X
(2 k − 5) (−3) 2k+1
1 4
x
.
1 − 2 x + x + a1 x +
(2 k + 1)!
3
2
k=1
(33)
Mar. 07, 2012
5
Remark 14. For this problem it is clear that the radius of convergence is ∞.
3. Notes and Comments
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Whether the power series converges at x = x0 ± ρ is tricky to determine. Different approaches are
needed for different power series.
•
In the textbook “analytic
at x0” is defined as “if, in an open interval about x0, this function is the sum
P∞
of a power series n=0 an (x − x0)n that has a positive radius of convergence.” The last part “that has
a positive radius of convergence” is redundant. If a power series sums up to a well-defined function
in an interval containing x0, then the radius of convergence is positive by definition.
•
Note that, “f (x) is analytic at x0” is not the same as “the Taylor expansion of f at x0 has positive
radius of convergence”. A classical example is the following:
(
−1/x
x>0 .
f (x) = e
(34)
0
x60
Turns out we have
f (n)(0) = 0
(35)
for all n. Consequently its Taylor expansion is
0 + 0x + 0 x2 + 0 x3 + (36)
which has radius of convergence ∞. But obviously this power series sum up to 0 which is different
from f .
Note that it is easy to modify the example so that the power series looks less artificial. For example
if we let
(
f (x) =
ex + e−1/x x > 0 .
ex
x60
(37)
then the Taylor expansion is
x
which of course sum up to e , not f (x).
∞
X
xn
n!
n=0
(38)