7.E: Power series methods (Exercises)
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7.1: Power Series
Is the power series ∑∞k=0ekxk convergent? If so, what is the radius of convergence?
Is the power series ∑∞k=0kxk convergent? If so, what is the radius of convergence?
Is the power series ∑∞k=0k!xk convergent? If so, what is the radius of convergence?
Is the power series ∑∞k=01(2k)!(x−10)k convergent? If so, what is the radius of convergence?
Determine the Taylor series for sinx around the point x0=π.
Determine the Taylor series for lnx around the point x0=1, and find the radius of convergence.
Determine the Taylor series and its radius of convergence of 11+x around x0=0.
Determine the Taylor series and its radius of convergence of x4−x2 around x0=0. Hint: You will not be able to use the ratio test.
Expand x5+5x+1 as a power series around x0=5.
Suppose that the ratio test applies to a series ∑∞k=0akxk. Show, using the ratio test, that the radius of convergence of the differentiated series is the same as that of the original series.
Suppose that f is an analytic function such that f(n)(0)=n. Find f(1).
Is the power series ∑∞n=1(0.1)nxn convergent? If so, what is the radius of convergence?
- Answer
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Yes. Radius of convergence is 10.
Is the power series ∑∞n=1n!nnxn convergent? If so, what is the radius of convergence?
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Yes. Radius of convergence is e.
Using the geometric series, expand 11−x around x0=2. For what x does the series converge?
- Answer
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11−x=−11−(2−x) so 11−x=∞∑n=0(−1)n+1(x−2)n, which converges for 1<x<3.
Find the Taylor series for x7ex around x0=0.
- Answer
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∞∑n=71(n−7)!xn
Imagine f and g are analytic functions such that f(k)(0)=g(k)(0) for all large enough k. What can you say about f(x)−g(x)?
- Answer
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f(x)−g(x) is a polynomial. Hint: Use Taylor series.
7.2: Series solutions of linear second order ODEs
In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the kth coefficient.
Use power series methods to solve y″+y=0 at the point x0=1.
Use power series methods to solve y″+4xy=0 at the point x0=0.
Use power series methods to solve y″−xy=0 at the point x0=1.
Use power series methods to solve y″+x2y=0 at the point x0=0.
The methods work for other orders than second order. Try the methods of this section to solve the first order system y′−xy=0 at the point x0=0.
Chebyshev’s equation of order p:
- Solve (1−x2)y″−xy′+p2y=0 using power series methods at x0=0.
- For what p is there a polynomial solution?
Find a polynomial solution to (x2+1)y″−2xy′+2y=0 using power series methods.
- Use power series methods to solve (1−x)y″+y=0 at the point x0=0.
- Use the solution to part a) to find a solution for xy″+y=0 around the point x0=1.
Use power series methods to solve y″+2x3y=0 at the point x0=0.
- Answer
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a2=0, a3=0, a4=0, recurrence relation (for k≥5): ak=−2ak−5k(k−1), so y(x)=a0+a1x−a010x5−a115x6+a0450x10+a1825x11−a047250x15−a199000x16+⋯
We can also use power series methods in nonhomogeneous equations.
- Use power series methods to solve y″−xy=11−x at the point x0=0. Hint: Recall the geometric series.
- Now solve for the initial condition y(0)=0, y′(0)=0.
- Answer
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- a2=12, and for k≥1 we have ak=ak−3+1k(k−1), so y(x)=a0+a1x+12x2+a0+16x3+a1+112x4+340x5+a0+230x6+a1+242x7+5112x8+a0+372x9+a1+390x10+⋯
- y(x)=12x2+16x3+112x4+340x5+115x6+121x7+5112x8+124x9+130x10+⋯
Attempt to solve x2y″−y=0 at x0=0 using the power series method of this section (x0 is a singular point). Can you find at least one solution? Can you find more than one solution?
- Answer
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Applying the method of this section directly we obtain ak=0 for all k and so y(x)=0 is the only solution we find.
7.3: Singular points and the method of Frobenius
Find a particular (Frobenius-type) solution of x2y″+xy′+(1+x)y=0.
Find a particular (Frobenius-type) solution of xy″−y=0.
Find a particular (Frobenius-type) solution of y″+1xy′−xy=0.
Find the general solution of 2xy″+y′−x2y=0.
Find the general solution of x2y″−xy′−y=0.
In the following equations classify the point x=0 as ordinary, regular singular, or singular but not regular singular.
- x2(1+x2)y″+xy=0
- x2y″+y′+y=0
- xy″+x3y′+y=0
- xy″+xy′−exy=0
- x2y″+x2y′+x2y=0
In the following equations classify the point x=0 as ordinary, regular singular, or singular but not regular singular.
- y″+y=0
- x3y″+(1+x)y=0
- xy″+x5y′+y=0
- sin(x)y″−y=0
- cos(x)y″−sin(x)y=0
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- ordinary,
- singular but not regular singular,
- regular singular,
- regular singular,
- ordinary.
Find the general solution of x2y″−y=0.
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y=Ax1+√52+Bx1−√52
Find a particular solution of x2y″+(x−34)y=0.
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y=x3/2∞∑k=0(−1)−1k!(k+2)!xk (Note that for convenience we did not pick a0=1.)
Find the general solution of x2y″−xy′+y=0.
- Answer
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y=Ax+Bxln(x)