Assignment with Solutions for Advanced Calculus | MATH 360, Assignments of Mathematics

Material Type: Assignment; Class: ADVANCED CALCULUS; Subject: Mathematics; University: University of Pennsylvania; Term: Fall 2002;

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Pre 2010

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Homework for Section 5.4
Mathematics 360
due Monday, December 9
Read Elementary Classical Analysis, Section 5.4.
1. Exercises 2–7, pp. 267–268 of ECA.
2. The Bessel function Jis defined by the power series
J(x) =
X
n=0 1
4nx2n
n!2
Prove that Jis twice differentiable and that Jsatisfies the differential equation
d
dx xdJ
dx +xJ = 0
3. Using the differential equation, show that
d
dx J0(x)2+J(x)2=2
xJ0(x)2
Use this to prove carefully that both J0and Jare bounded for x0. What happens
for x0?
4. Using an argument similar to the one used in class to study C(x) (the cosine function),
show that Jhas a positive zero. Hint: first show that xJ0(x) is decreasing if x > 0 and
J(x)>0.
5. Show that Jhas infinitely many positive zeroes. Hint: Suppose we have already found
a zero x1, with J(x1) = 0 and J0(x1)<0. Show that Jis concave up whenever x > 0,
J(x)<0, and J0(x)<0. From this, show that Jhas a local minimum at some point
x > x1. Then use the argument from the previous problem to get the next zero. The
case where J0(x1)>0 is very similar.

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Homework for Section 5.

Mathematics 360

due Monday, December 9

Read Elementary Classical Analysis, Section 5.4.

  1. Exercises 2–7, pp. 267–268 of ECA.
  2. The Bessel function J is defined by the power series

J(x) =

∑^ ∞

n=

)n x^2 n n!^2

Prove that J is twice differentiable and that J satisfies the differential equation

d dx

x

dJ dx

  • xJ = 0
  1. Using the differential equation, show that

d dx

J′(x)^2 + J(x)^2

x

J′(x)^2

Use this to prove carefully that both J′^ and J are bounded for x ≥ 0. What happens for x ≤ 0?

  1. Using an argument similar to the one used in class to study C(x) (the cosine function), show that J has a positive zero. Hint: first show that xJ′(x) is decreasing if x > 0 and J(x) > 0.
  2. Show that J has infinitely many positive zeroes. Hint: Suppose we have already found a zero x 1 , with J(x 1 ) = 0 and J′(x 1 ) < 0. Show that J is concave up whenever x > 0, J(x) < 0, and J′(x) < 0. From this, show that J has a local minimum at some point x > x 1. Then use the argument from the previous problem to get the next zero. The case where J′(x 1 ) > 0 is very similar.