Problem Set #7 Part B Official Solutions Total Points: 20 (+6)

Inna Zakharevich

November 17, 2003

(10) B1. We will first find the limit of (n + 1) ∫_{0}^{1} x^{n}f(x) dx. Since the ratio between the function that we want and this goes to 1 its limit will be the same as the limit of our function. Let M = sup_{[0,1]} |f|. Fix ϵ > 0. Since f is uniformly continuous we know that there exists δ > 0

such that if |x − x′| < δ then |f(x) − f(x′)| < ϵ. Consider the two integrals ∫_{0}^{1−δ} x^{n}f(x) dx and ∫_{1−δ}^{1} x^{n}f(x) dx. Then

(n + 1) |∫_{0}^{1−δ} x^{n}f(x) dx| ≤ (n + 1) ∫_{0}^{1−δ} |x^{n}||f(x)| dx ≤ (n + 1) ∫_{0}^{1−δ} x^{n}M dx = M(1 − δ)^{n+1}.

On the other hand,

(n + 1) ∫_{1−δ}^{1} x^{n}f(x) dx ≥ (n + 1) ∫_{1−δ}^{1} x^{n}(f(1) − ϵ) dx = (f(1) − ϵ) − (1 − δ)^{n+1}(f(1) − ϵ).

As n → ∞ the first of these integrals goes to 0, and the second integral goes to f(1) − ϵ. Thus we see that the limit of the total integral as n → ∞ is at least f(1). On the other hand,

(n + 1) ∫_{1−δ}^{1} x^{n}f(x) dx ≤ (n + 1) ∫_{1−δ}^{1} x^{n}(f(x) + ϵ) = (f(1) + ϵ) − (1 − δ)^{n+1}(f(1) + ϵ).

Thus the second integral is also less than or equal to f(1). So the limit of the integral is going to be f(1). Now we want to write I_{n} as f(1) + g(n) where g(n) = o(1). Note

∫_{0}^{1} nx^{n}f(x) = \frac{n}{n + 1} (f(x)x^{n+1}|_{0}^{1} − ∫_{0}^{1} x^{n+1}f(x) dx) = \frac{n}{n + 1} (f(1) − ∫_{0}^{1} x^{n+1}f′(x) dx) .

Then

n (n ∫_{0}^{1} x^{n}f(x) dx − f(1)) = \frac{n^{2}}{n + 1}f(1) − \frac{n^{2}}{n + 1} ∫_{0}^{1} x^{n+1}f′(x) dx − nf(1).

Let C(n) = n ∫_{0}^{1} x^{n+1}f′(x) dx. We know that lim_{n→∞} C(n) = f′(1). Then the above simplifies to

−\frac{n}{n + 1}(C(n) + f(1))

which goes to −f(1) − f′(1) as n → ∞. Thus we see that lim_{n→∞} n(I_{n} − f(1)) exists, so

I_{n} ∼ f(1) − \frac{1}{n}(f(1) + f′(1)) + o(1/n).

(10 (+6)) B2.

(5) (1) Note that the integrand is a C^{∞} function of z which is integrated on a compact set. Thus, by a

simple induction on m, we know that

\frac{d^{m}}{dz^{m}} J_{n}(z) = \frac{1}{π} \frac{d^{m}}{dz^{m}} ∫_{0}^{π} cos(nθ − z sin θ) dθ = \frac{1}{π} ∫_{0}^{π} \frac{∂^{m}}{∂z^{m}} cos(nθ − z sin θ) dθ

which will exist and be continuous for all m.

(5) (2) Note that

J_{n−1}(z) − J_{n+1}(z) = \frac{1}{π} ∫_{0}^{π} cos((n − 1)θ − z sin θ) dθ − \frac{1}{π} ∫_{0}^{π} cos((n + 1)θ − z sin θ) dθ
= \frac{1}{π} ∫_{0}^{π}(cos((n − 1)θ − z sin θ) − cos((n + 1)θ − z sin θ)) dθ
= \frac{1}{π} ∫_{0}^{π} 2 sin(nθ − z sin θ) sin θ dθ.

Also

J′_{n}(z) = \frac{1}{π} ∫_{0}^{π} \frac{∂}{∂z} cos(nθ − z sin θ) dθ = \frac{1}{π} ∫_{0}^{π} sin(nθ − z sin θ) sin θ dθ.

Thus we see that

J_{n−1}(z) − J_{n+1}(z) = 2J′_{n}(z).

Note that for n = 0 the first formula doesn’t work. However, note that

J_{1}(z) = \frac{1}{π} ∫_{0}^{π} cos(θ − z sin θ) dθ
= \frac{1}{π} ∫_{0}^{π} cos θ cos(z sin θ) + \frac{1}{π} ∫_{0}^{π} sin θ + sin(z sin θ) dθ
= \frac{1}{π} ∫_{−π/2}^{π/2} cos(θ − π/2) cos(z sin(θ − π/2)) dθ − J′_{0}(z)
= −\frac{1}{π} ∫_{−π/2}^{π/2} sin θ cos(z cos θ) − J′_{0}(z)
= −J′_{0}(z)

because the first integral is a symmetric integral of an odd function.