MATH 543: METHODS OF APPLIED MATHEMATICS I

First Midterm Exam: 31 October 2001 Wednesday 18.00-20.00 SAZ:13

QUESTIONS: (25)1. Generating function g(x, t) for the Legendre polynomials P_{n}(x) is given by

\frac{1}{\sqrt{t^{2} − 2xt + 1}} = ∑_{n=0}^{∞} t^{n}P_{n}(x)

(i) Find P_{i}(x) for i = 0 − 3, (ii) Prove that P′_{n+1} − xP′_{n} − (n + 1)P_{n} = 0, (iii) Find ||P_{n}(x)||, (iv) Expand \frac{1}{\sqrt{5−4x}} where x ∈ [−1, 1] interms of the Legendre polynomials. Find the first three terms and discuss the convergence of the series expansion.

Solution: (i) The function g(x, t) is differentiable , using the Taylor theorem we obtain

\frac{1}{n!} \frac{d^{n} g(x, t)}{dx^{n}}|_{t=0} = P_{n}(x)

Hence letting n = 0, 1, 2, 3 we obtain P_{0} = 1, P_{1} = x, P_{2} = \frac{1}{2}(3x^{2} − 1), P_{3} = \frac{1}{2}(5x^{3} − 3x) (ii) Differentiating both sides of the generating function relation above with respect to the variable x we obtain

\frac{1 − xt}{(t^{2} − 2xt + 2)^{(3/2)}} = ∑ t^{n}(P′_{n+1} − xP′_{n})

Now taking the derivative of the same equation wrt the t variable in this case we ger

\frac{1 − xt}{(t^{2} − 2xt + 1)^{(3/2)}} = ∑ t^{n}(n + 1)P_{n}

This completes the proof. (iii) Taking the square of the above equation and integration in [−1, 1] we obtain

∫_{−1}^{1} \frac{dx}{t^{2} − 2xt + 1} = ∑ t^{2n}||P_{n}||^{2}

here we used the orthogonality property of P_{n}. The untegral on the left hand sude can be easily evaluated . Then

\frac{1}{t} ln \frac{1 + t}{1 − t} = ∑ t^{2n}||P_{n}||^{2}

The left hand side can be expanded as

2 ∑ \frac{t^{2n}}{2n + 1}

Hence ||P_{n}|| = \frac{2}{2n+1}. If you don’t remember the expansion of ln \frac{1+t}{1−t} you can use ln(1−t) = − ∫_{0}^{t} \frac{dx}{1−x} = − ∫_{0}^{t} [1+x+x^{2} +x^{3} +⋯]dx = −(x+ \frac{x^{2}}{2} + \frac{x^{3}}{3} +⋯) (iv) Letting t = 1/2 in the above formula we get

\frac{1}{\sqrt{1/4 − x + 1}} = ∑_{n=0}^{∞} \frac{1}{2^{n}} P_{n}(x)

or after some simplifications

\frac{1}{\sqrt{5 − 4x}} = ∑ \frac{1}{2^{n+1}} P_{n}(x)

. This is a convergent series in [−1, 1] due to the Weierstrass theorem. Or by using |P_{n}(x)| < 1 qone can show that the series at the right hand side converges absolutely. Hence it converges uniformly.

(25)2. The Rodriguez formula for the orthogonal polynomials C_{n}(x) is given by

C_{n}(x) = \frac{1}{w} \frac{d^{n}}{dx^{n}} (ws^{n})

where

(1) C_{1}(x) is a first degree polynomial in x, (ii) s(x) is polynomial in x of degree ≤ 2 , with real roots, (iii) w(x) is real , postive and integrable in the interval [a, b] and satisfies the boundary conditions w(a)s(a) = w(b)s(b) = 0. Prove the following Solution For parts (a) and (b) See the lecture notes and DK pages 203-204. For part (c) see the lectures.

(a) Let us denote p_{(≤k)}(x) as any arbitrary polynomial in x of degree ≤ k. Prove te following for all m , n and for all k

\frac{d^{m}}{dx^{m}} (ws^{n} p_{(≤k)}) = ws^{(n−m)}p_{(≤k+m})

(b) C_{n}(x) is a polynomial in x of degrre n which is orthogonal on [a, b] with weight w(x) to any polynomial p_{m}(x) of degree m less then n

(c) Prove that C_{n}(x) with x ∈ I = [a, b] has n zeros in I. (25)3. (a) If (e_{1}, e_{2}, ⋯, e_{n}) is a finite orthonormal family of functions in L_{w}^{2}(a, b) and f ∈ L_{w}^{2}(a, b), then

|||f > − ∑_{k=1}^{n} f^{k} |e_{k} > ||

has its minimum value when f^{k} =< f|e_{k} > and the Bessel’s inequality holds

∑_{k=1}^{n} | < f|e_{k} > |^{2} ≤ ||f||^{2}

b) Prove that the set of orthonormal vectors |e_{i} >, (i = 1, 2, ⋯) form a basis of L_{w}^{2}(a, b) if and only if the Fourier coefficients with respect to |e_{i} > satisfy the Parseval’s relation.

Solution (a) Define the vector

|f_{λ} >= |f > −λ ∑_{k=0}^{n} f^{k} |e_{k} >

where λ is a real varaible (≠ 0). Find the norm of |f_{λ} >

||f_{λ}||^{2} = ||f||^{2} − ∑ | < f|e_{k} > |^{2} + ∑ |f^{k}λ− < f|e_{k} > |^{2}

This norm has minimum when the second term on the right hand side vanishes (why?) Hence for λ = 1 we have

f^{k} =< f|e_{k} >

Since the left hand side is positive the remaining term at the right hand side must be also positive, then

||f||^{2} ≥ ∑_{k=1}^{n} | < f|e_{k} > |^{2}

(b) See the lecture notes and also DK page 195

(25)4. (a) Prove that the sequence of functions f_{n}(x) defined by

f_{n}(x) = \{ \begin{matrix}n & if 0 < x < \frac{1}{n} \\ 0 & otherwise \end{matrix}

defines the generalized function δ(x) (Dirac delta function)

(b) Let f(x) be a good function, prove that δ(x) f(x) = f(0) δ(x),

(c) Let a ≠ 0 then prove that δ(ax) = \frac{1}{|a|}δ(x)

Solution (a) Let g(x) be a good function then

lim_{n→∞} ∫_{∞}^{∞} f_{n}(x) g(x)dx = lim_{n→∞} ∫_{0}^{1/n} n g(x)dx

the right hand side is equal to

= g(0) + lim_{n→∞} ∫_{0}^{1/n} n(g(x) − g(0))dx

the second term on the right hand side vanishes (using the MVT for the good function g(x).

(b) In the first part we showed that

∫_{∞}^{∞} δ(x)g(x)dx = g(0)

Letting g̃(x) = f(x)g(x) where f(x) is a fairly good function and g(x) is good function. Their product is a good function. then using the above property of the δ-function for the function g̃(x) we get

∫ δ(x)f(x)g(x)dx = f(0)g(0) = f(0) ∫ δ(x)g(x)dx

hence we can conclude that f(x)δ(x) = f(0)δ(x) (c)

lim_{n→∞} ∫ h_{n}(ax)g(x)dx = ∫ δ(ax)g(x)dx

assuming a > 0 in the left hand side we get

lim_{n→∞} ∫_{0}^{1/an} ng(x)dx = \frac{1}{a}g(0)

assuming a < 0 in the left hand sude we get

lim_{n→∞} ∫_{1/an}^{0} ng(x)dx = −\frac{1}{a}g(0)

this concludes the proof.