MAT 128A - Practice Midterm Exam

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1 MAT 8A - Practice Midterm Exam Karry Wong October 3, 08 Problem (True or False) Given that f : r, s Ñ R is a continuous function, and that ta n u are its Chebyshev coefficients. Also, for N P N, p N pxq N a n T n pxq. statements is true or false. No need to justify your answers. Determine whether each of the following (I) If f is continuously differentiable, then }p N f} 8 Ñ 0 as N Ñ 8. (II) if f is k-times continuously differentiable, then a n Op n k q. (III) If f is infinitely differentiable, then there exists an r 0 such that a n expp rnq. (IV) If a n n for all nonnegative integers n, then fpxq p N pxq N positive integers N and all x P r, s. for all (V) For each positive integer N, p N is the unique polynomial of degree N which interpolates f at the points j cos N π, j 0,,,, N Ans: (I) TRUE (Not graded - My reason): for continuously differentiable f : r, s Ñ C, its Chebyshev series converges absolutely and uniformly on the interval r, s. In particular, uniform convergence means exactly lim NÑ8 }f p N} 8 0 (II) TRUE

2 (Not graded - My reason): In class, we learned that given f being k-times continuously differentiable, its Chebyshev coefficients a n Op n k q, which in turn implies that, a n Op n k q Remark. In general, given two real-valued functions f, g, it is clear that f Opgq ñ f Opgq fpxq since lim xñ8 gpxq 0 8 (III) FALSE (Not graded - My reason): exponential decay of Chebyshev coefficients is a consequence when f is analytic over an ellipse. Given that f is infinitely differentiable (C 8 ), it does not imply that f is analytic. Indeed the class of analytic functions is a proper subset of C 8 -function. Here is an example of a real-valued function that is C 8 but not analytic $ & expp fpxq q, x x 0 % 0, x 0 We can show that f is infinitely differentiable but its Taylor s expansion at x 0 converges to zero and hence fails to converge to fpxq for x 0. See for more details. Remark. For those of you who have taken complex analysis, the picture is fundamentally different for complex-valued functions. function is analytic and vice versa! (IV) TRUE (Not graded - My reason): first, note that T n pxq T n pxq cospn cos pxqq is basically a cosine function. Second, fpxq p n pxq kn Replace n with N, we are done. (V) TRUE a k T k pxq Indeed, any holomorphic (= complex differentiable) kn a k T k pxq for all n and all x P r, s since kn a k kn n n (Not graded - My reason): This is a theorem shown in lecture 5. In addition, the Chebyshev coefficient is computed by the quadrature rule (which is exact in this case) a n N N j0 fpx jqt n px j q, where j x j cos N π, j 0,,,, N

3 Problem (Fourier Series) Compute fptq dt where f : r, πs Ñ C is the function defined by the Fourier series fptq n e int Ans: As far as I can tell, there is only one way to get to the right answer. (if you have a better solution, please let me know!) It is tempting for some of you to interchange the integral and the infinite sum. But it is wrong! (Correct). Since the given form of fptq is already a Fourier series, $ & 0, n 0 fptq e it 3 e it a n e int where a n % n, n 0 n 8 The integral ³ π fptq dt we want to compute is part of the definition of a 0, therefore a 0 : π fptq dt ñ fptq dt πa 0 πpq 4π (Incorrect). It is tempting to write down the following answer: fptq dt n e int dt p q n e int dt e 0 dt n e int dt 4π n 0 But the step p q is wrong! I hope that you learned in your analysis class that it is NOT always possible to interchange the integral and the infinite sum. Here is a counterexample: Let tf n pxq cospnxqu. Then consider N cospnxq, sinp n q sinp x q sinp 3x q sinpx q Since N cospnxqsinp x q 8 sinp x sinp x q q sinppn xq xq sinp n xq sinp pn q pn q xq sinp xq 3

4 So we have Ņ cospnxq sinp x q pn q sinp xq sinp x q Now for 8 cospnxq is divergent (try to let N Ñ 8 in the above expression.) Therefore, ³ π N cospnxq is undefined! But at the same time, for all integers n, ³ π cospnxq 0 and ³ π Therefore, N ³ π cospnxq π. So we see in this example that generally» X» f n pxq f n pxq X cosp0xq π. Remark 3. Some of you might recall that Fubini s theorem justifies the interchange of integrals for an integrable function. As a special case, Fubini s theorem also justifies the interchange of the integral and the infinite sum since infinite sum can be identified as the integration with respect to the counting measure: Theorem. For a sequence of measurable functions tf n u 8 with f n : X Ñ R and µ the Lebesgue measure on R, if»» either f n dµ 8 or f n dµ 8, X X Then we can interchange the integral and infinite sum over tf n u 8, i.e.»» f n dµ f n dµ X X But this theorem is NOT applicable in the above example, since f n pxq n e int,» f n dµ X 8 n e int 8 4

5 Problem 3 (Condition Number) Let κ f pxq denote the condition number of evaluation of the function f at the point x. Find a function f which is infinitely differentiable on the interval p0, q (but which may have singularities at x 0) such that lim xñ0 κ f pxq 8 Note: Since κ f 0 by definition, therefore 8 here is clearly 8 Ans: fpxq expp q, then x κ f pxq xf fpxq x expp x x q expp x q x Therefore, lim xñ0 κ f pxq lim xñ0 x 8. (Idea - Not graded:) First recall the formula for the condition number of a function fpxq: κ f pxq xf fpxq If we look at this formula carefully, any fpxq as some power of x will NOT give lim xñ0 κ f 8, since assume fpxq x α for some α P R: κ f xαxα α α 8 This simply wiil not work. How about we draw inspiration from example fpxq e x presented in class: κ f xex x e x That is actually insightful since we learn that exponential growth will lead to blow-up of the condition number, i.e. lim xñ 8 κ f pxq 8 Can we produce exponential growth around x 0? expp x q. Yes! Simplest answer: f pxq By the same token, we can produce infinitely many functions as such, i.e. expp x q, expp x 3 q,... 5

6 Problem 4 (Error of interpolant) Suppose that f : r0, s Ñ R is twice differentiable function such that f pxq for all 0 x. Let p be the unique polynomial of degree which interpolates f at the points 0 and. Show that ppxq fpxq 8 for all x P r0, sq. Ans: we apply the interpolation error formula in this case, i.e. x 0 0, x where ξ x P p0, q. fpxq ppxq f pξ x q! px x 0 qpx x q f pξ x q xpx q! In order to find an upper bound for fpxq ppxq, we look at the function gpxq : xpx q. Using calculus, g pxq x 0 ñ x g pxq ñ gp q 0 So gpxq has an absolute maxium at x and gp q. 4 Using f pxq, we have fpxq ppxq 4 8 for all x P r0, s. 6

7 Problem 5 (Chebyshev coefficients) Suppose that f : r, s Ñ R is an infinitely differentiable function, and that ta n u is the sequence of Chebyshev coefficients of f that is, ta n u defined via the formula a n π» fpxqt n pxq? x Suppose also that N is a positive integer, and p is the polynomial, p N a n T n pxq Show that if q is any polynomial of degree N, then a N 4 fpxq qpxq π for all x P r, s. Hint: You might want to use the fact that ³ T n pxq? x for all n without proving it. Ans: The desired statement as above is NOT correct. Choosing fpxq T 0 pxq T pxq x which is infinitely differentiable. Now choose qpxq to be a polynomial of degree 0, namely a constant, let say qpxq. Then L.H.S a but R.H.S. 4 π p x q 4 π x. Taking x 0, we have L.H.S The desired statement should be where the N R.H.S.. a N 4 π sup fpxq qpxq x -coefficients provides a lower bound for the distance between f and any polynomials of degree N. The proof s idea is as follows: π a» N» fpxqt N pxq? x rfpxq qpxq qpxqs T N pxq? x»» pfpxq qpxqqt N pxq? x qpxqt N pxq? x» fpxq qpxq T N pxq? x xqpxq, T N pxqy wpxq The reason for ³ qpxqt N pxq? x 0 is that since qpxq is a polynomial of degree N, qpxq can be expressed as a sum using T 0 pxq, T pxq, T pxq,, T N pxq, i.e. qpxq Ņ a n T n pxq a 0T 0 pxq a T pxq a N T N pxq 7 0

8 And T N pxq is orthogonal to all T 0 pxq, T pxq, T N pxq with respect to the weighted inner product, i.e. Therefore,»»» qpxqt N pxq? x T n pxqt m pxq? 0, n m x Ņ Ņ a n T n pxqt N pxq? x Now going on with the upper bound for a N, we have π a N sup fpxq qpxq x sup fpxq qpxq x»» T N pxq? x ñ a N 4 π sup fpxq qpxq x a n T n pxq? 0 x 8

MATH 360 Final Exam Thursday, December 14, a n 2. a n 1 1

MATH 360 Final Exam Thursday, December 14, a n 2. a n 1 1 MATH 36 Final Exam Thursday, December 4, 27 Name. The sequence ta n u is defined by a and a n (a) Prove that lim a n exists by showing that ta n u is bounded and monotonic and invoking an appropriate result.