Math 5310 Fall 2010 Homework set #3, solutions Problem 1 Part (a) The Laguerre polynomials are differentiable so we can use the Wronskian. In this case, the Wronskian matrix is upper triangular with diagonal elements (1, −1, 1, −1). (You might ask yourself how you can conclude that without actually constructing the whole Wronskian matrix.) The determinant of a triangular matrix is the product of the diagonal elements (if this is not obvious to you, you should try expanding the determinant of a triangular matrix in minors) so the determinant is 1 for all x. Because 9x such that W(x) 6= 0 the functions are linearly independent. Part (b) 3 n Any cubic polynomial written in the Vandermonde basis, p(x) = ∑ anx can be represented as a column n=0 vector 0 1 a0 B a C B 1 C B . C . @ . A aN By reading off the coefficients of the Laguerre polynnomials it should be easy to see that the Laguerre polyno- mials (through degree three) expressed in the Vandermonde basis have the column vector representations 0 1 1 B 0 C (L ) = B C 0 V @ 0 A 0 0 1 1 B −1 C (L ) = B C 1 V @ 0 A 0 0 1 1 B −2 C (L2)V = B 1 C @ 2 A 0 0 1 1 B −3 C (L3)V = B 3 C . @ 2 A 1 − 6 3 Now, suppose we have a cubic written in the Laguerre basis, p(x) = ∑ bn Ln(x). Expressed as a matrix-vector n=0 muliply, that summation is 0 1 0 1 1 1 1 1 b0 B 0 −1 −2 −3 C B b1 C B 1 3 C B C , @ 0 0 2 2 A @ b2 A 1 b 0 0 0 − 6 3 1 and the result of the multiplication is the vector of coefficients of p in the Vandermonde basis. So pV = TVL pL with 0 1 1 1 1 1 B 0 −1 −2 −3 C TVL = B 1 3 C . @ 0 0 2 2 A 1 0 0 0 6 Part (c) −1 To find the transformation from the Vandermonde basis to the Laguerre basis, compute TVL . I used Mathe- matica’s Inverse[] function to find 0 1 1 2 6 1 − B 0 −1 −4 −18 C T = T 1 = B C . LV VL @ 0 0 2 18 A 0 0 0 −6 Part (d) The polynomial p(x) = x3 + 2x2 + x + 5 is, using the Vandermonde basis, the column vector 0 5 1 B 1 C p = B C . V @ 2 A 1 To convert to the Laguerre basis, multiply by TLV 0 1 1 2 6 1 0 5 1 0 16 1 B 0 −1 −4 −18 C B 1 C B −27 C p = T p = B C B C = B C . L LV V @ 0 0 2 18 A @ 2 A @ 22 A 0 0 0 −6 1 −6 As a check, compute p(x) = 16 L0(x) − 27 L1(x) + 22 L2(x) − 6 L3(x) 22 6 = 16 − 27 (1 − x) + x2 − 4x + 2 − −x3 + 9x2 − 18x + 6 2 6 = (16 − 27 + 22 − 6) + (27 − 44 + 18) x + (11 − 9) x2 + x3 = 5 + x + 2x2 + x3 which is the original p as expressed in the Vandermonde basis. Problem 2 Part (a) The Maclaurin series for ex, sin x, and cos x are ¥ xn x2 x3 x4 x5 ex = ∑ = 1 + x + + + + ··· n=0 n! 2 6 24 120 ¥ 2n+1 3 5 + x x x sin x = (−1)n 1 = x − + − · · · ∑ ( + ) n=0 2n 1 ! 6 120 ¥ x2n x2 x4 cos x = (−1)n = 1 − + − · · · . ∑ ( ) n=0 2n ! 2 24 2 + Observe that i2n = (−1)n and i2n+1 = (−1)n 1 i. When we write eix using the Maclaurin series for the expo- nential, all the even powers of ix will be real while all the odd powers of ix will be imaginary. Plug ix into the series, ¥ inxn eix = ∑ n=0 n! and split the sum into odd and even terms, ¥ i2nx2n ¥ i2n+1x2n+1 eix = + ∑ ( ) ∑ ( + ) n=0 2n ! n=0 2n 1 ! ¥ 2n ¥ 2n+1 x + x eix = (−1)n + i (−1)n 1 ∑ ( ) ∑ ( + ) n=0 2n ! n=0 2n 1 ! = cos x + i sin x which is Euler’s formula. Note: splitting the series into two series with odd and even powers might seem like a trivial step, but it isn’t. Some of you may know that rearranging the terms in a series is allowed only when the series converges absolutely; if it doesn’t, different rearrangements of the same series can converge to different values! Series that converge but don’t converge absolutely must be used very carefully. All series used in this problem do converge absolutely, so this calculation is safe and sound. Interestingly, convergence of series was not understood in Euler’s time, and indeed Euler himself manipulated series in the fast-and-loose style of this solution. As a consequence he occasionally ran into paradoxical results that weren’t understood until decades later. We’ll worry more about convergence issues next semester. Part (b) cos x sin x The Wronskian for fcos x, sin xg is = 1 6= 0 so the functions are LI. − sin x cos x Part (c) − eix e ix The Wronskian for eix, e−ix is = −2i 6= 0 so the functions are LI. ieix −ie−ix Part (d) We’ll need the identities eix + e−ix cos x = 2 eix − e−ix sin x = 2i which are straightforward to derive from Euler’s formula. Equate the trig and exponential representations of f A cos x + B sin x = ae−ix + be−ix and then use the above identities to express the trig functions in the exponential basis, A B eix + e−ix + eix − e−ix = aeix + be−ix 2 2i A B A B + eix + − e−ix = aeix + b−ix. 2 2i 2 2i 3 Because the complex exponentials are linearly independent, we can equate coefficients. Therefore, a 1 −i A = 1 b 2 1 i B so the transformation matrix Tet is 1 1 −i . 2 1 i Part (e) 1 1 T = T−1 = . te et i −i Part (f) If f 2 F1 then we can represent it in any convenient basis, e.g., f (x) = A cos x + B sin x. Then f 0 = −A sin x + B cos x which is a linear combination of the same basis functions, so f 0 2 F1. Part (g) 0 If ft = A cos x + B sin x then ft = −A sin x + B cos x. In vector form, A B f = , f 0 = t B t −A 0 so the matrix Dtt such that ft = Dtt ft is 0 1 D = . tt −1 0 Part (h) ix −ix 0 ix −ix If fe = ae + be then fe = iae − ibe . In vector form, a ia f = , f 0 = e b e −ib 0 so the matrix Dee such that fe = Dee fe is i 0 D = . ee 0 −i Notice that Dee = TetDttTte as expected. 4.