1 Polynomial approximation of the gamma function

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1 AM05: Solutions to take-home midterm exam 1 1 Polynomial approximation of the gamma function Part (a) One way to answer this question is by constructing the Vandermonde linear system Solving this matrix system yields g g g =. (1) g g 4 (g 0, g 1, g, g, g 4 ) = ( , 16.58, ,.14667, ) () from which an approximation to the gamma function is given by g(x) = 4 k=0 g kx k. Part (b) The coefficients of the polynomial p(x) = 4 k=0 p kx k are given by Solving this system yields p p p = log. () p log p 4 log 4 (p 0, p 1, p, p, p 4 ) = (1.1507, , 0.880, , ) (4) from which the second gamma function approximation is given by h(x) = exp(p(x)). Parts (c) and (d) The program MidtermSol1.py constructs the two gamma function approximations. Figure 1 shows g(x) and h(x) in comparison to Γ(x). The program also calculates the maximum absolute relative error of the two approximations. To ensure that the desired level of accuracy is achieved, the program first scans over a coarse grid G α of equally spaced points in x, and computes the maximum relative error M α. The program then considers a finer grid G α+1 and computes the corresponding M α+1. This process is continued until the ratio M α+1 M α /M α drops below a threshold of ɛ = 10 4, whereupon it is assumed that M α+1 is a good approximation to the global 1 Solutions written by Henry Wilkin and Chris Rycroft. Thanks to Yuexia Luna Lin for help with the solution to question. 1

2 Figure 1: The gamma function Γ(x) and the two interpolating functions g(x) and h(x) considered in question 1. maximum. The maximum relative error of g(x) (defined as max x [1,5] Γ(x) g(x) /Γ(x)) was found to be , and the maximum relative error of h(x) was found to be The approximation h(x) is better, likely because the logarithm of a rapidly growing positive function typically has smaller n-th order derivatives, so the error bounds associated with the formula of Cauchy for polynomial interpolation considered in lecture are more restrictive. Least-squares spline fitting Parts (a) and (b) The program MidtermSol.py computes the coefficents of the least-squares spline fit. In Python the s k basis functions can be defined using scipy.interpolate.splmake, with the kind argument set to natural in order to match the specified boundary conditions. The spline can also be constructed manually, following the same procedures described in the spline notes on the website. The best fit coefficients for the points in the sdata1.txt data file are b = (0.491, , 0.859, 0.5, 0.596, 0.50, 1.661,.974, 1.46) (5) and the best fit coefficients for the points in the sdata.txt data file are b = (0.47, 0.174, 0.795,.50, 8.711, 1.77, 1.881,.850, 1.576). (6) Plots of the two best-fit splines are shown in Figs. 4 & 5.

3 Figure : A plot of the absolute relative error between g(x) and Γ(x) over the range [1, 5]. Figure : A plot of the absolute relative error between h(x) and Γ(x) over the range [1, 5].

4 1 spline curve sdata Figure 4: The data points from question (a), and the least-squares fit of the spline function. 0 spline curve sdata Figure 5: The data points from question (b), and the least-squares fit of the spline function. 4

5 Part (c) Of the two splines, the second one looks unreasonable since there is a large negative bump in the curve around x = 4 that does not appear consistent with the provided data points. This occurs because the spline fitting procedure does not work well when there is a region without data points. Consider the spline basis function s 4 (x): it is largest near x = 4. For all of the given data points, which lie in the ranges x < and x > 5, this function is very small. When performing the fitting procedure, the coefficient b 4 will be set based on the contribution of these very small values to the square residual, potentially creating an unreasonably large coefficient on s 4. In other words, the coefficient b 4 is poorly constrained by the data. Note that this problem would not occur for least-squares fitting based on polynomials, since the monomial basis functions all vary significantly at the given data points, and thus all of the fitted coefficients will be well-constrained. However spline-based fitting could be useful in some situations, provided the data is evenly distributed. Stability of a scheme for the heat equation Part (a) The Taylor expansion of f to second order is f (x + h) = f (x) + h f (x) + h f (x) + O(h ). (7) Note also that if f (z) has a convergent power series expansion near z = x, then f (x + h) + f (x h) = f (x) + f (x)h + O(h 4 ). This is because the left hand side is even in h, so only even powers of h appear in the power series expansion. To fourth order, the expression a f (x) + b( f (x + h) + f (x h)) + c( f (x + h) + f (x h)) is given by which simplifies to (a + b) f (x) + bh f (x) + c f (x) + 4ch f (x), (8) (a + b + c) f (x) + (b + 4c)h f (x). (9) For this to equal h f (x) up to error terms of order h 4, it is necessary that a + b + c = 0 and b + 4c = 1. Together, these imply that b = c a and c = 1 + a. Hence c(a) = 1 + a 6 (10) and therefore b(a) = 1 a. (11) Part (b) Substituting in the expressions from part (a) shows that discretization is given by U n+1 j = U n j + t h ( auj n 1 + a (U n j+1 + Un j 1 ) + + a 6 (Un j+ + Un j ) ). (1) 5

6 Setting Uj n = [λ(k)] n e ijkh and using the result e i(j+δ)kh + e i(j δ)kh = e ijkh cos(δkh)) yields the expression λ(k) = 1 + t ( h a + 4a cos(kh) + + a ) cos(kh) (1) for the amplification factor λ(k). Since cos θ = 1 sin θ, cos(kh) = 1 sin (kh) In additon cos(kh) = 1 s and therefore = 1 (1 + cos(kh))(1 cos(kh)) ( = 1 sin kh ) sin kh = 1 8s (1 s ). (14) ( λ(k) = 1 + µ a + 4a (1 s(k)) + + a [ ] ) 1 8s(k)(1 s(k)) ) = 1 + 4µ( s(k) (a + ) s(k). (15) Part (c) Setting a equal to gives b = = 1 and c = 1 6 = 0, so the method is equivalent to ordinary forward Euler integration with the standard second-order centered-difference stencil as discussed in the lectures. The eigenvalue λ(k) in this case is given by 1 4µs(k). Since s = sin hk 0, and s 1, λ(k) is stable as long as 1 4µ < 1. This is achieved at 1 4µ = 1, where µ = 1. Hence stability requires µ 1. Part (d) The method is unstable if there is any value of k for which λ(k) > 1. Using the definition s = sin hk, searching over arbitrary k is equivalent to searching over the range s [0, 1]. From Eq. 15, the case of λ > 1 corresponds to ( ) (a + )s 4µs 1 > 0. (16) which simplifies to and hence (a + )s 1 > 0, (17) (a + )s > (18) If a > a where a = 1, then Eq. 18 will be true for s = 1 regardless of the choice of µ, and thus the method will be unconditionally unstable. If a 1, then it follows that (a + )s 1 0 (19) 6

7 for all s [0, 1]. Hence for sufficiently small µ, it follows that ( ) (a + )s µs 1 1 (0) and thus λ(k) 1 for all k and the method is stable. Thus the method is unconditionally unstable if and only if a > a. Part (e) Equation 15 can be rewritten as λ(s) = 8(a + )µ s 4µs + 1, (1) where λ is now viewed as a function of s. This is a parabola that passes through the point (s, λ) = (0, 1). The parabola has a stationary point at s = s e, which is given by 0 = λ (s e ) = 16(a + )µs e 4µ () where the derivative is taken with respect to s, and hence s e = /4(a + ). The stability properties are given by considering the parabola over the range s [0, 1] and there are several cases: 1. If a < then λ < 0 and the parabola curves downward. In addition the parabola has a maximum at s e < 0. Hence, λ is monotonically decreasing over s [0, 1] and λ(1) = 8(a + )µ 4µ + 1 = 4(a + 1)µ + 1 () For stability, it is necessary for λ(1) 1 since this will ensure that all values in the range of s [0, 1] satisfy λ(s) 1. This will occur when µ (1 + a). (4). If < a 5 4 then λ > 0 and the parabola curves upward. It has a minimum at s e 1, and is therefore monotonically decreasing over s [0, 1]. Therefore for stability it is necessary to have λ(1) 1, which corresponds to as in the previous case. µ (1 + a). If 5 4 < a 1, then the parabola curves upward but has a minimum value at s e [ 1, 1). Since the parabola is symmetric about its extremum point in [ 1, 1) and λ(0) = 1, it follows that λ(1) < 1. Thus to ensure stability it suffices to check that λ(s e ) 1, which corresponds to 8(a + )µ 9 1 λ(s e ) = 16(a + ) µ a = 1 µ (a + ). (6) This will be true if µ 7 (5) 4(a + ). (7)

8 In summary, the range of permissible µ for stability is { 4(a+) if 5 4 µ(a) < a 1, if 5 (1+a) 4 a. (8) Part (f) This maximum step size of µ = is achieved for a = 1 = a. For this value b(a) = 0 and c(a) = 1 4, and thus the scheme is given by U n+1 j = U n j + t h ( 1 Un j (Un j+ + Un j ) ). (9) Note that in this case, the value at gridpoint j only depends on the values at gridpoints j, j, and j +. If the scheme is run on the initial data Uj 0 = ( 1) j, which alternates between 1 and 1 at successive gridpoints, then the solution will remain invariant as Uj n = ( 1) j for all timesteps. All second derivative calculations will evaluate to zero, since they either sample all 1 values, or all 1 values. This is formally stable, and is consistent with the calculations in part (e). However, it would not be good in practice physcially, the heat equation should damp out local variations, and thus it is undesirable to have the..., 1, 1, 1, 1,... solution remain invariant with time. Note that this is true regardless of the choice of µ; decreasing the timestep will not help. Solving an elliptic problem on an irregular domain Part (a) Since f appears eight times with weight β and once with weight α, matching with f to maximal order requires α = 8β. Meanwhile, since f (x + h) = f (x) + h f (x) + h f (x) +..., f (x + h) + f (x h) + f (x + h) + f (x h) = 4 f (x) + h f (x) + (h) f (x) = 4 f (x) + 5h f (x) (0) and the sum multiplying β is 8v(i, j) + 5h xv(i, j) + 5h yv(i, j). Hence, β = 1, and α = 8. 5h 5h Alternatively, this result can also be derived from your expressions for b(a), c(a) from problem (a) by solving for the value of a at which b = c. This requires 1 a = 1 + a 6 and hence 5a/6 = /, so a = 4/5. The stencil in this question can be made by superposing the question (a) stencil in the x and y directions. Hence (1) α = a = 8 5 () and β = b = c = 1 5. () 8

9 Part (b) If (j, k) is an interior site, let C = {s n} N C n=1 be the set of nearest neighbor sites of (j, k) in Ω\ Ω, and let C = {s l }N C l=1 be the set of next nearest neighbor stencil sites that lie in the interior. If there are N e stencil sites in the exterior (i.e. not in Ω or Ω), then the discrete approximation at (i, j) can be written as 8 + N e 5h V j,k + 1 ( ) 5h V s + V s. (4) s C s C For example, when (j + 1, k) is a boundary site and (j +, k) is exterior, N e = 1 and the explicit discrete Laplacian at (j, k) is Parts (c) and (d) 9 5h V j,k + 1 5h ( V j,k+1 + V j,k 1 + V j 1,k + V j,k + V j,k+ + V j,k ). (5) The program MidtermSol.py solves the elliptic problem using the given stencils. To define a map from lattice sites to vector indices (and vice versa), the program first sweeps over rows and columns of the discrete approximation to Ω and assigns a vector index to the field value on each lattice site in the order visited. Thus, (v 0, v 1, v,...) = (V (0,18), V (0,19), V (0,0),..., V (1,18), V (1,19),...]). (6) In order to build a representation of the discrete Laplacian with this labeling scheme, it helps to introduce some auxiliary structures. First, a len(v) array vi ls that gives the -dimensional lattice site of each linear vector index, and a 5 5 dimensional array ls vi that maps interior lattice sites to their corresponding vector indices. Also, a discrete approximation ω M (5,5) for Ω and its boundary is constructed according to: 1 if (hi, hj) is on the interior of Ω, ω ij = 0 if (hi, hj) is on the boundary Ω, 1 if (hi, hj) is in the complement of Ω. These structures make it easier to construct adjacency networks for the nodes v (with respect to the stencil), and to count the number of stencil sites at each node that lie in the complement of Ω. The Laplacian can then be defined in a straightforward way from these. The Cholesky factorization can be found either using numpy.linalg.cholesky in Python or chol in MATLAB. The algorithm described in lecture seven can also be used. Plots of the solution are shown in Fig. 6. The maximum occurs at site [11, 8], where the field value is to seven significant figures. (7) 9

10 Figure 6: Three-dimensional plot (left) and contour plot (right) of the two-dimensional solution to the elliptic problem in question 4, solved via the Cholesky factorization. 10