The answer in each case is the error in evaluating the taylor series for ln(1 x) for x = which is 6.9.

Transcription

1 Brad Nelson Math 26 Homework #2 /23/2. a MATLAB outputs: >> a=(+3.4e-6)-.e-6;a- ans = 4.449e-6 >> a=+(3.4e-6-.e-6);a- ans = 2.224e-6 And the exact answer for both operations is 2.3e-6. The reason why way I is so inaccurate is that the number +3.4e-6 is stored on the computer as ( )( + ɛ mach ) which corresponds to an error of magnitude 6 in double precision. Then the floating point operation ( ). 6 is accurate to (( ). 6 )( + ɛ mach ), which also introduces an error of magnitude 6. The final step, a, doesn t introduce significant error into this calculation, since the resulting number is on magnitude of 6 so the error is of the magnitude 32. The two operations which introduce error of magnitude 6 result in an inaccurate answer since the real answer is of the same magnitude. Way II is more accurate since ( = ( + ɛ mach ) only has an error 32, which is not significant for this calculation. Then = ( )( + ɛ mach ) introduces an error of magnitude 6 which is significant since the final answer is of this magnitude. Once again, the final operation, a does not introduce significant error. While the returned answer is inaccurate because the introduced error was of the same magnitude as the answer, there was only one such step that introduced such a large error, which compared to two such steps in way I, is more accurate. b MATLAB outputs: >> x =.999; a = ; for j=:6, a = a-x^j/j; end; a-log(-x) ans = 3.553e-3 >> x =.999; a = ; for j=6:-:, a = a-x^j/j; end; a-log(-x) ans = e-6 The answer in each case is the error in evaluating the taylor series for ln( x) for x =.999 which is For way I, the series is evaluated as a = j For way II, a = j 6 j. 6 j j= j= Looking at way I, the first term evaluated is.999 and the second term is.5. Adding these two terms together introduces an error of.5ɛ mach which is of the magnitude 6. Since the partial sum monotonically approaches , it is always of magnitude. This means that adding each term to the series introduces an error of 6, which compounded over 6 terms turns into a total error of 2. The actual observed error

2 is close to this, at 3, and may be smaller than the prediction due to errors in opposite directions from the true answer canceling each other out. As for way II, the first term evaluated is of order 3. Performing operations on numbers of this magnitude will introduce a relative error of 6, which is an absolute error of 47. Since these errors are so small, the introduced error doesn t reach a magnitude of 6 until towards the end of the calculation, so the compounded error is still 6. From these observations, we can conclude that when summing together a list of numbers using floating-point arithmetic, it is best to start with the smallest numbers first. This way the total error of the summation stays small relative to the size of the answer until the last terms are added, which prevents a large relative error from compounding over many iterations. 2. a f (x) = 2x computed via x x. On a machine, this algorithm is f (x) = f l(x) f l(x). To prove backwards stability, we want to show that x X, f x x (x) = f ( x) for some x : = O(ɛ x mach ). To see if this is possible, we set f (x) = (x( + ɛ ) + x( + ɛ ))( + ɛ 2 ) = x( + ɛ 3 ) + x( + ɛ 3 ) where ɛ,2 ɛ mach. It follows that ɛ 3 = ɛ + ɛ 2 + ɛ ɛ 2. Thus, ɛ 3 2ɛ mach + ɛ 2 mach, and ɛ 2 = O(ɛ mach ). Thus, the algorithm is backwards stable. b f (x) = + x computed via x. Set: f (x) = ( + x( + ɛ ))( + ɛ 2 ) = + x( + ɛ 3 ) = f ( x) where ɛ,2 ɛ mach. Ignoring high-order terms (because of their small size), we have ɛ 3 = ɛ + ɛ 2 + ɛ 2 x Since for choices of x, ɛ 3 will blow up, the algorithm is not backwards stable. However to test stability, consider f (x) and f ( x). If we force ɛ 3 < ɛ mach, then f (x) f ( x) f (x) = ɛ 2 + x(ɛ + ɛ 2 ɛ 3 ɛ ɛ 2 ) + x( + ɛ 3 ) We see that for x, the algorithm is backwards stable, so focusing on x <, if x is still close to, the numerator is O(ɛ mach ), while the denominator 2, thus f (x) f ( x) f = (x) O(ɛ mach ). For x, the numerator is ɛ 2 and the denominator is, so f (x) f ( x) f = (x) O(ɛ mach ). Thus, while the algorithm is not backwards stable, it is stable. 3. Computing the eigenvalues of a symmetric matrix via the characteristic polynomial: i Restricting to 2x2 diagonal matrices with diagonal entries a and b, the characteristic polynomial is λ 2 (a + b)λ + ab. If we choose A = I, then we have λ 2 2λ +. Since the coefficients of this polynomial are stored with error ɛ mach, the polynomial can be written ( + ɛ a )λ 2 2( + ɛ b )λ + ( + ɛ c ) where ɛ a,b,c ɛ mach. To find the roots of this polynomial using the quadratic formula, we have, λ = 2( + ɛ b) ± 4( + ɛ b ) 2 4( + ɛ a )( + ɛ c ) 2( + ɛ a ) By expanding the inside of the square root, we see that the largest order error terms will be O( ɛ mach ). 2

3 ii While the above was done for A = I, it is easy to see that for any quadratic polynomial ax 2 + bx + c, perturbation of O(ɛ mach ) in the coefficients will lead to a perturbation of O( ɛ mach ) in the roots of a quadratic polynomial by placing the coefficients in the quadratic formula. Thus if ɛ mach = 6, then there will be an error of O( 8 ) in the roots of the polynomial. In general, for a degree n polynomial, we would expect to see O( n ɛ mach ) if a similar method is used. Since n ɛ goes to slower than ɛ, f (x) f ( x) O(ɛ mach ) for any x x f (x) x : = O(ɛ x mach ), and the algorithm is not stable. Since stability is a weaker condition than backwards stable, it is also not backwards stable. Thus, the algorithm is unstable. 4. Code: lagrange.m, lagrange eq.m. The following plot was produced: 5 lagrange basis functions for 2 equally spaced points on [,] Note how for each node, all polynomials except for one go to, and the one that does not goes to. 5. Code: interp error.m, hw2fns.m The two functions in the following questions look like: 3

4 f (x) = sin(2e x ) f (x) = ( + 25x 2 ) The following two plots used 25 equally spaced points on [, ] to create an interpolated polynomial for each function. The interpolation error E n (x) := f (x) (L n f )(x) is plotted. 4

5 .5 x 9 interpolation error for sin(2e x ) (a) Note that the error is at most 9. f (x) = sin(2e x ) 3 interpolation error for (+25*x 2 ) (b) f (x) = ( + 25x 2 ) Note that the error blows up to 25 at edges. 5

6 interpolation error as function of n points Sup norm of the interpolation error for f (x) = sin(2e x ) as a function of n equally spaced points on [, ]. Note how line appears to be roughly linear between n= and 27. Since this is on a semilog plot, this indicates exponential convergence in this region. To explain the reverse trend when n > 27, we note that sup x [,],k {...n} l k (x) grows exponentially: 9 sup of n lagrange basis functions on [,] Sup norm over all n Lagrange basis functions on [, ] as a function of n. Since the supremum is growing exponentially, the supremum in the error of the sum over the 6

7 these functions is going to grow exponentially as ɛ mach sup x [,],k {...n} l k (x), following the rules of floating point arithmetic. Note that at n 27, the supremum is 6, so the supremum of the error will be if ɛ mach = 6. This is where the graph of the sup norm of the interpolation error above suddenly reverses course, and begins to be less accurate, as the increase in error due to the sup norm of the basis functions overtakes the decrease in error due to adding more terms to the interpolant. Code: lbasis sup.m 6. Code: interp error cheby.m, lagrange cheby.m After updating the codes to use Chebychev nodes instead of equally spaced nodes, the following plots were produced:.2 lagrange basis functions for 2 Chebychev points on [,] Lagrange basis functions for 2 Chebychev nodes on [, ]. By inspecting the graph, it is clear that for each l k (x), sup x [,] l k (x). 7

8 5 x 2 interpolation error for sin(2e x ) interpolation error for f (x) = sin(2e x ) plotted for 25 Chebychev points on [, ]. Maximum error is 2, which is better than for equally-spaced points. x 3 interpolation error for (+25*x 2 ) interpolation error for f (x) = ( + 25x 2 ) plotted for 25 Chebychev points on [, ]. The maximum error for this function is now 9 3, which is a drastic improvement over the scenario with equally-spaced nodes. 8

9 interpolation error as function of n points Sup norm of the interpolation error for f (x) = sin(2e x ) as a function of n Chebychev points on [, ]. interpolation error as function of n points Sup norm of the interpolation error for f (x) = ( + 25x 2 ) as a function of n Chebychev points on [, ]. Note that the interpolation error is converging exponentially to. Thus, the Runge phenomenon does not persist. Comparing the graphs for the sup norm of the interpolation error of f (x) = sin(2e x ) for Chebychev points and equally spaced points, it appears that the Chebychev points allow a better achievable interpolation error. For equally spaced points, the best possible error is 9

10 about while Chebychev points have a best possible error of 5. We can t reasonably expect the error to get much better, as this is close to ɛ mach, since the maximum for this function is.