Question

3. Consider the function f(x) = -0.1.24 – 0.15x3 – 0.522 – 0.25x + 1.2. (a) Obtain the analytical expression (i.e. True or Exact Solution) for the first derivative, Eval- uate its value at 1 =0.5. Box your answer and label it as fexact- (b) Now assume the function is discretized on a grid with uniform spacing of h. Evaluate your finite difference approximation at x = 0.5 using central differencing with step sizes starting at 1 and re- duced by a factor of 10, up to 10-10, i.e. (Use h = 1,0.1,0.01, 0.001, 0.0001,0.00001, ...., 10-10). Now compute the error e = || fum- fexact || for for each grid spacing. Here fexact is the value obtained from the continuous expression and || || is the absolute value. Create a table with columns for h, f(Pi), f(li+1), f(li-1), fCentered: fExact, and e for different values of h. Note that you could do these calculations on a calculator and only use Matlab only for plotting the values in table. However, if you want to explore Matlab, you may write a program in Matlab as well). Irrespective of how you do it, show all steps i.e. manual calculations) for one particular step size; say h = 0.01. 2 (c) In MATLAB, plot e versus the grid spacing (i.e. e on vertical axis and h on the horizontal axis) on a log – log plot. (A log – log plot does not mean taking the logarithm of these values, but the scales are logarithmic.). i. What happens to the error as step size decreases? ii. Does the error always decrease with decreasing h? If not, why do you think it doesn't decrease? iii. Find the order of accuracy of the formula using the slope of the curve in the region of decreasing error with step size.

Answer 1

Solution The given function is, Muse -0.251 +12 f(x) = -0 1x" _ 0.15x² -0.5x² - a) Towe/Exact Solution of df dx df doc -0.4x² - 0 45 x - x -0.25 - -0.05 - 0.1125 -0.5-025 n=0.5 dh dx s borah =-0.9125 exact x=0.5 fexact 09125 b) Using Central difference rethod formula by approximating Tayloe's series expansion in the order of Error (h), f (x.) = (&xit - f(x; -1) 2h. a) step size h=1, x=0.5 $(xi+ 1) = (x+h) = 15 (1-5)=-1-3125 $(x) = (x) = 0 5 f(05)= 0.925 $(x,-1) =6e-=0.5->$(-0.5) - 1.2125 (1:5), (0.1)(1-5)40 15(15) - 0.5(1:5)? o 25(1 5)#1:2 = f(0:5)= -(0:1)(0-5)-0.15(0.5) -0.50:5)?-0.25(0-5)+(2 3 f(-0.5) = -(01)1-05)_0 151-015) 205(-0.5). 0:251-0.5.be

s'(num) - $(15)-f(-0.5) 2(h) = -1.3125 - (1 2125) 2 f (num) -- 1.2625 Error, E = Il f (num) - flexad 11 € - ||- 1.2625-(-0.9125)|| €=0.35 -> $(0-5) =0-925 In the same way step size are reduced by factor of 10 and coversponding f (num) and error values are calculated. b) step size, h=0.1, X=0 51 d) h-0.001, x=0.5 $(xi+1)= 0.6-7 flo-b)=0-824 Xiti = o 501 > $(0-501) - 0.92409 $(xi) = 0.5 $(0-5)=0.925xi 4(x,-1) = 0.4-) $100=1004 Xi-, 30-499 -> $(0499) = 0.92591 f (num) = 0 fllnum) = -0.91250035 1=0.824-1007 2 (0:1) 16:3.5e") f (num) = -0.916 eh-0.0001 X=0.5 G = || -0.916-(-0.9125) || Xit -05001 f(0 5001) = 0:92491 LE=0.0035 =0.5 → $105)-0.925 = 0.49993 $(04999) = c) h = 0.01, =0.5 -0.91250000 3 titl -0.51 - $10.50)=0 9157 xt > $(0-5) = 0.925 = 3.5e X; -1=0.497 f (0.49) 20:134 f (num) E: f'(num) - f'(exact)- 35e xi =092509 f (num) =0.5 = -0.91 2535

f)h= 0.00001) x=0.5 Xiti in=0.00oooool x=0.5 xi 0.50001 » (0.92499) = 0'5 +0.925) 0.49999760-92501) xirl =-0.912500603 f (num) 3.3609 с & ' (num) = =-0.9125 3.3 e 11 j) h = 0.000000ool, x=0.5 g) h=0.000ool, x=0.5 f' (num) = -0.925 00002 Xiti 0.5oooo I € = 1.998940 X;-1= -1=0.499999 f' (num) = -0.9125 5.42 2-2 hlh=0.0000001, x=0.5 Xiti 0.5000001 xt X;-1 0.4999999 f'(num) = -0.921499999 E = 5.49689 e -10

f(x) TRUE(f'(exact)) h 1 0.49 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001 0.00000001 0.000000001 (-(0.1*(x^4))-(0.15*(x^3))-(0.5*(x^2))-(0.25*x)+1.2) -0.9125 xi xi+1 xi-1 f(xi) f(xi+1) f(xi-1) f'(num)/f'(centered) Error(epsilon) 0.5 1.5 -0.5 -1.3125 1.2125 -1.2625 0.35 0.5 0.6 0.4 0.925 0.82464 1.00784 -0.916 0.0035 0.5 0.51 0.925 0.91579 0.93404 -0.912535 3.5E-05 0.5 0.501 0.499 0.925 0.92409 0.92591 -0.91250035 3.5E-07 0.5 0.5001 0.4999 0.925 0.92491 0.92509 -0.912500003 3.49985E-09 0.5 0.50001 0.49999 0.925 0.92499 0.92501 -0.9125 3.31783E-11 0.5 0.500001 0.499999 0.925 0.925 0.925 -0.9125 5.42288E-12 0.5 0.5000001 0.4999999 0.925 0.925 0.925 -0.912499999 5.49689E-10 0.5 0.50000001 0.49999999 0.925 0.925 0.925 -0.912500003 3.33609E-09 0.5 0.500000001 0.499999999 0.925 0.925 0.925 -0.91250002 1.99894E-08

Step size(h) 1 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001 15-08 1E-09 1 0.1 0.01 E r 0.001 r 0.0001 0 0.00001 r 0.000001 e 0.0000001 р 1E-08 S . 1E-09 1 1E-10 1E-11 o =) 1E-12

the step size h decreases . Later, C) (i) Error value decreases initially it slightly increases after certain point a certain limit error value increases. This is due to domination of "Round-off" Erdor at a very low step sine (111) Order of Accur (1) No, after curacy 3.5e 3-3.5e of formula, Slope -5 0: ol -0.0385 Slope the order is oliserued as 0 (6²) for error wirt step size