Math Homework 3

Transcription

1 Math 0 - Homework 3 Due: Friday Feb. in class. Write on your paper the lab section you have registered for.. Staple the sheets together.. Solve exercise 8. of the textbook : Consider the following data: x f(x) Find the first derivative f (.3) using the following methods: (a) three-point forward difference formula. (b) three-point backward difference formula. (c) two-point central difference formula. Solution for part (a) f (x) = 3f(x) + 4f(x + h) f(x + h) h f (.3) = 3f(.3) + 4f(.4) f(.5) (0.) = ( 3)(0.97) + 4(.84) =.9955 Solution for part (b) f (x) = f(x h) 4f(x h) + 3f(x) h

2 f f(.) 4f(.) + 3f(.3) (0.78) + 3(0.97) (.3) = = =.995 (0.) 0. Solution for part (c) f (x) = f(x + h) f(x h) h f (.3) = f(.4) f(.) (0.) = =.990

3 . (i) We want to approximate f (x) by the four-point forward difference formula f(x) 5f(x + h) + 4f(x + h) f(x + 3h) h Use the unterminated form of the Taylor expansion to find the order of truncation error. (ii) Write a Matlab program to approximate the second derivative f () of the function f(x) = x by the difference formula in part (i). In you program you start with h = 0.5 and then using a while loop you reduce h in each step by You continue until your h is larger than or equal to 0.0. Your output must have three columns: the first column for h, the second one for the approximate values, and the third column shall show the absolute error between the approximate value and the true value f () =. In each iteration print the three values (h, approximate value, and the error) in the their columns using the fprintf function of Matlab. Use the formatting %3.f for the first column, and %.f for columns two and three. Supply your code and the output. Solution for part (i) f(x + h) = f(x) + hf (x) + h! f (x) + h3 3! f (x) + h4 4! f (4) (x) + f(x + h) = f(x) + hf (x) + h f (x) + h3 f (x) + h4 4 f (4) (x) + () In this equality, replace h by h f(x + h) = f(x) + hf (x) + (h) f (x) + (h)3 f (x) + (h)4 4 f (4) (x) + 3

4 f(x + h) = f(x) + hf (x) + 4h f (x) + 8h3 f (x) + h4 4 f (4) (x) + () In (), replace h by 3h f(x + 3h) = f(x) + 3hf (x) + (3h) f (x) + (3h)3 f (x) + (3h)4 4 f (4) (x) + f(x + 3h) = f(x) + 3hf (x) + 9h f (x) + 7h3 f (x) + 8h4 4 f (4) (x) + (3) So we have had these: f(x + h) = f(x) + hf (x) + h f (x) + h3 f (x) + h4 4 f (4) (x) + () f(x + h) = f(x) + hf (x) + 4h f (x) + 8h3 f (x) + h4 4 f (4) (x) + () f(x + 3h) = f(x) + 3hf (x) + 9h f (x) + 7h3 f (x) + 8h4 4 f (4) (x) + (3) By multiplying these by the numbers -5, 4, and - we get: 5f(x + h) = 5f(x) 5hf (x) + 5h f (x) + 5h3 f (x) + 5h4 4 f (4) (x) + (4) 4f(x + h) = 4f(x) + 8hf (x) + h f (x) + 3h3 f (x) + 4h4 4 f (4) (x) + (5) f(x + 3h) = f(x) 3hf (x) + 9h f (x) 7h3 f (x) + 8h4 4 f (4) (x) + () Upon adding up these values, we get: 5f(x + h) + 4f(x + h) f(x + 3h) = f(x) + h f (x) + h4 f (4) (x) + 4 By taking the term f(x) to the left-hand side we get: f(x) 5f(x + h) + 4f(x + h) f(x + 3h) = h f (x) + h4 f (4) (x) + 4 4

5 f(x) 5f(x + h) + 4f(x + h) f(x + 3h) = h f (x) + O(h 4 ) Dividing by h gives: f(x) 5f(x + h) + 4f(x + h) f(x + 3h) h = f (x) + O(h ) f (x) = f(x) 5f(x + h) + 4f(x + h) f(x + 3h) h + O(h ) So: f (x) f(x) 5f(x+h)+4f(x+h) f(x+3h) h truncation error of O(h ) 5

6 3. Find the coefficients a and b in the model y = bx x + a that best fits the following data: x y Show all the values to four decimal digits. Show the details of your calculations; i.e. calculate the matrix X X, the matrix (X X), and so on. Once you find the values a and b, then plot the three things in one graph: (i) the observed values of the table, (ii) the fitted values, (iii) the curve y = bx x + a for the range x = 0.5 : 0.0 : 0.5 Use different symbols for the observed values and the fitted values so as to make them distinguishable in the graph. Solution We first write the model as y = x + a bx = b + a b x which is a linear model. The column Y consists of the values y. The first column of the matrix X is ones(5,) and its second column consists of the values x : X = x x x 3 x 4 x 5 = Y = y y y 3 y 4 y 5 =

7 Then X X = = (X X) = d b ad bc c a (5)(.85) (.77)(.77) = = X Y = = (X X) X Y = = = b a b So: b = a b = 0.00 b = = 9.0 a = ( a b )(b) = (0.00)(9.0) =.870 7

8 So, the best model is: y = bx x + a = 9.0 x x To display the observed values we must plot the data points (, 3.) (3, ) (5,.5) (7, 7.) (0, 7.) We display these point in red color with the circle symbol. If we put the x values into the right-hand side of the equation y = 9.0 x x+.870 we get the fitted y values x = y = 9.0 x x+.870 = x = 3 y = 9.0 x x+.870 = 0.77 x = 5 y = 9.0 x x+.870 = x = 7 y = 9.0 x x+.870 = x = 7 y = 9.0 x x+.870 = 0.99 To display the fitted values we must plot the data points (, 0.304) (3, 0.77) (5, 0.499) (7, 0.385) (0, 0.99) 8

9 We display these point in black color with the asterisk symbol. Finally we plot the graph of the function y = 9.0 x x+.870 Matlab code for plotting this graph is: x = 0.5 : 0.0 : 0.5; y = (9.0. x)./ (x +.870) ; plot(x, y) over the range x = 0.5 : 0.0 : 0.5. The piece of 8 model curve 7 observed points fitted points