Math 476 Homework - Answer Key

Homework: Assigned Monday, Sep 4 Due Friday, September 8 1. In Matlab, find the smallest value of p for which the expression: tan(x) − x x3 when evaluated in double precision at x = 10−p has no correct significant digits. (Hint: First we need the exact value- Take the limit (by hand) as x → 0. Then let p = −1, −2, etc. until you find the right p value). Solution: First we find the limit using L’Hospital’s rule (by the way, did you check that L’Hospital was valid at each step?) tan(x) − x sec2 (x) − 1 2 sec2 (x) tan(x) = = lim = lim x→0 x→0 x→0 x3 3x2 6x lim 1 4 sec2 (x) tan2 (x) + 2 sec4 (x) = x→0 6 3 In Matlab, we could type: lim format long x=10.^[-1 -2 -3 -4 -5 -6 -7 -8 -9]; y=(tan(x)-x)./(x.^3); %Why is this ./ and .^? You should see that the expression, evaluated at 10−8 returns 0. 2. Use the Bisection Method (in Matlab) to find the root to six correct decimal places: 3x3 + x2 = x + 5. Solution: We first need to rewrite this as f (x) = 0: f (x) = 3x3 + x2 − x − 5 Now we need an initial interval, so we do a quick plot: x=linspace(-3,3); y=3*x.^3+x.^2-x-5; plot(x,y) In my case, I’ll take the interval [0, 2]. Note that I can now predict how many iterations I’ll need to get the right accuracy: |b − a| 1 ≤ × 10−6 n+1 2 2 2 ⇒ 2n+1 ≤ 1 × 10−6 2 ⇒ Using the Bisection function from class, in Matlab I would type: 1 n = 21 >> format long >> f=inline(’3*x.^3+x.^2-x-5’); >> y=bisect(f,0,2,0.5e-6); Finished after 21 iterates >> y y = 1.16972589492798 3. Use Bisection to find two real numbers x, correct to within 6 decimal places, that make the determinant of the matrix:    A=  1 2 3 x 4 5 x 6 7 x 8 9 x 10 11 12      equal to 1000. Hint: A little preprocessing in Maple might make this an easier problem. Solution: First, use Maple: with(linalg): A:=matrix(4,4,[1,2,3,x,4,5,x,6,7,x,8,9,x,10,11,12]); f:=det(A); 2 4 -2475 - 202 x + 1404 x + x plot(f-1000,x=-20..15); We’ll be using −3475 for our constant, since we’re trying to find where the determinant is 1000. We see that the zeros are between −20, −15 and 5, 15. Therefore, I’ll use Matlab with intervals [2, 4] and [4, 6] (many choices possible): format long f=inline(’x.^4-202*x.^2+1404*x-3475’); y=bisect(f,5,15,0.5e-6); Finished after 24 iterates y = 9.70829933881760 y=bisect(f,-20,-15,0.5e-6); Finished after 23 iterates y = -17.18849807977676 4. You have a spherical tank with radius 1 meter. You pour 1 cubic meter of water into the tank. Find the height of the water to within 1 millimeter. Hint: The volume of the bottom H meters of a hemisphere of radius R is πH 2 (R − 31 H). 2 Solution: We are solving the following equation for H: 1 1 = πh2 (1 − h) 3 where we want h to be accurate to the thousandths place, or to within 0.0005. Make this into a root finding problem and find an initial interval (from physical conditions, we know the answer must be between 0, 1, but we can plot it anyway): >> format long >> f=inline(’pi*x.^2.*(1-(1/3)*x)-1’); >> x=linspace(0,1); >> y=f(x); >> plot(x,y); >> y=bisect(f,0,1,0.0005); Finished after 10 iterates >> y y = 0.63525390625000 5. Use fixed point iteration to find a root of cos(x) = sin(x). In order to compare your results with other people, let us get our iteration method by adding x to both sides, and start with x0 = 0, with 19 iterations (before using Matlab, what is the solution?). Have Matlab compute ei+1 /ei . Does this number look right? (Hint: Recall our proof of convergence). Solution: First we note that the solution is π/4. Next re-write the equation to a fixed point form. One way to do this would be to write: cos(x) = sin(x) ⇔ cos(x) − sin(x) + x = x Does this work? Let’s be sure that the derivative is less than 1 (this wasn’t a requirement, but it is a good idea): √ f 0 (π/4) = −2/ 2 + 1 ≈ −0.414... Now perform fixed point iteration: >> f=inline(’cos(x)-sin(x)+x’); >> y=fixpt(f,0,19); >> format long >> Error=abs(y-pi/4); >> Z=Error(2:20)./Error(1:19); >> Z’ ans = 0.27323954473516 0.40338351109981 0.41244791418838 0.41391310875287 (middle terms deleted) 3 0.41421356224673 0.41421356252052 0.41421356222495 0.41421356323653 If we let f (x) = cos(x) − sin(x) + x, we see that the limit is |f 0 (π/4)|, as we used in our proof of convergence! That was, |xn+1 − r| = |f (xn ) − f (r)| ≤ |f 0 (cn )||xn − r| 4 ⇒ |xn+1 − r| ≤ |f 0 (cn )| → |f 0 (r)| |xn − r|