MATH-3800 Lecture 13: 3.2 The Euler Phi Function

- MAA MATHEMATICAL ASSOCIATION OF AMERICA webwork/math102-all 2017w1/ assignment 11/6 Assignment 11: Problem 6 Previous Problem List Next 2 points) Consider the differential equation dy 5x dx with Initlal condition yO) - 3. A. Use Euler's method with two steps to estimate y when x = 1 : (1) 6.75 Be sure not to round your calculations at each step) Now use four steps: y(1)6.125 Be sure not to round your calculations at each step) B. What is the solution to this differential equation (with the given initial condition)? C. What is the magnitude of the eror in the two Euler approximations you found? Magnitude of error in Euler with 2 steps Magnitude of error in Euler with 4 steps = Note that in this example, the right hand side of the differential equation is a function of the independent variable x instead of the dependent variable y. Nonetheless, you can apply Euler's method in the similar way here. Note: You can earn partial credit on this problem Preview My Answers Submit Answers MacBook Air

Which (if any) of the following vector fields are conservative: F rightarrow (x, y) = exy i rightarrow + ex+y j rightarrow F rightarrow (x, y) = 2x cos y i rightarrow - x2 sin y j rightarrow Determine if the field F rightarrow (x, y, z) = cos xz, sin yz, xy sin z is conservative. Let F rightarrow (x, y) = xy2 i rightarrow + x2y j rightarrow. Evaluate C F rightarrow · ds rightarrow where C is the upper half of the circle of radius 1 centered at the origin orientated counterclockwise. Find a potential function for Frightarrow (x, y, z) = yz2, xz2 - z, 2xyz - y . Use part (a) to evaluate C F rightarrow · ds rightarrow where C is any path in space from P = (1, 2, 3) to Q = (3, 2, 1). Let F rightarrow (x, y, z) = z2/x i rightarrow + x2/y j rightarrow + 2z ln xy k rightarrow. Evaluate C F rightarrow · d s rightarrow Where C is the path of straight line segments from P = (1, 2, 1) to Q = (4, 1, 7) to R = (5, 11, 7), and then back to P. (Bonus question: worth 10 points. Total points for assignment not to exceed 100.) Suppose T nabla phi · d s rightarrow = 1 where T is the path from P = (-2, 1) to Q = (2, 1) along the top half of the ellipse 3x2 + 11y2 = 23. Determine the value of int B nabla phi · d s rightarrow where B is the path from P to Q along the bottom half of that ellipse.

Here we return to polynomial interpolation, let's assume that we arc given "nodes' x0.x1.x2 = {-1,0.2}. and function value fo.f1.f2 = {1.2.1} corresponding to the three points (-1,1), (0.2) and (2.1). We will try and build a polynomial interpolant through these three points. First, let's try and pass a parabolic interpolant p(x) = p0 + P 1 x -f p1x2 through these three points. Show that such parabolic interpolant must satisfy p(xo) = f0. p(xi) = f1 and p(x1) = f2, which corresponds to p(-l) = 1. p(0) = 2. and p(2) = 1. which corresponds to p0 + (-l)pi+(-lfp2=1 phi. + (+0)phi +{+0fpi=2 Po + (+2) phi + (+2)2 P2=1 Solve this equation using row-reduction and find the vector p (i.e.„ find p0.p1.p2). Plot p(x) and show that p(x) induce passed through the points (-1,1), (0.2) and (2,1). Find the QR decomposition of A. Can you solve the equation Ap - b using the QR decomposition? Hopefully you get th same answer as you did above. Now let's try and pass a linear interpolant I (x) = lo + '11 through these three points. Show that, ideally, such a linear interpolant should satisfy l (x0) = f0, l(x1) = f 1 and l (x2) = f2, which corresponds to l (-1) = 1, l (0) = 2. and l (2) = 1. which corresponds to l0 + (-l) l1 = l l0 + (+0) l1 = 2 l0 + (+2)l1, = 1 which corresponds to Can you solve this equation exactly? Why or why not? Can you obtain an approximate solution / such that Al is 'close' to b? This is what is referred to as the least-square solution to this problem. Can you show that, when Al is as close to b as possible, the residual f - Al - b is perpendicular to each column of .'1? If we assume such a residual condition, then in what sense is Al close to 6? Can you show that when f is perpendicular to each column of A then f is as short as possible (i.e., the length is as small as possible).