September, we found that integral^1/2_0 1/1 - x^2 dx = 1/2 in 3. Towards the end

Question

September, we found that integral^1/2_0 1/1 - x^2 dx = 1/2 in 3. Towards the end of that lecture, we also found that In 3 is a little bit more than 1. To get a better approximation to the true value of In 3, we might try approximating the function f(x) = 1/x by a quadratic function q(x) = ax^2 + bx + c and calculating the integral integral^3_1 q(x) dx. A reasonable choice for q(x) would be the quadratic polynomial that has the same values as f(x) at each of x = 1, 2, 3. The equations q(1) = 1, g(2) = 1/2, q(3) = 1/3 give a system of three linear equations in the unknowns a, b, c. Write down this system, and solve it using Gauss-Jordan elimination. Having found a, b, c in (a), calculate integral^3_1 (ax^2 + bx + c) dx using the Fundamental Theorem of Calculus, and compare the result with what your calculator gives for In 3.

Explanation / Answer

Ans-

The following demonstrates the first six iterations of Müller's method in Matlab. Suppose we wish to find a root of the same polynomial

p(x) = x7 + 3x6 + 7x5 + x4 + 5x3 + 2x2 + 5x + 5

starting with the same three initial approximations x0 = 0, x1 = -0.1, and x2 = -0.2.

The first formula in red is the root of the quadratic polynomial which is added onto the middle approximation x(2).

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