Question

Using Newton method, find the value of t that give a maximum value at an interval of [0 10] for the following function: 2 sin (- y (2) Use initial guess of t = 0.1 with stopping error of &s = 0.01%. Apply centered finite-difference formulas with step size of h 0.01 to calculate the derivatives For all calculation, use at least 5 significant figures for better accuracy.

Answer 1

Steps:

The following algorithm implements Newton’s method to determine the maximum or minimum of a function.

Initialization

Determine a reasonably good estimate x₀ for the maxima or the minima of the function f(x).

Step 1

Determine f'(x) and f''(x) .

Step 2

Substitute x_i+1, the initial estimate x₀ for the first iteration, f'(x) and f''(x) into Eqn. 1 to determine x_i and the function value in iteration i.

Step 3

If the value of the first derivative of the function is zero, then you have reached the optimum (maxima or minima), otherwise repeat Step 2 with the new value of x_i until the absolute relative approximate error is less than the pre-specified tolerance.

$f(t)=\frac{2sin(\sqrt t)-t}{e^{3t}}\\\Rightarrow f'(t)=\frac{cos(\sqrt t)+\sqrt t(-1+3t-6sin(\sqrt t))}{e^{3t}\sqrt t}\\ and,\ f''(t)=\frac{-(1+12t)cos(\sqrt t)-\sqrt t(6t(-2+3t)+(1-36t)sin(\sqrt t))}{e^{3t}t^{3/2}}$

to 0.1 Es =0.01% h 0.01

f (to) ti tof(to) 0.325646 = 0.1 -17.7185 0.11838 f (th) t2= tfti) 0.047211 0.11838 12.8938 - 0.12204 f(t2) t3 t2f"(t2) f'(0.12204) f"(0.12204) = 0.12204 0.122158 0.12216

f (t3) ta = ta-f"(t3) 0.12216 (t3) f" (t3) f' 0.122158 0.12216

Now,

f(0.12216)0.390062