1) Use two repetitions of the Newton-Raphson algorithm to approximate the follow

Question

1) Use two repetitions of the Newton-Raphson algorithm to approximate the following. (Use x0 = 3. For best results, do not round intermediate values in your calculations. Give your answers correct to five decimal places.)

x1 =
x2 =

Now use a calculator to determine the exact value. (Give your answer correct to five decimal places.)

____________________________________________________________________

2)    Use two repetitions of the Newton-Raphson algorithm to approximate the zero of x2 + 3x - 11 between -5 and -6. (Use x0 = -5. For best results, do not round intermediate values in your calculations. Give your answer correct to five decimal places.)

_____________________________________________________________________

3)     Use the Newton-Raphson algorithm to find an approximate solution to the following. There are two roots. (Use x0 = 2 for the smaller value and x0 = 6 for the larger value. For best results, do not round intermediate values in your calculations. Use two repetitions. Give your answers correct to five decimal places.)

e3 - x = 6 - x

_______(smaller value)
_______(larger value)

Explanation / Answer

This does not mean that you cannot calculate roots using the Newton-Raphson procedure, but that the function f(x) must have the desired root as the solution (or one of the solutions) of the condition f(x) = 0. For example, to obtain the cube root of 30 you could use the function f(x) = (x³ - 30). The solution of f(x) = 0 is then x = 30^(1/3), and the required derivative f '(x) = 3.x².

This would give the expression for the (n+1)th approximation in terms of xn, the preceding approximation, as

x(n+1) = xn - [xn³ - 30]/(3.xn²)

I have carried out the procedure for this equation, with a target accuracy of 5 decimal places and using x(0) = 3 and 5 as the initial guesses. The results are

x(0) = 3 . . . . . . . . .x(0) = 5
x(1) = 3.11111 . . . .x(1) = 3.73333
x(2) = 3.10724 . . . .x(2) = 3.20636
x(3) = 3.10723 . . . .x(3) = 3.11027
x(4) = 3.10723 . . . .x(4) = 3.10724
x(5) = 3.10723 . . . .x(5) = 3.10723

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.