f(x)=0 f(x)= x^4 - 8.5 x^3 - 35.5 x^2 + 465 x - 1000 x0 = 7 x1= 9 3 it ==> 3 approx errors The Secant Method using the f
Posted: Mon May 09, 2022 10:54 am
f(x)=0
f(x)= x^4 - 8.5 x^3 - 35.5 x^2 + 465 x - 1000
x0 = 7
x1= 9
3 it ==> 3 approx errors
The Secant Method using the formulas and steps:
Its algorithm is similar to that of the Newton-Raphson method, only the iterative equation being different and the need for a second initial value: 1. Two initial values are established Xo and X1 2. Calculate the approximate root: Xi+1 = x; - f(x;)(Xi_1 – x;) f(xi-1)-f(xi) 3. It is calculated a Xi+1 - Xi ..100. Xi+1 . If ε it is greater than an allowable value &ą, the calculation return from point 2 If ε less than or equal to &a , Xi+] will be the root obtained by the secant method .
f(x)= x^4 - 8.5 x^3 - 35.5 x^2 + 465 x - 1000
x0 = 7
x1= 9
3 it ==> 3 approx errors
The Secant Method using the formulas and steps:
- F X 0 F X X 4 8 5 X 3 35 5 X 2 465 X 1000 X0 7 X1 9 3 It 3 Approx Errors The Secant Method Using The F 1 (56.2 KiB) Viewed 27 times