Newton's method

[CyberAngel]

(11-06-2019 04:42 PM)Wes Loewer Wrote:  

(11-06-2019 06:48 AM)hazimrassam Wrote:  The formula is simply Xn-[f(Xn)/f'(Xn)]
Basically I need to plug in different x-values and solve quickly. Is there a way to plug in X values into the same equation without having to rewrite the equation each time I plug in a new x-value?

Are you wanting just to solve the equation using Newton's Method? If so, then the online help that the others mentioned will give you that info. (Look for the newton() and fsolve() functions.)

However, it sounds like you might be asking how to easily step through Newton's Method so that you can see the intermediate results. If so, what I find helpful with my students is to use the following:

To find the zeros of x^3+x+1, enter an initial guess and then use
X-(X^3+X+1)/(3*X^2+1)|(X=Ans)
and press Enter repeatedly till the result starts to repeat.


You could also use
Ans-(Ans^3+Ans+1)/(3*Ans^2+1)
but that's not as easy to type, nor is it as aesthetically pleasing.

I use RE for computing and adding i to the (complex) x
and store it away - Starting value is an approximate 0. (zero with a dot)
Kind of something like: Settings
(I'm required to give the needed steps for say, five decimals)
Number Format [Fixed]v [5]v
Complex [(a,b) ]v [V]
[HOME] 0. [Sto|>] [ALPHA] [Shift] x
x - (RE(x)^3+RE(x)+1) / (3*RE(x)^2+1) + i [Sto|>] x
[ Enter ] [ Enter ] [ Enter ] [ Enter ] [ Enter ] [ Enter ]
no changes, therefore, the answer is ~ (-0.68233, 5.00000)
5 steps to get -0.68233