Newton's method

11052019, 09:09 PM
Post: #1




Newton's method
How do I perform Newton's method on my HP prime calculator?


11062019, 01:12 AM
Post: #2




RE: Newton's method
Press the HELP button.
Once in the tree, press the "SEARCH" and look for "newton" (check the "find in content" if you want to find any instance of it). After a bit, you will find a command that I suspect is the one you want... TW Although I work for the HP calculator group, the views and opinions I post here are my own. 

11062019, 06:48 AM
Post: #3




RE: Newton's method
The formula is simply Xn[f(Xn)/f'(Xn)]
Basically I need to plug in different xvalues 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 xvalue? 

11062019, 01:39 PM
Post: #4




RE: Newton's method
As I gave directions, you will find it. There is a built in command that does exactly what you want.
The goal was to teach about the search capability, the built in help, and make it easier for you to find things next time. TW Although I work for the HP calculator group, the views and opinions I post here are my own. 

11062019, 04:14 PM
Post: #5




RE: Newton's method
Try the [Toolbox] key, (then [Help] key, if needed). It's a great resource ...
Dale 

11062019, 04:38 PM
Post: #6




RE: Newton's method
Hello,
the built in routine 'newton' (you find a complete description following Tim's advice) can be used to study the convergence of the iterationmethod (may be that is what you want?) step by step by looking at the result after only one iteration per functioncall: change to CAS , then example: f(x)= x^23 wanted: one root of f , estimated value could be 1.3 first input newton(x^23,x,1.3,1) enter key output: 1.8038... (this is x_1 in the newton's series) input newton(x^23,x,Ans,1) enter key output: 1.73347958012 (this is x_2) enter key output: 1.73347958012 (this is x_3) and so on hope that helps Thomas 

11062019, 04:42 PM
(This post was last modified: 11062019 04:44 PM by Wes Loewer.)
Post: #7




RE: Newton's method
(11062019 06:48 AM)hazimrassam Wrote: The formula is simply Xn[f(Xn)/f'(Xn)] 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. 

11062019, 06:29 PM
Post: #8




RE: Newton's method
(11062019 04:42 PM)Wes Loewer Wrote:(11062019 06:48 AM)hazimrassam Wrote: The formula is simply Xn[f(Xn)/f'(Xn)] 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 

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