How to run this code on any inputs?Newton

1 vue (au cours des 30 derniers jours)
cakey
cakey le 21 Sep 2014
Commenté : cakey le 23 Sep 2014
function [p,k] = newton(f,df,p0,tol)
for k=1:100
p = p0 - f(p0)/df(p0);
if abs(p-p0)<tol, break; end
p0 = p;
end
I have that but where do I put the tolerance, the function and everything else for it to run. I'm really new at all this.
  4 commentaires
cakey
cakey le 22 Sep 2014
Can you help me understand why I need to do this function handle business? Why do I also need to hit disp? I'm completely new to this all.
Stephen23
Stephen23 le 22 Sep 2014
Read about function handles, it might help. And use the contents browser, which is on the left-hand side of the help window, to discover related topics... it really is very handy!

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Matt J
Matt J le 22 Sep 2014
Modifié(e) : Matt J le 22 Sep 2014
When you encounter commands or expressions that you don't understand, you should be reading their documentation pages to see what they do,
>> doc disp
>> docsearch 'function handle'
Even more importantly, you should be playing with these commands individually at the command line to test your understanding of the documentation. It should have been very easy for you to discover that disp() is a way to display results in the command window just by trying it,
>> disp([1,2])
1 2
>> disp([3,4,5])
3 4 5
It should have been similarly easy to learn that constructing function handles like
f=@(x) (x.^3+x+2);
allow your newton routine to evaluate your f(x) at different points, as the algorithm requires:
>> f(0)
ans =
2
>> f(1)
ans =
4

Plus de réponses (1)

Stephen23
Stephen23 le 22 Sep 2014
What you have is a function, which can be run to create some outputs. It could be good to revise MATLAB's comprehensive help on calling functions . If you are learning MATLAB, you will have to get used to doing a lot of reading of help pages. But don't worry, MATLAB's help pages are really quite good, with plenty of examples to try, and explanations of concepts that you need to understand. You should learn to navigate using the help contents browser too. Good luck!
  1 commentaire
cakey
cakey le 23 Sep 2014
You were helpful too. Thank you.

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