Newton Raphson!but with multiple initial values :/

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alexp10
alexp10 2020-5-5
编辑： Michael Soskind 2020-5-5
This works for one initial value. Now I want to do the exact thing but for 2 more initial values. Where they all are x0= [0.5,1.5,2.5]; and the outputs will be r = [r1 , r2 , r3]. Any help would be appreciated, thanks.
r1 = newtonr()
function[r]= newtonr
i = 0;
x0=0.5;
n = 5;
while i < 5
f = x0^4 - 9*x0^3+29*x0^2-39*x0+18;
df= 4*x0^3 - 27*x0^2 +58*x0 -39;
xnew= x0 - f/df;
i= i+1
x0=xnew;
disp(x0);
if i== 5
r= xnew;
disp('root is')
disp(x0)
break
end
end
end

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采纳的回答

David Hill
David Hill 2020-5-5
i = 0;
x=[0.5,1.5,2.5];
n = 5;
f = @(x)x.^4 - 9*x.^3+29*x.^2-39*x+18;
df= @(x)4*x.^3 - 27*x.^2 +58*x -39;%functions should not be in loop
while i < 5
x= x - f(x)./df(x);
i= i+1
disp(x);
end
disp('roots are: ');
disp(x);

更多回答（1 个）

Michael Soskind
Michael Soskind 2020-5-5
编辑：Michael Soskind 2020-5-5
Hi Alex,
Edited: Please note, David Hill's answer is the one I recommend.
As a note:
To make calculations work on multiple values within an array, the operator should be preceded with a '.'
This is the case for multiplication, division, and raising to a power.
This is the main modification that David makes in his code, as well as implementing functions for the function and its derivative. This is in general better practice.

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