can anyone help me to find the error ?

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diadalina
diadalina 2020-2-28
评论： diadalina 2020-2-28
syms x1
esp=10^-10;
x0=0.5;
nmax=100;
df=@(x1)(diff(f(x1)))
[ xapp,it] = newton(x0,esp,f,df,nmax)
where:
function [ xapp,it ] = newton(x0,esp,f,df,nmax)
it=0;
i=1;
x=[];
x(1)=x0;
x(2)=x(1)-f(x(1))/df(x(1));
while abs(x(i+1)-x(i))>esp && it>=nmax
if df(x(i))==0
disp('division par zero est impossible')
break
end
i=i+1;
it=it+1;
x(i)=x(i-1)-f(x(i-1))./df(x(i-1));
end
xapp=x(it );
and
function y=f(x)
y=(x+2).^(2/5)
end
  2 个评论
Guillaume
Guillaume 2020-2-28
If you get an error, give us the full text of the error message you get, without any modification
If you don't get an error but the code doesn't behave as you expected, then please explain what it does and how it differs from what you expected.
diadalina
diadalina 2020-2-28
df =
function_handle with value:
@(x1)(diff(f(x1)))
Not enough input arguments.
Error in f (line 2)
y=(x+2).^(2/5)
Error in tp4 (line 12)
[ xapp,it] = newton(x0,esp,f,df,nmax)

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回答（1 个）

Guillaume
Guillaume 2020-2-28
[ xapp,it] = newton(x0,esp,@f,df,nmax)
should probably fix the problem.
  3 个评论
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Guillaume
Guillaume 2020-2-28
编辑：Guillaume 2020-2-28
Oh yes, your df function does not work at all. Since it calls diff it's guaranteed to return one less element than the input so you're always going to get this error and since df is given a scalar, diff is always going to return [].
It's not clear what that df function is trying to do.

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