# can anyone help me to find the error ?

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Commented: diadalina on 28 Feb 2020
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

Guillaume on 28 Feb 2020
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.
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)

Guillaume on 28 Feb 2020
[ xapp,it] = newton(x0,esp,@f,df,nmax)
should probably fix the problem.

In an assignment A(:) = B, the number of elements in A and B must be the same.
Error in newton (line 6)
x(2)=x(1)-f(x(1))./df(x(1));
Error in tp4 (line 12)
[ xapp,it] = newton(x0,esp,@f,df,nmax)
>>
Guillaume on 28 Feb 2020
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.