I'm unsure of why there is an error on F=@x

function Xs= newtonM(fun,FunDer,Xest,err)
for i=1:100
    Xs= Xest-Fun(Xest)/FunDer(Xest);
    error =abs((Xs-Xest)/Xs)*100;
    fprintf('%3i %11.6f %11.6\n', i, Xs, error)
    if error < err
        break;
    end
    Xest= Xs;
    
end
end
f=@(x)exp(-0.5*x)*(4-x)-2;
df=@(x)exp(-0.5*x)*(-3+0.5*x);

Xs = newtonM(f,df,5);

Answer 1

[Star Strider]

The statement order is reversed from what it should be —

f=@(x)exp(-0.5*x)*(4-x)-2;
df=@(x)exp(-0.5*x)*(-3+0.5*x);
 
Xs = newtonM(f,df,5,0.001)

  1  -45.729976   2  -43.807300   3  -41.887610   4  -39.971139   5  -38.058150   6  -36.148939   7  -34.243841   8  -32.343235   9  -30.447556  10  -28.557302  11  -26.673052  12  -24.795477  13  -22.925367  14  -21.063656  15  -19.211459  16  -17.370128  17  -15.541316  18  -13.727084  19  -11.930062  20  -10.153724  21   -8.402890  22   -6.684771  23   -5.011257  24   -3.404173  25   -1.907058  26   -0.607889  27    0.340095  28    0.795644  29    0.882955  30    0.885706  31    0.885709 

Xs = 0.8857

function Xs= newtonM(fun,FunDer,Xest,err)
    for i=1:100
        Xs= Xest-fun(Xest)/FunDer(Xest);
        error =abs((Xs-Xest)/Xs)*100;
        fprintf('%3i %11.6f %11.6\n', i, Xs, error)
        if error < err
            break;
        end
        Xest= Xs;
     
    end
end

Also, ‘newtonM’ needed an ‘err’ argument.

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