Finding roots using Newton raphson method

Here is my code and my output is a function and not a numerical value as I expected. Can anyone debug this code?

syms f y x df
f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);


output: 

- (sin((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3) + exp(- exp(-3)/(pi/3 + exp(-3)) - 3))/(exp(- exp(-3)/(pi/3 + exp(-3)) - 3) - (pi*cos((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3))/3) - exp(-3)/(pi/3 + exp(-3)) - 3

Answer 1

[Alan Stevens]

Just get rid of the first line, you don't need it

f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);
disp(root)

[R Abhinandan]

Still im getting the output as a function and not the root

output:

- (sin((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3) + exp(- exp(-3)/(pi/3 + exp(-3)) - 3))/(exp(- exp(-3)/(pi/3 + exp(-3)) - 3) - (pi*cos((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3))/3) - exp(-3)/(pi/3 + exp(-3)) - 3

expected output:

-3.0844

[Alan Stevens]

This is what I get:

>> f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);
disp(root)
   -3.0454

[R Abhinandan]

Thank you, im sorry i didnt know that we have to save before running the code.