Please help with Taylor Series Expansion for x^n

Please help with Taylor series expansion for x^(1/2). I have tried to use the general taylor series expansion formula to find the solution, But I am not able to get the right answer. I am trying to find the square root of 5 about the expansion point 4.(x=5; a=4)

 root_five = vpa(sqrt(5));
  n = input ('Enter number of intervals: ');
  syms a
  func = (a)^(0.5);
  a = 4;
  x = 5;
  f = subs (func,a);

for i = 1:n
    f = f + ((diff(func,i)*(x-a)^i)/factorial (i));
    root_five_4(i) = vpa(subs(f,a));
end

err = abs(root_five-root_five_4)

Answer 1

[Birdman]

Actually, your answer is correct. At each step of n, you become closer to zero, which means the real value of square root of 5 and estimated from Taylor expansion are becoming closer as the expansion degree increases and that is actually what we expect from Taylor expansion. For instance when n is 10 in your code, the answer is

0.00000000313464

which is nearly zero and this is what we want to see.

[Swati Umamaheshwaran]

Thank you.

[Birdman]

Actually, I did not receive any erratic result. The convergence is better for every step of n.

[Swati Umamaheshwaran]

I got the answer too. I had made a silly mistake. Thank you.