Please help with Taylor Series Expansion for x^n
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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)
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采纳的回答
  Birdman
      
      
 2017-12-5
        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.
3 个评论
  Birdman
      
      
 2017-12-5
				Actually, I did not receive any erratic result. The convergence is better for every step of n.
更多回答(1 个)
  A Jall
 2025-4-3
        - Find the root of the function using the Taylor series expansion at up to the fourth order. Implement the method using MATLAB.
- Analyze the convergence of the series to approximate the root.
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