Please help with Taylor Series Expansion for x^n
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Swati Umamaheshwaran
2017 年 12 月 5 日
コメント済み: Swati Umamaheshwaran
2017 年 12 月 5 日
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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Birdman
2017 年 12 月 5 日
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 件のコメント
Birdman
2017 年 12 月 5 日
Actually, I did not receive any erratic result. The convergence is better for every step of n.
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