Taylor expansion calculation of exp(x^2)

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aroj bhattarai
aroj bhattarai 2020 年 12 月 30 日
コメント済み: aroj bhattarai 2021 年 1 月 1 日
Dear Matlab users and experts,
I am aware that the exponential function is standarized as "exp" in Matlab . However, I need to calculate the function value exp(x^2) adjusting the (N) terms in the power series. Can anyone recommend the correct method to compute the function exp(x^2)?
My approach:
x = -3.0:0.1:3.0;
N = 12;
Taylor_p2 = 0;
for n = 0:N
Taylor_p2 = Taylor_p2 + (x.^(2.0.*n))./(factorial(n)); % Taylor_p2 = exp(x^2)
end
isn't giving me the desired value. I am using R2020b Matlab version.
Many thanks in advance.
Bhattarai

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 12 月 30 日
編集済み: Ameer Hamza 2020 年 12 月 30 日
The formula for taylor series is correct. Just increase the number of terms.
x = -3.0:0.1:3.0;
N = 12;
Taylor_p2 = 0;
for n = 0:N
Taylor_p2 = Taylor_p2 + (x.^(2.0.*n))./(factorial(n)); % Taylor_p2 = exp(x^2)
end
p2 = exp(x.^2);
err = norm(p2-Taylor_p2);
plot(x, p2);
hold on
plot(x, Taylor_p2, '*')
Result
>> err
err =
9.1544e-05
  3 件のコメント
aroj bhattarai
aroj bhattarai 2021 年 1 月 1 日
Dear John D'Errico,
Should you not expect a complex number as a result? The bug is in your code, and how you handle complex results.
Yes, you are right, complex number results are expected which leads to slower convergent for abs(x)>1. For such situation, is there any mathematical solution to get a proper result? Or the real number exponent in the exponential function is fundamentally non-sense and unstable? My apologies for directing the original question into a mathematical discussion rather than the Matlab problem.
Bhattarai

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