テイラー級数およびマクローリン級数（べき級数）計算機

Your input: calculate the Taylor (Maclaurin) series of $$$e^{x}$$$ up to $$$n=3$$$

A Maclaurin series is given by $$$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}x^k$$$

In our case, $$$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}x^k=\sum\limits_{k=0}^{3}\frac{f^{(k)}\left(a\right)}{k!}x^k$$$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$$

Evaluate the function at the point: $$$f\left(0\right)=1$$$

1. Find the 1st derivative: $$$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(e^{x}\right)^{\prime}=e^{x}$$$ (steps can be seen here).

   Evaluate the 1st derivative at the given point: $$$\left(f\left(0\right)\right)^{\prime }=1$$$

2. Find the 2nd derivative: $$$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(e^{x}\right)^{\prime}=e^{x}$$$ (steps can be seen here).

   Evaluate the 2nd derivative at the given point: $$$\left(f\left(0\right)\right)^{\prime \prime }=1$$$

3. Find the 3rd derivative: $$$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(e^{x}\right)^{\prime}=e^{x}$$$ (steps can be seen here).

   Evaluate the 3rd derivative at the given point: $$$\left(f\left(0\right)\right)^{\prime \prime \prime }=1$$$

Now, use the calculated values to get a polynomial:

$$$f\left(x\right)\approx\frac{1}{0!}x^{0}+\frac{1}{1!}x^{1}+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}$$$

Finally, after simplifying we get the final answer:

$$$f\left(x\right)\approx P\left(x\right) = 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}$$$

Answer: the Taylor (Maclaurin) series of $$$e^{x}$$$ up to $$$n=3$$$ is $$$e^{x}\approx P\left(x\right)=1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}$$$