We have to find the Taylor Polynomial of $f(x)=e^x$ up to the third degree, centered at $x=0$.
The Taylor Polynomial is given by:
(x)=∑
k=0
n
k!
f
(k)
(a)
(x−a)
k
where $f^{(k)}(a)$ is the $k$th derivative of $f(x)$ evaluated at $a$.
Using this formula, we can find the Taylor Polynomial as follows:
\begin{align*}
f(x) &= e^x \
f'(x) &= e^x \
f''(x) &= e^x \
f'''(x) &= e^x \
\end{align*}
Evaluating each derivative at $x=0$, we get:
\begin{align*}
f(0) &= e^0 = 1 \
f'(0) &= e^0 = 1 \
f''(0) &= e^0 = 1 \
f'''(0) &= e^0 = 1 \
\end{align*}
Substituting these values into the formula for the Taylor Polynomial, we get:
\begin{align*}
P_3(x) &= \frac{f(0)}{0!}(x-0)^0 + \frac{f'(0)}{1!}(x-0)^1 + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 \
&= 1 + x + \frac{x^2}{2} + \frac{x^3}{6}
\end{align*}
Therefore, the Taylor Polynomial of $f(x)=e^x$ up to the third degree, centered at $x=0$, is $P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$.