1. The problem is to understand the relationship between the nth Taylor polynomial of a function $f(x)$ and the Maclaurin series. 2. The Taylor polynomial of degree $n$ for a function $f(x)$ centered at $a$ is given by the formula: $$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k$$ where $f^{(k)}(a)$ is the $k$th derivative of $f$ evaluated at $a$. 3. The Maclaurin series is a special case of the Taylor series where the center $a=0$. Thus, the nth Taylor polynomial at $a=0$ is: $$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$$ 4. Therefore, the nth Taylor polynomial of $f(x)$ at $a=0$ is exactly the nth partial sum of the Maclaurin series of $f(x)$. 5. In simple terms, the Maclaurin series is the Taylor series centered at zero, so the nth Taylor polynomial at zero equals the nth Maclaurin polynomial. Final answer: The nth Taylor polynomial of $f(x)$ at $a=0$ is the nth Maclaurin polynomial of $f(x)$.