To approximate the value of a function using a polynomial, especially useful when the function is difficult to compute directly.

Because they are centered at $x=0$, which often simplifies calculations.

The function $f(x)$ evaluated at $x=a$, i.e., $f(a)$.

Generally, a higher degree polynomial provides a more accurate approximation, especially closer to the center $x=a$.

A Taylor polynomial is a truncated Taylor series, taking only the first few terms of the infinite series.

Derivatives determine the coefficients of the polynomial terms, reflecting the function's rate of change at the center.

1. Find derivatives of $f(x)$. 2. Evaluate derivatives at $x=a$. 3. Plug values into Taylor series formula. 4. Simplify.

1. Find derivatives of $f(x)$. 2. Evaluate derivatives at $x=0$. 3. Plug values into Maclaurin series formula. 4. Simplify.

1. Find $f'(x)$, $f''(x)$, $f'''(x)$. 2. Evaluate at $x=1$. 3. Use the Taylor polynomial formula up to the third degree. 4. Simplify: $(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3$.

1. Find derivatives up to order 5. 2. Evaluate at $x=0$. 3. Use Maclaurin polynomial formula. 4. Simplify: $1 - \frac{x^2}{2} + \frac{x^4}{24}$.

$\sum_{n=0}^infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

$f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n$

$\frac{f^{(n)}(a)}{n!}(x-a)^n$

$\sum_{n=0}^infty \frac{f^{(n)}(0)}{n!}x^n$

$1+5x+\frac{25}{2}x^2+\frac{125}{6}x^3$