A long time ago, I wrote about differentials. That estimates the function as a line at a specific point.

**Taylor polynomial**.

**Theorem:**If f has n derivatives at c, then the polynomial $$P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + ... + \frac{f^{(n)}(c)}{n!}(x-c)^n$$ is the nth Taylor polynomial for f at c. Recall that this is an n-degree polynomial approximation for f(x).

**Maclaurin polynomials**come up a lot, and those are just Taylor polynomials centered at the origin, so c = 0.

## Example: Estimating e

# Remainder of a Taylor Polynomial

A taylor polynomial is just an approximation, so it isn't equal to f(x), but it's close. So, you can say this:

There's a theorem that gives a general procedure for estimating the remainder. This is called **Taylor's Theorem**, and the remainder is called the **Lagrange Error Bound. **

## Example- Approximating to a Desired Accuracy

$$\text{We know } R_n = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$$ So, we need to figure out the (n+1)st derivative $$f'(x) = x^{-1}, f''(x) = -x^{-2}, f'''(x) = 2x^{-3}, f^{(4)} = -6x^{-24}$$ So, there's an evident pattern: $$f^{(n)}(x) = (-1)^{n - 1} * \frac{(n-1)!}{x^n}$$ $$f^{(n+1)}(x) + -1^{n} * \frac{n!}{x^{n+1}}$$ We can plug that in $$|R_n| = |\frac{n!}{x^{n+1}}\times\frac{1}{(n+1)!}\times(1.2-1)^{n+1}|$$ $$= \frac{(0.2)^{n+1}}{z^{n+1}(n+1)}$$ z > 1, so that's less than $$\frac{(0.2)^{n+1}}{n+1} < 0.001$$ Plug in values for n, and the smallest value that works is n = 3.

**Note: I didn't use the equation involving the max (n+1)st derivative. Sometimes that's not necessary**

David Witten