# How do you find the Remainder term in Taylor Series?

There are lots of different ways this can be thought about, though probably the simplest and most common way is to use the inequality |R_{k}(x)|\leq \frac{M\cdot r^{k+1}}{(k+1)!}, where the $\left(k + 1\right)$st derivative ${f}^{\left(k + 1\right)} \left(x\right)$ of $f$ satisfies $| {f}^{\left(k + 1\right)} \left(x\right) | \setminus \le q M$ over the interval $\left[c - r , c + r\right]$ (assuming sufficient differentiability/smoothness of $f$ over the interval).
The function ${R}_{k} \left(x\right)$ is the "remainder term" and is defined to be ${R}_{k} \left(x\right) = f \left(x\right) - {P}_{k} \left(x\right)$, where ${P}_{k} \left(x\right)$ is the $k$th degree Taylor polynomial of $f$ centered at $x = a$: P_{k}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots.
For example, if $f \left(x\right) = {e}^{x}$, $a = 0$, and $k = 4$, we get ${P}_{4} \left(x\right) = 1 + x + \setminus \frac{{x}^{2}}{2} + \setminus \frac{{x}^{3}}{6} + \setminus \frac{{x}^{4}}{24}$. Moreover, if we consider this over the interval $\left[- 2 , 2\right]$ so that $r = 2$, then the inequality above becomes |R_{4}(x)|\leq \frac{e^{2}\cdot 2^{5}}{5!}=\frac{4}{15}e^{2}\approx 1.97. So we can expect ${P}_{4} \left(x\right)$ to approximate ${e}^{x}$ to within this amount over the interval $\left[- 2 , 2\right]$ (actually, the approximation is even better than the error bound gives...the point is that the error bound gives a guarantee ).