Once you have found the infinite Taylor Series, you will often be asked to approximate the original function with a finite number of terms from the Taylor Series. Doing so introduces error since the finite Taylor Series does not exactly represent the original function. To handle this error we write the function like this.

\(\displaystyle{ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . . + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) }\)

where \(R_n(x)\) is the remainder. Notice we are cutting off the series after the n-th derivative and \(R_n(x)\) represents the rest of the series.

Lagrange's formula for this remainder term is

\(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} }\)

This looks very similar to the equation for the Taylor series terms . . . and it is, except for one important item. Notice that in the numerator, we evaluate the \(n+1\) derivative at \(z\) instead of \(a\). So, what is the value of \(z\)? \(z\) takes on a value between \(a\) and \(x\), but, and here's the key, we don't know exactly what that value is. So this remainder can never be calculated exactly. However, since we know that \(z\) is between \(a\) and \(x\), we can determine an upper bound on the remainder and be confident that the remainder will never exceed this upper bound. So how do we do that?

We usually consider the absolute value of the remainder term \(R_n\) and call it the upper bound on the error, also called Taylor's Inequality.

\(\displaystyle{ \abs{R_n(x)} \leq \frac{\abs{f^{(n+1)}(z)(x-a)^{n+1}} }{(n+1)!} }\)

Choosing z
To find an upper bound on this error, we choose the value of \(z\) using these rules.
- If the \(n+1\) derivative contains a sine or cosine term, we replace the sine or cosine term with one, since the maximum value of sine or cosine is one. This seems somewhat arbitrary but most calculus books do this even though this could give a much larger upper bound than could be calculated using the next rule. [ As usual, check with your instructor to see what they expect. ]
- If we do not have a sine or cosine term, we calculate \(\abs{f^{(n+1)}(z)}\) and then choose the value of \(z\) between \(a\) and the \(x\)-value that we are estimating that makes this term a maximum. Sometimes, we need to find the critical points and find the one that is a maximum. Since we have a closed interval, either \([a,x]\) or \([x,a]\), we also have to consider the end points. Many times, the maximum will occur at one of the end points, but not always.

Okay, so what is the point of calculating the error bound? The point is that once we have calculated an upper bound on the error, we know that at all points in the interval of convergence, the truncated Taylor series will always be within \(\abs{R_n(x)}\) of the original function \(f(x)\). This \(\abs{R_n(x)}\) is a mathematical 'nearness' number that we can use to determine the number of terms we need to have for a Taylor series.

Here is a great video clip explaining the remainder and error bound on a Taylor series.

Instructions: For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers. However, for these problems, use the techniques above for choosing z, unless otherwise instructed. Give all answers in exact form, if possible.