Radius and Interval of Convergence

Once the Taylor series or power series is calculated, we use the ratio test to determine the radius convergence and other tests to determine the interval of convergence. Without knowing the radius and interval of convergence, the series is not considered a complete function (This is similar to not knowing the domain of a function. As we say in the page on domain and range, a fully defined function always contains information about the domain, either implicitly or explicitly stated.) In this case, we can't leave off information about where the series converges because the series may not hold for all values of \(x\).

The ratio test looks like this. We have a series \( \sum{a_n} \) with non-zero terms. We calculate the limit \( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} = L} \).
There are three possible cases for the value \(L\).
\( L < 1 \): The series converges absolutely.
\( L = 1 \): The ratio test is inconclusive.
\( L > 1 \): The series diverges.

So we use the first case ( \(L<1\), since we want convergence ) and we set up the inequality \( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} < 1} ~~~~~ [ 1 ]\)

Key: Do not drop the absolute values.

To find the radius of convergence, we need to simplify the inequality [1] to the point that we have \( \left| x-a \right| < R \). This gives the radius of convergence as \(R\). This seems very simple but you need to be careful of the notation and wording your textbooks. Some textbooks use a small \(r\). Some textbooks ask for the ratio of convergence in which case you need to give the answer as \( \rho = 1/R \). This has always seemed kind of strange to me but there must be some reason for it. However, on this site, whenever we talk about the radius of convergence in this context, we will use \(R\) as defined above.

There are three possible cases for the radius of convergence.

The interesting thing is that we have a strict inequality in \( 0 < R < \infty \) and, because of the definition of the ratio test, we have no idea what happens when \( \left| x-a \right| = R \). The series could converge or diverge. The ratio test doesn't give us a clue on what happens in that case. That's where we need to find the interval of convergence, which we discuss next.

We use the radius of convergence, \(R\), to calculate the interval of convergence as follows
\( \begin{array}{rcccl} & & \left| x-a \right| & < & R \\ -R & < & x-a & < & R \\ -R + a & < & x & < & R + a \end{array} \)
So now we have an open interval \( (-R+a, R+a) \) in which the series converges. Now we need to look at the endpoints to determine what goes on. To do that we substitute each endpoint individually for \(x\) into the series and then use the other series test to determine convergence or divergence.

Notice when we substitute \(x=-R+a\) into the \((x-a)^n\) term, we end up with \((-R)^n\) which can be simplified to \((-1)^n R^n\). Now we have an alternating series. So often, the alternating series test can be used to determine convergence or divergence. The point is that, using other tests, we need to definitively determine convergence or divergence at each endpoint. The result will be an open interval, a half-open interval or a closed interval. We call this interval, the interval of convergence.

Notice the difference between the terms radius of convergence and interval of convergence. The radius of convergence gives information about the open interval but says nothing about the endpoints. The interval of convergence includes the radius of convergence AND information about convergence or divergence of the endpoints.

Remainder and Error Bounds

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.