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Radius and Interval of Convergence for Power and Taylor Series |
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In order to have a complete picture of what is going on with a power series (and Taylor series, since Taylor series is a special case of power series), we need to know the radius and interval of convergence. This is analogous to knowing the domain of a function. |
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Difference Between Radius and Interval of Convergence |
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The difference between the radius of convergence and the interval of convergence lies in what information we have about the endpoints.
radius of convergence | interval of convergence | |
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gives convergence information only between x-values | gives convergence information inside the interval AND convergence/divergence information at the endpoints | |
if \((5,7)\) is the radius of convergence | if \((5,7)\) is the interval of convergence | |
the series converges for \(5 < x < 7\) | the series converges for \(5 < x < 7\) | |
says nothing about the endpoints | AND we know that the series diverges at \(x = 5\) and \(x = 7\) | |
contains the same information as the radius of convergence as well as what is going on at the endpoints |
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 \(x\).
Using The Ratio Test |
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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.
Radius of Convergence |
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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.
\(R = 0\) | series converges only at the point \(x = a\) |
\( 0 < R < \infty \) | series converges in the interval |
\(R = \infty \) | series converges for all \(x\) |
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.
Interval of Convergence |
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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.
Here is a video clip that explains how to show that a series converges for all x.
Dr Chris Tisdell - What is a Taylor series? | |
Sometimes you are given the interval of convergence and asked to determine the radius convergence. Here is a short video explaining how to do that.
PatrickJMT - Radius of Convergence for a Power Series | |
Search 17Calculus
Practice Problems |
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Instructions - - Unless otherwise instructed, determine the radius and interval of convergence for each series.
Level A - Basic |
Practice A01 | |
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\(\displaystyle{\sum_{n=0}^{\infty}{\frac{x^n}{n!}}}\) | |
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solution |
Practice A02 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}(x-5)^n}}\) | |
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solution |
Practice A03 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{nx^n}}\) | |
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solution |
Practice A04 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{x^n}{\sqrt{n}}}}\) | |
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solution |
Practice A05 | |
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\(\displaystyle{e^{x^2}=\sum_{n=0}^{\infty}{\frac{x^{2n}}{n!}}}\) | |
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solution |
Practice A06 | |
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\(\displaystyle{\sum_{n=0}^{\infty}{n^3(x-5)^n}}\) | |
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solution |
Practice A07 | |
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\(\displaystyle{\sum{\left(\frac{x}{2}\right)^n}}\) | |
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solution |
Practice A08 | |
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\(\displaystyle{\sum{\frac{x^n}{n~2^n}}}\) | |
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solution |
Practice A09 | |
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\(\displaystyle{\sum{\frac{x^{2n}}{(2n)!}}}\) | |
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solution |
Practice A10 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{(-1)^nnx^n}}\) | |
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solution |
Practice A11 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^nx^n}{n}}}\) | |
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solution |
Practice A12 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+1)^{2n}}{n!}}}\) | |
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solution |
Practice A13 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{5^n x^n}{n!} } }\) | |
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solution |
Practice A14 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{\frac{x^n 2^n}{n!}} }\) | |
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solution |
Practice A15 | |
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Maclaurin series for \(\cos(x^2)\) | |
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solution |
Level B - Intermediate |
Practice B01 | |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(x-2)^n}{(n+1)3^n}}}\) | |
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solution |
Practice B02 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-4)^n}{\sqrt[3]{n}} } }\) | |
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solution |
Practice B03 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3(x+5)^n}{6^n}} }\) | |
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solution |
Practice B04 | |
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\( \displaystyle{\sum_{n=0}^{\infty}{ \frac{(x+2)^n}{n+3}}}\) | |
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solution |
Practice B05 | |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{x^n 5^n}{n^n} } }\) | |
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solution |
Practice B06 | |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{(x-2)^n}{\ln(n+4)} }}\) | |
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solution |
Practice B07 | |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{(x+3)^n }{n^2+2n } }}\) | |
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solution |
Practice B08 | |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n (x+1)^n }{ 5^n \sqrt{n} } } }\) | |
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solution |
Practice B09 | |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+2)^n }{ (n+1)\ln(n+1) } } }\) | |
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solution |
