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Taylor Series

Through the following posts, we will be looking at different series.  Our goal will be to determine if the series converges or diverges.  Furthermore, if the series converges and we can find the actual value, we will.  These posts are meant to help as you study series in Calculus 2.  As you work through these problems with us, we hope you will make use of the Series Convergence Test Decision Tree that we have made available.  You can find more series and other calculus study help posts at Calculus study help. Furthermore, there are videos meant to accompany many of these posts on YouTube.

Taylor Series

Here we will be finding the Taylor Series for the function \(f(x)=x^{-1}\) centered at the point \(x=1\).

First we need to remember that the Taylor Series of a function centered at \(x_{0}\) is given by
\begin{align*}
\sum_{k=0}^{\infty}\frac{f^{(k)}(x_{0})*(x-x_{0})^{k}}{k!}.
\end{align*}
Therefore, we will need to find the general derivative at the point \(x_{0}=1\). While doing so, I will make a table so that we can see if an patterns emerge.

\(k\) \(f^{(k)}(x)\) \(f^{(k)}(1)\)
0 \(x^{-1}\) \(1\)
1 \(-x^{-2}\) \(-1\)
2 \(2x^{-3}\) \(2\)
3 \(-3*2x^{-4}\) \(-3*2\)
4 \(4*3*2x^{-5}\) \(4*3*2\)

Now that we’ve created a table for our first few derivatives, hopefully we can determine what the pattern will be. For another example of determining patterns and defining rules for sequences, visit the post Sets: Finding elements or defining rules. When working with this example, we break the problem into two parts.

  • We first look at the negative values of each derivative.
    • The terms are alternating between positive and negative.
    • We get something of the form \((-1)^{k}\).
    • Check that this lines up properly.
    • \((-1)^{0}=1\), \((-1)^{1}=-1\), \((-1)^{2}=1\), and so on. Note that the positive and negatives line up with the table we made above, so we leave the term as \((-1)^{k}\).
  • We next look at the product of positive numbers.
    • The numbers look like \(1, 1*2, 1*2*3, 1*2*3*4\).
    • This tells us that something like \(k!\) will be in the terms.
    • Again, we make sure that these line up. Note that for \(k=0\), we have \(1\), for \(k=1\) we have \(1\), for \(k=2\) we have \(2*1\).
    • Therefore, we will have exactly \(k!\).

Combining these together, we now find that \(f^{(k)}(1)=(-1)^{k}k!\). We then get that the Taylor Series will be
\begin{align*}
\sum_{k=0}^{\infty}\frac{(-1)^{k}k!(x-1)^{k}}{k!}=\sum_{k=0}^{\infty}(-1)^{k}(x-1)^{k}.
\end{align*}

Interval and Radius of Convergence

We now need to determine where this series will converge. In order to do this, we will again go through our decision tree to find an appropriate test to use. Here, we need to be able to do these tests with an arbitrary \(x\).

We will again start with the first question in our decision tree,

  • Is \(\lim\limits_{k \to \infty}a_{k}=0\)?
    In this case, the answer would depend on our choice of \(x\). If \(|x-1| \geq 1\), then this is not true, so the series will diverge by the divergence test. However, this limit will be zero for \(-1 < x < 1\). We now continue our process assuming \(|x-1| <1\).
  • Is this an alternating series?
    Note that this could be an alternating series if \(1< x < 2\). In these cases, we will get a convergent series, but we would need to start over looking at the absolute value to determine if the series was absolutely or conditionally convergent. We, therefore, continue assuming we do not have an alternating series.
  • Does the series look like a geometric series?
    In this case, it does look like we multiply by something in order to find each new term.
  • Can you rewrite the terms so that they look exactly like \(ar^{k}\)?
    Yes, we already have the series in this form.
  • Is \(|r| <1\)?
    Since we have already stated that if \(|x-1| \geq 1\), we have a divergent series, we will have that \(|r|=|(-1)(x-1)|=|x-1| <1\).
  • Therefore, the series will converge for \(|x-1|<1\).

Because we have that the Taylor Series will converge if and only if \(|x| <1\), we get that the radius of convergence is \(1\). Furthermore, the interval of convergence is interval that satisfies
\begin{align*}
|x-1| &< 1 \\
-1 < x-1 &< 1 \\
0< x &< 2 .
\end{align*}
This is just the interval \((0,2)\), hence this is the interval of convergence.

In general it is worth taking a moment to consider the endpoints of the interval. However, we have already considered these points since the \(|r|=1\) implies a series will diverge for a geometric series.

Approximating \(\frac{1}{1.1}\)

Now that we have a Taylor Series for \(\frac{1}{x}\), we can approximate values of the function using Taylor polynomials of different degrees. Here we will approximate \(f(1.1)=\frac{1}{1.1}\) with a polynomial of degree 2. We can then find that
\begin{align*}
\frac{1}{1.1} & \approx 1-(1.1-1)+(1.1-1)^{2} \\
&=1-.1+.01 \\
&=.91.
\end{align*}

Conclusion

We were able to find the Taylor Series for the function \(\frac{1}{x}\) centered at \(x=0\). In order to do so, we constructed the pieces of the series, one by one, and determined any patterns that emerged that would allow us to describe what the terms looked like in general. Once we had the Taylor Series, we used our convergence tests to determine the interval and radius of convergence. Since we knew that \(1.1\) was in this interval, it meant we could use Taylor polynomial to approximate \(\frac{1}{1.1}\).

Make sure to continue working through the problems presented in the other posts so that you can work on more types of series. Also, if you find these helpful, please like the posts and share them with anyone else that may be studying series.

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