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Power Series 
Difference Between Power Series and Taylor Series
Taylor series are a specific case of Power series where the constants ( usually functions of n ) are related to the derivative of the function. Many of the same techniques that work for one will work for the other.
A Power Series is based on the Geometric Series using the equation
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1r} }\) which converges for \( r < 1 \), where r is a function of x. We can also use the ratio test and other tests to determine the radius and interval of convergence.
Power Series are discussed on this page.
A Taylor Series is based on the recurring formula
\(\displaystyle{
\sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(xa)^n} = f(a) + \frac{f'(a)}{1!}(xa) + \frac{f''(a)}{2!}(xa)^2 + . . .
}\)
We use the Ratio Test to determine the radius of convergence.
Taylor Series are discussed on a separate page.

The idea with power series is to use the fact that with a geometric series, we know it converges under certain circumstances AND we know what it converges to. So what we do is use algebra and, sometimes calculus, to write a function in the form of a geometric power series. 


The geometric series equation is \(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1r} }\). We will have equations where r is a function of x.
We then match the values of a and r to get the series form of the function.
We also need to specify the values of r for which this series converges, i.e. \( \abs{r} < 1 \).
Why would we do this? Well, one reason is that we can convert a somewhat more complicated function into a polynomial (which happens to be infinite) and it is much easier (almost trivial) to find derivatives and integrals of polynomials.
Sometimes, you can just use algebra, to get the equation in the right form. But other times, you need to do a little more than basic algebra. If you can either take the derivative or integrate the original function to get something in the form \(\displaystyle{ \frac{a}{1r} }\), then you can do the opposite (integrate or take the derivative) of the sum to get the original function. I'm sure by now you need an example to get your head around this. Let's watch some videos that explain these ideas in more detail with examples.
Here are several very good videos introducing and explaining power series. It will help you to watch all of them, in this order, so that you can get several different perspectives on the same topic.
This first video introduces power series and its relationship to Taylor/Maclaurin series. He also uses the example \(\displaystyle{ \sum_{ n=0 }^{\infty}{ \frac{1}{2^n}x^n } }\) to talk about under what conditions this series converges. If you have time to watch only one video, this is the one to watch.
 Dr Chris Tisdell  What is a power series? 

This next video clip is more theoretical than the previous one but it will help you understand why power series work.
 MIT OCW  Lec 38  MIT 18.01 Single Variable Calculus, Fall 2007 

The next video is very practical, giving you the basics and showing you how to work with power series.
 PatrickJMT  Power Series Representation of Functions 

Manipulating Power Series 
Okay, now that you know how to set up power series, you can do all kinds of manipulations of them to find power series of other functions. In this video, he discusses how to manipulate power series, by taking derivatives and integrating, in order to find power series representation of functions that are not already in geometric series form.
 Dr Chris Tisdell  What is a power series? 

Finding a Power Series Using The Binomial Series
The binomial series can be used to expand a special class of functions into power series.
If k is any real number and \( x < 1 \), then
\(\displaystyle{ (1+x)^k = 1 + kx + \frac{k(k1)}{2!} x^2 + \frac{k(k1)(k2)}{3!}x^3 + . . .
= \sum_{n=0}^{\infty}{ \left(
\begin{array}{c}
k \\
n
\end{array}
\right) x^n }
}\)
where \(\displaystyle{
\left(
\begin{array}{c}
k \\
n
\end{array}
\right)
= \frac{k!}{n!(kn)!}
}\)
You can also use Taylor Series Expansion to expand these kinds of functions. There are a couple of practice problems below that use this series.
Radius and Interval of Convergence
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 nonzero 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
\( \left xa \right < R \). 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 \( \left xa \right = R \). 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
\(
\begin{array}{rcccl}
& & \left xa \right & < & R \\
R & < & xa & < & R \\
R + a & < & x & < & R + a
\end{array}
\)
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 \((xa)^n\) 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 halfopen 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.
See the radius and interval of convergence page page for more detail, videos and practice problems.
List of Common Power Series
This panel contains a list of common power series. These are often used to build other power series. See the practice problems below for examples. You can find more discussion on the Taylor Series page.
function 
power series 
convergence interval 

\(\displaystyle{ \frac{1}{1x} }\) 
\(\displaystyle{ \sum_ {n=0}^{\infty}{x^n} }\) 

\( 1 < x < 1 \) 
power series 
\(e^x\) 
\(\displaystyle{ \sum_{n=0}^{x^n}{\frac{x^n}{n!}} }\) 
\(\displaystyle{ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots }\) 
\(\infty < x < \infty\) 
Maclaurin series 
\(\ln(x)\) 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(1)^{n+1} (x1)^n}{n} } }\) 
\(\displaystyle{ (x1)  \frac{(x1)^2}{2} + \frac{(x1)^3}{3}  \frac{(x1)^4}{4} + \cdots }\) 

Taylor series about \( x = 1 \) 
\(\sin(x)\) 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(1)^n x^{2n+1}}{(2n+1)!} } }\) 
\(\displaystyle{ x  \frac{x^3}{3!} + \frac{x^5}{5!}  \frac{x^7}{7!} + \cdots }\) 
\(\infty < x < \infty\) 
Maclaurin series 
\(\cos(x)\) 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(1)^n x^{2n}}{(2n)!} } }\) 
\(\displaystyle{ 1  \frac{x^2}{2!} + \frac{x^4}{4!}  \frac{x^6}{6!} + \cdots }\) 
\(\infty < x < \infty\) 
Maclaurin series 
Instructions   Unless otherwise instructed,
 if you are given a power series, determine the function that the sum represents. Assume that the values of x are such that the series converges.
 if you are given a function, build the power series of the function at the given point (if no point is given, use \(x=0\)), and determine the radius of convergence.
Practice A01 
\(\displaystyle{g(x)=\frac{1}{1x^3}}\) 


\(\displaystyle{\frac{1}{1x^3}=\sum_{n=0}^{\infty}{x^{3n}}~~~~~R=1}\) 
Below the video is a detail solution of this practice problem.
This is already in the form of a geometric series.
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1r} }\).
We want \(\displaystyle{ \frac{a}{1r} = \frac{1}{1x^3} }\).
By direct comparison, we know \( a=1 \) and \( r = x^3 \).
Plugging these into the geometric series, we have
\(\displaystyle{ \sum_{n=0}^{\infty}{(1)(x^3)^n} = \sum_{n=0}^{\infty}{x^{3n}} }\)
Now we need to determine the range of xvalues for which this series converges, called the radius of convergence \(R\). We know from the geometric series that as long as \( \abs{r}<1 \) the series will converge. In this problem,
\( \abs{r} = \abs{x^3} < 1 \) and so \( \abs{x} < 1 \).
Practice A01 Final Answer 
\(\displaystyle{\frac{1}{1x^3}=\sum_{n=0}^{\infty}{x^{3n}}~~~~~R=1}\) 
Practice A02 
\(\displaystyle{h(x)=\frac{1}{x6}}\) 


\(\displaystyle{ \frac{1}{x6} = \frac{1}{6} \sum_{n=0}^{\infty}{\frac{x^n}{6^n}}~~~~~R = 6 }\) 
In order to determine the power series of this function, we use the geometric power series
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1r} }\).
We need to determine a and r. Looking at the equation \(\displaystyle{ \frac{a}{1r} = \frac{1}{x6} }\)
we need to rewrite the denominator \( x6 \) to make it look like \(1r\).
\( x6 = 6+x = (6)(1x/6)\)
So now we have \(\displaystyle{ \frac{1}{x6} = \frac{1}{(6)(1x/6)} = \frac{1/6}{1x/6} }\)
When we compare this last expression to the geometric series we get
\(\displaystyle{ \frac{1/6}{1x/6} = \frac{a}{1r} }\) where \(a=1/6\) and \(r=x/6\). Now we can write the sum.
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \sum_{n=0}^{\infty}{(1/6)(x/6)^n} = \frac{1}{6} \sum_{n=0}^{\infty}{\frac{x^n}{6^n}} }\)
Next, we need to determine the radius of convergence, \( R \) . By the definition of the geometric series, \(\abs{r} < 1\) for convergence. In our case, \(\abs{r} = \abs{x/6} < 1 ~~\to~~ \abs{x} < 6\)
Practice A02 Final Answer 
\(\displaystyle{ \frac{1}{x6} = \frac{1}{6} \sum_{n=0}^{\infty}{\frac{x^n}{6^n}}~~~~~R = 6 }\) 
Practice A03 
\(\displaystyle{\frac{6x^2}{7+15x}}\) 


\(\displaystyle{\frac{6x^2}{7+15x}=\sum_{n=0}^{\infty}{\frac{(15)^n 6 x^{n+2}}{7^{n+1}}}~~~~~R=7/15}\) 
First, we need to get the function in the form \(\displaystyle{ \frac{a}{1r} }\).
\(\displaystyle{ \frac{6x^2}{7+15x} = \frac{6x^2}{7(1(15x/7))} = \frac{6x^2/7}{1(15x/7)} }\)
Comparing the last expression to \(\displaystyle{\frac{a}{1r} }\) we have \(a=6x^2/7\) and \(r=15x/7\). Now we can write the sum
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \sum_{n=0}^{\infty}{\frac{6x^2}{7} \left( \frac{15x}{7} \right)^n} = \sum_{n=0}^{\infty}{\frac{(15)^n 6 x^{n+2}}{7^{n+1}}} }\)
Next, we need to determine the radius of convergence. This is given by the geometric series definition as \(\abs{r} < 1\).
In our case, \(\abs{r}=\abs{15x/7} < 1 ~~\to~~ \abs{x} < 7/15\)
Practice A03 Final Answer 
\(\displaystyle{\frac{6x^2}{7+15x}=\sum_{n=0}^{\infty}{\frac{(15)^n 6 x^{n+2}}{7^{n+1}}}~~~~~R=7/15}\) 
Practice A04 
\(\displaystyle{f(x)=\frac{x^5}{8+x^2}}\) 


\(\displaystyle{ \frac{x^5}{8+x^2} = \sum_{n=0}^{\infty}{ \frac{(1)^nx^{2n+5}}{8^{n+1}}}~~~~~R = 2 \sqrt{2} }\) 
Our goal here is to get \(f(x)\) in the form \(\displaystyle{\sum_{n=0}^{\infty}{ar^n}=\frac{a}{1r}}\). We also require \(\abs{r}< 1\) for series convergence.
\(\displaystyle{\frac{x^5}{8+x^2}=\frac{x^5}{8[1(x^2/8)]}=\frac{x^5/8}{1(x^2/8)}}\)
Comparing this last equation to \( \displaystyle{ \frac{a}{1r} }\) we have \(\displaystyle{ r = \frac{x^2}{8} }\) and \(\displaystyle{a = \frac{x^5}{8}}\) giving us
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{\frac{x^5}{8} \left( \frac{x^2}{8} \right)^n } = \sum_{n=0}^{\infty}{\frac{(1)^nx^{2n+5}}{8^{n+1}}} }\)
Now we need to find the radius of convergence.
\( \displaystyle{
\begin{array}{rcl}
\abs{r} & < & 1 \\
\abs{ \frac{x^2}{8} } & < & 1 \\
x^2 & < & 8 \\
\abs{x} & < & \sqrt{8} = 2 \sqrt{2}
\end{array}
}\)
Practice A04 Final Answer 
\(\displaystyle{ \frac{x^5}{8+x^2} = \sum_{n=0}^{\infty}{ \frac{(1)^nx^{2n+5}}{8^{n+1}}}~~~~~R = 2 \sqrt{2} }\) 
Practice A05 
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{x^{n+2}}{n!}}}\) 


\(\displaystyle{\sum_{n=0}^{\infty}{\frac{x^{n+2}}{n!}}=x^2e^x}\) 
Practice A05 Final Answer 
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{x^{n+2}}{n!}}=x^2e^x}\) 
Practice A06 
\(\displaystyle{\sum_{n=2}^{\infty}{x^n}}\) 


\(\displaystyle{\sum_{n=2}^{\infty}{x^n}=\frac{1}{1x}1x}\) 
Practice A06 Final Answer 
\(\displaystyle{\sum_{n=2}^{\infty}{x^n}=\frac{1}{1x}1x}\) 
Practice A07 
\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{x^n}{n!}+x^n\right]}}\) 


\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{x^n}{n!}+x^n\right]}=e^x+\frac{1}{1x}}\) 
Practice A07 Final Answer 
\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{x^n}{n!}+x^n\right]}=e^x+\frac{1}{1x}}\) 
Practice A08 
\(\displaystyle{\sum_{n=1}^{\infty}{x^{n+1}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{x^{n+1}}=\frac{1}{1x}}\) 
Practice A08 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{x^{n+1}}=\frac{1}{1x}}\) 
Practice A09 
\(g(x)=(x+1)e^x\) 


The power series for \(g(x)=(x+1)e^x\) about \(x=0\) is \(\displaystyle{ g(x) = \sum_{k=0}^{\infty}{ \frac{(x+1)x^k}{k!} } }\)
for \(1 < x < 1\) 
Practice A09 Final Answer 
The power series for \(g(x)=(x+1)e^x\) about \(x=0\) is \(\displaystyle{ g(x) = \sum_{k=0}^{\infty}{ \frac{(x+1)x^k}{k!} } }\)
for \(1 < x < 1\) 
Practice A10 
\(\displaystyle{\frac{1}{1+x^2}}\) 


The power series for \(\displaystyle{\frac{1}{1+x^2} }\) about \(x=0\) is \(\displaystyle{\sum_{k=0}^{\infty}{ (1)^k x^{2k}}}\)
for \(1 < x < 1\) 
Practice A10 Final Answer 
The power series for \(\displaystyle{\frac{1}{1+x^2} }\) about \(x=0\) is \(\displaystyle{\sum_{k=0}^{\infty}{ (1)^k x^{2k}}}\)
for \(1 < x < 1\) 
Practice A11 
\(\displaystyle{f(x)=x^2\sin(x^3)}\) 


The power series for \(\displaystyle{f(x)=x^2\sin(x^3)}\) about \(x=0\) is \(\displaystyle{f(x)=\sum_{k=0}^{\infty}{\frac{(1)^k x^{6k+5}}{(2k+1)!}}}\) for \(1 < x < 1\) 
There is more to this problem than you were asked to do in the problem statement. Here is what he asks for in the video.
Suppose that \(f(x)=x^2 \sin(x^3)\).
a) By using the Maclaurin series for sine, find the Maclaurin series for f.
b) Hence show that zero is a stationary point for f.
c) Is zero a local maximum point, local minimum point or a horizontal point of inflection? Explain.
Practice A11 Final Answer 
The power series for \(\displaystyle{f(x)=x^2\sin(x^3)}\) about \(x=0\) is \(\displaystyle{f(x)=\sum_{k=0}^{\infty}{\frac{(1)^k x^{6k+5}}{(2k+1)!}}}\) for \(1 < x < 1\) 
Practice A12 
\(\displaystyle{\frac{1}{1+x}}\) 


The power series for \(\displaystyle{\frac{1}{1+x}}\) about \(x=0\) is \(\displaystyle{g(x)=\sum_{k=0}^{\infty}{(x)^k}}\) for \(1 < x < 1\) 
Practice A12 Final Answer 
The power series for \(\displaystyle{\frac{1}{1+x}}\) about \(x=0\) is \(\displaystyle{g(x)=\sum_{k=0}^{\infty}{(x)^k}}\) for \(1 < x < 1\) 
Practice A13 
\(\displaystyle{\frac{1}{1+9x^2}}\) 


The power series for \(\displaystyle{\frac{1}{1+9x^2}}\) about \(x=0\) is \(\displaystyle{\sum_{k=0}^{\infty}{(9x^2)^k}}\) for \(1/3 < x < 1/3\) 
Practice A13 Final Answer 
The power series for \(\displaystyle{\frac{1}{1+9x^2}}\) about \(x=0\) is \(\displaystyle{\sum_{k=0}^{\infty}{(9x^2)^k}}\) for \(1/3 < x < 1/3\) 
Practice A14 
\(\displaystyle{\frac{x}{4x+1}}\) 


The power series for \(\displaystyle{\frac{x}{4x+1}}\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{(4)^n x^{n+1}}}\)
for \(1/2 < x < 1/2\) 
Practice A14 Final Answer 
The power series for \(\displaystyle{\frac{x}{4x+1}}\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{(4)^n x^{n+1}}}\)
for \(1/2 < x < 1/2\) 
Practice A15 
\(\displaystyle{\frac{x}{9+x^2}}\) 


The power series for \(\displaystyle{\frac{x}{9+x^2}}\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{\frac{(1)^nx^{2n+1}}{9^{n+1}}}}\) for \(3 < x < 3\) 
Practice A15 Final Answer 
The power series for \(\displaystyle{\frac{x}{9+x^2}}\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{\frac{(1)^nx^{2n+1}}{9^{n+1}}}}\) for \(3 < x < 3\) 
Practice A16 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^{n+1}x^{2n}}{(2n1)!}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^{n+1}x^{2n}}{(2n1)!}}=x\sin(x)}\) 
In this problem, he integrates the series for \(\cos(x)\) to get the series for \(\sin(x)\). This is not necessary unless you are told to start with \(\cos(x)\) to get the answer.
Practice A16 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^{n+1}x^{2n}}{(2n1)!}}=x\sin(x)}\) 
Practice A17 
\(x\cos(x)\sin(x)\) 


\(\displaystyle{x\cos(x)\sin(x)=\left(\frac{x^3}{3!}\frac{x^3}{2!}\right)+\left(\frac{x^5}{4!}\frac{x^5}{5!}\right)+\left(\frac{x^7}{7!}\frac{x^7}{6!}\right)+\left(\frac{x^9}{8!}\frac{x^9}{9!}\right)+\cdots}\) 
In this problem, he integrates the series for \(\sin(x)\) to get the series for \(\cos(x)\). This is not necessary unless you are told to start with \(\sin(x)\) to get the answer.
Practice A17 Final Answer 
\(\displaystyle{x\cos(x)\sin(x)=\left(\frac{x^3}{3!}\frac{x^3}{2!}\right)+\left(\frac{x^5}{4!}\frac{x^5}{5!}\right)+\left(\frac{x^7}{7!}\frac{x^7}{6!}\right)+\left(\frac{x^9}{8!}\frac{x^9}{9!}\right)+\cdots}\) 
Practice A18 
\(\displaystyle{f(x)=x^2e^{4x}}\) 


\(\displaystyle{f(x)=x^2e^{4x}=\sum_{n=0}^{\infty}{\frac{4^nx^{n+2}}{n!}}}\) 
Practice A18 Final Answer 
\(\displaystyle{f(x)=x^2e^{4x}=\sum_{n=0}^{\infty}{\frac{4^nx^{n+2}}{n!}}}\) 
Practice A19 
For \(\displaystyle{f(x)=\sum_{n=0}^{\infty}{(5x1)^n}}\), evaluate\(\int{f(x)~dx}\) 


\(\displaystyle{\int{f(x)~dx}=C+\sum_{n=0}^{\infty}{\frac{(5x1)^{n+1}}{5(n+1)}}}\) 
Practice A19 Final Answer 
\(\displaystyle{\int{f(x)~dx}=C+\sum_{n=0}^{\infty}{\frac{(5x1)^{n+1}}{5(n+1)}}}\) 
Practice A20 
\(\displaystyle{f(x)=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}}\), find \(\displaystyle{h(x)=x^3\int_{0}^{x}{f(t)~dt}}\) 


\(\displaystyle{h(x)=\sum_{n=0}^{\infty}{\frac{x^{n+4}}{(n+1)!}}}\) 
Practice A20 Final Answer 
\(\displaystyle{h(x)=\sum_{n=0}^{\infty}{\frac{x^{n+4}}{(n+1)!}}}\) 
Practice A21 
\(\displaystyle{\frac{2}{x+4}}\) 


\(\displaystyle{\frac{2}{x+4}=\sum_{n=0}^{\infty}{\frac{(1)^n}{2}\left(\frac{x}{4}\right)^n}}\) 
Practice A21 Final Answer 
\(\displaystyle{\frac{2}{x+4}=\sum_{n=0}^{\infty}{\frac{(1)^n}{2}\left(\frac{x}{4}\right)^n}}\) 
Practice A22 
\(\displaystyle{\cos(x^2)}\) 


\(\displaystyle{\cos(x^2)=\sum_{n=0}^{\infty}{\frac{(1)^nx^{4n}}{(2n)!}}}\) 
Practice A22 Final Answer 
\(\displaystyle{\cos(x^2)=\sum_{n=0}^{\infty}{\frac{(1)^nx^{4n}}{(2n)!}}}\) 
Practice B01 
\(\ln(1+x)\) 


The power series for \(\ln(1+x)\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{\frac{(1)^n x^{n+1}}{n+1}}; ~~~ 1 < x < 1}\) 
Practice B01 Final Answer 
The power series for \(\ln(1+x)\) about \(x=0\) is \(\displaystyle{\sum_{n=0}^{\infty}{\frac{(1)^n x^{n+1}}{n+1}}; ~~~ 1 < x < 1}\) 
Practice B02 
\(\displaystyle{g(x)=\frac{1}{(x1)^2}}\) 


The power series for \(\displaystyle{g(x) = \frac{1}{(x1)^2}}\) about \(x=0\) is \(\displaystyle{g(x) = \sum_{k=1}^{\infty}{kx^{k1}}}\) for \(1 < x < 1\) 
Practice B02 Final Answer 
The power series for \(\displaystyle{g(x) = \frac{1}{(x1)^2}}\) about \(x=0\) is \(\displaystyle{g(x) = \sum_{k=1}^{\infty}{kx^{k1}}}\) for \(1 < x < 1\) 
Practice B03 
\(\displaystyle{\ln\left(\frac{1x}{1+x}\right)}\) 


The power series for \(\displaystyle{\ln\left(\frac{1x}{1+x}\right)}\) about \(x=0\) is \(\displaystyle{2\sum_{n=0}^{\infty}{\frac{x^{2n+1}}{2n+1}}}\) for \(1 < x < 1\) 
This is a great video showing how to use integration and differentiation to get a power series for a function that initially looks nothing like a geometric series. The key here is to notice \(\displaystyle{ \frac{d}{dx}[\ln(1x)] = \frac{1}{1x} }\) and \(\displaystyle{ \frac{d}{dx}[\ln(1+x)] = \frac{1}{1+x} }\)
Now the two fractions are in the right form to be able to find the power series from the geometric series equation. The last step is to integrate the series to get the natural log equations. Notice how he takes into account the need for a constant in the integration.
Practice B03 Final Answer 
The power series for \(\displaystyle{\ln\left(\frac{1x}{1+x}\right)}\) about \(x=0\) is \(\displaystyle{2\sum_{n=0}^{\infty}{\frac{x^{2n+1}}{2n+1}}}\) for \(1 < x < 1\) 
Practice B04 
\(\displaystyle{\frac{x^2}{(1+x)^3}}\) 


\(\displaystyle{\frac{x^2}{(1+x)^3}=\frac{1}{2}\sum_{n=0}^{\infty}{(1)^{n}(n+2)(n+1)x^{n+2}}}\) 
Practice B04 Final Answer 
\(\displaystyle{\frac{x^2}{(1+x)^3}=\frac{1}{2}\sum_{n=0}^{\infty}{(1)^{n}(n+2)(n+1)x^{n+2}}}\) 
Practice B05 
\(\displaystyle{f(x)=\cos^2(x)}\) 


\(\displaystyle{f(x)=\cos^2(x)=1+\sum_{n=0}^{\infty}{\frac{(1)^n2^{2n1}x^{2n}}{(2n)!}}}\) 
Practice B05 Final Answer 
\(\displaystyle{f(x)=\cos^2(x)=1+\sum_{n=0}^{\infty}{\frac{(1)^n2^{2n1}x^{2n}}{(2n)!}}}\) 
Practice B06 
\(\displaystyle{\frac{x^2}{(12x)^2}}\) 


\(\displaystyle{\frac{x^2}{(12x)^2}=\sum_{n=0}^{\infty}{2^n(n+1)x^{n+2}}}\) 
Practice B06 Final Answer 
\(\displaystyle{\frac{x^2}{(12x)^2}=\sum_{n=0}^{\infty}{2^n(n+1)x^{n+2}}}\) 
Practice B07 
\(\displaystyle{e^{x^2}\cos(x)}\) 


\(\displaystyle{e^{x^2}\cos(x)=1\frac{3x^2}{2}+\frac{25x^4}{24}+\cdots}\) 
Practice B07 Final Answer 
\(\displaystyle{e^{x^2}\cos(x)=1\frac{3x^2}{2}+\frac{25x^4}{24}+\cdots}\) 
Practice B08 
\(\displaystyle{\frac{x}{\sin(x)}}\) 


\(\displaystyle{\frac{x}{\sin(x)}=1+\frac{x^2}{6}+\frac{7x^4}{300}+\cdots}\) 
Practice B08 Final Answer 
\(\displaystyle{\frac{x}{\sin(x)}=1+\frac{x^2}{6}+\frac{7x^4}{300}+\cdots}\) 
Practice B09 
evaluate \(\displaystyle{\int{\frac{e^{x^2}}{x}dx}}\) as an infinite series 


\(\displaystyle{\int{\frac{e^{x^2}}{x}dx}=\ln\abs{x}+\sum_{n=1}^{\infty}{\frac{x^{2n}}{n!(2n)}}+C}\) 
Practice B09 Final Answer 
\(\displaystyle{\int{\frac{e^{x^2}}{x}dx}=\ln\abs{x}+\sum_{n=1}^{\infty}{\frac{x^{2n}}{n!(2n)}}+C}\) 
Practice B10 
\(\displaystyle{\frac{6}{2x+1},~x=1}\) 


\(\displaystyle{\frac{6}{2x+1}=\sum_{n=0}^{\infty}{\frac{(1)^n 2^{n+1}\abs{x1}^n}{3^n}}
}\) 
Practice B10 Final Answer 
\(\displaystyle{\frac{6}{2x+1}=\sum_{n=0}^{\infty}{\frac{(1)^n 2^{n+1}\abs{x1}^n}{3^n}}
}\) 
Practice B11 
\(\displaystyle{f(x)=\frac{x+3}{1x^2}}\) 


\(\displaystyle{f(x)=\sum_{n=0}^{\infty}{(2+(1)^n)(x^n)}}\) 
Practice B11 Final Answer 
\(\displaystyle{f(x)=\sum_{n=0}^{\infty}{(2+(1)^n)(x^n)}}\) 
Practice B12 
\(\displaystyle{f(x)=e^x\sin(x)}\) 


\(\displaystyle{f(x)=x+x^2+\frac{x^3}{3}\frac{x^5}{30}+\cdots}\) 
Practice B12 Final Answer 
\(\displaystyle{f(x)=x+x^2+\frac{x^3}{3}\frac{x^5}{30}+\cdots}\) 
Practice B13 
\(\displaystyle{f(x)=\tan(x)}\) 


\(\displaystyle{f(x)=x+\frac{x^3}{3}+\frac{2x^4}{15}+\cdots}\) 
Practice B13 Final Answer 
\(\displaystyle{f(x)=x+\frac{x^3}{3}+\frac{2x^4}{15}+\cdots}\) 
Practice B14 
use the binomial series to expand \(\displaystyle{\frac{1}{\sqrt[5]{32x}}}\) as a power series 


\(\displaystyle{\frac{1}{2}+\sum_{n=1}^{\infty}{\frac{1\cdot 6 \cdot \cdots \cdot (5n4)}{5^n2^{5n+1}n!}}}\) 
Practice B14 Final Answer 
\(\displaystyle{\frac{1}{2}+\sum_{n=1}^{\infty}{\frac{1\cdot 6 \cdot \cdots \cdot (5n4)}{5^n2^{5n+1}n!}}}\) 
Practice C01 
\(\displaystyle{g(x)=\tan^{1}(x)}\) 


The power series for \(\displaystyle{g(x) = \tan^{1}(x)}\) about \(x=0\) is \(\displaystyle{g(x) = \sum_{k=0}^{\infty}{ \frac{(1)^k x^{2k+1}}{2k+1}}}\) for \(1 < x < 1\) 
Practice C01 Final Answer 
The power series for \(\displaystyle{g(x) = \tan^{1}(x)}\) about \(x=0\) is \(\displaystyle{g(x) = \sum_{k=0}^{\infty}{ \frac{(1)^k x^{2k+1}}{2k+1}}}\) for \(1 < x < 1\) 
Practice C02 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{5^n}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{5^n}}=\frac{5}{16}}\) 
Practice C02 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{5^n}}=\frac{5}{16}}\) 
Practice C03 
\(\displaystyle{f(x)=\arctan(2x)}\) 


\(\displaystyle{f(x)=\arctan(2x)=\sum_{n=0}^{\infty}{\frac{(1)^n2^{2n+1}x^{2n+1}}{2n+1}}}\) 
Practice C03 Final Answer 
\(\displaystyle{f(x)=\arctan(2x)=\sum_{n=0}^{\infty}{\frac{(1)^n2^{2n+1}x^{2n+1}}{2n+1}}}\) 
Practice C04 
use a power series to approximate \(\displaystyle{\int_{0}^{0.5}{\frac{\sin(x)}{x}dx}}\) with error \(< 0.001\) 


\(\displaystyle{\int_{0}^{0.5}{\frac{\sin(x)}{x}dx}\approx 0.493}\) 
Practice C04 Final Answer 
\(\displaystyle{\int_{0}^{0.5}{\frac{\sin(x)}{x}dx}\approx 0.493}\) 
Practice C05 
find the maclaurin series using the binomial series for \(\arcsin(x)\) 


\(\displaystyle{\arcsin(x)=x+\sum_{n=1}^{\infty}{\frac{1\cdot 3\cdot 5\cdots (2n1)}{(2n+1)2^n\cdot n!}x^{2n}}}\) 
Practice C05 Final Answer 
\(\displaystyle{\arcsin(x)=x+\sum_{n=1}^{\infty}{\frac{1\cdot 3\cdot 5\cdots (2n1)}{(2n+1)2^n\cdot n!}x^{2n}}}\) 