You CAN Ace Calculus
related topics on other pages |
---|
external links you may find helpful |
Single Variable Calculus |
---|
Multi-Variable Calculus |
---|
Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
---|
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
| |
Help Keep 17Calculus Free |
---|
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.
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
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r} }\) 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
\[\displaystyle{
\sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n} = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . .
}\]
We use the Ratio Test to determine the radius of convergence.
Taylor Series are discussed on a separate page.
The geometric series equation is \(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r} }\). 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 \(\displaystyle{ \frac{a}{1-r} }\), 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.
video by Dr Chris Tisdell
This next video clip is more theoretical than the previous one but it will help you understand why power series work.
video by MIT OCW
The next video is very practical, giving you the basics and showing you how to work with power series.
video by PatrickJMT
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.
video by Dr Chris Tisdell
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
\(\displaystyle{ (1+x)^k = 1 + kx + \frac{k(k-1)}{2!} x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + . . .
= \sum_{n=0}^{\infty}{ \left(
\begin{array}{c}
k \\
n
\end{array}
\right) x^n }
}\)
where \(\displaystyle{
\left(
\begin{array}{c}
k \\
n
\end{array}
\right)
= \frac{k!}{n!(k-n)!}
}\)
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.
This section is part of the discussion found on the radius and interval of convergence page, where you will find more information, videos and practice problems. |
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 |
---|
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 |
---|
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 within 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 |
---|
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.
See the radius and interval of convergence page page for more detail, videos and practice problems.
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.
\(\displaystyle{ \frac{1}{1-x} = \sum_ {n=0}^{\infty}{x^n} }\) |
---|
converges \( -1 < x < 1 \) |
power series |
\(\displaystyle{ e^x = \sum_{n=0}^{x^n}{\frac{x^n}{n!}} }\) |
---|
converges \(-\infty < x < \infty\) |
Maclaurin series |
\(\displaystyle{\ln(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^{n+1} (x-1)^n}{n} } }\) |
---|
Taylor series about \( x = 1 \) |
\(\displaystyle{ \sin(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{2n+1}}{(2n+1)!} } }\) |
---|
converges \(-\infty < x < \infty\) |
Maclaurin series |
\(\displaystyle{ \cos(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{2n}}{(2n)!} } }\) |
---|
converges \(-\infty < x < \infty\) |
Maclaurin series |
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
---|
Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
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.
Basic Problems |
---|
\(\displaystyle{ g(x) = \frac{1}{1-x^3} }\)
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{1-x^3} }\)
Final Answer |
---|
\(\displaystyle{ \frac{1}{1-x^3} = \sum_{n=0}^{\infty}{ x^{3n} } ~~~~~ R=1 }\) |
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{1-x^3} }\)
Solution |
---|
This is already in the form of a geometric series.
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r} }\).
We want \(\displaystyle{ \frac{a}{1-r} = \frac{1}{1-x^3} }\).
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 x-values 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 \( \abs{x} < 1 \).
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \frac{1}{1-x^3} = \sum_{n=0}^{\infty}{ x^{3n} } ~~~~~ R=1 }\) |
close solution |
\(\displaystyle{ h(x) = \frac{1}{x-6} }\)
Problem Statement |
---|
\(\displaystyle{ h(x) = \frac{1}{x-6} }\)
Final Answer |
---|
\(\displaystyle{ \frac{1}{x-6} = \frac{-1}{6} \sum_{n=0}^{\infty}{\frac{x^n}{6^n}} ~~~~~ R = 6 }\) |
Problem Statement |
---|
\(\displaystyle{ h(x) = \frac{1}{x-6} }\)
Solution |
---|
In order to determine the power series of this function, we use the geometric power series
\(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r} }\).
We need to determine a and r. Looking at the equation \(\displaystyle{ \frac{a}{1-r} = \frac{1}{x-6} }\)
we need to rewrite the denominator \( x-6 \) to make it look like \(1-r\).
\( x-6 = -6+x = (-6)(1-x/6)\)
So now we have \(\displaystyle{ \frac{1}{x-6} = \frac{1}{(-6)(1-x/6)} = \frac{-1/6}{1-x/6} }\)
When we compare this last expression to the geometric series we get
\(\displaystyle{ \frac{-1/6}{1-x/6} = \frac{a}{1-r} }\) 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, \(\abs{r} = \abs{x/6} < 1 ~~\to~~ \abs{x} < 6\)
Final Answer |
---|
\(\displaystyle{ \frac{1}{x-6} = \frac{-1}{6} \sum_{n=0}^{\infty}{\frac{x^n}{6^n}} ~~~~~ R = 6 }\) |
close solution |
\(\displaystyle{ \frac{6x^2}{7+15x} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{6x^2}{7+15x} }\)
Final Answer |
---|
\(\displaystyle{ \frac{6x^2}{7+15x} = \sum_{n=0}^{\infty}{ \frac{(-15)^n 6 x^{n+2}}{7^{n+1}} } ~~~~~ R=7/15 }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{6x^2}{7+15x} }\)
Solution |
---|
First, we need to get the function in the form \(\displaystyle{ \frac{a}{1-r} }\).
\(\displaystyle{ \frac{6x^2}{7+15x} = \frac{6x^2}{7(1-(-15x/7))} = \frac{6x^2/7}{1-(-15x/7)} }\)
Comparing the last expression to \(\displaystyle{\frac{a}{1-r} }\) 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, \( \abs{r} = \abs{-15x/7} < 1 ~~ \to ~~ \abs{x} < 7/15 \)
Final Answer |
---|
\(\displaystyle{ \frac{6x^2}{7+15x} = \sum_{n=0}^{\infty}{ \frac{(-15)^n 6 x^{n+2}}{7^{n+1}} } ~~~~~ R=7/15 }\) |
close solution |
\(\displaystyle{ f(x) = \frac{x^5}{8+x^2} }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = \frac{x^5}{8+x^2} }\)
Final Answer |
---|
\(\displaystyle{ \frac{x^5}{8+x^2} = \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{2n+5}}{8^{n+1}}} }\) \(R = 2 \sqrt{2}\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = \frac{x^5}{8+x^2} }\)
Solution |
---|
Our goal here is to get \(f(x)\) in the form \(\displaystyle{ \sum_{n=0}^{\infty}{ ar^n } = \frac{a}{1-r} }\). 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 \( \displaystyle{ \frac{a}{1-r} }\) 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.
\(\begin{array}{rcl}
\abs{r} & < & 1 \\
\displaystyle{ \abs{ \frac{-x^2}{8} } } & < & 1 \\
x^2 & < & 8 \\
\abs{x} & < & \sqrt{8} = 2 \sqrt{2}
\end{array}\)
Final Answer |
---|
\(\displaystyle{ \frac{x^5}{8+x^2} = \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{2n+5}}{8^{n+1}}} }\) \(R = 2 \sqrt{2}\) |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^{n+2}}{n!} } }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^{n+2}}{n!} } }\)
Final Answer |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^{n+2}}{n!} } = x^2e^x }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^{n+2}}{n!} } }\)
Solution |
---|
video by MIT OCW
Final Answer |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^{n+2}}{n!} } = x^2e^x }\) |
close solution |
\(\displaystyle{ \sum_{n=2}^{\infty}{x^n} }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=2}^{\infty}{x^n} }\)
Final Answer |
---|
\(\displaystyle{ \sum_{n=2}^{\infty}{x^n} = \frac{1}{1-x}-1-x }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=2}^{\infty}{x^n} }\)
Solution |
---|
video by MIT OCW
Final Answer |
---|
\(\displaystyle{ \sum_{n=2}^{\infty}{x^n} = \frac{1}{1-x}-1-x }\) |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} + x^n \right] } }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} + x^n \right] } }\)
Final Answer |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} + x^n \right] } = e^x + \frac{1}{1-x} }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} + x^n \right] } }\)
Solution |
---|
video by MIT OCW
Final Answer |
---|
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} + x^n \right] } = e^x + \frac{1}{1-x} }\) |
close solution |
\(\displaystyle{ \sum_{n=-1}^{\infty}{ x^{n+1} } }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=-1}^{\infty}{ x^{n+1} } }\)
Final Answer |
---|
\(\displaystyle{ \sum_{n=-1}^{\infty}{ x^{n+1} } = \frac{1}{1-x} }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=-1}^{\infty}{ x^{n+1} } }\)
Solution |
---|
video by MIT OCW
Final Answer |
---|
\(\displaystyle{ \sum_{n=-1}^{\infty}{ x^{n+1} } = \frac{1}{1-x} }\) |
close solution |
\( g(x) = (x+1)e^x \)
Problem Statement |
---|
\( g(x) = (x+1)e^x \)
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 \( -1 < x < 1 \) |
Problem Statement |
---|
\( g(x) = (x+1)e^x \)
Solution |
---|
video by Dr Chris Tisdell
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 \( -1 < x < 1 \) |
close solution |
\(\displaystyle{ \frac{1}{1+x^2} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{1}{1+x^2} }\)
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 \(-1 < x < 1\) |
Problem Statement |
---|
\(\displaystyle{ \frac{1}{1+x^2} }\)
Solution |
---|
video by Dr Chris Tisdell
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 \(-1 < x < 1\) |
close solution |
\(\displaystyle{ f(x) = x^2\sin(x^3) }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = x^2\sin(x^3) }\)
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 \( -1 < x < 1 \) |
Problem Statement |
---|
\(\displaystyle{ f(x) = x^2\sin(x^3) }\)
Solution |
---|
There is more to this video solution 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.
video by Dr Chris Tisdell
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 \( -1 < x < 1 \) |
close solution |
\(\displaystyle{ g(x) = \frac{1}{1+x} }\)
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{1+x} }\)
Final Answer |
---|
The power series for \( g(x) \) about \( x=0 \) is \(\displaystyle{ g(x) = \sum_{k=0}^{\infty}{ (-x)^k } }\) for \(-1 < x < 1\) |
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{1+x} }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
The power series for \( g(x) \) about \( x=0 \) is \(\displaystyle{ g(x) = \sum_{k=0}^{\infty}{ (-x)^k } }\) for \(-1 < x < 1\) |
close solution |
\(\displaystyle{ \frac{1}{1+9x^2} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{1}{1+9x^2} }\)
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 \( -1/3 < x < 1/3 \) |
Problem Statement |
---|
\(\displaystyle{ \frac{1}{1+9x^2} }\)
Solution |
---|
video by PatrickJMT
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 \( -1/3 < x < 1/3 \) |
close solution |
\(\displaystyle{ \frac{x}{4x+1} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{x}{4x+1} }\)
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 \( -1/2 < x < 1/2 \) |
Problem Statement |
---|
\(\displaystyle{ \frac{x}{4x+1} }\)
Solution |
---|
video by PatrickJMT
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 \( -1/2 < x < 1/2 \) |
close solution |
\(\displaystyle{ \frac{x}{9+x^2} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{x}{9+x^2} }\)
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 \( -3 < x < 3 \) |
Problem Statement |
---|
\(\displaystyle{ \frac{x}{9+x^2} }\)
Solution |
---|
video by PatrickJMT
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 \( -3 < x < 3 \) |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}x^{2n}}{(2n-1)!} } }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}x^{2n}}{(2n-1)!} } }\)
Final Answer |
---|
\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}x^{2n}}{(2n-1)!} } = x\sin(x) }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}x^{2n}}{(2n-1)!} } }\)
Solution |
---|
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.
video by PatrickJMT
Final Answer |
---|
\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}x^{2n}}{(2n-1)!} } = x\sin(x) }\) |
close solution |
\( x \cos(x) - \sin(x) \)
Problem Statement |
---|
\( x \cos(x) - \sin(x) \)
Final Answer |
---|
\(\displaystyle{ x\cos(x) - \sin(x) = \left( \frac{x^3}{3!} - \frac{x^3}{2!} \right) + \left( \frac{x^5}{4!} - \frac{x^5}{5!} \right) + \left( \frac{x^7}{7!} - \frac{x^7}{6!} \right) + \left( \frac{x^9}{8!} - \frac{x^9}{9!} \right) + \cdots }\) |
Problem Statement |
---|
\( x \cos(x) - \sin(x) \)
Solution |
---|
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.
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ x\cos(x) - \sin(x) = \left( \frac{x^3}{3!} - \frac{x^3}{2!} \right) + \left( \frac{x^5}{4!} - \frac{x^5}{5!} \right) + \left( \frac{x^7}{7!} - \frac{x^7}{6!} \right) + \left( \frac{x^9}{8!} - \frac{x^9}{9!} \right) + \cdots }\) |
close solution |
\(\displaystyle{ f(x)=x^2e^{4x} }\)
Problem Statement |
---|
\(\displaystyle{ f(x)=x^2e^{4x} }\)
Final Answer |
---|
\(\displaystyle{ f(x)=x^2e^{4x} = \sum_{n=0}^{\infty}{ \frac{4^nx^{n+2}}{n!} } }\) |
Problem Statement |
---|
\(\displaystyle{ f(x)=x^2e^{4x} }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ f(x)=x^2e^{4x} = \sum_{n=0}^{\infty}{ \frac{4^nx^{n+2}}{n!} } }\) |
close solution |
For \(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{(5x-1)^n} }\), evaluate\( \int{f(x)~dx} \)
Problem Statement |
---|
For \(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{(5x-1)^n} }\), evaluate\( \int{f(x)~dx} \)
Final Answer |
---|
\(\displaystyle{ \int{f(x)~dx} = C + \sum_{n=0}^{\infty}{ \frac{(5x-1)^{n+1}}{5(n+1)} } }\) |
Problem Statement |
---|
For \(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{(5x-1)^n} }\), evaluate\( \int{f(x)~dx} \)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \int{f(x)~dx} = C + \sum_{n=0}^{\infty}{ \frac{(5x-1)^{n+1}}{5(n+1)} } }\) |
close solution |
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\), find \(\displaystyle{ h(x) = x^3\int_{0}^{x}{ f(t)~dt } }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\), find \(\displaystyle{ h(x) = x^3\int_{0}^{x}{ f(t)~dt } }\)
Final Answer |
---|
\(\displaystyle{ h(x) = \sum_{n=0}^{\infty}{ \frac{x^{n+4}}{(n+1)!} } }\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\), find \(\displaystyle{ h(x) = x^3\int_{0}^{x}{ f(t)~dt } }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ h(x) = \sum_{n=0}^{\infty}{ \frac{x^{n+4}}{(n+1)!} } }\) |
close solution |
\(\displaystyle{ \frac{2}{x+4} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{2}{x+4} }\)
Final Answer |
---|
\(\displaystyle{ \frac{2}{x+4} = \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2} \left( \frac{x}{4} \right)^n } }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{2}{x+4} }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ \frac{2}{x+4} = \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2} \left( \frac{x}{4} \right)^n } }\) |
close solution |
\(\displaystyle{ \cos(x^2) }\)
Problem Statement |
---|
\(\displaystyle{ \cos(x^2) }\)
Final Answer |
---|
\(\displaystyle{ \cos(x^2) = \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{4n}}{(2n)! } } }\) |
Problem Statement |
---|
\(\displaystyle{ \cos(x^2) }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ \cos(x^2) = \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{4n}}{(2n)! } } }\) |
close solution |
Intermediate Problems |
---|
\( \ln(1+x) \)
Problem Statement |
---|
\( \ln(1+x) \)
Final Answer |
---|
The power series for \(\ln(1+x)\) about \(x=0\) is \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{n+1}}{n+1} }; }\) \( -1 < x < 1 \) |
Problem Statement |
---|
\( \ln(1+x) \)
Solution |
---|
video by Dr Chris Tisdell
Final Answer |
---|
The power series for \(\ln(1+x)\) about \(x=0\) is \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{n+1}}{n+1} }; }\) \( -1 < x < 1 \) |
close solution |
\(\displaystyle{ g(x) = \frac{1}{(x-1)^2} }\)
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{(x-1)^2} }\)
Final Answer |
---|
The power series for \(\displaystyle{ g(x) = \frac{1}{(x-1)^2} }\) about \(x=0\) is \(\displaystyle{ g(x) = \sum_{k=1}^{\infty}{ kx^{k-1} } }\) for \(-1 < x < 1\) |
Problem Statement |
---|
\(\displaystyle{ g(x) = \frac{1}{(x-1)^2} }\)
Solution |
---|
video by Dr Chris Tisdell
Final Answer |
---|
The power series for \(\displaystyle{ g(x) = \frac{1}{(x-1)^2} }\) about \(x=0\) is \(\displaystyle{ g(x) = \sum_{k=1}^{\infty}{ kx^{k-1} } }\) for \(-1 < x < 1\) |
close solution |
\(\displaystyle{ \ln\left( \frac{1-x}{1+x} \right) }\)
Problem Statement |
---|
\(\displaystyle{ \ln\left( \frac{1-x}{1+x} \right) }\)
Hint |
---|
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 \(\displaystyle{ \frac{d}{dx}[\ln(1-x)] = \frac{-1}{1-x} }\) 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.
Problem Statement |
---|
\(\displaystyle{ \ln\left( \frac{1-x}{1+x} \right) }\)
Final Answer |
---|
The power series for \(\displaystyle{ \ln\left(\frac{1-x}{1+x}\right) }\) about \(x=0\) is \(\displaystyle{ -2\sum_{n=0}^{\infty}{ \frac{x^{2n+1}}{2n+1} } }\) for \(-1 < x < 1\) |
Problem Statement |
---|
\(\displaystyle{ \ln\left( \frac{1-x}{1+x} \right) }\)
Hint |
---|
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 \(\displaystyle{ \frac{d}{dx}[\ln(1-x)] = \frac{-1}{1-x} }\) 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.
Solution |
---|
video by PatrickJMT
Final Answer |
---|
The power series for \(\displaystyle{ \ln\left(\frac{1-x}{1+x}\right) }\) about \(x=0\) is \(\displaystyle{ -2\sum_{n=0}^{\infty}{ \frac{x^{2n+1}}{2n+1} } }\) for \(-1 < x < 1\) |
close solution |
\(\displaystyle{ \frac{x^2}{(1+x)^3} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{x^2}{(1+x)^3} }\)
Final Answer |
---|
\(\displaystyle{ \frac{x^2}{(1+x)^3} = \frac{1}{2} \sum_{n=0}^{\infty}{ (-1)^{n}(n+2)(n+1)x^{n+2} } }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{x^2}{(1+x)^3} }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \frac{x^2}{(1+x)^3} = \frac{1}{2} \sum_{n=0}^{\infty}{ (-1)^{n}(n+2)(n+1)x^{n+2} } }\) |
close solution |
\(\displaystyle{ f(x)=\cos^2(x) }\)
Problem Statement |
---|
\(\displaystyle{ f(x)=\cos^2(x) }\)
Final Answer |
---|
\(\displaystyle{ f(x) = \cos^2(x) = 1 + \sum_{n=0}^{\infty}{ \frac{(-1)^n2^{2n-1}x^{2n}}{(2n)!} } }\) |
Problem Statement |
---|
\(\displaystyle{ f(x)=\cos^2(x) }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ f(x) = \cos^2(x) = 1 + \sum_{n=0}^{\infty}{ \frac{(-1)^n2^{2n-1}x^{2n}}{(2n)!} } }\) |
close solution |
\(\displaystyle{ \frac{x^2}{(1-2x)^2} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{x^2}{(1-2x)^2} }\)
Final Answer |
---|
\(\displaystyle{ \frac{x^2}{(1-2x)^2} = \sum_{n=0}^{\infty}{ 2^n(n+1)x^{n+2} } }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{x^2}{(1-2x)^2} }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \frac{x^2}{(1-2x)^2} = \sum_{n=0}^{\infty}{ 2^n(n+1)x^{n+2} } }\) |
close solution |
\(\displaystyle{ e^{-x^2}\cos(x) }\)
Problem Statement |
---|
\(\displaystyle{ e^{-x^2}\cos(x) }\)
Final Answer |
---|
\(\displaystyle{ e^{-x^2}\cos(x) = 1 - \frac{3x^2}{2} + \frac{25x^4}{24} + \cdots }\) |
Problem Statement |
---|
\(\displaystyle{ e^{-x^2}\cos(x) }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ e^{-x^2}\cos(x) = 1 - \frac{3x^2}{2} + \frac{25x^4}{24} + \cdots }\) |
close solution |
\(\displaystyle{ \frac{x}{\sin(x)} }\)
Problem Statement |
---|
\(\displaystyle{ \frac{x}{\sin(x)} }\)
Final Answer |
---|
\(\displaystyle{ \frac{x}{\sin(x)} = 1 + \frac{x^2}{6} + \frac{7x^4}{300} + \cdots }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{x}{\sin(x)} }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \frac{x}{\sin(x)} = 1 + \frac{x^2}{6} + \frac{7x^4}{300} + \cdots }\) |
close solution |
Evaluate \(\displaystyle{ \int{ \frac{e^{x^2}}{x}dx } }\) as an infinite series.
Problem Statement |
---|
Evaluate \(\displaystyle{ \int{ \frac{e^{x^2}}{x}dx } }\) as an infinite series.
Final Answer |
---|
\(\displaystyle{ \int{ \frac{e^{x^2}}{x}dx} = \ln\abs{x} + \sum_{n=1}^{\infty}{ \frac{x^{2n}}{n!(2n)} } + C }\) |
Problem Statement |
---|
Evaluate \(\displaystyle{ \int{ \frac{e^{x^2}}{x}dx } }\) as an infinite series.
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \int{ \frac{e^{x^2}}{x}dx} = \ln\abs{x} + \sum_{n=1}^{\infty}{ \frac{x^{2n}}{n!(2n)} } + C }\) |
close solution |
\(\displaystyle{ \frac{6}{2x+1}, ~ x=1 }\)
Problem Statement |
---|
\(\displaystyle{ \frac{6}{2x+1}, ~ x=1 }\)
Final Answer |
---|
\(\displaystyle{ \frac{6}{2x+1} = \sum_{n=0}^{\infty}{ \frac{(-1)^n 2^{n+1} \abs{x-1}^n}{3^n} } }\) |
Problem Statement |
---|
\(\displaystyle{ \frac{6}{2x+1}, ~ x=1 }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ \frac{6}{2x+1} = \sum_{n=0}^{\infty}{ \frac{(-1)^n 2^{n+1} \abs{x-1}^n}{3^n} } }\) |
close solution |
\(\displaystyle{ f(x) = \frac{x+3}{1-x^2} }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = \frac{x+3}{1-x^2} }\)
Final Answer |
---|
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{ (2+(-1)^n)(x^n) } }\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = \frac{x+3}{1-x^2} }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{ (2+(-1)^n)(x^n) } }\) |
close solution |
\(\displaystyle{ f(x) = e^x \sin(x) }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = e^x \sin(x) }\)
Final Answer |
---|
\(\displaystyle{ f(x) = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} + \cdots }\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = e^x \sin(x) }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ f(x) = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} + \cdots }\) |
close solution |
\(\displaystyle{ f(x) = \tan(x) }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = \tan(x) }\)
Final Answer |
---|
\(\displaystyle{ f(x) = x + \frac{x^3}{3} + \frac{2x^4}{15} + \cdots }\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = \tan(x) }\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ f(x) = x + \frac{x^3}{3} + \frac{2x^4}{15} + \cdots }\) |
close solution |
Use the binomial series to expand \(\displaystyle{ \frac{1}{ \sqrt[5]{32-x} } }\) as a power series.
Problem Statement |
---|
Use the binomial series to expand \(\displaystyle{ \frac{1}{ \sqrt[5]{32-x} } }\) as a power series.
Final Answer |
---|
\(\displaystyle{ \frac{1}{2} + \sum_{n=1}^{\infty}{ \frac{1\cdot 6 \cdot \cdots \cdot (5n-4) }{ 5^n2^{5n+1}n! } } }\) |
Problem Statement |
---|
Use the binomial series to expand \(\displaystyle{ \frac{1}{ \sqrt[5]{32-x} } }\) as a power series.
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \frac{1}{2} + \sum_{n=1}^{\infty}{ \frac{1\cdot 6 \cdot \cdots \cdot (5n-4) }{ 5^n2^{5n+1}n! } } }\) |
close solution |
Advanced Problems |
---|
\(\displaystyle{ g(x) = \tan^{-1}(x) }\)
Problem Statement |
---|
\(\displaystyle{ g(x) = \tan^{-1}(x) }\)
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 \(-1 < x < 1\) |
Problem Statement |
---|
\(\displaystyle{ g(x) = \tan^{-1}(x) }\)
Solution |
---|
video by Dr Chris Tisdell
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 \(-1 < x < 1\) |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } }\)
Problem Statement |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } }\)
Final Answer |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } = \frac{5}{16} }\) |
Problem Statement |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } = \frac{5}{16} }\) |
close solution |
\(\displaystyle{ f(x) = \arctan(2x) }\)
Problem Statement |
---|
\(\displaystyle{ f(x) = \arctan(2x) }\)
Final Answer |
---|
\(\displaystyle{ f(x) = \arctan(2x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n2^{2n+1}x^{2n+1}}{2n+1} } }\) |
Problem Statement |
---|
\(\displaystyle{ f(x) = \arctan(2x) }\)
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ f(x) = \arctan(2x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n2^{2n+1}x^{2n+1}}{2n+1} } }\) |
close solution |
Find the Maclaurin series using the binomial series for \(\arcsin(x)\).
Problem Statement |
---|
Find the Maclaurin series using the binomial series for \(\arcsin(x)\).
Final Answer |
---|
\(\displaystyle{ \arcsin(x) = x + \sum_{n=1}^{\infty}{\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n+1)2^n\cdot n!}x^{2n} } }\) |
Problem Statement |
---|
Find the Maclaurin series using the binomial series for \(\arcsin(x)\).
Solution |
---|
video by PatrickJMT
Final Answer |
---|
\(\displaystyle{ \arcsin(x) = x + \sum_{n=1}^{\infty}{\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n+1)2^n\cdot n!}x^{2n} } }\) |
close solution |
Use a power series to approximate \(\displaystyle{ \int_{0}^{0.5}{ \frac{\sin(x)}{x}dx } }\) with error \(< 0.001\)
Problem Statement |
---|
Use a power series to approximate \(\displaystyle{ \int_{0}^{0.5}{ \frac{\sin(x)}{x}dx } }\) with error \(< 0.001\)
Final Answer |
---|
\(\displaystyle{ \int_{0}^{0.5}{ \frac{\sin(x)}{x} dx } \approx 0.493 }\) |
Problem Statement |
---|
Use a power series to approximate \(\displaystyle{ \int_{0}^{0.5}{ \frac{\sin(x)}{x}dx } }\) with error \(< 0.001\)
Solution |
---|
video by MIP4U
Final Answer |
---|
\(\displaystyle{ \int_{0}^{0.5}{ \frac{\sin(x)}{x} dx } \approx 0.493 }\) |
close solution |