Check my solution to a power series/taylor series problem

In summary, we used the ratio test to find the radius of convergence for the series, which is 1. Then, by plugging in x=0 and x=2, we found that the series converges conditionally at both points. It also converges absolutely on the interval 0<x<2. The fact that this series is the Taylor series for arctan(x) centered at x=1 does not provide a quicker solution.
  • #1
miglo
98
0

Homework Statement


[tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(x-1)^{2n+1}}{2n+1}[/tex]

Find the series' radius and interval of convergence. Then identify the values of x for which the series converges absolutely and conditionally.


Homework Equations


ratio test
absolute convergence test
alternating series test

The Attempt at a Solution


[tex]\lim_{n\rightarrow\infty}\vert\frac{(x-1)^{2n+3}}{2n+3}\frac{2n+1}{(x-1)^{2n+1}}\vert=\lim_{n\rightarrow\infty}(x-1)^{2}\frac{2n+1}{2n+3}[/tex]
[tex](x-1)^{2}\lim_{n\rightarrow\infty}\frac{2n+1}{2n+3}=(x-1)^{2}[/tex]
[tex](x-1)^{2}<1[/tex]
[tex]\sqrt{(x-1)^{2}}<1[/tex]
[tex]\vert x-1\vert<1[/tex]
[tex]0<x<2[/tex]
so my radius of convergence is 1
then plugging in x=0 and x=2
i get [tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(-1)^{2n+1}}{2n+1}[/tex] and
[tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(1)^{2n+1}}{2n+1}[/tex]
i found that at both x=0 and x=2 the series converges conditionally
so then my interval of convergence is [tex]0\leq x \leq 2[/tex] with absolute convergence on 0<x<2 and conditional convergence at x=0 and 2
i have a test tomorrow on sequences, infinite series, power series, and taylor series so I am working on the problems in my book but this one is an even numbered problme so i can't check my solution in the back of the book
also i noticed that this series is the taylor series for arctanx centered at x=1, could i have used this to get the solution much quicker than what i did?
 
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  • #2
That looks fine to me if you are clear on why x=0 and x=2 cases converge by the alternating series test. Knowing it's the series for arctan(x) doesn't help unless you know something about complex analysis (maybe). I think you are doing it the simplest way possible.
 
  • #3
well at x=0 the sequence [tex]\frac{1}{2n+1}[/tex] is positive for all n, its decreasing because 2n+1 is an increasing sequence and the reciprocal of an increasing sequence is decreasing and the sequence goes to zero as n goes to infinity so by the alternating series test the series converges but it diverges when applying the absolute convergence test to the series at x=0 and limit comparison with 1/n
at x=2 the sequence is the same as the above so it converges by alternating series
but it fails the absolute convergence test in the same way the first one failed so at both points they only converge conditionally
ohh i see, i was hoping that the taylor series would make things quicker
 

1. How do I know if my solution to a power series problem is correct?

To check if your solution to a power series problem is correct, you can plug in the values of x into the original function and compare it to the value you calculated using your power series. If they are close or equal, then your solution is likely correct. You can also use a graphing calculator to plot both the original function and the power series solution and see if they match up.

2. What is the difference between a power series and a Taylor series?

A power series is a series of terms with increasing powers of x, while a Taylor series is a special type of power series that is centered at a specific value of x, usually 0. The coefficients of a Taylor series can be found using derivatives of the original function at the center value.

3. How do I find the radius of convergence for a power series solution?

The radius of convergence for a power series solution can be found by using the ratio test. Take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If the limit is less than 1, then the series converges. The radius of convergence is the distance from the center value where the series converges.

4. Can I use a power series to approximate any function?

No, not all functions can be accurately approximated using a power series. The success of a power series approximation depends on the radius of convergence and the behavior of the function near the center value. Some functions, such as those with discontinuities or vertical asymptotes, cannot be accurately approximated using a power series.

5. What is the purpose of using a Taylor series to approximate a function?

The purpose of using a Taylor series to approximate a function is to simplify the original function into a more manageable form. This allows for easier calculations and can also provide insight into the behavior of the function near the center value. Taylor series can also be used to approximate values of a function that may be difficult to calculate directly.

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