Stuck on power series, need a refreshment Diff EQ converge?

In summary: The series is convergentfrom x = , left end included (enter Y or N):to x = , right end included (enter Y or N):The series converges for -1<\frac{x}{6}-1\leq 1 or 0<x\leq 12.
  • #1
mr_coffee
1,629
1
Hello everyone! I remember doing these in calc II, but forgot 99% of it. Here is the question:
Find the interval of convergence for the given power series.
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/5e/119a4cd07ce5d3f6f673477ed169641.png [Broken]
The series is convergent
from x = , left end included (enter Y or N):
to x = , right end included (enter Y or N):
Here is my work:
http://suprfile.com/src/1/4fiy34/lastscan.jpg [Broken]

Now I'm confused on what i do now, also u see where i have
|x-6|? Is that right or should it be |x+6| ? Alos is it suppose to be the limit as x-> or n->? Thanks!
I'm using the ratio test btw.
:biggrin:
 
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  • #2
Well you have the ratio test wrong there is no |x - x0| tacked on to the fron of the limit, and the limit should be as "n" goes to infinity, then in order for the series to converge the limit must be less than or equal to 1.
 
  • #3
thanks for the responce d_leet I'm looking at this professors website and following his example, and for some reason he tacked on a |x-xo|.

Scroll down to example one if u want to know where I got that |x-xo|
http://tutorial.math.lamar.edu/AllBrowsers/3401/PowerSeries.asp [Broken]
 
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  • #4
Grrrr... I lost what I just typed.. Ok well I see what he did. What you did is not quite the same, you tried to doboth methods simultaneously, if you had left out that factor then you would get the right answer doing what I suggested. The reason that it is like that on the other website is that he knows that factor will come out of the limit if he takes it as part of the a_n so he just takes it out right away. Tell me if any of that made any sense at all, I can try to exxplain better if it didn't.
 
  • #5
I really didn't understand but I think i'll just ignore that factor of |x-6| and just leave it as
[(x-6)*n]/(n+1)(-6)

okay if i take this as n goes to infinity, wouldn't i get infinity/infinity if i plug in infinity for n?
I remeber there was a rule for this but I forgot. Do i divide everything by n then plug in infity and see waht I'm left with? kinda sounds familiar?
 
  • #6
Okay i still didn't figure out how to take the limit of that, but the TI-89 did, and it got:
-6(x-6), now i know the rule is:
http://tutorial.math.lamar.edu/AllBrowsers/3401/PowerSeries_files/eq0013M.gif [Broken]
So do i set -6(x-6) = 1?
so I'm getting x = 5.833 which is > 1, so this means the series diverges. But how do i know what intervals? Because it wants:
The series is convergent
from x = , left end included (enter Y or N):
to x = , right end included (enter Y or N):

So I know the radius of convergence is 35/6, but what do i do with that? I entered it in, +/- and it didn't like it
 
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  • #7
Okay I redid this problem, and i think i did it right, but not, lies!
But I'm way closer to it then i was before, check my work out, it looks legit hah:
Note: ignore the First series up there, where i have the 4 scribbled out.
http://img233.imageshack.us/img233/7356/lastscan4ge.jpg [Broken]
 
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  • #8
Note that since [tex]|x-6|\geq 0,[/tex] for all x by definition of absolute value, it follows that [tex]|x-6|\geq 0 >-6[/tex] for all x (so you can also include x=0 above.)
 
  • #9
You have, at one point,
[tex]\left|\frac{(x-6)n}{-6(n+1)}\right|[/tex]
and then immediately after
[tex]\frac{\left|x-6\right|\frac{n}{n}}{-6(\frac{n}{n}+\frac{1}{n})}[/tex]
What happened to the absolute value in the denominator? You are aware that |-6|= 6 aren't you? Shouldn't it be
|x- 6|< 6?
 
  • #10
[tex]\sum_{n=1}^{\infty} \frac{(x-6)^n}{n(-6)^n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\left( \frac{x-6}{6}\right) ^n = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\left( \frac{x}{6}-1\right) ^n = -\log \left| \frac{x}{6}\right| [/tex]

which converges for [itex]-1<\frac{x}{6}-1\leq 1[/itex] or [itex]0<x\leq 12[/itex] per the usual Taylor series

[tex]\log |1+x| = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\mbox{, valid for }-1<x\leq 1[/tex]

and which agrees with HallsofIvy's post (and that is not bad for the confidence.)
 
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  • #11
Oo my bad, i do'nt know why i just said f the abs sign in the denominator, that made it work out perfect. And thanks for the explanation benorin! I'm e-mailing u both milk and cookies. Enjoy. :biggrin:
 

1. What is a power series in terms of differential equations?

A power series is a mathematical series that represents a function as a sum of terms, where each term is a polynomial multiplied by a constant raised to a power. In terms of differential equations, power series are often used to approximate the solutions to non-linear differential equations.

2. How do you determine if a power series solution converges?

To determine if a power series solution converges, you can use the ratio test or the root test. The ratio test involves taking the limit of the absolute value of the ratio of the (n+1)th term to the nth term as n approaches infinity. If this limit is less than one, the series will converge. The root test involves taking the nth root of the absolute value of the nth term and if this value is less than one, the series will converge.

3. What is the interval of convergence for a power series solution?

The interval of convergence for a power series solution is the range of values for which the series will converge. This can be determined by using the ratio or root test and finding the values of x that satisfy the convergence criteria.

4. Can a power series solution diverge at certain points within its interval of convergence?

Yes, it is possible for a power series solution to diverge at certain points within its interval of convergence. This can happen when the series converges for some values of x and diverges for others. This is known as a radius of convergence and can be determined by using the convergence tests.

5. How can power series solutions be used to solve differential equations?

Power series solutions can be used to solve differential equations by substituting the series into the equation and solving for the coefficients. This allows for a non-linear differential equation to be approximated by a polynomial function, making it easier to solve. However, it is important to check the convergence of the series and to check if the solution satisfies the boundary conditions of the differential equation.

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