1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove the following Power Series is monotonic

  1. Nov 24, 2011 #1
    [tex]f(x) = \sum_{n=1}^{\infty} \frac{(-1)^nx^{n}}{(n)^{\frac{3}{2}}}[/tex]


    Prove f(x) is strictly monotonic (where f is defined) and that there exists one solution to
    f(x)=1.5
    and f(x)=-0.5


    First, how do I show that the radius of convergence is between [tex]-1 \le x \le 1[/tex]?

    And then, how do I proceed to show the series is strictly monotonic ?
     
  2. jcsd
  3. Nov 24, 2011 #2
    Strictly monotonic is when the function is either strictly increasing or decreasing.
    r=limn→∞ lan/an+1l
     
  4. Nov 24, 2011 #3
    I don't think at all that I can work with that definition in that case.

    Maybe work around showing f is single value, I really don't know.
     
  5. Nov 24, 2011 #4
    Not too sure why you can't use that limit definition. Alternatively, the max values for f are for even n and min values are for odd n. And, it's easy to see that the sequence is decreasing. (this can be easily proven by ratios of coeffs). So now you can deduce what the max and min values for the series is. These will be in [-1,1].
     
  6. Nov 24, 2011 #5
    You are speaking a language I don't understand :(

    Why does "the max values for f are for even n and min values are for odd n"?
    And what are "ratios of coeffs", how specifically should I show the sequence is decreasing?

    And from that, how I deduce the max and min values?


    I'm sorry for all the questions but unfortunately I know very little of the subject and I'm not on that level yet.
     
  7. Nov 24, 2011 #6
    Shaon0 is right in that you can often use the ratio test to find the radius of convergence of a series--however, the formula he gave is wrong, so check out the link I supplied.

    The usual way to see if a function is monotonic is to examine its derivative.

    Show us your attempt and I'll be happy to give more hints.
     
  8. Nov 24, 2011 #7
    I have the ratio test:
    [tex]\displaystyle \frac{(-1)^{n+1}x^{n+1}}{(n+1)^{\frac{3}{2}}}\cdot \frac{n^{\frac{3}{2}}}{(-1)^{n}x^{n}}[/tex]

    [tex]=\displaystyle \lim_{n\to \infty}\frac{n^{\frac{3}{2}}|x|}{(n+1)^{\frac{3}{2}}}=|x|[/tex]

    why does r=[-1,1] and not (-1,1)?

    And the derivative:

    [tex]f'(x) = \sum_{n=1}^\infty\frac{(-1)^n x^{n-1}}{n^{1/2}} = -1 + \frac x{\sqrt2} - \frac{x^2}{\sqrt3} + \frac{x^3}{\sqrt 4} - \frac{x^4}{\sqrt5} + \ldots .[/tex]

    How from here how do I show f is monotonic and that there exists one solution to
    f(x)=1.5?
     
  9. Nov 24, 2011 #8
    Read again what conclusions you can draw from the ratio test. You probably will have to check the endpoints of the interval of convergence separately, as they usually correspond to the case where the ratio test is inconclusive.

    Your derivative is wrong. You made a silly mistake--with respect to what variable are you differentiating?
     
  10. Nov 24, 2011 #9
    What do you mean by "have to check the endpoints of the interval of convergence separately", how precisely do I do that? Sorry but I dodn't remeber the according tool. I though the ratio test was all there is.

    And the derivative looks good to me... I double checked.

    d/dx.
     
  11. Nov 24, 2011 #10
    The derivative does look right to me. To check the end points, substitute x=1 and x=-1 explicitly and look at the convergence of the resulting function. That's the traditional way end points are checked. Remember this when working with power series: the tests for convergence such as the ratio or root test never give information about the convergence of the endpoints, they always have to be checked explicitly.

    For the derivative, group the terms like this
    [tex]\left(-1 + \frac{x}{\sqrt{2}}\right) + \left(-\frac{x^2}{\sqrt{3}} + \frac{x^3}{\sqrt{4}}\right) + ...[/tex]
    since [itex]\|x\|\le 1[/itex] what can you tell me about the sign of this sum?

    Finally, if you only need to show that the function exists at 1.5 and -0.5, then that's automatic because the series converges. Explicitly, the function is analytic therefore it is continuous on the interval it is defined.
     
  12. Nov 24, 2011 #11
    Just plug the endpoints in for x and see if the series converges or not. But first, what did you get for the radius of convergence?

    And there are many tests to determine the convergence/divergence of series.
    Ah, okay sorry I didn't notice the cancellation.
     
  13. Nov 24, 2011 #12
    Thanks for all your help.
    I didn't think checking the endpoints was as easy as just plugging in.


    One lastquestion,
    "
    Finally, if you only need to show that the function exists at 1.5 and -0.5, then that's automatic because the series converges. Explicitly, the function is analytic therefore it is continuous on the interval it is defined."

    I don't understand this, why is this automatic? Even if f is continuous I don't see how that shows that f(x)=1.5 and f(x)=-0.5 and the solotuin is single?
     
  14. Nov 24, 2011 #13
    It is single simply because it's monotonous. Suppose there were two solutions, from Rolle's theorem there must be a point where the derivative is 0 which is a contradiction.
     
  15. Nov 24, 2011 #14
    Okay, I see, but what guarantees that the solution even exists?
     
  16. Nov 24, 2011 #15
    Like I've mentioned, the solutions exist because the series is convergent at those two values. The limit it converges to is defined as the value of the function. The monoticity of the function guarantees that the solutions are unique.
     
  17. Nov 24, 2011 #16
    But how do you know the series is convergent where f(x)=1.5?
    It is convergent in [-1,1], how do I take it that there is indeed an x_0 in [-1,1] that satisfies f(x_0)=1.5 and that x_0 is single?

    Can i take just any random number 123456 and say that there is some x in [-1,1] that satisfy f(x)=123456?
     
  18. Nov 25, 2011 #17
    Ah, true. I missed that point.

    The simplest way is to check the end points and verify that f(-1) >= 1.5 and f(1) <= -0.5. It should not be too difficult to do that. After which the fact that f(x)=1.5 and f(x)=-0.5 exists follows from the intermediate value theorem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook