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Homework Help: When a Taylor Series Converges

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    For what values of x do you expect the following Taylor series to converge?


    2. Relevant equations
    I'm not too sure

    3. The attempt at a solution

    Well quite frankly I have no idea what to do. If someone can push me in the right direction I'll get the rest done.
  2. jcsd
  3. Nov 23, 2009 #2
    Taylor expansions are an approximation of a function based on expansion around a point. They use polynomials to give a good analogue and are useful for very small variables. You would expect universal convergence from something like this, but this is not the case.

    For this particular problem, you just need to know that you cannot divide by 0. Figure out at which values the derivative of your function goes to a fraction with a 0 at the bottom, and the series should diverge at those points.
  4. Nov 23, 2009 #3
    Well the problem states the convergence about [tex]x = 1/3[/tex].

    If I substitute this number I get a negative root which is wrong. Now I would suggest to myself to construct a taylor series for this equation, but I don't know how. I'm able to do a Maclaurin series with my eyes closed, but this Taylor crap just wasn't explained properly >_>

    Should I try and construct one? And if I do, what do I do with it to find the convergence?
  5. Nov 23, 2009 #4
    Ok, I'm sorry, you can disregard my comment above.

    Here's your Taylor series formula. I assume that you have covered series convergence rules in your class?

    [tex] \sum_{n=0} ^ {\infty } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n} [/tex]

    A Maclaurin series is actually just a Taylor series with a=0.
  6. Nov 23, 2009 #5
    So I'm just going to use a Ratio test after I have found at least the 1st 2 terms of the Taylor series? And where does this [tex]x=1/3[/tex] come into play?
  7. Nov 23, 2009 #6


    Staff: Mentor

    You didn't include this information in the first post in this thread. You are supposed to post "The problem statement, all variables and given/known data"

    Furthermore, what you did post seems incorrect. You said
    What you showed there isn't a Taylor's series. For sqrt(x2 - x - 2) to even be defined (in the reals), x has to satisfy x2 -x - 2 >= 0, and x = 1/3 does not.
    A Maclaurin series is a Taylor's series in powers of x - 0. You won't be able to construct a Taylor's series in powers of (x - 1/3), because your function and all of its derivatives are not defined at x - 1/3.

    Please give us the exact problem description and we can go from there.
  8. Nov 23, 2009 #7
    For what values of x do you expect the following Taylor series to converge? Do not work out the series.

    a) [tex]sqrt(x^2-x-2)[/tex] about [tex]x=1/3[/tex]

    .........that is the question....
  9. Nov 23, 2009 #8


    Staff: Mentor

    If you let f(x) = sqrt(x^2 - x - 2), the Taylor series for f about a = 1/3 is
    [tex]f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ... + \frac{f^{(n)}}{n!}(x - a)^n + ...[/tex]

    If a = 1/3, neither f nor any of its derivatives is defined, so for no values of x would the Taylor series for the given function converge.
  10. Nov 23, 2009 #9
    I'm doing it now and I'm finding the derivatives. Since I aim to prove what you just said, what would [tex]x[/tex] equal?
  11. Nov 23, 2009 #10


    Staff: Mentor

    What's the point? You can find f'(x), f''(x), etc., but f(1/3) is undefined (as a real number), f'(1/3) is undefined, f''(1/3) is undefined, etc.
  12. Nov 23, 2009 #11
    Heres another:

    [tex]sin(1-\theta^2)[/tex] around [tex]\theta[/tex] = 0.

    Since the derivatives = 0 when a = 0, I can say the sequence converges at [tex]sin(1)[/tex] ?
  13. Nov 23, 2009 #12
    Ok I understood that, what about this sin taylor series?
  14. Nov 23, 2009 #13


    Staff: Mentor

    Which Taylor series? sin(1 - [itex]\theta^2[/itex]) isn't a Taylor series. It has a Taylor series around [itex]\theta[/itex] = 0, which is the same as saying it has a Maclaurin series.

    If you know the Maclaurin series for sin(x), you can get the Maclaurin series for sin(1 - [itex]\theta^2[/itex]) by replacing [itex]\theta[/itex] in the first series by (1 - [itex]\theta^2[/itex]).

    What sequence are you talking about? A series converges iff its sequence of partial sums converges.
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