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Need Help finding Radius & Interval of convergence

  1. Nov 29, 2012 #1
    Ʃ1. The problem statement, all variables and given/known data

    Determine the radius of convergence and the interval of convergence for the follwing function expanded about the point a=2.

    f(x)= ln(3-x)


    2. Relevant equations

    ln(1-x) = Ʃ (x^n+1)/n+1 n=0 which has radius of convergence at |x|<1

    3. The attempt at a solution

    I have figured it out ignoring the a=2.

    so, ln(3-x) = Ʃ (x^n+1)/(n+1)(3^n+1) comes out to |x/3|<1

    interval of convergence (-3,3) radius of convergence = 3


    but i feel like these arent the correct answers for what was asked, so my question is where does the a=2 come in place?

    any suggestions?
     
  2. jcsd
  3. Nov 29, 2012 #2

    Mute

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    Many functions can be expanded about a point "a" in a taylor series of the form

    $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$$

    The series you used was for the expansion of ##\ln(1-x)## about the point a = 0. So, there are a couple of ways you could proceed. You could try to find the general form of the nth derivative of ##\ln(3-x)## evaluated at x=2 and just plug everything into the general sum formula above. I'll tell you right now that that's the cumbersome way to do this problem.

    The problem is actually set up nicely so that you can use the series for ##\ln(1-x)## to quickly get the series you want. Hint: The series for ##\ln(1-y(x))##, for any y(x) whose range is in the radius of convergence of the series, can be expanded using the series you wrote down for ##\ln(1-x)##, with x replaced by y(x). Can you find a y(x) that basically solves your problem?
     
  4. Nov 29, 2012 #3
    so i should make a taylor series of f(a) for f(x)=ln(3-x) and solve it that way?

    im confused, im not sure what to do
     
    Last edited: Nov 29, 2012
  5. Nov 29, 2012 #4
    do I simply plug in the a=2 for n to get the results i want?

    Ʃ (x^n+1)/(n+1)(3^n+1) where n=0 before, but n=2 now.

    this would give an interval of ±(81)^1/3 cuberoot of 81

    is this right?
     
  6. Nov 29, 2012 #5
    Ive been working on this problem for a while now and this is what I have gotten so far:

    f(x) = ln(3-x) f(2)= 0
    f`(x) = (3-x)^-1 f`(2)= 1
    f''(x) = -(3-x)^-2 f''(2)= -1
    f'''(x) = 2(3-x)^-3 f'''(2)= 2

    taylor series f(x)= 0 + (x-2) - (x-2)^2/2! + 2(x-2)^3/3! +....

    cleaned up we get f(x) = (x-2) - 1/2(x-2)^2 + 1/3(x-2)^3 +.....

    in sigma notation I got Ʃ [(x-2)^n]/n then Ʃ [(x-2)^n+1]/n+1 when n=0 gets me

    |x-2|<1 so, -∞ < x < 3


    am I doing this right?
     
  7. Nov 29, 2012 #6

    Mute

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    You're doing it the hard way! I said in my post that this way was cumbersome. If you want to do it this way, it's best if you can identify a pattern for the derivatives of ##\ln(3-x)## and write down a general expression for ##d^n(\ln(1-x))/dx^n##, which you could then evaluate at ##x=2## and plug into your series. Again, however, that's the hard way!

    You already know the series for ##\ln(1-x)## about x=0. Can you think of anything you can do to ##\ln(3-x)## so that you can use the series for ##\ln(1-x)##. Maybe a change of variables?
     
  8. Nov 29, 2012 #7
    Im not following. Where do I get the interval and radius of convergence from that?
     
  9. Nov 29, 2012 #8

    Mute

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    Do you know the radius of convergence for the ##\ln(1-x)## series? If not, do you know how you would find it?

    I'm trying to guide you towards the finding the series for ##\ln(3-x)## using the ##\ln(1-x)## series without outright giving away the answer. Maybe I should let you solve it the "hard" way first, then I can show you the quick way. The "hard" way is not really all that hard, per se, it's just that if you want to do it fully rigorously you have to prove the general form for the nth derivative of ##\ln(3-x)## using induction or something similar.

    Your attempt to derive the series the "hard" way was a good start, but you made a few small mistakes, and your grader may say you have left out some steps. You wrote

    When taking the derivatives, you are forgetting that there is an extra minus sign that comes from the fact you have -x in the denominator. So, it turns out that all derivative terms have a minus sign in front of them. Otherwise, these derivatives look good.

    Your first expression for the sigma notation is correct except that you are missing an overall minus sign (note that if your original attempt at the derivatives which gave you the alternating minus signs had been correct, you would be missing a factor of (-1)^n, but it coincidentally got dropped).

    I'm not sure what you are asking about the n=0 case. You don't need an n=0 term in your series:

    $$-\sum_{n=1}^\infty \frac{(x-2)^n}{n}$$
    is a perfectly good Taylor series.

    Okay, so that series looks good. The only thing missing from the derivation, which your grader might take off points for depending on your class, is that you didn't prove that

    $$\frac{d^n}{dx^n}\ln(3-x) = - \frac{(k-1)!}{(3-x)^k}.$$
    You guessed the general pattern, so you could prove it with induction. Do you know how to do that?

    This is not correct for the radius of convergence. It's called a 'radius' because it has to be symmetric: |x| < R or -R < x < R. Also, how do you know that you need ##|x-2| < 1##? This is correct, but you haven't explained where it came from or how you know it. Do you know how to calculate the radius of convergence for a given series? (Look up the ratio test if you don't).

    As a last aside - if you look at the series you derived and the series for ##\ln(1-x)##, what do you notice?
     
    Last edited: Nov 29, 2012
  10. Nov 29, 2012 #9
    [QUOTE
    This is not correct for the radius of convergence. It's called a 'radius' because it has to be symmetric: |x| < R or -R < x < R. Also, how do you know that you need ##|x-2| < 1##? This is correct, but you haven't explained where it came from or how you know it. Do you know how to calculate the radius of convergence for a given series? (Look up the ratio test if you don't).
    QUOTE]


    I used taylor series to get that. then put it into sigma notation and took the limit n→∞

    f(x) = (x - 2) - 1/2(x -2)^2 + 1/3(x - 2)^3 +...

    Ʃ [(x - 2)^n] / n lim n→∞ [(x - 2)^n+1] / n+1 = |x-2|< 1 i was comparing the |x-2| to the radius of convergence to the ln(1-x) which is |x|<1
     
  11. Nov 30, 2012 #10

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    Ok! So, there are basically two ways to get the radius of convergence in this problem. One is to perform the ratio test with the series you derived, which it looks like you tried to do. If you use the ratio test you will indeed find that you need ##|x-2| < 1##, as you have said. Or, as you also suggest, you can compare the series to the series for ##\ln(1-x)##, for which you need ##|x| < 1##, and you deduced that since the series are basically the same that you need ##|x-2|<1##. Both ways are fine, I just want to make sure you understand how you got to them.

    Ok, so now that we know we need ##|x-2|<1##, what are the radius and interval of convergence? You still haven't explicitly written it down.

    Also, since you have pretty much solved the problem at this point, I want to point out now that the reason you could use the series for ##\ln(1-x)## to deduce that ##|x-2| <1## for your series for ##\ln(3-x)## is that they are the same series! The only difference is that x has been shifted in one of them. Notice that ##\ln(3-x) = \ln(1 + 2 - x) = \ln(1 - (x-2))##. This was what I was trying to get you to notice before. The problem was set up in a way that all you needed to do was realize that the series for ##\ln(3-x)## about x = 2 is exactly the same as the series for ##\ln(1-y)## about y = 0, where you take y to be x-2. Then you can use everything you knew about the series for ##\ln(1-y)## to find things like the radius or interval of convergence.

    Since the problem was basically set up to get you to exploit that fact, that's why I was trying to get you to do it this way instead of deriving the series from scratch.
     
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