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2 Homework Problems

  1. Aug 11, 2005 #1
    Infinite Series Calculus 2 hw problem

    Hello. I need help finding the answer on two of these.

    #1. Find a power series representation for [tex]f(x)=ln(1+x^2)[/tex]

    I have
    So I factor out 2x and have [tex]\frac{1}{(1+x^2)}[/tex] which is a geometric series =
    [tex]\sum(-1)^{n}x^{2n}[/tex] then multiply 2x times the series
    2x * [tex]\sum(-1)^{n}x^{2n}[/tex] = [tex]\sum(-1)^{n}2x^{2n+1}[/tex]

    Then, [tex]\int\sum(-1)^{n}2x^{2n+1}[/tex] = [tex] \sum\frac{(-1)^{n+1}2x^{2n+2}}{2n+2}[/tex]

    Did I do that right??

    Ok Problem #2 Find the radius of convergence and interval of convergence of the series.
    for R I get 1/2 and I am having trouble figuring out whether the endpoints (5/2)<x<(7/2), are convergent.

    Last edited: Aug 11, 2005
  2. jcsd
  3. Aug 11, 2005 #2


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    I'm not sure what you're doing:

    [tex]f(x) = \sum _{n=0} ^{\infty } \frac{f^{(n)}(0)}{n!}x^n[/tex]

    what you're doing looks nothing like this, so unless I made a mistake or am missing something, what you did looks wrong. Here's a quick check:

    ln(1 + 1²) = ln(2) = 0.69314718055994530941723212145818 ~ 0.69

    On the other hand:

    [tex] \sum\frac{(-1)^{n+1}2(1)^{2n+2}}{2n+2}[/tex]

    [tex] = \sum\frac{(-1)^{n+1}}{n+1}[/tex]

    [tex] = -\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} < 0 < 0.69[/tex]

    so that's definitely wrong.
  4. Aug 11, 2005 #3
    Ok say I want to find the power series representation for ln(5-x)
    it will be equal to [tex]ln(5)-\sum\frac{x^{n+1}}{(n+1)5^{n+1}}[/tex]
  5. Aug 11, 2005 #4
    i think

    I think what you did on the first one is correct, or at least i got the same result.
    I think AKG counterexample is not valid, because the power series you gave:
    only converges for -1<x<1 (This comes from the interval of convergence of the geometric where you started from).

    Problem 2: Your endpoints are wrong. The power series converges absolutely for -7/2<x<-5/2 and you should check for x=-7/2 and x=-5/2. For one you should use leibnitz and for the other one is just a p-series.

    Maybe i am saying something stupid, because i dont know a lot about math.
    I hope this helps, Paul.
  6. Aug 11, 2005 #5


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    Well if you take x very close to 1, but less than 1, then your series will still be negative. On the other hand ln(1 + x²) will still be close to 0.69.
  7. Aug 11, 2005 #6
    I found the series to be divergent at both endpoints. Is that what you got pbialos?

    AKG is he right about the first one?

    Thanks for the help
  8. Aug 11, 2005 #7


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    I haven't touched series in over a year, but the formula I gave you in the second post does give the McLaurian expansion (which is just the Taylor expansion). If that's what you're doing (if that's what you mean by power series) then what you're doing looks nothing like what I've seen before. Also, if pbialos is right in that the interval of convergence is (-1,1) then the modification of my counteraxmple which I gave in post 5 should hold. Plug in x=1 and you'll see that they're very unequal: the series is negative and ln(1 + x²) = ln(2) is approximately 0.69. If 1 is not a valid counterexample, then it's still pretty clear that x = 0.9999 will give ln(1 + x²) close to 0.69 (at very least, it will be positive) and the series will still be negative.
  9. Aug 11, 2005 #8

    For the second one, i got divergent when x=-7/2 and convergent for x=-5/2.
    On the first one, i am not sure, maybe AKG or someone else can show us what is wrong with your procedure. I am just studying this subject myself, and i am not really sure about a lot of things.

  10. Aug 11, 2005 #9
    How did you get convergent for x=-5/2?
    [tex]\frac{(-2)^{n}(-\frac{1}{2})^{n}}{\sqrt{n}}[/tex] = [tex]\frac{(-1)^n}{\sqrt{n}}[/tex] is a divergent p-series p<1
  11. Aug 11, 2005 #10


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    I'm not sure what you're procedure is. What does your textbook say? Have you seen a formula like this?:

    [tex]f(x) = \sum _{n=0} ^{\infty } \frac{f^{(n)}(0)}{n!}x^n[/tex]

    I don't know if there are various power series expansions of functions, but the Taylor series expansion does express f(x) as a power series, and it is given by the formula above. Actually, the above is a special case where you expand about 0, since it's easiest to do so, but in general, the expansion is given by:

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

    Could you tell me how you're getting what you're getting, and why you're doing it. What process does your book give you? Why are you factoring out 2x? Why aren't you computing higher order derivatives, and why is there no "n!" in your summand?
  12. Aug 11, 2005 #11

    [tex]\sum\frac{(-1)^n}{\sqrt{n}}[/tex] is not a p-series, it is an alternating series. You should use the alternating series test.
  13. Aug 11, 2005 #12


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    I you integrate the series term-wise correctly, you get:

    [tex]\int\sum_{n=0}^\infty (-1)^{n}2x^{2n+1} = \sum_{n=0}^\infty \frac{(-1)^n 2x^{2n+2}}{2n+2} + C=\sum\frac{(-1)^n x^{2n+2}}{n+1} + C[/tex]
    So your answer had the wrong sign and you had to account for the constant of integration (which you can find by plugging in x=0 for example).

    You could also get it from the power series of ln(1+x) (gotten by integrating 1/(1+x)):

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

    So replacing x by x^2:

    [tex]\ln(1+x^2)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+2}}{n+1}[/tex]
  14. Aug 11, 2005 #13
    Thanks I see my mistakes!
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