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Homework Help: Simplifying a/(2^a)

  1. Oct 29, 2009 #1
    Is it possible to simplify this?


    It's actually part of a limit:


    The limit I'm hoping tends to 3/4 as n tends to infinity. It certainly appears to do so when I plotted a graph of a/(2^a).

    Any little hints? Tried logs but got nowhere: I still end up with an 'a' on the top and bottom. I'm assuming that I need to elliminate it from either the numerator or the denominator.
    Thanks in advance.
  2. jcsd
  3. Oct 29, 2009 #2
    Okay, I got:

    Where [tex]C=log{2}[/tex].
    I don't know what to do with this.
    Last edited: Oct 29, 2009
  4. Oct 29, 2009 #3
    Does this help:

    [tex] \frac{d}{da} \left ( \frac{-1}{2^a} \left ) = \frac{a}{2^{a + 1}} [/tex]
  5. Oct 29, 2009 #4
    I couldn't get the same result. I got:
    [tex]\frac{d}{da} \left (\frac{-1}{2^a} \left ) = \frac {d}{da} (-1 * 2^{-a}) = \frac{\ln {2}}{2^a}[/tex]
    [tex]f(a)= \frac{-1}{2^a} = -1 * 2^{-a}[/tex]

    [tex]\ln {f(a)} = -1 * -a * \ln {2}[/tex]

    [tex]\frac{d}{da} \left (\ln {f(a)} \left ) = \ln {2} * \frac{d}{da} (a)[/tex]

    [tex]\frac {1}{f(a)} * \frac{d}{da} f(a) = 1 * \ln {2}[/tex]

    [tex]\frac{d}{da} f(a) = f(a) * \ln {2} = \frac{-1}{2^a} * \ln {2} = \frac{-\ln {2}}{2^a}[/tex]

    Is my calculus out?

    Am I trying to establish the following?:
    [tex]\int_4^\infty (\frac{a}{2a}) da[/tex]

    Any help much appreciated, thanks so much for the responses.
  6. Oct 29, 2009 #5


    User Avatar

    Staff: Mentor

    Wolfram alpha says you are right:

  7. Oct 29, 2009 #6


    Staff: Mentor

    You could approximate
    [tex]2 \sum_{a=4}^{\infty}\frac{a}{ 2^a}[/tex]
    by this integral:
    [tex]2 \int_4^{\infty} \frac{x dx}{2^x}[/tex]
    assuming that both converge.
  8. Oct 29, 2009 #7
    Thanks for the replies guys!
    I'm really sorry, I don't understand.
    I thought
    [tex]\sum_4^\infty \left (\frac{a}{2^a} \left )[/tex] would be a less accurate approximation of [tex]\int_4^\infty \left (\frac{a}{2^a} \left ) da[/tex];
    and that
    [tex]\int_4^\infty \left (\frac{a}{2^a} \left ) da = \int_4^\infty \left (\frac{x}{2^x} \left ) dx[/tex] when [tex]a = x[/tex]
    That is, are they not the same?
    I'm assuming they do as you have suggested it! But how do I make a safe assumption that this is the case?
    Thanks again!
    Last edited: Oct 29, 2009
  9. Oct 29, 2009 #8


    Staff: Mentor

    There's the Integral Test for infinite series...

    BTW, this is not a Precalculus problem. It should be in the Calculus and Beyond section.
  10. Oct 29, 2009 #9
    Thanks, I'll look it up. Might need to wish me luck!.....

    Apologies, but I didn't know that it would entail calculus.

    Once again, thanks for all your help!
  11. Oct 29, 2009 #10
    How do I go about doing this? It's difficult to integrate the function. I tried the easy way out by using the calculator but that didn't work.
  12. Oct 29, 2009 #11


    Staff: Mentor

    Change to [tex]2 \int_4^{\infty} x e^{-x ln 2}dx}[/tex]
    This can probably be done by integration by parts, with u = x, and dv = e-x ln2dx. That's how I would start.
  13. Oct 29, 2009 #12


    Staff: Mentor

    Also, since this is an improper integral, you would need to take a limit:
    [tex]2 \lim_{b \rightarrow \infty} \int_4^b x e^{-x ln 2}dx}[/tex]
  14. Oct 29, 2009 #13
    Thanks Mark, it seems you always come to my rescue.
    Do I integrate first with the limit present for b?
    i.e. [tex]\int_4^b x e^{-x ln 2}dx}[/tex], then employ the limit?
    I am not familiar with this, it's more 'advanced' than what I have come across so far.

    I'll go away and have a crack at it tomorrow, start afresh. (It's fairly late in Merry Old England.)
    Once again, thanks for the continuing help.
  15. Oct 29, 2009 #14


    Staff: Mentor

    Yes, integrate first, which will give you an expression involving b. Then take the limit.
  16. Oct 29, 2009 #15
    lol sorry about that nobahar, my mistake.
  17. Oct 30, 2009 #16
    No problem!

    I attempted to integrate the function:
    [tex]\int xe^{-x \ln{2}} dx = \frac{x * e^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \frac{e^{-x \ln{2}}}{(\ln{2})^2} = \left - \left \frac{(\ln {2})^2 * x * e^{-x \ln{2}} - e^{-x \ln{2}} * \ln{2}}{(\ln{2})^3} = \left - \left \frac{\ln{2} * x * e^{-x \ln{2}} - e^{-x \ln{2}}}{(\ln{2})^2}[/tex]
    (Including +C)
    Is this correct so far?
  18. Oct 30, 2009 #17
    Attempt to 'simplify':

    [tex]\left - \left \frac{e^{-x \ln{2}}(\ln{2} * x + 1)}{(\ln{2})^2}[/tex]
  19. Oct 31, 2009 #18


    Staff: Mentor

    Assuming that you did integration by parts correctly (I didn't check), and that you ended up with this, about the only thing I can see to do to simplify it is replace e-xln2 with 1/2x.
  20. Oct 31, 2009 #19
    I'll post the workings.

    [tex]\int xe^{-x \ln{2}} dx[/tex]

    [tex]u = x \left \left and \left \left \frac{dv}{dx} = e^{-x \ln{2}} \left \left then \left \left \frac{du}{dx} = 1 \left \left and \left \left v = \int e^{-x \ln{2}} dx = \frac{e^{-x \ln{2}}}{- \ln{2}}[/tex]

    [tex]\int xe^{-x \ln{2}} dx = \frac{xe^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \int \frac{e^{-x \ln{2}}}{- \ln{2}} dx = \frac{xe^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \frac{e^{-x \ln{2}}}{(\ln{2})^2}[/tex]
    Last edited: Oct 31, 2009
  21. Nov 2, 2009 #20
    Okay, so:
    [tex]\frac{x \ln{2} + 1}{2^x(\ln{2})^2}[/tex]

    Now I need:
    [tex]\lim_{b \rightarrow \infty}[\frac{x \ln{2} + 1}{2^x(\ln{2})^2}]^{x=b}_{x=4}][/tex]

    I don't know how to do this. Where would I start?
    Last edited: Nov 2, 2009
  22. Nov 2, 2009 #21
    It seems to me that JG89's comment is in the right direction of the usual way to work this problem. It does involve a trick that you'll need to see spelled out in detail. Remember it and use it to solve similar problems. Consider for some positive integer [itex]n[/itex] the function

    [tex]f(x) = \frac{x^n}{2^n}[/tex]​

    It is everywhere continuous and differentiable, and its derivative is

    [tex]f'(x) = \frac{n}{2^n}x^{n-1}[/tex]​

    Next, consider the sum:

    [tex]F_n(x) = \sum_{a=4}^n \frac{x^a}{2^a}[/tex]​

    so that we have a geometric series, i.e.

    [tex]F_n(x) = \sum_{a=4}^n \frac{x^a}{2^a} = \frac{x^4/16 - x^5/2^{n+1}}{1-x/2}[/tex]​

    The final step is to take the derivative:

    [tex]F'_n(x) = \sum_{a=4}^n \frac{a}{2^a}x^{a-1} = \frac{(x^3/4-5x^4/2^{n+1})(1-x/2) + 1/2(x^4/16-x^5/2^{n+1})}{(1-x/2)^2}[/tex]​

    Notice that

    [tex]\lim_{n \rightarrow \infty} F'_n(1) = \sum_{a=4}^\infty \frac{a}{2^a} = 5/4[/tex]​

    is the limit that you are looking for.
  23. Nov 3, 2009 #22
    That's truly amazing. Many, many thanks Tedjn. Initially, it's a little hard to follow because the derivitive is of the variable x, but then you take the sum of a. But I think I understand it. I'll work through it again.
    Once again, many thanks.
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