Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interesting question - Not homework

  1. Nov 24, 2005 #1
    Interesting question - Not homework!!

    Show that the sum of the infinite series:
    [tex] log_2 e - log_4 e + log_{16} e + {(-1^n)} log_{2^{2n}} e .... [/tex]
    equals [tex] \frac{1}{ln2\sqrt2} [/tex]
    Any ideas?!
     
    Last edited: Nov 24, 2005
  2. jcsd
  3. Nov 24, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    what is the relationship between log base A and log base B? converting all those logs into base e would be the place to start
     
  4. Nov 24, 2005 #3
    are you sure [tex] {(-1^n)} log_{2^{2n}} e .... [/tex] is correct ?
     
  5. Nov 24, 2005 #4
    Well perhaps there should be ..... before it but stilll.....any ideas?!! on how to proove?
     
  6. Nov 24, 2005 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I told you, change base in logs so that they are all base e.
     
  7. Nov 24, 2005 #6

    mezarashi

    User Avatar
    Homework Helper

    matt has been referring to the identity: [tex]log_2 e = \frac{ln e}{ln 2}[/tex]


    [tex] log_2 e - log_4 e + log_{16} e + ... {(-1)^n} log_{2^{2n}} e[/tex]

    [tex] \frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^n}[/tex]


    [tex] \frac{1}{ln2} ( 1 - \frac{1}{2} + ... {(-1)^n} \frac{1}{2^n})[/tex]


    Geometric series formula still at hand?
     
    Last edited: Nov 24, 2005
  8. Nov 24, 2005 #7

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    Your nth term can't be correct. Do you know the term right after [tex]\log_{16} e[/tex]?
     
  9. Nov 24, 2005 #8
    mezarashis nth term isnt correct either
     
  10. Nov 24, 2005 #9

    dx

    User Avatar
    Homework Helper
    Gold Member

    The nth term is [tex](-1)^{n+1}log_{2^n}{e}[/tex]
     
  11. Nov 24, 2005 #10

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    That won't sum to [tex]\frac{1}{\log 2^{3/2}}[/tex] either, and it doesn't match the 3rd term of the given series.

    For that matter the nth term NewScientist supplied doesn't match the first term. No need to look to the fourth to see it's not correct, though the 4th should give more insight on what the nth term is.
     
    Last edited: Nov 24, 2005
  12. Nov 24, 2005 #11
    No dx, the nth term is
    [tex] log_{2^{2n}} e [/tex]
    The step : [tex] \frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^n}[/tex]
    is correct
    Then one must obsewrve this is a GP with r = -1/2 and a 1/In 2^2n
    Sum to infinity using a / (1-r)
    Which gives
    [tex] \frac{\frac{1}{ln2}}{1+\frac{1}{2}} [/tex]
    This simplifies to
    [tex] \frac{1}{1.5 ln2} [/tex]
    [tex] {1.5 ln2} = ln 2^\frac{3}{2} [/tex]
    which gives
    [tex] ln 2\sqrt{2}[/tex]
     
  13. Nov 24, 2005 #12

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    If this is your nth term then:

    is wrong. [tex]\log_{2^{2n}} e=\frac{\log e}{\log 2^{2n}}[/tex]

    In any case, the sum that you wrote that I just quoted is Not a geometric series. It's a constant times [tex]\sum_{n=1}^{\infty}\frac{(-1)^n}{n}[/tex]
     
    Last edited: Nov 24, 2005
  14. Nov 24, 2005 #13

    but the first term, n=1 is positive though
     
  15. Nov 24, 2005 #14

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    "a constant times"
     
  16. Nov 24, 2005 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it isn't. For one thing, if the general term is [itex]log_{2^{2n}} e[/itex] as you keep insisting, then the first term, with n= 0 would be [itex]log_1 e[/itex] which does not exist. If we take n= 1 as the first term then it would be [itex]log_{4} e[/itex], not [itex]log_{2} e[/itex]. However, let's assume that the general form is [itex]log_{2^{2n}} e[/itex], starting with n= 1 and add on [itex]log_2 e[/itex] at the end.
    Since ln e= 1 (I don't know why everyone kept writing ln e.) This series is
    [tex]\frac{1}{2 ln 2}\Sigma_{n=1}^\infnty \frac{(-1)^n}{n}[/itex]
    except for the factor [itex]\frac{1}{2 ln 2}[/itex] this is the "alternating harmonic series" which is well known to converge to ln 2. In other words that series converges to 1/2! Adding on that first term, [itex]log_2 2= \frac{1}{ln 2}[/tex] which did not conform to the given general term, we have
    [tex]\frac{1}{2}+ \frac{1}{ln 2}= \frac{1+ ln 2}{2 ln 2}[/tex]
     
    Last edited: Nov 24, 2005
  17. Nov 24, 2005 #16
    Okay, I have been jumping backwards and forwards to this question all day so here is my full solution - one i have thought about properly! I hope I can make some of you eat your words!!
    Lets go
    [tex] log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e .... [/tex]
    This is given in the question and yes the 'fourth' term (the general term IS correct, let n= 0,1,2,3....
    Now, the 'end result' is in ln so lets convert to base e and see where thatgets us
    [tex] log_a x = \frac{log_b x}{log_b a} [/tex]
    Let
    [tex] x = e = b [/tex]
    [tex] a = 2^{2n} [/tex]

    This gives

    [tex] log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e .... = \frac{ln e}{ln 2} - \frac{ln e}{ln 4} + ... + {(-1^n)} \frac{ln e}{ln 2^{2n}} [/tex]

    [tex] log_x x = 1 \rightarrow lne = 1 [/tex]

    Which gives

    [tex] log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e .... = \frac{1}{ln 2} - \frac{1}{ln 4} + ... + \frac{(-1^n)}{ln 2^{2n}} [/tex]

    It is clear that the common ratio of the terms is -1/2. The first term is 1/ln2. The series is infinite so n = infinity! Let us sum this GP.

    [tex] Sn = \frac{a}{1-r} [/tex]

    [tex] \frac{\frac{1}{ln2}}{1+\frac{1}{2}} [/tex]

    [tex] \frac{1}{1.5 ln2} [/tex]

    [tex] {1.5 ln2} = ln 2^\frac{3}{2} [/tex]

    [tex] ln 2\sqrt{2}[/tex]

    QED
     
    Last edited: Nov 24, 2005
  18. Nov 24, 2005 #17

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    As Halls pointed out, n=0 in your general term does not exist and hence is not [tex]\log_2 e[/tex].

    Try [tex](-1)^n \log_{2^{(2^n)}} e[/tex] as your general term, starting at n=0.

    edit- again, even using your general term you DO NOT get a geometric series. With your general term your n=2 and n=3 terms are [tex]\frac{1}{\log{16}}-\frac{1}{\log{64}}=\frac{1}{4\log{2}}-\frac{1}{6\log{2}}[/tex]... the n=3 term is not -1/2 times the n=2 term.
     
    Last edited: Nov 24, 2005
  19. Nov 24, 2005 #18
    n can = 0

    edit : [tex] \frac{1}{ln2} = 2\frac{1}{ln4} [/tex]

    thus we have a geometric series - don't we?!

    -NS
     
    Last edited: Nov 24, 2005
  20. Nov 24, 2005 #19

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    [tex]\log_{2^{2(0)}}e=\log_{2^0}e=\log_{1}e[/tex] which is undefined, unless you can tell me what power to raise 1 to to get e...
     
  21. Nov 24, 2005 #20
    How do i correct proof then?! As my result is correct?! And indeed i never claimed log of base 2^2n worked for the first term :P
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Interesting question - Not homework
  1. Interesting question (Replies: 8)

  2. An Interesting Question (Replies: 27)

  3. Interesting Question (Replies: 0)

Loading...