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Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & Weber

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    With n>1, show that (a) [tex]\frac{1}{n}[/tex]-ln[tex]\frac{n}{n-1}[/tex]<0
    and (b) [tex]\frac{1}{n}[/tex]-ln[tex]\frac{n+1}{n}[/tex]>0

    Use these inequalities to show that the Euler-Mascheron constant (eq. 5.28 - page330) is finite.

    2. Relevant equations
    This is in the chapter on infinite series, in the section on Taylor Expansion, so I guess Taylor, Maclaurin, and Binomial theorem are fair game.



    3. The attempt at a solution
    I first wrote the logarithm as a difference of logs and then tried to expand them in the Maclaurin series. But that apparently doesn't work since ln(0) and 1/0 are undefined...

    I also don't understand the statement at the end. Is that supposed to be a hint or a third part to the problem?

    Any help would be great, thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 4, 2008 #2

    lurflurf

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    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    For the first two part change the log to log(1+x) so that we want to show
    x/(1+x)<log(1+x)<x
    for |x|<1
    this can be done by using an integral or series representation for log
    The third bit is indeed a part of the problem and not a hint, infact one could say the first two parts are a hint for the third
    The constant in question is
    lim_{n->infinity} [1/1+1/2+1/3+...+1/n-log(n)]
    use the given inequalities to bound the constant
     
  4. Dec 4, 2008 #3
    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    I'm not exactly sure how to go about changing the log to log(1+x). Do you mean to factor an x out of the demoninator for the first one and then have log(1-1/x), and for the second one log(1+1/x)?
     
  5. Dec 4, 2008 #4

    HallsofIvy

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    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    I think he means let x= n/(n-1) so that ln(n/(n-1)) becomes just ln x. Of course, then x(n-1)= xn- x= n so n(x-1)= x and n= x/x-1 so that 1/n= (x-1)/x= 1- 1/x. Then your inequality 1/n- ln(n/(n-1))<0 becomes 1- 1/x- ln(x)< 0.
     
  6. Dec 4, 2008 #5
    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    Ok, but I'm not seeing how the two approaches are connected since lurflurf said to change the logarithm to ln(1+x).
     
  7. Dec 4, 2008 #6

    lurflurf

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    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    find that log(1+x)~x (x~0)is a simple form to work with
    log(x)~-1+x (x~1) amounts to the same thing, matter of taste

    we have
    1/n-log(n/(n-1))
    and
    1/n-log((n+1)/n)

    n/(n-1)=1+1/(n-1)
    let
    x=1/(n-1)
    n/(n-1)=1+1/(n-1)->1+x
    1/n=1/(n-1+1)=[1/(n-1)]/[1+1/(n-1)]->x/(1+x)
    1/n-log(n/(n-1))->x/(1+x)-log(1+x)

    (n+1)/n=1+1/n
    let x=1/n
    (n+1)/n=1+1/n->1+x
    1/n-log((n+1)/n)->x-log(1+x)

    we desire to show
    x-log(1+x)>0
    and
    x/(1+x)-log(1+x)<0
    or
    x/(1+x)<log(1+x)<x
    |x|<1
    This is easily done by any number of methods including using series
    log(1+x)=x-x/2+x^3/3-x^4/4+...
    or integrals
    log(1+x)=int(1/(1+t),t,0,x)

    once the hint inequalitites are verified use them to bound the constant
     
  8. Dec 5, 2008 #7
    Re: Taylor expansion problem 5.6.9 in Math Methods for Physicists, 6th ed Arfken & We

    Ok, that helps a lot. Thanks!
     
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