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A very difficult problem

  1. Feb 4, 2010 #1
    i have thought about that for serval days= =

    please help


    Attached Files:

  2. jcsd
  3. Feb 4, 2010 #2
    It is 35.565955205923352

    But I can't tell you about [tex]\sum_{2}^{n} \frac{1}{\sqrt{n}}.[/tex]
  4. Feb 4, 2010 #3
    how about steps?
  5. Feb 4, 2010 #4
    Just entered into the calculator.
  6. Feb 4, 2010 #5

    i want solving it without cal
  7. Feb 4, 2010 #6
    In what form do u want the answer.
    you can approximate the sum by integral 1/sqrt(n) between 1 and 19.
    notice that 19^2 is 361
    that is 2(19-1) = 36
    The answer is close enough
  8. Feb 4, 2010 #7
    >that is 2(19-1) = 36

    HOW come to this step?
  9. Feb 4, 2010 #8
    [tex] \sum_{1}^{361} \frac{1}{\sqrt{n}} [/tex]
    is approximately equal to
    [tex] \int_{1}^{361} \frac{1}{\sqrt{n}}. [/tex] =[tex]2(\sqrt{a}-\sqrt{b})[/tex]
    Last edited: Feb 4, 2010
  10. Feb 4, 2010 #9
    Why do you think there is any answer better than 35.565955205923352 ??
  11. Feb 4, 2010 #10
    I don't think there's an analytical solution, but you can approximate it.
    First approximation is integral,

    Next approximations would involve some play with operators.

    Let's define [tex]D\equiv \frac{d}{dx}[/tex] as the derivative operator, and observe the operator [tex]\sigma \equiv e^{D}[/tex]

    Functions of operators are defined via their Taylor Series, so:

    [tex]e^{D} f=\sum^{\infty}_{k=0}\frac{D^{k}f(x)}{k!}=\sum^{\infty}_{k=0}\frac{1}{k!}\frac{d^{k}f(x)}{dx^{k}}(x+1-x)^k=f(x+1)[/tex]

    (Notice that the last sum is the Taylor Series of f(z) around the point x)

    So [tex]\sigma[/tex] is some sort of a delay\shift operator.

    Let's now observe the operator S defined as


    You can notice that [tex]\sigma Sf=\sum^{n+1}_{k=1}f(k)[/tex]

    So [tex]\sigma S f-Sf=f(n+1)=\sigma f[/tex]

    And conclude [tex]S(\sigma -1)=\sigma[/tex]

    And if you recall the definition of [tex]\sigma[/tex] then:



    Knowing the expansion of the exponential function, you can approximate the Taylor expansion of S. The interesting fact is that S has a simple pole at "D=0", which means that in the expansion of S you will have a negative power of D:


    Which means that integration (the inverse of D) is the first approximation to discrete summation.

    A better approximation would be


    Operate S on [tex]\frac{1}{\sqrt{n}}[/tex]
    And get:


    And after some tiresome calculation for n=361


    Which I think is a pretty good approximation, if only to show that the operator thing is correct.
    Taking further derivatives would make a better approximation.
  12. Feb 6, 2010 #11
    You can also use the first 5 terms of the http://planetmath.org/encyclopedia/EulerMaclaurinSummationFormula.html" [Broken] (i.e. the integral, the terms involving the first derivative, and the terms involving the second derivative) to approximate the sum as follows :

    S_n = \sum_{i=1}^n \frac{1}{\sqrt{i}}\approx 2\sqrt{n} +\frac{1}{2\sqrt{n}} - \frac{1}{24n\sqrt{n}} -

    The result is then :

    S_{361} \approx 36.56797638

    The error relative to the true result is about 0.0055%.
    Last edited by a moderator: May 4, 2017
  13. Feb 6, 2010 #12
    As stated in http://en.wikipedia.org/wiki/Bernoulli_numbers" [Broken] if the first Bernoulli number is chosen according to the convention [tex]B_1 = -1/2[/tex] then the Euler–MacLaurin formula is :

    \sum\limits_{a\leq k<b}f(k)=\int_a^b f(x)\,dx \ + \sum\limits_{k=1}^m \frac{B_k}{k!}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m).

    whereas if we choose the convention [tex]B_1 = 1/2[/tex], then the Euler–MacLaurin formula is :

    \sum\limits_{a<k\leq b} f(k)=\int_a^b f(x)\,dx + \sum\limits_{k=1}^m \frac{B_k}{k!} \left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m). \
    Last edited by a moderator: May 4, 2017
  14. Feb 6, 2010 #13
    i can't understand

    S_n = \sum_{i=1}^n \frac{1}{\sqrt{i}}\approx 2\sqrt{n} +\frac{1}{2\sqrt{n}} - \frac{1}{24n\sqrt{n}} -


    can you explain it please?
  15. Feb 7, 2010 #14
    All I am doing is substituting [tex]f(n) = 1/\sqrt{n}[/tex] into the Euler-Maclaurin formula. This time from Mathworld (since the formula is expanded in full), the http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html" [Broken] formula can be written as :

    \sum_{k=1}^{n-1}f(k)=\int_0^n f(k) dk-1/2[f(n)-f(0)]+1/(12)[f^{'}(n)-f^{'}(0)]-1/(720)[f^{'''}(n)-f^{'''}(0)]+1/(30240)[f^{(5)}(n)-f^{(5)}(0)]-1/(1209600)[f^{(7)}(n)-f^{(7)}(0)]+...

    Note that i have written [tex]1/2[f(n)-f(0)][/tex] instead of [tex]1/2[f(n)+f(0)][/tex]. The former is consistent with http://en.wikipedia.org/wiki/Bernoulli_number" [Broken].

    Then I make the approximation :

    \sum_{k=1}^{n-1}f(k)\approx \int_0^n f(k)dk-1/2[f(n)-f(0)]+1/(12)[f^{'}(n)-f^{'}(0)]

    Then I substitute [tex]f(k) = 1/\sqrt{k}[/tex], and then add [tex]f(n)[/tex] to the result since the summation in the formula is up to [tex](n-1)[/tex] where as we want it up to [tex]n[/tex]

    All you have to do is to subsitute [tex]f(k) = 1/\sqrt{k}[/tex] into the formula above, and verify my result.
    Last edited by a moderator: May 4, 2017
  16. Feb 7, 2010 #15
    Last edited by a moderator: Apr 24, 2017
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