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Homework Help: Find formula for sum of square roots

  1. Apr 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the asymptotic formula for

    [tex]\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{3}+...+\sqrt{n}[/tex]

    in the form of [tex]c \ast n^{\alpha}[/tex]
    Identify c and alpha.
    (Do NOT use the fundamental theorem of calculus)

    2. Relevant equations

    Area under curve [tex]y = \sqrt{x}[/tex] in [0, 1]
    is [tex]\frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i}[/tex]

    3. The attempt at a solution
    ...I'm really not sure from where to start...
    I know [tex]c = \frac{2}{3}[/tex] and [tex]\alpha = \frac{3}{2}[/tex], but do not know how to prove it.
     
    Last edited: Apr 11, 2008
  2. jcsd
  3. Apr 11, 2008 #2

    Dick

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    Go with the area under the curve idea. The sum from 1 to n of sqrt(n) is approximated by the area under the curve sqrt(x) from x=0 to n. Set that up as an integral and evaluate it.
     
  4. Apr 11, 2008 #3
    Sorry I forget to say that the fundamental theorem of calculus cannot be used, otherwise it is pretty straight forward.
    Thanks for the quick reply.
     
  5. Apr 11, 2008 #4

    Dick

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    Ok, then. If c is the area under the curve sqrt(x) on [0,1] and I(n) is your sum of square roots, then you have c=limit n->infinity (1/n^(3/2))*I(n) from your Riemann integral type expression (were you just given that?). So for large n, I(n)~c*n^(3/2). There's the 3/2. The easy way to get c is to integrate to get the area. But not allowed? Ok, then look at I(n)-I(n-1)~c*(n^(3/2)-(n-1)^(3/2)). Separate the n's from the c and take the limit as n -> infinity. Is that allowed? (All of this is really just doing calculus the HARD WAY).
     
  6. Apr 11, 2008 #5
    Well, all the fundamental theorem of calculus says is that definite integrals can be written in terms of antiderivatives, so as long as you don't use that fact, it should be a problem, no? So, as Dick suggests, set up the problem as an integral, and then evaluate the Riemann sums directly and take the limit, rather than plugging in known antiderivatives.
     
  7. Apr 11, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi zillac! Welcome to PF! :smile:

    Hint: when n -> ∞, cn^α + √n -> c(n+1)^α. :smile:
     
  8. Apr 12, 2008 #7
    I tried using Riemann sums and it will go back to finding the sum of the sequence ([tex]\frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i}[/tex])

    I(n)-I(n-1)~c*(n^(3/2)-(n-1)^(3/2)) will work when alpha is found.
    So,
    c = [tex]\frac{\sqrt{n}}{n^\frac{3}{2}-(n-1)^\frac{3}{2}}[/tex]
    ...
    = [tex]\frac{1+(1-1/n)^\frac{3}{2}}{3-3/n-3/n^2}
    [/tex]
    lim c = 2/3
    ---------
    then for the first half of question
    lim Area = lim 1/(n^3/2) * ∑√i ( lim n -> infinity)
    -> Area * n^3/2 = ∑√i = c * n^a
    -> alpha = 3/2..

    Is this precise enough to conclude the value of alpha?
     
    Last edited: Apr 12, 2008
  9. Apr 12, 2008 #8

    Dick

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    Sure, if you presume that the area is a well defined constant, the riemann sum immediately gives you the n^(3/2) dependence. You just have to work a little harder to get the c.
     
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