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Find equation for nth value.

  1. May 3, 2014 #1
    Sn = ∑ni = 1 (Sn - i)-1
    S0 = 1

    I want to know how to find the general equation for Sn (An example of what I mean by "general equation" would be
    Sn = ∑ni = 1i = n(n+1)/2).
    Here's S0 though S5:

    S0 = 1
    S1 = 1
    S2 = 1 + 1
    S3 = 1 + 1 + 1/2
    S4= 1 + 1 + 1/2 + 2/5
    S5 = 1 + 1 + 1/2 + 2/5 + 10/29

    Thanks in advance, sorry if anything is unclear.
     
  2. jcsd
  3. May 3, 2014 #2
    I dunno about an exact formula, but it seems that for large ##n##, ##S_n \approx \sqrt{2 n}##.
     
  4. May 3, 2014 #3
    That's neat! How did you get that result?
     
  5. May 3, 2014 #4

    jbunniii

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    We can view the sum as being a Riemann sum (using rectangles of width 1) corresponding to the following integral equation:
    $$f(x) = \int_{0}^{x}\frac{dt}{f(t)}$$
    Differentiating both sides, we get
    $$f'(x) = \frac{1}{f(x)}$$
    By the product rule, this is equivalent to
    $$ \frac{d}{dx}(f(x)\cdot f(x)) = 2$$
    Thus
    $$(f(x))^2 = 2x + c$$
    and
    $$f(x) = \sqrt{2x + c}$$
    The integral equation evaluated at ##x=0## forces ##c=0##.

    So ##f(x) = \sqrt{2x}## is a solution to the integral equation, and therefore, ##S_n = \sqrt{2n}## is an approximate solution to the original problem, when ##n## is large.
     
  6. May 3, 2014 #5
    Thanks a lot for explaining that, there's something I don't understand though:
    The integral equation evaluated at ##x=0## forces ##c=0##.
    Why is c=0 forced? Why isn't it c=1?
     
  7. May 3, 2014 #6

    jbunniii

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    We have
    $$f(x) = \int_{0}^{x}\frac{dt}{f(t)}$$
    Evaluating this at ##x=0## gives us
    $$f(0) = \int_{0}^{0}\frac{dt}{f(t)}$$
    The right hand side is an integral over an interval of zero width, so the result is zero. Note that this forces the integrand to have zero denominator at ##t=0##, but that's OK: the integral from 0 to 0 is still 0, and ##1/\sqrt{2t}## is (improperly) integrable over ##[0,x]## despite the singularity at ##t=0##.

    Don't be concerned that ##f(0) = 0## does not match ##S_0 = 1## - no one is claiming that ##\sqrt{2n}## is an exact fit, just that it is a good approximation for large ##n##. Note that even if we had ##c=1##, the difference between ##\sqrt{2n+1}## and ##\sqrt{2n}## is negligible for large ##n##.
     
  8. May 3, 2014 #7
    Okay thanks again, I understand now.
     
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