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Value of a sum

  1. Nov 15, 2005 #1
    Given the sum
    [tex]{_\lim {i} \rightarrow 0} \sum_{k=0}^{\frac{x}{i} - 1} i\sqrt{1 + i^{2n-2}((k+1)^{n} - k^{n})^{2}}[/tex]
    I want to know how to derive to the value of this sum exactly. This is actually the value of the lenghts of a curve from a point to the origin of the form f(x) = x^n... I thought the binominal theorem can be used, but i can't develop on this further more. Anyone is capable of showing to what value this converges?
     
    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 18, 2005 #2
    I'll give this a go. The added bonus is if I'm wrong one of the more math-y people in here is bound to catch my mistake :)
    What I first did is rewrote your equation like this:
    [tex]{_\lim {i} \rightarrow 0} \sum_{k=0}^{\frac{x}{i} - 1} i\sqrt{1 + (i^{n-1}((k+1)^{n} - k^{n}))^{2}}[/tex]
    Now look at [tex]i^{n-1}((k+1)^{n} - k^{n})[/tex]. What value does this converge to as i-> 0 and if [tex]k = \frac{x}{i}-1[/tex]? Do a binomial expansion and you will get an open form of something like [tex]i^{n-1}(((\frac{x}{i})^{n} - ((\frac{x}{i})^{n} - n(\frac{x}{i})^{n-1} - o(n-2))))[/tex] where o(n-2) is something that isn't particularly relevant to the answer that is of order n-2. I think you will find that for this particular value of k (and in fact all k) if you work out say [tex]k = \frac{x}{i}-2[/tex] and so on that you will find the limit always converges to the same value: [tex]nx^{n-1}[/tex], so the sum simplifies somewhat to [tex]{_\lim {i} \rightarrow 0} \sum_{k=0}^{\frac{x}{i} - 1} i\sqrt{1 + (nx^{n-1})^{2}}[/tex]

    Edit: It certainly doesn't converge. I eliminated the iterated variable k entirely, I don't really know what to make of that sum to be honest. I'll re-edit this tomorrow with maybe something.
     
    Last edited by a moderator: Nov 19, 2005
  4. Nov 23, 2005 #3

    benorin

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    Shouldn't [tex]\frac{x}{i} - 1[/tex] be an integer?
     
  5. Nov 23, 2005 #4
    it can be an integer

    certainly not (unless it is not mentioned that x is complex).it can be proved .by solving we get x-1\i.further xi + 1\-1. if x is not complex the expression will never be integer . x has to be an complex which is raised to an odd no: (which is quiet easy to understand )it also depends on the coefficient of x(x=ai ) then if a is any real number then then it depends on what a is .i think i am right .if wrong please send me a reply and also tell how u tpye equations in forums
     
  6. Nov 25, 2005 #5

    benorin

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    We use latex code to type equations in the forums: to see the latex code that generated a particular math expression, double click on it. Oh, and here is a link to the ever handy LaTeX code reference.

    And as to my question, it was rhetorical.
     
  7. Nov 26, 2005 #6
    i am sorry. i did not see what u wrote before but ur last line caught my attention. thanks for guidelines to type equations.
     
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