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Riemann sum of derivative (something like that)

  1. Jul 15, 2006 #1

    quasar987

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    Hi, maybe someone can help. When I think about it, I'm pretty sure that the following is true: Let c be a curve parametrized by [itex]t\in [a,b][/itex], let [itex]\sigma = \{t_0,...,t_N\}[/itex] be a partition of [a,b] and [itex]\delta_{\sigma}=\max_{0\leq k \leq N-1}(t_{k+1}-t_k)[/itex]. Also define [itex]\Delta t_k=t_{k+1}-t_k[/itex] Then,

    [tex]\lim_{\delta_{\sigma}\rightarrow 0}\sum_{k=0}^{N-1}\frac{|c(t_k+\Delta t_k)-c(t_k)|}{\Delta t_k}\Delta t_k=\int_a^b |\frac{dc}{dt}(t)|dt[/tex]

    Proving this would also amount to proving

    [tex]\lim_{\delta_{\sigma}\rightarrow 0}\sum_{k=0}^{N-1}\frac{|c(t_k+\Delta t_k)-c(t_k)|}{\Delta t_k}\Delta t_k=\lim_{\delta_{\sigma}\rightarrow 0}\sum_{k=0}^{N-1} |\frac{dc}{dt}(t_k)|\Delta t_k[/tex]

    Is there a way to do this using a finite succession of arguments?
     
  2. jcsd
  3. Jul 15, 2006 #2

    Hurkyl

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    If it can be done directly, it looks like a differential approximation should be the obvious thing to do! Remember that for any differentiable f, there is a remainder r such that:

    [tex]
    f(x + k) = f(x) + k f'(x) + k r(x, k)
    [/tex]

    and, for each x,

    [tex]
    \lim_{k \rightarrow 0} r(x, k) = 0
    [/tex]


    Of course, what you want to prove is trivial over any interval where c is monotonic.
     
    Last edited: Jul 15, 2006
  4. Jul 15, 2006 #3

    quasar987

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    Hi Hurkyl,

    If I make that substitution in the riemann integral, I get

    [tex]\lim_{max(k_i)\rightarrow 0} \sum_i^N \frac{f(x_i+k_i)-f(x_i)}{k_1}k_i = \int_a^b f'(x)dx + \lim_{max(k_i)\rightarrow 0} \sum_i^N r(x_i,k_i)k_i[/tex]

    So the problem is essentially the same: In my OP, I knew that the differential ratio was going to the derivative as [itex]\delta_{\sigma} \rightarrow 0[/itex] so the limit of the riemann sum should be [itex]\int_a^b |\frac{dc}{dt}(t)|dt[/itex], but did not know how to prove it. Now I know that as max(k)-->0, r(x,k)-->0, so the limit of the riemann sum should be [itex]\int_a^b 0dx[/itex], but still don't know how to prove it. :grumpy:
     
  5. Jul 15, 2006 #4

    Hurkyl

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    If you take off the absolute value signs, then it's very easy. Your sum is a telescoping series, and your integral is easily integrated.

    If you don't want to use that... do you know about uniform convergence?


    Anyways... *bonks self* forget about differential approximation. This is a job for the mean value theorem!
     
  6. Jul 15, 2006 #5

    quasar987

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    Hurray for Hurkyl and the mean value theorem! :D
     
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