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Upper and lower sum

  1. Apr 26, 2012 #1
    Let [itex]f:[0,1]->ℝ[/itex] and [itex]f(x) = \frac{1}{n}[/itex] when [itex]x=\frac{1}{n²}, n=1,2,...[/itex] and 0 in other case.

    Define such spacing/interval [itex]D[/itex] that [itex]S_D-s_D < \frac{1}{100}[/itex]. Here [itex]S_D[/itex] refers to the upper sum and [itex]s_D[/itex] the the lower sum.


    Now, i'm not sure how to approach this really since the sum is defined so that the values of x jump a different interval every time. Any help would be appreciated. Hopefully i managed to explain the problem clearly enough.
     
    Last edited: Apr 26, 2012
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  3. Apr 26, 2012 #2

    SammyS

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    That looks like a typo.

    Is [itex]\displaystyle f(x) = \frac{1}{x}[/itex], when [itex]\displaystyle x=\frac{1}{n^2}, n=1,2,...[/itex] and 0 in otherwise.

    ... or perhaps ... [itex]\displaystyle f(x) = x[/itex], when [itex]\displaystyle x=\frac{1}{n^2}, n=1,2,...[/itex] and 0 in otherwise.

    Are you doing this with Riemann sums ? ... and ... Do the intervals need to be equal in size?
     
  4. Apr 26, 2012 #3
    It actually is [itex]f(x)=\frac{1}{n}[/itex]. Yea, i'm trying to approach this with Riemann sums and the intervals don't need to be equal in size as far as i understand.
     
  5. Apr 26, 2012 #4
    Right, I gave it a wild shot and this is what i came up with.

    The upper sum is defined as
    [itex]\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x))[/itex]

    Now let's choose such width for the Riemann sum quadrilateral that
    [itex]0.5*(\frac{1}{n} - \frac{1}{n+1})= \frac{1}{2n(n+1)} = r_n[/itex].

    Lets place each quadrilateral so that the supremus (=the point 1/n) is in the middle. This way the infimum will always be 0 and the lower sum [itex]s_D=0[/itex].

    Calculating the upper sum gives
    [itex]\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x)) = \sum_{k=1}^{n}(2*r_n*\frac{k}{n})[/itex]
    because [itex]\frac{k}{n}, k=1,...,n[/itex] goes through all the supremums.
    So
    [itex]S_D = \sum_{k=1}^{n}(2*r_n*\frac{k}{n}) = 2*r_n\sum_{k=1}^{n}k[/itex]
    which with a little work leads into
    [itex]S_D = \frac{1}{2n}[/itex] since [itex]\sum_{k=1}^{n}k=\frac{n(n+1)}{2}[/itex]

    Now [itex]S_D-s_d=\frac{1}{2n}-0=\frac{1}{2n}<\frac{1}{100}[/itex] when [itex]n\geq 50[/itex].

    Whether that's right or wrong is beyond me...
     
  6. Apr 26, 2012 #5
    I'm actually starting to wonder whether it might be a typo by my professor.

    Since it seems the function would only get one value if defined the way i stated in the original post. [itex]\frac{1}{n}[/itex] right?... or?
     
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