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Thread: Riemann Sum for 1/x

  1. #1
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    Riemann Sum for 1/x

    How to find the integral 1/x from 1 to 2 using riemann sum?

    ∫ 1/x dx = lim as n goes infinity Σ(i = 1 to n)f(xi)Δx

    Δx = 2-1/n = 1/n
    xi = 1 + iΔx = 1 + (i/n)

    = lim as n goes infinity Σ(i = 1 to n)f(1 + (i/n))(1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (1 + (i/n)) ] (1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (n + i) ]

    NOW, how to get rid of i?
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  2. #2
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    Re: Riemann Sum for 1/x

    Quote Originally Posted by joshuaa View Post
    How to find the integral 1/x from 1 to 2 using riemann sum?

    ∫ 1/x dx = lim as n goes infinity Σ(i = 1 to n)f(xi)Δx

    Δx = 2-1/n = 1/n
    You mean Δx= (2- 1)/n. That had me puzzled for a minute!

    xi = 1 + iΔx = 1 + (i/n)

    = lim as n goes infinity Σ(i = 1 to n)f(1 + (i/n))(1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (1 + (i/n)) ] (1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (n + i) ]

    NOW, how to get rid of i?
    Your do the sum! But there is no "simple" formula for that sum, if that is what you are asking. Do you know the McLaurin series for ln(x)?
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  3. #3
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    Re: Riemann Sum for 1/x

    Here's a link to Euler's constant:https://en.wikipedia.org/wiki/Euler%...eroni_constant. I'll leave it to you to see why the defining limit exists.
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  4. #4
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    Re: Riemann Sum for 1/x

    I don't know the McLaurin series. It seems a tough problem because I don't know how to use Mclaurin series or Euler's constant.
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  5. #5
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    Re: Riemann Sum for 1/x

    -ln(2n) should = -ln n - ln2
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    Re: Riemann Sum for 1/x

    Quote Originally Posted by joshuaa View Post
    -ln(2n) should = -ln n - ln2
    sorry I have just seen the brackets
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    Re: Riemann Sum for 1/x

    Quote Originally Posted by johng View Post
    Here's a link to Euler's constant:https://en.wikipedia.org/wiki/Euler%...eroni_constant. I'll leave it to you to see why the defining limit exists.
    I understood those. thanks. What about Mclaurin series?
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  8. #8
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    Re: Riemann Sum for 1/x

    Have you not taken a Calculus course? A "McLaurin" series is a Taylor's series about a= 0. Specifically, the McLaurin Series for f(x) is
    f(0)+ f'(0)x+ (f''(0)/2)x^2+ \cdot\cdot\cdot+ f^{(n)}(0)x^n+ \cdot\cdot\cdot.

    ( f^{(n)} is the nth derivative of f.)
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  9. #9
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    Re: Riemann Sum for 1/x

    but zero is undefined in the function 1/x
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    Re: Riemann Sum for 1/x

    Quote Originally Posted by joshuaa View Post
    but zero is undefined in the function 1/x
    Hallsofivy in reply told #2 told you the truth. The fact is that you nor anyone else is going to find a neat way to use simple Riemann Sums to find $\int_1^2 {\frac{{dx}}{x}}$. All of the other replies rely on advanced facts and/or techniques.

    Some calculus textbooks develop the integral using an area function and never mention Riemann sums.
    In those texts, the logarithm is defined as $\displaystyle\log (x) = \int_1^x {\frac{{dt}}{t},\;x > 0}$, again no sums!
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  11. #11
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    Re: Riemann Sum for 1/x

    thanks for the clarification
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