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Math Help - Reimann sum of the logarithm function?

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation [SOLVED] Reimann sum of the logarithm function?

    Dear friends,

    I tried to evaluate the integral \int_{1}^{t}\ln(s)\mathrm{d}s, where t\geq1 by making the usage of the Reimann sum, but i failed.
    More precisely, how should I pick \xi_{k} and \Delta_{k} below?
    \int_{1}^{t}\ln(s)\mathrm{d}s=\ln\bigg(\lim_{n\to\  infty}\prod_{j=1}^{n}\xi_{k}^{\Delta_{k}}\bigg)
    Last edited by bkarpuz; September 21st 2009 at 05:08 AM.
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  2. #2
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    Quote Originally Posted by bkarpuz View Post
    Dear friends,

    I tried to evaluate the integral \int_{1}^{t}\ln(s)\mathrm{d}s, where t\geq1 by making the usage of the Reimann sum, but i failed.
    More precisely, how should I pick \xi_{k} and \Delta_{k} below?
    \int_{1}^{t}\ln(s)\mathrm{d}s=\ln\bigg(\lim_{n\to\  infty}\prod_{j=1}^{n}\xi_{k}^{\Delta_{k}}\bigg)
    You can evaluate \int{\ln{s}\,ds} without using Riemann sums.

    Have you ever used Integration by Parts before?

    \int{u\,dv} = uv - \int{v\,du}.


    In this case, u = \ln{s} and dv = 1.

    So du = \frac{1}{s} and v = s.


    Therefore

    \int{\ln{s}\,ds} = \int{\ln{s}\cdot 1\,ds}

     = s\ln{s} - \int{s\cdot \frac{1}{s}\,ds}

     = s\ln{s} - \int{1\,ds}

     = s\ln{s} - s + C.


    Can you work out

    \int_1^t{\ln{s}\,ds} now?
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Prove It View Post
    You can evaluate \int{\ln{s}\,ds} without using Riemann sums.

    Have you ever used Integration by Parts before?

    \int{u\,dv} = uv - \int{v\,du}.


    In this case, u = \ln{s} and dv = 1.

    So du = \frac{1}{s} and v = s.


    Therefore

    \int{\ln{s}\,ds} = \int{\ln{s}\cdot 1\,ds}

     = s\ln{s} - \int{s\cdot \frac{1}{s}\,ds}

     = s\ln{s} - \int{1\,ds}

     = s\ln{s} - s + C.


    Can you work out

    \int_1^t{\ln{s}\,ds} now?
    Many thanks my friend Prove It, you really made me feel like a kid .
    I already knew that \big[t\ln(t)-t\big]^{\prime}=\ln(t) but I just wanted to show it via Remiann sum and the integration by parts tool is not allowed for use.
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  4. #4
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    Quote Originally Posted by bkarpuz View Post
    Many thanks my friend Prove It, you really made me feel like a kid .
    I already knew that \big[t\ln(t)-t\big]^{\prime}=\ln(t) but I just wanted to show it via Remiann sum and the integration by parts tool is not allowed for use.
    Why would Integration by Parts not be allowed to be used?

    What level maths are you taking?


    Anyway, incase you were wondering, you can prove Integration by Parts using the Product Rule.

    If you can prove this relationship works, I'm sure you'd be allowed to use it.

    If u = u(x) and v = v(x), by the product rule...

    \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.

    So \int{\frac{d}{dx}(uv)\,dx} = \int{\left(u\frac{dv}{dx} + v\frac{du}{dx}\right)\,dx}

    \int{u\frac{dv}{dx}\,dx} + \int{v\frac{du}{dx}\,dx} = uv

    \int{u\,dv} + \int{v\,du} = uv

    \int{u\,dv} = uv - \int{v\,du}.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Prove It View Post
    Why would Integration by Parts not be allowed to be used?

    What level maths are you taking?


    Anyway, incase you were wondering, you can prove Integration by Parts using the Product Rule.

    If you can prove this relationship works, I'm sure you'd be allowed to use it.

    If u = u(x) and v = v(x), by the product rule...

    \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.

    So \int{\frac{d}{dx}(uv)\,dx} = \int{\left(u\frac{dv}{dx} + v\frac{du}{dx}\right)\,dx}

    \int{u\frac{dv}{dx}\,dx} + \int{v\frac{du}{dx}\,dx} = uv

    \int{u\,dv} + \int{v\,du} = uv

    \int{u\,dv} = uv - \int{v\,du}.
    I myself tried to solve it by Reimann sum but I could not pick \xi_{k},\Delta_{k} in a suitable way to solve it easily.
    Btw, I am a PhD student.

    In the line below I tried to make it look similar to the Taylor's formula of the exponential function but not worked.
    <br />
\int_{1}^{t}\ln(s)\mathrm{d}s=\lim_{n\to\infty}\su  m_{j=1}^{n}\ln\big(\xi_{j}^{\Delta_{j}}\big)=\lim_  {n\to\infty}\sum_{j=0}^{n-1}\frac{\Big[\ln\Big(t\big(\ln(t)-1\big)+1\Big)\Big]^{j}}{j!}.<br />
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    If I let x_0=1, x_1=1+\Delta x,\cdots,x_{n-1}=1+(n-1)\Delta x and \Delta x=\frac{t-1}{n}. Then:

    \int_{1}^{t} \ln(s)ds=\lim_{n\to\infty}\sum_{i=1}^{\infty} f(x_{n-1}) \Delta x=\lim_{n\to\infty}\sum_{i=1}^{\infty} \ln\big(1+(i-1)\Delta x\big)\Delta x

    =\lim_{n\to\infty}\sum_{i=1}^{\infty} \ln\left(1+\frac{(i-1)(t-1)}{n}\right)\frac{t-1}{n}

    and I can't go further with it.
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  7. #7
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    I think it's enough to provide a proof, using Riemann sums, that

    \int_{a}^{b}{f(x)\,dx} = F(b) - F(a), where \frac{d}{dx}[F(x)] = f(x)

    and then to directly evaluate \int_1^t{\ln{x}\,dx}.


    Technically speaking, if you're evaluating a definite integral, you ARE using a Riemann sum.

    I think you're making the problem harder than it needs to be...
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  8. #8
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Prove It View Post
    I think it's enough to provide a proof, using Riemann sums, that

    \int_{a}^{b}{f(x)\,dx} = F(b) - F(a), where \frac{d}{dx}[F(x)] = f(x)

    and then to directly evaluate \int_1^t{\ln{x}\,dx}.


    Technically speaking, if you're evaluating a definite integral, you ARE using a Riemann sum.

    I think you're making the problem harder than it needs to be...
    Because what I want is to show that it can be proved by Reimann sum by picking the terms appropriately, my way is not only calculating that the solution of the integral is t\ln(t)-t+1.

    May be the relation
    t\ln(t)-t+1=(t-1)^{2}\sum_{j=0}^{\infty}\frac{(-1)^{j+1}(t-1)^{j}}{(j+2)!}
    should be used. :S
    Last edited by bkarpuz; July 7th 2009 at 12:34 PM.
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  9. #9
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    OK, I've done it, but I don't think you'll like the solution...


    For this to work, we need to make use of the Mean Value Theorem.

    i.e. for some function F(x), which is continuous on x \in [a, b], we have

    f(c) = \frac{F(b) - F(a)}{b - a}, where a < c < b and \frac{d}{dx}[F(x)] = f(x).

    We can rearrange the formula so that it is of the form

    (b - a)f(c) = F(b) - F(a).


    For this function, f(x) = \ln{x}, suppose we take some rectangular strip, between x_{k - 1} and x_k.

    Then this strip's area is

    (x_k - x_{k - 1})f(c), for some c \in (x_{k - 1}, x_k).

    Using the Mean Value Theorem, we can rewrite this as

    F(x_k) - F(x_{k - 1}), where \frac{d}{dx}[F(x)] = f(x).

    Clearly, since f(x) = \ln{x}, this means F(x) = x\ln{x} - x.


    So the area of one strip is...

    x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})

     = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}.


    If we have n strips, then the Area between x_0 and x_n is

    A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}

     = x_n\ln{x_n} - x_n - x_0\ln{x_0} + x_0, since this is a telescopic series.


    Since x_0 = 1 and x_n = t, you have

    A = t\ln{t} - t - 1\ln{1} + 1

     = t\ln{t} - t + 1.



    To fully complete this, you would need to note that you make n \to \infty and then take the limit.


    Can you evaluate

    \lim_{n \to \infty}(t\ln{t} - t + 1)?
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  10. #10
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    Quote Originally Posted by Prove It View Post
    OK, I've done it, but I don't think you'll like the solution...


    For this to work, we need to make use of the Mean Value Theorem.

    i.e. for some function F(x), which is continuous on x \in [a, b], we have

    f(c) = \frac{F(b) - F(a)}{b - a}, where a < c < b and \frac{d}{dx}[F(x)] = f(x).

    We can rearrange the formula so that it is of the form

    (b - a)f(c) = F(b) - F(a).


    For this function, f(x) = \ln{x}, suppose we take some rectangular strip, between x_{k - 1} and x_k.

    Then this strip's area is

    (x_k - x_{k - 1})f(c), for some c \in (x_{k - 1}, x_k).

    Using the Mean Value Theorem, we can rewrite this as

    F(x_k) - F(x_{k - 1}), where \frac{d}{dx}[F(x)] = f(x).

    Clearly, since f(x) = \ln{x}, this means F(x) = x\ln{x} - x.


    So the area of one strip is...

    x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})

     = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}.


    If we have n strips, then the Area between x_0 and x_n is

    A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}

     = x_n\ln{x_n} - x_n - x_0\ln{x_0} + x_0, since this is a telescopic series.


    Since x_0 = 1 and x_n = t, you have

    A = t\ln{t} - t - 1\ln{1} + 1

     = t\ln{t} - t + 1.



    To fully complete this, you would need to note that you make n \to \infty and then take the limit.


    Can you evaluate

    \lim_{n \to \infty}(t\ln{t} - t + 1)?
    Many thanks Prove It, but as you mentioned I did not like it, cuz I dont see any Reimann sum

    By the way, is it your style to talk like that all the time?
    Quote Originally Posted by Prove It
    Can you evaluate
    \lim_{{\color{red}{n \to \infty}}}({\color{blue}{t\ln{t} - t + 1}})?
    Sorry, forgive me since I am not very used to it...
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    Something similar was posed a good while back. Here is the link if you're interested.

    http://www.mathhelpforum.com/math-he...tml#post248858
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  12. #12
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    Quote Originally Posted by bkarpuz View Post
    Many thanks Prove It, but as you mentioned I did not like it, cuz I dont see any Reimann sum

    By the way, is it your style to talk like that all the time?


    Sorry, forgive me since I am not very used to it...
    Yes it is my style.

    I think you might need to refresh your memory as to what a Riemann sum actually is, as what I provided is, in fact, a Riemann sum.
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    Quote Originally Posted by Prove It View Post
    Yes it is my style.

    I think you might need to refresh your memory as to what a Riemann sum actually is, as what I provided is, in fact, a Riemann sum.
    You are actually right that I need to refresh my mind about Reimann sum.
    I think you did not read what I wrote in the topic, I dont want to calculate the integral with formulas, I want to show that the Reimann sum converges to t\ln(t)-t+1.
    If you dont have any answer, please dont mess the topic, but if you really want to help then you are welcome.
    thanks.
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    Bkarpuz . . . that's a nice Riemann sum above. He wrote:

    Quote Originally Posted by Prove It View Post

    Then this strip's area is

    (x_k - x_{k - 1})f(c), for some c \in (x_{k - 1}, x_k).

    Using the Mean Value Theorem, we can rewrite this as

    F(x_k) - F(x_{k - 1}), where \frac{d}{dx}[F(x)] = f(x).

    Clearly, since f(x) = \ln{x}, this means F(x) = x\ln{x} - x.


    So the area of one strip is...

    x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})

     = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}.


    If we have n strips, then the Area between x_0 and x_n is

    A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}
    Also, I personally saw nothing wrong with how it was written. That's just math-speak. We all do it and it's nice, brief, concise, and to the point. Maybe you can reconsider.
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by shawsend View Post
    Bkarpuz . . . that's a nice Riemann sum above. He wrote:



    Also, I personally saw nothing wrong with how it was written. That's just math-speak. We all do it and it's nice, brief, concise, and to the point. Maybe you can reconsider.
    @shawsend: It is not a Reimann sum it is just a (telescopic) sum.
    If someone knows F which is the antiderivative of f, then one can easily write
    F(x_{n})-F(x_{0})=\sum_{i=0}^{n-1}\big[F(x_{i+1})-F(x_{i})\big] and then let x_{n}=t and x_{0}=s.

    He did not write anything wrong but it is not related to the solution.
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