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Math Help - Integration of greatest integer function

  1. #1
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    Integration of greatest integer function

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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by champrock View Post
    Separate the integral :
    \int_0^2 [x]^n f'(x) ~dx=\int_0^1 [x]^n f'(x) ~dx+\int_1^2 [x]^n f'(x)~dx
    if x is between 0 and 1, then [x]=0
    if x is between 1 and 2, then [x]=1

    so your integral is actually \int_1^2 f'(x) ~dx, if n>0

    if n=0, then the integral is \int_0^2 f'(x) ~dx
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    but its actually [X]^N whcih is most confusing. (mod raised to Nth power).

    so, it cant be seperated that easily.
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    Moo
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    Quote Originally Posted by champrock View Post
    but its actually [X]^N whcih is most confusing. (mod raised to Nth power).

    so, it cant be seperated that easily.
    It is... Because x varies from 0 to 1 and from 1 to 2.


    x in [0,1[ => [x]=0 => [x]^n=0^n=0, if n is not equal to 0.
    x in [1,2[ => [x]=1 => [x]^n=1^n=1.
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  5. #5
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    hmm.. lets just take an example. assume that n=5 and x=1.5, then 1.5^5 = 7.59..
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    Let k\ge2, be an integer then \int_{0}^{k}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=\sum\limits_{m=0}^{k-1}{\int_{m}^{m+1}{m^{n}f'(x)}\,dx}=\sum\limits_{m=  1}^{k-1}{m^{n}\big(f(m+1)-f(m)\big)}.

    Hence, for k=2 it's \int_{0}^{2}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=f(2)-f(1), which is the value of the integral.
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    Moo
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    Quote Originally Posted by champrock View Post
    hmm.. lets just take an example. assume that n=5 and x=1.5, then 1.5^5 = 7.59..
    But you're working with the greatest integer function !

    It's [x]^n, not [x^n], that's certainly not the same...

    what I showed you gives the exact same result as Krizalid. (apart from the case n=0)
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