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Thread: Greatest integer function

  1. #1
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    Post Greatest integer function

    Prove that
    $[x]+[x+\dfrac{1}{n}]+[x+\dfrac{2}{n}]+.....[x+\dfrac{n-1}{n}]=[nx]$
    Where [.] Represents greatest integer function and n$\in$N
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  2. #2
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    Re: Greatest integer function

    I would proceed by using induction. The base case is trivial, and suppose that the statement is true for n where $\displaystyle n \leq p$. Then proceed by proving the statement is true for p+1 using the induction step.
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    Re: Greatest integer function

    Quote Originally Posted by MacstersUndead View Post
    I would proceed by using induction. The base case is trivial, and suppose that the statement is true for n where $\displaystyle n \leq p$. Then proceed by proving the statement is true for p+1 using the induction step.
    I thought about this, too, but how do you relate $\lfloor x+\dfrac{1}{p+1}\rfloor$ to $\lfloor x+\dfrac{1}{p}\rfloor$? We know that the LHS is less than or equal to the RHS. We know it is equal to the RHS or the RHS minus one. But, without more information about $p$, we will not be able to make any assessment.

    Instead, consider $\overline{x} = x-\lfloor x \rfloor$. Obviously, $0 \le \overline{x} < 1$ and $nx = n\lfloor x \rfloor + n\overline{x}$.
    $\lfloor n\overline{x} \rfloor$ is the number of terms of $x, x+\dfrac{1}{n}, x+\dfrac{2}{n}, \ldots, x+\dfrac{n-1}{n}$ such that $\lfloor x+k\rfloor > \lfloor x \rfloor$ where $k \in \left\{0,\dfrac{1}{n},\dfrac{2}{n},\ldots, \dfrac{n-1}{n}\right\}$.
    But, I have not thought how to prove that. Intuitively, it seems obvious, though.
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  4. #4
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    Re: Greatest integer function

    Quote Originally Posted by Sidd07 View Post
    Prove that
    $[x]+[x+\dfrac{1}{n}]+[x+\dfrac{2}{n}]+.....[x+\dfrac{n-1}{n}]=[nx]$
    Where [.] Represents greatest integer function and n$\in$N
    Hermite's identity
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