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Thread: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

  1. #1
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    Post Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this.
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    Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    Quote Originally Posted by vivek View Post
    Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this.
    What does [x] mean?

    Either way, we can cancel the [x]'s on each side, leading to

    $\displaystyle \left [ \frac{x + 1}{n} \right ] + \text{ ... } + \left [ \frac{x + n - 1}{n} \right ] = 0$
    (Or perhaps [0] ?)

    Let's say that n = 4. Then we have
    $\displaystyle \left [ \frac{x + 1}{4} \right ] + \left [ \frac{x + 2}{4} \right ] + \left [ \frac{x + 3}{4} \right ] = 0$

    This doesn't seem right.

    -Dan
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    Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    Its my bad, on L.H.S , it was [x/n] instead of [x],and [x] is greatest integer function
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    Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    Quote Originally Posted by topsquark View Post
    What does [x] mean?

    Either way, we can cancel the [x]'s on each side, leading to

    $\displaystyle \left [ \frac{x + 1}{n} \right ] + \text{ ... } + \left [ \frac{x + n - 1}{n} \right ] = 0$
    (Or perhaps [0] ?)

    Let's say that n = 4. Then we have
    $\displaystyle \left [ \frac{x + 1}{4} \right ] + \left [ \frac{x + 2}{4} \right ] + \left [ \frac{x + 3}{4} \right ] = 0$

    This doesn't seem right.
    Lets consider $P(n,x)]=\sum\limits_{k = 0}^{n - 1} {\left\lfloor {\dfrac{{x + k}}{n}} \right\rfloor }$

    Now look at $P(10,4)$
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    Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    first assume $\displaystyle 0\leq x<n$

    then show that

    $\displaystyle \left\lfloor \frac{x+k}{n}\right\rfloor =$

    $\displaystyle 0$ if $\displaystyle 0\leq k\leq n-\lfloor x\rfloor -1$

    $\displaystyle 1$ otherwise
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  6. #6
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    Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

    $x=nq+r$ where $0\le r < n$

    So, $$\left\lfloor \dfrac{x}{n}\right\rfloor =q$$ but $$\left\lfloor \dfrac{x+n}{n}\right\rfloor = q+1$$

    For all $0\le k <n-r$, you have:

    $$\left\lfloor\dfrac{x+k}{n}\right\rfloor=q$$

    For all $n-r \le k < n$, you have

    $$\left\lfloor\dfrac{x+k}{n}\right\rfloor=q+1$$
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