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October 18th, 2019, 10:38 AM   #1
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need explanation for integer part

Is this statement true ? $\displaystyle \lfloor n/2 \rfloor \approx \frac{n}{2}+ \frac{1-(-1)^n }{4}$. $\displaystyle \; n\in \mathbb{N}$ and $\displaystyle n\neq 1$.

Last edited by idontknow; October 18th, 2019 at 10:51 AM.
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October 18th, 2019, 10:54 AM   #2
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not quite

$\left \lfloor \dfrac n 2 \right\rfloor = \dfrac n 2 - \dfrac{1-(-1)^n}{4}$
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October 19th, 2019, 02:50 AM   #3
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Also $\displaystyle (-1)^n =e^{i\pi n} =4\lfloor n/2 \rfloor -2n+1$.
Now the integer part can be expressed as infinite series .
Floor Function -- from Wolfram MathWorld

Last edited by idontknow; October 19th, 2019 at 03:03 AM.
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October 19th, 2019, 06:17 AM   #4
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Quote:
Originally Posted by idontknow View Post
Also $\displaystyle (-1)^n =e^{i\pi n} =4\lfloor n/2 \rfloor -2n+1$.
Now the integer part can be expressed as infinite series .
Floor Function -- from Wolfram MathWorld
My essence was to express (-1)^n in one single equality using integer parts .
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