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Why is the following true?

$$\left[\frac{N - it}{N}\right]^{j+1} = \exp\left(-\frac{ijt}{N}\right)$$ i,j - integers less than N.

Is there any theorem which allows me to get this result?

I tried with $$N=2^{32}, i=j=2^8, t=2^{16}$$ and its almost true.

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are $i$ and $j$ complex numbers? –  dato datuashvili Jul 27 '12 at 13:20
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Put $N = 2$, $j = 0$ and you have ${2 - it\over 2} = 1$; this would not appear to be an identity. The asserted result is false. –  ncmathsadist Jul 27 '12 at 13:23
    
But its stated here caislab.kaist.ac.kr/lecture/2010/spring/cs548/basic/B01.pdf on page 3? at the beginning –  Yola Jul 27 '12 at 13:30
    
Yap, there point above = sign. May be its upper bound? –  Yola Jul 27 '12 at 13:34
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I see a dot over the equal sign, I don't know this formalism, but maybe it means "when $N$ is large", so that $(1-it/N)^{j+1}$ tends to $e^{-i(j+1)t/N}$. Moreover, when $N$ is large $j+1$ and $j$ are almost the same when divided by $N$. –  enzotib Jul 27 '12 at 13:43
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1 Answer 1

up vote 1 down vote accepted

As has been made clear in the comments, the result is false as stated. However, it is approximately true in the limit as $it/N\to0$ (and $j$ is not too large).

Here is one way to get a rigorous estimate of this nature. First, note that $\ln(1-x)=-x+O(x^2)$ as $x\to0$ (you can easily give rigorous upper bounds on the $O(x^2)$ term if needed). With $x=it/N$, we take exponentials, then raise both sides to the $j$th power to get $$\Bigl(1-\frac{it}{N}\Bigr)^j=\exp\biggl(-\frac{ijt}{N}+jO\Bigl(\Bigl(\frac{it}{N}\Bigr)^2\Bigr)\biggr)$$ where the big O refers to the limit $it/N\to0$. The missing factor $1-it/N$ on the left hand side (because of $j$ replacing $j+1$) is clearly close to $1$ in this case.

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