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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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    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 $N$.2012-07-27

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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.