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3

I have to agree with Prometheus, as this is probably bunk. However, I think there is an interesting interpretation to be gleaned from this Gauss-attributed comment. I've had the distinct pleasure of talking with UChicago's Professor Benson Farb, an esteemed geometer and one of my personal idols, on two occasions. The first time I talked with him was ...


21

The Wikipedia entry, and a few other places that mention this, give as reference the popular book of John Derbyshire on the Riemann hypothesis. The book itself mentions this in the spirit of an apocryphal tale, with the passing remark that "I wouldn't put it past him", and gives no reference at all. It's probably best to treat this story with utmost ...


7

I fail to understand why we are looking at this so hard and not seeing it. We do not anymore have the series expansion of e^x, sin x, and cos x at our immediate reach* but at the time they did. The claim which to us sounds preposterous merely makes him a man of his day. * It would take us several seconds to recall each one of them.


15

If you look at the formula geometrically (and draw the unit circle), you'll see immediately that $e^{\pi i}=-1$, since it's half a circle from the origin. I don't see any other way for this to be all too apparent.


6

I imagine that since Gauss spent most of his life studying complex numbers, it would have been immediately apparent to him that $e^{\pi i}=\cos(\pi)$.


7

Your conjecture is correct for $k$ sufficiently large and, almost surely, the only counterexample is the case $k=4$ you have already noted. This inequality is insanely tight: In the end, it comes down to the fact that $e/6 < 1/2$ and some very fortunate coincidences concerning the continued fraction of $e$. We put $n = \lfloor k/e \rfloor$. We have ...


3

Not an answer, but an approach: Fix a $k$ and let $K := \lceil \frac{k}{e} \rceil$. Suppose $\left\{ \frac{k}{e} \right\} \ge \frac12$. We want to show that \begin{align*} \left(\frac{k}{\lfloor \frac ke \rfloor}\right)^{\lfloor \frac ke \rfloor} \le \left(\frac{k}{\lceil \frac ke \rceil}\right)^{\lceil \frac ke \rceil} \end{align*} Basic transformations ...


4

Not an answer, just an observation - hence CW. The plot of $\text{max} \begin{cases} \dfrac{e}{k}\log\left(\dfrac{k}{\lfloor k/e\rfloor}^{\lfloor k/e\rfloor}\right)\\ \\ \dfrac{e}{k}\log\left(\dfrac{k}{\lceil k/e\rceil}^{\lceil k/e\rceil}\right)\\ \end{cases}$ against background $\mod \dfrac{e}{2}$ shows that what you conjecture is not strictly true, ...



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