In  a paper by S. W. Golomb and A. W. Hales, "Hypercube Tic-Tac-Toe" [It's about the generalization of Tic-Tac-Toe game to n dimensional cube] there is an interesting passage about the following equality

\[\left\lceil\frac{2}{2^{1/n}-1}\right\rceil = \left\lfloor \frac{2n}{\log 2} \right\rfloor \]

the authors said that this equality is false but they gave a miss-calculated counterexample (due to the error in the computer multi-precision system). 

The equality is indeed very interesting! 

In 18th century when the computer wasn't good enough, this might be a kind of problem that might fool those who rely only on "trial and error method" [some profession joke usually refer it to engineer :p ]

Let's use it on the following joke :D

Suppose that the energy explosion risk given by the $n$ connected reactors devices is given by the formula:

\[E_n := n^2 \left(\left\lceil\frac{2}{2^{1/n}-1}\right\rceil - \left\lfloor \frac{2n}{\log 2} \right\rfloor\right) \]

(In TJ)

Some amateur engineers who ignore the power of mathematical proof asserts that $E_n=0$, They conclude this after performing a trial method i.e  inputting several values of $n$. 

As they already check that $E_1=E_2= \cdots = E_{10^{13}}=0$. Quite convincing for the assertion that $E_n=0$ for all $n$.  

Furthermore, by direct computation $E_{n}=0$ for $10^{20}\leq n \leq 10^{23}$. So what else? 

By this observation, the engineer said that the device is safe because the risk is always zero, and so they proceed by activating the device.

Suppose that at a very unlucky day, only 777451915729369 reactors on this device are fully functional, nevertheless we still have $E_{777451915729369}=0$, and the risk is still zero.

But then there is a damned cat goes to the lab and accidentally malfunctioning one reactor, resulting only $777451915729368$ reactors are fully functional.

At that time the world will be destroyed by 

\[E_{7774519157293698} = 604431481271265877115249138161\]

TJ explosion. End Of The Story.

The nature of $E_n$ is very interesting, $E_n=0$ for a very long time  then it takes a nonzero value  instantly for the first time when $n=7774519157293698$ (very big), after that  $E_n=0$ again for a very long time until it instantly nonzero again for the second time when $n=140894092055857794$ (and of course bigger). 

So be careful on the trial-error conclusion :p

The sequence $\frac{E_n}{n^2}$ numbers on OEIS is A1299935. See http://oeis.org/A129935

[UnSA] The Condition for existence of minimum in Survival-Related Function.

Today, I was asked about the proof  stated on this paper  (theorem 2.1 page 98), (the proof is on page 110), we are dazed with the argument given there, and started to think that it's probably an errata (mistyped but not serious) or we just missed the point.

Let $S_X(t)$ be a survival function, define for each $d>0$ and $0<\alpha < S_X(0)$ the function

\[\text{VaR}_T(d,\alpha) := \left\{\begin{array}{lr} d+\delta(d)\qquad &  \text{for }d \in (0,S_X^{-1}(\alpha)\\ S_X^{-1}(\alpha)+\delta(d) \qquad & \text{for } d > S_X^{-1}(\alpha)
\end{array}\right.\]

Where $\delta(d)=(p+1)\int_d^{\infty} S_X(t) \, dt$ and $p^*=\frac{1}{1+p}$.

The value $d^*>0$ such that $\text{VaR}_T(d^*,\alpha)=\min_{d>0}(\text{VaR}_T(d,\alpha))$ exists if and only if 

\[S_X^{-1}(\alpha)\geq S_X^{-1}(p^*)+\delta(S_X^{-1}(p))\]

and 

\[\alpha < p^* < S_X(0)\]

Proof:

Recall that, the survival function is monotone decreasing function positive. Also it can be proved that $\text{VaR}_T(d,\alpha)$ is continuous. For $d>S_X^{-1} (\alpha)$, the derivative of $\text{VaR}_T(d,\alpha)$ is 

\[\delta'(d)=-(1+p) S_X(d) < 0\] 

(note that the differentiation under integral sign is verified since $S_X(d)$ is bounded, hence clearly dominated).
Therefore $\text{VaR}_T(d,\alpha)$ is decreasing on $(S_X^{-1}(\alpha),\infty)$, with limit $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$. Also,  for $p^*>S_X(0)$ we have the derivative of $d+\delta(d)$ is

\[1-(p+1)S_X(d)<0\]

thus $d+\delta(d)$ is decreasing when $p^* > S_X(0)$ until $d=S_X^{-1}(p^*)$. Therefore the function $d+\delta(d)$ attains minimum value on $d_0=S_X^{-1}(p^*)$. If the both conditions hold then $d_0 \in (0,S_X^{-1}(\alpha))$  and the minimum point is now also the minimum point of $\text{VaR}_T(d,\alpha)$, therefore $d^*=d_0>0$.

Conversely, If $\alpha\geq p^*$ then since $S_X(t)$ is decreasing its inverse function is also  decreasing, it follows that $S^{-1}_X(\alpha) \leq S_X^{-1}(p^*)=d_0$, and for all $d\leq S^{-1}_X(\alpha)$ we have $S_X(d) \geq p^*$, therefore the derivative of the function $\text{VaR}_T(d,\alpha)$ is equal to $1-(p+1)S_X(d)$ on $d\in (0,S_X^{-1}(\alpha))$ and equal to $-(p+1)S_X(d)$ when $d>S_X^{-1}(\alpha)$ which is consistently negative. Thus $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ and as $d\rightarrow \infty$ it has limiting value $S_X^{-1}(\alpha)$, while no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, so no such $d^*$ exists.


Now suppose that $p^* \geq S_X(0)$, then we have $d=S_X^{-1}(p^*) \leq 0$ thus $d_0 =0$. Since $d_0$ is the only stationery point of the function $d+\delta(d)$, it follows that this function does not have a stationery point when $d>0$,  therefore on $0< d\leq S_X^{-1}(\alpha)$ the derivative of the function is either strictly positive or strictly negative (if it ever been positive in one point and then negative in other point, then since the derivative is continuous by Intermediate Value Theorem it has zero, thus it has stationery point.)
From the above argument it follows that $d+\delta(d)$ is either monotone decreasing or monotone increasing on $0< d \leq S_X^{-1}(\alpha)$.
[In the paper it is stated only the decreasing case] 
It it is increasing, then $VaR_T(d,\alpha)$ is increasing on $(0,S_X^{-1}(\alpha))$ and eventually starts decreasing on $(S_X^{-1}(\alpha),\infty)$ with limited value $S_X^{-1}(\alpha)$, while it has no minimum value on $(0,S_X^{-1}(\alpha))$ the limited value $S_X^{-1}(\alpha)$ is the smallest the function $\text{VaR}_T(d,\alpha)$ can approach, but as before no such $d^*$ with  $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$. If $\delta(d)+d$ is decreasing, then $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ with limiting value $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$, thus no such $d^*$ exists.

If condition (1) does not hold while condition (2) holds then $\inf_{d>0} \text{VaR}_T(d,\alpha)=S_X^{-1}(\alpha)$, in order to have the infimum a minimum, there should be exists $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, but this is clearly impossible, and the proof is completed.

Remark: Because I don't know much about the behavior of Survival Function, I tend to avoid the second derivative test for stationery point $d_0$, if we employ this test, we will get that this point is a minima, thus $\text{VaR}_T  (d,\alpha)$ is not always decreasing when $d \in (0,\infty)$, this is  where we think the argument in the paper is fishy.

Following the algebraic geometry post I have in this blog, I create a note, in which I can't complete (maybe I'm busy :D) The file was named zariskiv2.pdf, because I was intentionally wrote this note for introduction to zarizki topology, but it turned out to be larger than that.
I create a pdf file of introductory algebraic geometry,  a small intro which is prepared for a seminal course for undergraduate students who still doesn't familiar with schemes and sheaves. I wish to continue the discussion and example in projective variety and cohomology if I have more time completing it.

download here