TITLE: Solving $x! = x^2$ (or $\Gamma(x) = x)$
QUESTION [1 upvotes]: I want to solve $x^{2x!} = (x!)^{x^2}$.
Aside for $1$ and $2$, the solution is $x! = x^2$.
When dividing by $x$, we get $\Gamma(x) = x$.
I don't know how to continue from that. I tried replacing $\Gamma(x)$ by its integral form, but I'm not able to solve it.
Edit: I already plotted the equations even before I tried solving, and I know the result is about 3.55. I just want to know where it comes from. I am not interested in the value, but in the equation and the way to solve it.

REPLY [1 votes]: For the equation, there is a trivial root $x=1$.
For the other one, by inspection, you know that the solution is between $x=3$ and $x=4$ and Newton method would converge very fast; being myself quite lazy, I shall start with $x_0=3.5$ and get the following iterates  shown below
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 3.5000000 \\
 1 & 3.5662556 \\
 2 & 3.5623964 \\
 3 & 3.5623823
\end{array}
\right)$$


TITLE: Hodge standard conjecture in positive characteristic
QUESTION [8 upvotes]: In the Wikipedia article on the Hodge Standard Conjecture it is written (note [Oct. 2015]: it has since been fixed):

In characteristic zero the Hodge standard conjecture holds, being a consequence of Hodge theory. In positive characteristic the Hodge standard conjecture is known only for surfaces and abelian varieties.

I have three questions:
(1) Is the characteristic zero version the Hodge Index Theorem?
(2) If so, what is a good reference for an algebraic geometry proof? I know it can be proved for a surface using the Riemann-Roch Theorem. Does this continue to be true for higher dimensions?
(3) Is the conjecture really only known in positive characteristic for surfaces and abelian varieties? Surely it should be possible to at least compute a result for projective $n$-space.

REPLY [4 votes]: (1) No, the Hodge index theorem, valid in arbitrary characteristic, is the Hodge standard conjecture for divisors on surfaces. Taking 2-dimensional linear  sections, one can deduce the Hodge standard conjecture for divisors on arbitrary smooth projective varieties.
The Hodge standard conjecture over $\mathbf{C}$ is not a consequence of the Hodge index theorem but rather of the fact that the space of algebraic cycles is contained in the space of Hodge classes $H^{p,p}(X)\cap H^{2p}(X, \mathbf{Q})$, and by Hodge theory (Riemann bilinear relations), the Lefschetz pairing is definite on the primitive parts of those a priori bigger spaces.
(2) I do not think there is an algebraic proof, since the above arguments requires some "positivity" (a definite pairing on a $\mathbf{Q}$-vector space remains definite after restricting to any subspace).
Speculation: Maybe the Weil cohomology theory for varieties over $\overline{\mathbf{F}}_p$ with values in the Kottwitz category ${\rm Kt}_{\mathbf{R}}$ whose existence was recently conjectured by Scholze could make such an argument possible in positive characteristic.
(3) Tetsushi Ito in "Weight-monodromy conjecture for $p$-adically uniformized varieties" (Invent. Math. 2005) crucially proved the Hodge standard conjecture for the varieties obtained from $\mathbf{P}^n$ over $\mathbf{F}_q$ by successively blowing up all $\mathbf{F}_q$-points, the strict transforms of the lines between those points etc. The cohomology of these varieties is generated by algebraic cycles, but they do not lift to characteristic zero.
(Obviously, if a variety in characteristic $p$ whose cohomology is generated by algebraic cycles lifts to characteristic zero "together with all cycles", then the Hodge standard conjecture holds.)


TITLE: An orientable line bundle is always trivial?
QUESTION [0 upvotes]: In general, there exist non-trivial orientable bundles. But if we only consider line bundles, then orientable bundles are always trivial bundles. (Similarly, non-trivial bundles are always non-orientable bundles.)
Is this statement true?
I got this conclusion from the following question and answer:
If a line bundle admits a non-vanishing section then it is trivial. If a line bundle admits a non-vanishing section, then it conforms to the definitions of trivial bundle and orientable bundle at the same time.

REPLY [2 votes]: Let $(E,B,L)$ be an orientable line bundle, this means there exists trivializations $U_{i}$ such that $\forall i; \phi_{U_{i}}:\pi^{-1}(U_{i}) \mapsto   U\times \mathbb{R}$ is fiberwise orientation preserving, denote by $g_{i}$ the inverse of this map, pick a partition of unity $P_{i}$ of the base space subbordinated to the trivializations (such a partition exists whenever $B$ is assumed to be paracompact). Define the map $s:B\mapsto E$ by the formula $s(b)=\sum P_{i}(b) g_{i}(b,1)$ where the sum is (necessarily finite) and is over the $U_{i}$ to which $b$ belongs.
Remark that $\forall b\in B,s(b)$ is never zero because preserving orientation between a one dimensional vector space and $\mathbb{R}$ means that $g_{i}(b,1)$ is always one "the same side" for every $b$ and every $I$.
This proves that an orientable line bundle is necessarly trivial.the other direction is easy.


TITLE: Prove uniformly continuity at $\infty$ to continuous function
QUESTION [2 upvotes]: Say $f:[0,\infty) \to \mathbb{R}$ is a continuous function. Assume $\lim_{x \to \infty}[f(x)-ax]=b$ for some $a,b \in \mathbb{R}$ and prove $f$ is uniformly continuous in $[0, \infty)$

So if the set was closed than Cantor will do the trick. I've tried proving $\lim_{x \to \infty}f(x)$ exists, or using the definition of uniform continuity and using the continuity of $f(x)$ but to no ends. Any hints?

REPLY [2 votes]: For any $\epsilon>0$ there exists some $K>0$ such that $|f(x)-ax-b|<\epsilon$ for $x>K$.
Now show the uniform continuity on $[0,K]$ and on $(K,\infty)$.


TITLE: Why does this work in this type of function?
QUESTION [3 upvotes]: I was just thinking about this: In a function like y = 3x you can find the correct slope even if you were to add two coordinates together.
Example: Find the slope by using x = 2 and x = 4 So, correct way is (12-6) / (4-2) which equals a slope of 3.
But you will always get correct slope even if you were to add as follows: (12+6) / (4+2) = 3.
However this doesn't work in a function like y = 3x+1 (and most other functions I assume) So can someone clarify why it always works in the first function?

REPLY [2 votes]: The equation $y=a x$ imposes a multiplicative relation on $x$ and $y$ that's uniquely compatible with the divisive computation of slope.
Of course, given two points $(x_1,y_1)$ and $(x_2,y_2)$ on any line, we define
$$\text{slope} = \frac{\text{change in } y}{\text{change in }x} = \frac{y_1-y_2}{x_1-x_2}$$
For points on a line defined by $y=ax$ (which, note, is specifically a line through the origin),
$$y_1 = a x_1 \qquad\text{and}\qquad y_2 = a x_2$$
so that
$$\text{slope} = \frac{a x_1 - a  x_2}{x_1 - x_2} = \frac{a(x_1-x_2)}{x_1-x_2}= a$$
Also, as you observe, you can add coordinates:
$$\frac{y_1+y_2}{x_1+x_2} = \frac{ax_1 + ax_2}{x_1+x_2}= \frac{a(x_1+x_2)}{x_1+x_2} = a$$
But, even more generally, you can take any "linear combination" of the coordinates:
$$\frac{Py_1+Qy_2}{Px_1+Qx_2} = \frac{Pax_1+Qax_2}{Px_1+Qx_2} = \frac{a(Px_1+Qx_2)}{Px_1+Qx_2} = a$$
The point here is that, because $y$-values are just multiples of the $x$-values, we can factor-out the multiplier in the numerator, and then cancel the denominator. The simple multiplicative nature of $y=ax$ just happens to get un-done by the divisive definition of slope. It's a nice property (and part of why we like when quantities are directly proportional), but it's too nice to expect to hold beyond this special case.


TITLE: algebra solving equation problem (10th grade standard)
QUESTION [1 upvotes]: Solve this equation in the set of whole numbers:
$$2x^{3}+x+8=y^2$$ Please help!

REPLY [0 votes]: I love Peter's answer, but I do not know many 10th graders that can come up with such a strategy. Another strategy would be to graph the two functions. Let $f(x) = 2x^3 + x + 8$ and let $g(x)=x^2$ and find the point(s) of intersection (where $f(x)=g(x)$).  A modern graphing utility shows a single point of intersection at roughly (-1.3, 1.8). This is clearly not an integer solution.
However, the question asked for solutions in the set of whole numbers (positive integers). We note that when $x=0$, $f(0) = 8$ and $g(0) = 0$. In other words, $f(x) > g(x)$. As $x$ increases, $f(x)$ increases more rapidly than $g(x)$, maintaining the inequality. Thus $f(x) > g(x)$ for all $x > 0$, meaning the two functions will never intersect.