@JM It's okay, I think his comment actually answers my question. But I can't bring myself to do any more of this. At least not today. Will look at it again tomorrow.
@JonasTeuwen Set theory and logic are wrong. In fact, I don't even remember why I'm doing this to myself at all, studying maths that is.
You're only alloted a certain amount of neurons at birth.
From that point, it's a steady decline. The only good thing is that we have quite a lot to start with, and extant neurons are remarkably capable of picking up slack.
@JonasTeuwen There was a big jump in the number of functions from version five to version six. It seemed at the time that they wanted it to cook and launder for you as well... :D
@J.M. I think I have an answer to your question. (Which I'm ripping off of one of robjohn's answers which he got more votes on than I did. Payback time.)
At least the LHS of Gautschi's inequality follows from what robjohn has. I'm having trouble seeing it for the RHS.
For $0<s<1$, we have $$\Gamma(x+s)\le\Gamma(x)^{1-s}\Gamma(x+1)^s=\Gamma(x)x^s.$$ If we multiply both sides of this by $x^{1-s}\,\Gamma(x+s)^{-1}$ and use $x\Gamma(x)=\Gamma(x+1)$ one obtains the left-hand side of the Gautschis inequality.
(Actually to go from weak to strong inequality you might have to do some kind of endpoint discussion, but that's just book-keeping.) Thoughts anyone?
Wow, that looks pretty cool! I think it's stretched a lot because it's not an isometric realization, but there's really no way around that because the parallelogram has a flat metric.
Now imagine slowly shifting the fundamental domain (parallelogram) by a continuous parameter multiple (time) of the first period ($t=0$ to $t=1$) and then of the second period, and making it an animation! Also, smoother coloration. Probably not feasible on a PC but one can dream.
I like how twisty it is.
Crap, my pen exploded everywhere on my bed. This isn't looking very pretty.
I never liked pens. I used to write everything in ball point pen but a few months ago I discovered that I much prefer pencil. Especially to write proofs. : )
Yo robjohn. Scroll up a bit and look at the gamma function inequality J.M. and I were talking about; he has a question about it on MSE that links to your answer.
@Daniil: Yes, n=0 in a ring of characteristic n. Problem?
@Daniil I'm a bit late and anon has already answered but let me say something anyway: If $e$ is a zero divisor then there is an element $a \neq 0$ such that $e \cdot a = 0$. But $e \cdot a = a \neq 0$ which would be a contradiction.
$\mathbb{F}_{p^n}$ can be constructed as $\mathbb{F}_{p}[X]$ modulo an irreducible monic polynomial of degree $n$ (it doesn't matter which such polynomial). Now if you add any n polynomials in this ring, each coefficient vanishes separately.
@robjohn No, quite a bit of additional work. I meant: copy-paste with no mental effort. I typed out the paragraph from the textbook. I copy-pasted from the pdf, which obviously gets the text right but not the math; so I did a pass correcting it.
@Ilya Can you do me a favour? Go to any thread with answers posted: click on the "history" link under the question and under any of the answers. Can you tell me what you get?
@Skullpatrol "Snipping tool", if I remember right. Just google for screenshot in windows.
To solve this: $\left|\frac{2-x}{x+3}\right|\ge-\cos\left(4\right)$ I have to set up a system like this: $\begin{cases}\frac{2-x}{x+3}\le\cos\left(4\right)\\\frac{2-x}{x+3}\ge-\cos\left(4\right)\end{cases}$
(It's so funny MathJax when you know the commands xD)
"the proof may be found or adapted from your textbook therefore I'll vote to close as too localized" This rationale doesn't make sense to me - that paintbrush touches a large percentage of meaningful MSE questions.
