just because i used the f-word or such , though I used stars .... I find it lame that when using stars you can still get kicked ... There was no fight either ... lame
maybe someone took it to serious and personal ... i dont know who kicked me
@robjohn I might be proposed for a contest like Putnam (especially for the fact that it offers the possibility to finish it without using special functions, and no need for series).
@robjohn I've always thought that the analytic continuation of the beta function using the Pochhammer contour was pretty cool (even though it admittedly took me a while to understand it).
@mick Ah. Phase plot of Jacobi's $\vartheta_2^4(q)$ over the unit disk. Green is purely imaginary. (As opposed to the usual scheme where red is purely real.)
enough nonsense from me ... here is the link to my lastest question http://math.stackexchange.com/questions/457867/asymptotic-expansion-of-ln-left-fracxax-a-right-in-form-of-sum-limit/457887?noredirect=1#comment2020048_457887
@DanielFischer you need to do the dishes alone ? :p
"The feeling you get when you're trying to writing up all you have thought about on a problem" $\cong$ "The feeling you get when doing dishes after the eating"
Let $x>0$.
Let $J(x) = \sum \limits_{n=0}^{\infty} \dfrac{x^n}{2^{n(n-1)/2} n!}$
Let $J^{-1}(x)$ be the functional inverse of $J(x)$.
How do I show that for all $x$ there exists a fixed positive real constant $C$ such that:
$\left(J^{-1}(J(x)-J(x-1))-\dfrac{x}{2}\right)^2 < C $
edit :
I be...
If we have an equation that looks like $|a| = b|c|$ for $a,b,c>0$, can we use the fact that $|a| = \sqrt{a^2}$ and the same fact for $|c|$ to square our equality, then take the square root to get $a = |b|c$?
@mick, you can let $z_1 = a+bi$ and $z_2 = c+di$, then show that $\left| \frac{z_1}{z_2} \right|$ and $\frac{\left|z_1\right|}{\left|z_2\right|}$ are equal.
@mick and I challenge you to prove this result :D $$\int_0^{\pi/2} \frac{\displaystyle \log\left(\frac{2 \sin(x)}{\sin(x)+\cos(x)}\right)} {\log(\tan(x))} \ dx = \frac{\pi}{4}$$
Let $x>0$.
Let $J(x) = \sum \limits_{n=0}^{\infty} \dfrac{x^n}{2^{n(n-1)/2} n!}$
Let $J^{-1}(x)$ be the functional inverse of $J(x)$.
How do I show that for all $x$ there exists a fixed positive real constant $C$ such that:
$\left(J^{-1}(J(x)-J(x-1))-\dfrac{x}{2}\right)^2 < C $
edit :
I be...
Anyway, that's irrevelant. I need some clarification. You said that the inverse exists because the derivative is always positive for positive $x$. Why don't you keep that restriction and conjecture the inequality for all $x$ then?
@Chris'ssis Now note that symmetry says $$\int_0^{\pi/2}\frac{\log(\sqrt2\sin(x))}{\log(\tan(x))}\mathrm{d}x =-\int_0^{\pi/2}\frac{\log(\sqrt2\cos(x))}{\log(\tan(x))}\mathrm{d}x$$
@Chris'ssis Add the left side to both sides to get $$2\int_0^{\pi/2}\frac{\log(\sqrt2\sin(x))}{\log(\tan(x))}\mathrm{d}x =\int_0^{\pi/2}\frac{\log(\tan(x))}{\log(\tan(x))}\mathrm{d}x=\frac\pi2$$
@robjohn you mean that $$\int_0^{\pi/2} \frac{\displaystyle \log\left(\frac{2 \sin(x)}{\sin(x)+\cos(x)}\right)} {\log(\tan(x))} \ dx=\int_0^{\pi/2}\frac{\log(\sin(x))}{\log(\tan(x))}\mathrm{d}x$$ ?
@robjohn using that result, one can also compute $$\int_0^{\pi/2} \frac{\displaystyle \log\left(\tan(x)+1\right)} {\log(\tan(x))} \ dx$$ that is not that nice at first sight.
@mick Here is a problem already. I thought it was the bessel function and that you were trying to prove the series. This is because the question is hard to read.
so for example, if you take a set with $n$ elements, and you associate with it the set of all possible graphs on that set, then that's a combinatorial species (the map is)
long story short it gives you new ways of working with power series that can help you solve problems like this using counting, you might want to look into it
if you don't like that example there's lots of other species, maybe the most familiar one is the "power set" species $\mathcal{P}$, that sends a set $S$ to the set of subsets of $S$
@mick the thing that makes me think it could work is that you're looking at $J(x)-J(x-1)$... the discrete differnce there happens a lot in species theory
What's the most elementary/concrete example of an indefinite integral that it's anti-derivative cannot be found(or can be but it is nearly impossible) but we don't have a problem evaluating that integral?
@MikeMiller crossed homomorphisms are defined with respect to an $f:Q\rightarrow \operatorname{Aut}(N)$ (so I guess they should be called "$f$-crossed homomorphisms")
@UserX cause trapezoid rule is pretty easy to explain, and easy to extend to composite trapezoid. And you could try Simpson's rule too if she liked that, though it is a bit harder.
i think they generate all $f$-crossed homomporhisms, or some type of reduction like that. I never really see any of them discussed that aren't principle.
@TedShifrin How can I show to a high-schooler that evaluating a definite integrals doesn't always require finding the antiderivative and using the FTC? I'm confident you have the right example.
@TedShifrin I proved that a point $x$ and a closed set $A$ not containing $x$ on a metric space can be separated by open sets like this : $A$ is closed $A'$ is open hence there is an open sphere $S_r(x)$ around $x$ not overlapping $A$. Take the union $\cup S_{r_i}(x_i)$ around $x_i$s on $A$ such that $r_i + r < d(x, x_i)$. This is our desired open set around $A$.
^ that turned out to be false
Take a point outside the real line and an open sphere with boundary touching the real line
Let $x>0$.
Let $J(x) = \sum \limits_{n=0}^{\infty} \dfrac{x^n}{2^{n(n-1)/2} n!}$
Let $J^{-1}(x)$ be the functional inverse of $J(x)$.
How do I show that for all $x$ there exists a fixed positive real constant $C$ such that:
$\left(J^{-1}(J(x)-J(x-1))-\dfrac{x}{2}\right)^2 < C $
edit :
I be...
