For $t,T>0$, let $f(t):=-t^{frac{8}{3}}+t^{frac{2}{3}}T^2-1$. By analyzing the derivative $f'(t)=-frac{8}{3}t^{frac{5}{3}}+frac{2}{3}t^{-frac{1}{3}}T^2$, we see on $(0,infty)$, $f(t)$ reaches its maximum $cT^{frac{8}{3}}$ at $t=frac{T}{2}$. If $T$ is large enough, we can also see $f(t)$ has two positive roots, denoted $a<b$, both depending on $T$ and satisfying
$$0<a<frac{T}{2}<b<T.$$
It is easy to see that $f(t)$ is positive on $(a,b)$.
As $Tto infty$, we see $ato 0$ and $bto T$. Moreover, for $a$, since $-a^{frac{8}{3}}+a^{frac{2}{3}}T^2-1=0$ ($a$ is a root!),
$$a^{frac{2}{3}}=frac{1}{T^2-a^2}.$$
Since $0<a<T$ from above, we see from this identity that $a asymp frac{1}{T^3}$ as $T>>1$.
Here is my question (arise from research): how to estimate (no need to compute explicitly)
$$A(T):=int_a^b frac{[f(t)]^{frac{3}{2}}}{t^5}dt$$
in terms of $T$? This integral is hard since there is a "fractional polynomial" under the square root.
The only thing I could do is to perform a Cauchy-Schwarz inequality on the integral, which yields
$$sqrt{int_a^b frac{[f(t)]^{3}}{t^{10}}dt int_a^b 1 dt }.$$
This integrand has explicit anti-derivative and can be computed. By expanding $[f(t)]^{3}$, using the facts that $a asymp frac{1}{T^3}$ and $bto T$, we have
$$sqrt{int_a^b frac{[f(t)]^{3}}{t^{10}}dt int_a^b 1 dt}asymp T^{12}.$$
This gives an upper bound for $A(T)$. Now I want to show that actually
$$A(T)asymp T^{12}.$$
Does anyone have anything to suggest for the more precise estimate?
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What is the pushout in the category of commutative unital $C^{ast}$-algebras? Is it the tensor product? Is it the same as in the category of noncommutative unital $C^{ast}$-algebras?
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Suppose I have a polynomial that is given a form
$$
f(x)=x^n – a_{n-1}x^{n-1} – ldots – a_1x – 1
$$
where each $a_k$ can be either $0,1$.
I’ve tried a bunch of examples and found that the maximum real root seems to be between $1,2$, but as for specifics of a polynomial of this structure I am not aware.
Using IVT, we can see pretty simply that $f(1)leq0$ and $f(2)> 0$ so there has to be a root on this interval, but thats a pretty wide range was wondering if this was previously studied
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Could anyone please clarify to me what active particles, in particular active deformable particles are? I have never heard of them, and I am quite curious
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Let $X$ be a smooth projective variety over an algebraically closed field $k$ of characteristic zero.
Is there a description of $operatorname{Aut}(operatorname{Coh}(X))$, i.e. the autoequivalences of the category $operatorname{Coh}(X)$?
Clearly it contains $operatorname{Aut}(X)ltimesoperatorname{Pic}(X)$ as a subgroup.
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$$int_{n+1}^{infty}f(x)dx le R_n le int_{n}^{infty}f(x)dx$$
What does this actually mean?
Let’s use n=5.
The $R_5$ is the error of the partial sum $S_5$
That error is less than the sum of the remaining terms from 5 to $infty$ ?
Why is that?
Also, why is it greater than sum of the remaining terms from 6 to $infty$ ?
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Let $n$ be a positive integer and $p$ be a prime. Let $v_p(n)$ be the $p$-adic order of $n$, i.e., the exponent of the highest power of $p$ that divides $n$. I would like to know if there is a quick and simple proof for the following congruence relation.
$$
sum_{j=1}^{lfloor log_{p} n rfloor} {n-1 choose p^j-1} equiv v_p(n) ;mbox{(mod }pmbox{)}.
$$
Key ideas involved in a ‘not so simple proof’ can be found in http://math.colgate.edu/~integers/w61/w61.pdf
Best wishes.
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Considering $(-Delta)^s: H^s(Omega) to L^2(Omega)$, is it possible to show that this operator is closed (continuous)?
For instance, taking a sequence $(u_n) subset H^s(Omega)$ with $u_n to u$ in $H^s(Omega)$, we need to show that $(-Delta)^su_n to (-Delta)^su$ in $L^2(Omega)$.
Attempt:
Consider
$$lim_{n to +infty}int_Omega |(-Delta^s)u_n(x)|^2 dx = lim_{n to +infty} int_Omega Big(int_Omega frac{u_n(x) – u_n(y)}{|x – y|^{N + 2s}} dyBig)^2 dx.$$
So my idea would be to use some inequality (perhaps something similar to the Poincaré inequality) to obtain the norm of $u_n$ in $H^s(Omega)$ on the right side. But, I don’t know how to get rid of this one on the right side of the expression above.
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I am studying ribbon graphs on surfaces. I am reading this paper (page 5).
But I have confused about how to make surfaces using the ribbon graphs. Actually, we call this surfaces as associated ribbon surfaces.
The definition of ribbon graph is given in definition 1.5 of the above mentioned paper.
Consider the following ribbon graph
To get its associated ribbon surface, first we do its edge refinement by adding each edge a degree two vertex. Then consider the half edges incident to each vertices of the graph.
Then replace the half edges with thin strips with the orientation on boundaries (this orientation of the boundaries follows the orientation of each vertex), which is as follows:
My question is how we connect the strips corresponding to the two half edges following the orientation of their boundaries to form ribbons?
Please help.
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Let $Z$ be a $[0, infty)$-valued random variable satisfying $lambda E[g(Z+1)] = E[Zg(Z)]$ for all indicator functions $g$ of Borel subsets of $[0, infty)$.
Prove that $mathcal{L}(Z) = Poisson(lambda)$.
Hint: Consider $1_{(n,n+1)}$ for $n in mathbb{N}_0$.
Proof:
Let $n in mathbb{N}_0$ and g = 1_{(n,n+1)}, then holds:
begin{align*}
lambda E[g(Z+1)] = lambda P[Z+1 in (n,n+1)] = lambda (P[Z+1 = n+1] – P[Z+1 = n]) = lambda P[Z = n] – lambda P[Z = n-1]).
end{align*}
But what is
begin{align*}
E[Zg(Z)] = E[Z cdot 1_{(n,n+1)}(Z)] = ?
end{align*}
Have I got everything right so far? If so, how should I proceed now?
Thanks for the help!
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Bézout’s theorem states that a linear diophantine equation $ax+by=c$ has integer solutions if and only if $gcd(a,b)|c$. Then is there a theorem regarding the existence of a unique integer solution for a system of two linear equations $ax+by=c$ and $px+qy=m$, where $a,b,c,p,q,m$ are positive integers?
Additionally, can the theorem be generalized to the condition where there are more than two unknowns?
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