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Special functions, orthogonal polynomials, harmonic analysis, ordinary differential equations (ODE's), differential relations, calculus of variations, approximations, expansions, asymptotics.
4
votes
Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x +...
Yes. Suppose otherwise and let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ be two linearly independent points on the hyperplane which do not intersect the quintic. Then some po …
16
votes
Accepted
How to estimate the growth of a recurrence sequence
A very powerful way to estimate the growth of a recurrence is to look at the analytic properties of the generating function that it implies. In this case we should take the exponential generating fun …
3
votes
An identity for the cosine function
The standard way of doing problems like these is to look at the coefficients of the Chebyshev polynomials. The polynomial $T_n$ of degree $n$ such that $T_n(2 \cos \theta) = 2 \cos n \theta$ has lead …
2
votes
How small can the set of $p$ such that the $L^p$ norms are different for two fixed functions?
No. You can take $f, g$ to be sums of smooth bump functions with disjoint support. As long as $f, g$ each have bumps of the same size and shape you can put them in arbitrary locations (and orientati …
4
votes
Is there a topology on growth rates of functions?
Every totally ordered set naturally gives rise to a topology; the basis of the topology is the set of open intervals and open rays, just as in the order definition of the topology on R. See the Wikip …
10
votes
What is the term analogous to "Wronskian" for difference equations?
You can think of difference equations as "discretized differential equations," or alternately you can think of differential equations as "difference equations in the limit as the difference goes to ze …
7
votes
How to see meromorphicity of a function locally?
Checking whether a function is rational locally is straightforward, since the Taylor coefficients satisfy a very strong structure theorem. There must exist complex numbers $\alpha_1, ... \alpha_k$ an …
4
votes
Chebyshev-like polynomials with integral roots
I can satisfy conditions 1, 3 and almost satisfy condition 2. Letting $f_n(t) = {t+n-1 \choose n}$ we have the well-known generating function
$\displaystyle \sum_{n \ge 0} f_n(t) x^n = \frac{1}{(1 - …
7
votes
Subtle distinction in "completeness"?
This is related to something which has always annoyed me: some authors define, say, the group algebra $k[G]$ of a group $G$ as the set of functions $G \to k$ with finite support, equipped with convolu …
25
votes
Taylor's theorem and the symmetric group
One way is to use a combinatorial definition of the derivative. Let $A(z) = \sum a_n z^n$ be a power series. In combinatorics, where $A$ is likely to be an ordinary generating function, $a_n$ is likel …
13
votes
Connectifications?
The two-point discrete space already doesn't have a (universal) connectification, in the sense that two points don't have a coproduct in the category of connected spaces. If $X$ were such a coproduct, …
15
votes
Why do functions in complex analysis behave so well? (as opposed to functions in real analysis)
To expand on Dinakar's comment about boundary value problems, the physical intuition one should have here is that the real and complex parts of a complex differentiable functions are harmonic function …
6
votes
Accepted
Are there Generalisations of a Limit (for Just-divergent Sequences)?
Another common technique is Abel summation, which works a little better than Cesaro summation. Zeta regularization is also important in physics.
You might enjoy reading these posts at The Everything …
43
votes
6
answers
2k
views
Are there "natural" sequences with "exotic" growth rates? What metatheorems are there guaran...
A thing that consistently surprises me is that many "natural" sequences $f(n)$, even apparently very complicated ones, have growth rates which can be described by elementary functions $g(n)$ (say, to …
1
vote
Positivity of sequences via generating series
Just a quick observation. It is not hard to see that this is impossible if there exists some $i$ such that for all $j \ge i$, the coefficients of $p(x) r(x)^j$ are all positive after the first positi …