21

No. The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials
$$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!}$$
and as such if
$$\displaystyle f(x) = \sum a_n {x \choose n}, a_n \in \mathbb{Z}$$
is an integer-valued polynomial, then, for example, $f \left( - \frac{1}{2} \right)$ must have ...

answered Aug 18 '14 at 7:36

Qiaochu Yuan

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10

I assume you mean nondecreasing, i.e., monotonically increasing. If so, the answer is "no." For example, interpolate (0,0), (1,10) and (2,11) with a natural cubic spline:

7

The ellipsoid method applies to this problem: once you fix a degree, the space of polynomials threading through your corridor is a convex set. Given any polynomial, if there is a point on the graph which lies outside the corridor then that point provides a linear inequality (i.e. hyperplane) separating your polynomial from your desired convex set. You can ...

6

As noted by Dejan Govc, the set $E$ should also contain those points where the right limit of $F(x)$ equals the left limit of $G(x)$, because any continuous function bounded between $F$ and $G$ will be forced to the unique limit in these points.
Let's first describe a construction for a family of continuous functions, which will later be refined to give a ...

6

Looking at the same issue i found this paper :
D’Andrea, Carlos, and Luis Tabera. "Tropicalization and irreducibility of generalized Vandermonde determinants." Proceedings of the American Mathematical Society 137.11 (2009): 3647-3656. It's a bit above my algebra skills for now but maybe it could help.

6

Viewing $X_{i,d}$ as an $N \times d+1$ matrix, $det(A)$ can be written as a polynomial in the determinants of the $d+1\times d+1$ minors. This is just because it only depends on $X_{i,d}$ up to right-multiplication by $SL_{d+1}$, and the invariants of that action are generated by the determinants of the minors. (This is just the system of Plucker coordinates ...

6

Wolfram contains the following formula which should make your calculations easy:
If $\ell_1+\ell_2+\ell_3=2g$ then $$\begin{pmatrix}
\ell_1 &\ell_2 &\ell_3\\
0&0&0
\end{pmatrix}=$$
$$(-1)^g\sqrt{\frac{(2g-2\ell_1)!(2g-2\ell_2)!(2g-2\ell_3)!}{(2g+1)!}}\frac{g!}{(g-\ell_1)!(g-\ell_2)!(g-\ell_3)!}$$
and if $\ell_1+\ell_2+\ell_3=2g+1$ then ...

answered Oct 28 '15 at 20:41

Gjergji Zaimi

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6

This isn't working. Let's look at the simplest case, $N=1$ (but the example works in general). Then the identity you are hoping for becomes
$$
\det \begin{pmatrix} f(a) & g(a) \\ f(b) & g(b) \end{pmatrix} = (b-a)\det \begin{pmatrix} f(c) & g(c) \\ f'(c) & g'(c) \end{pmatrix} ,
$$
for some $a\le c\le b$. However, taking $f(a)=f(b)=0$ won't ...

6

The result that you mention in the first part of your question is a classical result by Faber
G. Faber, Uber die interpolatorsche Darstellung stetiger Funktionen, Jahresber. der deutschen Math. Verein. 23 (1914), 190-210.
Of course, this result does not exclude pointwise convergence. This question was negatively
answered by S. Bernstein in
S. ...

6

Let $n \ge 2$. Given any points $x_0 < x_1 < \dots < x_n$, there is a quadratic function positive at all those points, but negative somewhere in $[x_0,x_n]$.
Indeed, let $c \in [x_0,x_n]$ be any point other than those $n+1$ points. There is a quadratic $\phi(x) = -1+m(x-c)^2$ that is positive at those points. Simply take $m$ large enough.
...

6

Yes, interpolation on product spaces works componentwise, so $$\Bigl(\prod_{i=1}^n X_i,\prod_{i=1}^n Y_i\Bigr) = \prod_{i=1}^n (X_i,Y_i)$$ for any interpolation functor $(\cdot,\cdot)$ even with equal norms for a fixed choice of $\ell_p$ norm on the product spaces. This follows from the restriction/corestriction theorem in Section 1.2.4 in the bible book of ...

5

A much more efficient approach can be based on semidefinite programming feasibility testing. Namely, it is well-known that global nonnegativity of a
degree $2d$ polynomial $g(x)$ is equivalent to existence of a decomposition of $g$ into a sum of squares of polynomials, which in turn is equivalent to existence of a postitive semidefinite $(d+1)\times (d+1)$ ...

5

The answer is yes if the $x_i$ are integers. Write
$$
B(X):=\prod_{i=1}^n (X-x_i),
$$
so that $B(X)$ is a monic polynomial in $\mathbf{Z}[X]$. For any $P(X)\in\mathbf{Z}[X]$, we can write
$$
P(X) = B(X) Q(X) + R(X)
$$
where $Q,R\in\mathbf{Z}[X]$ and $\deg(R)<n$. Here $R(x_i)=P(x_i)=y_i$ for every $i$. Since there is a unique polynomial $f(X)\in\mathbf{...

5

We interpolate $f(x) = \frac{x^2}{\sin(\arccos(x))^2} = \frac{x^2}{1-x^2}$ on the Chebyshev-nodes in the $(-1,1)$ interval.
Lemma 1. [1, 6.4 lemma]. If $x_1, x_2, \ldots, x_{n+1}$ are the roots of the polynomial $\omega_{n+1}(x)$, then the $\ell_i(x)$ Lagrange-polynomials can be written in the following form
$$\ell_i(x) = \frac{\omega_{n+1}(x)}{(x-x_i)\cdot ...

5

There isn't a formula for what you're looking for. At least, the formula can't make sense for all initial data and depend continuously on that initial data.
Your initial data is two points in the Stiefel manifold $V_{n,d}$, together with two points in $\mathbb R^n$. If you were to find an embedding $B^d \to \mathbb R^n$ whose derivatives agreed at two ...

5

Notice that $e^x$ does not have an inverse on the whole real line.
Extension of iterates is possible if $f$ has a fixed point $x_0$. Suppose for example, that this fixed point is repelling that is $f(x_0)=x_0$ and $\lambda=f'(x_0)>1.$
I assume that $f$ is analytic, strictly increasing on $R$ and maps $R$ onto
itself. The Poincare equation
$$F(\lambda y)=...

answered Nov 19 '17 at 20:46

Alexandre Eremenko

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5

The following discussion is based on the book Gradient Flows by Ambrosio, Gigli, and Savare (2008).
Consider $p$-Wasserstein distance with $p>1$ on a Hilbert space (for the sake of uniqueness). Let $\gamma$ be the optimal transport plan between $\mu$ and $\nu$ under the $p$-Wasserstein distance. Denote by $\pi^i$ be the $i$-th projection ($i=1,2$) and by ...

pr.probability probability-distributions geometric-measure-theory interpolation optimal-transportation

4

A cubic bezier defined by $p_1, p_2, p_3, p_4$ has parametric equation $$B(t) = (1-t)^3p_1 + 3(1-t)^2tp_2 + 3(1-t)t^2p_3 + t^3p_4.$$
The setup here also defines $A(t) = (1-t) p_2 + tp_3$.
The way $C$ is defined, there are some real $s(t)$ and $u(t)$, both possibly depending on $p_1,\ldots,p_4$ such that $C = sA + (1-s)B = up_1 + (1-u)p_4$.
So $B - C = B - ...

4

The answer is no. I assume that the degree of a rational function is the maximum of degrees of its numerator and denominator (in the irreducible form).
Consider any interval $(x_i+j,x_i+j+1)$ with $0\leq i\leq g-1$ and $0\leq j\leq g-2$. If the denominator of $R(x)$ has no roots on this interval, then $R'(x)$ should have a root on it. Let $d$ be the degree ...

4

We have with $\chi\in C^\infty_c(\mathbb R)$, equal to $1$ near 0,
$$
\frac{1}{\sinh t}=\frac{\chi(t)}{\sinh t}+\frac{1-\chi(t)}{\sinh t}.
$$
The second function belongs to $L^1$ and Young's inequality implies boundedness on $L^p$ for all $p\in [1,+\infty]$.
We write,
$$
\frac{\chi(t)}{\sinh t}=\frac1t\frac{t\chi(t)}{\sinh t}=
\frac{\psi(t)}{t}
$$
where $\...

4

For a finite non-empty set $\Omega=\{\omega_1,\dots,\omega_{N+1}\}\subset \mathbb{R}$, denote by $\Phi_{\Omega}$ the linear functional which maps a function $g:\Omega\mapsto \mathbb{R}$ to $$\Phi_{\Omega}(g)=\sum_{\omega\in \Omega} \frac{g(\omega)}{\prod_{\tau\in \Omega\setminus \omega} (\omega-\tau)}.$$
This expression may be viewed as a coefficient of $x^{...

3

There is an excellent explanation of this in Chapter 4 of L. N. Trefethen's Approximation Theory and Approximation Practice (henceforth ATAP; the first 6 chapters are available for free online). I will try to summarize it in your notation, but I recommend reading his explanation.
Chebyshev interpolation
Interpolation at the Chebyshev points gives your $\...

3

My guess is, no. The reason is that there is a positive answer to the following slightly different question, which I blieve is the right one [at least, it is the right one for my project...]
Given a matrix $X = (X_{i,j})$, where $i < m$ and $j \leq n$, can one express the product of the minors of $X$ of order $n+1$ as a single determinant in which the ...

3

Yes. This is a known inequality, called e.g. "Gagliardo-Nirenberg inequality" in Brezis' book, Comment 1 to Chapter 8, page 233.

3

The answer to your question highly depends on the domain on which you plan to approximate your function.
For products of compact intervals, e.g. rectangles in the plane, some kind of tensor product construction can be used. You should take a look at [1], which describes how bivariate functions are handled in Chebfun [2], a Matlab package for representing ...

3

Let $D$ be the minimum distance between $x$'s, merging all the $x$'s into one list of length $N$.
Let $k$ be an integer $\ge \max(16\log(8/D^2),10N)\ /\ D^2$.
Then a polynomial of degree of $6(k+1)(N-1)$ suffices.
Proof:
Let $p(x) = \frac{1}{2}(3 q(x) - q^3(x))$, where $q(x) = \sum_i \pm \Pi_{j \neq i} r_{ij}(x)$ and
$$r_{ij}(x) = \left(1-\frac{(x-x_i)^2}...

3

You might look at Carleson's 1958 paper "An interpolation problem for bounded analytic functions". A modern treatment is given in Agler and McCarthy's Pick Interpolation and Hilbert Function Spaces, Chapter 9.

3

See pp.87~107 of
Prenter, Paddy M. Splines and variational methods. Courier Corporation, 2008. especially p.100 where "uniqueness theorem" is proved and spline is defined as minimizer to $\displaystyle{\int}_{K}[f^{(m)}(u)]^2 du$ among $C^m$ functions over a set $K=[a,b]\subset\mathbb{R}$. The basic technique there is to realize $\int_K (Lf)^2=\int_K (L(f-g)...

3

Can we use the number of local extremum points of $f$ to bound the number of local extremum points of $s$?
No. See https://en.m.wikipedia.org/wiki/Monotone_cubic_interpolation for a counterexample and for examples of alternative conditions that give the kind of cubic interpolation you seek.

3

From the properties of integrals it is not hard to derive an upper bound on the $L^2$ norm from the upper bound on the $L^\infty$ norm.
Since you have (from Hall & Meyer) a bound
$\left| f(x)-s(x) \right| \leq C h^4 = M$, then:
$\sqrt{\int_a^b (f(x)-s(x))^2 dx} \leq \sqrt{M^2(b-a)} = C h^4 \sqrt{b-a} $.
So, $\|f(x)-s(x)\|_2 \leq \sqrt{b-a} \|f(x)-s(x)\|...

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