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actual evaluation – What are methods to compute polynomials that converge from above and under to a steady and bounded duty in $[0,1]$?
While the next doesn’t absolutely respond my query, I make the next notes.
The following inequality from Nacu and Peres 2005 is beneficial: $$mathbb{E}[f(X/n)]f(okay/(2n)) le kappa_f(n), tag{1}$$ the place—
 $kappa_f(n)$ is a duty that relies upon solely on $f$ and $n$ and makes the inequality maintain undoubted for all $f$ belonging to a given class of features (resembling Lipschitz steady or twicedifferentiable features), and
 $X$ is a hypergeometric($2n$, $okay$, $n$) random variable.
Notably, if a duty $f$ is such that (1) holds undoubted, then in common, for all $n$ which are powers of two—
 the $okay$th Bernstein coefficient for the higher polynomial of $n$th diploma is $f(okay/n) + delta(n)$, and
 the $okay$th Bernstein coefficient for the scowl polynomial of $n$th diploma is $f(okay/n) – delta(n)$,
the place $n$ is the polynomial’s diploma and $delta(n)$ is an answer to the next useful equation:$$delta(n) = delta(n*2) + kappa_f(n),$$or equivalently, the linear recurrence $delta(n) = delta(n+1) + kappa_f(2^n)$.
For instance—
 if $f$ is Lipschitz steady with Lipschitz ceaseless $C$—
 $kappa_f(n) = C/sqrt{2*n}$ , so
 $delta(n) = (1+sqrt{2})C/sqrt{n}$ , and
 if $f$ is twice differentiable and $M$ isn’t lower than the best worth of $f′′(x)$ for any $x$ in [0, 1]—
 $kappa_f(n) = M/(4*n)$, so
 $delta(n) = M/(2*n)$
(Nacu and Peres 2005, Proposition 10). (Perhaps the twice differentiable illustration may breathe improved to $M/(8*n)$, as urged in Powell 1981, however I’m not positive if it might proper the fifth requirement of my query in that illustration.)
For the twice differentiable illustration, the next Python code makes use of the SymPy laptop algebra library to compute $M$ and higher and scowl bounds of the polynomials for a given $n$, given a duty $f(x)$ with two or extra steady derivatives:
i=Interval(0, 1)
d=diff(diff(func))
m=Max(most(d, x, i),most(d,x,i))
bound1=minimal(func,x,i)m/(n*8)
bound2=most(func,x,i)+m/(n*8)
(Unfortunately, although, discovering the utmost can fail for some features d
; a much less well various is to employ d.subs(x, nsolve(diff(d), 0.5),(0,1),solver="bisect")+0.1
.)
Of passage, this isn’t the one route to construct polynomials that converge to a duty within the method requested for by my query, and this respond does not decipher all the problems I point out in my query. Notably, the process above does not mask features that want two steady derivatives, resembling $min(lambda, c)$. Also, the process shifts the all approximation by a ceaseless, whereas it ought to breathe workable to supply completely different approximations to the identical duty which are optimized for explicit chances of heads, as urged in Thomas and Blanchet 2012.
Other customers are inspired so as to add different solutions to my query.
Moreover:

If $f$ is understood to breathe concave in $[0, 1]$ (which roughly signifies that its charge of progress there by no means goes up), the Bernstein coefficients for the scowl polynomials are merely $f(okay/n)$, because of Jensen’s inequality.

If $f$ is understood to breathe convex in $[0, 1]$ (which roughly signifies that its charge of progress there by no means goes down), the Bernstein coefficients for the higher polynomials are merely $f(okay/n)$, because of Jensen’s inequality.
REFERENCES:
 Powell, M.J.D., Approximation Theory and Methods, 1981.
 Nacu, Şerban, and Yuval Peres. “Fast simulation of new coins from old“, The Annals of Applied Probability 15, no. 1A (2005): 93115.
EDIT (Jan. 7): There is one point to I ought to make.
Assume $f(lambda)$ is actual analytic on [0, 1] and has a Taylor succession of the kind $sum_{j=0}^{infty} a_j lambda^j,$ the place the $a_j$ are all 0 or larger and sum to 1 or much less. (An instance is cosh($lambda$) − 1.) Then the scowl polynomial of $n$th diploma is just the $n$th Taylor polynomial of $f$, specifically $sum_{j=0}^{n} a_j lambda^j.$ However, plane although this might breathe transformed to a Bernsteinform polynomial, the Bernstein coefficients is not going to often equivocate in [0, 1] except its diploma is elevated sufficient instances, and plane then, it might breathe difficult to proper the requirement given in my query that “the difference between two consecutive polynomials in each sequence [must have] nonnegative Bernstein coefficients”, because the Thomas and Łatuszyński algorithms would fail in any other case.
EDIT (Jan. 20): Edited typically.
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