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You would do well to study the idea of Lagrange interpolation. There is a unique polynomial of the minimal reasonable degree fitting data. In your setting, $t^4=t$ so any of $x+y,x^4+y,x+y^4,x^4+y^4$ or $(x+y)^4=x^4+x^3y+xy^4+x^4$ agree at all points. Here are a few observations about your example: Consider the $9$ function values you specify, but as ...

1

(Extended comments in reply to Matt's comment to me.) The answer to your MO question was provided in the MSE question couched in terms of polynomials expressed as truncated power series, or ordinary generating functions (o.g.f.s) and their reciprocals. Here you use truncated Taylor series, or exponential generating functions (e.g.f.s). The series for the ...

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I figured it out. here is the code. from scipy.interpolate import CubicSpline def P(Cm): Ps = [] Cmb = [i for i in Cm] c0 = Cmb if c0 == 0.0 or c0 == 0: pass else: Cmb.insert(0, 0.0) for i in range(len(Cmb)-1): ys = [0.0 for i in range(len(Cmb))] ...

3

For associative algebras, as your required, see Plotkin, Algebras with the same (algebraic) geometry, Israel J. Math., 96 (2) (1996), 511–522. This is, being more precise, part of this nice relatively new field of Universal Algebraic Geometry which discuss such things. For a survey I recommend A. Shevlyakov, Lectures notes in universal algebraic geometry, ...

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Another equivalent characterization of Sheffer sequences is that they fit into a generating function of the form $$\sum_{n=0}^{\infty}\frac{p_n(x)}{n!}t^n=f(t)e^{xg(t)}.$$ Most of the results on Sheffer sequences apply to a more general setting where we work with a function $\Psi(x)=\sum_{n\geq 0}x^n/c_n$ and define $\Psi$-Sheffer sequences as those which ...

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While it may be, in general it is not. Consider as an example, $u=x+y+yp(x), v=x+yp(x)$ where$\deg p(x)\geq 2$. Then $(u,v)=(x,y)$ and so maximal. Notice that $u=y+v$. So, $$(u-a, v-b)=(u-v+b-a, v-b)=(y+b-a, v-b)=(y+b-a, x-b+(a-b)p(x))$$ and so most pairs of values of $a,b$, it is not maximal, as long as $\deg (x+(a-b)p(x))\geq 2$.

6

The sense of this inequality is that it turns out into equality when one of rectangles is inscribed in another as on the picture. We get $ax-by=a(m+n)-b(p+q)=am-bp=a^2\cos \varphi-b^2\cos \varphi=(a^2-b^2)\cos \varphi$, analogously $ay-bx=a(p+q)-b(m+n)=aq-bn=(a^2-b^2)\sin \varphi$, and your relation $(a^2-b^2)^2=(ax-by)^2+(ay-bx)^2$ reads as $\cos^2\varphi+\... 2 Let$r_i := p_i - q_i$. $${\bf A} (x) := \begin{bmatrix} p_1 (x) & q_1 (x) & \ldots & q_1 (x)\\ q_2 (x) & p_2 (x) & \ldots & q_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q_n (x) & q_n (x) & \ldots & p_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}_n^\top$$ Using the matrix ... 1 The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the$N$. We proceed by induction on$k$, the number of polynomials. For$k=1$, either$p_1$has only rational coefficients or it has at least one irrational. If it has an ... 5 Francesco Polizzi's idea is enough to solve the problem: Lemma. Let$f = ax^3 + bx^2 + cx + d \in \mathbf Z_{\geq 0}[x]$nonconstant with$d > 0$, such that$f(1)$is a prime$p > 2$. Then$f$is irreducible in$\mathbf Z[x]$. Proof. Suppose$f = gh$for$g,h \in \mathbf Z[x]$; we must show that$g \in \{\pm 1\}$or$h \in \{\pm 1\}$. First assume$\...

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