Unsolvability of Some Problems about Piecewise-Polynomial Functions

Abstract

The algorithmic unsolvability is proved for some problems concerning piecewise polynomials of one variable with infinite number of nodes.

Alfred Tarski proved in 1948 (see, e.g. [1]) that, for any system $S$ of rational equations and inequalities with real unknowns $x_{1},\ldots,x_{n}$ and parameters $a_{1},\ldots,a_{m}$, using some algorithm, we can construct systems $T_{1},\ldots,T_{k}$ of polynomial equations and inequalities which include just the parameters $a_{1},\ldots,a_{m}$ (but not the variables $x_{1},\ldots,x_{n}$) such that the system $S$ has a solution for $x_{1},\ldots,x_{n}$ iff the parameters $a_{1},\ldots,a_{m}$ satisfy at least one of the systems $T_{1},\ldots,T_{k}.$ This leads to the existence of an algorithm determining the solvability of any algebraic system of equations and inequalities with parameters. In other words, the elementary theory of the field of real numbers (and the theory of real closed fields) is solvable. An example of a real closed field is the field of all real algebraic numbers. If a polynomial with coefficients from this field has a real root, then this root is an algebraic number.

The Tarski algorithm is not feasible. In 1987 Grigor’ev proposed the best algorithm among the known ones nowadays.

The statement about the algorithmic solvability of real polynomial equations can easily be extended to the class of algebraic equations written with not only arithmetic operations, but also radicals. In particular, there exists an algorithm determining the solvability of real rational equations written by means of the modulus sign (because $|x|=\sqrt{x^{2}}$). Any such equation can be written in form of equality of two piecewise-rational functions composed of a finite number of rational functions. It is convenient to represent such functions of one variable by piecewise-monotonic, that is, by rational functions composed of monotonic ones (on their segments). It is clear that the class of such functions of one variable is closed with respect to arithmetic operations and superposition operation and includes all continuous functions obtained from $x,|x|$ and all constants from $\mathbb{R}$ by means of arithmetic operations and superposition, as well as all continuous piecewise-polynomial functions composed of a finite number of ‘‘pieces’’, that is, all functions generated by the basis $\mathbb{R}\cup\{x+y,xy,|x|\}$ by means of superposition operations. This class includes the class $\mathcal{A}$ of all piecewise-polynomial functions composed of a finite number of polynomials with integer coefficients defined on segments with the ends from the set $\mathbb{A}$. By the Tarski theorem, for real closed fields there exists an algorithm determining the solvability of any equation of form $f=0$, $f\in\mathcal{A}$ in the field $\mathbb{A}$ (and in the field $\mathbb{R}$).

A continuous piecewise-linear function $||x||=\min\{\{x\},1-\{x\}\}$ (where $\{x\}=x-\lfloor x\rfloor$ is the function ‘‘fractional part’’) equal to the distance from the point $x$ to the nearest integer number. It is clear that the function $||x||=1/2$ does not belong to the class $\mathcal{A}$, because it is composed of an infinite number of ‘‘pieces’’. The function $||x||$ can be expressed through $\sin,\arcsin,\pi,1$ and arithmetic operations by the formula

$$\frac{\arcsin\left(\sin\left(\dfrac{\pi}{2}(4x-1)\right)\right)}{2\pi}+\frac{1}{4};$$

therefore, it is an elementary function. Let $\mathfrak{A}$ be a class of all functions of one variable generated by the basis $\{1,x-y,xy,||x||\}$ by means of superposition operations. This includes all polynomials with integer coefficients. It is easy to see that all other functions from this class are piecewise-polynomial, the polynomial have integer coefficients, are defined on segments whose ends belong to $\mathbb{A}$, and, at that, partitioning of the straight line $\mathbb{R}$ into these segments may include an infinite number of them. Because the points of local extrema of an integer polynomial belong to $\mathbb{A}$, any such function can be considered a piecewise-monotonic on segments with the ends from $\mathbb{A}$. Indeed, suppose that $f,g\in\mathfrak{A}$. Each function corresponds to the algebraic partitioning of the straight line $\mathbb{R}$ into segments (this is the name of the partitioning by the points from the field $\mathbb{A}$). Intersection of these partitionings is obviously algebraic, and on each segment of the intersection both functions are equal to integer polynomials. Because the sum, difference, and multiplication of integer polynomials also are integer polynomials, $f\pm g,gf\in\mathfrak{A}$. Suppose that $h=f(g)$ is a superposition of $f$ and $g$. At each segment $I$ of its own partitioning $g(x)$ is a monotonic polynomial. Then $g(I)$ is a segment with ends from $\mathbb{A}$, because the value of the integer polynomial also is algebraic at the point from $\mathbb{A}$. We divide the segment $J=g(I)$ into segments $J_{i}$ by the points $x_{i}$ from the partitioning of the function $f$ (if they exist), obtain the algebraic partitioning, and consider the corresponding partitioning of the segment $I$ into the segments $I_{i}$ by the points $y_{i}=g^{-1}(x_{i})$. Because $x_{i}\in\mathbb{A}$, $y_{i}\in\mathbb{A}$ as a root of the equation $g(y_{i})=x_{i}$ due to the real closedness of the field $\mathbb{A}$. On the segment $I_{i}$ the superposition $h(x)=f(g(x))$ is an integer polynomial. The constructed polynomials and segments show that $h\in\mathfrak{A}.$

The Matiyasevich–Davis–Putnam–Robinson theorem [2] about nonsolvability of the problem about determining the solvability of Diophantine equations implies the following

Theorem. The problem of determining the solvability $($ in $\mathbb{R}$ and $\mathbb{A})$ of the equation $f(x)=0$ and inequality $f(x)\geqslant 0$ of one variable at $f(x)\in\mathfrak{A}$ is algorithmically unsolvable.

Proof. We modify and simplify the reasoning from [2–5], in which the class of functions generated by the basis $\{1,x\pm y,xy,\sin\}$ by means of superposition operations is considered. This class is more complex than $\mathfrak{A}$ and includes transcendent elementary functions.

Any Diophantine equation $D(y_{1},\ldots,y_{n})=0$ (below, we consider its solvability not in integer, but in natural numbers; zero is assumed to be a natural number) can be transformed into an equivalent equation in the field $\mathbb{R}$:

$$D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||=0.$$

If by $\lfloor x\rceil$ we denote the integer point closest to a number $x$ (this is the denotation by Khrapchenko; for semiinteger numbers $n+1/2$ we can take both $n$ and $n+1$), then for any polynomial $D\in\mathbb{Z}[x_{1},\ldots,x_{n}]$ the following inequality is obvious:

$$|D(x_{1},\ldots,x_{n})-D(\lfloor x_{1}\rceil,\ldots,\lfloor x_{n}\rceil)|\leqslant\sum_{i=1}^{n}D_{i}(x_{1},\ldots,x_{n})||x_{i}||,$$

where $x_{i}=\lfloor x_{i}\rceil\pm||x_{i}||$, $x_{i}\geqslant 0$ and $D_{i}$ are suitable polynomials from $\mathbb{Z}[x_{1},\ldots,x_{n}]$. Therefore,

$$|D(x_{1}^{2},\ldots,x^{2}_{n})-D(\lfloor x^{2}_{1}\rceil,\ldots,\lfloor x^{2}_{n}\rceil)|\leqslant P\delta,\quad\delta=\max_{i}||x^{2}_{i}||,\quad P=\sum_{i=1}^{n}D_{i}(x^{2}_{1},\ldots,x^{2}_{n}).$$

Suppose that $M(x_{1},\ldots,x_{n})=1+P(x_{1},\ldots,x_{n})$. Then, if

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1,$$

then $\delta<1/M^{2},$ $|D(x^{2}_{1},\ldots,x^{2}_{n})|<1/M$, which means that

$$|D(\lfloor x^{2}_{1}\rceil,\ldots,\lfloor x^{2}_{n}\rceil)|\leqslant|D(x^{2}_{1},\ldots,x^{2}_{n})|+|D(\lfloor x^{2}_{1}\rceil,\ldots,\lfloor x^{2}_{n}\rceil)-D(x^{2}_{1},\ldots,x^{2}_{n})|$$

$${}<1/M+\delta P<1/M+P/M^{2}=(M+P)/M^{2}=(1+2P)/(1+P)^{2}<1;$$

consequently, $D(\lfloor x^{2}_{1}\rceil,\ldots,\lfloor x^{2}_{n}\rceil)=0$, because it is an integer number. Therefore, the Diophantine equation

$$D(y_{1},\ldots,y_{n})=0$$

is solvable in natural numbers iff the inequality

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1$$

is solvable in real numbers. Indeed, if $D(y_{1},\ldots,y_{n})=0,y_{i}\in\mathbb{N}$, then at $x_{i}=\sqrt{y_{i}}$

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)$$

$${}=M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(y_{1},\ldots,y_{n})+\sum_{i=1}^{n}||y_{i}||\right)=0<1,$$

and, conversely, if

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1,$$

then, as we have proved above, $D(\lfloor x^{2}_{1}\rceil,\ldots,\lfloor x^{2}_{n}\rceil)=0$, and, hence, the equation $D(y_{1},\ldots,y_{n})=0$ has a natural solution.

To obtain an inequality of one variable from the inequality

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1,$$

we use

Lemma. A plane continuous curve $\mathbb{R}_{+}\rightarrow r(x)$ , $r(x)=(f(x),g(x))=(x||x^{3}||,x||x||)$ is everywhere dense on the octant $\mathbb{R}_{+}^{2}$ . A continuous curve $\mathbb{R}_{+}\rightarrow r^{n}(x)$ ,

$$r^{n}(x)=(r^{n}_{1}(x),\ldots,r^{n}_{n}(x))=(f(x),f(g(x)),f(g(g(x))),\ldots,\underbrace{g(g(\ldots g(x))\ldots)}_{n-1})$$

is everywhere dense on the octant $\mathbb{R}_{+}^{n}.$

Proof. Let $(a,b)\in\mathbb{R}_{+}$. Consider the inequality $g(x)\in[a-\epsilon,a]\subset\mathbb{R}_{+},\epsilon<1/2$. At $n\in\mathbb{N}$ on the segment $[n,n+1/2]$ the function $g(x)$ is equal to $x(x-n)$ and $g(x)\in[a-\epsilon,a]$ if for $n>2a+2$

$$x\in I_{n}=[a_{n},b_{n}]=[n+(a-\epsilon)/n,n+a/(n+1/2)].$$

If $x\in I_{n}$, then for $n>(a+1)/\epsilon>2a+2$ the length of the segment $|I_{n}|>(\epsilon-a/(2n+1))/n>\epsilon/(2n)$; consequently, $x^{3}\in J_{n}=[a_{n}^{3},b_{n}^{3}]$ and the length of the segment $J_{n}$

$$|J_{n}|>3|I_{n}|a_{n}^{2}>3\epsilon n/2>3(a+1)/2>3/2;$$

therefore, it includes the segment $[N,N+1/2],N\in\mathbb{N},$ $N\geqslant a_{n}^{3}$, and, when $x$ runs over the segment $[\sqrt[3]{N},\sqrt[3]{N+1/2}]\subset I_{n}$, then $f(x)=x||x^{3}||=x(x^{3}-N)$ increases from $0$ to $\sqrt[3]{N+1/2}/2$. For $n>2b$, by the inequality $\sqrt[3]{N}\geqslant a_{n}>n$ there exists an $x\in I_{n}$ such that $g(x)\in[a-\epsilon,a]$, $f(x)=b$; therefore, the plane curve $r(x)=(f(x),g(x))$ is everywhere dense on $\mathbb{R}_{+}^{2}$. The second statement can be proved by induction. The base is already proved.

The induction step. Suppose that $(a_{1},\ldots,a_{n})\in\mathbb{R}^{n}_{+}$ and a number $x$ is such that

$$r^{n-1}(x)=(r^{n}_{2}(x),\ldots,r^{n}_{n}(x))\in\prod_{i=2}^{n}[a_{i},a_{i}+\epsilon/2].$$

Due to the continuity of the curve, there exists a number $\delta>0$ such that for any $x\in[X,X+\delta]$

$$r^{n-1}(x)\in\prod_{i=2}^{n}[a_{i}-\epsilon,a_{i}+\epsilon].$$

For the curve $r(x)=(f(x),g(x))$ there exists a number $x$ such that $f(x)=a_{1},g(x)\in[X,X+\delta]$. Then

$$(r^{n}_{2}(x),\ldots,r^{n}_{n}(x))=r^{n-1}(g(x))\in\prod_{i=2}^{n}[a_{i}-\epsilon,a_{i}+\epsilon],\quad r^{n}_{1}(x)=f(x)=a_{1};$$

therefore, the curve $r^{n}$ is everywhere dense in $\mathbb{R}_{+}^{n}$. $\Box$

Further, instead of $r^{n}_{i}$ we write $r_{i},$ and $r^{2}_{i}$ denotes $(r_{i})^{2}$. The note on the class $\mathfrak{A}$ made before the formulation of the theorem implies that the functions $r_{1}(x),\ldots,r_{n}(x)\in\mathfrak{A}$ and for any $M,D\in\mathbb{Z}[x_{1},\ldots,x_{n}]$

$$h(x)=M^{2}(r_{1}(x),\ldots,r_{n}(x))\left(D^{2}(r^{2}_{1}(x),\ldots,r^{2}_{n}(x))+\sum_{i=1}^{n}||r^{2}_{i}(x)||\right)\in\mathfrak{A}.$$

If the inequality

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1$$

is solvable in the field $\mathbb{R}$, then, choosing a number $x>0$ such that $||r^{2}_{i}(x)||\rightarrow 0$ for all $i$ and the condition

$$D(\lfloor r^{2}_{1}(x)\rceil,\ldots,\lfloor r^{2}_{n}(x)\rceil)=0$$

holds, we obtain the fact that there exists an $x>0$ such that $0\leqslant h(x)<1/2$.

Choosing a number $x>0$ such that $||r^{2}_{i}(x)||\rightarrow 1/2$, we obtain the fact that there exists an $x>0$ such that $h(x)>1/2$ (in the case $n=1$ we should choose a semiinteger $y_{1}=x^{2}_{1}$ so that $D(y_{1})\neq 0$). Therefore, in the case of solvability of the Diophantine equation $D(y_{1},\ldots,y_{n})=0$ in natural numbers, the equation $2h(x)=1$ has roots in $\mathbb{R}_{+}$. Due to the note before the formulation of the theorem, they belong to the field $\mathbb{A}.$ If the equation $D=0$ has no solutions in $\mathbb{N}$, then the inequality

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)<1$$

has no solutions in $\mathbb{R}$ (otherwise, there would be solutions in $\mathbb{N}$ to the equation $D=0$), which means that the inequality

$$M^{2}(x_{1},\ldots,x_{n})\left(D^{2}(x^{2}_{1},\ldots,x^{2}_{n})+\sum_{i=1}^{n}||x^{2}_{i}||\right)\geqslant 1$$

always holds; therefore, $h(x)\geqslant 1$ and the equation $2h(x)=1$ has no roots in $\mathbb{R}$.

Because the problem of solvability of Diophantine equations in $\mathbb{N}$ is unsolvable, an equivalent problem of solvability (in the fields $\mathbb{R},\mathbb{A}$) of the equations of form $2h=1$, and, therefore, of any equations written by means of the functions from $\mathfrak{A}$, is unsolvable. The problem of solvability of the inequalities of form $h(x)<1$ also is unsolvable, because it is equivalent to the problem of the Diophantine equation $D=0$ in $\mathbb{N}$. Therefore, the problem of determining whether an arbitrary equation $f(x)=0$ or inequality $f(x)<0$, $f\in\mathfrak{A}$, has solutions in the fields $\mathbb{R}$ and $\mathbb{A}$ is also unsolvable.

If, instead of the class $\mathfrak{A}$ we take a class composed of the restrictions of the functions of the class $\mathfrak{A}$ to a given finite segment, then the theorem hypothesis becomes false, because the Tarski algorithm allows solving the problems considered in it.

Below, similarly to [2], we derive Corollaries 1–3.

Corollary 1. If, in addition to the function $||x||$ , we admit the use of the function $|x|$ , then the problem of testing the validity of an arbitrary identity $f(x)=g(x)$ is also unsolvable.

Proof. Indeed, the identity $f(x)+|f(x)|=0$ holds iff the inequality $f>0$ has not real solutions and the problem of determining the solvability of the inequalities written by means of the function $||x||$ is unsolvable. $\Box$

Corollary 2. In the class of functions of one variable generated by the basis $\{1,1/x,x-y,xy,||x||\}$ using the superposition operations, the problem of determining whether an improper integral converges is unsolvable.

Proof. Let $h_{D}(x)$ be a function constructed in the proof of the theorem by an arbitrary Diophantine equation $D=0$. Consider the integral

$$\int\limits_{-\infty}^{+\infty}\frac{dx}{(x^{2}+1)(2h_{D}(x)-1)^{2}}.$$

If the equation $D=0$ is unsolvable in $\mathbb{N}$, then $h_{D}\geqslant 1$ and the integral converges. Otherwise, there exists a root $x_{0}$ of the equation $2h_{D}(x_{0})-1=0$ and, for instance, to the left from $x_{0}$ the equality by the order $2h_{D}(x)-1=\Theta(x-x_{0})^{k}$ holds (the function $2h_{D}(x)-1$ is polynomial and nonzero on some segment). Then the integral

$$\int\limits_{-\infty}^{x_{0}}\frac{dx}{(x^{2}+1)(2h_{D}(x)-1)^{2}}$$

diverges, because the following integral diverges:

$$\int\limits_{-\infty}^{x_{0}}\frac{dx}{(x-x_{0})^{2k}}.$$

$\Box$

Corollary 3. In the class of functions of one variable generated by the basis $\{1,x-y,xy,\sqrt{|x|},||x||\}$ using the superposition operations, the problem of determining whether a given integral can be computed in elementary functions is unsolvable.

Proof. Let $f(x)=\sqrt{1+x^{6}}$. This differential binomial cannot be integrated in elementary functions. It is clear that the function $f$ belongs to the considered case, as does the function $g(x)=1+|4x-4|-|4x-3|$. Because $g(x)=0$ for $x\geqslant 1$ and $g(x)=2$ for $x\leqslant 3/4$, the function $s_{D}(x)=g(h_{D}(x))f(x)$ belongs to the given class; in the case of a solvable Diophantine equation $D=0$ it is identically equal to $2f(x)$ in some neighborhood of the root $x_{0}$ of the equation $h_{D}(x)=1/2$, and, in the case of an unsolvable equation $D=0$ the inequality $h_{D}(x)\geqslant 1$ and the function $s(x)=0$ hold identically. In the first case it cannot be integrated in elementary functions, whereas in the second case it can be integrated in elementary functions. $\Box$

We point out several further corollaries.

We call a finite system of continuous functions the complete basis if any continuous function can on any compact be arbitrarily accurately approximated in the uniform metric by the functions that can be expressed in form of superpositions of basis functions. For instance, the Weierstrass theorem implies the completeness of the basis $\{1/2,x-y,xy\}$. It is clear that the basis $\{1,x-y,xy\}$ is incomplete, because, by the Bernshtein theorem, on segments of length larger than $4$, we can arbitrarily accurately approximate polynomials just by the same polynomials with integer coefficients. Suppose that $\mathcal{K}$ is the class of all polynomial and piecewise-polynomial functions whose polynomials have integer or semiinteger coefficients and are defined on segments with ends belonging to the field $\mathbb{A}$, and the partitioning of the straight line $\mathbb{R}$ into these segments can contain an infinite number of them.

Corollary 4. The problem of determining the completeness for the bases expressible through the functions from $\mathcal{K}$ is algorithmically unsolvable.

Proof. The problem of determining the solvability of Diophantine equations $D=0$ whose free term of polynomial is less than $2$ in magnitude is clearly unsolvable (otherwise, this problem would be solvable for arbitrary Diophantine equations). Let $h_{D}(x)$ be a function defined in the proof of the theorem. Because for the functions from the lemma $f(m)=g(m)=0$ at $m\in\mathbb{Z}$, the functions $r^{n}_{i}$ are also zero at integer points. From the arguments in the proof of the theorem it follows that, if an equation $D=0$ is solvable, then $h_{D}(x)<1$, and if it is not, then $h_{D}(x)\geqslant 1$, where

$$h_{D}=M^{2}(r_{1}(x),\ldots,r_{n}(x))\left(D^{2}(r^{2}_{1}(x),\ldots,r^{2}_{n}(x))+\sum_{i=1}^{n}||r^{2}_{i}(x)||\right)\in\mathcal{K}.$$

In the case $h_{D}(x)\geqslant 1$ we then have $h_{D}(m)=M^{2}(0,\ldots,0)D^{2}(0,\ldots,0)\geqslant 4$ at $m\in\mathbb{Z}$. Suppose that

$$s(x)=(1+|x-1|-|x-2|)/4,$$

then $s(x)=0$ for $x\leqslant 1$, $s(x)=(x-1)/2$ for $1\leqslant x\leqslant 2$, $s(x)=1/2$ for $x\geqslant 2$, and the function $H_{D}(x)=s(h_{D}(x))\in\mathcal{K}.$ In the case of solvability of equation $D=0$ it is clear that $H_{D}$ is identically zero; otherwise, $H_{D}(m)=1/2$ at $m\in\mathbb{Z}$. In the first case the basis $\{H_{D}(x),1,x-y,xy\}$ is incomplete, and in the second case it is complete, because $H_{D}(1)=1/2$. Therefore, the problem of determining the completeness of the bases of form $\{H_{D}(x),1,x-y,xy\}$ is unsolvable. The basis $\{s(x),1,x-y,xy\}$ is clearly complete, because $s(2)=1/2$, whereas the basis $\{h_{D}(x),1,x-y,xy\}$ is incomplete, because $h_{D}(m)\in\mathbb{Z}$, which means that all functions of the basis preserve the set $\mathbb{Z}$; therefore, the constant $1/2$ cannot be approximated with an accuracy less than $1/2$. $\Box$

Below, we consider the bases consisting of piecewise-polynomial functions complete in the Boolean sense, that is, such functions in which we can express any Boolean function by formulas (a formula expresses a Boolean function if, at substitution into it zeros or unities instead of variables, the value of the formula becomes zero or unity). For an arbitrary Boolean function $f$ the minimal complexity of the formula implementing it in a given basis $B$ is denoted by $L_{B}^{\Phi}(f)$ and is called complexity of implementation of this function $f$ by formulas over $B$, and by $L^{\Phi}_{B}(n)$ we denote $\max_{f(x_{1},\ldots,x_{n}}\{L^{\Phi}_{B}(f)\}$. Similarly, we define $L_{B}(n)$ in the case of its implementation not by formulas, but by circuits (nonbranching programs). For finite bases it is well-known [6] that $L_{B}(n)=\Theta(2^{n}/n)$ and $L^{\Phi}_{B}(n)=\Theta(2^{n}/\log n)^{C_{B}}$, where $C_{B}$ is some constant. Of interest are also the bases containing, in addition to a finite number of functions, a continuum of constants. For instance, for a basis $B=\{(x-y)/2,|x|,x/2,\min(x,y),1-x\}\cup[0,1]$ it is known [6] that $L_{B}(n)=\Theta(2^{n}/n)$, and for the basis $B=\{x-y,|x|,x/2,\min(x,y),1-x\}\cup[0,1]$ it is known [6–88] that $L_{B}(n)=\Theta(2^{n/2})$.

From [6] we can conclude that for the basis $B=\{x-y,xy,||x||,|x|,x^{2}\}\cup[0,1]$ the order equality holds: $L_{B}(n)=\Theta(n)$ (it holds also without the function $x^{2}$), $L_{B}^{\Phi}(n)=\Theta(n)$, and for the basis $B=\{x-y,xy,||x||,|x|\}\cup[0,1]$ the order equality $L_{B}(n)=\Theta(n)$ holds, and from [6–8] we can conclude that $L_{B}^{\Phi}(n)=\Theta(2^{n/2})$.

Corollary 5. In the class of Boolean complete piecewise-polynomial finite bases with a continuum of constants, the problems of determining the solvability of any of the following relations are unsolvable:

$$L_{B}(n)=\Theta(L^{\Phi}_{B}(n)),\quad L^{\Phi}_{B}(n)>\Theta(2^{n/2}),\quad L_{B}(n)>n2^{n/2}.$$

Proof. We take the function defined in the proof of Corollary 4 $H_{D}(x)=s(h_{D}(x))\in\mathcal{K}$. If an equation $D=0$ is solvable, this function is identically zero; otherwise, $H_{D}(m)=1/2$ at $m\in\mathbb{Z}$. Suppose that $F_{D}(x,y)=2H_{D}(y)x^{2}$. Then, in the case of solvability of an equation $D=0$, the function $F_{D}(x,y)=0$; otherwise, $F_{D}(x,0)=x^{2}$. Consider a class of bases $B=\{x-y,xy,||x||,|x|,F_{D}(x,y)\}\cup[0,1]$. In the case of solvability of an equation $D=0$ we have $B=\{x-y,xy,||x||,|x|,0\}\cup[0,1]$, which means that $L_{B}^{\Phi}(n)=\Theta(2^{n/2})$, $L_{B}(n)=\Theta(n)$; otherwise, in the basis $B$ the function $x^{2}$ can be represented as a formula $F_{D}(x,0)$ and, therefore, $L_{B}(n)=\Theta(n),$ $L_{B}^{\Phi}(n)=\Theta(n)$, because these equalities hold for $B=\{x-y,xy,||x||,|x|,x^{2}\}\cup[0,1]$. Therefore, the problem of determining, by a given basis, whether the equality $L_{B}(n)=\Theta(L^{\Phi}_{B}(n))$ holds is unsolvable. By an analogous reason the problem of determining whether the equality $L^{\Phi}_{B}(n)\geqslant\Theta(2^{n/2})$ holds is unsolvable, as are the other similar problems.

For instance, let us take a function $F_{D}(x,y,z)=(H_{D}(z)+1/2)*(x-y)$. Then, if an equation $D=0$ is solvable, we have $F_{D}(x,y,z)=(x-y)/2$; otherwise, $F_{D}(x,y,0)=x-y$. Consider the class of bases

$$B=\{xy,|x|,\min(x,y),1-x,x/2,F_{D}(x,y,z)\}\cup[0,1].$$

Then, in the case of solvability of the equation $D=0$ we have $L_{B}(n)=\Theta(2^{n}/n)$; otherwise, $L_{B}(n)=O(2^{n/2})$. Hence, the problem of determining whether, for a given basis, the inequality $L_{B}(n)>n2^{n/2}$ holds is unsolvable. $\Box$

For an arbitrary complete basis $B$ and a continuous function $f$, for any $\epsilon>0$ we can determine the complexity of its $\epsilon$-approximation $L_{B}(f,\epsilon)$ as the least number of elements in the circuit constructed from the functions of the basis $B$ and computing the function uniformly deviating from $f$ on its domain by no more than $\epsilon$. For any compact functional class $K$ we can define $L_{B}(K,\epsilon)$ as $\max_{f\in K}L_{B}(f,\epsilon)$. For instance, for the class $K=W(1/8,1/8,[0,1])$ consisting of the functions defined on the segment $[0,1]$, bounded in their magnitude by the constant $1/8$, and satisfying the Lipschitz condition with the same constant and for the basis $B=\left\{\dfrac{x-y}{2},\max(x,y),|x|,-x,1\right\}\bigcup\{cx:0\leqslant c\leqslant 1\}$, it was proved in [6] that $L_{B}(K,\epsilon)=\Theta\left(\dfrac{1}{\epsilon\log_{2}\dfrac{1}{\epsilon}}\right)$ (the lower bound is valid for any Lipschitz class $W(M,N,[a,b])$), while in [9] it was proved that for any Lipschitz class $W(M,N,[a,b])$ and for the basis

$$B=\{x-y,\max(x,y),|x|,1\}\bigcup\{cx:0\leqslant c\leqslant 1\}\},$$

the order equality is valid: $L_{B}(W,\epsilon)=\Theta\left(\dfrac{1}{\sqrt{\epsilon}}\right)$.

Corollary 6. In the class of complete piecewise-polynomial bases containing a finite number of piecewise-polynomial functions and a continuum of linear functions $\{cx:0\leqslant c\leqslant 1\}$ , for any function $L(\epsilon)$ such that

$$\frac{1}{\sqrt{\epsilon}}\log\frac{1}{\epsilon}<L(\epsilon)<\dfrac{1}{\epsilon\log^{2}_{2}\dfrac{1}{\epsilon}},$$

the problems of determining the solvability of the following equalities are unsolvable for Lipschitz classes $K$ :

$$L_{B}(K,\epsilon)>L(\epsilon),\quad L_{B}(K,\epsilon)<L(\epsilon).$$

Proof. Assume the contrary. Suppose that $D=0$ is an arbitrary Diophantine equation. Let us take a function $F_{D}(x,y,z)=(H_{D}(z)+1/2)*(x-y)$ from the proof of Corollary 5. Consider the class $W(1/8,1/8,[0,1]$ and the basis $B=\{F_{D}(x,y,z),\max(x,y),|x|,-x,1\}\bigcup\{cx:0\leqslant c\leqslant 1\}.$ According to the proof of Corollary 5, when the equation $D=0$ is solvable, we have $F_{D}(x,y,z)=(x-y)/2$; in the opposite case, $F_{D}(x,y,0)=x-y$. In the first case the basis $B=\{(x-y)/2,\max(x,y),|x|,-x,1\}\bigcup\{cx:0\leqslant c\leqslant 1\}$, and, according to the remark made before the formulation of Corollary 6, $L_{B}(K,\epsilon)=\Theta\left(\dfrac{1}{\epsilon\log_{2}\dfrac{1}{\epsilon}}\right)>L(\epsilon)$. In the second case, using the considered basis, we can express the basis

$$B=\{x-y,\max(x,y),|x|,-x,1\}\bigcup\{cx:0\leqslant c\leqslant 1\}$$

for which $L_{B}(K,\epsilon)=\Theta\left(\dfrac{1}{\sqrt{\epsilon}}\right)$, and, therefore, for the considered basis $L_{B}(K,\epsilon)<L(\epsilon)$. If the studied problem of determining the solvability of inequalities would be solvable, then the problem about Diophantine equations would be also solvable. $\Box$