COMPUTATION OF POLYNOMIALS, THE LEAST DEVIATING FROM ZERO IN L1 METRICS BY SOLVING EXTREMAL PROBLEMS Текст научной статьи по специальности «Математика»

MATHEMATICAL SCIENCES

COMPUTATION OF POLYNOMIALS, THE LEAST DEVIATING FROM ZERO IN Li METRICS BY

SOLVING EXTREMAL PROBLEMS

Suleykin A.,

A teaching assistant at the Lomonosov Moscow State University Faculty of Mechanics and Mathematics Mammahajiyev R.,

A graduate student at the Lomonosov Moscow State University

Faculty of Space Research https://doi.org/10.5281/zenodo.7258432

Abstract

In this paper, we consider polynomials that deviate least from zero in the Li metric. Then we explicitly calculate Chebyshev polynomials of the second kind by solving extremal problems.

Keywords: Chebyshev polynomials, symmetric polynomials, extremal problems, normed spaces, Li space, Weierstrass theorem, Jackson theorem, polynomials deviating least from zero, Waring's inverse formula, Chebyshev polynomials of the second kind, approximation theory, optimization methods.

1. Introduction

Consider normed spaces Li([-1;1]) of functions x(t) measurable on the interval [-1;1] such that the value is finite:

The general statement of the problem of polynomials deviating least from zero is as follows:

Among the polynomials pn-i £ Pn-i, among polynomials of the form pn-i(t) = ^xttk, k=0 find the one that delivers the minimum in the task

71— 1

fnl(v) =

tn+P„-l

Theorem 3.1 ([4]). Function

£i([-i;i]) fnl (#) =

tr

k=o

xktk

¿i([-i;i])

—> min.

(P)

tn+Pn-1

11 £i{[—i;l]) is convex, continuous, and grows at infinity.

The function f„i() is convex, continuous in x, and the solution in the problem exists according to the Weierstrass theorem and its corollary.

Theorem 3.2 (Weierstrass). A continuous function on a non-empty bounded closed subset of a fnite-dimen-sional space (compact set) reaches its absolute maximum and minimum.

We single out a simple consequence of this theorem. Corollary 1. If the function f is continuous on the space R11 and

then it reaches its absolute minimum on any closed subset ofRn (not necessarily bounded). A similar statement for the absolute maximum holds if limfx) =

Let us prove that when a continuous function is approximated by algebraic polynomials of best approximation in the corresponding metrics of the spaces Li and C is unique. Definition 1. Consider a system of real functions

fo(t),fi(t),...,fn(t),

continuous on the segment [a;b]. Consider

f(t) = ft) + f) + ... + ft)

for arbitrary coefficients fi0,...,^„. Let each f(t) have at most n distinct zeros on (a,b). Such a system will be called a Chebyshev system with respect to the interval (a;b).

Theorem 3.3 (Jackson). If any non-zero polynomial in a system of linearly independent continuous ones on the segment [a;b] offunctions has at most n changes of sign on the interval (a;b), then for any function f £ C([a;b]) there exists a unique polynomial of best approximation in the metric of the space Li([a;b]) in this system.

Theorem 3.4 (Haar). For any function f £ C([a;b]) there is a unique polynomial of best approximation in the metric of the space C([a;b]) with respect to the system of linearly independent continuous ones on the interval [a; b] functions if and only if any non-zero polynomial in this system of functions has on [a;b] at most n different zeros. In other words, a system offunctions is Chebyshev if and only if it satisfies the condition Haar.

Example 1. The system of algebraic polynomials {1 ,t,...,f} satisfies the Haar condition on any inten'al \a:b |,

since any non-trivial polynomial Ek—o has at most n different zeros on the inten'al (a:b), and, hence, is Chebyshevskaya.

It follows from the above example and the theorems of Jackson and Haar that when a continuous function is approximated by algebraic polynomials, the element of best approximation in the corresponding metrics of the spaces Li and C is unique. Let us denote the solution of the problem by x, the value by and the corresponding polynomial by T,i(-).

In this way.

71— 1

Tnl(t) -tn + jykt\ k=0

fnl

Tnl(')

M[-i;i]>

The polynomial T„i( ) is called the polynomial that deviates least from zero in the metric Li([-1;1]). 2. Polynomials deviating least from zero

We introduce several auxiliary theorems that we will need to calculate polynomials.

Lemma 1. Let ft) be a piecewise continuous function on the segment [a;b] and b

tkf(t)dt = 0,

k = 0,...,« - 1.

Then the function f/) changes sign on the interval (a;b) at least n times.

Lemma 2. The polynomial Tnq(^) for 1 < q < œ has exactly n zeros on the interval (—1 ;1)

Theorem 4.1. In the space Lp([-a;a]), 1 < p < an even function is approximated by an even polynomial, and an odd function is approximated by odd polynomial.

Proof. Take a function /(0- Let the polynomial /,.—o is a polynomial of best

approximation for it, i.e.,

= inf

m-m

P&Vn

m-p(t)

Due to the symmetry with respect to zero of the segment [-a;a] for an even function ft) we have

m-m

m-p(-t)

n-t)-p(-t)

Hence, due to the uniqueness of the best approximation polynomial p(-t) = p(t). Hence the polynomial p(t) is even. Similarly, due to the symmetry with respect to zero of the

segment [—a;a] for the odd functionf(t) we have ft) — p(t)llLp =lf(—t) — p(—t)ILp = I — f(t) — p(—t)llLp = f(t) — (—p(—t))ILp.

Hence, due to the uniqueness of the best approximation polynomial p(t) = -p(-t) ^^ p(-t) = -p(t). Hence, the polynomial p(t) odd.

□

Let us formulate a necessary condition for an extremum of I order in a finitedimensional unconstrained problem, which is an analog of Fermât's theorem.

Theorem 4.2. //' ~ (xl> • ■ • > xn) £ locextr f .-point oflocal extremum ofthe function ofn variables f(x1,...,xn) and the function f £ D(x) is differentiable at the point x, then

fix) - 0

r)f(x)

df(x) dxn

= 0

dxi

3. Symmetric polynomials

Definition 2. Polynomial P(xi,x2,...,x„) is called symmetric, if for any permutation x1i,x,2,...,xi„ variables xi ,x2,...,xn

P(Xii,Xi2,...,Xin) = P(Xi,X2,...,Xn).

a

J\

Definition 3. Polynomials

h, J k = 0,

Gk(xi,x2,...,x„) = II X xil ■■■xik, 1 < k < n,

l<il<-.<ik<n

^0, k > n,

are called elementary symmetric polynomials of the kth degree.

Let's write out elementary symmetric polynomials:

Oi(Xi,X2,...,Xn) = X1 + X2 + ■■■ + Xn; CT2(Xl,X2,---,Xn) = X1X2 + X1X3 + ■■■ + X^^;

cr3(xl,x2.....xn) = xlx2x3 + xlx2x4 + ... + xn—2xn—\xir„...

Definition 4. For k £ N, a polynomial of the form^k t^lt • ■ • ixn) + • • • + Xn js called the power

sum of the kth degree ■ For example,

S1 (Xl, X2, .. ., X;7) = Xl +... + x„:

S3 (Xj, X2, • . • , = ^Cj + . - . + X^

1S4 (X] j Xq j ... ^ Xn) — Xj I - - h. I Xp

Obviously, the polynomials 0•k(xl,x2,■■■,xn) and S^XbX^^Xn) are symmetric polynomials.

Theorem 5.1 (Waring's formula). For anv A" £ N the formula

sk = k J2

Oil,.",On

where the sum is taken over all sets of non-negative integers a^^asatisfying the equality a1 + 2a2 + ■■■ + nan = k Here Sk = Skx^^^Xn), ai = стi(x1,■■■,x„), i = l^^n^ For example, Si = ai;

-a2

Î - 2^2;

s3 — of - 3ci o"2 + 3cr3; 34 = af - 4crfa2 + 2crf + 4cr1cr3 - 4a4

Theorem 5.2 (Waring's inverse formula). For any k £ N the formula

where the sum is taken over all sets of non-negative integers satisfying ^ +2fi2 + ■■■ + nft„ = k Here ak

= ak(xl,■■■,x„), Si = S^Xi^^Xn), i = 1,■■■,n■

For example, a1 = x1 + ■■■ + x„; a2 =x1x2+■■■+xn-1xn; a3 = x1x2x3 + ■■■ + xn-2xn-1xn; a4 = x1x2x3x4 + ■■■ + xn-3xn-2xn-1xn.

4. Computation of Chebyshev polynomials of the second kind

Let's solve the problem of finding Chebyshev polynomials of the second kind and express it in terms of symmetric polynomials.

Example 2.

Solution. Since an even function t2 in the space Z([-l;l]) is approximated by an even

polynomial, then xi = 0. Function "1 is convex and differentiable. A

necessary and sufficient condition for the minimum of such a function is:

1

dfix) dx n

= 0

J sign(i2 + x0)dt =

0

-1

(1)

By Lemma 2, the polynomial ^21 = t2, + Xq changes sign twice on the interval (-1:1). Let's denote the sign change points .

Then, by virtue of relation (1), we get:

-ii 6 1

dt --1

dt + dt = 0 ^^

-6 6

Hence

Answer. 4.

Example 3.

Solution. Since the odd function t3 in the space Z([-l;l]) is approximated by an odd polynomial, then x; = x, l

/(Xl) = f \t3 + Xlt\dt = 0. Function -l is convex and differentiable.

A necessary and sufficient condition for the minimum of such a function is:

(2)

By Lemma 2, the polynomial^! = + -'-l'on the interval (-1:1) changes sign tliree times: at the points and zero at the point. Then, by virtue of relation (2), we get:

Answer. 2

Example 4

Solution. Since an even function f in the space Z([-l;l]) is approximated by an even

polynomial, then = x\ = 0. Function -1 is convex and differ-

entiable. A necessary and sufficient condition for the minimum of such a function is:

Of (:

x

f'(x) = 0

dx0 df(x) dxo.

0, 0,

f sign(i4 + x2t2 + x0)dt = 0, -l

l

f t2 sign(i4 + x2t2 +x0)dt = 0

(3)

By Lemma 2, the polynomial^ li — t1 + x2t2 + Xo on the interval (-1:1) changes sign four times: at the points ±ii,±&.

41

V I 0 © (b i ►

1 (i) (p -i V6 î A & J \

Figure 3:

Then, using the fact that the subplay expression is an even function, then from (3) we get:

2. We use

the well-known Waring formulas and findM £2 andfr+ei From'S's — <71 3<Ji<T2 wg have tliat:

1 / l\3 o / 1\ 3 3 . , . , 1 1

By Waring's formula^ we have

T41 = t4 - S2t2 + {S*^S2)2 = t4- (a2 - 2a2)t2 + a2 = t4 - ^t2 + ^

Thus, we get

1 16.

Example 5.

Solution. Since the odd function t5in the space !([-!;!]) is approximated by an odd

f(x) 0

df(x)

dxi df(x) dxz

polynomial, then xi = x: = xn = 0 . Function and differentiable. A necessary and sufficient condition for (4)

= 0, f isign(i5 + x3t3 + x1t)dt = 0,

= 0, f t3 sign(fF' + x3t3 + Xit)dt = 0

is convex he minimum of such a function is:

By Lemma 2, the polynomial^5l = t + + X\t on the interval (-1:1) changes sign five times: at the points ±£,±£2 and at zero.

Then, by virtue of relation (4), we get:

-£2

-£1

£1

Figure 4: 1

- tdt + tdt - tdt + tdt - inl££2itdt +tdt = 0 ^^

-1

"6

-£1

-(s -1)+i? - s+ÎÎ+8 - fe2 - m +1-Î22

We

Hence

3_ 16

T51 - i5 - 52i3 + i^i-^Ïi = ¿5 _ (a2 _ ¿3 + ^ = ¿5 _ t3 + __t

Example 6.

0

0

Solution. Let n = 2k. Since by Theorem 2.5 an even function t2kin the space Z([-1;1]) is approximated by an even polynomial, then x = X3 = ... = Xn-3 = xn-i = 0. 1

Function/(xo,X2,...,Xn-2) : -1

|tn + Xn-2tn 2 + ... + xo1 Idt is convex and differentiable.

A necessary and sufficient condition for the minimum of such a function is:

C,j V i/ Vi, f.y 1

-i \ ' /

Figure 5:

By Lemma 2, thepolynomialonthe interval (-1:1) changes sign;? times. Let us denote the sign change points.

Y

= 0;

fr

-1 -Çk -<ä—i -Çk Çk-1 1

t2dt -

-1 -Çk

t2dt + ... -

-&-1

Il \-Çktn-2dt - $k-1tn-2dt + ... - jÇk tn-2dt + /1 tn-2dt = 0.

1-1 -Çk -Çk-1

( -Çk + 1 - Çk-1 - Çk + ... - Çk- Çk-1 + 1 - Çk = 0;

= 0;

Çk

t2dt +

Çk

t2dt = 0;

1

-iT1 +1 - - er1 + - ■ - - er1 - Ck:i +1 - er = o.

Cn— 1 I ¿n—\ I I fn-1 I ftl— J c

Çfc + +Çi — ^n-i — — 2'

= (t - m+6)... (t - m+&) = p - ¿42f-2 + EEe2^-4+

¿=1 i=l j=l

i (-1 y-'EE ■ ■ ■ E^ • • • e.i2 + (- DaEE ■ ■ ■ EE^J • • • -

(=1 j = l m=l i=l j—1 m=lp=l

fc-1

= tn + E^1)^"2'^2 - + +

/=1

3 _

where

/=1,2.....A -land

And the sum is taken over all sets of non-negative integers fi\...../;„ satisfying the equality fi\ + 202 +... + njJ>„

= k.

Let n = 2k + 1. Since by Theorem 25 the odd function t2k+1 in the space I([-1;1]) is approximated by an odd polynomial, then xo = X2 = ■■■ = x„-3 = x„-1 = 0. Function 1

f(Xl,X3,...,Xn-2) = -1

necessary and sufficient condition for the minimum of such a function is:

' df(x)

\tn + xn-2tn 2 + ... + X3t3 + xit ' |dt is convex and differentiable. A

dxi df(x)

dm

, dxn-2

0,

= 0,

= 0,

r i

f t sign (P + xn-otn~2 + ... + x0) dt = 0,

l

/ i3 sign (tn + xn-2tn~2 + ... + xo)dt = 0, -1

f tn~2 sign (tn + xn-2tn~2 + • • - + Xo) dt = 0

By Lemma 2, the polynomial^ni = tn + %n-2tn " + ...+ x$t + it'if0n the interval (-1:1) changes sign n times. Denote the sign change points ±£b—,±£(„-1y2 and £ = 0. Then, by virtue of the resulting system, we have:

f

-1

- tdt + tdt - ... --1 -&-1 -$ $-1 1

- t3dt +t3dt - ... --1 -$

tdt = 0;

$

t3dt +

$

-£kT1

t3dt = 0;

$.

j-$

$k-1 j$

tn-2dt + tn-2dt - ... -

tn-2dt +

tn-2dt = 0.

-$

-$-1

$

-$2 + 1 + $2-1 - $2 + ... - $2 + $2-1 + 1 - $2 = 0;

= 0;

-$77-1 - 1 - $77--11 - $77-1 + ... - $77-1 - $77--11 — 1 — $77-1 = 0.

ift _ ifc-i + ■ • • + ~~ = 2

1

¿■n-l _ ¿71 1 J_ , ¿"Ti-1 _ CXI-! _ _

Sfc Sfc-1 T • ■ ■ -t" ?2 si — 2 ■

fe k

Tnl = t(t m+eo = r - Ee>2f7-2+EE +

£=i ¿=i j=i

• <-i;ft- ;EE■ ■ ■ E^• • • e*2 + (- DaEE■ ■ ■ EE-n • • • -

i=l j—1 m=l i=l y = l m=lp=l

fc-l

= tn + E(-l)^n"2iK - 2(7;_i(T;+I + <T;+2) + (-l)fcaft, /=1

d - + ■ ■ ■ + e24 - = J

3ft/ Jft,-± 31 i)

k k

erf - 2<jl_1al+1 + 2^+2 = E • • • E ^ • • •

, where / = 1,2,...,k - 1

n=1 l

n=1

<7*: = E

...../3:!.../3n!

= k.

and the sum is taken over all sets of non-negative integers ft\,...,p„, satisfying the equality fii + 2^2 + ... + For example,

in = t;

Tu = t4

T5l = t5

(,aI - 2a2)t2 + a2 = t4 - S2t2 + ( [<j\ - 2a2)t3 + a2t = t5- S2t3 +

' Ç-2 C N J2

~2 2 J

( S\ S2

= ¿4 _ 3 2 1 4 16.

t = p - tö + —t 16 .

1

T61 = t6 - (a2 - 2a2)t4 + (a22 - 2a^)t2 - a23 = te - S2t4 + ^ - SA)t2 - a2 =

= t»~ -14 + It2 - -

References:

4. E.M. Galeev Approximation theory. Baku: Publishing house of the Baku branch of Lomonosov

1. E.M. Galeev Optimization: Theory, examples, Moscow State University, 2016.

problems. Moscow: Nauka, 2010.

2. E.M. Galeev Optimization methods. Baku, Petersburg, 1903.

5. Oblomievsky D. D. Symmetric functions, St.

2016.

6. N.I. Akhiezer Lectures on approximation the-

3. E.M. Galeev Approximation of classes of pe- ory, second edition. Moscow: Nauka, 1965.

riodic functions (Part 1).

7. Vinberg E. B. Algebra of Polynomials, Pros-veshchenie, Moscow, 1980.

MAXIMISATION OF CONDITIONAL MUTUAL INFORMATION

Suleykin A.,

A teaching assistant at the Lomonosov Moscow State University Faculty of Mechanics and Mathematics Mammahajiyev R.

A graduate student at the Lomonosov Moscow State University

Faculty of Space Research https://doi.org/10.5281/zenodo.7258567

Abstract

In this paper we consider mutual information for a pair of random variables and find a third variable (condition) that maximise conditional mutual information of three of them.

Keywords: Mutual information, conditional mutual information, entropy, information inequality, Lagrange theorem, conditional extremum.

1 Solution

Let A, B and X take two values for convenience. We will solve the problem using methods from the optimal control course. In entropy, we take the natural logarithm for ease of differentiation. Consider I(A;B\X) as a function

of Pvk-

Here the values A have the index i, the values B have the index j, the values X have the index k. Now let's express everything in terms of pjk: