On a question by D. I. Mendeleev A. Markov read in the session of the Physicomathematical Section on 24 October 1889

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1 On a question by D. I. Mendeleev A. Markov read in the session of the Physicomathematical Section on 4 October 88 In the present work, we will consider the set of all polynomials [lit. entire functions] f(z) = p 0 z n + p z n + p z n + + p n z + p n whose degree does not exceed a given whole number n, whose absolute values do not exceed another given value L for all values of the variable z lying between the two given values a and b > a. Thus, L < f(z) < L for a < z < b. The question is which bound the absolute value of the derivative f (x) = np 0 x n + (n )p x n + + p n x + p n of f(x) with respect to x does not exceed. This question was posed by D.I.Mendeleev, for n =, in his paper The analysis of water solutions by specific weight ( 86). The answer depends on how much the number x is pinned down. We distinguish two cases: ) x is a given number, ) x is an arbitrary number between a and b. Correspondingly, we consider two problems. Problem No. To find, for particular x, the biggest absolute value of f (x). Solution. We denote by y that function f(z) of the ones considered by us for which f (x) in absolute value is biggest. By the statement of the question, L y +L for all values z that lie between a and b. From all these values z, we single out those at which y equals ±L. Let us write them in a sequence Denoting by α, α,..., α i, α i+,..., α s. y(α i ) the value of y at z = α i, equal to ±L, we notice that the sequence of s ratios y(α ) y(α ), y(α 3) y(α ),..., y(α s ) y(α s ) must contain at least n numbers equal to. Indeed, in the contrary case it would not be difficult to find, among the polynomials of degree n, an infinite set of those whose ratios over y at z = α, α,...,α s

2 are negative numbers. If then, having multiplied one of these ϕ(z) by (z x) and by a sufficiently small positive number ǫ, we add the product to y, then we obtain the polynomial ǫ(z x) ϕ(z) Y = y + ǫ(z x) ϕ(z) of degree n in z and such that, for a < z < b, Y < L and, for z = x, dy = dy. Finally, if we multiply Y by the ratio of the number L over the biggest absolute value of Y on a < z < b, then the new function obtained this way will belong to the functions f(z) considered by us and, at z = x, its derivative is bigger in absolute value than dy. Hence, s is no smaller than n, and the sequence of ratios y(α ) y(α ), y(α 3) y(α ),..., y(α s ) y(α s ) () contains no less than n numbers equal to. If occurs n times in the sequence (), then it is known that y is reduced to ±L cosnarccos x a b b a = ±f 0 (z). In that case, dy = ±nl sin n arccos x a b = ±f (z a)(b z) b a 0(z). f 0 We study the condition under which the biggest absolute value of f (x) equals the absolute value of (x)ṡince we concern ourselves with absolute values, then among all functions f(z) we can consider only those for which f (x) has the same sign as f 0 (x). We take, for brevity, and a b cos iπ n + b + a Considering the value of f(z) and f 0 (z) at = ξ i, i = 0,,,..., n, f(z) f 0 (z) = ϕ(z). z = ξ 0, ξ, ξ,..., ξ n, we find Hence, the equation f 0 (ξ n) = +L and therefore ϕ(ξ n ) 0 f 0(ξ n ) = L and therefore ϕ(ξ n ) 0 f 0 (ξ n ) = +L and therefore ϕ(ξ n ) f 0 (ξ 0) = ( ) n L and therefore ( ) n ϕ(ξ 0 ) 0. ϕ(z) = 0

3 must have one root between ξ 0 and ξ, between ξ and ξ,..., between ξ n and ξ n. In other words, the function ϕ(z) must decompose into real factors of first degree in z, where and ϕ(z) = q(z η )(z η ) (z η n ), a = ξ 0 η ξ η ξ ξ n η n ξ n = b. Concerning the coefficient q, it must be negative. With this, we have since f 0(z) vanishes at and the leading term of the polynomial f 0 (z) is ( f (x) = f 0 (x) ) ϕ(x) x η x η x η n f 0(x) = n nl (b a) n (x ξ )(x ξ ) (x ξ n ) z = ξ, ξ,..., ξ n n nl (b a) n. We start with the case when x lies outside the bounds a and b. Then each of the expressions ϕ(x) ϕ(x) ϕ(x),,..., x η x η x η n has its sign opposite the sign of f 0 (x) and so f (x) < f 0(x). Thus, if x lies outside the bounds a and b, then the biggest value of f (x) equals f 0 (x). Now we assume that x lies between ξ i and ξ i. Then ϕ(x) x η i = q(x η )(x η ) (x η i )(x η i+ ) (x η n ) has its sign the opposite of the sign of f 0 (z). This leaves us to consider the sign of the sum x η i x η + x η i x η + + x η i x η i + x η i x η i + + x η i x η n = Σ, which, for brevity, we denote by the single letter Σ. We now denote by f(z) some arbitrary polynomial of nth degree in z that satisfies the conditions L < f(z) < +L for a < z < b and Then the numbers f (x) > 0. (x) f 0 η, η,..., η n 3

4 can take on arbitrary values, subject only to the inequalities ξ 0 η ξ η ξ ξ n η n ξ n and the coefficient q is sufficiently small in absolute value. Taking into account this remark, it is easy to see that the smallest (extreme) value of the sum Σ is equal to the smaller of the numbers and x ξ i x ξ 0 + x ξ i x ξ x ξ { i f 0 = (x ξ i ) (x) x ξ n f 0 (x) + } x a x ξ i + x ξ i x ξ { i f 0 = (x ξ i ) (x) x ξ x ξ x ξ n f 0 (x) + }. x b If the smallest value of Σ is positive, then also all values of Σ are positive and the sign of the expression ( ) ϕ(x) x η x η x η n is the opposite of the sign of f 0(x); in addition, surely, f (x) < f 0 (x). But if the smallest value of Σ is negative, then the undetermined numbers η, η,..., η n, can be so chosen that f (x) exceeds f 0 (x) From this we conclude that the biggest value of f (x) equals the biggest value of f 0 (x) if and only if x lies outside the bound a and b, or else a < x < b, f 0 (x) f 0 (x) + x a f 0 > 0, and (x) f 0 (x) + < 0. () x b Instead of the rational expressions f 0 (x) f 0 (x) + x a and f 0 (x) f 0 (x) + x b it is possible to consider (x a)f 0 (x) + f 0(x) and (x b)f 0 (x) + f 0(x), since, firstly, in obedience to the inequality (), the expressions (x a)f 0 (x) + f 0 (x) and (x b)f 0 (x) + f 0 (x) (3) have the same sign and, secondly, our inequalities () hold in case the signs of the expressions (3) are the same and a < x < b. Having considered in this fashion the case y = f 0 (z), we turn to the others. If y is not = f 0 (z), then, by the above, the sequence of ratios y(α ) y(α ), y(α 3) y(α ),..., y(α s ) y(α s ) 4

5 contains n numbers equaling. Also, s = n and, of the two differences α a, b α n at least one must be zero. We take one of the functions f(z) that satisfy our conditions. The equation f(z) y = 0 of nth or lower degree in z has a root where between α and α, between α and α 3,..., between α n and α n. In other words, the difference f(z) y must decompose into real factors of first degree in z: Also, we have f (x) = f(z) y = ψ(z) = (qz r)(z η )(z η ) (z η n ) α η η α n η n α n, r q α n or α. ( ) ( dy ) ψ(x), x η x η x η n x η n with η n = r q. It is not hard to see also that the sign of the difference qz r is the opposite of the sign of y(α n ) for all values of z that lie between α and α n. Under the conditions pointed out above, the numbers η, η,..., η n, can be given arbitrary values, subject only to the condition that q be sufficiently small. We assume, to begin with, that x is bigger than α n. Then, for η n > x, we get the inequalities 0 < x η + x η + + and the undeterminedness of the numbers 0 > x η n < x η n > η, η,..., η n x α + + can be made use of in such a way that the quantity ( ) ψ(x) x η x η x η n x α n will have an arbitrary sign. It follows that the case is impossible. x > α n 5

6 One shows similarly that the case x < α is impossible. So assume that x lies between α i and α i+. Then, the sign of ψ(x) x η i is the opposite of the sign of ( ) n i y(α n ) and so, in order for f (x) to be, in absolute value, smaller than must be the same as the sign of Now, the expression x η i x η + x η i x η + + x η i x η i + + x η i x η n ( ) n i y(α n ) ( ) n i y(α n ) ( ) dy. ( ) dy ( ) dy, the sign of the sum is a positive number, hence the sign of ( ) n i y(α n ) ( ) is the same as that of y(α i+ ) and of dy. On the other hand, it is not hard to see that the smallest value of the sum equals the smaller of the numbers x η i x η + x η i x η + + x η i x η i + x η i x η i + + x η i x η n + x η i x η n { (x α i ) x α x α x α i x α n { (x α i+ ) x α x α 3 and therefore cannot be greater nor less than zero. For that reason, we arrive at the following condition x α x α x α i+ + + x α n + Our considerations also show that, excluding the case that simultaneously n =, α = a, α n = b, x = a + b, x α n }, + } x α n x α x α n = 0. (4) the derivative f (x) takes its biggest absolute value only for two functions f(z) and these two only differ by a sign. But if n = and x = a + b, 6

7 then the biggest absolute value of f (x) equals L b a and there exists an infinite set of different functions f(z): namely, all functions of the form with We recall that of the two differences { } z a b L + q(z a)(z b) b a (b a) < q < (b a). α a, b α n at least one must be zero, and correspondingly distinguish three cases: ) α = a, α n < b; ) α > a, α n = b; 3) α = a, α n = b. If then we can adjoin to the numbers also some number α = a and α n < b, α, α,..., α n α n+ which is bigger than b and satisfies the condition y(α n+ ) = y(α n ), since, as z increases continuously from α n to +, the ratio y y(α n ) also changes continuously from to +. Therefore we have y = ±L cosn arccos z a α n+ = ±f (z). α n+ a The unknown α n+, in accordance with equation (4), must satisfy the equation i=,,...,n x a+αn+ αn+ a cos iπ n = 0, i.e., f (x) f (x) + x a = 0 and also the inequalities whence Therefore, for the case α n+ > b > a + α n+ b a sin π n cos π n + α n+ a > α n+ > b. α = a, α n < b really to occur, one of the values α n+ satisfying the equation cos π n, (x a)f (x) + f (x) = 0 (5) 7

8 must lie between b a sin π n cos π n and b. And only one, since in the contrary case the sought-for biggest value of f (x) would be taken on by several different functions f(z), but the preceding considerations show this to be impossible. Considering then the sum x a+αn+ αn+ a cos iπ n i=,,...,n as a function of α n+, we note this function increases continuously with α n+ with the exception of those values α n+ at which it is infinite. Therefore, equation (5) cannot have multiple roots. From this, it is not hard to conclude that the case α = a, α n < b occurs exactly when during the passage of α n+ from b to b asin π n cos π n the expression changes its sign. We also note that, for α n+ = b, the expression reduces to (x a)f (x) + f (x) (x a)f (x) + f (x) (x a)f 0 (x) + f 0 (x). In exactly the same way, with the introduction of the variable α 0 and taking L cosn arccos z α 0 b b α 0 = f (z), we see that the case α > a, α n = b occurs exactly when during the passage of α 0 from a bsin π n cos π n to a the expression changes its sign. Then where α 0 must satisfy the equation and the inequalities We now address the case (x b)f (x) + f (x) y = ±f (z), (x b)f (x) + f (x) = 0 α 0 < a < α 0 + b + b α 0 α = a, α n = b, which occurs exactly when neither of the preceding cases occurs. If α = a, α n = b, 8 cos (n )π. n

9 then the equation of (n )st degree in z has the n roots dy = 0 α, α 3,..., α n between a and b and one root outside these bounds. We denote this last root by the letter β and assume for definiteness that β > b. In this case, y, as z moves from b to β, grows but, as z grows even larger, first diminishes to zero and then grows without bound. Also, the equation y L = 0 of nth degree in z certainly has the n double roots and the two simple roots α, α 3,..., α n and also two roots which we will denote by the letters and These last two roots are bigger than β. It follows that a, b γ and δ. y L = p 0(z α ) (z α 3 ) (z α n ) (z a)(z b)(z γ)(z δ) dy = np 0(z α )(z α 3 ) (z α n )(z β), from which we derive the first order differential equation y L = (z a)(z b)(z γ)(z δ) n (z β) ( ) dy. (6) E. I. Zolotarev, in his work The application of elliptic functions to questions concerning functions that deviate least and most from zero, expressed the solution of that last equation in terms of elliptic functions. Without relying on E. I. Zolotarev s formulas, we show how it is possible to reduce our problem to three algebraic equations. For this, we obtain from the equation (6) by differentiation n (z β) 3 y = (z a)(z b)(z γ)(z δ)y + { (z a)(z b)(z γ)(z δ)(z β) z a + + z δ } y. z β (7) Taking now y = p 0 (z β) n + p (z β) n + + p n (z β) + p n and rewriting equation (7) in powers of z β and comparing coefficients, we arrive at a system of n + equations in the n + unknowns p, p,..., p n, p n, β, γ, δ. p 0 p 0 p 0 p 0

10 It is not hard to eliminate the unknowns p p 0, p p 0,..., p n p 0, p n p 0 which depend linearly on the remaining three β, γ, δ. By eliminating p p 0, p p 0,..., p n p 0, p n p 0 we arrive at two algebraic equations for the unknowns β, γ, δ. Now, condition (4) gives the third equation ( ) y y + x a + x b + x β = 0. (8) As concerns the coefficient p 0, it is determined from the condition y(a) = ±L. We arrive at the same results also in case β is less than a except that, for β < a, γ and δ must be less than β. For what is to follow it is important to notice that in each case the expression (z γ)(z δ) (z β) is greater than unity for all values of z lying between a and b. We now show that equation (6) may be transformed into two first-order linear differential equations with two unknown polynomials. Here, for definiteness, we ll take Let n be even. Then, denoting the products y(a) = L; a < b < β < γ < δ. (z α )(z α 4 ) (z α n ) and (z α 3 )(z α 5 ) (z α n ) respectively by we obtain U and V, y L =p 0 (z a)(z δ)v y + L =p 0 (z b)(z γ)u y =p 0 {(z a)(z δ)v + (z a δ)v }V =p 0 {(z b)(z γ)u + (z b γ)u}u =np 0 (z β)uv and thereby arrive at the desired two linear first-order differential equations (z a)(z δ)v + (z a δ)v = n(z β)u 0

11 In similar fashion, for odd n, denoting the products (z b)(z γ)u + (z b γ)u = n(z β)v. (z α )(z α 4 ) (z α n ) and (z α 3 )(z α 5 ) (z α n ) respectively by we obtain U and V, (z a)(z b)(z γ)v + {3z + (a + b + γ)z + ab + aγ + bγ}v = n(z β)u (z δ)u + U = n(z β)v. I n =. In this case, Hence, for f 0 (z) = f 0 (z) = Examples. L (b a) { 8(z a)(z b) + (b a) }, 8L (z a b), f (b a) 0 = 6L (b a) (x a)f 0 (x) + f 0(x) = 8L (4x 3a b) (b a) (x b)f 0 (x) + f 0 8L (x) = (4x 3b a). (b a) x > 3b + a 4 the greatest value of f (x) equals the absolute value of f 0 (x) = or x < 3a + b, 4 8L (x a b). (b a) Turning now also to the functions f (z) and f (z), we find f (z) = L (α 3 a) { 8(z a)(z α3 ) + (α 3 a) }, (x a)f (x) + f (x) = f (z) = 8L (α 3 a) (4x 3a α 3) L (b α 0 ) { 8(z α0 )(z b) + (b α 0 ) }, (x b)f (x) + f (x) = 8L (b α 0 ) (4x 3b α 0). α 3 = 4x 3a, α 0 = 4x 3b from which we conclude that the absolutely largest value of f (x) for for 3a+b 4 < x < a+b 8L equals (α 3 a) (x α 3 a) = L x a a+b < x < 3b+a 8L 4 equals (b α 0) (x α 0 b) = L b x.

12 As to the function y determined by the differential equation (6), for n = it plays no role in our question. II n = 3. Taking for simplicity of the results we find where Hence, for a = and b = +, f 0 (z) = L(4z 3 3z), f 0(z) = 3L(4z ), f 0 (z) = 4Lz (x a)f 0 (x) + f 0(x) = 3L(x + 8x ) = 36L(x ω )(x ω ) (x b)f 0 (x) + f 0(x) = 3L(x 8x ) = 36L(x ω )(x ω ), ω = 7 6 < ω = 7 6 < ω = x < ω, ω < x < ω or x > ω, the absolutely largest value of f (x) equals the absolute value of f 0 (x) = 3L(4x ). Turning now to the functions f (z) and f (z), we find { ( ) } 3 z α4 f (z) = L 4 3 z + α 4 α 4 + α 4 + < ω = (x a)f (x) + f (x) = 6L (α 4 + ) 3 [ 6(x + α4 )(x + ) + 4(x + α 4 ) (α 4 + ) ] (x b)f (x) + f (x) = The expression for α 4 = turns into 6L ( α 0 ) 3 { 6(x α0 )(x ) + 4(x α 0 ) ( α 0 ) } 6(x + α 4 )(x + ) + 4(x + α 4 ) (α 4 + ) 48x 3 + 3x 4 = 48(x ω )(x ω ), and for α 4 = +sin π 6 cos π 6 = 5 3, it turns into where 3(x 3 )(x + ) + 6(x 3 ) 64 = 48x x 6 = ǫ = 8 = 48(x ǫ )(x ǫ ), and ǫ = + 8. From this we conclude that the biggest absolute value of f (x) equals the absolute value of f (x) in those cases when ω < x < ǫ or ω < x < ǫ. Here, the number α 4 must be determined from the equation 6(x + α 4 )(x + ) + 4(x + α 4 ) (α 4 + ) = 0.

13 In order to give the expression f (x) a possibly simple form, we take α 4 = + ξ and x + = t. Then f (x) = tf (x) { f (x) = (t ξ) ξ 3 L = 6 ( t ξ ) 3 ( ) } t L ξ t 48t 3tξ + 3ξ = 0, t ξ = 4 ± 7 ( ) 3 t ξ ( ) t = 0 ± 7 7 ξ 44.6 f (x) = 0 ± 7 7 L x +. Of the two signs ± for 7 one must take the one for which lies between and 5 3. And the inequalities are equivalent to α 4 = + 4 ± (x + ) 7 ± > α 4 > > x > ± 7. 6 Comparing these last inequalities with the ones found earlier we see that the absolutely largest value of f (x) Similarly, taking ω < x < ǫ or ω < x < ǫ, 7 for ω < x < ǫ equals and for ω < x < ǫ equals we find that the absolutely largest value of f (x) If now x lies = ǫ and + 8 = ǫ for ǫ < x < ω equals but for ǫ < x < ω equals L x+ L x+. L x L x. between ǫ and ǫ or between ǫ and ǫ then the biggest absolute value of f (x) coincides with that of the function y which is determined by the equations (6) and (8) for n = 3, a =, b = +. 3

14 In our example, the differential equation (6) can be changed into the two equations y L = p 0 (z )(z γ), y L = p 0 (z α ) (z δ), from which we deduce that γ = δ + α, = α + α δ, L = p 0 (α δ + γ) δ = + α α, γ = 3α α, p 0 = 4α L ( α ). Now, equation (8) becomes Hence x α + x + + x = 0. x α = x x, α = 3x x α = + x 3x x = ( x)( + 3x), γ = ( α )( + 3α ) x α + α = 3x + x x = ( + x)(3x ), + γ = ( + α )(3α ) x 3α + 3α = x + x 3 x = (x ǫ )(x ǫ ) x 3α = x x 3 = (x ǫ )(x ǫ ) x x } ( ) dy = 4α { L ( α 3x 3α x ) α = Now it is not hard to see that, for 6x 3 L ( x )( x ). ǫ < x < ǫ or ǫ < x < ǫ, = 4(x α )(3xα + ) ( α ) L the function y constructed by us satisfies all the aforementioned conditions and the absolutely largest value of f (x) equals the absolute value of 6x 3 L ( x )( x ). Problem No. To find the biggest absolute value of f (x) for all x lying between a and b. 4

15 Solution. In solving the previous problem, we found all those functions f(z) for which f (x) takes on its largest absolute value. One of our results is the fact that, for the absolutely largest value of f (x) equals (x b)f 0 (x) + f 0(x) (x a)f 0 (x) + f 0 > 0, (x) f 0(x) = nl sinn arccos x a b b a (x a)(b x). Assuming now we find that and If 0 < ϕ < π n x = a + b f 0 (x) = L cosnϕ, f 0 (x) = + b a f 0(x) = cosϕ, nl sinnϕ (b a)sin ϕ, 4nL{sinnϕcosϕ n cosnϕsin ϕ} (b a) sin, ϕ (x b)f 0 (x) + f 0 (x) cosϕ sin nϕ + n cosnϕsin ϕ (x a)f 0 (x) + f 0 = (x) + cosϕ sin nϕ n cosnϕsin ϕ. or π > ϕ > π π n, then On the other hand, from the formula sin nϕ > n cosnϕsinϕ (x b)f 0 (x) + f 0 (x) (x a)f 0 (x) + f 0 > 0. (x) f 0 nl sinnϕ (x) = (b a)sin ϕ, it is evident that, for a x b, the largest absolute value of f 0 (x) equals and occurs when x = a and x = b. Therefore, for all values of x lying n L b a between a and a + b b a the absolutely largest value of f (x) equals We assume now that x lies between cos π n a + b or between + b a cos π n n L b a. and b, a + b b a cos π n and a + b 5 + b a cos π n.

16 In this case, ( ) ( b a b + a (x a)(b x) = x) > ( b a ) sin π n > ( b a ) n. The derivative f (x) takes on its absolutely largest value at one of the aforementioned functions f 0 (x), f (x), f (x) or for the function y that satisfies the differential equation (6). But, by the above observed f 0(x) < n L b a, and, in the same way, we see that and Also, from equation (6) and for the inequality results, and therefore a + b b a ( ) dy < f (x) < n L α n+ a < n L b a f (x) < n L b α 0 < n L b a. cos π n < x < a + b + b a cos π n, n (x a)(b x) L < 4n4 (b a) L ( ) dy < n L b a. All these results show that the sought-for biggest absolute value of f (x) equals n L b a. 6