Định Lý Gauss và ứng dụng

DOI 10.1007/s10888-009-9120-9Gauss’ theorem concerning the center of gravity and its application Miodrag Tomi´c Received: 20 July 2009 / Accepted: 20 July 2009 © Springer Science + Busin

DOI 10.1007/s10888-009-9120-9

Gauss’ theorem concerning the center

of gravity and its application

Miodrag Tomi´c

Received: 20 July 2009 / Accepted: 20 July 2009

© Springer Science + Business Media, LLC 2009

Keywords Gauss’ theorem · Center of gravity · Complex plane

Section 1

If a νare arbitrary points in the complex variable plane andα νare positive numbers (weights), then the complex number, represented by the expression:

ξ n=

n



ν=1 α ν a ν n



ν=1 α ν

, n = 2, 3, ,

which is the center of gravity of a ν, as noted by Gauss,1is situated either inside or on

the boundary of the smallest convex polygon drawn around the points a ν.

The exact geometric interpretation of this theorem is given later by H A Schwarz, and particularly by L Fejér [1]

1 Gauss [2] See also Lucas [7] 2 e S.

Translated from Serbian by Ivica Urban from volume 1, pages 31–40, of Vesnik Društva

matematiˇcara i fiziˇcara Narodne Republike Srbije, with the kind permission of the editor

of Matematiˇcki vesnik and of the Society of Mathematicians of Serbia.

Section 2

Let there be n points a νin the complex plane (Fig.1), and n positive numbers α ν Then, from:

ξ2= α1a1+ α2 a2

α1+ α2 = a1+

α2

α1+ α2 (a2− a1 ) = a2− α1

α1+ α2 (a2− a1 ) ,

it follows thatξ2is situated on the line a1 a2between points a1 and a2, for all possible

choices of numbers α1> 0 and α2> 0, i.e that ξ2 lies inside the smallest convex

polygon drawn around the points a ν and a2.

From:

ξ n= α1a1+ α2a2+ · · · + α n a n

α1+ α2 + · · · + α n = (α1+ α2+ · · · + α n−1) ξ n−1+ α n a n

(α1+ α2 + · · · + α n−1) + α n

and from the preceding remark it follows that the point ξ n is situated on the line

ξ n−1a n−1, i.e inside the smallest convex polygon around the a ν, because the point

ξ n−1is also situated inside or on the boundary of that polygon

Section 3

From the latter, a well-known geometric construction of the center of gravityξ nis obtained

To obtain:

ξ2=α1a1+ α2 a2

α1+ α2

we need to draw through the points a1 and a2, in any direction (different from a1 a2),

two parallel line segments in the mutually opposite directions towards a1 a2, one

Fig 1

beginning at a1of lengthα2, the other beginning at a2of lengthα1 The intersection

of the line going through these two new points with a1 a2givesξ2 As:

ξ n= A n−1ξ n−1+ αn a n

A n−1+ α n where An−1= α1 + α2 + · · · + α n−1, the preceding construction leads, after n steps,

to the center of gravityξ n.

Section 4

Among many applications of this theorem, we mention Jensen’s well known theorem about convex functions

Jensen2has shown in 1905 that for every convex functionϕ(t) on an interval (a, b)

the following inequality is satisfied:

ϕ

⎛

⎜

⎝

n



ν=1 α ν x ν

n



ν=1 α ν

⎞

⎟

⎠ ≤

n



ν=1 α ν ϕ (x ν )

n



ν=1 α ν

(A)

where x ν are numbers in the interval (a, b ) and the α νare any positive numbers Jensen’s proof of this inequality is completely analogous to Cauchy’s proof3that the geometric mean is smaller than the arithmetic mean Cauchy’s first inequality:

√

ab ≤12(a + b)

i.e

n(a) + n(b)

2 ≤ n

a + b

2

corresponds to Jensen’s inequality:

ϕ

x + y

2 ≤ϕ (x) + ϕ (y)2 that serves for defining functions which are convex from above In this way Jensen introduces the most general class of functions that satisfy inequalityA

Section 5

As Jensen noted, this inequality generalizes some well-known and in analysis often used inequalities

For example, for:

ϕ (x) = x p , p > 1

we have:

n

ν=1

α ν x ν

p

≤

n

ν=1

α ν

p−1 n

ν=1

α ν x ν p

which is Buniakowsky’s inequality, usually called the Cauchy–Schwarz inequality ([4], p 16)

For p = 2 and α2

ν x ν = βνα νwe obtain Cauchy’s inequality:

n

ν=1

α ν β ν

2

≤

n

ν=1

α2

ν·

n

ν=1

β2

ν

Replacing x νin the inequality preceding the last one with

β ν

α ν1/p

and raising both

sides to the power 1/ p, one obtains:

n

ν=1

α1 −1/p

ν ·β1/p

n

ν=1

α ν

1 −1/p

·

n

ν=1

β ν

1/p

which is Hölder’s inequality (and so on)

Section 6

Jensen’s inequalityAfollows directly from Gauss’s theorem about the center of

grav-ity, reflecting the obvious fact that every polygon inscribed into a convex continuous

curve is convex.

Let y = f (x) be any convex line and let a1, a2, , anbe points lying on that line,

with real parts (abscissae) x1, x2, , xn(Fig.2), and let a be the point on the curve

whose real part (abscissa) is:

x∗=

n



ν=1 α ν x ν

n



ν=1 α ν

Fig 2

Since ak = xk + if (xk ), it follows that:

ξ n=

n



ν=1 α ν x ν

n



ν=1 α ν + i

n



ν=1 α ν f (x ν )

n



ν=1 α ν

and:

x∗= Re {a} = Re {ξn} = α1x1+ α2 x2+ · · · + αn x n

α1+ α2 + · · · + αn

Gauss’s theorem obviously implies:

Im {a} ≤ Im {ξn}

and this yields Jensen’s inequality:

f

α

1x1+ α2 x2+ · · · + αn x n

α1+ α2 + · · · + αn



≤

n



ν=1 α ν f (x ν )

n



ν=1 α ν

.

Section 7

M Petrovich [9] has provided an upper limit for the expression:

n

ν=1

α ν f (x ν )

α ν

From the above mentioned geometric interpretation we immediately obtain another upper bound that is achieved when allα ν = 0, except α1andα n It is obvious

(from Fig.2) that this upper bound is the imaginary part (ordinate) of the point b ,

i.e that:

n



ν=1 α ν f (x ν )

n



ν=1 α ν

≤ f (x1 ) + f (x1) − f (x n )

x1− xn

n



ν=2 α ν x ν

n



ν=1 α ν

Expanding the theorem of Petrovich, J Karamata has in the same issue of Publications offered one of the most general formulations of this theorem, that as

a special case contains the following theorem:

Theorem 1 Let x ν and X ν , ν = 1, 2, , n, be two sequences of numbers in an interval (a, b) sorted in increasing order, i.e.

In order that the inequality:

n

ν=1

f (x ν ) ≤

n

ν=1

is satisfied for every convex function f (t), it is necessary and sufficient that:

k

ν=1

X ν≤

k

ν=1

x ν , for every k = 1, 2, , n − 1 (2) and

n

ν=1

X ν=

n

ν=1

Several proofs4 of this theorem are analytical in nature; moreover, Karamata’s general theorem is stated in the form of a Stieltjes integral However, this proof can

be derived also in an elementary way on the basis of the above mentioned geometric observation With this approach, the theorem can in fact be stated in a more precise way by replacing the equality sign in Eq.3 with≤ under the assumption that the

function f (t) is convex and monotone.

Theorem 2 Let x ν and X ν , ν = 1, 2, , n, be the two sequences of numbers in an interval (a, b) sorted in increasing order, i.e.

X ν ≤ Xν+1 and x ν ≤ xν+1

and let f (t) be a convex function on (a, b); then the inequality:

n

ν=1

f (x ν ) ≤

n

ν=1

holds:

(a) if the function f (t) is decreasing and if :

k

ν=1

X ν≤

k

ν=1

x ν , for every k = 1, 2, , n (4)

or :

(b) if the function f (t) is increasing and if :

n

ν=k

X ν≥

n

ν=k

x ν , for every k = 1, 2, , n. (5)

4 Karamata [6], see also Hardy et al [3], L Fuchs, A new proof of an inequality of Hardy, Littlewood, Polya, and J Aczél, A generalization of a notion convex functions, and loc cit 1, p 83.

Although this theorem assumes more on the function f (t), i.e that it must

be monotonic, it still in a certain sense contains Theorem 1, as we will show in Section9, applying it separately to the decreasing and increasing parts of a convex function

Section 8

The proof of part (b) of Theorem 2 is obtained directly from (a) by replacing an

increasing function f (t) by the function f (b − t) and the sequences x ν and X νby the

sequences b − x n +1−ν and b − X n +1−ν, respectively

Thus, it is sufficient to prove part (a) of Theorem 2

By Eq.4, for k= 1, we have:

X1≤ x1,

and from monotonicity of the function f (t) it follows that:

f (X1) ≥ f (x1)

Case 1 If x2≤ X2 and if we denote by (1, 1) and (ξ1,ϕ1) the coordinates of the

center of gravity T1 of the line segment M1 M2, and the center of gravity t1of the line

Fig 3

segment m1 m2respectively (see Fig.3), then from condition4for k= 2, that can be written as:

1= X1+ X2

2 ≤ x1+ x2

2 = ξ1

it follows that:

1= f (X1) + f (X2)

2 ≥ f (x1) + f (x2)

2 = ϕ1,

because in this case the line segment M1 M2is above m1 m2, and the abscissa1of

the center of gravity T1is to the left of the abscissaξ1of the center of gravity t1, and also the center of gravity T1 is above t1.

Case 2 If x2≥ X2, then, because of monotonicity of the function f (t), we have that:

f (X2) ≥ f (x2) ,

which, summing with:

f (X1) ≥ f (x1)

gives:

f (X1) + f (X2) ≥ f (x1) + f (x2)

Thus, in both cases this inequality is fulfilled

Let us now apply the same argument to the points T1, M3, t1 and m3.

Case 1 If x3≤ X3 then the line segment t1 m3lies under the line segment T1 M3, and according to condition4with k= 3 we have:

2= X1+ X2 + X3

3 ≤ x1+ x2 + x3

3 = ξ2

Thus, we have:

2= f (X1) + f (X2) + f (X3)

3 ≥ f (x1) + f (x2) + f (x3)

3 = ϕ2 Case 2 If x3≥ X3, then because of monotonicity of the function f (t) we have

f (X3) ≥ f(x3), which, adding to:

f (X1) + f (X2) ≥ f (x1) + f (x2)

gives:

f (X1) + f (X2) + f (X3) ≥ f (x1) + f (x2) + f (x3)

It is clear that this procedure can be extended to any number of points; hence, Theorem 2 is proved

Section 9

Let us show now that Theorem 1 is a direct consequence of Theorem 2 Assume

therefore that the function f (t) is convex in an interval (a, b), monotonically

de-creasing for a < t < ¯x, and monotonically increasing for ¯x < t < b It is clear that,

because of convexity of the function f (t), there can exist only one point ¯x at which

the curve changes the sense of monotonicity

If to the left of the point¯x there are p points Xν and q points xν, and if p = q, then the sequence X ν and the sequence x ν can always be extended with r = |p − q| points

¯x, such that the newly obtained sequences:

X ν and xν , ν = 1, 2, , n + r

have the same number of points in the interval (a, x ), and therefore, also in the

interval (x , b)

These two, new sequences, are:

Case 1 if p < q, r = q − p:

X ν =

⎧

⎨

⎩

X ν forν = 1, 2, , p,

¯x forν = p + 1, p + 2, , p + r = q,

X ν−r forν = q + 1, q + 2, , n + r,

and:

xν=

⎧

⎨

⎩

x ν forν = 1, 2, , q,

¯x forν = q + 1, q + 2, , q + r,

x ν−r forν = q + r + 1, q + r + 2, , n + r;

Case 2 if p < q, r = p − q:

X ν =

⎧

⎨

⎩

X ν forν = 1, 2, , p,

¯x forν = p + 1, p + 2, , p = r,

X ν−r forν = p + r + 1, p + r + 2, , n + r,

and:

xν=

⎧

⎨

⎩

x ν forν = 1, 2, , q,

¯x forν = q + 1, q + 2, , q + r = p,

x ν−r forν = p + 1, p + 2, , n + r.

Since by the assumption of Theorem 1, the sequences x ν and X νsatisfy conditions

2and3, it is easy to see:

Case 1 that:

n +r

ν=1

X ν =

n +r

ν=1

xν ,

Case 2 that the points of the sequences X ν and xν lying in the interval (a, ¯x) satisfy

condition4, and those points which belong to the interval (¯x, b) satisfy condition4

In other words, if we denote by s the maximum of p and q, then:

k

ν=1

X ν ≤

k

ν=1

xν , for k = 1, 2, , s

and:

n +r

ν=k

X ν≥

n +r

ν=k

xν , for k = s + 1, s + 2, , n + r.

On the other hand, the function f (t) is decreasing on the interval (a, ¯x), and

increasing on the interval (¯x, b); therefore according to Theorem 1, parts (a) and

(b), we have:

s

ν=1

f

xν

≤

s

ν=1

f

X ν and:

n +r

ν=s+1

f

xν

≤

n +r

ν=s+1

f

X ν Summing these two inequalities we obtain:

n +r

ν=1

f

xν

≤

n +r

ν=1

f

X ν

,

and this inequality reduces to inequality1, because on its left and right sides r equal members f (¯x) cancel out, which proves Theorem 1.

References

1 Fejér, L.: Über die Wurzel von kleinsten absoluten Betrage einer algebraischen Gleichung Math.

Annalen, Bd 65 (1908)

2 Gauss, C.F.: Werke Bd 3 Ed (1866)

3 Hardy, G.H., Littlewood, J.F., Pólya, G.: Some simple inequalities satisfied by convex functions.

Mess Math LVIII, 149–152 (1929)

4 Hardy, G.H., Littlewood, J.F., Pólya, G.: Inequalities Cambridge Univ Press (1934)

5 Jensen, J.L.W.V.: Sur les fonctions convexes et les inégalités ente les valeurs moyennes Acta

Math T 30, 175 (1905)

6 Karamata, J.: Sur une inégalité relative aux fonctions convexes Publ Math T I (1932)

7 Lucas, C.H.F.: C R de l’Acad De Paris (1866)

8 Pólya, G., Szegö, G.: Aufgaben und Lehrsätze aus der Analysis I, pp 51–53 Berlin (1925)

9 Petrovich, M.: Sur une fonctionnelle Publ Math Beograd, T L (1932)

Section 6

Jensen’s inequalityAfollows directly from Gauss? ??s theorem about the center of

grav-ity, reflecting the obvious fact that every polygon... · · + αn x n

α1+ α2 + · · · + αn

Gauss? ??s theorem obviously implies:

Im {a} ≤ Im {ξn}

and this yields Jensen’s... algebraischen Gleichung Math.

Annalen, Bd 65 (1908)

2 Gauss, C.F.: Werke Bd Ed (1866)

3 Hardy, G.H., Littlewood, J.F., Pólya, G.: