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Topic: Analogy between Roots of Polynomial Equqtions and Centroid Problem
Replies: 1   Last Post: Mar 21, 2013 10:34 AM

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Doctor Nisith Bairagi

Posts: 23
From: Uttarpara, West Bengal, India
Registered: 3/2/13
Analogy between Roots of Polynomial Equqtions and Centroid Problem
Posted: Mar 20, 2013 2:34 PM
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From: Doctor Nisith Bairagi
[Retired Professor of Structural Engineering,
Indian Institute of Technology-Bombay, India]
Uttarpara, West Bengal, India.
Subject : Analogy between Roots of Polynomial Equations and Centroid Problem
sci.math.independent
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The problem of locating centroid of regular geometrical lamina of uniform thickness can be easily done by utilising the analogy between the average of the roots of a polynomial equation of degree n, and the location of the point of action of the resultant of parallel forces.
This analogy is proposed as follows:

(A) For parallel forces of equal magnitudes:
Consider the expression for moments (with, a = force, and x = distance):
a<1>.x<1> + a<2>.x<2> +?+a<n>.x<n> = R .x<g> ?????????Eq(1)
If: a<1> = a<2> =?= a<n> = 1, and
then: R (resultant) = a<1> + a<2> +?+a<n> = 1 x n = n,
Eq(1) yields:
x<g> = [x<1> + x<2> +?+ x<n>] / n = = [sum of x<i>, (i = 1 to n)] / n) ?.Eq(2)
This expression speaks for the location x<g>, of the resultant R of the n number of a<i>, (i = 1 to n), forces, each of unit magnitude.

Similar property of the average of the roots of any polynomial equation of any degree (say, n) exists, as is imminent here:

Consider the basic polynomial equation of degree n as:
a<n> x^n + a<n-1> x^(n-1) + a<n-2> x^(n-2) +?+a<1> x + a<0> = 0 ??...Eq(3)
[It has number of roots = n, and it is in 0-th stage of differentiation (k = 0), i.e., no differentiation is performed on it.]

In terms of the roots, x<1>, x<2>, ?, x<n>, the n-th degree polynomial equation, alternatively, takes the form as:
(x ? a<1>).(x ? a<2>).(x ? a<3>). ??(x ? a<n>) = 0 ?????????.Eq(4)

The 1-st, 2-nd, 3-rd,?k-th successive stages of differentiation of the polynomial equation with respect to x, produce polynomial equations of degree (n-1), (n-2), (n-3),?(n-k)-th, respectively. Such differentiation is possible up to k = (n-1), when a linear equation in x, results, which when solved, provides a single value of the root.

Observe that, the average of the roots x<av>, of each of the polynomials equations, of degree m (say), considered at any stage of differentiation, k = 0,1, 2, 3, ?, (n -1) results in:
x<av> = [sum of x<i>, (i = 1 to m)] / m ???????????????Eq(5)
and its value, x<av> = ? (a<m-1> / a<m>) / m
[Take m = n for the basic polynomial equation for which, k = 0.]
The form of Eq(5) and Eq(2) are identical.

This feature when used analogously, to locate of the resultant R, of parallel force systems simply by considering that, if through each of the root points, x<i>, corresponding to any given polynomial equation, a force of unit magnitude parallel to one another acts, the resultant R (of magnitude = degree of the polynomial, so also = total number of roots), of which then acts through the average of the root points x<av>, which actually is the centroid of these parallel force system, x<g>.

Based on the analogy, two RULES are proposed:
RULE - 1 :
The value of x derived from the (n-1) ? th derivative of a polynomial equation is equal to the average of the n- number of roots of the given polynomial equation. This average of the roots is equal to those of each of the polynomial equation derived at any k-th successive stage of differentiation, [where, k is less than or equal to (n-1)], including the basic given polynomial equation (for which, k = 0)].

RULE - 2 :
If n number of like parallel forces, each of unit magnitude, acting normal to a certain straight line, is expressed by a polynomial equation of degree n, which is formed by the position of its individual roots, x<i>, (i = 1 to n), then the centroid, that is, the location through which the line of action the resultant R of magnitude n, of this force system passes, is obtained analogously, either : (1) by averaging the roots, or, (2) by solving the (n ? 1)-th derivative, [this always yields linear equation], set to zero.

The ANALOGOUS EQUIVALENCE between the system of parallel forces and the polynomial equation are :
(1). System of parallel forces (each of unit magnitude) = polynomial equation with constant coefficients.
(2). Number of parallel forces, n = degree of polynomial, n.
(3). Value of the resultant, R = degree of polynomial, n.
(4). Location of the action of the resultant
(= centroid of forces), x<g> = average of roots, x<av>.

(B) For parallel forces of unequal magnitudes:
If each of the acting parallel forces, a<1>, a<2>, a<3>,?, a<n> is not equal to1, but of different from one another, acting at the corresponding root point x<1>, x<2>, x<3>, ?x<n>, take all the applied forces, a<i>, (i = 1 to n), as if = 1, only that, the root points, through which these individual unit forces, each of unit magnitude, would act, need modification.
The modified root points are: [a<i>.x<i> /C], where, C = [sum of the given a<i>] /n, (i = 1 to n). [Note: (given a<i>) means actual value of the force that is acting]

In terms of the modified roots, the required n-th degree polynomial equation, in this case, alternatively, becomes:
(x ? a<1>.x<1> /C).(x ? a<2>.x<2> /C). ??(x ? a<n>.x<n> /C) = 0 ???..Eq(6)

(C) For centroid of uniform laminas (x<g>, y<g>):
For actual centroid problem of uniform laminas, two sets of polynomial equations of the Eq(6) type need to be written, one for x- directional, and the other for y- directional forces (actually, they are the number of areas a<i>, of the total area when divided into parts).

To locate the x<g>, for the y- directional forces, get x<av>, by differentiating this type of equation up to (n ? 1)- th stage, and set to zero. Similarly, write this type of equation in terms of y, and get y<g> from the x- directional forces.

Example 1. An L-shaped rectangular lamina ABCFHGEDA is described in (x,y) coordinate system. The coordinates are : A(0, 0), B(4, 0), C(8, 0), F(8, 2), H(8, 6), G(4, 6), E(4, 2) and D(0, 2). Find the centroid of the lamina.

Solution by dividing the lamina into three areas:
n = 3, a<1> = area ABED = 8, x<1> = 2, y<1> = 1, a<2> = area BCFE = 8, x<2> = 6, y<2> = 1, a<3> = area GHFE = 16, x<3> = 6, y<3> = 4, [a<1> + a<2> + a<3>]= 32, C = 10.33.
(a<1> x<1>) / C = 1.549, (a<2> x<2>) / C = 4.647, (a<3> x<3>) / C = 9.293,
(a<1> y<1>) / C = 0.774, (a<2> y<2>) / C = 0.774, (a<3> y<3>) / C = 6.196
Polynomial equation along x : (x ? 1.549) (x ? 4.647) (x ? 9.293) = 0,
or, x^3 ? 15.489 x^2 + 64.777 x ? 66.891 = 0.
Value of x from the second successive differentiation, x = + 5.163 (accepted as = 5),
also x <av> = x<g> = (15.489 / 1) / 3 = + 5.163.
Polynomial equation along y : (y ? 0.774) (y ? 0.774) (y ? 6.196) = 0,
or, y^3 ? 7.744 y^2 + 10.490 y ? 3.711 = 0.
Value of y from the second successive differentiation, y = + 2.581 (accept = 2.5),
also y <av> = y<g> = (7.744 / 1) / 3 = + 2.581.
So the centroid of the given lamina is at (5.0, 2.5).
[The round off error in decimal places has crept in the calculator calculation].

My question/enquiries to my Readers:
I have verified this analogy and understand that, it works perfectly. Please send me your valuable comments, and constructive criticism and suggestions for improvement on this topic. ?????????????????????????????
For detailed treatment, Please refer to My Books:
1) Appendix B: ?Advanced Trigonometric Relations through Nbic Functions? by Nisith K Bairagi, New Age International publishers, New Delhi (2012),[ISBN: 978-81-224-3023-3], and
2) Chapter 12: Advanced Solid Mechanics? by Nisith K Bairagi, Khanna Publishers, Delhi (1985).

Thanks to all,
Doctor Nisith Bairagi



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