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### Polynomial Function with Matching Maxima

Date: 11/17/2008 at 00:43:17
From: Chris
Subject: Finding a polynomial's "natural" Minimax curve

Let c(t) be a polynomial of degree 2n, n>=2 being a positive integer
for which all roots are unique, exist and are located between [0, 1].
I further specify that the first and last root are at t=0 and t=1.  As
such, the polynomial can be defined with the equation:

c(t) = t * (t - a_1) * ... * (t - a_(n-1)) * (1-t)

I would like to find every root of the polynomial that makes it such
that every maximum of the polynomial are equal.

I cannot think of a way to get a system that I can compute the answer
to.  Maple also doesn't seem to be able to solve this one.  I found a
solution for a polygon of degree 4.  I did so by noticing the symmetry
that will arise at t=1/2 and thereby defining the equation as follows:
c(t) = t * (t-1) * (t-u) * (t-(1-u)).

I can then solve my equation by solving {dc(a)/dt = 0, c(a) =
-c(1/2)}, restricting my solution to those that fit {a < 1/2, a > 0, u
< 1/2}.  I can find that {a =0.1173165676, u =0.2928932188} and can
verify that this is indeed what was expected and that the properties
are in fact respected.

I tried to tackle the same problem with a polynomial of degree 6 but
failed to get anywhere with that one.

Date: 11/17/2008 at 05:19:59
From: Doctor Jacques
Subject: Re: Finding a polynomial's

Hi Chris,

The usual solution to these "equal ripple" problems makes use of the
Chebyshev polynomials.

The idea behind this is that cos(n*u) is a rational function of
cos(u).  We define the polynomials T_n(x) by:

x = cos(u)     0 <= u <= pi
T_n(x) = cos(n*u)

As the maxima and minima of cos(n*u) are all equal to 1 or -1,
respectively, we get indeed the desired property.

Using basic trigonometric identities, it is possible to compute these
polynomials by means of a recurrence formula.  You can see a plot of
at:

Wolfram Mathworld: Chebyshev Polynomial of the First Kind
http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html

Look in particular at the equations (9) and (26).

As we will see, we don't really need to compute the polynomials
themselves.

For your particular problem, we cannot use these polynomials
directly, because:

* The normal domain of the T_n(x) is the interval [-1,+1], and you
want your polynomial defined over [0,1].  This can be easily taken
care of by a linear transformation of the variable.

* The T_n(x) take values of 1 or -1 at the endpoints of the
interval.  We can solve this problem by limiting the interval to
the outermost two roots; this will only imply the need for other
coefficients in the linear transformation above.

We can compute the roots of T_n(x) explicitly, because the
corresponding values of u are the solutions of the equation:

cos(n*u) = 0

and these values are given by:

u = (2k+1)*pi/(2n)     k = 0, ..., n-1

If the corresponding values of x = cos(u) are x[1],...,x[n] in
ascending order, we define the new variable t by:

t = (x - x[1])/(x[n] - x[1])

and the desired polynomial c(t) = T_n(x) will have the desired
properties.

If we don't care about the value of the maxima, we may write down
the roots of c(t) directly, as:

t[i] = (x[i] - x[1])/(x[n] - x[1])

and we have c(t) equal to the product of the factors (t - t[i]).

For example, for n = 4, the solutions of cos(4*u) = 0 are (2k+1)*pi/8.

We can make the following table:

u[i]     x[i]      t[i]
--------------------------
7*pi/8 -0.923880  0.000000
5*pi/8 -0.382683  0.292893
3*pi/8  0.382683  0.707107
1*pi/8  0.923880  1.000000

Notes:

* The u[i] are in descending order, since cos is a decreasing
function on the interval considered, and we want the x[i] in
ascending order.

* The second column is computed as x[i] = cos(u[i])

* The third column contains the roots of c(t) and is computed as:
t[i] = (x[i] - x[1])/(x[4] - x[1])
= (x[i] + 0.923880)/1.847759

The resulting polynomial is:

c(t) =  t*(t-0.292893)*(t-0.707107)*(t-1)

Note that this also works for polynomials of odd degree.

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
College Calculus

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