Roots of a Random Polynomial

Calculus Level 4

a 2016 x 2016 + a 2015 x 2015 + + a 1 x + a 0 = 0 a_{2016}x^{2016} + a_{2015}x^{2015} + \cdots + a_1x + a_0= 0

The coefficients of the equation above are independently sampled from a uniform distribution over [ 0 , 1 ] , [0,1], and the 2016 roots are plotted in the complex plane . Which of the following graphs is most likely to depict these roots?

Bonus: Explain this phenomenon, extend it to polynomials of arbitrary degree, and/or discuss what happens if the distribution of coefficients is changed.

C B A

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Pranshu Gaba
Feb 17, 2016

Since the coefficients of given polynomial are all real, the non-real complex roots will appear in pairs along with their corresponding conjugates. Thus when we plot the roots of the polynomial, the plot will be symmetrical about the real axis. Out of the given graphs, only B \boxed{B} is symmetric about the real axis.


On a serious note:

We want to prove that for a random polynomial whose coefficients are randomly sampled from the uniform distribution [ 0 , 1 ] [ 0, 1] , the absolute value of the roots is close to 1 1 . I have been able to prove this for a linear polynomial.

Let the polynomial be a 1 x + a 0 = 0 a_{1} x + a_{0} = 0 . The root of the polynomial is x = a 0 / a 1 x = -a_{0}/a_{1} . The distribution of a 0 / a 1 a_{0}/a_{1} will be uniform ratio distribution (equation 8) .

P ( u ) = { 0 u = 0 1 2 0 u 1 1 2 u 2 u > 1 P(u) = \begin{cases} 0 & u = 0\\ \frac{1}{2} & 0 \leq u \leq 1\\ \frac{1}{2u^{2}} & u > 1 \end{cases}

On solving the definite integral m P ( u ) d u = 1 2 \displaystyle \int _{-\infty} ^{m } P(u) \, du = \frac{1}{2} , we get the median value of P ( u ) P(u) is m = 1 m = 1 . Thus, we can say

The median value of a 0 a 1 \dfrac{a_{0}}{a_{1}} is 1, ( ) ~~~~~~~(*)

Therefore the median value of the root of the linear polynomial is 1 -1 and the root of the linear polynomial is most likely to lie close to the complex circle z = 1 |z| = 1 .


Using the fact in ( ) (*) , we will try to extend this result for polynomials of higher degrees. We will proceed and convert the given polynomial to a monic polynomial. Consider the following polynomial:

a n x n + a n 1 x n 1 + + a 3 x 3 + a 2 x 2 + a 1 x 1 + a 0 = 0 a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{3}x^{3} + a_{2}x^{2} + a_{1} x^{1} + a_{0} = 0

We will divide throughout by a n a_{n} to obtain a monic polynomial. Each coefficient of this monic polynomial has median value equal to 1 1 since the median value of a i / a n a_{i}/a_{n} is 1 for all i. Therefore the median polynomial is

x n + x n 1 + x n 2 + + x 3 + x 2 + x + 1 = 0 x^{n} + x^{n-1} + x^{n-2 } + \cdots + x^{3} + x^{2} + x + 1 = 0

The roots of this equation are all the ( n + 1 ) th (n+1)^{\text{th}} roots of unity except 1. The roots of the median polynomial lie on the unit circle z = 1 |z| = 1 . Hence we can say that the roots of the randomly chosen polynomial of any degree n n are most likely to lie close the unit circle. _\square

Note: In this solution I have assumed that the median is close to the most likely values (mode). This is true when the standard deviation of a distribution is low.

Moderator note:

Good point about the symmetry.

As you point out, your result is not complete. For example, the fact that the median root for the linear polynomial is 1 -1 does not directly tell us that the roots should have magnitude close to 1. However, your general intuition is on the right track (and a full rigorous solution will likely involve Complex Analysis and some more technical details).

To get closer to a more convincing argument, try thinking about the effect of the degree of each term and the fact that the coefficients are uniformly distributed with the same distribution.

Alternatively, to understand this phenomenon, suppose instead that the roots were chosen from U [ 0 , 1 ] . U[0,1]. What would the distribution of the coefficients be like? Hint: They definitely would not be linear, and they would be a function of the degree of their term.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...