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Finding the Roots of a Function


Date: 11/4/96 at 12:29:14
From: Scott Thompson
Subject: Roots of a Function

I need help with the following 2 questions:

Descartes' "Rule of signs" states: Let f(x)=0 be a polynomial equation 
with real coefficients arranged in descending powers of x. The number 
of positive roots of the equation is either equal to the number of 
variations in sign presented by the coefficients of f(x), or less than 
this number of variations by a positive even number. 

1) Find the positive roots of x^9 + 3x^8 - 5x^3 + 4x + 6 = 0   
My answer: I see 1 variation in sign presented by the coefficients: a 
minus sign before 5x^3.  But I think there are 4 positive roots, which 
doesn't fit the rule.  Can somebody explain this to me?

2) Show that x^n - 1 = 0 has exactly 2 roots if n is even, and only 1 
real root if n is odd.
My answer: If n is even, then there would only be 2 roots because you 
would have x^even# - 1^2 for any even number.  Any suggestions as to 
how I would show this?


Date: 11/05/96
From: Doctor Pete
Subject: Re: Roots of a Function

1)  Given the equation x^9 + 3x^8 - 5x^3 + 4x + 6 = 0, you should 
first notice that there are 2 variations of sign.  If you write down 
the signs of the coefficients, you get ++-++; the first variation is 
+-, and the second is -+. Second, there is one real root to this 
equation, and it is negative. You can see this by noting that f(-3) 
and f(-2) have opposite signs, so f crosses the x-axis somewhere in 
between these two numbers. Note this does not violate Descartes' Rule 
of Signs because 0 is an even number (2) minus the number of 
variations, which is 2.

It should be mentioned that you can't really say anything about 
positivity for complex numbers.

2) To show that x^n - 1 = 0 has exactly 2 roots if n is even, and only 
1 real root if n is odd, we note that clearly, x = 1 is always a root.  
To complete the argument, show that there are no additional roots when 
n is odd by considering the first derivative. If n is even, note x = -
1 is another root.  In this case, consider both the first and second 
derivative.


Date: 11/05/96 at 12:00:17
From: Scott Thompson
Subject: Re: (no subject)

Thanks for the help. However, I'm still not clear on these questions. 

In order for something to be a "ROOT", would it have to be able to 
satisfy the equation f(x)=0?  If this is the case, I guess there 
wouldn't be an easy way of determining the root other than trial and 
error. Please clarify for me.

Thanks,
Scott


Date: 11/06/96 at 10:23:52
From: Doctor Pete
Subject: Re: (no subject)

By definition, r is a root of f(x) if and only if f(r) = 0. 
Furthermore, if f is a polynomial of degree n, then there are exactly 
n roots (counting multiplicities) according to the Fundamental Theorem 
of Algebra. But these n roots are not necessarily real. In general, 
they will be complex numbers of the form a+bI, where I is the 
imaginary number Sqrt[-1]. The point is that Descartes' Rule of Signs 
only applies to the *real* roots of f(x). After all, is 3-2I positive, 
or negative? (Neither!) As I pointed out previously, the polynomial in 
problem 1 has only one real root, and it is negative. Thus, there are 
zero positive real roots.  The remaining 8 roots come in 4 pairs of 
complex conjugates.

If your goal is to find the roots of f(x), this is not easy because 
f(x) is degree 9. If deg(f) < 4, then it is possible to find the 
*exact* value of the roots. In general, it is not always possible to 
find the exact value of the roots when deg(f) is 5 or greater.

For Problem 2, note that the derivative of x^n - 1 is nx^(n-1).  
If n is odd, then the power on x is even, and hence the derivative 
is always nonnegative. What this means is that x^n - 1 is a strictly 
increasing function for odd n. Since it is clear that x = 1 is a 
root, it follows that this must be the only real root, because a 
strictly increasing function can only cross the x-axis once.

To do the case where n is even, the argument is similar but one should 
also consider the second derivative.  Alternatively, show that the 
derivative has only one extremum (maximum or minimum; why would this 
help?)

In both cases, it helps to think graphically.  A recurring theme 
throughout both these problems is the idea that a polynomial of degree 
n has n roots, but *at most* n real roots. Often the number of real 
roots is less.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Calculus
High School Functions
High School Polynomials

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