Finding the Roots of a FunctionDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/