Extremes without Calculus

There was an interesting question last week on the AP Calculus Community Bulletin Board. A teacher was working with polynomial functions in her pre-calculus class and the class had learned how to find the roots, and how the roots and their multiplicities affected the graph. Her question was whether there is a way to determine the number of extreme values (maximums and minimums) without using calculus.

You might want to work this yourself before reading on.

Before we go into more detail recall two facts about polynomial functions with real coefficients:

  • As a corollary to the Fundamental Theorem of Algebra, we know that a polynomial function of degree n has exactly n factors some of which may be the same. Therefore the polynomial has exactly n roots, again not necessarily different. The multiplicity of a root is the number of times the corresponding factor appears in the factorization.
  • Between any two consecutive roots there is exactly one turning point. (You may need some calculus to justify this. If there were more than one turning point between two consecutive roots, there must be at least three of them. Then the derivative would have more zeros than the original polynomial, which of course cannot happen.

One reader suggested a method for dealing with cubics which I think generalizes to any polynomial. The method is this: Subtract the constant term from the polynomial. This translates the graph so that it will contain the origin. The translated graph is congruent to the original and therefore will have the same number of extreme values.

Red Graph: y={{x}^{3}}-4{{x}^{2}}+3x+4
Blue Graph: y={{x}^{3}}-4{{x}^{2}}+3x

Find the Real roots of the translated polynomial.  Then the number of turning points for both polynomials can be found this way (assuming, for the moment, that all the roots are Real numbers):

  1. For each distinct root, count 1
  2. For each root of even multiplicity, count 1
  3. Each root of odd multiplicity greater than one, ( 3, 5, 7, …), count 0

The number of turning points will be one less than the total.

Example P(x)=x{{\left( x-1 \right)}^{4}}{{\left( x-2 \right)}^{3}}{{\left( x-3 \right)}^{8}}, the count is 4 + 2 + 0 = 6. Turning points = 5

The reasoning goes something like this (for a polynomial with Real coefficients and constant term of zero):

  1. A linear polynomial has one root and no turning points.
  2. Each time you multiply by a different linear factor you add one root and one turning point.
  3. Each time you multiply by several of the same factor you already used to end with an even power for that factor, you add no roots, but you add one turning point (on the x-axis at the root).
  4. Each time you multiply by several of the same factor you already used to end with an odd power for that factor, you add no roots, and no turning points, since the graph will cross the axis at such a value.
  5. Then the count scheme above will then give the number of turning points.

Another approach is this: consider a polynomial with n distinct roots and (n – 1) turning points spread out along the number line. Now move two adjacent roots together (resulting in a root of multiplicity 2).  The turning point between them moves onto the x-axis. You have lost a root but not a turning point; thus, in the count item 1 decreases by one, so to compensate you increase item 2 by one to get the same total. Now move a third adjacent root to the place of the other two to get a root of multiplicity 3: this time you lose another root and a turning point so decreasing item 1 will keep the number of turning points given by the count correct.

Did I miss anything? Yes. I did not consider translated polynomials with Complex roots. That’s because I have not (yet) figured that out.  I’m pretty sure that a unfactorable quadratic factor (i.e.  one with Complex conjugate roots) does not add any turning points to the graph, but I haven’t quite convinced myself.  Any suggestions?

 

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2 thoughts on “Extremes without Calculus

  1. Yes, an unfactorable quadratic factor can add turning points. For an example, see the graph in red above. In that case, translating it downwards helps, but there’s no reason in general why translating it so that it passes through the origin will be helpful. (For example, suppose that you translate the red graph to the right instead so that it passes through the origin … not that you have any reason to do that, bit Tiny suppose that you were initially given that function. Then it still has turning points from an irreducible quadratic factor, and yet it already goes through the origin.)

    In fact, it may be that no translation whatsoever can make all of the turning points visible from the x-axis. For example, consider 3x^5 – 30x^4 + 95x^3 – 90x^2 + 23 (which I thought of by antidifferentiating x(x-1)(x-3)(x-4), with the gap intended to give space between the turning points).

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  2. The following was sent by Jim Hobelman. It is the “calculus” proof of the formula developed in this post. His result (2e + o – 1) is the same as mine (number of roots) + e – 1 since the number of roots is (e + o).

    Thank you, Jim – Lin Mc.

    Let p(x) be a polynomial with real coefficents of degree n having no non-real roots.
    Let k be the number of distinct roots.
    Let r1 < r2 < .. < rk be these roots
    Let the multiplicities of these roots be m1,m2,..,mk
    So n = m1 + m2 + … +mk
    Let o be the number of roots of odd multiplcity and e the number of roots of even multiplicity.
    So k = o + e
    Claim – Under these hypotheses the number of extreme points of p(x) is 2e + o -1
    Proof –
    The extreme points of p(x) are exactly the roots of p'(x) of odd multiplicity.

    Now the points r1,r2,…,rk give roots of p'(x) of total multiplicity

    (m1-1) + (m2-1) + … + (mk-1) = n – k

    Such roots of p'(x) which are extreme values of p(x) are exactly the even roots of p(x)
    so their number is e.

    Now the total multiplicity of all roots of p'(x) is n-1.

    So in addition to the roots of p'(x) among r1,r2,..,rk we have additional roots
    with a total multiplicity of (n-1) – (n-k) = k-1.

    But there is at least one root of p'(x) between consecutive roots of p(x) which
    gives at least k-1 additional roots. Thus these give all additional roots and each
    with multiplicity one. So these additional roots of p'(x) are all extreme valurs of p(x)
    and number k-1.

    So the total number of extreme points of p(x) is e + k -1

    But k = o + e

    So the total number or extreme points of p(x) is 2e + o -1 as claimed.

    This rule does not generally work if there are non-real roots.

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