Friday 20 November 2009

What Can You Say About a Polynomial with All Coefficients Equal

My lunch math kids and I came across a problem about a function with all its coefficients equal (hereafter called PWEC, polynomials with equal coefficients). I had to admit to them that I had never thought about such a function, so we developed an interactive graph on geogebra and played around with it, looking for what we might say about such a function.. So here is your chance to show how clever you are...

I took a couple of the kids observations about PWEC functions and intentionally violate at least one of them in each graph. Each of the functions is NOT a function with all coefficients equal; and your task is to say WHY? Some have more than one reason...

Game on.. here we go:

Can you get it? If not, scroll down a little to find at least one reason...
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Ok, observation number one, Joe S picked up on this one right away... Every odd powered PWEC will have a zero at x=-1... easy to confirm because the alternating powers will be negative and positive. Descarte's rule of signs proves that there can be no real zeros that are positive. My experience is that it always has just one; the one at x=-1.

Did you get it?? Well, how about another...

Well? What about this one?....
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This time you can't get much from Descarte, but it seems from observation that an even PWEC can never have a zero. I played with this a little and haven't succeeded in proving it analytically for anything but a few simple polynomials.. any offers?

Ok..just one more... another even function... what do you think...


OK, no easy stuff on this one... what do you think?
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This time there are no tricks with the zeros... Ok, maybe one more.... notice that the horizontal tangent occurs with x>0...but that is a no-no as well. The derivative will also have all positive coefficients, so by Descarte's rule of signs again, it can not have a zero when x>0.

I admit I had never thought of any of these things, and noticed them all by working with interactive graphs on Geogebra.. one reason I am thankful for the power of modern computing... I know more math because of the technology available.. and I want my kids to see that it can be more than an easy way to get the answer, it can help you learn more about the ideas of mathematics.

1 comment:

r. r. vlorbik said...

complexify; divide out the coefficient.

z^n + z^(n-1) + ... + z + 1 = 0

and multiply through by z -1:

z^n - 1.

which has n distinct complex roots
by the usual "graphical" proof.