I was just reminded of a beautiful polynomial relationship in Dan Kalman's wonderful, Unusual Mathematical Excursions, Polynoimia and Related Realms (below).It is not that I'm so clever I had figured this out on my own, but Prof. Dan had showed me this a while before.

If you are teaching quadratics, or polynomials in general right now; and even if you aren't, here is a beautiful little polynomial puzzle that will dazzle your students, and even may teach them something.

Ask one to pick a quadratic or cubic (The idea works with any polynomial but calculations get a little freaky with too many terms) with all positive integer coefficients. Then tell them if they answer two polynomial evaluations of their function, you will tell them the coefficients. I will use a cubic to keep my life simple: suppose they pick f(x)= 2x

^{3}+ 3x

^{2}+5x+4.

You first ask them to tell you f(1). This will tell you the sum of the coefficients, in this case 14. Now ask them to tell you f(14). When they do, you can convert their answer to base 14 and get the coefficients of the polynomial. This turns out to be the incredibly large number 6150. Now to earn their wows, you need to turn this into a base 14 number, but beautifully, 6150 base ten works out to be 2(14

6150 / 14 is 439 with rem. of 4. This first remainder is the constant term of the original polynomial.

^{3}) + 3(14^{2}) + 5(15) + 4. And the coefficients of the powers are the coefficients of the original polynomial. Now dividing by 14 and recording the successive remainders gives us the coefficients.....6150 / 14 is 439 with rem. of 4. This first remainder is the constant term of the original polynomial.

439/14 is 31 R 5, so 5 becomes the linear term.

31/14 is 2 R 3 , so 3 becomes the quadratic term

and 0/14 is 0 R 2 so our last, and leading, coefficient is 2.

Student's who find this method of converting to base 14 (or any other base) mysterious can be shown the relationship between it and synthetic division by the intermediate of Horner's Method of writing a polynomial Ax

^{3}+Bx^{2}+ Cx + D in the form ((Ax+B)x+C)x+D with which we can convert any number from any base, x, to it's decimal equivalent. It should be quickly obvious that to go back our first remainder will be D, with a quotient of (Ax+B)x+C .
There is much mystery in this method for the young who are less experienced with dividing polynomials, and I would follow it with the decimal approach shown farther below for young students. For more mature students, I think it would be wise to show them the decimal method (Which they will usually pick up quickly as trivial), but then show them how to convert the coefficients from the base f(1) and encourage them to prove that it is always true with some simple algebra.

Before I show the decimal method, here is the simple explanation of why the above works.

If we start with a cubic, Ax

^{3}+ Bx^{2}+Cx +D and let f(1) = p, then f(p) will be
Ap

^{3}+ Bp^{2}+Cp +D with all of A, B, C and D less than p (since p was the sum of all the coefficients and hence no messy carry over problems) and in base p, the coefficients will be the same as the originals. The unknowing student gives you this sum in base ten, and you simply convert to base p by repeated division of each quotient by p.
If prefer you can use a base ten approach that also works. After you get f(1) and know the sum of the coefficients, you can pick the smallest power of 10 that is greater than f(1). In the above cubic example we could use 100 since that is greater than f(1). Now when we ask for f(100) you will get 2030504. This would work even if one or more of the original coefficients had two digits as long as f(1) was smaller than 100. My preference was to always tell them that all the coefficients had to be a single digit, and then f(100) would also work. With that example I could have every student pick his own example of a cubic or quadratic (or for that matter any thing up to a tenth power) and enter them as functions on their calculators, then when they evaluate f(100) they instantly realize their number has been revealed. For newer algebra students just studying polynomials, this helps them to start to see why the arithmetic they learn with polynomials is so like the arithmetic they learned in elementary school.

Professor Kalman's book is full of beautiful and interesting ideas regarding polynomials. I think it would be an excellent addition to the library of any teacher of HS math.

The image at top is what is commonly called a zeotrope, a device that produces the illusion of a moving image by showing different pictures in rapid succession. It was invented by William George Horner in 1834, for whom Horner's method is named, (but which he did NOT invent and which was used 600 years earlier, by the Chinese mathematician Qin Jiushao). Horner called it 'Daedalum' that translates to 'The wheel of the devil'. The American developer William F. Lincoln names his the 'Zoetrope', which means 'wheel of life'.

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