Of course, I could just give them a huge number of polynomials to multiply, but, aside from being boring, that would require me to work out the answers myself in order to check their answers!

I’ve found a simple alternative that goes back to the great eighteenth-century mathematician Lagrange. “Lagrange interpolation” is actually interesting and useful in itself (although even well-educated technical people seem often to be ignorant of it nowadays), it happens to require quite a lot of multiplying of polynomials, checking of signs, etc. so that it is good algebra practice, and, best of all, it is automatically self-checking.

Here in a nutshell is how it works:

You are given a table of values for the variables x and y, and you want to find a polynomial that gives exactly the correct values of y when you plug in the values for x.

Suppose, for example, you are given the following values:

x yand you want to find a polynomial

1 1

2 4

3 9

5 9

y = a xthat goes through those points.^{3}+ b x^{2}+ c x + d

There are various ways to solve this problem -- for example, you can use linear algebra if you view (a,b,c,d) as a vector in a four-dimensional space.

The method published by Lagrange uses a much simpler idea.

What we do is find four separate polynomial, each of which vanishes at all but one of the values of x.

For example, expression A:

(x - 2) * (x - 3) * (x - 5)obviously vanishes when x is 2, 3, or 5, but obviously does not vanish when x is 1.

What is the value of expression A when x equals 1? Well, just plug 1 in for x and you find that the value is -8.

Now, when x is 1, according to our table, we need y to have a value of 1, not -8. So, we will just divide expression A by -8 and multiply it by 1, getting expression B:

1 * (x - 2) * (x - 3) * (x - 5) / (-8)If you do the same thing for the case where x is 2, you get expression C:

4 * (x - 1) * (x - 3) * (x - 5) / (3)Run the same trick for x equal to 3, and you get expression D:

9 * (x - 1) * (x - 2) * (x - 5) / (-4)Finally, for x equals 5, you get expression E:

9 * (x - 1) * (x - 2) * (x - 3) / (24)Now, expression B gives the right value for y when x is 1, and, it is created so that it will vanish at the other three values of x, so it will not mess up the values of y there. Similarly, expression C is created so that it gives the right value for y when x is 2, and it is zero at the other three values of x.

So, if we simplify expressions B, C, D, and E by multiplying each one out, and then add them all together, combining like terms, we will get a polynomial that gives the right values of y for each of the four values of x.

I don't need practice on algebra, so I had my kids do this. Their answer is expression F:

y = (-2/3) xHow do they (and I) know that they did the algebra right?^{3}+ 5 x^{2}+ (-22/3) x + 4

Simple – they plugged into expression F the values 1, 2, 3, and 5 for x. They should find that the values of y will then be 1, 4, 9, and 9 as planned. If they do not get the right values of y, they need to find their algebra error!

Note that Lagrange interpolation always works: you can choose any real numbers, positive, negative or zero, integral or fractional, for x and for y. (You can even use complex numbers if you wish.)

In particular, the values of x do not have to be evenly space: in my example, I “skipped” 4, and it worked fine. Nor do the values of y have to be in any pattern: 1, 4, and 9 in my example seemed to be starting a pattern, but I wrecked that pattern by using 9 twice instead of using the “obvious” choice of 16.

The values of x do have to all be different (although, you can play some interesting tricks by letting two values of x get “infinitely close” – basically, you can then control the slope at x as well as the value of y).

You can in fact prove that this is the only polynomial of degree three or lower that goes through our four points. (In general, if you have n points, with different values of x, there is always a unique polynomial of degree n-1 or lower that goes through those points.)

The more points you use, the more complicated the algebra gets. I'd start with only two or three points for someone who is just learning algebra.

Isn't this too hard for kids in first-year algebra? No, my kids have learned it without too much trouble: there is no division of polynomials here, no quadratic formula, no trig functions, etc. This really is just first-year algebra.

But, isn't it complicated?

A little.

But this is the kind of complication that you get in real math applied to real problems. No one in real life (not even in science or engineering) ever faces the problem of multiplying ( x - 2 ) times ( x- 3) just for the fun of it. And, it is very, very rare that anyone ever faces the familiar textbook sort of algebra problem: “A train leaves Albuquerque going towards Santa Fe at 70 mph and a train leaves Santa Fe...” or “Jane had five times as many dolls as Ginger, but after Jane got two more dolls....”

Algebra is an abstract science; traditional algebra is about understanding the abstract properties of the four basic arithmetic operations: what does and can happen when you use the operations of addition, subtraction, multiplication and division in a systematic way? (Modern university algebra is about the properties of more general systems of mathematical operations that can operate on elements very different from ordinary real numbers.)

Mathematics is not really about balancing your checkbook or calculating the amount of tile you need to re-tile the kitchen – we have electronic calculators to do that for us.

Mathematics is about the possible abstract structures that can logically exist.

Those structures are often based on arithmetic and geometry, so you do need to know traditional math to understand modern mathematics.

But really learning math means trying not just to learn to get the right answer but actually exploring the universe of mathematics much as traditional explorers explored newly-discovered continents.

Lagrange interpolation is a very simple example of such exploration. Actually graphing the polynomials you get through Lagrange interpolation can also be enlightening: while the method always works, it can give some pretty “snaky” curves if you fit more than three points (for three points, the resulting curve is much nicer).

Is it too complicated? Well, someone unwilling to tackle topics such as Lagrange interpolation is not really learning math. If you want to know what math is really about, rather than just working you way through the watered-down, over-simplified picture of mathematics portrayed in American public-school textbooks, you need to try to wrap you mind around ideas such as Lagrange interpolation.

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* From time to time, I plan on posting a “Math Interlude,” in which I’ll try to explain some significant idea in math not known to most educated American adults, that I have in fact taught to my own kids in grade school or middle school, and that a bright middle-school student should be able to grasp.