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Importance of Algebra, Prof. Askey
Posted:
May 27, 1997 10:25 AM


Dear AP Calculus Teachers,
A discussion has evolved in this forum on the need for basic techniques of algebra in order to learn calculus and to pursue higher mathematics. Richard Sisley, and no doubt many others, maintain that less attention should be given to basic algebra skills because of the availability of computer algebra systems to students.
Richard Sisley even claims that such skills as factoring quadratic polynomials over the integers need no longer be taught.
The following message comes from Richard (Dick) Askey, John Bascom Professor of Mathematics, at the University of Wisconsin, Madison (one of the premier institutions for mathematics in the world). Dick is a fellow of the Indian Academy of Sciences, a fellow of The American Academy of Arts and Sciences, and currently serves on the American Mathematical Society committee to advise NCTM about a revised version of their Standards.
Dick has kindly given examples of algebra problems he would like good students to be able to solve. The last one he gives in his message below is particularly interesting because he suggests using a computer algebra system for the initial stages of a problem. The completion of the problem is up to you with whatever tools you choose.
I STRONGLY ENCOURAGE THOSE WHO DON'T THINK BASIC ALGBRA SKILLS ARE IMPORTANT TO CAREFULLY READ DICK'S MESSAGE AND TO WORK THROUGH THE EXERCISES HE PROVIDES (especially the last one in his message).
I also strongly encourage any math teacher to take advantage of Dick's generous offer and request a copy of the algebra notes he and Prof Wu wrote. I have a copy. The notes are clear, concise, and beautifully written. They are valuable for any teacher of mathematics.
David Klein Math Dept. California State University, Northridge
_______________________ Dear David, Pass this on to the APcalc group. I have been getting some of the messages about what algebra students should learn in high school. HungHsi Wu at Univ. of CaliforniaBerkeley and I have written a set of notes about what we think students should have learned in high school mathematics before they start calculus. I can send a copy to people who are interested in seeing these notes. Send a message asking for them to askey@math.wisc.edu I may have to stop sending if there are more than 100 requests, and will have to have more printed in any case, so it may be a little while before these are send. Send a US postal address. One thing which has not been mentioned in the messages sent to me is work with equations with parameters. This is very important in both mathematics and physics, but little is done with it in high school mathematics according to what students tell me and from what I read in school texts. One very nice problem is to find the rational points on the unit circle this way. Consider the unit circle and the straight line through (1,0) with slope t. To find the points of intersection, replace y^2 in x^2 + y^2 = 1 by y = t(x+1) to get x^2 + t^2(x+1)^2 =1. At this point, I ask students if this is a hard or an easy problem to do. They should be able to say it is easy, for it is clearly a quadratic equation and all of these can be solved via the quadratic formula. However, very few students are able to do see this. They do not understand what polynomials are, or even recognize quadratic polynomials when the quadratic is not given in the form ax^2+bx+c or as something like 6x^2+4x5. Students need much more experience seeing polynomials in different forms. To show that all rational points are found this way takes a little more work, but finding the formulas for Pythagorean triples, which is what is being done here, is enough to interest most students. There were questions about whether or not higher degree polynomials ever arise which one needs to factor. Here is an important example of one. You all know that binomial coefficients count things. One thing they count is a refinement of the number of ways of putting 0s and 1s into n places. C(n,k)=n!/k!(nk)! counts the number of ways of putting k 0s and nk 1s into n spots. This is a refinement since the sum on k from 0 to n of C(n,k)=2^n, which is the number of ways of putting zeros and ones into n spots. There is a further refinement. Consider the following numbers
0011 0 0101 1 0110 2 1001 2 1010 3 1100 4
This is a listing of the ways of putting 2 zeros and 2 ones into four spots. The numbers on the right give the number of inversions needed to switch 0s and 1s to get back to the original pattern of zeros on the left and 1s on the right. Form a generating function of the numbers in the right hand column. Write this as 1q^0 + 1q^1 + 2q^2 + 1q^3 + 1q^4 = 1 + q + 2q^2 + q^3 +q^4. When q=1, this is 6, as it should be. However, the binomial coefficient C(4,2) is not only 6, it is better written as 4!/2!2!, so we want to write the fourth degree polynomial above in a similar form. This can be done by factoring it first. One way to do this is to write it as 1+q+q^2 +q^2+q^3+q^4 = 1+q+q^2 +q^2(1+q+q^2) = (1+q+q^2)(1+q^2). This is analogous to 3*2 rather than 4*3/1*2. To get the better form, sum the geometric series to get (1q^3)(1q^4)/(1q)(1q^2), and then put in the extra factors to get (1q)(1q^2)(1q^3)(1q^4)/ (1q)(1q^2)(1q)(1q^2), which starts to look like 4!/2!2!. To make it look even more like this, divide each factor by 1q and expand via the sum of a finite geometric series. If n!_q=1(1+q)...(1+q+...+q^(n1)), then this is 4!_q  2!_q2!_q
This is probably new to all or almost all of you. The idea of having products like those found above is old, and a version of this was found in the early part of the nineteenth century. The qextensions of the binomial coefficients are called Gaussian binomial coefficients. The interpretation of the coefficients in the expansion of these coefficients as polynomials in q that was given above was found by MacMahon about 1915. There is a version of this due to Schutzenberger in 1953 in the form of a noncommutative version of the binomial theorem. Consider (x+y)^n and expand it, but use the strange multiplication rule that yx=qxy, qx=xq, qy=yq. The result is sum of C_q(n,k)x^(nk)y^k, where the Gaussian binomial coefficient has what should be the obvious extension of C(n,k), using the case n=4, k=2 as a model. It is a nice exercise to show this is true by induction. Detailed ways of counting which lead to such beautiful formulas will certainly have important uses, and this now does. There are objects called quantum groups, which first arose in some work in mathematical physics, and now occur in many areas of mathematics. This noncommutative version of the binomial theorem is central in this subject. Toward the end of his active career, George Polya wrote a couple of papers about inversions and Gaussian polynomials. These were written after Schutzenberger introduced the noncommutative version of the binomial theorem, but Polya was unaware of it. His version was more complicated to understand. Even so, he suggested that this would be a nice topic for high school mathematics. The commentator on these two papers in Polya's "Collected Papers" expressed reservations about this. I have done this in a high school mathematics club, and the students could follow and help with some of the work, and found it interesting. One thing which came up in the argument above was the finite geometric series. Students need to know this, and very few of them do when they get to college. The first day in first semester calculus, I ask the students to turn a repeating decimal into a fraction, say .454545... At most 20% are able to do this even in an honor section, and only about half of these know how to do it by calling the number x, and multiplying by the appropriate power of 10. The other half now do it by pattern matching, saying it is 45/99, but have no idea why. The first method is important, for it leads to the sum of the general finite geometric series. The other method is a dead end, since it leads nowhere. Unfortunately, there were a number of letters in Mathematics Teacher praising a student Joey for discovering this pattern matching, and none pointing out that the method is a dead end, and so should not be encouraged. There is a lot of symbolic manipulation which students should learn in algebra. The book "Algebra" by I.M. Gelfand and A. Shen has some very nice problems. This is published by BirkhauserBoston, and the paperback version costs under $20. Both Tom Romberg, who chaired the committee which wrote the NCTM Standards, and Hyman Bass, chair of the Mathematical Sciences Education Board, have written that this book "corresponds to the NCTM Standards". I can not make that claim, since I do not understand many things in the "Standards", but claim that it is a very good book, full of interesting problems and explanations for why things work as they do. Here is one problem from this book, along with a solution showing how one can do the factoring, and a couple of things which follow from the factoring. It is a cubic, and the argument is not obvious, but one learns things from it. Factor
x^3 + y^3 + z^3  3xyz.
I will skip a screen so that those of you who want to work on this can stop reading, and scroll down once you have either solved it or decided you spent enough time.
MORE BELOW
Observe that this polynomial vanishes when x=y=z. Set z=x+a, y=x+b. Then the polynomial is
x^3 + x^3 + 3x^2b + 3xb^2 + b^3 + x^3 + 3x^2a + 3xa^2 + a^3  3x(x^2 + ax + bx + ab) = 3x(a^2ab+b^2) + a^3 + b^3 =(3x + a + b)(a^2  ab + b^2) =(x+y+z)(x^2+y^2+z^2xyxzyz).
One thing which follows from this is the arithmeticgeometric mean for three numbers. For, x^2+y^2+z^2xyxzyz =1/2[(xy)^2 + (yz)^2 + (zx)^2] which is nonnegative. The general solution of a cubic also follows, since the quadratic just given above can be factored into two linear factors, when treated as a function of x. Consider the general cubic equation. ax^3+bx^2+cx+d=0. We can assume a=1 by division, and then that b=0 by shifting x by a constant to remove the x^2 term. Thus we can consider x^3 + cx + d =0 = x^3  3yzx + y^3 + z^3. Set 3yz=c and y^3+z^3=d, and solve for y and z, by finding y from the first equation, putting it in the second and solving the resulting equation which is a quadratic in y^3. Can all students do this? Probably not given the weak background they have. Should many students see this? Yes. If we look at the results of the eighth grade TIMSS test, we see that we have 25% of our students in the bottom 25% of the students internationally. However, we only have 18% of our students in the top 25% internationally, and only 5% of our students in the top 10%. This seems to say that we are cheating our better students more than our poorer students, if we measure by what we should expect them to accomplish. Contrast this with results from Singapore, where only 1% of the students are in the bottom 25% and 45% are in the top 10%. Yes, all students can learn mathematics, and real mathematics, not the watered down version we have been giving to our students. For those of you who like to use computer algebra systems, try the following problem. Let
f(x) = 8(1(1x)^(1/2))^3  x^2
Compute f(f(x)) and ask the computer algebra system to simplify it. Then graph it and get a surprise. Finally, try to prove that the surprise is true. Something which has been outlined in this message will help. Dick Askey Dept. of Mathematics Univ. of WisconsinMadison 480 Lincoln Drive Madison, WI 53706
askey@math.wisc.edu



