Saturday, June 13, 2009

An Equation Which May Be More 'Complex' Than It Appears!

Maybe I should rename this blog to Saturday 'Morning' Post. After all, no one reads that either anymore!

As the school year comes to a close (and I'm assuming it's already over for some), here's an innocent-looking equation which might be worth discussing with your advanced algebra/precalculus students now or next year. I might have considered saving this for our next online math contest but it's complex nature makes it more suitable for discussion in the classroom than on a test. Have you seen exercises like this in your Algebra or Precalculus texts? Do students often delve beneath the surface of these? It's kind of like a black box. We often feel we simply cannot reveal too much of the mystery here or we will not finish required content. Well, you know my philosophy of 'less is more' and I don't even live in Westport, CT. (Ok, that's a post for another day!).

SOLVE (by at least two different methods):

2a-3/2 - a-1/2 - a1/2 = 0

Preliminary Comments/Questions/Issues

  • Is the term solve ambiguous here, i.e., should we always specify the domain to be over the reals or over the complex numbers or is that understood in the context of the problems? I'm guessing that most advanced algebra students learn that the domain of the variable or solve instructions may impact on the result, but, that is precisely one of the objectives of this problem.
  • Should students immediately change all fractional exponents to radical form? OR use the gcf approach (which requires strong skill)?
  • It's not hard to guess that 1 is a solution but is it the only solution? Can we make a case for -2 being the other solution? The graph doesn't reveal this and surely, -2 doesn't make sense or does it....
  • Is there ambiguity in raising a negative real number to a fractional exponent (never mind raising i to the i)? Why? Isn't there a principal value for such an expression? How is it defined? This problem raises fundamental and sophisticated issues about numbers which can be taken as far as one chooses to go Just how complex can complex numbers get?
  • What is the role of the graphing calculator here? Mathematica? Wolfram Alpha? In addition to verifying solutions or determining answers, can these tools also be useful in clarifying ideas or raising new questions?
  • Students (and the rest of us) are now capable of quickly filling in the gaps in their knowledge base by visiting Wolfram's MathWorld or Wikipedia for more background. Should this impact on how we present material? Typically, in the pre-web days teachers would avoid opening up a can of worms like complex solutions here, but, with your more capable groups, the sky's the limit now IMO...


Sue VanHattum said...

I bet you have more readers than you think, Dave!

Dave Marain said...

That was very nice of you to say, Sue! I am hoping this post generates some discussion. Complex numbers and exponentiation are non-trivial!

Eric Jablow said...

You left out two other approaches students may have learned:

1. Get rid of a^{1/2} by substituting another variable for it.

2. Remove the negative exponents by multiplying by a suitable power of a (knowing that a cannot equal 0).

Approach 1 does not work well. Approach 2 leads to an immediate solution.

Dave Marain said...

Agreed, Eric. Many students, who have the skill, would opt for one of those, particularly since factoring often proves difficult for them (as in, 'fast becoming a lost art').

However, I wrote this problem to have another dimension. The algebra is challenging enough, but, after obtaining the quadratic 2-a-a^2 = 0, which leads to a = 1 or a = -2, there is the issue of discarding the extraneous solution '-2'. As suggested in the post, if we permit complex solutions, this opens up a can of worms. It would be interesting to ask students (who have a knowledge of complex numbers) to substitute -2 mechanically and see what happens.How would they evaluate (-2)^(-3/2) for example, as an imaginary number. Is it merely the square root of -1/8, which produces the principal value, 1/(2i√2)?

Then store the original equation in Y1 on their graphing calculator and evaluate Y1(-2) when in real vs. complex mode. Also, look at the result Wolfram Alpha gives for the solution to the equation. Too far beyond most high schoolers or something to pique their interest...

Eric Jablow said...

I didn't mention the extraneous solutions from the quadratic because I didn't want to give the entire problem away. The other issue that entering the complex domain leads to is that of whether one may choose different values of the square root for the three terms. And that leads to the old paradox about 1 = -1:

1 = (-1)·(-1)

√1 = √[(-1)·(-1)]

√1 = √(-1)·√(-1)

1 = i·i

1 = -1.

Dave Marain said...

It may be a classic paradox, Eric, but I'm willing to bet there are a few out there who have not seen it! It's one of my favorites and the fallacy is subtle. There are other similar ones based on the false assumption that laws of exponents for reals can be extended to the complex domain. I felt that the multifaceted nature of this problem could be appreciated on several levels. At first glance the advanced algebra/precalc student would see this just as an algebra manipulation or one they could load into their graphing calculator. When the instructor suggests changing the mode to a+bi form, there should be some surprise -- at least as much as when we ask students what ln(-1) is (or how it could have infinitely many values!).

Cecil Kirksey said...

Hi Dave:
Interesting problem. If we consider this as a function and ask for the zeros then we must first look at the domain of the function. I do not believe that complex analysis is covered in any high school math class so we should assume that the domain is the real numbers. But for this function a must be > 0. A zero can be complex but it can never be outside the real domain of the function.

However there is an intersting problem regarding solutions for the cubic equation. The real root or zero can be defined as a function of a real variable but the domain becomes restricted because of square roots of negative numbers. This I find is a really deep issue. How do you calculate a real valued function whose domain and range a re all real numbers but the calculation must use complex numbers? BTW there is different function for the real root that does not have this prpblem or so I believe.

Why I got really interested in this type of problem was the following. I was helping a student solving radical equations. (I am a tutor at a CC.) The resulting answers came out to be radicals. i.e., a + sqrt(b). Now the standard text book procedure is to check each answer to be sure it solves the original equation. But I was sure that this would not work in all cases. For some examples that I have looked at the correct procedure is to first check to see if the radical expression must always be nonnegative (works for some cases). I have not arrived at a general conclusion on this yet. Obviously graphing (using the real domain) can be used but doesn't help in an algebraic solution.

BTW I am talking about simple radical equations involving two square roots and a constant.

I always look at your blog for interesting math problems. Keep up the great work.