Solving linear and nonlinear equations through the method of iteration (Picard's method) has been enormously facilitated by scientific calculators. The topic is approached here through graphical analysis which gives a clear visual representation of the process involved. It has ready application in problems for which algebraic solutions are messy or not possible, or where working backwards proves useful.
Though the actual work with the calculator described here will be similar to that used in the paper by Harrow and Straley in this volume, the overall approach will be quite different. Transformational geometry lay at the heart of their method, the visual technique of web diagrams forms the core of this technique.
Let's begin with a system of two linear equations, say f(x) = .5x+8 and g(x) = -3x+1. The graph of these two functions is given below in Figure 1.
Figure 1 Figure 2
On the right is the identical graph with the first step of graphical analysis sketched in; an initial value for x, xo, is arbitrarily selected; the first of the two functions is then evaluated at xo, giving f(xo); in Figure 2, this is shown by the vertical line from xo on the x-axis to the function f(x) at the point (xo, f(xo)); this value of f(x) is then set equal to g(x) on the second curve, a process denoted by the horizontal line from f(x) to g(x) at the value f(xo). In Figure 3 below, this process is continued by moving vertically to the f(x) curve, then horizontally to g(x), and so on toward the solution to this pair of simultaneous equations.
Algebraically, the move from g(x) vertically to f(x) requires our finding the value of x, call it x1, for which g(x1) = f(xo). By definition, finding the value of x given the value of g(x) means finding the inverse of g(x). And since every linear function has an easily obtained inverse, this poses no obstacle. That is to say that x1 = g-1(f(xo)). In which case the vertical move from g(x) to f(x) amounts to evaluating f(x1); and the next value of x, x2, is g-1(f(x1)). Continuing this iterative process will lead to the value of x that is the solution to this pair of simultaneous equations.
On the calculator, this means selecting any initial x (the closer to the final solution, the fewer steps required to home in on the solution) and iterating the function g-1(f(x)), where f(x) is given above, and g-1(x) = (-1/3)(x - 1). The function to be iterated is therefore, g-1(f(x)) = (-1/3)(.5x + 7). Let's try a seed of -4. Within several iterations, the fixed point value of -2 appears on the calculator screen. Note that the function iterated here is just what would have resulted had we set the two functions equal to each other, and solved for x on one side in terms of x on the other. Our result is just Picard's method arrived at graphically.
-3x + 1 = .5x + 8
-3x = .5x + 7
x = (-1/3)(.5x + 7)
The TI-82 enables us to carry out this process in steps one-to-one with the web diagram:
store a seed to x; type in f(x) in terms of x, store to x; colon; type in g-1(x) in terms of x; press ENTER, ENTER, ....
In the example given above, the screen should look something like
-4 --> x
.5x+8 --> x:(x-1)/-3 -->x
Let us suppose now that we had chosen g(x) as our "first" function, f(x) second. The composite function to be iterated now is f-1(g(x)) = 2(-3x - 7). Try this, again with a seed of -4. Rapidly our iterations fly away from the solution we seek. Graphically this amounts to doing what we had done originally, only now in reverse. Moving backwards on the web diagram means spiraling out from the fixed point. It is now a repelling fixed point where before it was attracting (see Figure 4).
Comparing Figure 4 with Figure 3, we can generalize that if we pick the line with the slope closer to zero as our "first" function, we will have success. If we choose the steeper slope first, we will fail to converge on our sought-for solution.
1. Using the same two functions given above, iterate the composite function g-1(f(x)) thirteen times with a seed value of -4. Now, without clearing the screen, enter the function f-1(g(x)) (this can be done on the TI-82 by hitting 2nd ENTRY a few times to recover this function if you used it not long ago) and iterate, effectively going backwards from the final iterate of the original thirteen iterations. What did you find?
Try the same exercise, only iterate g-1(f(x)) twenty times this trial, then go backwards with f-1(g(x)). What do you find now? What does this tell you about the workings of your calculator?
2. Try some other functions. What happens as the values of the slopes get closer and closer to one another? What happens in the case of oppositely signed slopes as the absolute values of the slopes get closer and closer to one another? What happens when one slope is m, the other -m? Make web diagrams of these various possibilities. Does your position relative to the x and y-axes make any difference to your graphical analysis?
Though the interplay of geometry and algebra leads to some interesting insights while exploring linear systems, the real power of this method comes in the non-linear realm. Suppose you would like to solve the equation 8x2 +.5 = 2cosx. Sketching the graphs of f(x) = 8x2 +.5 and g(x) = 2cosx, we see that there are two solutions, and the curve y = 2cosx has the slope closer to 0 in each case (Figure 5).
Using the method outlined above, evidently we need the inverse of h(x) = 8x2 +.5. For x 0 in the original function, the inverse is given by h-1(x) = ((x - .5)/8).5, and for x < 0 in the original function, the inverse is given by h-1(x) = -((x - .5)/8).5. The composite functions thus become:
for solution 0, iterate h-1(y) = ((2cosx - .5)/8).5
for solution 0, iterate h-1(y) = -((2cosx - .5)/8).5
Iterating each of these with a seed of 1 yields the two solutions, .40856... and -.40856....
Note that we could have solved the original equation for x on the left side, in terms of cosx on the right side, and arrived at the same equations given above, to be iterated as in Picard's method, just as for the linear case.
8x2 +.5 = 2cosx
x2 = (2cosx -.5)/8,
x = +/- ((2cosx -.5)/8).5
Note also that for a poor choice of seed, our iteration would have come to a screeching halt. Try a seed value of 2, for example. The web diagram (Figure 5) shows the problem.
This then, is a benefit of looking at such problems using graphical analysis. A quick sketch by hand or with a graphing calculator allows one to determine such matters as appropriate seed values and for which function the inverse will be necessary; or equivalently, in solving for x in Picard's method, from which side one should proceed to isolate x so that the function set equal to x will cut through the diagonal y = x less steeply than the diagonal itself. In general, solving for x in the manner of Picard will be more direct in arriving at solutions, but both functions should be sketched to gain some insight about the functions' relationship to one another. Moreover, the concept of inverse and moving backwards is instructive and has application in other areas, such as in the pictorial creation of Julia Sets on a graphing calculator by working backwards (see Pettus and Taylor in this volume, for example).
Finally, consider the set of problems one encounters in financial math in which a mortgage, for example, is to be knocked back by a payment of $5000/year for 20 years, with an effective annual interest rate of 8%. What is the value of the initial balance that will yield a 0 balance after 20 years? Various methods are open to us, for example:
1) Using the calculator, one can iterate the function f(x) = 1.08x - 5000 twenty times making refined guesses at the seed value. The value that results in 0 after the required twenty iterations is the solution.
2) One could solve analytically, using either geometric series or the fixed point form of the function, if either is familiar.
However, by employing the graphical analysis argument discussed above, you could start with a value of zero and iterate backwards (i.e. with the inverse function) and arrive directly on the answer after twenty iterations. f-1(x) = (x + 5000)/1.08, x1 = 0, and after twenty iterations the result is 49,090.737.... Therefore an initial amount borrowed of $49,090.74 will lead to a 0 balance after 20 years at the interest and payment rates given above. The first several steps of the graphical analysis for this problem are shown below.
1. Given 2x = x10. A sketch of the curves, f(x) = 2x and g(x) = x10, shows that there are three solutions, two of which are at points where the slope of f(x) is smaller than that of g(x), and one the other way around. The solutions to this and other such problems of the reader's own creation are left as exercises.
2. Suppose your state must pay off a lottery winner with 10 equal installments of $100,000/year, starting today and ending nine years from now. How much must the state set aside now to make these payments if its unpaid balance earns 6% effective annual interest, paid at the end of each year (just before the next $100,000 is paid out)? At the time of the last payment, the balance should be reduced to 0.
3. You are 45 years old and have determined that you would like to establish a retirement account at age 65 to be worth $250,000. You have decided to put an initial lump sum into an interest bearing account which you will add to at the end of each year to the tune of $5000. You assume the minimum annual interest you will earn to be 6%. What is the minimum lump sum you must start with to insure $250,000 at retirement?
Contemporary Precalculus through Applications, The North Carolina School of Science and Mathematics, Janson Publications, Inc., Dedham, MA
Chaos, Fractals, and Dynamics, Robert L. Devaney, Addison-Wesley Publishing Co.