     TOPIC: ROOT and SOLVER 
     PRODUCT:  TI-85 and TI-82 
     DATE: October 6, 1993 
 
     A question has been asked about action of the interactive ROOT 
     function on the graph of y1 = 3x^3 - 2x^2 - 13x - 8 in the viewing 
     rectangle of [-5,5] by [-30,15].  The question is why does the 
     function want to return the root at x = 8/3, even when the cursor 
     is pointed to the root at x = -1? 
 
     The TI-85 root finder or solver is a purely numeric method for 
     finding zeros of a function.  It does not use any information other 
     than function evaluations in its search for a root.  The basic 
     concept is to work between two "end points" and draw a secant line 
     through them, projecting this line to its intersection with the 
     x-axis.  If the intersection is between these end points, the 
     function value there is used to determine how to reduce the x-range 
     while maintaining a sign change (if it exists) or following lower 
     function values.  If the secant intersection is out of bounds or 
     will not reduce the x-range as much as bisection, then the x-range 
     is bisected instead.  This continues until the x-range has been 
     reduced to a very small range. 
 
     Prior to getting into the routine described above, the solver attempts 
     through some heuristics to get two "good" end points to start with. 
     The guess you provide is one of these, the other depends on a lot 
     of factors, but the solver prefers a sign change if it can find it. 
     In the case of this example, the calculator spends some time "going 
     downhill" in the neighborhood of the even multiple root at -1, then 
     "sees" the function begin increasing.  It then checks the function 
     values at the the bounds, continuing to try to find a sign change. 
     It finds one at the right bound and then begins the routine above 
     using your guess and the right hand bound as the end points.  Since it 
     won't lose the sign change, this forces it to the root at 8/3. 
     Because an even multiple root doesn't have a sign change, you sort 
     of have to "force" the solver into doing the bisections needed to 
     reduce the range around this "minimum" that (to it) may or may not 
     be a root.  In fact, for the calculator to call this a root, it 
     requires a function evaluation of opposite sign or exactly zero. 
     For this equation, in exact arithmetic we should not get a change 
     of sign in the neighborhood of -1 and we may not be likely to 
     evaluate the function at -1 exactly.  But due to roundoff error, 
     we may (and usually do) get either a function evaluation of either 
     exactly zero for an argument other than -1 or a function evaluation 
     of opposite sign (positive in this case). 
 
     If you are interested, you can observe this effect of subtractive 
     cancellation, which is a fact of life for fixed precision arithmetic 
     in evaluating polynomials at multiple roots.  Plot the equation 
     y2=sign(y1)*abs log y1 in the window [-1.000001 -.999999] by [-18 15]. 
     (You may want to set the graph format to DrawDot).  You see that 
     near the outside of this range, the function evaluates to about 
     -1E-11 (-11 on the graph) and close to x=-1 in the center, the 
     function evaluates to about -3E-13 (-12.5 on the graph).  However, 
     in the range of about +/- 2E-7 either side of x=-1, the function 
     evaluation becomes erratic, with exact zero evaluations where pixels 
     are blank and evaluations with positive sign at the top of the graph. 
     The width of this erratic region can be roughly predicted by the 
     formula: 
            2*R*10^-(14/M)   where R is the value of the root and M is 
                             the multiplicity of the root.  The 14 
                             represents the 14 digit arithmetic of the 
                             TI-85. 
     This is one of the fundamental reasons why it is numerically 
     difficult to compute multiple roots of polynomials accurately. 
     We realize that limitations such as this are usually not what you 
     are teaching, but they are commonplace on most computing platforms. 
     And as the above demonstration shows, the graphing calculator is well 
     suited to studying itself. 
 
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