Chapter 2: Answers 4 Jack K. Cohen Colorado School of Mines

  1. Divide the equation by a to get an equation of the form x3 + px2 + qx + r = 0. Let f (x) = x3 + px2 + qx + r. Note that for sufficiently large positive x, f (x) is as large as you like—hence we can find B > 0 such that f (B) > 0. Similarly, considering large negative x, we see that we can find A < 0 such that f (A) < 0. Since f (x) is continuous in the closed interval [A, B] and f (A) < 0 < f (B), the intermediate value theorem guarantees that there is a C in [A, B] such that f (C) = 0.

    Discussion: the division by a is not essential, it just lets us avoid considering two cases (positive a and negative a). Likewise, insisting on a positive endpoint B and a negative endpoint A was just one way of guaranteeing A < B to avoid having to consider the possibility that the interval is really [B, A]. Most importantly, note that a rough sketch of f (x) makes clear the existence of the root and is a guide to the formal proof.

  2. Each bisection cuts the interval in half. So 10 bisections reduces it to 2-10-th of its original size. The approximation 210 = 1024 $\approx$ 103 justifies the folklore. Ahem. Remembering that 210 $\approx$ 103 may well be more important than anything else on this worksheet!

  3. r $\approx$ 50008.332.

  4. r $\approx$ 50008.331912. An appropriate input is: 
    Bisect[Cos[1000/r] - r/(r + h), {r, 50008.3215, 50008.3325},  0.0000005]

  5. Our interval has length on the order of 106, so after about 30 iterations we'd have 3 places after the decimal point (do you know why?). This might be considered practical by some, but expert opinion is that bisection alone is an inadequate equation solving method. Stay tuned.