166 5. Approximation and Location of ZerosExplorations
E.42. Convergence of Newton Approximations. It is not always the
case that Newton’s Method approximates the intended zero of a polyno-
mial. The difference between successive approximations to a zero of p(t) is
given by an expression of the form p(u)/p’(u), which may be large if p’(u)
is small (the graph is quite flat at the approximation) or p(u) is large (the
approximation is too far from the zero).
Consider, for example the polynomial t3 - llt2 + 24t = t(t - 3)(t - 8).
Sketch the graph of this polynomial. Examine the sequence of approxima-
tions which begin respectively with 4, 5, 5.5, 5.7, 6.
One may well wonder what happens if the polynomial has no real zeros.
Experiment with the polynomial t2 + 1. Show that if we begin with the
approximation cot 0, then the next approximation is cot 28. You might also
observe that if the first approximation is Re w/Im w for some w E C, then
the next one is Re w2/Im w2.
E.43. Newton’s Method According to Newton. In his monograph,
Analysis by equations of an infinite number of terms, Isaac Newton pro-
vided a method of showing how a function can be written as a power series.
As an illustration, he approximated a solution of a numerical equation.
Study the following passage and compare it to the procedure described in
Exercise 5:
Let the Equation ~~-29-5 = 0 be proposed to be resolved:
and let 2 be a number which differs from the Root sought, by
less than a tenth Part of itself. Then I put 2 + p = y, and I
substitute this Value in Place of it in the Equation, and thence
a new Equation arises, viz. p3 + 6p2 + lop - 1 = 0, whose Root
p is to be sought for, that it may be added to the Quotient:
viz. thus (neglecting p” + 6p2 upon the Account of their small-
ness) lop - 1 = 0, or p = 0,l is near the Truth; therefore
I write 0, 1 in the Quotient, and then suppose 0,l + q = p,
and this it’s [sic!] value I substitute, as formerly, whence results
q3 + 6, 3q2 + 11, 23q + 0,061 = 0.
And since 11,23q + 0,061 = 0 comes near to the Truth, or
since q is almost equal to -0,0054 (viz. by dividing until as
many Figures arise as there are places betwixt the first Figures
of this and the principal Quotient) I write -0,0054 in the lower
part of the Quotient, since it is negative.
And then supposing -0,0054 + r = q, I substitute this as
formerly, and thus the Operation is continued as far as you
please. But if I desire to continue the Work only to twice as
many Figures as there are in the Quotient except one, instead
of q I substitute -0,0054+r into this 6, 3q3+11, 23q+O, 061, viz.
neglecting its first Term (q3) upon the Account of it’s Smallness,