Unknown

5.1. Approximation of Roots 167

and there arises 6, 3r2+ll, 16196r+O, 000541708 = 0 almost, or

(rejecting 6, 3r2) r = -0,00004853 almost, which I write in the

negative part of the Quotient. Finally, subducting the negative

Part of the Quotient from the affirmative, I have 2,09455147

the Quotient sought.

y3--y-5=0

2+p=y

0,1+q=p

-0.0054 + r = q

+Y3

+2Y

-5

Sum

+P”

+6p2

+1op

-1

Sum

+6, 3q2

+11 23q

+O, 061

Sum

+2,10000009

-0,00544853

+2,09455147 = y

+8+12p+6p2+p3

-4-2~

-5

-I+ lop + 6p2 + p”

+o, 001+ 0,03q + 0,3q2 + q3

+0,06+1,2+6,0

+1, +10

-1

+O, 061+ 11,23q + 6, 3q2 + q3

+0,000183708 - 0,06804r + 6, 3r2

-0,060642 + 11,23

+O, 061

SO, 000541708 + 11,16196r + 6, 3r2

-0,00004854 + s = r

Verify that the method given by Newton is essentially the method given
in the Exercises, and work Newton’s example in the modern way. Interpret
Newton’s table.

E.44. Newton’s Method and Hensel’s Lemma. Refer to Explorations
E.33 and E.43. In Newton’s Method, we start with an approximation u
and move to a new approximation v = u - f(u)/f’(u). If we are lucky, we
will end up with an approximation which is closer to the desired root, in
the sense that If( will b e smaller than if(~ Thus, the absolute value is
used to give a measure of closeness. To make this more precise, we introduce
into R a distance function d defined by

d(a, b) = la - bl.

The smaller the value of d, the closer the points a and b. This distance
function has three fundamental properties:

(i) d(a, b) = 0 if and only if a = b;