Question #4cf41

1 Answer
Nov 8, 2017

(Using the Newton Method (see below)
#(0.462)^(-1)~~2.16450216#

Explanation:

Newton Method:
We want to find the value of #x# for which
#color(white)("XXX")x=(0.462)^(-1)#
or, assuming #x!=0#,
#color(white)("XXX")1/x=0.462#

This is equivalent to finding the root of
#color(white)("XXX")f(x)=1/x-0.462=0#

The Newton Method tells us to start with some initial guess, #x_0#
(We've been told to use #x_0=2#).
and
to developing increasingly better approximations #x_(n+1)# using the recursive relation:
#color(white)("XXX")x_(n+1)=x_n-(f(x_n))/(f'(x_n))#

Since
#color(white)("XXX")f(x_n)=1/(x_n)-0.462=(x_n)^-1-0.462#
using the exponent rule for derivatives, we have
#color(white)("XXX")f'(x_n)=-1(x_n^(-2))=-1/(x_n)^2#

Entering all of this into a spread sheet:
enter image source here
It is normal to stop this iteration once the difference between successive approximations is less than 8 digits. (In fact, going beyond the last line displayed here has reached the limits for accuracy in my spreadsheet, so no further improvement is possible).

Note: for this case, if we were doing the calculations by hand, performing the long division would be easier.