Question

Question #4cf41

Answer 1

(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:

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.