How do you use Newton's approximation method with #f(x) = x^2 - 2# to iteratively solve for the positive zero of #f(x)# accurately to within 4 decimal places using at least #6# iterations?

1 Answer
Truong-Son N.
Oct 7, 2016

It's basically a recursion problem to repeatedly guess until you converge onto your answer. You take the result you got previously and utilize it in the next iteration, getting closer each time.

The fractions get ugly, but that's what Wolfram Alpha is for.

#1)# Using #f(x) = x^2 - 2#, if we go to #n = 6#, we utilize #x_1, x_2, . . . , x_5# and ultimately acquire #x_6#.

We also use #f(x_1), f(x_2), . . . , f(x_5)# and the derivatives of #f# evaluated at #x_1, x_2, . . . , x_5#. It's easiest if you have a TI calculator (say, TI-83, TI-84?) and you set up the following input:

#color(blue)("("x - "(("x")"^2 - 2")/("2"("x")))"->x)#

which reads as...

#(x - ((x)^2 - 2)/(2(x)))->x#

i. Evaluate #(x - ((x)^2 - 2)/(2(x)))# for the #(n+1)#th guess, #x_(n+1)#.
ii. Store the result into #x# for next time, i.e. for #x_(n+1)#.

where #f = x^2 - 2# and #f' = 2x#.

Now, each time you press Enter, you evaluate for #x_(n+1)#. So...

  • The first time you enter that, you evaluate #x_1#.
  • Press Enter, and you evaluate #x_2#.
  • Press Enter again, and you evaluate #x_3#. And so on.

That's how you evaluate it numerically. But algebraically, it works out like this. Let #x_1 = 7#. Then:

#color(green)(x_2) = 7 - ((7)^2 - 2)/(2(7))#

#= 98/14 - 47/14#

#= color(green)(51/14 ~~ 3.643)#

#color(green)(x_3) = 51/14 - ((51/14)^2 - 2)/(2(51/14))#

#= 51/14 - (2601/196 - 2)/(51/7)#

#= 51/14 - (2601/196 - 392/196)(7/51)#

#= 36414/9996 - (18207/9996 - 2744/9996)#

#= (20951)/(9996)#

#= color(green)(2993/1428 ~~ 2.096)#

#color(green)(x_4) = 2993/1428 - ((2993/1428)^2 - 2)/(2(2993/1428))#

#= 2993/1428 - (8958049/2039184 - 4078368/2039184)/(2993/714)#

#= 2993/1428 - (8958049/2039184 - 4078368/2039184)(714/2993)#

#= 17916098/8548008- 4879681/8548008#

#= color(green)(13036417/8548008 ~~ 1.525)#

#2)# Continue on with #x_5# and #x_6# (using Wolfram Alpha, like I've been using) to get:

#color(green)(x_5 = 316085049734017/222870793614672 ~~ 1.418)#

#color(green)(x_6 = 199252539958223443417674291457/140892251767906875111177394848 ~~ 1.414)#

In summary, we went through the following iterations, with approximate decimals:

#x_1 -> x_2" " -> " "x_3 -> x_4 " "-> x_5 " "-> x_6#
#7 -> 3.643 -> 2.096 -> 1.525 -> 1.418 -> 1.414#

Thus, at #x_6#, we have convergence onto #sqrt2# with an error of about #5.725xx10^(-6)#. We are actually accurate to the 4th decimal place before rounding gets us a noticeably different comparison.

Our convergence was upon #1.414219cdots#, while #sqrt2 = 1.414213cdots#.

Again, doing this on a TI-83, TI-84, some calculator of that sort, would make visualizing this a bit easier.

I would evaluate the fractions using Wolfram Alpha, but do it step by step like I show for #x_2#, #x_3#, and #x_4#.

Related questions
See all questions in Using Newton's Method to Approximate Solutions to Equations
Impact of this question
1894 views around the world
You can reuse this answer
Creative Commons License