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?

Question

Ezekiel Alexander · Accepted Answer

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... 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#.

Paisley Johnson · Answer

undefined To use Newton's approximation method to iteratively solve for the positive zero of $ f(x) = x^2 - 2 $ accurately to within 4 decimal places using at least 6 iterations, follow these steps:

1. Choose an initial guess $ x_0 $ close to the root.

2. Use the formula for Newton's iteration:

$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

3. Compute the derivative of $ f(x) $, which is $ f'(x) = 2x $.

4. Iteratively apply the formula to find $ x_{n+1} $ from $ x_n $ until the desired accuracy is achieved.

5. Continue iterating until the difference between successive approximations is less than $ 0.0001 $ (4 decimal places).

Here is an example of the iterative process:

Initial guess: $ x_0 = 2 $ (since $ f(2) = 2^2 - 2 = 2 > 0 $ and we're looking for the positive zero).

Iteration 1:
$$ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{2^2 - 2}{2 \cdot 2} = 2 - \frac{2}{4} = 2 - 0.5 = 1.5 $$

Iteration 2:
$$ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.5 - \frac{1.5^2 - 2}{2 \cdot 1.5} = 1.5 - \frac{0.25}{3} \approx 1.41667 $$

Iteration 3:
$$ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.41667 - \frac{(1.41667)^2 - 2}{2 \cdot 1.41667} \approx 1.41422 $$

Continue this process until you reach the desired accuracy or until you have completed at least 6 iterations.