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
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.
We also use
#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
- 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
#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)#
#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
Our convergence was upon
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