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# help

0
374
3

Idk how to workout these questions

Mar 6, 2019

#1
+6203
+1

$$x_{n+1} = \dfrac{(x_n)^3-1}{4},~x_1 = -1\\ x_2 = \dfrac{(x_1)^3-1}{4} = \dfrac{(-1)^3-1}{4} = -\dfrac 1 2\\ x_3 = \dfrac{(x_2)^3-1}{4} = \dfrac{\left(-\dfrac 1 2\right)^3 - 1}{4} = -\dfrac{9}{32}$$

$$\text{The solution occurs when the recursion has stabilized on a single value}\\ \text{so just run the iteration until }x_{n+1} = x_n \text{ to 6 decimal places}\\ \text{Use a spreadsheet or something}\\ \text{You should get }x = -0.254101$$

Now just copy this method for problem 2 which has a slightly different recursion relation.

Mar 6, 2019
#2
0

I didn't get the last one right

Mar 6, 2019
#3
+114254
+1

Second one

[(-1)^3 - 5] / 10 =  -6/ 10 = -.6

[ (-.6)^3 - 10 ] / 5   = -.5216

[ (-.5216)^3 - 5] / 10  = -0.5141909

[( -0.5141909)^3 - 5 ] / 10   = -0.5135948

[(-0.5135948)^3 - 5 ] / 10  = -0.5135475

[(-0.5135475)^3 -  5 ] / 10   = -0.5135438

{ (-0.5135438)^3 - 5 ] / 10    =  -0.513543

Mar 6, 2019