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Idk how to workout these questions

 

 Mar 6, 2019
 #1
avatar+6046 
+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
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0

I didn't get the last one right

 Mar 6, 2019
 #3
avatar+106533 
+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                

 

 

cool cool cool

 Mar 6, 2019

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