We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
141
3
avatar

Idk how to workout these questions

 

 Mar 6, 2019
 #1
avatar+6042 
+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
avatar
0

I didn't get the last one right

 Mar 6, 2019
 #3
avatar+103948 
+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

5 Online Users