+0  
 
0
667
3
avatar

Set   a=6 - 4sqrt(2),     y=Sqrt(2) - 1,   n=0, 1, 2, 3.........etc.    Iterate the following..........

 

y=[1 - (1 - y^4)^1/4] / [ 1 - (1 + y^4)^1/4], then:

a=a(1 + y)^4 - [2^(2n+3). y(1 + y +y^2)]..........converges to what?.

advanced
 Jan 16, 2016

Best Answer 

 #2
avatar+118723 
+5

I just want to rewrite your question.....

 

Set  

 

\(a=6 - 4\sqrt{2}, \qquad   y=\sqrt2 - 1, \qquad  n=0, 1, 2, 3.........etc.   \)

 

 Iterate the following..........

 

\(y=\frac{1 - (1 - y^4)^{1/4}}{ 1 - (1 + y^4)^{1/4}}\)

then:

 

\(a=a(1 + y)^4 - [2^{(2n+3)}* y(1 + y +y^2)]\)

 

..........converges to what?.

 Jan 17, 2016
 #1
avatar
0

SORRY, SEE CORRECTED FORMULA ABOVE.

 Jan 16, 2016
 #2
avatar+118723 
+5
Best Answer

I just want to rewrite your question.....

 

Set  

 

\(a=6 - 4\sqrt{2}, \qquad   y=\sqrt2 - 1, \qquad  n=0, 1, 2, 3.........etc.   \)

 

 Iterate the following..........

 

\(y=\frac{1 - (1 - y^4)^{1/4}}{ 1 - (1 + y^4)^{1/4}}\)

then:

 

\(a=a(1 + y)^4 - [2^{(2n+3)}* y(1 + y +y^2)]\)

 

..........converges to what?.

Melody Jan 17, 2016
 #3
avatar+118723 
+5

Asked again here

 

http://web2.0calc.com/questions/sorry-correction

 

I have no idea if the questions are identical of which one is the one the guest wants answered.     angry

 Jan 17, 2016

3 Online Users