Question 1161304
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No.  Your entire premise is faulty.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 1\ =\ 0]


Let *[tex \Large x\ =\ j]


Ok, but the *[tex \Large j] substitution has no value that I can see.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ j^2\  -\  j\  -\  1\ =\ 0]


Isolate *[tex \Large j^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ j^2\ =\ 1\ +\ j]  That's ok


Take the square root of both sides


Here is where you go high and right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ j\ \neq\ \sqrt{1}\ +\ j]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ j\ =\ \pm\sqrt{1\ +\ j}\ \neq\ 1\ +\ j]


The idea that *[tex \LARGE j\ =\ 1\ +\ j] is ludicrous.  If that were true, you could subtract j from both sides and get 0 = 1.


The rest of your post is nonsense.


I think what you were trying to get to was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{1\,+\,\sqrt{1\,+\,\sqrt{1\,+\,\sqrt{1\,+\,\sqrt{1\,+\,...}}}}}]


In which case you need at least 10 iterations to be correct to the 4th decimal place. This method converges very slowly.


The 19th Fibonacci number divided by the 18th Fibonacci number gets you 5 decimal place accuracy with a crapload less arithmetic.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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