
#1
Jul2107, 03:25 PM

P: 83

Hi, my question is given the recurrence relation for the convergents, could we construct a continued fraction so..
[tex] \alpha = a_{0}+ \frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}}+.... [/tex] all the coefficients a's and b's are equal to a certain integer ? for example if all the coefficients (numerators and denomiators) * are one we have just the Fibonacci (Golden ratio) constant [tex] \frac{2}{\sqrt 5 1} [/tex] * are two we have exactly [tex] \sqrt 2 +1 [/tex] i the sense that expanding the 2 numbers above their continued fraction is made only by 1 or 2, but can we construct a general continued fraction with all the numbers equal to 3,4,5,6,...... 



#2
Jul2607, 11:26 AM

P: 891

Postscript. On the other hand: It seemed to me that any similar manner of construction a continued fraction will give an irrational number of some fixed value which would be a multiple of the continued fraction comprising only ones, but then n/n = 1 so I guess I made a mistake. Anyway at least the numbers are completely predefined. 


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