# continued fractions

by Klaus_Hoffmann
Tags: continued, fractions
 P: 83 Hi, my question is given the recurrence relation for the convergents, could we construct a continued fraction so.. $$\alpha = a_{0}+ \frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}}+....$$ 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 $$\frac{2}{\sqrt 5 -1}$$ * are two we have exactly $$\sqrt 2 +1$$ 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,......
P: 891
 Quote by Klaus_Hoffmann Hi, my question is given the recurrence relation for the convergents, could we construct a continued fraction so.. $$\alpha = a_{0}+ \frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}}+....$$ 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 $$\frac{2}{\sqrt 5 -1}$$ * are two we have exactly $$\sqrt 2 +1$$ 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,......
Something doesn't seem right. If making all the a's and b's 1 gives the Fibonacci (golden ratio) constant, wouldn't making all the a's and b's 2 simply be 2 times the Golden ration rather than $$\sqrt 2 + 1$$.

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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