Can All Coefficients in a Continued Fraction Be the Same Integer?

[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,...

 

[ramsey2879]

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.

 

[tacman]

?

It is possible to construct a continued fraction where all the coefficients are equal to a certain integer. This type of continued fraction is known as a regular continued fraction. In fact, the examples you have provided (Fibonacci constant and \sqrt 2 +1) are both regular continued fractions with all coefficients equal to 1 and 2, respectively.

To construct a regular continued fraction with all coefficients equal to a certain integer, we can use the recurrence relation for convergents. This recurrence relation is given by:

\frac{p_{n}}{q_{n}} = a_{n}+\frac{1}{\frac{q_{n-1}}{p_{n-1}}}

where p_{n} and q_{n} denote the numerator and denominator of the nth convergent, and a_{n} is the nth coefficient.

In order for all the coefficients to be equal to a certain integer, we can set each a_{n} equal to that integer. For example, if we want all the coefficients to be equal to 3, we would have:

\frac{p_{n}}{q_{n}} = 3+\frac{1}{\frac{q_{n-1}}{p_{n-1}}}

This would give us a regular continued fraction with all coefficients equal to 3. Similarly, we can construct regular continued fractions with all coefficients equal to 4, 5, 6, and so on.

However, it is important to note that not all regular continued fractions with equal coefficients will have a closed form expression like the examples you have provided. In fact, most regular continued fractions with equal coefficients will not have a closed form expression. This is because regular continued fractions with equal coefficients often represent irrational numbers, and irrational numbers do not have finite decimal expansions or closed form expressions.

In conclusion, it is possible to construct a continued fraction with all coefficients equal to a certain integer using the recurrence relation for convergents. However, not all of these continued fractions will have a closed form expression, as most of them will represent irrational numbers.