# continued fraction of $\log_y x$ where $x>y$

I am looking to find a continued fraction evaluation for $$\frac{log3}{log2}$$. Originally, I was following Hollos' (personal website) directions, but got worried when I read Shank (original paper). Shank says that his algorithm works when $$x (for $$\log_y x$$), which isn't the case here ($$\frac{log3}{log2}=\log_23$$).

I saw Somos outline a very nice algorithm (on MSE) for evaluating $$\frac{\log x}{\log y}$$. He includes a sidenote that briefly states what to do when $$x>y$$, but it isn't clear to me what he means here. $$\dagger$$. Second, it seems like the results he is getting are very nice approximations, that give very nice answers, but I am trying to find the (long winded) continued fraction.$$\dagger\dagger$$

All in all though, it seems there is a fairly straightforward way to manipulate Shank's algorithm (in a very similar manner to what Somos did) so that we can find a continued fraction for $$\log_y x$$ when $$x>y$$.

$$\dagger$$ Somos states

First, initial values are $$x_0 = x>1, y_0 = y>1.$$ In step $$n$$ we have $$x_n = x^{a_n}/y^{b_n},\; y_n = y^{d_n}/x^{c_n}.$$ If $$x_n=y_n$$ then we stop since $$x^{a_n+c_n}=y^{b_n+d_n}$$ and $$\log x/\log > y=(b_n+d_n)/(a_n+c_n).$$ Otherwise, suppose $$x_n Then let $$x_{n+1}=x_n,\; y_{n+1}=y_n/x_n,\;$$ and $$\;y_{n+1}=y^{d_n+b_n}/x^{a_n+c_n}.$$ So now we get that $$a_{n+1}=a_n,\; b_{n+1}=b_n,\; c_{n+1}=c_n+a_n,\; d_{n+1}=d_n+b_n.$$ Similarly if $$x_n>y_n,$$ with the roles of $$x$$ and $$y$$ reversed. For all $$n$$, $$\;(b_n+d_n)/(a_n+c_n)\approx \log x/\log y=\log_y x.$$

"$$x_n>y_n$$, with the roles of $$x$$ and $$y$$ reversed" is a little unclear for me. If I swap x's and y's in line 4 I get $$y_{n+1}=y_2$$, $$x_{n+1}=\frac{x_n}{y_n}$$, and $$x_{n+1}=x^{a_n+c_n}/y^{d_n+b^n}$$. Which I don't think is what he means.

$$\dagger\dagger$$ Somos writes

For an example, let $$x=2,y=10.$$ Then, $$x_4=1.6=2^4/10^1,\; > y_4=1.25=10^1/2^3$$ giving $$\log_{10}2 \approx (1+1)/(4+3)=2/7$$ and next $$x_5=1.28=2^7/10^2,\;y_5=1.25=10^1/2^3$$ giving $$\log_{10}2\approx > (2+1)/(7+3)=3/10=.300.$$ Next approximation is $$4/13=.307\dots\;$$ and so on.

Here, he is getting an approximation (ie 4/13). But I am looking for a continued fraction. Something along the lines of $$\frac{a}{b+\frac{c}{d+\frac{f}{g+\cdots}}}$$. It is very possible I am misinterpreting, though

• I am only aware of a numerical method to compute the simple continued fraction , unless we have a quadratic surd. Commented Jul 10 at 10:45
• I have no idea about Hollos & Shank & Somos. When the method is not suitable for $x=3,y=2$ , you can easily take $x=3,y=4$ , then Divide that Output by $2$.
– Prem
Commented Jul 10 at 11:15
• Multiply by $2$ , it was a typo.
– Prem
Commented Jul 10 at 11:48
• Hollos' link for finding a continued fraction of logx/logy is the most accessible.
– ness
Commented Jul 10 at 11:55
• It just seems like multiplying the number before and dividing it back out at the end would give me incorrect numbers in the algorithm here? Possibly this is a completely incorrect assumption on my part --- it just doesn't seem like it would work through the part where you pick the exponent on the bounds.
– ness
Commented Jul 10 at 12:00