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 $x<y$ (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 $x_n<y_n.$ 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

  • $\begingroup$ I am only aware of a numerical method to compute the simple continued fraction , unless we have a quadratic surd. $\endgroup$
    – Peter
    Commented Jul 10 at 10:45
  • $\begingroup$ 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 $2$. $\endgroup$
    – Prem
    Commented Jul 10 at 11:15
  • $\begingroup$ Multiply by $2$ , it was a typo. $\endgroup$
    – Prem
    Commented Jul 10 at 11:48
  • $\begingroup$ Hollos' link for finding a continued fraction of logx/logy is the most accessible. $\endgroup$
    – ness
    Commented Jul 10 at 11:55
  • $\begingroup$ 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. $\endgroup$
    – ness
    Commented Jul 10 at 12:00


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