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amistre64
Continued fractions ...
a continued fraction can be constructed by taking subsequent recpiricals
\[\frac pq=\frac{1}{q/p}\] i was wondering how they got it for like sqrt(2)
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is this a tutorial or a question?
\[\sqrt{2}=1+(\sqrt{2}-1)=1+\cfrac{1}{\frac{1}{\sqrt{2}-1}}\] \[1+\cfrac{1}{\frac{1}{\sqrt{2}-1}*\frac{\sqrt{2}+1}{\sqrt{2}+1}}=1+\cfrac{1}{\frac{\sqrt{2}+1}{2-1}}\] a little of both
i like how the bump timer fakes you out; it shows the button but says you cant when you hit it lol
lol that happens with me sometimes
how do we find the continued fraction of pi?
not that these things are unique, but i still have to wonder
hmn for that you will have to wait just for 1 min.. please
1+pi -1 = pi 1 + 1/(1/pi) -1 = pi correct?
\[\large{1+\pi -1=\pi}\] \[\large{1+\cfrac{1}{\frac{1}{\pi}}-1=\pi}\]
that is correct so far, but im unsure how that helps us proceed
you would most likely want to reciprocate the last 2 terms together
hmm\[4+\pi-4\] \[4+\frac{1}{\pi-4}\] i cant see it from there either
i spose in teh end i could just google it :) but what fun is there in that
\[3+\frac{1}{\pi-3}\] \[3+\frac{1}{.14159...}\frac{10}{10}\] \[3+\frac{10}{1+(1.4159...)-1}\] \[3+\cfrac{10}{1+\cfrac{1}{(1.4159...)-1}}\] \[3+\cfrac{10}{1+\cfrac{1}{(.4159...)}\cfrac{10}{10}}\] \[3+\cfrac{10}{1+\cfrac{10}{4+(4.159...)-4}}\] that seems to be a run on it, but you would have to know the value of pi to begin with id assume
:) that is great work...