Question

\[\large [3;6,\overline{1,4}] = 3+\dfrac{1}{6+\dfrac{1}{1+\dfrac{1}{4 + \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} =?\]

Answer 1

\[{{\dfrac{1}{4 + \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} := ~ ~ t \]

\[{{{4 + \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} = ~ ~ \frac{1}{t}\]
\[{{{ \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} =\frac{1}{t} - 4 = t\]And so,\[1 - 4t = t^2\Rightarrow t^2 + 4t-1 =0 \Rightarrow t = \sqrt{5} - 2 \]

Answer 2

Half of our job is done.

Answer 3

\[3 + \frac{1}{6+t} = 3 + \frac{1}{\sqrt{5} + 4 }\]

Answer 4

Neat !

Answer 5

Is there some sort of trick when it comes to continuous fractions like these? Because I'm pretty sure what I showed above is just a standard-old-boring kind of solution.

Answer 6

why didn't u let
\[1+{{\dfrac{1}{4 + \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} := ~ ~ t\]

Answer 7

these continued fractions always converge to an irrational number

Answer 8

yeah they always converge

Answer 9

is this correct?
$$
\large {{{ \dfrac{1}{\color{red}1+\dfrac{1}{4+\cdots}}}}} =\frac{1}{t} - 4 = t

$$

Answer 10

Well, I suppose we could've done that, and substituted\[\frac{1}{6 + \frac{1}{t}}\] in the end.

Answer 11

why is that equal to t, when t had a 4

Answer 12

Ah, I read the fraction wrong.

Answer 13

Thanks for pointing out. Let's make the modifications.\[{{{ \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} =\frac{1}{t} - 4 \]\[\dfrac{1}{1 + t} = \frac{1}{t} - 4\]I hope I'm right this time.

Answer 14

\[t = t+1 - 4t(t+1)\]\[4t^2 + 4t -1=0\]

Answer 15

\[t = \frac{\sqrt 2 -1}{2}\]

Answer 16

im getting the same..

Answer 17

looks good now :)

Answer 18

\[\large [3;6,\overline{1,4}] = 3+\dfrac{1}{6+\dfrac{1}{1+\dfrac{1}{4 + \dfrac{1}{1+\dfrac{1}{4+\cdots}}}}} = 3+\dfrac{1}{6+\dfrac{1}{1+t}}\]

Answer 19

should end up with
\[\dfrac{14-\sqrt{2}}{4}\]

Answer 20

the reverse of this problem is more interesting :

represent \(\sqrt{2}\) as infinite continued fraction

Answer 21

there seems to be a nice method to write any irrational number as a continued fraction
\[\sqrt{2} = [1;\overline{2}]=1+\dfrac{1}{2+\dfrac{1}{2+\cdots}}\]

Answer 22

http://gyazo.com/43e10aaa20413f2ba38cd9f2e8b6b8a6

Answer 23

The continued fraction of \(\sqrt[3]{2}\), on the other hand, doesn't really seem to help.

Answer 24

It's like an "irrational" continued fraction expansion, y'know.

Answer 25

i mean the terms in continued fraction form dont seem to be periodic

Answer 26

http://gyazo.com/48db46211dcfe4fe5dc9d7115f773841

Answer 27

wolfram seem to know how to work these continued fraction forms quickly !

Answer 28

this doesnt look periodic too
http://gyazo.com/0c07cf4c190f6f341b9fff30c33ea2b9

Answer 29

see
http://gyazo.com/512d974365626a6baa125e12ce413d5b

Answer 30

Yeah, I can understand that since \(\pi \) isn't algebraic.

Answer 31

\(e\) has some pattern to it though

Answer 32

Surprising...

Answer 33

\[e = [2;\overline{1,2n,1}]\]

where \(n=1,2,3,\ldots\)

Answer 34

its not exactly periodic but it has that surprising pattern !

Answer 35

and phi has a very nice pattern

Answer 36

very simple i mean

Answer 37

what is the continued fraction for cube root of 2

Answer 38

\(\sqrt[3]{2}\) doesn't look periodic
http://www.wolframalpha.com/input/?i=continued+fraction+2%5E%281%2F3%29

Answer 39

may be not all irrational numbers are periodic in contined fraction representation

Answer 40

right, but at least the square root of integers are

Answer 41

\[\phi = [1;\overline{1}] = 1+\dfrac{1}{1+\dfrac{1}{1+\cdots}}\]

i think this is the most nice looking continued fraction

Answer 42

yes its surprising, and e is nice as well

Answer 43

interesting.. is it easy to prove square roots of integers are periodic in continued fraction representation

Answer 44

all the examples i saw so far are periodic