Question

\[1/1, 3/2, 7/5, 17/12, 41/29, ?\]

Answer 1

what is the question?

Answer 2

I see some fibinocci stuff going on

Answer 3

or however you spell it

Answer 4

1,1,2,3,5,8,

oops nevermind

Answer 5

I thought I see some adding of previous terms in there

Answer 6

interesting... thats possible i guess but my knowledge in this topic limited so not so sure how to link this sequence to fibonacci..

Answer 7

1/1
3/2 1+1 , 1+2
7/5 3+2, 5+2
17/12 7+5, 5+12
41/29 17+12, 12+29

Answer 8

so we have 41+29, 29+(41+29)

Answer 9

\[\frac{29+(41+29)}{41+29}\]

Answer 10

one sec checking my pattern

Answer 11

the next bottom should be 41+29
the next top should be 29+(new bottom)

Answer 12

that works nicely xD

Answer 13

\[\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\]

Answer 14

that looks simple compared to the method i am using! :

\[a_n = 1 + \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\cdots\text{(n times)}} }}\]

Answer 15

\[a_1 = 1 = 1/1\]
\[a_2 = 1 = 1 + \dfrac{1}{2} = \dfrac{3}{2}\]
\[a_3=1+\dfrac{1}{2+\dfrac{1}{2}} = 1+\dfrac{2}{5} = \dfrac{7}{5}\]
\[\cdots \]

Answer 16

easy to see.. keep doing this forever gives us the pythaogrean number \(\sqrt{2}\)

Answer 17

I was eating sorry
I'm trying currently to write my thingy in like a general form
It kinda seems like it might be tricky

Answer 18

getting a recurrence relation is easy, but an explicit relation is like finding a formula for nth digit of of sqrt(2) right ?

Answer 19

like how do you say this in a pretty way:

\[\text{ term }=\frac{ \text{ previous numerator } +2 \text{ previous denominator }}{ \text{ previous numerator }+ \text{ previous denominator }}\]

Answer 20

\(a_n = \dfrac{p_n}{q_n};~~p_1 = 1, q_1 = 1,~p_2=3,~q_2=2\)


\(p_n = 2*p_{n-1} + p_{n-2}\)
\(q_n = 2*q_{n-1} + q_{n-2}\)

Answer 21

not completely sure if that works... still checking.. .

Answer 22

well p3 is 7
and q3 is 5
so an is 7/5

Answer 23

oops a3*

Answer 24

seems like it is working

Answer 25

awesome question
and nice work on the explicit and non-explicit forms

Answer 26

this might be hard to show
\[\lim_{n \rightarrow \infty}1 + \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\cdots\text{(n times)}} }} =\sqrt{2}\]

Answer 27

I don't know enough about limits to show that is what I maen

Answer 28

not realy that hard, at least not for you im pretty sure

Answer 29

here is another interesting limit
\[\lim_{n \rightarrow \infty}1 + \dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\cdots\text{(n times)}} }} =\phi\]

\(\phi = \dfrac{\sqrt{5}+1}{2}\)

Answer 30

turns out every irrational number has this infinite continued fraction form... so we can use this to approximate irrational numbers to any decimal place

Answer 31

kinda neat
I think I remember the golden number being somehow related to Fibonacci sequence
but I can't remember

Answer 32

ratio of two consecutive fibonacci numbers approaches golden ratio as \(n\to \infty\)

Answer 33

\[\lim\limits_{n\to \infty}\dfrac{F_{n+1}}{F_n} = \phi\]

this one right ?

Answer 34

oh yeah you are right

Answer 35

read about them in the morning haha
yeah that limit is related to infinite continued fraction representaiton of \(\phi\)

Answer 36

According to this one site if I can somehow show that new ugly thing you wrote can come from
\[\phi=1+\frac{1}{\phi} \\ \text{ then I can solve this for } \phi \\ \ \phi^2-\phi-1=0 \\ \phi=\frac{1 \pm \sqrt{1+4}}{2}=\frac{1 \pm \sqrt{5}}{2}\]
when we can show that our sequence is increasing and not negative so we can throw out the negative solution

Answer 37

I called it ugly not because you wrote it in a ugly fashion but because it a seriously ugly looking fraction :p

Answer 38

That is it! the same trick works for \(\sqrt{2}\) too !

Answer 39

lol the word "ugly" is well taken :)

Answer 40

great news