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Tagged with pi continued-fractions
33 questions
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Is there a lower bound of $| \pi/4 - a/b|$ dependent only on b?
I was wondering if there exists a lower bound for $| \pi/4 -a/b|$, where $b$ is an integer and $a$ is chosen such that $a/b$ is the best approximation to $\pi/4$, which only depends on $b$. I am not ...
- 1
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1
answer
126
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Deriving the continued fraction for Pi [closed]
So I was searching online for methods to approximate Pi and found this
continued fraction that supposedly approximates to Pi when continued infinitely. I've tried searching all over the internet for ...
- 31
1
vote
0
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200
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How to explain it ? Difference between two continued fraction very small
Well I find it as a coincidence but how to explain :
$$\frac{1}{1+\frac{a}{1+\frac{a^{2}}{1+\frac{a^{3}}{1+\frac{\cdot\cdot\cdot}{a^{2n}}}}}}-\frac{1}{1+\frac{b}{1+\frac{b^{2}}{1+\frac{b^{3}}{1+\frac{\...
- 3,742
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1
answer
203
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Request for proof: Regularity of pi's continued fractions
Notice
In this post, a continued fraction is represented as follows
$$ a + \cfrac{1^2}{b+\cfrac{3^2}{b+\cfrac{5^2}{\ddots}}} = a +K^\infty_{k=1}\frac{(2k-1)^2}{b} $$
When I was checking the ...
- 426
12
votes
2
answers
451
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An uncommon continued fraction of $\frac{\pi}{2}$
I'm currently stuck with the following infinite continued fraction:
$$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$
There is an obscure clue on this: ...
- 123
10
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1
answer
411
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355/113 and small odd cubes
An important approximation to $\pi$ is given by the convergent $\frac{355}{113}$.
The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.
$$\frac{...
- 5,335
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votes
2
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270
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Irrationality of ${\pi}$ and $e^{x/y}$
I attempt to prove that ${\pi}$ is an irrational number. For this, I use the beautiful continued fraction given by Brouncker who rewrote Wallis' formula as a continued fraction, which Wallis and later ...
4
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1
answer
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Ramanujan's $\sqrt{\frac{\pi e}{2}}$ formula [duplicate]
The following identity is due to Ramanujan:
$$\DeclareMathOperator{\k}{\vphantom{\sum}\vcenter{\LARGE K}} \sqrt{\frac{\pi e}{2}}=\frac{1}{1+\k_{n=1}^\infty \frac{n}{1}}+\sum_{n=0}^\infty\frac{1}{(2n+1)...
- 405
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6
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779
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Is $22/7$ the closest to $\pi$, among fractions of denominator at most $50$?
Is $22/7$ the closest to $\pi$, among fractions of denominator at most $50$?
I am currently studying continued fractions, while I know that for all denominators at most $Q_n$, $\frac{P_n}{Q_n}$ is the ...
- 873
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1
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172
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What is the simple continued fraction of $τ$ ($2π$)?
I cannot find any information on Google or Wolfram Mathworld to answer this question. I also don't have the skills to calculate it myself so I thought it would be good if someone with this knowledge ...
- 111
14
votes
0
answers
333
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Why Is $\ln 23+\cfrac{1}{\color{red}{163}+\cfrac{1}{1+\cfrac{1}{\color{red}{41}}}}\approx\pi$
I know from reading that the Heegner number 163 yields the prime generating or Euler Lucky Number 41. Now apparently $\ln23<\pi$ and this can be shown without calculators. I noticed that
$$
\pi-\...
- 3,938
1
vote
1
answer
282
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Continued fraction of $π$ using sums of cubes
Recently I came across this identity:
$$\pi=3+\cfrac1{6+\cfrac{1^3+2^3}{6+\cfrac{1^3+2^3+3^3+4^3}{6+\cfrac{1^3+2^3+3^3+4^3+5^3+6^3}{6+\ddots}}}},\tag1$$
thus
$$\pi=3+\cfrac{1}{6+\cfrac{(1\cdot3)^2}{6+\...
- 528
7
votes
3
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418
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On expressing $\frac{\pi^n}{4\cdot 3^{n-1}}$ as a continued fraction.
It is a celebrated equation that $$\frac{\pi}{4}=\cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}$$
However, there are two other conjectured equations that I found which, if true (...
- 9,595
2
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0
answers
81
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A conjectured continued fraction involving non-polynomial patterns
After having been devoting some time for many years to experimental mathematics, I am thinking to publish the details of some of my most fruitful computing workflows for discovering identities. I ...
- 1,013
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109
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How to prove, that this chain fraction equals $\frac{4}{\pi}$?
I would like to ask how do I prove this chain fraction equality?
\begin{equation}
\frac{4}{\pi} = 1+{\raise{-0.8ex}\mathop{\huge\mathrm{K}}_{n=1}^{\infty}}\left[\frac{(2n-1)^2}{2}\right]= 1+\cfrac{1}...
user675788
5
votes
2
answers
1k
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Proof of this formula for $\sqrt{e\pi/2}$ and similar formulas.
\begin{align}
\sqrt{\frac{e\pi}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}
\end{align}
as seen here.
...
- 2,811
17
votes
1
answer
891
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The most complex formula for the golden ratio $\varphi$ that I have ever seen. How was it achieved?
I am fascinated by the following formula for the golden ratio $\varphi$: $$\Large\varphi = \frac{\sqrt{5}}{1 + \left(5^{3/4}\left(\frac{\sqrt{5} - 1}{2}\right)^{5/2} - 1\right)^{1/5}} - \frac{1}{e^{2\...
- 9,595
4
votes
1
answer
191
views
$\sqrt{\frac{\pi e}{2}}=\frac{1}{1+\mathrm{K}_{i=1}^{\infty}{\frac{i}{1}}}+\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!!}}$ implies $\sqrt{\pi e/2}\notin Q$?
On the OEIS Wiki immediately after the formula
$$\sqrt{\frac{\pi e}{2}}=\frac{1}{1+\mathrm{K}_{i=1}^{\infty}{\frac{i}{1}}}+\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!!}}$$
(where I am using $\mathrm{K}$ as ...
- 3,196
0
votes
1
answer
167
views
Infinite fraction
I wasn't sure if this belongs to programming but the problems seems more mathematical so i posted here.
I was looking at pi and made a function(Haskell) similar to it without squaring
...
- 23
4
votes
0
answers
107
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A conjectured continued fraction related to a 2F1 hypergeometric function (and a formula for π)
In a previous question, some months ago, I described a continued fraction for which I was trying to find an identity. I finally found what I was looking for, and I will publish an answer to this ...
- 1,013
1
vote
2
answers
341
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Is there infinite $1$ in the continued fraction of $\pi$?
I asked this question just for curiosity! I guess that it's an unsolved problem, but I can't find any reference that mentions that.
You can see http://oeis.org/A001203 and How to find continued ...
- 485
1
vote
1
answer
340
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Intuition behind irrationality of $\pi$
Would the existence of arbitrarily large terms in the continued fraction expansion of $\pi$ imply its irrationality?
Edit: I completely changed the question to remove speculation on my part.
- 779
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400
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Infinitely nested radical expansions for real numbers
Conjecture. For any real number $x \in (0,1]$ there exists a unique expansion in the form $x=-2+\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\cdots}}}$ with $a_k$ being natural numbers from the set $(2,3,4,5,6)$. ...
- 32.2k
16
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2
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528
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What's the formula for this series for $\pi$?
These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}
= \frac{4}{1} - \...
- 56.2k
14
votes
1
answer
668
views
Continued fraction estimation of error in Leibniz series for $\pi$.
The following arctan formula for $\pi$ is quite well known
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\tag{1}$$ and bears the name of Madhava-Gregory-Leibniz series after ...
- 89.7k
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765
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"Bizarre" continued fraction of Ramanujan! But where's the proof?
$$\frac{e^\pi-1}{e^\pi+1}=\cfrac\pi{2+\cfrac{\pi^2}{6+\cfrac{\pi^2}{10+\cfrac{\pi^2}{14+...}}}}$$
"Bizarre" continued fraction of Ramanujan! But where's the proof? i have no training in continued ...
- 1,804
2
votes
1
answer
541
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Continued fraction expansion of Pi (oeis A001203). [duplicate]
I would like to understand how you get the numbers
$$3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+...}}}}$$
i.e. $\{3,7,15,1,292,...\}$ (A001203).
In the comments of A046965 is explained a method ...
- 7,926
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1
answer
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Chinese estimate for $\pi$. Were they lucky?
The famous Chinese estimate $\pi\approx\frac{355}{113}$ is good. I think that is too good. As a continued fraction,
$$\pi=[3; 7,15,1,292,\ldots].$$
That $292$ is a bit too big. Is there a reason for ...
- 66.3k
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votes
4
answers
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How to find continued fraction of pi
I have always been amazed by the continued fractions for $\pi$. For example some continued fractions for pi are:
$\pi=[3:7,15,1,292,.....]$ and many others given here.
Similarly some nice continued ...
- 6,208
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5
answers
1k
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Is π unusually close to 7920/2521?
EDIT: One can look at a particular type of approximation to $\pi$ based on comparing radians to degrees. If you try to approximate $\pi$ by fractions of the form $180n/(360k+1)$, you can find that $\...
- 1,891
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1
answer
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A strange "pattern" in the continued fraction convergents of pi?
From the simple continued fraction of $\pi$, one gets the convergents,
$$p_n = \frac{3}{1}, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102}, \frac{104348}{33215}, \frac{208341}{...
- 56.2k
11
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1
answer
356
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Request for a proof of the following continued-fraction identity
I have been poring over many texts about continued fractions, but none of them seem to be helping me to prove the following beautiful continued-fraction identity (I am nowhere close):
$$
\cfrac{1}{\...
- 19.9k
3
votes
2
answers
346
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Unique continued fraction
If $x$ is a uniformly random number in $[0,1]$, what distribution should the $n$-th term in its continued fraction expansion follow?
What is the expected vale of $a_n$ in $[a_0;a_1,a_2,\dots]$?
Here ...
user25597