129
votes
Is there any integral for the Golden Ratio?
Potentially interesting:
$$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$
Perhaps also worthy of consideration:
$$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{...
- 10.6k
86
votes
Is there any integral for the Golden Ratio?
In this answer, it is shown that
$$
\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi}
$$
- 328k
80
votes
Accepted
Why does this process map every fraction to the golden ratio?
Instead of representing $\frac{a}{b}$ as a fraction, represent it as the vector $\left( \begin{array}{c} a \\ b \end{array} \right)$.
Then, all you are doing to generate your sequence is repeatedly ...
- 1,190
58
votes
Is there any integral for the Golden Ratio?
An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $\phi$ factor:
$$ \frac{1}{(\sqrt{\phi\sqrt{5}})e^{2\pi/5}} = 1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}...
- 6,204
47
votes
Accepted
Is there any integral for the Golden Ratio?
For $k>0$, we have
$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty \ln \left( \frac{x^2-2kx+k^2}{x^2+2kx\cos \sqrt{\pi^2-\phi}+k^2}\right) \;\frac{\mathrm dx}{x}=\phi}}$$
I ...
- 10.5k
43
votes
Is there any integral for the Golden Ratio?
$$\int_{-1}^1 dx \frac1x \sqrt{\frac{1+x}{1-x}} \log{\left (\frac{2 x^2+2 x+1}{2 x^2-2 x+1}\right )} = 4 \pi \operatorname{arccot}{\sqrt{\phi}}$$
- 134k
42
votes
Prove that the golden ratio is irrational by contradiction
Here's one idea that works directly without showing anything about $\sqrt 5$:
We know $\varphi > 1$ so if it is rational, we could write
$$ \varphi = \frac{a}{b},$$
where $a > b > 0$ are ...
- 2,918
41
votes
How many pairs of numbers are there so they are the inverse of each other and they have the same decimal part?
So we want values of $0<x<1$ such that $x+k= \frac{\large 1}{\large x}$ for positive integer $k$, meaning $x^2+kx-1 =0$. This has a positive solution in the range for every $k$.
- 38.6k
36
votes
Accepted
A golden ratio series from a comic book
Using the series
$$
(1-4x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}x^n\tag{1}
$$
we get
$$
\begin{align}
f(x)
&=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!}x^{n+2}\\
&=\frac12\sum_{n=1}^\...
- 328k
35
votes
Is there any integral for the Golden Ratio?
Here's a series:
$$
\phi = 1 + \sum_{n=2}^\infty \frac{(-1)^{n}}{F_nF_{n-1}}
$$
where $F_n$ is the $n$th Fibonacci number.
To see this, rewrite the numerator using the identity $(-1)^n=F_{n+1}F_{n-...
- 36.4k
31
votes
Is there any integral for the Golden Ratio?
Based on the fact that $\varphi = \frac{1+\sqrt{5}}{2}$:
$$\varphi = \int_4^5 \frac32+\frac1{4\sqrt{x}} \mathrm{d}x$$
Based on the fact that $\varphi = 2\cos(\frac{\pi}{5})$:
$$\varphi = \int_{\...
- 24.4k
30
votes
Is there any integral for the Golden Ratio?
$$\int_0^{\infty} \frac{x^2}{1+x^{10}} \, \mathrm{d}x = \frac{\pi}{5 \phi}.$$
- 301
29
votes
Why does every "fibonacci like" series converge to $\phi$?
Let's define a Fibonacci-like sequence to be a sequence satisfying:
$$s_n=s_{n-1}+s_{n-2}.$$
Notice that if we take two Fibonacci-like sequences and add them together, we get another Fibonacci-like ...
- 58.9k
26
votes
A pattern appearing in the powers of $\phi$
This can be seen from the following formula:
$$L_n = \varphi^n + \frac{1}{(-\varphi)^n}$$
Where $L_n$ are the Lucas numbers, which are integers. Because the term $\dfrac{1}{(-\varphi)^n}$ alternates ...
- 10.1k
25
votes
Accepted
Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.
We can prove by induction that
if $x+\dfrac1x$ is an integer, $x^n+\dfrac1{x^n}$ will be an integer
as $$\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)=x^{n+1}+\frac1{x^{n+1}}+x^{n-1}+\frac1{x^...
- 269k
24
votes
23
votes
How to compute $\int_0^\infty \frac{1}{(1+x^{\varphi})^{\varphi}}\,dx$?
Since $\frac1\varphi=\varphi-1$,
$$
\begin{align}
\int_0^\infty\frac1{(1+x^\varphi)^\varphi}\,\mathrm{d}x
&=(\varphi-1)\int_0^\infty\frac{x^{\varphi-2}}{(1+x)^\varphi}\,\mathrm{d}x\tag{1}\\[6pt]
&...
- 328k
22
votes
Accepted
Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$
We will use a well-known$^{[1]}$ formula for sum of arctangents:
$$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$
The exact equality holds for $uv<1$, for other ...
- 45.5k
22
votes
A golden ratio series from a comic book
First Approach: Catalan Numbers
Some straightforward manipulations of the sum brings it to the form
$$
S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2(n+1))!}{(n+2)!(n+1)! 4^{2n+3}}
$$
Using the ...
- 12k
22
votes
Is there any integral for the Golden Ratio?
All the following is based on the simple fact that:
$$\phi=2 \cos \left( \frac{\pi}{5} \right)=2 \sin \left( \frac{3\pi}{10} \right)$$
These integrals are the small sample of what we can build using ...
- 30.4k
22
votes
Accepted
A Series For the Golden Ratio
First of all note that
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$
Lets rewrite your sum as the following
$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\...
- 15.3k
22
votes
Accepted
Fibonacci sequence and other metallic sequences emerged in the form of fractions
Answer to question 1) :
The generating function for the Fibonacci numbers $F_n$ is known to be
$$\dfrac{1}{1-(x+x^2)}=\underbrace{1}_{F_0}+\underbrace{1}_{F_1}x+\underbrace{2}_{F_2}x^2+\underbrace{3}_{...
- 66.5k
22
votes
Why does this process map every fraction to the golden ratio?
Let $f$ be the map that takes $a/b$ to $(a+b)/(a+2b)$. We can prove inductively that the $n$th iteration of this process gives
$$f^n(a/b) = \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b},$$
where $F_n$ ...
- 39.5k
21
votes
Accepted
Verifying a long polynomial equation in (the reciprocal of) the Golden Ratio
$$\begin{align}
&\quad 4 s^{12} + 11 s^{11} + 11 s^{10} + 9 s^9 + 7 s^8 + 5 s^7 + 3 s^6 +
s^5 + s^4 + s^3 + s^2 - s - 1 \\
=&\quad 4 s^{12}+ 4s^{11}-4 s^{10} \\
&\quad \phantom{4 s^{10}}...
- 70.2k
17
votes
Is there any integral for the Golden Ratio?
How about this one:
$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\frac{2}{\phi}-\ln \phi$$
There is an infinitely nested radical in the denominator.
A finite one is also possible:
$$...
- 30.4k
17
votes
Accepted
A conjectured continued fraction for $\phi^\phi$
Set $C_1=2-1/\phi$, then your CF can be writen as ($\phi^2=\phi+1$):
$$
\phi^{\phi}=2+\textbf{K}^{\infty}_{n=1}\left(\frac{(n+1)\left(1-n/\phi\right)/\phi}{(n+1)C_1}\right)=\ldots=2+C_1\textbf{K}^{\...
- 7,012
17
votes
Accepted
Golden ratio mod 1 distribution
Part 1.
As mentioned in the comments, the equidistribution theorem, states that in the limit any irrational value will produce an equidistributed sequence. That is, in the limit as $n \rightarrow \...
- 1,629
16
votes
How to compute $\int_0^\infty \frac{1}{(1+x^{\varphi})^{\varphi}}\,dx$?
Hint: Make $x \mapsto \dfrac{1}{x}$ and use $\phi^2 = \phi + 1$ to further simplify. The final result should be $1$.
$$\int_0^{\infty} \frac{1}{(1+x^{\phi})^{\phi}}\,dx = \int_0^{\infty} \frac{x^{\...
- 17.5k
16
votes
Fibonacci sequence and other metallic sequences emerged in the form of fractions
Following on from Jean Marie's answer, the metallic sequence
$$M_{n,k}=nM_{n,k-1}+M_{n,k-2}$$
Has the generating function
$$G_n(x)=M_{n,0}+M_{n,1}x+M_{n,2}x^2+\dots$$
Such that
$$xG_n(x)=M_{n,0}x+M_{n,...
- 19.8k
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