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{...
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  • 10.6k
86 votes

Is there any integral for the Golden Ratio?

In this answer, it is shown that $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi} $$
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  • 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 ...
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  • 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}...
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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 ...
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  • 10.5k
43 votes

Is there any integral for the Golden Ratio?

$$\int_{-1}^1 dx \frac1x \sqrt{\frac{1+x}{1-x}} \log{\left (\frac{2 x^2+2 x+1}{2 x^2-2 x+1}\right )} = 4 \pi \operatorname{arccot}{\sqrt{\phi}}$$
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  • 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 ...
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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 $k$.
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  • 38.6k
36 votes
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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}^\...
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  • 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-...
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  • 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_{\...
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  • 24.4k
30 votes

Is there any integral for the Golden Ratio?

$$\int_0^{\infty} \frac{x^2}{1+x^{10}} \, \mathrm{d}x = \frac{\pi}{5 \phi}.$$
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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 ...
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  • 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 ...
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25 votes
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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^...
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24 votes

Is there any integral for the Golden Ratio?

$$\int_0^{\infty} \frac{dx}{(1+x^\phi)^\phi}=1$$
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  • 3,685
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] &...
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  • 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 ...
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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 ...
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  • 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 ...
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  • 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\...
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  • 15.3k
22 votes
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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}_{...
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  • 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 $F_n$ ...
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  • 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}}...
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  • 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: $$...
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  • 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}^{\...
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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 \...
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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^{\...
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  • 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,...
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