# Tag Info

• 6,204
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

### 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

### 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

### 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
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• 24.4k

### 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

### 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

### 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
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• 328k
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### 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

### 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

### 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
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• 66.5k

### 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
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• 30.4k
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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
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,...