User:Curiosity123

Radical
$$x=\sqrt{1+\left(x-\frac{1}{x}\right)\sqrt{1+\left(x-\frac{1}{x}\right) \sqrt{1+\left(x-\frac{1}{x}\right)\sqrt{1\cdots}}}}$$

$$x=\sqrt{-1+\left(x+\frac{1}{x}\right) \sqrt{-1+\left(x+\frac{1}{x}\right)\sqrt{-1+\left(x+\frac{1}{x}\right)\sqrt{-1\cdots}}}}$$

Tanh(1/x)
$$\tanh\left(\frac{y}{x}\right)=\frac{e^{\frac{2y}{x}}-1}{e^{\frac{2y}{x}}+1}= \frac{y}{1x+\frac{y^2}{3x+\frac{y^2}{5x+\frac{y^2}{7x+\cdots}}}}$$

$$\tanh\left(\frac{\pi}{2}\right)=\frac{e^{\pi}-1}{e^{\pi}+1}= \frac{\pi}{2+\frac{\pi^2}{6+\frac{\pi^2}{10+\frac{\pi^2}{14+\cdots}}}}$$

Which gives Ramanujan equation

ln4
$$\ln4=1+\frac{1}{3}+\frac{1}{3^3-3}+\frac{1}{6^3-6}+\frac{1}{9^3-9}+\frac{1}{12^3-12}+\cdots= 1+\frac{1}{3}+\sum_{k=1}^{\infty}\frac{1}{(3k)^3-3k}$$

Nested Radical
$$2=\sqrt[3]{3!+\sqrt[3]{3!+\sqrt[3]{3!+\sqrt[3]{3!+\cdots}}}}$$

$$3=\sqrt[3]{4!+\sqrt[3]{4!+\sqrt[3]{4!+\sqrt[3]{4!+\cdots}}}}$$

$$5=\sqrt[3]{5!+\sqrt[3]{5!+\sqrt[3]{5!+\sqrt[3]{5!+\cdots}}}}$$

$$9=\sqrt[3]{6!+\sqrt[3]{6!+\sqrt[3]{6!+\sqrt[3]{6!+\cdots}}}}$$

Strange pattern!

$$3!+2=2^3\cdots(1)$$

$$4!+3=3^3\cdots(2)$$

$$5!+5=5^3\cdots(3)$$

$$6!+9=9^3\cdots(4)$$

$$x!+y=y^3\cdots(5)$$

$$y=\sqrt[3]{x!+\sqrt[3]{x!+\sqrt[3]{x!+\sqrt[3]{x!+\cdots}}}}$$

Is there more of this type

Nested Radical
$$n-1=\sqrt[3]{n!+\sqrt[3]{n!+\sqrt[3]{n!+\sqrt[3]{n!+\cdots}}}}$$

Infinite product
p = {2,3,5,7,11,13,...}

$$\frac{7}{6}=\prod_{p}^{\infty}\frac{p^4+1}{p^4-1}$$

$$\frac{123}{122}=\prod_{p}^{\infty}\frac{p^8+1}{p^8-1}$$

Arctan(π/4)
$$\frac{\pi}{2}=\sum_{n=1}^{\infty}Arctan\left(\frac{1}{n^2}\right)$$

$$\frac{\pi}{4}=\sum_{n=1}^{\infty}Arctan\left(\frac{1}{2n^2}\right)$$

$$\frac{\pi}{4}=\sum_{n=1}^{\infty}Arctan\left(\frac{1}{n^2+n+1}\right)$$

Cube of Fibonacci series
$$1^3+1^3+2^3+3^3+5^3+\cdots{+}F_{n}^3= \frac{F_{n}F_{n+1}^2+(-1)^{n+1}F_{n-1}+1}{2}$$

π
$$\frac{\pi}{2^3\cdot{3^2}}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{s} \left[\frac{1}{e^{k\pi}-1}-\frac{1}{3(e^{2k\pi}-1)}+\frac{1}{6(e^{4k\pi}-1)}\right]$$

Logarithms of 2
$$\ln2=12\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(e^{k\pi}-1)}+ 4\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(e^{k\pi}+1)}$$

$$\ln2=\sum_{k=1}^{\infty}\frac{1}{k}\left[n\zeta(2k)-\sum_{s=1}^{n}\frac{n}{s^{2k}}- \sum_{s=1}^{\infty}\frac{n-s}{(n+s)^{2k}}\right]$$

$$\ln2=8\sum_{k=0}^{\infty}\frac{1}{2k+1}\left[\frac{1}{e^{(2k+1)\pi}-1} +\frac{1}{e^{(2k+1)+1}}\right]$$

$$\ln2=\frac{1}{2^k-(1+k)}\cdot\sum_{n=1}^{\infty}\frac{1}{n} \left[\sum_{s=1}^{2^k-2}\frac{2^k-(1+s)}{(1+s)^{2n}}\right]$$

Where k ≥ 2

$$\ln2=\frac{1}{k}\cdot\sum_{n=1}^{\infty}\frac{1}{n} \left[(2^k-1)\zeta(2n)-\sum_{s=1}^{2^k-1}\frac{2^k-s}{s^{2n}}\right]$$

Where k ≥ 1

Euler's Constant
$$1=\lim_{x \to \infty}\left[\frac{\gamma}{x}+e^{-\frac{\gamma}{x}}\prod_{y=1}^{\infty}\left(\frac{xy}{1+xy}\right) e^{\frac{1}{xy}}\right]$$

Alternating symmetric formula
$$2\times{\frac{4}{\pi}}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)^{(-1)^{n+1}}} {\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}$$

Square Root of 2
Algorithms

$$F_{n+1}=\frac{F_{n}}{a}+\frac{b}{F_{n}}=\sqrt{\frac{ab}{a-1}}$$

Setting a = 2 and b = 1 

$$F_{n+1}=\frac{F_{n}}{2}+\frac{1}{F_{n}}=\sqrt{2}$$

$$F_{n+1}=\frac{F_{n}}{a}+\frac{b}{F_{n}}+\frac{F_{n}}{c}=\sqrt{\frac{abc}{ac-(a+c)}}$$

Fo is an estimate

$$ \sqrt{2}+ \frac{1}{\sqrt{2}}= \sqrt{2+\frac{\sqrt{2}}{2^0} \sqrt{2+\frac{\sqrt{2}}{2^1} \sqrt{2+\frac{\sqrt{2}}{2^2} \sqrt{2+\frac{\sqrt{2}}{2^3} \sqrt{2+\frac{\sqrt{2}}{2^4} \sqrt{2}+\cdots}}}}}$$

$$ \frac{1}{\sqrt{2}}= \sqrt{2-\frac{\sqrt{2}}{2^0} \sqrt{2-\frac{\sqrt{2}}{2^1} \sqrt{2-\frac{\sqrt{2}}{2^2} \sqrt{2-\frac{\sqrt{2}}{2^3} \sqrt{2-\frac{\sqrt{2}}{2^4} \sqrt{2}-\cdots}}}}}$$

$$\sqrt{2}=\prod_{s=1}^{\infty}\frac{(4s)^2-1^2}{(4s)^2-2^2}$$

'''K = 0.9159655... (Catalan's Constant) '''

$$\sqrt{2}=e^{1-\frac{2K}{\pi}}\prod_{s=1}^{\infty}\left(\frac{4s-1}{4s+1}\right)^{4s}e^2$$

Continued fraction
$$\sqrt[16]p=\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+...}}}$$

$$\sqrt{\sqrt{4}+\sqrt{3}}-\sqrt{2}=\frac{1}{\sqrt{2}+\frac{1}{\sqrt{2}+\frac{1}{\sqrt{2}+...}}}$$

$${\phi^2}=\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

Continued fraction of phi
$$\phi=\frac{\sqrt{5}+1}{2}$$

$$\frac{1}{\phi^0}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}+ \frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{0}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi^2}=\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{0}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$\frac{1}{\phi^3}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi^4}=\frac{2}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$\frac{1}{\phi^5}=\frac{2}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{3}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}} $$

$$\frac{1}{\phi^6}=\frac{5}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{3}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$\frac{1}{\phi^7}=\frac{5}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{8}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}} $$

and so on ...

Unity
$$1=\phi^2-1\phi\cdots(1)$$

$$1=2\phi^2-1\phi^3\cdots(2)$$

$$1=2\phi^4-3\phi^3\cdots(3)$$

$$1=5\phi^4-3\phi^5\cdots(4)$$

$$1=5\phi^6-8\phi^5\cdots(5)$$

$$1=13\phi^6-8\phi^7\cdots(6)$$

$$1=13\phi^8-21\phi^7\cdots(7)$$

$$1=34\phi^8-21\phi^9\cdots(8)$$

'''and so on ... '''

Integers
$$\phi=\frac{\sqrt{5}+1}{2}$$

$$1=\phi^2-\phi\cdots(1)$$

$$2=-\phi^3+3\phi^2-\phi\cdots(2)$$

$$3=2\phi^4-4\phi^3+3\phi^2-\phi\cdots(3)$$

$$4=-3\phi^5+7\phi^4-4\phi^3+3\phi^2-\phi\cdots(4)$$

$$5=5\phi^6-11\phi^5+7\phi^4-4\phi^3+3\phi^2-\phi\cdots(5)$$

and so on ...

$$n=\left(\sum_{m=1}^n(-\phi)^mL_{m+1}\right)+(-\phi)^{n+1}F_n$$

Approximation
$$a\phi^{\frac{3}{2}}+b\phi^{\frac{1}{2}}-c\phi\approx{d}$$

$$(a+c)\phi^{\frac{3}{2}}+(b+d)\phi^{\frac{1}{2}}-(a+b+c)\phi\approx(a+d)$$

Example

$$1\phi^{\frac{3}{2}}+3\phi^{\frac{1}{2}}-3\phi\approx1$$

$$4\phi^{\frac{3}{2}}+4\phi^{\frac{1}{2}}-7\phi\approx2$$

$$11\phi^{\frac{3}{2}}+6\phi^{\frac{1}{2}}-15\phi\approx6$$

$$26\phi^{\frac{3}{2}}+12\phi^{\frac{1}{2}}-32\phi\approx17$$

'''and so on ... '''

Square root
$$\frac{\sqrt{2}}{\sqrt{1}+\sqrt{3}}=\sqrt{\sqrt{4}-\sqrt{3}}$$

Prime number
$$5^2+5^4=17^2+19^2\cdots(1)$$

Is there any more of this kind?

$$p{_1}^2+p{_1}^4=p_{2}^2+p_{3}^2$$

$$\sqrt{4n^2+4n-7}=2\sqrt{-1+\frac{2n-1}{2^0}\sqrt{1+\frac{2n-1}{2^1}\sqrt{1+\frac{2n-1}{2^2}\sqrt{1+...}}}}$$

$$\sqrt{4n^2+4n-7}$$

is prime from n = 2 to 12 only

n = 2, gives

$$\sqrt{17}=2\sqrt{-1+\frac{3}{2^0}\sqrt{1+\frac{3}{2^1}\sqrt{1+\frac{3}{2^2}\sqrt{1+...}}}}$$

2
$$2=\frac{\left[n^p+(n+r)^p\right]^2+\left[n^p+(n+r)^p+2n\right]^2} {n^2+\left[n^p+(n+r)^p+n\right]^2}$$

0
$$0=1^8+2^6-3^4+4^2\cdots(1)$$

Ramanujan's problem
$$2^N-7=X^2\cdots(1)$$

Solution N = 3, 4 , 5 , 7 , 15

Sum

$$3^2+4^2+5^2+7^2+15^2=18^2\cdots(1)$$

$$3+4+5+7+15=34\cdots(2)$$

$$2^{34}-2^{18}=131071^2\cdots(3)$$

$$2^{34}+2^{18}=131073^2\cdots(4)$$

$$2^{18}=\frac{131071^2+131073^2}{131072}\cdots(5)$$

Continued fraction

$$\sqrt[16]p=\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+...}}}$$

$$\frac{1}{\sqrt[16]p}=\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+...}}}$$

$$\sqrt[16]{37634}=\sqrt{\sqrt{3}+2}=\sqrt{2}+\frac{1}{\sqrt{2}+\frac{1}{\sqrt{2}+\frac{1}{\sqrt{2}+...}}}$$

$$\sqrt[16]{277727}=\sqrt{3}+\frac{1}{\sqrt{3}+\frac{1}{\sqrt{3}+\frac{1}{\sqrt{3}+...}}}$$

$${\phi^2}=\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\sqrt[16]{38925119}=\sqrt{7}+\frac{1}{\sqrt{7}+\frac{1}{\sqrt{7}+\frac{1}{\sqrt{7}+...}}}$$

phi

$$\frac{1}{\phi^0}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}+ \frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{0}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi^2}=\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{0}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$\frac{1}{\phi^3}=\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}$$

$$\frac{1}{\phi^4}=\frac{2}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$\frac{1}{\phi^5}=\frac{2}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}- \frac{3}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}} $$

$$\frac{1}{\phi^6}=\frac{5}{\sqrt{5}+\frac{1}{\sqrt{5}+\frac{1}{\sqrt{5}+...}}}- \frac{3}{\sqrt{1}+\frac{1}{\sqrt{1}+\frac{1}{\sqrt{1}+...}}}$$

$$a\phi^{\frac{3}{2}}+b\phi^{\frac{1}{2}}-c\phi\approx{d}$$

$$(a+c)\phi^{\frac{3}{2}}+(b+d)\phi^{\frac{1}{2}}-(a+b+c)\phi\approx(a+d)$$

$$\sqrt{n}=continued_fraction$$

Approximate to root

$$\frac{a}{b},\frac{2b+a}{b+a},\frac{4b+3a}{3b+2a},\frac{10b+7a}{7b+5a},\cdots=\sqrt{2}$$

$$\frac{a}{b},\frac{3b+a}{b+a},\frac{6b+4a}{4b+2a},\frac{18b+10a}{10b+6a},\cdots=\sqrt{3}$$

$$\frac{a}{b},\frac{nb+a}{b+a},\frac{2bn+a(n+1)}{b(n+1)+2a}, \frac{nb(n+3)+a(3n+1)}{b(3n+1)+a(n+3)},\cdots=\sqrt{n}$$

Recursive

$$F_{n+1}=\frac{F_{n}}{a}+\frac{b}{F_{n}}=\sqrt{\frac{ab}{a-1}}$$

$$F_{n+1}=\frac{F_{n}}{a}+\frac{b}{F_{n}}+\frac{F_{n}}{c}=\sqrt{\frac{abc}{ac-(a+c)}}$$

Fo is an estimate

$$ \sqrt{2}+ \frac{1}{\sqrt{2}}= \sqrt{2+\frac{\sqrt{2}}{2^0} \sqrt{2+\frac{\sqrt{2}}{2^1} \sqrt{2+\frac{\sqrt{2}}{2^2} \sqrt{2+\frac{\sqrt{2}}{2^3} \sqrt{2+\frac{\sqrt{2}}{2^4} \sqrt{2}+\cdots}}}}}$$

$$ \frac{1}{\sqrt{2}}= \sqrt{2-\frac{\sqrt{2}}{2^0} \sqrt{2-\frac{\sqrt{2}}{2^1} \sqrt{2-\frac{\sqrt{2}}{2^2} \sqrt{2-\frac{\sqrt{2}}{2^3} \sqrt{2-\frac{\sqrt{2}}{2^4} \sqrt{2}-\cdots}}}}}$$

$$\frac{\sqrt{2}}{\sqrt{1}+\sqrt{3}}=\sqrt{\sqrt{4}-\sqrt{3}}$$

Prime square

$$5^2+5^4=17^2+19^2$$

$$7^2-7^4=13^2+599^2-(11^2+601^2)$$

$$13^2+11^4=83^2+89^2$$

$$277^2+11^4=199^2+193^2$$

Question

$$p^2+p^4=(p_1)^2+(p_2)^2$$

$$5^2+5^4=17^2+19^2$$

$$p{_1}^2+p{_1}^4=p_{2}^2+p_{3}^2$$

Is there any more of this kind?

$$(p_1)^2+(p_2)^2=(p_3)^2+(p_4)^2$$

(p_3) if odd and not prime then it has to be a^4 for the above equation to be equal.

Example

$$47^2+43^2=63^2+89$$

$$13^2+121^2=83^2+89^2$$

$$121^2=11^4$$

True or fale

if

$$(p_1)^2+(p_2)^2=(p_3)^2+(p_4)^4$$

then

$$p_2-p_1=6$$

$$p_3=prime$$

p_3 ≠ p_4

example

$$89^2+83^2=13^2+11^4$$

$$79^2+73^2=107^2+11^4$$

$$199^2+193^2=277^2+11^4$$