Mr Pie
• Member for 5 years, 11 months
• Last seen more than 1 year ago
• Melbourne VIC, Australia
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I may be found in a very cool chatroom.

Some of my favourite original identities:

$$(a\cdot x+b\cdot y)^2+(a\cdot \overline{a_1+p}+b\cdot\overline{b_1+q})^2+(a\cdot \overline{a_2+p}+b\cdot\overline{b_2+q})^2$$ $$+(a\cdot\overline{a_1+a_2+x}+b\cdot\overline{b_1+b_2+y})^2$$ $$=$$ $$(a\cdot p+b\cdot q)^2+(a\cdot\overline{a_1+x}+b\cdot\overline{b_1+y})^2 +(a\cdot\overline{a_2+x}+b\cdot\overline{b_2+y})^2$$ $$+(a\cdot\overline{a_1+a_2+p}+b\cdot\overline{b_1+b_2+q})^2$$ $$\bowtie$$ If $ab+bc+ac=0$, then $$(2a+b)^2+(2b+c)^2+(2a+c)^2=(2a-b)^2+(2b-c)^2+(2a-c)^2$$ $$(-a+b+c)^2+(a-b+c)^2+(a+b-c)^2=3(a+b+c)^2$$ $$\frac{({-3a}+b+c)^5+(a-3b+c)^5+(a+b-3c)^5+(a+b+c)^5}{({-3a}+b+c)^3+(a-3b+c)^3+(a+b-3c)^3+(a+b+c)^3}=10(a+b+c)^2$$ $$\bowtie$$ For all $q\notin [-1,1]$, it follows for the $n^{\text{th}}$ prime $p_n$ that $$\sum_{n=1}^{\infty}\bigg(\frac 1{q^{p_n}-1}-\sum_{k=1}^{n-1}\frac 1{q^{p_np_{n-k}}-1}\bigg)=\frac 1{q(q-1)}$$ $$\bowtie$$ $$\sqrt[3]{\frac 52\bigg(\frac{\sqrt[5]2+4}{\sqrt[5]4+2}-1\bigg)}-\sqrt[3]{\frac 7{\sqrt[5]8}-\frac{2-\sqrt[5]2}{2+\sqrt[5]8}-4}=\sqrt[5]2-\sqrt[5]{\frac 14}$$ $$\frac{7-\sqrt 5+\sqrt{6\big(5-\sqrt 5\big)}}{7+\sqrt 5-\sqrt{6\big(5+\sqrt 5\big)}}=\frac 14\bigg(9-\sqrt 5+\sqrt{6\cdot\overline{5+\sqrt 5}}\bigg)$$ $$\sqrt[4]{3\bigg(1-\frac 1{\sqrt[3]4}\bigg)}\Bigg(\sqrt{\frac 3{\sqrt[3]2}-1}+\sqrt{\frac 13\cdot\overline{1+\frac 1{\sqrt[3]2}}}\Bigg)=2$$ $$\bowtie$$ Denote by $\Phi_n(x)$ the $n^{\text{th}}$ cyclotomic polynomial of x, then for all odd $pn$ given $p$ is prime, it follows that $$\cfrac{\Phi_{pn}(x)+\Phi_{pn}(-x)}{\Phi_{pn}(x)-\Phi_{pn}(-x)}=\cfrac{\Phi_{pn}\big(\frac 1x\big)+\Phi_{pn}\big({-\frac 1x}\big)}{\Phi_{pn}\big(\frac 1x\big)-\Phi_{pn}\big({-\frac 1x}\big)}$$ $$\bowtie$$ If $t\neq 0$ is a root of $(b-at)u^2+2(c-bt)u+3(d-ct)=0$, then $\dfrac 1t$ is a root of $$\frac 14\big(a+bv+cv^2+dv^3\big)=a+(b-at)v+(c-bt)v^2+(d-ct)v^3$$ $$\bowtie$$ For $a\neq 0$, let $r=\sqrt[3]{\dfrac{b^2-ac}{a^2}}$ and $t=\dfrac ba+1$. If $$d=-\frac 1a(c-bt)^2+ct$$ then $$ax^3+3bx^2+3cx+d=0$$ with $x=r^2+r-\dfrac ba$. $$\bowtie$$ If $x=\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}$ then, for any $a$ and $b$, it follows that $$(a_2x^2-a_1x+a)(b_2x^2-b_1x+b)=(a_2b+a_1b_1+ab_2)x^2-(ab_1+a_1b)x+ab$$ $$\bowtie$$ If $\varphi$ is the golden ratio and $\psi$ is its conjugate, then $$\frac{\pi}5\sqrt{\frac{2-\psi}{2+\psi}}$$ $$=\bigg(1-\frac{2-\varphi}{2+\varphi}\bigg)\bigg(1-\cfrac{2-\sqrt{2+\varphi}}{2+\sqrt{2+\varphi}}\bigg)\Bigg(1-\cfrac{2-\sqrt{2+\sqrt{2+\varphi}}}{2+\sqrt{2+\sqrt{2+\varphi}}}\Bigg)\cdots$$ $$\bowtie$$ $$2\arctan\frac 1{\sqrt[3]2}-\arctan\frac 5{3+4\sqrt[3]2}=\frac{\pi}4$$

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