## Wednesday, 7 July 2010

### More MathJax testing

$\tfrac12\left(a \pm \sqrt{a^2-b^2c}\right)$

$\sqrt[4]{\frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}}} = \frac{ \sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}=\tfrac12\left(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125}\right)$

$\sqrt{ \sqrt[3]{28} - \sqrt[3]{27}} = \tfrac13\left(\sqrt[3]{98} - \sqrt[3]{28} -1\right)$

$\sqrt[3]{ \sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}} } = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}$

Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':

$? = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \,$

This can be solved by noting a more general formulation:

$? = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt\mathrm{\cdots}}} \,$

Setting this to F(x) and squaring both sides gives us:

$F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt\mathrm{\cdots}} \,$

Under certain conditions infinitely nested square roots such as

$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$

represent rational numbers. This rational number can be found by realizing that x also appears under the radical sign, which gives the equation

$x = \sqrt{2+x}.$

If we solve this equation, we find that x = 2 (the second solution x = −1 doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if n > 0, then:

$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \tfrac12\left(1 + \sqrt {1+4n}\right).$

The same procedure also works to get

$\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}} = \tfrac12\left(-1 + \sqrt {1+4n}\right).$

This method will give a rational x value for all values of n such that

$n = x^2 + x. \,$

 Cube roots

In certain cases, infinitely nested cube roots such as

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

can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation

$x = \sqrt[3]{6+x}.$

If we solve this equation, we find that x = 2. More generally, we find that

$\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots}}}}$

is the real root of the equation x3 − x − n = 0 for all n > 0.

The same procedure also works to get

$\sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\cdots}}}}$

as the real root of the equation x3 + x − n = 0 for all n and x where n > 0 and |x| ≥ 1.

$\left [\begin{smallmatrix} \frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\\ -1&1&0&0&0&0&0&0 \\ 0&-1&1&0&0&0&0&0 \\ 0&0&-1&1&0&0&0&0 \\ 0&0&0&-1&1&0&0&0 \\ 0&0&0&0&-1&1&0&0 \\ 0&0&0&0&0&-1&1&0 \\ 1&1&0&0&0&0&0&0 \\ \end{smallmatrix}\right ].$

$g_{\mu \nu}~~=[S1] \times \operatorname{diag}(-1,+1,+1,+1)$

$\binom nk_q = \frac{[n]_q!}{[n-k]_q! [k]_q!}.$

$\Gamma_0(N) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbf{Z}) : c \equiv 0 \pmod{N} \right\}.$

\begin{align} &\text{Let }s = 1 + r + r^2 + r^3 + \cdots. \\[4pt] &\text{Then }rs = r + r^2 + r^3 + r^4 + \cdots. \\[4pt] &\text{Then }s - rs = 1,\text{ so }s(1 - r) = 1,\text{ and thus }s = \frac{1}{1-r}. \end{align}

$c = \operatorname{st}(x) = \operatorname{st}(y).\,$

$\mathcal L \left\{\sum_{k=0} a_k J_{\nu+k} \right\}(s)= \frac 1 \sqrt{1+s^2} \sum_{k=0} \frac{a_k}{(s+\sqrt{1+s^2})^{\nu+k}}$

$\operatorname{Der}_K(A,M)\subset \operatorname{Der}_k(A,M),\,$

$\ \ \matrix{ &&0&&0&&0 \cr&&\downarrow&&\downarrow&&\downarrow \cr &&A_1 & & B_1& &C_1 \cr &&\downarrow & &\downarrow&&\downarrow \cr &&A_2 & \to & B_2 & \to & C_2 & \cr &&\downarrow &&\downarrow&&\downarrow \cr 0&\to&A_3 & \to & B_3 & \to & C_3 }\ \$

$\sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n},$