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. \,$

[edit] 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
}\ \$

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

MathJax testing

The following equations are represented in the HTML source code as LaTeX expressions.

The Lorenz Equations

\[\begin{aligned}

\dot{x} & = \sigma(y-x) \\


\dot{y} & = \rho x - y - xz \\

\dot{z} & = -\beta z + xy

\end{aligned} \]

The Cauchy-Schwarz Inequality

\[ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) \]

A Cross Product Formula

\[\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}


\mathbf{i} & \mathbf{j} & \mathbf{k} \\

\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\

\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0


\end{vmatrix} \]

The probability of getting \(k\) heads when flipping \(n\) coins is:

\[P(E) = {n \choose k} p^k (1-p)^{ n-k} \]

An Identity of Ramanujan

\[ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =

1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}

{1+\frac{e^{-8\pi}} {1+\ldots} } } } \]

A Rogers-Ramanujan Identity

\[ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =


\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},

\quad\quad \text{for} |q|<1. \]

Maxwell’s Equations

\[ \begin{aligned}

\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\

\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\


\nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}

\]

Finally, while display equations look good for a page of samples, the ability to mix math and text in a paragraph is also important. This expression \(\sqrt{3x-1}+(1+x)^2\) is an example of an inline equation. As you see, MathJax equations can be used this way as well, without unduly disturbing the spacing between lines.