'; SDV($GUIButtons['ASCIIsvg'],array(1000, '\\\begin{graph} ', ' \\\end{graph}', 'width=300; height=200; xmin=-6.3; xmax=6.3; xscl=1; plot(sin(x));', '$ASCIIMathMLUrl/graph.gif"$[ASCIIsvg-graph]"')); SDV($GUIButtons['ASCIImath'],array(1000, '`', '`', 'sqrtn', '$GUIButtonDirUrlFmt/math.gif"$[Math formula (ASCIIMath)]"')); Math etc - ASCII Math ML Sandbox

ASCII Math ML Sandbox

a``graph plot(sin(x)) plot(cos(x)) end``graph

`a_b^c` \begin{graph} width=300; height=200; xmin=-1; xmax=1; xscl=1; plot(x*sin(1/x)); \end{graph}




amath int_0^3 x^2dx endamath

Testing a new automathrecognize mode. This mode is activated on any text surrounded by ama``th...enda``math amath $\overset{\mathbf E}{\simeq}$ $\simeq^{\mathbf E}$

decimal numbers: 11.00+12.0=23.00 \ \ 11.00 + 12.0 = 23.00

Solving the quadratic equation. Suppose a x^2+b x+c=0 and a!=0. We first divide by \a to get x^2+b/a x+c/a=0.

Then we complete the square and obtain x^2+b/a x+(b/(2a))^2-(b/(2a))^2+c/a=0 . The first three terms factor to give (x+b/(2a))^2=(b^2)/(4a^2)-c/a. Now we take square roots on both sides and get x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a).

Finally we move the b/(2a) to the right and simplify to get the two solutions: x_(1,2)=(-b+-sqrt(b^2-4a c))/(2a)

A set X, can I have that p in NN?

int x^2dx=x^3/3+C and sqrtx = x^(1//2)

q is a matrix ((1,2),(3,4)) and set {x | x != 1} `{x | x != 1}` $\{x | x \ne 1\}$ and |x|={(x,\if x>=0),(-x,\if x<0):}.


Theorem 1: In any right-angled triangle with sides a, b and hypotenuse c we have a^2 + b^2 = c^2.

Proof: Let h be the height drawn from the right angle to the hypotenuse. ... QED

Some standard LaTeX: $\int_0^1 x^2\,dx = \frac{1}{3}$ and as a display $\int_0^1 x^2\,dx = \frac{1}{3}$

The graph requires Firefox 2.0+ (or WinXP IE+Adobe SVGview).


 xmin=-1; xmax=3; xscl=1; ymin=-.5;
 text((1.2,0.5),"$a=\\int_0^1 x^2\\,dx=\\frac{1}{3}$","right")


theorem 2: Every positive number has a unique prime factorization.

Example: example 1

Example 2: example 2

Proof: Some would say this is obvious but a rigorous proof requires a bit of thought. QED

\begin{definition} A number p>1 is prime if p is only divisible by 1 and p. \end{definition}

\begin{lemma} For any natural numbers a,b and any prime p, if p divides a*b then p divides a or p divides b. \end{lemma}

Pythagorean theorem: 0+a^2+b^2=c^2

AA x in CC (sin^2x+cos^2x=1)

sum_(i=1)^n i^3=(sum_(i=1)^n i^2)^2

Definition of the Riemann integral: If f is continuous on the interval (a,b), except perhaps at finitely many points, then int_a^b f(x)dx=lim_(n->oo)sum_[i=1]^n f(x_i^(**))Delta x where Delta x=(b-a)/n, x_i=a+i Delta x and x_i^(**)in[x_[i-1],x_i].

And then after extending the Riemann integral to improper integrals (unbounded domains) you get the magic formula: int_0^oo e^{-x^2}dx = 1/2sqrt{pi}.

x/x=(1 if x!=0)

int_0^pi sin x dx=-cos x]_0^pi=-cos pi-(-cos0)=-(-1)-(-1)=2

Decimal numbers (right-click on the expression to see the MathML code): epsilon=.001 quad h=-.01 quad pi~~3.14159 quad weird number -0.123.456 and dot product u.v

Test: RR = uuu_{n=0}^oo[-n,n] and {0} = nnn_{n=1}^oo(- 1/n,1/n)

^^^_{i=1}^n phi_i = phi_1 ^^ phi_2 ^^ \cdots ^^ phi_n and

vvv_{i=1}^n phi_i = phi_1 vv phi_2 vv \cdots vv phi_n


int_-1^1 sqrt(1-x^2)dx = pi/2

lim_(x->a) f(x)=L <=> AA epsi > 0 EE delta > 0 AAx[0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon]

Chain fractions: 1/(1+1/(1+...))

f\left( x \right) = x_0 + a \times x



Testing Arrows and Quantum Mechanics

Singlet State in Driac Notation

This dosn't work:

( |uarr darr:) - |uarr darr:) ) / sqrt(2)

You get:

`( |uarr darr:) - |uarr darr:) ) / sqrt(2)`

Use this instead:

( |{:uarr darr:) - |{:uarr darr:) ) / sqrt(2)

`( |{:uarr darr:) - |{:uarr darr:) ) / sqrt(2)`