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== My Goal ==
My goal is to make mathematics more accessible and fun for everyone, and a big part of that is to explain mathematics using "easy language", but this requires a balancing act between precision and comprehension.

Let me explain: there is an educational concept called the spiral, which roughly means that a subject comes around again and again, always at a higher level. For example, a young person is taught that multiplication is just repeated addition. But then a year later the subject is revisited and multiplying by negatives is taught, then decimals come along ... 

[[image:multiply-p2n3.gif|frame|
This is an illustration of 2 times -3. Observe that our toddler is (according to him) moving '''forward''' two paces at a time, but he does this three times in a negative direction. If he were stepping backwards two paces at a time while facing forwards, that would be -2 times 3. Have a look at [[http://www.mathsisfun.com/multiplying-negatives.html Multiplying by Negatives]] for a longer description.]]

== The Website ==
And that is why I have developed ([http://www.mathsisfun.com Math is Fun], or "'''Maths''' is Fun" in British English), to be a place where mathematics can be explained in a more "user-friendly" manner.

And like all people who embark on explaining Science to the general public I must at times leave out details which would only confuse, but it can be very hard to know where to draw the line.

So please forgive me, fellow Wikipedians, when I over-simplify! And correct me gently, but do correct me!

== Contact Details ==
Use this [http://www.mathsisfun.com/contact.php Contact Form]
or leave a message on the [http://www.mathisfunforum.com Math is Fun Forum]

== Test Area Stats ==

<math> \chi^2 = \sum \frac{(O - E)^2}{E} </math>

== Test Area Taylor ==

<math>f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots. </math>

<math>e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots</math>

<math>e^{x} = \sum^{\infin}_{n=0} \frac{x^n}{n!}</math>

<math>\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots</math>

<math>\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}</math>

<math>\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots</math>

<math>\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} </math>

<math>\tan x = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \cdots\quad\text{ for }|x| < \frac{\pi}{2}\!</math>

<math>\tan x = \sum^{\infin}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}\quad\text{ for }|x| < \frac{\pi}{2}\!</math>

<math>\frac{1}{1-x} = 1+x+x^2+x^3+\cdots\quad\text{ for }|x| < 1\!</math>

<math>\frac{1}{1-x} = \sum^\infin_{n=0} x^n\quad\text{ for }|x| < 1\!</math>



<math>e^{x} = \sum^{\infin}_{n=0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots</math>

<math>e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots</math>

<math>e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \cdots</math>

<math>e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right)</math>

<math>e^{ix} = \cos x +  i\sin x</math>

== Test Area Scratch ==

[[Help:Displaying_a_formula]]

<math>
r_{xy}=\frac{n\sum x_iy_i-\sum x_i\sum y_i}
{\sqrt{n\sum x_i^2-(\sum x_i)^2}~\sqrt{n\sum y_i^2-(\sum y_i)^2}}.
</math>

:<math>f(ta+(1-t)b)\geq t f(a)+(1-t)f(b)</math>


<math> 
|A| 
= a \cdot \begin{vmatrix} e & f \\ h & i \end{vmatrix} 
- b \cdot \begin{vmatrix} d & f \\ g & i \end{vmatrix} 
+ c \cdot \begin{vmatrix} d & e \\ g & h \end{vmatrix} 
</math>

<math> 
|A| 
= a \cdot \begin{vmatrix} f & g & h \\ j & k & l \\ n & o & p \end{vmatrix} 
- b \cdot \begin{vmatrix} e & g & h \\ i & k & l \\ m & o & p \end{vmatrix} 
+ c \cdot \begin{vmatrix} e & f & h \\ i & j & l \\ m & n & p \end{vmatrix} 
- d \cdot \begin{vmatrix} e & f & g \\ i & j & k \\ m & n & o \end{vmatrix} 
</math>

== Test Area Symbols ==
<math>\Rightarrow \iff \approx</math>

== Test Area Stats ==
<math>
r_{xy}=\frac{\sum\limits_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}
            {\sqrt{\sum\limits_{i=1}^n (x_i-\bar{x})^2 \sum\limits_{i=1}^n (y_i-\bar{y})^2}},
</math>

<math>
\begin{align}
\text{Variance: } \sigma^2 & = \frac{206^2 + 76^2 + (-224)^2 + 36^2 + (-94)^2}{5} \\ 
                     & = \frac{42,436 + 5,776+ 50,176+ 1,296 + 8,836}{5} \\
                     & = \frac{108,520}{5} = 21,704\\
\end{align}
</math>

== Test Area Sigma ==

<math>\sum f = 15+27+8+5 = 55</math>

<math>\sum fx = 15 \times 1 + 27 \times 2 + 8 \times 3 + 5 \times 4 = 113</math>

<math>\bar{x} = \frac{\sum fx}{\sum f} = \frac{15 \times 1 + 27 \times 2 + 8 \times 3 + 5 \times 4}{15+27+8+5} = 2.05...</math>

<math>\sum_{k=m}^n ca_k = c\sum_{k=m}^n a_k</math>

<math>\sum_{k=m}^n 6k^2 = 6\sum_{k=m}^n k^2</math>

<math>\sum_{k=m}^n (a_k+b_k) = \sum_{k=m}^n a_k + \sum_{k=m}^n b_k</math>

<math>\sum_{k=m}^n (a_k-b_k) = \sum_{k=m}^n a_k - \sum_{k=m}^n b_k</math>

<math>\sum_{k=m}^n (k+k^2) = \sum_{k=m}^n k + \sum_{k=m}^n k^2</math>


<math>\sum_{n=1}^{4} ( 2n+1) = 3+5+7+9=24</math>

<math>\sum_{i=2}^{14} \left ( \$7 \times 4 (i-1) + \$11 \times (i-2)^2  \right )</math>

<math>\sum_{i=2}^{14} \$7 \times 4 (i-1) + \sum_{i=2}^{14} \$11 \times (i-2)^2  </math>

<math>\$7 \times 4 \sum_{i=2}^{14} (i-1) + \$11 \times \sum_{i=2}^{14} (i-2)^2  </math>

<math>\$7 \times 4 \sum_{j=1}^{13} j + \$11 \sum_{k=1}^{12} k^2  </math>

<math>\$7 \times 4 \times \frac{13 \times 14}{2} + \$11 \times \frac{12 \times 13 \times 25}{6}  </math>

<math>\$7 \times 4 \times 91 + \$11 \times 650 </math>

<math>\$7 \times 364 + \$11 \times 650 </math>

<math>\$2548 + \$7150 = \$9698 \,</math>

== Test Area Partial Sums ==

<math>\sum_{k=1}^n 1 = n</math>

<math>\sum_{k=1}^n c = nc</math>

<math>\sum_{k=1}^n k = \frac{n(n+1)}{2}</math>

<math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>

<math>\sum_{k=1}^n k^3 = \left (\frac{n(n+1)}{2} \right )^2</math>

<math>\sum_{k=1}^n k^3 = \left (\sum_{k=1}^n k \right )^2</math>

<math>\sum_{k=1}^n (2k-1) = n^2</math>




<math>\sum_{k=1}^{14} k^2 = \frac{14(14+1)(2\cdot 14+1)}{6} = 1015</math>

<math>\sum_{k=0}^{n-1} (a+kd) = \frac{n}{2} (2a+(n-1)d)</math>

<math>\sum_{k=0}^{10-1} (1+k \cdot 3) = \frac{10}{2} (2 \cdot 1 +(10-1)\cdot 3)</math>

<math>\sum_{k=0}^{n-1} (ar^k) = a \left (\frac{1-r^n}{1-r}\right )</math>

<math>\sum_{k=0}^{4-1} (10 \cdot 3^k) = 10 \left (\frac{1-3^4}{1-3}\right ) = 400</math>

<math>\sum_{k=0}^{10-1} \tfrac{1}{2}(\tfrac{1}{2})^k = \frac{1}{2} \left (\frac{1 - (\frac{1}{2})^{10}}{1-\frac{1}{2}}\right ) = \frac{1}{2} \left (\frac{1 - \frac{1}{1024}}{\frac{1}{2}}\right ) = 1 - \frac{1}{1024} </math>

<math>\sum_{k=0}^{64-1} (1 \cdot 2^k) = 1 \left (\frac{1-2^{64}}{1-2}\right )</math>

<math>\sum n</math>

<math>\sum_{n=1}^4 n</math>

<math>\sum_{n=1}^4 n = 1 + 2 + 3 + 4 = 10</math>

<math>\sum_{n=1}^4 n^2 = 1^2 + 2^2 + 3^2 + 4^2 = 30</math>

<math>\sum_{i=1}^3 i(i+1) = 1\cdot 2 + 2\cdot 3 + 3\cdot 4 = 20</math>

<math>\sum_{i=3}^5 \frac{i}{i+1} = \frac{3}{4} + \frac{4}{5} + \frac{5}{6}</math>

== Test Area Binomial ==

<math>(a+b)^n=\sum_{k=0}^n{n \choose k}a^{n-k}b^{k}</math>


<math>{n \choose k}(\tfrac{1}{n})^{k} = \frac{n!}{k!(n-k)!} \cdot \frac{1}{n^k}</math>


<math>
\begin{align}
\sum_{k=0}^\infty{\frac{1}{k!}} &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}  + \frac{1}{4!} + ... \\
                                &= 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + ... \\
\end{align}
</math>






<math>
\begin{align}
(a+b)^3 & = \sum_{k=0}^3{3 \choose k}a^{3-k}b^{k} \\
        & = {3 \choose 0}a^{3-0}b^{0} + {3 \choose 1}a^{3-1}b^{1} + {3 \choose 2}a^{3-2}b^{2} + {3 \choose 3}a^{3-3}b^{3} \\
        & = 1 \cdot a^3b^0 + 3 \cdot a^2b^1 + 3 \cdot a^1b^2 + 1 \cdot a^0b^3 \\
        & = a^3 + 3a^2b + 3ab^2 + b^3 \\
\end{align}
</math>


<math>
\begin{align}
(1+\tfrac{1}{n})^n & = \sum_{k=0}^n{n \choose k}1^{n-k}(\tfrac{1}{n})^{k} \\
                   & = \sum_{k=0}^n{n \choose k}(\tfrac{1}{n})^{k} \\
                   & = \sum_{k=0}^n~\frac{n!}{k!~(n-k)!} \cdot \frac{1}{n^k} \\
\end{align}
</math>



<math>
\begin{align}
(x+5)^4 & = \sum_{k=0}^4{4 \choose k}x^{4-k}5^{k} \\
        & = {4 \choose 0}x^{4-0}5^{0} + {4 \choose 1}x^{4-1}5^{1} + {4 \choose 2}x^{4-2}5^{2} +  {4 \choose 3}x^{4-3}5^{3} +  {4 \choose 4}x^{4-4}5^{4} \\
        & = 1 \cdot x^45^0 + 4 \cdot x^35^1 + 6 \cdot x^25^2 + 4 \cdot x^15^3 + 1 \cdot x^05^4 \\
        & = x^4 + 4x^35 + 6x^25^2 + 4\cdot 5^3 + 5^4 \\
\end{align}
</math>


<math>
\begin{align}
\sum_{k=0}^{10-1} \tfrac{1}{2}(\tfrac{1}{2})^k & = \frac{1}{2}  \left (\frac{1 - (\frac{1}{2})^{10}}{1-\frac{1}{2}}\right )  \\
        & = \frac{1}{2} \left (\frac{1 - \frac{1}{1024}}{\frac{1}{2}}\right )  \\
        & = 1 - \tfrac{1}{1024}  \\
        & = 0.9990234375 \\
\end{align}
</math>

== Test Area Sigma 2 ==

<math>\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2}</math>

<math>s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}</math>


<math>\sum_{k=0}^{n-1} (ar^k) = a \left (\frac{1-r^n}{1-r}\right )</math>

<math>\sum_{k=0}^{\infty} (ar^k) = a \left (\frac{1}{1-r}\right )</math>

<math>\sum_{k=0}^{\infty} \left (\frac{1}{2} \cdot (\frac{1}{2})^k \right ) = \frac{1}{2} \left (\frac{1}{1-\frac{1}{2}}\right )</math>

<math>\sum_{k=0}^{\infty} (\tfrac{1}{2} \cdot (\tfrac{1}{2})^k ) = \tfrac{1}{2} \left (\frac{1}{1-\frac{1}{2}}\right )</math>

<math>
\begin{align}
0.999... & = 0.9 + 0.09 + 0.009 + ... \\
        & = 0.9 \cdot 0.1^0 + 0.9 \cdot 0.1^1 + 0.9 \cdot 0.1^2 + ... \\
        & = \sum_{k=0}^{\infty} 0.9 \cdot 0.1^k \\
\end{align}
</math>

<math>\sum_{k=0}^{\infty} 0.9 \times 0.1^k = 0.9 \left (\frac{1}{1-0.1}\right ) = 0.9 \left (\frac{1}{0.9}\right ) = 1</math>

== Test Area Trig ==

<math>\frac{\sqrt{a^2-b^2}}{a}</math>

<math>\frac{\sqrt{a^2+b^2}}{a}</math>

<math> \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}</math>

<math> \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}</math>

<math>c^2 = a^2 + b^2 - 2ab\cos(C) \, </math>

<math>a^2 = b^2 + c^2 - 2bc\cos(A) \, </math>

<math>b^2 = a^2 + c^2 - 2ac\cos(B) \, </math>

<math>\tan \theta = \frac{\sin \theta}{\cos \theta}</math>

<math>\frac{\sin \theta}{\cos \theta} = \frac{Opposite / Hypotenuse}{Adjacent / Hypotenuse} = \frac{Opposite}{Adjacent} = \tan \theta</math>

<math>\cot \theta = \frac{\cos \theta}{\sin \theta}</math>


<math>\sin \frac{\theta}{2} =  \pm\, \sqrt{\frac{1 - \cos \theta}{2}}</math>

<math>\cos \frac{\theta}{2} =  \pm\, \sqrt{\frac{1 + \cos\theta}{2}}</math>

<math>\tan \frac{\theta}{2} = \pm\, \sqrt{1 - \cos \theta \over 1 + \cos \theta} = \frac{\sin \theta}{1 + \cos \theta} = \frac{1-\cos \theta}{\sin \theta} = \csc \theta - \cot \theta </math>

<math>\cot \frac{\theta}{2} = \pm\, \sqrt{1 + \cos \theta \over 1 - \cos \theta} = \frac{\sin \theta}{1 - \cos \theta} = \frac{1 + \cos \theta}{\sin \theta} = \csc \theta + \cot \theta </math>

== My Test Area Other ==


<math>c = \sqrt{x^2 + y^2}</math>

<math>c = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} </math>

<math>c = \sqrt{(3-9)^2 + (2-7)^2}</math>

<math>c = \sqrt{(-6)^2 + (-5)^2} = \sqrt{36+25}= \sqrt{61} = 7.81...</math>


<math>c = \sqrt{(-3-7)^2 + (5-(-1))^2}</math>

<math>c = \sqrt{(-10)^2 + (6)^2} = \sqrt{100+36}= \sqrt{136} = 11.66...</math>


<math>c = \sqrt{(9-3)^2 + (7-2)^2}</math>

<math>c = \sqrt{6^2 + 5^2} = \sqrt{36+25}= \sqrt{61} = 7.81...</math>


<math>c = \sqrt{(x_\text{A}-x_\text{B})^2 +(y_\text{A}-y_\text{B})^2}</math>

<math>
\begin{align}
c &= \sqrt{(9-4)^2 + (2-8)^2 + (7-10)^2} \\
  &= \sqrt{25 + 36 + 9} = \sqrt{70} = 8.37... \\
\end{align}
</math>


<math>\frac{2x^2-5x-1}{x-3} = 2x + 1 + \frac{2}{x-3}</math>

<math>\frac{1}{\sqrt{2}}</math>

<math>\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} </math>

<math>\sqrt[n]{a^m} = (\sqrt[n]{a})^m</math>

<math>\sqrt[3]{27^2} = (\sqrt[3]{27})^2 = 3^2 = 9</math>

<math>\sqrt[3]{4^6} = (4)^\frac{6}{3} = 4^2 = 16</math>

<math>\sqrt[n]{a^m} = a^\frac{m}{n}</math>

<math>\sqrt[n]{a} = a^\frac{1}{n}</math>


<math>\sqrt[3]{2^3} = 2</math>

<math>\sqrt[3]{-2^3} = -2</math>

<math>\sqrt[4]{-2^4} = |-2| = 2</math>


<math>5^4=625 \ \ so \ \ 5 = \sqrt[4]{625}</math>


<math>\sqrt[n]{ab} = \sqrt[n]{a}\cdot\sqrt[n]{b}</math>

<math>\sqrt[3]{128} = \sqrt[3]{64\cdot2} = \sqrt[3]{64}\cdot\sqrt[3]{2} = 4\sqrt[3]{2}</math>


<math>\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}</math>

<math>\sqrt[3]{\frac{1}{64}} = \frac{\sqrt[3]{1}}{\sqrt[3]{64}} = \frac{1}{4}</math>


<math>\sqrt[n]{a+b} \neq \sqrt[n]{a}+\sqrt[n]{b}</math>

<math>\sqrt[n]{a-b} \neq \sqrt[n]{a}-\sqrt[n]{b}</math>

<math>\sqrt[n]{a^n+b^n} \neq a+b</math>


<math>\sqrt[n]{a^n} = a</math>

<math>\sqrt{a} \times \sqrt{a} = a</math>

<math>\sqrt[3]{a} \times \sqrt[3]{a} \times \sqrt[3]{a} = a</math>

<math>\underbrace{\sqrt[n]{a} \times \sqrt[n]{a} \times ... \times \sqrt[n]{a}}_{n\ of\ them} = a</math>

== Ellipse a and b ==

<math>h = \frac{(a-b)^2}{(a+b)^2}</math>

<math>p = \pi (a+b) \left( 1 + \sum_{n=1}^\infty  {0.5 \choose n}^2 \cdot h^n \right)\!\,</math>

<math>p = \pi (a+b) \sum_{n=0}^\infty  {0.5 \choose n}^2 h^n \!\,</math>

<math>p = \pi (a+b) \left( 1 + \frac{1}{4}h + \frac{1}{64}h^2 + \frac{1}{256}h^3 + ...\right)\!\,</math>

Ellipse perimeter, simple formula:

<math>p \approx 2 \pi \sqrt{\frac{a^2 + b^2}{2}}\!\,</math>

A better approximation by Ramanujan is:

<math>p \approx \pi \left[3(a+b) - \sqrt{(3a+b)(a+3b)}\right]\!\,</math>

<math>p\approx\pi\left(a+b\right)\left(1+\frac{3h}{10+\sqrt{4-3h}}\right)</math>

<math>e = \frac{\sqrt{a^2-b^2}}{a}</math>

<math>p = 2 a \pi \left( 1 - \sum_{i=1}^\infty  \frac{(2i)!^2}{(2^i \cdot i!)^4 } \cdot \frac{e ^{2i}}{2i-1} \right)</math>

<math>p = 2 a \pi \left[1 - \left(\frac{1}{2}\right)^2 e ^2 - \left(\frac{1 \cdot 3}{2 \cdot 4}\right)^2 \frac{e ^4}{3} - \left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^2 \frac{e ^{6}}{5} - \dots \right]</math>

<math>p = 2 a \pi \left[1 - \left(\frac{1}{2}\right)^2 e ^2 - \left(\frac{1 \cdot 3}{2 \cdot 4}\right)^2 \frac{e ^4}{3} - \dots - \left(\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n}\right)^2 \frac{e ^{2n}}{2n-1} - \dots \right]</math>


<math>Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0\;</math>

== Ellipse r and s ==

<math>h = \frac{(r-s)^2}{(r+s)^2}</math>

<math>p = \pi (r+s) \left( 1 + \sum_{n=1}^\infty  {0.5 \choose n}^2 \cdot h^n \right)\!\,</math>

<math>p = \pi (r+s) \sum_{n=0}^\infty  {0.5 \choose n}^2 h^n \!\,</math>

<math>p = \pi (r+s) \left( 1 + \frac{1}{4}h + \frac{1}{64}h^2 + \frac{1}{256}h^3 + ...\right)\!\,</math>

Ellipse perimeter, simple formula:

<math>p \approx 2 \pi \sqrt{\frac{r^2 + s^2}{2}}\!\,</math>

A better approximation by Ramanujan is:

<math>p \approx \pi \left[3(r+s) - \sqrt{(3r+s)(r+3s)}\right]\!\,</math>

<math>\varepsilon= \frac{\sqrt{r^2-s^2}}{r}</math>

<math>p = 2 r \pi \left( 1 - \sum_{i=1}^\infty  \frac{(2i)!^2}{(2^i \cdot i!)^4 } \cdot \frac{\varepsilon^{2i}}{2i-1} \right)</math>

<math>p = 2 r \pi \left[1 - \left(\frac{1}{2}\right)^2 \varepsilon^2 - \left(\frac{1 \cdot 3}{2 \cdot 4}\right)^2 \frac{\varepsilon^4}{3} - \left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^2 \frac{\varepsilon^{6}}{5} - \dots \right]</math>

<math>p = 2 r \pi \left[1 - \left(\frac{1}{2}\right)^2 \varepsilon^2 - \left(\frac{1 \cdot 3}{2 \cdot 4}\right)^2 \frac{\varepsilon^4}{3} - \dots - \left(\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n}\right)^2 \frac{\varepsilon^{2n}}{2n-1} - \dots \right]</math>


<math>Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0\;</math>

== My Test Exponents ==

<math>x^\frac{1}{n} = \sqrt[n]{x}</math>

<math>27^\frac{1}{3} = \sqrt[3]{27} = 3</math>

<math>x^\frac{m}{n} = \sqrt[n]{x^m}</math>

<math>\sqrt[n]{625} = 5</math>
<math>\sqrt[4]{625} = 5</math>
<math>\sqrt[n]{a^n} = a</math>


<math>
\begin{align}
x^\frac{m}{n} & = \sqrt[n]{x^m} \\
              & = (\sqrt[n]{x})^m \\
\end{align}
</math>

<math>x^\frac{m}{n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m </math>


<math>
\begin{align}
x^\frac{2}{3} & = \sqrt[3]{x^2} \\
              & = (\sqrt[3]{x})^2 \\
\end{align}
</math>




<math>x^\frac{m}{n} = x^{(m \times \frac{1}{n})} = (x^m)^{\frac{1}{n}} = \sqrt[n]{x^m}</math>

<math>x^\frac{m}{n} = x^{(\frac{1}{n} \times m)} = (x^{\frac{1}{n}})^m = (\sqrt[n]{x})^m</math>

== My Test Area ==

<math>\int_a^b f(x)\,dx</math>

<math>2x^2+5x+3=0\,</math>
<math>x^2-3x=0\,</math>
<math>5x-3=0\,</math>

<math>\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1</math>

<math>\sqrt{1-e^2}</math>

<math>\frac{\sqrt{2}}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = </math><math> \frac{\sqrt{2} \times \sqrt{3}}{3 \times 3} = </math><math> \frac{\sqrt{6}}{9} </math>

<math>\sqrt{\tfrac{1}{2}} = \sqrt{\tfrac{2}{4}} = \frac{\sqrt{2}}{\sqrt{4}} = \frac{\sqrt{2}}{2}</math>

<math>\sqrt{\tfrac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}</math>

<math>x^\frac{2}{3} = \sqrt[3]{x^2}</math>

<math>10^{\,\!10^{100}}</math>

<math>10^{\,\!10^{10^{1000}}}</math>

<math>{{n!} \over {(n - r)!}} \times {{1} \over {r!}} = {{n!} \over {r!(n - r)!}}</math>

<math>{{n!} \over {r!(n - r)!}} = {n \choose r}</math>

<math>f(k;n,p)={n\choose k}p^k(1-p)^{n-k}</math>

for <math>k=0,1,2,\dots,n</math> and where

<math>{n\choose k}=\frac{n!}{k!(n-k)!}</math>

<math>f(3;10,0.5)={10\choose 3}0.5^3(1-0.5)^{(10-3)}={10\choose 3}0.5^30.5^7</math>

<math>{10\choose 3}=\frac{10!}{3!(10-3)!}=\frac{10!}{3!7!}=120</math>

<math>{4\choose 2}=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1}=6</math>

<math>f(3;10,0.5)=120\times0.5^30.5^7=0.1171875</math>

<math> P(n,r) = {}^n\!P_r = {}_n\!P_r = \frac{n!}{(n-r)!}</math>

<math> C(n,r) = {}^n\!C_r = {}_n\!C_r = {n\choose r}=\frac{n!}{r!(n-r)!}</math>

<math> P(n,r) = \frac{n!}{(n-r)!}. </math>

== Test Area 2 ==
<math>\vec{i} \times \vec{j}=\vec{k}</math>

[[Image:Hexadecimal_multiplication_table.svg|right|thumb|250px|A hexadecimal [[multiplication table]]]]

<math>nC_{r}=\frac{n!}{(n-r)!(r!)}</math>

<math>\sum_{k=1}^\infty 10^{-k!}</math> = 0.110001000000000000000001000...

<math>0 < |x-\frac{p}{q}| < \frac{1}{q^n}</math>

<math>A = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\,</math>

<math>C = cos^{-1}(\frac{a^2+b^2-c^2}{2ab})</math>


<math>\varphi = \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}</math>

<math>\varphi = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}} = 1.618...</math>

<math>\frac{1}{3-\sqrt{2}}</math>

<math>\frac{1}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} = \frac{3+\sqrt{2}}{3^2-(\sqrt{2})^2} = \frac{3+\sqrt{2}}{7}</math>

<math>\frac{2-\sqrt{x}}{4-x}</math>

<math>\frac{2-\sqrt{x}}{4-x} \times \frac{2+\sqrt{x}}{2+\sqrt{x}} = \frac{2^2-(\sqrt{2})^2}{(4-x)(2+\sqrt{x})} = \frac{(4-x)}{(4-x)(2+\sqrt{x})}  = \frac{1}{2+\sqrt{x}} </math>
== Test Area Comb Perm ==
<math>n^r \,</math>

<math>{n!} \over {(n - r)!} \,</math>

<math>{n!} \over {r!(n - r)!} \,</math>

<math>{{(r + n - 1)!} \over {r!(n - 1)!}}</math>

<math>{{r + n - 1} \choose {r}} = {{(r + n - 1)!} \over {r!(n - 1)!}}</math>

<math>{{r + n - 1} \choose {r}} = {{r + n - 1} \choose {n - 1}} = {{(r + n - 1)!} \over {r!(n - 1)!}}</math>

== Test Area Sets ==

<math>
f(t)=\begin{cases}
\$50&\text{if } t\leq6\\
\$80&\text{if } t>6 \text{ and } t\leq 15\\
\$80+\$5(t-15)&\text{if } t>15\end{cases}
</math>

<math>
f(x)=\begin{cases}
x^2&\text{if } x<2\\
6&\text{if } x=2\\
10-x&\text{if } x>2 \text{ and } x\leq 6 \text{ .}\end{cases}
</math>

<math>f(x)=|x| = \begin{cases} x, & \mbox{if }  x \ge 0  \\ -x,  & \mbox{if } x < 0 \text{ .} \end{cases} </math> 

<math>h(x)= \begin{cases} 2, & \mbox{if }  x \le 1  \\ x,  & \mbox{if } x > 1 \text{ .} \end{cases} </math> 

<math>\frac{1}{x}</math> <math>\frac{1}{y}</math> <math>x^2\,</math> <math>\sqrt{y}\,</math> <math>x^n\,</math> <math>\sqrt[n]{y}</math> <math>y^\frac{1}{n}</math>

<math>e^x\,</math> <math>ln(y)\,</math> <math>a^x\,</math> <math>log_a(y)\,</math>

<math>sin(x)\,</math> <math>arcsin(y)\,</math> <math>sin^{-1}(y)\,</math>

<math>cos(x)\,</math> <math>arccos(y)\,</math> <math>cos^{-1}(y)\,</math>

<math>tan(x)\,</math> <math>arctan(y)\,</math> <math>tan^{-1}(y)\,</math>

[[Help:Displaying_a_formula]]

<math>f\colon \mathbb{N}\rightarrow\mathbb{N}</math>

<math>f\colon \{1,2,3,...\}\rightarrow\{1,2,3,...\}</math> 

<math>f\colon \mathbb{R}\rightarrow\mathbb{R}</math> 

<math>f\colon\,x\mapsto x^2</math>

<math>f(x) = x^2 \,</math>

From [[Set-builder notation]]

Examples:
* <math>\{x \mid x = x^2 \}</math> is the set <math>\{0, 1\}</math>,
* <math>\{x : x \in \mathbb{R} \land x > 0\}</math> is the set of all [[positive number|positive]] [[real numbers]],
* <math>\{k : n \in \mathbb{N} \land k = 2n\}</math> is the set of all [[even number|even]] [[natural numbers]],
* <math>\{a : \exists \ p, q \in \mathbb{Z}, q \ne 0 : a = p/q\}</math> is the set of [[rational number]]s, or numbers that can be written as the ratio of two [[integers]].

<math>\sum_{n=1}^4 (2n+1)</math>
-

== Test Area Limits ==

<math>\lim_{x\to1} \frac{x^2-1}{x-1} = 2</math>

<math>\lim_{x\to1} \frac{x^2-1}{x-1} = \lim_{x\to1} \frac{(x-1)(x+1)}{x-1} = \lim_{x\to1} (x+1)</math>

<math>\lim_{x\to1} (x+1) = 1+1 = 2</math>

<math>\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1</math>

<math>\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = e</math>

<math>\lim_{x\to10} \frac{x}{2}= 5</math>

<math>\lim_{x\to4} \frac{2-\sqrt{x}}{4-x}</math>

<math>\frac{2-\sqrt{x}}{4-x} \times \frac{2+\sqrt{x}}{2+\sqrt{x}} = \frac{2^2-(\sqrt{x})^2}{(4-x)(2+\sqrt{x})} = \frac{(4-x)}{(4-x)(2+\sqrt{x})}  = \frac{1}{2+\sqrt{x}} </math>

<math>\frac{2-\sqrt{x}}{4-x} \times \frac{2+\sqrt{x}}{2+\sqrt{x}}</math>

<math>\frac{2^2-(\sqrt{x})^2}{(4-x)(2+\sqrt{x})}</math>

<math>\frac{(4-x)}{(4-x)(2+\sqrt{x})}</math>

<math>\frac{1}{2+\sqrt{x}}</math>

<math>\lim_{x\to4} \frac{2-\sqrt{x}}{4-x} = \lim_{x\to4} \frac{1}{2+\sqrt{x}} = \frac{1}{2+\sqrt{4}} = \frac{1}{4}</math>

== Test Area Derivatives ==

<math>\frac{2x\Delta x + \Delta x^2}{\Delta x}</math>

<math>\lim_{\Delta x\to0} \frac{2x\Delta x + \Delta x^2}{\Delta x} = \lim_{\Delta x\to0} 2x + \Delta x</math>

<math>\lim_{\Delta x\to0} 2x + \Delta x = 2x</math>

<math>\lim_{\Delta x\to0} \frac{2x\Delta x + \Delta x^2}{\Delta x} = \lim_{\Delta x\to0} \frac{2x\Delta x}{\Delta x} = \lim_{\Delta x\to0} 2x = 2x</math>

<math>\lim_{\Delta x\to0} \frac{3x^2\Delta x + 3x\Delta x^2 + \Delta x^3}{\Delta x} = \lim_{\Delta x\to0} 3x^2 + 3x\Delta x + \Delta x^2 = 3x^2</math>

<math>f^\prime(x)\ = \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}</math>

<math>\frac{\Delta y}{\Delta x} = \frac{f(x+\Delta x)-f(x)}{\Delta x}</math>

<math>\frac{dy}{dx} = \frac{f(x+dx)-f(x)}{dx}\,</math>

<math>\frac{dy}{dx}\,</math>

<math>= \frac{f(x+dx)-f(x)}{dx}\,</math>

<math>= \frac{(x+dx)^2 - x^2}{dx}\,</math>

<math>= \frac{x^2 + 2x \cdot dx + dx^2 - x^2}{dx}\,</math>

<math>= \frac{2x \cdot dx + dx^2}{dx}\,</math>

<math>= 2x + dx\,</math>

<math>= 2x\,</math>

== Test Area Integrals ==

<math>\int_1^2 2x\,dx = 2^2 - 1^2 = 3</math>

<math>\int_{0.5}^1 cos(x)\,dx = sin(1) - sin(0.5) = 0.841... - 0.479... = 0.362...</math>

<math>\int_1^3 cos(x)\,dx = sin(3) - sin(1) = 0.141... - 0.841... = -0.700...</math>

<math>\int_0^1 sin(x)\,dx = -cos(1) - (-cos(0)) = -0.540... - (-1) = 0.460...</math>


<math>\int_\frac{\pi}{4}^\frac{\pi}{2} cos(x)\,dx</math>