User:MathsIsFun

Wikipedia:Babel

Other

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 ...

File:Multiply-p2n3.gif

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 [Multiplying by Negatives] for a longer description.

The Website

And that is why I have developed (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 Contact Form or leave a message on the Math is Fun Forum

Test Area Stats

${\chi }^2=\sum \frac{(O-E{)}^2}{E}$

Test Area Taylor

$f(x)=f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+\cdots .$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

$e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$

$\sin x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$

$\cos x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n)!}{x}^{2n}$

$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots \text{ for }|x|<\frac{\pi }{2}$

$\tan x=\sum _{n=1}^{\mathrm{\infty }}\frac{B_{2n}(-4{)}^n(1-4^n)}{(2n)!}{x}^{2n-1}\text{ for }|x|<\frac{\pi }{2}$

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots \text{ for }|x|<1$

$\frac{1}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{x}^n\text{ for }|x|<1$

$e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots$

$e^{ix}=1+ix+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+\frac{(ix{)}^4}{4!}+\frac{(ix{)}^5}{5!}+\cdots$

$e^{ix}=1+ix-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}-\cdots$

$e^{ix}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots )+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots )$

$e^{ix}=\cos x+i\sin x$

Test Area Scratch

Help:Displaying_a_formula

$r_{xy}=\frac{n\sum x_i{y}_i-\sum x_i\sum y_i}{\sqrt{n\sum x_i^2-(\sum x_i{)}^2}\sqrt{n\sum y_i^2-(\sum y_i{)}^2}}.$

$f(ta+(1-t)b)\ge tf(a)+(1-t)f(b)$

$|A|=a\cdot \left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\cdot \left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\cdot \left|\begin{array}{cc}d& e\\ g& h\end{array}\right|$

$|A|=a\cdot \left|\begin{array}{ccc}f& g& h\\ j& k& l\\ n& o& p\end{array}\right|-b\cdot \left|\begin{array}{ccc}e& g& h\\ i& k& l\\ m& o& p\end{array}\right|+c\cdot \left|\begin{array}{ccc}e& f& h\\ i& j& l\\ m& n& p\end{array}\right|-d\cdot \left|\begin{array}{ccc}e& f& g\\ i& j& k\\ m& n& o\end{array}\right|$

Test Area Symbols

$\Rightarrow ⟺\approx$

Test Area Stats

$r_{xy}=\frac{\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sqrt{\sum _{i=1}^n(x_i-\overline{x}{)}^2\sum _{i=1}^n(y_i-\overline{y}{)}^2}},$

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

Test Area Sigma

$\sum f=15+27+8+5=55$

$\sum fx=15\times 1+27\times 2+8\times 3+5\times 4=113$

$\overline{x}=\frac{\sum fx}{\sum f}=\frac{15\times 1+27\times 2+8\times 3+5\times 4}{15+27+8+5}=2.05...$

$\sum _{k=m}^{n}c{a}_k=c\sum _{k=m}^n{a}_k$

$\sum _{k=m}^{n}6k^2=6\sum _{k=m}^n{k}^2$

$\sum _{k=m}^n(a_k+b_k)=\sum _{k=m}^n{a}_k+\sum _{k=m}^n{b}_k$

$\sum _{k=m}^n(a_k-b_k)=\sum _{k=m}^n{a}_k-\sum _{k=m}^n{b}_k$

$\sum _{k=m}^n(k+k^2)=\sum _{k=m}^{n}k+\sum _{k=m}^n{k}^2$

$\sum _{n=1}^4(2n+1)=3+5+7+9=24$

$\sum _{i=2}^{14}(\mathrm{\$}7\times 4(i-1)+\mathrm{\$}11\times (i-2{)}^2)$

$\sum _{i=2}^{14}\mathrm{\$}7\times 4(i-1)+\sum _{i=2}^{14}\mathrm{\$}11\times (i-2{)}^2$

$\mathrm{\$}7\times 4\sum _{i=2}^{14}(i-1)+\mathrm{\$}11\times \sum _{i=2}^{14}(i-2{)}^2$

$\mathrm{\$}7\times 4\sum _{j=1}^{13}j+\mathrm{\$}11\sum _{k=1}^{12}{k}^2$

$\mathrm{\$}7\times 4\times \frac{13\times 14}{2}+\mathrm{\$}11\times \frac{12\times 13\times 25}{6}$

$\mathrm{\$}7\times 4\times 91+\mathrm{\$}11\times 650$

$\mathrm{\$}7\times 364+\mathrm{\$}11\times 650$

$\mathrm{\$}2548+\mathrm{\$}7150=\mathrm{\$}9698$

Test Area Partial Sums

$\sum _{k=1}^{n}1=n$

$\sum _{k=1}^{n}c=nc$

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

$\sum _{k=1}^n{k}^3={\left(\frac{n(n+1)}{2}\right)}^2$

$\sum _{k=1}^n{k}^3={\left(\sum _{k=1}^{n}k\right)}^2$

$\sum _{k=1}^n(2k-1)=n^2$

$\sum _{k=1}^{14}{k}^2=\frac{14(14+1)(2\cdot 14+1)}{6}=1015$

$\sum _{k=0}^{n-1}(a+kd)=\frac{n}{2}(2a+(n-1)d)$

$\sum _{k=0}^{10-1}(1+k\cdot 3)=\frac{10}{2}(2\cdot 1+(10-1)\cdot 3)$

$\sum _{k=0}^{n-1}(a{r}^k)=a\left(\frac{1-r^n}{1-r}\right)$

$\sum _{k=0}^{4-1}(10\cdot 3^k)=10\left(\frac{1-3^4}{1-3}\right)=400$

$\sum _{k=0}^{10-1}\frac{1}{2}(\frac{1}{2}{)}^k=\frac{1}{2}\left(\frac{1-(\frac{1}{2}{)}^{10}}{1-\frac{1}{2}}\right)=\frac{1}{2}\left(\frac{1-\frac{1}{1024}}{\frac{1}{2}}\right)=1-\frac{1}{1024}$

$\sum _{k=0}^{64-1}(1\cdot 2^k)=1\left(\frac{1-2^{64}}{1-2}\right)$

$\sum n$

$\sum _{n=1}^{4}n$

$\sum _{n=1}^{4}n=1+2+3+4=10$

$\sum _{n=1}^4{n}^2=1^2+2^2+3^2+4^2=30$

$\sum _{i=1}^{3}i(i+1)=1\cdot 2+2\cdot 3+3\cdot 4=20$

$\sum _{i=3}^5\frac{i}{i+1}=\frac{3}{4}+\frac{4}{5}+\frac{5}{6}$

Test Area Binomial

$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$

$(\genfrac{}{}{0ex}{}{n}{k})(\frac{1}{n}{)}^k=\frac{n!}{k!(n-k)!}\cdot \frac{1}{n^k}$

$\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}& =\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...\\ & =1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+...\\ \end{array}$

$\begin{array}{rl}(a+b{)}^3& =\sum _{k=0}^3(\genfrac{}{}{0ex}{}{3}{k})a^{3-k}{b}^k\\ & =(\genfrac{}{}{0ex}{}{3}{0})a^{3-0}{b}^0+(\genfrac{}{}{0ex}{}{3}{1})a^{3-1}{b}^1+(\genfrac{}{}{0ex}{}{3}{2})a^{3-2}{b}^2+(\genfrac{}{}{0ex}{}{3}{3})a^{3-3}{b}^3\\ & =1\cdot a^3{b}^0+3\cdot a^2{b}^1+3\cdot a^1{b}^2+1\cdot a^0{b}^3\\ & =a^3+3a^{2}b+3a{b}^2+b^3\\ \end{array}$

$\begin{array}{rl}(1+\frac{1}{n}{)}^n& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})1^{n-k}(\frac{1}{n}{)}^k\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(\frac{1}{n}{)}^k\\ & =\sum _{k=0}^n\frac{n!}{k!(n-k)!}\cdot \frac{1}{n^k}\\ \end{array}$

$\begin{array}{rl}(x+5{)}^4& =\sum _{k=0}^4(\genfrac{}{}{0ex}{}{4}{k})x^{4-k}{5}^k\\ & =(\genfrac{}{}{0ex}{}{4}{0})x^{4-0}{5}^0+(\genfrac{}{}{0ex}{}{4}{1})x^{4-1}{5}^1+(\genfrac{}{}{0ex}{}{4}{2})x^{4-2}{5}^2+(\genfrac{}{}{0ex}{}{4}{3})x^{4-3}{5}^3+(\genfrac{}{}{0ex}{}{4}{4})x^{4-4}{5}^4\\ & =1\cdot x^4{5}^0+4\cdot x^3{5}^1+6\cdot x^2{5}^2+4\cdot x^1{5}^3+1\cdot x^0{5}^4\\ & =x^4+4x^{3}5+6x^2{5}^2+4\cdot 5^3+5^4\\ \end{array}$

$\begin{array}{rl}\sum _{k=0}^{10-1}\frac{1}{2}(\frac{1}{2}{)}^k& =\frac{1}{2}\left(\frac{1-(\frac{1}{2}{)}^{10}}{1-\frac{1}{2}}\right)\\ & =\frac{1}{2}\left(\frac{1-\frac{1}{1024}}{\frac{1}{2}}\right)\\ & =1-\frac{1}{1024}\\ & =0.9990234375\\ \end{array}$

Test Area Sigma 2

$\sigma =\sqrt{\frac{1}{N}\sum _{i=1}^N(x_i-\mu {)}^2}$

$s=\sqrt{\frac{1}{N-1}\sum _{i=1}^N(x_i-\bar{x}{)}^2}$

$\sum _{k=0}^{n-1}(a{r}^k)=a\left(\frac{1-r^n}{1-r}\right)$

$\sum _{k=0}^{\mathrm{\infty }}(a{r}^k)=a\left(\frac{1}{1-r}\right)$

$\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{2}\cdot (\frac{1}{2}{)}^k)=\frac{1}{2}\left(\frac{1}{1-\frac{1}{2}}\right)$

$\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{2}\cdot (\frac{1}{2}{)}^k)=\frac{1}{2}\left(\frac{1}{1-\frac{1}{2}}\right)$

$\begin{array}{rl}0.999...& =0.9+0.09+0.009+...\\ & =0.9\cdot 0.1^0+0.9\cdot 0.1^1+0.9\cdot 0.1^2+...\\ & =\sum _{k=0}^{\mathrm{\infty }}0.9\cdot 0.1^k\\ \end{array}$

$\sum _{k=0}^{\mathrm{\infty }}0.9\times 0.1^k=0.9\left(\frac{1}{1-0.1}\right)=0.9\left(\frac{1}{0.9}\right)=1$

Test Area Trig

$\frac{\sqrt{a^2-b^2}}{a}$

$\frac{\sqrt{a^2+b^2}}{a}$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$c^2=a^2+b^2-2ab\cos (C)$

$a^2=b^2+c^2-2bc\cos (A)$

$b^2=a^2+c^2-2ac\cos (B)$

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

$\frac{\sin \theta }{\cos \theta }=\frac{Opposite/Hypotenuse}{Adjacent/Hypotenuse}=\frac{Opposite}{Adjacent}=\tan \theta$

$\cot \theta =\frac{\cos \theta }{\sin \theta }$

$\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$

$\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

$\tan \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\frac{\sin \theta }{1+\cos \theta }=\frac{1-\cos \theta }{\sin \theta }=\csc \theta -\cot \theta$

$\cot \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\frac{\sin \theta }{1-\cos \theta }=\frac{1+\cos \theta }{\sin \theta }=\csc \theta +\cot \theta$

My Test Area Other

$c=\sqrt{x^2+y^2}$

$c=\sqrt{(-2{)}^2+3^2}=\sqrt{4+9}=\sqrt{13}$

$c=\sqrt{(3-9{)}^2+(2-7{)}^2}$

$c=\sqrt{(-6{)}^2+(-5{)}^2}=\sqrt{36+25}=\sqrt{61}=7.81...$

$c=\sqrt{(-3-7{)}^2+(5-(-1){)}^2}$

$c=\sqrt{(-10{)}^2+(6{)}^2}=\sqrt{100+36}=\sqrt{136}=11.66...$

$c=\sqrt{(9-3{)}^2+(7-2{)}^2}$

$c=\sqrt{6^2+5^2}=\sqrt{36+25}=\sqrt{61}=7.81...$

$c=\sqrt{(x_{\text{A}}-x_{\text{B}}{)}^2+(y_{\text{A}}-y_{\text{B}}{)}^2}$

$\begin{array}{rl}c& =\sqrt{(9-4{)}^2+(2-8{)}^2+(7-10{)}^2}\\ & =\sqrt{25+36+9}=\sqrt{70}=8.37...\\ \end{array}$

$\frac{2x^2-5x-1}{x-3}=2x+1+\frac{2}{x-3}$

$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$\sqrt[n]{a^m}=(\sqrt[n]{a}{)}^m$

$\sqrt[3]{27^2}=(\sqrt[3]{27}{)}^2=3^2=9$

$\sqrt[3]{4^6}=(4{)}^{\frac{6}{3}}=4^2=16$

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt[n]{a}=a^{\frac{1}{n}}$

$\sqrt[3]{2^3}=2$

$\sqrt[3]{-2^3}=-2$

$\sqrt[4]{-2^4}=|-2|=2$

$5^4=625so5=\sqrt[4]{625}$

$\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$

$\sqrt[3]{128}=\sqrt[3]{64\cdot 2}=\sqrt[3]{64}\cdot \sqrt[3]{2}=4\sqrt[3]{2}$

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$\sqrt[3]{\frac{1}{64}}=\frac{\sqrt[3]{1}}{\sqrt[3]{64}}=\frac{1}{4}$

$\sqrt[n]{a+b}\ne \sqrt[n]{a}+\sqrt[n]{b}$

$\sqrt[n]{a-b}\ne \sqrt[n]{a}-\sqrt[n]{b}$

$\sqrt[n]{a^n+b^n}\ne a+b$

$\sqrt[n]{a^n}=a$

$\sqrt{a}\times \sqrt{a}=a$

$\sqrt[3]{a}\times \sqrt[3]{a}\times \sqrt[3]{a}=a$

$\underset{nofthem}{\underbrace{\sqrt[n]{a}\times \sqrt[n]{a}\times ...\times \sqrt[n]{a}}}=a$

Ellipse a and b

$h=\frac{(a-b{)}^2}{(a+b{)}^2}$

$p=\pi (a+b)(1+\sum _{n=1}^{\mathrm{\infty }}{(\genfrac{}{}{0ex}{}{0.5}{n})}^2\cdot h^n)$

$p=\pi (a+b)\sum _{n=0}^{\mathrm{\infty }}{(\genfrac{}{}{0ex}{}{0.5}{n})}^2{h}^n$

$p=\pi (a+b)(1+\frac{1}{4}h+\frac{1}{64}{h}^2+\frac{1}{256}{h}^3+...)$

Ellipse perimeter, simple formula:

$p\approx 2\pi \sqrt{\frac{a^2+b^2}{2}}$

A better approximation by Ramanujan is:

$p\approx \pi [3(a+b)-\sqrt{(3a+b)(a+3b)}]$

$p\approx \pi (a+b)(1+\frac{3h}{10+\sqrt{4-3h}})$

$e=\frac{\sqrt{a^2-b^2}}{a}$

$p=2a\pi (1-\sum _{i=1}^{\mathrm{\infty }}\frac{(2i){!}^2}{(2^i\cdot i!{)}^4}\cdot \frac{e^{2i}}{2i-1})$

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

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

$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Ellipse r and s

$h=\frac{(r-s{)}^2}{(r+s{)}^2}$

$p=\pi (r+s)(1+\sum _{n=1}^{\mathrm{\infty }}{(\genfrac{}{}{0ex}{}{0.5}{n})}^2\cdot h^n)$

$p=\pi (r+s)\sum _{n=0}^{\mathrm{\infty }}{(\genfrac{}{}{0ex}{}{0.5}{n})}^2{h}^n$

$p=\pi (r+s)(1+\frac{1}{4}h+\frac{1}{64}{h}^2+\frac{1}{256}{h}^3+...)$

Ellipse perimeter, simple formula:

$p\approx 2\pi \sqrt{\frac{r^2+s^2}{2}}$

A better approximation by Ramanujan is:

$p\approx \pi [3(r+s)-\sqrt{(3r+s)(r+3s)}]$

$\epsilon =\frac{\sqrt{r^2-s^2}}{r}$

$p=2r\pi (1-\sum _{i=1}^{\mathrm{\infty }}\frac{(2i){!}^2}{(2^i\cdot i!{)}^4}\cdot \frac{{\epsilon }^{2i}}{2i-1})$

$p=2r\pi [1-{\left(\frac{1}{2}\right)}^2{\epsilon }^2-{\left(\frac{1\cdot 3}{2\cdot 4}\right)}^2\frac{{\epsilon }^4}{3}-{\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)}^2\frac{{\epsilon }^6}{5}-\dots ]$

$p=2r\pi [1-{\left(\frac{1}{2}\right)}^2{\epsilon }^2-{\left(\frac{1\cdot 3}{2\cdot 4}\right)}^2\frac{{\epsilon }^4}{3}-\dots -{\left(\frac{1\cdot 3\cdot 5\cdot \dots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \dots \cdot 2n}\right)}^2\frac{{\epsilon }^{2n}}{2n-1}-\dots ]$

$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

My Test Exponents

$x^{\frac{1}{n}}=\sqrt[n]{x}$

$27^{\frac{1}{3}}=\sqrt[3]{27}=3$

$x^{\frac{m}{n}}=\sqrt[n]{x^m}$

$\sqrt[n]{625}=5$ $\sqrt[4]{625}=5$ $\sqrt[n]{a^n}=a$

$\begin{array}{rl}{x}^{\frac{m}{n}}& =\sqrt[n]{x^m}\\ & =(\sqrt[n]{x}{)}^m\\ \end{array}$

$x^{\frac{m}{n}}=\sqrt[n]{x^m}=(\sqrt[n]{x}{)}^m$

$\begin{array}{rl}{x}^{\frac{2}{3}}& =\sqrt[3]{x^2}\\ & =(\sqrt[3]{x}{)}^2\\ \end{array}$

$x^{\frac{m}{n}}=x^{(m\times \frac{1}{n})}=(x^m{)}^{\frac{1}{n}}=\sqrt[n]{x^m}$

$x^{\frac{m}{n}}=x^{(\frac{1}{n}\times m)}=(x^{\frac{1}{n}}{)}^m=(\sqrt[n]{x}{)}^m$

My Test Area

$\underset{a}{\overset{b}{\int }}f(x)dx$

$2x^2+5x+3=0$ $x^2-3x=0$ $5x-3=0$

$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$

$\sqrt{1-e^2}$

$\frac{\sqrt{2}}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=$$\frac{\sqrt{2}\times \sqrt{3}}{3\times 3}=$$\frac{\sqrt{6}}{9}$

$\sqrt{\frac{1}{2}}=\sqrt{\frac{2}{4}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{2}$

$\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}$

$x^{\frac{2}{3}}=\sqrt[3]{x^2}$

$10^{10^{100}}$

$10^{10^{10^{1000}}}$

$\frac{n!}{(n-r)!}\times \frac{1}{r!}=\frac{n!}{r!(n-r)!}$

$\frac{n!}{r!(n-r)!}=(\genfrac{}{}{0ex}{}{n}{r})$

$f(k;n,p)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

for $k=0,1,2,\dots ,n$ and where

$(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!}$

$f(3;10,0.5)=(\genfrac{}{}{0ex}{}{10}{3})0.5^3(1-0.5{)}^{(10-3)}=(\genfrac{}{}{0ex}{}{10}{3})0.5^{3}0.5^7$

$(\genfrac{}{}{0ex}{}{10}{3})=\frac{10!}{3!(10-3)!}=\frac{10!}{3!7!}=120$

$(\genfrac{}{}{0ex}{}{4}{2})=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}=6$

$f(3;10,0.5)=120\times 0.5^{3}0.5^7=0.1171875$

$P(n,r)={}^n{P}_r={}_n{P}_r=\frac{n!}{(n-r)!}$

$C(n,r)={}^n{C}_r={}_n{C}_r=(\genfrac{}{}{0ex}{}{n}{r})=\frac{n!}{r!(n-r)!}$

$P(n,r)=\frac{n!}{(n-r)!}.$

Test Area 2

$\overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k}$

File:Hexadecimal multiplication table.svg

A hexadecimal multiplication table

$n{C}_r=\frac{n!}{(n-r)!(r!)}$

$\sum _{k=1}^{\mathrm{\infty }}10^{-k!}$ = 0.110001000000000000000001000...

$0<|x-\frac{p}{q}|<\frac{1}{q^n}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$C=co{s}^{-1}(\frac{a^2+b^2-c^2}{2ab})$

$\phi =\frac{1}{2}+\frac{\sqrt{5}}{2}=\frac{1+\sqrt{5}}{2}$

$\phi =1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots }}}=1.618...$

$\frac{1}{3-\sqrt{2}}$

$\frac{1}{3-\sqrt{2}}\times \frac{3+\sqrt{2}}{3+\sqrt{2}}=\frac{3+\sqrt{2}}{3^2-(\sqrt{2}{)}^2}=\frac{3+\sqrt{2}}{7}$

$\frac{2-\sqrt{x}}{4-x}$

$\frac{2-\sqrt{x}}{4-x}\times \frac{2+\sqrt{x}}{2+\sqrt{x}}=\frac{2^2-(\sqrt{2}{)}^2}{(4-x)(2+\sqrt{x})}=\frac{(4-x)}{(4-x)(2+\sqrt{x})}=\frac{1}{2+\sqrt{x}}$

Test Area Comb Perm

$n^r$

$\frac{n!}{(n-r)!}$

$\frac{n!}{r!(n-r)!}$

$\frac{(r+n-1)!}{r!(n-1)!}$

$(\genfrac{}{}{0ex}{}{r+n-1}{r})=\frac{(r+n-1)!}{r!(n-1)!}$

$(\genfrac{}{}{0ex}{}{r+n-1}{r})=(\genfrac{}{}{0ex}{}{r+n-1}{n-1})=\frac{(r+n-1)!}{r!(n-1)!}$

Test Area Sets

$f(t)=\{\begin{array}{ll}\mathrm{\$}50& \text{if }t\le 6\\ \mathrm{\$}80& \text{if }t>6\text{ and }t\le 15\\ \mathrm{\$}80+\mathrm{\$}5(t-15)& \text{if }t>15\end{array}$

$f(x)=\{\begin{array}{ll}{x}^2& \text{if }x<2\\ 6& \text{if }x=2\\ 10-x& \text{if }x>2\text{ and }x\le 6\text{ .}\end{array}$

$f(x)=|x|=\{\begin{array}{ll}x,& \text{if }x\ge 0\\ -x,& \text{if }x<0\text{ .}\end{array}$

$h(x)=\{\begin{array}{ll}2,& \text{if }x\le 1\\ x,& \text{if }x>1\text{ .}\end{array}$

$\frac{1}{x}$ $\frac{1}{y}$ $x^2$ $\sqrt{y}$ $x^n$ $\sqrt[n]{y}$ $y^{\frac{1}{n}}$

$e^x$ $ln(y)$ $a^x$ $lo{g}_a(y)$

$sin(x)$ $arcsin(y)$ $si{n}^{-1}(y)$

$cos(x)$ $arccos(y)$ $co{s}^{-1}(y)$

$tan(x)$ $arctan(y)$ $ta{n}^{-1}(y)$

Help:Displaying_a_formula

$f:\mathbb{\mathbb{N}}\to \mathbb{\mathbb{N}}$

$f:\{1,2,3,...\}\to \{1,2,3,...\}$

$f:\mathbb{ℝ}\to \mathbb{ℝ}$

$f:x\mapsto x^2$

$f(x)=x^2$

From Set-builder notation

Examples:

• $\{x\mid x=x^2\}$ is the set $\{0,1\}$,
• $\{x:x\in \mathbb{ℝ}\wedge x>0\}$ is the set of all positive real numbers,
• $\{k:n\in \mathbb{\mathbb{N}}\wedge k=2n\}$ is the set of all even natural numbers,
• $\{a:\mathrm{\exists }p,q\in \mathbb{\mathbb{Z}},q\ne 0:a=p/q\}$ is the set of rational numbers, or numbers that can be written as the ratio of two integers.

$\sum _{n=1}^4(2n+1)$ -

Test Area Limits

$\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}=2$

$\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}=\underset{x\to 1}{\lim }\frac{(x-1)(x+1)}{x-1}=\underset{x\to 1}{\lim }(x+1)$

$\underset{x\to 1}{\lim }(x+1)=1+1=2$

$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1$

$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n=e$

$\underset{x\to 10}{\lim }\frac{x}{2}=5$

$\underset{x\to 4}{\lim }\frac{2-\sqrt{x}}{4-x}$

$\frac{2-\sqrt{x}}{4-x}\times \frac{2+\sqrt{x}}{2+\sqrt{x}}=\frac{2^2-(\sqrt{x}{)}^2}{(4-x)(2+\sqrt{x})}=\frac{(4-x)}{(4-x)(2+\sqrt{x})}=\frac{1}{2+\sqrt{x}}$

$\frac{2-\sqrt{x}}{4-x}\times \frac{2+\sqrt{x}}{2+\sqrt{x}}$

$\frac{2^2-(\sqrt{x}{)}^2}{(4-x)(2+\sqrt{x})}$

$\frac{(4-x)}{(4-x)(2+\sqrt{x})}$

$\frac{1}{2+\sqrt{x}}$

$\underset{x\to 4}{\lim }\frac{2-\sqrt{x}}{4-x}=\underset{x\to 4}{\lim }\frac{1}{2+\sqrt{x}}=\frac{1}{2+\sqrt{4}}=\frac{1}{4}$

Test Area Derivatives

$\frac{2x\mathrm{\Delta }x+\mathrm{\Delta }{x}^2}{\mathrm{\Delta }x}$

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{2x\mathrm{\Delta }x+\mathrm{\Delta }{x}^2}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }2x+\mathrm{\Delta }x$

$\underset{\mathrm{\Delta }x\to 0}{\lim }2x+\mathrm{\Delta }x=2x$

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{2x\mathrm{\Delta }x+\mathrm{\Delta }{x}^2}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{2x\mathrm{\Delta }x}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }2x=2x$

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{3x^2\mathrm{\Delta }x+3x\mathrm{\Delta }{x}^2+\mathrm{\Delta }{x}^3}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }3x^2+3x\mathrm{\Delta }x+\mathrm{\Delta }{x}^2=3x^2$

$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$

$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$

$\frac{dy}{dx}=\frac{f(x+dx)-f(x)}{dx}$

$\frac{dy}{dx}$

$=\frac{f(x+dx)-f(x)}{dx}$

$=\frac{(x+dx{)}^2-x^2}{dx}$

$=\frac{x^2+2x\cdot dx+d{x}^2-x^2}{dx}$

$=\frac{2x\cdot dx+d{x}^2}{dx}$

$=2x+dx$

$=2x$

Test Area Integrals

$\underset{1}{\overset{2}{\int }}2xdx=2^2-1^2=3$

$\underset{0.5}{\overset{1}{\int }}cos(x)dx=sin(1)-sin(0.5)=0.841...-0.479...=0.362...$

$\underset{1}{\overset{3}{\int }}cos(x)dx=sin(3)-sin(1)=0.141...-0.841...=-0.700...$

$\underset{0}{\overset{1}{\int }}sin(x)dx=-cos(1)-(-cos(0))=-0.540...-(-1)=0.460...$

$\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}cos(x)dx$