Wikipedia:Babel en This user is a native speaker of the English language. de-1 Dieser Benutzer hat grundlegende Deutschkenntnisse. es-1 Este usuario puede contribuir con un nivel básico de español. da-1 Denne bruger har et grundlæggende kendskab til dansk.

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prog-4 This user is an expert programmer. This user's favourite colour is blue.

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 ...

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

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

Test Area Taylor

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Test Area Scratch

Help:Displaying_a_formula

${\displaystyle r_{xy}={\frac {n\sum x_{i}y_{i}-\sum x_{i}\sum y_{i}}{{\sqrt {n\sum x_{i}^{2}-(\sum x_{i})^{2}}}~{\sqrt {n\sum y_{i}^{2}-(\sum y_{i})^{2}}}}}.}$

${\displaystyle f(ta+(1-t)b)\geq tf(a)+(1-t)f(b)}$

${\displaystyle |A|=a\cdot {\begin{vmatrix}e&f\\h&i\end{vmatrix}}-b\cdot {\begin{vmatrix}d&f\\g&i\end{vmatrix}}+c\cdot {\begin{vmatrix}d&e\\g&h\end{vmatrix}}}$

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

Test Area Symbols

${\displaystyle \Rightarrow \iff \approx }$

Test Area Stats

${\displaystyle r_{xy}={\frac {\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sqrt {\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})^{2}\sum \limits _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}},}$

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

Test Area Sigma

${\displaystyle \sum f=15+27+8+5=55}$

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

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

${\displaystyle \sum _{k=m}^{n}ca_{k}=c\sum _{k=m}^{n}a_{k}}$

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

${\displaystyle \sum _{k=m}^{n}(a_{k}+b_{k})=\sum _{k=m}^{n}a_{k}+\sum _{k=m}^{n}b_{k}}$

${\displaystyle \sum _{k=m}^{n}(a_{k}-b_{k})=\sum _{k=m}^{n}a_{k}-\sum _{k=m}^{n}b_{k}}$

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

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

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

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

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

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

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

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

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

${\displaystyle \$2548+\$7150=\$9698\,}$

Test Area Partial Sums

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \sum n}$

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

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

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

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

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

Test Area Binomial

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

${\displaystyle {n \choose k}({\tfrac {1}{n}})^{k}={\frac {n!}{k!(n-k)!}}\cdot {\frac {1}{n^{k}}}}$

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

${\displaystyle {\begin{aligned}(a+b)^{3}&=\sum _{k=0}^{3}{3 \choose k}a^{3-k}b^{k}\\&={3 \choose 0}a^{3-0}b^{0}+{3 \choose 1}a^{3-1}b^{1}+{3 \choose 2}a^{3-2}b^{2}+{3 \choose 3}a^{3-3}b^{3}\\&=1\cdot a^{3}b^{0}+3\cdot a^{2}b^{1}+3\cdot a^{1}b^{2}+1\cdot a^{0}b^{3}\\&=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}(1+{\tfrac {1}{n}})^{n}&=\sum _{k=0}^{n}{n \choose k}1^{n-k}({\tfrac {1}{n}})^{k}\\&=\sum _{k=0}^{n}{n \choose k}({\tfrac {1}{n}})^{k}\\&=\sum _{k=0}^{n}~{\frac {n!}{k!~(n-k)!}}\cdot {\frac {1}{n^{k}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}(x+5)^{4}&=\sum _{k=0}^{4}{4 \choose k}x^{4-k}5^{k}\\&={4 \choose 0}x^{4-0}5^{0}+{4 \choose 1}x^{4-1}5^{1}+{4 \choose 2}x^{4-2}5^{2}+{4 \choose 3}x^{4-3}5^{3}+{4 \choose 4}x^{4-4}5^{4}\\&=1\cdot x^{4}5^{0}+4\cdot x^{3}5^{1}+6\cdot x^{2}5^{2}+4\cdot x^{1}5^{3}+1\cdot x^{0}5^{4}\\&=x^{4}+4x^{3}5+6x^{2}5^{2}+4\cdot 5^{3}+5^{4}\\\end{aligned}}}$

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

Test Area Sigma 2

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

${\displaystyle s={\sqrt {{\frac {1}{N-1}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}}$

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

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

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

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

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

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

Test Area Trig

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

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

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

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

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos(C)\,}$

${\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos(A)\,}$

${\displaystyle b^{2}=a^{2}+c^{2}-2ac\cos(B)\,}$

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

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

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

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

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

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

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

My Test Area Other

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle {\sqrt[{n}]{a+b}}\neq {\sqrt[{n}]{a}}+{\sqrt[{n}]{b}}}$

${\displaystyle {\sqrt[{n}]{a-b}}\neq {\sqrt[{n}]{a}}-{\sqrt[{n}]{b}}}$

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

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

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

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

${\displaystyle \underbrace {{\sqrt[{n}]{a}}\times {\sqrt[{n}]{a}}\times ...\times {\sqrt[{n}]{a}}} _{n\ of\ them}=a}$

Ellipse a and b

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

${\displaystyle p=\pi (a+b)\left(1+\sum _{n=1}^{\infty }{0.5 \choose n}^{2}\cdot h^{n}\right)\!\,}$

${\displaystyle p=\pi (a+b)\sum _{n=0}^{\infty }{0.5 \choose n}^{2}h^{n}\!\,}$

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

Ellipse perimeter, simple formula:

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

A better approximation by Ramanujan is:

${\displaystyle p\approx \pi \left[3(a+b)-{\sqrt {(3a+b)(a+3b)}}\right]\!\,}$

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

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

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

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

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

${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\;}$

Ellipse r and s

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

${\displaystyle p=\pi (r+s)\left(1+\sum _{n=1}^{\infty }{0.5 \choose n}^{2}\cdot h^{n}\right)\!\,}$

${\displaystyle p=\pi (r+s)\sum _{n=0}^{\infty }{0.5 \choose n}^{2}h^{n}\!\,}$

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

Ellipse perimeter, simple formula:

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

A better approximation by Ramanujan is:

${\displaystyle p\approx \pi \left[3(r+s)-{\sqrt {(3r+s)(r+3s)}}\right]\!\,}$

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

${\displaystyle p=2r\pi \left(1-\sum _{i=1}^{\infty }{\frac {(2i)!^{2}}{(2^{i}\cdot i!)^{4}}}\cdot {\frac {\varepsilon ^{2i}}{2i-1}}\right)}$

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

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

${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\;}$

My Test Exponents

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

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

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

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

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

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

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

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

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

My Test Area

${\displaystyle \int _{a}^{b}f(x)\,dx}$

${\displaystyle 2x^{2}+5x+3=0\,}$ ${\displaystyle x^{2}-3x=0\,}$ ${\displaystyle 5x-3=0\,}$

${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1}$

${\displaystyle {\sqrt {1-e^{2}}}}$

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

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

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

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

${\displaystyle 10^{\,\!10^{100}}}$

${\displaystyle 10^{\,\!10^{10^{1000}}}}$

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

${\displaystyle {{n!} \over {r!(n-r)!}}={n \choose r}}$

${\displaystyle f(k;n,p)={n \choose k}p^{k}(1-p)^{n-k}}$

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

${\displaystyle {n \choose k}={\frac {n!}{k!(n-k)!}}}$

${\displaystyle f(3;10,0.5)={10 \choose 3}0.5^{3}(1-0.5)^{(10-3)}={10 \choose 3}0.5^{3}0.5^{7}}$

${\displaystyle {10 \choose 3}={\frac {10!}{3!(10-3)!}}={\frac {10!}{3!7!}}=120}$

${\displaystyle {4 \choose 2}={\frac {4!}{2!(4-2)!}}={\frac {4!}{2!2!}}={\frac {4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}}=6}$