User:MathsIsFun

Wikipedia:Babel

enThis user is a native speaker of the English language.

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Other

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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 ...

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 Scratch

Help:Displaying_a_formula

${\begin{aligned}\log _{a}(m\times n)&=\log _{a}(a^{log_{a}m}\times a^{log_{a}n})\\&=\log _{a}(a^{log_{a}m\ +\ log_{a}n})\\&=\log _{a}m+\log _{a}n\\\end{aligned}}$

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

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

Test Area Symbols

$\Rightarrow \iff \approx$

Test Area Sigma

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\$7\times 4\times 91+\$11\times 650$

$\$7\times 364+\$11\times 650$

$\$2548+\$7150=\$9698\,$

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

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

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

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

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

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

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

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

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

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

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

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

$\sum n$

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

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

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

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

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

Test Area Binomial

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

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

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

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

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

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

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

Test Area Sigma 2

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

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

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

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

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

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

Test Area Trig

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

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

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

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

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

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

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

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

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

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

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

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

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

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

My Test Area Other

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ellipse a and b

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

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

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

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

Ellipse perimeter, simple formula:

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

A better approximation by Ramanujan is:

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

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

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

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

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

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

Ellipse r and s

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

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

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

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

Ellipse perimeter, simple formula:

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

A better approximation by Ramanujan is:

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

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

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

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

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

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

My Test Exponents

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

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

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

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

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

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

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

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

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

My Test Area

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

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

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

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

${\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}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$P(n,r)={}^{n}\!P_{r}={}_{n}\!P_{r}={\frac {n!}{(n-r)!}}$

$C(n,r)={}^{n}\!C_{r}={}_{n}\!C_{r}={n \choose r}={\frac {n!}{r!(n-r)!}}$

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

Test Area 2

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

A hexadecimal multiplication table

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

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

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

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

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

${{n+r-1} \choose {r}}={{(n+r-1)!} \over {r!(n-1)!}}$

${{n+r-1} \choose {r}}={{n+r-1} \choose {n-1}}={{(n+r-1)!} \over {r!(n-1)!}}$

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

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

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

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

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

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

Test Area Sets

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

$f(x)=|x|={\begin{cases}x,&{\mbox{if }}x\geq 0\\-x,&{\mbox{if }}x<0{\text{ .}}\end{cases}}$

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

$e^{x}\,$ $ln(y)\,$ $a^{x}\,$ $log_{a}(y)\,$

$sin(x)\,$ $arcsin(y)\,$ $sin^{-1}(y)\,$

$cos(x)\,$ $arccos(y)\,$ $cos^{-1}(y)\,$

$tan(x)\,$ $arctan(y)\,$ $tan^{-1}(y)\,$

Help:Displaying_a_formula

$f\colon \mathbb {N} \rightarrow \mathbb {N}$

$f\colon \{1,2,3,...\}\rightarrow \{1,2,3,...\}$

$f\colon \mathbb {R} \rightarrow \mathbb {R}$

$f\colon \,x\mapsto x^{2}$

$f(x)=x^{2}\,$

From Set-builder notation

Examples:

• $\{x\mid x=x^{2}\}$ is the set $\{0,1\}$,
• $\{x:x\in \mathbb {R} \land x>0\}$ is the set of all positive real numbers,
• $\{k:n\in \mathbb {N} \land k=2n\}$ is the set of all even natural numbers,
• $\{a:\exists \ p,q\in \mathbb {Z} ,q\neq 0:a=p/q\}$ 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

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

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

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

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

$\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

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

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

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

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

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

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

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

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

Test Area Derivatives

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

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

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

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

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

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

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

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

${\frac {dy}{dx}}\,$

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

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

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

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

$=2x+dx\,$

$=2x\,$