User:MathsIsFun

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1 My Goal
2 The Website
3 Contact Details
4 My Test Area Other
5 My Test Area Geometry
6 My Test Area
7 Test Area 2
8 Test Area Sets
9 Test Area Limits
10 Test Area Derivatives

[edit] 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.

[edit] 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!

[edit] Contact Details

Use this Contact Form or leave a message on the Math is Fun Forum

[edit] My Test Area Other

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sqrt[3]{128} = \sqrt[3]{64\cdot2} = \sqrt[3]{64}\cdot\sqrt[3]{2} = 4\sqrt[3]{2}$

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

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

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

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

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

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

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

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

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

[edit] My Test Area Geometry

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

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

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

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

Ellipse perimeter, simple formula:

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

A better approximation by Ramanujan is:

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

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

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

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

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

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

[edit] My Test Area

$\int_a^b f(x)\,dx$

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

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

$\sqrt{1-e^2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$f(3;10,0.5)={10\choose 3}0.5^3(1-0.5)^{(10-3)}={10\choose 3}0.5^30.5^7$

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

$f(3;10,0.5)=120\times0.5^30.5^7=0.1171875$

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

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

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

[edit] 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...

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

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

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

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

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

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

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

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

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

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

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

[edit] Test Area Sets

Help:Displaying_a_formula

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

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

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

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

From Set-builder notation

Examples:

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

$\sum_{n=1}^4 (2n+1)$

[edit] Test Area Limits

$\lim_{x\to1} \frac{x^2-1}{x-1} = 2$

$\lim_{x\to1} \frac{x^2-1}{x-1} = \lim_{x\to1} \frac{(x-1)(x+1)}{x-1} = \lim_{x\to1} (x+1)$

$\lim_{x\to1} (x+1) = 1+1 = 2$

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

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

$\lim_{x\to10} \frac{x}{2}= 5$

$\lim_{x\to4} \frac{2-\sqrt{x}}{4-x}$

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