Other
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 ...
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!
Use this Contact Form
or leave a message on the Math is Fun Forum
Test Area Scratch
Help:Displaying_a_formula
log
a
(
m
×
n
)
=
log
a
(
a
l
o
g
a
m
×
a
l
o
g
a
n
)
=
log
a
(
a
l
o
g
a
m
+
l
o
g
a
n
)
=
log
a
m
+
log
a
n
{\displaystyle {\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}}}
y
=
log
4
(
1
4
)
{\displaystyle y=\log _{4}({\frac {1}{4}})\,}
4
y
=
4
log
4
(
1
4
)
{\displaystyle 4^{y}=4^{\log _{4}({\frac {1}{4}})}\,}
log
3
(
x
)
=
5
{\displaystyle \log _{3}(x)=5\,}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle 3^{\log_{3} (x)} = x \,}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle 3^{\log_{3} (x)} = 3^5 \,}
log
a
(
a
x
)
=
x
{\displaystyle \log _{a}(a^{x})=x\,}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle a^{\log_{a} (x)} = x \,}
2
x
x
−
3
+
3
=
6
x
−
3
{\displaystyle {\frac {2x}{x-3}}+3={\frac {6}{x-3}}}
2
x
+
3
(
x
−
3
)
=
6
{\displaystyle 2x+3(x-3)=6\,}
2
x
+
3
(
x
−
3
)
−
6
=
0
{\displaystyle 2x+3(x-3)-6=0\,}
2
x
+
3
x
−
9
−
6
=
0
{\displaystyle 2x+3x-9-6=0\,}
5
x
−
15
=
0
{\displaystyle 5x-15=0\,}
5
(
x
−
3
)
=
0
{\displaystyle 5(x-3)=0\,}
x
−
3
=
0
{\displaystyle x-3=0\,}
x
=
3
{\displaystyle x=3\,}
x
≠
3
{\displaystyle x\neq 3\,}
(
x
−
5
)
(
x
+
2
)
=
x
2
+
2
x
−
5
x
−
10
=
x
2
−
3
x
−
10
{\displaystyle {\begin{aligned}(x-5)(x+2)&=x^{2}+2x-5x-10\\&=x^{2}-3x-10\\\end{aligned}}}
1
x
+
1
{\displaystyle {\frac {1}{x+1}}}
x
x
3
−
1
{\displaystyle {\frac {x}{x^{3}-1}}}
x
2
−
1
x
+
1
{\displaystyle {\frac {x^{2}-1}{x+1}}}
4
x
3
−
3
5
x
3
+
1
{\displaystyle {\frac {4x^{3}-3}{5x^{3}+1}}}
Test Area Symbols
⇒
⟺
≈
{\displaystyle \Rightarrow \iff \approx }
Test Area Sigma
∑
k
=
m
n
c
a
k
=
c
∑
k
=
m
n
a
k
{\displaystyle \sum _{k=m}^{n}ca_{k}=c\sum _{k=m}^{n}a_{k}}
∑
k
=
m
n
6
k
2
=
6
∑
k
=
m
n
k
2
{\displaystyle \sum _{k=m}^{n}6k^{2}=6\sum _{k=m}^{n}k^{2}}
∑
k
=
m
n
(
a
k
+
b
k
)
=
∑
k
=
m
n
a
k
+
∑
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}}
∑
k
=
m
n
(
a
k
−
b
k
)
=
∑
k
=
m
n
a
k
−
∑
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}}
∑
k
=
m
n
(
k
+
k
2
)
=
∑
k
=
m
n
k
+
∑
k
=
m
n
k
2
{\displaystyle \sum _{k=m}^{n}(k+k^{2})=\sum _{k=m}^{n}k+\sum _{k=m}^{n}k^{2}}
∑
n
=
1
4
(
2
n
+
1
)
=
3
+
5
+
7
+
9
=
24
{\displaystyle \sum _{n=1}^{4}(2n+1)=3+5+7+9=24}
∑
i
=
2
14
(
$
7
×
4
(
i
−
1
)
+
$
11
×
(
i
−
2
)
2
)
{\displaystyle \sum _{i=2}^{14}\left(\$7\times 4(i-1)+\$11\times (i-2)^{2}\right)}
∑
i
=
2
14
$
7
×
4
(
i
−
1
)
+
∑
i
=
2
14
$
11
×
(
i
−
2
)
2
{\displaystyle \sum _{i=2}^{14}\$7\times 4(i-1)+\sum _{i=2}^{14}\$11\times (i-2)^{2}}
$
7
×
4
∑
i
=
2
14
(
i
−
1
)
+
$
11
×
∑
i
=
2
14
(
i
−
2
)
2
{\displaystyle \$7\times 4\sum _{i=2}^{14}(i-1)+\$11\times \sum _{i=2}^{14}(i-2)^{2}}
$
7
×
4
∑
j
=
1
13
j
+
$
11
∑
k
=
1
12
k
2
{\displaystyle \$7\times 4\sum _{j=1}^{13}j+\$11\sum _{k=1}^{12}k^{2}}
$
7
×
4
×
13
×
14
2
+
$
11
×
12
×
13
×
25
6
{\displaystyle \$7\times 4\times {\frac {13\times 14}{2}}+\$11\times {\frac {12\times 13\times 25}{6}}}
$
7
×
4
×
91
+
$
11
×
650
{\displaystyle \$7\times 4\times 91+\$11\times 650}
$
7
×
364
+
$
11
×
650
{\displaystyle \$7\times 364+\$11\times 650}
$
2548
+
$
7150
=
$
9698
{\displaystyle \$2548+\$7150=\$9698\,}
∑
k
=
1
n
1
=
n
{\displaystyle \sum _{k=1}^{n}1=n}
∑
k
=
1
n
c
=
n
c
{\displaystyle \sum _{k=1}^{n}c=nc}
∑
k
=
1
n
k
=
n
(
n
+
1
)
2
{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}}
∑
k
=
1
n
k
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
{\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}
∑
k
=
1
n
k
3
=
(
n
(
n
+
1
)
2
)
2
{\displaystyle \sum _{k=1}^{n}k^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}}
∑
k
=
1
14
k
2
=
14
(
14
+
1
)
(
2
⋅
14
+
1
)
6
=
1015
{\displaystyle \sum _{k=1}^{14}k^{2}={\frac {14(14+1)(2\cdot 14+1)}{6}}=1015}
∑
k
=
0
n
−
1
(
a
+
k
d
)
=
n
2
(
2
a
+
(
n
−
1
)
d
)
{\displaystyle \sum _{k=0}^{n-1}(a+kd)={\frac {n}{2}}(2a+(n-1)d)}
∑
k
=
0
10
−
1
(
1
+
k
⋅
3
)
=
10
2
(
2
⋅
1
+
(
10
−
1
)
⋅
3
)
{\displaystyle \sum _{k=0}^{10-1}(1+k\cdot 3)={\frac {10}{2}}(2\cdot 1+(10-1)\cdot 3)}
∑
k
=
0
n
−
1
(
a
r
k
)
=
a
(
1
−
r
n
1
−
r
)
{\displaystyle \sum _{k=0}^{n-1}(ar^{k})=a\left({\frac {1-r^{n}}{1-r}}\right)}
∑
k
=
0
4
−
1
(
10
⋅
3
k
)
=
10
(
1
−
3
4
1
−
3
)
=
400
{\displaystyle \sum _{k=0}^{4-1}(10\cdot 3^{k})=10\left({\frac {1-3^{4}}{1-3}}\right)=400}
∑
k
=
0
10
−
1
1
2
(
1
2
)
k
=
1
2
(
1
−
(
1
2
)
10
1
−
1
2
)
=
1
2
(
1
−
1
1024
1
2
)
=
1
−
1
1024
{\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}}}
∑
k
=
0
64
−
1
(
1
⋅
2
k
)
=
1
(
1
−
2
64
1
−
2
)
{\displaystyle \sum _{k=0}^{64-1}(1\cdot 2^{k})=1\left({\frac {1-2^{64}}{1-2}}\right)}
∑
n
{\displaystyle \sum n}
∑
n
=
1
4
n
{\displaystyle \sum _{n=1}^{4}n}
∑
n
=
1
4
n
=
1
+
2
+
3
+
4
=
10
{\displaystyle \sum _{n=1}^{4}n=1+2+3+4=10}
∑
n
=
1
4
n
2
=
1
2
+
2
2
+
3
2
+
4
2
=
30
{\displaystyle \sum _{n=1}^{4}n^{2}=1^{2}+2^{2}+3^{2}+4^{2}=30}
∑
i
=
1
3
i
(
i
+
1
)
=
1
⋅
2
+
2
⋅
3
+
3
⋅
4
=
20
{\displaystyle \sum _{i=1}^{3}i(i+1)=1\cdot 2+2\cdot 3+3\cdot 4=20}
∑
i
=
3
5
i
i
+
1
=
3
4
+
4
5
+
5
6
{\displaystyle \sum _{i=3}^{5}{\frac {i}{i+1}}={\frac {3}{4}}+{\frac {4}{5}}+{\frac {5}{6}}}
Test Area Binomial
(
a
+
b
)
n
=
∑
k
=
0
n
(
n
k
)
a
n
−
k
b
k
{\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k}}
(
n
k
)
(
1
n
)
k
=
n
!
k
!
(
n
−
k
)
!
⋅
1
n
k
{\displaystyle {n \choose k}({\tfrac {1}{n}})^{k}={\frac {n!}{k!(n-k)!}}\cdot {\frac {1}{n^{k}}}}
∑
k
=
0
∞
1
k
!
=
1
0
!
+
1
1
!
+
1
2
!
+
1
3
!
+
1
4
!
+
.
.
.
=
1
+
1
+
1
2
+
1
6
+
1
24
+
.
.
.
{\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}}}
(
a
+
b
)
3
=
∑
k
=
0
3
(
3
k
)
a
3
−
k
b
k
=
(
3
0
)
a
3
−
0
b
0
+
(
3
1
)
a
3
−
1
b
1
+
(
3
2
)
a
3
−
2
b
2
+
(
3
3
)
a
3
−
3
b
3
=
1
⋅
a
3
b
0
+
3
⋅
a
2
b
1
+
3
⋅
a
1
b
2
+
1
⋅
a
0
b
3
=
a
3
+
3
a
2
b
+
3
a
b
2
+
b
3
{\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}}}
(
1
+
1
n
)
n
=
∑
k
=
0
n
(
n
k
)
1
n
−
k
(
1
n
)
k
=
∑
k
=
0
n
(
n
k
)
(
1
n
)
k
=
∑
k
=
0
n
n
!
k
!
(
n
−
k
)
!
⋅
1
n
k
{\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}}}
(
x
+
5
)
4
=
∑
k
=
0
4
(
4
k
)
x
4
−
k
5
k
=
(
4
0
)
x
4
−
0
5
0
+
(
4
1
)
x
4
−
1
5
1
+
(
4
2
)
x
4
−
2
5
2
+
(
4
3
)
x
4
−
3
5
3
+
(
4
4
)
x
4
−
4
5
4
=
1
⋅
x
4
5
0
+
4
⋅
x
3
5
1
+
6
⋅
x
2
5
2
+
4
⋅
x
1
5
3
+
1
⋅
x
0
5
4
=
x
4
+
4
x
3
5
+
6
x
2
5
2
+
4
⋅
5
3
+
5
4
{\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}}}
∑
k
=
0
10
−
1
1
2
(
1
2
)
k
=
1
2
(
1
−
(
1
2
)
10
1
−
1
2
)
=
1
2
(
1
−
1
1024
1
2
)
=
1
−
1
1024
=
0.9990234375
{\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
∑
k
=
0
n
−
1
(
a
r
k
)
=
a
(
1
−
r
n
1
−
r
)
{\displaystyle \sum _{k=0}^{n-1}(ar^{k})=a\left({\frac {1-r^{n}}{1-r}}\right)}
∑
k
=
0
∞
(
a
r
k
)
=
a
(
1
1
−
r
)
{\displaystyle \sum _{k=0}^{\infty }(ar^{k})=a\left({\frac {1}{1-r}}\right)}
∑
k
=
0
∞
(
1
2
⋅
(
1
2
)
k
)
=
1
2
(
1
1
−
1
2
)
{\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)}
∑
k
=
0
∞
(
1
2
⋅
(
1
2
)
k
)
=
1
2
(
1
1
−
1
2
)
{\displaystyle \sum _{k=0}^{\infty }({\tfrac {1}{2}}\cdot ({\tfrac {1}{2}})^{k})={\tfrac {1}{2}}\left({\frac {1}{1-{\frac {1}{2}}}}\right)}
0.999...
=
0.9
+
0.09
+
0.009
+
.
.
.
=
0.9
⋅
0.1
0
+
0.9
⋅
0.1
1
+
0.9
⋅
0.1
2
+
.
.
.
=
∑
k
=
0
∞
0.9
⋅
0.1
k
{\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}}}
∑
k
=
0
∞
0.9
×
0.1
k
=
0.9
(
1
1
−
0.1
)
=
0.9
(
1
0.9
)
=
1
{\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
a
2
−
b
2
a
{\displaystyle {\frac {\sqrt {a^{2}-b^{2}}}{a}}}
a
2
+
b
2
a
{\displaystyle {\frac {\sqrt {a^{2}+b^{2}}}{a}}}
a
sin
A
=
b
sin
B
=
c
sin
C
{\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}}
sin
A
a
=
sin
B
b
=
sin
C
c
{\displaystyle {\frac {\sin A}{a}}={\frac {\sin B}{b}}={\frac {\sin C}{c}}}
c
2
=
a
2
+
b
2
−
2
a
b
cos
(
C
)
{\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos(C)\,}
a
2
=
b
2
+
c
2
−
2
b
c
cos
(
A
)
{\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos(A)\,}
b
2
=
a
2
+
c
2
−
2
a
c
cos
(
B
)
{\displaystyle b^{2}=a^{2}+c^{2}-2ac\cos(B)\,}
tan
θ
=
sin
θ
cos
θ
{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}}
sin
θ
cos
θ
=
O
p
p
o
s
i
t
e
/
H
y
p
o
t
e
n
u
s
e
A
d
j
a
c
e
n
t
/
H
y
p
o
t
e
n
u
s
e
=
O
p
p
o
s
i
t
e
A
d
j
a
c
e
n
t
=
tan
θ
{\displaystyle {\frac {\sin \theta }{\cos \theta }}={\frac {Opposite/Hypotenuse}{Adjacent/Hypotenuse}}={\frac {Opposite}{Adjacent}}=\tan \theta }
cot
θ
=
cos
θ
sin
θ
{\displaystyle \cot \theta ={\frac {\cos \theta }{\sin \theta }}}
sin
θ
2
=
±
1
−
cos
θ
2
{\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}}
cos
θ
2
=
±
1
+
cos
θ
2
{\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}}
tan
θ
2
=
±
1
−
cos
θ
1
+
cos
θ
=
sin
θ
1
+
cos
θ
=
1
−
cos
θ
sin
θ
=
csc
θ
−
cot
θ
{\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 }
cot
θ
2
=
±
1
+
cos
θ
1
−
cos
θ
=
sin
θ
1
−
cos
θ
=
1
+
cos
θ
sin
θ
=
csc
θ
+
cot
θ
{\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
c
=
(
3
−
9
)
2
+
(
2
−
7
)
2
{\displaystyle c={\sqrt {(3-9)^{2}+(2-7)^{2}}}}
c
=
(
−
6
)
2
+
(
−
5
)
2
=
36
+
25
=
61
=
7.81...
{\displaystyle c={\sqrt {(-6)^{2}+(-5)^{2}}}={\sqrt {36+25}}={\sqrt {61}}=7.81...}
c
=
(
−
3
−
7
)
2
+
(
5
−
(
−
1
)
)
2
{\displaystyle c={\sqrt {(-3-7)^{2}+(5-(-1))^{2}}}}
c
=
(
−
10
)
2
+
(
6
)
2
=
100
+
36
=
136
=
11.66...
{\displaystyle c={\sqrt {(-10)^{2}+(6)^{2}}}={\sqrt {100+36}}={\sqrt {136}}=11.66...}
c
=
(
9
−
3
)
2
+
(
7
−
2
)
2
{\displaystyle c={\sqrt {(9-3)^{2}+(7-2)^{2}}}}
c
=
6
2
+
5
2
=
36
+
25
=
61
=
7.81...
{\displaystyle c={\sqrt {6^{2}+5^{2}}}={\sqrt {36+25}}={\sqrt {61}}=7.81...}
c
=
(
x
A
−
x
B
)
2
+
(
y
A
−
y
B
)
2
{\displaystyle c={\sqrt {(x_{\text{A}}-x_{\text{B}})^{2}+(y_{\text{A}}-y_{\text{B}})^{2}}}}
2
x
2
−
5
x
−
1
x
−
3
=
2
x
+
1
+
2
x
−
3
{\displaystyle {\frac {2x^{2}-5x-1}{x-3}}=2x+1+{\frac {2}{x-3}}}
1
2
{\displaystyle {\frac {1}{\sqrt {2}}}}
1
2
×
2
2
=
2
2
{\displaystyle {\frac {1}{\sqrt {2}}}\times {\frac {\sqrt {2}}{\sqrt {2}}}={\frac {\sqrt {2}}{2}}}
a
m
n
=
(
a
n
)
m
{\displaystyle {\sqrt[{n}]{a^{m}}}=({\sqrt[{n}]{a}})^{m}}
27
2
3
=
(
27
3
)
2
=
3
2
=
9
{\displaystyle {\sqrt[{3}]{27^{2}}}=({\sqrt[{3}]{27}})^{2}=3^{2}=9}
4
6
3
=
(
4
)
6
3
=
4
2
=
16
{\displaystyle {\sqrt[{3}]{4^{6}}}=(4)^{\frac {6}{3}}=4^{2}=16}
a
m
n
=
a
m
n
{\displaystyle {\sqrt[{n}]{a^{m}}}=a^{\frac {m}{n}}}
a
n
=
a
1
n
{\displaystyle {\sqrt[{n}]{a}}=a^{\frac {1}{n}}}
2
3
3
=
2
{\displaystyle {\sqrt[{3}]{2^{3}}}=2}
−
2
3
3
=
−
2
{\displaystyle {\sqrt[{3}]{-2^{3}}}=-2}
−
2
4
4
=
|
−
2
|
=
2
{\displaystyle {\sqrt[{4}]{-2^{4}}}=|-2|=2}
5
4
=
625
s
o
5
=
625
4
{\displaystyle 5^{4}=625\ \ so\ \ 5={\sqrt[{4}]{625}}}
a
b
n
=
a
n
⋅
b
n
{\displaystyle {\sqrt[{n}]{ab}}={\sqrt[{n}]{a}}\cdot {\sqrt[{n}]{b}}}
128
3
=
64
⋅
2
3
=
64
3
⋅
2
3
=
4
2
3
{\displaystyle {\sqrt[{3}]{128}}={\sqrt[{3}]{64\cdot 2}}={\sqrt[{3}]{64}}\cdot {\sqrt[{3}]{2}}=4{\sqrt[{3}]{2}}}
a
b
n
=
a
n
b
n
{\displaystyle {\sqrt[{n}]{\frac {a}{b}}}={\frac {\sqrt[{n}]{a}}{\sqrt[{n}]{b}}}}
1
64
3
=
1
3
64
3
=
1
4
{\displaystyle {\sqrt[{3}]{\frac {1}{64}}}={\frac {\sqrt[{3}]{1}}{\sqrt[{3}]{64}}}={\frac {1}{4}}}
a
+
b
n
≠
a
n
+
b
n
{\displaystyle {\sqrt[{n}]{a+b}}\neq {\sqrt[{n}]{a}}+{\sqrt[{n}]{b}}}
a
−
b
n
≠
a
n
−
b
n
{\displaystyle {\sqrt[{n}]{a-b}}\neq {\sqrt[{n}]{a}}-{\sqrt[{n}]{b}}}
a
n
+
b
n
n
≠
a
+
b
{\displaystyle {\sqrt[{n}]{a^{n}+b^{n}}}\neq a+b}
a
n
n
=
a
{\displaystyle {\sqrt[{n}]{a^{n}}}=a}
a
×
a
=
a
{\displaystyle {\sqrt {a}}\times {\sqrt {a}}=a}
a
3
×
a
3
×
a
3
=
a
{\displaystyle {\sqrt[{3}]{a}}\times {\sqrt[{3}]{a}}\times {\sqrt[{3}]{a}}=a}
a
n
×
a
n
×
.
.
.
×
a
n
⏟
n
o
f
t
h
e
m
=
a
{\displaystyle \underbrace {{\sqrt[{n}]{a}}\times {\sqrt[{n}]{a}}\times ...\times {\sqrt[{n}]{a}}} _{n\ of\ them}=a}
Ellipse a and b
h
=
(
a
−
b
)
2
(
a
+
b
)
2
{\displaystyle h={\frac {(a-b)^{2}}{(a+b)^{2}}}}
p
=
π
(
a
+
b
)
(
1
+
∑
n
=
1
∞
(
0.5
n
)
2
⋅
h
n
)
{\displaystyle p=\pi (a+b)\left(1+\sum _{n=1}^{\infty }{0.5 \choose n}^{2}\cdot h^{n}\right)\!\,}
p
=
π
(
a
+
b
)
∑
n
=
0
∞
(
0.5
n
)
2
h
n
{\displaystyle p=\pi (a+b)\sum _{n=0}^{\infty }{0.5 \choose n}^{2}h^{n}\!\,}
p
=
π
(
a
+
b
)
(
1
+
1
4
h
+
1
64
h
2
+
1
256
h
3
+
.
.
.
)
{\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:
p
≈
2
π
a
2
+
b
2
2
{\displaystyle p\approx 2\pi {\sqrt {\frac {a^{2}+b^{2}}{2}}}\!\,}
A better approximation by Ramanujan is:
p
≈
π
[
3
(
a
+
b
)
−
(
3
a
+
b
)
(
a
+
3
b
)
]
{\displaystyle p\approx \pi \left[3(a+b)-{\sqrt {(3a+b)(a+3b)}}\right]\!\,}
e
=
a
2
−
b
2
a
{\displaystyle e={\frac {\sqrt {a^{2}-b^{2}}}{a}}}
p
=
2
a
π
(
1
−
∑
i
=
1
∞
(
2
i
)
!
2
(
2
i
⋅
i
!
)
4
⋅
e
2
i
2
i
−
1
)
{\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)}
p
=
2
a
π
[
1
−
(
1
2
)
2
e
2
−
(
1
⋅
3
2
⋅
4
)
2
e
4
3
−
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
2
e
6
5
−
…
]
{\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]}
p
=
2
a
π
[
1
−
(
1
2
)
2
e
2
−
(
1
⋅
3
2
⋅
4
)
2
e
4
3
−
⋯
−
(
1
⋅
3
⋅
5
⋅
⋯
⋅
(
2
n
−
1
)
2
⋅
4
⋅
6
⋅
⋯
⋅
2
n
)
2
e
2
n
2
n
−
1
−
…
]
{\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]}
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\;}
Ellipse r and s
h
=
(
r
−
s
)
2
(
r
+
s
)
2
{\displaystyle h={\frac {(r-s)^{2}}{(r+s)^{2}}}}
p
=
π
(
r
+
s
)
(
1
+
∑
n
=
1
∞
(
0.5
n
)
2
⋅
h
n
)
{\displaystyle p=\pi (r+s)\left(1+\sum _{n=1}^{\infty }{0.5 \choose n}^{2}\cdot h^{n}\right)\!\,}
p
=
π
(
r
+
s
)
∑
n
=
0
∞
(
0.5
n
)
2
h
n
{\displaystyle p=\pi (r+s)\sum _{n=0}^{\infty }{0.5 \choose n}^{2}h^{n}\!\,}
p
=
π
(
r
+
s
)
(
1
+
1
4
h
+
1
64
h
2
+
1
256
h
3
+
.
.
.
)
{\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:
p
≈
2
π
r
2
+
s
2
2
{\displaystyle p\approx 2\pi {\sqrt {\frac {r^{2}+s^{2}}{2}}}\!\,}
A better approximation by Ramanujan is:
p
≈
π
[
3
(
r
+
s
)
−
(
3
r
+
s
)
(
r
+
3
s
)
]
{\displaystyle p\approx \pi \left[3(r+s)-{\sqrt {(3r+s)(r+3s)}}\right]\!\,}
ε
=
r
2
−
s
2
r
{\displaystyle \varepsilon ={\frac {\sqrt {r^{2}-s^{2}}}{r}}}
p
=
2
r
π
(
1
−
∑
i
=
1
∞
(
2
i
)
!
2
(
2
i
⋅
i
!
)
4
⋅
ε
2
i
2
i
−
1
)
{\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)}
p
=
2
r
π
[
1
−
(
1
2
)
2
ε
2
−
(
1
⋅
3
2
⋅
4
)
2
ε
4
3
−
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
2
ε
6
5
−
…
]
{\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]}
p
=
2
r
π
[
1
−
(
1
2
)
2
ε
2
−
(
1
⋅
3
2
⋅
4
)
2
ε
4
3
−
⋯
−
(
1
⋅
3
⋅
5
⋅
⋯
⋅
(
2
n
−
1
)
2
⋅
4
⋅
6
⋅
⋯
⋅
2
n
)
2
ε
2
n
2
n
−
1
−
…
]
{\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]}
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\;}
My Test Exponents
x
1
n
=
x
n
{\displaystyle x^{\frac {1}{n}}={\sqrt[{n}]{x}}}
27
1
3
=
27
3
=
3
{\displaystyle 27^{\frac {1}{3}}={\sqrt[{3}]{27}}=3}
x
m
n
=
x
m
n
{\displaystyle x^{\frac {m}{n}}={\sqrt[{n}]{x^{m}}}}
625
n
=
5
{\displaystyle {\sqrt[{n}]{625}}=5}
625
4
=
5
{\displaystyle {\sqrt[{4}]{625}}=5}
a
n
n
=
a
{\displaystyle {\sqrt[{n}]{a^{n}}}=a}
x
m
n
=
x
m
n
=
(
x
n
)
m
{\displaystyle {\begin{aligned}x^{\frac {m}{n}}&={\sqrt[{n}]{x^{m}}}\\&=({\sqrt[{n}]{x}})^{m}\\\end{aligned}}}
x
m
n
=
x
m
n
=
(
x
n
)
m
{\displaystyle x^{\frac {m}{n}}={\sqrt[{n}]{x^{m}}}=({\sqrt[{n}]{x}})^{m}}
x
2
3
=
x
2
3
=
(
x
3
)
2
{\displaystyle {\begin{aligned}x^{\frac {2}{3}}&={\sqrt[{3}]{x^{2}}}\\&=({\sqrt[{3}]{x}})^{2}\\\end{aligned}}}
x
m
n
=
x
(
m
×
1
n
)
=
(
x
m
)
1
n
=
x
m
n
{\displaystyle x^{\frac {m}{n}}=x^{(m\times {\frac {1}{n}})}=(x^{m})^{\frac {1}{n}}={\sqrt[{n}]{x^{m}}}}
x
m
n
=
x
(
1
n
×
m
)
=
(
x
1
n
)
m
=
(
x
n
)
m
{\displaystyle x^{\frac {m}{n}}=x^{({\frac {1}{n}}\times m)}=(x^{\frac {1}{n}})^{m}=({\sqrt[{n}]{x}})^{m}}
My Test Area
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\,dx}
2
x
2
+
5
x
+
3
=
0
{\displaystyle 2x^{2}+5x+3=0\,}
x
2
−
3
x
=
0
{\displaystyle x^{2}-3x=0\,}
5
x
−
3
=
0
{\displaystyle 5x-3=0\,}
lim
x
→
0
sin
(
x
)
x
=
1
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1}
1
−
e
2
{\displaystyle {\sqrt {1-e^{2}}}}
2
3
3
×
3
3
=
{\displaystyle {\frac {\sqrt {2}}{3{\sqrt {3}}}}\times {\frac {\sqrt {3}}{\sqrt {3}}}=}
2
×
3
3
×
3
=
{\displaystyle {\frac {{\sqrt {2}}\times {\sqrt {3}}}{3\times 3}}=}
6
9
{\displaystyle {\frac {\sqrt {6}}{9}}}
1
2
=
2
4
=
2
4
=
2
2
{\displaystyle {\sqrt {\tfrac {1}{2}}}={\sqrt {\tfrac {2}{4}}}={\frac {\sqrt {2}}{\sqrt {4}}}={\frac {\sqrt {2}}{2}}}
3
4
=
3
4
=
3
2
{\displaystyle {\sqrt {\tfrac {3}{4}}}={\frac {\sqrt {3}}{\sqrt {4}}}={\frac {\sqrt {3}}{2}}}
x
2
3
=
x
2
3
{\displaystyle x^{\frac {2}{3}}={\sqrt[{3}]{x^{2}}}}
10
10
100
{\displaystyle 10^{\,\!10^{100}}}
10
10
10
1000
{\displaystyle 10^{\,\!10^{10^{1000}}}}
n
!
(
n
−
r
)
!
×
1
r
!
=
n
!
r
!
(
n
−
r
)
!
{\displaystyle {{n!} \over {(n-r)!}}\times {{1} \over {r!}}={{n!} \over {r!(n-r)!}}}
n
!
r
!
(
n
−
r
)
!
=
(
n
r
)
{\displaystyle {{n!} \over {r!(n-r)!}}={n \choose r}}
f
(
k
;
n
,
p
)
=
(
n
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle f(k;n,p)={n \choose k}p^{k}(1-p)^{n-k}}
for
k
=
0
,
1
,
2
,
…
,
n
{\displaystyle k=0,1,2,\dots ,n}
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
{\displaystyle {n \choose k}={\frac {n!}{k!(n-k)!}}}
f
(
3
;
10
,
0.5
)
=
(
10
3
)
0.5
3
(
1
−
0.5
)
(
10
−
3
)
=
(
10
3
)
0.5
3
0.5
7
{\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}}
(
10
3
)
=
10
!
3
!
(
10
−
3
)
!
=
10
!
3
!
7
!
=
120
{\displaystyle {10 \choose 3}={\frac {10!}{3!(10-3)!}}={\frac {10!}{3!7!}}=120}
(
4
2
)
=
4
!
2
!
(
4
−
2
)
!
=
4
!
2
!
2
!
=
4
⋅
3
⋅
2
⋅
1
2
⋅
1
⋅
2
⋅
1
=
6
{\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}
f
(
3
;
10
,
0.5
)
=
120
×
0.5
3
0.5
7
=
0.1171875
{\displaystyle f(3;10,0.5)=120\times 0.5^{3}0.5^{7}=0.1171875}
P
(
n
,
r
)
=
n
P
r
=
n
P
r
=
n
!
(
n
−
r
)
!
{\displaystyle P(n,r)={}^{n}\!P_{r}={}_{n}\!P_{r}={\frac {n!}{(n-r)!}}}
C
(
n
,
r
)
=
n
C
r
=
n
C
r
=
(
n
r
)
=
n
!
r
!
(
n
−
r
)
!
{\displaystyle C(n,r)={}^{n}\!C_{r}={}_{n}\!C_{r}={n \choose r}={\frac {n!}{r!(n-r)!}}}
P
(
n
,
r
)
=
n
!
(
n
−
r
)
!
.
{\displaystyle P(n,r)={\frac {n!}{(n-r)!}}.}
Test Area 2
i
→
×
j
→
=
k
→
{\displaystyle {\vec {i}}\times {\vec {j}}={\vec {k}}}
A hexadecimal multiplication table
n
C
r
=
n
!
(
n
−
r
)
!
(
r
!
)
{\displaystyle nC_{r}={\frac {n!}{(n-r)!(r!)}}}
∑
k
=
1
∞
10
−
k
!
{\displaystyle \sum _{k=1}^{\infty }10^{-k!}}
0
<
|
x
−
p
q
|
<
1
q
n
{\displaystyle 0<|x-{\frac {p}{q}}|<{\frac {1}{q^{n}}}}
A
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
{\displaystyle A={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}\,}
C
=
c
o
s
−
1
(
a
2
+
b
2
−
c
2
2
a
b
)
{\displaystyle C=cos^{-1}({\frac {a^{2}+b^{2}-c^{2}}{2ab}})}
(
n
+
r
−
1
r
)
=
(
n
+
r
−
1
)
!
r
!
(
n
−
1
)
!
{\displaystyle {{n+r-1} \choose {r}}={{(n+r-1)!} \over {r!(n-1)!}}}
(
n
+
r
−
1
r
)
=
(
n
+
r
−
1
n
−
1
)
=
(
n
+
r
−
1
)
!
r
!
(
n
−
1
)
!
{\displaystyle {{n+r-1} \choose {r}}={{n+r-1} \choose {n-1}}={{(n+r-1)!} \over {r!(n-1)!}}}
φ
=
1
2
+
5
2
=
1
+
5
2
{\displaystyle \varphi ={\frac {1}{2}}+{\frac {\sqrt {5}}{2}}={\frac {1+{\sqrt {5}}}{2}}}
φ
=
1
+
1
1
+
1
1
+
1
1
+
⋯
=
1.618...
{\displaystyle \varphi =1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+\cdots }}}}}}=1.618...}
1
3
−
2
{\displaystyle {\frac {1}{3-{\sqrt {2}}}}}
1
3
−
2
×
3
+
2
3
+
2
=
3
+
2
3
2
−
(
2
)
2
=
3
+
2
7
{\displaystyle {\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}}}
2
−
x
4
−
x
{\displaystyle {\frac {2-{\sqrt {x}}}{4-x}}}
2
−
x
4
−
x
×
2
+
x
2
+
x
=
2
2
−
(
2
)
2
(
4
−
x
)
(
2
+
x
)
=
(
4
−
x
)
(
4
−
x
)
(
2
+
x
)
=
1
2
+
x
{\displaystyle {\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
)
=
{
x
2
if
x
<
2
6
if
x
=
2
10
−
x
if
x
>
2
and
x
≤
6
.
{\displaystyle 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
|
=
{
x
,
if
x
≥
0
−
x
,
if
x
<
0
.
{\displaystyle f(x)=|x|={\begin{cases}x,&{\mbox{if }}x\geq 0\\-x,&{\mbox{if }}x<0{\text{ .}}\end{cases}}}
1
x
{\displaystyle {\frac {1}{x}}}
1
y
{\displaystyle {\frac {1}{y}}}
x
2
{\displaystyle x^{2}\,}
y
{\displaystyle {\sqrt {y}}\,}
x
n
{\displaystyle x^{n}\,}
y
n
{\displaystyle {\sqrt[{n}]{y}}}
y
1
n
{\displaystyle y^{\frac {1}{n}}}
e
x
{\displaystyle e^{x}\,}
l
n
(
y
)
{\displaystyle ln(y)\,}
a
x
{\displaystyle a^{x}\,}
l
o
g
a
(
y
)
{\displaystyle log_{a}(y)\,}
s
i
n
(
x
)
{\displaystyle sin(x)\,}
a
r
c
s
i
n
(
y
)
{\displaystyle arcsin(y)\,}
s
i
n
−
1
(
y
)
{\displaystyle sin^{-1}(y)\,}
c
o
s
(
x
)
{\displaystyle cos(x)\,}
a
r
c
c
o
s
(
y
)
{\displaystyle arccos(y)\,}
c
o
s
−
1
(
y
)
{\displaystyle cos^{-1}(y)\,}
t
a
n
(
x
)
{\displaystyle tan(x)\,}
a
r
c
t
a
n
(
y
)
{\displaystyle arctan(y)\,}
t
a
n
−
1
(
y
)
{\displaystyle tan^{-1}(y)\,}
Help:Displaying_a_formula
f
:
N
→
N
{\displaystyle f\colon \mathbb {N} \rightarrow \mathbb {N} }
f
:
{
1
,
2
,
3
,
.
.
.
}
→
{
1
,
2
,
3
,
.
.
.
}
{\displaystyle f\colon \{1,2,3,...\}\rightarrow \{1,2,3,...\}}
f
:
R
→
R
{\displaystyle f\colon \mathbb {R} \rightarrow \mathbb {R} }
f
:
x
↦
x
2
{\displaystyle f\colon \,x\mapsto x^{2}}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}\,}
From Set-builder notation
Examples:
{
x
∣
x
=
x
2
}
{\displaystyle \{x\mid x=x^{2}\}}
{
0
,
1
}
{\displaystyle \{0,1\}}
{
x
:
x
∈
R
∧
x
>
0
}
{\displaystyle \{x:x\in \mathbb {R} \land x>0\}}
positive real numbers ,
{
k
:
n
∈
N
∧
k
=
2
n
}
{\displaystyle \{k:n\in \mathbb {N} \land k=2n\}}
even natural numbers ,
{
a
:
∃
p
,
q
∈
Z
,
q
≠
0
:
a
=
p
/
q
}
{\displaystyle \{a:\exists \ p,q\in \mathbb {Z} ,q\neq 0:a=p/q\}}
rational numbers , or numbers that can be written as the ratio of two integers .
∑
n
=
1
4
(
2
n
+
1
)
{\displaystyle \sum _{n=1}^{4}(2n+1)}
Test Area Limits
lim
x
→
1
x
2
−
1
x
−
1
=
2
{\displaystyle \lim _{x\to 1}{\frac {x^{2}-1}{x-1}}=2}
lim
x
→
1
x
2
−
1
x
−
1
=
lim
x
→
1
(
x
−
1
)
(
x
+
1
)
x
−
1
=
lim
x
→
1
(
x
+
1
)
{\displaystyle \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
→
1
(
x
+
1
)
=
1
+
1
=
2
{\displaystyle \lim _{x\to 1}(x+1)=1+1=2}
x
2
−
1
x
−
1
=
(
x
−
1
)
(
x
+
1
)
x
−
1
=
x
+
1
{\displaystyle {\frac {x^{2}-1}{x-1}}={\frac {(x-1)(x+1)}{x-1}}=x+1}
lim
n
→
∞
(
1
+
1
n
)
n
=
e
{\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e}
lim
x
→
10
x
2
=
5
{\displaystyle \lim _{x\to 10}{\frac {x}{2}}=5}
lim
x
→
4
2
−
x
4
−
x
{\displaystyle \lim _{x\to 4}{\frac {2-{\sqrt {x}}}{4-x}}}
2
−
x
4
−
x
×
2
+
x
2
+
x
=
2
2
−
(
x
)
2
(
4
−
x
)
(
2
+
x
)
=
(
4
−
x
)
(
4
−
x
)
(
2
+
x
)
=
1
2
+
x
{\displaystyle {\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}}}}}
2
−
x
4
−
x
×
2
+
x
2
+
x
{\displaystyle {\frac {2-{\sqrt {x}}}{4-x}}\times {\frac {2+{\sqrt {x}}}{2+{\sqrt {x}}}}}
2
2
−
(
x
)
2
(
4
−
x
)
(
2
+
x
)
{\displaystyle {\frac {2^{2}-({\sqrt {x}})^{2}}{(4-x)(2+{\sqrt {x}})}}}
(
4
−
x
)
(
4
−
x
)
(
2
+
x
)
{\displaystyle {\frac {(4-x)}{(4-x)(2+{\sqrt {x}})}}}
1
2
+
x
{\displaystyle {\frac {1}{2+{\sqrt {x}}}}}
lim
x
→
4
2
−
x
4
−
x
=
lim
x
→
4
1
2
+
x
=
1
2
+
4
=
1
4
{\displaystyle \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
2
x
Δ
x
+
Δ
x
2
Δ
x
{\displaystyle {\frac {2x\Delta x+\Delta x^{2}}{\Delta x}}}
lim
Δ
x
→
0
2
x
Δ
x
+
Δ
x
2
Δ
x
=
lim
Δ
x
→
0
2
x
+
Δ
x
{\displaystyle \lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{\Delta x}}=\lim _{\Delta x\to 0}2x+\Delta x}
lim
Δ
x
→
0
2
x
+
Δ
x
=
2
x
{\displaystyle \lim _{\Delta x\to 0}2x+\Delta x=2x}
lim
Δ
x
→
0
2
x
Δ
x
+
Δ
x
2
Δ
x
=
lim
Δ
x
→
0
2
x
Δ
x
Δ
x
=
lim
Δ
x
→
0
2
x
=
2
x
{\displaystyle \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
Δ
x
→
0
3
x
2
Δ
x
+
3
x
Δ
x
2
+
Δ
x
3
Δ
x
=
lim
Δ
x
→
0
3
x
2
+
3
x
Δ
x
+
Δ
x
2
=
3
x
2
{\displaystyle \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
′
(
x
)
=
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
{\displaystyle f^{\prime }(x)\ =\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}
Δ
y
Δ
x
=
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}}
d
y
d
x
=
f
(
x
+
d
x
)
−
f
(
x
)
d
x
{\displaystyle {\frac {dy}{dx}}={\frac {f(x+dx)-f(x)}{dx}}\,}
d
y
d
x
{\displaystyle {\frac {dy}{dx}}\,}
=
f
(
x
+
d
x
)
−
f
(
x
)
d
x
{\displaystyle ={\frac {f(x+dx)-f(x)}{dx}}\,}
=
(
x
+
d
x
)
2
−
x
2
d
x
{\displaystyle ={\frac {(x+dx)^{2}-x^{2}}{dx}}\,}
=
x
2
+
2
x
⋅
d
x
+
d
x
2
−
x
2
d
x
{\displaystyle ={\frac {x^{2}+2x\cdot dx+dx^{2}-x^{2}}{dx}}\,}
=
2
x
⋅
d
x
+
d
x
2
d
x
{\displaystyle ={\frac {2x\cdot dx+dx^{2}}{dx}}\,}
=
2
x
+
d
x
{\displaystyle =2x+dx\,}
=
2
x
{\displaystyle =2x\,}
![{\displaystyle {\sqrt[{n}]{a^{m}}}=({\sqrt[{n}]{a}})^{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0a3f37dda326e4d2eafb747e7858fb9ea0b0974)
![{\displaystyle {\sqrt[{3}]{27^{2}}}=({\sqrt[{3}]{27}})^{2}=3^{2}=9}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d887a387d1aa91ad64eee24297a68edca816c259)
![{\displaystyle {\sqrt[{3}]{4^{6}}}=(4)^{\frac {6}{3}}=4^{2}=16}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9b43e9a56af7859ab6959eb1f65a76c19e7cb61)
![{\displaystyle {\sqrt[{n}]{a^{m}}}=a^{\frac {m}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c714f8be840b1ed81134c7179d0a72b3c0a76c9)
![{\displaystyle {\sqrt[{n}]{a}}=a^{\frac {1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2211e22214f87e6a2303d3d4c2465a3229984bde)
![{\displaystyle {\sqrt[{3}]{2^{3}}}=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bcd43ce83fea48005fdaca8d05fa3fd760d446e)
![{\displaystyle {\sqrt[{3}]{-2^{3}}}=-2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb04630cab013e894078837bb77099e6ca028d47)
![{\displaystyle {\sqrt[{4}]{-2^{4}}}=|-2|=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b232efa6bc22e73dcfe6c3033aed8df5068a85f1)
![{\displaystyle 5^{4}=625\ \ so\ \ 5={\sqrt[{4}]{625}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae0e9c195cdad618d25d858682b3e00f62364e03)
![{\displaystyle {\sqrt[{n}]{ab}}={\sqrt[{n}]{a}}\cdot {\sqrt[{n}]{b}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9135a0a2544783e81640f65650fd2e6b6c3c928a)
![{\displaystyle {\sqrt[{3}]{128}}={\sqrt[{3}]{64\cdot 2}}={\sqrt[{3}]{64}}\cdot {\sqrt[{3}]{2}}=4{\sqrt[{3}]{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dd827e35a20baaf9fd41ff1d997b14c604e2a5e)
![{\displaystyle {\sqrt[{n}]{\frac {a}{b}}}={\frac {\sqrt[{n}]{a}}{\sqrt[{n}]{b}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/372ddb7d13541806e35a6053ba614df98a87b655)
![{\displaystyle {\sqrt[{3}]{\frac {1}{64}}}={\frac {\sqrt[{3}]{1}}{\sqrt[{3}]{64}}}={\frac {1}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4474280018585533c60505dff15977c735ec1f87)
![{\displaystyle {\sqrt[{n}]{a+b}}\neq {\sqrt[{n}]{a}}+{\sqrt[{n}]{b}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61620a4858d7f93838fa5b30da60215a44ced271)
![{\displaystyle {\sqrt[{n}]{a-b}}\neq {\sqrt[{n}]{a}}-{\sqrt[{n}]{b}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bace7a8dabfa46243daa2560ef74cc5d8c064207)
![{\displaystyle {\sqrt[{n}]{a^{n}+b^{n}}}\neq a+b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93fee168a3c6ef7166a7bd1b2c3efe8063ee303a)
![{\displaystyle {\sqrt[{n}]{a^{n}}}=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bac4d6692e40c78859274a2b5c34e8cc8f5d3bc6)
![{\displaystyle {\sqrt[{3}]{a}}\times {\sqrt[{3}]{a}}\times {\sqrt[{3}]{a}}=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e88e05bc08f7afeba226c6d02034f1e8d5e718b)
![{\displaystyle \underbrace {{\sqrt[{n}]{a}}\times {\sqrt[{n}]{a}}\times ...\times {\sqrt[{n}]{a}}} _{n\ of\ them}=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac91a111b2e7d68ff80ad1b1bea5604693dabcad)
![{\displaystyle p\approx \pi \left[3(a+b)-{\sqrt {(3a+b)(a+3b)}}\right]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a1df9a75a6bf7008ab0eb473f0416ba97749a59)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1becc7e40139a18a6b95828db4235c41153321b1)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b70eb9199bf749e224a11405ea5eeed81c2656df)
![{\displaystyle p\approx \pi \left[3(r+s)-{\sqrt {(3r+s)(r+3s)}}\right]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd61b4c0e1129e4294b6067c7efa2a3f25110d5e)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d963070dc24c5e20351cc7b7cd2567723edde4e7)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e58e054ef293c52e588d274652fd0af7e891ca4)
![{\displaystyle x^{\frac {1}{n}}={\sqrt[{n}]{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5af185e5a580b4e83edd1c512c2563cb481bcd25)
![{\displaystyle 27^{\frac {1}{3}}={\sqrt[{3}]{27}}=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb900682768103f384a349465653b268144ba2ff)
![{\displaystyle x^{\frac {m}{n}}={\sqrt[{n}]{x^{m}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c52238caa2a931ae81e11a0d2628e4f8e751aa32)
![{\displaystyle {\sqrt[{n}]{625}}=5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c68fb65ebae790711e6e2505ae7d9d9952feecc9)
![{\displaystyle {\sqrt[{4}]{625}}=5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3742e7f5b46dc783da7504a025cf3d4591f93366)
![{\displaystyle {\begin{aligned}x^{\frac {m}{n}}&={\sqrt[{n}]{x^{m}}}\\&=({\sqrt[{n}]{x}})^{m}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e4b7e57245f3cff5d98e160ec10029c69c3a2c7)
![{\displaystyle x^{\frac {m}{n}}={\sqrt[{n}]{x^{m}}}=({\sqrt[{n}]{x}})^{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8297bbdfb863abf9583cda19d6422ebc98635cc)
![{\displaystyle {\begin{aligned}x^{\frac {2}{3}}&={\sqrt[{3}]{x^{2}}}\\&=({\sqrt[{3}]{x}})^{2}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2149e2636e53c39972d0b6cdd17f1bb497bc4d26)
![{\displaystyle x^{\frac {m}{n}}=x^{(m\times {\frac {1}{n}})}=(x^{m})^{\frac {1}{n}}={\sqrt[{n}]{x^{m}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20a8ba46e9d9efe0e770779c4fb33011ee31a918)
![{\displaystyle x^{\frac {m}{n}}=x^{({\frac {1}{n}}\times m)}=(x^{\frac {1}{n}})^{m}=({\sqrt[{n}]{x}})^{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d379ec811a09bd1e6154487c003056b1c436d527)
![{\displaystyle x^{\frac {2}{3}}={\sqrt[{3}]{x^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30bafddab7d6faf08ea43dd4d3285a0a1b00f511)
![{\displaystyle {\sqrt[{n}]{y}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50a063a32c43a9627099cdc262110bd579221381)
