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Balancing a circle Let the triple
(
θ
i
,
d
i
,
w
i
)
{\displaystyle (\theta _{i},d_{i},w_{i})}
Let P be some particles on a unit disk such that a given diameter axis
d
1
{\displaystyle d_{1}}
d
1
{\displaystyle d_{1}}
d
2
{\displaystyle d_{2}}
Prove that all
d
{\displaystyle d}
Consider two axis orthogonal to each other and a set of weights with the above property (which can be achieved by calculating the position of the
n
t
h
{\displaystyle n^{th}}
n
−
1
{\displaystyle n-1}
Note that the perpendiculars to a new rotated orthogonal pair of axes from a given weight all lie on the circumference of the circle centred halfway between the origin and the weight, radius half the full distance from O to W.
This applies to each weight, and so there are a set of circles each with a common point on the circumference of the origin and the new axes intersect with these circles at the new distances.
From trigonometric identities , we have:
sin
(
α
+
β
)
=
sin
α
cos
β
+
cos
α
sin
β
cos
(
α
+
β
)
=
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle {\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}}
We know:
∑
w
i
sin
α
i
=
0
∑
w
i
cos
α
i
=
0
{\displaystyle {\begin{aligned}\sum w_{i}\sin \alpha _{i}&=0\\\sum w_{i}\cos \alpha _{i}&=0\end{aligned}}}
If
α
i
{\displaystyle \alpha _{i}}
β
{\displaystyle \beta }
w
i
sin
(
α
i
+
β
)
=
w
i
sin
α
i
cos
β
+
w
i
cos
α
i
sin
β
w
i
cos
(
α
i
+
β
)
=
w
i
cos
α
i
cos
β
−
w
i
sin
α
i
sin
β
{\displaystyle {\begin{aligned}w_{i}\sin(\alpha _{i}+\beta )&=w_{i}\sin \alpha _{i}\cos \beta +w_{i}\cos \alpha _{i}\sin \beta \\w_{i}\cos(\alpha _{i}+\beta )&=w_{i}\cos \alpha _{i}\cos \beta -w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}}
and summing:
∑
w
i
sin
(
α
i
+
β
)
=
∑
w
i
sin
α
i
cos
β
+
∑
w
i
cos
α
i
sin
β
∑
w
i
cos
(
α
i
+
β
)
=
∑
w
i
cos
α
i
cos
β
−
∑
w
i
sin
α
i
sin
β
{\displaystyle {\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\sum w_{i}\sin \alpha _{i}\cos \beta +\sum w_{i}\cos \alpha _{i}\sin \beta \\\sum w_{i}\cos(\alpha _{i}+\beta )&=\sum w_{i}\cos \alpha _{i}\cos \beta -\sum w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}}
∑
w
i
sin
(
α
i
+
β
)
=
cos
β
∑
w
i
sin
α
i
+
sin
β
∑
w
i
cos
α
i
=
0
+
0
=
0
∑
w
i
cos
(
α
i
+
β
)
=
cos
β
∑
w
i
cos
α
i
−
sin
β
∑
w
i
sin
α
i
=
0
−
0
=
0
{\displaystyle {\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\sin \alpha _{i}+\sin \beta \sum w_{i}\cos \alpha _{i}=0+0=0\\\sum w_{i}\cos(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\cos \alpha _{i}-\sin \beta \sum w_{i}\sin \alpha _{i}=0-0=0\end{aligned}}}
The resultant weighted vector at the origin is therefore zero, and this is invariant over axis rotation.
Or, the lines midway between two are balanced, and this process continues ad infinitum.
Convex function proofs
edit
Consider
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
x
≤
y
,
0
≤
t
≤
1
{\displaystyle x\leq y,0\leq t\leq 1}
f
(
t
x
+
(
1
−
t
)
y
)
−
t
f
(
x
)
−
(
1
−
t
)
f
(
y
)
{\displaystyle f(tx+(1-t)y)-tf(x)-(1-t)f(y)}
=
t
2
x
2
+
2
t
(
1
−
t
)
x
y
+
(
1
−
t
)
2
y
2
−
t
x
2
−
(
1
−
t
)
y
2
{\displaystyle =t^{2}x^{2}+2t(1-t)xy+(1-t)^{2}y^{2}-tx^{2}-(1-t)y^{2}}
=
t
(
t
−
1
)
x
2
+
2
t
(
1
−
t
)
x
y
+
(
1
−
t
)
(
−
t
)
y
2
{\displaystyle =t(t-1)x^{2}+2t(1-t)xy+(1-t)(-t)y^{2}}
=
−
t
(
1
−
t
)
(
x
2
−
2
x
y
+
y
2
)
{\displaystyle =-t(1-t)\left(x^{2}-2xy+y^{2}\right)}
=
−
t
(
1
−
t
)
(
x
−
y
)
2
{\displaystyle =-t(1-t)(x-y)^{2}}
≤
0
{\displaystyle \leq 0}
Consider
f
(
x
)
=
log
(
x
)
{\displaystyle f(x)=\log(x)}
0
<
x
≤
y
,
0
≤
t
≤
1
{\displaystyle 0<x\leq y,0\leq t\leq 1}
f
(
t
x
+
(
1
−
t
)
y
)
−
t
f
(
x
)
−
(
1
−
t
)
f
(
y
)
{\displaystyle f(tx+(1-t)y)-tf(x)-(1-t)f(y)}
=
log
(
t
x
+
(
1
−
t
)
y
)
−
(
t
log
(
x
)
+
(
1
−
t
)
log
(
y
)
)
{\displaystyle =\log(tx+(1-t)y)-(t\log(x)+(1-t)\log(y))}
=
log
(
t
x
+
(
1
−
t
)
y
)
−
log
(
x
t
y
1
−
t
)
{\displaystyle =\log(tx+(1-t)y)-\log(x^{t}y^{1-t})}
=
log
(
t
x
+
(
1
−
t
)
y
x
t
y
1
−
t
)
{\displaystyle =\log({\frac {tx+(1-t)y}{x^{t}y^{1-t}}})}
So need to prove:
t
x
+
(
1
−
t
)
y
x
t
y
1
−
t
≥
1
{\displaystyle {\frac {tx+(1-t)y}{x^{t}y^{1-t}}}\geq 1}
or
t
x
+
(
1
−
t
)
y
≥
x
t
y
1
−
t
{\displaystyle tx+(1-t)y\geq x^{t}y^{1-t}}
or, dividing by
y
{\displaystyle y}
t
(
x
y
−
1
)
+
1
≥
(
x
y
)
t
{\displaystyle t\left({\frac {x}{y}}-1\right)+1\geq \left({\frac {x}{y}}\right)^{t}}
So:
t
(
x
y
−
1
)
+
1
−
(
x
y
)
t
{\displaystyle t\left({\frac {x}{y}}-1\right)+1-\left({\frac {x}{y}}\right)^{t}}
=
(
x
y
−
1
)
(
t
−
(
x
y
)
t
−
1
x
y
−
1
)
{\displaystyle =\left({\frac {x}{y}}-1\right)\left(t-{\frac {\left({\frac {x}{y}}\right)^{t}-1}{{\frac {x}{y}}-1}}\right)}
=
(
x
y
−
1
)
(
t
−
(
1
+
x
y
+
⋯
+
(
x
y
)
t
−
1
)
)
{\displaystyle =\left({\frac {x}{y}}-1\right)\left(t-(1+{\frac {x}{y}}+\dots +\left({\frac {x}{y}}\right)^{t-1})\right)}
≥
0
{\displaystyle \geq 0}
Catalan q-polynomials
edit
The Catalan q-polynomials
C
k
(
q
)
{\displaystyle C_{k}(q)}
k , and start
C
0
(
q
)
=
1
{\displaystyle C_{0}(q)=1}
C
1
(
q
)
=
1
{\displaystyle C_{1}(q)=1}
C
2
(
q
)
=
1
+
q
{\displaystyle C_{2}(q)=1+q}
C
3
(
q
)
=
1
+
q
+
2
q
2
+
q
3
{\displaystyle C_{3}(q)=1+q+2q^{2}+q^{3}}
C
4
(
q
)
=
1
+
q
+
2
q
2
+
3
q
3
+
3
q
4
+
3
q
5
+
q
6
{\displaystyle C_{4}(q)=1+q+2q^{2}+3q^{3}+3q^{4}+3q^{5}+q^{6}}
The sum of the coefficients of
C
k
(
q
)
{\displaystyle C_{k}(q)}
C
k
{\displaystyle C_{k}}
To generate the next level, we add a horizontal and vertical step. The horizontal step is always placed at the origin, and the vertical step can be placed anywhere off the bounding diagonal, and is the first time the path touches the diagonal.
The first path from the convolution is inserted along the (k-1) -th diagonal created by the two endpoints of the new steps, and the second is placed along the k -th diagonal.
With Dyck words, where the new template is
X
(
k
)
Y
(
n
−
k
)
{\displaystyle X(k)Y(n-k)}
(
k
)
{\displaystyle (k)}
k . The X step is always first, and the Y step comes after a k -length Dyck word.
The new polynomial is therefore the heights of the previous polynomials, plus the rectangle created by the insertion.
C
n
(
q
)
=
∑
i
=
0
n
−
1
q
i
(
n
−
i
)
C
i
(
q
)
C
n
−
i
(
q
)
{\displaystyle C_{n}(q)=\sum _{i=0}^{n-1}q^{i(n-i)}C_{i}(q)C_{n-i}(q)}
For example,
C
4
(
q
)
=
q
0
(
1
)
(
1
+
q
+
2
q
2
+
q
3
)
+
q
3
(
1
)
(
1
+
q
)
+
q
4
(
q
+
1
)
(
1
)
+
q
3
(
1
+
q
+
2
q
2
+
q
3
)
(
1
)
{\displaystyle C_{4}(q)=q^{0}(1)(1+q+2q^{2}+q^{3})+q^{3}(1)(1+q)+q^{4}(q+1)(1)+q^{3}(1+q+2q^{2}+q^{3})(1)}
12 34 56
13 26 45
14 25 36
15 23 46
16 24 35 12 34 58 67
13 24 57 68
14 25 36 78
15 26 37 48
16 27 38 45
17 28 35 46
18 23 47 56
1
+
4
x
+
x
2
(
1
−
x
)
4
=
1
+
8
x
+
27
x
2
+
64
x
3
+
125
x
4
+
…
{\displaystyle {\frac {1+4x+x^{2}}{(1-x)^{4}}}=1+8x+27x^{2}+64x^{3}+125x^{4}+\dots }
https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E4
The general relation is given by the Eulerian numbers .
The GF's for
1
,
x
,
x
2
{\displaystyle 1,x,x^{2}}
1
1
−
x
,
x
(
1
−
x
)
2
,
x
2
+
x
(
1
−
x
)
3
{\displaystyle {\frac {1}{1-x}},{\frac {x}{(1-x)^{2}}},{\frac {x^{2}+x}{(1-x)^{3}}}}
Therefore
3
(
x
2
+
x
)
(
1
−
x
)
3
+
3
x
(
1
−
x
)
2
+
1
1
−
x
{\displaystyle {\frac {3(x^{2}+x)}{(1-x)^{3}}}+{\frac {3x}{(1-x)^{2}}}+{\frac {1}{1-x}}}
generates
3
x
2
+
3
x
+
1
{\displaystyle 3x^{2}+3x+1}
This equals
3
(
x
2
+
x
)
+
3
x
(
1
−
x
)
+
(
1
−
2
x
+
x
2
)
(
1
−
x
)
3
{\displaystyle {\frac {3(x^{2}+x)+3x(1-x)+(1-2x+x^{2})}{(1-x)^{3}}}}
=
1
+
4
x
+
x
2
(
1
−
x
)
3
{\displaystyle ={\frac {1+4x+x^{2}}{(1-x)^{3}}}}
and generates the sequence
1
,
7
,
19
,
37
,
…
{\displaystyle 1,7,19,37,\dots }
https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E3
Multiplying by
1
1
−
x
{\displaystyle {\frac {1}{1-x}}}
1
+
4
x
+
x
2
(
1
−
x
)
4
{\displaystyle {\frac {1+4x+x^{2}}{(1-x)^{4}}}}
generates the sequence
1
,
8
,
27
,
64
,
…
{\displaystyle 1,8,27,64,\dots }
Again gives
1
+
4
x
+
x
2
(
1
−
x
)
5
{\displaystyle {\frac {1+4x+x^{2}}{(1-x)^{5}}}}
generates the sequence
1
,
9
,
36
,
100
,
…
{\displaystyle 1,9,36,100,\dots }
https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E5
Prove
(
m
j
)
(
n
k
)
≤
(
m
+
n
j
+
k
)
{\displaystyle {\binom {m}{j}}{\binom {n}{k}}\leq {\binom {m+n}{j+k}}}
=
m
!
n
!
j
!
k
!
(
m
−
j
)
!
(
n
−
k
)
!
≤
(
m
+
n
)
!
(
j
+
k
)
!
(
m
+
n
−
j
−
k
)
!
{\displaystyle ={\frac {m!n!}{j!k!(m-j)!(n-k)!}}\leq {\frac {(m+n)!}{(j+k)!(m+n-j-k)!}}}
Consider
(
m
+
n
)
!
m
!
n
!
j
!
k
!
(
j
+
k
)
!
(
m
−
j
)
!
(
n
−
k
)
!
(
m
+
n
−
j
−
k
)
!
{\displaystyle {\frac {(m+n)!}{m!n!}}{\frac {j!k!}{(j+k)!}}{\frac {(m-j)!(n-k)!}{(m+n-j-k)!}}}
where
(
a
+
b
)
!
a
!
b
!
=
(
a
+
b
a
)
=
(
a
+
1
)
⋯
(
a
+
b
)
b
!
=
∏
i
=
1
b
(
1
+
a
i
)
{\displaystyle {\frac {(a+b)!}{a!b!}}={\binom {a+b}{a}}={\frac {(a+1)\cdots (a+b)}{b!}}=\prod _{i=1}^{b}\left(1+{\frac {a}{i}}\right)}
so
∏
i
=
1
m
(
1
+
n
i
)
∏
i
=
1
j
(
1
+
k
i
)
∏
i
=
1
m
−
j
(
1
+
n
−
k
i
)
{\displaystyle {\frac {\prod _{i=1}^{m}\left(1+{\frac {n}{i}}\right)}{\prod _{i=1}^{j}\left(1+{\frac {k}{i}}\right)\prod _{i=1}^{m-j}\left(1+{\frac {n-k}{i}}\right)}}}
and
(
m
+
n
m
)
(
j
+
k
j
)
(
m
+
n
−
j
−
k
m
−
j
)
{\displaystyle {\frac {\binom {m+n}{m}}{{\binom {j+k}{j}}{\binom {m+n-j-k}{m-j}}}}}
The meaning of logic
edit
TRUTH
0000 (0) FALSE
0110 (6) XOR
1001 (9) NXOR
1111 (F) TRUE IDEMPOTENT
0001 (1) AND
0011 (3) A
0101 (5) B
0111 (7) OR INJECTIVE
0100 (4) LT
1100 (C) NB
1101 (D) LTE
1110 (E) NAND SURJECTIVE
0010 (2) GT
1000 (8) NOR
1011 (B) GTE
1010 (A) NA Partial sums of binomials
edit
∑
i
=
0
k
(
n
−
k
−
1
+
i
i
)
2
n
−
k
−
i
{\displaystyle \sum _{i=0}^{k}{\binom {n-k-1+i}{i}}2^{n-k-i}}
n
=
6
,
k
=
3
:
1
+
6
+
15
+
20
=
42
{\displaystyle n=6,k=3:1+6+15+20=42}
(
2
0
)
2
3
+
(
3
1
)
2
2
+
(
4
2
)
2
+
(
5
3
)
=
8
+
12
+
12
+
10
=
4
{\displaystyle {\binom {2}{0}}2^{3}+{\binom {3}{1}}2^{2}+{\binom {4}{2}}2+{\binom {5}{3}}=8+12+12+10=4}
GF for Catalan numbers
edit
Catalan number
Start with the GF for the central binomial coefficient .
1
1
−
4
x
=
∑
k
=
0
∞
(
2
k
k
)
x
k
{\displaystyle {\frac {1}{\sqrt {1-4x}}}=\sum _{k=0}^{\infty }{\binom {2k}{k}}x^{k}}
Integrating and setting the constant to
1
2
{\displaystyle {\frac {1}{2}}}
x
=
0
{\displaystyle x=0}
1
−
1
−
4
x
2
=
∑
k
=
0
∞
1
k
+
1
(
2
k
k
)
x
k
+
1
{\displaystyle {\frac {1-{\sqrt {1-4x}}}{2}}=\sum _{k=0}^{\infty }{\frac {1}{k+1}}{\binom {2k}{k}}x^{k+1}}
Divide by x to get
1
−
1
−
4
x
2
x
=
∑
k
=
0
∞
C
k
x
k
{\displaystyle {\frac {1-{\sqrt {1-4x}}}{2x}}=\sum _{k=0}^{\infty }C_{k}x^{k}}
Primitive root modulo n
The order of an element is the smallest k where
a
k
≡
1
(
mod
n
)
{\displaystyle a^{k}\equiv 1{\pmod {n}}}
k divides
ϕ
(
n
)
{\displaystyle \phi (n)}
k returns negative if and only if the candidate is a primitive root.
a/k
1
2
3
4
5
6
1
1
1
1
1
1
1
2
2
4
1
2
4
1
3
3
2
6
4
5
1
4
4
2
1
4
2
1
5
5
4
6
2
3
1
6
6
1
6
1
6
1
Fermat's theorem on sums of two squares
(5) can be proved by Euler's criterion . It also tells us that if
p
=
4
k
+
1
{\displaystyle p=4k+1}
p
|
∏
i
=
1
4
k
(
i
2
k
+
(
i
−
1
)
2
k
)
{\displaystyle p|\prod _{i=1}^{4k}\left(i^{2k}+(i-1)^{2k}\right)}
which is the sum of two squares by (1). (2), (3) and (4) can then be used to find it.
Finding primitive roots for primes
edit
For each prime p of p-1, remove k^p from 1,..,p-1 as k runs over the same range. Only primitive roots remain.
edit
a
k
≡
1
(
mod
p
)
{\displaystyle a^{k}\equiv 1{\pmod {p}}}
k
|
p
−
1
{\displaystyle k|p-1}
For example,
x
7
−
1
{\displaystyle x^{7}-1}
13 , 17 , etc..., but is by 29 .
Chinese Remainder Theorem
edit
Coordinates of circumcentre
edit
Let
△
A
B
C
{\displaystyle \triangle ABC}
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
,
(
x
3
,
y
3
)
{\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})}
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle \left({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}}\right)}
(
x
1
+
x
3
2
,
y
1
+
y
3
2
)
{\displaystyle \left({\frac {x_{1}+x_{3}}{2}},{\frac {y_{1}+y_{3}}{2}}\right)}
The perpendiculars are
(
y
2
−
y
1
x
1
−
x
2
)
{\displaystyle {\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}}
(
y
3
−
y
1
x
1
−
x
3
)
{\displaystyle {\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}}
and the equations of the perpendiculars through the midpoints are
(
x
y
)
=
(
x
1
+
x
2
2
y
1
+
y
2
2
)
+
s
(
y
2
−
y
1
x
1
−
x
2
)
{\displaystyle {\binom {x}{y}}={\binom {\frac {x_{1}+x_{2}}{2}}{\frac {y_{1}+y_{2}}{2}}}+s{\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}}
(
x
y
)
=
(
x
1
+
x
3
2
y
1
+
y
3
2
)
+
t
(
y
3
−
y
1
x
1
−
x
3
)
{\displaystyle {\binom {x}{y}}={\binom {\frac {x_{1}+x_{3}}{2}}{\frac {y_{1}+y_{3}}{2}}}+t{\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}}
with equality when
x
1
+
x
2
2
+
s
(
y
2
−
y
1
)
=
x
1
+
x
3
2
+
t
(
y
3
−
y
1
)
{\displaystyle {\frac {x_{1}+x_{2}}{2}}+s(y_{2}-y_{1})={\frac {x_{1}+x_{3}}{2}}+t(y_{3}-y_{1})}
y
1
+
y
2
2
+
s
(
x
1
−
x
2
)
=
y
1
+
y
3
2
+
t
(
x
1
−
x
3
)
{\displaystyle {\frac {y_{1}+y_{2}}{2}}+s(x_{1}-x_{2})={\frac {y_{1}+y_{3}}{2}}+t(x_{1}-x_{3})}
so
(
x
2
−
x
3
)
+
2
s
(
y
2
−
y
1
)
=
2
t
(
y
3
−
y
1
)
{\displaystyle (x_{2}-x_{3})+2s(y_{2}-y_{1})=2t(y_{3}-y_{1})}
(
y
2
−
y
3
)
+
2
s
(
x
1
−
x
2
)
=
2
t
(
x
1
−
x
3
)
{\displaystyle (y_{2}-y_{3})+2s(x_{1}-x_{2})=2t(x_{1}-x_{3})}
and then
(
x
2
−
x
3
)
(
x
1
−
x
2
)
+
2
s
(
x
1
−
x
2
)
(
y
2
−
y
1
)
=
2
t
(
x
1
−
x
2
)
(
y
3
−
y
1
)
{\displaystyle (x_{2}-x_{3})(x_{1}-x_{2})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{2})(y_{3}-y_{1})}
(
y
2
−
y
3
)
(
y
2
−
y
1
)
+
2
s
(
x
1
−
x
2
)
(
y
2
−
y
1
)
=
2
t
(
x
1
−
x
3
)
(
y
2
−
y
1
)
{\displaystyle (y_{2}-y_{3})(y_{2}-y_{1})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{3})(y_{2}-y_{1})}
so
(
x
2
−
x
3
)
(
x
1
−
x
2
)
−
(
y
2
−
y
3
)
(
y
2
−
y
1
)
=
2
t
(
x
1
y
3
−
x
2
y
3
+
x
2
y
1
−
x
1
y
2
+
x
3
y
2
−
x
3
y
1
)
{\displaystyle (x_{2}-x_{3})(x_{1}-x_{2})-(y_{2}-y_{3})(y_{2}-y_{1})=2t(x_{1}y_{3}-x_{2}y_{3}+x_{2}y_{1}-x_{1}y_{2}+x_{3}y_{2}-x_{3}y_{1})}
Hamiltonian cycles
Generate the permutations of all vertices. Remove all those with non-adjacent vertices.
Two cuts are adjacent if the corresponding partitions differ by only 1 vertex.
An integer in base b can be written as
n
=
∑
i
=
0
N
a
i
b
i
{\displaystyle n=\sum _{i=0}^{N}a_{i}b^{i}}
and the same integer in base d is
n
=
∑
i
=
0
M
c
i
d
i
{\displaystyle n=\sum _{i=0}^{M}c_{i}d^{i}}
If
d
+
e
=
b
{\displaystyle d+e=b}
n
=
∑
i
=
0
N
a
i
(
d
+
e
)
i
{\displaystyle n=\sum _{i=0}^{N}a_{i}(d+e)^{i}}
Expand this using the binomial formula to get
n
=
∑
i
=
0
N
a
i
∑
k
=
0
i
(
i
k
)
d
k
e
i
−
k
{\displaystyle n=\sum _{i=0}^{N}a_{i}\sum _{k=0}^{i}{\binom {i}{k}}d^{k}e^{i-k}}
and collect by like terms in d to get the second equation.
n
=
∑
k
=
0
N
d
k
∑
i
=
k
N
a
i
(
i
k
)
e
i
−
k
{\displaystyle n=\sum _{k=0}^{N}d^{k}\sum _{i=k}^{N}a_{i}{\binom {i}{k}}e^{i-k}}
because
∑
i
=
0
N
∑
k
=
0
i
=
∑
k
=
0
N
∑
i
=
k
N
{\displaystyle \sum _{i=0}^{N}\sum _{k=0}^{i}=\sum _{k=0}^{N}\sum _{i=k}^{N}}
using
i
+
k
=
N
{\displaystyle i+k=N}
(
i
,
k
)
{\displaystyle (i,k)}
We want to add two bytes A and B . We use auxiliary bytes C and D to store the carries and answer, and a carry flag for the final bits.
Let C be initially zero, and a ,b ,c ,d be aligned bits.
Then
d
=
a
⊗
b
⊗
c
{\displaystyle d=a\otimes b\otimes c}
c
′
=
(
a
∧
b
)
∨
(
a
∧
c
)
∨
(
b
∧
c
)
{\displaystyle c'=(a\land b)\lor (a\land c)\lor (b\land c)}
where
⊗
{\displaystyle \otimes }
XOR and c' is the next carry bit.
If f and g are continuous functions on
[
a
,
b
]
{\displaystyle [a,b]}
h
(
x
,
λ
)
=
λ
f
(
x
)
+
(
1
−
λ
)
g
(
x
)
{\displaystyle h(x,\lambda )=\lambda f(x)+(1-\lambda )g(x)}
If x is fixed,
h
(
x
,
λ
+
d
λ
)
−
h
(
x
,
λ
)
=
d
λ
(
f
(
x
)
+
g
(
x
)
)
{\displaystyle h(x,\lambda +d\lambda )-h(x,\lambda )=d\lambda (f(x)+g(x))}
which is continuous as
f
(
x
)
+
g
(
x
)
{\displaystyle f(x)+g(x)}
If
λ
{\displaystyle \lambda }
h
(
x
+
d
x
,
λ
)
−
h
(
x
,
λ
)
=
λ
(
f
(
x
+
d
x
)
−
f
(
x
)
)
+
(
1
−
λ
)
(
g
(
x
+
d
x
)
−
g
(
x
)
)
{\displaystyle h(x+dx,\lambda )-h(x,\lambda )=\lambda (f(x+dx)-f(x))+(1-\lambda )(g(x+dx)-g(x))}
which is continuous because
f
(
x
)
,
g
(
x
)
{\displaystyle f(x),g(x)}
3-sphere
The notion of gluing can be extended from the 2d 'above/below' into 3d with inside (the bounding sphere)/outside.
As a 3-sphere can be stereographically projected to the one point compactification of
R
3
{\displaystyle \mathbb {R^{3}} }
Note that the removal of the 'one-point' creates a hole.
The stereographic projection of the 2-sphere onto the 3-sphere matches to the inverse projection of the 4-sphere.
A torus isn't locally Euclidean as the inner ring necessarily has a higher measure (point density) than the outer ring.
Client-Vendor-Bank model
edit
The Client-Vendor-Bank model (CVB) involves the oAuth model.
Only the Authentication and Authorization layers are used, along with vendorId and vendorSecret. These are registered with the Bank.
The steps involved are:
1. C sends V transaction request
2. V asks C for Authentication with vendorId and transaction details from B
3. B replies to V with the AuthCode for the transaction
4. V posts AuthCode, vendorId and vendorSecret to B
5. B commits the transaction and sends receipt to V
6. V sends receipt to C
This model has minimal external interference, and is secure as B has two potentially conflicting items for the same transaction.
Catalan interpretation
edit
The 15 Bell numbers for n = 4 with the Catalan equivalent:
1234 - (((())))
1 234 - ()((()))
2 134 - (()(()))
3 124 - ((()()))
4 123 - ((()))()
12 34 - (())(())
13 24 - crossed
14 23 = ((())())
12 3 4 - (())()()
13 2 4 - (()())()
14 2 3 - (()()())
23 1 4 - ()(())()
24 1 3 - ()(()())
34 1 2 - ()()(())
1 2 3 4 - ()()()()
The congruent number problem asks which square numbers are in an arithmetic progression of 3.
Are there higher powers such that 3 lie in an AP?
That is
a
n
+
x
=
b
n
{\displaystyle a^{n}+x=b^{n}}
b
n
+
x
=
c
n
{\displaystyle b^{n}+x=c^{n}}
or
a
n
−
b
n
=
b
n
−
c
n
{\displaystyle a^{n}-b^{n}=b^{n}-c^{n}}
so
a
n
+
c
n
=
2
b
n
{\displaystyle a^{n}+c^{n}=2b^{n}}
A random walk on a 2d grid arrives at coordinate
(
x
,
y
)
{\displaystyle (x,y)}
z steps, where
R
(
x
,
y
,
z
)
=
R
(
x
−
1
,
y
,
z
−
1
)
+
R
(
x
+
1
,
y
,
z
−
1
)
+
R
(
x
,
y
−
1
,
z
−
1
)
+
R
(
x
,
y
+
1
,
z
−
1
)
{\displaystyle R(x,y,z)=R(x-1,y,z-1)+R(x+1,y,z-1)+R(x,y-1,z-1)+R(x,y+1,z-1)}
with
R
(
0
,
0
,
0
)
=
1
{\displaystyle R(0,0,0)=1}
This leads to the definition
R
(
x
,
y
,
z
)
=
∑
k
=
0
z
−
x
−
y
z
!
(
x
+
k
)
!
k
!
(
y
+
z
−
x
−
y
−
k
)
!
(
z
−
x
−
y
−
k
)
!
{\displaystyle R(x,y,z)=\sum _{k=0}^{z-x-y}{\frac {z!}{(x+k)!k!(y+z-x-y-k)!(z-x-y-k)!}}}
ln
(
(
1
−
x
)
(
1
−
x
)
)
=
(
1
−
x
)
ln
(
1
−
x
)
{\displaystyle \ln((1-x)^{(1-x)})=(1-x)\ln(1-x)}
=
lim
x
→
1
(
1
−
x
)
(
−
x
−
x
2
2
−
x
3
3
−
…
)
{\displaystyle =\lim _{x\to 1}(1-x)\left(-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\dots \right)}
=
lim
x
→
1
(
x
−
1
)
∑
i
=
1
∞
x
i
i
{\displaystyle =\lim _{x\to 1}(x-1)\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}}
=
lim
x
→
1
∑
i
=
1
∞
x
i
+
1
i
−
∑
i
=
1
∞
x
i
i
{\displaystyle =\lim _{x\to 1}\sum _{i=1}^{\infty }{\frac {x^{i+1}}{i}}-\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}}
=
lim
x
→
1
∑
i
=
1
∞
x
i
+
1
−
x
i
i
{\displaystyle =\lim _{x\to 1}\sum _{i=1}^{\infty }{\frac {x^{i+1}-x^{i}}{i}}}
=
∑
i
=
1
∞
1
−
1
i
{\displaystyle =\sum _{i=1}^{\infty }{\frac {1-1}{i}}}
=
0
{\displaystyle =0}
Consider the function
x
n
{\displaystyle x^{n}}
[
0
,
y
]
{\displaystyle [0,y]}
0
,
h
,
2
h
,
…
,
k
h
=
y
{\displaystyle 0,h,2h,\dots ,kh=y}
The area of each rectangle at point
q
h
{\displaystyle qh}
(
q
h
)
n
⋅
h
{\displaystyle (qh)^{n}\cdot h}
h
∑
j
=
0
k
(
j
h
)
n
=
h
n
+
1
∑
j
=
0
k
j
n
{\displaystyle h\sum _{j=0}^{k}(jh)^{n}=h^{n+1}\sum _{j=0}^{k}j^{n}}
which is
h
n
+
1
{\displaystyle h^{n+1}}
sum of powers given by
S
n
(
k
)
=
1
n
+
1
∑
r
=
0
n
(
n
+
1
r
)
B
r
k
n
−
r
+
1
{\displaystyle S_{n}(k)={\frac {1}{n+1}}\sum _{r=0}^{n}{\binom {n+1}{r}}B_{r}k^{n-r+1}}
so
lim
h
→
0
h
n
+
1
S
n
(
k
)
=
1
n
+
1
(
h
k
)
n
+
1
{\displaystyle \lim _{h\to 0}h^{n+1}S_{n}(k)={\frac {1}{n+1}}(hk)^{n+1}}
and as
h
k
=
y
{\displaystyle hk=y}
∫
0
y
x
n
d
x
=
y
n
+
1
n
+
1
{\displaystyle \int _{0}^{y}x^{n}dx={\frac {y^{n+1}}{n+1}}}
Square pyramid numbers
edit
Imagine the sample space
000
001
010
011
100
101
110
111
If you pick a random entry from each row, you will have picked both a 0 and a 1 from at least one column.
Say you pick column c from row r , and it is b . Then all the rows with
¬
b
{\displaystyle \lnot b}
c have this column removed from the options, which halves the size of the sample space. After n turns therefore the sample space is reduced from
2
n
{\displaystyle 2^{n}}
1 , but the final row has no available moves.
determinant
det
(
a
+
b
,
c
+
d
,
e
+
f
)
=
det
(
a
,
c
,
e
)
+
det
(
a
,
c
,
f
)
+
det
(
a
,
d
,
e
)
+
det
(
a
,
d
,
f
)
+
det
(
b
,
c
,
e
)
+
det
(
b
,
c
,
f
)
+
det
(
b
,
d
,
e
)
+
det
(
b
,
d
,
f
)
{\displaystyle \det(a+b,c+d,e+f)=\det(a,c,e)+\det(a,c,f)+\det(a,d,e)+\det(a,d,f)+\det(b,c,e)+\det(b,c,f)+\det(b,d,e)+\det(b,d,f)}
Eccentricity yields
x
2
+
(
y
−
f
)
2
=
e
(
y
+
f
)
2
{\displaystyle x^{2}+(y-f)^{2}=e(y+f)^{2}}
(from Parabola#Axis of symmetry parallel to the y axis ).
so the equations for the conic sections are
x
2
=
(
e
−
1
)
(
y
2
+
f
2
)
+
2
(
e
+
1
)
y
f
{\displaystyle x^{2}=(e-1)(y^{2}+f^{2})+2(e+1)yf}
or
x
2
=
(
e
−
1
)
(
y
+
f
)
2
+
4
y
f
{\displaystyle x^{2}=(e-1)(y+f)^{2}+4yf}
We wish to add an arbitrary (finite) list of positive integers with a given maximum size, e.g. 4,12,2,7,0. The 0 indicates the end of the list.
A ruple is executed in the following order:
1. Write
2..Move
3. State
There are no 'if' statements (hence 'P' - an 'if' statement is non-deterministic to a qubit ). Movement is always R, and no symbol is written except for at 0.
The transition function is a basically hand-written sum table, e.g.
symbol 1 2 3 4 ...
state 0: 1 2 3 4 ...
state 1: 2 3 4 5...
...
state k: k+1 k+2 k+3 k+4...
Symbol 0 writes the current state down as a symbol.
state k, symvol 0, write k .
Complex exponentiation
edit
(
r
e
i
θ
)
i
=
e
−
θ
e
i
ln
r
,
(
r
e
i
θ
)
−
i
=
e
θ
e
−
i
ln
r
{\displaystyle \left(re^{i\theta }\right)^{i}=e^{-\theta }e^{i\ln r},\left(re^{i\theta }\right)^{-i}=e^{\theta }e^{-i\ln r}}
(
(
r
e
i
θ
)
i
)
i
=
(
r
e
i
θ
)
−
1
=
1
r
e
−
i
θ
{\displaystyle \left(\left(re^{i\theta }\right)^{i}\right)^{i}=\left(re^{i\theta }\right)^{-1}={\frac {1}{r}}e^{-i\theta }}
Symmetric group
If
[
a
,
b
]
{\displaystyle [a,b]}
a -th element with the b -element, then
r
=
[
1
,
2
]
[
1
,
3
]
…
[
1
,
n
]
{\displaystyle r=[1,2][1,3]\dots [1,n]}
1 element, hence
r
n
=
I
{\displaystyle r^{n}=I}
Cross product
(
a
b
c
)
×
(
d
e
f
)
=
(
0
−
c
b
c
0
−
a
−
b
a
0
)
(
d
e
f
)
{\displaystyle {\begin{pmatrix}a\\b\\c\end{pmatrix}}\times {\begin{pmatrix}d\\e\\f\end{pmatrix}}={\begin{pmatrix}0\,\,\,-c\,\,\,b\\c\,\,\,0\,\,\,-a\\-b\,\,\,a\,\,\,0\end{pmatrix}}{\begin{pmatrix}d\\e\\f\end{pmatrix}}}
Navier-Stokes equations
edit
If the boundary is a sphere then the equations are harmonic , i.e. the Laplacian is zero. The outside of a sphere is unbounded, and therefore the pressure is zero.
ρ
(
∂
u
∂
t
+
u
⋅
∇
u
)
=
−
∇
p
+
∇
⋅
[
μ
(
∇
u
+
(
∇
u
)
T
)
]
+
ρ
g
{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+\rho \mathbf {g} }
Derivation of the Navier–Stokes equations#Incompressible Newtonian fluid
If we ignore the 'extra' bit, there are therefore two PDE's to solve, namely, no pressure
ρ
(
∂
u
∂
t
+
u
⋅
∇
u
)
=
∇
⋅
[
μ
(
∇
u
+
(
∇
u
)
T
)
]
{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]}
and no movement
ρ
(
∂
u
∂
t
+
u
⋅
∇
u
)
=
−
∇
p
{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p}
Analyticity of holomorphic functions
f
(
z
)
=
∑
n
=
0
∞
c
n
(
z
−
a
)
n
{\displaystyle f(z)=\sum _{n=0}^{\infty }c_{n}(z-a)^{n}}
Let
z
=
a
+
r
e
i
θ
{\displaystyle z=a+re^{i\theta }}
a
∈
C
,
r
,
θ
∈
R
{\displaystyle a\in \mathbb {C} ,r,\theta \in \mathbb {R} }
Then, if f is holomorphic,
|
f
(
a
+
r
)
|
≥
|
f
(
a
+
r
e
i
θ
|
{\displaystyle |f(a+r)|\geq |f(a+re^{i\theta }|}
i.e.
θ
=
0
{\displaystyle \theta =0}
f .
Diffferentiation of theta function
edit
Area of a parallelogram
edit
From the picture, the area of the parallelogram is
(
a
+
c
)
(
b
+
d
)
−
a
b
−
c
d
−
2
b
c
=
a
d
−
b
c
{\displaystyle (a+c)(b+d)-ab-cd-2bc=ad-bc}
Random walks on a 2d lattice are governed by
R
(
d
,
x
,
y
)
=
R
(
d
−
1
,
x
−
1
,
y
)
+
R
(
d
−
1
,
x
+
1
,
y
)
+
R
(
d
−
1
,
x
,
y
−
1
)
+
R
(
d
−
1
,
x
,
y
+
1
)
{\displaystyle R(d,x,y)=R(d-1,x-1,y)+R(d-1,x+1,y)+R(d-1,x,y-1)+R(d-1,x,y+1)}
After
2
d
{\displaystyle 2d}
(
x
,
y
)
{\displaystyle (x,y)}
x
+
y
{\displaystyle x+y}
Let
a
=
x
+
y
2
,
b
=
x
−
y
2
{\displaystyle a={\frac {x+y}{2}},b={\frac {x-y}{2}}}
(
0
,
0
,
0
)
{\displaystyle (0,0,0)}
(
2
d
,
x
,
y
)
{\displaystyle (2d,x,y)}
R
(
2
d
,
x
,
y
)
=
(
2
d
d
+
a
)
(
2
d
d
+
b
)
{\displaystyle R(2d,x,y)={\binom {2d}{d+a}}{\binom {2d}{d+b}}}
(
k
a
)
(
k
b
)
{\displaystyle {\binom {k}{a}}{\binom {k}{b}}}
=
(
(
k
−
1
a
−
1
)
+
(
k
−
1
a
)
)
(
(
k
−
1
b
−
1
)
+
(
k
−
1
b
)
)
{\displaystyle =\left({\binom {k-1}{a-1}}+{\binom {k-1}{a}}\right)\left({\binom {k-1}{b-1}}+{\binom {k-1}{b}}\right)}
=
(
k
−
1
a
−
1
)
(
k
−
1
b
−
1
)
+
(
k
−
1
a
)
(
k
−
1
b
−
1
)
+
(
k
−
1
a
−
1
)
(
k
−
1
b
)
+
(
k
−
1
a
)
(
k
−
1
b
)
{\displaystyle ={\binom {k-1}{a-1}}{\binom {k-1}{b-1}}+{\binom {k-1}{a}}{\binom {k-1}{b-1}}+{\binom {k-1}{a-1}}{\binom {k-1}{b}}+{\binom {k-1}{a}}{\binom {k-1}{b}}}
Vertex cover
Let V be a subset of vertices and V' be its 1-adjacent neighbourhood.
If
V
∩
V
′
=
∅
{\displaystyle V\cap V'=\emptyset }
v
,
v
′
∈
V
⟹
N
(
v
)
∩
N
(
v
′
)
=
∅
{\displaystyle v,v'\in V\implies N(v)\cap N(v')=\emptyset }
V
∪
V
′
=
V
G
{\displaystyle V\cup V'=V_{G}}
then V is a perfect cover in G.
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle [0,y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b72d202f0d734e94c466e0b6bc8b34b5a83d155b)
![{\displaystyle r=[1,2][1,3]\dots [1,n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79eead25f43e5339f914902a4758fa5f05f38275)
![{\displaystyle \int _{-\infty }^{\infty }S(t)e^{iTx}dx=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }s(x)e^{-ix(t-T)}dx\right]dy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/028e87e5a024ff68660aa553bc379b2826f504eb)
![{\displaystyle \int _{-\infty }^{\infty }S(t)e^{iTx}dx=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }s(x)e^{-ix(t-T)}dy\right]dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53ef8096354995b9c1421487b9e0e2f45b6db29e)
![{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+\rho \mathbf {g} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6d944b99d8ad4c39438d58e4e964ec1eb04b6b6)
![{\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2500ed4bc4e92cca13499d3ef9cc59fb65f051d1)
