User:Darcourse/sandbox

Balancing a disc

Balancing a circle

Let the triple $(\theta _{i},d_{i},w_{i})$ be the angle, distance and weight of a particle from the centre of the disc.

Let P be some particles on a unit disk such that a given diameter axis $d_{1}$ means that the sum of perpendiculars from P to $d_{1}$ is zero. Let $d_{2}$ be a distinct other axis with the same property.

Prove that all $d$ have this property.

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 $n^{th}$ weight after the first $n-1$ weights have been placed).

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:

${\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}$

We know:

${\begin{aligned}\sum w_{i}\sin \alpha _{i}&=0\\\sum w_{i}\cos \alpha _{i}&=0\end{aligned}}$

If $\alpha _{i}$ is the original angle, and $\beta$ is the rotated amount, then we have:

${\begin{aligned}w_{i}\sin(\alpha _{i}+\beta )&=w_{i}\sin \alpha _{i}\cos \beta +w_{i}\cos \alpha _{i}\sin \beta \\w_{i}\cos(\alpha _{i}+\beta )&=w_{i}\cos \alpha _{i}\cos \beta -w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}$

and summing:

${\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\sum w_{i}\sin \alpha _{i}\cos \beta +\sum w_{i}\cos \alpha _{i}\sin \beta \\\sum w_{i}\cos(\alpha _{i}+\beta )&=\sum w_{i}\cos \alpha _{i}\cos \beta -\sum w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}$

${\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\sin \alpha _{i}+\sin \beta \sum w_{i}\cos \alpha _{i}=0+0=0\\\sum w_{i}\cos(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\cos \alpha _{i}-\sin \beta \sum w_{i}\sin \alpha _{i}=0-0=0\end{aligned}}$

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

Consider $f(x)=x^{2}$, $x\leq y,0\leq t\leq 1$. Then:

$f(tx+(1-t)y)-tf(x)-(1-t)f(y)$

$=t^{2}x^{2}+2t(1-t)xy+(1-t)^{2}y^{2}-tx^{2}-(1-t)y^{2}$

$=t(t-1)x^{2}+2t(1-t)xy+(1-t)(-t)y^{2}$

$=-t(1-t)\left(x^{2}-2xy+y^{2}\right)$

$=-t(1-t)(x-y)^{2}$

$\leq 0$

---

Consider $f(x)=\log(x)$, $0<x\leq y,0\leq t\leq 1$. Then:

$f(tx+(1-t)y)-tf(x)-(1-t)f(y)$

$=\log(tx+(1-t)y)-(t\log(x)+(1-t)\log(y))$

$=\log(tx+(1-t)y)-\log(x^{t}y^{1-t})$

$=\log({\frac {tx+(1-t)y}{x^{t}y^{1-t}}})$

So need to prove:

${\frac {tx+(1-t)y}{x^{t}y^{1-t}}}\geq 1$

or

$tx+(1-t)y\geq x^{t}y^{1-t}$

or, dividing by $y$:

$t\left({\frac {x}{y}}-1\right)+1\geq \left({\frac {x}{y}}\right)^{t}$

So:

$t\left({\frac {x}{y}}-1\right)+1-\left({\frac {x}{y}}\right)^{t}$

$=\left({\frac {x}{y}}-1\right)\left(t-{\frac {\left({\frac {x}{y}}\right)^{t}-1}{{\frac {x}{y}}-1}}\right)$

$=\left({\frac {x}{y}}-1\right)\left(t-(1+{\frac {x}{y}}+\dots +\left({\frac {x}{y}}\right)^{t-1})\right)$

$\geq 0$

Catalan q-polynomials

The Catalan q-polynomials $C_{k}(q)$ count the number of blocks present in the diagram under the diagonal height k, and start

• $C_{0}(q)=1$
• $C_{1}(q)=1$
• $C_{2}(q)=1+q$
• $C_{3}(q)=1+q+2q^{2}+q^{3}$
• $C_{4}(q)=1+q+2q^{2}+3q^{3}+3q^{4}+3q^{5}+q^{6}$

The sum of the coefficients of $C_{k}(q)$ give $C_{k}$.

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 $X(k)Y(n-k)$, where $(k)$ is a Dyck word of length 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.

$C_{n}(q)=\sum _{i=0}^{n-1}q^{i(n-i)}C_{i}(q)C_{n-i}(q)$

For example, $C_{4}(q)=q^{0}(1)(1+q+2q^{2}+q^{3})+q^{3}(1)(1+q)+q^{4}(q+1)(1)+q^{3}(1+q+2q^{2}+q^{3})(1)$.

Cubic polynomial

Let

$f(x)=x^{3}+cx+d$

The local minima and maxima are given by the zeroes of the differential of $f(x)$.

$f'(x)=3x^{2}+c$

If $x={\sqrt {\frac {-c}{3}}}$ then (c is negative)

$f(x)={\sqrt {\frac {c}{3}}}^{3}-c{\sqrt {\frac {-c}{3}}}+d$

and $f(x)>0$ means

${\sqrt {\frac {c}{3}}}^{3}\left({\frac {1}{3}}-1\right)>d$

${\sqrt {\frac {-c}{3}}}^{3}\left({\frac {2}{3}}\right)>d$

$-4c^{3}>27d^{2}$

Round Robin

6

• 12 34 56
• 13 26 45
• 14 25 36
• 15 23 46
• 16 24 35

8

• 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

GF for cubes

${\frac {1+4x+x^{2}}{(1-x)^{4}}}=1+8x+27x^{2}+64x^{3}+125x^{4}+\dots$

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.

Proof

The GF's for $1,x,x^{2}$ are ${\frac {1}{1-x}},{\frac {x}{(1-x)^{2}}},{\frac {x^{2}+x}{(1-x)^{3}}}$.

Therefore

${\frac {3(x^{2}+x)}{(1-x)^{3}}}+{\frac {3x}{(1-x)^{2}}}+{\frac {1}{1-x}}$

generates $3x^{2}+3x+1$.

This equals

${\frac {3(x^{2}+x)+3x(1-x)+(1-2x+x^{2})}{(1-x)^{3}}}$

$={\frac {1+4x+x^{2}}{(1-x)^{3}}}$

and generates the sequence $1,7,19,37,\dots$.

https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E3

Multiplying by ${\frac {1}{1-x}}$ gives the partial sums, and so

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

generates the sequence $1,8,27,64,\dots$.

Again gives

${\frac {1+4x+x^{2}}{(1-x)^{5}}}$

generates the sequence $1,9,36,100,\dots$.

https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E5

Binomials

Prove

${\binom {m}{j}}{\binom {n}{k}}\leq {\binom {m+n}{j+k}}$

$={\frac {m!n!}{j!k!(m-j)!(n-k)!}}\leq {\frac {(m+n)!}{(j+k)!(m+n-j-k)!}}$

Consider

${\frac {(m+n)!}{m!n!}}{\frac {j!k!}{(j+k)!}}{\frac {(m-j)!(n-k)!}{(m+n-j-k)!}}$

where

${\frac {(a+b)!}{a!b!}}={\binom {a+b}{a}}={\frac {(a+1)\cdots (a+b)}{b!}}=\prod _{i=1}^{b}\left(1+{\frac {a}{i}}\right)$

so

${\frac {\prod _{i=1}^{m}\left(1+{\frac {n}{i}}\right)}{\prod _{i=1}^{j}\left(1+{\frac {k}{i}}\right)\prod _{i=1}^{m-j}\left(1+{\frac {n-k}{i}}\right)}}$

and

${\frac {\binom {m+n}{m}}{{\binom {j+k}{j}}{\binom {m+n-j-k}{m-j}}}}$

The meaning of logic

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

$\sum _{i=0}^{k}{\binom {n-k-1+i}{i}}2^{n-k-i}$

$n=6,k=3:1+6+15+20=42$

${\binom {2}{0}}2^{3}+{\binom {3}{1}}2^{2}+{\binom {4}{2}}2+{\binom {5}{3}}=8+12+12+10=4$2.

GF for Catalan numbers

Catalan number

Start with the GF for the central binomial coefficient.

${\frac {1}{\sqrt {1-4x}}}=\sum _{k=0}^{\infty }{\binom {2k}{k}}x^{k}$

Integrating and setting the constant to ${\frac {1}{2}}$ from the case $x=0$ yields

${\frac {1-{\sqrt {1-4x}}}{2}}=\sum _{k=0}^{\infty }{\frac {1}{k+1}}{\binom {2k}{k}}x^{k+1}$

Divide by x to get

${\frac {1-{\sqrt {1-4x}}}{2x}}=\sum _{k=0}^{\infty }C_{k}x^{k}$

Primitive roots

Primitive root modulo n

The order of an element is the smallest k where $a^{k}\equiv 1{\pmod {n}}$.

k divides $\phi (n)$. Testing a candidate against each possible 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 $p=4k+1$,

$p|\prod _{i=1}^{4k}\left(i^{2k}+(i-1)^{2k}\right)$

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

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.

Fermat's little power test

$a^{k}\equiv 1{\pmod {p}}$ if and only if $k|p-1$

For example, $x^{7}-1$ is never divisible by 13, 17, etc..., but is by 29.

Chinese Remainder Theorem

Say $x\equiv a{\pmod {p}}$, and also $x\equiv b{\pmod {q}}$ with p,q primes, and a,b least residues.

Note that $p^{q-1}$ is $0{\pmod {p}}$ and $1{\pmod {q}}$. Similarly $q^{p-1}$ is $1{\pmod {p}}$ and $0{\pmod {q}}$.

Therefore $aq^{p-1}+bp^{q-1}\equiv a{\pmod {p}}$ and $aq^{p-1}+bp^{q-1}\equiv b{\pmod {q}}$, and so $x\equiv aq^{p-1}+bp^{q-1}{\pmod {pq}}$.

Also $p^{q-1}\equiv p(p\#q)^{q-2}{\pmod {pq}}$, where $p\#q=p\mod q$

So the solution is ${\frac {p}{p\#q}}{\pmod {pq}}$

e.g. ${\frac {5}{-2}}\equiv 15{\pmod {35}}$.

Coordinates of circumcentre

Let $\triangle ABC$ be coordinates $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$. Then the midpoints of AB and AC are

$\left({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}}\right)$

$\left({\frac {x_{1}+x_{3}}{2}},{\frac {y_{1}+y_{3}}{2}}\right)$

The perpendiculars are

${\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}$

${\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}$

and the equations of the perpendiculars through the midpoints are

${\binom {x}{y}}={\binom {\frac {x_{1}+x_{2}}{2}}{\frac {y_{1}+y_{2}}{2}}}+s{\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}$

${\binom {x}{y}}={\binom {\frac {x_{1}+x_{3}}{2}}{\frac {y_{1}+y_{3}}{2}}}+t{\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}$

with equality when

${\frac {x_{1}+x_{2}}{2}}+s(y_{2}-y_{1})={\frac {x_{1}+x_{3}}{2}}+t(y_{3}-y_{1})$

${\frac {y_{1}+y_{2}}{2}}+s(x_{1}-x_{2})={\frac {y_{1}+y_{3}}{2}}+t(x_{1}-x_{3})$

so

$(x_{2}-x_{3})+2s(y_{2}-y_{1})=2t(y_{3}-y_{1})$

$(y_{2}-y_{3})+2s(x_{1}-x_{2})=2t(x_{1}-x_{3})$

and then

$(x_{2}-x_{3})(x_{1}-x_{2})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{2})(y_{3}-y_{1})$

$(y_{2}-y_{3})(y_{2}-y_{1})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{3})(y_{2}-y_{1})$

so

$(x_{2}-x_{3})(x_{1}-x_{2})-(y_{2}-y_{3})(y_{2}-y_{1})=2t(x_{1}y_{3}-x_{2}y_{3}+x_{2}y_{1}-x_{1}y_{2}+x_{3}y_{2}-x_{3}y_{1})$

Graphs

Hamiltonian cycles

Hamiltonian cycles

Generate the permutations of all vertices. Remove all those with non-adjacent vertices.

Cuts

Two cuts are adjacent if the corresponding partitions differ by only 1 vertex.

Base conversion

An integer in base b can be written as

$n=\sum _{i=0}^{N}a_{i}b^{i}$

and the same integer in base d is

$n=\sum _{i=0}^{M}c_{i}d^{i}$

If $d+e=b$, then the first equation becomes

$n=\sum _{i=0}^{N}a_{i}(d+e)^{i}$

Expand this using the binomial formula to get

$n=\sum _{i=0}^{N}a_{i}\sum _{k=0}^{i}{\binom {i}{k}}d^{k}e^{i-k}$

and collect by like terms in d to get the second equation.

$n=\sum _{k=0}^{N}d^{k}\sum _{i=k}^{N}a_{i}{\binom {i}{k}}e^{i-k}$

because

$\sum _{i=0}^{N}\sum _{k=0}^{i}=\sum _{k=0}^{N}\sum _{i=k}^{N}$

using $i+k=N$. This only changes the ordering of the $(i,k)$ pairs and not the function.

Add

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 $d=a\otimes b\otimes c$ and $c'=(a\land b)\lor (a\land c)\lor (b\land c)$

where $\otimes$ is XOR and c' is the next carry bit.

Homotopy

If f and g are continuous functions on $[a,b]$, let $h(x,\lambda )=\lambda f(x)+(1-\lambda )g(x)$.

If x is fixed,

$h(x,\lambda +d\lambda )-h(x,\lambda )=d\lambda (f(x)+g(x))$

which is continuous as $f(x)+g(x)$ is constant.

If $\lambda$ is fixed,

$h(x+dx,\lambda )-h(x,\lambda )=\lambda (f(x+dx)-f(x))+(1-\lambda )(g(x+dx)-g(x))$

which is continuous because $f(x),g(x)$ are.

3-sphere

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 $\mathbb {R^{3}}$ so can every manifold homeomorphic to it.

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.

Client-Vendor-Bank model

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

• Bell numbers
• Catalan numbers

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 - ()()()()

Congruent number

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

$a^{n}+x=b^{n}$

$b^{n}+x=c^{n}$

or

$a^{n}-b^{n}=b^{n}-c^{n}$

so

$a^{n}+c^{n}=2b^{n}$

Lattice walks

A random walk on a 2d grid arrives at coordinate $(x,y)$ after z steps, where

$R(x,y,z)=R(x-1,y,z-1)+R(x+1,y,z-1)+R(x,y-1,z-1)+R(x,y+1,z-1)$

with $R(0,0,0)=1$.

This leads to the definition

$R(x,y,z)=\sum _{k=0}^{z-x-y}{\frac {z!}{(x+k)!k!(y+z-x-y-k)!(z-x-y-k)!}}$

Cubic equation

If

$a+b+c=0$

then

$a^{3}+b^{3}+c^{3}=3abc$

e.g.

$(-3,-4,7):-27-64+343=252=3\cdot 3\cdot 4\cdot 7$

because

$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b+c)(ab+bc+ca)-3abc$

Also, with $a+b+c=0,$

$a^{3}+b^{3}+c^{3}$

$=-(b+c)^{3}+b^{3}+c^{3}$

$=-3b^{2}c-3bc^{2}$

$=-3bc(b+c)$

$=3abc$

Distributivity

OR is $\lor$, AND is $\land$. They are commonly written as $x+y$ and xy where $\mathbb {F_{2}}$ is assumed.

In $\mathbb {Z}$, they are $x+y-xy,xy$.

OR distributes, so $x\lor (y\land z)=(x\lor y)\land (x\lor z)$. This might seem unusual as it looks like $x+(yz)=(x+y)(x+z)$.

Proof 1

$(x+y-xy)(x+z-xz)=x+yx-xyz$

Proof 2

$x\lor (y\land z)={\overline {\overline {x\lor (y\land z)}}}={\overline {x'\land {\overline {(y\land z)}}}}={\overline {x'\land (y'\lor z')}}={\overline {(x'\land y')\lor (x'\land z')}}={\overline {(x'\land y')}}\land {\overline {(x'\land z')}}=(x\lor y)\land (x\lor z)$

0^0=1

$\ln((1-x)^{(1-x)})=(1-x)\ln(1-x)$

$=\lim _{x\to 1}(1-x)\left(-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\dots \right)$

$=\lim _{x\to 1}(x-1)\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}$

$=\lim _{x\to 1}\sum _{i=1}^{\infty }{\frac {x^{i+1}}{i}}-\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}$

$=\lim _{x\to 1}\sum _{i=1}^{\infty }{\frac {x^{i+1}-x^{i}}{i}}$

$=\sum _{i=1}^{\infty }{\frac {1-1}{i}}$

$=0$

Integration

Consider the function $x^{n}$ on $[0,y]$ at the points $0,h,2h,\dots ,kh=y$.

The area of each rectangle at point $qh$ is $(qh)^{n}\cdot h$ and the sum of all the rectangles is

$h\sum _{j=0}^{k}(jh)^{n}=h^{n+1}\sum _{j=0}^{k}j^{n}$

which is $h^{n+1}$ times the sum of powers given by

$S_{n}(k)={\frac {1}{n+1}}\sum _{r=0}^{n}{\binom {n+1}{r}}B_{r}k^{n-r+1}$

so

$\lim _{h\to 0}h^{n+1}S_{n}(k)={\frac {1}{n+1}}(hk)^{n+1}$

and as $hk=y$

$\int _{0}^{y}x^{n}dx={\frac {y^{n+1}}{n+1}}$

Square pyramid numbers

Square pyramidal numbers

1222222333333
1222222333333
1222222333333
1222222333333
1222222333333
1222222333333
444444555555

The 1,4 and 5 blocks are vertical $n\cdot n$ squares. Block 6 is split into $n\cdot 1$ and $n\cdot (n-1)$

1222222333336
1222222333336
1222222333336
1222222333336
1222222333336
1222222333336
4444445555556

This is an $n\cdot (n+1)\cdot (2n+1)$ box with an $(n-1)\cdot n\cdot (2n-1)$ 'interior'.

Find a pair

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.

Proof

Say you pick column c from row r, and it is b. Then all the rows with $\lnot b$ in column 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 $2^{n}$ to 1, but the final row has no available moves.

Determinants

determinant

$\det(a+b,c+d,e+f)=\det(a,c,e)+\det(a,c,f)+\det(a,d,e)+\det(a,d,f)+\det(b,c,e)+\det(b,c,f)+\det(b,d,e)+\det(b,d,f)$

Conic sections

Eccentricity yields

$x^{2}+(y-f)^{2}=e(y+f)^{2}$

(from Parabola#Axis of symmetry parallel to the y axis).

so the equations for the conic sections are

$x^{2}=(e-1)(y^{2}+f^{2})+2(e+1)yf$

or

$x^{2}=(e-1)(y+f)^{2}+4yf$

Harmonic numbers

Harmonic number

Prove

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

e.g.

$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}={\frac {50}{24}}$

${\frac {5}{2}}\left({\frac {1}{4}}+{\frac {1}{6}}+{\frac {1}{6}}+{\frac {1}{4}}\right)={\frac {5}{2}}{\frac {5}{6}}$

Proof

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

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

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

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

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

Addition is P

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

$\left(re^{i\theta }\right)^{i}=e^{-\theta }e^{i\ln r},\left(re^{i\theta }\right)^{-i}=e^{\theta }e^{-i\ln r}$

$\left(\left(re^{i\theta }\right)^{i}\right)^{i}=\left(re^{i\theta }\right)^{-1}={\frac {1}{r}}e^{-i\theta }$

Symmetric group

Symmetric group

If $[a,b]$ swaps the a-th element with the b-element, then $r=[1,2][1,3]\dots [1,n]$ left-shifts a permutation by 1 element, hence $r^{n}=I$.

Cross product

Cross product

${\begin{pmatrix}a\\b\\c\end{pmatrix}}\times {\begin{pmatrix}d\\e\\f\end{pmatrix}}={\begin{pmatrix}0\,\,\,-c\,\,\,b\\c\,\,\,0\,\,\,-a\\-b\,\,\,a\,\,\,0\end{pmatrix}}{\begin{pmatrix}d\\e\\f\end{pmatrix}}$

Exact sequence

The first isomorphism theorem states that the kernel of H is a normal subgroup of G.

Therefore a group homomorphism must take a group to a group with a normal subgroup for the sequence to be an exact sequence.

Fourier series

Fourier series

Fourier transform

It is assumed that function $S(t)$ can be expressed as the integral of a complex linear combination of complex exponentials, where $S(t)$ is not necessarily periodic.

$S(t)=\int _{-\infty }^{\infty }s(x)e^{-itx}dx$

Multiplying both sides by $e^{iTx}$ gives

$S(t)e^{iTx}=\int _{-\infty }^{\infty }s(x)e^{-ix(t-T)}dx$

$\int _{-\infty }^{\infty }S(t)e^{iTx}dx=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }s(x)e^{-ix(t-T)}dx\right]dy$

$\int _{-\infty }^{\infty }S(t)e^{iTx}dx=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }s(x)e^{-ix(t-T)}dy\right]dx$

$\int _{-\infty }^{\infty }S(t)e^{iTx}dx=s(t)$

Navier-Stokes equations

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.

$\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+\rho \mathbf {g}$

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

$\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=\nabla \cdot \left[\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathsf {T}}\right)\right]$

and no movement

$\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p$

Analytic functions

Analyticity of holomorphic functions

$$f(z)=\sum _{n=0}^{\infty }c_{n}(z-a)^{n}$$

Let $z=a+re^{i\theta }$, with $a\in \mathbb {C} ,r,\theta \in \mathbb {R}$.

Then, if f is holomorphic,

$|f(a+r)|\geq |f(a+re^{i\theta }|$

i.e. $\theta =0$ naximizes f.

Diffferentiation of theta function

$x^{\frac {1}{2}}\left(2\psi (x)+1\right)=2\psi ({\frac {1}{x}})+1$

$x^{\frac {1}{2}}2\psi '(x)+{\frac {1}{2}}x^{-{\frac {1}{2}}}\left(2\psi (x)+1\right)=-2{\frac {1}{x^{2}}}\psi '({\frac {1}{x}})$

At $x=1$

$2\psi '(1)+{\frac {1}{2}}\left(2\psi (1)+1\right)=-2\psi '(1)$

${\frac {1}{2}}+\psi (1)+4\psi '(1)=0$

Second derivative

$x^{\frac {1}{2}}2\psi '(x)+{\frac {1}{2}}x^{-{\frac {1}{2}}}\left(2\psi (x)+1\right)=-2{\frac {1}{x^{2}}}\psi '({\frac {1}{x}})$

$x^{\frac {1}{2}}2\psi ''(x)+{\frac {1}{2}}x^{-{\frac {1}{2}}}2\psi '(x)+{\frac {1}{2}}x^{-{\frac {1}{2}}}2\psi '(x)-{\frac {1}{4}}x^{-{\frac {3}{2}}}\left(2\psi (x)+1\right)=2{\frac {1}{x^{4}}}\psi ''({\frac {1}{x}})+4{\frac {1}{x^{3}}}\psi '({\frac {1}{x}})$

At $x=1$

$2\psi ''(1)+\psi '(1)+\psi '(1)-{\frac {1}{4}}\left(2\psi (1)+1\right)=2\psi ''(1)+4\psi '(1)$

${\frac {1}{2}}+\psi (1)+4\psi '(1)=0$

!!!

Area of a parallelogram

From the picture, the area of the parallelogram is

$(a+c)(b+d)-ab-cd-2bc=ad-bc$

2d random walks

Random walks on a 2d lattice are governed by

$R(d,x,y)=R(d-1,x-1,y)+R(d-1,x+1,y)+R(d-1,x,y-1)+R(d-1,x,y+1)$

After $2d$ moves, we are at position $(x,y)$ where $x+y$ is even.

Let $a={\frac {x+y}{2}},b={\frac {x-y}{2}}$. Then the number of paths from $(0,0,0)$ to $(2d,x,y)$) is

$R(2d,x,y)={\binom {2d}{d+a}}{\binom {2d}{d+b}}$

Proof

${\binom {k}{a}}{\binom {k}{b}}$

$=\left({\binom {k-1}{a-1}}+{\binom {k-1}{a}}\right)\left({\binom {k-1}{b-1}}+{\binom {k-1}{b}}\right)$

$={\binom {k-1}{a-1}}{\binom {k-1}{b-1}}+{\binom {k-1}{a}}{\binom {k-1}{b-1}}+{\binom {k-1}{a-1}}{\binom {k-1}{b}}+{\binom {k-1}{a}}{\binom {k-1}{b}}$