text:
# Sports ranking methods, 4

This is the fourth of a series of expository posts on matrix-theoretic sports ranking methods. This post discusses the Elo rating.

This system was originally developed by Arpad Elo (Elo (1903-1992) was a physics professor at Marquette University in Milwaukee and a chess master, eight-time winner of the Wisconsin State Chess Championships.) Originally, it was developed for rating chess players in the 1950s and 1960s. Now it is used for table tennis, basketball, and other sports.

We use the following version of his rating system.

As above, assume all the $n$ teams play each other (ties allowed)
and let $r_i$ denote the rating of Team $i$, $i=1,2,\dots,n$.

Let $A=(A_{ij})$ denote an $n\times n$ matrix of score results:

$A_{ij}= \left\{ \begin{array}{rr} -1,& {\rm if\ team\ } i {\rm \ lost\ to\ team\ } j,\\ +1,& {\rm if\ team\ } i {\rm\ beat\ team\ } j,\\ 0, & {\rm if}\ i=j. \end{array} \right.$

Let $S_{ij}=(A_{ij}+1)/2$.

As in the previous post, the matrix $A$ associated to the example of the Patriot league is the adjacency matrix of a diagraph.

1. Initialize all the ratings to be $100$: ${\bf r}=(r_1,\dots,r_n) = (100,\dots,100)$.
2. After Team $i$ plays Team $j$, update their rating using the formula

$r_i = r_i+K(S_{ij}-mu_{ij}),$

where $K=10$ and

$\mu_{ij} = (1+e^{-(r_i-r_j)/400})^{-1}.$

In the example of the Patriot league, the ratings vector is

${\bf r}=(85.124, 104.79, 104.88, 85.032, 94.876, 124.53).$

This gives the ranking

Lafayette $<$ Army $<$ Lehigh $<$ Bucknell $<$ Holy Cross $<$ Navy.

This gives a prediction failure rate of $13.3\%$.

Some SageMath code for this:

def elo_rating(A):
"""
A is a signed adjacency matrix for a directed graph.

Returns elo ratings of the vertices of Gamma = Graph(A)

EXAMPLES:
sage: A = matrix(QQ,[
[0 , -1 , 1  , -1 , -1 , -1 ],
[1,   0 ,  -1,  1,  1,   -1  ],
[-1 , 1 ,  0 ,  1 , 1  , -1  ],
[1 , -1 , -1,  0 ,  -1 , -1  ],
[1 , - 1 , - 1 , 1 , 0 , - 1  ],
[1 ,  1  ,  1  , 1  , 1  , 0 ]
])
sage: elo_rating(A)
(85.124, 104.79, 104.88, 85.032, 94.876, 124.53)

"""
n = len(A.rows())
RR = RealField(prec=20)
V = RR^n
K = 10
r0 = 100 # initial rating
r = n*[r0]
for i in range(n):
for j in range(n):
if ij and A[i][j]==1:
S = 1
elif ij and A[i][j]==-1:
S = 0
else:
S = 1/2
mu = 1/(1+e^(-(r[i]-r[j])/400))
r[i] = r[i]+K*(S-mu)
return V(r)

# Sports ranking methods, 1

This is the first of a series of expository posts on matrix-theoretic sports ranking methods. This post, which owes much to discussions with TS Michael, discusses Massey’s method.

Massey’s method, currently in use by the NCAA (for football, where teams typically play each other once), was developed by Kenneth P. Massey
while an undergraduate math major in the late 1990s. We present a possible variation of Massey’s method adapted to baseball, where teams typically play each other multiple times.

There are exactly 15 pairing between these teams. These pairs are sorted lexicographically, as follows:

(1,2),(1,3),(1,4), …, (5,6).

In other words, sorted as

Army vs Bucknell, Army vs Holy Cross, Army vs Lafayette, …, Lehigh vs Navy.

The cumulative results of the 2016 regular season are given in the table below. We count only the games played in the Patriot league, but not including the Patriot league post-season tournament (see eg, the Patriot League site for details). In the table, the total score (since the teams play multiple games against each other) of the team in the vertical column on the left is listed first. In other words, ”a – b” in row $i$ and column $j$ means the total runs scored by team $i$ against team $j$ is $a$, and the total runs allowed by team $i$ against team $j$ is $b$. Here, we order the six teams as above (team $1$ is Army (USMI at Westpoint), team $2$ is Bucknell, and so on). For instance if X played Y and the scores were $10-0$, $0-1$, $0-1$, $0-1$, $0-1$, $0-1$, then the table would read $10-5$ in the position of row X and column Y.

 X\Y Army Bucknell Holy Cross Lafayette Lehigh Navy Army x 14-16 14-13 14-24 10-12 8-19 Bucknell 16-14 x 27-30 18-16 23-20 10-22 Holy Cross 13-14 30-27 x 19-15 17-13 9-16 Lafayette 24-14 16-18 15-19 x 12-23 17-39 Lehigh 12-10 20-23 13-17 23-12 x 12-18 Navy 19-8 22-10 16-9 39-17 18-12 x

Win-loss digraph of the Patriot league mens baseball from 2015

In this ordering, we record their (sum total) win-loss record (a 1 for a win, -1 for a loss) in a $15\times 6$ matrix:

$M = \left(\begin{array}{cccccc} -1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & -1 & 1 \end{array}\right).$

We also record their total losses in a column vector:

${\bf b}= \left(\begin{array}{c} 2 \\ 1 \\ 10 \\ 2 \\ 11 \\ 3 \\ 2 \\ 3 \\ 14 \\ 4 \\ 14 \\ 10 \\ 11 \\ 22 \\ 6 \\ \end{array}\right).$

The Massey ranking of these teams is a vector ${\bf r}$ which best fits the equation

$M{\bf r}={\bf b}.$

While the corresponding linear system is over-determined, we can look for a best (in the least squares sense) approximate solution using the orthogonal projection formula

$P_V = B(B^tB)^{-1}B^t,$

valid for matrices $B$ with linearly independent columns. Unfortunately, in this case $B=M$ does not have linearly independent columns, so the formula doesn’t apply.

Massey’s clever idea is to solve

$M^tM{\bf r}=M^t{\bf b}$

by row-reduction and determine the rankings from the parameterized form of the solution. To this end, we compute

$M^tM= \left(\begin{array}{rrrrrr} 5 & -1 & -1 & -1 & -1 & -1 \\ -1 & 5 & -1 & -1 & -1 & -1 \\ -1 & -1 & 5 & -1 & -1 & -1 \\ -1 & -1 & -1 & 5 & -1 & -1 \\ -1 & -1 & -1 & -1 & 5 & -1 \\ -1 & -1 & -1 & -1 & -1 & 5 \end{array}\right)$

and

$M^t{\bf b}= \left(\begin{array}{r} -24 \\ -10 \\ 10 \\ -29 \\ -10 \\ 63 \\ \end{array}\right).$

Then we compute the rref of

$A= (M^tM,M^t{\bf b}) = \left(\begin{array}{rrrrrr|r} 5 & -1 & -1 & -1 & -1 & -1 & -24 \\ -1 & 5 & -1 & -1 & -1 & -1 & -10 \\ -1 & -1 & 5 & -1 & -1 & -1 & 10 \\ -1 & -1 & -1 & 5 & -1 & -1 & -29 \\ -1 & -1 & -1 & -1 & 5 & -1 & -10 \\ -1 & -1 & -1 & -1 & -1 & 5 & 63 \end{array}\right),$

which is

$rref(M^tM,M^t{\bf b})= \left(\begin{array}{rrrrrr|r} 1 & 0 & 0 & 0 & 0 & -1 & -\frac{87}{6} \\ 0 & 1 & 0 & 0 & 0 & -1 & -\frac{73}{6} \\ 0 & 0 & 1 & 0 & 0 & -1 & -\frac{53}{6} \\ 0 & 0 & 0 & 1 & 0 & -1 & -\frac{92}{3} \\ 0 & 0 & 0 & 0 & 1 & -1 & -\frac{73}{6} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right).$

If ${\bf r}=(r_1,r_2,r_3,r_4,r_5,r_6)$ denotes the rankings of Army, Bucknell, Holy Cross, Lafayette, Lehigh, Navy, in that order, then

$r_1=r_6-\frac{87}{6},\ \ r_2=r_6-\frac{73}{6},\ \ r_3=r_6-\frac{53}{6},\ \ r_4=r_6-\frac{92}{6},\ \ r_5=r_6-\frac{73}{6}.$

Therefore

Lafayette $<$ Army = Bucknell = Lehigh $<$ Holy Cross $<$ Navy.

If we use this ranking to predict win/losses over the season, it would fail to correctly predict Army vs Holy Cross (Army won), Bucknell vs Lehigh, and Lafayette vs Army. This gives a prediction failure rate of $20\%$.

# The minimum upset ranking problem

Suppose n teams play each other, and let Team $r_1 <$ Team $r_2 < \dots <$ Team $r_n$ denote some fixed ranking (where $r_1,\dots,r_n$ is some permutation of $1,\dots,n$). An upset occurs when a lower ranked team beats an upper ranked team. For each ranking, ${\bf r}$, let $U({\bf r})$ denote the total number of upsets. The minimum upset problem is to find an “efficient” construction of a ranking for which $U({\bf r})$ is as small as possible.

In general, let $A_{ij}$ denote the number of times Team i beat team $j$ minus the number of times Team j beat Team i. We regard this matrix as the signed adjacency matrix of a digraph $\Gamma$. Our goal is to find a Hamiltonian (undirected) path through the vertices of $\Gamma$ which goes the “wrong way” on as few edges as possible.

1. Construct the list of spanning trees of $\Gamma$ (regarded as an undirected graph).
2. Construct the sublist of Hamiltonian paths (from the spanning trees of maximum degree 2).
3. For each Hamiltonian path, compute the associated upset number: the total number of edges transversal in $\Gamma$ going the “right way” minus the total number going the “wrong way.”
4. Locate a Hamiltonian for which this upset number is as large as possible.

Use this sagemath/python code to compute such a Hamiltonian path.

def hamiltonian_paths(Gamma, signed_adjacency_matrix = []):
"""
Returns a list of hamiltonian paths (spanning trees of
max degree <=2).

EXAMPLES:
sage: Gamma = graphs.GridGraph([3,3])
sage: HP = hamiltonian_paths(Gamma)
sage: len(HP)
20
sage: A = matrix(QQ,[
[0 , -1 , 1  , -1 , -1 , -1 ],
[1,   0 ,  -1,  1,  1,   -1  ],
[-1 , 1 ,  0 ,  1 , 1  , -1  ],
[1 , -1 , -1,  0 ,  -1 , -1  ],
[1 , - 1 , - 1 , 1 , 0 , - 1  ],
[1 ,  1  ,  1  , 1  , 1  , 0 ]
])
sage: HP = hamiltonian_paths(Gamma, signed_adjacency_matrix = A)
sage: L = [sum(x[2]) for x in HP]; max(L)
5
sage: L.index(5)
21
sage: HP[21]
[Graph on 6 vertices,
[0, 5, 2, 1, 3, 4],
[-1, 1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1]]
sage: L.count(5)
1

"""
ST = Gamma.spanning_trees()
HP = []
for X in ST:
L = X.degree_sequence()
if max(L)<=2:
#print L,ST.index(X), max(L)
HP.append(X)
return HP
HP = []
for X in ST:
L = X.degree_sequence()
if max(L)<=2:
#VX = X.vertices()
EX = X.edges()
if EX[0][1] != EX[-1][1]:
ranking = X.shortest_path(EX[0][0],EX[-1][1])
else:
ranking = X.shortest_path(EX[0][0],EX[-1][0])
signature = [A[ranking[i]][ranking[j]] for i in range(len(ranking)-1) for j in range(i+1,len(ranking))]
HP.append([X,ranking,signature])
return HP

Wessell describes this method in a different way.

1. Construct a matrix, $M=(M_{ij})$, with rows and columns indexed by the teams in some fixed order. The entry in the i-th row and the j-th column is defined by$m_{ij}= \left\{ \begin{array}{rr} 0,& {\rm if\ team\ } i {\rm \ lost\ to\ team\ } j,\\ 1,& {\rm if\ team\ } i {\rm\ beat\ team\ } j,\\ 0, & {\rm if}\ i=j. \end{array} \right.$
2. Reorder the rows (and corresponding columns) to in a basic win-loss order: the teams that won the most games go at the
top of $M$, and those that lost the most at the bottom.
3. Randomly swap rows and their associated columns, each time checking if the
number of upsets has gone down or not from the previous time. If it has gone down, we keep
the swap that just happened, if not we switch the two rows and columns back and try again.

An implementaiton of this in Sagemath/python code is:

def minimum_upset_random(M,N=10):
"""
EXAMPLES:
sage: M = matrix(QQ,[
[0 , 0 , 1  , 0 , 0 , 0 ],
[1,   0 ,  0,  1,  1,   0  ],
[0 , 1 ,  0 ,  1 , 1  , 0  ],
[1 , 0 , 0,  0 ,  0 , 0  ],
[1 , 0 , 0 , 1 , 0 , 0  ],
[1 ,  1  ,  1  , 1  , 1  , 0 ]
])
sage: minimum_upset_random(M)
(
[0 0 1 1 0 1]
[1 0 0 1 0 1]
[0 1 0 0 0 0]
[0 0 1 0 0 0]
[1 1 1 1 0 1]
[0 0 1 1 0 0], [1, 2, 0, 3, 5, 4]
)

"""
n = len(M.rows())
Sn = SymmetricGroup(n)
M1 = M
wins = sum([sum([M1[j][i] for i in range(j,6)]) for j in range(6)])
g0 = Sn(1)
for k in range(N):
g = Sn.random_element()
P = g.matrix()
M0 = P*M1*P^(-1)
if sum([sum([M0[j][i] for i in range(j,6)]) for j in range(6)])>wins:
M1 = M0
g0 = g*g0
return M1,g0(range(n))

# Splitting fields of representations of generalized symmetric groups, 7

In this post, we discover which representations of the generalized symmetric group $G = S_n\ wr\ C_\ell = C_\ell^n\, >\!\!\lhd \, S_n$ can be realized over a given abelian extension of ${\mathbb{Q}}$.

Let $\theta_{\mu,\rho}\in G^*$ be the representation defined previously, where $\rho\in ((S_n)_\mu)^*$.

Let $K\subset {\mathbb{Q}}(\zeta_\ell)$ be a subfield, where $\zeta_\ell$ is a primitive $\ell^{th}$ root of unity. Assume $K$ contains the field generated by the values of the character of $\theta_{\mu,\rho}$. Assume $K/{\mathbb{Q}}$ is Galois and let $\Gamma_K=Gal({\mathbb{Q}}(\zeta_\ell)/K)$. Note if we regard $C_\ell$ as a subset of ${\mathbb{Q}}(\zeta_\ell)$ then there is an induced action of $\Gamma_K$ on $C_\ell$,

$\sigma:\mu \longmapsto \mu^\sigma, \ \ \ \ \ \ \ \ \ \mu\in (C_\ell)^*,\ \ \sigma\in \Gamma_K,$

where $\mu^\sigma(z)=\mu(\sigma^{-1}(z))$, $z\in C_\ell$. This action extends to an action on $(C_\ell^n)^*=(C_\ell^*)^n$.

Key Lemma:
In the notation above, $\theta_{\mu,\rho}\cong\theta_{\mu,\rho}^\sigma$ if and only if $\mu$ is equivalent to $\mu^\sigma$ under the action of $S_n$ on $(C_\ell^n)^*$.

Let

$n_\mu(\chi)=|\{i\ |\ 1\leq i\leq n,\ \mu_i=\chi\}|,$

where $\mu=(\mu_1,...,\mu_n)\in (C_\ell^n)^*$ and $\chi\in C_\ell^*$.

Theorem: The character of $\theta_{\mu,\rho}\in G^*$ has values in $K$ if and only if $n_\mu(\chi)=n_\mu(\chi^\sigma)$,
for all $\sigma\in \Gamma_K$ and all $\chi\in C_\ell^*$.

This theorem is proven in this paper.

We now determine the splitting field of any irreducible character of a generalized symmetric group.

Theorem: Let $\chi=tr(\theta_{\rho,\mu})$ be an irreducible character of $G=S_n\ wr\ C_\ell$. We have

$Gal({\mathbb{Q}}(\zeta_\ell)/{\mathbb{Q}}(\chi))= Stab_\Gamma(\chi).$

This theorem is also proven in this paper.

In the next post we shall give an example.

# Linear systems of graphs in Sage

Let $\Gamma$ be a graph. A divisor on $\Gamma$ is an element of the free group generated by the vertices $V$, ${\mathbb{Z}}[V]$.

We say that divisors $D$ and $D^\prime$ are linearly equivalent and write $D \sim D^\prime$ if $D^\prime-D$ is a principal divisor, i.e., if $D^\prime = D + \text{div}(f)$ for some function $f : V \rightarrow {\mathbb{Z}}$. Note that if $D$ and $D^\prime$ are linearly equivalent, they must have the same degree, since the degree of every principal divisor is 0. Divisors of degree 0 are linearly equivalent if and only if they determine the same element of the Jacobian. If $D$ is a divisor of degree 0, we denote by $[D]$ the element of the Jacobian determined by $D$. A divisor $D$ is said to be effective if $D(v) \geq 0$ for all vertices $v$. We write $D \geq 0$ to mean that $D$ is effective. The linear system associated to a divisor $D$ is the set

$|D| = \{ D^\prime \in \text{Div}(\Gamma ) : D^\prime \geq 0 \text{ and } D^\prime \sim D\},$

i.e., $|D|$ is the set of all effective divisors linearly equivalent to $D$. Note that if $D_1 \sim D_2$, then $|D_1| = |D_2|$. We note also that if $\text{deg}(D)<0$, then $|D|$ must be empty.

Sage can be used to compute the linear system of any divisor on a graph.

def linear_system(D, Gamma):
"""
Returns linear system attached to the divisor D.

EXAMPLES:
sage: Gamma2 = graphs.CubeGraph(2)
sage: Gamma1 = Gamma2.subgraph(vertices = ['00', '01'], edges = [('00', '01')])
sage: f = [['00', '01', '10', '11'], ['00', '01', '00', '01']]
sage: is_harmonic_graph_morphism(Gamma1, Gamma2, f)
True
sage: PhiV = matrix_of_graph_morphism_vertices(Gamma1, Gamma2, f); PhiV
[1 0 1 0]
[0 1 0 1]
sage: D = vector([1,0,0,1])
sage: PhiV*D
(1, 1)
sage: linear_system(PhiV*D, Gamma1)
[(2, 0), (1, 1), (0, 2)]
sage: linear_system(D, Gamma2)
[(0, 2, 0, 0), (0, 0, 2, 0), (1, 0, 0, 1)]
sage: [PhiV*x for x in linear_system(D, Gamma2)]
[(0, 2), (2, 0), (1, 1)]

"""
Q = Gamma.laplacian_matrix()
CS = Q.column_space()
N = len(D.list())
d = sum(D.list())
#print d
lin_sys = []
if d < 0:
return lin_sys
if (d == 0) and (D in CS):
lin_sys = [CS(0)]
return lin_sys
elif (d == 0):
return lin_sys
S = IntegerModRing(d+1)^N
V = QQ^N
for v in S:
v = V(v)
#print D-v,v,D
if D-v in CS:
lin_sys.append(v)
return lin_sys

# Differential calculus using Sagemath

Granville’s classic text book Elements of the Differential and Integral Calculus fell into the public domain and then much of it (but not all, at the time of this writing) was scanned into wikisource primarily by R. J. Hall. Granville’s entire book contains material on differential, integral, and even multivariable calculus. The material in the subset here is restricted to differential calculus topics, though contains some material which might properly belong to an elementary differential geometry course. The above-mentioned wikisource document uses mathml and latex and some Greek letter fonts.

In particular, the existence of this document owes itself primarily to three great open source projects: TeX/LaTeX, Wikipedia, and Sagemath (http://www.sagemath.org). Some material from Sean Mauch’s public domain text on Applied Mathematics, was also included.

The current latex document is due to David Joyner, who is responsible for re-formatting, editing for readability, the correction (or introduction) of typos from the scanned version, and any extra material added (for example, the hyperlinked cross references, and the Sagemath material). Please email corrections to wdjoyner@gmail.com.

Acknowledgements: I thank the following readers for reporting typos: Mario Pernici, Jacob Hicks.

Now available from amazon.com for \$20 (not including shipping).

# Discrete Fourier transforms using Sagemath

Here are some Sagemath examples for DFTs, DCTs, and DST’s. You can try copying and pasting them into the Sagemath cloud, for example.

The Sagemath dft command applies to a sequence S indexed by a set J computes the un-normalized DFT: (in Python)

[sum([S[i]*chi(zeta**(i*j)) for i in J]) for j in J]Here are some examples which explain the syntax:

sage: J = range(6)
sage: A = [ZZ(1) for i in J]
sage: s = IndexedSequence(A,J)
sage: s.dft(lambda x:x^2)
Indexed sequence: [6, 0, 0, 6, 0, 0]
indexed by [0, 1, 2, 3, 4, 5]
sage: s.dft()
Indexed sequence: [6, 0, 0, 0, 0, 0]
indexed by [0, 1, 2, 3, 4, 5]
sage: G = SymmetricGroup(3)
sage: J = G.conjugacy_classes_representatives()
sage: s = IndexedSequence([1,2,3],J) # 1,2,3 are the values of a class fcn on G
sage: s.dft()   # the "scalar-valued Fourier transform" of this class fcn
Indexed sequence: [8, 2, 2]
indexed by [(), (1,2), (1,2,3)]
sage: J = AbelianGroup(2,[2,3],names='ab')
sage: s = IndexedSequence([1,2,3,4,5,6],J)
sage: s.dft()   # the precision of output is somewhat random and arch. dependent.
Indexed sequence: [21.0000000000000, -2.99999999999997 - 1.73205080756885*I, -2.99999999999999 + 1.73205080756888*I, -9.00000000000000 + 0.0000000000000485744257349999*I, -0.00000000000000976996261670137 - 0.0000000000000159872115546022*I, -0.00000000000000621724893790087 - 0.0000000000000106581410364015*I]
indexed by Multiplicative Abelian Group isomorphic to C2 x C3
sage: J = CyclicPermutationGroup(6)
sage: s = IndexedSequence([1,2,3,4,5,6],J)
sage: s.dft()   # the precision of output is somewhat random and arch. dependent.
Indexed sequence: [21.0000000000000, -2.99999999999997 - 1.73205080756885*I, -2.99999999999999 + 1.73205080756888*I, -9.00000000000000 + 0.0000000000000485744257349999*I, -0.00000000000000976996261670137 - 0.0000000000000159872115546022*I, -0.00000000000000621724893790087 - 0.0000000000000106581410364015*I]
indexed by Cyclic group of order 6 as a permutation group
sage: p = 7; J = range(p); A = [kronecker_symbol(j,p) for j in J]
age: s = IndexedSequence(A,J)
sage: Fs = s.dft()
sage: c = Fs.list()[1]; [x/c for x in Fs.list()]; s.list()
[0, 1, 1, -1, 1, -1, -1]
[0, 1, 1, -1, 1, -1, -1]

The DFT of the values of the quadratic residue symbol is itself, up to a constant factor (denoted c on the last line above).

Here is a 2nd example:

sage: J = range(5)
sage: A = [ZZ(1) for i in J]
sage: s = IndexedSequence(A,J)
sage: fs = s.dft(); fs
Indexed sequence: [5, 0, 0, 0, 0]
indexed by [0, 1, 2, 3, 4]
sage: it = fs.idft(); it
Indexed sequence: [1, 1, 1, 1, 1]
indexed by [0, 1, 2, 3, 4]
age: it == s
True
sage: t = IndexedSequence(B,J)
sage: s.convolution(t)
[1, 2, 3, 4, 5, 4, 3, 2, 1]

Here is a 3rd example:

sage: J = range(5)
sage: A = [exp(-2*pi*i*I/5) for i in J]
sage: s = IndexedSequence(A,J)
sage: s.dct()    # discrete cosine
Indexed sequence: [2.50000000000011 + 0.00000000000000582867087928207*I, 2.50000000000011 + 0.00000000000000582867087928207*I, 2.50000000000011 + 0.00000000000000582867087928207*I, 2.50000000000011 + 0.00000000000000582867087928207*I, 2.50000000000011 + 0.00000000000000582867087928207*I]
indexed by [0, 1, 2, 3, 4]
sage: s.dst()        # discrete sine
Indexed sequence: [0.0000000000000171529457304586 - 2.49999999999915*I, 0.0000000000000171529457304586 - 2.49999999999915*I, 0.0000000000000171529457304586 - 2.49999999999915*I, 0.0000000000000171529457304586 - 2.49999999999915*I, 0.0000000000000171529457304586 - 2.49999999999915*I]
indexed by [0, 1, 2, 3, 4]

Here is a 4th example:

sage: I = range(3)
sage: A = [ZZ(i^2)+1 for i in I]
sage: s = IndexedSequence(A,I)
sage: P1 = s.plot()
sage: P2 = s.plot_histogram()

P1 and P2 are displayed below:

The plot of P1

The plot of P2


text:
• A DURGA DEVI

Articles written in Pramana – Journal of Physics

• Symmetries and integrability of the modified Camassa–Holm equation with an arbitrary parameter

We study the symmetry and integrability of a modified Camassa–Holm equation (MCH), with an arbitrary parameter $k$, of the form $u_t + k(u − u_{xx})^{2}u_x − u_{xxt} + (u^2 − u_x^2)(u_x − u_{xxx} ) = 0.$ The commutator table and adjoint representation of the symmetries are presented to construct one-dimensional optimal system. By using the one-dimensional optimal system, we reduce the order or number of independent variables of the above equation and also we obtain interesting novel solutions for the reduced ordinary differential equations. Finally, we apply the Painlevé test to the resultant nonlinear ordinary differential equation and it is observed that the equation is integrable.

• # Pramana – Journal of Physics

Volume 96, 2022
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019

format: plain


text:
# chord angle formula

a = \frac{1}{2} \cdot (140 ^{\circ}) Theorem: The measure of the angle formed by 2 chords that intersect inside the circle is 1 2 the sum of the chords' intercepted arcs. Real World Math Horror Stories from Real encounters. Click here for the formulas used in this calculator. Note: The chord of a circle is a straight line that connects any two points on the circumference of a circle. \\ Chords were used extensively in the early development of trigonometry. Circle Calculator. Multiply this root by the central angle again to get the arc length. The measure of the angle formed by 2 chords Cloudflare Ray ID: 616a1c69e9b4dc89 Statement: The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangent through one of the endpoints of the chord is equal to the angle in the alternate segment. D represents the perpendicular distance from the cord to the center of the circle. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Hence the sine of the angle BCM is (c/2)/r = c/(2r). The chord length formula in mathematics could be written as given below. The measure of the arc is 160. If you know radius and angle you may use the following formulas to calculate remaining segment parameters: The first has the central angle measured in degrees so that the sector area equals π times the radius-squared and then multiplied by the quantity of the central angle in degrees divided by 360 degrees. Chords $$\overline{JW}$$ and $$\overline{LY}$$ intersect as shown below. Please enable Cookies and reload the page. 1. The outputs are the arclength s, area A of the sector and the length d of the chord. Let R be the radius of the circle, θ the central angle in radians, α is the central angle in degrees, c the chord length, s the arc length, h the sagitta (height) of the segment, and d the height (or apothem) of the triangular portion. \angle \class{data-angle-label}{W} = \frac 1 2 (\overparen{\rm \class{data-angle-label-0}{AB}} + \overparen{\rm \class{data-angle-label-1}{CD}}) The first step is to look at the chord, and realize that an isosceles triangle can be made inside the circle, between the chord line and the 2 radius lines. \angle A= \frac{1}{2} \cdot (38^ {\circ} + 68^ {\circ}) Radius and central angle 2. Theorem: This calculation gives you the radius. Choose one based on what you are given to start. Diagram 1. Now if we focus solely on this isosceles triangle that has been formed. The triangle can be cut in half by a perpendicular bisector, and split into 2 smaller right angle triangles. $$\text{m } \overparen{\red{JKL}}$$ is $$75^{\circ}$$ $$\text{m } \overparen{\red{WXY}}$$ is $$65^{\circ}$$ and What is the value of $$a$$? Hence the central angle BCA has measure. The chord length formulas vary depends on what information do you have about the circle. These two other arcs should equal 360° - 140° = 220°. Show that the angles of Intersecting chords are equal to half the sum of the arcs that the angle and its opposite angle subtend, m∠α = ½(P+Q). \\ A chord that passes through the center of the circle is also a diameter of the circle. $$\\ Namely,$$ \overparen{ AGF }$$and$$ \overparen{ CD }$$. The formulas for all THREE of these situations are the same: Angle Formed Outside = $$\frac { 1 }{ 2 }$$ Difference of Intercepted Arcs (When subtracting, start with the larger arc.) \\ . Calculate the height of a segment of a circle if given 1. 220 ^{\circ} =\overparen{TE } + \overparen{ GR } (Whew, what a mouthful!) \class{data-angle}{89.68 } ^{\circ} = \frac 1 2 ( \class{data-angle-0}{88.21 } ^{\circ} + \class{data-angle-1}{91.15 } ^{\circ} ) radius = So far everything is fine. The problem with these measurements is that if angle AEC = 70°, then we know that$$\overparen{ ABC }$$+$$\overparen{ DF }$$should equal 140°. Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. Use the theorem for intersecting chords to find the value of sum of intercepted arcs (assume all arcs to be minor arcs).$$ This theorem applies to the angles and arcs of chords that intersect anywhere within the circle. . \angle A= \frac{1}{2} \cdot (\overparen{\red{HIJ}} + \overparen{ \red{KLM } }) Chord Length and is denoted by l symbol. This particular formula can be seen in two ways. \\ Angle Formed by Two Chords. 110^{\circ} = \frac{1}{2} \cdot (\overparen{TE } + \overparen{ GR }) The first known trigonometric table, compiled by Hipparchus, tabulated the value of the chord function for every 7.5 degrees. The length a of the arc is a fraction of the length of the circumference which is 2 π r. In fact the fraction is . \angle A= \frac{1}{2} \cdot (106 ^{\circ}) C represents the angle extended at the center by the chord. Circular segment. $$For example, in the above figure, Using the figure above, try out your power-theorem skills on the following problem: If$$ \overparen{\red{HIJ}}= 38 ^{\circ} $$,$$ \overparen{JK} = 44 ^{\circ} $$and$$ \overparen{KLM}= 68 ^{\circ} $$, then what is the measure of$$ \angle $$A? We must first convert the angle measure to radians: Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle. However, the measurements of$$ \overparen{ CD }$$and$$ \overparen{ AGF }$$do not add up to 220°. Special situation for this set up: It can be proven that ∠ABC and central ∠AOC are supplementary. Formula for angles and intercepted arcs of intersecting chords. = 2 × (r2–d2. The value of c is the length of chord.$$, $$m \angle AEB = m \angle CED$$ CED since they are vertical angles. The chord radius formula when length and height of the chord are given is. We also find the angle given the arc lengths. You may need to download version 2.0 now from the Chrome Web Store. Then a formula is presented that we will use to meet this lesson's objectives. \angle AEB = \frac{1}{2} (\overparen{ AB} + \overparen{ CD}) m \angle AEC = 70 ^{\circ} \\ \\ Chord and central angle $$. Multiply this result by 2.  Chord Length when radius and angle are given calculator uses Chord Length=sin (Angle A/2)*2*Radius to calculate the Chord Length, Chord Length when radius and angle are given is the length of a line segment connecting any two points on the circumference of a circle with a given value for radius and angle. \\ Note:$$ \overparen { NO } $$is not an intercepted arc, so it cannot be used for this problem. Enter the radius and central angle in DEGREES, RADIANS or both as positive real numbers and press "calculate". If the radius is r and the length of the chord is c then triangle CMB is a right triangle with |BC| = r and |MB| = c/2. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Radius of circle = r= D/2 = Dia / 2 Angle of the sector = θ = 2 cos -1 ((r – h) / r) Chord length of the circle segment = c = 2 SQRT[ h (2r – h) ] Arc Length of the circle segment = l = 0.01745 x r x θ$$ \angle Z= 40 ^{\circ} Chord Length = 2 × √ (r 2 − d 2) Chord Length Using Trigonometry. \\ \overparen{AGF}= 170 ^{\circ } Find the measure of the angle t in the diagram. The units will be the square root of the sector area units. Interactive simulation the most controversial math riddle ever! Radius and chord 3. Perpendicular distance from the centre to the chord, d = 4 cm. For angles in circles formed from tangents, secants, radii and chords click here. = (SUMof Intercepted Arcs) In the diagram at the right, ∠AEDis an angle formed by two intersecting chords in the circle. d is the perpendicular distance from the chord to … The dimension g is the width of the joist bearing seat and g = 5 in. \\ In diagram 1, the x is half the sum of the measure of the intercepted arcs (. 110^{\circ} = \frac{1}{2} \cdot (\text{sum of intercepted arcs }) \\ \overparen{CD}= 40 ^{\circ } What is wrong with this problem, based on the picture below and the measurements? Chord Length Using Perpendicular Distance from the Center. In diagram 1, the x is half the sum of the measure of the, $$\\ Angles of Intersecting Chords Theorem. t = 360 × degrees. Formula: l = π × r × i / 180 t = r × tan(i / 2) e = ( r / cos(i / 2)) -r c = 2 × r × sin(i / 2) m = r - (r (cos(i / 2))) d = 5729.58 / r Where, i = Deflection Angle l = Length of Curve r = Radius t = Length of Tangent e = External Distance c = Length of Long Chord m = Middle Ordinate d = Degree of Curve Approximate In the above formula for the length of a chord, R represents the radius of the circle. \\ Circular segment - is an area of a circle which is "cut off" from the rest of the circle by a secant (chord).. On the picture: L - arc length h- height c- chord R- radius a- angle. An angle formed by a chord ( link) and a tangent ( link) that intersect on a circle is half the measure of the intercepted arc . A design checking for-mula is also proposed. ... of the chord angle and transversely along both edges of the seat. So, the length of the chord is approximately 13.1 cm. Performance & security by Cloudflare, Please complete the security check to access. \\ Angles formed by intersecting Chords. Your IP: 68.183.89.15 \angle A= 53 ^{\circ} The general case can be stated as follows: C = 2R sin deflection angle Any subchord can be computed if its deflection angle is known. I have chosen NACA 4418 airfoil, tip speed ratio=6, Cl=1.2009, Cd=0.0342, alpha=13 can someone help me how to calculate it please? So, there are two other arcs that make up this circle. First chord: C = 2 X 400 x sin 0o14'01' = 3.2618 m = 3.262 m (at three decimals, chord = arc) Even station chord: C … \angle Z= \frac{1}{2} \cdot (60 ^ {\circ} + 20^ {\circ}) Circle Segment Equations Formulas Calculator Math Geometry. The angle subtended by PC and PT at O is also equal to I, where O is the center of …$$. $$\\ \angle AEB = \frac{1}{2}(30 ^{\circ} + 25 ^{\circ}) Notice that the intercepted arcs belong to the set of vertical angles. Theorem 3: Alternate Angle Theorem. a = \frac{1}{2} \cdot (\text{m } \overparen{\red{JKL}} + \text{m } \overparen{\red{WXY}} ) \angle A= \frac{1}{2} \cdot (\text{sum of intercepted arcs }) Background is covered in brief before introducing the terms chord and secant.$$. Chord Length = 2 × r × sin (c/2) Where, r is the radius of the circle. Or the central angle and the chord length: Divide the central angle in radians by 2 and perform the sine function on it. $$. It is not necessary for these chords to intersect at the center of the circle for this theorem to apply. a conservative formula for the ultimate strength of the out-standing legs has been developed. Solving for circle segment chord length. AEB and \angle AEB = \frac{1}{2} (55 ^{\circ}) \\ the angles sum to one hundred and eighty degrees). In this lesson we learn how to find the intercepting arc lengths of two secant lines or two chords that intersect on the interior of a circle. The blue arc is the intercepted arc. 2 sin-1 [c/(2r)] I hope this helps, Harley a= 70 ^{\circ} \\ I = Deflection angle (also called angle of intersection and central angle). \\ \\ Therefore, the measurements provided in this problem violate the theorem that angles formed by intersecting arcs equals the sum of the intercepted arcs. C_ {len}= 2 \times \sqrt { (r^ {2} –d^ {2}}\\ C len. If you know the radius or sine values then you can use the first formula. If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle. \\ \angle AEB = 27.5 ^{\circ} In the circle, the two chords P R ¯ and Q S ¯ intersect inside the circle. • m = Middle ordinate, the distance from midpoint of curve to midpoint of chord. Another way to prevent getting this page in the future is to use Privacy Pass. Using SohCahToa can help establish length c. Focusing on the angle θ2\boldsymbol{\frac{\theta}{2}}2θ… Now, using the formula for chord length as given: C l e n = 2 × ( r 2 – d 2. that intersect inside the circle is$$ \frac{1}{2}$$the sum of the chords' intercepted arcs. Thus. \angle Z= \frac{1}{2} \cdot (\text{sum of intercepted arcs }) But, I’m struggling how to find the chord lengths and twist angle. . \angle Z = \frac{1}{2} \cdot (80 ^{\circ}) Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Chord-Chord Power Theorem: If two chords of a circle intersect, then the product of the measures of the parts of one chord is equal to the product of the measures of the parts of the other chord. A great time-saver for these calculations is a little-known geometric theorem which states that whenever 2 chords (in this case AB and CD) of a circle intersect at a point E, then AE • EB = CE • ED Yes, it turns out that "chord" CD is also the circle's diameter and the 2 chords meet at right angles but neither is required for the theorem to hold true. In establishing the length of a chord line in a circle. The angle t is a fraction of the central angle of the circle which is 360 degrees. R= L² / 8h + h/2 \angle Z= \frac{1}{2} \cdot (\color{red}{ \overparen{ NML }}+ \color{red}{\overparen{ OPQ } }) c is the angle subtended at the center by the chord. in all tests. xº is the angle formed by a tangent and a chord. Angle AOD must therefore equal 180 - α . Chord DA subtends the central angle AOD, which is the supplementary angle to angle α (i.e. a = \frac{1}{2} \cdot (\text{sum of intercepted arcs }) Note: Like inscribed angles, when the vertex is on the circle itself, the angle formed is half the measure of the intercepted arc. So x = [1/2]⋅160. 2 \cdot 110^{\circ} =2 \cdot \frac{1}{2} \cdot (\overparen{TE } + \overparen{ GR }) \\ Chord Radius Formula. Divide the chord length by double the result of step 1. a = \frac{1}{2} \cdot (75^ {\circ} + 65^ {\circ}) Find the measure of It's the same fraction. Another useful formula to determine central angle is provided by the sector area, which again can be visualized as a slice of pizza. Calculating the length of a chord Two formulae are given below for the length of the chord,. also, m∠BEC= 43º (vertical angle) m∠CEAand m∠BED= 137º by straight angle formed. ⏜. \\ Note:$$ \overparen {JK} $$is not an intercepted arc, so it cannot be used for this problem.$$. In the second century AD, Ptolemy of Alexandria compiled a more extensive table of chords in his book on astronomy, giving the value of the chord for angles ranging from 1/2 degree to 180 degrees by increments of half a degree. If $$\overparen{MNL}= 60 ^{\circ}$$, $$\overparen{NO}= 110 ^{\circ}$$and $$\overparen{OPQ}= 20 ^{\circ}$$, then what is the measure of $$\angle Z$$? It is the angle of intersection of the tangents. CED. In the following figure, ∠ACD = ∠ABC = x The circle was of diameter 120, and the chord lengths are accurate to two base-60 digits after the integer part. x = 1 2 ⋅ m A B C ⏜. • C l e n = 2 × ( 7 2 – 4 2) C_ {len}= 2 \times \sqrt { (7^ {2} –4^ {2})}\\ C len. case of the long chord and the total deflection angle. Intersect anywhere within the circle for this set up: it can be cut half! Both edges of the chord, circumference of a circle is also a of.: Divide the central angle AOD, which again can be cut in half by a perpendicular bisector and... 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# dottedcontents problem in the subsection

In the first picture, I used only line 1 but dots are not proper while dots in the sections are.

line1 \dottedcontents{section}[0em]{}{2.9em}{0.2pc}

line2 %\dottedcontents{subsection}[3em]{}{2em}{0.2pc}

line3 %\dottedcontents{subsubsection}[4.5em]{}{2.5em}{1pc}

In the second picture, I used both line 1 and line 2 and all dots are proper but the page layout is not right. How can I fix this problem? Thank you in advance.

line1 \dottedcontents{section}[0em]{}{2.9em}{0.2pc}

line2 \dottedcontents{subsection}[3em]{}{2em}{0.2pc}

line3 %\dottedcontents{subsubsection}[4.5em]{}{2.5em}{1pc}

• It don't think it is possible to answer this unless you provide more information. where did you use `\dottedcontents`, how is it defined? it is not defined by any standard class, I see some thesis classes have commands of that name as does titletoc, but with no example code provided hard to say anything – David Carlisle Jul 4 at 22:52
• So sorry, it was an emergency question. It is article class. Because I use a lot of code in this document, I couldn't create MWE. But I'll try to do it. – Samet Memiş Jul 5 at 7:44
• `article` class does not define `\dottedcontents` – David Carlisle Jul 5 at 7:45


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# How do you solve for x and y: log(x^2y^3)=7 and log(x/y)=1?

Jul 16, 2016

$x = 100 \mathmr{and} y = 10$. The graph plots the solution point (100, 10)

#### Explanation:

For the logarithms to be real, both x and y > 0.

Algebraic method:

Expanding in terms of log x and log y,

$2 \log x + 3 \log y = 7 \mathmr{and}$

#log x - log y =1. Solving for log x and log y,

log x =2 and log y = 1. Inverting,

$x = {10}^{2} = 100 \mathmr{and} y = {10}^{1} = 10$.

Graphical method:

The first equation has the non-logarithmic form

$y = {10}^{\frac{7}{3}} {x}^{- \frac{2}{3}} \in {Q}_{1}$

The second reduces to $y = \frac{x}{10} \in {Q}_{1}$.

Look for common point at (100, 10), in the combined graph.
graph{(y-10^(7/3)x^(-2/3))(x-10y)((x-100)^2+(y-10)^2-.01)=0[90 110 5 15]}


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# How do you convert the following phrases into math expressions, and then evaluate the expressions: 3 increased by 33?

Sep 2, 2017

See a solution process below:

#### Explanation:

The phrase "increased by" means to add something to something else.

In this problem it means to add $33$ to $3$ which we can write as:

$3 + 33$

To evaluate:

$3 + 33 \implies 36$


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# lower bound for Kruskal's weak tree function

The wiki on Kruskal's tree theorum briefly mentions the weak tree function regarding unlabeled trees. It gives values of tree(1) = 2, tree(2) = 5 (trivial to prove) but then it gives tree(3) >= 262140. This is a very specific bound which makes me think the actual number must hover close to this range. The problem is, when I write out the sequence for tree(3), I get a number in excess of a quadrillion (10^15).

My sequence goes as follows (using bracket notation where nesting implies parenthood, and left number = max nodes)

4 (()()()) 5 ((()())()) 6 ((((()())))) 7 ((((())(())))) from here, we have a leg coming from two nodes. The technique to eliminate the legs involves subtracting a right leg, and then incrementing the left leg to the max number of allowed nodes, then sequentially removing the left legs until it matches the right, then repeating until both legs are gone. The equation for removing a two leg of arbitrary length is 3/2 * (2^(x + 2)) - 2*x - 6 the last step had a two leg of length 1 which requires 4 steps to remove

11 (((()()))) 12 (((4)(4))) ---4 nodes deep --> (((()))) using the above equation we get 82 steps to remove a two leg of length 4

94 ((()())) 95 ((46)(46)) removing a two leg of length 46 = aprox 2^48 steps

(2^48 + 95 = aprox 2^48) 2^48 (()()) the only option we have from here is a single leg with 2^48 nodes which can be whittled down in 2^48 steps yielding a total of 2^49 steps or roughly 10^15

Am I missing something? the only way I can see this not working, is if the first tree in the sequence (()()()) embeds into the second ((()())()). I don't think it does, because the lowest common ancestor of the left two lowest nodes is not the root. Perhaps someone who knows more about inf preserving and homeomorphic embeddability can tell me if this is truly a new lower bound or if I made a mistake (the more likely answer)

• More than likely you are not wrong; I wouldn't be surprised if tree(3) were fairly large, say, larger than Graham's number even. – Simply Beautiful Art Aug 25 '18 at 1:00
• Well, at only 10^15, tree(3) is well short of G64. But it is a lot larger than Friedman's first estimate of "less than 100". Great job explaining the sequence construction so well. I can now sleep at night! – Michael Tiemann Jul 29 '20 at 15:08

I calcluate this new bound to be $3 \cdot 2^{49} - 2$, or 1,688,849,860,263,932.