Chaitanya's Random Pages

June 30, 2017

The ballot problem and Catalan’s triangle

Filed under: Uncategorized — ckrao @ 10:15 pm

The ballot problem asks for the probability that candidate A is always ahead of candidate B during a tallying process if they respectively end up with p and q votes where p > q. For example if p = 2, q = 1 there are 3 ways in which the three votes are counted (AAB, ABA, BAA) but the only favourable outcome in which A remains ahead throughout is occurs if the tally appears as AAB. Hence the probability A remains ahead is 1/3.

If there are no restrictions, the number of ways the votes are tallied is the binomial coefficient \binom{p+q}{p}. The number of favourable outcomes (the numerator of the desired probability) in which A remains ahead can counted recursively in a similar way to Pascal’s triangle (each number the sum of the two neighbours above it) except no number may appear to the left of the vertical midline, as illustrated below. For example, the second element of the fifth row (3) corresponds to the case p = 3, q = 1 (AAAB, AABA, ABAA). More generally, dividing into the cases where the final vote is A or B, the number of ways N_{p,q} in which A remains ahead of B is equal to N_{p-1,q} + N_{p,q-1} where N_{p,q} = 0 if q \geq p. This sequence appears as A008313 in the OEIS and is the reversed form of Catalan’s triangle.

Catalan_tri

A way of generating the general term is making use of a beautiful reflection principle that gives a 1-1 correspondence between the number of tallies leading to a tie at some point and the number of tallies in which the first vote goes to candidate B: simply interchange A with B for all votes up to and including that tie. This amounts to reflecting the random walk about the midline, as illustrated below with the blue path corresponding to ABAA and the the red path BAAA.

pathreflection

Since p > q, the probability candidate A always leads is 1 minus the probability the sequence ties at some point. But the bijection above shows an equal number of these start with A and with B, so our desired probability is

\displaystyle 1 - 2 \text{Pr(sequence starts with B)} = 1 - 2\frac{q}{p+q} = \frac{p-q}{p+q}.

The numbers in the triangle are also formed by differences of adjacent entries of Pascal’s triangle, namely row p+q has terms of the form

\displaystyle \begin{aligned} N_{p,q} &= \binom{p+q}{p}\frac{p-q}{p+q}\\ &= \frac{(p+q-1)!(p-q)}{p!q!}\\&= \binom{p+q-1}{q}-\binom{p+q-1}{p}.\end{aligned}

This can be interpreted as the number of unrestricted sequences with p As and q Bs of length (p+q) that start with A minus the corresponding number that start with B, again following from the reflection principle.

As an aside, looking at the bottom row above we see N_{8,6} = N_{10,4} = 429, or equivalently

\displaystyle \binom{13}{4} - \binom{13}{3} = \binom{13}{6} - \binom{13}{5} = 429.

Finally we note that the Catalan numbers arise from the following parts of the triangle above:

  • as entries in the first column (counting Dyck paths)
  • as the sum of squares of each row
  • as the sum of entries in NE-SW diagonals

Catalan’s triangle can be generalised to a trapezium in which we count the number of strings consisting of n As and k Bs such that in every initial segment of the string the number of Bs does not exceed the number of As by m or more.

April 9, 2017

Highest aggregates and averages after n test matches/innings

Filed under: cricket,sport — ckrao @ 9:14 am

Soon after the 2017 India-Australia test series I noticed that among players who have played 54 tests, nobody has scored more than Steven Smith’s 5251 runs. Here is a list of the top three aggregates and averages after 54 tests and 100 innings.

54 tests 5251 SPD Smith (AUS) 5210 SM Gavaskar (INDIA) 4991 L Hutton (ENG) 4840 JB Hobbs (ENG)
54 tests 61.06 SPD Smith (AUS) 60.73 H Sutcliffe (ENG) 59.51 GS Sobers (WI) 59.02 JB Hobbs (ENG)
100 innings 5354 JB Hobbs (ENG) 5345 GS Sobers (WI) 5279 WR Hammond (ENG) 5251 SPD Smith (AUS)
100 innings 61.06 SPD Smith (AUS) 60.74 GS Sobers (WI) 60.68 WR Hammond (ENG) 58.48 L Hutton (ENG)

A more complete list of the top 10 scorers in these categories after n tests/innings is below. Statistics are from ESPN Cricinfo and are current to 9 April 2017. Corrections are welcome.

The following players have been ranked first for some n (as of 9 April 2017):

  • highest aggregate after n innings: Foster, Rowe, Javed Miandad, Kambli, Weekes, Bradman, Hobbs, Sobers, Hammond, Hutton, Sehwag, Tendulkar, Sangakkara, Lara
  • highest aggregate after n tests: Rowe, Foster, Javed Miandad, Gavaskar, Bradman, Smith, Sobers, Hutton, Sehwag, Sangakkara, Younis Khan, Lara, Ponting, Kallis, Dravid, Tendulkar
  • highest average after n innings: Foster, Rowe, Bell, Trott, Kambli, Gavaskar, Harvey, Bradman, Voges, Sutcliffe, Hobbs, Smith, Barrington, Sobers, Hammond, Hutton, Dravid, Tendulkar, Ponting, Sangakkara, Kallis
  • highest average after n tests: Rowe, Rudolph, Bell, Gavaskar, Kambli, Samaraweera, Harvey, Bradman, Voges, Sutcliffe, Smith, Hobbs, Hammond, Sobers, Dravid, Tendulkar, Ponting, Sangakkara, Kallis

March 19, 2017

Two similar geometry problems based on perpendiculars to cevians

Filed under: mathematics — ckrao @ 7:18 am

In this post I wanted to show a couple of similar problems that can be proved using some ideas from projective geometry.

The first problem I found via the Romantics of Geometry Facebook group: let M be the point of tangency of the incircle of \triangle ABC with BC and let E be the foot of the perpendicular from the incentre X of the \triangle ABC to AM. Then show EM bisects \angle BEC.

 

perpendicularcevian1

The second problem is motivated by the above and problem 2 of the 2008 USAMO: this time let AM be a symmedian of ABC and E be the foot of the perpendicular from the circumcentre X of \triangle ABC to AM. Then show that EM bisects \angle BEC.

perpendicularcevian2

Here is a solution to the first problem inspired bythat of Vaggelis Stamatiadis. Let the line through the other two points of tangency P, Q of the incircle with ABC intersect line BC at the point N as shown below. Note that since AP and AQ are tangents to the circle, line NPQ is the polar of A with respect to the incircle.

perpendicularcevian1a

Since N is on the polar of A, by La Hire’s theorem, A is on the polar of N. The polar of N also passes through M (as NM is a tangent to the circle at M). We conclude that the polar of N is the line through A and M.

Next, let MN intersect PQ at R. By theorem 5(a) at this link, the points (N, R, P, Q) form a harmonic range. Since the cross ratio of collinear points does not change under central projection,  considering the projection from A, (N,M,B,C) also form a harmonic range. (Alternatively, this follows from the theorems of Ceva and Menelaus using the Cevians intersecting at the Gergonne point and transveral NPQ). Also, NE \perp EM as both NI and IE are perpendicular to polar AM of N.

Considering a central projection from E of line NMBC to a line N', M, P', C' parallel to NE through M, we see that (N', M, P', C') form a harmonic range. Since N' is a point at infinity, this implies M is the midpoint of B'C' and so triangles B'EM and C'EM are congruent (equality of two pairs of sides and included angle is 90^{\circ}). Hence EM bisects \angle BEC as was to be shown.

perpendicularcevian1b

For the second problem, we use the following characterisation of a symmedian: AM extended concurs with the lines of tangency of the circumcircle at B and C. (For three proofs of this see here.)

perpendicularcevian2a

Define N as the intersection of XE with BC and D as the intersection of AM with the tangents at B, C. Note that line NBMC is the polar of D with respect to the circumcircle. By La Hire’s theorem, D must be on the polar of N. This polar is perpendicular to NX (the line joining N to the centre of the circle) and as ED \perp EX by construction of E, it follows that line AEMD is the polar of N. Again by theorem 5(a) in reference (2), (N, M, B, C) form a harmonic range. Following the same argument as the previous proof, this together with NE \perp EM imply EM bisects \angle BEC as required.

By similar arguments, one can prove the following, left to the interested reader. If X is the A-excentre of \triangle ABC, M the ex-circle’s point of tangency of BC, and E the foot of the perpendicular from X to line AM, then EM bisects \angle BEC.

perpendicularcevian3

References

(1) Alexander Bogomolny, Poles and Polars from Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml, Accessed 19 March 2017

(2) Poles and Polars – Another Useful Tool! | The Problem Solver’s Paradise

(3) Yufei Zhao, Lemmas in Euclidean Geometry

December 24, 2016

Kohli’s 2016

Filed under: Uncategorized — ckrao @ 9:18 pm

Here is a list of the scores Virat Kohli made in Test, ODI, T20 and IPL cricket during 2016. Stunning numbers.

Test ODI T20I IPL
200 (283) 91 (97) 90* (55) 75 (51)
44 (90) 59 (67) 59* (33) 79 (48)
3 (8) 117 (117) 50 (36) 33 (30)
4 (17) 106 (92) 7 (12) 80 (63)
9 (10) 8 (11) 49 (51) 100* (63)
18 (40) 85* (81) 56* (47) 14 (17)
9 (28) 9 (13) 41* (28) 52 (44)
45 (65) 154* (134) 23 (27) 108* (58)
211 (366) 45 (51) 55* (37) 20 (21)
17 (28) 65 (76) 24 (24) 7 (7)
40 (95) 82* (51) 109 (55)
49* (98) 89* (47) 75* (51)
167 (267) 16 (9) 113 (50)
81 (109) 54* (45)
62 (127) 0 (2)
6* (11) 54 (35)
235 (340)
15 (29)
1215 @ 75.9 739 @ 92.37, SR 100 641 @ 106.8, SR 140.3 973 @ 81.2, SR 152.0

December 19, 2016

Some special functions and their applications

Filed under: mathematics — ckrao @ 9:55 am

Here are some notes on special functions and where they may arise. We consider functions in applied mathematics beyond field (four arithmetic operations), composition and inverse operations applied to the power and exponential functions.

1. Bessel and related functions

Bessel functions of the first (J_{\alpha}(x)) and second (Y_{\alpha}(x)) kind of order \alpha satisfy:

\displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - \alpha^2)y = 0.

Solutions for integer \alpha arise in solving Laplace’s equation in cylindrical coordinates while solutions for half-integer \alpha arise in solving the Helmholtz equation in spherical coordinates. Hence they come about in wave propagation, heat diffusion and electrostatic potential problems. The functions oscillate roughly periodically with amplitude decaying proportional to 1/\sqrt{x}. Note that Y_{\alpha}(x) is the second linearly independent solution when \alpha is an integer (for integer n, J_{-n}(x) = (-1)^n J_n(x)). Also, for integer n, J_n has the generating function

\displaystyle  \sum_{n=-\infty}^\infty J_n(x) t^n = e^{(\frac{x}{2})(t-1/t)},

the integral representations

\displaystyle J_n(x) = \frac{1}{\pi} \int_0^\pi \cos (n \tau - x \sin(\tau)) \,d\tau = \frac{1}{2 \pi} \int_{-\pi}^\pi e^{i(n \tau - x \sin(\tau))} \,d\tau

and satisfies the orthogonality relation

\displaystyle \int_0^1 x J_\alpha(x u_{\alpha,m}) J_\alpha(x u_{\alpha,n}) \,dx = \frac{\delta_{m,n}}{2} [J_{\alpha+1}(u_{\alpha,m})]^2 = \frac{\delta_{m,n}}{2} [J_{\alpha}'(u_{\alpha,m})]^2,

where \alpha > -1, \delta_{m,n} Kronecker delta, and u_{\alpha, m} is the m-th zero of J_{\alpha}(x).

Modified Bessel functions of the first (I_{\alpha}(x)) and second (K_{\alpha}(x)) kind of order \alpha satisfy:

\displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - (x^2 + \alpha^2)y = 0

(replacing x with ix in the previous equation).

The four functions may be expressed as follows.

\displaystyle J_{\alpha}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}

\displaystyle I_\alpha(x) = \sum_{m=0}^\infty \frac{1}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}

\displaystyle Y_\alpha(x) = \frac{J_\alpha(x) \cos(\alpha\pi) - J_{-\alpha}(x)}{\sin(\alpha\pi)}

\displaystyle K_\alpha(x) = \frac{\pi}{2} \frac{I_{-\alpha} (x) - I_\alpha (x)}{\sin (\alpha \pi)}

(In the last formula we need to take a limit when \alpha is an integer.)

Note that K and Y are singular at zero.

The Hankel functions H_\alpha^{(1)}(x) = J_\alpha(x)+iY_\alpha(x) and H_\alpha^{(2)}(x) = J_\alpha(x)-iY_\alpha(x) are also known as Bessel functions of the third kind.

The functions J_\alphaY_\alpha, H_\alpha^{(1)}, and H_\alpha^{(2)} all satisfy the recurrence relations (using Z in place of any of these four functions)

\displaystyle \frac{2\alpha}{x} Z_\alpha(x) = Z_{\alpha-1}(x) + Z_{\alpha+1}(x),
\displaystyle 2\frac{dZ_\alpha}{dx} = Z_{\alpha-1}(x) - Z_{\alpha+1}(x).

Bessel functions of higher orders/derivatives can be calculated from lower ones via:

\displaystyle \left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ x^\alpha Z_{\alpha} (x) \right] = x^{\alpha - m} Z_{\alpha - m} (x),
\displaystyle \left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ \frac{Z_\alpha (x)}{x^\alpha} \right] = (-1)^m \frac{Z_{\alpha + m} (x)}{x^{\alpha + m}}.

In particular, note that -J_1(x) is the derivative of J_0(x).

The Airy functions of the first (Ai(x)) and second (Bi(x)) kind satisfy

\displaystyle \frac{d^2y}{dx^2} - xy = 0.

This arises as a solution to Schrödinger’s equation for a particle in a triangular potential well and also describes interference and refraction patterns.

2. Orthogonal polynomials

Hermite polynomials (the probabilists’ defintion) can be defined by:

\displaystyle \mathit{He}_n(x)=(-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-\frac{x^2}{2}}=\left (x-\frac{d}{dx} \right )^n \cdot 1,

and are orthogonal with respect to weighting function w(x) = e^{-x^2} on (-\infty, \infty).

They satisfy the differential equation

\displaystyle \left(e^{-\frac{x^2}{2}}u'\right)' + \lambda e^{-\frac{1}{2}x^2}u = 0

(where \lambda is forced to be an integer if we insist u be polynomially bounded at \infty)

and the recurrence relation

\displaystyle {\mathit{He}}_{n+1}(x)=x{\mathit{He}}_n(x)-{\mathit{He}}_n'(x).

The first few such polynomials are 1, x, x^2-1, x^3-3x, \ldots. The Physicists’ Hermite polynomials H_n(x) are related by H_n(x)=2^{\tfrac{n}{2}}{\mathit{He}}_n(\sqrt{2} \,x) and arise for example as the eigenstates of the quantum harmonic oscillator.

Laguerre polynomials are defined by

\displaystyle L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}\left(e^{-x} x^n\right) =\frac{1}{n!} \left( \frac{d}{dx} -1 \right) ^n x^n = \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k,

and are orthogonal with respect to e^{-x} on (0,\infty).

They satisfy the differential equation

\displaystyle  xy'' + (1 - x)y' + ny = 0,

recurrence relation

\displaystyle L_{k + 1}(x) = \frac{(2k + 1 - x)L_k(x) - k L_{k - 1}(x)}{k + 1},

and have generating function

\displaystyle \sum_n^\infty  t^n L_n(x)=  \frac{1}{1-t} e^{-\frac{tx}{1-t}}.

The first few values are 1, 1-x, (x^2-4x+2)/2. Note also that L_{-n}(x)=e^xL_{n-1}(-x).

The functions come up as the radial part of solution to Schrödinger’s equation for a one-electron atom.

Legendre polynomials can be defined by

\displaystyle P_n(x) = {1 \over 2^n n!} {d^n \over dx^n } \left[ (x^2 -1)^n \right]

and are orthogonal with respect to the L^2 norm on (-1,1).

They satisfy the differential equation

\displaystyle {d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0,

recurrence relation

and have generating function

\sum_{n=0}^\infty P_n(x) t^n = \displaystyle \frac{1}{\sqrt{1-2xt+t^2}}.

The first few values are 1, x, (3x^2-1)/2, (5x^3-3x)/2.

They arise in the expansion of the Newtonian potential 1/|x-x'| (multipole expansions) and Laplace’s equation where there is axial symmetry (spherical harmonics are expressed in terms of these).

Chebyshev polynomials of the 1st kind T_n(x) can be defined by

T_n(x) =\begin{cases} \cos(n\arccos(x)) & \ |x| \le 1 \\ \frac12 \left[ \left (x-\sqrt{x^2-1} \right )^n + \left (x+\sqrt{x^2-1} \right )^n \right] & \ |x| \ge 1 \\ \end{cases}

and are orthogonal with respect to weighting function w(x) = 1/\sqrt{1-x^2} in (-1,1).

They satisfy the differential equation

\displaystyle (1-x^2)\,y'' - x\,y' + n^2\,y = 0,

the relations

\displaystyle T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)

\displaystyle (1 - x^2)T_n'(x) = -nx T_n(x) + n T_{n-1}(x)

and have generating function

\displaystyle \sum_{n=0}^{\infty}T_n(x) t^n = \frac{1-tx}{1-2tx+t^2}.

The first few values are 1, x, 2x^2-1, 4x^3-3x, \ldots. These polynomials arise in approximation theory, namely their roots are used as nodes in piecewise polynomial interpolation. The function f(x) = \frac1{2^{n-1}}T_n(x) is the polynomial of leading coefficient 1 and degree n where the maximal absolute value on (-1,1) is minimal.

Chebyshev polynomials of the 2nd kind U_n(x) are defined by

\displaystyle  U_n(x)  = \frac{\left (x+\sqrt{x^2-1} \right )^{n+1} - \left (x-\sqrt{x^2-1} \right )^{n+1}}{2\sqrt{x^2-1}}

and are orthogonal with respect to weighting function w(x) = \sqrt{1-x^2} in (-1,1).

They satisfy the differential equation

\displaystyle  (1-x^2)\,y'' - 3x\,y' + n(n+2)\,y = 0,

the recurrence relation

\displaystyle U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x)

and have generating function

\displaystyle \sum_{n=0}^{\infty}U_n(x) t^n = \frac{1}{1-2 t x+t^2}.

The first few values are 1, 2x, 4x^2-1, 8x^3-4x, \ldots. (There are also less well known Chebyshev  polynomials of the third and fourth kind.)

Bessel polynomials y_n(x) may be defined from Bessel functions via

\displaystyle y_n(x)=\sqrt{\frac{2}{\pi x}}\,e^{1/x}K_{n+\frac 1 2}(1/x)  = \sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\,\left(\frac{x}{2}\right)^k.

They satisfies the differential equation

\displaystyle x^2\frac{d^2y_n(x)}{dx^2}+2(x\!+\!1)\frac{dy_n(x)}{dx}-n(n+1)y_n(x)=0.

The first few values are 1, x+1, 3x^2+3x+1,\ldots.

3. Integrals

The error function has the form

\displaystyle \rm{erf}(x) = \frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\,\mathrm dt.

This can be interpreted as the probability a normally distributed random variable with zero mean and variance 1/2 is in the interval (-x,x).

The cdf of the normal distribution $\Phi(x)$ is related to this via \Phi(x) = (1 + {\rm erf}(x/\sqrt{2})/2. Hence the tail probability of the standard normal distribution Q(x) is Q(x) = (1 - {\rm erf}(x/\sqrt{2}))/2.

Fresnel integrals are defined by

\displaystyle S(x) =\int_0^x \sin(t^2)\,\mathrm{d}t=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+3}}{(2n+1)!(4n+3)}
\displaystyle C(x) =\int_0^x \cos(t^2)\,\mathrm{d}t=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+1}}{(2n)!(4n+1)}

They have applications in optics.

The exponential integral {\rm Ei}(x) (used in heat transfer applications) is defined by

\displaystyle {\rm Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt.

It is related to the logarithmic integral

\displaystyle {\rm li} (x) =   \int_0^x \frac{dt}{\ln t}

by \mathrm{li}(x) = \mathrm{Ei}(\ln x) (for real x).

The incomplete elliptic integral of the first, second and third kinds are defined by

\displaystyle F(\varphi,k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}

\displaystyle E(\varphi,k) =  \int_0^\varphi \sqrt{1-k^2 \sin^2\theta}\, d\theta

 \displaystyle \Pi(n ; \varphi \setminus \alpha) = \int_0^\varphi  \frac{1}{1-n\sin^2 \theta} \frac {d\theta}{\sqrt{1-(\sin\theta\sin \alpha)^2}}

Setting \varphi = \pi/2 gives the complete elliptic integrals.

Any integral of the form \int_{c}^{x} R \left(t, \sqrt{P(t)} \right) \, dt, where c is a constant, R is a rational function of its arguments and P(t) is a polynomial of 3rd or 4th degree with no repeated roots, may be expressed in terms of the elliptic integrals. The circumference of an ellipse of semi-major axis a, semi-minor axis b and eccentricity e = \sqrt{1-b^2/a^2} is given by 4aE(e), where E(k) is the complete integral of the second kind.

(Some elliptic functions are related to inverse elliptic integral, hence their name.)

The (upper) incomplete Gamma function is defined by

\displaystyle \Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t.

It satisfies the recurrence relation \Gamma(s+1,x)= s\Gamma(s,x) + x^{s} e^{-x}. Setting s= 0 gives the Gamma function which interpolates the factorial function.

The digamma function is the logarithmic derivative of the gamma function:

\displaystyle \psi(x)=\frac{d}{dx}\ln\Big(\Gamma(x)\Big)=\frac{\Gamma'(x)}{\Gamma(x)}.

Due the relation \psi(x+1) = \psi(x) + 1/x, this function appears in the regularisation of divergent integrals, e.g.

\sum_{n=0}^{\infty} \frac{1}{n+a}= - \psi (a).

The incomplete Beta function is defined by

\displaystyle B(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,\mathrm{d}t.

When setting x=1 this becomes the Beta function which is related to the gamma function via

\displaystyle B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}.

This can be extended to the multivariate Beta function, used in defining the Dirichlet function.

\displaystyle B(\alpha_1,\ldots,\alpha_K) = \frac{\Gamma(\alpha_1) \cdots \Gamma(\alpha_K)}{\Gamma(\alpha_1 + \ldots + \alpha_K)}.

The polylogarithm, appearing as integrals of the Fermi–Dirac and Bose–Einstein distributions, is defined by

\displaystyle {\rm Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} = z + \frac{z^2}{2^s} + \frac{z^3}{3^s} + \cdots

Note the special case {\rm Li}_1(z) = -\ln (1-z) and the case s=2 is known as the dilogarithm. We also have the recursive formula

\displaystyle {\rm Li}_{s+1}(z) = \int_0^z \frac {{\rm Li}_s(t)}{t}\,\mathrm{d}t.

4. Generalised Hypergeometric functions

All the above functions can be written in terms of generalised hypergeometric functions.

\displaystyle {}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}

where (a)_n = \Gamma(a+n)/\Gamma(a) = a(a+1)(a+2)...(a+n-1) for n > 0 or (a)_0 = 1.

The special case p=q=1 is called a confluent hypergeometric function of the first kind, also written M(a;b;z).

This satisfies the differential equation (Kummer’s equation)

\displaystyle \left (z\frac{d}{dz}+a \right )w = \left (z\frac{d}{dz}+b \right )\frac{dw}{dz}.

The Bessel, Hankel, Airy, Laguerre, error, exponential and logarithmic integral functions can be expressed in terms of this.

The case p=2, q=1 is sometimes called Gauss’s hypergeometric functions, or simply hypergeometric functions. This satisfies the differential equation

\displaystyle \left (z\frac{d}{dz}+a \right ) \left (z\frac{d}{dz}+b \right )w =\left  (z\frac{d}{dz}+c \right )\frac{dw}{dz}.

The Legendre, Hermite and Chebyshev, Beta, Gamma functions can be expressed in terms of this.

Further reading

The Wolfram Functions Site

Wikipedia: List of mathematical functions

Wikipedia: List of special functions and eponyms

Wikipedia: List of q-analogs

Wikipedia Category: Orthogonal polynomials

Weisstein, Eric W. “Laplace’s Equation.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/LaplacesEquation.html

July 29, 2016

Distribution of Melbourne’s length of day

Filed under: geography — ckrao @ 10:34 pm
According to timeanddate.com, Melbourne, Australia in 2016 has a minimum daylength of 9 hours 32 minutes and 32 seconds, and a maximum daylength of 14 hours 47 minutes and 19 seconds (the asymmetry is due to the way day length is calculated). Here is a look at the distribution of day length through the year.
Duration of daylength (hrs) Dates Frequency
< 10 19 May-24 July 67
10-10.5 25 July-10 August, 2-18 May 34
10.5-11 11-24 August, 18 April-1 May 28
11-11.5 25 August-6 September, 5-17 April 26
11.5-12 6-19 September, 24 March-4 April 25
12-12.5 20 September-1 October, 12-23 March 24
12.5-13 1-13 October, 28 February-11 March 25
13-13.5 14-26 October, 16-27 February 25
13.5-14 9 Nov, 2-15 Feb 28
14-14.5 10 Nov-27 Nov, 16 Jan-1Feb 35
>14.5 28 Nov-15 Jan 49

What surprised me the most about this was that only 100 days of the year have daylength between 11 and 13 hours and we have a good 84 days with light longer than 14 hours.

June 26, 2016

2016 has many factors

Filed under: mathematics — ckrao @ 11:16 am

The number 2016 has at least as many factors (36) as any positive integer below it except 1680 = 2^4\times 3\times 5\times 7 (which has 40 factors). The next time a year will have more factors is 2160 = 2^4\times 3^3\times 5, also with 40 factors.

Here are the numbers below 2160 also with 36 factors:

  • 1260 = 2^2 \times 3^2 \times 5 \times 7
  • 1440 = 2^5 \times 3^2 \times 5
  • 1800 = 2^3 \times 3^2 \times 5^2
  • 1980 = 2^2 \times 3^2 \times 5 \times 11
  • 2016 = 2^5 \times 3^2 \times 7
  • 2100 = 2^2 \times 3 \times 5^2 \times 7

The first integer with more than 40 factors is $2520 = 2^3 \times 3^2 \times 5 \times 7$ (48 factors).

References

[1] N. J. A. Sloane and Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from N. J. A. Sloane), A000005 – OEIS.

[2] Highly composite number – Wikipedia, the free encyclopedia

March 28, 2016

Recent months of global warmth

Filed under: climate and weather — ckrao @ 5:02 am

According to both NASA’s Goddard Institute for Space Studies [1] and the NOAA National Centers for Environmental Information [2], February 2016 set another record of the highest deviation of global temperatures above the monthly mean. In fact NASA’s dataset has seen the past five months record the largest five monthly global warm anomalies [3]. Some plots of global temperatures from recent months can be seen at Makiko Sato’s page here. One case in point is Longyearbyen, Svalbard (78°N) whose temperatures have barely been below average for the past six months (data from [4-5]).

longyearbyen

February set the record of greatest anomaly from mean monthly temperatures, beating the previous record (set only the previous month) by more than 0.2°C. The map here shows that the vast majority of the planet had above-average temperatures, with the greatest deviation in the arctic region. As an example, check out the temperatures of Salekhard, Russia on the arctic circle during this time (this is a place that registers temperatures below -40 during winters). Over the month its average was 12.5°C above the mean! Data is from [6].

salekhard

More reading and references

[1] Record Warmth in February : Image of the Day

[2] NOAA National Centers for Environmental Information, State of the Climate: Global Analysis for February 2016, published online March 2016, retrieved on March 27, 2016 from http://www.ncdc.noaa.gov/sotc/global/201602.

[3] February 2016 Was the Most Abnormally Warm Month Ever Recorded, NOAA and NASA Say | The Weather Channel

[4] Ogimet: Synop report summary for Svalbard airport

[5] Longyearbyen February Weather 2016 – AccuWeather

[6] Погода и Климат – Климатический монитор: погода в Салехарде (pogodaiklimat.ru)

[7] Reliable, official numbers now in for February 2016 show that it smashed the previous record for the month – ImaGeo

[8] Record-Shattering February Warmth Bakes Alaska, Arctic 18°F Above Normal | ThinkProgress

[9] February Smashes Earth’s All-Time Global Heat Record by a Jaw-Dropping Margin | Dr. Jeff Masters’ WunderBlog

Save

March 26, 2016

Applying AM-GM in the denominator after flipping the sign

Filed under: mathematics — ckrao @ 8:44 pm

There are times when solving inequalities that one has a sum of fractions in which applying the AM-GM inequality to each denominator results in the wrong sign for the resulting expression.

For example (from [1], p18), if we wish to show that for real numbers x_1, x_2, \ldots, x_n with sum n that

\displaystyle \sum_{i = 1}^n \frac{1}{x_i^2 + 1}\geq \frac{n}{2},

we may write x_i^2 + 1 \geq 2x_i (equivalent to (x_i-1)^2 \geq 0), but this implies \frac{1}{x_i^2 + 1} \leq \frac{1}{2x_i} and so the sign goes the wrong way.

A way around this is to write

\begin{aligned}  \frac{1}{x_i^2 + 1} &= 1 - \frac{x_i^2}{x_i^2 + 1}\\  &\geq 1 - \frac{x_i^2}{2x_i}\\  &= 1 - \frac{x_i}{2}.  \end{aligned}

Summing this over i then gives \sum_{i=1}^n \frac{1}{x_i^2 + 1} \geq n - \sum_{i=1}^n (x_i/2) = n/2 as desired.

Here are a few more examples demonstrating this technique.

2. (p9 of [2]) If a,b,c are positive real numbers with a + b + c = 3, then

\dfrac{a}{1 +b^2} + \dfrac{b}{1 +c^2} + \dfrac{c}{1 +a^2} \geq \dfrac{3}{2}.

To prove this we write

\begin{aligned}  \frac{a}{1 + b^2} &= a\left(1 - \frac{b^2}{1 + b^2}\right)\\  &\geq a\left(1 - \frac{b}{2}\right) \quad \text{(using the same argument as before)}\\  &=a - \frac{ab}{2}.  \end{aligned}

Next we have 3(ab + bc + ca) \leq (a + b + c)^2 = 9 as this is equivalent to (a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0. This means ab + bc + ca \leq 3. Putting everything together,

\begin{aligned}  \frac{a}{1 + b^2} + \frac{b}{1 + c^2} + \frac{c}{1 + a^2}&\geq \left( a - \frac{ab}{2} \right) + \left( b - \frac{bc}{2} \right) + \left( c - \frac{ca}{2} \right)\\  &= (a + b + c) - (ab + bc + ca)/2\\  &\geq 3 - 3/2\\  &=\frac{3}{2},  \end{aligned}

as required.

3. (based on p8 of [2]) If x_i > 0 for i= 1, 2, \ldots, n and \sum_{i = 1}^n x_i^2 = n then

\displaystyle \sum_{i=1}^n \frac{1}{x_i^3 + 2} \geq \frac{n}{3}.

By the AM-GM inequality, x_i^3 + 2 = x_i^3 + 1 + 1 \geq 3x_i, so

\begin{aligned}  \frac{1}{x_i^3 + 2} &= \frac{1}{2}\left( 1 - \frac{x_i^3}{x_i^3 + 2} \right)\\  &\geq \frac{1}{2}\left( 1 - \frac{x_i^3}{3x_i} \right)\\  &= \frac{1}{2}\left( 1 - \frac{x_i^2}{3} \right).  \end{aligned}

Summing this over i gives

\begin{aligned}  \sum_{i=1}^n \frac{1}{x_i^3 + 2} &\geq \frac{1}{2} \sum_{i=1}^n \left( 1 - \frac{x_i^2}{3} \right)\\  &= \frac{1}{2}\left( n - \frac{n}{3} \right)\\  &= \frac{n}{3}.  \end{aligned}

4. (from [3]) If x, y, z are positive, then

\dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } \geq \dfrac {x + y + z}{2}.

Once again, focusing on the denominator,

\begin{aligned}  \dfrac {x ^ 3}{x ^ 2 + y ^ 2} &= x\left(1 - \dfrac {y ^ 2} {x ^ 2 + y ^ 2} \right)\\  &\geq x \left(1 -\dfrac{xy^2}{2xy} \right)\\  &= x-\dfrac{y}{2}.  \end{aligned}

Hence,

\begin{aligned}  \dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } &\geq x-\dfrac{y}{2} + y-\dfrac{z}{2} + z-\dfrac{x}{2}\\  &= \dfrac {x + y + z}{2},  \end{aligned}

as desired.

5. (from the 1991 Asian Pacific Maths Olympiad, see [4] for other solutions) Let a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n be positive numbers with \sum_{i = 1}^n a_i = \sum_{i = 1}^n b_i. Then

\displaystyle\sum_{i=1}^n\frac{a_i^2}{a_i + b_i} \geq \frac{1}{2}\sum_{i=1}^n a_i.

Here we write

\begin{aligned}  \sum_{i=1}^n\frac{a_i^2}{a_i + b_i} &= \sum_{i=1}^n a_i \left(1 - \frac{b_i}{a_i + b_i} \right)\\  &\geq \sum_{i=1}^n a_i \left(1 - \frac{b_i}{2\sqrt{a_i b_i}} \right) \\  &= \frac{1}{2} \sum_{i=1}^n \left( 2a_i - \sqrt{a_i b_i} \right) \\  &= \frac{1}{4} \sum_{i=1}^n \left( 4a_i - 2\sqrt{a_i b_i} \right)\\  &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i +a_i - 2\sqrt{a_i b_i} + b_i \right) \quad \text{(as } \sum_{i=1}^n a_i = \sum_{i=1}^n b_i\text{)}\\  &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i + \left(\sqrt{a_i} - \sqrt{b_i}\right)^2 \right)\\  &\geq \frac{1}{4} \sum_{i=1}^n 2a_i\\  &= \frac{1}{2} \sum_{i=1}^n a_i,  \end{aligned}

as required.

References

[1] Zdravko Cvetkovski, Inequalities: Theorems, Techniques and Selected Problems, Springer, 2012.

[2] Wang and Kadaveru, Advanced Topics in Inequalities, available from http://www.artofproblemsolving.com/community/q1h1060665p4590952

[3] Cauchy Reverse Technique: https://translate.google.com.au/translate?hl=en&sl=ja&u=http://mathtrain.jp/crt&prev=search

[4] algebra precalculus – Prove that \frac{a_1^2}{a_1+b_1}+\cdots+\frac{a_n^2}{a_n+b_n} \geq \frac{1}{2}(a_1+\cdots+a_n). – Mathematics StackExchange

February 28, 2016

The race up the charts for two recent movies

Filed under: movies and TV — ckrao @ 7:59 pm

In recent times Jurassic World and Star Wars VII (The Force Awakens) have respectively become the fourth and third biggest movies of all time worldwide (behind Avatar and Titanic). Here is how they ranked in the all-time US/Canada charts day by day (using data from boxofficemojo.com).

Day Jurassic World Star Wars: The Force Awakens
1 786 464
2 275 183
3 143 96
4 108 68
5 79 40
6 69 29
7 54 22
8 37 11
9 28 6
10 18 5
11 15 5
12 11 5
13 10 4
14 9 3
15 7 2
16-19 5 2
20-21 5 1
22-44 4 1
45+ 3 1

I was amazed by Jurassic World’s summer run and then that of The Force Awakens simply blew my mind. 🙂

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