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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. 🙂

February 27, 2016

Cutting a triangle in half

Filed under: mathematics — ckrao @ 9:40 pm

Here is a cute triangle result that I’m surprised I had not known previously. If we are given a point on one of the sides of a triangle, how do we find a line through the triangle that cuts its area in half?

Clearly if that point is either a midpoint or one of the vertices, the answer is a median of the triangle. A median cuts a triangle in half since the two pieces have the same length side and equal height.

median

So what if the point is not a midpoint or a vertex? Referring to the diagram below, if P is our desired point closer to A than B, the end point Q of the area-bisecting segment would need to be on side BC so that area(BPQ) = area(ABC)/2.

areabisector

In other words, we require area(BPQ) = area(BDQ), or, subtracting the areas of triangle BDQ from both sides,

\displaystyle |DPQ| = |DCQ|.

Since these two triangles share the common base DQ, this tells us that we require them to have the same height. In other words, we require CP to be parallel to DQ. This tells us how to construct the point Q given P on AB:

  1. Construct the midpoint D of AB.
  2. Draw DQ parallel to AP.

See [1] for an animation of this construction.

In turns out that the set of all area-bisecting lines are tangent to three hyperbolas and enclose a deltoid of area (3/4)\ln(2) - 1/2 \approx 0.01986 times the original triangle. [2,3,4]

References

[1] Jaime Rangel-Mondragon, “Bisecting a Triangle” http://demonstrations.wolfram.com/BisectingATriangle/ from the Wolfram Demonstrations Project Published: July 10, 2013

[2] Ismail Hammoudeh, “Triangle Area Bisectors”  http://demonstrations.wolfram.com/TriangleAreaBisectors/ from the Wolfram Demonstrations Project

[4] Henry Bottomley, Area bisectors of a triangle, January 2002

January 31, 2016

Most wickets after n test matches from debut

Filed under: cricket,sport — ckrao @ 12:18 am

Here is a list of the leading test cricket wicket takers after playing n matches. A few current-day players feature and I hope to update this list over time (last updated January 31 2016). It was created largely manually using ESPNCricinfo’s Statsguru starting from [1] and [2], and using lists of the fastest to multiples of 50 wickets here.  Corrections are more than welcome (updated: February 26 2017).

1 Hirwani, Massie (16) F Martin (Eng), Krejza (12) several tied on 11
2 Hirwani (24) Bedser (22) Massie (21)
3 Hirwani (31) Turner (29) Hogg (27)
4 Turner (39) Hirwani (36) Hogg, Valentine (33)
5 Turner (45) Hogg (40) Valentine (39)
6 Turner (50) Philander (45) Valentine (43)
7 Richardson (53) Philander, Turner (51) Valentine (49)
8 Richardson (66) Turner (56) Valentine (55)
9 Richardson (66) Turner (63) Ferris (61)
10 Richardson (71) Turner (69) Tate (65)
11 Richardson, Turner (72) Yasir Shah (69) Grimmett, Peel, Roberts, Tate, Valentine (65)
12 Turner (80) Richardson, Yasir Shah (76) Grimmett (71)
13 Yasir Shah (86) Turner (81) Richardson (78)
14 Richardson (88) Yasir Shah (87) Turner (83)
15 Yasir Shah (90) Lohmann (89) Barnes, Grimmett, Philander (87)
16 Lohmann (101) Yasir Shah (95) Barnes, Turner (94)
17 Lohmann (109) Yasir Shah (102) Barnes, Grimmett, Turner (101)
18 Barnes, Lohmann, Yasir Shah (112) Grimmett (105) Ashwin (104)
19 Yasir Shah (116) Barnes (112) Grimmett (109)
20 Barnes (122) Yasir Shah (116) Grimmett (112)
21 Barnes (122) Grimmett (120) Yasir Shah (119)
22 Barnes (135) Grimmett (128) Botham (118)
23 Grimmett (142) Barnes (140) Botham (122)
24 Barnes (150) Grimmett (142) Waqar Younis (134)
25 Barnes (167) Grimmett, Waqar Younis (143) Botham (139)
26 Barnes (175) Waqar Younis (148) Grimmett (144)
27 Barnes (189) Waqar Younis (154) Grimmett (147)
28 Waqar Younis (159) Grimmett (155) Botham (149)
29 Waqar Younis (166) Grimmett (156) Ashwin, Saeed Ajmal (153)
30 Waqar Younis (169) Saeed Ajmal (159) Ashwin, Grimmett (157)
31 Waqar Younis (180) Ashwin (169) Grimmett (164)
32 Waqar Younis (187) Ashwin (176) Grimmett (172)
33 Waqar Younis (190) Ashwin (183) Grimmett (177)
34 Waqar Younis (191) Ashwin (189) Grimmett (183)
35 Waqar Younis (194) Grimmett (193) Ashwin (192)
36 Grimmett (203) Waqar Younis (196) Ashwin (193)
37 Grimmett (216) Ashwin (203) Waqar Younis (199)
38 Ashwin (207) Lillee (206) Waqar Younis (200)
39 Ashwin (220) Waqar Younis (208) Lillee (206)
40 Ashwin (223) Lillee (206) Steyn (205)
41 Ashwin (231) Waqar Younis (216) Steyn (211)
42 Ashwin (235) Waqar Younis (217) Lillee (214)
43 Ashwin (247) Lillee, Waqar Younis (222) Steyn (217)
44 Ashwin (248) Lillee (229) Waqar Younis (227)
45 Ashwin (254) Steyn (232) Lillee (230)
46 Ashwin (261) Steyn (238) Lillee (237)
47 Steyn (244) Lillee (243) Donald (237)
48 Lillee (251) Steyn (249) Donald (241)
49 Lillee (259) Steyn (255) Donald (248)
50 Lillee (262) Steyn (260) Donald (251)
51 Lillee (269) Steyn (263) Donald, Waqar Younis (254)
52 Lillee (273) Steyn (265) Muralitharan (261)
53 Lillee (279) Steyn (270) Muralitharan (265)
54 Lillee (290) Steyn (272) Donald, Muralitharan (265)
55 Lillee (296) Steyn (279) Muralitharan (278)
56 Lillee (305) Muralitharan (283) Steyn (282)
57 Lillee (305) Muralitharan (291) Steyn (287)
58 Lillee (315) Muralitharan (302) Marshall (290)
59 Lillee (321) Muralitharan (303) Marshall (296)
60 Lillee (321) Muralitharan (310) Hadlee, Marshall, Steyn (299)
61 Lillee (321) Muralitharan (315) Steyn (304)
62 Lillee (325) Muralitharan (317) Steyn (312)
63 Lillee (328) Muralitharan (325) Steyn (323)
64 Lillee (332) Muralitharan (329) Steyn (327)
65 Muralitharan (340) Lillee (335) Hadlee, Steyn (332)
66 Muralitharan (350) Lillee, Steyn (336) Hadlee (334)
67 Muralitharan (361) Steyn (340) Hadlee, Lillee (336)
68 Muralitharan (371) Hadlee, Lillee (342) Steyn (341)
69 Muralitharan (374) Hadlee (351) Steyn (350)
70 Muralitharan (382) Steyn (356) Hadlee, Lillee (355)
71 Muralitharan (395) Steyn (361) Hadlee (358)
72 Muralitharan (404) Hadlee (363) Steyn (362)
73 Muralitharan (412) Hadlee (373) Steyn (371)
74 Muralitharan (417) Steyn (375) Hadlee (373)
75 Muralitharan (420) Steyn (383) Hadlee (378)
76 Muralitharan (430) Steyn (389) Hadlee (388)
77 Muralitharan (433) Hadlee (391) Steyn (389)
78 Muralitharan (437) Steyn (396) Hadlee (395)
79 Muralitharan (442) Steyn (399) Hadlee (396)
80 Muralitharan (450) Hadlee (403) Steyn (402)
81 Muralitharan (455) Hadlee (406) Steyn (402)
82 Muralitharan (459) Hadlee (408) Steyn (406)
83 Muralitharan (470) Hadlee (415) Steyn (408)
84 Muralitharan (478) Hadlee (419) Steyn (416)
85 Muralitharan (485) Hadlee (423) Steyn (417)
86 Muralitharan (496) Hadlee (431) Kumble (415)
87-101 Muralitharan Kumble McGrath
102-126 Muralitharan Kumble Warne
127 Muralitharan (770) Warne (611) Kumble (608)
128 Muralitharan (777) Warne (623) Kumble (608)
129 Muralitharan (783) Warne (629) Kumble (613)
130 Muralitharan (786) Warne (634) Kumble (616)
131 Muralitharan (788) Warne (638) Kumble (616)
132 Muralitharan (792) Warne (645) Kumble (619)
133 Muralitharan (800) Warne (651) Kallis (258)
134-145 Warne Kallis Waugh
146-166 Kallis Waugh Tendulkar
167-168 Waugh Tendulkar Ponting
169-200 Tendulkar

References

[1] Most wkts in consec Tests from debut – Google Groups

[2] Top 10 bowlers with most wickets in 10 Tests or less (crictracker.com)

January 30, 2016

Patterns early in the digits of pi

Filed under: mathematics — ckrao @ 8:59 pm

Recently when taking a look at the early decimal digits of \pi I made the following observations:

 3.141592653589793238462643383279502884197169399375105820974944592307816406286…

  • The first run of seven distinct digits (8327950, shown underlined) appears in the 26th to 32nd decimal place. Curiously the third such run (5923078, also underlined) in decimal places 61 to 67 contains the same seven digits. (There is also a run of seven distinct digits in places 51 to 57 with 5820974.)
  • Decimal digits 60 to 69 (shown in bold) are distinct (i.e. all digits are represented once in this streak). The same is true for digits 61 to 70 as both digits 60 and 70 are ‘4’.

Assume the digits of \pi are generated independently from a uniform distribution. Firstly, how often would we expect to see a run of 7 distinct digits? Places k to k+6 are distinct with probability

\displaystyle \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} \times \frac{6}{10} \times \frac{5}{10} \times \frac{4}{10} = \frac{9!}{3.10^6} = \frac{189}{3125}.

Hence we expect runs of 7 distinct digits to appear 3125/189 \approx 16.5 places apart. This includes the possibility of runs such as 12345678 which contain two runs of 7 distinct digits that are only 1 place apart.

How often would we expect to see the same 7 digits appearing in a run as we did in places 26-32 and 61-67? Furthermore let’s assume the two runs have no overlap, so we discount possibilities such as 12345678 which have a six-digit overlap. We expect a given sequence (e.g. 1234567, in that order) to appear 1/10^7 of the time. There are 7! = 5040 permutations of such a sequence, but of these 1 has overlap 6 (2345671), 2! has overlap 5 (3456712 or 3456721), 3! has overlap 4, …, 6! has overlap 1 with the original sequence. This leaves us with 5040 - (1 + 2 + 6 + 24 + 120 + 720) = 4167 possible choices of the next run to have the same 7 digits but non-overlapping (or to appear in precisely the same order – overlap 7). Hence we expect the same 7 digits to recur (no overlap with the original run) after approximately 10^7/4167 \approx 2400 places apart so what we saw in the first 100 places was remarkable.

Now let us turn to runs of all ten distinct digits. Repeating the argument above, such runs occur every 10^{10}/10! \approx 2756 places. According to [1] the next time we see ten distinct digits is in decimal places 5470 to 5479.

To answer the question of when we would expect to see the first occurrence of ten distinct digits, we adopt an argument from renewal-reward theory based on Sec 7.9.2 of [2] (there also exist approaches based on setting up recurrence relations, or martingale theory (modelling a fair casino), see [3]-[4]). Firstly we let T be the first time we get 10 consecutive distinct values – we wish to find E[T] where E denotes the expected value operator. Note that this will be more than the 2756 answer we obtained above since we make no assumption of starting with a run of ten distinct digits – there is no way T could be 1 for example, but we could have two runs of ten distinct digits that are 1 apart.

From a sequence of digits we first define a renewal process in which after we get 10 consecutive distinct values (at time T) we start over and wait for the next run of 10 consecutive distinct values without using any of the values  up to time T. Such a process will then have an expected length of cycle of E[T].

Next, suppose we earn a reward of $1 every time the last 10 digits of the sequence are distinct (so we would have obtained $1 at each of decimal places 69 and 70 in the \pi example). By an important result in renewal-reward theory, the long run average reward is equal to the expected reward in a cycle divided by the expected length of a cycle.

In a cycle we will obtain

  • $1 at the end
  • $1 at time 1 in the cycle with probability 1/10 (if that digit is the same as ten digits before it)
  • $1 at time 2 in the cycle with probability 2/100 (if the last two digits match those ten places before it)
  • $1 at time 9 in the cycle with probability 9!/10^9 (if the last nine digits match those ten places before it)

Hence the expected reward in a cycle is given by

\displaystyle 1 + \sum_{i=1}^9 \frac{i!}{10^i} = \sum_{i=0}^9 \frac{i!}{10^i}.

We have already seen that the long run average reward is 10!/10^{10} at each decimal place. Hence the expected length of a cycle E[T] (i.e. the expected number of digits before we expect the first run of ten consecutive digits) is given by

\frac{10^{10}}{10!}\sum_{i=0}^9 \frac{i!}{10^i} \approx 3118.

Hence it is pretty cool that we see it so early in the decimal digits of \pi. 🙂

References

[1] A258157 – Online Encyclopedia of Integer Sequences

[2] S. Ross, Introduction to Probability Models, Academic Press, 2014.

[3] Combinatorics Problem on Expected Value – Problem Solving: Dice Rolls – Daniel Wang | Brilliant

[4] A Collection of Dice Problemsmadandmoonly.com

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