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October 30, 2014

Mountain ranges of the world

Filed under: geography — ckrao @ 1:06 pm

As a geography lesson for myself, here is a list of major mountain ranges of the world, where those that are either long or have a significantly high peak are shown (hence several near Tibet are given). Maps are shown below the table.


Continent Name of Range Countries spanned Length (km) Name of Highest Peak Elevation of highest peak (m)
Antarctica Transantarctic 3500 Kirkpatrick 4528
Oceania Southern Alps New Zealand 350 Aoraki (Cook) 3724
Great Dividing Range Australia 3000 Kosciuszko 2228
New Guinea Highlands Indonesia, Papua New Guinea 1600 Puncak Jaya 4884
Africa Drakensberg South Africa, Lesotho 1000 Thabana Ntlenyana 3482
Ethiopian Highlands Ethiopia, Eritrea 1500 Ras Dashan 4533
Atlas Morocco, Algeria, Tunisia 2400 Jbel Toubkal 4167
Asia Himalayas Bhutan, China, India, Nepal, Pakistan 2400 Everest 8849
Karakoram Pakistan, India, China 500 K2 8611
Hindu Kush Afghanistan, Pakistan 1200 Tirich Mir 7690
Pamir Tajikistan, Krygyzstan, Afghanistan, Pakistan, China ~500 Ismoil Somoni Peak 7495
Tian Shan China, Pakistan, India, Kazakhstan, Kyrgyzstan, Uzbekistan 1500 Jengish Chokusu 7439
Kunlun China, India 3000 Kongur Tagh 7649
Altun China 800 Sulamutag Feng 6245
Altai Russia, China, Mongolia, Kazakhstan 2000 Belhuka 4506
Sayan Mongolia, Russia 1500 Mounkou 3492
Verkhoyansk Russia 1200 Mus-Khaya 2959
Sredinny Russia 900 Ichinsky 3607
Hengduan China, Myanmar 800 Gongga 7556
Qin China 500 Taibai 3767
Taihang China 400 Wutai 3061
Arakan Myanmar 1000 Victoria 3094
Annamite Laos, Vietnam, Cambodia 1100 Ngoc Pan 2598
Barisan Indonesia 1700 Kerinci 3800
Western Ghats India 1600 Anamudi 2695
Eastern Ghats India 1300 Arma Konda 1680
Zagros Iran, Iraq, Turkey 1500 Zard Kuh 4548
Alborz Iran 600 Damavand 5671
Caucasus Russia, Georgia, Azerbaijan, Armenia, Turkey 1100 Elbrus 5642
Yemen Highlands Yemen, Saudi Arabia 1500 Jabal an Nabi Shu’ayb 3666
Pontic Turkey 1000 Kackar Dagi 3492
Taurus Turkey 600 Demirkazik 3756
Asia/Europe Ural Russia, Kazakhstan 2500 Narodnaya 1895
Europe Kjølen Norway, Sweden, Finland 1700 Galdhøpiggen 2469
Alps Germany, Austria, France, Italy, Liechtenstein, Monaco, Slovenia, Switzerland 1200 Blanc 4811
Pyrenees France, Spain, Andorra 430 Aneto 3404
Apennines Italy, San Marino 1200 Corno Grande 2912
Carpathians Austria, Slovakia, Poland, Czech Republic, Hungary, Romania, Ukraine, Serbia 1500 Gerlachovsky stit 2655
North America Arctic Canada 1000 Barbeau Peak 2616
Brooks USA, Canada 1100 Mount Chamberlin 2749
Aleutian USA 1000 Redoubt 2788
Alaskan USA 650 McKinley 6194
Rocky USA, Canada 4800 Elbert 4401
Coast USA, Canada 1600 Waddington 4019
Cascade USA, Canada 1100 Rainier 4392
Sierra Nevada USA 650 Whitney 4421
Appalachian USA, Canada 2400 Mitchell 2037
Sierra Madre Occidental Mexico 1250 Cerro Mohinora 3250
Sierra Madre Oriental Mexico 1250 Cerro San Rafael 3700
Sierra Madre del Sur Mexico 1000 Teotepec 3703
Sierra Madre Chiapas Mexico, Guatemala, El Salvador, Honduras 600 Tajumulco 4220
 South America Andes Venezuela, Colombia, Ecuador, Peru, Bolivia, Chile, Argentina 7000 Aconcagua 6962
Serra do Mar Brazil 1500 Pico Maior de Friburgo 2316



October 25, 2014

An integral related to Bessel polynomials

Filed under: mathematics — ckrao @ 12:46 am

In this post I want to share this cute result that I learned recently:

\displaystyle \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}\ dx = \int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^2}\ dx = \frac{\pi}{e} \quad\quad (1)

Let us see how the first integral is derived and then generalised. The integrand has poles at \pm i in the complex plane and we may apply contour integration to proceed. Note that as \sin x /(x^2+1) is an odd function, \int_{-\infty}^{\infty} \sin x /(x^2+1)\ dx = 0 and so

\int_{-\infty}^{\infty}\frac{\cos x}{x^2+1}\ dx = \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}.\quad\quad(2)

It therefore suffices to consider the integrand f(z) = e^{iz}/(z^2+1). We consider the semicircular contour of radius R (to be traversed anticlockwise) in the upper half plane centred at 0. It encloses the pole at i.

semicircular contour

Along this closed contour we use the residue theorem to compute

\begin{aligned}  \oint f(z)\ dz &= 2\pi i \text{Res}[f(z)]_{z=i}\\  &= 2\pi i \lim_{z \rightarrow i} (z-i)\frac{e^{iz}}{z^2 +1}\\  &= 2\pi i \lim_{z \rightarrow i} \frac{e^{iz}}{z +i}\\  &= 2\pi i \frac{e^{-1}}{2i}\\  &= \frac{\pi}{e}.\quad\quad (3)  \end{aligned}

On the semicircular arc z = R(\cos \theta + i\sin \theta) = Re^{i\theta} for 0 \leq \theta \leq \pi and so

\begin{aligned}  |f(z)| &= \left| \frac{e^{iz}}{z^2+1} \right|\\  &= \frac{|e^{iR(\cos \theta + i\sin \theta)}|}{|R^2 e^{i2\theta} + 1|}\\  &= \frac{|e^{-R \sin \theta}|}{|R^2 e^{i2\theta} + 1|}\\  &\leq \frac{e^{-R\sin \theta}}{R^2-1}\\  &\leq \frac{1}{R^2-1} \quad \text{for } 0 \leq \theta \leq \pi,\quad\quad (4)  \end{aligned}

where in the second last step we use the fact that distance to the origin of a circle of radius R^2 centred at -1 is at least R^2-1. Then

\begin{aligned}  \lim_{R \rightarrow \infty} \int_C f(z)\ dz &\leq \lim_{R \rightarrow \infty} \int_C |f(z)| \ |dz| \\  &\leq \lim_{R \rightarrow \infty} \pi R \sup_{z \in C} |f(z)|\\  &\leq \lim_{R \rightarrow \infty} \pi R \frac{1}{R^2-1} \quad \text{(by (4))}\\  &= 0.\quad\quad(5)  \end{aligned}

It follows that \lim_{R \rightarrow \infty} \int_C f(z)\ dz = 0 and we are left with

\begin{aligned}  \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}\ dx &= \lim_{R \rightarrow \infty} \int_{-R}^R f(z)\ dz \\  &= \lim_{R \rightarrow \infty} \oint f(z)\ dz - \int_C f(z) \ dz\\  &= \lim_{R \rightarrow \infty} \oint f(z)\ dz\\  &= \frac{\pi}{e} \quad\text{(by (3)).}\quad \quad(6)  \end{aligned}

To generalize this result to integrals of the form \int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^{n+1}}\ dx = \int_{-\infty}^{\infty} \frac{e^{ix}}{(x^2+1)^{n+1}}\ dx for non-negative integers n, we choose the same contour as above and use the residue limit formula for poles of order (n+1):

\displaystyle \text{Res} [f(z)]_{z=z_0} = \frac{1}{n!} \lim_{z \rightarrow z_0} \left[\frac{d^n}{dz^n} (z-z_0)^{n+1} f(z) \right]. \quad\quad (7)

To apply (7) we make use of the General Leibniz rule for the n‘th derivative of a product:

\displaystyle (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)} g^{(n-k)}.\quad \quad (8)


\begin{aligned}  \oint \frac{e^{iz}}{(z^2+1)^{n+1}}\ dz &= 2\pi i \text{Res} \left[ \frac{e^{iz}}{(z^2+1)^{n+1}} \right]_{z=i}\\  &= \frac{2\pi i}{n!} \left[\frac{d^n}{dz^n}\frac{e^{iz}}{(z+i)^{n+1}} \right]_{z=i}\\  &= \frac{2\pi i}{n!} \sum_{k=0}^n \left[\binom{n}{k}\left(e^{iz}\right)^{(n-k)} \left(\frac{1}{(z+i)^{n+1}}\right)^{(k)}\right]_{z=i} \quad\text{(applying (8))}\\  &= \frac{2\pi i}{n!} \sum_{k=0}^n \binom{n}{k} (i)^{n-k} e^{-1} (-n-1)(-n-2)\ldots (-n-k) \frac{1}{(2i)^{n+k+1}}\\  &= \frac{2\pi i}{n!e} \sum_{k=0}^n \binom{n}{k}\frac{1}{i^{2k+1}}\frac{(n+k)!}{n!}(-1)^k\\  &= \frac{\pi}{n!e} \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{(n+k)!}{n!}\frac{1}{2^{n+k}}\\  &= \frac{\pi}{e} \frac{1}{2^n n!}\sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!}\frac{1}{2^k}\\  &= \frac{\pi}{e} \frac{1}{2^n n!} y_n(1), \quad\quad(8)\end{aligned}

where y_n(x) is the n‘th order Bessel polynomial defined by

\displaystyle y_n(x) = \sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k.\quad\quad(9)

For example, for n = 1, the integral is given by

\begin{aligned} 2\pi i \left[ \frac{d}{dz} \frac{e^{iz}}{(z+i)^2}\right]_{z=i} &= \left[ \frac{-2e^{iz}}{(z+i)^3} + \frac{ie^{iz}}{(z+i)^2}\right]_{z=i}\\ &= 2\pi i\left(\frac{-2e^{-1}}{8i^3} + \frac{ie^{-1}}{-4}\right)\\ &= \frac{\pi}{e}. \quad\quad (10)\end{aligned}

The general sum in (8) can be also be written as

\displaystyle \sum_{k=0}^n \frac{(k+n)!}{2^k k!(n-k)!} = e \sqrt{\frac{2}{\pi}}K_{n+1/2}(1),\quad \quad (11)

where K_{\alpha}(x) is the modified Bessel function of the second kind.

Proceeding similarly to (4), \left| e^{iz}/(1+z^2)^{n+1} \right| \leq 1/(R^2-1)^{n+1} and so similar to (5)  the integral on the arc converges to 0 as R \rightarrow \infty. Hence following the same argument as in (6) the desired line integral along the real axis is equal to the contour integral in (8).

Evaluating (8) for n = 0, 1, 2,3,4,5,\ldots the first terms of  (e/\pi)\int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^2}\ dx are given by

\displaystyle \frac{1}{1}, \frac{2}{2}, \frac{7}{8}, \frac{37}{48}, \frac{266}{384}, \frac{2431}{3840}, \cdots\quad\quad(12)

In this sequence a_n/b_n, the denominators b_n are related to the previous ones by multiplication by 2n, while curiously the numerators are related by the second order recurrence

a_{n} = (2n-1)a_{n-1} + a_{n-2}.\quad\quad(13)

This follows from the following recurrence relation for the Bessel polynomials:

y_n(x) = (2n-1)xy_{n-1}(x) + y_{n-2}(x).\quad\quad (14)

This can be proved using (9). We have

\begin{aligned}  (2n-1)xy_{n-1}(x) + y_{n-2}(x) &= (2n-1)x\sum_{k=0}^{n-1} \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k + \sum_{k=0}^{n-2} \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k\\  &=(2n-1)\sum_{k=1}^n \frac{(n+k-2)!}{(k-1)!(n-k)!}2\frac{x^k}{2^k}  + \sum_{k=0}^{n-2} \frac{(n+k-2)!}{k!(n-2-k)!}\left(\frac{x}{2}\right)^k\\  &= \frac{(n-2)!}{0!(n-2)!} + \sum_{k=1}^{n-2} (n+k-2)! \left[  \frac{2(2n-1)}{(k-1)!(n-k)!} + \frac{1}{k!(n-k-2)!}\right] \left(\frac{x}{2}\right)^k \\  & \quad \quad + (2n-1) \left[\frac{n + n -1-2)!}{(n-2)!1!}2 \left(\frac{x}{2}\right)^{n-1} + \frac{(2n-2)!}{(n-1)!0!}2\left(\frac{x}{2}\right)^n\right]\\  &= 1 + \sum_{k=1}^{n-2} (n+k-2)! \left[\frac{2(2n-1)k + (n-k-1)(n-k)}{k!(n-k)!}\right]\left(\frac{x}{2}\right)^k\\  & \quad \quad + (2n-1) \left[ \frac{(2n-3)!}{(n-2)!}2\left(\frac{x}{2}\right)^{n-1} + \frac{(2n-2)!}{(n-1)!}2\left(\frac{x}{2}\right)^n\right].\quad\quad(15)  \end{aligned}


\begin{aligned}  2(2n-1)k + (n-k-1)(n-k) &= 2k(2n-1) + (n + k - 1 - 2k)(n-k)\\  &= 2k(2n-1-n+k) + (n+k-1)(n-k)\\  &= (n+k-1)(2k + n-k)\\  &= (n+k-1)(n+k).\quad\quad(16)  \end{aligned}

Substituting this into (15),

\begin{aligned}  (2n-1)xy_{n-1}(x) + y_{n-2}(x) &= 1 + \sum_{k=1}^{n-2}\frac{(n+k)!}{k!(n-k)!}\left(\frac{x}{2}\right)^k + \frac{(2n-1)!(n-1)2}{(n-1)!(2n-2)}\left(\frac{x}{2}\right)^{n-1} + \frac{(2n)!n2}{n! 2n}\left(\frac{x}{2}\right)^{n}\\  &= \sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!}\left(\frac{x}{2}\right)^k\\  &= y_n(x),  \end{aligned}

thus verifying (14).

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