Chaitanya's Random Pages

February 28, 2015

Large US+Canada box office openings

Filed under: movies and TV — ckrao @ 1:38 pm

The site boxofficemojo.com lists the opening weekend grosses of movies in the US and Canada dating back to the early 1980s. Via this page on top opening weekends, I worked out movies that at their time of release attained the n’th highest grossing opening weekend where n ranges from 1 to 10 (all dollar amounts in $US). It gives a perspective on how big some movies were at the time. It also shows how movie grosses have grown through inflation and more frontloading over the years. Note that only opening weekends are shown here – for example Superman’s third weekend was once the largest grossing weekend at the time, but is not listed here.

 

n=1 (i.e. current and previous record-breaking openings):

Title Opening Date (mm/dd/yyyy)
Marvel’s The Avengers $207,438,708 5/4/2012
Harry Potter and the Deathly Hallows Part 2 $169,189,427 7/15/2011
The Dark Knight $158,411,483 7/18/2008
Spider-Man 3 $151,116,516 5/4/2007
Pirates of the Caribbean: Dead Man’s Chest $135,634,554 7/7/2006
Spider-Man $114,844,116 5/3/2002
Harry Potter and the Sorcerer’s Stone $90,294,621 11/16/2001
The Lost World: Jurassic Park $72,132,785 5/23/1997
Batman Forever $52,784,433 6/16/1995
Jurassic Park $47,026,828 6/11/1993
Batman Returns $45,687,711 6/19/1992
Batman $40,489,746 6/23/1989
Ghostbusters II $29,472,894 6/16/1989
Indiana Jones and the Last Crusade $29,355,021 5/24/1989
Beverly Hills Cop II $26,348,555 5/20/1987
Indiana Jones and the Temple of Doom $25,337,110 5/23/1984
Return of the Jedi $23,019,618 5/25/1983
Star Trek II: The Wrath of Khan $14,347,221 6/4/1982
Superman II $14,100,523 6/19/1981
Star Trek: The Motion Picture $11,926,421 12/7/1979
Every Which Way But Loose $10,272,294 12/20/1978

 

n=2 (i.e. the second largest opening at the time)

Title Opening Date
Iron Man 3 $174,144,585 5/3/2013
Star Wars: Episode III – Revenge of the Sith $108,435,841 5/19/2005
Shrek 2 $108,037,878 5/19/2004
The Matrix Reloaded $91,774,413 5/15/2003
Planet of the Apes (2001) $68,532,960 7/27/2001
The Mummy Returns $68,139,035 5/4/2001
Star Wars: Episode I – The Phantom Menace $64,820,970 5/19/1999
Independence Day $50,228,264 7/3/1996
Lethal Weapon 3 $33,243,086 5/15/1992
Terminator 2: Judgment Day $31,765,506 7/3/1991
Rocky III $12,431,486 5/28/1982
The Cannonball Run $11,765,654 6/19/1981
Smokey and the Bandit II $10,883,835 8/15/1980
The Empire Strikes Back $10,840,307 6/20/1980

 

n=3:

Title Opening Date
The Dark Knight Rises $160,887,295 7/20/2012
The Hunger Games $152,535,747 3/23/2012
The Twilight Saga: New Moon $142,839,137 11/20/2009
Shrek the Third $121,629,270 5/18/2007
Harry Potter and the Prisoner of Azkaban $93,687,367 6/4/2004
Harry Potter and the Chamber of Secrets $88,357,488 11/15/2002
Star Wars: Episode II – Attack of the Clones $80,027,814 5/16/2002
Hannibal $58,003,121 2/9/2001
Mission: Impossible II $57,845,297 5/24/2000
Toy Story 2 $57,388,839 11/24/1999
Austin Powers: The Spy Who Shagged Me $54,917,604 6/11/1999
Men in Black $51,068,455 7/2/1997
The Lion King $40,888,194 6/24/1994
Rambo: First Blood Part II $20,176,217 5/22/1985
Star Trek III: The Search for Spock $16,673,295 6/1/1984

 

n=4:

Title Opening Date
X-Men: The Last Stand $102,750,665 5/26/2006
Harry Potter and the Goblet of Fire $102,685,961 11/18/2005
X2: X-Men United $85,558,731 5/2/2003
Austin Powers in Goldmember $73,071,188 7/26/2002
Rush Hour 2 $67,408,222 8/3/2001
Pearl Harbor $59,078,912 5/25/2001
Mission: Impossible $45,436,830 5/22/1996
Twister $41,059,405 5/10/1996
Back to the Future Part II $27,835,125 11/22/1989
Rocky IV $19,991,537 11/27/1985
Beverly Hills Cop $15,214,805 12/5/1984
Jaws 3-D $13,422,500 7/22/1983
Superman III $13,352,357 6/17/1983

 

n=5:

Title Opening Date
The Twilight Saga: Breaking Dawn Part 1 $138,122,261 11/18/2011
Iron Man 2 $128,122,480 5/7/2010
Pirates of the Caribbean: At World’s End $114,732,820 5/25/2007
How the Grinch Stole Christmas $55,082,330 11/17/2000
Interview with the Vampire $36,389,705 11/11/1994
Home Alone 2: Lost in New York $31,126,882 11/20/1992
Bram Stoker’s Dracula $30,521,679 11/13/1992
Star Trek IV: The Voyage Home $16,881,888 11/26/1986
The Best Little Whorehouse in Texas $11,874,268 7/23/1982
E.T.: The Extra-Terrestrial $11,835,389 6/11/1982

 

n=6:

Title Opening Date
The Hunger Games: Catching Fire $158,074,286 11/22/2013
Harry Potter and the Deathly Hallows Part 1 $125,017,372 11/19/2010
Alice in Wonderland (2010) $116,101,023 3/5/2010
The Passion of the Christ $83,848,082 2/25/2004
Monsters, Inc. $62,577,067 11/2/2001
X-Men $54,471,475 7/14/2000
Ace Ventura: When Nature Calls $37,804,076 11/10/1995
Robin Hood: Prince of Thieves $25,625,602 6/14/1991
Total Recall $25,533,700 6/1/1990
Teenage Mutant Ninja Turtles $25,398,367 3/30/1990
Ghostbusters $13,578,151 6/8/1984
Staying Alive $12,146,143 7/15/1983

 

n=7:

Title Opening Date
Transformers: Revenge of the Fallen $108,966,307 6/24/2009
Spider-Man 2 $88,156,227 6/30/2004
Batman and Robin $42,872,605 6/20/1997
Lethal Weapon 2 $20,388,800 7/7/1989
Star Trek V: The Final Frontier $17,375,648 6/9/1989

 

n=8:

Title Opening Date
The Twilight Saga: Breaking Dawn Part 2 $141,067,634 11/16/2012
The Lord of the Rings: The Return of the King $72,629,713 12/17/2003
Godzilla $44,047,541 5/20/1998
The Flintstones $29,688,730 5/27/1994
Gremlins $12,511,634 6/8/1984

 

n=9:

Title Opening Date
Finding Nemo $70,251,710 5/30/2003
The Mummy $43,369,635 5/7/1999
Deep Impact $41,152,375 5/8/1998

 

n=10:

Title Opening Date
Toy Story 3 $110,307,189 6/18/2010
Indiana Jones and the Kingdom of the Crystal Skull $100,137,835 5/22/2008
Iron Man $98,618,668 5/2/2008
Dick Tracy $22,543,911 6/15/1990

(To create the above lists the movie lists in decreasing order of gross were pasted into Excel and the opening weekend date was converted to a number by creating a new column with formula =–TEXT(,”mm/dd/yyyy”). This was then converted to a rank by a countif formula to count the number of occurrences with higher gross that predated each movie. Finally a filter was applied to select ranks 1 to 10.)

February 27, 2015

Cross sections of a cube

Filed under: mathematics — ckrao @ 9:59 pm
When a plane intersects a cube there is a variety of shapes of the resulting cross section.
  • a single point (a vertex of the cube)
  • a line segment (an edge of the cube)
  • a triangle (if three adjacent faces of the cube are intersected)
  • a parallelogram (if two pairs of opposite faces are intersected – this includes a rhombus or rectangle)
  • a trapezium (if two pairs of
  • a pentagon (if the plane meets all but one face of the cube)
  • a hexagon (if the plane meets all faces of the cube)

The last five of these (the non-degenerate cases) are illustrated below and at http://cococubed.asu.edu/images/raybox/five_shapes_800.png . Some are demonstrated in the video below too.

cs1 cs2 cs3 cs4 cs5

One can use [1] to experiment interactively with cross sections given points on the edges or faces, while [2] shows how to complete the cross section geometrically if one is given three points on the edges.

Let us be systematic in determining properties of the cross sections above. Firstly, if the plane is parallel to an edge (any of four parallel edges), the cross section can be seen to be a line or rectangle with the longer dimension of length at most \sqrt{2} times the other. That rectangle becomes a square if the plane is parallel to a face.

If the plane is not parallel to a face, we may set up a coordinate system where a unit cube is placed in the first octant aligned with the coordinate axes and the normal to the plane has positive x, y and z coordinates. In other words, we may assume the plane has equation ax + by + cz = 1 and intercepts at (1/a,0,0), (0,1/b,0) and (0,0,1/c), where a, b, c are positive.

coordsystem

The cross section satisfies ax + by + cz = 1 and the inequalities 0 \leq x \leq 1, 0 \leq y \leq 1 and 0 \leq z \leq 1. This can be considered the intersection of the two regions

\displaystyle ax + by + cz = 1, 0 \leq x, 0 \leq y, 0 \leq z,

\displaystyle ax + by + cz = 1, x \leq 1, y \leq 1, z \leq 1,

each of which is an acute-angled triangle in the same plane (acute because one can show that the sum of the squares of any two sides is strictly greater than the square of the third side). Note that the triangles have parallel corresponding sides, being bounded by the pairs of parallel faces of the cube x = 0, x=1, y = 0, y= 1, z=0, z=1. Hence the two triangles are oppositely similar with a centre of similarity.

The following diagram shows the coordinates of the vertices of the two triangles, which in this case intersect in a hexagon.

2triangles

The centre of similarity of the two triangles is the intersection of two lines joining corresponding sides – this can be found to be the point (1/(a+b+c), 1/(a+b+c), 1/(a+b+c)), which is the intersection of the unit cube’s diagonal from the origin (to (1,1,1)) and the plane ax + by + cz = 1.

Side lengths of the triangles and distances between corresponding parallel sides may be found by Pythagoras’ theorem and are shown below for one pair of corresponding sides (the remaining lengths can be found by cyclically permuting a,b,c).

lengthsofsides

To sum up, all of the possible cross sections of a cube where the plane is not parallel to an edge can be described by the intersection of two oppositely similar triangles with corresponding sides parallel.

The type of polygon obtained depends on which vertices of the figure below are selected, as determined by the values of a,b,c.constraints

In this figure a vertex for the cross-sectional polygon is chosen if the constraint associated with it is satisfied. A red vertex has a conflicting constraint with its neighbouring two blue vertices, so either a red point or one or more blue points in this area can be chosen. Note that for the plane to intersect the cube at all we require (1,1,1) to be on the different side of the plane from the origin, or in other words, a + b + c \geq 1.

Let us look at a few examples. Firstly, if a, b, c are all greater than 1 we choose the following triangle.

tricase

Similarly if a+b, b+c, c+a all are less than 1, the oppositely similar triangle on the red vertices would be chosen.

Next, if c > 1, a < 1, a+b < 1 we obtain the following parallelogram.

parallelogramcase

 

 

If c > 1, a < 1, a+b > 1 we obtain either a pentagon (parallelogram truncated at a vertex) or a trapezium depending on whether b < 1 or b \geq 1 respectively.

b < 1:

pentagoncaseb\geq 1:
trapeziumcase

Finally, if a, b, c are less than 1 and a+b, b+c, c+a are greater than 1, we obtain a hexagon.

hexagoncase

For details on calculating the areas of such polygons refer to [3], especially the method applying the area cosine principle that relates an area of a figure to its projection. For calculating volumes related to regions obtained by the cross section refer to [4].

References

[1] Cross Sections of a Cube: http://www.wou.edu/~burtonl/flash/sandbox.html

[2] Episode 16 – Cross sections of a cube: http://sectioneurosens.free.fr/docs/premiere/s02e16s.pdf

[3] calculus – Area of the polygon formed by cutting a cube with a plane – Mathematics Stack Exchange

[4] integration – Volume of cube section above intersection with plane – Mathematics Stack Exchange

January 30, 2015

AB de Villiers’ fastest ODI century

Filed under: cricket,sport — ckrao @ 12:51 pm

Recently we witnessed the fastest century in one-day cricket history, with AB de Villiers coming in during the 39th over at the fall of South Africa’s first wicket (1/247) and proceeding to blast an incredible 149 off 44 balls in just 59 minutes propelling the team to 2/439 (Amla also made 153* and Rossouw 128). The match produced all types of records including many instances of most runs in n consecutive overs, documented in other blog post here.

Here is the ball-by-ball breakdown of his innings followed by a graph of runs versus balls.

4 2 1 | 1 4 6 4 6 | 6 . 2 2 LB | 1 6 6 6 4 | 6 1 4 1 | 1 . 4 4 | . 6 | 4 6 6 6 | LB 6 | 1 | 6 6 4 6 6 2 | 2 . out

ABbattinggraph

We see that three times he hit 28 runs in 5 balls and 26 runs in another space of 5 balls – that’s 110 runs in just 20 balls with 15 6s and 5 4s right there! He was already 82 off 27 balls (7 6s, 7 4s) and then made 63 off his next 13 balls including 9 6s and 2 4s to surge to 145 off 40 balls!! I have never seen such a concentration of 6 hitting. More analysis of his strike rate is at this @dualnoise post.

Here is what he scored (with balls faced) against the four bowlers who were up against him:

Taylor: 30 (13)
Holder: 45 (9)
Russell: 35 (12)
Smith: 39 (10)

Some other amazingly fast centuries in limited over cricket are in this blog post. Here is the breakdown of the previous fastest ODI century (36 balls) by Corey Anderson in this match:

1 | . 4 1 | . 1 | 6 1 . 2 1 | 4 6 1 | 6 6 . 6 . 6 | 1 . . 6 | 6 6 6 6 1 | . 1 | 4 4 1 1 | 6 6 4 1 | 2 4 1 | 6 1 | 2 2 1

January 26, 2015

Affine Transformations of Cartesian Coordinates

Filed under: mathematics — ckrao @ 11:43 am

A very common exercise in high school mathematics is to plot transformations of some standard functions. For example, to plot y = 2\sin (5x + \frac{2\pi}{3}) - 1 we may start with a standard sine curve and apply the following transformations in turn:

  • squeeze it by a factor of 5 in the x-direction
  • shift it left by \frac{2\pi}{3}
  • stretch it by a factor of 2 in the y-direction
  • shift it down by 1

This leads to the plot shown.

2sin_5x2pi3_-1

For sine and cosine graphs an alternative is to plot successive peaks/troughs of the curve and interpolate accordingly. For example, to plot y = 2\sin (5x + \frac{2\pi}{3}) - 1 we may proceed as follows.

  • Since \sin(x) has a peak at \frac{\pi}{2}, solve 5x + \frac{2\pi}{3} = \frac{\pi}{2} to find x = -\frac{\pi}{30} as a point where there is a peak at y = 2\times 1 - 1 = 1. Hence plot the point (-\frac{\pi}{30},1).
  • Since the angular frequency is 5, the period is \frac{2\pi}{5} and we may plot successive peaks spaced \frac{2\pi}{5} apart from the point (-\frac{\pi}{30},1).
  • Troughs will be equally spaced halfway between the peaks at y = 2\times (-1) - 1 = -3 (at x = -\frac{\pi}{30} + \frac{\pi}{5} + k\frac{2\pi}{5}). Then join the dots with a sinusoidal curve.
  • Additionally x- and y- intercepts may be found by setting y = 0 and x = 0 respectively. We find that the x-intercepts are at x = \frac{2\pi k}{5} - \frac{\pi}{10}, \frac{2\pi k}{5} + \frac{\pi}{30}\ (k \in \mathbb{Z}) and y-intercept is y = 2\sin (\frac{2\pi}{3}) - 1 = \sqrt{3}-1.

2sin_5x2pi3_-1-interpolated

The first approach is more generalisable to plotting other functions. Instead of thinking of the graph transforming, we also may consider it as a change of coordinates. For example, if we translate the parabola y = x^2 so that its turning point is at (2,1), this is equivalent to keeping the parabola fixed and shifting axes so that the new origin is at (-2,-1) with respect to the old coordinates. This is illustrated below where the black coordinates are modified to the red ones. The parabola has equation y = x^2 under the black coordinates and y - 1 = (x-2)^2 or y = (x-2)^2 + 1 under the red coordinates.

graphshift

As another example suppose we take a unit circle and stretch it by a factor of 2 in the x direction and a factor of 3 in the y direction. This is the equivalent of changing scale so that the x-axis is squeezed by 2 and the y-axis is squeezed by 3.

graphsqueeze

graphsqueeze2

Under this stretching of the circle or squeezing of axes, the unit circle equation x^2 + y^2 = 1 transforms to that of the ellipse \left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1.

More generally, by stretching a Cartesian graph f(x,y) = 0 by a in the x-direction and b in the y-direction, then shifting it along the vector (h,k), we obtain the equation

\displaystyle f\left(\frac{x-h}{a}, \frac{y-k}{b}\right) = 0.

This uses the fact that \frac{x-h}{a} and \frac{y-k}{b} are the inverses of ax+h and by+k respectively. Note that if |a| or |b| are less than 1, the stretch becomes a squeezing of the graph, while a < 0 or b < 0 correspond to a reflection in the x or y axes.

We can extend this idea to the rotation of a graph. Suppose for example we wish to rotate the hyperbola y = 1/x by 45 degrees anti-clockwise. This is equivalent to a rotation of the axes by 45 degrees clockwise and the matrix corresponding to this linear transform is

\displaystyle \left[ \begin{array} {c} x' \\ y' \end{array} \right] = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1\\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x \\ y \end{array} \right] = \left[ \begin{array} {c} \frac{x+y}{\sqrt{2}} \\ \frac{-x+y}{\sqrt{2}}\end{array} \right].

(Here the columns of the change of basis matrix correspond to where the basis vectors (1,0) and (0,1) map to under a 45 degree clockwise rotation.)

In other words we replace x' with (x+y)/\sqrt{2} and y' with (-x+y)/\sqrt{2} in the equation y' = 1/x' and obtain (-x+y)/\sqrt{2} = 1/( (x+y)/\sqrt{2}) or y^2 - x^2 = 2.

hyperbola

Here is the same transformation applied to the parabola y = x^2 to obtain \frac{-x+y}{\sqrt{2}} = \frac{(x+y)^2}{2} or x^2 + y^2 + 2xy + \sqrt{2}(x-y) = 0:

rotated_parabola

If a graph is affinely transformed (by an invertible map) so that (1,0) maps to (a,b) and (0,1) maps to (c,d) followed by a shift along the vector (h,k), then this is equivalent to the coordinates shifting by (-h,-k) and then transforming under the inverse mapping \displaystyle \left[ \begin{array}{cc} a & c\\ b & d\end{array} \right]^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -c\\ -b & a\end{array} \right]:

\displaystyle \boxed{\begin{aligned} x' &= (x-h) \frac{d}{ad-bc} - (y-k) \frac{c}{ad-bd}\\ y' &= -(x-h) \frac{b}{ad-bc} + (y-k) \frac{a}{ad-bd}\end{aligned}}

Here are some special cases of this formula:

  • rotation of the graph by \theta anti-clockwise: \displaystyle x' = x\cos \theta + y \sin \theta, y' = - x \sin \theta + y\cos \theta (the example \theta = \pi/4 was done above)
  • reflection of the graph in y = x: x' = y, y' = x
  • reflection of the graph in y = mx where m = \tan \theta: verify that a = \cos 2\theta, b = \sin 2\theta, c = \sin 2\theta, d = -\cos 2\theta so \displaystyle x' = x \cos 2\theta + y \sin 2\theta, y' = x \sin 2\theta - y \cos 2\theta
  • reflection of the graph in y = mx + c where m = \tan \theta: this is equivalent to a reflection in the line y = mx followed by a shift along the vector (h,k) = (-c \sin 2\theta, c (1+\cos 2 \theta) ) so
    \displaystyle \begin{aligned} x' &= (x+c \sin 2\theta)\cos 2\theta + (y-c(1+\cos 2\theta))\sin 2\theta\\ &= x \cos 2\theta + (y-c)\sin 2\theta,\\ y' &= (x+c \sin 2\theta)\sin 2\theta - (y-c(1+\cos 2\theta))\cos 2\theta\\ &= x \sin 2\theta - (y-c) \cos 2\theta + c \end{aligned}
  • reflection of the graph in y = x + c (special instance of the previous case with \sin 2\theta = 1, \cos 2\theta = 0): x'= y - c, y' = x+c

December 15, 2014

Types of -saurs that are not dinosaurs

Filed under: nature — ckrao @ 12:06 pm

Below is a reference for myself of types of (mostly) prehistoric animals that are not dinosaurs but have names ending in -saur (sauria means lizard but most of these are not that closely related to lizards).

Group Prefix meaning When it lived Brief description
Aetosaur eagle late Triassic heavily armoured
Anteosaur Antaeus (son of Poseidon and Gaia) 272-260 Ma large Dinocephalians (therapsid)
Cotylosaur cup late Carboniferous-Permian basal reptile (also known as Captorhinids)
Ichthyosaur fish 245-90 Ma dolphin-like marine reptile
Ictidosaur ferret late Triassic – mid Jurassic mammal-like cynodonts, also known as tritheledontids
Mesosaur middle 299-280 Ma like a small aligator
Mosasaur Meuse River late Cretaceous marine reptile similar to monitor lizards
Nothosaur false/hybrid Triassic slender marine reptile
Pachypleurosaur thick-ribbed Triassic like an aquatic lizard
Pareiasaur shield 270-250 Ma large anapsid
Pelycosaur axe or bowl 320-251 Ma non-therapsid synapsids (e.g. Dimetrodon)
Phytosaur plant 228-200 Ma long-snouted archosauriforms
Plesiosaur close to/near 210-65 Ma marine reptile with broad flat body and short tail
Pliosaur closely 200-89 Ma short-necked plesiosaur
Poposaur discovered on Popo Agie River (ref) late Triassic carnivorous paracrocodylomorphs
Protorosaur early Permian-Triassic long-necked archosauromorphs
Pterosaur winged 228-65 Ma closest relatives to dinosauromorphs
Rhynchosaur beaked Triassic beaked archosauromorphs
Teleosaur end/last early Jurassic – early Cretaceous marine crocodyliforms
Thalattosaur ocean Triassic marine reptile with long flat tail
Trilophosaur three ridged late Triassic lizard-like archosauromorphs
Xenosaur strange present (Cenozoic) knob-scaled lizards

December 14, 2014

Basic combinatorics results

Filed under: mathematics — ckrao @ 8:11 pm

The following lists most of the introductory combinatorics formulas one might see in a first course expressed in terms of the number of arrangements of letters in which repetition or order matters.

number of letters alphabet size letters repeated? order matters? formula comments
k n yes yes n^k samples with replacement
k  n no yes \begin{aligned} & n(n-1)\ldots (n-k+1)\\ &= \frac{n!}{(n-k)!} = P(n,k) = (n)_k\end{aligned} samples without replacement (permutations)
n  n no  yes n! if letters in a line

(n-1)! if letters in a ring and rotations are considered equivalent

(n-1)!/2 if letters in a ring and rotations & reflections are considered equivalent

 k n no no \frac{n!}{k!(n-k)!} = \binom{n}{k} = C(n,k) binomial coefficient

(combinations)

n 2 yes: k of type 1, n-k of type 2 yes \binom{n}{k}
n m yes: k_1 of type i for i = 1,\ldots, m yes \frac{n!}{k_1! k_2! \ldots k_m!} = \frac{(k_1 + k_2 + \ldots + k_m)!}{k_1! k_2! \ldots k_m!} = \binom{n}{k_1, k_2, \ldots, k_m} multinomial coefficient (multiset permutations)

e.g. number of arrangements of “BANANA” is \frac{6!}{3!2!1!}

n k yes no \binom{n+k-1}{k-1} = \binom{n+k-1}{n} = \left(\!\!{n \choose k}\!\!\right) multiset coefficient (combinations with repetition):

number of non-negative integer solutions to n_1 + n_2 + \ldots + n_k = n

n 2 no two consecutive letters of type 1 yes F_{n+2} Fibonacci number where F_{n+2} = F_{n+1} + F_n and F_1 = F_2 = 1

e.g. number of sequences of 6 coin tosses with no two consecutive heads is F_8 = 21

November 23, 2014

Fastest to n test centuries for n sufficiently large

Filed under: cricket,sport — ckrao @ 12:04 am

My blog entry about fewest innings to n ODI centuries has encountered much traffic in recent times due to the consistent century-scoring feats of Amla, Kohli and de Villiers. As requested there, here is a similar table for test cricket.

As expected Bradman dominates the table leading the pack for n = 7 to 29 – he reached n=13, 27 and 28 roughly twice as fast as anyone else! It’s interesting to see how other players feature. Headley is in the list for every century until his last while Arthur Morris raced to 10 centuries in 36 innings before stalling. Sutcliffe, Harvey and Walcott jostle for positions after that and then Hayden and Gavaskar arrive. Tendulkar is somewhere in the top 3 from his 21st century onwards (except when Mohammad Yousuf narrowly takes his spot at n=23), while Ponting, Kallis and Sangakkara are present in the tail end of the list (updated: 31 January 2015).

 

n Fastest to n test centuries (innings required)
3 Azharuddin (4) Headley, Morris, Kambli (6) Gavaskar, Harvey, Hunte (7)
4 Gavaskar, Headley, Kambli (8) Sutcliffe, Weekes (9) Harvey (10)
5 Weekes (10) Harvey, Sutcliffe (12) Bradman, Headley (13)
6 Harvey (13) Bradman (15) Headley (17)
7 Bradman (18) Headley (21) Sutcliffe, Morris (22)
8 Bradman (19) Morris (28) Headley (30)
9 Bradman (22) Morris (30) Headley (31)
10 Bradman (23) Headley (32) Morris (36)
11 Bradman (25) Harvey (46) Sutcliffe, Walcott (49)
12 Bradman (26) Sutcliffe, Walcott (50) Harvey (52)
13 Bradman (28) Walcott (55) Sutcliffe (56)
14 Bradman (40) Walcott (56) Sutcliffe (58)
15 Bradman (41) Sutcliffe (59) Harvey (67)
16 Bradman (48) Sutcliffe (62) Hayden (77)
17 Bradman (50) Gavaskar (81) Hayden (82)
18 Bradman (51) Gavaskar (82) Hayden (87)
19 Bradman (53) Gavaskar (85) Hayden (94)
20 Bradman (55) Gavaskar (93) Hayden (95)
21 Bradman (56) Gavaskar (98) Tendulkar (110)
22 Bradman (58) Gavaskar (101) Tendulkar (114)
23 Bradman (59) Gavaskar (109) Mohammad Yousuf (122)
24 Bradman (66) Tendulkar (125) Gavaskar (128)
25 Bradman (68) Tendulkar (130) Gavaskar (138)
26 Bradman (69) Tendulkar (136) Gavaskar (144)
27 Bradman (70) Tendulkar (141) Gavaskar (154)
28 Bradman (72) Tendulkar (144) Gavaskar (160)
29 Bradman (79) Tendulkar (148) Gavaskar, Hayden (166)
30 Tendulkar (159) Hayden (167) Ponting (170)
31 Tendulkar (165) Ponting (174) Gavaskar (192)
32 Ponting (176) Tendulkar (179) Gavaskar (195)
33 Ponting (178) Tendulkar (181) Sangakkara (199)
34 Tendulkar (192) Ponting (193) Gavaskar (206)
35 Ponting (194) Tendulkar (200) Sangakkara (209)
36 Ponting (200) Sangakkara (210) Tendulkar (218)
37 Ponting (212) Sangakkara (218) Tendulkar (220)
38 Ponting (222) Sangakkara (224) Tendulkar (232)
39 Tendulkar (236) Ponting (239) Kallis (245)
40 Kallis (246) Tendulkar (251) Ponting (273)
41 Kallis, Tendulkar (254) Ponting (275)
42 Kallis (256) Tendulkar (257)
43 Kallis (258) Tendulkar (263)
44 Kallis (263) Tendulkar (266)
45 Tendulkar (268) Kallis (280)
46-51 Tendulkar (270,271,274,
279,286,289)

 

 

Here are some notable streaks observed while preparing the table:

  • Sir Donald Bradman scored 13 centuries in 25 innings in two separate streaks! He scored 26 of his 29 centuries between his 4th and 28th and later between his 48th and 72nd innings inclusive.
  • Mohammad Yousuf entered the table during a streak of 10 centuries in 20 innings (103rd to 122nd innings).
  • Sir Clyde Walcott scored 10 centuries in 23 innings (34-56).
  • Sir Garfield Sobers had no centuries in his first 28 innings and then 11 in his next 31 (29-59).
  • Denis Compton scored 11 centuries in 35 innings (23-57).
  • Mahela Jayawardene scored 12 centuries in 37 innings (129-165).
  • Sunil Gavaskar scored 13 centuries in 41 innings (61-101) and already had 19 centuries after his first 45 tests.
  • Matthew Hayden scored 17 centuries in 59 innings (37-95) in his most productive patch.
  • Inzamam-ul-Haq scored only 6 centuries in his first 90 innings and then 19 in his next 87 (91-177).
  • Ricky Ponting had two separate amazing stretches of 11 centuries in 30 innings (87-116) and later 10 in 24 (155-178).
  • Sachin Tendulkar scored 11 centuries in 34 innings (81-114) and 18 in 68 (81-148). His last 12 centuries were in the space of 39 innings (251-289).
  • Kumar Sangakkara arrived late in the list while scoring 27 centuries in 110 innings (101-210).
  • Jacques Kallis scored 14 centuries in 43 innings (221-263) after 20 centuries in 88 innings earlier in his career (102-189).

Candidate players were found using the best all-time ratings for batsmen on cricketratings.com. Data for the above was sourced from the original version of Statsguru, by displaying the career cumulative averages of each player. Corrections/additions are welcome!

November 22, 2014

A cute sum of Ramanujan

Filed under: mathematics — ckrao @ 3:09 am

Here is a beautiful sum I found in [1], apparently due to Ramanujan.

\displaystyle \frac{1}{1^3\times 2^3} + \frac{1}{2^3 \times 3^3} + \frac{1}{3^3 \times 4^3} + \cdots = \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3} = 10-\pi^2\quad\quad(1)

Note that the result also demonstrates that \pi^2 is slightly less than 10, an alternative to the approaches in [2].

Many of his results require advanced number theory to prove, but this one is not too tricky, as long as we know the following similarly attractive result that I had previously mentioned in this blog post.

\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\quad\quad(2)

To prove (1), the idea is to write \frac{1}{k^3(k+1)^3} as a sum of partial fractions and then sum a telescoping series. You might wish to try this yourself before reading further.

We write

\displaystyle \frac{1}{k^3(k+1)^3} = \frac{a(k)}{k^3} + \frac{b(k)}{(k+1)^3} = \frac{a(k)(k+1)^3 + b(k)k^3}{k^3(k+1)^3},\quad\quad(3)

where a(k) and b(k) are quadratic polynomials. One way of finding a(k) and b(k) would be to compare coefficients of 1, k, k^2, k^3 and solve a system of equations. Another approach, the Extended Euclidean Algorithm, does so via finding the greatest common divisor of k^3 and (k+1)^3.

The following manipulations verify that the greatest common divisor of k^3 and (k+1)^3 is 1, where the next line computes quotients and remainders based on the previous line.

\displaystyle  \begin{aligned}  (k+1)^3 &= k^3 + (3k^2 + 3k + 1) & \quad (4)\\  k^3 &= \frac{k}{3}(3k^2 + 3k + 1) - \left(k^2 + \frac{k}{3}\right) & \quad (5)\\  3k^2 + 3k+ 1 &= 3\left(k^2 + \frac{k}{3}\right) + (2k + 1) & \quad (6)\\  k^2 + \frac{k}{3} &= \frac{k}{2}(2k+1) - \frac{k}{6} & \quad (7)\\  2k+1 &= 12 \frac{k}{6} + 1 & \quad (8)  \end{aligned}

Reversing the steps of (4)-(8) gives us 1 as quadratic polynomial combinations of k^3 and (k+1)^3, thus providing us with a(k) and b(k).

\displaystyle  \begin{aligned}  1 &= (2k+1) - 12\left[\frac{k}{2}(2k+1) - \left(k^2 + \frac{k}{3}\right)\right]\\  &= (2k+1)(1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\  &= \left[3k^2 + 3k + 1 - 3\left(k^2 + \frac{k}{3}\right) \right](1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\  &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left(k^2 + \frac{k}{3}\right)\\  &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left[\frac{k}{3}(3k^2+3k+1) - k^3\right]\\  &= (3k^2 + 3k + 1)\left[1-6k + 4k - k(1-6k)]\right] + \left[12 - 3(1-6k)\right]\left(-k^3\right)\\  &= (k+1)^3\left(6k^2 - 3k+ 1\right) - k^3\left(6k^2 - 3k+ 1 + 12 - 3 + 18k\right)\\  &= (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10).\quad\quad(9)  \end{aligned}

Using (9) we then have

\displaystyle  \begin{aligned}  \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3}  &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{1}{k^3(k+1)^3}\\  &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{ (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10)}{k^3(k+1)^3}\\  &= \lim_{N \rightarrow \infty} \left(\sum_{k=1}^{N} \frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6k^2 + 15k+10}{(k+1)^3}\right)\\  &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6(k+1)^2 + 3(k+1) + 1}{(k+1)^3}\right)\\  &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{6}{k} - \frac{3}{k^2} + \frac{1}{k^3}\right) - \sum_{k=2}^{N+1} \left(\frac{6}{k} + \frac{3}{k^2} + \frac{1}{k^3}\right)\\  &= \left(\frac{6}{1} - \frac{3}{1} + \frac{1}{1}\right) - \lim_{N \rightarrow \infty} \sum_{k=2}^{N} \frac{6}{k^2} - \lim_{N \rightarrow \infty}\left(\frac{6}{N+1} + \frac{3}{(N+1)^2} + \frac{1}{(N+1)^3} \right)\\  &= 4 - (\pi^2 - 6) - 0 \quad\text{(using (2))}\\  &= 10-\pi^2,  \end{aligned}

thus verifying (1).

The interested reader might like to find other identities in a similar manner. :)

References

[1] Clifford A. Pickover, A Passion for Mathematics: Numbers, Puzzles, Madness, Religion, and the Quest for Reality, John Wiley & Sons, 2005.

[2] Noam D. Elkies, Why is \pi^2 so close to 10?

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

Maps

References

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)

Hence

\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}

Now

\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).

Next Page »

The Rubric Theme. Create a free website or blog at WordPress.com.

Follow

Get every new post delivered to your Inbox.

%d bloggers like this: