Mathematical Topics

09/29/2018

   
   
   
   
 

 

FRACTALS  See http://www.youtube.com/results?search_query=Mandelbrot+fractal+tour+guide  Also see http://www.scientificamerican.com/article/mathematics-ramanujan/  and http://en.wikipedia.org/wiki/Self-similarity   and also http://en.wikipedia.org/wiki/Fractal and http://en.wikipedia.org/wiki/Chaos_theory

In October 1992, when I was President of the Society of Actuaries, I had a video presentation of pictures of fractals and some background at the Annual Meeting of the Society of Actuaries. Robert Devaney developed the script.  See Chaos video, but start about 40% of the way through. (OOPs is copywrite protected so will only play on my computer)  If interested, read Don's Presidential Address.

See Robert Devaney and chaotic systems and fractals at                                 https://www.youtube.com/watch?v=6QIhaDvTHXk  and http://math.bu.edu/people/bob/

Harlan Brothers articles on Mandelbrot and Fractal Music.                              https://www.youtube.com/watch?v=s-SJvJK3r3I

Also info from David

David Richeson: Division by Zero  

Math Crafts: Salt Designs, Newton Snowflakes, Fractal Christmas Trees, Paper Pentagons, and Flip Books     Posted by Dave Richeson on December 18, 2017

I have had a crafty late fall and early winter. Ive been good about posting my crafts on Twitter, but not so good at blogging about them. So, Ive collected them all and will share them all here in one blog post.

The Geometry of Salt

I came across this neat pdf by Troy Jones about using salt to do geometry. So, over Thanksgiving break I got my kids and their cousins together to do a little mathematics. We cut various shapes out of paper, propped them up on glasses, and poured salt over them. The salt is a natural bisector. The ridges can be used to bisect angles and to find the locus of points equidistant from two curves. We had fun making triangle centers, Voronoi diagrams, and conic sections. I had a good time thinking about why this worked (it is a fun exercise to see why these ridges form the various conic sections).

.DPklrweXkAA_tg3.jpg  DPkltFIWsAAjn1w.jpg

Paper Pentagon                                                                               In a recent paper by John Sharp I learned about tying a strip of paper into a regular pentagon. It goes back to Tom Tit, which was the pen-name of Arthur Good (18531928).

DLudvRfUMAAAiHc.jpg_large.jpeg 

Newton Snowflake                                                                      The American Institute of Physics gave templates to make physics-related snowflakes. I used their template to make this Isaac Newton snowflake. (As cool as it is, it doesnt have six-fold symmetry like a true snowflake.) They also have a crystallography and a Nikola Tesla snowflake.

DQiYZReWAAAUT34.jpg_large.jpeg

Flip Book                                                                                       I have been wanting to make a mathematical flip book for a long time. Yesterday was the day that I made it happen. My 13-year old son and I figured out how to do it using the Adobe suite (my son is becoming a self-taught Adobe wiz). I started by creating the first four stages of the Koch snowflake using Illustrator. My son imported them into After Effects and had them morph from one to the other. He used Media Encoder to export them as an animated gif:

Koch.gif

Then we exported the frames of the gif as 90 separate images. I printed them on card stock with nine per page (heres a pdf). I cut them out and used a binder clip to hold them together. It seemed like 90 frames was perhaps too many, so I took every other frame and made two books of 45 frames. Those worked well. Lastly, I put the two books back to back and took following video of the triangle turning into a snowflake turning back into a triangle. Im excited to make more flip books.

Fractal Christmas Trees I found this blog post, which has instructions on how to make a fractal Christmas tree. I made one and thought it looked great. In fact, it folds up nicely into a card. So my daughter made one to give to her teacher.

DRCDZADVoAAriWM.jpg_large                                                                           After that, I tried making one of my own design. Rather than cutting the paper in half at each stage, I cut it into thirds. The result is below. It looks cool, but when it is folded as a card, part of the inside sticks out.

DRF7ByoVwAAqMOb.jpg_large

Dave Richeson   Professor of Mathematics    Dickinson College
Contact Information    richesod@dickinson.edu   Tome Scientific Building Room 244     717.245.1744

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https://divisbyzero.com/

David Richeson

Contact Information    richesod@dickinson.edu                                                      Tome Scientific Building Room 244     717.245.1744
http://divisbyzero.com  Buy this

Bio

His research involves the study of dynamical systems from a topological viewpoint, the history of mathematics (especially the history of geometry and topology), and recreational mathematics. He is interested in promoting expository mathematical writing.

Education

bulletB.A., Hamilton College, 1993
bulletM.S., Northwestern University, 1994
bullet Ph.D., 1998   https://divisbyzero.com/    Read about at https://divisbyzero.com/research/eulers-gem/

 Euler's Gem      April 19, 2016

"Euler's Gem is one of the most elegant popular mathematics books I have read, a lovely exposition of some fascinating material which could be enjoyed by everyone from school students and lay readers to potential mathematicians. . . . Every school library should have a copy of Euler's Gem and it would be a great present for an otherwise curious student under-challenged or unexcited by the mathematics they are doing.

Mandelbrot Sets See http://www.youtube.com/watch?v=8ma6cV6fw24   Also see discussion by Mandelbrot at http://www.youtube.com/watch?v=ay8OMOsf6AQ   http://en.wikipedia.org/wiki/Mandelbrot_set  and https://www.youtube.com/watch?v=56gzV0od6DU

Julia Sets  See http://www.youtube.com/watch?v=2AZYZ-L8m9Q   and                                             https://www.youtube.com/watch?v=mg4bp7G0D3s

Newton Fractal  See  http://vimeo.com/9770779

Serpenski and other results See https://www.youtube.com/watch?v=DO8yFGbbGmg

 FIBONACCI SERIES See http://www.youtube.com/watch?v=SjSHVDfXHQ4

The first two terms are 1 and 1.  Then use the formula

F_n = F_{n-1} + F_{n-2},\!\,

The first 21 Fibonacci numbers Fn for n = 0, 1, 2, ..., 20 are:

F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765

F_{n}=\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} \tbinom {n-k-1} k.

These numbers also give the solution to certain enumerative problems.  The most common such problem is that of counting the number of compositions of 1s and 2s that sum to a given total n: there are Fn+1 ways to do this. For example F6 = 8 counts the eight compositions:

1+1+1+1+1 = 1+1+1+2 = 1+1+2+1 = 1+2+1+1 = 2+1+1+1 = 2+2+1 = 2+1+2 = 1+2+2,

all of which sum to 6−1 = 5.

Golden Ratio  is the only number that is one more than its resiprocal. See http://en.wikipedia.org/wiki/Golden_ratio

The Fibonacci numbers have a closed-form solution. It is known as Binet's formula:

F_n = \frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}  where  \varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\cdots\,

Coin Game   You have a pile of coins, and you can take up to and including twice what your opponent took last timeuses a winning strategy of leaving your opponent on a Fibonacci number or on the sum of two or more Fibonacci numbers.  The person who picks up the last coin wins.  On your first turn you cant take all. 

Start with 11 coins.  Take 3, leaving 8 and you win, as 8 is a Fibonacci number. 

Start with 12 coins.  Take 4 and the other guy can take 8 and you lose.  So reduce the pile to the sum of two Fibonacci numbers 8 +3 = 11 and take 1 reducing the 12 to 11.  Your opponent can now take 1 or 2.  If he takes 1, you take 2 and he has 8 to choose from.  If he takes 2, you take 1 and he has 8 to choose from.  Both are bad.

Whatever he chooses, say 2, you take 1 and leave him with 5.  He can take 1 or 2.  If he takes 1, you take 1 leaving him with 3.  He can take 1 or 2and you take whats left.  

BINARY NUMBERS

The modern binary number system was invented by Gottfried Leibniz in 1679.

Decimal pattern Binary numbers
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001

 

Nim  The traditional game of Nim used three rows of 3, 4, and 5 coins.  On your turn, you can take any number from only one row.  The person to pick the last coin wins.  The winning strategy uses the Binary System to  matched up pairs of powers of 2's.

 3 = 2 + 1, 4 = 4,  5 = 4 +1   The person who goes first will win by  taking 2 from the 3 pile to create matches.

If the game was four rows of  8, 13, 17, and 20:

20 = 16 + 4, 17 = 16 + 1,  13 = 8 + 4 + 1,   8   = 8   The person going first loses as we have matches

If the game was five rows of  5, 6, 7, 8, 10:

10 = 8 + 2,   8 =  8,  7 = 4 + 2 + 1,  6 = 4 + 2,  5 =  4 + 1 The person who goes first will win  Take 2 from 5 pile, leaving 2 and 1.

The name is probably derived from German nimm meaning "take [imperative]", or the obsolete English verb nim of the same meaning.

Wythoff's game is a two-player mathematical game of strategy, played with two piles of counters. Players take turns removing counters from one or both piles; in the latter case, the numbers of counters removed from each pile must be equal. The game ends when one person removes the last counter or counters, thus winning.

Martin Gardner claims that the game was played in China under the name 捡石子 jiǎn shzǐ ("picking stones"). The Dutch mathematician W. A. Wythoff published a mathematical analysis of the game in 1907.

The Last Biscuit game  is played by removing cookies from two jars, either from a single jar, or the same number from both jars. This game is also called the Puppies and Kittens game.

The strategy of winning Wythoff's Nim is to reduce the piles to a number pair.. If the starting pile numbers are safe, the first player loses. He is certain to leave an unsafe pair of piles, which his opponent can always reduce to a safe pair on his next move. If the game begins with unsafe numbers, the first player can always win by reducing the piles to a safe pair and continuing to play to safe pairs.

 Let us take the safe pairs in sequence, starting with the pair nearest 0/0, and arrange them in a row with each smaller number above its partner, as in the table below. Above the pairs write their "position numbers." The top numbers of the safe pairs form a sequence we shall call A. The bottom numbers form a sequence we shall call B.

Position (n) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
A 1 3 4 6 8 9 11 12 14 16 17 19 21 22 24
B 2 5 7 10 13 15 18 20 23 26 28 31 34 36 39

You cannot win with 21, 34.   a)  If you take 6 from the 34 pile leaving 21, 28  your opponent will take 4 from the 21 pile leaving you with 17, 28.  b)  If you take 7 from both piles leaving 14, 27 your opponent will take 4 from the 27 pile leaving you with 14, 23. 

You cannot win with 21, 34.  a) If you take 6 from both piles leaving 15, 28 your opponent will take 19 from the 28 pile leaving you with 9, 15.  b) If you take 7 from the 34 pile leaving your opponent with 21, 27 your opponent will take 12 from each pile leaving you with 9, 15.    

The two sequences, each one strictly increasing, have so many remarkable properties that dozens of technical papers have been written about them. Each B number is the sum of its A number and its position number. If we add an A number to its B number, the sum is an A number that appears in the A sequence at a position number equal to B. (An example is 8 + 13 = 21. The 13th number of the A sequence is 21.)

Can we generate the sequences by a recursive algorithm that is purely numerical?  Yes.

Start with 1 as the top number of the first safe pair. Add this to its position number to obtain 2 as the bottom number. The top number of the next pair is the smallest positive integer not previously used. It is 3. Below it goes 5, the sum of 3 and its position number. For the top of the third pair write again the smallest positive integer not yet used. It is 4. Below it goes 7, the sum of 4 and 3. Continuing in this way will generate series A and B.

There is a bonus. We have discovered one of the most unusual properties of the safe pairs. It is obvious from our procedure that every positive integer must appear once and only once somewhere in the two sequences.

Is there a way to generate the two sequences nonrecursively? Yes. Wythoff was the first to discover that the numbers in sequence A are simply multiples of the golden ratio rounded down to integers! (He wrote that he pulled this discovery "out of a hat.")

Also see: http://www.fq.math.ca/Scanned/17-3/hoggatt.pdf

SYMMETRY and ITERATION  See   www. rmmsmsp.ucdenver.edu/instructormaterial/geometry-daisies.pps

Scientists strive to find mathematical patterns. However, Sir Francis Bacan said: "There is no beauty that hath not some strangeness in the proportion."  Since beauty lies in the eyes of the beholder, are the most beautiful creations short lyrics or long symphonies? 

David Wells said we might conclude that the beauty of the Mandelbrot set is "romantic"

Hardy said:  "A mathematician, like a painter or poet, is a maker of patterns.  If his patterns are more permanent than theirs, it is because they are made with ideas."  Hardy also hoped that "nothing he had ever discovered would have any practical use. 

Paul Erdős  said, "Why are numbers beautiful? It's like asking why is Beethoven's Ninth Symphony beautiful. If you don't see why, someone can't tell you. I know numbers are beautiful. If they aren't beautiful, nothing is".

Some see beauty in mathematical results that establish connections between two areas of mathematics that at first sight appear to be unrelated. These results are often described as deep. One example are often cited is Euler's identity:

\displaystyle e^{i \pi} + 1 = 0\, .

This is a special case of Euler's formula which states that, for any real number x,

e^{ix} = \cos x + i\sin x \
 
 
Physicist Richard Feynman called Euler's formula "our jewel" and "the most remarkable formula in mathematics".

Modern examples include the modularity theorem, which establishes an important connection between elliptic curves and modular forms (work on which led to the awarding of the Wolf Prize to Andrew Wiles and Robert Langlands), and "monstrous moonshine", which connects the Monster group to modular functions via string theory for which Richard Borcherds was awarded the Fields Medal.

An example of mathematical elegance is Leibnitz's series:

\sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} \;=\; \frac{\pi}{4}.\! 1 \,-\, \frac{1}{3} \,+\, \frac{1}{5} \,-\, \frac{1}{7} \,+\, \frac{1}{9} \,-\, \cdots \;=\; \frac{\pi}{4}.\!

BIG NUMBERS  See http://www.guardian.co.uk/world/2009/mar/25/trillion-dollar-rescue-plan

MATH TERMS  See http://www.cut-the-knot.org/glossary/stop.shtml

Repunits are numbers where all digits are the same, and were named repunits by Albert Beiler.

In particular, the decimal (base-10) repunits that are often referred to as simply repunits are defined as

R_n=R_n^{(10)}={10^n-1\over{10-1}}={10^n-1\over9}\qquad\mbox{for }n\ge1.
The problem of discovering which repunits are prime, and factoring others, is similar to the analyzing Mersenne numbers (named after the French monk Marin Mersenne who studied them in the early 17th century)  of the form M_n=2^n-1\, . The first four Mersenne primes are 3, 7, 31 and 127.  If n is a composite number then so is 2n − 1. The definition is therefore unchanged when written M_p=2^p-1 where p is assumed prime.  More generally, numbers of the form M_n=2^n-1\, without the primality requirement, are called Mersenne numbers.

Repunits R_n therefore have exactly n decimal digits.

 

Amazingly, the squares of the repunits  R_n^2  give the Demlo numbers, 1^2=1, 11^2=121, 111^2=12321  Ramchandra Dattaraya Kaprekar (19051986) studied the Demlo numbers, named after a train station 30 miles from Bombay on the then G. I. P. Railway where he had the idea of studying them.  These are the numbers 1, 121, 12321, , which are the squares of the repunits 1, 11, 111, .   He was an Indian recreational mathematician who described several classes of natural numbers including the Kaprekar, Harshad and Self numbers and discovered the Kaprekar constant, named after him. Despite having no formal postgraduate training and working as a schoolteacher, he published extensively and became well known in recreational mathematics circles.  The following table contains some repunits.

 

 3 x 37 = 111
 6 x 37 = 222
 9 x 37 = 333
12 x 37 = 444
15 x 37 = 555
18 x 37 = 666
21 x 37 = 777
24 x 37 = 888
27 x 37 = 999
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111=12345678987654321
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111
2519 Mod 2 = 1
2519 Mod 3 = 2
2519 Mod 4 = 3
2519 Mod 5 = 4
2519 Mod 6 = 5
2519 Mod 7 = 6
2519 Mod 8 = 7
2519 Mod 9 = 8
2519 Mod 10 = 9
0 x 9 + 8 = 8
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
987654321 x 9 - 1 = 8888888888
9876543210 x 9 - 2 = 88888888888
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
                     

 

 

  142857 x 2 = 285714                               
142857 x 3 = 428571
142857 x 4 = 571428
142857 x 5 = 714285
142857 x 6 = 857142

91  

times 1 = 0 9 1 7x7=49
91 times 2 = 1 8 2 67x67=4489
91 times 3 = 2 7 3 667x667=44889
91 times 4 = 3 6 4 6667x6667=44448889
91 times 5 = 4 5 5 66667x66667=4444488889
91 times 6 = 5 4 6 666667x666667=444444888889
91 times 7 = 6 3 7 6666667x6666667=44444448888889
91 times 8 = 7 2 8 etc.
91 times 9 = 8 1 9  
1x9+2 = 11 9 x 9 + 7 = 88  9 x 9 = 81   6 x 7 = 42
12x9+3 = 111 98 x 9 + 6 = 888 99 x 99 = 9801 66 x 67 = 4422
123x9+4 = 1111 987 x 9 + 5 = 8888  999 x 999 = 998001 666 x 667 = 444222
 1234x9+5 = 11111 9876 x 9 + 4 = 88888  9999 x 9999 = 99980001 6666 x 6667 = 44442222
 12345x9+6 = 111111 98765 x 9 + 3 = 888888 etc etc
 123456x9+7 = 1111111 987654 x 9 +2 = 8888888    
 1234567x9+  = 11111111 9876543x9+1= 88888888    
 12345678x9+9=111111111 98765432x9 = 888888888    
1x1=1 4x4=16
11x11=121 34x34=1156
111x111=12321 334x334=111556
1111x1111=1234321 3334x3334=11115556
11111x11111=123454321 33334x33334=1111155556
111111x111111=12345654321 etc.
1111111x1111111=1234567654321
11111111x11111111=123456787654321
111111111x111111111=12345678987654321

Santa's Options  Assuming Rudolph was in front, there are 40,320 ways to arrange the other eight reindeer.

Geometry  Are the eight balls moving in a circle or a straight line?
http://showyou.com/v/y-pNe6fsaCVtI/crazy-circle-illusion?u=multimotion

Roman Numerals  The original Roman year had 10 named months Martius "March", Aprilis "April", Maius "May", Junius "June", Quintilis "July", Sextilis "August", September "September", October "October", November "November", December "December", and probably two unnamed months in the dead of winter when not much happened in agriculture. The year began with Martius "March". Numa Pompilius, the second king of Rome circa 700 BC, added the two months Januarius "January" and Februarius "February". He also moved the beginning of the year from Marius to Januarius and changed the number of days in several months to be odd, a lucky number. After Februarius there was occasionally an additional month of Intercalaris "intercalendar". This is the origin of the leap-year day being in February. In 46 BC, Julius Caesar reformed the Roman calendar (hence the Julian calendar) changing the number of days in many months and removing Intercalaris.

The numerals are: 1=I (unus); 5=V (quinque); 10=X (decem); C=100 (centum), and M=1000 (mille).   They also used 50=L (quinquaginta); and 500=D (quingenti). 

When a small number comes before a larger number, the smaller number is subtracted.  4 = IV or 5-1. When a smaller number follows a larger one, the two are added together: 7 = VII, or 5 + 2 and 19 = XIX, or 10 + 9.  Although the Romans used a decimal system for whole numbers, they used a duodecimal system for fractions because the divisibility by 12 made it easier to handle the common fractions of 1/3 and 1/4.

Chinese Multiplication      See http://www.youtube.com/watch?v=8iIU9EDC2GQ  and http://www.youtube.com/watch?v=maRN2fUOF0o

Leonhard Euler (1707-1783):  One of his many contributions was called "Euler's Formula".  The formula states that, for any real number  xe^{ix} = \cos x + i\sin x \!   where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions, with the argument x given in radians. The formula is still valid if x is a complex number.   Richard Feynman called Euler's formula "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics".   

Euler Line

In the 18th century, the Swiss mathematician Leonhard Euler noticed that three of the  centers of a triangle are always collinear (they always lie on a straight line). The three centers that have this surprising property are the triangle's centroid (where the three medians of the triangle's sides meet), circumcenter (where the perpendicular bisectors of the triangle's sides meet) and the orthocenter (where the three altitudes to the vertices of the triangle meet).  The distance from the orthocenter to the centroid is two times the distance from the centroid to the circumcenter.  (Another center, the incenter, where the bisectors of the three angles meet, is not on this line.)

Acute Triangle
Obtuse Triangle
Equilateral Triangle

Triangle centers

Incenter
                                  Located at intersection of the angle bisectors.

 
Centroid
                                  Located at intersection of the medians

Circumcenter
                                  Located at intersection of the perpendicular bisectors of the sides
 
Orthocenter
                                  Located at intersection of the altitudes
 
Exeter point.svg The medians  meet the circumcircle of triangle ABC at A' , B' and C' . Let DEF be the triangle formed by the tangents at A, B, and C to the circumcircle of triangle ABC.  The lines through DA' , EB' and FC' meet at the Exeter point MorleysTheorem The points of intersection of the adjacent angle trisectors of the angles of any triangle DeltaABC are the polygon vertices of an equilateral triangle DeltaDEF known as the first Morley triangle. .

 

The nine-point circle is a circle through nine points:

The midpoint of each side of the triangle.

The foot of each altitude

The midpoint of the line segment from each vertex of the triangle to the orthocenter.

 

Nine Point Circle above

 
The Euler line in any non equilateral triangle passes through five points:

The orthocenter, the circumcenter, the centroid, the Exeter point and the center of the nine-point circle of the triangle.

The Exeter Point is not shown on the left.

Orthocenter (Blue), Incenter (Orange), Circumcircle (Green), Center of the Nine point circle (Red)

Euler Line in red

Euler and the Nine Point Circle

The nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant concyclic points defined from the triangle. These nine points are:

bulletThe midpoint of each side of the triangle
bulletThe foot of each altitude
bulletThe midpoint of the line segment from each vertex of the triangle to the orthocenter (where the three altitudes meet; these line segments lie on their respective altitudes).

The nine-point circle is also known as Feuerbach's circle, Euler's circle,  and Terquem's circle.

To construct the nine point circle of a triangle, see http://jwilson.coe.uga.edu/EMT668/EMT668.Folders.F97/Anderson/geometry/geometry1project/construction/construction.html

.1. Draw a triangle ABC and construct the midpoints of the three sides. Label them as L, M, N.

2. Construct the feet of the altitudes of the triangle ABC. Label them as D, E, F. Label the point of intersection of the three altitudes as H. This is also called the orthocenter.

 

3. Construct the midpoints of the segments AH, BH, CH. Label them as X, Y, Z.

4. Notice the nine points, L,M,N,D,E,F,X,Y, Z, lie in a circle called the Nine-Point Circle..

5. Construct the circumscribed circle for triangle LMN. Label the center of that circle U.

The center U of the circumscribed circle for triangle LMN will also be the center of the Nine-Point Circle.

More on Nine-Point Circle at http://en.wikipedia.org/wiki/Triangle_center where we learn  the following:

Let A, B, C denote the vertex angles of the reference triangle, and let x : y : z be a variable point in trilinear coordinates; then an equation for the Euler line is

\sin 2A \sin(B - C)x + \sin 2B \sin(C - A)y + \sin 2C \sin(A - B)z = 0.\,
Center Trilinear Coordiates On Euler Line?
Orttocenter Secant A: Secant B : Secant C Yes
Centroid Cosecant A: Cosecant B: Cosecant C Yes
Circumcenter Cosine A: Cosine B: Cosine C Yes
Nine Point Circle Cosine (B - C): Cosine (C - A): Cosine (A - B) Yes
In Center 1:1:1 Only if Isosceles

Euler and polyhedrons

A platonic solid is a regular, convex polyhedron with congruent faces of regular polygons and the same number of faces meeting at each vertex.  There are five regular polyhedrons that  meet those criteria, and each is named after its number of faces.:

Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron
Tetrahedron.svg Hexahedron.svg Octahedron.svg POV-Ray-Dodecahedron.svg Icosahedron.svg
4 Triangles 4 Squares 8 Triangles 12 Pentagons 20 Triangles

Euler's formula for polyhedrons is:   V - E + F = 2  That is the number of vertices, minus the number of edges, plus the number of faces, is equal to two.

More on Euler

In analytical mathematics, Euler's identity (also known as Euler's equation) is the equality:

eiπ + 1 = 0
e is Euler's number, the base of natural logarithms
i is the imaginary unit, which satisfies i2 = −1
π is pi, the ratio of the circumference of a circle to its diameter
    

Euler's  number e is an important mathematical constant, approximately equal to 2.71828, that is the base of the natural logarithm.  It is the limit of (1 + 1/n)n as n becomes large, an expression that arises in the study of compound interest, and can also be calculated as the sum of the infinite series:

e =  \displaystyle\sum\limits_{n = 0}^{ \infty} \dfrac{1}{n!} = 1 + \frac{1}{1} + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \cdots

e is the unique real number such that the value of the derivative (slope of the tangent line) of the function f(x) = ex at the point x = 0 is equal to 1. The function ex so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base e.

The number e is of eminent importance in mathematics, alongside 0, 1, π and i. All five of these numbers play important and recurring roles across mathematics, and are the five constants appearing in one formulation of Euler's identity. Like the constant π, e is irrational: it is not a ratio of integers; and it is transcendental: it is not a root of any non-zero polynomial with rational coefficients. The numerical value of e truncated to 50 decimal places is 2.71828182845904523536028747135266249775724709369995...

e and compound interest 

Let P be the principal (initial investment), rbe the annual compounded rate, i^((n))the "nominal rate," nbe the number of times interest is compounded per year (i.e., the year is divided into nconversion periods), and tbe the number of years (the "term"). The interest rate per conversion period is then

 r=(i^((n)))/n.

If interest is compounded ntimes at an annual rate of r(where, for example, 10% corresponds to r=0.10), then the effective rate over 1/nthe time (what an investor would earn if he did not redeposit his interest after each compounding) is

 (1+r)^(1/n).

The total amount of holdings Aafter a time twhen interest is re-invested is then

 A=P(1+(i^((n)))/n)^(nt)=P(1+r)^(nt).

Note that even if interest is compounded continuously, the return is still finite since

 lim_(n->infty)(1+1/n)^n=e,

where e is the base of the natural logarithm.

The time required for a given principal to double (assuming n=1 conversion period) is given by solving

 2P=P(1+r)^t, or

 t=(ln2)/(ln(1+r)),

where ln is the natural logarithm. This function can be approximated by the so-called rule of 72:

 t approx (0.72)/r.

NUMBER THEORY

Numbers are classified according to type. The first type of number is the first type you ever learned about: the counting, or "natural" numbers:

1, 2, 3, 4, 5, 6, ...

The next type is the "whole" numbers, which are the natural numbers together with zero:

0, 1, 2, 3, 4, 5, 6, ...

Then come the "integers", which are zero, the natural numbers, and the negatives of the naturals:

... -6, -5, -4, -3, -2, -1, 9, 1, 2, 3, 4, 5, 6, ...

The next type is the "rational", or fractional, numbers, which are technically regarded as ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer.

Sums of Natural Numbers  The sum of the first n natural numbers is:  1 + 2 + 3 + 4 ....... + n  =  n(n+1)/2      Gauss developed this formula when in primary school: the average of the first number and the last number times the number of numbers!                                                                                                                                                        If n = 6: 1 + 2 + 3 + 4 + 5  + 6 = 21   = (6x7)/2

Sums of Even Numbers  The sum of the first k even natural numbers is: 2 + 4 + 6 ..... + 2k = k(k+1).                      If k = 3:     2 + 4 + 6 = 12    = (3x4)

Sums of Odd Numbers  The sum of the first k odd natural numbers is:  1 + 3 + 5 ..... + (2k - 1) =  k2                       If k = 3:      1 + 3 + 5 = 9     = (3x3)         

n^2 = \sum_{k=1}^n(2k-1)

Sums of Squares  The sum of the squares of the first n natural numbers is:
sumSquaresNatNumbersFormulae.gif

30 =  (12 + 22 + 32 + 42) = 1 + 4 + 9 + 16

Also see:   http://www.takayaiwamoto.com/Sums_and_Series/sumsqr_1.html  and http://www.math.utah.edu/~palais/sums.html

365 = ( 102 + 112 + 122)  = (132 + 142)

    

Sums of Cubes  The sum of the cubes of the first n natural numbers is:

1^3+2^3+\dots+n^3 = (1+2+\dots+n)^2=\left(\frac{n(n+1)}{2}\right)^2.
 
As you can see, the sum of the cubes of the first n natural numbers is equal to the square of the sum of the first n natural numbers.
 

      Divisor Function: D(x) = number of divisors in a number including 1 and x.  If m and n are relatively prime, then D(mn) = D(m) x D(n)

m

n

mn

D(m)

D(n)

D(mn) = D(m) x D(n)

Divisors

2

3

6

2

2

4

1, 2, 3, 6

4

9

36

3

3

9

1, 2, 3, 4, 6, 9, 12,18, 36
15 1 15 4

1

4

1, 3, 5, 15

28

1

28

6

1

6

1, 2, 4, 7, 14, 28
15 28 420 6 4 24 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420

Take any integer. Examples: 15 and 28.  Write its factors. Underneath each factor write the D(x) for each factor:

Factors of 15     15 5 3 1
D(Factors of 15)     4 2 2 1
Factors of 28 28 14 7 4 2 1
D(Factors of 28) 6 4 2 3 2 1

      Liouville's Theorem says: The sum of the cubes of the numbers in the second line equals the square of the sum of those same numbers:   64 + 8 + 8 + 1 = 81.     4 + 2 + 2 + 1 = 9, and 9 squared = 81.                                 The sum of the cube of the numbers in the fourth line equals the square of the sum of those same numbers:     216 + 64 + 8 + 27 + 8 + 1 = 324.       6 + 4 + 2 + 3 + 2 + 1 = 18, and18 squared = 324. 

      Conjecture is:   Pick small numbers and the sum of their cubes is less than the square of their sums. But if you pick large numbers the converse is true.  See examples using four numbers are below:

          Their Sum Square of their Sum Sum of their Cubes    Square of their Sum minus the Sum of their Cubes
A 1 1 1 3 6 36 30 6
B 1 1 2 2 6 36 18 18
C 2 2 3 4 11 121 107 14
D 1 2 2 4 9 81 81 0
E 1 1 2 5 9 81 135 -54
F 1 2 6 8 17 289 737 -448
G 3 6 7 8 24 576 1098 -522
 
Sum of 1 - 2 + 3 - 4 + 5 - 6  ... =
 Summation from n equals 1 to m of the series n * (-1)^(n-1)
http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7
 

In a 1749 report, Leonhard Euler admits that the series diverges but prepares to sum it anyway:

...when it is said that the sum of this series 1−2+3−4+5−6 etc. is 14, that must appear paradoxical. For by adding 100 terms of this series, we get −50, however, the sum of 101 terms gives +51, which is quite different from 14 and becomes still greater when one increases the number of terms. But I have already noticed at a previous time, that it is necessary to give to the word sum a more extended meaning....

Euler proposed a generalization of the word "sum" several times; see Euler on infinite series. In the case of 1 − 2 + 3 − 4 + ..., his ideas are similar to what is now known as Abel summation:

...it is no more doubtful that the sum of this series 1−2+3−4+5 + etc. is 14; since it arises from the expansion of the formula 1(1+1)2, whose value is incontestably 14. The idea becomes clearer by considering the general series 1 − 2x + 3x2 − 4x3 + 5x4 − 6x5 + &c. that arises while expanding the expression 1(1+x)2, which this series is indeed equal to after we set x = 1.

There are many ways to see that, at least for absolute values |x| < 1, Euler is right in that

1-2x+3x^2-4x^3+\cdots = \frac{1}{(1+x)^2}.

 

One can take the Taylor expansion of the right-hand side, or apply the formal long division process for polynomials. Starting from the left-hand side, one can follow the general heuristics above and try multiplying by (1+x) twice or squaring the geometric series 1 − x + x2 − .... Euler also seems to suggest differentiating the latter series term by term.

In the modern view, the series 1 − 2x + 3x2 − 4x3 + ... does not define a function at x = 1, so that value cannot simply be substituted into the resulting expression. Since the function is defined for all |x| < 1, one can still take the limit as x approaches 1, and this is the definition of the Abel sum:

\lim_{x\rightarrow 1^{-}}\sum_{n=1}^\infty n(-x)^{n-1} = \lim_{x\rightarrow 1^{-}}\frac{1}{(1+x)^2} = \frac14.

Leonhard Euler is most famous for the "Euler Identity":  e^{ix} = \cos x + i\sin x \                                                           The special case, with x = π  gives the beautiful identity:  e^{i \pi} + 1 = 0, which involves 0, 1, i, e and π. 

We can easily see this result by looking at the diagram of a circle. 

The x axis contains the real numbers (Re) and the y axis contains the imaginary numbers (Im). 

The radius of the circle is i

The identity is when the angle equals  π

Triangular Numbers 

The numbers which can be arranged in a compact triangular pattern are termed as triangular numbers. The triangular numbers are formed by partial sum of the series 1+2+3+4+5+6+7......+n. So

T1 = 1
T2 = 1 + 2 = 3
T3 = 1 + 2 + 3 = 6
T4 = 1 + 2 + 3 + 4 = 10

So the nth triangular number can be obtained as Tn = n(n+1)/2, where n is any natural number. In other words triangular numbers form the series 1,3,6,10,15,21,28.....

n2  = the sum of two consecutive triangular numbers, because Tn +  Tn-1  = n(n+1)/2  +  (n-1)(n)/2  = n2

Cubics, Quartics, and Quintics

Niccolo Tartaglia, who solved the cubic, failed miserably for the rest of his life (mainly because he spent it trying to discredit Cardano).  See  http://www.storyofmathematics.com/16th_tartaglia.html
Giralamo Cardano, who stole Tartaglia's solution, is also credited with solving the cubic. 
Lodovico Ferrara, solved the general quartic, was poisoned, probably by his sister, over an inheritance dispute.
Evariste Galois, who showed the general quintic was unsolvable, died in a duel at the age of 29.
Niels Henrik Abel, who duplicated and extended Galois' proof independently, finally managed to receive his first faculty  position. The notification letter arrived a few days after Abel had died of pneumonia. He was 29.
 

Brocard's Problem:  N factorial + 1 = X squared.  This is true  for X = 5, 11, and 71, but that may be all.                Pierre Rene Jean Baptiste Henri Brocard: (1845 - 1922).  Bruce Berndt and William Galway used a computer in 2000 to show there are no other solutions up to N = one billion.

Primes, other than 2 or 3 are either of the form 6n + 1 or 6n - 1.     

Lychrel Numbers.  Most numbers become a palindrome by reversing their digits and adding repeatedly.  (349 + 943 = 1292, 1292 + 2921 = 4213, 4213 + 3124 = 7337 a palindrome.  Those that do not convert, are Lychrel Numbers. The name "Lychrel" was coined by Wade Van Landingham: a rough anagram of his girlfriend's name Cheryl.

Catalan's conjecture (occasionally now referred to as Mihăilescu's theorem)  was conjectured by the mathematician Eugene Charles Catalan  in 1844 and proven in 2002 by Preda Mihăilescu.

To understand the conjecture, notice that 23 and 32 (i.e. 8 and 9) are two powers of natural numbers, whose values 8 and 9 respectively are consecutive. The conjecture states that this is the only case of two consecutive powers. That is to say, that the only solution in the natural numbers of

xayb = 1

for x, a, y, b > 1 is x = 3, a = 2, y = 2, b = 3.

SQUARE ROOT  The square root of the number 81 is 9.  81 is the only number whose square root is the sum of its digits.

What did Pythagoras say when he was first confronted with the square root of 2?  "There has to be a rational explanation for this."

 PARTITIONS See http://en.wikipedia.org/wiki/Partition_(number_theory)

In number theory and combinatorics, a partition of a positive integer n, also called an integer partition, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. If order matters, the sum becomes a composition. For example, 4 can be partitioned in five distinct ways:

4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1

The order-dependent composition 1 + 3 is the same partition as 3 + 1, while 1 + 2 + 1 and 1 + 1 + 2 are the same partition as 2 + 1 + 1.

A summand in a partition is also called a part. The number of partitions of n is given by the partition function p(n). So p(4) = 5. The notation λ n means that λ is a partition of n.

Partitions can be graphically visualized with Young diagrams (boxes) or Ferrers diagrams (dots). They occur in a number of branches of mathematics and physics, including the study of symmetric polynomials, the symmetric group and in group representation theory in general.

Young diagrams associated to the partitions of the positive integers 1 through 8. They are arranged so that images under the reflection about the main diagonal of the square are conjugate partitions. (below)

File:Ferrer partitioning diagrams.svg

 GAMMA FUNCTION

The (complete) gamma function Gamma(n) is defined to be an extension of the factorial to complex and real number arguments. It is related to the factorial by  Gamma(n)=(n-1)!, a slightly unfortunate notation due to Legendre which is now universally used instead of Gauss's simpler Pi(n)=n!  There are no points z at which Gamma(z)=0.

The gamma function can be defined as a definite integral for R[z]>0(Euler's integral form) as

  Gamma(z)=int_0^1[ln(1/t)]^(z-1)dt. or int_0^inftyt^(z-1)e^(-t)dtor 2int_0^inftye^(-t^2)t^(2z-1)dt,

The complete gamma function Gamma(x) can be generalized to the upper incomplete gamma function Gamma(a,x)and lower incomplete gamma function gamma(a,x).

GammaReImAbs

Plots of the real and imaginary parts of Gamma(z)in the complex plane are illustrated above.

Below we see the gamma function along part of the real axis: 

GammaFunction

Integrating   Gamma(z)2int_0^inftye^(-t^2)t^(2z-1)dt, by parts for a real argument, it can be seen that Gamma(x)(x-1)Gamma(x-1).

If  x is an integer  Gamma(n) = (n-1)!, so the gamma function reduces to the factorial for a positive integer argument.

A beautiful relationship between Gamma(z)and the Riemann zeta function zeta(z)is given by:

 zeta(z)Gamma(z)=int_0^infty(u^(z-1))/(e^u-1)du   for R[z]>1

The gamma function can also be defined by an infinite product form:

 Gamma(z)=[ze^(gammaz)product_(r=1)^infty(1+z/r)e^(-z/r)]^(-1),  where gammais the Euler-Mascheroni constant

The Euler limit form is  Gamma(z)=1/zproduct_(n=1)^infty[(1+1/n)^z(1+z/n)^(-1)], so, Gamma(z) = lim_(n->infty)(n!)/((z)_(n+1))n^z

The reciprocal of  the gamma function 1/Gamma(z) is an entire function  expressed as

  1/(Gamma(z))=zexp[gammaz-sum_(k=2)^infty((-1)^kzeta(k)z^k)/k],  

where gammais the Euler-Mascheroni constant and zeta(z)is the Riemann zeta function 

 

 

 

 

 

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