Some of My Favorite Reads

Astronomy Magazine. 

This is the one magazine that I get every month on a regular basis.  Pictured is the June 2016 issue.

Star Names: Their Lore and Meaning - Richard Hinckley Allen (ISBN 978-0-486-21079-7, 2014/1963)

Quadrivium - various authors, published by Wooden Books and Bloomsbury  (ISBN 978-0-8027-7813-0, 2010)

Applied Mathematical Physics with Programmable Pocket Calculators - Robert M. Eisberg  (ISBN 0-07-019109-3, 1976)

Math for Mystics - Renna Shesso (ISBN 978-1-57863-383-8, 2007)

Power-Up: Unlocking the Hidden Mathematics in Video Games - Matthew Lane (ISBN 9780691161518, 2017)

The Jazz of Physics: The Secret Link Between Music and the Structure of the Universe - Stephon Alexander (ISBN 9780465034994, 2016)

The Magic of Math: Solving for x and Figuring Out Why - Arthur Benjamin (ISBN 978-0-465-05472-5, 2015)

(not pictured)

The Milky Way: An Insider's Guide - William H. Waller (ISBN 978-0-691-12224-3, 2013)

Eddie

This blog is property of Edward Shore, 2017.  

Pascal’s Triangle: Polynomials and TI-84 Plus

Pascal’s Triangle:  Polynomials and TI-84 Plus

Pascal's Triangle

Powers of 2

Pascal’s Triangle holds a great number of properties.   For instance, the sum of each row (R) is a power of two (2^R).  The top row is referred to as row 0.  Hence:

1 = 2^0

1 + 1 = 2 = 2^1

1 + 2 + 1 = 4 = 2^2

1 + 3 + 3 + 1 = 8 = 2^3

1 + 4 + 6 + 4 + 1 = 16 = 2^4

1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^5

1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^6

and so on…

Binomial Expansion

Take a look what happens when you expand the binomial (x + y)^R:

(x + y)^0 = 1

(x + y)^1 = 1*x + 1*y

(x + y)^2 = 1*x^2 + 2*x*y + 1*y^2

(x + y)^3 = 1*x^3 + 3*x^2*y + 3*x*y^2 + 1*y^3

(x + y)^4 = 1*x^4 + 4*x^3*y + 6*x^2*y^2 + 4*x*y^3 + 1*y^4

(x + y)^5 = 1*x^5 + 5*x^4*y + 10*x^3*y^2 + 10*x^2*y^3 + 5*x*y^4 + 1*y^5

(x + y)^6 = 1*x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5 + 1*y^6

and so on…

Notice the coefficients (in blue)?  They represent rows of the Pascal’s Triangle. 

Combinatorics

The formula for find the number of combinations of N items from R is:

COMB(R, N) = R nCr N = R! / ((R – N)! * N!)

If you R stand for a row of the Pascal’s Triangle and N stand for an entry (starting from 0), you can get R nCr N from the triangle. 

For instance, for the 4^th row (R = 4 with entries 1, 4, 6, 4, 1), 2^nd entry (N = 2, with left most entry designated as 0), Pascal’s Triangle will state that 4 nCr 3 = 4! / (2! * 2!) = 6.

For the 5^th Row (R = 5):

N = 0, 5 nCr 0 = 1

N = 1, 5 nCr 1 = 5

N = 2, 5 nCr 2 = 10

N = 3, 5 nCr 3 = 10

N = 4, 5 nCr 4 = 5

N = 5, 5 nCr 5 = 1

Sierpinksi Triangle

One of the books I got for Christmas is the book “The Magic of Math: Solving for x and Figuring Out Why” written by Arthur Benjamin.  The book is well written and if you want a good read I recommend this book.  It has something for everyone. One of things I learned from Benjamin’s book is that if you mark all the odd numbers, and you take many rows, you get the famous fractal the Sierpinski Triangle.

Take a look at the diagram below:

Making the Sierpinski Triangle

TI-84 Plus:  Generating a Row of Pascal’s Triangle

A short program to generate a row of Pascal’s Triangle.  The result is stored in list L6.  The first entry is the 0^th entry.

Input "ROW:",R

R+1→dim(L₆)

For(K,0,R)

R nCr K→L₆(K+1)

End

Disp "L₆:"

Pause L₆

I think this is self-explanatory.

Eddie

HAPPY NEW YEAR!

Source:

Benjamin, Arthur.  The Magic of Math: Solving for x and Figuring Our Why.  Basic Books:  New York.  2015

This blog is property of Edward Shore.   2015.