A blog is that is all about mathematics and calculators, two of my passions in life.
Wednesday, December 31, 2014
Sunday, December 28, 2014
TI84+: Doppler Effect, Finding an Equation of a Line with 2 Points, Arc Length of f(x), Orbit around the Sun
Planet/Celestial Object

Mass (pounds)

Avg. Radius (miles)

Orbit (years)

Earth

1.317E25

92.954E6

1.000009092

Jupiter

4.186E27

483.682E6

11.86410888

Pluto

3.244E22

3.67005E9

248.0906781

Sunday, December 21, 2014
Happy December Solstice!
I want to wish everyone a Happy, Healthy, Safe, and Celebratory Solstice, and Happy Holidays (Christmas, Hanukkah, and all the holidays that are being celebrated worldwide).
I thank everyone who follows this blog, comments, and readers. Thank you for making this blog the success that it is.
I'll probably speak to you next week  until then, CHEERS!
Take care of each other.
Eddie
Tuesday, December 16, 2014
HP Prime and HP 50g: Using the LSQ Function to find the MoorePenrose Inverse of a Matrix
Sunday, December 14, 2014
Bingo: Number of Cards and Odds of Winning
A standard BINGO board has a 5 x 5 matrix. Each column is dedicated to a letter, mainly B, I, N, G, and O. 75 numbers are assigned to each number as follows:
B: 1 to 15
I: 16 to 30
N: 31 to 45
G: 46 to 60
O: 61 to 75
Each column has 5 of 15 possible numbers, except the N column, which only has 4. The N column has a Free Space.
How Many Different Bingo Cards Are There?
A bingo card has five rows, five columns , and two diagonals. The middle row, middle column, and the two diagonals contain the Free Space.
Each of columns have the following number of arrangements:
B Column: nPr(15,5) = 15!/(155)! = 360,360
I Column: 360,360 (see B column)
N Column: nPr(15,4) = 15!/(154)! = 32,760 (remember the Free Space!)
G Column: 360,360 (like the B and I columns)
O Column: 360,360 (like the B, I, and G columns)
Note: nPr is the Permutation function n!/(nr)!, sometimes labeled PERM or (n r) shown vertically. Simply put, permutation means arrangement: "How many ways can we arrange r out of n objects?
Since each of the permutations of B column can have each of the permutations of the I column; which in turn, each of those permutations can contain each permutation of the N column, and so on; the number of total number of bingo boards is:
360,360 * 360,360 * 32,760 * 360,360 * 360,360
= 360,360^4 * 32,760
= 552,446,474,061,128,648,601,600,000
≈ 5.52446 x 10^26
We can get the number of different cards by using rows as well.
1st Row: 15 * 15 * 15 * 15 * 15 = 15^5 =759,375 (15 numbers available per column)
2nd Row: 14 * 14 * 14 * 14 * 14 = 14^5 = 537,824 (14 numbers available per column)
3rd Row: 13 * 13 * 1 * 13 * 13 = 28,561 (don't forget about the Free space)
4th Row: 12 * 12 * 13 * 12 * 12 = 269,568
5th Row: 11 * 11 * 10 * 11 * 11 = 175,692
And the number of cards is:
759,375 * 537,824 * 28,561 * 269,568 * 175,692
= 552,446,474,061,128,648,601,600,000
≈ 5.52446 x 10^26
We arrived at the same destination.
What The Odds of Having a Winning Card in Bingo?
Even though there are 2,0711,126,800 ways to pick 5 numbers out of 75; thus making you very lucky to get a Bingo if the you win after just five draws.
Despite this, we look at the number of cards that are played in the game rather than the numbers themselves when answering this question.
So instead of having a long calculation, which I once thought was the case, the odds of having the winning card boils down to this:
1/(number of cards in play)
This is true if you are playing BINGO by a line or blackout bingo.
Don't forget to you yell BINGO! when you win. :)
Have a great day! Now I am off to errands and maybe wrap some presents.
Eddie
This blog is property of Edward Shore. 2014
P.S. I am going to make it one of my resolutions to check on the comments timely. I appreciate and thank everyone who leaves a comment.  Eddie (12/15/14)
Thursday, December 11, 2014
Annie Jump Cannon
Google dedicated their home page to Annie Jump Cannon today. The Henry Draper Award winner classified and observed 500,000 stars, while breaking ground for women in science.
She also has a crater named after her on our Moon, and her stellar classification is standard (as determined by the IAU).
http://en.m.wikipedia.org/wiki/Annie_Jump_Cannon
Saturday, December 6, 2014
Some Science Stocking Stuffings
Einstein's Time Dilation
Recall that equation for time dilation is:
Δt = Δt0 / √(1  u^2/c^2)
Where:
Δt = time observed by the person standing still
Δt0 = time observed by the traveler (in the observer's frame of reference)
u = speed of the moving object , which contains the traveler
c = speed of light in a vacuum = 299,792,458 m/s
In short, the traveler will note observe and note that time Δt0 has passed, which the person standing still observes that time Δt has passed. The closer someone goes to the speed of light, that less person she/he experiences. You will need to go super fast. Driving at 65 mph (29.0756 m/s) on the freeway won't cut it and here's why:
For 1 unit of time to pass for the person driving 65 mph (Δt0 = 1), the change of time for the observer (Δt) is:
Δt = 1/√(1  29.0756^2/299,792,458^2) ≈ 1.0000000000000047 (that is fourteen zeroes between the decimal point and the 4) (The Wolfram Alpha app was used for this calculation).
Virtually the same time passes.
Same deal regarding observing an airplane flying at 600 mph (268.224 m/s). For the person watching the plane, for a plane's passenger, pilot, or cocktail and peanut server to observe one unit of time, us watchers observe 1.0000000000004 (twelve zeroes) units of time.
If you want to be on an object where the people observing you experience twice the time (Δt = 2) you do (Δt0 = 1), you will need to travel (c*√3)/2 or 259,627,884.491 m/s (580,771,037.247 mph). That is 86% the speed of light!
Archer's Paradox (E.J. Rendtroff, 1913)
In archery, archers aim their arrows slightly to the side, instead of directly at the target. On the surface, that seems crazy. This is where the Archer's Paradox comes into effect.
Basically, when the arrow is shot, it travels in an "S" curve. Drawing a string makes the arrow bend such that he tip is pointed away from the target. As the arrow is fired and the string returns to the bow, the arrow bends the other way, turning the arrow back to the target.
Other factors to making accurate shots include the stiffness of the arrow, brace height, and wind conditions. I found this video by Billgsgate helpful:
http://youtu.be/bNlx6MBlymw
Galactic Coordinates
Galactic Coordinates are spherical coordinates that are set such as:
* The center is our sun.
* The coordinates are (l°, b°), where l is the longitude (0° to 360°, flat angle) and the latitude (90° to 90°, height).
* Pointing "due east", l = 0° and b = 0° is pointing towards the center of the Milky Way. "Due west", l = 180° and b = 0° points away from the center of the Galaxy.
Using Wolfram Alpha ( http://m.wolframalpha.com ) and an online coordinator converter ( http://ned.ipac.caltech.edu/forms/calculator.html ), here are the 30° longitude markers when latitude is 0° (b = 0°):
* 0° points towards the constellation Sagittarius (as it should, it having the Milky Way center)
* 30° points towards Aquila the Eagle
* 60° points towards the Vulpecula the Fox
* 90° points towards Cygnus the Swan
* 120° points towards Cassiopeia the Vain Queen
* 150° points towards Perseus the Hero
* 180° points towards Auriga the Charioteer
* 210° points towards Monoceros the Unicorn
* 240° points towards Puppis (a ship's poop deck)
* 270° points towards Vela (a ship's sails)
* 300° points towards Crux, The Southern Cross
* 330° points towards Norma (a carpenter's square (measuring tool))
Spherical Lenses (see figure 2 below)
Variables:
P = place of object (with distance s)
P' = place of image (with distance s')
C = center of curvature
V = vertex of the lens
General relations:
tan α = h/(s  δ)
tan β = h/(s'  δ)
tan Φ = h/(R  δ)
However, if α < π/2 ( α < 45° ), we are dealing with paraxial rays:
1/S + 1/S' = 2/R
Surprisingly, the object is far from the vertex.
Forgive me if I repeat things. Instead of going out to clobber everyone last Black Friday for that extra 10% off, I stayed home and cracked open some books that I have been meaning to look at for months. I hope you find this enjoyable and insightful.
The best always,
Eddie
P.S. I was thinking about adding a section (or blog entry) about the age of Aquarius and some of the head scratchers I have about it. I am not sure if this subject would be appropriate for this blog. If you have any thoughts about this, or anything else, go and ahead and comment!  E.S.
This blog is property of Edward Shore. 2014
Wednesday, November 26, 2014
Astronomy: Distance and Midpoint Between Two Celestial Objects
Astronomy: Distance and Midpoint Between Two Celestial Objects
Stars and celestial objects have the equatorial coordinates ( α, δ, R ) where:
α = right ascension in hours, minutes, and seconds
δ = declination in degrees, minutes, and seconds
R = distance to the celestial about (in light years, astronomical units, etc)
Convert α and δ to decimal degrees:
α = (hours + minutes/60 + seconds/3600) * 15
δ = sign(δ) * ( degrees + minutes/60 + seconds/3600 )
However, calculating distance between two points in space requires that we have cartesian coordinates (x, y, z). The conversion formulas to go from equatorial to cartesian coordinates are:
x = R * cos δ * sin α
y = R * cos δ * cos α
z = R * sin δ
For completeness, here are the conversion formulas to from cartesian to equatorial coordinates:
R = √(x^2 + y^2 + z^2)
δ = asin(z/R)
α = atan(y/x) (pay attention to the quadrant!). Alternatively: α = arg(x + i*y)
The distance formula between two points in space is:
D = √( (x2  x1)^2 + (y2  y1)^2 + (z2  z1)^2 )
Source: http://en.m.wikipedia.org/wiki/Equatorial_coordinate_system
Example
Find the distance between Sagittarius A* (center of the Milky Way Galaxy) and the Andromeda Galaxy (M31).
Equatorial Coordinates:
From Wolfram Alpha: ( http://m.wolframalpha.com )
Sagittarius A*:
R = 24,824 light years
α = 17h45m40s = 266.42°
δ = 29°0'28" = 29.008°
Andromeda Galaxy:
R = 2,571,000 light years
α = 42m40s = 10.68°
δ = 41°16'8" = 41.269°
Converting to the cartesian system:
Sagittarius A*:
x = 21,667.514038
y = 1,355.611275
z = 12,037.945401
Andromeda Galaxy:
x = 358,122.65578
y = 1,898,943.67736
z = 1,695,818.99789
Calculation:
D = √( (21,667.514038  358,122.65578)^2 + (1,355.611275  1,898,943.67736)^2
+ (12,037.945401  1,695,818.99789)^2 ) ≈ 2,583,051.16059
The distance between Sagittarius A* and the Andromeda Galaxy is about 2.583 million light years.
Happy Thanksgiving, or Day of Thanks, wherever you are. Please be safe and use good judgement. We need all the positive we can get on Earth, and in the cosmos. Cheers!
Eddie
This blog is property of Edward Shore. 2014
Tuesday, November 25, 2014
Complex Analysis: Line Integral
Contour Integrals
Let the complex function f(z) be defined as f(z) = u(x,y) + i*v(x,y)
Using the definition of the integral:
∫ f(z) dz = lim n → ∞ [ Σ f(z_k) * Δz_k from k = 1 to n ]
= lim n → ∞ [ Σ ( u(x_k, y_k) + i*v(x_k, y_k) ) * ( Δx_k + i*Δy_k ) from k = 1 to n ]
= lim n → ∞ Σ [ (u(x_k, y_k) * Δx_k  v(x_k, y_k) * Δy_k) + i*(v(x_k, y_k) * Δx_k
+ u(x_k, y_k) * Δy_k) ]
= ( ∫ u(x,y) dx  ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )
In summary:
∫ f(z) dz = ( ∫ u(x,y) dx  ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )
What is needed:
1. A contour curve f = y(x). This is where you get your end points.
2. We need f(z) to integrate. Separate f(z) into its parametric parts.
Let's look at a couple examples.
Examples
1. Contour: y = x + 1 from (3,4) to (4,5). Integrate ∫ z^2 dz.
Since y = x + 1, x = y  1
z^2 = (x + i*y)^2 = (x^2  y^2) + i*(2*x*y)
u = x^2  y^2
v = 2*x*y
u = x^2  y^2
u = x^2  (x + 1)^2
u = 2*x  1
∫ u dx from 3 to 4 = 8
v = 2 * x * y
v = 2 * (y  1) * y
∫ v dy from 4 to 5 = 95/3
u = (y  1)^2  y^2
u = 2 * y + 1
∫ u dy from 4 to 5 = 8
v = 2 * x * y
v = 2 * x * (x + 1)
∫ v dx from 3 to 4 = 95/3
∫ z^2 dz from (3 + 4*i) to (4 + 5*i) = (8  95/3) + i*(8 + 95/3) ≈ 39.666667 + 23.666667*i
2. Contour: y = x^2  1 from (0, 1) to (2, 1). Integrate ∫ z^2 + 1 dz.
y = x^2  1, x = √(y + 1)
z^2 + 1 = (x + i*y)^2 + 1 = x^2 + 2*i*x*y  y^2 + 1
u = x^2  y^2 + 1
v = 2 * x * y
u = x^2  y^2 + 1
u = x^2  (x^2  1)^2 + 1
∫ u dx from 0 to 2 = 8/5
v = 2 * x * y
v = 2 * √(y + 1) * y
∫ v dy from 1 to 1 = 8/15 * √2
u = x^2  y^2 + 1
u = y + 1  y^2 + 1
u = y^2 + y + 2
∫ u dy from 1 to 1 = 10/3
v = 2 * x * y
v = 2 * x * (x^2  1)
∫ v dx from 0 to 2 = 4
∫ z^2 + 1 dz from i to 2+i = (8/5  8/15 * √2) + i*(4 + 10/3) ≈ 0.845753 + 7.333333*i
This is the basics of line integrals.
Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. AddisonWesley Publishing Company. 1994.
Take care, Eddie
This blog is property of Edward Shore. 2014
Saturday, November 22, 2014
Complex Analysis: CauchyReimann Equations
Continuing my blogging adventures at the Last Drop Cafe in Claremont, CA:
CauchyReimann Equations
Let the complexvalued function f(z), where z = x+i*y, be defined in the parametric form:
f(z) = u(x,y) + i*v(x,y) where u(x,y) = Re(f(z)) and v(x,y) = Im(f(z)).
Then by the definition of the derivative:
f(z) = lim x0 → x [(f(z) + x0)  f(z))/(x  x0)]
= lim x0 → x [(u(x +x0, y)  u(x, y) + i*v(x + x0, y)  i*v(x, y))/(x  x0)]
= du/dx + i*dv/dx (I)
Also:
f(z) = lim y0 → y [(f(z + y0)  f(z))/(i*y  i*y0)]
= lim y0 → y [(u(x, y + y0)  u(x, y) + i*v(x, y + y0)  i*v(x, y))/(i*(y  y0))]
= i*(lim y0 → y [(u(x, y + y0)  u(x, y) + i*v(x, y + y0)  i*v(x, y))/(y  y0)])
= i*du/dy + dv/dy (II)
Taking the real and imaginary parts of (I) and (II):
du/dx = dv/dy and du/dy = dv/dx
The CauchyRiemann equations can be used to determine whether f(z) is differentiable and if so, where.
Two Quick Examples
1. z^2 = (x^2  y^2) + 2*i*x*y
u=x^2  y^2
v=2*i*x*y
du/dx = 2*x, dv/dy = 2*x
du/dy = 2*y, dv/dx = 2*y (note the negative sign in front of du/dy!)
Since the CauchyRiemann equations hold, and without restriction, then z^2 is differentiable for all z. And:
df/dz = 2*x + 2*i*y = 2*(x+i*y) = 2*z
2. e^z = e^(x+i*y) = e^x*cos y + i*e^x*sin y
u=e^x*cos y
v=e^x*sin y
du/dx = e^x*cos y, dv/dy = e^x*cos y
du/dy = e^x*sin y, dv/dx = e^x*sin y
Since the CauchyRiemann equations hold, and without restriction, then e^z is differentiable for all z. And:
df/dx = e^x*cos y + i*e^x*sin y = e^z
I used (I), but using (II) will garner the same result.
That is the CauchyRiemann equations in a nutshell!
Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. AddisonWesley Publishing Company. 1994.
As always, have a great day! Take care. Eddie
This blog is property of Edward Shore. 2014
Complex Analysis: The Conjugate, the Modulus, and its Properties
Blogging today from Last Drop Cafè in Claremont, CA. I think I found a new favorite drink: mint mocha made with soy milk.
Here are some basics of the conjugate and modulus of complex numbers.
Let z = x+i*y, with i=√1
Conjugate
Usually labeled "zbar" (z with a line over it), the conjugate is also labeled conj(z) and z*.
conj(z) = x  i*y
Modulus or Absolute Value
Not surprisingly, the modulus, also called the absolute value of the complex number z is defined as:
z = √(x^2 + y^2)
Properties
Let's explore some properties of the conjugate and modulus.
z + conj(z) = (x + i*y) + (x  i*y) = 2*x
z  conj(z) = (x + i*y)  (x  i*y) = 2*i*y
(conj(z))^2 = (x  i*y)^2 = x^2  2*i*x*y + (i*y)^2 = x^2  y^2  2*i*x*y
conj(z)^2 + z^2 = (x  i*y)^2 + (x + i*y)^2 = x^2  2*i*x*y  y^2 + x^2 + 2*i*x*y  y^2
= 2*(x^2  y^2)
conj(z) * z = (x  i*y) * (x + i*y) = x^2 + i*x*y  i*x*y  i^2*y^2 = x^2 + y^2 = z^2
which easily leads to: z = √(z * conj(z)) and
z1 * z2 = √(z1 * conj(z1) * z2 * conj(z2)) = √(z1 * conj(z1)) * √(z2 * conj(z2)) = z1 * z2
and:
z1/z2 = √((z1 * conj(z1))/(z2 * conj(z2))) = √((z1 * conj(z1))/√((z2 * conj(z2)) = z1/z2
Source: Wuncsh, David A. Complex Numbers with Applications. 2nd Edition. AddisonWesley Publishing Company. Reading, MA. 1994
Wednesday, November 12, 2014
Casio fx9750gII, fx9860gII, Prizm: Prime and Complex Number Prime (Gaussian Prime)
“N”?>N
Int N>N
For 2>K To S
If Frac (N÷K)=0
Then
“NOT PRIME” /rtri
Stop
IfEnd
Next
“PRIME”
36 is NOT PRIME
a+bi
“Z”?>Z
(Abs Z)^2>T
If ReP Z=0 or ImP Z=0
Then
Goto 1
IfEnd
For 2>K To Abs Z
If Frac(T÷K)=0
Then Goto 2
IfEnd
Next
Goto 3
Lbl 1
If MOD(Abs Z, 4)≠3
Then Goto 2
IfEnd
Next
Goto 3
Lbl 2
“NOT PRIME” /rtri
Stop
Lbl 3
“PRIME”
7 is (complex) prime
5+4i is (complex) prime
Tuesday, November 11, 2014
Darts: The Probability of Getting the Best Score
Area of a Circular Ring
= (π * (R3^2 – R2^2 + R5^2 – R6^2)/(π * R^2)
= (R3^2 – R2^2 + R5^2 – R6^2)/R^2
= (R6^2 – R5^2)/R^2
= (R4^2 – R3^2)/R^2
= (π * R2^2)/(π * R^2)
= R2^2/R^2
Scoring 60
RCL 40 Book On Sale
RCL 40 Book On Sale The fantastic book, RCL 40: Recollection, Reinvention, & HP Calculators , edited by W.A.C. MierJedrezejowic, Ph. D,...