Can't wait for Nashville!

I am going to a calculator conference in Nashville, TN next weekend. The conference is for Hewlett Packard calculator enthusiasts. Can't wait!

Eddie

Cartesian Coordinates to Pixel (Screen) Coordinates

Converting Cartesian Coordinates to Screen Coordinates

When working with computer or calculator graphics, sometimes we have to work with screen coordinates. The screen coordinate system has the following:

1. The upper corner pixel (point) is (0,0).
2. The lower corner pixel is (A, B).
3. The x axis increases in the right direction.
4. The y axis increases in the down direction, the opposite direction of the Cartesian plane.

In order to to convert a point (x, y) in the Cartesian coordinates to point (xp, yp) in Screen coordinates, first observe the following:

1. Picture the plane you are working with as a screen. If you are working with a calculator or computer, this is fairly easy.
2. Let Xmin be the least-valued x of the screen (left edge) and Xmax be the most-valued x (right edge). Similarly, let Ymin be the least-valued y of the screen (bottom) and Ymax be the most-valued y of the screen (top).

You can check with the window settings on a graphing calculator to verify Xmax, Xmin, Ymax, and Ymin.

For the Hewlett Packard 28 series, 48 series and 50g, the variables XRNG and YRNG list the screen's dimensions, {Xmin, Xmax} for XRNG and {Ymin, Ymax} for YRNG.

To transform Cartesian coordinates to screen coordinates, we can use transformation matrices for scaling and translation. The general form

show the transformation from coordinates (x, y) to (x', y') where:

s_x = scale of the new x axis. If s_x < 0, the x point is reflected with respective to the x axis.

s_y = scale of the new y axis. If s_y < 0, the y point is reflected with respected to the y axis. In going from Cartesian coordinates to screen coordinates, s_y < 0.

x_s = Is the translation (shift to the new center) in the x-direction. If x_s > 0, the new center is to the right of the original center. Similarly, if x_s < 0, the new center is to the left of the original center.

y_s = Is the translation (shift to the new center) in the y-direction. If y_s > 0, the new center is to the above the original center. Similarly, if y_s < 0, the new center is to the below the original center.

For the transformation from Cartesian coordinates to screen coordinates:

s_x = A / (Xmax - Xmin)

s_y = -B / (Ymax - Ymin)

Since the new center will be at (Xmin, Ymax):

x_s = -Xmin

y_s = -Ymax

Therefore by matrix multiplication, with x' = xp and y' = yp:

xp = (x - Xmin) * A / (Xmax - Xmin)

yp = (y - Ymax) * -B / (Ymax - Ymin)

On the HP 48 series and 50g, the translation from Cartesian to screen coordinates can be accomplished with the C→PX function.

Example:

Using the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, transform the Cartesian coordinate (1, 0) to screen coordinates. The screen has pixel size 130 × 79.

In this case, A = 130 and B = 79.

Then:

xp = 130 / (5 - (-5)) * (1 - (-5)) = 78
yp = -79 / (4 - (-4)) * (0 - 4) = 39.5

The screen coordinate of (1, 0) is {78, 39.5}.

HP 50g Programs

DPIX

This program figures the distance in pixels between two points stated in Cartesian coordinates.

Input:
2: coord-y
1: coord-x

Output:
1: Distance

Program:
<< C→PX SWAP C→PX - DUP * ΣLIST B→R √ R→B >>

Example:

Using the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, find the pixel distance between (0,1) and (-1,-2)? (HP 50g pixel dimension 130 × 79)

2: (0,1)
1: (-1,-2)
DPIX

Answer in Decimal mode: #32d. (Approxmiately 32)

ARCEZ

This draws the arc using Cartesian coordinates and the radius in Cartesian scale (the way one would think if drawing on a Cartesian plane). This program does all the necessary transformation for you. You can use it in a program (requiring a PICTURE command later in a program) or alone (press the left arrow button to show the graph screen).

Input:
4: Cartesian Coordinates (x,y)
3: Radius in Cartesian form (the way you think)
2: Beginning Angle
1: Ending Angle

<< → P R A B
<< P DUP R +
C→PX SWAP C→PX - DUP * ΣLIST B→R √ R→B
P C→PX SWAP A →NUM B →NUM ARC >> >>

Example:

Using the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, draw a circle with radius 1.5 and center (1,0). (HP 50g pixel dimension 130 × 79)

In Radians Mode:
4: (1,0)
3: 1.5
2: 0
1: 2 * π
ARCEZ

Pressing the left arrow button will show the picture below (with no functions selected and additional drawing):

This concludes today's blog entry. As always, thank you for all your comments and support. Have a great day!


This blog is property of Edward Shore. © 2012

September 11

Taking some time from my math blog to speak about today.

This is a day we in the United States will never forget. Blessings to the families who suffered the losses on that horrible day in 2001.

May this world find peace soon.

Time Value of Money Tips and Tricks (Real Estate)

Tips and Tricks - Time Value of Money

Here is a way to solve the following financial problems with TVM (Time Value of Money) functions:

* Finding the balance of a loan
* Finding the total amount of interest paid
* Finding the APR
* Finding the maximum loan amount based on the 28/36 ratio

Conventions and Assumptions

Variables:
N = number of months
I/YR = annual interest
i = periodical interest. This is the interest rate divided by 12. This key is present on some financial calculators instead of the I/YR key. The most famous example is the HP-12.
PV = present value.
PMT = periodic payment
FV = future value

We are assuming the following;
1. Payments are made on a monthly basis
2. The payments are due at the end of each month.
3. P/Y (payments are year) and if applicable, C/Y (compounding periods per year), are set to 12.
4. If a calculator or app is used, the normal cash flow convention applies. In our examples today, we will use the borrower's point of view. PV is positive (loan is "received"), and PMT and FV (paid out).

Let's begin!

Finding the Balance after M months.

Steps:
1. Establish the loan information. You will need, interest rate, loan amount, and payment.
2. Enter M for N and solve for FV.

Example: Find the balance on a 30-year, 5%, $214,000 loan after 5 years (60 payments).

Step 1: Establish the loan information. At this time, payment is not known, so let's solve for it first.

N = 360
I/YR = 5
PV = 214,000
FV = 0
Solve for PMT. (-1,148.80)

Step 2: Find the balance after 60 months.

N = 60
I/YR, PV, and PMT remain unchanged.
Solve for FV. (-196,513.49)

After 60 months, the balance is $196,513.40.

Finding the total amount of interest paid

Steps - Interest paid after the entire loan is paid:
1. Establish the loan information: term, interest rate, payment, and loan amount.
2. The interest paid is found by the formula: N × PMT + PV

A suggested keystroke sequence: RCL N × RCL PMT + RCL PV
(Remember in loans PMT is negative, PV is positive from the point of view of the borrower.)

Example: A company finances a machine costing $5,412.50 which is financed for 3 years at 6%. The loan is carried full term. What is the total interest paid?

Step 1: Establish loan information.
N = 36
I/YR = 6
PV = 5412.50
FV = 0
Solve for PMT (-164.66)

Step 2: Find the total interest paid over the life of the loan.
PMT × N + PV = -164.66 × 36 + 5412.50 = -515.21

The company will pay a total of $515.21 over the life of the loan.

Finding the APR

The APR, known as the annual percentage rate, is the true interest rate of the mortgage when fees, insurance, and closing costs are considered. The APR of the loan will be higher than the loan's stated rate. By law (U.S.), lending institutions must explicitly state the loan's APR.

Steps:
1. Figure out the total payment when factoring in the loan fees, insurance, and closing costs.

N = number of payments
I/YR = original rate
PV = loan amount plus loan fees plus insurance plus closing costs
FV = 0
Calculate PMT

2. Figure APR (interest rate, I/Y) with the loan amount only set as PV.

N = N from Step 1
PV = loan amount
FV = FV from Step 1
PMT = PMT from Step 1
Calculate I/Y

Example: A couple is purchasing a home costing $300,000. The couple put 20% and will finance the balance. The loan is 30 years, financed at the rate of 4.25%. The loan charges 1 point (1% of the price) in fees and $1,500 in mortgage insurance.

$300,000 less the 20% down payment is $240,000. This will be treated as the loan amount.

Step 1: Payment when factoring the fees
PV = (300,000 - 20%) + (300,000 × 1%) + 1,500.00
= 240,000 + 3,000 + 1,500 = 244,500
N = 360
I/Y = 4.25
FV = 0
Calculate PMT (-1,202.79)

The payment is $1,202.79.

Step 2: Finding the APR
PV = 240,000
Calculate I/Y (4.41)

The APR is 4.41%.

Finding the maximum loan amount based on the 28/36 ratio.

This is how a way to find the qualifying amount for a potential borrower.

The 28/36 ratio refers to ability to pay monthly bills. On the front end, your monthly mortgage payments, which includes property taxes and insurance must be no greater than 28% of your monthly gross income. On the back end, total monthly debt payments, including the mortgage, child support, credit card, student loans, and automobile loans, must be no greater than 36%. If both ratios are met, your monthly payment is deemed qualified - you can afford it.


Using this information, we can find the maximum allowable loan amount.

Step 1: Take the lesser of:

X = .36 × monthly gross income - (monthly property insurance + monthly property taxes + other monthly debt payments)

or

X = .28 × monthly gross income - (monthly property insurance + monthly property taxes)

Step 2: Set up your TVM variables as such:
N = number of payments
I/YR = stated rate of the loan
FV = 0
PMT = -X (note the negative sign)
Solve for PV. This is your maximum loan amount.

Example: A couple wants to know the maximum price of a house they can afford in Math City. The following data is presented:
1. The couple's gross income each month is $7,600.
2. The average property taxes and mortgage in Math City is approximately $314 a month.
3. The couple is currently paying $500 in credit card debt. In addition, they have financed one of their cars with a monthly payment of $355 a month.
4. The Square Root Bank, the mortgage company the couple is currently working with, is offering a 30 year loan at 5.12%.

Step 1:
At the 36% limit, X = .36 × 7600 - (314 + 500 + 355) = 1,567
At the 28% limit, X = .28 × 7600 - 314 = 1,814
Let X = 1,567.

Step 2:
N = 360
I/YR = 5.12
FV = 0
PMT = -1567
Solve for PV (287,956.53)

The maximum loan the couple can afford is $287,956.53.

I hope you find this helpful. Until next time,

Eddie


This blog is property of Edward Shore. © 2012

The Largest Shape That Can Fit

We are finding the dimensions of the largest object that can fit in another geometric shape.

Largest Circle in a Square

Known: The radius, which is represented by line segment

AN

.
Goal: Find the dimension of one side of the square. Let the measure of the line segment AC be represent the measure of the side.

Let:

AB

represent the line segment which length is the diameter of the circle.

By the Pythagorean Theorem:

AC^2 + CB^2 = AB^2

Since AC = CB and AB = 2 AN,

2 AC^2 = (2 AN)^2
AC^2 = 2 AN^2
AC = AN √2

Take the positive root, as the negative root has negligible meaning.

The measure of the square's side is the radius times √2. (s = r √2)

Largest Circle in a Square

Known: Radius of the circle, R.
Goal: Find the length of a side of the square, S.

Observe that the largest circle will fit when it's center is in the square's center. As the picture above indicates, some of the edge of the circle touches the square.

Therefore, S = 2 R

Largest Triangle in a Square

Known: A square with each side measuring S.
Goal: The dimensions of the largest triangle.

If we want the largest triangle to fit in the square, it would make sense to make one of the sides to be of length S. It would also make sense to have the vertex of the triangle's two other sides at the other side of the square. This can be accomplished by a 45°-45°-90° triangle, not an equilateral triangle. (See the diagram above)

To solve for B, use trigonometry.

Observe that sin 45° = cos 45° = √2 / 2

Then:

√2 / 2 = B / S
B = (S √2) / 2

So our triangle would have the dimensions S × (S √2) / 2 × (S √2) / 2

Largest Square in an Equilateral Triangle

Known: An equilateral triangle with sides of length B.
Goal: The length of the inscribed square, S.

Observe, by inspection, that the inscribed square is the largest when one of it's sides is touching the triangle's base. This allows for the largest S possible.

The measurement of the base is (B - S)/2 + S + (B - S)/2 = B.

We can find S by working with one of the two "mini right triangles" formed at the bottom of the triangle. A diagram of one of the "mini right triangles" is shown below:

Using trigonometry:

tan 60° = S / ((B - S) / 2)
√3 = (2S) / (B - S)
B √3 - S √3 = 2S

S = (B √3) / (2 + √3)

which is the desired result.

Largest Circle in an Equilateral Triangle

I admit I have some difficulty with this one. In order for the circle to fit, the following have to be true:

1. The area of the triangle must be greater than the area of the circle.
2. The length of the triangle's side must be greater than twice the circle's radius.
3. The height of the triangle must be greater than twice the circle's radius.

Even obeying these rules did not guarantee an answer.

For me this is an open question. This may be one where numerical methods must be used. Any ideas, as always, are welcome.


Until then,

Eddie

This blog is property of Edward Shore. © 2012

Labor Day: Reference Angle, Construction Calculators, Why a Negative Times a Negative is Positive

Reference Angle

Each angle can be represented by a reference angle. The reference angle is determined by finding the smallest angle between the terminal ("ending side") side of the angle and the x-axis.

The reference angle is always positive and has the range 0° ≤ θ ≤ 90°.

Examples:
The reference angle of 135° is 45°.
The reference angle of 220° is 30°.
The reference angle of 329° is 31°.

This page by www.analyzemath.com gives examples and a method of finding the reference angle.

Math warehouse also gives a good explanation of reference angle.

Another way to get a reference angle is to use a scientific calculator. For most, if not all scientific calculators and math programs, the arcsin function (usually labeled sin^-1) gives the angle between -90° and 90°. On a scientific calculator, we can use the following keystroke sequence:

Angle [SIN] [SIN^-1] [ABS]

Or the command abs(asin(sin(x)). A graph of the reference angle is shown below. The angle in the graph is measure of radians.

Construction Calculators

I usually accompany my dad to trips to Home Depot and Lowes whenever he gets ideas for home improvement. The latest one was to build a fence. My dad and I often talk about construction and home improvement plans.

Calculated Industries is a company that makes specialized calculators that cover construction, real estate, hot rods, electronics, and even cooking and quilting. While I dare say the math can be done with a scientific and finance calculator, these calculators are good because (1) the keys are specialized and (2) they focus on a specialized area.

I managed to buy all three of calculators the stores have, which two of them are pictured below. The Materials Estimator is just what it says, it's main focus is estimate the materials needed, such as of tiles needed, gallons of paint, or my favorite, the fence materials. The Construction Master 5 focuses on roofs, stairs, and circular structures. My favorite feature of the construction calculators is the ability to calculate dimensional math in yards, feet, and inches with fractions (precision can be set to 1/64).

Whether a home improvement project is actually going to happen I am not sure (cost is a factor), but at the least these calculators are nice additions to my collection.

Why a Negative Times a Negative is a Positive

We all heard the rule "A negative number times a negative number is a positive number". This line reminds me of the 1988 movie "Stand and Deliver". And yes, I had to look up the title because I remember the wrong name, but I remember Edward James Olmos as a teacher getting his classroom to chant the rule.

Confession: I am not a movie person, at all This may be strange considering I was born in Los Angeles blocks away from Hollywood. Sitting a dark room for two or more hours looking at a screen just doesn't do anything for me.

Anyway, I feel like for students to understand and fully grasp this concept, a nice short, concise explanation is needed.

Some resources tend to use an example such as:

-4 × -5
= -1 × 4 × -1 × 5
Using the Commutative Property for multiplication:
= -1 × -1 × 4 × 5
= -1 × -1 × 20
= 20.

Yet I feel this explanation is incomplete. There is no explanation of why -1 × -1 = 1.

I think a better way to show that a negative times a negative is positive is to consider the equation:

S = A × B + -A × B + -A × -B

Where S is the sum of the right hand expression, and A and B are positive numbers. (A > 0, B > 0).

We can find S in two ways. First, factor out a B in the first and second term of the right hand side:

S = A × B + -A × B + -A × -B
= (A + -A) × B + -A × -B
= 0 + -A × -B
= -A × -B

A second way is to factor -A on the second and third term on the right hand side:

S = A × B + -A × B + -A × -B
= A × B + -A × (B + -B)
= A × B + 0
= A × B

Note the following:
A > 0 and B > 0.
-A < 0 and -B < 0.
And A × B > 0, and S > 0.

Also, note S = A × B = -A × -B, and since S > 0, this implies -A × -B > 0.

There are many explanations that can be found on a web search either on a search engine or on YouTube. This one is my favorite because it is the most logical and it introduces logic and mathematical thinking to students.

That concludes this entry for today. Have a great Labor Day and I will talk to you soon!

Eddie

This blog is property of Edward Shore. © 2012

Obtaining More Digits with the HP 50g

Intro:

I was reading the Fun With Num3ers blog by Ben Vitalis. The August 18, 2012 entry talked about certain fractions produced various patters.

Link to Ben's 10/81 post here. (8/18/12)

Examples include:

10. / 81 = 0.123456790123456790...

10 / 81 - ((3340 / 3267 * 10^-9)) = 0.12345678910111213141516... (and the number continues a counting fashion)

It was the latter fraction that inspired me to attempt to extract a larger number of decimal points. On most calculators, the most decimal points you can get is 12 in any numeric calculations. However with a calculator with CAS capabilities, we may be are able to extract more decimal places. (HP 50g, HP 49g+, TI-89, TI nSpire CAS are three models with CAS capabilities)


Steps for Fractions and Expressions - HP 50g:
1. Set the HP 50g to the Exact Mode.
2. Enter the expression. Remember, do not include a decimal point, use fractions.
4. Press EVAL.
5. Enter an integer that represents the number of decimal places. This number must be an integer!
6. Press [left shift], [EEX] (10^x) (ALOG function). Multiply.
7. Find PROPFRAC (soft key label PROPF) in the Arithmetic menu. Execute it twice. The result you want is an integer plus a fraction. The integer part represents number * 10^n. Ignore the "remainder".

Examples:

22 / 7 to 24 decimal places

n = 24

Note that 22 / 7 ≈ 3.14285714286

Keystrokes:
22 ENTER 7 / 24 10^x * PROPFRAC PROPFRAC

Press the up key and then F2 to view the answer. The result is:

3141285714285714285714285714 + 1 / 7

Hence 22 / 7 to 24 places is 3.141285714285714285714285714.

10 / 81 - ((3340 / 3267 * 10^-9)) to 24 decimal places

n = 24
10 / 81 - ((3340 / 3267 * 10^-9)) ≈ 0.12345689101

Keystrokes:
10 ENTER 81 / 3340 ENTER 3267 / 9 +/- 10^x * - 24 10^x * PROPFRAC PROPFRAC

Result:

123456789101112131415161 + (7039 / 9801)

So to 24 places, 10 / 81 - ((3340 / 3267 * 10^-9)) is 0.1234567890101112131415161.

This method can be extended to include sums and series. Finite series and sums work well. Infinite sums and series, that's another story. Sometimes a program will work better.

Steps for a program is used:
1. Set the HP 50g to the Exact Mode.
2. Enter 0. (or an appropriate starting sum)
3. Enter a program that represents a loop. The format is like this: << start end FOR var required expression + EVAL (we want exact answers) NEXT (or increment STEP >>.
4. Press EVAL twice.
5. Enter an integer that represents the number of decimal places. This number must be an integer!
6. Press [left shift], [EEX] (10^x) (ALOG function). Multiply.
7. Find PROPFRAC (soft key label PROPF) in the Arithmetic Menu. You may have to execute it twice. The result you want is an integer plus a fraction. The integer is the number * 10^n. Ignore the "remainder".

Notes:
* This method works best fast converging series. Series for π may not converge as desired (please see the example below).
* The loop can take a long time if the ending value is high.
* You have to make a mental point of where the decimal point should be. It is best to get an approximate value of the number you want to try to get more decimal points for before beginning.
* This method is not perfect! Remember, we get the result of the calculation, which may not be the desired number.


Σ (X^-1, X, 1, 20) to 24 places

Σ (X^-1, X, 1, 20) ≈ 3.5773965714

RPN syntax for Σ : variable lower_limit upper_limit expression Σ

Keystrokes, with X cleared (purged):
X ENTER 1 ENTER 20 ENTER X 1/X Σ 24 10^x * PROPFRAC PROPFRAC

Result: 10^24 * Psi(21) - 10^24 * Psi(1). Clearly, this is not what we want. Let's try a program.

1. Enter 0, then press ENTER.
2. Enter << 1 20 FOR X X INV + EVAL NEXT>>. Press ENTER, then EVAL. INV is the same as 1/X.
3. Key 24 ENTER 10^x * PROPFRAC PROPFRAC

Result:

3597739657143681911483769 + (66839 / 969969)

To 24 decimal places, Σ (X^-1, X, 1, 20) = 3.597739657143681911483769.

The difficulty for π

There are many infinite series for π. The very slow converging Gregory-Leibiniz series is the classic example:

π / 4 = 1 - 1 / 3 + 1 / 5 - 1 / 7 + 1 / 9 + ...

And the series from Nilakantha:

Let's use the series from Nilakantha. Remember the integer part divided by 10^24 is our "real" answer.

Key (I combine all the steps in a single program)
<< 3
2 50 FOR K
4
K
DUP 1 +
DUP 1 +
* * INV *
-1 K 2 / 1 + ^
* + EVAL 2 STEP
24 ALOG * PROPFRAC PROPFRAC >>

Result:
3141606851347550068996630 + 1606620270334125710 / 373805630307495883

Matches π * 10^24 for the first four digits.

Changing the upper limit garners these results (ignoring the fraction):
Limit of 100: 314160685134755006899630
Limit of 250: 3141590769849795180512794
Limit of 500 (after about 20 minutes): 3141592637780581078269478

int(π *10^24) = 31415926535897932384626433


This blog entry is a technique to "trick" the calculator into giving more decimal places beyond 12

Eddie


This blog is property of Edward Shore. © 2012.

Repeated Operations - Revisited

Repeated Operations with the TI nSpire

What happens when you repeat an operation or a function on a number? I am sure we have all entered a number on a standard calculator and pressed the square root function a number of times. Eventually, each non-zero number goes towards 1.

The following graphs show are functions that are repeated. The curve in blue represents the function, the "curve" (really a Scatterplot) in red demonstrates what happens to a certain interval when I repeat a function 10 times. I used a TI nSpire CX CAS for this. That Ans function (last answer) comes in handy!

Some graphs "stabilize" after 10 iterations, and the result becomes a constant for each point in the domain. I will point that out when applicable.

Enjoy!

Eddie

This blog is property of Edward Shore. © 2012

The first set uses the domain x ∈ {0, 5}. The Scatterplot has points in this interval in increments of Δx = 0.25

The next set has the domain x ∈ [0, 10] with the Scatterplot in increments Δx = 0.1. The function is in blue, and the Scatterplot in red represents the function repeated ten times.

Santa Barabra

I am on vacation this week and loving it. I visited UCSB and found the math department. Below is a picture to remember my visit by.

Look forward to blogging again in September!

Eddie

Chaos and a War Against Set Theory?

This is Eddie, at the Friends Cafe at Monrovia, CA, enjoying lunch and a great Friday.

Chaos

Below are three pages from a beautiful book, "An Eye For Fractals" by Michael McGuire (1991). I got this from Book Alley in Pasadena, CA. Book Alley has every book ever written, or at least it felt like it when the aisles literally full with books.

My understanding of chaos theory and fractals is limited at this point. Making fractals involve applying repeating operations or calculations a lot (and yes this can include an infinite) of times and evaluating the results. Famous examples of fractals include the Koch Snowflake, Sierpniski Triangle (the TI-82 manual had a program to generate this), the Cantor Set, and the Barnsley Fern.

I have tried to plot the Madelbrot Set on a TI-84+, unsuccessfully.

Set Theory Is Evil!

Or this is what the publishers at the Christian publisher A Beka want us to think. I first heard of the publisher while I listening to episode #108 of Math/Maths Podcast. Clicking this sentence takes you to the Math/Maths podcast page. You can also download their podcast through iTunes.

They referred us to an article written by Maggie Koerth-Baker, which you can read by clicking here.

I agree with Koerth-Barker. The fundamentalist books are out of touch with reality and is a disguised as an attack on anyone and everyone who does not belong to the fundamentalist tribe. To exclude or twist facts to fit an ancient text and to teach it to children can pry the door to another Dark Age.

Mathematically, without set theory, we would be without, among other things, we would be without the ability to qualitatively compare groups of items, vector spaces, computing, and how to order objects of set. I would think that with A Beka's objection, we would not be able to tell which one of two members is greater or less - which would really screw up subtraction and the number line. However, while I was writing this blog entry, I was informed by Samuel Hansen (of the Math/Maths podcast) Twitter that their objection is to modern set - so ordering of numbers and basic math is OK. Thanks, Samuel.



Eddie


This blog is property of Edward Shore. © 2012

Applications and Programming: Rotation of a Graph

Rotation of a Function - TI-84+

The following program rotates the function y1(x) at an angle θ. The angle θ is measured counterclockwise, with θ = 0 indicating due east (positive x direction).

Let the coordinates (x', y') be e new coordinates after rotating the original coordiante, (x, y) at an angle θ. Then (x', y') can be calculated by the following equations:

x' = x cos θ - y sin θ
y' = x sin θ + y cos θ

The rotated coordinates are compiled into two lists: one for x', and the other for y'. A stat plot is made of the resulting lists.

Steps (TI-84+):

1. Store Y1. Two ways to do this:
* Press Y=, then edit Y1, or
* Type the function in quotes and store it in the variable Y1.

2. Run ROTATEY1 (see below). The original Y1 is graphed along its rotated sibling.

Two screen shots of examples are shown below. Have fun,


Eddie

This blog is property of Edward Shore. © 2012

Program ROTATEY1 - TI-84+


: Func
: FnOff
: PlotsOff
: FnOn 1
: PlotsOn 1
: Radian
: Disp "TYPE ° FOR DEGREES"
: Input "θ:", θ
: Xmin → X
: {X cos( θ ) - Y1 sin(θ → L1
: {X sin( θ ) - Y1 cos(θ → L2
: 1→ C
: For(K,Xmin,Xmax - ΔX, ΔX)
: K + ΔX → X
: augment(L1, {X cos( θ ) - Y1 sin(θ → L1
: augment(L2, {X sin( θ ) - Y1 cos(θ → L2
: C + 1 → C
: ClrHome
: Output(1,1,"PROGRESS:
: Output(2,1,"round(100C/95,2
: End
: Plot(Scatter,L1,L2, square or marker of your choice
: DispGraph

The Blue HP 50g Part 2 (Pictures)

First of all, this is where I bought the blue HP 50g:

The Calculator Store (Barcelona, Spain)

Now some pictures. I posted a picture of the calculator in its packaging last week. You can check that out here.

I will also be showing my black HP 50g, which I bought in 2006, and has been a companion when I was getting my Master's Degree in Mathematics.

Enjoy!

Eddie

This blog is property of Edward Shore. © 2012

Applications and Programming: Ellipses

The following program works with several properties with ellipses. The general equation for an ellipse with center (x0, y0) is:

(x - x0)^2 / a^2 + (y -y0)^2 / b^2 = 1

For today's blog, I will center the ellipse at the origin (0,0).

With a and b as the length of the semi-axes:

* min(a,b) is the length of the semi-minor axis. min is the Minimum function, the least of a and b.
* max(a,b) is the length of the semi-major axis. max is the Maximum function, the greater of a and b.
* The eccentricity of the ellipse is e = √ (1 - [min(a,b) / max(a,b)]^2). In this sense, e is not the constant 2.7182818285... . If e = 0, then the ellipse is actually a circle. Further more, you can determine where the foci are, which is the distance e * max(a,b) along the semi-major axis.
* The area of an ellipse is easy enough: A = π a b
* The circumference is a different story. A good estimate, suggest by the NCEES, if a and b are close together is C ≈ 2 π √ ( [a^2 - b^2] / 2)

Finding the true circumference will require some calculus. Let the ellipse be defined by the parametric equations:

x(t) = a cos t
y(t) = b sin t
With 0 ≤ t ≤ 2 π

The arc length of a parametric equation is

∫ ( √ ( (dx/dt)^2 + (dy/dt)^2 ) dt, t0, t1)

Which means that the true circumference of an ellipse is

∫ ( √ ( a^2 sin(t)^2 + b^2 cos(t)^2 ) dt, 0, 2 π )

Any calculator with integration can handle this.

The ELLIPSE Program, TI-84+, 184 bytes

The following program will:
1. Prompt for a and b. (A, B)
2. Calculate eccentricity (E).
3. Calculate area (A).
4. Calculate circumference (C).
5. Draw the ellipse. The program leaves the user on the graph screen. The ellipse can be traced and used for further analysis if desired.


ELLIPSE - TI-84+

: Param
: Radian
: Disp "X^2/A^2+Y^2/B^2 =1"
: Prompt A,B
: "A cos(T)"→X1T
: "B sin(T)" →Y1T
: FnOff
: PlotsOff
: FnOn 1
: 0 →Tmin
: 2 π →Tmax
: π / 128 →Tstep
: √(1-(min(A,B)/max(A,B))² → E
: π A B → R
: fnInt(√(A² sin(T)² + B² cos(T)² , T, 0, 2 π) → C
: Disp "ECC.=", E
: Pause
: Disp "AREA=", R
: Pause
: Disp "CIRC.=",C
: Pause
: ZoomFit


Here is the ELLIPSE program executed with A=2 and B=3:

I thank you once again. Thank you to the followers of my blog, love the comments, I appreciate it.

Until next time,

Eddie


This blog is property of Edward Shore. © 2012

The Blue HP 50g Has Arrived

Thanks to the Calculator Store in Spain! :)

This is a HP 50g, only in a blue case.

Eddie