Logic Operators and the TI-36X Pro

Introduction

This blog entry is in response to a email regarding the logic operations I received from jcroot on the Texas Instruments TI-36X Pro Review: (link here)

Hello Eddie!

How are you today? I was just wondering if you could please give us an example of how to use the "Logic function" on this calculator, I am trying to follow the manual but I still can't understand in which situations this could be useful.

jcroot, I hope that this blog entry (at least in part) answers your question. I had to research a bit because I must confess that I do not use logic operators much on calculators.

Logic Operators

The most common logic operators are: and, or, not, 2's compliment and xor. The TI-36X has two additional operators xnor and nand. The latter two are not common on calculators.

I tend to think binary numbers as the prime example for these logic operators. It is my understanding that the operators compare the binary form (base 2 integers) of an integer and if necessary, convert the answer back into the required form (decimal, hexadecimal, or octal).

Binary Numbers

Binary numbers are numbers consisting of 0 and 1, with each digit being a bit. Often, the bit 0 is referred to "off" or "false", while the bit 1 is referred to "on" or "true". Each bit resembles a power of 2, with powers of 2 increasing from right to left:

...(2^n)...(2^5)(2^4)(2^3)(2^2)(2^1)(2^0)

For example, if the first (rightmost bit) is 1, then the value "1" is turned on. If the second bit is 1, then the value "2" is turned on, the third bit gives a value of "4", fourth bit gives a value of "8", and so on. Binary numbers are often written with a subscript of 2. On calculators, they are often designated with b or a "#b".

On the TI-36X Pro

Change to BINary Mode: Press [mode], arrow down five times, select BIN, hit [enter], then [2nd] [mode] (quit). You can designate DECimal (normal - base 10), HEXadecimal (base 16), and OCTal (base 8) modes in a similar way.

Designate a binary number in any mode: type your binary number, [2nd] [ 9 ] (base n), select TYPE, select 2: b. This procedure is not necessary if you are in Binary Mode.

[2nd] [ 9 ] (base n) gives you access to conversions (CONV), designations (TYPE), and operations (LOGIC).

One Argument Logic Operators

not: This operator basically switches a bit. In binary numbers, if the bit is 0, then not 0 = 1. Similarly, if the bit is 1, then not 1 = 0.

2's Complement: This complement is a way to manipulate binary numbers such that the arithmetic, specifically addition and subtraction of binary numbers, is accurate. More information is found by clicking on this sentence.

Two Argument Logic Operators

The following logic operations answer different types of questions:

and: Are both bit 1 true and bit 2 true?

or (inclusive or): Is either bit 1 or bit 2 true? Bit 1 and bit 2 can both be true.

xor (exclusive or): Is either bit 1 true or bit 2 true - without both bits being true?

xnor (not exclusive or): The opposite of xor. Are both bits true or both bits false?

nand (not and): The opposite of and. Is at least one of the bits false?

The following table summaries and, or, xor, xnor, and nand.

How the TI-36X Works With Binary Numbers

All binary numbers work with 10 bit integers. Attempting to work integers greater than 10 bits causes a SYNTAX Error. This error is not mentioned in the TI-36X Manual.

The binary number of a TI-36X looks like this:

##########

where # is a 0 or 1. If you enter less than 10 bits, the calculator "fills" the leftmost bits with 0. Note that I designated the leftmost bit (10th bit) red and bold. This is known as the sign bit. If this bit is 0, then the integer is positive.

(sign) (2^8) (2^7) (2^6) (2^5) (2^4) (2^3) (2^2) (2^1) (2^0)
0 [ 1 = "on", 0 = "off"]

However, if the leftmost bit (10th bit) is 1, then the integer is negative. I am not 100% sure how the rest of the bits are designated.

Note in all examples, I designate binary numbers with the "b", like the calculator does.

10001b = 0000010001b = 17

1111101111b = -17

For the TI-36X Pro, the boundary for binary integers in the interval [-511, 511].
(2^9 - 1 = 511) Remember, the 10th bit is the sign bit on this calculator.

When using logic operators, the TI-36X compares each bit, including the 10th bit.

By the way, binary integers can be stored and recalled in variables.

Examples:

Let x = 0000011101b (5th, 4th, 3rd, and 1st bits are on, x = 29)
and y = 0000010100b (5th and 3rd bits are on, y = 20)

(decimal equivalents in parenthesis)
x and y = 0000010100b (29 and 20 = 20)
x or y = 0000011101b (29 or 20 = 29)
x xor y = 0000001001b (29 xor 20 = 9)
x xnor y = 1111110110b (29 xnor 20 = -10)
x nand y = 1111101011b (29 nand 20 = -21)

So, when I think of logic operations, I think of comparing binary numbers (or binary equivalent of integers) bit by bit, given a designated amount of bits.

Hope this helps,

Eddie.



This blog is property of Edward Shore. © 2012

Calculus Revisited #11: Integration by Substitution

Welcome to Part 11 of our 21-part Calculus Revisited journey! Today we are continuing with integrals - and starting the first of four blog entries concerning techniques of integration. I will try my best to explain each one but I feel the best way to demonstrate these techniques is by example.

Integration by Substitution

We have the integral ∫ f(x) dx.

Find a function u(x) such that

du = u'(x) dx and

∫ f(u) * du is easier to integrate than ∫ f(x) dx.

Then ∫ f(u) du = F(u) + C

For the definite integral

b
∫ f(x) dx
a

du = u'(x) dx

and the integral is re-written as

u(b)
∫ f(u) du = F(u(b)) - F(u(a))
u(a)

Let's get to the solved problems to demonstrate this technique.

Problems
Indefinite Integrals

1. ∫ √(4 + 5x) dx

Let u(x) = 4 + 5x
Then du = 5 dx

Note that √u = √(4 + 5x)
and dx = du/5

Rewriting the integral:

∫ √(4 + 5x) dx
= ∫ √u * 1/5 du
= 1/5 * ∫ √u du
= 1/5 * (u^(3/2))/(3/2) + C
= 1/5 * 2/3 * u^(3/2) + C
= 2/15 * u^(3/2) + C
When substitution is used, please don't not forget to state the answer in the original variable, in this case, x. u = 4 + 5x
= 2/15 * (4 + 5x)^(3/2) + C

2. ∫ sin x / √(1 + 2 cos x) dx

Let u = 1 + 2 cos x
Then du = - 2 sin x dx

Note - du/2 = sin x dx (which matches the numerator)

Rewriting the integral:
∫ sin x / √(1 + 2 cos x) dx
= ∫ 1/√u * -1/2 du
= -1/2 * ∫ u^(-1/2) du
= -1/2 * u^(1/2) * 2 + C
= -u^(1/2) + C
= -√u + C
Remember... the answer must contain the original variable!
= -√(1 + 2 cos x) + C

The right substitutions make integration much easier. The next two problems illustrate this.

3. ∫ sin^4 x * cos x dx

Let u = sin x
Then du = cos dx (note the cos x in the integral)

∫ sin^4 x * cos x dx
= ∫ u^4 du
= u^5/5 + C
= sin^5 x/5 + C

4. ∫ e^x / (e^(2x) + 1) dx

Let u = e^x
Then du = e^x dx (matches the numerator)

∫ e^x / (e^(2x) + 1) dx
= ∫ du/(u^2 + 1)
= atan u + C
= atan (e^(2x)) + C

Definite Integrals

If you use the substitution technique with indefinite integrals, do NOT forget to account for the limits! For definite integrals, you will recalculate the limits to account for the substitution. The next two problems illustrate this point.

5. Calculate

1
∫ cos (2x) dx
-1

Let u = 2x
Then du = 2 dx

Note du/2 = dx (try to get the term with the dx to match some part of the integral)

Accounting for the limits:
New Upper Limit = 2 * 1 = 2
New Lower Limit = 2 * -1 = -2

Putting it together:

1
∫ cos (2x) dx
-1

2
= ∫ cos u * 1/2 du
-2

= 1/2 * (sin(2) - sin(-2))

Recall sin(-x) = -sin x

= 1/2 * (sin(2) + sin(2)) = sin(2) ≈ 0.90930


Finding the right substitution may be tricky at times. Algebraic manipulation may be necessary. Here is a more difficult problem.

6. Calculate

9
∫ √t / (√t - 1) dt
4

Let u = √t - 1
Then du = 1/(2√t) dt

To help match the integral, (2√t) du = dt

Note that 2√t = 2(u + 1) = 2u + 2 (Look at the substitution and solve for √t).

So (2u + 2) du = dt

We can't forget about the limits, since we are working with a definite integral.
u = √t - 1

New Upper Limit: √9 - 1 = 4 - 1 = 3
New Lower Limit: √4 - 1 = 2 - 1 = 2

Putting this together:

9
∫ √t / (√t - 1) dt
4

2
= ∫ (u + 1)/u * (2u + 2) du
1

2
= ∫ (1 + 1/u) * (2u + 2) du
1

2
= ∫ 2u + 2 + 2 + 2/u du
1

2
= ∫ 2u + 4 + 2/u du
1

The antiderivative is u^2 + 4u + 2 ln u

= ( (2)^2 + 4(2) + 2 ln (2) ) - ( (1)^2 + 4(1) + 2 ln (1) )

= ( 4 + 8 + 2 ln 2 ) - ( 1 + 4 + 2 * 0 )

= 7 + 2 ln 2 ≈ 8.38629


Best thing to do if you are in a calculus course is to practice, practice, practice!

I hope these six example problems give you an understanding of the substitution technique. Mastery of this technique will come with time and practice - and will speed up integral evaluation.

Next time, we look at the famous (or maybe infamous) Integration by Parts technique.

See you soon,

Eddie


This blog is property of Edward Shore. © 2012

Link to Part 10: Calculus Revisited

Here is the link to Part 10 of our Calculus Revisited Series: Integration Basics

http://edspi31415.blogspot.com/2012/04/calculus-revisited-10-integration.html

(The original post is buried in April 2012)

First Look: Casio FC-200V Financial Calculator (Updated 4/6/2014)

Hi everyone! In tonight's blog entry we will look at the Casio FC-200V solar financial calculator. I purchased this calculator from Amazon on Super Saver shipper and it came one day earlier than scheduled. It is a rare calculator in terms that I believe it is the only financial calculator besides the Texas Instruments BA-35 Solar that is to be solar-powered. (I could be wrong and if I am feel free to let me know.)

Below is the picture of the calculator in it's packaging:

Packaging
The packaging came with the calculator, a hard case with a cheat sheet, and three instruction manuals: one in Spanish, one in French, and one in English.

Features

From the packaging, the FC-200V is said to have the following:

* SMPL (Simple Interest)
* CMPD (Compound Interest)
* CASH (Cash Flows: NPV, NFV, PBP, and IRR)**
* AMRT (Amortization)
* DPRC (Depreciation: SL, Fixed %, SYD, DB)
* Bonds
* BEVN (Break Even)
* DAYS (Days & Days Between Dates)
* COST (Cost Selling Margin)

Comparing the FC-200V against other financial calculators (I'll use the Hewlett Packard HP 10bII+ for comparison tonight - picture shown below), it seems that the FC-200V breaks away from the traditional financial calculator set up.

Time Value of Money
Most financial calculators sport five keys for using TVM (Time Value of Money) calculations:

N = number of periods
I/YR = yearly interest rate (some have I for periodic interest rate instead)
PV = present value
PMT = periodic payment
FV = future value

The FC-200V breaks away from this model and instead presents TVM calculations in two modules:
1. SMPL for Simple Interest
2. CMPD for Compound Interest

SMPL

Pressing SMPL gives the following options:
Set: Toggle between 365 or 360 days a year
Days: Number of Days
I%: Annual Interest Rate
PV: Present Value

This mode solves for SI (Simple Interest) and SFV (Simple Future Value).

Note: All my examples, unless stated otherwise, use 365 days per year.

Example: If you deposit $900.00 in the bank today in an aggressive 90-day term CD that pays 5% a year, how much will you have in 90 days?

SMPL Mode:
Set: 365
Days = 90
I% = 5
PV = -900

Results:
SI = 11.10
SFV = 911.10

Traditional Set Up:
You could use the TVM keys, however, you answers match better if you use the formula:

SFV = -(PV + Days/365 * I%/100 * PV) (denominator can be 360)

CMPD Mode
This is the traditional TVM set up:
Set*: End or Beginning
n = Number of compounding periods
I% = Annual Interest Rate
PV = Present Value
PMT = Periodic Payment
FV = Future Value
P/Y = Number of payments per year (affects n)
C/Y = Number of compounding periods per year (affects interest) - set separately from P/Y

Use the SOLVE key to solve for the desired variable. Use the cash flow: inflows (receipts) are positive, outflows (disbursements) are negative.

FC-200V vs. HP 10bII+

Example 1: Solving for Payment - END Mode
N = 48, I%YR = 9%, PV = 60,000, FV = 0, P/Y = C/Y = 12

PMT = -1,493.10 (HP 10bII+ is instant, beating out FC-200V by a microsecond)

Example 2: Solving for Interest - END Mode
N = 60, PV = -10,000, PMT = -250, FV = 30,000, P/Y = C/Y = 12

I%YR = 5.15% (The HP 10bii+ being operated by battery only, returns this result instantly. However, the FC-200V returns the result within 3 seconds, so it is not a slow poke - not bad for a solar calculator)

Cash Flow**

The FC-200V stores up to 80 single cash flows, or 40 when frequency is involved. By contrast, the HP 10bII+ can store up to 45 cash flows.

Like the HP10bII+, the FC-200V can solve for NPV (Net Present Value), NFV (Net Future Value), and IRR (Internal Rate of Return). The FC-200V also solves for PBP (Payback Period).

Example:
Cash Flows for a 5 year project:
Initial Investment (x1): -$35,000
Year 1 (x2): $10,000
Year 2 (x3): $12,000
Year 3 (x4): $16,600
Year 4 (x5): $17,500
Year 5 (x6): $24,100

Use a discount rate of 10%

To enter cash flows, hit the [CASH] key and select Csh row. It will take you to the stats screen. Yes, the FC-200V shares the statistic list with the cash flow list.

NPV = $23,397.03
NFV = $37,681.15
IRR (takes a few seconds on the FC-200V) = 29.96%

** On the HP 10bii+: don't forget rate goes I/YR and P/YR = 1

<hr>UPDATE:  4/6/2014

It doesn't look like the FC-200V can handle certain types of cash flow problems correctly.  We run into trouble when at least one flow CFn has the same sign as the initial flow (designated as CF1 on the FC-200V).  Please see the comments below for examples. 

Hence, if you are looking for a reliable cash flow function, pass on the FC-200V. 

I apologize for any inconvenience. - Eddie 

<hr>

Other Features
CNVR: Convert between EFF (effective) and APR (nominal percentages) with compounding

n: Payments per year
I%: Interest to be converted, select to EFF or to APR

The [COMP] key returns you to the home screen.

COST: Cost-Sell-Margin

DAYS: Days between dates using 360 day years or 365 day years. Use the format for dates: mmddyyyy (USA) or ddmmyyyy (International) (no decimal point)

Bonds and Depreciation

An Expansive Breakeven Module, including Degree of Operating Leverage, Degree of Financial Leverage, Degree of Combined Leverage, and Margin of Safety

Statistics

The FC-200V has the same statistical regressions of its scientific sisters fx-115ES and fx-300ES: linear, logarithmic, two types of exponential, power, inverse, quadratic.

Scientific Functions

On recent financial calculators, there has been a growing number of scientific functions. The FC-200V is no exception. Trigonometry (inverse functions have to be accessed by the catalog), polar and rectangular conversions (via the catalog), powers, roots (real numbers only), logarithms, exponentials, hyperbolic functions (via the catalog), and random numbers. No normal distribution functions though. However, room for two shortcuts and formula memories.

I will looking through the manual over the next week. This is just a quick overview of the FC-200V.

Thanks for reading!

This blog is property of Edward Shore. © 2012

Links to the First Nine Parts of Calculus Revisited

#1: Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-1-functions.html

#2: Trig Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-2-working-with.html

#3: Logarithm and Exponential Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-3-logarithmic-and.html

#4: Limits
http://edspi31415.blogspot.com/2012/04/calculus-revisited-4-limits.html

#5: Derivatives
http://edspi31415.blogspot.com/2012/04/calculus-revisited-5-derivatives.html

#6: Chain Rule
http://edspi31415.blogspot.com/2012/04/calculus-revisited-6-chain-rule.html

#7: Finding Extrema
http://edspi31415.blogspot.com/2012/04/calculus-revisited-7-finding-extrema.html

#8: Related Rates
http://edspi31415.blogspot.com/2012/04/calculus-revisited-8-related-rates.html

#9: Newton's Method
http://edspi31415.blogspot.com/2012/05/calculus-revisited-9-newton-method.html

Calculus Revisited #9: Newton's Method

Welcome to blog entry #9 of 21 of our Calculus Revisited. Today we will cover Newton's Method: an effective and powerful way of finding roots of functions.

Process: Newton's Method

Goal: Find roots of the function f(x). That is, solve for x: f(x) = 0

1. Start with an initial guess. Let this be x_n.

2. Calculate x_(n+1) = x_n - f(x_n)/f'(x_n) (function over its derivative)

3. If x_n+1 matches x_n, you are done, with the root being x_n+1. You determine how accurate the answer is by selecting how many decimal points to match. Calculator accuracy (what you see on the screen) varies from 10 to 12 decimal points.

Caution: The choice of initial value is important. Usually, selecting initial guesses close to the root works the best, but this does not always work.

Also, this method is not 100% in finding roots. If you get x_n and x_n+1 flopping between two distinct values, you are caught in a loop - a root will not be found.

I recommend that you use a calculator when working with Newton's Method. On scientific calculators, you may be able to take advantage of the last answer feature, setting x_n = ans.

Example: (TI -36X Pro, TI-30 Multiview, Casio fx-300, Casio fx-115ES, most graphing calculators)
2 ENTER (2 is stored in ans)
ans - f(ans)/f'(ans) (calculates x_1 with x_n = 2)
Keep pressing ENTER

Problems
Newton's Method will often be used to estimate nth roots (square root, cube root, etc). The first problem will be an example of such.

1. Estimate √6 to 5 decimal places.

Observe that √4 = 2 and √9 = 3, which means √6 lies in between 2 and 3. Let's make an initial guess 2.5.

Use the function f(x) = x^2 - 6. Why?

x^2 = 6
x^2 - 6 = 0

Finding the root to the above equation leads to √6.

Then f'(x) = 2x and

x_n+1 = x_n - (x^2 - 6)/(2x)

Having x_0 as the initial guess,
x_0 = 2.50000
x_1 = 2.45000
x_2 = 2.44949
x_3 = 2.44949

(x_1 = 2.5 - (2.5^2 - 6)/(2*2.5) = 2.45,
x_2 = 2.45 - (2.45^2 - 6)/(2*2.45) = 2.44949, and so on)

So to five decimal places, √6 ≈ 2.44949

2. Find a root of the equation e^x - x = 5 for any x > 0. Let x_0 = 2. (the initial guess)

Get the equation in form of f(x) = 0:
e^x - x = 5
e^x - x - 5 = 0

Then f(x) = e^x - x - 5, f'(x) = e^x - 1, and

x_n+1 = x_n - (e^x_n - x_n - 5)/(e^x_n - 1)

By calculation, we get:

x_0 = 2.00000
x_1 = 1.93911
x_2 = 1.93685
x_3 = 1.93685

Then a root of f(x) is x≈1.93685.

3. Solve x^4 - 4x - 4 = 0 with the condition -1 < x < 1. Find the root to five decimal places.

Hint: You may want to graph f(x) to estimate an appropriate initial guess.

Let x0 = -.5

Then f(x) = x^4 - 4x - 4, f'(x) = 4x^3 - 4, and

x_n+1 = x_n - (x_n^4 - 4x_n - 4)/(4x_n^3 - 4)

By calculation, we get:
x_0 = -0.50000
x_1 = -0.93056
x_2 = -0.86520
x_3 = -0.86198
x_4 = -0.86198

A root of f(x) where -1
Here is a situation where Newton's Method does not work for three picks. Not all f(x) will be successful.
4. Find a root of f(x) = √(x^2 - 4) Fortunately, we can find them using algebraic methods (x = ± 2).

f(x) = √(x^2 - 4)
f'(x) = 1/(2 * √(x^2 - 4) * 2x = x/√(x^2 - 4)
x_n+1 = x_n - (x^2 - 4)/x

If x_0 = 1, then subsequent calculations result in a loop between 1 and 4.
If x_0 = 1.5, then subsequent calculations result in a loop between 1.5 and 2.66667.
If x_0 = 1.75, then subsequent calculations result in a loop between 1.75 and 2.828571.

You may try this with other x_0 and run into similar trouble.

Just be aware.

That wraps it up on Newton's Method. Next we head into integration! - Eddie


This blog is property of Edward Shore. © 2012

Calculus Revisited #7: Finding Extrema

We have arrived at the 7th in a 21 blog entry in our Calculus Revisited series.

Today - we will use the derivative to find local extrema. Extrema include the maximum and minimum values of a function f(x). Finding extrema determines where the function f(x) "turns".

First Derivative Test: A process for finding extreme of the function f(x):

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Compute f'(x) where x is near c but not at x = c.

(A) If f'(x) < 0 when x < c and f'(x) > 0 when x > c, then f(x) has a minimum at x = c.

(B) If f'(x) > 0 when x < c and f'(x) < 0 when x > c, then f(x) has a maximum at x = c.

(C) If (A) or (B) do not apply, then f(x) has an inflection point at x = c.

Second Derivative Test: Another process, this one involving the second derivative:

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Calculate f''(c) (second derivative).

(A) If f''(c) >0, then f(x) has a strict local maximum.

(B) If f''(c)<0, then f(x) has a strict local minimum.

(C) If (A) or (B) do not apply, the second derivative test is inconclusive.

Some observations:

1. If f'(x) > 0, f(x) is said to be increasing at x.
2. If f'(x) < 0, f(x) is said to be decreasing at x.
3. If f''(x) > 0, f(x) is said to be concave up at x. (think of a bowl holding water)
4. If f''(x) < 0, f(x) is said to be concave down at x. (think of a bowl turned upside down)

Problems
1. Find extrema of the function y(x) = -x^3 + 2x + 1

Find the critical points
y'(x) = 0
d/dx (-x^3 + 2x + 1) = 0
-3x^2 + 2 = 0
-3x^2 = -2
x^2 = 2/3
x = ±√(2/3)

The critical points are x = √(2/3) and x = -√(2/3).

By the second derivative test:
y''(x) = -6x

y''(√(2/3)) = -6 * √(2/3) < 0. f(x) has a maximum at x = √(2/3).

y''(-√(2/3)) = -6 * -√(2/3) = 6 * √(2/3) > 0. f(x) has a minimum at x = √(2/3).

√(2/3) ≈ 0.81650

2. An open box will be made out of a 12" x 10" sheet of cardboard. (see drawing below, not drawn to scale). A square of length x inches will be cut from each corner of the sheet. What should be the dimensions to construct a box with maximum volume?

x is going to be restricted. First of all x > 0. Second, x < 5. (because 10 - 2 * 5 = 0).

The volume of the box is:
V = (12 - (x + x)) * (10 - (x + x)) * x
= (12 - x^2) * (10 - 2x) * x
= 4x^3 - 44x^2 + 120x

To find the maximum volume, find the critical points. Use the derivative of V in terms of x.

dV/dx = 0
12x^2 - 88x + 120 = 0
By the quadratic formula:
x = (88 ± √(88^2 - 4*12*120))/(2*12)
where:
x ≈ 1.8107 and x ≈ 5.5226.

The latter measurement does not make sense since 10 - 2*5.5226 < 0, and negative dimensions do not make sense.

Test point x ≈ 1.8107.

The second derivative:
d^2V/dx^2 = 24x- 88

Then 24(1.8107) - 88 < 0. The test point is a local maximum.

The dimensions of this box are approximately 8.3786'' x 6.3786'' x 1.8107'' with a volume of about 96.7706 in^3.

3. Minimize the distance between the point (3,1) and the line y(x) = 2x + 5. (see diagram below).

The distance formula between two points (x1, y1) and (x2, y2).

D = (x2 - x1)^2 + (y2 - y1)^2

Find the distance between (3,1) and (x, 2x+5).

D = (x - 3)^2 + (2x + 5 - 1)^2
= (x - 3)^2 + (2x + 2)^2
= 5x^2 + 2x + 13

Taking the derivative D'(x)

D'(x) = 0
10x + 2 = 0
10x = -2
x = -1/5

Since D''(x) = 10 > 0, the point x is a strict local minimum.

The minimum distance is 5 * (-1/5)^2 + 2 * (-1/5) + 13 = 68/5 = 13.6

Next time we deal with problems involving related rates. Until next time, Eddie.


This blog is property of Edward Shore. © 2012

Calculus Revisited #6: The Chain Rule

Today we cover a very useful rule in Calculus: The Chain Rule. The rule deals with so many applications in higher mathematics. This is entry #6 in a series of 21.

The Chain Rule
d/dx f(g(x)) = f'(g(x)) * g'(x) = df/dg * dg/dx

If you learn nothing else in calculus - learn this rule! We will do some practice problems with chain rule today.

Problems
1. d/dx (3x^2 + 1)^2

Using the rule above, let f = g^2 and g = 3x^2 + 1.

Then d/dx f(g) = df/dg * dg/dx,
f = g^2
df/dg = 2g
g = 3x^2 + 1
dg/dx = 6x

Putting it all together:
d/dx (3x^2 + 1)^2 = 2g * 6x = 2(3x^2 + 1) * 6x = 36x^3 + 12x

2. d/dx cos(2x^3 - 1)

Let f = cos g and g = 2x^3 -1
Then:
df/dg = -sin g
dg/dx = 6x^2

Put it together:
d/dx cos(2x^3 - 1) = -sin g * 6x^2 = -sin(2x^3 -1) * 6x^2 = -6x^2 * sin(2x^3 - 1)

3. d/dx √(e^x + 1)

Note √x = x^(1/2)

Let f = √g and g = e^x + 1
Then:
df/dg = 1/2 * 1/√g
dg/dx = e^x

So:
d/dx √(e^x + 1) = 1/2 * 1/√g * e^x = e^x / (2 * √(e^x + 1))

Now that you get the though process, let's try another one.

4. d/dx cos^3 x = d/dx (cos x)^3

d/dx (cos x)^3 = 3 * (cos x)^2 * d/dx cos x = 3 * (cos x)^2 * -sin x = - 3 cos^2 x sin x

5. Evaluate the slope of the function y(x) = ln sin x at x = π/4

d/dx ln sin x
= 1 / sin x * d/dx (sin x)
= 1 / sin x * cos x
= cot x

To find the slope at x = π/4, find cot π/4 = 1.

6. The Chain Rule can be applied more than once. This problem illustrates it. Find d/dx ln (cos √x)

d/dx ln (cos √x)
= 1/(cos √x) * d/dx (cos √x)
= 1/(cos √x) * (-sin √x) * d/dx (√x)
= (-sin √x)/(cos √x) * 1/(2√x)
= (-tan √x)/(2√x)

Next time, we will find the extremes of functions.

This blog is property of Edward Shore. © 2012

Calculus Revisited #5: Derivatives

Welcome to entry #5 of 21 in our Calculus Revisited series. Today, we tackle derivatives!

First the formal definition.

Derivative:

df/dx = f'(x) =

lim ( f(x + δx) - f(x) ) / δx
δx → 0

Note that both symbols for derivatives, the quotient-like symbol and the use of the prime symbol are used interchangeably.

Use the derivative to:
1. Find the instantaneous rate of change at point x0 using the function f(x).
2. Find the slope at given point.
3. When given an equation dealing with the position of an object, f(t), you can find the velocity of an object by calculating f'(t).

But we are going to jump into doing the derivatives. Here is a basic of derivatives:

d/dx ( f(x) + g(x) ) = f'(x) + g'(x)
d/dx ( f(x) * g(x) ) = f'(x) * g(x) + f(x) * g'(x)
d/dx ( f(x) / g(x) ) = (g(x) * f'(x) - g'(x) * f(x))/(g(x)^2)
d/dx a = 0 (a is a constant)
d/dx x = 1
d/dx x^n = n * x^(n-1)
d/dx sin x = cos x
d/dx cos x = -sin x
d/dx tan x = sec^2 x
d/dx e^x = e^x
d/dx a^x = a^x ln a (a is a constant)
d/dx ln x = 1/x
d/dx asin x = 1/√ (1 - x^2)
d/dx acos x = -1/√ (1 - x^2)
d/dx atan x = 1/(x^2 + 1)

TIP: For d/dx ( f(x) / g(x) ), let N = f(x) (numerator) and D = g(x) (denominator). Then d/dx N / D = ( D * N' - D' * N)/(D^2)

Higher Order Derivative: Repeat the derivative operation on f(x).

f''(x) = d^2/dx^2 is the second derivative: take the derivative of f(x), twice.
f'''(x) = d^3/dx^3 is the third derivative: take the derivative of f(x), thrice (three times).
f^(n)(x) = d^n/dx^n is the nth derivative: take the derivative of f(x) n times.

Problems
1. Find the derivative of f(x) = x^2 + 2x + 3.

We can use the addition property to help us.

d/dx (x^2 + 2x + 3)
= d/dx (x^2) + d/dx (2x) + d/dx (3)
= 2x + 2 + 0
= 2x + 2

Remember, the derivative of a constant is 0.

2. Find the slope of f(x) = -x^5 + 2x - 1 at x = 1

First find the derivative
d/dx (-x^5 + 2x - 1)
= d/dx (-x^5) + d/dx(2x) - d/dx(1)
= -5x^4 + 2 - 0
= -5x^4 + 2
The slope is determined by calculating f'(1).
f'(1) = -5(1)^4 + 2 = -3

The slope of f(x) at x = 1 is -3.

3. Find the derivative of f(x) = x^2 * ln x.

Use the product rule: x^2 multiplied by ln x.

d/dx (x^2 * ln x)
= d/dx (x^2) * ln x + x^2 * d/dx (ln x)
= 2x * ln x + x^2 * 1/x
= 2x * ln x + x

4. Find the derivative of f(x) = (x^2 -1)/(x + 2)

Use the division rule: numerator is x^2 -1 and denominator is x + 2.

d/dx ((x^2 - 1)/(x + 2))
= ((x + 2) * d/dx(x^2 - 1) - (x^2 - 1) * d/dx(x + 2))/(x + 2)^2
= ((x + 2) * 2x - (x^2 - 1) * 1)/(x + 2)^2
= (2x^2 + 4x - x^2 + 1)/(x + 2)^2
= (x^2 + 4x + 1)/(x + 2)^2

5. Higher order derivatives: Find the first, second, and third derivative of f(x) = 3x^6 + 7x and g(x) = sin x respectively.


f(x) = 3x^6 + 7x
First Derivative
f'(x) = 18x^5 + 7
Second Derivative
f''(x) = 90x^4
Third Derivative
f'''(x) = 360x^3

g(x) = sin x
g'(x) = cos x
g''(x) = -sin x
g'''(x) = -cos x

6. An interesting property: d/dx( asin x + acos x)

d/dx(asin x + acos x)
= d/dx asin x + d/dx acos x
= 1/√(1-x^2) + (-1)/√(1-x^2)
= 0

This implies that the function asin x + acos x for all x. Bonus question: what is that constant?

Thanks for joining us. Next time we will work with an important rule in calculus: The Chain Rule.

This blog is property of Edward Shore. © 2012

By the way, the answer to the bonus question: asin x + acos x = π/2

Calculus Revisited #4: Limits

Welcome to Calculus Revisited! This is blog entry #4 of a 21 blog entry series. Today, we will cover the basics of limits.

Limit: The limit of a function f(x) is a value of which a function approaches as x creeps closer and closer to it's target x=a. This is not the same as f(a).

In notation:

lim f(x) = L
x→a

The limit may or may not exist.

Left Side Limit: The limit of a function f(x) as x approaches a from the left side. Hence, x < a.

In notation:

lim f(x) = L
x→ a-

Right Side Limit: The limit of a function f(x) as x approaches a from the right side. Hence, x > a.

In notation:

lim f(x) = L
x→ a+

If the left side limit is equal to the right side limit, then the general limit exists at L.

Continuous Function: A function f(x) is continuous if a limit exists for each x in the domain (or specified interval), and that limit is the same as the function's value. It has been often said that you can graph continuous functions without lifting a pencil.

Common continuous functions include:
f(x) = p(x) (polynomials a_n * x^n + ... + a_0)
f(x) = e^x
f(x) = ln x (for x>0 only)
f(x) = sin x
f(x) = cos x

Properties of Limits

lim (f(x) + g(x)) = lim f(x) + lim g(x)

lim c *f(x) = c * lim f(x), c is a constant


Common Ways to Attack Limits

1. If f(x) is continuous, then

lim f(x) = L
x →a

for all x.

2. "Calculator Method": plug in various x_i as x approaches closer and closer to x = a (but not at x = a), observe the results and make a educated conclusion.

3. Graph the function. If graphing calculators are allowed, this is the time to use them.

Problems
1. Find

lim x^2 + 1
x→2

Using the "calculator method":

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .01) = 4.9601
f(2 - .001) = 4.99601
f(2 - .0001) = 4.99960001
f(2 - 10^-9) = 4.999999996
The value is getting close to 5.

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .01) = 5.0401
f(2 + .001) = 5.004001
f(2 + .0001) = 5.00040001
f(2 + 10^-9) = 5.000000004
The value is getting close to 5.

We can reasonably conclude that

lim x^2 + 1 = 5 as x → 2

We could have also observed that x^2 + 1 is continuous everywhere and figured the limit out by plugging in 2 for x.

Now let's go to a case where f(x) is not continuous everywhere.

2. Find

lim 1/(x-2)
x→2

f(2) = 1/0. So plugging in x=2 does not work here.

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .001) = -1,000
f(2 - .0001) = -10,000
f(2 - .00001) = -100,000
f(2 - 10^-9) = -1,000,000,000
Note that f(x) is getting to be a very large negative number, towards -∞

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .001) = 1,000
f(2 + .0001) = 10,000
f(2 + .00001) = 100,000
f(2 + 10^-9) = 1,000,000,000
Note that f(x) is getting to be a very large positive number, towards ∞

But -∞ ≠ ∞

3. Find

lim (sin x)/x
x→ 0

Again, f(0) = 0/0, the plugging it won't work.

Left Side Limit:
f(-.001) = .9999998333
f(-.0001) = .9999999983
f(-.00001) = 1 (calculator returns 1)

Right Side Limit:
f(.001) = .9999998333
f(.0001) = .9999999983
f(.00001) = 1 (calculator returns 1)

From the "Calculator Method":

lim (sin x)/x = 1
x → 0

The next time we will working with derivatives. See you next time, Eddie.

This blog is property of Edward Shore. © 2012

Calculus Revisited #3: Logarithmic and Exponential Functions

Welcome to Part 3 of the 21 part series: Calculus Revisited. Today will we talk about exponential and logarithmic functions.

Exponential and Logarithmic Functions

The function y(x) = e^x is a function where the constant e is taken to the power of x. To 25 decimal places, e = 2.7182818284590452353602874. e is known as the Euler's Number or Napier's Constant. (Wikipedia Source)

The function y(x) = ln x is known as the natural logarithmic function (base e), and is the inverse function of the exponential function. In computer mathematical software and Microsoft Excel ln x is referred to as log x.

Note: On calculators, the function log x refers to a common logarithmic function, not natural. log x uses base 10.

Graphs of both e^x and ln x follow.

Common Properties of the Exponential and Logarithmic Functions

e^x * e^y = e^(x + y)
e^x / e^y = e^(x- y)
(e^x)^y = e^(x * y)

ln(x * y) = ln x + ln y
ln(x / y) = ln x - ln y
ln(x^y) = y * ln x

e^(ln x) = x
ln (e^x) = x

log_n θ = ln θ / ln n (logarithm to base n)

Problems
1. Solve 2^x = 68

2^x = 68
Take the logarithm of both sides
ln (2^x) = ln 68
x ln 2 = ln 68
x = ln 68/ln 2 ≈ 6.08746

2. Solve e^(2x) = 2 * e^(3x)

e^(2x) = 2 * e^(3x)
e^x ≠ 0 for all x, so we can divide.
Divide by e^(2x)
1 = 2 * e^(3x) / e^(2x)
1 = 2 * e^(3x - 2x)
1 = 2 * e^x
1/2 = e*^x
x = ln (1/2) ≈ -.69315

3. Solve e^(2x + 1) = 2^(2x - 3)

e^(2x + 1) = 2^(2x - 3)
ln e^(2x + 1) = ln 2^(2x - 3)
ln e = 1
2x + 1 = (2x - 3) * ln 2
2x + 1 = (2 ln 2)x - (3 ln 2)
(2 - 2 ln 2)x = -3 ln 2 - 1
(2 - 2 ln 2)x = -(3 ln 2 + 1)
x = -(3 ln 2 + 1)/(2 - 2 ln 2) ≈ -5.01778

Next time, we tackle limits. Until next time, take care! Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #2: Working with Trigonometric Functions

Welcome to Part 2 of a series of the 21 Part Series Calculus Revisited, where we cover some of the major topics. Today's and tomorrow's blog entries will be cover the various transcendental functions: today trigonometric functions, tomorrow exponential and logarithmic functions.

Trigonometric Functions

Trigonometric Functions include sine, cosine, and tangent. (abbreviated sin, cos, and tan, respectively) They have corresponding reciprocal functions, cosecant, secant, and cotangent (abbreviated csc, sec, and cot respectively).

The relationship between the trigonometric and their reciprocal functions:

csc x = 1 / sin x
sec x = 1 / cos x
cot x = 1 / tan x

The inverse functions of sine, cosine, and tangent are the arcsine, arccosine, and arctangent functions, respectively. (abbreviated asin or sin^-1, acos or cos^-1, atan or tan^-1 respectively)

sin (asin x) = x
asin (sin x) = x
cos (acos x) = x
acos (cos x) = x
tan (atan x) = x
atan (tan x) = x


The graphs of sin x, cos x, and tan x follow.

Remember: sin^2 x = (sin x)^2. sin^-1 x = asin x. This applies for all trigonometric functions.

More Common Trig Identities

tan x = sin x/cos x
cot x = cos x/tan x

sin^2 x + cos^2 x = (sin x)^2 + (cos x)^2 = 1
tan^2 x + 1 = sec^2 x
1 + cot^2 x = csc^2 x

sin(π - x) = sin x
cos(π - x) = -cos x

sin 2x = 2 sin x cos x
cos 2x = 1 - 2 sin^2 x

sin(a + b) = cos a sin b + sin a cos b
cos(a + b) = cos a cos b - sin a sin b

Problems

1. Let y(x) = a sin (bx), where a and b are any real numbers. What is the significance of a and b?

a is the amplitude of the sine function, where a is the maximum value of y(x) and -a is the minimum value of y(x).

b involved in the period of y(x). The period is π/b.

Note: The variables of a and b have the same significance in the function y(x) = a cos (bx).

2. Is sin x an even or odd function? What about cos x and tan x?

The function f(x) is an even function if for every x, f(-x) = f(x). The function f(x) is an odd function if for every x, f(-x) = -f(x).

Using the graphs above as a reference, sin(-x) = -sin x and cos(-x) = cos x. Note that tan x = sin x/cos x. Then tan(-x) = sin(-x)/cos(-x) = - sin x/cos x = - tan x.

Therefore, sin x is an odd function, cos x is an even function, and tax x is an odd function.

3. Solve cos^2 x - cos x = sin^2 x.

Recall sin^2 x + cos^2 x = 1
Then sin^2 x = 1 - cos^2 x

cos^2 x - cos x = sin^2 x
cos^2 x - cos x = 1 - cos^2 x
2 cos^2 x - cos x - 1 = 0
2 (cos x)^2 - (cos x) - 1 = 0
This is quadratic equation in terms of cos x. This implies that
(cos x - 1) * (2 cos x + 1) = 0
Hence:
cos x - 1 = 0
cos x = 1
x = acos 1 = 0
and
2 cos x + 1 = 0
2 cos x = -1
cos x = -1/2
x = acos (-1/2) = 2π/3

The solutions are x = 0, 2π/3

4. Solve 2 sin x = sin 2x

Recall sin 2x = 2 sin x cos x

2 sin x = sin 2x
2 sin x = 2 sin x cos x
Since sin x can be zero, don't divide by sin x. Instead
2 sin x - 2 sin x cos x = 0
Factor out 2 sin x...
(2 sin x)(1 - cos x) = 0
Then
2 sin x = 0
sin x = 0
x = asin 0 = 0
and
1 - cos x = 0
1 = cos x
x = acos 1 = 0

The solution is x = 0.

Next time we will work exponential and logarithmic functions. Until then, take care! Eddie


This blog is property of Edward Shore. © 2012

Calculus Revisited #1: Functions

Welcome to Calculus Revisited

Greetings. Over the next 21 blog entries (Monday - Friday), Eddie's Math and Calculator Blog will revisit and review some of the major topics in one-variable calculus. I may not get to everything, but I am going to hit a lot of the major topics. Let's start off with some pre-calculus and then head into derivatives and integrals. This blog is going to serve as a reference: whether you are a student or a professional. Enjoy, and let's get started!

Sources Used for this Series

Ross, Debra Anne. "Master Math: Calculus" Career Press: Franklin Lakes, NJ 1998

Silverman, Richard A. Ph.D "Calculus with Analytic Geometry" Prentice-Hall Inc., Englewood Cliffs, NJ 1985


With the preliminairies out of the way...

Function Basics

Function: Is an expression (formula or equation) that associates each element of the domain to a corresponding element of the range. A true function is an expression where every element in the domain corresponds to 1 and only 1 element in a range. So no one element in domain can correspond to two (or more) elements in the range.

Domain: The initial set to which a function is applied. The variable that describes the domain is the independent variable.

Range: The set of results. The variable that describes the range is the dependent variable.

So...

domain → Function → range
x → y = f(x) → y

Note: For the purpose of this series, I will use x, t, θ (the Greek letter theta) for independent variables and y, f, and g for dependent variables. Obviously any letter or description can be used for variables.

The set of real numbers: that is all numbers including integers, decimals, fractions, and irrational numbers. Real numbers just don't include any complex or imaginary number. So, if √-1 (labeled i or j) is involved, the number is not a real number.

This series will focus exclusively on the set of real numbers.

---------------


1. Find the domain and range of the function y(x) = 2/(x-4).

Note that y(x) is not defined at x=4. Why? y(4) = 2/(4-4) = 2/0. We all know that you can't divide by zero. You can calculate y(x) at various points to verify, however y(x) returns answers that cover the set of real numbers.

Domain: The set of real numbers where x ≠ 4
Range: The set of real numbers

2. Find the domain and range of the function y(x) = √ (16 - x^2).

At first glance, y(x) is defined for all x that make 16 - x^2 nonnegative (positive or zero). To calculate the appropriate domain:

Set:
16 - x^2 ≥ 0
16 ≥ x^2
Take the square root of both sides.
x ≥ -4 and x ≤ 4
The proper domain is
-4 ≤ x ≤ 4

Observe that y(x) hits its maximum point when x = 0
y(0) = √ (16 - 0^2) = √ 16 = 4
Minimum points:
y(4) = √ (16 - 4^2) = 0 and
y(-4) = √ (16 - (-4)^2) = 0

Domain: -4 ≤ x ≤ 4
Range: 0 ≤ y ≤ 4

Composite Functions: Basically, a function of a function. Often noted with a raised circle.

Let f(x) and g(x) be two functions. Then f(g(x)) and g(f(x)) are two composite function. In f(g(x)), the g(x) is treated as the independent variable, and vice versa.

3. Let f(x) = x^2 and g(x) = sin x. Find the composite functions f(g(x)) and g(f(x)).

f(g(x)) = f(sin x) = (sin x)^2 = sin^2 x

Caution: Calculators and most mathematical (if not all) software will not allow the user to enter expressions such as sin^2 x. Use the expression (sin x)^2 instead.

g(f(x)) = g(x^2) = sin (x^2)

Inverse Functions: A function that describes a rule from an element of the range to the element of the domain. If y = f(x), then x = f^-1(y).

4. Let f(x) = 7x + 1. Find the inverse function.

Set y as y = 7x + 1. Solve for x.
y = 7x + 1
y - 1 = 7x
(y - 1)/7 = x

Hence the inverse function is f^-1(y) = (y - 1)/7

Tomorrow: Working with Trigonometric Functions

This blog is property of Edward Shore. © 2012

Review: Casio fx-115 ES PLUS Review

Welcome to blog entry #73. Just an announcement:

Starting Monday on April 23, 2012, I begin my Calculus Revisited Series, where we revisit some of the major topics of calculus. Each blog entry is going to contain a quick summary of the topic, followed by a few example problems. Hopefully this serves as a review for students in calculus in time for finals or perhaps you want to revisit some of the topics. It will be one blog each day, Monday through Thursday except on Friday, where you get three for the price of one. (don't worry, it's free).

Onto today's topic... the Casio fx-115ES Plus Calculator Review!

The fx-115ES PLUS is the successor to the 2006(?) [when was this calculator first released?] fx-115 ES. Outside of the United States, these calculators are called fx-991ES PLUS and fx-991 ES, respectively.

I purchased the fx-115ES PLUS at a WalMart in Glendora, CA. Apparently, this is the only major store at this time that carries the new fx-115ES PLUS outside of ordering online. It cost me $17.48, around the same price the classic fx-115 ES cost.

The picture below is a comparison between both models. The new fx-115 is on the left.

Features

The fx-115ES PLUS retains the Natural-V.P.A.M. (Visually Perfect Algebraic Method), which means that you can enter equations and expressions exactly as written. The MathIO method allows the user to accomplish just that. Fractions, radicals, expressions with π, and calculus function templates are all there. Exact answers can include fractions, square roots, and coefficients of π (MathIO).

For those who want a older style, one-line entry method, the LineIO method has been retained. Personally, I prefer the MathIO method.

All of the other features that were present in the classic fx-115 ES are here to:
* Integrals of f(x)
* Numeric Derivatives: d/dx
* Sums of a function: ∑ f(x)
* Base modes Decimal, Octal, Binary, Hexadecimal
* Numeric solver, of equations and roots of expressions. Yes, the variable to solve for is X. (see below)
* CALC button allows for calculating expressions repeated amount of times.
* Statistics including 1-variable, linear regression (a+bx), quadratic regression (a+bx+cx^2), cubic regression (a+bx+cx^2+dx^3), 2 types of exponential (a + b * e^x and a x^b), power (b a^x), logarithmic (a + b ln x), and inverse (a + b/x).
* Equations - 2x2 and 3x3 simultaneous equations, quadratic, and cubic equation
* Matrices: functions include transpose, inverse, and determinant
* Vectors
* Multi Line Statements with the colon (:)
* Complex Number Mode - unfortunately it's still limited to arithmetic, x^2, 1/x.

Hint: To clear calculator history, while in COMP mode, press the [ ON ] key.

Hint: To force approximate answers, press [SHIFT] [ = ]. This works on both versions of the fx-115 ES.

What's New?
* The number of available memories have increased from 7 to 9. (A, B, C, D, X, Y, M, and now E, F). Previously E and F were available only for the Hexadecimal mode.
* You now have the ability to calculate using repeated numbers. For example, you can type the decimal form of 1/3 using 0.3 with the bar above the three. I believe that this is first line of calculators that has this ability.
* New number functions are: GCD, LCM, Integer Part, Fractional Part, Random Integers, Integer Division (÷R) that gives quotient and remainder, and Prime Factorization (up to three digit factors). To factor a number, enter it, press [ = ], then [SHIFT], [ º ' '' ].
* Products of function f(x): ∏ f(x).
* In Table Mode you can include two functions f(x) and g(x). This is turned on in SET UP.
* The [ALPHA] key acts as both as an variable key and a second shift key.
* The rref and ref functions are added to the Matrix Mode (but not eigenvalues).
* The fx-115ES PLUS has a curve design, and boasts a faster processor.

New Modes
* Inequality Solver of quadratic and cubic equations
* Verification mode, used for compare expressions (i.e. Does π/4 < π/2? Does 1 = 9/9 = e^0?)
* Distribution mode: Normal Distribution (CDF, PDF, and Inverse (Yes!)), Binomial Distribution (CDF, PDF), and Poisson Distribution (CDF, PDF).

Final Thoughts
If the fx-115 ES was solid, then the fx-115ES PLUS is taking a great calculator and making it better. At about $18, you get a huge bang for the buck.

Rating: ***** (5 out of 5)

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UPDATE-CORRECTION

Many thanks to jcroot for pointing this out to me:

The Solve feature can solve for more than just X. You can specify the variable as so:

equation, variable to solve for

If there is no second argument, then the variable to solve for defaults to X.

Example:
X - 2Y = 1, X solves for X
X - 2Y = 1, Y solves for Y
-------------------------------------------------------------------------




This blog is property of Edward Shore. © 2012