LINEAR PROGRAMMING----- FORM FOUR.

LINEAR    PROGRAMMING-----FORM  FOUR

Sections: Optimizing linear systems, Setting up word problems

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INTRODUCTION:

Linear programming is the process of taking various linear inequalities relating to some situation, and finding the "best" value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions.
In "real life", linear programming is part of a very important area of mathematics called "optimization techniques". This field of study (or at least the applied results of it) are used every day in the organization and allocation of resources. These "real life" systems can have dozens or hundreds of variables, or more. In algebra, though, you'll only work with the simple (and graphable) two-variable linear case.
The general process for solving linear-programming exercises is to graph the inequalities (called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility region"). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the "optimization equation") for which you're trying to find the highest or lowest value.

• Find the maximal and minimal value of z = 3x + 4y subject to the following constraints:
The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula "z = 3x + 4y" is the optimization equation. I need to find the (x, y) corner points of the feasibility region that return the largest and smallest values of z.
My first step is to solve each inequality for the more-easily graphed equivalent forms:

It's easy to graph the system:   Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved

To find the corner points -- which aren't always clear from the graph -- I'll pair the lines (thus forming a system of linear equations) and solve:


y = –( 1/2 )x + 7y = 3x	y = –( 1/2 )x + 7y = x – 2	y = 3x y = x – 2
–( 1/2 )x + 7 = 3x–x + 14 = 6x14 = 7x 2 = x y = 3(2) = 6	–( 1/2 )x + 7 = x – 2 –x + 14 = 2x – 4 18 = 3x 6 = x y = (6) – 2 = 4	3x = x – 2 2x = –2x = –1 y = 3(–1) = –3
corner point at (2, 6)	corner point at (6, 4)	corner pt. at (–1, –3)

So the corner points are (2, 6), (6, 4), and (–1, –3).
Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, I only need to plug these three points into "z = 3x + 4y".

(2, 6):      z = 3(2)   + 4(6)   =   6 + 24 =   30
(6, 4):      z = 3(6)   + 4(4)   = 18 + 16 =   34
(–1, –3):  z = 3(–1) + 4(–3) = –3 – 12 = –15
Then the maximum of z = 34 occurs at (6, 4),
and the minimum of z = –15 occurs at (–1, –3).

• Given the following constraints, maximize and minimize the value of z = –0.4x + 3.2y.
First I'll solve the fourth and fifth constraints for easier graphing:

The feasibility region looks like this:

From the graph, I can see which lines cross to form the corners, so I know which lines to pair up in order to verify the coordinates. I'll start at the "top" of the shaded area and work my way clockwise around the edges:


y = –x + 7y = x + 5	y = –x + 7x = 5	x = 5y = 0
–x + 7 = x + 5 2 = 2x 1 = x y = (1) + 5 = 6	y = –(5) + 7 = 2	[nothing to do]
corner at (1, 6)	corner at (5, 2)	corner at (5, 0)

y = 0y = –( 1/2 )x + 2	y = –( 1/2 )x + 2x = 0	x = 0y = x + 5
–( 1/2 )x + 2 = 0 2 = (1/2)x 4 = x	y = –( 1/2 )(0) + 2 y = 0 + 2 y = 2	y = (0) + 5 = 5
corner at (4, 0)	corner at (0, 2)	corner at (0, 5)

Now I'll plug each corner point into the optimization equation, z = –0.4x + 3.2y:

(1, 6):  z = –0.4(1) + 3.2(6) = –0.4 + 19.2 = 18.8
(5, 2):  z = –0.4(5) + 3.2(2) = –2.0 + 6.4   =   4.4
(5, 0):  z = –0.4(5) + 3.2(0) = –2.0 + 0.0   = –2.0
(4, 0):  z = –0.4(4) + 3.2(0) = –1.6 + 0.0   = –1.6
(0, 2):  z = –0.4(0) + 3.2(2) = –0.0 + 6.4   =   6.4
(0, 5):  z = –0.4(0) + 3.2(5) = –0.0 + 16.0 = 16.0
Then the maximum is 18.8 at (1, 6) and the minimum is –2 at (5, 0).

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Given the inequalities, linear-programming exercise are pretty straightforward, if sometimes a bit long. The hard part is usually the word problems, where you have to figure out what the inequalities are. So I'll show how to set up some typical linear-programming word problems.

• At a certain refinery, the refining process requires the production of at least two gallons of gasoline for each gallon of fuel oil. To meet the anticipated demands of winter, at least three million gallons of fuel oil a day will need to be produced. The demand for gasoline, on the other hand, is not more than 6.4 million gallons a day.
If gasoline is selling for $1.90 per gallon and fuel oil sells for $1.50/gal, how much of each should be produced in order to maximize revenue?
The question asks for the number of gallons which should be produced, so I should let my variables stand for "gallons produced".


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x: gallons of gasoline producedy: gallons of fuel oil produced
Since this is a "real world" problem, I know that I can't have negative production levels, so the variables can't be negative. This gives me my first two constraints: namely, x > 0 and y > 0.
Since I have to have at least two gallons of gas for every gallon of oil, then x > 2y.
For graphing, of course, I'll use the more manageable form "y < ( ¹/₂ )x".
The winter demand says that y > 3,000,000; note that this constraint eliminates the need for the "y > 0" constraint. The gas demand says that x < 6,400,000.

I need to maximize revenue R, so the optimization equation is R = 1.9x + 1.5y. Then the model for this word problem is as follows:

R = 1.9x + 1.5y, subject to:
x > 0
x < 6,400,000   Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved
y > 3,000,000
y < ( ¹/₂ )x
Using a scale that counts by millions (so "y = 3" on the graph means "y is three million"), the above system graphs as follows:

Taking a closer look, I can see the feasibility region a little better:

When you test the corner points at (6.4m, 3.2m), (6.4m, 3m), and (6m, 3m), you should get a maximal solution of R = $16.96m at (x, y) = (6.4m, 3.2m).

PROBABILITY ----- FORM FOUR ----- BY. MWL. JAPHET MASATU

1. Introduction
2. Basic Concepts
3. Conditional Probability Demo
4. Gambler's Fallacy Simulation
5. Permutations and Combinations
6. Birthday Simulation
7. Binomial Distribution
8. Binomial Demonstration
9. Poisson Distribution
10. Multinomial Distribution
11. Hypergeometric Distribution
12. Base Rates
13. Bayes' Theorem Demonstration
14. Monty Hall Problem Demonstration
15. Statistical Literacy
16. Exercises

Probability is an important and complex field of study. Fortunately, only a few basic issues in probability theory are essential for understanding statistics at the level covered in this book. These basic issues are covered in this chapter.

The introductory section discusses the definitions of probability. This is not as simple as it may seem. The section on basic concepts covers how to compute probabilities in a variety of simple situations. The Gambler's Fallacy Simulation provides an opportunity to explore this fallacy by simulation. The Birthday Demonstration illustrates the probability of finding two or more people with the same birthday. The Binomial Demonstration shows the binomial distribution for different parameters. The section on base rates discusses an important but often-ignored factor in determining probabilities. It also presents Bayes' Theorem. The Bayes' Theorem Demonstration shows how a tree diagram and Bayes' Theorem result in the same answer. Finally, the Monty Hall Demonstration lets you play a game with a very counterintuitive result.

 

PROBABILITY ----- FORM FOUR BY. MWL. JAPHET MASATU.

INTRODUCTION    TO    PROBABILITY.

BASI  CONCEPTS.

Learning Objectives

1. Compute probability in a situation where there are equally-likely outcomes
2. Apply concepts to cards and dice
3. Compute the probability of two independent events both occurring
4. Compute the probability of either of two independent events occurring
5. Do problems that involve conditional probabilities
6. Compute the probability that in a room of N people, at least two share a birthday
7. Describe the gambler's fallacy

Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called favorable outcomes. Given that all outcomes are equally likely, we can compute the probability of a one or a six using the formula:

In this case there are two favorable outcomes and six possible outcomes. So the probability of throwing either a one or six is 1/3. Don't be misled by our use of the term "favorable," by the way. You should understand it in the sense of "favorable to the event in question happening." That event might not be favorable to your well-being. You might be betting on a three, for example.

The above formula applies to many games of chance. For example, what is the probability that a card drawn at random from a deck of playing cards will be an ace? Since the deck has four aces, there are four favorable outcomes; since the deck has 52 cards, there are 52 possible outcomes. The probability is therefore 4/52 = 1/13. What about the probability that the card will be a club? Since there are 13 clubs, the probability is 13/52 = 1/4.

Let's say you have a bag with 20 cherries: 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet? There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is 14/20 = 7/10. There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.

Here is a more complex example. You throw 2 dice. What is the probability that the sum of the two dice will be 6? To solve this problem, list all the possible outcomes. There are 36 of them since each die can come up one of six ways. The 36 possibilities are shown below.

Die 1	Die 2	Total		Die 1	Die 2	Total		Die 1	Die 2	Total
1	1	2		3	1	4		5	1	6
1	2	3		3	2	5		5	2	7
1	3	4		3	3	6		5	3	8
1	4	5		3	4	7		5	4	9
1	5	6		3	5	8		5	5	10
1	6	7		3	6	9		5	6	11
2	1	3		4	1	5		6	1	7
2	2	4		4	2	6		6	2	8
2	3	5		4	3	7		6	3	9
2	4	6		4	4	8		6	4	10
2	5	7		4	5	9		6	5	11
2	6	8		4	6	10		6	6	12

You can see that 5 of the 36 possibilities total 6. Therefore, the probability is 5/36.

If you know the probability of an event occurring, it is easy to compute the probability that the event does not occur. If P(A) is the probability of Event A, then 1 - P(A) is the probability that the event does not occur. For the last example, the probability that the total is 6 is 5/36. Therefore, the probability that the total is not 6 is 1 - 5/36 = 31/36.

Probability of Two (or more) Independent Events

Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs. Let's take a simple example. A fair coin is tossed two times. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss. The two events are (1) first toss is a head and (2) second toss is a head. So these events are independent. Consider the two events (1) "It will rain tomorrow in Houston" and (2) "It will rain tomorrow in Galveston" (a city near Houston). These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.

Probability of A and B

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. More formally, if events A and B are independent, then the probability of both A and B occurring is:

P(A and B) = P(A) x P(B)

where P(A and B) is the probability of events A and B both occurring, P(A) is the probability of event A occurring, and P(B) is the probability of event B occurring.

If you flip a coin twice, what is the probability that it will come up heads both times? Event A is that the coin comes up heads on the first flip and Event B is that the coin comes up heads on the second flip. Since both P(A) and P(B) equal 1/2, the probability that both events occur is

1/2 x 1/2 = 1/4

Let's take another example. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). Therefore, the probability of both events occurring is 1/2 x 1/6 = 1/12.

One final example: You draw a card from a deck of cards, put it back, and then draw another card. What is the probability that the first card is a heart and the second card is black? Since there are 52 cards in a deck and 13 of them are hearts, the probability that the first card is a heart is 13/52 = 1/4. Since there are 26 black cards in the deck, the probability that the second card is black is 26/52 = 1/2. The probability of both events occurring is therefore 1/4 x 1/2 = 1/8.

See the section on conditional probabilities on this page to see how to compute P(A and B) when A and B are not independent.

Probability of A or B

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

In this discussion, when we say "A or B occurs" we include three possibilities:

1. A occurs and B does not occur
2. B occurs and A does not occur
3. Both A and B occur

This use of the word "or" is technically called inclusive or because it includes the case in which both A and B occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:

1. A-and-B happens
2. A-and-not-B happens
3. not-A-and-B happens.

The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the simple event B happens if either A-and-B happens or not-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B) + P(not-A-and-B). We can make these two sums equal by subtracting one occurrence of P(A-and-B) from the first. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B).

Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,

P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

P(6 or head) = P(6) + P(head) - P(6 and head)
             = (1/6) + (1/2) - (1/6)(1/2)
             = 7/12

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.

Conditional Probabilities

Often it is required to compute the probability of an event given that another event has occurred. For example, what is the probability that two cards drawn at random from a deck of playing cards will both be aces? It might seem that you could use the formula for the probability of two independent events and simply multiply 4/52 x 4/52 = 1/169. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as:

P(ace on second draw | an ace on the first draw)

The vertical bar "|" is read as "given," so the above expression is short for: "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw." What is this probability? Since after an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the probability that one of these aces will be drawn is 3/51 = 1/17.

If Events A and B are not independent, then P(A and B) = P(A) x P(B|A).

Applying this to the problem of two aces, the probability of drawing two aces from a deck is 4/52 x 3/51 = 1/221.

One more example: If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card? There are two ways you can satisfy this condition: (a) You can get the Ace of Diamonds first and then a black card or (b) you can get a black card first and then the Ace of Diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case B: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case B is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case A. Recall that the probability of A or B is P(A) + P(B) - P(A and B). In this problem, P(A and B) = 0 since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.

Birthday Problem

If there are 25 people in a room, what is the probability that at least two of them share the same birthday. If your first thought is that it is 25/365 = 0.068, you will be surprised to learn it is much higher than that. This problem requires the application of the sections on P(A and B) and conditional probability.

This problem is best approached by asking what is the probability that no two people have the same birthday. Once we know this probability, we can simply subtract it from 1 to find the probability that two people share a birthday.

If we choose two people at random, what is the probability that they do not share a birthday? Of the 365 days on which the second person could have a birthday, 364 of them are different from the first person's birthday. Therefore the probability is 364/365. Let's define P2 as the probability that the second person drawn does not share a birthday with the person drawn previously. P2 is therefore 364/365. Now define P3 as the probability that the third person drawn does not share a birthday with anyone drawn previously given that there are no previous birthday matches. P3 is therefore a conditional probability. If there are no previous birthday matches, then two of the 365 days have been "used up," leaving 363 non-matching days. Therefore P3 = 363/365. In like manner, P4 = 362/365, P5 = 361/365, and so on up to P25 = 341/365.

In order for there to be no matches, the second person must not match any previous person and the third person must not match any previous person, and the fourth person must not match any previous person, etc. Since P(A and B) = P(A)P(B), all we have to do is multiply P2, P3, P4 ...P25 together. The result is 0.431. Therefore the probability of at least one match is 0.569.

Gambler's Fallacy

A fair coin is flipped five times and comes up heads each time. What is the probability that it will come up heads on the sixth flip? The correct answer is, of course, 1/2. But many people believe that a tail is more likely to occur after throwing five heads. Their faulty reasoning may go something like this: "In the long run, the number of heads and tails will be the same, so the tails have some catching up to do." The flaws in this logic are exposed in the simulation in this chapter.

Question 2 out of 12.
You have a bag of marbles. There are 3 red marbles, 2 green marbles, 7 yellow marbles, and 3 blue marbles. What is the probability of drawing a yellow or red marble?

TEN WAYS TO SURVIVE THE MATH BLUES ---- BY. MWL. JAPHET MASATU.

TEN   WAYS   TO  SURVIVE   THE   MATH   BLUES.

1. Figure out the Big Picture: Find out why you are doing this math. How does it fit with your other courses (science, geography, English, engineering)? You could do some Internet searches on the math you are studying and include "application". Get a sense of where you are going and why you are doing this. Mathematics is compulsory in most of the world – there has to be a reason…
2. Get on top of it before it gets on top of you. Yep, mathematics is one of those things that builds on prior knowledge. Yet many students learn things only for an examination and then promptly forget it, setting themselves up for later difficulties. Learn for the future, not for tomorrow’s test.
3. Read Ahead. It is strongly advised that you read over next week’s math right now. You won’t understand it all, but you will have a better sense of what is coming up and how it fits with what you are doing this week. Then, when your class goes through it later, your doubts and uncertainties will reduce – and you will understand and remember it better.
4. Use more than one resource. It often happens that you can’t follow the teacher’s explanation and your textbook is very confusing. Borrow 2 or 3 textbooks similar to your own from your library and read what they have to say about the topic. Often they will have a diagram, a picture or an explanation that gives you the "Ahhh – I get it!" that you desire.
5. Don’t join the Blame Game. Teaching mathematics is tough. Teachers really have to work hard to make math fun, interesting and engaging. It is easy to blame a teacher for a bad grade, but who is really responsible for your future?
6. Practice makes Perfect. You don’t expect to be able to play guitar or drive a car without practice. Well, learning mathematics (unfortunately) involves some slogging away and doing exercises. Don’t get bogged down, though – use your other resources to help you through the homework.
7. Time Management. Start homework assignments as soon as you get them. There may be some things on there that you haven’t done in class yet (because maybe it is not due for a few weeks). That’s good – it helps to focus your thoughts so that when you are doing that section in class, you know that it is important and you’ll know what you don’t know. Nobody plans to fail – but many fail to plan…
8. Don’t fall into the trap of copying from a friend to survive. They probably have the wrong answer anyway. Besides, a lot of students resent being asked for their assignments for copying – they are too afraid of a ruined relationship to say no. Hey, you can do it – have the confidence in your own ability.
9. Never, never give up. Math uses a different part of the brain than most other things in school. It can be stressful when you can’t figure out something. Work on something else for a while and come back to it later.
10. Keep a sense of humour! Don’t lose the ability to laugh at yourself and your own mistakes. Mistakes are not the end of the world – they are the beginning of real learning!

HOW TO UNDERSTAND MATHEMATICS FORMULAS ----- BY. MWL. JAPHET MASATU.

HOW    TO  UNDERSTAND  MATHEMATICS   FORMULAS. INTRODUCTION.

In a recent IntMath Poll, many readers reported that they find math difficult because they have trouble learning math formulas and an almost equal number have trouble understanding math formulas.
I wrote some tips on learning math formulas here: How to learn math formulas.
Now for some suggestions on how to understand math formulas. These should be read together with the “learning” tips because they are closely related.
a. Understanding math is like understanding a foreign language: Say you are a native English speaker and you come across a Japanese newspaper for the first time. All the squiggles look very strange and you find you don’t understand anything.
If you want to learn to read Japanese, you need to learn new symbols, new words and new grammar. You will only start to understand Japanese newspapers (or manga comics ^_^) once you have committed to memory a few hundred symbols & several hundred words, and you have a reasonable understanding of Japanese grammar.
When it comes to math, you also need to learn new symbols (like π, θ, Σ), new words (math formulas & math terms like “function” and “derivative”) and new grammar (writing equations in a logical and consistent manner).
So before you can understand math formulas you need to learn what each of the symbols are and what they mean (including the letters). You also need to concentrate on the new vocabulary (look it up in a math dictionary for a second opinion). Also take note of the “math grammar” — the way that it is written and how one step follows another.
A little bit of effort on learning the basics will produce huge benefits.
b. Learn the formulas you already understand: All math requires earlier math. That is, all the new things you are learning now depend on what you learned last week, last semester, last year and all the way back to the numbers you learned as a little kid.
If you learn formulas as you go, it will help you to understand what’s going on in the new stuff you are studying. You will better recognize formulas, especially when the letters or the notation are changed in small ways.
Don’t always rely on formula sheets. Commit as many formulas as you can to memory — you’ll be amazed how much more confident you become and how much better you’ll understand each new concept.
c. Always learn what the formula will give you and the conditions: I notice that a lot of students write the quadratic formula as

But this is NOT the quadratic formula! Well, it’s not the whole story. A lot of important stuff is missing — the bits which help you to understand it and apply it. We need to have all of the following when writing the quadratic formula:

The solution for the quadratic equation
ax² + bx + c = 0
is given by

A lot of students miss out the “x =” and have no idea what the formula is doing for them. Also, if you miss out the following bit, you won’t know how and when to apply the formula:
ax² + bx + c = 0
Learning the full situation (the complete formula and its conditions) is vital for understanding.
d. Keep a chart of the formulas you need to know: Repetition is key to learning. If the only time you see your math formulas is when you open your textbook, there is a good chance they will be unfamiliar and you will need to start from scratch each time.
Write the formulas down and write them often. Use Post-It notes or a big piece of paper and put the formulas around your bedroom, the kitchen and the bathroom. Include the conditions for each formula and a description (in words, or a graph, or a picture).
The more familiar they are, the more chance you will recognize them and the better you will understand them as you are using them.
e. Math is often written in different ways — but with the same meaning: A lot of confusion occurs in math because of the way it is written. It often happens that you think you know and understand a formula and then you’ll see it written in another way — and panic.
A simple example is the fraction “half”. It can be written as 1/2, and also diagonally, as ½ and in a vertical arrangement like a normal fraction. We can even have it as a ratio, where the ratio of the 2 (equal) parts would be written 1:1.
Another example where the same concept can be written in different ways is angles, which can be written as capital letters (A), or maybe in the form ∠BAC, as Greek letters (like θ) or as lower case letters (x). When you are familiar with all the different ways of writing formulas and concepts, you will be able to understand them better.
Every time your teacher starts a new topic, take particular note of the way the formula is presented and the alternatives that are possible.

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Do you have any tips to add? How do you figure out your math formulas? Which formulas are hardest to understand?
Good luck with understanding math formulas!

How to learn math formulas

In a recent IntMath Poll, readers indicated that the hardest thing about math was learning the formulas.
Here are 10 things you can do to improve your memory for math formulas.

1. Read ahead

Read over tomorrow’s math lesson today. Get a general idea about the new formulas in advance, before your teacher covers them in class.
As you read ahead, you will recognize some of it, and other parts will be brand new. That’s OK – when your teacher is explaining them you already have a “hook” to hang this new knowledge on and it will make more sense — and it will be easier to memorize the formulas later.
This technique also gives you an overview of the diagrams, graphs and vocabulary in the new section. Look up any new words in a dictionary so you reduce this stumbling block in class.
This step may only take 15 minutes or so before each class, but will make a huge difference to your understanding of the math you are studying.
I always used to read ahead when I was a student and I would be calm in class while all my friends were stressed out and confused about the new topic.

2. Meaning

All of us find it very difficult to learn meaningless lists of words, letters or numbers. Our brain cannot see the connections between the words and so they are quickly forgotten.
Don’t just try to learn formulas by themselves — it’s just like learning that meaningless list.
When you need to learn formulas, also learn the conditions for each formula (it might be something like “if x > 0″).
Also draw a relevant diagram or graph each time you write the formula (it might be a parabola, or perhaps a circle). You will begin to associate the picture with the formula and then later when you need to recall that formula, the associated image will help you to remember it (and its meaning, and its conditions).
During exams, many of my students would try to answer a question with the wrong formula! I could see that they successfully learned the formula, but they had no idea how to apply it. Diagrams, graphs and pictures always help.
Most of us find it difficult to learn things in a vacuum, so make sure you learn the formulas in their right context.
When you create your summary list of formulas, include conditions and relevant pictures, graphs and diagrams.

3. Practice

You know, math teachers don’t give you homework because they are nasty creatures. They do it because they know repetition is a very important aspect of learning. If you practice a new skill, the connections between neurons in your brain are strengthened. But if you don’t practice, then the weak bonds are broken.
If you try to learn formulas without doing the practice first, then you are just making it more difficult for yourself.

4. Keep a list of symbols

Most math formulas involve some Greek letters, or perhaps some strange symbols like ^ or perhaps a letter with a bar over the top.
When we learn a foreign language, it’s good to keep a list of the new vocabulary as we come across it. As it gets more complicated, we can go back to the list to remind us of the words we learned recently but are hazy about. Learning mathematics symbols should be like this, too.
Keep a list of symbols and paste them up somewhere in your room, so that you can update it easily and can refer to it when needed. Write out the symbol in words, for example: ∑ is “sum”; ∫ is the “integration” symbol and Φ is “capital phi”, the Greek letter.
Just like when learning whole formulas, include a small diagram or graph to remind you of where each symbol came from.
Another way of keeping your list is via flash cards. Make use of dead time on the bus and learn a few formulas each day.

5. Absorb the formulas via different channels

I’ve already talked about writing and visual aids for learning formulas. Also process and learn each one by hearing it and speaking it.
An example here is the formula for the derivative of a fraction involving x terms on the top and bottom (known as the “Quotient Rule”). Then in words, the derivative is:
dy/dx = bottom times derivative of top minus top times derivative of bottom all over bottom squared.
The formula is actually as follows, if we let u = numerator and v = denominator of the fraction, then:

6. Use memory techniques

Most people are capable of learning lists of unrelated numbers or words, as long as they use the right techniques. Such techniques can be applied to the learning of formulas as well.
One of these techniques is to create a story around the thing you need to learn. The crazier the story, the better it is because it is easier to remember. If the story is set in some striking physical location, it also helps to remember it later.

7. Know why

In many examinations, they give you a math formula sheet so why do you still need to learn formulas? As mentioned earlier, if students don’t know what they are doing, they will choose a formula randomly, plug in the values and hope for the best. This usually has bad outcomes and zero marks.
I encourage you to learn the formulas, even if they are given to you in the exam. The process of learning the conditions for how to use the formula and the associated graphs or diagrams, means that you are more likely to use the correct formula and use it correctly when answering the question. This is also good for future learning, because you have a much better grasp of the basics.

8. Sleep on it

Don’t under-estimate the importance of sleep when it comes to remembering things. Deep sleep is a phase during the night where we process what we thought about during the day and this is when more permanent memories are laid down. During REM (rapid eye movement) sleep, we rehearse the new skills and consolidate them.
Avoid cramming your math formulas the night before an exam until late. Have a plan for what you are going to learn and spread it out so that it is not overwhelming.

9. Healthy body, efficient brain

The healthier you are, the less you need to worry about sickness distracting from your learning. Spend time exercising and getting the oxygen flowing in your brain. This is essential for learning.

10. Remove distractions

This one is a problem for those of us that love being on the Internet, or listening to music, or talking to our friends. There are just so many things that distract us from learning what we need to learn.
Turn off all those distractions for a set time each day. You won’t die without them. Concentrate on the formulas you need to learn and use all the above techniques.
When you are done, reward yourself with some media time — but only after you have really accomplished something.

UNDERSTANDING MATHEMATICS LEARNING PROBLEMS --- BY. MWL. JAPHET MASATU

UNDERSTANDING   MATHEMATICS   PROBLEMS.

What Are They?

Students who experience significant problems learning and applying mathematics manifest their math learning problems in a variety of ways. Research indicates that there are a number of reasons these students experience difficulty learning mathematics (Mercer, Jordan, & Miller, 1996; Mercer, Lane, Jordan, Allsopp, & Eisele, 1996; Mercer & Mercer, 1998; Miller & Mercer, 1997.) The following list includes these research-based math disability characteristics.

Characteristics of Students Who Have Learning Problems

Learned Helplessness - Students who experience continuous failure in math expect to fail. Their lack of confidence compels them to rely on assistance from others to complete tasks such as worksheets. Assistance that only helps the student "get through" the current set of problems or tasks and does not include re-teaching the concept/skill, only reinforces the student's belief that he cannot learn math.

Passive Learners - Students who have learning problems often are not "active" learners. They do not actively make connections between what they already know and what they are presently learning. When presented with a problem-solving situation, they do not employ strategies or activate prior knowledge to solve the problem. For example, students may learn that 8 x 4 = 32, but when presented with 8 x 5 = ___, they do not actively connect the process of multiplication to that of repeated addition. They do not think to add eight more to thirty-two in order to solve the problem. Students that have learning problems often believe that students who are successful in math just know the answers. They do not understand that students who are successful in math are good at employing strategies to solve problems.

Memory Problems - Memory deficits play a significant role in these students' math learning problems. Memory problems are most evident when students demonstrate difficulty remembering their basic addition, subtraction, multiplication, & division facts. Memory deficits also play a significant role when students are solving multi-step problems and when problem-solving situations require the use of particular problem solving strategies. A common misconception about the memory problems of these students is that it is an information storage problem; that somehow, these students just never get it stored properly. This belief probably arises because one day the student can do a math task but then the next day they can't. Teachers then re-teach the skill only to have the same experience repeated. In contrast to an information storage problem, these memory deficits are often a result of an information retrieval problem. For these students, instruction should include teaching students strategies for accessing and retrieving the information they have stored.

Attention Problems - Math requires a great deal of attention, particularly when multiple steps are involved in the problem solving process. During instruction, students who have attention problems often "miss" important pieces of information. Without these important pieces of information, students have difficulty trying to implement the problem solving process they have just learned. For example, when learning long division, students may miss the "subtract" step in the "divide, multiply, subtract, bring down" long division process. Without subtracting in the proper place, the student will be unable to solve long division problems accurately. Additionally, these students may be unable to focus on the important features that make a mathematical concept distinct. For example, when teaching geometric shapes, these students may attend to features not relevant to identifying shapes. Instead of counting the number of sides to distinguish triangles from rectangles, the student may focus on size or color. Using visual, auditory, tactile (touch), and kinesthetic (movement) cues to highlight the relevant features of a concept is helpful for these students.

Cognitive/Metacognitive Thinking Deficits - Metacognition has to do with students' ability to monitor their learning: 1.) Evaluating whether they are learning; 2.) Employing strategies when needed; 3.) Knowing whether a strategy is successful; and, 4.) Making changes when needed. These are essential skills for any problem solving situation. Because math is problem solving, students who are not metacognitively adept will have great difficulty being successful with mathematics. These students need to be explicitly taught how to be metacognitive learners. Teachers who model this process, who teach students problem solving strategies, who reinforce students' use of these strategies, and who teach students to organize themselves so they can access strategies, will help students who have metacognitive deficits become metacognitive learners.

Low Level of Academic Achievement - Students who experience math failure often lack basic math skills. Because it takes students with math disabilities a longer time to process visual and auditory information than typical learners, they often never have enough time or opportunity to master the foundational concepts/skills that make learning higher level mathematics possible. Providing these students many opportunities to respond to math tasks and providing these practice opportunities in a variety of ways is essential if these students are to ever master the math concepts/skills we teach. Additionally, teachers need to plan periodic review and practice of concepts/skills that students have previously mastered.

Math Anxiety - These students often approach math with trepidation. Because math is difficult for them, "math time" is often an anxiety-ridden experience. The best cure for math anxiety is success. Providing success starts first with the teacher. By understanding why students are having the difficulties they are having, we are less inclined to place "blame" on the students for their lack of math success. These students already feel they are not capable. The attitude with which we approach these students can be a crucial first step in rectifying the math problems they are having. Providing these students with non-threatening, risk-free opportunities to learn and practice math skills is greatly encouraged. Celebrating both small and great advances is also important. Last, if we provide instruction that is effective for these students, we will help them learn math, thereby helping them to experience the success they deserve.

Math Instruction Issues That Impact Students Who Have Math Learning Problems

Although it is very important to understand the learning characteristics of students with math learning problems, it is also important to understand how math instruction/curriculum issues negatively affect these students (Mercer, Jordan, & Miller, 1996; Mercer, Lane, Jordan, Allsopp, & Eisele, 1996; Mercer & Mercer, 1998; Miller & Mercer, 1997). The following list includes these instruction/curriculum issues as well as how they impact the students described above.

Spiraling Curriculum - Within a spiraling curriculum, students are exposed to a number of important math concepts the first year. The next year, students return to those math concepts, expanding on the foundation established the year before. This cycle continues with each successive year. While the purpose of this approach is logical and may be appropriate for students who are average to above average achievers, the spiraling curriculum can be a significant impediment for students who have math learning problems. The primary problem for these students is the limited time that is devoted to each concept. Students who have math learning problems are never able to truly master the concept/skill being taught. For these students, "exposure" to foundational skills is not enough. Without an appropriate number of practice opportunities, these students will only partially acquire the skill. When the concept/skill is revisited the next year, the student is at a great disadvantage because the foundation they are expected to have is incomplete. After several years, the student has not only "not mastered" basic skills, but has also not been able to make the important connections between those basic skills and the higher level math skills being taught as the students moves through the elementary, middle, & secondary grades.

Teaching Understanding/Algorithm Driven Instruction - Although the National Council on Teaching Mathematics (NCTM) strongly encourages teaching mathematical understanding and reasoning, the reality for students with math learning problems is that they spend most of their math time learning and practicing computation procedures. Because of their memory problems, attention problems, and metacognitive deficits, these students have difficulty accurately performing multi-step computations. Therefore, instructional emphasis for these students is often placed on procedural accuracy rather than on conceptual understanding. This emphasis on algorithm (procedure) proficiency supersedes emphasis on conceptual understanding. An example of this is the process of multiplication. Students who only are taught the procedure of multiplication through drill and practice often do not really understand what the process represents. For example, consider the relationship of the following two multiplication problems: 2 x 4 = 8 and ½ x ¼ = 1/8. When students are asked why the answer in the first problem is greater than its multipliers but the answer to the second problem is less than each of its multipliers, the students are unable to answer why. They have never really understood that the multiplication sign really means "of" and that "2 x 4 = 8" means two groups of fours objects, while " ½ x ¼ = 1/8" means one-half of one-fourth. Teaching understanding of the math processes as well as teaching the algorithms (procedures) for computing solutions is critical for students with math learning problems.

Teaching to Mastery - As described under "Spiraling Curriculum," students with learning problems need many opportunities to respond to specific math tasks in order to master them. Teaching to mastery requires that both the teacher and the student monitor the student's learning progress on a daily basis. Mastery is indicated only when the student is able to perform a math task at 100% accuracy for at least three consecutive days. In situations where student progress is assessed only by unit tests, it is very difficult to determine whether a student has really mastered the skills covered in that unit. Even if the student performs well on the unit test, a teacher cannot be certain that the student actually has reached mastery. Because of the learning characteristics common for these students, it is possible that the student would not score as well if given the same test the next day. Mastery can only be inferred when the student demonstrates consistent mastery performance over time. Such continuous assessment is rare in math classrooms. When evaluation of student progress occurs only by unit tests and the students with learning problems do not perform well, the teacher is left with a difficult dilemma. Does the teacher take additional class time to re-teach the skill, thereby falling behind the mandated curriculum's instructional pace? Conversely, does the teacher instead move on to new material, knowing that these students have not mastered the preceding skills, making it less likely the student will have the prerequisite skills to learn the new information? This no-win situation can be avoided if continuous daily assessment is implemented for these students. It is easier and more time efficient to re-teach an individual math skill the same day of initial instruction, or on the following day. Attempting to re-teach multiple math skills many days after initial instruction is much more difficult and time consuming. Due to the hierarchical nature of mathematics, if students do not master prerequisite skills, it is likely that they will not master future skills.

Reforms That Are Cyclical in Nature - The cyclical nature of mathematics curriculum/instruction reforms creates changing instructional practices that confuse students with learning problems. Like reforms for reading instruction, math instruction swings from primarily skills-based emphasis to primarily meaning-based emphasis dependent on the philosophical and political trends of the day. Most students experience at least one of these shifts as they move through grades K to 12. While students who are average to above average achievers are able to manage these changes in instruction, students who have learning problems do not adjust well to such change.

Application of Effective Teaching Practices for Students who have Learning Problems - Research has identified math instructional practices that are effective for students who have learning problems, but these instructional practices are not always implemented in our schools. These instructional practices are described and modeled in this CD-ROM program. Descriptions also include how the particular characteristics of each instructional strategy complement the learning characteristics of students with learning problems. Guidance is also provided which will help you implement these instructional practices in an organized and systematic way.

How Does This Information Help Me?

Teachers who understand the learning needs of their students are more empowered to provide the kind of instruction their students need. Knowing why a student is struggling to learn math provides a basis for understanding why particular instructional strategies/approaches are effective for him/her. Each of the instructional strategies included in this program has unique characteristics that positively impact the learning characteristics of students who have math learning problems. As you learn about each strategy, you are encouraged to refer often to the learner characteristics described in this section. While reading about each instructional strategy and then watching a teacher model the strategy, note how the specific instructional characteristics of the strategy complement or "match" the learning characteristics of students with math learning problems. The text descriptions for each instructional strategy found in this manual clarify these relationships. The elaborated video clips in the CD-ROM also emphasize how the specific characteristics for each instructional strategy positively impact students who have math learning problems.