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Home < Volume 2 Three Skills For Algebra << Chapter 14. Forward and Backward Use of a Formula

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Chapter 14 
Compound Interest Calculations
(Forwards and Backwars!)

Volume 2, Three Skills for Algebra

This chapters uses the compound interest formula to introduce the idea of using formulas directly and indirectly, that is forwards and backwards, and also to introduce and compare arithmetic and algebraic solutions to problems. The ideas identify, put into words,  a unifying theme for teen and adult education in mathematics & science.   Every formula you meet will be used forwards and backwards. And if you understand the algebraic solution method for one formula, you will be able to understand it for all.  

You are now going to meet the compound interest formula. When you meet a formula for the first time, you should wonder what it does or means. You should wonder where it came from or how it was obtained. In this discussion of the compound interest formula, I just want to show you how to use it and how to manipulate or change it to extract other formulas from it. Examples with and without numbers will be given. You should regard the justification or origin of the formula as a problem, one that you should solve by finding an explanation for it.

You need to do more than read and to use formulas as written. You should be able to change them or modify them into another form that might be useful. The authors of mathematics books write for people with this manipulative ability. For such people, modification of formulas as needed is routine. The aim is to provide that ability. 

1  Compound Interest - Numerical Introduction

When you place (or invest) money in a bank account, a bank pays you money for keeping your money with it, a form of rent for its use of it. The bank is using your money to make loans or investments. The money you are paid is called interest. The amount of interest paid and how often depends on the type of account.

In a compound interest account, a bank adds the interest to your account at the end of a period. This period may be a day, a month, a quarter year, a half-year or a full-year. In each period, all the money in your account is now earning interest. So you now receive interest or rent not only for your original deposit, but also for interest previously added to the account on the completion of each period. Here interest paid at the end of one period will earn interest in future periods. Your money is said to be earning compound interest, or more briefly, compounding. The following table (not in the paperback version) is included to review or introduce how and why the compound interest formulas works.

2  The Compound Interest Formula

The compound interest formula describes the total amount (including the initial amount) which you find in your bank account when all interest earned is put back (reinvested) to earn further interest, rather than being sent to you or being put aside - provided the interest rate is constant.

When you place an initial amount P into an account, it is called the principal. In a compound interest account the following happens. The money in your account grows to an amount A after n periods. (The number n here identifies the number of periods your money stays in the account without any withdrawals, or deposits, except for interest payments at the end of each period.) The amount A is given by the compound interest formula

A = P(1+i)n
In this formula, the interest rate per period is given by the quantity i. The formula should only be used when interest is compounded. Again, compounded means the interest is reinvested at the end of each period with no other deposits or withdrawals, Each interest payment deposited in your account then earns interest (rent from the bank) in the following periods.

Here is a numerical examples with  i = 5% and P = 1000 to show  how or why the formula works. Observe how the amount at the end of a period is the same as the amount at the start of the next period.

Period
n 		Amount at
Start of Period 		Amount of Interest Amount at end of Period 103(1.05)n
1 1000.00 50.00 1050.00 1050.00
2 1050.00 52.50 1102.50  
3 1102.50      
4        
5        
  Observe how the amout at the end of a period is equals 100% of  the initial amount plus 5% of the initial amount. So the amount at the end each period is 105% of 1.05 times the initial amount.
Fill in this table with the aid of a calculator to the nearest penny (two decimal places). Observe the formula use shortens the calculation. Note how the amount at the end of one period becomes the amount at the start of the next.  If you do not like to work with interest calculations, turn this whole chapter into a compound population growth model using the values of  A = P(1+i)n to nearest whole number as an approximation to the whole number of individuals present in the population.

The compound interest formula gives an example of a calculation described in algebraic shorthand notation. To use the compound interest formula someone has to explain or show to you the role of each piece of the shorthand. That is done next.

The final compounded amount A on the left-hand side of the compound interest formula can be computed when three numbers are given, namely

  1. the initial amount, also called the principal P,
  2. the interest rate i, and
  3. the number n of compounding periods, possibly months, in which interest is compounded.
When these numbers or quantities are not given, we can only talk or write about compound interest calculations and not do them. Examples in which calculations are done and numbers appear are given below.
  • Other people thinking perhaps of the word rate rather than the word interest in the phrase interest rate, use the letter r instead of i. The shorthand selected does not matter. Like a play, only the plot is important. The actors or letters can be changed.

We could try to describe the compound interest calculation in words alone. This description might be a good essay assignment in a language course alongside the essay of describing in words alone how to tie a shoelace. The task is formidable. The task should persuade you that the algebraic shorthand notation has a few space-saving advantages, even if it may be difficult to read aloud in an understandable way. Formulas like pictures need to be seen to be fully appreciated. Often, mathematics is better written and not spoken.

Examples of how to use the formula directly and indirectly follow. Try to understand both the numerical (arithmetic) and algebraic solutions.

3  Using The Formula

The compound interest formula A = P(1+i)n involves four quantities, namely A, P, i and n. When any three are known, the fourth can be found. Properties of arithmetic and algebra say how this is done. Read on. The easiest quantity to find is A. In the following examples, we consider the cases where the fourth quantity is A or P or i. We can also consider the case where the fourth quantity is n.1

1A formula for this case will be stated at the end of this chapter. How that formula is obtained or used will not be explained here. Another intellectual debt is created.

3.1  Direct Use: Example 1

Problem:   Find the final amount A of an investment, if the initial amount invested is $1500, the interest rate per year is 8% and the interest is compounded for 4 years.

ARITHMETIC SOLUTION. Here the compounding period is one year. In the compound interest formula A = P(1+i)n we then have

  • the interest rate i = 8% = 8 ×0.01 = 0.08 since 1% = 0.01 = [1/100].
  • the number n of compounding periods is 4 and,
  • the principal P = $1500.
Therefore substitution or replacement yields
A = P(1+i)n = $1500 (1 +.08)4
So the final amount (maturity value)
A = $1500 ×1.36049 = $ 2040.73
to the nearest penny or [1/100]th of a dollar.

Suggestion: check the above calculations (and those done below) by hand or with the help of a calculator.

Note: (1+.08)4 = (1.08)4 = 1.08×1.08 ×1.08×1.08 is the shorthand for the product of the number 1.08 with itself, 4 times. This four-fold product was obtained with the aid of a calculator. 2

2This product can be regrouped. It equals (1.08×1.08)2 and so its calculation involves only two multiplications. Aside: how many multiplications does the computation of (1.08)16 require? The answer is 15 or 4 depending on how this product is computed. Hint: (1.08)16 = (1.08)8 ·(1.08)8 = [ (1.08)8 ]2.

Note: Rates of interests can be written as percentages, fractions or decimals. The percentage form can be changed to a decimal form by replacing the percent sign % by one of its equals 0.01 or ([1/100]). The fraction or decimal form can also be changed into a percentage by multiplying by 100% = the percentage representation of the number 1 = [100/100].

3.2  Direct Use: Example 2

Problem:   The principal amount $1200 is invested for 31/2 years in a compound interest account paying 8% compounded monthly. Find the final amount in the account. (See the solutions below for the meaning of this phrase: 8% compounded monthly.)


ARITHMETIC SOLUTION: Note that the interest rate per month is not 8%. It is instead i = [8%/12] = 2/3% per month. Also the number of periods (here months) is n = 3.5 ×12 = 42 = the number of months in 3.5 years. So we can use all this in the compound interest formula to get by replacement (or substitution)

A = P(1+i)n = $1200 (1 + .08
12 		)(3.5×12) = $1200 (1 + .08
12 		)42
Remember to do calculations inside parentheses before those outside.3

3Suggestion: When you replace an expression by another put the other in parentheses.

With the help of a calculator, the final amount in the account is

A = $1200 (1.00666667)42 = $1200 ×1.321919 = $1586.30
Suggestion: check this with the help of a calculator.

3.2  Direct Use: Example 2

Problem:   The principal amount $1200 is invested for 31/2 years in a compound interest account paying 8% compounded monthly. Find the final amount in the account. (See the solutions below for the meaning of this phrase: 8% compounded monthly.)


ARITHMETIC SOLUTION: Note that the interest rate per month is not 8%. It is instead i = [8%/12] = 2/3% per month. Also the number of periods (here months) is n = 3.5 ×12 = 42 = the number of months in 3.5 years. So we can use all this in the compound interest formula to get by replacement (or substitution)

A = P(1+i)n = $1200 (1 + .08
12 		)(3.5×12) = $1200 (1 + .08
12 		)42
Remember to do calculations inside parentheses before those outside.3

3Suggestion: When you replace an expression by another put the other in parentheses.

With the help of a calculator, the final amount in the account is

A = $1200 (1.00666667)42 = $1200 ×1.321919 = $1586.30
Suggestion: check this with the help of a calculator.

  Using The Formula Backwards

The compound interest formula A = P(1+i)n involves four quantities, namely A, P, i and n. When any three are known, the fourth can be found. Properties of arithmetic and algebra say how this is done. Read on. The easiest quantity to find is A. In the following examples, we consider the cases where the fourth quantity is A or P or i. We can also consider the case where the fourth quantity is n.1

3.3  Indirect Use: Example 3

Problem (Finding the principal): Tom Oublier, lucky Tom, finds he has $1350 in an account today. For the past 21/2 years, the account has been paying Tom 9% compounded monthly. Tom Oublier has forgotten the initial amount he had in the account. What was the initial amount (principal) that he placed or deposited in the account?

There are two ways of getting the result. Both will be given. The advantages of each will be noted.


ARITHMETIC SOLUTION. The principal P is unknown. The principal is the initial amount in the account which we want to find. Now in the compound interest formula A = P(1+i)n,

  • the final amount A = $1350 is given or known,
  • the interest rate i = 9% ×12 = .75% = .0075 is known, and
  • the number of compound interest periods (months) in 2.5 years n = 2.5 ×12 = 30.
Therefore replacement or substitution in the formula A = P(1+i)n gives
$1350 = P(1+.75%)30 = P(1+0.75× 1
100 		)30
Calculation yields first
$1350 = P(1.0075)30
and then
$1350 = P×1.251272
Here (1.0075)30 = 1.251272 represents the number 1.0075 multiplied by itself 30 times. Suggestion: Use a calculator to check this multiplication result.

Now multiplying a quantity P by a nonzero number4 (or quantity) and then dividing by the same number yields the quantity P, no matter what P equals. In our situation,

P = (P×1.251272) × 1.251272 = ($1350) × 1.251272 = $1078.90
The second equality follows from the replacement of P×1.251272 by its equal $1350. The initial amount invested was $1078.90 to the nearest penny or cent. This yields the solution of one problem. In the next solution method, we will see how to algebraically describe the arithmetic solution of all similar problems.


ALGEBRAIC SHORTHAND SOLUTION. There is a pattern in the first solution. This pattern can be followed if we had the same problem again, but with different numbers. For instance, how would we solve the above problem if the 9% was replaced by 8%? We will try to capture the pattern using shorthand notation. This approach is given next. It requires a little more work. But it will give a formula for solving many similar problems.

On your first reading of the shorthand solution you may assume the letter have the values given above. On your second reading, pretend A, i and n have values not known to you - for instance they might be hidden in an sealed enveloped. This second viewpoint is a key to algebra.

First, note the following simple idea. The compound interest formula A = P(1+i)n says the symbol A and the expression P(1+i)n both stand for, represent or give the final amount in the account.5

5More precisely we can say:

  • The symbol A is shorthand for a quantity.
  • The expression P(1+i)n when computed gives the same quantity.
Here the symbol A and the more complicated expression or symbol P(1+i)n both represent the same quantity. So we take the liberty of using one in place of the other, as convenient.

Second, we can assume and use the rule: when B is a nonzero number or quantity, then

P ×B
B 		= (P ×B)×B = P
This rule holds no matter what the number B might be, provided this number B is nonzero. We are now ready (after all this preparation) for a one-line shorthand solution.

We now use the above ideas. We will apply this rule with B = (1+i)n. That is, the value of B is given by (1+i)n whenever the latter is computed. So

P = P×B
B 		= P×(1+i)n
(1+i)n 		= A
(1+i)n 		.
The second equality follows by replacing P×(1+i)n by its equal A. This gives us the formula
P = A
(1+i)n
for P in terms of the other three quantities A, i and n. So we can compute P whenever the values of A, i and n are given. This formula solves many forgotten principal problems at once. This formula for the principal P, the initial amount in a compound interest account, is known as the present value formula. It gives the value Pnow of getting the amount A in n periods from now if you investment Pnow compounds at the rate i for each of these n periods.

using the formula - number substitution. The substitutions A = $1305 , i = [.09/12] = 0.0075, and n = 30 leads to

P = A
(1+i)n 		= $1350
(1+.75%)30
This gives
P = $1350
(1.0075)30 		= $1350
1.251272
or
P = ($1350) × 1.251272 = $1078.90
The result is the same as before. The arithmetic is also the same as before.

Remark (Algebraic Viewpoint). The formula we have found, namely

P = A
(1+i)n
for P in terms of the other three quantities A, i and n can be used whenever the last three quantities are given. We can further use the formula without us repeating the above reasoning. The above reasoning with shorthand notation solves many problems at once. Shorthand notation helps us describe calculations. It also leads us to new ones.

3  Using The Formula Backward, More Examples

The compound interest formula A = P(1+i)n involves four quantities, namely A, P, i and n. When any three are known, the fourth can be found. Properties of arithmetic and algebra say how this is done. Read on. The easiest quantity to find is A. In the following examples, we consider the cases where the fourth quantity is A or P or i. We can also consider the case where the fourth quantity is n.1

3.4  Indirect Use: Example 4

Problem:   Joan places $500 dollars in an investment. Four years later, the investment was worth $645.34. What interest rate, compounded yearly, would give her this amount?

The arithmetic solution given next will be followed by an algebraic solution. The algebraic solution again captures a pattern present in the arithmetic solution.

ARITHMETIC SOLUTION. (Suggestion: look at the shorthand solution first). Here we are given the final amount A = $645.34, the initial amount (principal) P = $500 and the number n = 4 of periods (here years) that the money is earning interest. The compound interest formula says

A = P(1+n)n
With the values given, we may write
$645.34 = $500(1+i)4
From the previous equation, we get two more equalities:
$645.34
$500 		= $500(1+i)4
$500 		= (1+i)4
The first equality is justified since we can replace $645.34 by its equal $ 500(1+i)4 . The second follows since the number (1+i)n is both multiplied and divided by the quantity $500.

The foregoing implies

$645.34
$500.00 		= (1+i)4
Now canceling the dollar units $ from top and bottom6
645.34
500.00 		= (1+i)4
We could replace the number [645.34/500.00] by its decimal form 1.29068 now or later. We will do this replacement later because keeping the fractional form helps us follow each number and its role in our reasoning and manipulations. The aim here is to identify a solution pattern which can be followed in other examples. To isolate the unknown i even further, we raise each side to the ¼ power.7

7When the symbol a denotes a positive number, b = a[(1)/(4)] stands for the number b > 0 which satisfies b4 = a. Here b = 1+i satisfies bn = a when a = (1+i)n.

This yields

|
|
| 		645.34
500 		|
|
| 		¼

 
 = [(1+i)4]¼ = (1+i)
since (B4)[(1)/(4)] = B for any positive number B. So we conclude
|
|
| 		645.34
500 		|
|
| 		¼

 
 		= 1+i
Finally we complete the isolation of the (forgotten) rate i by using the last equation and the equation8 i = (1+i)-1.

8An equation which always holds is called an identity.

 From both together, we get

i = (1+i)-1 = |
|
| 		645.34
500 		|
|
| 		¼

 
 -1
This tells us how to compute the interest rate i.

Next we use a calculator to see

i = [1.29068][¼]-1 = 1.065870997 -1 = .065780997
approximately. But this is a number. To express the answer as a percentage we note
0.065870997
=
(.065870997 ×100)× 1
100
=
6.5870997×0.01
=
6.5870997×1.0%
since 1.0% = .01 = [1/100]. The answer is that the previously unknown and forgotten interest rate i = 6.5870997%.

ALGEBRAIC SHORTHAND SOLUTION. The shorthand solution is as follows. The first part of this solution does not care what values have been acquired by the symbols A, n and P. We are about to solve many problems at once. This shows the power of shorthand notation in manipulating equations.

We will start with the compound interest formula

A = P(1+i)n
Again, if you do not like dealing with letters, suppose A, n and P are shorthand for the numbers given in the arithmetic solution (or any other numbers you want). On your second reading, imagine or give other values for the three quantities in question. As a first step to isolate the interest rate i, we note
A
P 		= P(1+i)n
P
upon replacing the quantity A by its equal P(1+i)n. Now multiplying and dividing by a nonzero quantity P ¹ 0 is the same as multiplying by 1. Therefore the right hand side is just (1+i)n. This observation suggests the intermediate result
A
P 		= (1+i)n
This gives us two expressions whose values when computed should be equal. Now we take the [1/(n)]-th power (n-th root) of the left-handside [(A)/(P)], and then replace it by its equal. This yields9
|
|
| 		A
P 		|
|
| 		1/n

 
 = [(1+i)n]1/n = 1+i

9When a is a positive number, b = a[1/(n)] stands for the number b > 0 which satisfies bn = a.

Therefore

i = (1+i)-1 = |
|
| 		A
P 		|
|
| 		1/n

 
 -1
since adding and then subtracting 1 from i gives i, no matter what quantity or number i is. More briefly,
i = |
|
| 		A
P 		|
|
| 		1/n

 
 -1
like before, except now we have shorthand in place of numbers. I will call this formula, the forgotten interest rate formula.

To return to arithmetic and the use of numbers, we substitute A = $645.34, P = $500  dollars, and n = 4 into the forgotten interest rate formula. This yields

i = |
|
| 		$645.34
$500 		|
|
| 		¼

 
 -1 = é
ê
ë 		645.34
500 		|
|
| 		¼

 
 -1
You have seen the rest of the calculation. So with the aid of a calculator we see as before that i = .06587710 = 6.58771%.

Up to the point of substitution of the given numbers, our reasoning did not depend on the numbers themselves. Any others could be used instead. We can use or employ the forgotten interest rate formula to find i from the equation

A = P(1+i)n
when the quantities A, P and n are given - or can be found first. This again shows the power of the shorthand notation to describe many possible calculations and solutions all at once.

The examples above show the role of algebraic shorthand notation in describing and in changing the way calculations are done. The compound interest formula has been used and manipulated but the origin of this formula has not been described. Ask a mathematics instructor for an explanation.

4  Review and Further Notes

We will review what we have met. We will also state a formula for the exponent n in the compound interest formula A = P(1+i)n. How this formula for n is obtained from the compound interest formula will not be shown here - another intellectual IOU is created.

In money matters dealing with the compound interest formula, we can ask for the final compounded amount given by the direct use of the formula, but we can also ask for the other three quantities. That is, we may solve for the principal, for the interest rate or for the number of compounding periods. The compound interest formula can be viewed as one relationship between four quantities, anyone of which can be solved for or expressed in terms of the other three. In particular, the compound interest formula and equation A = P(1+i)n involves four quantities. When any three are known, the fourth can be found. The easiest quantity to find is A. Given the three numbers and quantities P, i and n, you can find the final amount A by the direct use of the formula. But by indirect use of the compound interest formula, that is by changing or manipulating it, given any three of the four quantities A, P, i and n, we can calculate the fourth. From the compound interest formula

A = P(1+i)n
in its usual form, we can obtain formulas for P, i and n. Their description follows.

  •  

  • The so-called present value formula
    P = A
    (1+i)n
    This present value formula says what amount (or principal) P will grow to the amount A in n periods time if the interest rate is i. Vocabulary: the amount P is called the present value of the final amount A. Further the amount A is called the future or maturity value of P at the end of the n-th period.
  • the interest rate formula
    i = |
    |
    |     		A
    P     		|
    |
    |     		1/n

     
    		                [n] /
     -1 =       /
                \/
    		 __
    A
    P

         		  -1.

  •  

  • A nameless formula for the exponent (or power) n. From the compound interest formula, we can also get or find a expression for n, the number of compound periods in terms of the other three quantities P, A and i. The expression is
    n =   
    log A
    P
    log(1+i)
    Understanding this requires familiarity with logarithms. Using it requires say a calculator with a log button. Again, why or how this last formula is obtained is left as an intellectual IOU.

You have seen the derivation of the first two of the above formulas from the compound interest formula. Explanation of the third is left as anintellectual debt. This chapter has shown the usefulness of algebra and shorthand notation in dealing with the compound interest formula. The further study of powers, roots and logarithms is left to another text.

Further Readings

  1. Mathematics of Finance, 3rd Edition by P. Zima & R. Brown, McGraw-Hill Ryerson Ltd, IBSN: 0-07-549491-4,
  2. The chapter Money Computations below.
 

Postscript: Derivation of formula for n assuming a knowledge of logarithms.

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Parent Center: Help your child or teen learn:

Parent-friendly Work Booklets for ages 3+ to 13 Use these or others to check or build skills. Other booklets are available but these booklets allow parents unsure of themselves in mathematics to help their children. The selection acquired in Canada is published in the USA. So it has a US orientation. In retrospect, the selection shows parents what to check with the booklets or by other ways, the choice is theirs. But in retrospect, the selection does not cover integral and fractions liquid weights and measures - ask the publishers to correct that! For ages 9 to 12 say, parents may compensate by showing boys and girls how to use weights or mass, and further measures in food preparation. Beyond that children may be shown how to measure and calculate angles, lengths and areas [proportional amounts too] directly or by using maps and plans drawns to scale. Learning how to gather and measure all the ingredients, pots and pans for a dish or a meal, along with cleaning up sets the stage for like activities or experiments in science courses, and in developing organizational skills, gives boys and girls a head start. Good luck. At the other extreme, more comprehensive than light, if your motto is McCainian: drill, drill, drill then Toronto mathematician and actor John Mighton's jump math organization has jump math workbooks for at least grades 3 to 8 for at-home and in-school use - training sessions for teachers available. Jump math has been expanding to cover older students. Jump Math Samples: plus Fractions for Grades 3-4 & Grades 5-6 [Read] Free Resources grades 1 to 8 [unread - likely to be good]. and

Mathematics Skills For Ages 3 to 14 - technical!

Skills with take home value - A few ideas

Basic skills include time-date-calendar Matters; money matters; map, plan and scale diagram matters;counting, measuring and figuring; decision making with logic and likelyhood; being careful and being aware of the domino effect of mistakes; reading and writing with precision.

Is your child able to add, subtract and multiply amounts of money, work with fractions, work with clocks and calendars, work with maps and plans, and measure length, weight-mass and volume? Schools may promote your son or daughter without providing basic skills in reading, writing and arithmetic.

Arithmetic and Number Theory Skills

Algebra Starter Lessons

1 Working With Sets
2 Formula Forward Use - Evaluation
3 Solving Linear Equations - Skip first step with students able to solve 1 eqn in 1 unknown.
4 Computation Rules and Function Notation
5 Real Numbers
6 More Less Greater Than Inequalities and Comparison
7 Axioms Logic and Equivalent Equations
8 Unifying Theme For Algebra
9 Proportionality Backwards and Forwards
10 Examples of Algebraic Reasoning
A Origins of Counting and Figuring Methods
B Real Numbers Extrinsic Development


Site coverage of formuala evaluation format, of computation rules and axioms, and of the forward and backward use of formulas and proportionality relations lessens the amount of natural talent needed to understand and explain algebra.

Geometry - maps plans trigonometry vectors

1 Maps Plans Measurement
2 Euclidean Geometry - Constructions + extras
3 Cartesian and Polar Coordinates
4 Lines and Slopes Take 1
5 What is Similarity
6 Trigonometry first steps
7 Complex Numbers
8 Unit-Circle Trigonometry
9 Lines and Slopes Take 2 with tangent function
10 Intersecting Straight Lines and Transversals
11 Parallel Straight Lines and Transversals
12 Function Translating and Rescaling
13 Vectors
14 Degrees to Radians and Radians to Degrees
15 Arc or Inverse Trigonometric Function

Pre-Teen and young teen mastery of skills and practices which should be common with map-plans-diagrams drawn to scale, contour interpretation included, has actual or potential take-home value for daily- and adult-life in solving routine problems. Elevating some practices to principles, axioms or postualates, provides a base for analytic and Euclidean geometry, an analytic view of similarity, and an efficient mastery of trigonometry and complex numbers. Right triangle trigonometry provide an analytic alternative to solving geometric problems by drawing diagrams to scale.

More Algebra

Natural-Logarithms Exponentials Powers Roots
Five Polynomial Operations
Quadratics Geometrically
Functions
5 Factored Polynomial Sign Analysis Examples
Rewriting algebraic substitution as function substitutions

The first topic leads to a full high school level theory for the forward and backward mastery of growth and decay models and for definition, range and domains of radicals, roots and powers. The next two topics make quadratics and polynomials easier to learn and teach. Site coverage of functions turns vertical and horizontal line rules into computation methods for evaluating functions.

70 Calculus Starter Lessons

Calculus Lessons Elsewhere:

  1. How to Ace Calculus: Street Wise Guide - Mostly Text.

  2. Flash Video for Calculus Phobics

They cover basic topics in ways likely to complement your notes, your textbooks and site material. When Goldilocks trespassed in the house of the three bears, she found three bowls of porridge, two not to her liking, and one just right. Different bears have different tastes. As invited guest here and elsewhere, if one or more explanations is not to liking, try another. It may be better or just right.

Unsolicited Advice

Learning to do and high marks if it comes to easy is often deceptive - light rather than deep. For that reason, students with learning difficulties determined not to let it get in their way may go deeper and farther than those with none. High marks, if the come easy, may be deceptive - provide a too light and not a deep mastery. That could have been your problem in secondary school, one that leads to comprehension shock or difficulties in calculus and more generally in the first year of college. Bon Appetite.

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Home < Volume 2 Three Skills For Algebra << Chapter 14. Forward and Backward Use of a Formula

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Logic-Reason for all
Careful Thinking
Chains of Reason
Mathematical Induction
Responsibility
Bodies-of-Knowledge

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science

Geometry
1 Maps + Plans Use
2 Euclidean Geometry
3 Rct +Polr Coordinates
4 Lines-Slopes [I]
5. What is Similarity 		Algebra Starters - the base
1. Better Work Format
2. Solve Linear Eqns
3. Computation Rules
4. Axioms, Item 3 Viewpnt
5. Formulas Backwards 		More Algebra
Logarithms-ax & m/nth roots
Five Polynomial Operations
Quadratics Geometrically
Functions || Vectors too
Arith. Skill Check+Answers 		Calculus Prep/Preview
What is a Variable
Why study slopes
Why factor polynomials
Complex Numbers
Limits + Continuity

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