Three Skills
For
Algebra
Volume 2
|
Chapters and Appendices
Book Entrance
14 The Formula
14. Direct Use - First Example
14. Direct Use, Second Example
14 Indirect Use - First Example I
14 Indirect Use - Second Example
14 Going Further
Foreword
1. Introduction
2. Implication Rules [4]
3. Chains of Reason [3]
4. Induction Mathematical
4. Romeo and Juliet
6 Old Language
5 Knowledge Islands [2]
7 Arith Skill Check [4 X 2]
Arith Webvideos
7. The Next Chapters
8 The Three Skills
8 VNR-Concise-Encyclopedia
PS. What is a Variable [8]
9. Algebra Talk [7]
10 Two More Skills[5]
11 Why Shorthand
12 Shorthand Usage [10]
13 What's Next
PS: The 4-th Skill For Algebra
14 Compound Interest [6]
15 Linear Equations [5]
16 Painless Proofs
17 Pythagoras
PS I. Distributive Law
PS II. Polynomials
18 Rules of Algebra [20]
19 Functions & Sets
20 Degrees & Radians
21 What's Next
22. Arith & Geometric Sums [2]
23 Summation Notation
24 Your Money [3]
25 Induction & Recursion [4]
26 What's Next
27 Pronouns in Logic
28 Occurrence Tables
29 Contrapositive
30 Truth Tables
31 Indirect Reason
Pathways for Learning
Would you like to show yourself or others how to be algebra
power users?
What is a Variable?
Introduction
Variation between Examples
Variation of Letters
A letter denotes a variable
Cases of Double Variation
Three Notions of a Variable
Constants, Parameters
& Variables
Talking about numbers
Dependent
or Independent
Variable, a Matter of Choice
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Chapter 14
Compound Interest Calculations
(Compound Growth Calculations)
Previous: 14. Direct Use of Compound Interest
Formula, Secoond Example
3 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
and then
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
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
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
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.
More Chapter Sections: [ Up ] [ 14 The Formula ] [ 14. Direct Use - First Example ] [ 14. Direct Use, Second Example ] [ 14 Indirect Use - First Example I ] [ 14 Indirect Use - Second Example ] [ 14 Going Further ]
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