## Interest

Interest mathematics.

Simple interest. You should know the formula I = Prt, interest equals principal * rate * time.

The amount at the end of the time period is A = P + Prt = P(1+rt).

A better deal is to get compound interest, in which the interest is tallied up on a regular basis, then added to the amount deposited.

We typically represent the amount deposited as P, the interest rate (per year) as r, the number of years as t, and the number of time periods per year as n.

Consider $1000 invested at 12% for a period of 10 years. For simple interest, the amount on deposit at the end of 10 years will be 1000(1+.12*10)=2200.00.

If the interest is compounded annually, one will get the interest at the end of the first year added in. After one year, one has 1000(1+.12). This figure is used for computing the amount at the end of the second year, which is then 1000(1+.12)^{2} =$1254.40.

Do you see where this is going? After 10 years, we have

1000(1+.12)^{10}. = 3105.85.

If the interest is compounded quarterly, we have

1000(1.03)^{40} = $3262.04.

Note that .12/4 = .03 and 10 * 4 = 40.

Suppose the interest is added in every month. After one month, one has

1000(1.01) =$1010.

After a year,

1000(1.01)^{12} = $1126.83.

After 10 years,

1000*(1.01)^{120} =$3300.39.

Some interest uses what is called continuous compounding. In continuous compounding, the interest is accumulated as it is earned. Continuous compounding does not do continued multiplying which is shown above. Instead of multiplying, we use the number e (named for Leonhard Euler). e = 2.7182818…. e has its own key on the TI: it is above the divide key.

We raise e to a power; this key is found to the left of the 4 key.

The theory behind the number e is properly the subject of calculus, as e is defined by using a limit process. Limits are defined and used in calculus. Here I just show you how to use e.

Some of you have already met what is called the pert formula: A = Pe^{rt}.

In this formula P is the amount invested at the beginning of the time period, e is Euler’s number, r is the annual interest rate (as a decimal), and t is the time (in years).

This formula is typically used in biology (population growth), physics (radioactive decay), and other subjects which encounter constant rates of growth or decay.

Using the same figures as above, we have, for $1000 invested at 12% for 10 years,

1000e^{.12*10} =$3320.12.

Note that one gains an additional $11 (approximately) in changing from monthly compounding to continuous compounding.

Regular deposits or repayments:

For those who are familiar with changing a repeating decimal into a fraction, we proceed as follows: Suppose one wishes to change s=0.363636363636… into a fraction.

One takes 100s = 36.3636363636… and subtracts s.

This gives us 99s = 36, or s = 36/99 = 4/9.

The above technique will always give a fraction for a repeating decimal, though not always in lowest terms.

The same method is used with any power series.

We begin with S = 1 + y + y^{2} + y^{3} + … + y^{n-1}.

In this expression y is the ratio between successive terms.

We add up the terms in the power series; then we multiply each term by y (the ratio between successive terms). All that is needed is the distributive law: a(b+c)=ab+ac. This law allows yS (the multiplication by y) to be distributed (spread out) among all of the terms of S. Subtracting results in removing of all the terms, except the first and the last.

Thus yS = y(1 + y + y^{2} + y^{3} + … + y^{n-1}) = y + y^{2} + y^{3} + … + y^{n}.

The same method is used to find S that is used to change a repeating decimal to a fraction.

yS = y + y^{2} + y^{3} + … + y^{n}

S = 1 + y + y^{2} + y^{3} + … + y^{n-1}

yS – S = y^{n} – 1, as all of the middle terms are subtracted out.

S(y – 1) = y^{n} – 1.

Should we wish to do this computation on the TI, there is a pair of functions which will do the terms of the series and the summation for us. The functions are

sum [LIST MATH 5] and seq [LIST OPS 5].

The syntax for seq is seq(expr, var, start, stop{, incr} ). var is the letter for the variable (ususally X). expr is the expression which is added up (in this case y^X). start is the beginning value. stop is the ending value. incr is an optional term for the increment (amount added to the exponent in each step). The assumed value for incr is 1, unless you say otherwise.

All interest is presumed to be annual. 12% interest means 1% per month. 1+.12/12=1.01.

The payment is assumed to be made monthly.

So 12%, on an amount of $1000 invested and compounded monthly, for two years, becomes

1000(sum(seq(X,1.01^X,0,23)=26973.46.

Or one can do the computation by use of the formula shown above and the ^ key:

1000(1.01^24-1)/(1.01-1)=26973.46. This is much easier.

Or we can use the TVM solver, with N = 24, I% = 12, PV = 0, PMT = -1000, P/Y=12, and solve for FV. This is the easiest of all.

We want to know about the amount owed on an account, where money is borrowed. This might be a house purchase, a car purchase, or any other type of loan which is repaid in constant amounts.

Let A represent the amount borrowed, let i represent the interest per time period, let P represent the amount paid, and let y=1+i. Each month the amount still owed is multiplied by y

(that is, A(1+i) = Ay), and P is deducted from the amount owed.

Thus (Ay-P)y=Ay^{2}-Py. Each month the entire amount owed is multiplied by y, and P is subtracted.

0: A

1: Ay – P

2: Ay^{2} – Py – P

3: Ay^{3} – Py^{2} – Py – P

…

n: Ay^{n} – Py^{n-1} – Py^{n-2} – … – Py – P = Ay^{n}– P(1 + y + y^{2} + y^{3} + … + y^{n-1})

Do you see where the power series in y comes in?

n:

If we wish to determine the amount of the regular payment, we assume that the balance is zero after n periods. Then

The same method is used to compute the amount which is accumulated in a savings account, where regular periodic payments are made.

1: P

2: Py+P

3: Py^{2}+Py+p

…

n: Py^{n-1} + Py^{n-2} + … + P^{2} + P = P(1 + y + y^{2} + y^{3} + … + y^{n-1})

=

Same method, same result. One need not compute all the intermediate values; one can do the whole thing at one pass.

This is sufficiently useful that the TI calculator has the formulas pre-coded in an APP. The APP is called Finance 1 1, or TVM Solver. On the old TI-83, it is in FINANCE.

APPS Finance 1 1 brings up the TVM Solver.

N=number of periods (deposits, payments)

I%=interest rate per year (as a percent, thus 6% becomes 6 )

PV = present value (how much do you start with)

PMT = amount paid in or paid out

FV = future value (how much do you end up with)

P/Y=number of payments per year

C/Y=number of compoundings per year (ususally the same as P/Y)

One fills in the known quantities, places the cursor on the unknown value, and does ALPHA ENTER [Solve]

Peculiarity: if PV=0, PMT and FV have OPPOSITE signs.

If FV=0, PV and PMT have opposite signs.

For the problem referenced above, set N=24, I%=12, PV=0, PMT=-1000, P/Y=12, C/Y=12, and place the cursor on FV. Then do SOLVE and the answer appears.

Or you can use the formulas above.

To use the TVM solver with continuous compounding, we resort to the following:

the continuous compounding interest can be converted to an equivalent monthly, quarterly or annual interest by using the pert formula.

8% computed monthly: find e^{.08/12}. Then find (e^{.08/12}-1)*12 = .0802672602.

Place I%=8.02672602 and proceed as before.

Thus $100, deposited monthly at 8%, with the interest compounded continuously, will produce $18,321.90 in 10 years.

This compares to $18,294.60 when the interest is compounded monthly.