We have to work with money every day. While balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, when we start saving, planning for retirement, or need a loan, we need more mathematics.
Discussing interest starts with the principal, or amount your account starts with. This could be a starting investment, or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.
\(I = P_0 r\)
\(A = P_0 +I = P_0 + P_0 r = P_0(1+r)\)
\(I\) is the interest
\(A\) is the end amount; principal plus interest
\(P_0\) is the principal (starting amount)
\(r\) is the interest rate in decimal form (example \(5%=0.05\)
A friend asks to borrow $300 and agrees to repay it in 30 days with 3% interest. How much interest will you earn?
\(P_0 = $300\text{,}\) the principal
\(r = 0.03\text{,}\) the 3% interest rate
\(I = $300(0.03) = $9.\)You will earn $9 interest.
One-time simple interest is only common for extremely short-term loans. For longer term loans, it is common for interest to be paid on a daily, monthly, quarterly, or annual basis. Therefore we will focus our attention on what is known as compound interest.
With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.
Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?
The 3% interest is an annual percentage rate (APR) the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn \(3%/12 = 0.25%\) per month.
In the first month,
\(P0 = $1000\)
\(I = $1000 (0.0025) = $2.50\)
\(A = $1000 + $2.50 = $1002.50\)
In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50
In the second month,
\(P_0 = $1002.50\)
\(I = $1002.50 (0.0025) = $2.51 \) (rounded)
\(A = $1002.50 + $2.51 = $1005.01\)
Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding of interest gives us.
Calculating out a few more months yields the following table:
Month
Starting balance
Interest earned
Ending Balance
1
1000.00
2.50
1002.50
2
1002.50
2.51
1005.01
3
1005.01
2.51
1007.52
4
1007.52
2.52
1010.04
5
1010.04
2.53
1012.57
6
1012.57
2.53
1015.10
7
1015.10
2.54
1017.64
8
1017.64
2.54
1020.18
9
1020.18
2.55
1022.73
10
1022.73
2.56
1025.29
11
1025.29
2.56
1027.85
12
1027.85
2.57
1030.42
To find an equation to represent this, if Pm represents the amount of money after m months, then we could write the recursive equation:
\(P_0 = $1000\)
\(P_m = (1+0.0025)P_{m-1}\)
You probably recognize this as the recursive form of exponential growth. If not, we could go through the steps to build an explicit equation for the growth:
\(P_0 = $1000\)
\(P_1 = 1.0025P_0 = 1.0025 (1000)\)
\(P_2 = 1.0025P_1 = 1.0025 (1.0025 (1000)) = 1.0025^2(1000)\)
\(P_3 = 1.0025P_2 = 1.0025 (1.0025^2(1000)) = 1.0025^3(1000)\)
\(P_4 = 1.0025P_3 = 1.0025 (1.0025^3(1000)) = 1.0025^4(1000)\)
Observing a pattern, we could conclude
\(P_m = (10025)^m($1000)\)
Notice that the $1000 in the equation was \(P_0\text{,}\) the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year. Generalizing our result, we could write
In this formula \(m\) is the number of compounding periods (months in our example), \(r\) is the annual interest rate, and \(k\) is the number of compounds per year.
While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If \(N\) is the number of years, then \(m = N k\text{.}\) Making this change gives us the standard formula for compound interest.
\(P_N\) is the balance in the account after \(N\) years.
\(P_0\) is the starting balance of the account (also called initial deposit, or principal)
\(r\) is the annual interest rate in decimal form
\(k\) is the number of compounding periods in one year.
If the compounding is done annually (once a year), \(k = 1.\) If the compounding is done quarterly, \(k = 4.\) If the compounding is done monthly, \(k = 12\text{.}\) If the compounding is done daily, \(k = 365\text{.}\)
The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.
A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?
In this example, \(P_0 = $3000 \) is the initial deposit; \(r = 0.06\) since the annual rate is 6%; \(k = 12\) since there are 12 months in 1 year; and \(N = 20\) since were looking for how much well have after 20 years.
So \(P_{20} = 3000 \left( 1+\frac{0.06}{12} \right)^{20 \times 12} = $9930.61\) (rounded to the nearest penny).
When we need to calculate something like \(5^3\) it is easy enough to just multiply \(555=125\text{.}\) But when we need to calculate something like \(1.005^{240}\text{,}\) it would be very tedious to calculate this by multiplying 1.005 by itself 240 times! So to make things easier, we can harness the power of our scientific calculators.
Most scientific calculators have a button for exponents. It is typically either labeled like:
\([\) ^ \(]\text{,}\) \([y^x]\text{,}\) or \([ x^y ]\)
To evaluate \(1.005^{240}\) we'd type [1.005] [ ^ ] [240] or [1.005] [ \(y^x\) ] [240]. Try it out - you should get something around 3.3102044758.
You know that you will need $40,000 for your childs education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?
In this example, were looking for \(P_0\text{.}\)
We know \(r = 0.04 \) (since the interest is 4%), \(k = 4\) (since there are 4 quarters in 1 year), \(N = 18\) since this is when we want to have achieved the desired balance, and \(P_{18} = $40,000\) since this is the amount we want in 18 years.
In this case, were going to have to set up the equation, and solve for \(P_0\text{:}\)
\(40000 = P_0 \left( 1+\frac{0.04}{4} \right)^{18 \times 4}\)
\(40000 = P_0 (2.0471)\)
\(P_0 = \frac{40000}{2.0471} = $19539.84\)
So you would need to deposit $19,539.84 now to have $40,000 in 18 years.
