Example

Build an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly.

Answer:

• P₀ = $1000
• P­₁ = 1.0025P­₀ = 1.0025 (1000)
• P­₂ = 1.0025P­₁ = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)
• P­₃ = 1.0025P­₂ = 1.0025 (1.00252(1000)) = 1.00253(1000)
• P­₄ = 1.0025P­₃ = 1.0025 (1.00253(1000)) = 1.00254(1000)

Observing a pattern, we could conclude

• P_m = (1.0025)^m($1000)

Notice that the $1000 in the equation was P₀, 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 [latex-display]{{P}_{m}}={{P}_{0}}{{\left(1+\frac{r}{k}\right)}^{m}}[/latex-display]   In this formula:

• m is the number of compounding periods (months in our example)
• r is the annual interest rate
• k is the number of compounds per year.

View this video for a walkthrough of the concept of compound interest. https://youtu.be/xuQTFmP9nNg

Example

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?

Answer: In this example,

P0 = $3000	the initial deposit
r = 0.06	6% annual rate
k = 12	12 months in 1 year
N = 20	since we’re looking for how much we’ll have after 20 years

So [latex]{{P}_{20}}=3000{{\left(1+\frac{0.06}{12}\right)}^{20\times12}}=\$9930.61[/latex] (round your answer to the nearest penny)

A video walkthrough of this example problem is available below. https://youtu.be/8NazxAjhpJw

Example

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?

Answer: In this example, we’re looking for P₀.

r = 0.04	4%
k = 4	4 quarters in 1 year
N = 18	Since we know the balance in 18 years
P18 = $40,000	The amount we have in 18 years

In this case, we’re going to have to set up the equation, and solve for P₀. [latex-display]\begin{align}&40000={{P}_{0}}{{\left(1+\frac{0.04}{4}\right)}^{18\times4}}\\&40000={{P}_{0}}(2.0471)\\&{{P}_{0}}=\frac{40000}{2.0471}=\$19539.84 \\\end{align}[/latex-display] So you would need to deposit $19,539.84 now to have $40,000 in 18 years.

Example

To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.

P0 = $1000	the initial deposit
r = 0.05	5%
k = 12	12 months in 1 year
N = 30	since we’re looking for the amount after 30 years

If we first compute r/k, we find 0.05/12 = 0.00416666666667 Here is the effect of rounding this to different values:

r/k rounded to:	Gives P­30­ to be:	Error
0.004	$4208.59	$259.15
0.0042	$4521.45	$53.71
0.00417	$4473.09	$5.35
0.004167	$4468.28	$0.54
0.0041667	$4467.80	$0.06
no rounding	$4467.74

If you’re working in a bank, of course you wouldn’t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough - $5 off of $4500 isn’t too bad. Certainly keeping that fourth decimal place wouldn’t have hurt. View the following for a demonstration of this example. https://youtu.be/VhhYtaMN6mo

Examples

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?

Answer: This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,

P0 = $2000	the initial deposit
r = 0.06	6% annual rate
k = 12	12 months in 1 year

So our general equation is [latex]{{P}_{N}}=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}[/latex]. We also know that we want our ending amount to be double of $2000, which is $4000, so we’re looking for N so that P_N = 4000. To solve this, we set our equation for P_N equal to 4000.

[latex]4000=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}[/latex]              Divide both sides by 2000

[latex]2={{\left(1.005\right)}^{12N}}[/latex]                                   To solve for the exponent, take the log of both sides

[latex]\log\left(2\right)=\log\left({{\left(1.005\right)}^{12N}}\right)[/latex]             Use the exponent property of logs on the right side

[latex]\log\left(2\right)=12N\log\left(1.005\right)[/latex]                     Now we can divide both sides by 12log(1.005)

[latex]\frac{\log\left(2\right)}{12\log\left(1.005\right)}=N[/latex]                              Approximating this to a decimal

N = 11.581 It will take about 11.581 years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.

Get additional guidance for this example in the following: https://youtu.be/zHRTxtFiyxc