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## SectionContinuous Growth

###### Supplemental Videos

The main topics of this section are also presented in the following videos:

Recall that in the previous section we found that with a principal $P$ and interest rate $r\text{,}$ if we compound this interest $n$-times a year the value, $A(t)\text{,}$ of the investment after $t$ years is given by the formula

\begin{equation*} A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}. \end{equation*}

As $n$ gets large (i.e. as $n$ approaches infinity) the growth factor in the above formula, $\left(1+\dfrac{r}{n}\right)^n\text{,}$ approaches $e^r$ where $e \approx 2.71828\text{.}$ So, for very large values of $n\text{,}$

\begin{equation*} P\left(1+\dfrac{r}{n}\right)^{nt}\approx Pe^{rt}. \end{equation*}

In general, we have the following definition:

###### Compounded Continuously

The amount, A(t), accumulated (principal plus interest) in an account bearing interest compounded continuously is

\begin{equation*} A(t) = P(e)^{rt} \end{equation*}

where \begin{align*} \amp P \amp\amp \text{is the principal invested,} \\ \amp r \amp\amp \text{is the continuous interest rate,} \\ \amp t \amp\amp \text{is the time period, in years}. \end{align*}

###### Caution200

In the formula above, $e$ is not a variable! It is a constant and does not depend on the given information. It will always be approximately $2.71828\text{.}$

### SubsectionToo Many Rates!

Throughout this chapter we have discussed several different rates which can be very confusing, especially since some of the different names mean the same thing! But remember that the context of the variable $r$ is what determines its label. Let's summarize the different labels we have for this $r$ variable.

Effective interest rate and annual interest rate mean the exact same thing. (These terms exist in finance and business.) If we label this interest rate as $r\text{,}$ then this would be the $r$ appearing in the formula

\begin{equation*} A(t)=a(1+r)^t \end{equation*}

for an interest which is compounded once a year (or annually).

Nominal rate is synonymous with stated (or given) rate. This means that in most problems, if you are told a rate then it is nominal. This means that nominal can apply everywhere! If we label this rate as $r \text{,}$ then this would be the $r$ appearing in each of the following formulas:

• $a(1+r)^t\text{,}$
• $a\left(1+\dfrac{r}{n}\right)^{nt}\text{,}$
• $P(e)^{rt}.$

Continuous interest rate is the interest rate appearing in the formula for interest which is compounded continuously. In other words, if we label this interest rate as $r$ then this would be the $r$ appearing in the formula

\begin{equation*} A(t)=P(e)^{rt}. \end{equation*}

The following example illustrates how to translate between some of these rates.

###### Example201

Let's say you are given a nominal rate of $5$% with a principal investment of one dollar. (Note that the principal investment won't affect our rates.)

1. If we compounded annually, what is the effective annual interest rate?

2. If we compounded monthly, what is the effective annual interest rate?

3. If we compounded continuously, what is the effective annual interest rate?

Solution
1. Looking at our compound interest formula

\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}, \end{equation*}

we know that we are compounding once a year, so $n=1\text{,}$ and the nominal rate is $5$%, so $r=0.05\text{.}$ The principal amount is one dollar, so $P=1\text{.}$ Using this information, we get the formula

\begin{equation*} A(t) = 1\left(1 + \dfrac{0.05}{1}\right)^{1t}= (1 + 0.05)^{t}. \end{equation*}

Since the principal value was one dollar, we can omit that from the multiplication. Since we rewrote the formula to be in the form $A(t)=P(1+r)^{t}\text{,}$ the context has changed to be for annual! So, we can read it from the formula, the effective (or annual) interest rate is $0.05$ or $5$%.

2. Part a is to demonstrate a relatively simple case, but we can follow the same process to find more complicated scenarios. This time, we compounded monthly and so $n=12\text{.}$ We get \begin{align*} A(t) \amp = 1\left(1 + \dfrac{0.05}{12}\right)^{12t} \\ \amp \approx \left[(1 + 0.0041667)^{\alert{12}}\right]^{t} \\ \amp \approx (1.05116189)^t \\ \amp \approx (1 + 0.05116189)^t \end{align*} Since we rewrote the formula to be in the form $A(t)=P(1+r)^{t}\text{,}$ we can read it from the formula, the effective (or annual) interest rate is about $5.116189$%.

3. This time, we compounded continuously and so we have to use the compounded continuously formula: $A(t)=P(e)^{rt}\text{.}$ We get \begin{align*} A(t) \amp = 1(e)^{0.05t} \\ \amp = (\alert{e^{0.05}})^{t} \\ \amp \approx (1.05127109)^t \\ \amp \approx (1 + 0.05127109)^t \end{align*} Since we rewrote the formula to be in the form $A(t)=P(1+r)^{t}\text{,}$ we can read it from the formula, the effective (or annual) interest rate is about $5.127109$%.

You may have noticed that what essentially doing here is finding the growth factor $b$ of the situation and subtracting by $1\text{.}$ That is another way to find the effective annual rate, but the above shows you why it works! You're rewriting the situation to be in the context of annual rates!

NOTE: In the next chapter, you will learn how to go from effective annual rate back to continuous interest rate or other rates!