# Exponential Growth

This is a Helper Page for:
Euler's Number
Bedsheet Problem
Taylor Series

# Basic Description

Exponential growth (or exponential decay if the growth rate is negative) is produced by a mathematical function with a variable exponent. As the variable changes, the value of the function increases (or decreases) in proportion to its current value.

An example of such a function is f (x) = 2x. Every time x increases by 1, the value of the function doubles, or in other words, grows by 100% of its existing value. For the function g (x) = 0.5x, each time x increases by 1, the value of the function halves, or decreases by 50%.

 x f1 (x) = 3x f2 (x) = 2x f3 (x) = 1.5x f4 (x) = 1.2x f5 (x) = 0.8x f6 (x) = 0.3x 0 1 1 1 1 1 1 1 3 2 1.5 1.2 0.8 0.3 2 9 4 2.25 1.44 0.64 0.09 3 27 8 3.375 1.728 0.512 0.027 4 81 16 5.0625 2.074 0.4096 0.0081 5 243 32 7.59375 2.488 0.32768 0.00243 ... ... ... ... ... ... ... 10 59049 1024 57.67 6.192 0.107 0.0000059 20 3486784401 1048576 3325.26 38.338 0.012 3.487 × 10-11

As shown in the table above for various exponential functions, for equally spaced x values, the value of the function grows or shrinks by a constant factor. For example, $\frac{f_1(x+1)}{f_1(x)}=\frac{3^{x+1}}{3^x}=3$

for all x values.

Exponential growth (decay) is often seen and used in the studies of world population, natural resources, half-life decay, economic growth, interest compounding and other domains.

# General Formula

## For the Discrete Case

The basic formula for discrete exponential growth is $x_t = x_0(1+r)^t$

where x0 is the initial value, r is a constant representing the growth rate, and xt is the value after t periods of time. Here, discrete means that in the function, the possible values for t are some distance apart from each other. For example, in some cases, the function is defined where t can only be integers, so numbers such as 1.5 are not allowed when we plug in t values.

When r > 0, the function experiences exponential growth. This is because the of the function, 1 + r, is greater than 1. For example, if r = 40%, it means the value of the function increases by 40% (becomes 1.4 times its current value) each time t increases by 1.

On the other hand, when r < 0, the function experiences exponential decay because the base is smaller than 1. If r = -40%, each time t increases by 1, the value of the function decreases by 40% (shrinks to 60% of its current value).

Let's look at two common real-world scenarios of discrete exponential growth and decay to understand this formula: the compounded interest problem and elimination of a drug from the body.

### Exponential Growth: Compound Interest

At the beginning of the year, you deposit $1000 in your bank account, and the annual interest rate is 5%. Assuming no other deposits or withdrawals and a constant interest rate, what will be the value of the account after t years? One way to look at this problem is to use a table. Another way is to look at the graph. Year Amount of Money Calculation Method Graph 0$1000.00
1 $1050.00 1000.00 × (1+5%) = 1050.00 2$1102.50 1050.00 × (1+5%) = 1102.50
3 $1157.63 1102.50 × (1+5%) ≈ 1157.63 4$1215.51 1157.63 × (1+5%) ≈ 1215.51
5 $1276.28 1215.51 × (1+5%) ≈ 1276.28 ... ... ... 15$2078.93 1000 × (1+5%)15 ≈ 2078.93
25 $3386.35 1000 × (1+5%)25 ≈ 3386.35 50$11467.40 1000 × (1+5%)50 ≈ 11467.40
80 $49561.44 1000 × (1+5%)80 ≈ 49561.44 From the table and graph we can see that, every year, the money you have grows by 50%, and the difference between two consecutive years becomes larger as t increases. ### Exponential Decay: Elimination of a Drug from the Body When people take medicine, the drug gets metabolized and eliminated at a constant rate. Suppose the initial amount of a drug in the body is 200mg and it is eliminated at a rate of 30% per hour. Let Q denote the amount of drug left in the body. Written as a function, $Q(t) = 200(1-30%)^t = 200(0.7)^t$ We can use the following table and graph to illustrate the process. Time Amount of Drug in Body (mg) Calculation Method Graph 0 200 1 140 200 × (1-30%) = 140 2 98 140 × (1-30%) = 98 3 68.6 98 × (1-30%) ≈ 68.6 4 48.0 68.6 × (1-30%) ≈ 48.0 5 33.6 48.0 × (1-30%) ≈ 33.6 ... ... ... 9 8.1 200 × (1-30%)9 ≈ 8.1 12 2.8 200 × (1-30%)12 ≈ 2.8 15 0.9 200 × (1-30%)15 ≈ 0.9 From above we can see that as time elapses, the amount of drug left in the body decreases, and the rate of the elimination slows down gradually. ## For the Continuous Case In the compounded interest rate problem, if the interest is paid more frequently than once a year, say every 3 months, the amount of interest you get at the end of the year will be different: • If paid once a year, you will get$1000 × (1+5%) = $1050 at the end of one year. • If the bank calculates interest every 3 months, the quarterly interest rate is 5% ÷ 4 = 1.25%. The amount of money you have at the end of the year is$1000 × (1+1.25%)4 = \$1050.95.

See below for a more detailed table showing the different results you would get if the interest is compounded in different frequencies:

Note:

The calculation method for a given compounding frequency is $x_t = 1000(1 + \frac{0.05}{n})^{nt}$

where xt represents the money you have at time t, and n represents the number of times the bank compounds interest in a year.

The last column of the table uses the formula $x_t=1000 e^{0.05t}$

which is the basic formula for continuous exponential growth functions and will be explained soon.

 Year Number of Times Interest is Compounded per Year Use e 1 2 5 10 50 500 0 1000 1000 1000 1000 1000 1000 1000 1 1050.00 1050.63 1051.01 1051.14 1051.12 1051.27 1051.27 2 1102.50 1103.81 1104.62 1104.90 1105.12 1105.17 1105.17 3 1157.63 1159.69 1160.97 1161.40 1161.75 1161.83 1161.83 4 1215.51 1218.40 1220.19 1220.79 1221.28 1221.39 1221.40 5 1276.28 1280.08 1282.43 1283.23 1283.87 1284.01 1284.03 ... ... ... ... ... ... ... ... 10 1628.89 1638.62 1644.63 1646.67 1648.31 1648.68 1648.72 20 2653.30 2685.06 2704.81 2711.52 2716.92 2718.15 2718.28

As shown in the table, if we compound the interest more often, the amount of money you have at a given time will be closer and closer to 1000e0.05t. What if we compound interest 10,000 times a year, or even more frequently? Then we are facing a continuously compounding problem, in which the change happens instantaneously.

In fact, Euler's Number, is closely connected to continuous compounding. Let r represent the annual growth rate, and n represent the number of times the compounding happens during one year: $\lim_{n\rightarrow \infty}(1+\frac{r}{n})^n = e^r$

Here is a link to an online calculator that shows this limit is reasonable.
(Note: I suggest not to input too big a value for n. The result may become inaccurate when n exceeds 1,000,000,000,000 because of the program)

Recall the formula for the discrete case that $x_t = x_0(1+r)^t$ where the base of the function, 1 + r, represents the ratio between the amount of money you have after one year and the amount you initially have.

If interest is compounded n times a year, this ratio is $(1+\frac{r}{n})^n$

Thus, the amount of money you have after t years is $x_t = x_0(1+\frac{r}{n})^{nt}$

If the frequency of compounding approaches infinity, $\lim_{n\rightarrow \infty}(1+\frac{r}{n})^{nt} = e^{rt}$

Therefore, the general formula used for continuous exponential growth problems is $x_t=x_0 e^{rt}$

where x0 is the initial value of the function, and r represents the growth rate.

Similar to the discrete case, when r is positive, we have exponential growth, and when r is negative, we have exponential decay. This is because as time passes, e is raised to an increasingly large power if the growth rate is positive, which leads the function to increase more. However, if the growth rate is negative, e is raised to an increasingly negative power. In other words, 1/e is raised to a higher power. This leads the function to decrease.

To derive this formula, we need to use differential equations.

Write the function in terms of x and t. Say the rate of growth is r, and the initial value of x is x0, a positive constant. Then in the first form of equation, the function can be written as

Eq.1 $x_t=x_0 (1+r)^t$

According to the definition of exponential growth, for each time interval, the value of the function increases proportionally to its current value. In this case, the proportion is r. Therefore, $\frac{dx}{dt}=rx$

Separating the variables, $\frac{1}{x} dx = r dt$

Integrate both sides: $\int {\frac{1}{x} dx} = \int{r dt}.$

We get $\ln x = rt + C$ where C is a constant

Solving for x, the equation becomes $x = e^{rt+C} = e^C e^{rt}$

We know from Eq.1 that when t = 0, x = x0.

Therefore, eC = x0.

Hence, $x = x_0 e^{rt}$.

# Half-Life Decay

Half-life, usually denoted as $t_{\frac{1}{2}}$, is defined as the amount of time needed for a system undergoing exponential decay to decrease to half of its initial value. The term is usually used in describing the characteristics of radioactive elements and testing pharmaceutical substances. For example, the half-life of carbon-14 is about 5730 years, which means that it takes about 5730 years for carbon-14 to decay to half of its original amount(the decayed half becomes nitrogen-14).
Half-life does not depend on the initial amount of the substance. Rather, it depends only on the rate of decay.
Use the general formula for continuous exponential decay: $x_t=x_0 e^{rt}$ with r < 0

We are looking for the point where $\frac{x_t}{x_0}=\frac{1}{2}$

that is the case when $e^{rt} = \frac{1}{2}$.

Solving this equation, we get <template>AlignEquals |e1l= \ln (e^{rt}) |e1r= \ln \frac{1}{2} |e2l= rt |e2r= \ln \frac{1}{2} |e3l= t |e3r= \frac{\ln \frac{1}{2} }{r} </template> Therefore, $t_{\frac{1}{2}}=\frac{\ln \frac{1}{2} }{r} \approx \frac{-0.6931}{r}$

We can use this formula to calculate the half-life of any system given the value of r.

# Doubling Time and the Rule of 70

How long does it take for a substance undergoing exponential growth to double? We can use the method above to find that $t_{double} = \frac{\ln 2}{r}$.

The rule of 70 states that the amount of time needed for a quantity to double when it is growing exponentially at a constant percentage rate can be approximated by dividing 70 by that percentage number(note: 70 is divided by the percentage number, not the percentage rate. If the percentage rate is x%, the percentage number is just x, not 0.01x.) For example, if the population is growing by 2% annually, its doubling time is approximately 70/2 = 35 years.

To understand the math behind the rule of 70, we use the formula for continuous exponential growth.

To understand the math behind the rule of 70, we use the formula for continuous exponential growth.

Let $x(t) = x_0 e^{rt}$ where x0 is the initial quantity, x(t) is the quantity after t periods of time, and r is the growth rate given as a percent. So to find the amount of time needed for the quantity to double, we need: $rt = \ln 2 \approx 0.69$

Then, the doubling time is $t = \frac{0.69}{r}$.

We can then multiply both the numerator and denominator by 100 and round up from 69 to 70. Therefore, the doubling time $t = \frac{70}{100r}$.

Note that 100r is the percentage number of the growth rate. Thus the rule of 70 holds as a valid approximation of the doubling time.

# Exponential vs. Polynomial Growth Exponential growth in contrast with linear and polynomial growth

Compared to linear and polynomial growth, exponential growth may have a relatively slow start, but as time elapses, a small amount can grow extremely rapidly and become astronomically large. Some of the very famous stories about such growth are the Bedsheet problem and King's problem, both showing that exponential growth becomes faster and faster. This is because, though the growth rate, or , is constant over time, the becomes bigger as the function grows.

We can also use calculus to explain this. For a function, its absolute rate of change is exactly its derivative. For a function that is growing linearly, its derivative is a constant, while for a function that is growing exponentially, its derivative increases as the variable grows. For example, if f(x) = 4x, then f′(x) = 4; if g(x) = ex, then g′(x) = ex, a value that grows as x does. Therefore, exponential growth gets faster as the variable increases. Soon, it is growing at a higher rate than polynomial growth, and eventually, the value of the exponential function exceeds that of the polynomial function.