The Natural Base e

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The number e,e, commonly called the natural base, is a mathematical constant named by the mathematician Leonhard Euler. It is an irrational number, meaning it can't be expressed as a fraction of two integers.

e=2.7182818284e = 2.7182818284\ldots

This constant appears in several areas of mathematics and has multiple uses. For instance, it is often used as the base of exponential functions.
Explanation

Deriving ee

The value of the number ee can be found in different ways. Here, compound interest will be used to find its value. The formula for compound interest is the exponential growth function y=P(1+rn)nt, y=P\left(1+\frac{r}{n}\right)^{nt}, where the constant rr is the interest rate in decimal form. If the interest rate was 100%,100\,\%, which is extremely profitable, the value of rr equals 1.1. y=P(1+1n)nt. y=P\left(1+\frac{1}{n}\right)^{nt}. The number nn is the amount of times the interest is compounded each year. That is, how often the accrued interest is added to the balance. The more often the interest is compounded, the higher the profit will be each year. What happens if the interest is compounded very often? That is, when nn is large. To examine this, the function will be rewritten using the the power of a power property. y=P(1+1n)nt=P((1+1n)n)t y=P\left(1+\frac{1}{n}\right)^{nt}=P\left(\left(1+\frac{1}{n}\right)^n\right)^t For the rewritten function, the expression inside the outer parentheses is a constant depending only on n.n. To analyze what happens when nn increases, larger and larger values will be substituted into the expression.

nn (1+1n)n\left(1+\dfrac{1}{n}\right)^n Expression value
1010 (1+110)10\left(1+\dfrac{1}{10}\right)^{10} 2.593742.59374\ldots
100100 (1+1100)100\left(1+\dfrac{1}{100}\right)^{100} 2.704812.70481\ldots
10001000 (1+11000)1000\left(1+\dfrac{1}{1000}\right)^{1000} 2.716922.71692\ldots
1000010\,000 (1+110000)10000\left(1+\dfrac{1}{10\,000}\right)^{10\,000} 2.718142.71814\ldots
100000100\,000 (1+1100000)100000\left(1+\dfrac{1}{100\,000}\right)^{100\,000} 2.718262.71826\ldots

The table shows that for high values of n,n, the expression seems to approach 2.718.\sim2.718. In fact, as n,n\rightarrow \infty, the value of the expression approaches the natural base ee: e=2.71828 e=2.71828\ldots Substituting ee for the expression gives the function y=Pet, y=Pe^{t},

which applies when the interest rate is 100%,100\,\%, compounded infinitely often. When interest is compounded infinitely often, it is said to be continuously compounded. Using this exponential function, instead of the original formula, makes this growth easier to work with.
Concept

Natural Base Exponential Function

Exponential functions are commonly expressed using the natural base ee, as they then exhibit useful characteristics. These functions are called natural base exponential functions and are written in the form f(x)=aerx. f(x) = ae^{rx}. If aa and rr are both positive, the function is an exponential growth function. If aa is positive and rr is negative, the resulting function is instead an exponential decay function.

Exercise


Determine whether the exponential function f(x)=0.5e0.3x f(x) = 0.5 e^{0.3x} shows growth or decay. Then, rewrite the function in the form f(x)=abxf(x) = ab^x and graph it.

Solution

The function is currently expressed as a natural base exponential function, f(x)=aerx. f(x) = ae^{rx}. Thus, identifying the signs of the constants aa and rr will help us decide whether it is an exponential growth or decay function. Both aa and rr are positive in this case. Thus, it models an exponential growth. To rewrite the function as f(x)=abx, f(x) = ab^x, we have to make sure that the exponent is nothing but x.x. This is achieved using the power of a power property and calculating the new base.

f(x)=0.5e0.3xf(x) = 0.5e^{0.3x}
f(x)=0.5(e0.3)xf(x) = 0.5\left(e^{0.3}\right)^x
f(x)=0.5(1.34985)xf(x) = 0.5 \cdot (1.34985\ldots)^x
f(x)0.51.35xf(x) \approx 0.5 \cdot 1.35^x

Now that the function is rewritten, we can graph it as usual, starting by plotting the initial value, 0.5.0.5.

Next, more points are found by repeatedly increasing the xx-value by 11 and multiplying the function value by the constant multiplier, 1.35.1.35.

Connecting the points with a smooth curve gives us the desired graph.


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Rule

Continuously Compounded Interest

When an interest rate of 100%100\,\% is compounded continuously, meaning it is compounded infinitely often, the resulting function is y=Pet. y = Pe^t. When the interest rate is something other than 100,100, the arbitrary rate rr is used. y=P(1+rn)nt. y = P \left( 1 + \dfrac r n \right)^{nt}. Suppose that the interest rate doubles, r=2.r = 2. Because interest is compounded continuously, this effectively leads to the same growth in half the time. Similarly, three times the interest leads to the same growth in a third of the time. This corresponds to a horizontal stretch or shrink, leading to the following function.

y=Perty = Pe^{rt}

Exercise


Carlos won the lottery! In total, his winnings are $50000.\$50\,000. He wants to buy a high quality miniature blimp, which costs $66499.\$66\,499. Since his winnings aren't enough, he's decided to deposit them into a savings account for 55 years, earning 6%6\,\% yearly interest compounded continuously. Assuming the price stays constant, will he have enough money in his account to buy the miniature blimp after the 55 years?

Solution

As the interest of the account is continuously compounded, the balance can be modeled with the function y=Pert, y = Pe^{rt}, where PP is the principal and rr is the interest rate in decimal form. In this case, the principal is $50000\$ 50\,000 and the interest rate is 0.06,0.06, resulting in the function y=50000e0.06t. y = 50\,000e^{0.06t}. By substituting t=5t = 5 into the function, we'll find the balance at 55 years.

y=50000e0.06ty = 50\,000e^{0.06t}
y=50000e0.065y = 50\,000e^{0.06 \cdot {\color{#0000FF}{5}}}
y=50000e0.3y = 50\,000e^{0.3}
y=67492.94038y = 67\,492.94038\ldots
y67493y \approx 67\,493

At 55 years, his account balance will be $67493.\$ 67\,493. Thus, he will be able to afford the miniature blimp at $66499.\$ 66\,499.

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Exercises

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