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Exponential functions are always increasing (for a > 0) and concave up.

> plot(2^x, x=-2..2);

> plot(2^x, x=-5..5);

> plot(2^x, x=-10..10);

- Plot the following on the same graph.
- 2
^{x}and 3^{x}between -2 and 2. - 2
^{x}and 2^{2x}between -2 and 2. - 2
^{x}and 2^{x+1}between -2 and 2. - 2
^{x}and 3*2^{x}between -2 and 2. - 2
^{x}and x*2^{x}between -2 and 2.

- 2

Let f(x) = a^{x} and consider f(*x+h*)/f(*x*)
= a^{x+h}/a^{x} = a^{h}.
Thus the ratio of a exponential functions at equal distance are equal.
That is, for fixed *h* f(*x+h*)/f(*x*) is the same regardless of
*h*.
Often data is presented in the form of points, this gives us a useful way of
checking whether a set of data is related by an exponential function.

Consider the following table which gives the worlds population in millions at 50 year intervals from 1750 to 1950:

Year | 1750 | 1800 | 1850 | 1900 | 1950 |
---|---|---|---|---|---|

Population | 700 | 900 | 1200 | 1600 | 2200 |

> L := [ 700, 900, 1200, 1600, 2200 ];

> evalf(L[i]/L[i+1] $ i = 1..4);

f(t) = 700 2

(2

> f := t -> 700 * 2^(.4344028242 * (t - 1750)/50);

> f(1800);

> f(1850);

> f(1900);

> f(1950);

Use this approximation to estimate the population in 2000.

In fact this estimate is somewhat low.

- Use the method above to check if the following data for the population of
the United States, in millions, has an exponential form. If it does try to
guess what this function might be and use it to estimate the population in
2000.
Year 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 Population 76 92 106 123 131 150 179 203 227 250

Where e = 2.718281828459...

Maple represents the exponential function by

> plot(exp(x), x=-3..3);

> evalf(exp(1));

- Plot the following graphs.
- exp(-x)
- exp(x) + exp(-x)
- exp(x^2)
- exp(-x^2)

The exponential function has the property that it is its own derivative, i.e.

D(exp)(x) = exp(x).

The exponetial function is the underlying function in many applications.

This is particularly useful in the context of differential equations of the
form

dy/dx = ky.

y = A*exp(kx) satisfies this equation for any constant value of A. (See section
6.5 of Stewart.)

Consider the graph of exp(-x^2):

> plot(exp(-x^2), x=-3..3);

This is the graph of the *normal distribution* in probability, with mean 0
and standard deviation 1.

In general exp(-((x - m)/s)^2) is the graph of the
normal distribution with mean m and standard deviation s.

- Plot the following normal distributions on the same graph as
exp(-x^2).
- mean 0, s.d. 2
- mean 0, s.d. 3
- mean 1, s.d. 2
- mean -1, s.d. 3

Over the complex plane the exponential function is related to the
trigonometric functions:

exp(I x) = cos(x) + I sin(x)

Substituting x and -x and solving gives

cos(x) = (exp(I x) + exp(-I x))/(2 I)

sin(x) = (exp(I x) - exp(-I x))/2.

Over the reals a similar expression relates the exponential function to the
hyperbolic sine and hyperbolic cosine functions, which are defined by

cosh(x) = (exp(x) + exp(-x))/2

sinh(x) = (exp(x) - exp(-x))/2

- Plot exp(x), exp(-x) and (exp(x) + exp(-x))/2 on the same graph.
- Plot exp(x), exp(-x) and (exp(x) - exp(-x))/2 on the same graph.

> exp(ln(x));

> plot(ln(x), x =-1..20);

- Plot exp(x), ln(x) and the line y = x on the same graph.

Logarithms are useful for placing one exponential in terms of another.

If *c* = ln(*a*) then *a ^{x} = e^{cx}*

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