y_{n} = y_{0} · m^{ n}.The geometric sequence is completely described by giving its starting value y_{0} and the multiplication factor m. For the above example y_{0} = 1 and m = 3. Another example is the geometric sequence {40, 20, 10, 5, 2.5, …} for which y_{0} = 40 and m = 0.5.
y = y_{0} · m^{ x}.It gets its name from the fact that the variable x is in the exponent. The “starting value” y_{0} may be any real constant but the base m must be a positive real constant to avoid taking roots of negative numbers.
y = y_{0} · e^{ x},with base e is the only one whose slope equals its own height everywhere. (Recall that the number e, which is approximately 2.718, was introduced in Section 12.2 as the base used for natural logarithms.) This property makes this exponential function an important standard in calculus where we study the slopes of different functions.
y = y_{0} · e^{ bx},is the only function whose slope is b times its own height everywhere and which goes through the point (x = 0 , y = y_{ 0} ). (b may be positive or negative.) We prove all these assertions below.
n (1 + 1 / n)^{ n} 1 2 2 2.25 10 2.59374 10^{3} 2.71692 10^{6} 2.71828
Summary: The function y = e^{ x} is the function whose slope equals its own height everywhere and which goes through the point (x = 0 , y = 1). Furthermore, the function: y = y_{0} · e^{ bx}is the function whose slope is b times its own height everywhere and which goes through the point (x = 0 , y = y_{ 0} ). (b may be positive or negative.) |
y = y_{0} · m^{ x}.From now on we will replace the variable x in the exponential function by the variable t to represent time. Unless stated otherwise t will be measured in seconds. The exponential function now looks like this:
y = y_{0} · m^{ t}.We already know that y has the value y_{0} when t = 0 and that y is multiplied by a factor m whenever time t is increased by 1 second. If m > 1 then y grows with time and if m < 1 then y decays with time (we call the growth or decay exponential because t is in the exponent). We have seen how we can change the base by using the exponential form of the change of base formula. In this section we show how using various bases or putting the exponent in various forms can illuminate various features of the exponential growth or decay.
y = y_{0} · m^{+ t},with positive exponent always means exponential growth, and:
y = y_{0} · m^{− t},with negative exponent always means exponential decay. The growth or decay is by the factor m every second. For example the function y = (¼)^{ t} describes exponential decay. Using the rules of exponents we can rewrite it as:
y = (¼)^{ t} = (4^{−1})^{ t},or simplified,
y = 4^{ − t}.These two forms of the function show that the statements “y is multiplied by a factor of 1/4 every second” and “y decays by a factor of 4 every second” are equivalent.
y = y_{0} · b^{+ t / T},where b is any desired base > 1, and the value of T depends on b. This form is useful because it shows that y grows by a factor b each time t is increased by T seconds as we can see from the following table of values of y (that is, of the expression y_{0} · b^{+ t / T} ) versus t:
t y 0 y_{0} T y_{0} · b^{+ T / T} = y_{0} · b 2 T y_{0} · b^{+ 2 T / T} = y_{0} · b^{ 2}
y = y_{0} · b^{ − t / T},showing that y decays by a factor b when t is increased by T seconds.
y = y_{0} · e^{+ t / τ},showing that y grows by a factor of e or approximately 2.7 every τ seconds. The decay formula reads:
y = y_{0} · e^{ − t / τ},showing that y decays by a factor of e or approximately 2.7 every τ seconds (or decays to e^{ − 1} ≈ 37% of its former value every τ seconds). This can be seen from the following table of values of y (that is, of the expression y_{0} · e^{ − t / τ} ) versus t:
t y 0 y_{0} τ y_{0} · e^{ − τ / τ} = y_{0} · e^{ − 1} ≈ 37% of y_{0} 5 τ y_{0} · e^{ − 5 τ / τ} = y_{0} · e^{ − 5} ≈ 0.7% of y_{0}
Summary: The time constant τ is the period of time that it takes for an exponentially decaying quantity to decay to a fraction e^{ − 1} of its initial value, or to about 37% of its initial value. After a period of 5 time constants it has decayed to less than 1% of its initial value and for practical purposes equals zero. |
y = y_{0} · e^{+ r t}.Comparing this with the time constant form we see that r = 1 / τ. Since τ has units of seconds, r has units of 1/seconds. r is called the instantaneous growth rate. Bankers and people interested in rates of growth prefer this form. The equation y = y_{0} · e^{+ r t} represents a quantity y whose initial value is y_{0} and whose rate of growth at any instant equals r times its value at that instant.
y = y_{0} · e^{ − r t},This equation represents a quantity y whose initial value is y_{0} and whose rate of decay at any time equals r times its value at that time.
a) to decay by a factor of 9, andSolution:
b) to decay by a factor of 100.
T = 4.19 seconds.Substituting this into the desired form gives: This form clearly shows that y decays by a factor of 100 every 4.19 seconds.
(a) What amount will be in the account at the end of 15 months?Solution:
(b) After how many years will there be $ 4000 in the account?
t = 12.26 yrs.Answers:
(a) After 15 months the account holds $ 1657.75.
(b) The account holds $ 4000 after 12.26 yrs.
Algebra Coach Exercises |
i = i_{0} · e^{ − t / τ}to describe the current.
ln(3.222) = 2.85 / τNow back-substitute this value of τ into, say, the first of the two equations to get i_{0}:
τ = 2.436
8.7 = i_{0} · e^{ − 1.25 / 2.436}Thus the equation is i = 14.5 · e^{ − t / 2.44}.
i_{0} = 14.5
2.7 = 8.7 · e^{ − 2.85 / τ}This equation can be solved for τ:
τ = 2.436Now get the actual i_{0} by substituting τ and one of the points on the curve, say (t = 1.25, i = 8.7), into the equation i = i_{0} · e^{ − t / τ} to get:
8.7 = i_{0} · e^{ − 1.25 / 2.436}Thus again we find the equation to be i = 14.5 · e^{ − t / 2.44}.
i_{0} = 14.5
Algebra Coach Exercises |
(a) y = 5 − 5 e^{ − t / 2}
(b) y = 5 + 3 e^{ − t / 2}
(c) y = 5 + 3 e^{ − t / 6}
(d) y = 5 − 7 e^{ − t / 6}
y = y_{∞} + a e^{ − t / τ},There are 3 parameters: y_{∞} , a and τ. If we write transpose y_{∞} to the left-hand-side and write the equation as
y − y_{∞} = a e^{ − t / τ},then we see that the right side is the familiar exponential decay in time constant form, and that the difference of y from y_{∞} equals a when t = 0 and that this difference decays with time constant τ. When t = ∞ then y = y_{∞}. Because there are 3 parameters, we must be given the value of y at 3 different times to fix them.
v = v_{∞} + a e^{ − t / τ},Calculate the values of a and τ.
0.3226 = e^{ −1.8 / τ}Solving gives τ = 1.591 and back-substituting gives a = −13.18.
2.0 = 6.2 e^{ − 1.8 / τ}Solving for τ gives:
τ = 1.591Now find the value of a by substituting τ, v_{∞} and the value of t and v at either one of the given points into the equation v = v_{∞} + a e^{ − t / τ}. This again gives a = −13.18.
v = v_{∞} + a e^{ − t / τ},Calculate the values of v_{∞} , a and τ accurate to 3 significant figures.
v = 116 − 153 e^{ − t / 8.15}.