Unformatted text preview: 25 1,006.60 0.276 2 1,000.27 0.274 26 1,006.87 0.276 3 1,000.55 0.274 27 1,007.15 0.276 4 1,000.82 0.274 28 1,007.42 0.276 5 1,001.10 0.274 29 1,007.70 0.276 6 1,001.37 0.274 30 1,007.98 0.276 After 30 days the total principal would be $1, 007.98 + 0.276 ≈ $1, 008.26 .!
What will happen if we compound the interest even more frequently? We can work out the
framework of Example 2.8 in abstract terms. If r is the annual interest rate and the time period is
represented as a fraction of the year, ∆t (which is 1 / 365 in the example), then the interest rate
during that time period will be r × ∆t . If P denotes the principal at time t then the interest earned
during the time ∆t from t to t + ∆t will be P × r ∆t . This interest accounts for the change in the
principal ∆P , so we have
∆P = P(r ∆t ) . (2.7) We want to let ∆t → 0 . Of course, this says that the time period goes to zero. Hence, so does the
rate and therefore also the interest earned - not very surprising in view of the interest listed in the
table above. In such situations we can learn something by considering the rate at which the
interest goes to zero, namely the ratio ∆P / ∆t . By (2.7) this ratio is rP . However according to
(2.6), as ∆t → 0 the ratio ∆P / ∆t approaches the derivative, the instantaneous rate of change of
the function P(t ) . Thus we arrive at the differential equation
= rP ,
dt (2.8) which we take as the definition of continuous compounding of interest.
Definition 2.1: Money earns interest at an annual rate r compounded continuously if the principal
satisfies equation (2.8).!
Although we can't physically compound the interest continuously, we can solve the differential
equation (2.8) and thereby figure out what the principal would be if that idealized process could be
achieved. The formula turns out to be quite simple and can be used to directly compute the
Example 2.9: Find the general formula for the principal P (t ) when money is compounded
continuously at an annual interest rate r , assum...
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