Using the exponential growth and decay formula for compound interest I have what seems to be a rather simple question but one that is confusing me a lot. When looking at standard exponential growth/decay models(such as the decay of Carbon-14 etc), we can use the formula A=P(1+r)^t in order to find things such as the half-life/rate. However, in these models, aren't the substances (such as Carbon-14) assumed to decay continuously? If so, why can we NOT use the formula Pe^(rt) when is this assumed to be the formula for continuous growth/decay? For instance, in a compound interest problem where interest is compounded continuously, we would have to use the Pe^(rt) formula right?(We couldn't use the formula A(1+r)^t)

sincsenekdq

sincsenekdq

Answered question

2022-09-03

Using the exponential growth and decay formula for compound interest
I have what seems to be a rather simple question but one that is confusing me a lot.
When looking at standard exponential growth/decay models(such as the decay of Carbon-14 etc),
we can use the formula A = P ( 1 + r ) t in order to find things such as the half-life/rate.
However, in these models, aren't the substances (such as Carbon-14) assumed to decay continuously?
If so, why can we NOT use the formula P e r t ,
when is this assumed to be the formula for continuous growth/decay?
For instance, in a compound interest problem where interest is compounded
continuously, we would have to use the P e r t formula right?
(We couldn't use the formula A ( 1 + r ) t

Answer & Explanation

Gracelyn Paul

Gracelyn Paul

Beginner2022-09-04Added 17 answers

Suppose that I am saving up for 20 years, with some sort of savings bond, and that the interest rate is locked in at 5% per year, for the whole 20 years. I am depositing 1 million dollars, and the compounding period is going to be one of the choices below.
For each of the following compounding periods, and P = 1, 000, 000, what you get is the following:
First column = Period of Compounding
Second column = Periods per Year m
Third column = Number of Periods n=mt
Fourth column = Interest per Period i=r/m
Fifth column = Amount A = P ( 1 + i ) n
[ A n n u a l l y 1 20 ( 0.05 / 1 = 0.05 ) 2 , 653 , 297.71 S e m i a n n u a l l y 2 40 ( 0.05 / 2 = 0.025 ) 2 , 685 , 063.84 Q u a r t e r l y 4 80 ( 0.05 / 4 = 0.0125 ) 2 , 701 , 484.94 B i m o n t h l y 6 120 ( 0.05 / 6 = 0.0833.. ) 2 , 707 , 041.49 M o n t h l y 12 240 ( 0.05 / 12 = 0.0416.. ) 2 , 712 , 640.29 B i w e e k l y 26 520 ( 0.05 / 26 = 0.00192307... ) 2 , 715 , 672.70 W e e k l y 52 1040 ( 0.05 / 52 = 0.000961538.. ) 2 , 716 , 976.11 D a i l y 360 7200 ( 0.05 / 360 = 0.00013888.. ) 2 , 718 , 093.08 H o u r l y ( 360 24 ) 172800 ( 5.78703.. 10 6 ) 2 , 718 , 273.96 M i n u t e l y ( 360 24 60 ) 10368000 ( 9.64506.. 10 6 ) 2 , 718 , 281.70 S e c o n d l y ( 360 24 60 60 ) 622080000 ( 1.60751.. 10 6 ) 2 , 718 , 281.92 ]
Now surely, the last three compounding periods are just fictional. No one, except possibly a mafia loan-shark, would compound interest hourly. They are printed here to prove a point: observe that as you go down the table, n is getting very large—but the amount, A, is going toward a fixed number. This fixed number is the value of the continuously compounded interest where m = .
Before you continue, you should verify the arithmetic.
Let’s verify the daily one together. As we said before, bankers believe that there are 360 days per year.
We know that i = r/m and since r = 0.05 in this case, our calculator tells us that i = 0.05/360 = 0.0001388... .
The principal is given to us as $ 1,000,000.00.
All we need now is n, and n = m * t = (360)(20) = 7200. Finally, we have
A = P ( 1 + i ) n
A = 1 , 000 , 000 ( 1 + 0.00013888 · · · ) 7200
A=(1,000,000)(2.71809⋅⋅⋅)
A=2,718,093.08

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