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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 $A(1+r)^t$)

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  • $\begingroup$ They are essentially the same formula. Let $R = 1+r$. Then $$(1+r)^t = R^t = \mathrm{e}^{\ln(R)t} = \mathrm{e}^{kt}, $$ where $r$, $R$, and $k$ are all constants that relate to the rate of decay. So $A = P(1+r)^t$ and $A = P\mathrm{e}^{rt}$ are telling you exactly the same thing, though the meaning of $r$ in the two formulæ is slightly different. $\endgroup$ – Xander Henderson Aug 18 '18 at 2:05
  • $\begingroup$ @XanderHenderson Hey thanks for the comment. Although I do understand your derivation of Pe^rt, I don't understand why can't the original formula be used in continuously compounded interest problems? (For instance, using an initial balance of 100 and 20% interest compounded continuously, we can clearly see that 100(1.2)^t is not the same as 100e^0.2t.) $\endgroup$ – Samvit Agarwal Aug 18 '18 at 2:39
  • $\begingroup$ But $100(1.2)^t$ is the same as $100e^{\ln(1.2)t}$. In the first case $r=0.2$ but in the second case $r=\ln(1.2)$. $\endgroup$ – John Wayland Bales Aug 18 '18 at 5:00
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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 $A = P(1 + i)^n$

$ \begin{bmatrix} {Annually} & 1 & 20 & (0.05/1 =0.05) & 2,653,297.71 \\ {Semiannually} & 2 & 40 & (0.05/2 =0.025) & 2,685,063.84 \\ {Quarterly} & 4 & 80 & (0.05/4 =0.0125) & 2,701,484.94 \\ {Bimonthly} & 6 & 120 & (0.05/6 =0.0833..) & 2,707,041.49 \\ {Monthly} & 12 & 240 & (0.05/12 =0.0416..) & 2,712,640.29 \\ {Biweekly} & 26 & 520 & (0.05/26 =0.00192307...) & 2,715,672.70 \\ {Weekly} & 52 & 1040 & (0.05/52 =0.000961538..) & 2,716,976.11 \\ {Daily} & 360 & 7200 & (0.05/360=0.00013888..) & 2,718,093.08 \\ {Hourly} & (360*24) & 172800 & (5.78703..*10^{−6)} & 2,718,273.96 \\ {Minutely} & (360*24*60) & 10368000 & (9.64506..*10^{−6)} & 2,718,281.70 \\ {Secondly} & (360*24*60*60) & 622080000 & (1.60751..*10^{−6)} & 2,718,281.92 \\ \end{bmatrix} $

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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