[Compound Interest] Layman way vs. Derivative way

[TadeusPrastowo]

My https://www.amazon.com/dp/0073532320/?tag=pfamazon01-20 (p. 176 Example 7.1) pointed out that an investment $p(t) = 100\,2^t$ ($t$ in year) that doubles the capital every year starting with an initial capital of $100, has an (instantaneous) rate-of-change $\frac{\text{d}}{\text{d}t} p(t) = p'(t) = 100\,2^t\,\ln 2$.

Layman usually uses the formula $p(t/12) = 100\,2^{t/12}$ if he wants to know his capital at the end of each month. And, I think that makes sense because if the capital is never cashed, at the end of the first year, the initial capital doubles exactly to $200.

Now, the textbook claims that $\frac{p'(t)}{p(t)} = \ln 2 \approx 69.3%$ is the percentage change per year, and that should be surprising to most people because a percentage rate of 69.3% will double the investment each year if compounded "continuously".

Why it should make sense at all? I think people simply use something like $p(t/12) = 100\,2^{t/12}$ to calculate their capital after certain months. Why would knowing the instantaneous rate of change be beneficial at all in business, and so, should surprise people by the percentage change of 69.3% per year? Is the example simply a thought exercise without any direct connection to the real world?

Thank you.

 

[TadeusPrastowo]

Why would knowing the instantaneous rate of change be beneficial at all in business, and so, should surprise people by the percentage change of 69.3% per year? Is the example simply a thought exercise without any direct connection to the real world?

Taking an analogy to instantaneous rate-of-change of car's position (i.e., car's speed), the rate-of-change of position is very beneficial because speed relates to the magnitude of the car's kinetic energy where the greater the energy is, the more fatal a car crash becomes.

So, I think the rate-of-change of a compound saving is not beneficial to calculate the amount of capital at the end of each month (analogously, the rate-of-change of a car's position is not beneficial to decide the final position of the car after certain miles). But, while there is a great advantage of knowing the car's speed as a car's driver (e.g., minimizing fatality in an accident), is there any advantage of knowing the rate-of-change of a compound saving as an investor?

Thank you.

 

[lurflurf]

The practical interest is in comparing investments.
The interest formula is
$$A=A_0 \, \left( 1+\tfrac{r}{n}\right) ^{n t}$$
If your are comparing several options each with different n and r you need to be able decide which is best.
In your example say you can choose 100% interest compounded annually or 70% interest compounded continuously. You would choose the later. Someone who does not understand interest might take the former as 100>70.

 

[TadeusPrastowo]

The practical interest is in comparing investments.
The interest formula is
$$A=A_0 \, \left( 1+\tfrac{r}{n}\right) ^{n t}$$
If your are comparing several options each with different n and r you need to be able decide which is best.

That makes sense.

In your example say you can choose 100% interest compounded annually or 70% interest compounded continuously. You would choose the later. Someone who does not understand interest might take the former as 100>70.

Ah, I see the point now.

Thank you very much.