Dividing $x$ by a small annual $r$-percent cumulative interest rate approximates the number of years needed to double your investment with a bank. Find $x$. Hint: The word “small” may be important.
1. For which real values of $a$ does $(1+a)^x=e$ yield real values of $x$?
2. If the number of bacteria in a sample doubles every hour, how many are there after 10 hours, if the initial population was 1?
3. Using Taylor expansion about 0 (Maclaurin), find $\sin(0.1)$ correct to 2 decimal places.

For $r$-percent cumulative interest every year, how much money will you have after $n$ years?

If the amount of money doubles after $n$ years, find an expression relating $n$ to $r$.

Can you use the fact that $r$ is small and the Taylor series for $\ln(1+x)$ to simplify your expression?

[Trivia: Luca Pacioli in 1494 said “72” (for slightly larger $r$), and this result was apparently already well known at that time.]

The question asks us to find $x$, which when divided by $r$ approximates the number of years needed to double the investment. This translates to $\frac{x}{r} = n.$ Let the initial investment amount be $a_0$, and the amount in the bank after $n$ years be $a_n$. Each year, the amount increases by $r$ percent, which gives us the recurrence $a_{n+1} = a_n\big(1+\frac{r}{100}\big).$ Unwinding this recursion, one finds that $a_n = a_0 \big(1+\frac{r}{100}\big)^n.$

We wish to find $n$ such that $a_n = 2a_0.$ Substitute above to obtain $2 = \big(1+\frac{r}{100}\big)^n,$ and apply log to get $n = \frac{\ln 2}{\ln(1+\frac{r}{100})}.$ Since $r$ is very small, we can approximate the denominator by using only the first term of the Taylor expansion of $\ln(1+\frac{r}{100}) \approx \frac{r}{100},$ which gives us $n = \frac{100 \ln 2} {r}.$ We now substitute into the previous equation to get $x = 100 \ln 2 \approx 69.3.$

The Taylor expansion of $\ln(1+x)$ around $x=0$ is $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$.