$x$ 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.
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- For which real values of
$a$does$(1+a)^x=e$yield real values of$x$? - If the number of bacteria in a sample doubles every hour, how many are there after 10 hours, if the initial population was 1?
- Using Taylor expansion about 0 (Maclaurin), find
$\sin(0.1)$correct to 2 decimal places.
- For which real values of
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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?
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[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$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$.