 # Practice Paper 1 Question 16

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.

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## Warm-up Questions

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.

## Hints

• Hint 1
For $$r$$-percent cumulative interest every year, how much money will you have after $$n$$ years?
• Hint 2
If the amount of money doubles after $$n$$ years, find an expression relating $$n$$ to $$r$$.
• Hint 3
Can you use the fact that $$r$$ is small and the Taylor series for $$\ln(1+x)$$ to simplify your expression?

## Solution

[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 $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$.

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