The Computer Laboratory

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.

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

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

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.footasc@sulp.ecitcarp. Please do not write to this address regarding general admissions or course queries.