Can we find an approximation to $(1.0099)^6$?

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The binomial theorem says that, if \(x\) is a real number and \(n\) is a positive integer, then \[ (1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i \] where \[\begin{align*} \binom{n}{0} &= 1, \\ \binom{n}{i} &= \frac{n(n-1) \dotsm (n-(i-1))}{i(i-1) \dotsm 1} \quad\text{for each integer $1 \le i \le n$.} \end{align*}\] By applying this to expression \((1+p)^6\) and ignoring terms beyond the fourth, we see that \[\begin{align*} (1+p)^6 &= \binom{6}{0} p^0 + \binom{6}{1} p^1 + \binom{6}{2} p^2 + \binom{6}{3} p^3 + \dotsb \\ &= 1 + \frac{6}{1} p + \frac{6 \times 5}{2 \times 1} p^2 + \frac{6 \times 5 \times 4}{3 \times 2 \times 1} p^3 + \dotsb \\ &= 1 + 6p + 15p^2 + 20p^3 + \dotsb. \end{align*}\] By taking \(p = x - x^2\) in the above expansion, we have that \[ (1+x-x^2)^6 = 1 + 6(x-x^2) + 15(x-x^2)^2 + 20(x-x^2)^3 + \dotsb. \] Ignoring the terms corresponding to \(p^3\), \(p^4\),… we have, up to and including the term featuring \(x^2\), \[\begin{align*} (1+x-x^2)^6 &= 1 + 6(x-x^2) + 15x^2 + \dotsb \\ &= 1 + 6x - 6x^2 + 15x^2 + \dotsb \\ &= 1 + 6x + 9x^2 + \dotsb. \end{align*}\]

If we intend to use a binomial expansion of \((1+x-x^2)^6\), it would be sensible to find some \(x\) such that \[ 1+x-x^2 = 1.0099. \] That is, \(x\) must be such that \[ x - x^2 = 0.0099. \] However, as \[ 0.0099 = 0.01 - 0.0001 = 0.01 - (0.01)^2, \] we can simply take \(x = 0.01\).

By applying this solution \(x=0.01\) to the approximation from the previous part \[ (1+x-x^2)^6 \approx 1 + 6x + 9x^2 \] which improves as \(x\) tends to zero, we get \[ (1.0099)^6 \approx 1 + 6 \times 0.01 + 9 \times (0.01)^2 = 1 + 0.06 + 0.0009 = 1.0609. \] In fact, \[ (1.0099)^6 = 1.0608897, \] and so the error in the approximation is tiny: it is less than \(0.001\)%.