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

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

Question

Obtain the first $4$ terms in the expansion of $(1 + p)^6$ in ascending powers of $p$.

By substituting $p = x - x^2$ obtain the expansion of $(1 + x - x^2)^6$ as far as the term in $x^2$.

Find a value for $x$ which would enable you to calculate $(1.0099)^6$.

Suggestion

How can we use our expansion from the first part of the question?

We want to use an expansion of $(1 + x - x^2)^6$ - can we find an equation that $x$ must satisfy?

Or can we see straight away what $x$ must be?

Solution

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

$$\begin{align*} \binom{n}{0} &= 1, \\ \binom{n}{i} &= \frac{n(n-1) \dots m (n-(i-1))}{i(i-1) \dots m 1} \quad\text{for each integer }1 \le i \le n\text{.} \end{align*}$$

By applying this to expression

$(1+p)^6$

and ignoring terms beyond the fourth, we see that

$$\begin{align*} (1+p)^6 &= \binom{6}{0} p^0 + \binom{6}{1} p^1 + \binom{6}{2} p^2 + \binom{6}{3} p^3 + \dots b \\ &= 1 + \frac{6}{1} p + \frac{6 \times 5}{2 \times 1} p^2 + \frac{6 \times 5 \times 4}{3 \times 2 \times 1} p^3 + \dots b \\ &= 1 + 6p + 15p^2 + 20p^3 + \dots b. \end{align*}$$

By taking

$p = x - x^2$

in the above expansion, we have that

$$(1+x-x^2)^6 = 1 + 6(x-x^2) + 15(x-x^2)^2 + 20(x-x^2)^3 + \dots b.$$

Ignoring the terms corresponding to

$p^3$

,

$p^4$

,… we have, up to and including the term featuring

$x^2$

,

$$\begin{align*} (1+x-x^2)^6 &= 1 + 6(x-x^2) + 15x^2 + \dots b \\ &= 1 + 6x - 6x^2 + 15x^2 + \dots b \\ &= 1 + 6x + 9x^2 + \dots b. \end{align*}$$

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 $x = 0.01$.

Alternatively, we could solve the quadratic equation $x^2 - x + 0.0099 = 0$.

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 $0.001$%.