Write down the first four terms in the expansion of \((1+x)^8\) in ascending powers of \(x\).
Alternatively, we can just use the fact that the 9th line of Pascal’s triangle starts 1, 8, 28, 56, … and just write the final line down.
Evaluate \((1.001)^8-(0.999)^8\) to eight significant figures.
We can expand to get \[(1-x)^8 = 1 - 8x + 28x^2 - 56x^3 + \dotsb + x^8.\] Thus \[(1+x)^8-(1-x)^8=16x+112x^3+ 112x^5 + 16x^7,\] with the even powers of \(x\) all cancelling out.
If we pick \(x=0.001\) and substitute it into the above equation we get precisely what we need: \[\begin{align} (1.001)^8-(0.999)^8&\approx16\times 0.001+112\times(0.001)^3 \\ &=0.016 + 0.000\,000\,112=0.016\,000\,112 . \end{align}\]So to eight significant figures \((1.001)^8-(0.999)^8\approx0.016\,000\,11\).
Do the higher-order terms make a contribution that we need to consider?
The \(x^5\) term does not contribute because \((0.001)^5=0.000000000000001\) and the coefficient is not of the order \(10^6\) that would be needed to bring it within 8 significant figures. The other terms are even smaller.
