Review question

# Can we estimate a difference of eighth powers? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6503

## Solution

Write down the first four terms in the expansion of $(1+x)^8$ in ascending powers of $x$.

We can use the binomial theorem for $n$ a positive integer, $(a+b)^n=\sum\limits_{k=0}^n \binom{n}{k}a^{n-k}b^k,$ with $n=8$, $a=1$ and $b=x$, to get \begin{align*} (1+x)^8 &= \binom{8}{0}x^0+\binom{8}{1}x^1+\binom{8}{2}x^2+\binom{8}{3}x^3 + \dotsb + x^8 \\ &= 1 + \frac{8}{1} x^0 + \frac{8 \cdot 7}{2 \cdot 1} x^2 + \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} x^3 + \dotsb x^8 \\ &= 1 + 8x + 28x^2 + 56x^3 + \dotsb + x^8. \end{align*}

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.