Binomial Series — Question 7

Question 7 – Solution

Step 1: Use the binomial expansion.

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\cdots.$$ Set $x=0.04$.

Step 2: Keep two terms.

$$\sqrt{1.04}\approx1+\frac{0.04}{2}=\boxed{1.02}.$$

Step 3: Bound the error.

Beginning with the quadratic term, the omitted terms alternate and decrease in magnitude at $x=0.04$. Therefore $$|R|\le\frac{(0.04)^2}{8}=\boxed{0.0002}.$$ The first correction is negative, so the linear approximation is an overestimate. More precisely, $$1.0198\le\sqrt{1.04}<1.02.$$ The actual value, $1.0198039\ldots$, is consistent with this bound.