Calculating error tolerance for approximate values
If $a$ and $b$ are nearly equal,show that
$$ \frac{a+b}{2}-\frac{(a-b)^2}{4(a+b)}$$
is the approximate value of $\sqrt{ab}$ with an error in excess less than
$\endgroup$ 3$$\frac{(a-b)^4}{4(a+b)^3}$$
1 Answer
$\begingroup$Without loss of generality assume $a,b>0$ and $b=a+x$, where $x$ is small. Let $$ \frac{a+b}{2}-\frac{(a-b)^2}{4(a+b)} = \frac{2a+x}{2}-\frac{x^2}{4(2a+x)} =: f(x) $$ and $$ \frac{(a-b)^4}{4(a+b)^3} = \frac{x^4}{4(2a+x)^3} =: g(x). $$ Then we have to show $$ \left|\frac{\sqrt{ab}-f(x)}{g(x)}\right| = \left| \frac{\sqrt{a(a+x)}-f(x)}{g(x)} \right| < 1 $$ After all these definitions this is the easy part, because with a CAS like Maple we get the first terms of the series for the error expression $$ \left| \frac{\sqrt{a(a+x)}-f(x)}{g(x)} \right| = \left|-\frac{1}{4} -\frac{1}{32}\Big(\frac{x}{a}\Big)^2 + O\left(\left(\frac{x}{a}\right)^3\right)\right| $$ And this actually shows that the error factor is less than $\frac{1}{4}$ for sufficient small $x$.
$\endgroup$ 1
