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Let $a,b,c,d \in \mathbb{R}$ with $c,d > 0$. Assume that $|a-b| < \epsilon_1$ and $|c-d| < \epsilon_2$.

What can be said about $|(a/c)^2 - (b/d)^2|$?

What I tried is the following:

\begin{align*} |(a/c)^2 - (b/d)^2| & \leq | a/c - b/d | \cdot | a/c + b/d | \\ & = | a/c - b/c + b/c - b/d | \cdot | a/c + b/d | \\ & \leq | (a-b)/c | + |b| |1/c - 1/d | \cdot | a/c + b/d | \\ & \leq \epsilon_1/c + |b|/(cd) |d-c | \cdot | a/c + b/d | \\ & \leq \epsilon_1/c + \epsilon_2 |b|/(cd) \cdot | a/c + b/d | \end{align*}

Is it possible to further improve this bound? Especially the last factor $| a/c + b/d | $ troubles me.. Thanks!

1 Answers 1

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It should bother you. The $cd$ in the denominator should bother you also. If $c,d$ are small enough, $cd$ can be huge. Let us take $\epsilon_1=\epsilon_2=.0001$ just for fun. Then we can have $a=b=1, c=.0001, d=10^{-12}$ Then $|(\frac ac)^2-(\frac bd)^2|=|10^8-10^{24}| \approx 10^{24}$ It should be clear that we can make this as large as we want even if the $\epsilon$s are very small.

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    In fact, the bound given is even worse; while the actual 'error' depends 'only' quadratically on $1/c, 1/d$, the bound does cubically. The estimated error with the numbers given in this answer is $10^4 + 10^{16}*(10^{12} + 10^4) \approx 10^{28}$. This happens in the second inequality. – 2017-02-02
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    thank you for pointing this out. is there a way to get the bound sharper? – 2017-02-02