Numerical computation of $\sqrt{x} - \sqrt{y}$ where $x$ and $y$ are almost equal

Question

A problem asks the following

Compute $\sqrt{4321} - \sqrt{4318}$ using 4 decimal digit arithmetic.

This comes out to $0.02,$ when the exact result should be $0.02282...$. It then asks the following,

Reformulate the expression $\sqrt{x} - \sqrt{y}$ with $x, y \in \Bbb{Z}$ very similar in size such that the answer produced is more accurate than in the above computation.

I'm unsure of what is even being asked here. Am I to do some analysis on the function $f(x,y) = \sqrt{x} - \sqrt{y}$ or am I missing something?

Thanks for any pointers.

score 4 · Answer 1 · answered Apr 19 '16 at 11:38

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Hint: $\sqrt{x} - \sqrt{y} = \dfrac{x-y}{\sqrt{x}+\sqrt{y}}$, which is much better because there is little cancellation.

answered Apr 19 '16 at 11:38

user21820

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Isn't there cancellation in $x-y$ ? – lhf Apr 19 '16 at 11:39
@lhf: That cancellation does not affect because it is absolutely precise since $x,y$ are integers. – user21820 Apr 19 '16 at 11:40
Thanks for reminding me, I'll edit into the question that x and y should be integers. – Edward Evans Apr 19 '16 at 11:41
That works for the numerical example but not for the reformulation. Let's see if the OP clarifies this. – lhf Apr 19 '16 at 11:41
Specifically, if you use fixed-point precision arithmetic throughout, the left-hand side suffers severe loss of precision while the right-hand side does not when $x,y$ are integers. – user21820 Apr 19 '16 at 11:41
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@lhf: Ok he has clarified it. – user21820 Apr 19 '16 at 11:42

score 3 · Answer 2 · answered Apr 19 '16 at 11:52

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$$\sqrt x-\sqrt y=\frac{x-y}{\sqrt x+\sqrt y}$$

In the given example one can do $$\sqrt{4321}-\sqrt{4318}=\frac {3}{\sqrt x+\sqrt y}\approx \frac{3}{2\sqrt{4321}}\approx 0.02281$$ Optionally $$\frac{3}{2\sqrt{4318}}\approx0.02282$$

answered Apr 19 '16 at 11:52

Piquito

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score 1 · Accepted Answer · answered Apr 19 '16 at 11:34

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Hint: Write $x=y+t$ and expand $\sqrt{y+t}$ in a Taylor series in $t$ around $t=0$.

The first term after $\sqrt y$ is $ \dfrac{t}{2 \sqrt y} $. Use more terms if you need.

answered Apr 19 '16 at 11:34

lhf

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Your method is no better than mine in terms of precision, since to find $t$ you will also have loss of precision if $x,y$ are not special such as integers. However, indeed Taylor expansion is the general way to systematically separate out significant terms that can be cancelled exactly, which in this case is $\sqrt{y}$. My method is ad-hoc and works better on this problem but does not extend to other problems. – user21820 Apr 19 '16 at 11:43
I like both answers but this seems more intuitive to me in terms of extension to other problems. Thanks for your answer though. – Edward Evans Apr 19 '16 at 11:50
@Ed_4434: In fact I recommend always using Taylor expansions to understand the asymptotic behaviour. My answer is more tailored to elementary level (like for high-school). =) – user21820 Apr 19 '16 at 11:56
Yeah that's fine, thanks anyway! – Edward Evans Apr 19 '16 at 12:00

Numerical computation of $\sqrt{x} - \sqrt{y}$ where $x$ and $y$ are almost equal

3 Answers3