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Relevant definitions:

If $\tilde a$ is an approximation of $a$, then the absolute error of $a$ is $$ |\tilde a - a| $$ and the relative error of $a$ is $$ \frac{| \tilde a - a|}{a} .$$

Statement of the Problem:

Consider the function $f(x) = \sqrt{x+1} - 1$. Assume that we have a good implementation of square root, so that if $y = \sqrt{x}$, then the relative error in $\sqrt{x}$ is no bigger than the relative error in $x$ or $y$. That is, $\sqrt{x}$ can be computed as accurately as possible, given the scheme we are using for representing numbers on the computer.

Using double precision, roughly how accurately can a computer calculate $f(1)$? Give the absolute and relative error.

Where I Am:

I'm not sure what is being asked of me here. We have that using double precision numbers is roughly equivalent to working with $16$ significant digits in base $10$. Now, computing $f(1)$ to $16$ significant digits is

$$ f(1) = \sqrt{1+1} - 1 = \sqrt{2} - 1 \approx 0.414213562373095. $$

So, am I just being asked to compute

$$ |\tilde a - a| = |0.414213562373095 - (\sqrt{2} - 1) | $$

for the absolute error, and

$$ \frac{|\tilde a - a|}{a} = \frac{|0.414213562373095 - (\sqrt{2} - 1) |}{\sqrt{2} - 1} $$

for the relative error?

If so, any method of computing it would involve, well, a computer that itself is likely only accurate enough to compute $f(1)$ to $16$ significant digits, so it seems somewhat pointless. Otherwise, just presenting the above results as a solution seems incomplete (and rather pointless in itself).

Perhaps someone more familiar with numerical analysis could help me out here.

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    $\begingroup$ dont forget parentheses when doing minus $\endgroup$ – mathreadler Oct 1 '17 at 21:39
  • $\begingroup$ @mathreadler Whoops! Thanks. Edited. $\endgroup$ – thisisourconcerndude Oct 1 '17 at 21:41
  • $\begingroup$ Other than that I think you are doing it right. It is probably just a warm up exercise to learn what absolute and relative errors are so you learn it. $\endgroup$ – mathreadler Oct 1 '17 at 21:43
  • $\begingroup$ @Hellen Is double precision not a type of floating point representation? (That I'm not sure should give you an idea of how little I know about this subject...) $\endgroup$ – thisisourconcerndude Oct 1 '17 at 21:55
  • $\begingroup$ Yes double precision is an IEEE floating point standard. 64 bits of which 52-53 something are mantissa. Slightly below 16 decimals precision since $\log_{10}(2^{53}) \approx 15.9$. $\endgroup$ – mathreadler Oct 1 '17 at 22:06
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The point is that you computed $\sqrt 2$ to $16$ decimal digits, so the relative error is about $10^{-16}$. When you subtracted $1$ the difference has only $15$ decimal digits, so the relative error is about $10^{-15}$. This is not much of a loss in this case, but if you were to subtract an exact $1.41421$ from $\sqrt 2$ you would lose six digits and the relative error would increase by a factor of about a million. Yes, you are being asked to compute just what you say. I suspect that there is another exercise coming with much worse cancellation. You will then be shown the advantage of removing the cancellation analytically.

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