0
$\begingroup$

enter image description here Suppose a machine with the floating-point system $\beta = 10$, $t = 8$, $L = -50$, and $U = 50$ is used to calculate the roots of a quadratic equation $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are given real coefficients. For each of the following, state the numerical difficulties that arise when using the standard formula for computing the root. Explain how to overcome these difficulties when possible.

I dont need an answer for all of them, I tried reading the book - I dont understand the concept of $\beta $, $t$, $L$, and $U$, so if you could just explain that in relation to this question, that'd be great. Thank you!

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Essentially you have a number system which retains $t$ correct decimal places (decimal because $\beta=10$) in each operation, and can represent positive numbers as large as $10^U=10^{50}$ and as small as $10^L=10^{-50}$. So for example, in either the second or third cases, $b^2$ overflows. (I'm not really sure why this exercise doesn't just use the actual IEEE double precision values of these numbers, which is to say $\beta=2,t=53,L=1024,U=-1023$ (I might have the 1024 and 1023 reversed). $\endgroup$ – Ian Aug 31 '16 at 23:47
  • $\begingroup$ @Ian: I think that should be an answer. It seems to address OP's concerns. $\endgroup$ – Ross Millikan Sep 1 '16 at 0:17
0
$\begingroup$

Essentially you have a number system which retains $t=8$ correct decimal places (decimal because $\beta=10$) in each operation, and can represent positive numbers as large as $10^U=10^{50}$ and as small as $10^L=10^{-50}$. So for example, in both the second and third situations, $b^2$ overflows when you attempt to compute it. As best I can tell the first situation is actually not a problem, though maybe I'm mistaken.

For reference, these days this sort of thing is mainly applicable in the case of IEEE double precision arithmetic, which uses $\beta=2,t=53,U=1024,L=-1023$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.