when I am subtracting two same values found out from different equations not giving 0, it is giving imaginary value

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BIKRAM KESHARI KHANDAI 2021-12-29

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编辑： David Goodmanson 2021-12-29

采纳的回答： David Goodmanson

n = 7;

t = 0.6;

tp = 0.05;

tc = 0.05;

x = (n-1)*(tp+tc);

delta = sqrt(t-x)

Ans:- delta = 0.0000e+00 + 1.0537e-08i

Here I am getting some imaginary value , How I can get 0 value?

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Answer 1

David Goodmanson 2021-12-29

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编辑：David Goodmanson 2021-12-29

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Hi Bikram,

this is all because of precision issues when dealing with floating point numbers. After running your code,

tminusx = t-x
       tminusx =   -1.1102e-16

Since double precision numbers have 16 digits, an expression differing from 0 by something on the order of 10^(-16) is generally as good as you can expect.

The answer you are getting looks a lot worse than it really is, because you are taking the square root of t-x. So 10^(-16) becomes 10^(-8), with an additional factor of i since it so happens (50-50 chance) that t-x is negative.

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BIKRAM KESHARI KHANDAI 2021-12-29

Thank you Torsten for your suggestion. I will keep this in my mind.

David Goodmanson 2021-12-29

编辑：David Goodmanson 2021-12-29

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Hi Bikram,

In this case you can get a better answer by scaling things appropriately. Suppose time is in seconds, and you know that t = .6 exactly and tp = tc = .05 exactly. Then t = 600 msec, tp = tc = 50 msec. In terms of milliseconds, all the times are integers, which are stored exactly. The calculation is also in integers, so if you do the same calculation you get delta = 0.

By the calculation being in integers I mean that there are, for example, no divisions of integers with a non-integer result, such as y = 1/3. Now y is a floating point quantity and you get back to the same issues. A well known example is