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

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?

Answer 1

[David Goodmanson]

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.

[BIKRAM KESHARI KHANDAI]

Hii David,

Thank you for your response

Can I round off it and use it to get my solution or any other way is there

[Torsten]

If you make a more complicated computation than the one above, you don't know the result in advance. So in this case, you can't cheat afterwards to manipulate the result according to your needs.

What I want to say: It is 100% legal to keep the result in the complex form as it is.

Of course, you can take the real part of the result (delta = real(delta)), but this is, as said, some kind of cheating (and maybe even wrong).

[BIKRAM KESHARI KHANDAI]

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

[David Goodmanson]

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

1/10 + 2/10 -3/10
  ans = 5.5511e-17