# Why this expression simplfies to 0.+0.I? [closed]

 Simplify[E^(-0.2 (1 - τ2)) (2^-(-21 + τ1)^2 Sin[
1.256 + π (-21 + τ1)] +
2^-(-20 + τ1)^2 Sin[
1.256 + π (-20 + τ1)]) (2^-(-21 + τ2)^2 Sin[
1.256 + π (-21 + τ2)] +
2^-(-20 + τ2)^2 Sin[1.256 + π (-20 + τ2)]) Sin[
0.9797958971132712 (1 - τ2)] (1/2 E^(-0.2 (1 - τ1))
Sin[0.9797958971132712 (1 - τ1)] +
E^(-0.2 (-τ1 + τ2))
Sin[0.9797958971132712 (-τ1 + τ2)])]


(mma11 win7 64bit)

• A related question. – J. M. will be back soon Oct 15 '16 at 2:51
• Use Simplify[expr] // Chop – Bob Hanlon Oct 15 '16 at 2:58
• @Bob Hanlon It may not be that problem. – WateSoyan Oct 15 '16 at 3:00
• @WateSoyan well, it is your job to state the problem well. So what did you expect, 12? – Kuba Dec 14 '16 at 9:26
• Probably precision-related. For any value of the parameters that I try, I get extremely small values, effectively zero. Remember that any number that has a decimal point is considered inexact, so the argument that "but it's not exactly zero!" doesn't apply. A counterargument is: Have you investigated how much precision would be lost to roundoff during such a calculation? Perhaps more than the magnitude of the result. – Szabolcs Dec 14 '16 at 9:33

I understand that you are not very sure about the purpose of Simplify[] function. Take it out of the expression and you will get the long full result.

    E^(-0.2 (1 - τ2)) (2^-(-21 + τ1)^2 Sin[
1.256 + π (-21 + τ1)] +
2^-(-20 + τ1)^2 Sin[
1.256 + π (-20 + τ1)]) (2^-(-21 + τ2)^2 Sin[
1.256 + π (-21 + τ2)] +
2^-(-20 + τ2)^2 Sin[1.256 + π (-20 + τ2)]) Sin[
0.9797958971132712 (1 - τ2)] (1/2 E^(-0.2 (1 - τ1)) Sin[
0.9797958971132712 (1 - τ1)] +
E^(-0.2 (-τ1 + τ2)) Sin[
0.9797958971132712 (-τ1 + τ2)])


Which results in a long expression that you may inspect. Also try substitute Simplify with N and also eliminate some precision from the code.