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Mathematics Wolfram Language Numerical Computation

Why Arg[z] is not always the same as Im[Log[z]]

Mikko Tommila

Posted 7 years ago

My question regards this calculation. First define: x := N[9999999966650702/10^16,16] y := N[-6618533197939224/10^28,16] z := x + I y Then, I want to compute the arg of z, and I can do this in three different ways, but the result is not always the same: Arg[z] gives -6.6185332200115710^-13 ArcTan[y / x] gives also -6.6185332200115710^-13 Im[Log[z]] gives -6.619*10^-13 i.e. otherwise the same result but only to four digits of precision. This puzzles me. Is the result of the Im[Log[z]] calculation correct (and is the precision also correct)? Why it is correct, as one would expect that the imaginary part of Log[z] is equal to the arg of z (and to the same precision, one might expect)? Is it simply because Arg[z] and Im[Log[z]] are implemented with a different algorithm? Are all of the results "correct"? Why?

POSTED BY: Mikko Tommila

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Sander Huisman

Sander Huisman, University of Twente

Posted 7 years ago

It is the same but with a different precision, which is more clear with more digits: x = N[9999999966650702/10^16, 50]; y = N[-6618533197939224/10^28, 50]; z = x + I y; Arg[z] Precision[%] ArcTan[y/x] Precision[%] ArcTan[x, y] Precision[%] Im[Log[z]] Precision[%] -6.61853322001156766770653239333525420366865484805310^-13 49.699 -6.61853322001156766770653239333525420366865484805310^-13 49.699 -6.61853322001156766770653239333525420366865484805310^-13 49.699 -6.618533220011567667706532393335254203710^-13 37.9713 as you can see the 'log'-route costs more precision. Why exactly? it depends on how Log is calculated internally. Log might be optimized for a different range of numbers compared to Arg and ArcTan. Note that the 'angle' is very very close to 0. Such that y/x = ArcTan(y/x) very closely which might aid in finding higher numbers of high precision. A series expansion of Log around {1,0} in complex plane gives a similar answer, but with higher precision: Series[Log[\[FormalX] + I \[FormalY]], {\[FormalX], 1, 5}, {\[FormalY], 0, 5}] // Normal Im[% /. {\[FormalX] -> x, \[FormalY] -> y}] -6.61853322001156766770653239333525420366865484805310^-13 Always try to use the function that is intended to do the job. If you go a 'long route' it might costs you precision in each step... Second thought: Log calculates both the Real and Imaginary part: ReIm[Log[z]] {-3.3349298055608781788025361368464813210166610^-9, -6.6185332200115676677065323933352542037*10^-13} Could it be that some precision is 'lost' in the real component? I'm not sure if that is possible, but I could imagine that there is a balance in precision for each component somehow...

POSTED BY: Sander Huisman

S M Blinder

S M Blinder, WRI and Univ of Michigan

Posted 7 years ago

Note that if z is accurate to 16 figures, Log[z] is accurate to Log[16] approx 3. Try this: In[1]:= x = N[9999999966650702/10^16, 40]; y = N[-6618533197939224/10^28, 40]; z := x + I y In[4]:= Arg[z] Out[4]= -6.6185332200115676677065323933352542036710^-13 In[5]:= ArcTan[y/x] Out[5]= -6.6185332200115676677065323933352542036710^-13 In[6]:= Im[Log[z]] Out[6]= -6.618533220011567667706532393*10^-13

Note that if z is accurate to 16 figures, Log[z] is accurate to Log[16] approx 3. Try this:

In[1]:= x = N[9999999966650702/10^16, 40];
y = N[-6618533197939224/10^28, 40];
z := x + I y

In[4]:= Arg[z]

Out[4]= -6.61853322001156766770653239333525420367*10^-13

In[5]:= ArcTan[y/x]

Out[5]= -6.61853322001156766770653239333525420367*10^-13

In[6]:= Im[Log[z]]

Out[6]= -6.618533220011567667706532393*10^-13

POSTED BY: S M Blinder

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