exp( ) gives unexpected answer for (2,2) input

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Jonathon Cheah
Jonathon Cheah 2022-9-2
编辑： Bruno Luong 2022-9-9
I have trouble understanding the results returned by exp():
given a= [1.4142 1.4121; 1.4142 1.4121] ;
exp(j*2*pi*a*10^3) = [0.2269 +0.9739i 0.8312+0.5560i ; 0.2269+0.9739i 0.8312+0.5560i]
but checking by elements:
exp(j*2*pi*1.4142*10^3)=0.3090+0.9511i
exp(j*2*pi*1.4121*10^3)=0.890+0.5878i
if a =[1.4142 1.4121];
then the answers agree with the element check. What am I missing?
  1 个评论
James Tursa
James Tursa 2022-9-2
编辑：James Tursa 2022-9-2
Please show the actual code ... i.e., copy & paste the code and actual MATLAB output. It could be that you hand typed in numbers based on MATLAB truncated display but the actual numbers used by MATLAB are different. E.g., what does 1.4142-a(1,1) and 1.4141-a(1,2) show? Or use format longg and display the "a" matrix again.

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John D'Errico
John D'Errico 2022-9-2
编辑：John D'Errico 2022-9-2
Ran in:
I would bet a large sum of money that a is not exactly what you think it is.
a_you_think = [1.4142 1.4121; 1.4142 1.4121];
However, the odds are instead very good that those numbers are only rounded to 5 significant digits, then they were displayed as such. Remember that MATLAB carries roughly 16 significant digits in a double precision number. So that value of 1.4142 for a(1,1) could be anything in the interval (1.41415,1.41425).
a11 = [1.41415*(1+eps);1.41425*(1- eps)]
a11 = 2×1
1.4142 1.4142
format long g
exp(j*2*pi*a11*10^3)
ans =
0.587785252291231 + 0.80901699437585i 1.67332990617082e-12 + 1i
So is your output unexpected? It is only unexpected for YOU. I'm not at all surprised.
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Jonathon Cheah
Jonathon Cheah 2022-9-9
Bruno was pointing out my 2 numbers (1.414213562373095-1.412093835408965) x10^3 being large enough for many "turns" (cycles?) which is true, but I was referring to each single number 1.414213562373095 with 1.4142 as an example. Knowing Euler identity exp(ix)=cos(x)+isin(x), how big x is, is moot by mod(x,2pi). I asked for answer because I was blindsighted by Matlab's 16 significant digits as John D'Errico pointed out, and I think James Tursa was going the same direction. Indeed, when mod(x,2pi) ~ 0 or pi for sine for exampe, tiniest delta will give relatively bigger swing in answers. I expected that, but not by not knowing the inherent Matlab number resolutions. The factor x10^3 is necessary but not sufficient effects in my "small" delta comment. I hope this clears up the confusions.
Bruno Luong
Bruno Luong 2022-9-9
编辑：Bruno Luong 2022-9-9
Ran in:
@Jonathon Cheah "The factor x10^3 is necessary but not sufficient effects in my ""small" delta comment.
The factor 10^3 + truncating digits of 1.414213562373095 are perfectly sufficient to explain. There is noting mysterious as you seem to imply.
Here are two numbers
a=1.414213562373095
a = 1.4142
ar=1.4141
ar = 1.4141
Now the exponential values with factor 10^3
z=exp(1i*2*pi*a*10^3)
z = 0.2269 + 0.9739i
zr=exp(1i*2*pi*ar*10^3)
zr = 0.8090 + 0.5878i
Q: How much is the difference in angles after scaling by 10e3?
theta=2*pi*(ar-a)*10^3 % in radian
theta = -0.7135
A: And the angle error in degree,
thetadeg=rad2deg(theta) % in degres
thetadeg = -40.8825
OK the error of angle si about -40 degres, not quite one turn (=360 degres) but it is not small by any mean due solely to the factor 10^3, nothing else.
Q: Can I retreive the angle from z and zr alone?
A: Yes:
atan2(imag(zr/z),real(zr/z)) % compare with theta above
ans = -0.7135
Q: Does the error of angle can explain the different between z and zr
z*exp(1i*theta)
ans = 0.8090 + 0.5878i
zr
zr = 0.8090 + 0.5878i
A: Perfectly
To summarize:
There is nothing surprise when you take time to decompose and check the bound of the error.
There is no lesson to be learnsed other then checking how your calculation chain propagate the error.

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Paul
Paul 2022-9-2
Ran in:
Seems to give the expected result?
a= [1.4142 1.4121; 1.4142 1.4121]
a = 2×2
1.4142 1.4121 1.4142 1.4121
exp(j*2*pi*a*10^3)
ans =
0.3090 + 0.9511i 0.8090 + 0.5878i 0.3090 + 0.9511i 0.8090 + 0.5878i

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