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

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?

[James Tursa]

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.

Answer 1

[John D'Errico]

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.

[Jonathon Cheah]

You are absolutely correct. Checked using format long, my numbers were 1.414213562373095 and 1.412093835408965. Surprising part is that for such a small delta in inputs, the answers are so wildly different. I learned 2 new things today; Matlab carries full resolutions even if FORMAT type is not set. Complex EXP(j) will yield results that are widely varied with small delta change in the input.

Sir, thank you.

[Bruno Luong]

"Surprising part is that for such a small delta in inputs, the answers are so wildly different."

What?

You give two angles different by

(1.414213562373095-1.412093835408965)*1e3

ans = 2.1197

as number of turns and you call it small delta?

[Jonathon Cheah]

The delta I meant was between 1.414213562373095 and 1.4142.

[Stephen23]

"The delta I meant was between 1.414213562373095 and 1.4142."

But as Bruno Loung correctly pointed out, those are not the input values to EXP. The values you actually used with EXP are around 1e3 times larger... so not really a very "small" delta at all.

[Jonathon Cheah]

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]

@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.

Answer 2

[Paul]

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