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Complex roots equality problem

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Bryce Inman
Bryce Inman on 21 Nov 2016
Commented: James Tursa on 22 Nov 2016
I want to find the real roots of a polynomial that are 0<=roots<=1, for example:
>> rts=roots([16 -32 116 0 -100])
rts =
0.9050 + 2.6262i
0.9050 - 2.6262i
1.0000 + 0.0000i
-0.8100 + 0.0000i
I should be able to find that rts(3) satisfies these requirements. However,
>> myrt=rts(logical(~imag(rts) & real(rts)>=0 & real(rts)<=1))
myrt =
Empty matrix: 0-by-1
This is because real(rts(3))<=1 yields false, which is a bug so far as I can tell since complex(1,0)<=1 yields true. If I open the variable rts and double click rts(3) (which shows that it really is stored just like complex(1,0)), I can then execute real(rts(3))<=1 and it produces the correct result. What gives?
How do I find that rts(3) is the correct answer when I write a script? I am using MATLAB R2015b.

Accepted Answer

James Tursa
James Tursa on 21 Nov 2016
Edited: James Tursa on 21 Nov 2016
R2016b:
>> rts=roots([16 -32 116 0 -100])
rts =
0.904997916303652 + 2.626226923351172i
0.904997916303652 - 2.626226923351172i
1.000000000000000 + 0.000000000000000i
-0.809995832607303 + 0.000000000000000i
>> rts(logical(~imag(rts) & real(rts)>=0 & real(rts)<=1))
ans =
1.000000000000000
>> num2hex(rts(3))
ans =
3feffffffffffffc
>> polyval([16 -32 116 0 -100],rts(3))
ans =
-8.526512829121202e-14
>> polyval([16 -32 116 0 -100],1)
ans =
0
R2011a:
>> rts=roots([16 -32 116 0 -100])
rts =
0.904997916303650 + 2.626226923351169i
0.904997916303650 - 2.626226923351169i
1.000000000000000
-0.809995832607303
>> rts(logical(~imag(rts) & real(rts)>=0 & real(rts)<=1))
ans =
Empty matrix: 0-by-1
>> num2strexact(rts(3))
ans =
1.000000000000000444089209850062616169452667236328125
>> num2hex(rts(3))
ans =
3ff0000000000002
>> polyval([16 -32 116 0 -100],rts(3))
ans =
8.526512829121202e-014
>> polyval([16 -32 116 0 -100],1)
ans =
0
So it looks like the roots function (which uses eig in the background) has different behavior depending on the version. The versions missed being exactly 1.0 by 2-3 least significant bits (one version less than and the other version greater than). You will have to adjust your code to account for this.
  2 Comments
James Tursa
James Tursa on 22 Nov 2016
Not sure what you are saying here. How do you get real(rts(3))<=1 to evaluate as true? You double clicked on rts in the Workspace list and then that brings up the Variable Editor. Then what did you do?

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More Answers (1)

Walter Roberson
Walter Roberson on 21 Nov 2016
You will find that real(rts(3))-1 is not 0.
When I test on my Mac, in R2015b, rts(3) is 1+2*eps but in R2016b, rts(3) is 1-2*eps
  1 Comment
Bryce Inman
Bryce Inman on 22 Nov 2016
Good point, I should have subtracted 1 to see the error.

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