ismembertol does not work as documented

Hello,

This function should work with absolute tolerance. Here is an example where it doesn't work as documented

B=[ 1.9500    1.0000];
A=[3.0000    2.0000    4.0000    2.5000    1.2000    1.1000];
[Loc1,Loc2]=ismembertol(B,A,0.1,'DataScale',1)

What should come out according to documentation is

Loc1 =
  1×2 logical array
   1   1
Loc2 =
     2     6

What does come out however is

Loc1 =
  1×2 logical array
   1   0
Loc2 =
     2     0

It seems someone forgot the absolute when comparing :)

Answer 1

[Matt J]

The documentation isn't wrong. You've set a tolerance that can only satisfied reliably at A(6) in infinite precision arithmetic. Observe:

>> [Loc1,Loc2]=ismembertol(B,A,0.1+eps,'DataScale',1)

Loc1 =

  1×2 logical array

   1   1


Loc2 =

     2     6

[Guillaume]

To clarify, 1.1 is not actually representable as a double. What's actual stored in memory is 1.100000000000000088817841970012523233890533447265625. Similarly 0.1 is not actually representable as double (exact value is 0.1000000000000000055511151231257827021181583404541015625), The difference between actual value of 1.1 and 1 is slightly larger than that:

>> (1.1 - 1) <= 0.1
ans =
  logical
   0

so, yes ismembertol is going to say that 1.1 is not within range.

[Matt J]

Hmmm. But 1.100000000000000088817841970012523233890533447265625 looks higher than double precision (more than 16 decimal points) ?

[Guillaume]

It's the complete expansion of the binary fraction. I used Jame Tursa's num2strexact for that. As James says on that page, Don't confuse the exact conversion with significance. These extra digits are just noise. The double before 1.1 is:

>> num2strexact(1.1-eps(1.1))
ans =
    '1.0999999999999998667732370449812151491641998291015625'

and the difference between the two is

>> eps(1.1)  %1.1-eps(1.1) has the same eps
ans =
   2.2204e-16

as you can see the difference is around 1e-16 so at the 16th digit.