There's a problem with the Range(3) function.

Here's an example:

1. Create a vector for a variable n using the range(3) function such as:

n:=0.03, 0.0325.. 0.05

This should produce a vector with 9 elements ranging from 0.03 to 0.05 in 0.0025 increments.

However, it actually produces an 8 element vector from 0.03 to 0.0475. It appears the last element (0.05) is truncated.

Why?

Thanks in advance,
Brett

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Bug in Range(3) function, SMath 0.95.sm (5 KiB) downloaded 46 time(s).

Hello Brett,

I think it is due to the machine floating point rounding error. Here is the possible explanation of it.

[MATH=eng]n:range(0.03,0.05+10^{-16},0.0325)[/MATH]


[MATH=eng]n=mat(0.03,0.0325,0.035,0.0375,0.04,0.0425,0.045,0.0475,0.05,9,1)[/MATH]

Adding a rather small number to the last number (order of magnitude of computing precision ~15 decimal place) might solve this.

Regards,
Radovan

Thanks Radovan! That worked (though I'm somewhat concerned that an error like this popped up in the first place).

Brett

I agree with Radovan that this feature is not exactly a bug. However, testing the loop exit condition with some tolerance halfway between the iterator increment and machine precision would be a default that might perhaps meet the intent of most users. Then ugly workarounds could be avoided.

I think the easiest (and perhaps cleaner) workaround is the alternate method I included in the attachment. By using a For loop with a dummy (integer) index variable (e.g., i) run through a range of 1 to 9, then under the For loop create the vector n in an equation using the index variable.

For i:=1..9
n:=0.03+(i-1)*0.0025

The same without for loop:

[MATH lang=ENG]0.03+0.0025*(range(0,8))=mat(0.03,0.0325,0.035,0.0375,0.04,0.0425,0.045,0.0475,0.05,9,1)[/MATH]

Both versions of the workaround obfuscate the upper loop limit, that's why I would like the range(3) function to be modified. However, the workarounds make the iterator increment more obvious. Would perhaps be fine to have an alternative range(3) with range(start,stop,step) displaying as

start..stop by step

Wrote

The same without for loop:

[MATH lang=ENG]0.03+0.0025*(range(0,8))=mat(0.03,0.0325,0.035,0.0375,0.04,0.0425,0.045,0.0475,0.05,9,1)[/MATH]

Both versions of the workaround obfuscate the upper loop limit, that's why I would like the range(3) function to be modified. However, the workarounds make the iterator increment more obvious. Would perhaps be fine to have an alternative range(3) with range(start,stop,step) displaying as

start..stop by step

I agree with Martin :d