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*To*: schlie@xxxxxxxxxxx*Subject*: Re: comparison operators and *typos*From*: Aubrey Jaffer <agj@xxxxxxxxxxxx>*Date*: Fri, 8 Jul 2005 22:25:15 -0400 (EDT)*Cc*: srfi-70@xxxxxxxxxxxxxxxxx*Delivered-to*: srfi-70@xxxxxxxxxxxxxxxxx*In-reply-to*: <BEF2B887.ABEC%schlie@xxxxxxxxxxx> (message from Paul Schlie on Thu, 07 Jul 2005 10:46:31 -0400)*References*: <BEF2B887.ABEC%schlie@xxxxxxxxxxx>

| Date: Thu, 07 Jul 2005 10:46:31 -0400 | From: Paul Schlie <schlie@xxxxxxxxxxx> | | > | Date: Wed, 06 Jul 2005 13:55:40 -0400 | > | From: Paul Schlie <schlie@xxxxxxxxxxx> | > | | > | Where absolute zero is designated as 0, and who's reciprocal is 0, as | > | the average value of it's -1/0 and +1/0 limits would be 0; as would 0/0, | > | > Then the FINITE? predicate becomes useless. | | - so? SRFI-70: This SRFI attempts to strike a balance between tolerance and signaling of range errors. Having #i+/0 and #i-/0 allows one level of calculation to be executed before bounds checking. That detection is via the FINITE? procedure. Subsequent numerical calculations on infinities can signal an error or return a non-real infinity: #i0/0. #i0/0 can be considered an error waiting to happen. I'm sorry that SRFI-70 can't accommodate your proposals. They are complicated and nonuniform; they don't significantly improve upon the capabilities of SRFI-70; and you have not grounded them in terms of standard mathematical constructs. | > | (limit (lambda (x) (tan x)) (pi/2 -0/1)) => +1/0 | > | (limit (lambda (x) (tan x)) (pi/2 +0/1)) => -1/0 | > | | > | (+ 4. (/ (abs 0) 0)) => 4.0 | > | (limit (lambda (x) (+ 4. (/ (abs x) x))) (0 -0/1)) => 3.0 | > | (limit (lambda (x) (+ 4. (/ (abs x) x))) (0 +0/1)) => 5.0 | > | > LIMIT already handles these cases correctly. But I am unconvinced | > that a procedure can automatically pick the evaluation points given no | > information about the test function. | | - it seems fairly straight foreword to for +-0/1 to imply the use | of the smallest value representable by an implementation as the | delta value in it's calculation of a limit value its argument's | value? That doesn't work in general. Intermediate quantities in a calculation can overflow or underflow if the delta is too small. (define (func x) (expt (+ (/ x) (exp (/ x)) (exp (/ 2 x))) x)) ;;The mathematically correct answer is e^2 = 7.3890560989306504: (limit func 0 1/40) ==> 7.3890560989306504 (limit func 0 1/41) ==> #i0/0 (limit func 0 1/42) ==> #f (limit func 0 1/43) ==> #i+/0 (limit func 0 1/444) ==> #i+/0 (limit func 0 1e-323) ==> #i+/0

**Follow-Ups**:**Re: limit function***From:*Paul Schlie

**References**:**Re: comparison operators and *typos***From:*Paul Schlie

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