Question is on the highlighted statement below:
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Does the argument (like the scan operation) become the previous output of the command i.e. the orange highlight becomes the parameter to f to yield the blue highlight ?
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In the same output mentioned in q1 above, does the program stop because machine because of 2 reasons:
1. 1e-256 squared leads a 0
2. two 0s are not printed because it would violate the rule in violet below
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Please confirm : “value matching the argument” is a subset of the “last seen is produced” clause , because “next value” is the " function applied to the previous value" and hence “previous value” is the argument for the “next value”. Better explained in question 4 below.
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In the yellow highlight below, it does not mention that the first output will be the argument itself
iterate
If f is a monadic function, f\ calls f on its argument repeatedly until a value matching the argument or the last seen is produced. The result is the argument followed by all the results except the last. For example:
q)(neg\)11 -1q)(rotate[1]\)"abcd"("abcd";"bcda";"cdab";"dabc")q)({x\*x}\)0.10.1 0.01 0.0001 1e-08 1e-16 1e-32 1e-64 1e-128 1e-256 0q){x\*x}\[0.1] / alternate syntax0.1 0.01 0.0001 1e-08 1e-16 1e-32 1e-64 1e-128 1e-256 0
W