CODEWITHSUNDEEP
maxima
Topologicalife
Topologicalife

Reputation: 15

Output of a function in maxima

I have programmed a function F which round the input following some specific rules. For example, if my input is F(0.1355) it gives as output 1.4b-1. The problem is that, when I do F(0.1355)^2, it gives me 1.96382999420166b-2, when it should give me 0.14^2, i.e, 1.96b-2. Why is this happening? and, how can I fix it?

This is the code of my function:

F(x):=(block ([log10,aaa,fpprec],log10(x):=entier(log(x)/log(10)), 
    vnorm(x):= if x/(10^(log10(x)))<2 then 1/(10^(log10(x)-1)) else 1/10^(log10(x)),
     aaa(x):= x*vnorm(x),
    fpprec:if aaa(x)<20 then 2 else 1,bfloat((round(aaa(x)))*1/vnorm(x)))); `

More info: I can't declare fpprec as global variable. I know that would solve my problem. Best regards,

Lievet.

Upvotes: 0

Views: 283

Answers (1)

Robert Dodier
Robert Dodier

Reputation: 17576

Well, the problem is that F(x)^2 is computed with the value of fpprec outside of the block, which is 16. First F(x) is computed with fpprec = 2; call that result foo. Then foo^2 is computed with fpprec = 16.

I don't know a way to limit the precision for bigfloat computations to the precision of the operands. It may be possible with simplification rules (tellsimp or tellsimpafter) but, if so, it is probably not easy.

Two ideas to make some progress here. (1) Consider using rational numbers instead of bigfloats. I.e. return 7/5 instead of 1.4b-1. (2) If you are just trying to limit the digits that are printed, use fpprintprec.

Maybe if you say more about what is your larger goal here, someone will have some advice.

Upvotes: 1

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