Re: A solution to calculation of odd roots of negative numbers with Calc

Hi,

Le 16/07/2012 19:14, marlon orlando barahona alvarez a écrit :
> when we try to calculate the cube root of -27 (using the formula =
> POWER (A1, 1/3) -3 Calc should give us but gives us the message # VALUE!
> I solved the problem by modifying the formula as follows:
> =-POWER (-A1, 1/3) and this modification Calc can give us the correct
> answer.
> Well I think to include this modification in Cal would take a decision
> block, suppose that the entries are "numbers, index". The terms of the
> decision would be:
> if (num <0 & 1/index% 2 = 1)
> The first one test if num is a negative number, the second one trys if
> index is an odd number, yet if an integer. If true both conditions
> would be used if the proposed amendment and was not used the normal
> function.
> If this helps please let me know. I hope not look like a fool. If Cal
> can do the calculation differently let me know as it does, thanks for
> everything.
I think it should be better to define a root() function specifically
dedicated to compute the root(s). Indeed we should not do confusion
between exponentiation and root(s) extraction. x^y is mathematically
defined for real numbers and only for x > 0 : x^y = exp(y.Ln(x)). Root
extraction is another problem : number of roots, make difference between
real and complex root, one argument is real and another is integer.

Best regards.
JBF

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