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<div class="moz-cite-prefix">Hi,<br>
<br>
Le 16/07/2012 19:14, marlon orlando barahona alvarez a écrit :<br>
</div>
<blockquote
cite="mid:CAAowQz8mqgXZ9mjKRHGm-tfAetF0+PSZYMj9_gwAVKdkYVU16Q@mail.gmail.com"
type="cite">
<div><font size="4">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!</font></div>
<div><font size="4">I solved the problem by modifying the formula
as follows:</font></div>
<div><font size="4">=-POWER (-A1, 1/3) and this modification Calc
can give us the correct answer.</font></div>
<div><font size="4">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:</font></div>
<div><font size="4">if (num <0 & 1/index% 2 = 1)</font></div>
<div><font size="4">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.</font></div>
<div><font size="4">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.</font></div>
</blockquote>
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.<br>
<br>
Best regards.<br>
JBF<br>
<pre class="moz-signature" cols="72">--
Seuls des formats ouverts peuvent assurer la pérennité de vos documents.
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