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This morning, Mathieu had a nice experience in his course on computational method in actuarial science. But let us start with some mathematical formal definitions.First, recall that  is - somehow - a standard expression. No one should be surprised to see such an expression. Generally (as explained in http://en.wikipedia.org/... ), this function is defined only when . The idea is that the definition of  is thatAnd it is a definition. Such a function exists only if  (maybe excluding ). This would be a standard definition in real-analysis.Now, this 'power' function appears also in complex analysis, when dealing with unit roots. From instance, if  , where  and , for some , then . Thus, in complex-analysis it might be more complex to define properly  since it might not be unique. But we can relate (sometimes, when  is the inverse of an integer, or maybe a rational number ?) with roots of polynomial functions. So far, nothing new...Let us get back to Mathieu's problem. Actually, in his course, he wanted to compute . With a French version of Excel, enteringyou do get . If you look at the 'help' window, you have some more detailsIt looks like this hat function can be used to define objects such as . But withyou get(meaning that this is a problem...). It is also possible to use the power (puissance in French) function of Excel,Here, you also getThe weird part here is that, in the 'help' window, you can read that this power function can be used with any number in .Another point... what about  ? Somehow, it is just the square of the previous one (with the fraction)... Here, typingyou get(similarly with the power function). So clearly, it is not that simple to use this power function. Now, if you use Google (which is now my new online calculator when I am in class, when I cannot use R), if the power is a fraction (or to be more specific the inverse of an integer), then it works as Excel  you get But if you type (which should be close, from a continuity property of the power function)  you getand similarlyOn Wolfram Mathworld, enterMathematica does recognize that we try to deal with unit roots: the result is herewith - as expected - a numerical approximationWith Matlab, Mathieu did obtain the same as Mathematica (its decimal approximation). And to conclude, with R, Mathieu did obtain&#62; (-8)^(1/3)[1] NaN&#62; (-8)^(.333333333333333)[1] NaNSo for R, you cannot use this hat function on negative numbers.Now, how can we interpret those outputs ?1) My understanding is that clearly, with MS Excel, sincewhich is problematic. For instance, in insurance, with monthly discounts, we do have functions like . What if2) The problem comes - probably (MS Excel is not an open software, so it might be hard to check) -  from the fact that  is interpreted as an inverse of a (possibly) bijective function. To be more specific,  means that . When  is an odd integer, then (in real-analysis) there is a unique inverse, and thus,  is uniquely defined, since  is a bijective  function. This is what MS Excel (and Google) is doing:  is a bijective  function, so  means that we need to find the unique (real) value  such that . Thus, somehow, it makes sense to return .3) There is still a problem with Google, and Mathematica. That is fine to return unit roots in . But how comes there is only one value ? I mean, yes  is a possible answer, sincebut one can also observe that , and similarly,  andOne can check withWith R, since we do not deal with power function here, but with roots, if we want to find  such that , the function is&#62; polyroot(c(8,0,0,1))[1]  1+1.732051i -2+0.000000i  1-1.732051iWhich is different... Weird isn't it ?