|i Factorial| You Won’t Believe The Outcome
|i Factorial| You Won’t Believe The Outcome

Hello,

I apologize in advance if I mess up some mathematical terms, English isn’t my native language. I tried my best to look up the English equivalents of terms I’m about to use, but please do correct me if I made some mistakes.

I’m not a mathematician by trade but I’ve always had a penchant for math and calculus (I had the more or less “standard” course of it in the university). I’ve recently stumbled onto an article about fractional calculus and it really interested me. While reading about it and calculating simple derivatives I came up with an interesting observation. Unfortunately there’s no way to simply ask this question without showing how I came to it, so please bear with me for the next few paragraphs.

Suppose we want to test that the 0.5th derivative of the 0.5th derivative of a function f(x) = x will give us the same result as the 1st derivative of the same function (that is, f0.5(f0.5(x)) = f1(x)). This statement makes sense since the powers of derivatives add up when we perform them one after another. For example, 2nd der. of the 3rd der. is the 5th der., etc.

Using the fairly obvious idea from this section of the Wikipedia article, we attempt to calculate the 0.5th derivative of f(x) = x. See image below:

The image (step 1: calculating the 0.5th derivative of the original function)

If we don’t know anything about the Gamma function, we have a problem here: this result has a factor of 1/(0.5)!, which is not, to the best of our knowledge, a meaningful number. But we can decide to continue with our calculation. We can simply leave this factor in this original form (as a constant of sorts) and see where it gets us. And so, we continue by calculating the 0.5th derivative of this result. See image below:

The image (step 2: calculating the 0.5th derivative of the result of the previous calculation)

The original idea that the property of the powers of the derivatives to add up is preserved with fractional derivatives stands true. Indeed, we got the result (= 1) which is the same as the result of the direct calculation of the 1st derivative.

But what’s more important is that the “bad”, “unknown” number (the factorial of 0.5) simply disappeared in the process and we got ourselves a result which is a real number. I’ve tried to do the same thing with other fractional powers of the derivatives (calculating the 1/3rd der. 3 times in a row, calculating the 1/4th der. 4 times in a row) and it works successfully every time. I’m sure there is a simple enough proof that it works for all such cases but honestly I haven’t bothered.

The problem that is left is to find out the “real” value of the factorial of 0.5. Here one might argue that it’s not actually necessary to know this value, since it “disappears” in the process of calculating the 0.5th der. two times. What I mean here is that we can imagine the factorial of 0.5 to be 432, 100500, 0.234 or any other arbitrary number – it will not affect the end result. Two 0.5th derivatives in a row would still give the correct result.

But since we want to know the numerical value of the first calculation (“square root of x times what?”), we have found a way of calculating the factorial of 0.5 by introducing the Gamma function. It “extends” the factorial function onto a set of real numbers; in particular, we can look up the value of Γ(0.5+1) (= (0.5)!) and see that it’s half of the square root of π. As a result, the 0.5th derivative of f(x) = x is approximately 1.129*sqrt(x).

And here we finally come to my question (which, unfortunately, I haven’t found a way to express without showing you my thought process above). Why have we chosen to use the Gamma function to calculate the values of the factorial for non-integer numbers? As shown above, it doesn’t matter what numerical value we use for the factorial of 0.5 – be it 1.772, 250 or -28 – the fundamental property of the derivative powers to add up would still hold true. If (hypothetically!) the process of calculating the 0.5th derivative two times in a row would give us something like this as a result (just as an example):

Image (hypothetical result for example!)

… then we would have something to work on. We would know that 0.5!/1.5! equals 1 (the real numerical value of the 1st der. of x), and from this proportion, we would derive some equations to find the numerical value of 0.5!. But since all such “questionable” numbers disappear from the equations as soon as we get to an integer power of the derivative, we have nothing to compare these “strange” numbers to.

So, once again, this is the question that confuses me: why have we chosen to use Gamma function to calculate factorials of non-integers? Even the Wiki article on the factorial states (in the very end of this section) that there are other ways to interpolate integer factorial values. Does it mean that if we decide to use some other function instead of Gamma to calculate factorials of non-integers, then we’ll get some other numerical result for the fractional derivatives? And is there a way to know “for sure” which of these numerical results would be the actual correct one?

P.S. The whole thing with “bad” numbers disappearing from the equations during the final act reminds me of cubic equations and how complex numbers can pop up and go away when you use the Cardano’s solution. Even if we know nothing about the numerical values of complex numbers, we can simply treat them as some constants and then they will simply go away in the end. However, we do have a way of calculating the numerical values of complex numbers from other mathematical fields. Here, however, it seems that we just trust the Gamma function that it does an okay job at interpolating the factorial, because we have no other way to “test” these numerical results. Is it true?

TL;DR: We seemingly have no way of “testing” the values of the Gamma function for being proper representations of the factorials of non-integers, yet we use these values in fractional calculus. Why do we use Gamma function and not some other arbitrary function that happens to be equal to N! for integer N’s?

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