The Square Root Function: Understanding the Difference

[TheDemx27]

I went to splash at MIT a while back, and I took a class on cesaro summation. We were promised to go over an interesting identity but we never did: $4(\frac{1}{2}!)^2=\pi$. Now, this doesn't make any sense to me, since I thought you could only do factorials with integers, like in the famous example of recursive code:

int fact(int n)
{
    int result;

    if(n==1)
    {
        return 1;
    }

    result = fact(n-1)* n;
    return result;
}

This was my only concept of factorials. How would one make a separate function to handle the fractions and what would the maths be?

 

[Stephen Tashi]

Factorials of fractions are (tranditionally) defined in terms of the values of the Gamma function.

As to why this is a good way to do things, I haven't seen any simple explanation. In particular, I haven't seen any explanation of why computing the gamma function is, in any sense, an extension of the algorithm used to compute factorials of postive integers.

 

[pwsnafu]

In particular, I haven't seen any explanation of why computing the gamma function is, in any sense, an extension of the algorithm used to compute factorials of postive integers.

The only property gamma has that other extensions do not is log-convexity (Bohr-Mollerup theorem). As to why that is related to the factorial, your guess is as good as mine.

 

[mathman]

Γ(n+1) = n!. Γ(z) is defined for all complex z, while factoral is only for integers.

 

[pwsnafu]

Γ(n+1) = n!. Γ(z) is defined for all complex z, while factoral is only for integers.

That is not true. Gamma has poles at 0, -1, -2 etc.

 

[Char. Limit]

Convexity is nice. Log-convexity is even nicer. Convexity basically means that the function in question has a second derivative that is never negative - that is to say, the function always curves "up". Log-convexity means that even the logarithm of the function is convex, and is a stronger condition than convexity. Thus, since the Gamma Function is equal to the factorials on all integers*, is analytic, and is log-convex, and is the only function that satisfies all three of these properties, it is considered the "best" extension of the factorials to all complex numbers excepting the negative reals.

*: Well, not quite true. The Gamma Function is shifted by 1. I share the opinion of Euler and many others that this shift was a dumb one.

 

[mathman]

That is not true. Gamma has poles at 0, -1, -2 etc.

You are quibbling. Yes it has poles. So does this mean it is not defined or is it defined by the poles?

 

[pwsnafu]

You are quibbling. Yes it has poles. So does this mean it is not defined or is it defined by the poles?

Gamma is not defined at 0 and negative integers. A pole is a type of singularity. You don't say $\frac{1}{x}$ is defined at x=0 by a pole for the same reason.

 

[mathman]

Gamma is not defined at 0 and negative integers. A pole is a type of singularity. You don't say $\frac{1}{x}$ is defined at x=0 by a pole for the same reason.

The quibble is in the definition of the word "define". Does it mean to have a specific value or does it mean it can be described (a pole at that point)?

 

[David Carroll]

The function switches concavity at each pole. The reason why it isn't defined at each pole is because on the right side it would approach negative infinity and on the left side it would approach positive infinity (or vice versa).

 

[mathman]

The description you gave is the definition. The quibble is over what it means to "define" something.

 

[HallsofIvy]

The quibble is in the definition of the word "define". Does it mean to have a specific value or does it mean it can be described (a pole at that point)?

To say that a function is or is not defined at a point has a very specific meaning- it means that there is a unique value for the function at that point.

 

[mathman]

To say that a function is or is not defined at a point has a very specific meaning- it means that there is a unique value for the function at that point.

Does this mean that square roots (two valued) are not defined?

 

[pwsnafu]

Does this mean that square roots (two valued) are not defined?

"square roots (two valued)" is not a function. That's irrelevant.

 

[mathman]

I guess the term "function" needs to be defined. So what do you call the square root of x?

 

[David Carroll]

Perhaps "function" could be defined as any operation whose outputs could be entered into an inverse form of that operation and have an output that was the same as the input of the original operation. Hence, both (-x) and (x) could be squared and you would get x², which was the input to the original operation of "taking the square root". If, however, you were to input "negative infinity" or "positive infinity" into an inverse gamma function, you would get an infinite number of possible outputs.

 

[pwsnafu]

I guess the term "function" needs to be defined.

A function is a triple $(A, B, \Gamma)$ where

1. A is a set called the domain,
2. B is a set called the co-domain,
3. $\Gamma \subset A \times B$ is called the graph,
4. for all $x \in A$ there exists $(x,y) \in \Gamma$,
5. if $(x,y) \in \Gamma$ and $(x,z) \in \Gamma$ then $x = z$.

In words, a function is a relation that is defined on its domain and satisfies the horizontal line test.

So what do you call the square root of x?

Do you understand that "the square root of x" is not the same thing as "square roots (two values)"?
And if you were wondering, the former is a function. The latter is not.