Monday Math 21: The Gamma Function (Part 4/?)

The Gamma Function Part 4: Euler Reflection Formula and the Sine Function:
(Part 1)
(Part 2)
(Part 3)

Let us now define the function , for non-integer x, as . Now, we remember that , and if we replace x with –x, we see that , and so . Thus, when we examine , we see:

So is periodic with a period of one.

Now, we examine the behavior of φ near zero (and thus near all integers due to periodicity). We can write as:

or thus

The gamma functions both have values of unity at x=0, and the singularity at the origin for the fraction is removable; expanding the sine in its Maclaurin series, we have:

Which gives , and thus we have for x an integer as well.

From part 3, we had the Legendre duplication formula:
.
If we replace x with 1-x, we obtain:
.
Using these, we see:
.

Now, define . Then g(x) is periodic, since φ is periodic. Now, we can apply the duplication formula for φ to see:


Now, since g(x) is continuous on the closed interval [0,1], it is bounded; there is a constant M such that for (see the ‘boundedness theorem’ here). As g(x) is periodic, we see it is thus bounded by M for all x. Applying our previous result for g(x):

And thus g(x) can be bounded by . Repeating this process, we obtain bound . We can continue repeating; the result is that the bound of g(x) goes to zero. Thus , which tells us that is linear. As it is periodic and continuous, it must then be a constant; and thus is a constant. Since , we thus see that for all x. Using the definition of , we have the Euler reflection formula:


Solving the reflection formula for , we have:
.
Now, using , we find that

In part 2, we found the Weierstrass form of the gamma function:

Plugging in –x, we obtain:

And thus

Plugging this into our formula for :
.

Thus, we have an expression for the sine function as an infinite product.

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6 Responses to “Monday Math 21: The Gamma Function (Part 4/?)”

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