What I just did is enough for one day... As I said... I'm tired...
ah come on...how can you get tired of math?
Alright, let's try another one then; how about a proof of the euler reflection formula,
G(z)
G(1-z)=
pcsc(
px), valid for all z
ÎC when z
¹{1,0,-1,-2,...}. We will start with the expression of the left side:
G(z)
G(1-z)
By the gamma functional identity, this can be written as
-z
G(z)
G(-z)
Expanding this into the weierstrass product definitions, we have:
-z×(ze
gz Õ¥k=1 (1+z/k)×e
-z/k)
-1×(-ze
-gz Õ¥k=1 (1-z/k)×e
z/k)
-1The factors e
gz, e
-gz cancel out each other, as do the factors e
-z/k and e
z/k. The two z's can be multiplied to get z², and the negative signs cancel out. Since the order in which the terms in the products are multiplied does not matter, we can merge the two products into one:
z×(z²
Õ¥k=1 (1+z/k)×(1-z/k) )
-1One of the pairs of z's outside the product cancels out, and the quantity inside the product can be expanded to get:
(-z
Õ¥k=1 (1-z²/k²) )
-1Now let u=
pz, so that z=u/
p. Substituting this in, we have:
(u/
p Õ¥k=1 (1-u²/(k²
p²) )
-1Notice that the product combined with the u on the outside is now exactly that of the sine function sin(u), and this can be verified by taking the logarithm of the product to transform into a sum and then exponentiating. Therefore this becomes:
(1/
p sin(u) )
-1We now substitute z back in and simply the expression to get the final result:
pcsc(
pz)
This completes the proof. Beautiful, isn't it?
I would put up something else too but I have to go at the moment, so see you in a little while.
