Using Stirling's Formula to calculate ratio I am working on this question:

For equidistant points $x_j=j, 0\leqslant j\leqslant n$, $n$ even, let $$\omega(x)=(x-x_0)(x-x_1)\ldots (x-x_n)$$ Use Stirling's formula to estimate the ratio $$\frac{\omega\left(\frac 12\right)}{\omega\left(\frac{n}{2}+\frac 12\right)}$$ for large $n$. Explain Runge phenomenon.

So far I have figured out that $$\omega(1/2)=\frac12·\left(-\frac12\right)·\left(-\frac32\right)\ldots\left(\frac12-n\right)=(-1)^n2^{-n-1}1·3·5\ldots(2n-1)
=(-1)^n2^{-2n-1}\frac{(2n)!}{n!}$$
Now, I know that I have to use Stirling's Formula. But how do I compute $\omega\left(\frac{n}{2}+\frac 12\right)$, to find the ratio? Also how does the Runge phenomenon play into this?
 A: Let us use the notation
$$\omega_q(x) = \prod_{i=0}^q(x-x_i)$$
and take the work already done to arrive at
$$\omega_n\left(\frac 12\right)=\frac12·\left(-\frac12\right)·\left(-\frac32\right)\ldots\left(\frac12-n\right)=(-1)^n2^{-n-1}1·3·5\ldots(2n-1)\\
=(-1)^n2^{-2n-1}\frac{(2n)!}{n!}$$
Given that we have $n=2m$ even, this becomes
$$\omega_{2m}\left(\frac12\right) = 2^{-4m-1}{(4m)!\over(2m)!}$$
We also wish to find the value of $\omega_n\left(\frac n2+\frac 12\right)$, and this value is $(\frac n2+\frac 12)(\frac n2-\frac 12)(\frac n2-\frac 32)\cdots(\frac 32)(\frac 12)(-\frac 12)(-\frac 32)\cdots(-\frac n2+\frac 32)(-\frac n2+\frac 12)$
It is worth noting that the terms $(\frac 12)(\frac 32)\cdots(\frac n2-\frac 12)$ have the same value as $\left|\omega_m\left(\frac 12\right)\right|=2^{-2m-1}{(2m)!\over m!}$, and therefore we have
$$\omega_n\left(\frac n2+\frac 12\right) = (-1)^m\left(\omega_m\left(\frac 12\right)\right)^2\left(\frac n2+\frac 12\right)\\
=(-1)^{m}2^{-4m-2}\left({(2m)!\over m!}\right)^2\left(m+\frac 12\right)$$
This leads us to conclude
$${\omega_n\left(\frac 12\right)\over\omega_n\left(\frac n2+\frac 12\right)}={2^{-4m-1}{(4m)!\over(2m)!}\over (-1)^{m}2^{-4m-2}\left({(2m)!\over m!}\right)^2\left(m+\frac 12\right)}\\
=(-1)^{m}{(4m)!(m!)^2\over 2((2m)!)^3\left(m+\frac 12\right)}$$
The next step would probably be to use an approximation for $(2k)!\over k!k!$ in a couple spots and see what the ratio turns out to be.  If $n=4$, this becomes $-1\cdot {8!4\over 2\cdot 64\cdot 27\cdot 8\frac 52}=-{8!\over 128\cdot 27\cdot 5}=-{7\over 3}\lt -2$, and it is reasonable to expect that this ratio will increase in absolute value.  By comparison, if $n=2$ we get result $1$.
