Wallis Product Proof

The answer for the first product is $2^{1+\frac m2-2^m}$ We will show this by induction on m. The base case of m=0 is straightforward. Using the double angle identity for sin(x) $\sin(\frac{\pi j}{2^{m+1}})=2\sin(\frac{\pi j}{2^{m+2}}) \cos(\frac{\pi j}{2^{m+2}})=2 \sin(\frac{\pi j}{2^{m+2}}) \sin(\frac{(2^{m+1}-j)\pi}{2^{m+2}})$ Notice that $\prod_{j=1}^{2^m} \sin(\frac{\pi j}{2^{m+1}}) = 2^{2^m-\frac 12}\prod_{j=1}^{2^{m+1}} \sin(\frac{\pi j}{2^{m+2}})$ (The extra $2^{\frac {-1}{2}}$ appears because $\sin\frac{\pi}4$ got included twice in the previous calculation. Also $\sin\frac{\pi}{2}$ did not get included but that doesn't matter because it is equal to 1) This completes the proof because $2^{1+\frac m2-2^m} \div 2^{2^m-\frac 12} = 2^{1+\frac {m+1}2-2^{m+1}}$

The answer for the second product is $2^{\frac 12-2^m}$ We will show this by induction on m. The base case of m=0 is straightforward. Using the double angle identity for sin(x) $\sin(\frac{2\pi j-\pi}{2^{m+2}})=2\sin(\frac{2\pi j-\pi}{2^{m+3}}) \cos(\frac{2\pi j-\pi}{2^{m+3}})=2 \sin(\frac{2\pi j-\pi}{2^{m+3}}) \sin(\frac{2^{m+2} \pi-(2\pi j-\pi)}{2^{m+3}})$ Notice that $\prod_{j=1}^{2^m} \sin(\frac{2\pi j-\pi}{2^{m+2}}) = 2^{2^m}\prod_{j=1}^{2^{m+1}} \sin(\frac{2\pi j-\pi}{2^{m+3}})$ This completes the proof because $2^{\frac 12-2^m} \div 2^{2^m} = 2^{\frac {1}2-2^{m+1}}$

Bonus: Consider the product $\prod_{j=1}^{2^m} \frac{\sin^2(\frac{\pi j}{2^{m+1}})}{\sin(\frac{2\pi j-\pi}{2^{m+2}})\sin(\frac{2\pi(j+1)-\pi}{2^{m+2}})}$ By the formulae that we just found, this is equal to $\frac{(2^{1+\frac m2-2^m})^2}{(2^{\frac 12-2^m})^2 \sin(\frac{2^{m+1}\pi+\pi}{2^{m+2}}) \div \sin(\frac{\pi}{2^{m+2}})}$ This simplifies to $2^{m+1} \tan \frac{\pi}{2^{m+2}}$ Taking the limit as m goes to infinity of each side of the equation gives us $\frac{\pi}{2} = \frac 21 \frac23 \frac 43 \frac 45 \frac 65 \frac 67 . . .$

Note: To take the limit of the product at the end we technically need to show that the limit of the product is equal to the product of the limits. To make this a little easier, let $n=2^m$ . We can simplify each term to $\frac{4j^2+e}{4j^2-1+f}$ where e and f are small error terms whose most significant term is proportional to $n^{-2}$ . The number of terms in which this error gets multiplied is n, so we need to take $\frac{4j^2+e}{4j^2-1+f}$ to the nth power. We can see that the error terms do go to zero even if they get multiplied n times, so the limit of the product is indeed equal to the product of the limits.

For more information see https://www.3blue1brown.com/sridhars-corner/2018/4/17/wallis-product-supplement-dominated-convergence