Show that $\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2$ for $k \in \mathbb{N}$.

Question

I want to prove $\Gamma(\frac1m) \dots \Gamma(\frac{m-1}{m})=(2 \pi)^{m-1} m^{-\frac12}$.

My attempts:

For $m=2l+1$, I used the reflection formula $\Gamma(\frac{k}{m}) \Gamma(1-\frac{k}{m})=\frac{\pi}{\sin \frac{\pi k}{m}}$ and then replaced $\lim_{x \to 1} \dfrac{x^m-1}{x-1}$ once with $m$ and once with multiplication of roots of unity which results in the requested equality to prove.

For $m=2l$ this approach fails because of appearance of $\Gamma(\frac12)$ in the middle of the list with even with many tricks I used it fails to prove the final result. Reducing it to prove this : $$\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2 $$ How can I approach from this line on? It is true for small numbers of $k$ but appearance of $k$ in th denominator makes by-induction-method useless.

Added : The following is from the text but I believe it works for $m$ an odd integer. Does it work for $m$ even and if so how?

Answer 1

First, notice that $j=k$ can be excluded from consideration as the corresponding factor is $\sin\frac{\pi k}{2k}=1$.

Second, notice that $\pm\sin\frac{\pi j}{2k}$ for $j=1,2,\dots, k-1$ are the roots of the Chebyshev polynomial $U_{2k-1}(x)$ divided by $x$. By Vieta theorem, their product $(-1)^{k-1}\left(\prod_{j=1}^{k-1} \sin\frac{\pi j}{2k}\right)^2$ equals the free term of $\frac1xU_{2k-1}(x)$, which is $(-1)^{k-1}2k$, divided by its leading coefficient, which is $2^{2k-1}$. That is, $$\left(\prod_{j=1}^{k-1} \sin\frac{\pi j}{2k}\right)^2 = \frac{k}{2^{2k-2}}.$$

Answer 2

It is too long for a comment, so I post it as answer.

I will present two solutions.

First solution:

The approach proposed by Max Alekseyev is correct.

Let us prove

$\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2$, for $k \in \mathbb{N}, k > 0$

Let us go step by step.

Step 1. Given $k \in \mathbb{N}, k > 0$, let us consider the Chebyshev polynomial of the second kind $U_{2k-1}$.

We have, by definition, that $$U_{2k-1}(\cos \theta)\sin \theta= \sin 2k \theta \tag{1}$$ and its explicit form is: $$U_{2k-1}(x) = \sum_{t=0}^{k-1} (-1)^t \binom{{2k-1}-t}{t}~(2x)^{{2k-1}-2t} \tag{2}$$

Step 2. From $(2)$, we see that:

1. the degree of $U_{2k-1}$ is $2k-1$;
2. the leading coefficient of $U_{2k-1}$ is $2^{2k-1}$
3. the term of lowest degree is $(-1)^{k-1}\binom{k}{k-1}~(2x)$, that is $(-1)^{k-1}2kx$

Step 3. Let us look at the roots of $U_{2k-1}$. From $(1)$, taking $\theta = \frac{r \pi}{2k}$ where $r= 1, \dots, 2k-1$, we have

$$U_{2k-1}\left (\cos \frac{r \pi}{2k} \right )\sin\frac{r \pi}{2k}= \sin 2k \frac{r \pi}{2k} = \sin r \pi =0$$ Since for each $r= 1, \dots, 2k-1$, $\sin\frac{r \pi}{2k}\ne 0$, we have that, for each $r= 1, \dots, 2k-1$, $$ U_{2k-1}\left (\cos \frac{r \pi}{2k} \right ) = 0$$ So we have that $U_{2k-1}$ has $2k-1$ distinct roots and they are $\cos \frac{r \pi}{2k}$ for $r= 1, \dots, 2k-1$.

Now, note that for each $r= 1, \dots, 2k-1$, $$ \cos \frac{r \pi}{2k} = \cos \left ( \frac{\pi}{2}+ \frac{(r -k) \pi}{2k} \right ) = -\sin \frac{(r -k) \pi}{2k}$$ So the roots of $ U_{2k-1}$ are $-\sin \frac{(r -k) \pi}{2k}$ for $r= 1, \dots, 2k-1$. In other words (taking $s = r-k$), the roots of $ U_{2k-1}$ are $-\sin \frac{s \pi}{2k}$ for $s= -(k-1), \dots, k-1$.

It means that the roots of $ U_{2k-1}$ are $0$ and $\pm \sin \frac{s \pi}{2k}$ for $s= 1, \dots, k-1$.

Step 4. Since the term of lowest degree in $ U_{2k-1}$ is $(-1)^{k-1}2kx$ (or, equivalently, $0$ is a simples root of $ U_{2k-1}$), we can divide $ U_{2k-1}$ by $x$. Let $P$ be the polynomial obtained by such division. It is immediate that

1. $P$ has degree $2k-2$;
2. the roots of $P$ are $\pm \sin \frac{s \pi}{2k}$ for $s= 1, \dots, k-1$;
3. the leading coefficient of $P$ is $2^{2k-1}$
4. the term of lowest degree is $(-1)^{k-1}2k$

It follows immediately that $$(-1)^{k-1}\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)=(-1)^{(2k-2)} \frac{(-1)^{k-1}2k}{2^{2k-1}}$$ that is $$\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)^2=\frac{k}{2^{2k-2}}$$ Since $\sin \frac{k\pi }{2k}= \sin \frac{\pi }{2} =1$, we can write: $$\left (\prod_{s=1}^k \sin \frac{s\pi }{2k} \right)^2=\frac{k}{2^{2k-2}}$$

$\square$

Second solution: Here is another way to prove, using complex numbers, that

$\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2$, for $k \in \mathbb{N}, k > 0$

Proof: For $k=1$, it is immediate that $$\dfrac{k}{2^{2k-2}}= 1 = \left (\sin \frac{\pi }{2} \right)^2 =\left (\prod_{j=1}^k \sin \frac{\pi j}{2k} \right )^2$$

Now, note that for $m \in \Bbb N$, $m \geqslant 1$
$$\frac{x^{m} - 1}{x-1} = \prod_{j=1}^{m-1} \left (x-e^\frac{2j\pi i} {m} \right )$$ On the other hand, $$\frac{x^{m} - 1}{x-1} = x^{m-1}+ x^{m-2}+ \cdots+ x +1$$ So $$ x^{m-1}+ x^{m-2}+ \cdots+ x +1 = \prod_{j=1}^{m-1} \left (x-e^\frac{2j\pi i} {m} \right )$$ Making $x=1$, we get $$ m = \prod_{j=1}^{m-1} \left (1-e^\frac{2j\pi i} {m} \right )$$ So, for any $k \in \Bbb N$, $k \geqslant 2$, we have (making $m=2k$), $$ 2k = \prod_{j=1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right )$$ Now, note that, for all $j \in \{1, \dots ,k-1\}$, $1-e^\frac{j\pi i} {k}$ is the conjugate of $1-e^\frac{(2k-j)\pi i} {k}$. So we have \begin{align*} 2k &= \prod_{j=1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right )= \\ &= \left [\prod_{j=1}^{k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] \left (1-e^\frac{k\pi i} {k} \right ) \left [\prod_{j=k+1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] = \\ &= 2 \left [\prod_{j=1}^{k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] \left [\prod_{j=k-1}^{1} \left (1-e^\frac{(2k-j)\pi i} {k} \right ) \right ] = \\ &= 2 \prod_{j=1}^{k-1} \left |1-e^\frac{j\pi i} {k} \right |^2 =\\ &= 2 \prod_{j=1}^{k-1} \left[ \left (1- \cos \frac{j\pi } {k} \right )^2 + \left (\sin \frac{j\pi } {k} \right )^2 \right] =\\ &= 2 \prod_{j=1}^{k-1} \left[ 2 \left (1- \cos \frac{j\pi } {k} \right ) \right] =\\ &= 2^k \prod_{j=1}^{k-1} \left (1- \cos \frac{j\pi } {k} \right ) = \\ &= 2^k \prod_{j=1}^{k-1} \left[ 2 \left (\sin \frac{j\pi } {2k} \right )^2 \right] =\\ &= 2^{2k-1} \prod_{j=1}^{k-1} \left (\sin \frac{j\pi } {2k} \right )^2 \end{align*} note that we used the fact that $1 -\cos \theta = 2 \left(\sin \frac{\theta}{2} \right)^2$.

So, we have $$ \frac{k}{2^{2k-2}}=\left (\prod_{j=1}^{k-1} \sin \frac{j\pi }{2k} \right )^2 $$ Since $ \sin \frac{k\pi }{2k} = \sin \frac{\pi }{2}=1$, we can write $$ \frac{k}{2^{2k-2}}=\left (\prod_{j=1}^{k} \sin \frac{j\pi }{2k} \right )^2 $$