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I have found an interesting family of infinite products. The most interesting one of them being:

$\sqrt{2}-1=\dfrac{1\cdot7\cdot9\cdot15\cdot17\cdot23\cdots}{3\cdot5\cdot11\cdot13\cdot19\cdot21\cdots}$

The numerators follow the pattern $+6,+2,+6,+2,\cdots$ and the denominators follow $+2,+6,+2,+6,\cdots$

The family of products was derived from assuming: $\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$

And subsequently replacing $x$ as $\dfrac{\pi}{2}$

The keyword here being "assumed", so I don't know if this product can be proven. The values do seem to be equal after a few iterations though.

However, I would greatly appreciate any proofs or alternate derivations.

Anton Vrdoljak
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LithiumPoisoning
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    Multiplying $10^8+1$ factors gives a value about $10^{-9}$ far away , so the product seems to converge (however slowly) to $\sqrt{2}-1$. Of course , this is not yet a proof. – Peter Sep 28 '23 at 06:32
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    I would write $\sin(x) - 1/\sqrt 2 = \sin(x) - \sin(\pi/4)$, apply the sum to product identity, and then write each factor as an infinite product https://math.stackexchange.com/q/674769/42969. – Martin R Sep 28 '23 at 06:58
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    Might be connected : https://math.stackexchange.com/questions/157372/proving-frac-sin-xx-left1-fracx2-pi2-right-left1-fracx22 : https://math.stackexchange.com/questions/674769/sinx-infinite-product-formula-how-did-euler-prove-it : https://math.stackexchange.com/questions/134870/how-was-euler-able-to-create-an-infinite-product-for-sinc-by-using-its-roots : https://math.stackexchange.com/questions/3899658/infinite-product-of-sqrt2 : "Weierstrass Factorization Theorem" & "Wallis Product" ?? Interesting Question !! – Prem Sep 28 '23 at 07:08
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    In the following paper by Sondow and Yi, a number of relevant infinite products are derived. In particular, I think it is worthwhile to look into equations (9) and (11). These give expressions for $\sqrt{ 2 + \sqrt{2}} $ and $\sqrt{ 2 - \sqrt{2}} $. The denominators also have the +2, +6 pattern. The numerators have a +2, +4 pattern: https://www.jstor.org/stable/pdf/10.4169/000298910x523399.pdf?casa_token=zMnO3gmsdHEAAAAA:_aQ3vfAFDNJvOcU_usaVXbQhMJuQyqX2_WfWkD_FskT1pzYx5ZqEMJd7pTHGyuGVRpWvFNwB3Pwig-jEz9RlnNblDDLBVu_ER_olfw1R8NVrOnPHU7pN – Max Muller Sep 28 '23 at 07:53
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    ^ correction: the numerators have a +4, +4 pattern. The denominators are correct as stated. – Max Muller Sep 28 '23 at 08:22
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    You can write this as $$\prod_{i=0}^\infty\frac{4i+2-(-1)^i}{4i+2+(-1)^i}$$ – Akiva Weinberger Sep 29 '23 at 05:33
  • You also can exclude $9$ from the fraction to have only prime numbers. I don't know if there are other composite numbers in the continuation of the fraction, but I guess they can be excluded too. –  Oct 01 '23 at 06:54
  • I suppose that this must have some connection with $L(\chi_8,1) = 1-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{15}+\cdots=\dfrac{\log(1+\sqrt{2})}{\sqrt{2}}$. Here $\chi_8=\left(\dfrac{\cdot}{2}\right)$ is a principal Dirichlet character modulo $8$, and $\left(\dfrac{\cdot}{\cdot}\right)$ is the Kronecker symbol. – Jianing Song Oct 05 '23 at 17:00

5 Answers5

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You wrote

The family of products was derived from assuming: $\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$

That assumption is indeed correct.

Proof: Using the “sum to product” identity for the sine we have $$ f(x) = \sin(x) - \frac{1}{\sqrt 2} = \sin(x) - \sin\left(\frac \pi 4\right) = -2 \sin\left(\frac x2 - \frac \pi 8\right) \sin\left(\frac x2 - \frac {3\pi}8 \right) \, . $$ Using the infinite product for the sine (see, e.g., here), the first factor can be written as $$ \sin\left(\frac x2 - \frac \pi 8\right) = \left(\frac x2 - \frac \pi 8\right)\prod_{n=1}^\infty \left(1-\frac{x}{2\pi n} + \frac{1}{8n}\right) \left(1+\frac{x}{2\pi n} - \frac{1}{8n}\right) \\ = (-\frac \pi 8) \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1 + \frac{1}{8n}\right)\left(1-\frac{4x}{(8n+1)\pi} \right) \left(1- \frac{1}{8n}\right)\left(1+\frac{4x}{(8n-1)\pi}\right) \\ = C_1 \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-1)\pi}\right)\left(1-\frac{4x}{(8n+1)\pi} \right) $$ with the constant $$ C_1 = (-\frac \pi 8)\prod_{n=1}^\infty \left(1- \left(\frac{1}{8n}\right)^2\right) \, . $$ In the same way we get for the second factor $$ \sin\left(\frac x2 - \frac {3\pi}8 \right) = C_2 \left(1 - \frac {4x}{3\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-3)\pi} \right) \left(1-\frac{4x}{(8n+3)\pi}\right) $$ with some constant $C_2$.

Combining these results we have $$ f(x) = C \left(1 - \frac {4x}{\pi} \right)\left(1 - \frac {4x}{3\pi} \right) \\ \times \prod_{n=1}^\infty\left(1+\frac{4x}{(8n-3)\pi} \right) \left(1+\frac{4x}{(8n-1)\pi} \right)\left(1-\frac{4x}{(8n+1)\pi} \right)\left(1-\frac{4x}{(8n+3)\pi} \right) $$ with some constant $C$.

Setting $x=0$ shows that $C= -1/\sqrt 2$, and that concludes the proof.

Martin R
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This is an alternative derivation, not a proof. Noting the pattern of $+8$ between alternating multiplicands. Your infinite product can be expressed as two separate products:

$$\prod_{k=0}^{\infty}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot \prod_{k=1}^{\infty}{\left( \dfrac{8k - 1}{8k - 3}\right)} \\= \prod_{k=0}^{\infty}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot 3\prod_{k=0}^{\infty}{\left( \dfrac{8k - 1}{8k - 3}\right)}$$

Now, consider the more general form:

$$\prod_{k=0}^{n}{\left( \dfrac{8k + 1}{8k + 3}\right)} \cdot 3\prod_{k=0}^{n}{\left( \dfrac{8k - 1}{8k - 3}\right)} \\= 3\cdot\left(\dfrac{\left(n + \frac{1}{8}\right)! \cdot \left(\frac{3}{8}\right)!}{3\cdot\left(n + \frac{3}{8}\right)! \cdot \left(\frac{1}{8}\right)!} \right) \cdot \left( \dfrac{\left(n - \frac{1}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(n - \frac{3}{8}\right)! \cdot \left(-\frac{1}{8}\right)!} \right)$$

As $n \to \infty$, the product converges to

$$\dfrac{\left(\frac{3}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(\frac{1}{8}\right)! \cdot \left(-\frac{1}{8}\right)!} $$

Recall that $(-z)!\cdot z! = \Gamma{(1 - z)}\cdot\Gamma{(1 + z)} = \Gamma{(1 - z)}\cdot z\cdot\Gamma{(z)}$. By the reflection identity, $\Gamma{(1 - z)}\cdot z\cdot\Gamma{(z)} = \dfrac{\pi z}{\sin{\pi z}}$

Hence,

$$\dfrac{\left(\frac{3}{8}\right)! \cdot \left(-\frac{3}{8}\right)!}{3\cdot\left(\frac{1}{8}\right)! \cdot \left(-\frac{1}{8}\right)!} \\= \dfrac{\frac{3\pi/8}{\sin{3\pi/8}}}{3\cdot\frac{\pi/8}{\sin{\pi/8}}} \\= \dfrac{\sin{\pi/8}}{\sin{3\pi/8}} \\= \sqrt{2} - 1$$

You can use the sine expansion formulas in the last step.

Dstarred
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    May I ask a silly question? There was a $3$ at the top. Why has it gone to the bottom? – Bob Dobbs Sep 28 '23 at 10:57
  • @BobDobbs You question is not silly :D. Thanks for hinting the error. I forgot to include the $\dfrac{1}{3}$ factors after evaluation of both products. Will correct now! – Dstarred Sep 28 '23 at 11:48
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Consider the partial product $$P_k=\frac{\prod _{n=1}^k a_n}{\prod _{n=1}^k b_n}$$ where $$a_n=4 n+(-1)^n-2\qquad \qquad\text{and} \qquad\qquad b_n=4 n-(-1)^n-2$$ that is to say $$P_k=\prod _{n=1}^k \left(1+\frac{2 (-1)^n}{4 n-(-1)^n-2}\right)$$ which is $$P_k=\frac{ \Gamma \left(\left\lfloor \frac{k-1}{2}\right\rfloor +\frac{9}{8}\right) \Gamma \left(\left\lfloor \frac{k}{2}\right\rfloor +\frac{7}{8}\right)}{\Gamma \left(\left\lfloor \frac{k-1}{2}\right\rfloor +\frac{11}{8}\right) \Gamma \left(\left\lfloor \frac{k}{2}\right\rfloor +\frac{5}{8}\right)}\,\,\tan \left(\frac{\pi }{8}\right)$$

The big front factor oscillates but converge very slowly to $1$. Then the result.

Asymptotically $$P_{2p}=\left( 1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)\,\,\tan \left(\frac{\pi }{8}\right)$$ $$P_{2p+1}=\left(1-\frac{1}{8 p}+\frac{9}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)\,\,\tan \left(\frac{\pi }{8}\right)$$ and $$\frac{P_{2p+2} }{P_{2p} }=1-\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ $$\frac{P_{2p+3} }{P_{2p+1} }=1+\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$

Claude Leibovici
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As an (overkill) alternative, consider Hurwitz's zeta function $$ \zeta(s,a)=\sum_{n\geq 0}(n+a)^{-s} $$ which satisfies (cf. 25.11.18 in https://dlmf.nist.gov/25, or, for a proof, Section 13.21 in Whittaker and Watson's Course of Modern Analysis) $$ \zeta(0,a)=\tfrac 12-a,\qquad \frac d{ds}\zeta(s,a)\big|_{s=0}=\log\Gamma(a)-\frac 12\log(2\pi).\qquad\qquad(\star) $$ Introduce the function \begin{align*} f(s)&=\sum_{n\geq 1}(-1)^n\bigl((4n-1)^{-s}+(4n+1)^{-s}\bigr) \\ &=8^{-s}\bigl[-\zeta (s,\tfrac{3}{8})-\zeta(s,\tfrac{5}{8})+\zeta(s,\tfrac{7}{8})+\zeta(s,\tfrac{9}{8})\bigr] \end{align*} We have $$ \sum_{n\geq 1}(-1)^n\bigl(\log(4n-1)+\log(4n+1)\bigr)=\log\biggl(\frac{7\cdot9\cdot15\cdot 17\cdots}{3\cdot 5\cdot 11\cdot 13\cdots}\biggr)=\frac d{ds}f(s)\big|_{s=0} $$ which can be evaluated thanks to $(\star)$ and Euler's reflection formula for the Gamma function.

Giulio R
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Not an answer by any means but a few interesting observations:

Since it has been proved by the other wonderful answers that:$\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$

One could take the logarithm on both sides:

$\log{(\dfrac{1}{\sqrt{2}}-\sin(x))}=\log{(\dfrac{1}{\sqrt{2}})}+\log\left(1-\dfrac{4x}{\pi}\right)+\log\left(1-\dfrac{4x}{3\pi}\right)+\log\left(1+\dfrac{4x}{5\pi}\right)+\log\left(1+\dfrac{4x}{7\pi}\right) \cdots$

And differentiate:

$\dfrac{-\cos(x)}{\frac{1}{\sqrt{2}}-0}=>\sqrt{2}=0+\dfrac{4}{\pi(1-\dfrac{4x}{\pi})}+\dfrac{4}{3\pi(1-\dfrac{4x}{3\pi})}+\dfrac{-4}{5\pi(1-\dfrac{4x}{5\pi})}+\dfrac{-4}{7\pi(1-\dfrac{4x}{7\pi})}+\cdots$

When $x=0,$ the resulting summation becomes:

$\dfrac{\pi}{2\sqrt{2}}=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots$, So one can derive a subset of the Dirichlet L-summations from the $\sin(x)-k$ representation

Additionally replace $x=-\dfrac{\pi}{4}$ in: $\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right) \cdots$

To get, $1=(1+\dfrac{1}{3})(1-\dfrac{1}{5})(1-\dfrac{1}{7})(1+\dfrac{1}{9})(1+\dfrac{1}{11})\cdots$

:)