1

The product can be transformed into:

$1=\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots$

The result after a few products seems to approach $1$ as expected. Here is my proof of the identity which uses trigonometry and the Weierstrass factorisation theorem: https://math.stackexchange.com/a/4777515/42969

However, I would like to know if the identity can be derived using an alternate method, without any reference to trigonometry. Additionally I would like to know if this identity has been observed before.

  • 6
    The product on the left is odd and the product on the right is even. If they are infinite then they both diverge. – John Douma Sep 30 '23 at 05:20
  • 1
    The equality in the title is (debatably) true but for silly reasons (as John says both sides diverge) and it is not equivalent to the $1 = \dots $ identity. For that identity you first need to check that the infinite product converges at all (which can be done by taking the logarithm and then applying the alternating series test). – Qiaochu Yuan Sep 30 '23 at 05:42
  • Ohh thanks for clarifying. I understand that this is a very silly question :) but the infinite product $(1+1/3)(1-1/5)\cdots$ does seem to approach $1$... Please try to prove this identity(and ignore my mistake of trying to equate diverging series) :$$1=(1+1/3)(1-1/5)\cdots$$ I have proved the identity using trigonometry but I would like a purely algebraic or non trigonometric approach :) –  Sep 30 '23 at 05:50
  • 2
    Maybe you could change the title and provide us your trigonometric proof? – garondal Sep 30 '23 at 08:01
  • 1
    You proved the identity here: https://math.stackexchange.com/a/4777515/42969, which looks correct to me. Since it is an infinite product, any proof would involve some limiting process. Can you clarify what kind of “algebraic proof” you are looking for? – Martin R Sep 30 '23 at 08:10
  • 1
    Yep, but I am obsessed with the Basel problem :) The way I proved it uses an extension of Weierstrass factorisation theorem. The Basel problem is a direct consequence of this theorem. I however want to prove the identity in the question with no reference to the factorisation, please use whatever method you like :) Proving this identity may help in an alternate derivation of the Basel problem –  Sep 30 '23 at 08:23
  • Are you familiar with https://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf which gives 14 proofs of $\zeta(2)=\pi^2/6$? – Gerry Myerson Sep 30 '23 at 10:19
  • So is the claimed equation $\prod_{k\ge0}\frac{(8k+4)^2(8k+6)(8k+10}{(8k+3)(8k+5)(8k+7)(8k+9)}=1$? – J.G. Sep 30 '23 at 10:29
  • Yes indeed the product is $\prod_{k\ge0}\frac{(8k+4)^2(8k+6)(8k+10)\cdots}{(8k+3)(8k+5)(8k+7)(8k+9)\cdots}=1$(As per my limited understanding of these notations)

    I'll expand the initial product: $$1=(1+\dfrac{1}{3})(1-\dfrac{1}{5})(1-\dfrac{1}{7})(1+\dfrac{1}{9})(1+\dfrac{1}{11})(1-\dfrac{1}{13})(1-\dfrac{1}{15})(1+\dfrac{1}{17})(1+\dfrac{1}{19})\cdots$$

     –  Sep 30 '23 at 10:37
  • @GerryMyerson Yes I do know of pdf. However I want to prove it in my own way. Since, by proving the identity in the question, it becomes easy to show that the square of the class number formula(L(1,-8)): $$\dfrac{\pi}{2\sqrt{2}}=1+1/3-1/5-1/7+1/9+1/11-\cdots$$ is simply, $$\dfrac{\pi^2}{8}=1^2+(1/3)^2+(1/5)^2+(1/7)^2+(1/9)^2+\cdots$$ –  Sep 30 '23 at 10:42
  • Follow-up question posted as https://math.stackexchange.com/questions/4778744/proof-verification-of-basel-problem – Gerry Myerson Oct 01 '23 at 12:26

1 Answers1

3

Consider the product:$(1+1)\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots$ , as:

$$\prod{\dfrac{(8n+2)(8n+4)(8n+4)(8n+6)}{(8n+1)(8n+3)(8n+5)(8n+7)}}--(1)$$

(Reference:New Wallis- and Catalan-Type Infinite Products, Jonathan Sondow, Huang Yi https://arxiv.org/abs/1005.2712) (This is their proof, not mine)

Rewrite as: $$\prod{\dfrac{(n+1/4)(n+1/2)(n+1/2)(n+3/4)}{(n+1/8)(n+3/8)(n+5/8)(n+7/8)}}--(2)$$

Euler's reflection formula:

$\Gamma(x)\Gamma(1-x)=\dfrac{x}{\sin(\pi x)}$

Weierstrass infinite product of gamma function gives us the general formula:

$\prod{\dfrac{(n+a_1)\cdots(n+a_k)}{(n+b_1)\cdots(n+b_k)}}=\dfrac{\Gamma(b_1)\cdots\Gamma(b_k)}{\Gamma(a_1)\cdots\Gamma(a_k)}$ (Where $a_1+a_2+\cdots+a_k=b_1+b_2+\cdots+b_k$)

Using these two formulae, $(2)$ becomes:

$$\dfrac{\Gamma(1/8)\Gamma(3/8)\Gamma(5/8)\Gamma(7/8)}{\Gamma(1/4)\Gamma(1/2)\Gamma(1/2)\Gamma(3/4)}=\dfrac{\sin(\pi/4)\sin(\pi/2)}{\sin(\pi/8)\sin(3\pi/8)}=2$$

Therefore, $$1=\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots$$ (End of referenced proof)

Consequently, since you're so obsessed with the Basel problem, you could substitute this result in the Dirichlet series $$L_{-8}(1)=\dfrac{\pi}{2\sqrt{2}}=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots$$

Squaring reveals, $$\dfrac{\pi^2}{8}=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+\cdots+2\left[\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots-1 \right]$$(The combining might have been a bit wonky tho)

which results in: $$\dfrac{\pi^2}{8}=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+\cdots$$

LithiumPoisoning
  • 1,164
  • 1
  • 16
  • I don't know too much about the mentioned Dirichlet class number formule derivation since it deals with advanced mathematics. However if the result $L_{-8}(1)$ can be derived without any reference to the Basel problem or the Zeta function, this could be an alternate approach to solving the Basel problem.... – LithiumPoisoning Oct 01 '23 at 06:12