Eulers product identity is as follows

\begin{align} \prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 \end{align}

How one can explicitly prove this identity?

Note here $q$ deonotes a complex number satisfying $|q|<1$

  • $\begingroup$ what's $q$?.... $\endgroup$ – voldemort Jan 25 '15 at 15:29
  • $\begingroup$ @voldemort, q is just variable $\endgroup$ – phy_math Jan 25 '15 at 15:30
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    $\begingroup$ How do you compute an identity? Do you mean proving that it holds? $\endgroup$ – Wojowu Jan 25 '15 at 15:31
  • $\begingroup$ @Wojowu, yes, it is known as Euler identity $\endgroup$ – phy_math Jan 25 '15 at 15:32
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    $\begingroup$ Rewrite it as: $$\prod(1+x^n)=\prod\frac{1}{1-x^{2n-1}}$$ Then the left side coefficient of $x^N$ is the number of ways to partition $N$ into distinct positive integers, and the right is the number of ways to partition $N$ into odd not-necessarily distinct integers. This reduces it to a combinatorial question, at least. $\endgroup$ – Thomas Andrews Jan 25 '15 at 15:37

For $\lvert q\rvert < 1$, define

$$f(q) = \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n).$$

The product is absolutely and locally uniformly convergent on the open unit disk, hence $f$ is holomorphic there, and the product can be reordered and regrouped as desired. It is clear that $f(0) = 1$, and by reordering and regrouping, we have

$$\begin{aligned} f(q) &= \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n)\\ &= \prod_{n=1}^\infty(1-q^{2n-1})(1+q^{2n-1})\cdot \prod_{n=1}^\infty (1+q^{2n})\\ &= \prod_{n=1}^\infty \bigl(1-(q^2)^{2n-1}\bigr)\bigl(1+(q^2)^n\bigr)\\ &= f(q^2). \end{aligned}$$

Since for $0 < \lvert q \rvert < 1$ the set $\{ q^{2^k} : k \in \mathbb{N}\}$ has an accumulation point in the unit disk (namely $0$), it follows that $f$ is constant.


We have \begin{align}\prod_{n = 1}^\infty (1 - q^{2n-1})(1 + q^n) &= \prod_{n = 1}^\infty \frac{(1 - q^{2n-1})(1 - q^{2n})}{1 - q^n}\\ &= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 - q^n)}\\ &= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 -q^{2n})(1 - q^{2n-1})}\\ &= 1\end{align}

  • $\begingroup$ Great! it helps me a lot $\endgroup$ – phy_math Jan 25 '15 at 15:55
  • $\begingroup$ Very slick. Might even be easier to see by starting with $$\prod (1-q^n) = \prod (1-q^{2n-1})(1-q^{2n}),$$ which seems obvious, and then divide. $\endgroup$ – Thomas Andrews Jan 25 '15 at 17:30
  • $\begingroup$ @Thomas thanks. I wanted to present a calculation that doesn't look like I'm working backwards. I did use that identity at the end though. As you know, it follows from breaking the product into even and odd $n$. $\endgroup$ – kobe Jan 25 '15 at 17:43

Another approach, similar to Daniel's, without an appeal to properties of holomorphic functions.

$$(1-q)(1+q)(1+q^2)\cdots (1+q^{2^n})\cdots = 1$$

since the partial product is $1-q^{2^{n+1}}$ .

Substituting $q^{2k-1}$, you get:

$$(1-q^{2k-1})(1+q^{2k-1})(1+q^{(2k-1)\cdot 2})\cdots\left(1+q^{(2k-1)\cdot 2^n}\right)\cdots = 1$$

So $$1=\prod_{k=1}^\infty \left((1-q^{2k-1})\prod_{n=0}^\infty \left(1+q^{(2k-1)2^n}\right)\right)$$

Now, ever positive integer $N$ can be written in exactly one way as $(2k-1)2^n$. So we can rearrange.


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