Let $x * y = \frac{x + y}{1 - xy}$. I want a single formula for $x_1 * x_2 * \ldots * x_n$, for all natural $n$.

In order to generate plausible candidates, let's see what happens at small values of $n$:

  • $x_1 * x_2 * x_3 = \dfrac{x_1 + x_2 + x_3 - x_1x_2x_3}{1 - x_1x_2 - x_1x_3 - x_2x_3}$
  • $x_1 * x_2 * x_3 * x_4 = \dfrac{x_1 + x_2 + x_3 + x_4 - x_1x_2x_3 - x_1x_2x_4 - x_1x_3x_4 - x_2x_3x_4}{1 - x_1x_2 - x_1x_3 - x_1x_4 - x_2x_3 - x_2x_4 - x_3x_4 + x_1x_2x_3x_4}$

Then I conjecture the following:

$x_1 * x_2 * \ldots * x_n = \dfrac {\sum_{I \in T_1} \prod_{i \in I} x_i - \sum_{I \in T_3} \prod_{i \in I} x_i} {\sum_{I \in T_0} \prod_{i \in I} x_i - \sum_{I \in T_2} \prod_{i \in I} x_i}$

Where $T_k = \{\, I \in \wp(\{1, 2 \ldots n\}) : k \equiv |I| \pmod 4 \, \}$

Is there a nice proof of the above not involving transcendental functions?

  • 2
    $\begingroup$ $$x*y=\tan(\arctan x+\arctan y)$$ $\endgroup$ – lab bhattacharjee Jan 24 '16 at 6:29
  • $\begingroup$ @labbhattacharjee: Is there a proof not involving transcendental functions? $\endgroup$ – pyon Jan 24 '16 at 6:30
  • $\begingroup$ Continuing @labbhattacharjee's brilliant observation $$ x_1 * x_2 * \cdots * x_n = \tan \left( \arctan x_1 + \arctan x_2 + \cdots + \arctan x_n \right) $$ $\endgroup$ – Kaster Jan 24 '16 at 6:42
  • $\begingroup$ The suggestion of @labbhattacharjee suggests looking at $\left. \frac{\Im t}{\Re t}\right|_{t = \prod_{j=1}^n (1+ \mathrm{i} x_j)}$. This just encodes the slopes (tangents) in the components of a complex number. $\endgroup$ – Eric Towers Jan 24 '16 at 7:14
  • 1
    $\begingroup$ I edited to try fulfilling your wishes. Feel free to edit again if you don't like it. Also, I changed your set notation to use ":", because usually "|" can be confusing especially if you have absolute value signs or divisibility signs. In any case if you want the original one use \mid which spaces nicely. $\endgroup$ – user21820 Jan 24 '16 at 7:53

Expanding on my comment...

Suppose all the $x_j$ are real. We proceed by induction. We write $\frac{\Im}{\Re}(t)$ to represent the ratio of the imaginary to real parts of the complex number $t$. Then $$\begin{align} x_1 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1) ) = x_1 \text{ and }\\ x_1 \ast x_2 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1)(1+\mathrm{i} x_2) ) = \frac{x_1 + x_2}{1 - x_1 x_2}, \end{align}$$ establishing the result for the smallest allowed $n$. (Actually, I've assumed this is what you want for $n=1$, since you didn't specify.) Suppose now $n >2$ and $\ast_{j=1}^{n-1} x_j = \frac{\Im}{\Re}\left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)$. Then $$\begin{align} \frac{\Im}{\Re}\left( \prod_{j=1}^{n} (1 + \mathrm{i}x_j) \right) &= \frac{\Im \left( (1 + \mathrm{i}x_n) \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)}{\Re\left( (1 + \mathrm{i}x_n) \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\ &= \frac{\Im \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) + x_n \Re \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) }{\Re \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) - x_n \Im \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\ &= \frac{ \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) + x_n }{1 - x_n \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\ &= \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) \ast x_n \\ &= \left( \ast_{j=1}^{n-1} x_j \right) \ast x_n \\ &= \ast_{j=1}^{n} x_j. \end{align}$$ (It's traditional to write this last display in reverse order, but I think the manipulations are easier to follow in this order.) The step yielding $\dots {} - x_n \Im \dots$ in the denominator may be less than obvious: we want the negative imaginary part here because (in the previous line) when it is multiplied by $\mathrm{i} x_n$, it must yield a positive real part. (Trying to stumble across this manipulation in the traditional order is why I think the order above is easier to follow.) This completes our induction.

Having done that, it should be no great challenge to show that $\frac{\Im}{\Re}\left( \prod_{j=1}^{n} (1 + \mathrm{i}x_j) \right)$ splits the even and odd degree terms into the denominator and numerator, respectively, and also has the signs you want for your $\pmod{4}$ representation.

  • $\begingroup$ $*$ has obvious identity and inverses, commutes just as obviously, and I had already separately proved that it's associative. But it would've never occured to me to represent each $x_j$ by $1 + ix_j$ (or, rather, its equivalence class modulo scaling by a nonzero real). Thanks! $\endgroup$ – pyon Jan 24 '16 at 8:19
  • $\begingroup$ I think the most progress came from @labbhattacharjee. I also recognized this as the sum of angles formula for tangent. Your "split into classes mod 4" makes me think of the powers of i, or of the Gaussian integers. (If your formula had been mod 6, I would have thought of the Eisenstein integers.) The rest was just figuring out $\alpha$ and $\beta$ in $\alpha + \beta x$ for the multiplicands. $\endgroup$ – Eric Towers Jan 24 '16 at 15:29

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