The following formula for a determinant is attributed to Karl Rohn (e.g. in Bruijn, de, N. G. (1955). On some multiple integrals involving determinants. Journal of the Indian Mathematical Society. New Series, 19, 133-151.):

$$ \det \left(\left(\frac{x_i-x_j}{x_i+x_j} \right)_{i, j=1, \ldots, n}\right) = \prod_{1 \le i < j \le n}\left(\frac{x_i-x_j}{x_i+x_j} \right)^2 \, . $$

In a recent answer I provided a proof of that formula via induction. A crucial role in the induction step played the following identity: $$ \tag{1} \frac{a-b}{a+b}\cdot\frac{c-d}{c+d} + \frac{a-c}{a+c}\cdot\frac{d-b}{d+b} + \frac{a-d}{a+d}\cdot\frac{b-c}{b+c}\\ = \frac{a-b}{a+b}\cdot\frac{a-c}{a+c}\cdot\frac{a-d}{a+d}\cdot\frac{b-c}{b+c}\cdot\frac{b-d}{b+d}\cdot\frac{c-d}{c+d} \, , $$ which came a bit surprising. $(1)$ holds for all real (or complex) numbers as long as it is well-defined, i.e. no denominator is zero. I verified it with a computer algebra system (Maxima).

One could also argue as follows:

  • For fixed $b, c, d$ are both sides of $(1)$ rational functions in $a$, of degree at most $3$.
  • These rational functions both have zeros at $a=b, c, d$ and poles at $a=-b, -c, -d$.
  • It follows that these rational functions differ only by a constant factor.

Therefore it suffices to show that $(1)$ holds for $a \to \infty$, i.e. that $$ \tag{2} \frac{c-d}{c+d} + \frac{d-b}{d+b} + \frac{b-c}{b+c} = \frac{b-c}{b+c}\cdot\frac{b-d}{b+d}\cdot\frac{c-d}{c+d}\, . $$ And that is a known identity, see for example Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$, where one can find some elegant proofs of $(2)$.

One could also repeat the argument:

  • For fixed $c, d$ are both sides of $(2)$ rational functions in $b$, of degree at most $2$.
  • These rational functions both have zeros at $b=c, d$ and poles at $b=-c, -d$.
  • It follows that these rational functions differ only by a constant factor.

Therefore it suffices to show that $(2)$ holds for $b \to \infty$, i.e. that $$ \tag{3} \frac{c-d}{c+d} = \frac{c-d}{c+d} \, , $$ and that is obviously true.

So $(1)$ implies $(2)$ and that implies $(3)$. On the other hand, $(3)$ can be used to prove $(2)$, and that can be used to prove $(1)$.

What I am asking for is:

  • Other proofs of the initial identity $(1)$.
  • If possible, some deeper insight how equations $(1)$, $(2)$, and $(3)$ are related.
  • If possible, further generalizations: What would the corresponding identity be for five numbers (if there is one)?
  • $\begingroup$ @José: Thanks for fixing the formula, you have a sharp eye! $\endgroup$
    – Martin R
    Mar 24, 2021 at 21:11
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    $\begingroup$ And I thank you for posting such a question. It's among the greatest questions that I have ever seen around here. $\endgroup$ Mar 24, 2021 at 21:14

1 Answer 1


Alternative proof: Let $z_1=a,z_2=b$, etc. Put the LHS over a common denominator so it reads $$\mathrm{LHS} = \frac{f(z)}{\prod_{i<j}(z_i + z_j)}$$ and note that its numerator $f$ must have degree six. But the LHS is totally antisymmetric in $z_i$, and the denominator is totally symmetric, and as such the numerator $f$ is antisymmetric and vanishes when any pair of $z$s coincide. Thus it must have a factor of a particular degree six polynomial: $$\prod_{i<j}(z_i - z_j) \mid f$$ But $f$ is degree six. Thus the LHS and RHS are identical up to an overall constant, which is easily checked e.g. by computing a single coefficient (e.g. that of $a^3b^2c^1d^0$) or evaluating at one suitable point.

Connection: This makes clear the connection between the identities you offer: the LHS is a totally antisymmetric function of the variables involved in each case, with the common denominator totally symmetric; hence the numerator has a factor of $\prod_{i<j}(z_i - z_j)$ and is of the same degree as this product (the so-called Vandermonde determinant).

Generalization: Off the top of my head, I can't think of a simple generalization to odd numbers of variables greater than 3 [Edit: See below.] -- the most obvious generalization is to Schur's Pfaffian identity (1911), for polynomials in $2n$ variables, relying on a Pfaffian-type antisymmetrization. Namely, let $P$ be the set of all $(2n)!/(2^n n!)$ distinct pairings of $z$s, so e.g. $$P=\{((1,2),(3,4)), ((1,3),(2,4)), ((1,4),(2,3))\}$$ for $2n=4$ variables. Then $$ \sum_{p\in P} \sigma(p) \prod_{i=1}^{n} \frac{z_{p_{i,1}} - z_{p_{i,2}}}{z_{p_{i,1}} + z_{p_{i,2}}} = \prod_{i<j}\frac{z_i-z_j}{z_i+z_j} $$ where $\sigma(p)$ is the signature of the permutation $(p_{1,1},p_{1,2},\ldots,p_{n,1},p_{n,2})$ of $(1,\ldots,2n)$.

Reference: This identity is more commonly expressed as $$ \mathrm{Pf}\left(\frac{z_j - z_i}{z_j + z_i}\right)_{i,j} = \prod_{i<j} \frac{z_j - z_i}{z_j + z_i} $$ but I've rewritten it to be more clearly in the spirit of your question.

This, and similar identities, are given in e.g. Generalizations of Cauchy’s Determinant Identity and Schur’s Pfaffian Identity (Soichi Okada).

Finally, note that squaring this Pfaffian identity immediately gives the result in the linked question.

Odd Numbers of Variables (Edit): As Martin points out in the comments, the technique of taking $z_{2n} \to \infty$ gives an identity for $2n-1$ variables from the one above. This identity has the same number of terms in the sum on the LHS as the parent identity, but one fewer term in each product there, and $2n-1$ fewer terms in the product on the RHS. Thus there is a five variable identity with 15 terms like $\frac{a-b}{a+b}\frac{c-d}{c+d}$ with various signs on the LHS, and the RHS being the ratio of $(a \mp b)(a \mp c)\cdots(d \mp e)$.

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    $\begingroup$ Recalling the origin (for me), I see myself going in circles ;) $\endgroup$
    – metamorphy
    Mar 24, 2021 at 22:56
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    $\begingroup$ @metamorphy Ah, but I hope this breaks the circle! The above approach gives a direct proof of the Pfaffian identity using only standard symmetry properties of the Pfaffian (antisymmetry under permutations of rows and columns together), right? $\endgroup$ Mar 24, 2021 at 23:27
  • $\begingroup$ Thank you for the detailed answer. I'll need some more time to digest all that, then I will come back to you. – One thought: Taking the limit $x_{2n}\to \infty$ in the identity for $(2n)$ variables should give an identity for $(2n-1)$ variables, since all fractions containing $x_{2n}$ converge to one. That's how I derived (2) from (1), and the same method would give an identity for any odd number of variables. $\endgroup$
    – Martin R
    Mar 25, 2021 at 8:44
  • $\begingroup$ Good point! So that just means leaving one variable unpaired and antisymmetrizing over which one it is in some appropriate way on the left hand side. Presumably this gives $(2n+1)!/(2^n n!)$ terms in the sum for $2n+1$ variables. $\endgroup$ Mar 25, 2021 at 9:28

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