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Recently I discovered the differential identity

$$ \frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2} = \frac{(1 \times 3 \times \dots \times k)^2}{(1+x^2)^{(k+2)/2}}$$

valid for any odd natural number $k$; for instance $\frac{d^6}{dx^6} (1+x^2)^{5/2} = \frac{225}{(1+x^2)^{7/2}}$. This identity was surprising at first, since usually the repeated application of the product rule and chain rule leads to far messier expressions than this, but there are now several proofs that adequately explain this identity (collected at this blog post of mine). There is also the more general identity

$$ |\frac{d}{dx}|^{2s-1} (1+x^2)^{s-1} = \frac{2^{2s-1}\Gamma(s)}{\Gamma(1-s)} (1+x^2)^{-s}$$

valid for any complex $s$ (if everything is interpreted distributionally), which is related to the isomorphisms between principal series representations of $PGL_2({\bf R})$.

The purpose of my question here is not to ask for more proofs of this identity (but you are welcome to visit the above-mentioned blog post to contribute another proof, if you wish). Instead, I am asking as to whether this identity (or something close to it) already appears in the literature - I find it hard to believe that such a simple identity has been missed for centuries, given that it could easily have been discovered and proven by (say) Euler. The closest match that I know of so far are the Rodrigues formulae for the classical orthogonal polynomials, but I was not quite able to place the above identity as a special case of these formulae (the exponents don't quite match up).

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  • $\begingroup$ Letting Q be $(1+x^2)^{1/2}$, I rewrite the identity using operator notation as Q^{k+1}D^{k+1}Q^k = C for a certain nonzero constant C. I then wonder if this is part of a larger total differential or perhaps the result of some inversion formula. Just some random musings, in case a random start toward an answer proves useful. $\endgroup$ – The Masked Avenger Jun 3 '15 at 16:21
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    $\begingroup$ The Fourier transforms of the distributions $(1+x^2)^\lambda$, $\lambda\in\mathbb{C}$, are studied in vol.1 of Gelfand and Shilov's book Generalized Functions, Chap.II, Sec. 2.6. The identity you discovered behaves nicely with respect to the Fourier transform and maybe it follows from the mountain of formulas in the above book. $\endgroup$ – Liviu Nicolaescu Jun 3 '15 at 16:35
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    $\begingroup$ If I'm not mistaken the formula as stated has a typo: $(k+1)/2$ in the denominator on the right-hand side should read $1+k/2$. (At least that's what the Rodrigues formula in my answer below would give.) $\endgroup$ – Carlo Beenakker Jun 3 '15 at 20:08
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    $\begingroup$ These steampunk identities are really great. You can prove Rodriguez formula using matrix elements of $SO(3)$. $\endgroup$ – john mangual Jun 4 '15 at 21:51
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It's the Rodrigues formula for Gegenbauer polynomials $C_n^{(\alpha)}(x)$, in the special case $\alpha=-n/2$ when the polynomial is just unity.

The general formula reads

$$C_n^{(\alpha)}(x)=\frac{(-2)^n}{n!}\frac{\Gamma(n+\alpha)\Gamma(n+2\alpha)}{\Gamma(\alpha)\Gamma(2n+2\alpha)}(1-x^2)^{-\alpha+1/2}\frac{d^n}{dx^n}\left[(1-x^2)^{n+\alpha-1/2}\right].$$

Substitution of $\alpha=-n/2$, with $C_n^{(-n/2)}(x)=1$ for $n$ even, gives

$$1=\frac{(-2)^n(n/2)!}{n!(n-1)!}\lim_{\alpha\rightarrow-n/2}\frac{\Gamma(n+2\alpha)}{\Gamma(\alpha)}(1-x^2)^{(n+1)/2}\frac{d^n}{dx^n}\left[(1-x^2)^{(n-1)/2}\right]$$

$$\qquad = (-1)^{n/2}\left(\frac{1}{(n-1)!!}\right)^2(1-x^2)^{(n+1)/2}\frac{d^n}{dx^n}\left[(1-x^2)^{(n-1)/2}\right],$$

or with $x\to ix$,

$$\frac{d^n}{dx^n}\left[(1+x^2)^{(n-1)/2}\right]=[(n-1)!!]^2(1+x^2)^{-(n+1)/2},$$

which is the desired expression.

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This can be considered as a case of the Lagrange reversion formula.

Consider the equation $v = x + y \sqrt{1+v^2}$ (where $v$ is a function of $x$ and $y$). It has an explicit solution

$$ v = \dfrac{\sqrt{1+x^2-y^2}\; y - x}{y^2-1} $$

Lagrange reversion (with $f(x) = (1+x^2)^{1/2}$ and $g = \arctan$) can be written as

$$ \eqalign{\arctan(v) - \arctan(x) &= \sum_{j=0}^\infty \dfrac{y^{j+1}}{(j+1)!} \left(\dfrac{\partial}{\partial x}\right)^{j} (f(x)^{j+1} g'(x))\cr &= \sum_{j=0}^\infty \dfrac{y^{j+1}}{(j+1)!} \left(\dfrac{\partial}{\partial x}\right)^{j}(1+x^2)^{(j-1)/2} } \tag{1} $$

Take the partial derivative with respect to $y$; it turns out that

$$ \eqalign{\dfrac{\partial }{\partial y} \arctan(v) &= \dfrac{1}{1+v^2} \dfrac{\partial v}{\partial y} = \dfrac{-1}{\sqrt{1+x^2-y^2}}\cr &= -\sum _{k=0}^{\infty } (1+x^2)^{-k-1/2}{\frac { \left( 2\,k \right) !}{ 4^k \;\left( k! \right) ^{2}}} y^{2k} }$$

Thus equating the coefficients of $y^{2k}$ here and in the partial derivative of the right side of (1), you get your formula.

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Here is a direct induction argument. Let $h_k(x):=(1+x^2)^{k/2}$, $g(x):=x$, $p_k:=(1\times3\times\dots\times k)^2$. The identity in question is $$D^{k+1}h_k=p_k/h_{k+2}$$ for positive odd $k$, where $D$ is the differentiation operator. It is easy to check it for $k=1$. Then for odd $k\ge3$ $$D^{k+1}h_k=k D^k(gh_{k-2})=kg D^k h_{k-2}+k^2 D^{k-1}h_{k-2}$$ $$=kg D(p_{k-2}/h_k)+k^2 p_{k-2}/h_k=p_k/h_{k+2},$$ as desired; for the third equality in the above display we use the induction, and for the second we use the Leibniz rule.

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    $\begingroup$ I thought the purpose was not to ask for more proofs (the OP states so), but to find out if and when the identity has been studied before, and in what context, no? $\endgroup$ – Suvrit Jun 7 '15 at 20:38
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    $\begingroup$ Your comment is correct, and I admit that I didn't read at first the last paragraph of the statement of the question. However, such a simple proof could still be relevant, since it seems to be no more complicated than a derivation from a nontrivial known result. In particular, this proof can be considered a derivation from the empty statement that states nothing, or maybe from the Leibniz rule. $\endgroup$ – Iosif Pinelis Jun 7 '15 at 21:17
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    $\begingroup$ I think posting such a simple proof is well justified here, even though it doesn't answer the exact question asked. $\endgroup$ – Deane Yang Jun 8 '15 at 2:15
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    $\begingroup$ I do like this proof (and the other two proofs too). I was just trying to respect Terry's request to instead contribute proofs to his blog post :-) $\endgroup$ – Suvrit Jun 8 '15 at 2:37

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