Here’s a cute identity I discovered by accident recently. Observe that

and so one can conjecture that one has

when is even, and

when is odd. This is obvious in the even case since is a polynomial of degree , but I struggled for a while with the odd case before finding a slick three-line proof. (I was first trying to prove the weaker statement that was non-negative, but for some strange reason I was only able to establish this by working out the derivative exactly, rather than by using more analytic methods, such as convexity arguments.) I thought other readers might like the challenge (and also I’d like to see some other proofs), so rather than post my own proof immediately, I’ll see if anyone would like to supply their own proofs or thoughts in the comments. Also I am curious to know if this identity is connected to any other existing piece of mathematics.

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30 May, 2015 at 8:54 am

Paata IvanishviliHere is an approach:

Let , and let be the following operator: it takes derivative with respect to x, then divides the result by , then again takes derivative with respect to and then again divides result by . Note that . Let , and . All we need to show is that . Note that . Now I think it should not be difficult to iterate this process and get the desired result.

30 May, 2015 at 11:17 am

Paata IvanishviliI am sorry for the typos: “All we need to show ”.

And by saying “to iterate this process”, of course I meant simple induction ( is a linear operator).

30 May, 2015 at 10:24 am

Ofir GoroI think Lagrange Inversion formula sheds light on this problem.

Let . If we denote (and note that satisfies , Lagrange Inversion says:

for any power series .

Choosing , we get:

. Now we recall and get the constant coefficient in your formula (replace by to see this more clearly).

By plugging such that , we can get the rest of the coefficients, and see that they agree with the formula. Not very slick, I agree, as the formula for the ‘s become more complicated.

30 May, 2015 at 11:35 am

AmieHere is a rather brute force proof.

Let .

Let , and let

let

Then

Inductively:

Thus:

We are left to verify the formula

(or perhaps there’s something slick with generating functions. But back to work).

30 May, 2015 at 11:38 am

Emmanuel AmiotIdentifying power expansions seems to work.

30 May, 2015 at 11:52 am

AnonymousIt seems that this identity (for complex argument ) follows from Rodrigues formula for Chebyshev polynomials . see e.g. table 18.5.1 in dlmf.nist.gov

30 May, 2015 at 12:00 pm

1 – Fields Medalist Terence Tao stumbles upon a “differentiation identity | Offer Your[…] : https://terrytao.wordpress.com/2015/05/30/a-differentiation-identity/ […]

30 May, 2015 at 12:00 pm

Ofir GorodetskyIt is worth noting that the straightforward brute force solution (comparing coefficients) actually works, and is relatively simple.

Namely, the coefficient of ( even) in ( is odd) is . After differentiating times, we get that the coefficient of is .

On the other hand, the coefficient of in the RHS is .

For the equality amounts to , which is verified by induction on odd ‘s, where the induction step follows from

(the identity is useful here)

To proceed, perform induction on even values of . The base case was done now, and the induction step follows from the following identity:

(the identity is useful here)

30 May, 2015 at 2:28 pm

Ofir GorodetskyHere’s a proof using Leibnitz’ law. Write as . We differentiate this times and get:

where is the falling factorial.

We need to prove this sum equals . Multiplying both sides by , the desired equality becomes:

At this point one notices the interesting-though-tautological identity . Plugging it, the equality reduces to

The sum is just (binomial theorem), hence everything boils down to the following:

This is again kind tautological, since the LHS is .

30 May, 2015 at 2:33 pm

Ofir Gorodetsky(last post for tonight) I just realized we could have used the mentioned identity right after we used Leibniz’ law, and in this way find the answer without even knowing it beforehand.

30 May, 2015 at 5:28 pm

AnonymousThis proof seems to be “from the book!”.

30 May, 2015 at 3:49 pm

Ivan Z.I bet there is some nice parsimonious modular forms thingy behind this!

30 May, 2015 at 4:20 pm

Paolo D'IsantoWe develope the binomial and so we have

Now, because we are interested to only values of exponent of whose integral is not zero we impose that

so being an integer is necessarily odd. In the other cases the Cauchy integral is zero. From this relation we have

and so posing we have:

Remembering that

and

we obtain finally

Best Regards

Paolo D’Isanto

30 May, 2015 at 5:19 pm

AnonymousThe contour of integration should be clearly defined (to avoid the branch line connecting the two branch points ).

30 May, 2015 at 6:10 pm

Zoltan ToroczkaiLet , , and , for odd. Using and (here prime denotes differentiation w.r.2 ) we find . It is easy to show the identity for odd: . Using this identity then follows for odd. Assuming that holds for all , odd, we can write: , where we used (*). The second term is computed from (**) and we find: The terms cancel and we find that the identity is valid for (odd) as well. There is some sort of duality that appears here, which could be exploited for a short proof.

30 May, 2015 at 6:22 pm

Zoltan ToroczkaiThere was a bit of a jump in my description above, this is more explicit: and then the rest easily follows.

30 May, 2015 at 7:20 pm

AnonymousIn the neighborhood of infinity, can be written as a series of decreasing powers of , starting from . If we differentiate times, the nonnegative powers disappear, and the negative powers give powers of exponent at most . Hence, the -th derivative is at infinity. On the other hand, an immediate induction implies that it is a polynomial times : the domination implies that the polynomial is constant. This constant can be computed by taking times the term in in the expansion of at zero.

[This is essentially the proof I eventually came up with also :-) – T.]30 May, 2015 at 10:09 pm

Iosif PinelisLet , ,

. Then for odd

as desired; for the third equality in the above display we use the induction, and for the second we use the Leibniz rule. Here is of course the differentiation operator.

31 May, 2015 at 12:43 am

Paolo D'IsantoThanks Anonymous, sorry for all this posts, i was very tired because of i had the idea of demonstration this night! So, the most general contour on which the integral is calculated is every closed curve containing in the -strip :

So, the Tao-conjecture is verified for every

.

Thanks a lot.

31 May, 2015 at 5:56 am

Terence TaoThanks for all the contributed proofs and proof strategies!

The proof of the identity that I originally had is essentially that found in this previous anonymous comment (the only difference being that I computed the constant through Taylor expansion around infinity, rather than around zero), but I just worked out a Fourier-analytic proof which has the advantage of extending to non-integer values of , and which is somewhat reminiscent of the proof of the functional equation for the zeta function via modularity of the theta function (confirming the suspicion of Ivan Z. above). For any , we scale the Gamma function

to express as the average of Gaussians:

This formula is valid for any . Taking (distributional) Fourier transforms, we conclude that

Making the change of variables we conclude

If we compare this with (*) with replaced by (assuming temporarily that ), we conclude that

which on inverting the Fourier transform leads to the distributional identity

This formula was only derived for , but using analytic (or meromorphic) continuation in the space of tempered distributions it is valid for all . Specialising to the case for odd , we obtain

As is an odd integer, , while from the basic formula one sees that

and the claim follows.

1 June, 2015 at 12:38 am

AnonymousThe above distributional identity implies also the case for even (in fact, the RHS vanishes whenever is a pole of – i.e. is a positive integer.)

2 June, 2015 at 4:37 am

AnonymousIt is not clear if is really special. Therefore it seems natural to consider all the solutions for the general class of nonlinear(!) fractional derivatives eigenproblems

where are real parameters, and is the corresponding “eigenvalue”.

2 June, 2015 at 8:58 pm

Terence TaoPaul Nelson (private communication) has informed me that the fact that is a scalar multiple of can be interpreted as a consequence of the isomorphism of the principal series representation of with , where one can express as the space of (suitably nice) functions obeying the homogeneity for , with the usual action of . One can view this principal series representation as the induced representation of the character on the Borel subgroup B of upper triangular matrices (I may have some signs incorrect here).

The homogeneous version of can be viewed as generating the spherical functions in this principal series representation, that is to say the functions invariant under the action of the maximal compact subgroup . Because the two representations are isomorphic, their Whittaker models (that is to say, G-maps to the representation induced from a character of the nilpotent group ) must agree up to scalar multiples, and in particular the two spherical functions mentioned above must become scalar multiples of each other after applying their respective Whittaker maps. On computing this map explicitly, this ends up showing that the functions and are scalar multiples of each other, giving the claim after inverting the Fourier transform.

So it seems the “high level” explanation of the identity stems from the isomorphism between the principal series representations and , but I don’t know of a simple way to establish this isomorphism (it seems to be a medium-length computation involving Frobenius reciprocity and Haar measure integrations). In any event, this seems to indicate that the similarity with the Riemann zeta functional equation is not entirely coincidental.

[Disclaimer: my grasp of the representation theory of Lie groups is a little shaky, and any inaccuracies in the above are due to myself and not to Nelson. -T.]3 June, 2015 at 9:53 pm

Gergely HarcosIn more classical (and more elementary) terms, this is equivalent to , i.e. to . See, for example, 8.432.5 in Gradshteyn-Ryzhik: Tables of integrals, series, and products (7th ed., 2007).

[Here denotes the modified Bessel function of the second kind – T.]4 June, 2015 at 8:55 am

Gergely HarcosJust to clarify my point a bit, the fifth display in your Fourier-analytic proof precisely states in the light of 8.432.5 in Gradshteyn-Ryzhik. This symmetry has been known classically for the -Bessel function, while later it has been realized to be a reflection of (or an equivalent form of) . So in a sense you re-discovered all these symmetries.

31 May, 2015 at 10:13 am

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31 May, 2015 at 2:30 pm

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1 June, 2015 at 5:00 pm

AnonymousBrute force induction also works: replacing by one has to prove that

Now, by differentiating twice the left-hand side becomes

Using the induction hypothesis twice gives

From this the proof is completed by straightforward computation.

2 June, 2015 at 9:16 am

Christopher D. Long (@octonion)This is equivalent to my solution. The algebra is nicely simplified by the substitution .

2 June, 2015 at 1:46 pm

Zoltan ToroczkaiSimilar to what I provided above; I also wrote out the induction steps in detail.

3 June, 2015 at 5:32 am

Iosif PinelisAbove one can find my direct induction proof, which is bit simpler and uses a more compact notation.

2 June, 2015 at 1:14 am

QuoraHow can one prove Terence Tao’s cute differentiation identity?X= sinh y (1 + x^2)^1/2 = cosh y = dx/dy I’ve now spotted that this solution has already been posted on the original web site. https://terrytao.wordpress.com/2015/05/30/a-differentiation-identity/ So I leave you to read it there

2 June, 2015 at 9:00 am

Christopher D. Long (@octonion)The standard substitution reduces this to differentiating times. After two differentiations we get . By induction, we end up with , which simplifies to .

3 June, 2015 at 12:30 am

Mr MathMathematics is so beautiful.

3 June, 2015 at 6:17 am

Shalosh B. Ekhad and Doron ZeilbergerIf you go to Maple, and download the Maple package

http://www.math.rutgers.edu/~zeilberg/tokhniot/EKHAD

and you type:

lprint(AZd((2*k)!*(1+z^2)^(k-1/2)/(z-x)^(2*k+1) ,z,k,K)[1]);

you would get, after one nano-second

-(2*k+1)^2+(1+x^2)*K

which means that the left side (with k replaced by 2*k-1), let’s call it A(k),

satisfies the first-order recurrence

-(2*k+1)^2*A(k)+(1+x^2)*A(k+1)=0,

that together with the trivial value for A(0) implies the original statement.

For a verbose version, type:

AZpapd((2*k)!*(1+z^2)^(k-1/2)/(z-x)^(2*k+1) ,z,k,K);

Note that the proof is fully rigorous. Procedure AZd is an implementaion

of the Almkvist-Zeilberger algorithm, part of WZ theory.

k does not have to be an integer, and the recurrence is valid for all k,

hence one can get Gamma functions.

Of course, AZd can give recurrences for ANY expression of the form

D_x^(k) ( P(x)^k*Q(x)),

for a very wide class of P(x) and Q(x), but, alas, the recurrence is

usually not first-order. All the Rodriguez formulas for the classical

orthogonal polynomials yield the respective three-term recurrences.

3 June, 2015 at 9:30 pm

Terence TaoSome miscellaneous remarks on the identity.

I posted a related question on MathOverflow at http://mathoverflow.net/questions/208341/is-this-differential-identity-known where it was pointed out by Carlo Beenakker that the identity follows from the Rodrigues formula for the Gegenbauer polynomials , together with the fact that when n is even.

A reformulation of the identity (in the more general Gamma function form) is as a two-dimensional identity

where . Amusingly, this identity is essentially its own Fourier transform!

Another amusing consequence of the identity is that the function is “very convex” on the positive real axis in the sense that the first k+1 derivatives are all non-negative there. (This follows from repeated integration of the differentiation identity together with a Taylor expansion at the origin.) I don’t know of any other way to prove this “convexity” without going through the differential identity.

4 June, 2015 at 2:44 am

AnonymousIs it possible to reformulate the two-dimensional identity on the sphere ? (perhaps by spherical harmonics?)

11 June, 2015 at 8:10 pm

AnonymousProbably a shot in the dark. Notice that the equation for k = n has, on the right side, a function of the left side of the equation for k = n+2. Inductively, test that k = 1 works. Suppose k = n works, then reciprocate and multiply by the double factorial squared. Then take the “k+2″nd derivative of the resulting equation. Looking to simplify it to the equation where k = n + 2.

I tried it for a half an hour and didn’t get anywhere.

12 June, 2015 at 12:17 am

antonioYou have forgotten the chain rule

23 June, 2015 at 11:18 am

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24 June, 2015 at 1:27 am

UbikHere is another answer that is not really connected to any other existing piece of beautiful mathematics…

The function is a solution of the first order ODE :

If we differentiate -times this equation (using the Leibniz’s Rule) we get

If and , we obtain that satisfies the ODE :

Hence

It remains to compute but the equation gives a recurrence formula for the derivative at of :

So that

For we get

30 June, 2015 at 10:45 pm

AnonymousThis is a good topic for classes. In the post, you said “Here’s a cute identity I discovered by accident recently”. Could you please mention a little about the context that motivated matter ? Thanks.

15 July, 2015 at 2:54 am

A differentiation identity — Thanks to Prof Terence Tao « Mathematics Hothouse[…] https://terrytao.wordpress.com/2015/05/30/a-differentiation-identity […]

22 August, 2015 at 2:04 am

AnonymousA trivial generalization of the given identity is obtained by scaling the variable to where and are independent of

The result so obtained is

25 November, 2015 at 7:03 pm

JunThis paper of Milnor: http://people.ucsc.edu/~lewis/Math208/hairyball.pdf remind me of this post, though I don’t know if there are any connections behind.