# A question regarding a claim of V. I. Arnold

In his Huygens and Barrow, Newton and Hooke, Arnold mentions a notorious teaser that, in his opinion, "modern" mathematicians are not capable of solving quickly. Then, he adds that the exception that proved the rule in this case of his was the German mathematician Gerd Faltings.

My question is whether any of you knows the complete story behind those lines in Arnold's book. I mean, did Arnold pose the problem somewhere (maybe Квант?) and Faltings was the only one that submitted a solution after Arnold's own heart? Is the previous conjecture totally unrelated to the actual development of things?

P.S. It seems that this teaser of Arnold eventually became a cult thingy in certain branches of the Russian mathematical community. Below you can find a photograph taken by a fellow of mine of one of the walls of IUM's cafeteria (where IUM stands for Independent University of Moscow). As the Hindu mathematician Bhāskara would say (or so the legend has it): BEHOLD!

• What was the notorius teaser? – Mariano Suárez-Álvarez Apr 8 '10 at 7:02
• I think the author meant the story also retold here: jstor.org/stable/2031461 – Igor Pak Apr 8 '10 at 7:06
• I'm sure I've seen this (especially given Angelo's comment below) somewhere on the internet, perhaps on a math(s) blog. – Yemon Choi Apr 8 '10 at 7:35
• Just for the sake of completeness (since Lyosha asked): it is page 28 in Birkhauser's English edition (1990), available via googlebooks here books.google.ba/… – Harun Šiljak Apr 8 '10 at 10:17
• The notorious teaser is to calculate $\mathop{\lim}\limits_{x \to 0} \frac{\sin \tan x - \tan \sin x}{\arcsin \arctan x-\arctan \arcsin x}$ – I. J. Kennedy Jun 10 '10 at 20:07

Here is a problem which I heard Arnold give in an ODE lecture when I was an undergrad. Arnold indeed talked about Barrow, Newton and Hooke that day, and about how modern mathematicians can not calculate quickly but for Barrow this would be a one-minute exercise. He then dared anybody in the audience to do it in 10 minutes and offered immediate monetary reward, which was not collected. I admit that it took me more than 10 minutes to do this by computing Taylor series.

This is consistent with what Angelo is describing. But for all I know, this could have been a lucky guess on Faltings' part, even though he is well known to be very quick and razor sharp.

The problem was to find the limit

$$\lim_{x\to 0} \frac { \sin(\tan x) - \tan(\sin x) } { \arcsin(\arctan x) - \arctan(\arcsin x) }$$

The answer is the same for $$\lim_{x\to 0} \frac { f(x) - g(x) } { g^{-1}(x) - f^{-1}(x) }$$ for any two analytic around 0 functions $f,g$ with $f(0)=g(0)=0$ and $f'(0)=g'(0)=1$, which you can easily prove by looking at the power expansions of $f$ and $f^{-1}$ or, in the case of Barrow, by looking at the graph.

End of Apr 8 2010 edit

Beg of Apr 9 2012 edit

Here is a computation for the inverse functions. Suppose $$f(x) = x + a_2 x^2 + a_3 x^3 + \dots \quad \text{and} \quad f^{-1}(x) = x + A_2 x^2 + A_3 x^3 + \dots$$

Computing recursively, one sees that for $n\ge2$ one has $$A_n = -a_n + P_n(a_2, \dotsc, a_{n-1} )$$ for some universal polynomial $P_n$.

Now, let $$g(x) = x + b_2 x^2 + b_3 x^3 + \dots \quad \text{and} \quad g^{-1}(x) = x + B_2 x^2 + B_3 x^3 + \dots$$

and suppose that $b_i=a_i$ for $i\le n-1$ but $b_n\ne a_n$. Then by induction one has $B_i=A_i$ for $i\le n-1$, $A_n=-a_n+ P_n(a_2,\dotsc,a_{n-1})$ and $B_n=-b_n+ P_n(a_2,\dotsc,a_{n-1})$.

Thus, the power expansion for $f(x)-g(x)$ starts with $(a_n-b_n)x^n$, and the power expansion for $g^{-1}(x)-f^{-1}(x)$ starts with $(B_n-A_n)x^n = (a_n-b_n)x^n$. So the limit is 1.

• A user called "VA" gives an authoritative account of a lecture by Vladimir Arnold. Surely it can't be... – Tom Leinster Apr 8 '10 at 19:45
• Arnold does not live in the US, VA does. – Noah Snyder Apr 8 '10 at 20:56
• Wolfram Alpha can do it: wolframalpha.com/input/?i=Limit[(Sin[Tan[x]]+-+Tan[Sin[x]])/… – F_G Apr 9 '10 at 20:08
• It's not that WA can do it, sir. WA knows it by heart... – José Hdz. Stgo. Apr 23 '10 at 8:26

I heard this story from Dinesh Thakur several years ago. This is what he told me. When Arnold posed this question, Faltings immediately said 1. After the talk somebody complimented Faltings on his quickness, and Faltings replied that what immediately came to mind was that the answer had to be either 0 or 1. Since 1 was a more interesting answer, he went with that.

The "inspecting the graph" comment might refer to something like this.

Consider two smooth curves $y = f(x)$ and $y = g(x)$ that are tangent to $y = x$ at $(0,0)$. For $x$ near 0, define $u(x)$ and $v(x)$ so that $g^{-1}(x) - f^{-1}(x) = u(x)$ and $f(x) - g(x) = v(x)$. In the picture, $v(x) = BC$ and $u(x) = AD$. But both curves have slope very close to 1, so $AD \approx BC$, i.e. $u(x) \approx - v(x)$, and $\frac{f(x) - g(x)}{g^{-1}(x) - f^{-1}(x)} \approx 1$.

• This is the most pellucid explanation I've seen yet on this thread (and from memory is substantially what is contained in the footnote given for this problem in Arnold's book, but it's explained better here IMO). – Todd Trimble May 9 '11 at 17:31
• (For "footnote", read "endnote".) – Todd Trimble May 9 '11 at 17:32
• Dear Robert, let me try to explain myself better. This is what I am saying: Denote the vertex of the right triangle by $P$ (it lies on the line $y=x$). Then it is easy to draw a triangle in which the slopes of AB and DC are very close to 1, but AD/BC=2, for as long as AD and BC are much much smaller than AP and BP. But this is indeed the case here. – VA. Apr 9 '12 at 19:04
• Once one can show that AD/BC is close to 1 for macroscopic perturbations (in which AD, BC ~ AP, BP), this implies that AD/BC is close to 1 for microscopic perturbations (AD, BC << AP, BP) as well, by smoothly varying one of the functions f, g and noting that all quantities involved are uniformly smooth (after rescaling by x). Basically, there isn't enough curvature available wrt the perturbation parameter to make the microscopic ratio too far away from the macroscopic ratios. (Or, if one wishes, one can use L'hopital's rule followed by numerical differentiation.) – Terry Tao Apr 9 '12 at 23:09
• Oh, I see what @VA is saying. Yes, you do need more than just smooth functions, you need analytic I guess. Otherwise consider e.g. $f(x) = x + x^2 + e^{-1/x} \cos(\pi/x)$, where for $x=1/n$ you get $D-A \approx -(B-C)$. – Robert Israel Apr 10 '12 at 14:06

I heard Arnold tell the story in a talk, twenty-odd years ago. He had presented the teaser during a seminar in Princeton (some limit involving tangent functions, I don't remember exactly), and Faltings immediately stated that the answer was 1. I could not do it quickly, not being Faltings, but thought a little afterwards, and it wasn't hard.

The limit $$\lim_{x\to 0}\frac { f(x) - g(x) } { f^{-1}(x) - g^{-1}(x)} = -\left(f'(0)\right)^6$$ appears in the Problems section of Mathematics Magazine, where it is calculated under the assumption that $f$ and $g$ are analytic in a neighborhood of $0$ odd functions such that $f'(0)=g'(0)\neq 0$, $f^{(3)}(0)= g^{(3)}(0)$, and $f^{(5)}(0)\neq g^{(5)}(0)$ (Problem 1672, vol. 77, No. 3, June 2004, pp. 234-235) .

In Arnold's example, we have that $$\sin(\tan x)= x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55x^7}{1008}+O(x^9)$$ $$\tan(\sin x)= x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}+O(x^9)$$ $$\arcsin(\arctan x)=x-\frac{x^3}{6}+\frac{13x^5}{120}-\frac{341x^7}{5040}+O(x^9)$$ $$\arctan(\arcsin x)=x-\frac{x^3}{6}+\frac{13x^5}{120}-\frac{173x^7}{5040}+O(x^9)$$ Now, $$\lim_{x\to 0} \frac { \sin(\tan x) - \tan(\sin x) } { \arcsin(\arctan x) - \arctan(\arcsin x) } = \lim_{x\to 0} \frac { -\frac{55x^7}{1008}+\frac{107x^7}{5040} +O(x^9)} { -\frac{341x^7}{1008}+\frac{173x^7}{5040}+O(x^9)}$$ $$=\lim_{x\to 0} \frac{-\frac{168x^7}{5040}+O(x^9)}{-\frac{168x^7}{5040}+O(x^9)}=1.$$

Given that one has to use the Taylor series expansion of $f$ and $g$ up to the seventh order, I find it somewhat difficult to see the result just by inspecting the graph.

Edit. And for the sake of completeness, here's the original argument from Arnold's book (Birkhauser Verlag 1990, P. 108).

If the graphs of non-coincident analytic functions $f$ and $g$ touch the line $y = x$ at the origin (Fig. 37), then the ratios $|AB|/|BC|$ and $|BC|/|ED|$ tend to one as $A$ tends to the origin. Therefore the required limit of the ratio $|AB|/|D'E'|$ is equal to one.

• Well, isn't your condition that $f$ and $g$ are different odd analytic functions ? And sure enough, $\sin$ and $\tan$ don't commute... – BS. May 9 '11 at 16:51
• Is the generalization supposed to be that that limit quotient is $-(f'(0))^{2n+2}$ if $n$ is the least integer where $f^{2n+1}(0)$ and $g^{2n+1}(0)$ differ? – Todd Trimble May 9 '11 at 17:01
• @Todd and Andrey : the limit is $-(f'(0))^{k+1}$, when $f(x)-g(x)=ax^k+O(x^{k+1})$, $k>1$, $a\neq 0$. The proof is to observe that replacing $f$, $g$ by $f_c(x)=f(cx)$, $g_c(x)=g(cx)$, the quotient for $f_c,g_c$ is then equivalent (at $0$) to $c^{k+1}$ times that for $f,g$ (note that $f_c^{-1}=c^{-1}f$). One is now reduced to the case $f'(0)=g'(0)=1$. – BS. May 9 '11 at 17:53
• of course I meant $f_c^{-1}=c^{-1}f^{-1}$ – BS. May 9 '11 at 17:54
• @BS and Todd Trimble: Thank you for the comments. – Andrey Rekalo May 9 '11 at 18:08

I imagine that Newton et al would have no trouble solving this teaser because they would say: let x be infinitesimal, in which case both numerator and denominator equal x-x; quotient equals 1 :)

• I have a feeling that this might have been Arnold's intended point, behind the jibe – Yemon Choi Apr 8 '10 at 8:31
• What? This just makes no sense whatsoever as written. Certainly Newton was more knowledgeable than that... – fedja Apr 8 '10 at 14:38
• You should read "Method of Fluxions ans Infinite Series" by Newton! You would be surprised ... – Paul Broussous Apr 8 '10 at 18:56
• Newton was more knowledgeable, especially about power series. But I think he also had enough feel for the behavior of sin x and tan x, for x small, to see that the ratio of numerator and denominator is arbitrarily close to 1. Having said that, I doubt that any such tricky limit was considered by 17th-century mathematicians. – John Stillwell Apr 8 '10 at 21:48