## 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 G. 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 (Квант Magazine?) and G. Faltings was the only one that submitted a solution after his own heart? Is the previous conjecture totally unrelated to the actual development of things?

I thank you in advance for you insightful replies.

P.S. Guess it'd be just great for MO if we somehow managed to get first-hand information about this query of mine.

-
What was the notorius teaser? – Mariano Suárez-Alvarez Apr 8 2010 at 7:02
I think the author meant the story also retold here: jstor.org/stable/2031461 – Igor Pak Apr 8 2010 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 2010 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 2010 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 2010 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 2010 at 19:45
Arnold does not live in the US, VA does. – Noah Snyder Apr 8 2010 at 20:56
Wolfram Alpha can do it: wolframalpha.com/input/… – FG Apr 9 2010 at 20:08
It's not that WA can do it, sir. WA knows it by heart... – J. H. S. Apr 23 2010 at 8:26

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 2011 at 17:31
(For "footnote", read "endnote".) – Todd Trimble May 9 2011 at 17:32
At first, this looks like a great geometric explanation. Unfortunately, I do not understand it. The problem is that AD and BC are much smaller than the catheti, i.e. the sides of the triangle (in the original example, AD and BD are on the order of $x^7$ and the sides are on the order of $x^3$). So you can easily draw a similar triangle like, with slopes very close to 1, in which AD/BD=2 or anything else you like. – VA Apr 9 2012 at 16:17
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 2012 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 2012 at 14:06

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 2011 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 2011 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 2011 at 17:53 of course I meant $f_c^{-1}=c^{-1}f^{-1}$ – BS May 9 2011 at 17:54 @BS and Todd Trimble: Thank you for the comments. – Andrey Rekalo May 9 2011 at 18:08

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.

-

You should read "Method of Fluxions ans Infinite Series" by Newton! You would be surprised ...

-
 Sorry I wanted to answer John Stillwell's remark. – Paul Broussous Apr 8 2010 at 18:59

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 2010 at 8:31
What? This just makes no sense whatsoever as written. Certainly Newton was more knowledgeable than that... – fedja Apr 8 2010 at 14:38
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 2010 at 21:48

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.

-