Question

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.

Answer 1

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

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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.

Answer 2

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 $\frac{f(x) - g(x)}{g^{-1}(x) - f^{-1}(x)} \approx 1$.

Answer 3

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.

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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.

Answer 4

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.

Answer 5

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

Answer 6

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 :)

Answer 7

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.