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
limxto0fracsin(tanx)−tan(sinx)arcsin(arctanx)−arctan(arcsinx)
The answer is the same for
limxto0fracf(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+a2x2+a3x3+dotsquadtextandquadf−1(x)=x+A2x2+A3x3+dots
Computing recursively, one sees that for nge2 one has
An=−an+Pn(a2,dotsc,an−1)
for some universal polynomial Pn.
Now, let
g(x)=x+b2x2+b3x3+dotsquadtextandquadg−1(x)=x+B2x2+B3x3+dots
and suppose that bi=ai for ilen−1 but bnnean. Then by induction one has Bi=Ai for ilen−1, An=−an+Pn(a2,dotsc,an−1) and Bn=−bn+Pn(a2,dotsc,an−1).
Thus, the power expansion for f(x)−g(x) starts with (an−bn)xn, and the power expansion for g−1(x)−f−1(x) starts with (Bn−An)xn=(an−bn)xn. So the limit is 1.
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