Hey Dave Renfro,

Here's the solution I sent to CMJ for problem #497, in 1993 I think. It is NOT the solution that was printed (way too long, for one thing).

I'll leave this page up for your use and/or abuse in conjunction with your post on the Math Forum.

I guess the problem given by Aboufadel had a reference to Arnol'd, but I am not absolutely certain what the book was. I believe it was the one titled Huygens and Barrow, Newton and Hooke, but I don't guarantee it. I'm at home and all my stuff is at the office. However, I am certain that you, Dave Renfro, with your legendary thoroughness and attention to detail, will provide the exact references!

Rick

 

Department of Mathematics
Louisiana State University
     in Shreveport
Shreveport, LA 71115
(318) 797-5282

Roger B. Nelsen
CMJ Problems
Department of Mathematics
Lewis and Clark College
Portland, OR 97219

 

This is to give an ``alternate'' solution (i.e., not the one given by V.I. Arnol'd) to problem #497.

 

497. Proposed by Edward Aboufadel, Rutgers

Evaluate


lim
x ® 0 
 sin(tanx)-tan(sinx)
arcsin(arctanx)-arctan(arcsinx)

 

Solution by Rick Mabry, LSUS, Shreveport, LA 71115

The limit is 1.

Since I have seen the (geometric) solution given by Arnol'd in the book mentioned by the proposer, I would like to offer a solution of a different type, but one that is very elementary and broadly applicable: solution by power series. Arnol'd even mentions in his book that Newton did most of his analysis using power series.

Perhaps the proposer assumes that most of us are too quick to apply L'Hôpital's rule to such problems. (He would most likely be correct.) In the given problem, even at first glance, the derivatives may already seem too unwieldy to apply L'Hôpital's rule, but in this particular instance, L'Hôpital's rule wouldn't even give an answer until it had been applied seven times, and by then things would be completely unmanageable.

The limit has the form


lim
x ® 0 
 f(x)-g(x)
g-1(x)-f-1(x)
 ,
where f(x) = sin(tanx) and g(x) = tan(sinx), but this limit is 1 regardless of the choice of f and g, so long as they are not identical functions and each is analytic about zero with f(0) = 0 = g(0), f¢(0) = 1 = g¢(0). (These are the same assumptions made by Arnol'd in his solution.) A function h(x) of this type will be expressible as
h(x) = x+ ¥
å
 n = N 
 cn xn,
for x in a neighborhood of zero, where N > 1 and cN ¹ 0. This function will be invertible near zero, and its inverse will have a power series of the form
h-1(x) = x+ ¥
å
 n = N 
 dn xn
in a neighborhood of zero, where dN = -cN. It is also true that each coefficient dn of h-1(x) does not depend on any of the coefficients cn+1,cn+2,¼, and in fact,
dn = -cn+(terms involving only c2,c3,¼cn-1).
So let
f(x) = x+ ¥
å
 n = K 
 an xn  and   g(x) = x+ ¥
å
 n = L 
 bn xn,
where aK ¹ 0 ¹ bL. Then if K > L, we have

lim
x ® 0 
 f(x)-g(x)
g-1(x)-f-1(x)
 =
lim
x ® 0 
 -bL xL + ( higher order terms )
-bL xL + ( higher order terms )
 = 1.
If K < L, then

lim
x ® 0 
 f(x)-g(x)
g-1(x)-f-1(x)
 =
lim
x ® 0 
 aK xK + ( higher order terms )
-(-aK) xK + ( higher order terms )
 = 1.
Lastly, if K = L, let M be the smallest integer for which aM ¹ bM (such an M must exist, given our assumptions). Then aM-bM ¹ 0, hence

lim
x ® 0 
 f(x)-g(x)
g-1(x)-f-1(x)
 =
lim
x ® 0 
 (aM-bM) xM + (h.o.t)
(-bM-(-aM)) xM + (h.o.t)
 = 1,
and we're done.

It seems worth noting that we could also have considered general functions of the same type as above and written the limit in the form


lim
x ® 0 
 f(g(x))-g(f(x))
f-1(g-1(x))-g-1(f-1(x))
 ,
In this case the series of f(g(x)) and g(f(x)) agree up to the x4 term, so L'Hôpital's rule would require at least four applications. If we assume further that f(x) and g(x) are odd, as is the case in the posed problem, then a minimum of seven application of L'Hôpital's rule are needed.

I wonder what Arnol'd and the proposer would suggest we do with a problem like


lim
x ® 0 
 cosxcoshx -1
(cosx - coshx)2.
The use of powers series serves up the answer in moments, whereas L'Hôpital's rule takes four applications. Even using a power series on the original problem

lim
x ® 0 
 sin(tanx)-tan(sinx)
arcsin(arctanx)-arctan(arcsinx)
is not all that bad. (That's how I first did it, I confess!) And I cannot even think of any other way that I'd want to compute

lim
x ® 0 
 sin(tanx)-tan(sinx) +x7/30
arcsin(arctanx)-arctan(arcsinx)+x7/30
 = -27
13
 .

File translated from TEX by TTH, version 2.32.
On 16 Apr 2000, 12:21.