[tan(sin x) - sin(tan x)]/(x^7) as x --> 0
Dave L. Renfro
"[tan(sin x) - sin(tan x)]/(x^7) as x --> 0"
[Feb. 13 and April 16, 2000] at
<http://forum.swarthmore.edu/epigone/sci.math/friblorkhul>
I briefly discussed the behavior of tan(sin x) - sin(tan x)
for x near zero. This past weekend I spent some time in a library
looking some things up, and while doing so I came across this
paper:
Arthur Cayley, "On the value of tan(sin theta) - sin(tan theta)",
Messenger of Mathematics 14 (1894-95), 191-192.
A biography of Cayley can be found at
<http://www-groups.dcs.st-and.ac.uk:80/~history/Mathematicians/Cayley.html>
Interestingly, Cayley died in on January 26, 1895, about the time
this paper appeared. Since the paper is quite short, I've reproduced
it below, modulo notational changes needed for this ASCII text
format. [NOTE: Every occurrence of "z" below is "$\theta$" in the
original.] I haven't checked the details of Cayley's argument,
but I did proof-read several times what I typed. Fixed font spacing
should be used when reading this--otherwise the expressions will
not line up very well from line to line.
Dave L. Renfro
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"On the value of tan(sin z) - sin(tan z)" by Arthur Cayley
The following equation is given p. 59 of the "Lady's and
Gentleman's Diary" for 1853:
tan(sin z) - sin(tan z) = (1/30)*z^7 + (29/756)*z^9 + &c.
Write in general
X = z + A*z^3 + B*z^5 + C*z^7 + D*z^9 + ...,
Y = z + A'*z^3 + B'*z^5 + C'*z^7 + D'*z^9 + ... .
Then as far as z^9 we have
X + A'*X^3 + B'*X^5 + C'*X^7 + D'*X^9
= z + A*z^3 + B*z^5 + C*z^7 + D*z^9
+ A'*[z^3 + 3A*z^5 + 3(A^2 + B)*z^7 + (A^3 + 6AB + 3C)*z^9]
+ B'*[ z^5 + 5A*z^7 + (10*A^2 + 5B)*z^9]
+ C'*[ z^7 + 7A*z^9]
+ D'*[ z^9]
= + z
+ (z^3)*(A + A')
+ (z^5)*(B + 3AA' + B')
+ (z^7)*(C + 3(A^2)A' + 3A'B + 5AB' + C')
+ (z^9)*(D + (A^3)A' + 6AA'B + 3A'C + 10(A^2)B' + 5BB' + 7AC + D'),
and hence
X + A'*(X^3) + B'*(X^5) + C'*(X^7) + D'*(X^9)
- Y - A*(Y^3) - B*(Y^5) - C*(Y^7) - D*(Y^9)
= (z^7)*[3AA'(A - A') + 2(AB' - A'B)]
+ (z^9)*[AA'(A^2 - A'^2) + 6AA'(B - B') + 4(AC' - A'C)
+ 10[(A^2)B' - (A'^2)B)] ],
X = sin z = (1/6)*z^3 + (1/120)*z^5 - (1/5040)*z^7 + ...;
A = -1/6, B = 1/120, C = -1/5040,
Y = tan z = (1/3)*z^3 + (2/15)*z^5 + (17/315)*z^7 - ...;
A' = 1/3, B' = 2/15, C' = 17/315,
we have therefore
AA' = -1/18, A - A' = -1/2, A + A' = 1/6, B - B' = -1/8,
AB' - A'B = -1/40, AC' - A'C = -1/112, (A^2)B' - (A'^2)B = 1/360.
Hence coeff. z^7 = 1/12 - 1/20, = 1/30,
coeff. z^9 = 1/216 + 1/24 - 1/28 + 1/36, = 29/756
(viz. multiplying by 1512; 7 + 63 - 54 + 42 = 58),
and the required equation is thus verified.
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ROGER BAGULA
Very good research!
General type of function of meromorphics as noncommutive (functionally),
maybe:
Let f(z) and g(z) be meromorphic functions, then
h(z)=f(g(z))-g(f(z))
Will h(z) always br meromorphic as well?
Another question?
Suppose we have
f(z)=(a*z+b)/(c*z+d)
g(z)=(e*z+f)/(g*z+h)
both in the special linear group SP(2,z).
Is there a case where:
f(g(z))=g(f(z))+K*f(z)*g(z)
or
f(g(z))=Mod(g(f(z)),K) ?
A modular forms Klein group.
"Dave L. Renfro" wrote:
--
Respectfully,
Roger L. Bagula
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