P R O B L E M S
By Pooya Farshim (P2572) on Tuesday,
July 11, 2000 - 07:42 pm :
Prove that:
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1
sin x
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1
x
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+ |
¥
å
n=1
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(-1)n 2x
x2-n2p2
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By Michael Doré (P904) on Thursday, July 20,
2000 - 12:12 am :
Here is a rather non-rigorous proof of the series you're
trying to show. I'll see what I can do to make it better. First
of all I'll post the dodgy proof...
By Michael Doré (P904) on Thursday, July 20,
2000 - 12:12 am :
See Dan Goodman's message on April 8th in this
thread . Here Euler's infinite product for sin x is
given:
sin x=...(1+x/(3p))(1+x/(2p))(1+x/p)x(1-x/p)(1-x/(2p))(1-x/(3p))...
(actually the x was left out but I'm pretty sure that was a
mistake - it doesn't make sense otherwise). According to the
message, this was guessed by Euler because sin has roots at
n*pi.
Now continuing on in the crazy spirit Euler started out in why
not differentiate both sides with respect to x using (and you're
going to love this) the product rule extended to infinite
products . (No doubt this isn't valid - I'll work on this
later.)
In general if y = ab (where y,a,b are functions of x) then:
y' = (a')(b) + (a)(b')
If y = abc then:
y' = (a')(b)(c) + (a)(b')(c) + (a)(b)(c')
So we try to extend this to infinite products. y =
abcdef....
y' = (a')bcdef.... + a(b')cdef... + ab(c')def... + ...
Now using Euler's infinite product and applying the difference of
two squares rule, sin x = ...f(-2)f(-1)f(0)f(1)f(2)...
where f(n) = (1-x/n*pi ) for n =/= 0, and f(0) = x. Agreed?
Now differentiating both sides of the equation it is easy to
see:
cos x = ... + g(-2) + g(-1) + g(0) + g(1) + g(2) + ...
where g(n) = ...f(n-2)f(n-1)f'(n)f(n+1)f(n+2)...
where f'(n) = derivative of f(n) with respect to x.
But g(n) = ...f(n-1)f(n)f(n+1)... * [f'(n)/f(n)] = sin x *
f'(n)/f(n)
(you simply replace the infinite product with sin x due to
Euler's infinite product formula).
Now for n =/= 0, f(n) = 1-x/n*pi, so f'(n)/f(n) =
(-1/n*pi)/(1-x/n*pi) = 1/(x-n*pi). And f'(0)/f(0) = 1/x, so in
fact f'(n)/f(n) = 1/(x-n*pi) for all n.
Therefore for all n:
g(n) = sin(x)/(x-n*pi ). And we know:
cos x = ... + g(-1) + g(0) + g(1) + ...
So:
cos x=...+sin(x)/(x-p)+sin(x)/(x)+sin(x)/(x+p)+...
Divide through by sin x:
cot x=...+1/(x-2×p)+1/(x-p)+1/x+1/(x+p)+1/(x+2×p)+... (*)
This is almost the right hand side of your equation but not
quite. We need to make the terms alternate in sign. To do this we
first consider cot(x/2)
cot(x/2)=...+1/(x/2-2×p)+1/(x/2-p)+1/(x/2)+1/(x/2+p)+1/(x/2+2× p)+...
So:
cot(x/2)=...+2/(x-2p)+2/x+2/(x-2×p)+...
Now subtract equation (*):
cot(x/2)-cot(x)=...-1/(x-3×p)+1/(x-2×p)-1/(x-p)+1/x-1/(x+p) +1/(x+2×p)-...
Now if you combine 1/(x-n×p) and 1/(x+n×p) pairwise you can
see that this is indeed the right hand side.
All that remains to be proven is that:
cot(x/2) - cot(x) = 1/sin(x)
LHS = 1/tan(x/2) - 1/tan(x) = 1/tan(x/2) - (1-tan2
(x/2))/(2tan(x/2)) = 1/2*(tan(x/2) + cot(x/2))
Multiply the top and bottom of the fraction by
sin(x/2)cos(x/2):
LHS = 1/2*(cos2 (x/2) + sin2
(x/2))/(sin(x/2)cos(x/2)) = 1/(2*sin(x/2)*cos(x/2)) = 1/sin x =
RHS.
Thus "proving" the equation.
I'd be extremely interested on anybody's ideas on making the
middle part of this a little more rigorous.
Yours,
Michael