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By Pooya Farshim (P2572) on Tuesday, July 11, 2000 - 07:42 pm:

Prove that:


 1
sinx
 =  1
x
 + ¥
å
n=1 
  (-1)n 2x
x2-n2p2
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*p) for n¹0, 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 n¹0, 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*p). 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-2*p) + 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