P R O B L E M S

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

Prove that:

\frac{1}{\sin x} = \frac{1}{x} + \sum_{n = 1}^{\infty} \frac{(- 1)^{n} 2 x}{x^{2} - n^{2} π^{2}}

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 / (3 π)) (1 + x / (2 π)) (1 + x / π) x (1 - x / π) (1 - x / (2 π)) (1 - x / (3 π)) . . .

(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 - π) + \sin (x) / (x) + \sin (x) / (x + π) + . . .

Divide through by sin x:

$\cot x = . . . + 1 / (x - 2 \times π) + 1 / (x - π) + 1 / x + 1 / (x + π) + 1 / (x + 2 \times π) + . . .$ (*)
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)

$\cot (x / 2) = . . . + 1 / (x / 2 - 2 \times π) + 1 / (x / 2 - π) + 1 / (x / 2) + 1 / (x / 2 + π) + 1 / (x / 2 + 2 \times π) + . . .$

So:

$\cot (x / 2) = . . . + 2 / (x - 2 π) + 2 / x + 2 / (x - 2 \times π) + . . .$

Now subtract equation (*):

$\cot (x / 2) - \cot (x) = . . . - 1 / (x - 3 \times π) + 1 / (x - 2 \times π) - 1 / (x - π) + 1 / x - 1 / (x + π) + 1 / (x + 2 \times π) - . . .$

Now if you combine $1 / (x - n \times π)$ and $1 / (x + n \times π)$ 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