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From: Peter Fairbrother <peter@tsto.co.uk>
Newsgroups: sci.math
Subject: Re: question
Date: Sat, 18 May 2024 04:14:05 +0100
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On 17/05/2024 13:56, Jim Burns wrote:
> On 5/16/2024 7:41 PM, Peter Fairbrother wrote:
> 
>> Is lim (cos pi/2n)^n = 1 as n -> infinity?
> 
> Yes, 1.

a) thanks

b) yikes!  I remember doing some of the below math at school and 
university, and thinking I'd probably never need it again. This is now 
the second time I have needed something like that, in 50 years, so maybe 
I should learn it to immediately usable status, or maybe not: but in any 
case I will have to look at the details which you have so kindly provided.


It's about a physics thing, passing light through polarising filters: eg 
when n=1 the expression =0, no light passes through a pair of polarisers 
which are crossed at 90 degrees.

Hmmm, when n=0 does the expression (cos pi/2n)^n = 1/2?

Insert another suitably orientated filter in between those filters 
(n=2), and some light passes. Oooh, quantum weirdness, recent Nobel 
prizes, Bell's theorem, spooky-action-at-a-distance, and so on.

Insert lots and lots of filters in a rotating pattern, and all (1/2) the 
light passes through.

Except - suppose it's all really simple instead, and light which passes 
through a filter just has its polarisation changed a little. Then all 
the Bell hidden variables are irrelevant, as they change depending on 
their particles' history.

Note, there is necessarily a measurement of the photons in a polarising 
filter - but this does not necessarily involve a complete wavefunction 
collapse.

On to entanglement... - but that's another story, for later.

> Take the logarithm of both sides,
> Swap log and lim, because continuity.
> Evaluate lim by L'Hôpital's rule .
> Take the exponential of both sides.
> 
> lim(cos(1/n)^n
> [n→inf])
> 
> exp(log(lim(cos(1/n)^n)))
> [n→inf])))
> 
> exp(lim(log(cos(1/n)^n)))
> [n→inf]))
> continuity
> 
> exp(lim(
> n*log(cos(1/n))
> [n→inf]))
> 
> exp(lim(
> log(cos(1/n))
> ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
> 1/n
> [n→inf]))
> 
> exp(lim(
> log(cos(x))
> ⎯⎯⎯⎯⎯⎯⎯⎯⎯
> x
> [x→0]))
> 
> exp(lim(
> -sin(x)/cos(x)
> ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
> 1
> [x→0]))
> L'Hôpital
> 
> exp(0)
> 
> 1
> 
>> Any formula for calculating it from a given n
>> (other than the obvious)?
> 
> What is it that is obvious?

just that F(n) =(cos pi/2n)^n and is easily calculated, (or perhaps F(x) 
=(cos pi/2x)^x) - can it be simplified?

Thanks again


Peter Fairbrother