Re: question

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