question

[Peter Fairbrother]

Is lim (cos pi/2n)^n = 1 as n -> infinity?

Any formula for calculating it from a given n (other than the obvious)?

Thanks

(not a homework question)

Peter F

[Jim Burns]

Post by Peter Fairbrother
Is lim (cos pi/2n)^n = 1 as n -> infinity?

Yes, 1.
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

Post by Peter Fairbrother
Any formula for calculating it from a given n
(other than the obvious)?

What is it that is obvious?

[Peter Fairbrother]

Post by Jim Burns

Post by Peter Fairbrother
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.

Post by Jim Burns
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

Post by Peter Fairbrother
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

[Jim Burns]

Post by Peter Fairbrother

Post by Jim Burns

Post by Peter Fairbrother
Is lim (cos pi/2n)^n = 1 as n -> infinity?

Yes, 1.

a) thanks

My pleasure.

Post by Peter Fairbrother
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,
but in any case
I will have to look at the details
which you have so kindly provided.

Sorry if my answer was a little too.much.information.ish.
It might have been shaped by my recent discussions,
in which every last #$%^ing.obvious detail
gets disputed.
That, of course, has nothing to do with you.

Expanding functions into series
and throwing away higher order terms
is a more physicsy presentation:

cos(x) ≅ 1 - x²/2 + ...

(1 - x)^n ≅ 1 - n⋅x + ...

cos(π/2n)^n ≅
(1 - (π/2n)²/2)^n ≅
1 - n⋅(π/2n)²/2 ≅
1 - (π²/8)/n ⟶ 1

Whew! Same answer! Yay me.

Post by Peter Fairbrother
It's about a physics thing,
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?

(cos π/0)^0 is undefined.
(cos π/2⋅2)^2 = 1/2

Post by Peter Fairbrother
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.

That effect would make a nice exhibit at
a science museum
https://cosi.org/
https://www.exploratorium.edu/
etc

I'm thinking of
two fixed polarizers at 90°, backlit,
and shaped polarizers: letters animals puppets
which can be inserted between them.

Where the shapes are, more light, not less.
Weirdness.
Budding Nobel.prize.winners galore!

Post by Peter Fairbrother
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.

I can do the math for all that,
but I struggle with the concepts.
What I read tells me I have illustrious company.

I imagine some future generation,
maybe not very far off in time,
being as comfortable with 21st cen. quantum weirdness
as we are with 19th cen. electromagnetic weirdness.

It seems to me that the de.weirding will be brought by
the widespread use of quantum technologies --
not very far off at all.

I imagine us today pecking out "Chopsticks" on
the grand piano of quantum physics, with
our quantum computing and our quantum key distribution.
There will come a Mozart, a Van Cliburn,
and, until they come,
we won't be able to imagine what they'll play.

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

My probably.out.of.date understanding of
quantum wave collapse
is that
pre.collapse, there are interference patterns
post.collapse, there aren't interference patterns.

I'd be surprised if the interference patterns
of multiply.polarized light goes away,
even partially.
My intuition gives it a thumbs.down,
but I haven't looked into this carefully.

Post by Peter Fairbrother
On to entanglement... -
but that's another story, for later.

I hope I'll get to see someone tell that story.