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    From Peter Fairbrother@21:1/5 to All on Fri May 17 00:41:44 2024
    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

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  • From Jim Burns@21:1/5 to Peter Fairbrother on Fri May 17 08:56:42 2024
    On 5/16/2024 7:41 PM, Peter Fairbrother wrote:

    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

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

    What is it that is obvious?

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  • From Peter Fairbrother@21:1/5 to Jim Burns on Sat May 18 04:14:05 2024
    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

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  • From Jim Burns@21:1/5 to Peter Fairbrother on Sat May 18 13:30:01 2024
    On 5/17/2024 11:14 PM, Peter Fairbrother wrote:
    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

    My pleasure.

    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.

    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.

    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?

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

    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!

    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.

    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.

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

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

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