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Path: ...!news.mixmin.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: Peter Fairbrother <peter@tsto.co.uk> Newsgroups: sci.math Subject: Re: question Date: Sat, 18 May 2024 04:14:05 +0100 Organization: A noiseless patient Spider Lines: 96 Message-ID: <v296du$2jukf$1@dont-email.me> References: <v265jo$1qg8g$1@dont-email.me> <401d3516-fca4-4b3c-a089-ddd40718d303@att.net> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sat, 18 May 2024 05:14:07 +0200 (CEST) Injection-Info: dont-email.me; posting-host="c56dd11ff0b5acd14d5c1e8195f55e84"; logging-data="2751119"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19OMDwMdXLv8NMFLAwv+GOI6M1snxfFvEg=" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:+klChtpUWHvf8wGaLWDbw6cZy8o= Content-Language: en-GB In-Reply-To: <401d3516-fca4-4b3c-a089-ddd40718d303@att.net> Bytes: 3537 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