An arithmetic-geometric limit

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[@MathFromAlphaToOmega]

As an interesting consequence, this shows that e^t ≥ t+1 from the AM-GM inequality, with equality if and only if t=0.

[@complexquestion3601]

There's a solution that's fairly simpler. The GM can be rewritten as (n!)^(t/n). Use Stirling's formula for n!, and because of the t/n exponent almost all factors have a limit of 1, and you're left with an asymptotic equivalent of (n/e)^t for the GM. Next, plug this equivalent into the GM/AM quotient, which is now (n^t / e^t )* (n/(sum of i^t from 1 to n)). Then divide both the numerator and the denominator by n^t, which transforms the AM term into (sum of (i/n)^t from 1 to n)/n, which in the limit is the definition of the Riemann integral of (x^t dx) from 0 to 1. All that's left is to compute the integral, which gives 1/(t+1), and plug that into the quotient, giving (t+1)/e^t.

[@pavlopanasiuk7297]

Same here. I am wondering however if I skipped part where we consider t+1

[@complexquestion3601]

@@pavlopanasiuk7297 Indeed, if t

[@ridefast0]

Speaking of AM and GM, I enjoyed finding the 'hybrid' AGM in the formula for the exact period of a simple pendulum. Would that be worth a look, Michael?

[@nirajmehta6424]

sounds very interesting!

[@BilalAhmed-wo6fe]

Sound interesting

[@ridefast0]

leapsecond.com/hsn2006/pendulum-period-agm.pdf

[@peilingLeslieLiu]

indeed gauss AGM yield pi fastest :) and also arttan

[@HagenvonEitzen]

11:04 Well, "approximating too small" only holds if the integrand is increasing, in other words, if t>0

[@user-en5vj6vr2u]

Good point. You can also squeeze thm the GM and AM alone then you won’t have to worry about flipping inequalities

[@adiaphoros6842]

Geomethic Arithmedian.

[@maxwelljennings4178]

bless you

[@blazeottozean469]

Can we apply the limit of GM and AM separately (get some kind of infinity) then combine together to get the answer? Like: lim GM = lim n^t/e^t lim AM = lim n^t/(t+1)

[@cmilkau]

Fun fact: the natural log of the geometric mean it's the arithmetic mean of the natural logs.

[@ianjlilly]

John Cook has done some articles on these topics recently.

[@minwithoutintroduction]

22:24 رائع جدا كالعادة

[@ruilopes6638]

Shouldn’t our assumptions about t happen when we are looking for the inequalities on the AM. t been lesse than one there flips the inequalities the other way around. I believe what was calculated holds only for t>1

[@faradayawerty]

this is awesome

[@nikhilprabhakar7116]

Where is @goodplacetostop ???

[@Ahmed-Youcef1959]

we really miss him

[@gp-ht7ug]

Nice video! ❤️

[@iGeen7]

But what about t

[@BongoFerno]

I'm never sure if this is serious math, or trolling non mathematicians people.

[@zachbills8112]

I just used Stirling's Approximation up top and approximated the bottom sum by it's integral, which is valid asymptotically. This is simple enough to do in your head.

[@stevenmellemans7215]

Why not the integral from 0 to n of x^t dt instead of adding the one?

[@peilingLeslieLiu]

❤interesting 😊

[@bndrcr82a08e349g]

Beautiful

[@Alan-zf2tt]

I think I may take up knitting 🙂

[@stefanalecu9532]

I fully encourage you in this endeavor

[@Alan-zf2tt]

@@stefanalecu9532Thank you! I do not know whether to thank you or curse as there seems playful ambiguity on "this" I will take a finer interpretation and double up on "Thanks dood!" 🙂

[@juniorcyans2988]

Amazing🎉🎉🎉So fun❤❤❤

[@eugen9454]

What was is used for? Because for a standalone publication is to easy, no?

[@EinsteinsHair]

It has been 30 years since I looked at Mathematics Magazine, but this is probably not too easy. The magazine is aimed at undergraduate teachers and students. As I remember they also included Problems to solve. Also Proofs Without Words showing simple geometric proofs. Not exactly a formal Journal.