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.
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?
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)
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
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.
@@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!" 🙂
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.
![The Most Useful Curve in Mathematics [Logarithms]](http://i.ytimg.com/vi/OjIwCOevUew/mqdefault.jpg)
As an interesting consequence, this shows that e^t ≥ t+1 from the AM-GM inequality, with equality if and only if t=0.
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.
Same here. I am wondering however if I skipped part where we consider t+1
@@pavlopanasiuk7297 Indeed, if t
Geomethic Arithmedian.
bless you
11:04 Well, "approximating too small" only holds if the integrand is increasing, in other words, if t>0
Good point. You can also squeeze thm the GM and AM alone then you won’t have to worry about flipping inequalities
Fun fact: the natural log of the geometric mean it's the arithmetic mean of the natural logs.
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?
sounds very interesting!
Sound interesting
leapsecond.com/hsn2006/pendulum-period-agm.pdf
indeed gauss AGM yield pi fastest :) and also arttan
John Cook has done some articles on these topics recently.
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)
I'm never sure if this is serious math, or trolling non mathematicians people.
22:24 رائع جدا كالعادة
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
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.
But what about t
this is awesome
Nice video! ❤️
Where is @goodplacetostop ???
we really miss him
I think I may take up knitting 🙂
I fully encourage you in this endeavor
@@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!" 🙂
Why not the integral from 0 to n of x^t dt instead of adding the one?
Beautiful
What was is used for? Because for a standalone publication is to easy, no?
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.
❤interesting 😊
Amazing🎉🎉🎉So fun❤❤❤