the Stolarsky mean
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- เผยแพร่เมื่อ 4 พ.ย. 2023
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The logarithmic mean is the limit as p→0, which checks out looking at it from a calculus perspective. Interestingly, the limit as p→1 also exists, and its value is 1/e * (b^b/a^a)^(1/(b-a)). That one was a pretty comprehensive exercise. Not really sure how to interpret the results though… idk what would make sense to fall between the arithmetic and logarithmic means.
This video dropped an hour early. Did you forget to turn your clock back? 😉
timezone changed a week ago😂.
@@nikos4677 no, it changed last night, at least here
@@pierreabbat6157 yeah propably each country changes their timezone at different dates
Wow, I independently discovered this after watching your video on the logarithmic mean... Cool! Here's something neat I noticed:
f and g generate the same mean iff f(x) = Ag(x) + Bx + C for constants A≠0,B,C. Pretty fun exercise to show.
Also the case as p -> 1 of the Stolarski mean is interesting.
The final conclusion is so awesome because it is very consistent with what we acknowledge as what a mean should be. Really cool!
I appreciate how the limit at the end produced the maximum, which is also what happens when you take the limit of the p-norm to infinity.
I would love to learn about what (if any) connections there are between the Stolarsky mean and the p-norm.
Could the Stolarsky mean be related to some sort of uncountable extension to the p-norm?
The p-norm comes with a Banach space L^P. Maybe Stolarsky means also have spaces associated with them? Also, L^1 is well defined (it's the Manhattan distance).
Edit: The first Stolarsky mean can be defined as the "identric mean". It's not the same as the Manhattan distance.
The general form of the mean presented at the start doesn't really generalise clearly to more than 2 numbers. On the other hand, Jensen's inequality (concerning convex functions) does give a large class of means that have nice properties, and leads to the concept of an Orlicz space. These function spaces include and generalise the L^p spaces; could be worth looking up!
@@thudso Good point. So the Stolarsky mean is kind of in it's own category and we can't expect interesting spaces to pop out.
As always best content ever!
The more I watch M|ichael's videos the more I see beauty in math. Here we see how discrete and continuous sort of merge and understand why Category Theory does attract interest.
A patterns of patterns in a pattern?
And now I have to dig what happens with tan/arctan or maybe also the "half derivative" by myself...
Really interesting! My first thought when seeing the different p-values corresponding to well-established means like the arithmetic or the geometric mean was to ask whether the arithmetic-geometric mean, which is usually defined by series, has a distinct p-value as well. After some fairly rough calculation in Excel it seems that the AGM of two numbers corresponds to the Stolarsky mean for p=0.5. Is this correct?
It seems
S0.5(a,b) = 1/4 (sqrt(a) + sqrt(b))^2
which certainly isn't AGM.
Is there a value of p that gives you the harmonic mean ?
my favorite means are the AGM, the arithmetic-geomentric mean of Gauss, and the means that come from the ratios of elementary symmetric functions. For some matrix versions of these, see some of my papers.
Hi,
1:08 : there can be several of them, specially if the function is not monotonous. The theorem just says that there is at least one.
It seems like the condition on f' is that it must be "strictly monotonous", IOW either increasing or decreasing, but not constant, and even "never" constant.
What whould be the function f to get (ab)/(a+b) ?
13:17 : good way to find interesting means.
The condition that f' is 1-1 is sufficient to guarantee the uniqueness of such point
Could there be a complex p-adic (C_p) analogue of this?
While this generalized mean is interesting, there's another one that I like better. It's called the generalized f-mean and has a formula of f⁻¹(arithmeticMean(f(a), f(b), f(c), ...)). It's also called the Kolmogorov-Nagumo-de Finetti mean and the quasi-arithmetic mean. So it's only a little bit different than the mean in the video, what without the derivative and using an arithmetic mean as part of the general formula, but those differences are pretty important. And also it's set up for averaging more than just two terms, which is nice. The best part is if you let f(x)=arctan(x). Then the mean has all kinds of neat properties. It can give meaningful values to averages including terms that are infinite, positive, negative, and zero. For example, the arctan mean of 0 and infinity is 1. The mean of zero, infinity and another infinity is sqrt(3). The mean of positive infinity and negative infinity is zero.
I learned about such means from a book by Bruno de Finetti
What do you get in the limit as p→1?
Oh I just solved this, it’s 1/e * (b^b/a^a)^(1/(b-a)) 😧
Not really sure what you could interpret it as.
why this mean shows same properties as Quasi-arithmetic mean ?
Edit:
Now ic why because they are equivalent.
Let's call this mean as MVT mean and Quasi-arithmetic mean as QA mean
In MVT mean if we consider D as derivative and S as slope operator on function f then MVT mean may stated as Df=Sf
If we replace f(x) by ∫(a to x)f(t) dt in Df=Sf
We get
f(c)=1/(b-a)[∫(a to b)f(t) dt]
Which is equivalent to QA mean.
Looks like S0 is the logarithmic mean again. S1 is something funny,
S1(a,b) = (a^(a/(a - b)) b^(-b/(a - b)))/e = (a^a/b^b)^(1/(a-b)) /e for positive a, b
according to Wolfram Alpha (and me).
Very nice video! L’Hopital is fundamental
here he could have factorized the b^p inside the ln to get the limit
Application-wise, if your data behaves quadratically, then the arithmetic mean would be the best pick? I am trying to interpret this and map it to applications (I will think it on my own, but I want to drop this question here while this video is still recent)
What do you mean by "if your data behaves quadratically"?
would be interesting to consider p*(x^p) and use the product logarithm W
is your pfp a covariance matrix, or some iterated bitwise operator related?
I Like the case for x^n
I like "nicely defined" functions :)
Consider a function f(x) and its derivative f'(x). If you have the derivative f'(x) and initial conditions, you can use integration to find the original function f(x). In this sense, the integral acts as the inverse operation to the derivative. -- ChatGPT3.5
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No, you cannot explode it.