What is the logarithmic mean, and why does it fit right in the middle??
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- Опубликовано: 8 авг 2023
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For the red trapezoid, you don't need to find the actual heights to find the average height. The average height is the height at the middle point, which is where the tangent line is referenced to.
@Keithfert490 Clever❕
I was going to comment the same. By finding the average height you need to simplify e^(½lnb+½lna)(lnb-lna) which is pretty simple
Right, the convexity holds true for all R.
yep, typical Calc II concept
WHY define the log mean thi s way --why notjust lna -ln b or lna + ln b or something else.
(sqrt(b) - sqrt(a))^2 >= 0
b + a - 2sqrt(ba) >= 0
(b+a)/2 >= sqrt(ab)
The log mean is ubiquitous in engineering, especially heat transfer and anything having to do with cylindrical shapes.
I knew I'd seen it before
Indeed, I noticed the format too!
I wanted to write exactly the same comment 👍
5:32 If you notice that the limit is the reciprocal of the derivative definition of the natural logarithm at x=a, which is why the limit is 1/(1/a)
That's why one could say it's almost like circular reasoning
I wonder how the log-average for a series of points works out? The other two are (a1*a2*a3...*an)^(1/n) and (a1+a2+...an)/n
Exactly what I was wondering
Log Mean's actually have a lot of application to modeling transport phenomena when you have heat exchangers or separation membranes where two fluids can exchange heat or mass according to Newton's Law of Cooling or Fick's Law of Diffusion as they flow parallel to the surface separating the two. I don't remember the exact derivation, but it involves the fact that both of these processes show an exponential decay in concentration/temperature differences over time so the arithmetic average of temperature/concentration isn't as the logarithmic mean.
Yep very familiar with the log mean as an engineer. Log mean temperature difference is used in sizing counter-current heat exchangers.
"Is there an in between mean?"
Ah yes, the mean mean.
*Central Limit Theorem has entered the chat*
Stop it! 😅
@@BikeArea That's mean! :-)
I like this kind of videos when geometric explanations are given in Cartesian axes
You should do the arctangent mean. It's my favorite of the generalized means. It's defined as tan((1/n)*Σarctan(aₖ)), where the sum runs over all elements aₖ that you want to average. Its virtues include that it's well-defined for all real numbers, not just positives, and for all hyperreal numbers (which are basically the reals plus infinity), although infinitesimals are treated as having the infinitely small part being zero. This means that you can average infinities with actual numbers and get meaningful results. For example, the average of infinity and 0 is 1. The average of infinity, infinity, and 0 is √3.
Nice! That's one I had not met before, and I like being able to take on infinity in a series when you are seeking a mean.
I wonder where it lies on the infinite scale of power means (harmonic
Just by checking Desmos, it looks like it's always
There shouldn’t be one. Thatd be like saying x^a behaves as logx for some a
@@user-en5vj6vr2u Well, in some sense it does, for a=0.
Another expression which always gives a result which lies between the geometric and the arithmetic mean is (a + b + sqrt(ab))/3.
It occurs e. g. when calculating the volume of a frustrum: That's given by its height times this "mean" of the area of the base and the area of the top.
And I just found out that there is actually a name for this: It's called the "Heronian mean".
Isn't that the arethmetic mean between the arethmetic mean and the geometric mean!?
@@minerscale Actually, it's not a simple arithmetic mean of both, but a _weighted_ mean: 2/3 times the arithmetic mean plus 1/3 times the geometric mean.
Numerically, here's the inequalities between the means : arithmetic ≥ Heronian ≥ logarithmic ≥ geometric ≥ harmonic
@@Nolord_ so the Heronian is always above the log mean? Is the Heronian the smallest convex combination of am and gm above the log mean?
@@josea748 Maybe, that can depend on your definition of combination.
You have heard of the elf on the shelf. Today we shall learn about the mean in-between.
"And we'll maybe derive this real quick just for completeness"
This is how my wife wins arguments.
As an engineer I'm really glad to see a video about the log mean.
What about the AGM (arithmetic-geometric mean)? Doesn't it also lie between the arithmetic and geometric means? It's also related to the elliptic integrals and surely deserves more attention at the HS and undergraduate level.
It of course does lie between the arithmetic and geometric means by construction. Looking at it on Desmos, it seems to lie between the gemoetric mean and the logarithmic mean.
Using mean value theorem gives us a more insight of how a value c within (a,b) is related with AM-GM inequality!
14:59
Could this definition somehow be extended to more inputs?
Like arithmetic mean is clearly (a+b+c)/3 and geometric mean is clearly cbrt(abc) but it doesn't seem obvious how this would work for the log mean
It can if the log mean is associative. Is it? Trying to solve that in my head but I'm getting logs of sums of logs ant it gets ugly.
@@JCCyC well no because AM3(a, b, c) ≠ AM2(a, AM2(b, c)) if you know what i mean
With 4 inputs, both the arithmetic and geometric means of a, b, c and d are mean(mean(a, b), mean(c, d)). For instance, ((a+b)/2 + (c+d)/2)/2 = (a+b+c+d)/4 and √(√(ab)√(cd)) = 4thRoot(abcd).
Let's try this with the logarithmic mean: Assume 0 < a < b < c < d. We have the two sub-means u := (b-a)/(ln(b)-ln(a)) and v := (d-c)/(ln(d)-ln(c)). Wait, was this the right thing to do? Maybe u should have been calculated from a and d, while v should be using b and c? Or maybe use a and c for u and b and d for v?? No matter what we do, we then need to figure out whether u or v is the larger one, and then calculate the mean of those two again. And now we have like, what, 6 possible cases to chew through? I feel like this doesn't generalize well...
One of my favorite videos. Not terrible complicated, but still really interesting.
An alternative way to show the initial inequality between the geometric and the arithmetic mean in by taking the logarithm and using Jensen's inequality.
Such a cool video, thanks for sharing Michael! I loved the geometric elegance of the inequality proof!
Here is a generalization of this mean I thought of after watching this:
L_a(x,y) = (a(x-y)/(x^a-y^a))^(1/(1-a)), taking appropriate limits when necessary.
It reduces to the arithmetic mean when a=2, the logarithmic when a=0 and the geometric when a=-1. This raises the question, what about a=1? This would in a sense lie exactly between the arithmetic and logarithmic means.
It turns out to become (x^x/y^y)^(1/(x-y)) / e, which I think could make for an interesting exercise.
Thanks Michael! I've been playing with various means you discussed (arithmetic, geometric, and harmonic) and also the RMS (root mean square). I always thought there must be an exponential or logarithmic mean and toyed with various ideas without success. Well you've answered that riddle with your video. I also like your proofs of the inequalities.
Very nice!
Wish I had known about this proof much earlier in my life. 🙂
I'm wondering, You can write the arithmetic mean like (t_1 + t_2 + ... + t_n) / (n) and the geometric mean as (t_1 t_2 ... t_n)^(1/n).
Is there a similar way to put the logarithmic mean or does it only work if you have only two numbers?
Have a look at generalised means (Wikipedia can get you started)
@@Francisco-vl5ub Thanks m8, will check it out
You should rename the video "What does the logarithmic mean mean?"
don't be mean
Hmmm it is as if there is a set of two-argument functions called the "set of means", the "mean" function that belongs to it have the same properties as the arithmetic mean a+b/2, and each "mean" function of two real arguments a and b can be ordered
Is it condition that the secant line is always "above" the tangent line a necessary and sufficient condition for convexity? Are there convex functions that don't have this property?
AM-LM has practical application in physics in heat transfer physics problem.
Amazing!
Lovely video! I'd never heard of the logarithmic mean and deriving the inequality using areas was very interesting, your uploads are a bright point of my day.
For completeness sake it might have been nice to first show that the log mean behaves as a mean (i.e. a
I don't think means actually need that property (think for example of the average bearing, you can't order bearings in that way yet the mean has meaning), but yeah it would have been nice to show.
@@MagicGonads You may be right, after (far too much) searching for a good definition, I couldn't actually find a well defined one for what "mean" actually refers to.
Most dictionaries just define it as the arithmetic mean, while the closest real definitions I could find were vague at best, such as "a measure of central tendency" and Wikipedia's explanation that a "mean serves to summarize a group of data, often to better understand the overall value."
I find it odd that it doesn't seem to have a well established mathematical definition.
Actually that's a consequence of the shown inequality: Since min(a,b)
In addition to an arithmetic mean and a geometric mean, any two numbers above 1 have what I call a double-geometric mean. If log_a (b)=log_b (c), b is the DGM of a and c. Means with other degrees of exponentiality are subjective to the base (although you could very well use e).
As usual, a very good video! FYI.. I have a very interesting little book from which I learned a lot called "Analytical Inequalities" by Nicholas D. Kazarinoff ; University of Michigan, HOLT RINEHART AND WINSTON 1961, but it doesn't seem to discuss the logarithmic inequality.
if am0 = arithmetic mean and gm0 = geometric mean, and am1 and gm1 the arithmetic and geometric mean of am0 and gm0, and repeat this a few times, then quickly am approaches gm and thus also the logarithmic mean, and this converges to something in the middle as well
While they approach each other, they don't approach the logarithmic mean. They approach the arithmetic-gemoetric eman, which is defined exactly by this limit.
Can we generalize the notion of a "mean" in another way?
We note:
i: x -> x is an increasing function, and the arithmetic mean, (a + b)/2 = inv i [ (i(a)+ i(b))/2 ]
In is also an increasing function and we may write the geometric mean as sqrt (ab) = inv ln [ (in(a) + ln(b))/2 ]
So let f be any increasing function (defined, say, on [0, inf), and define the "f-mean" of a, b in [0, inf), denoted M(f; a, b), to be inv f [ (f(a) + f(b)) / 2 ]
Thus, M(x -> x^3; a, b) = cube-rt [ (a^3 + b^3) / 2 ]
We note M(f; a, a) = a and for a < b, a < M(f; a, b) < b
Does a "mean" have to have any other properties?
And can the present mean be brought into this scheme? I.e., is there an increasing function, f, s.t.:
f [ (b - a) /(ln b - In a) ] = [ f(a) + f(b) ] / 2 ?
"Does a "mean" have to have any other properties?" Well, it has to be symmetric in its arguments
Oh, and f, above, only has to be strictly monotonic: the harmonic mean uses f: x -> 1/x
Actually a lot of means are defined that way. For example, the harmonic mean is obtained with i(x) = 1/x. The root-mean-square is obtained with i(x) = x^2.And you can even get the geometric mean this way, using i(x) = ln(x).
Great!
The log mean has other nice properties. It is homogeneous like other means: The mean of kx and ky is k times the mean of x and y. Thus you can scale the log mean to a function of one variable x/y.
Also, it isn't difficult to show that L(x,y) < ((³√x+³√y)/2)³, the power mean of exponent 1/3. (which is less than the arithmetic mean).
I've always been fascinated by the Harmonic Mean. Average Speed and Resistance in Parallel.
Is there a way to generalise the logarithmic mean that is useful and/or unambiguously “correct”, preserving both the inequality and that it should return “a” for the mean of arbitrarily many copies of “a”? 🤔
I don't think that complicated calculation of h1, h2 near the end was necessary. Because, (h1+h2)/2 (the arithmetic mean of the heights) is obviously just the middle point, so sqrt(ab). This is because they're connected with a straight line segment.
Since it's trapezoid, we can derive '(h1 + h2)/2 = e^(lnb+lna/2) = sqrt(ab)' direct
How can be extended this logarithmic mean to a set of numbers? It seems not obvious to me
How to expand the definition to the arbitrary number of variables? I.e., a1, a2, a3 etc. instead a and b
Awesome
Am I trying to understand this too soon? Because I'm struggling with it.
As a chemistry "high schooler", I actually wonder if this is stuff that's usually taught at university or high school. I watched this because a teacher said "use the logarithmic mean for this because it's more precise" (related to heat exchange) and did not really explain why and what it is.
12:24 Surely it's easier to integrate y between ln(b) and ln(a)?
How do you do log mean of 3 or more numbers?
But why did you take such a long and circuitous path to the pink area, rather than just writing it down on one or two lines from geometry? If (x, h) marks the midpoint of a straight line between points (ln(a), h1) and (ln(b), h2), then it follows from linearity that h = (h1 + h2)/2. Therefore the area of the trapezoid under the line is h(ln(b)-ln(a)). If h = exp[(ln(a) + ln(b))/2] (since h lies on the curve where the tangent is drawn), then immediately h = exp(ln(ab)/2) = sqrt(ab).
5:32 Using L'Hopital ? why
This is reciprocal of derivative of ln(x) at x=a
so l'Hopital rule may produce circular reasoning
Because it's an indeterminate form. If you plug in x = a, you get 0/0, which is indeterminate.
But for two numbers a and b, we get GM(a, b) = GM(HM(a, b), AM(a, b)). We don't get the same for LM, i.e. LM(a, b) ≠ LM(GM(a, b), AM(a, b)). And that's kind of sad. I wonder what it'll converge to though.
If by "it" you mean taking arithmetic and geometric means repeatedly, the limit is, by definition, the arithmetic-geometric mean. This by construction has the property that AGM(AM(a,b), GM(a,b)) = AGM(a,b).
What about the Heronic Mean?
Isn't this the application of Hermite-Hadamard inequality for f = exp(x) from ln(a) to ln(b) ? THis is the simplest proof I've seen from this. Thanks!
why you do the stuff only with 2 elements and not n elements ? cause it don't tell me if it's generalize to n-sqrt(sum(a_i)) =< sum(a_i)/n :/
Very nice and elegent proof
But can we use it in inequalities ?
Yo Michael, benjamin7853 down there wondered if log mean can be extended to more than two operands. I realized it can if it's associative, but I tried to find that out in my head and it got gnarly. Good subject for a follow-up video?
These are some dank means
Take a look at problem 5 of IMO 2022
And Problem 1 of IMO 2023
Both are number theory problems :)
That's interesting but I wonder if that log mean has any practical use.
So, is there a factorial mean?
Well, AFAIK the Gamma function (which generalizes the factorials) is strictly convex on the positive reals, which means the derivative is strictly increasing and thus invertible. Let's call the inverse function of that derivative I. Then according to the mean value theorem, the function I((Gamma(x)-Gamma(y))/(x-y)) is in between x and y. I guess one could call that the factorial mean of x and y,
For the area of the magenta area, you could also calculate the integral of the line: \sqrt{ab} \int_{\ln(a)}^{\ln(b)} x + 1 - \ln(\sqrt{ab}) dx = \sqrt{ab} [x^2/2 + x(1-\ln(\sqrt{ab})]_{\ln(a)}^{\ln(b)} = \sqrt{ab} ( \ln(b)^2/2 + \ln(b) - 1/2 \ln(b)(\ln(a) + \ln(b)) - \ln(a)^2/2 - \ln(a) + 1/2 \ln(a)(\ln(b) + \ln(a)) = \sqrt{ab}(\ln(b) - \ln(a)
Michael is just being mean today...
At 1:36 you didn't subtract (a-b) ^2.
You don't need to, because he's using an inequality and he's only makes the top of the fraction bigger, as (a-b)^2 is greater than 0
@@casdinnissen6032 What's the factorial equality and unequality?
@@zakiabg845 Im not sure what you mean with the factorial function.
The inequality says that (ab)^(1/2)
@@casdinnissen6032 I'm not talking about factorial fonction now.
In the pre-amble: WLOG, take a>=b
[sqrt(a) - sqrt(b)]^2 >= 0.
Expand and you are done. Short and sweet.