why is the power mean so important?

[@barutjeh]

The derivative at the end is wrong. Can't use the derivative of an exponential if the base is variable. Have to write it as exp(ln( )), then use the chain rule. Also, the exponent is 1/x, not x, resulting in more complications.

[@skylardeslypere9909]

Yeah that's what I thought as well. A little trick to differentiate f(x)^g(x). First consider the base f(x) to be a number and differentiate, then consider the exponent g(x) to be a number and differentiate. Add the two, you get your derivative.

[@deinauge7894]

yep. so g'(x) = (-log((1+c^x)/2)/x^2 + 2log(c)c^x/(x(1+c^x))) * g(x) now the first half is positive, but the second half is not... makes thing complicated i guess.

[@deinauge7894]

aaghh even the first part is positive only when x>0. Another error that's also made in the video at 17:49 ... the argument of the log etc 😅

[@r.maelstrom4810]

He should have used Jensen's Inequality for ((a^p + b^p)/2)^(1/p) < ((a^q + b^q)/2)^(1/q), when 0

[@stewartcopeland4950]

@@deinauge7894 no factor 2 on the second term

[@FranzBiscuit]

That the means could even be unified at all is remarkable in and of itself. Thanks for bringing to light such an interesting mathematical object. Kudos!

[@Demo-critus]

Very interesting! Note that for p=2 you get the root-mean-square (RMS) value (used in science quite a bit). We know that RMS is more biased towards larger values than the average, and this now makes sense. The greater p, the more biased towards larger values, with p to infinity resulting in only the largest value (max). Cool!

[@micayahritchie7158]

-1 is also interesting

[@skylardeslypere9909]

Am I tripping or is the derivative of g(x) around 17:00 totally wrong? The derivative of f(x)^g(x) is equal to g(x)*f(x)^{g(x)-1}*f'(x) + ln(f(x))*f(x)^g(x)*g'(x). Also plotting everything in Desmos shows that the derivative is wrong.

[@AntoshaPushkin]

At 17:53 I don't think ln((1+c^x)/2) is necessarily negative, for example, c = 1/2, x = -1, we get ln((1+(1/2)^(-1))/2) = ln(3/2) > 0 Also a previous step looks wrong, the derivative of g should be something different

[@richardjacobson3124]

Please do a video on Jensen's Inequality! It is the mother of all mean inequalities and right up your alley

[@brhayashi]

I encountered the power mean with p = 1/2 in the wild for the first time this week, so this video is very topical! It turns out that the p = 1/2 power mean on two positive numbers is the arithmentic mean of the arithmentic mean of the two numbers and the geometric mean of the two numbers.

[@plislegalineu3005]

You mean... AM( AM(a,b) , GM(a,b) ) ??

[@brhayashi]

@plislegalineu3005 Yes, M_{1/2}(a, b) follows the formula you gave. When I encountered that form for a circuit I was researching, I checked it against the p = 1/2 power mean and was surprised to see the identity appear. (Something similar happens for the Stolarsky means.) This seems to be an exceptional case for the power mean on two positive inputs. I didn't see anything elegant appear relating other power means to each other for other non-integer, rational numbers p.

[@m_stifeev]

Finally, we have "it's a good place to stop" at the end of the video. I missed it.

[@Bodyknock]

13:00 The power mean being increasing over p also proves all these averages are between the maximum and minimum values using the prior lemma that the limit as p goes to infinity of m is the maximum and as p goes to negative infinity is the minimum. (I.e. You could add Minimum and Maximum at both ends of that Pythagorean Inequality diagram.)

[@txikitofandango]

You also showed that all these means fall between the max and min of the values

[@xinpingdonohoe3978]

This may help people see why the supremum norm ||•||∞ is defined the way it is, because it's the limit as p→∞ of the Lp norm ||•||p.

[@r.maelstrom4810]

All the last part of the video is wrong (though the conclusion is right). Instead, you should have used Jensen's Inequality for ((a^p + b^p)/2)^(1/p) < ((a^q + b^q)/2)^(1/q), when 0

[@Alan-zf2tt]

I doubted Michael's claim as variable x was invoked without being introduced so ... @ 14:12 I exercised due diligence and plopped f(x) into a spreadsheet and tried various values for a and b and x taking positive and negative values on x @ 14:50 slapped forehead as the wlog assumption is far easier to put into a spreadsheet @15:52 I deserved another slap to forehead and resolved never to be impatient again (it did not last very long tho). g(x) is a dream to enter into a spreadsheet. Distracted by thoughts of a,b and x make for 3 variable and neat trick of tieing these to 2 variables by creating a ratio c on wlog b/a where a>b or a=b Pondered what would happen were x a complex number but variables c and x seem somewhat nicer and more pleasant to deal with analytically and spreadsheetally Impatience returned soon followed by frustration. Resolved to wait and see if Michael does a follow up on complex variables ... And yes, trials suggest original f is non-decreasing but I obtained a funny result at a=-4 b=-3 x=-1 but that may be my spreadsheet. If time allows I will look at it again and spend more time with it. EDIT: entered spreadsheet formula based on g(x) and did encounter problems I used x values -30 -2 -1 0 1 2 30 and of course experienced DIV by zero errors but that was expected for sure

[@AlecBrady]

Also for p = 2 we get the engineer's (and statistician's) old friends, the root mean square and the standard deviation. Are there any known applications of any higher values of p?

[@Adraria8]

Video on the generalized f mean next????

[@wanfuse]

Funny I was just working on a problem last week and serviced a similar technique but using different techniques, no idea if the results I got actually work, but in principle it’s similar, but instead of geometric , arithmetic, … I was deriving actual values at points on the distribution.

[@Alan-zf2tt]

Interim conclusion? to the inquiring mind everything is a learning experience Corollary: the non-inquiring mind is none the wiser

[@Nnbvcxfz]

Do a video on the hölder ineq pls

[@maxhagenauer24]

Wait so when p = 1 then it becomes the arithmetic mean byt when p = 0, it becomes the natural log if the geometry mean?

[@Xeroxias]

No, just the geometric mean. He showed that ln(m_0) = ln(GM).

[@maxhagenauer24]

@Xeroxias Oh yes my bad, you are correct.

[@skillhunter4804]

Isn't the last derivative inomplete?

[@deinauge7894]

and the argument at 17:49 is also only true for positive x. But this is of no relevance then ...😂

[@emanuellandeholm5657]

I'm thinking: What kind of means might we rediscover if we consider multivariate convex mean functions?

[@vuinshhsniuv]

Is there some (irrational? ) value of p for which Mp = AGM (arithmetic-geometric mean) ?

[@magicalgirlgleamingmoonlight]

m_p is continuous on p in (0,1], and m_1 >= AGM >= m_0, so by the intermediate value theorem there is a value of p in (0,1] s.t m_p = AGM as far as its value, i have no idea if its rational or not my gut says irrational, and probably only definable through the IVT but notably, a_1 in the definition of the AGM is m_(1/2)

[@chrisglosser7318]

Limit p-> \infty ~ max

[@marc-andredesrosiers523]

It would have been nice to explore probability implications. 🙂

[@Alan-zf2tt]

Probably 🙂

[@richardsandmeyer4431]

Nice, interesting generalization. Also, considering negative values for p, we get: for p = -1, the harmonic mean for p --> - infinity, the minimum.

[@Lucashallal]

kzbin.info/www/bejne/aaOUpWyGm5Z5nM0si=ZVsjXmXVk9J82H65 Does it relate with the mean you showed in this video?

[@dang-x3n0t1ct]

I think he showed 3 different generalisations of means already.

[@PATRICKZWIETERING]

Ok, so if f(x)=-1/x, then f'(x)=1/x^2>0, which means that f(x) is an increasing function on the whole set of real numbers?? This is a bit disappointing for a math. professor...😔

[@noicemaster5173]

:O

[@knivesoutcatchdamouse2137]

pretty sloppy ngl

[@一只没有名字的羊]

this video is so mean