Problem-Solving Trick No One Taught You: RMS-AM-GM-HM Inequality
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- Опубліковано 24 лип 2024
- This inequality is famous in math competitions and in theoretical proofs. But why is it true? The video presents a great geometric visualization and proof for two variables. Pay attention--I'll use this inequality in an upcoming video!
Desmos.com link
www.desmos.com/calculator/6kb...
Link to proofs of generalized case
artofproblemsolving.com/wiki/...
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Hey this is Presh Talkwalker
Identity crises can be brief but hard-hitting...
Because somebody in the comments misheard his name last video, I think.
Nah, people have been mis-hearing his name for ages & ages. He's just getting slower.
Or something.
Incidentally, you misspelled his name. There's no "e" in it. It ends with "ar."
Fred
Did you figure this out?
Since "Walker" is a common English-language last name, I'm sure many people thought his name was "Preshtal Walker".
I learned this for the first time when I was about 40 yrs old. This geometric proof is just so elegant. Such a shame I never encountered it at school or university.
*Thank you for the math lesson.* I've gained much more knowledge by watching this than those debatable puzzles.
first answer
How to get root(ab): Use proportions. Let c be the length of the red segmant. Then a/c = c/b, which implies c^2 = ab, since the triangles are similar.
how do you know both triangles are congruent?
Seriously, how have I never heard of this proof.
because this proof only works for two terms, which isn't the most useful form
@@fernandowong5799 we can then apply induction to prove it for n number of terms.
i know it is pretty off topic but does anyone know of a good site to stream newly released movies online?
@Brodie Enrique Lately I have been using FlixZone. You can find it by googling :)
@Kenzo Rowen definitely, I have been using FlixZone for months myself :)
Fresh Tall Water. Got it.
You mean Fresh Saltwater
Presh tall walker
@Sai Sasank presh talwalkar l, his sirname is indian
@@sahilsagwekar no he lives in USA
@@sahilsagwekar r/whoosh
This is the best video you've made thus far. A cool visualization / geometric proof of a useful theorem. Nice job
It took me quite a while to work out where you got sqrt(ab) and xAM=GM^2. So you could go into those more explicitly
Similar triangles have sides of equal ratios
GM/x = AM/GM
Yes, so from similar triangles from his diagram:
h/a = b/h
==> h^2 = ab
==> h = sqrt(ab) = GM
@@anishkrishnan9698 Thanks a lot!
@@adamwho9801 Aaah that seems easier I took wrote pythogoras equations for three triangles and equated them, 4 steps but yeah works
You have done everything elegantly. Nice immaculate work ! 👍
I have algebraic proofs of the inequalities using proof by contradiction:
First, suppose that RMS
Nevan Lowe u may just suggest that it will be proven by contradiction and leave it to the readers
But he wanted to provide the solution too. It is his right. So next time let your suggestions at home please.
Jaroslav Ševčík
The joke: y=1-x^2
You: (0,0)
Jaroslav Ševčík he is obviously just joking. In most math Olympiad books there's always statements like this.
Nevan Lowe ...Thanks for sharing that with us, a math lesson in itself, and for taking all the time that must have been necessary to present it so well with the math symbol limitations of a computer keyboard. Good job!!
IMO, this is seriously one of your best videos ever. This one gets to show us that Geometry, I believe, is intrinsically linked to all of Mathematics, even with the most seemingly unrelated topic. What a beautiful proof, Presh. Thank you!
get it, useful in IMO 😂
My mind gets blown everytime when I watch your stuff. I can barely keep up.
The video description of all the means merging together when a=b is wonderful.
Really helpful. Thank you very much. I am a twelth grader and have never seen such an interesting proof of this inequality.
amazing ...ua-cam.com/video/hdgvuORnEiA/v-deo.html
Wow, very amazing and elegant! We can even see other things from the graph, for example that AM/RMS >= HM/GM.
Explanation: These ratios are the cosines of the top angles. And the top angles are arctan((b-a)/2GM) and arctan((b-a)/2AM) respectively. Since arctan is ascending and AM and GM are in the denominator, the right top angle is smaller than the left top angle. And since cosine is a descending function on the interval [0, pi/2], the ratio AM/RMS is larger (or equal) than RM/GM.
Really nice prove! I would like to see more of this format.
Thumps up👍
Most useful contribution of the last weeks. Thank You.
Thats awesome man! Nice video !!!!
Great one! Hope to see more videos like these in the future ^_^
Great idea to pepper your videos from time to time with this kind of educational content!
A video on why each of these means is useful would be nice. I'm not a mathematician and sure I can go look it up myself, but it would be much easier for me if you did it. Jokes aside, an in depth explanation of these various means would make a video I would definitely watch.
en.m.wikipedia.org/wiki/Mean
The best description regarding means ever!
Beautifully demonstrated. Love it!
amazing ...ua-cam.com/video/hdgvuORnEiA/v-deo.html
Enjoyed very much. Waiting for such type of videos.
A nice visualised video.
One of your best videos so far.
Simply superb! Super excellent!
Awesomely simple.
What a beauty! That was awesome! Thank Sir!
Alternatively: derive the general power mean and explain the intuitions behind them (with the sum fixed, the greater power has more impact when elements are more spread out)
What about Gauss's arithmetic, geometric mean? (Useful for elliptic integrals.)
I had to learn about the root mean square when I was reading Descartes' Geometry but I never learned this. Fascinating
helpful indeed, a lot better than complicated ways, my goodness, thank you for this super cool way. loved it
Très belle démonstration géométrique ! Bravo.
I was searching for such geometric approach for proving it today I am glad to watch this video, thanks!
amazing ...ua-cam.com/video/hdgvuORnEiA/v-deo.html
Thank you, I always poor in math, but your lesson truly raises me up. I am in a tremendous excitment of handle some of these difficulties. Thanks again!💕💕💕
Real math! Hooray!
I should be similar with all these series. Arithmetic Means is used in Statistics. Geometric means is used in calculation of population and compound interest. Wonderfully you have proveded some associations with other two.
Nice. I am beginning to be addicted to your math problems.
Gajab Bhai..I heard first time about root mean square
Excellent proof dear. Zordaar zabardast zindabaad
You, indisputably have the best technical/math presentation platform on the web. I would be filthy rich if I had a buck for every comment you have received begging you to disclose how you pulled this off?
The animation and capacity to explain and erase stuff clearly is world class.
Cool explanation. Thanks, amigo!
Amazing video, please do more of such proofs
Me a 36 years old failure in both professional and personal life loves your video try to solve questions, watch them many times. They are lifeline for me.
Hi Mr Presh thanks a lot for this beautiful explanation and believe me nobody can do better than you did . Perfect just perfect .
Very nice, such a meaningful demonstration..
Very nice demonstration!
Thanks, I have to visualize mathematics geometrically to really understand, so this was cool
Whenever he said 'mean' I heard 'meme'
Beautiful demonstration.
What an elegant set of proofs
Very nice description.
you can also use concavity of graphs to extend it to infinite positive no.s
The RMS is also used in the kinetic theory of gases in thermodynamics
This video deserves 1M+ likes, bcz it is really super awesome, and super cool idea .. No one ever told me this thing
Sir,it is very useful video for all mathematics learner's:-CHVSN as a INDIAN mathematician
Interesting, beautiful and useful. Thanks
+ Presh: At 3m50s: You can also quickly verify that this altitude is the GM by similar triangles, because a/h = h/b
And I really like your geometric demo of that chain of inequalities!!
BTW, you might mention that all these means are related by being "functional transforms" of the simple (arithmetic) mean. A "transformed mean," TM, using a monotonic function f, is:
TM( ֿx ) = f⁻¹(AM(f( ֿx )))
where ֿx = x[1...n]; AM(f( ֿx )) = (1/n)∑ᵢ₌₁ⁿ f( xᵢ )
So:
• when f(x) = x², f⁻¹(x) = √x, and TM = RMS
• when f(x) = x, f⁻¹(x) = x, and TM = AM
• when f(x) = ln(x), f⁻¹(x) = eˣ, and TM = GM
• when f(x) = 1/x, f⁻¹(x) = 1/x, and TM = HM
Neat, huh? ;-)
PS: I suspect that some property of each function - maybe something involving the second derivative - can be used to arrive at those inequalities, but I haven't delved into that.
Actually, looking at the list, I'm getting a very strong hunch . . .
Fred
Fred: your hunch is correct. This all relies on Jensen’s inequality applied generally to the functions x^a. This leads to the so-called power means, which generalize the four special cases mentioned here.
Nice proof, I have never seen such proof although I have known about the inequality
This video was interesting. Thank you
Presh you re just too awesome!!
+MindYourDecisions I would state the geometric mean - even for two numbers - as ab^(1/2). Differently from RMS, where the use of square and square root would not change with the number of terms, the power (or root) order in a geometric mean will change.
I would put the segment at 7:00 at the front and use the fractional power notation for the root: this way it's clear one is always _dividing_ something by the numerosity of the data, then state this for n=2 and only last change the power notation to a root, if you think it's more familiar/easier to understand for people when looking at right triangles.
Other than that, nice video and animation; thank you!
Awesome....thanks...
Our sir taught us
He used it in many qns
This is a really important and interesting inequality
6:37 That animation seems to use way to much CPU for his PC (volume up)
Nice explanation sir....thank u...
What program do ypu use to illustrate your problems???
A beautiful proof. Thanks
what would a and b have to be in order for the four values to be positive numbers?
Please make similar type of video on circumcentre, orthocentre, incentre,
Finally a respectable video from this channel! How have I never seen this geometric argument
Thanks very much..... I was able to prove only A.M, G.M and H.M
before you said you made it in desmos, I thought to myself I should make this in desmos lol
Very interesting! This is one of my favorites! Thanks
Your content is amazing
thanks for wonderful geometry and you
Thank-you!
You're a genius... how you can do such things like that in math...
That was beautiful bro..
I really wanted this video....
Thnx
Excellent.
Great..Well done
Beautiful champ. I liked it and have no criticism.
How about adding the contraharmonic mean to this? It's always greater than or equal to the RMS for any given data set...essentially it's the arithmetic mean of the squares of all the values divided by the arithmetic mean of the values. So for (3,4) it's equal to 3.571, which is greater than the RMS of 3.536. Could that be interpreted geometrically?
Where can I send you my doubt ?
i have the algebric proof without using contradiction and I knew this inequalities a long before
andthe inequalities only work if both a and b are positive
archana rathore works also for negative but the signs flip
What about the median and the mode? They, too, are also types of arithmetic means.
is the hypotenuse length right doe?
In the two-variable case, at least, it's easy to show algebraically equality holds if a=b. Just set the two variables equal, and simplify all four expressions to a.
Hello Mr presh . Hello everyone . When i went further ive found that the intersecting point of HM and RMS ALWAYS LAYS DOWN ON AN EYE SHAPE AND VERY BUTTOM POINT OF THE HM LAYS ON AN OVOID SHAPE ( Yes egg shape not an ellipse neither anoval ) and of course the midpoint of the segment GM LAYS ON AN ELLIPSE .
at 3:48 where do we get the sqrt ab from?
Due to the chord chord power theorem.
Nice geometric proof i knew just how to prove it by algebra
Bring more like this
Visual proofs of algebraic theorems are always great.
This is really very helpful.....thnku...sir😁😁😁😁😁
Fisrt time i understood "Fresh water" !
Sir,, if we know the diference of 'a' and 'b', and there is a problem like R.M.S. = x(A.M.), can we define 'x'??
Beautifull proff.
thanks you very much.
Frank K.
There is also a rather beautiful generalization: Let t be any real number, and for any POSITIVE x1, x2, ..., xn, define M[t](x1,x2,...,xn) = (Sum[(xk)^t, {k,1,n}])^(1/t). Then if t1 < t2, we have M[t1] ≤ M[t2], with equality iff all xk are equal. In particular, M[-1] = HM, M[0] = GM, M[1] = AM and M[2] = RMS, giving the result in the video. It is also interesting to note that Limit(t -> -Inf) M[t] = min{x1,x2,...,xn} and Limit(t -> +Inf) M[t] = max{x1,x2,...,xn}.
Super sum