That is a very nice proof of AM-GM worth remembering.
@thedoublehelix5661
3 жыл бұрын
Totally agree
@aashsyed1277
3 жыл бұрын
@@thedoublehelix5661 I LIKE YOU BECAUSE OF THAT
@thedoublehelix5661
3 жыл бұрын
@@aashsyed1277 w- what?
@aashsyed1277
3 жыл бұрын
@@thedoublehelix5661 GOOD
@yoav613
3 жыл бұрын
I saw the proof forAMGM with regular induction and this forward backward induction is great !
@ProfOmarMath
3 жыл бұрын
Isn’t it? So much nicer!
@bastiana.n.4277
3 жыл бұрын
Wonderful! This makes me think that there has to be an even more powerful formulation of induction that enables to deduce other formulations (like the one you presented on the video) that might make solving a particular problem easier
@ProfOmarMath
3 жыл бұрын
There are several. We will see more in this series!
@bastiana.n.4277
3 жыл бұрын
@@ProfOmarMath Great, I'll wait patiently
@thedoublehelix5661
3 жыл бұрын
The backwards part did involve some pretty slick moves. But it makes sense to choose the average of the previous numbers as the nth term
@ProfOmarMath
3 жыл бұрын
Ya after filming I realized it wasn’t as straightforward as I had in my head
@SimchaWaldman
Жыл бұрын
WOW! Real art of induction proofs!
@hrs7305
3 жыл бұрын
Nice ! Looking forward to more videos on induction
@ProfOmarMath
3 жыл бұрын
The next one is fun!
@coldsoup49
3 жыл бұрын
How do you always find such cool topics??? I am in awe
@ProfOmarMath
3 жыл бұрын
Thanks coldsoup! Decades of experience! More to come 😍
@shouligatv
3 жыл бұрын
Awesome ! One thing I'd really love to see is a proof of how all the variations of classical induction are logically equivalent
@ProfOmarMath
3 жыл бұрын
You can actually deduce some on your own! For example the 2n could have been replaced with 3n and things would still work. It’s kind of a directed graph problem with vertices being the natural numbers
@georgedoran9299
3 жыл бұрын
@@ProfOmarMath I’d imagine that there’s an infinity of possible induction techniques, you just need to ensure you cover all the cases.
@physicsnovice656
3 жыл бұрын
I remember when my calculus professor showed us this example during exercise sessions in 2010. He mentioned that he knew about it from Knuth's book Concrete Mathematics that I bought later, but I didn't go over it 100%.
@ProfOmarMath
3 жыл бұрын
Cool you get to revisit it!
@goodplacetostop2973
3 жыл бұрын
As the saying almost goes, take a big jump forward to take a step back. 😝
@ProfOmarMath
3 жыл бұрын
This made me snort!!
@mathsandsciencechannel
3 жыл бұрын
Well done sir. Love the video. Thanks
@ProfOmarMath
3 жыл бұрын
Thanks!
@aashsyed1277
3 жыл бұрын
@@ProfOmarMath SAME HERE!
@danelrosen5461
3 жыл бұрын
Awesome proof! New subscriber :D
@ProfOmarMath
3 жыл бұрын
Thanks Santiago! Enjoy the channel 😍
@advaithkumar5966
3 жыл бұрын
could you make a video on how to use this technique on the problem 'IMO Shortlist 2016 A1'? Thanks again
@joseluishablutzelaceijas928
3 жыл бұрын
This was REALLY cool! Thank you very much!
@ProfOmarMath
3 жыл бұрын
Definitely!
@aashsyed1277
3 жыл бұрын
I like you and Micheal penn
@ghimpugelu5375
3 жыл бұрын
can someone explain why choosing an= (a1+a2+...+an-1)/(n-1) does not lose generality when proving p(n) -> p(n-1)?
@bigjazbo9217
Жыл бұрын
I thought the same, but consider this: The AM-GM inequality for n numbers--where n is a power of 2--is proven true for any set of n positive numbers by forward induction. This is true for an arbitrary set of n positive numbers, so it will still be true even if one of those n numbers was specially constructed. Omar then demonstrated that for a special set of n numbers (n-1 arbitrary numbers plus one special number), the subset of n-1 arbitrary numbers also satisfies AM-GM. That means that if given any set of numbers where the number of elements is one less than a power of 2, we can add a special element to construct a special set of numbers of cardinality that is a power of 2, which we already proved by forward induction satisfies AM-GM, and which also implies that our given set of arbitrary numbers satisfies AM-GM. Now we know that all sets of numbers having cardinality one less than a power of 2 satisfies AM-GM. We can repeat the backwards induction process as far as necessary until we have proven AM-GM for all cardinalities between n and the next lower power of 2.
@comuniunecuosho-campulbudi7611
9 ай бұрын
p(2ᵏ) true ⇔p(n) true, n=2ᵏ for any x₁,...,xₙ ⇒p(n) true even after we change xₙ (to a specially constructed one as we like) ⇒p(n-1) true so indeed p(n)⇒p(n-1)
@aashsyed1277
3 жыл бұрын
I love this channel
@kanewilliams1653
3 жыл бұрын
What software are you using Prof Omar? Love your red highlights that fade away.
@ProfOmarMath
3 жыл бұрын
Thanks Kane. I’m using GoodNotes on an iPad Pro!
@antormosabbir4750
3 жыл бұрын
you can try bdmo 2020 higher secondary category's one of the problem. it needs this idea
@ProfOmarMath
3 жыл бұрын
Neat 😀
@Mrpallekuling
Жыл бұрын
This proof was discovered by the great Cauchy (1789-1857)
@yassinezaoui4555
3 жыл бұрын
Nice approach 8)
@ProfOmarMath
3 жыл бұрын
Thanks!
@yoavshati
3 жыл бұрын
You should prove by induction that proving by forward-backward induction works
@ProfOmarMath
3 жыл бұрын
😅
@chandankar5032
2 жыл бұрын
Ah, the idea was to when one comes up with AM GM, it's easy to observe with assuming a+b/2 geq sqrt(ab) . We can do 4,6,,8... As any even number will be divided into to parts and base case can be applied. Similarly the other part, n positive numbers having am geq gm, well among thus n numbers, if we assume one is am of other n-1 s then we readily prove that for n-1 numbers. This the two statements. After the student observe these two, then you should tell them, the formal statement unless it's just checking and verifying. Another fun instance, say if I want to show identify permutation is even. Say if e has r cycles, it can be shown that it's essentially r-2 cycles, then r-4 cycles... I. E. P(r) implies P(r-2). So the crux of the matter is, there is a set at most countable, then by induction is just a way of saying one element generates other property s.t. all elements are covered. That property itself is equivalent to well ordering principle. Anyway, I'm glad I figure this chnanel (from Michael Penn)
@robertveith6383
4 ай бұрын
(a + b)/2
@sandorszabo2470
3 жыл бұрын
Clever.
@ProfOmarMath
3 жыл бұрын
It’s fun!
@petrotkach8757
3 жыл бұрын
Forward step can be any, right? ( For example, P(5k) )
@ProfOmarMath
3 жыл бұрын
Yup!
@tomkerruish2982
3 жыл бұрын
Will you be covering transfinite induction?
@ProfOmarMath
3 жыл бұрын
I’m unsure yet because it might involve a long discussion of ordinals
@tomkerruish2982
3 жыл бұрын
@@ProfOmarMath You could do it for sets in general and only rely on the axiom of foundation, like how Conway did for his "one-line" proofs for the surreals. Namely, that if a property P of sets is such that, for any set x, if P holds for every element of x implies that P holds for x itself, then P holds for every set. As I write this, it occurs to me that it might be more appropriate for a course on set theory. I find it interesting, though, that there is no need for a base case, as P would necessarily hold for the empty set. Thank you for your response. I appreciate it.
@aashsyed1277
3 жыл бұрын
Nyc
@advaithkumar5966
3 жыл бұрын
is this method also known as 'cauchy induction'?
@ProfOmarMath
3 жыл бұрын
Yes!
@advaithkumar5966
3 жыл бұрын
@@ProfOmarMath thanks for your response! Also, could you make a video on how to use this technique on the problem 'IMO Shortlist 2016 A1'?
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