Proofs by mathematical induction.

[@piotryjak5643]

It would be so great if you could make it easier to find previous parts of your courses. A playlist, or even an old-fashioned link in the description to a first video of you multi part content would be so helpful. Great content, love how fluent you are, just freaking badass, you mister are a rock star.

[@MichaelPennMath]

There are playlists, but maybe they are hard to find. I just hired a student today that is going to help me organize things!! So things should be improving.

[@tomatrix7525]

@@MichaelPennMath wow, sounds good! You’re channel is really becoming a gem

[@marienbad2]

Agree. I actually went through and made a playlist of some videos on here I wanted to watch and saved it so I could watch them all in order later. Was a pain to do though lol!

[@goodplacetostop2973]

@@MichaelPennMath Make sure this student knows where are the good places to stop.

[@buxeessingh2571]

I used to use that thumbnail in class. Very few admitted they got the joke. I want to emphasise the importance of remembering the base case in the induction step. Often, how you would perform your induction can be discovered from using the n = 1 case in a proof for n = 2. That is where you would use the base case in the induction step.

[@xcheese1]

A man deft in liquor production Runs stills of flawless construction. The alcohol boils Through magnetic coils. He says that it’s “proof by induction.”

[@heh2393]

Epic!

[@goodplacetostop2973]

23:34 Good Place To Sto-

[@yakov9ify]

I absolutely love proofs by induction and especially their variants such as the ones for graph theory, trees and ordinals. But my favorite has to be what I like to call Analytic Induction. It goes as follows: Let X be a connected topological space and P(x) is some property of every point x in X. Assume that there exists at least one element y in X such that P(y) is true. (base case). Also assume that if P(x) is true for some x then there exists an open neighborhood U of x in X such that P(z) is true for all z in U. Finally assume that if P(x_n) is true for some sequence of x_n's then P(z) is true for all limit points z of x_n. If you can show the above you have successfully proven that P(x) is true for all x in X. I love this because it gives me such a vivid image of what I am proving, the property P(x) spreading from point to point till it covers all of X. Absolutely beautiful.

[@sinecurve9999]

ElectroBOOM has entered the chat.

[@eleazaralmazan4089]

By far my favorite proof technique! #MathematicalInducation

[@tomatrix7525]

I just noticed the thumbnail is a reference to electrical induction

[@rockinroggenrola7277]

If I'm not mistaken, didn't you make another video about induction before?

[@MichaelPennMath]

You are right, but that was for a problem solving group that I was coaching and this is more tailored to a class that I am teaching...

[@MrRyanroberson1]

Interestingly, the angle sum rule doesn't require that the figure be convex, so long as it is euclidean (which is a bit of a circular definition, since euclidean space can be defined as that which obeys the angle sum for all polygons)

[@judysalazar0208]

Professor Penn, in the induction hypothesis we assume that there exists some natural number k such that p(k) implies p(k+1)?

[@lionking2192]

Try India's exam 'JEE ADVANCED' maths problems... U will find very good calculus problems out there!!

[@martinnyberg9295]

@MichaelPenn Please tell us more about all your pretty chalk! It looks very nostalgic, and soft and comfortable. I haven’t taught using chalk since the 1990s; all I get to use these days are stinkin’ whyteboard pens. 😁

[@MichaelPennMath]

I use Hagoromo chalk. It is great. Chalk talks are still quite common even at fancy international math conferences!!

[@tomasbeltran04050]

I left at 15:30. I'm writing down the examples.

[@AntoshaPushkin]

Isn't strong induction just a regular induction, but instead of P(n) we make a new predicate Q(n) which is Q(n) = for all k, k ≤ n => P(k) prove base and step for Q, and then we get Q(n) implies P(n) for all n?

[@michalbotor]

(3:55) all horses are the same color (all people are the same height)

[@sapientum8]

excellent content

[@Hobbit183]

I like to watch these videos even if i understand very little ;) its meditating

[@marcozarantonello2180]

This is a great video

[@poi_aithhkunnnRVC]

23:34 broken "good place to stop" first

[@prathmeshraut1616]

These is my first Sum of Exercise

[@gaborendredi8161]

In high school my math teacher once made an induction proof like this. He proved for n=2 as base case. Then he showed that P(n) => P(n^2) and P(n) => P(n-1). He claimed, that this way was still more easy than to show the usual P(n) => P(n+1). Unfortunately I don’t remember the statement he proved this way.

[@fullfungo]

IIRC it’s AM-GM

[@nathanisbored]

I like to imagine induction as like an infinite row of dominoes. For it to work, the dominoes need to be evenly spaced (which is why n and n+1 should be integers). Proving the induction step is like setting up the dominoes so that they are aligned, and showing the base case is like knocking over the first domino (though you can really do these in either order).

[@travisnell6849]

This really breaks down when we move to transfinite induction. You can do induction on a much broader class. In this context it's more important to think that if the property fails, it must have a first time it fails. The induction step(s) in this broader context are to make sure that there cannot be a first time it fails.

[@blazedinfernape886]

I think it as a staircase where you can only see the step you are on but you know that you can get on the next step. Now if both of these statements are true then you can climb up the staircase. If one of them or none of them are true then you can't climb up the staircase.

[@travisnell6849]

@@blazedinfernape886 These analogies break down immediately when one does induction on a well-founded partial order, rather than a well ordering.

[@prithujsarkar2010]

Awesome

[@tylershepard4269]

Hey now be careful when throwing shade at us Electrical Engineers. You are on the internet after all...

[@ozjapie]

principal? principle

[@martinnyberg9295]

The principal tool is the principle of mathematical induction. 😏

[@peaceandknowledge3668]

Hi!

[@mathunt1130]

You haven't proven the principle of induction is valid. You can find this proof in the book, Mathematical Analysis by K.G. Binmore.

[@Loots1]

heuristics and philosophy are not the same thing

[@aamierulharith5294]

"One is even" hmmmmm press x to doubt

[@thephysicistcuber175]

L.

[@f5673-t1h]

23:34 is a good place to stop

[@malawigw]

Proof by induction shows that one is even

[@davidbrisbane7206]

Hello World

[@CesarMaglione]

¡7! great! ;) Well done! ¿Do you sleep at any moment? :P

[@michalbotor]

honestly? i hate proofs by induction. the only proofs that i hate more are proofs by contradiction (as these are like: proof: "therefore this exists", me: "ok. but how does it look like?" proof: "i dunno. but this exists!") and proofs involving axiom of choice (as these tend to produce hairy monsters, like vitali sets). and my quarrel with proofs by inductions is that they can be dangerous, if you mess up the base case, and even if they work they leave you completely in the dark as to what that you proved actually means. i prefer geometric/visual, combinatorial or probabilistic proofs instead, if possible. like with your 1 + 3 + 5 + ... = n^2 claim. it has a very natural visual proof, like so: .. O X O X O .. O X O X X .. O X O O O .. O X X X X .. O O O O O .. .. .. .. .. ..

[@arch3866]

I too always would prefer a visual proof, but I think proof by induction is also pretty neat since it seems to have a pretty simple process. I think, (depending on the p(n), I'm not sure how hard some p(n) can get to evaluate ) it'd be pretty hard to mess up. not sure about this tho also, the thought process hopefully should not be that difficult to understand, as it basically goes like this, as Michael explained: 1) ok so this works for n=1 2) let us say it works for some integer k>=1 3) oooh look it works for k+1 4) so that means it works for 1, 1+1=2, 3, 4 ... which is the set of natural numbers!! boom finished. not sure tho, I haven't done many proofs so I don't know how common/rare a specific type of proof is in olympiads, but I'd thinked that completely visual proof is rare. thanks for reading my ted talk

[@michalbotor]

@@arch3866 here's a "proof": let a be a real number different than zero, then a^n = 1 for every natural number n. proof by induction: base case: a^0 = 1 by definition. inductive step: lets assume that a^n =1 for all natural k