HI

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Group Theory can be used to solve Rubik's cube, 15-puzzle and a lot of other permutation puzzles!!

Great video! Well-paced and understandable for beginners!

If we consider division as a stand alone operation, it is not associative. As an example, (4/6)/7 is not equal to 4/(6/7)

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For the exersize at 3:06 - A light hearted example of non associativity is rock-paper-scissors: (🪨 · 🗒️) · ✂️ → 🗒️ · ✂️ → ✂️ 🪨 · (🗒️ · ✂️) → 🪨 · ✂️ → 🪨

Amazing video! Had to subscribe after this one 😮😅

This lecture has a very pleasant form, with handwriting. Your enthusiasm are warm and promising of much more advanced math. I can actually comprehend the proof , in spite of my fragmented [youtube] education. Thank you very much. May this new Year the best Year [yet] of this Millennium, for You and for Everyone May I ask what software you use for the handwriting?

Excellent

This is exactly what i want more of on KZbin. What a delightful result. Thank you for publishing these videos. It really does feel like these sum = product identities are super useful for proofs along with chaining together inequalities. There is the R(s) = Σ(1/nˢ) = Π(1/(1 - 1/pˢ)) as a classic example of this. Also Πf(n) = e^[Σlog(f(x))] Id be really interested to explore more generalized inf sum = inf product identities like the one presented in this video in a number thory proofs context. Very fun stuff. Happy new year all!

Nice video, but I was a bit confused seeing a_j under the summation sign at 6:28 but nowhere else in the summation. I guess it means r summations over each of the exponents. I suppose one could use two summation signs with dots between them and each summation sign is for one of the exponents, with each summation going from 0 to K. Is that right?

Congratulations, great video 👍

Great proof. I found 2 proofs of the infinitude of primes, one when I was a sophomore in high school and another one around 5 years ago.

Genial la prueba y la explicación

What an amazing proof! Thanks for sharing.

Great video!!

Very nice proof. Thank you for the video.

There's close to a dozen problems in "Mathematical Methods for Physicists" by George B. Arfken where he points out the Euler-Mascheroni Constant.

The segment division part is confusing because i do not get the why. I think the clearest would be to have AG as a fraction of AB feom 0 to 100% AG(vector)= A(origin coords)+t*AB(vector with coordinates just differences of point coordinates) with t from 0 to 1 With coordinates it would be G=A+t(B-A)=A(1-t)+Bt so correlating to your notation t=n/(m+n). I see it is the same thing except with different parameterization when so there are some limite cases t=0<=>m=0 n=n G=A and when t=1 <=> n=0 m=m G=B

Well for the first problem I would take a different approach... As I see induction all cases up to and including P(K) are true and we want to show P(K+1) then P(k+1)=P(k-2)+$3 but this approach involves showing a larger base case which means P(8)=$5+$3 P(9)=3x$3 P(10)=2x$5 so for any k>=11 this will automatically imply P(K+1) by simply taking the arrangement for p(k-2) and adding a $3 on top. But I like your argument a lot. For second problem I like your inequality. I was thinking about the unit circle and inside it sin and cos are sides of a triangle with the other side 1 so by triangle inequality sin(x)+cos(x)>=1. For the induction step I would go on to see how the base case for k=2 pans out ... my idea remains linked to the unit circle. Let there be a triangle angled to centre a then |sin(a)| is the length of perpendicular to 0x with ends A and B . Draw the intersection of line angled a+b and the intersection to unit circle . The worst case is when a and b have the same sign. Then | cos(a+b) | is the projection from the last point drawn on unit circle down to 0x and | sin (a) | is a segment perpendicular to 0x, while |sin(b)| is a segment not perpendicular to 0x. You can see from drawing that the sum of segments algebraically does not account for them having different direction and will result in a longer line than the ordered one. Alternatively we can draw a projection from where the line sized | sinb | starts on the first triangle and its projection will have a length smaller than | sin a | , as similar triangles will be |sin a| |cos b|. The lines will form a right angled trapezoid with one base |sin (a+b) | the other |sin a| |cos b| and the slanted length |sin b|. by completing the rectangle we get leftover a right angle triangle of hypothenuse |sin b| which means |sin b|> |sin (a+b)|- |sin a| |cos b| |sin b| + |sin a| |cos b|>|sin(a+b)| We can use this so the thing to prove is | sin a| + |sin b| + |cos (a+b)| >= |sin a| |cos b|+|sin b| + | cos(a+b)| >= |sin (a+b)| + |cos(a+b) | >=1 from previous case I guess the key here is to show that |sin a1| + |sin a2| +...+ |sin an| >= | sin(a1+a2+...+an)| BC |sin a| >= | sin a| true consider true up until k checking for k+1 | sin(a1)|+|sin(a2)|+...+|sin(ak)| + | sin(ak+1)|>= | sin(a1+a2+...+ak)| + |sin(ak+1)|>= |sin(a1+...)||cos(ak+1)|+|sin(ak+1)||cos(a1+...)|>= |sin (a1+...)cos(ak+1)+sin(ak+1)cos(a1+...)|=| sin(a1+a2+...+ak+1)| I think there was something wrong with the direction of inequality in the video proof but will have to recheck video. Edit It is all cool Nice stuff with Fermat's theorem. I was just considering factoring out k+1 but that is a trap because you lose factors of p from the combinatorics which sucks

Merry Christmas, This looks like a typical problem. It is a bit similar to arcsin x + arccos x = Pi/2 I was thinking in a few simple ways since i am a simple man :) 1) using trigonomoetry if we do a change of variable and let u= tg(m) E(u)=Arctg(u)+arctg(1/u) = arctg(tg(m))+arctg(cos(m)/sin(m)) Now using a trig identity sin(x)=cos(Pi/2-x) and cos(x)=sin(pi/2-x) we will have the expression E(u)=m+arctg(sin(Pi/2-m)/cos(Pi/2-m))=m+arctg(tg(Pi/2-m))=m+Pi/2-m =Pi/2 2) using calculus let there be f(x)=arctg(x)+arctg(1/x) defined on R\{0} this function will be continuous and derivable on definition interval f'(x)=1/(1+x^2)+1/(1+1/x^2)*(-1)*1/x^2=1/(1+x^2)-1/(1+x^2)=0 on the whole interval as a consequence of Theorem of Lagrange the function must be constant so to find the value it takes just plug in a value for x f(1)=2*arctg(1)=Pi/2

Hello Tedszy, Nice video very much appreciated. When I saw the definition there were 2 things I thought about : 1) proving that the harmonic sum diverges using Theorem of Lagrange 2) it might just be the difference between the sum of rectangles of size 1/k * 1 when k=1..n and the area under the graph of 1/x from taken 1 to n I am somewhat fond of Theorem of Lagrange or theorem of finite growth as it is very useful in understanding calculus with the property that f'(x)>0 function is increasing as a consequence of this. Well theorem of Lagrange states that given f(x):[a,b]->I with f continuous on [a,b] and derivable on (a,b) then there will exist between a and b at least a point c where the tangent to the graph is parallel to the line passing trough points (a,f(a)) and (b,f(b)) or equivalent a<c<b such that f'(c)=(f(b)-f(a))/(b-a) Now looking at the harmonic series it would make sense to have f(x)=ln(x) then as an elementary function it satisfies Theorem of Lagrange for any interval [k,k+1] so for each such interval there will exist a point c such that f'(c)=(ln(k+1)-ln(k))/(k+1-k) or 1/c= ln(k+1)-ln(k) we also know that k<c<k+1 raise to power -1 1/k>1/c>1/(k+1) and substutitute indetity 1/k>ln(k+1)-ln(k)>1/(k+1) k=1 1> ln(2) -ln(1)> 1/2 k=2 1/2>ln(3)-ln(2)>1/3 ....... k=n 1/n>ln(n+1)-ln(n)>1/(n+1) adding up sum(1/i ; i=1..n)>ln(n+1)-ln(1) doing a limit as n goes to infinity shows that the harmonic series diverges When I was 11th grade learning this theorem and Th of Cauchy and applications there wasn't anything such as this in my school years that make you feel both smart and dumb at the same time. Smart because you see the graphical explanation and dumb because you can't see where you need to manipulate stuff to make the problems work

NIce video.

whats the proof for the lemma where 0≠r rational, s irrational then rs irrational?

Fermi principle evocation is a coup de theatre! bravo!

Aside from the Fermat numbers, can you find another infinite sequence of easily-generated numbers which are all coprime to each other? Like, comment, SUBSCRIBE!

Simply gorgeous...

Analysis is beautiful! Like, comment, SUBSCRIBE!

Woah, great voice 👍 You could be a radio anchor ⚓️

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you could also write (B^(x-A) / A^(x-B)) as (B/A)^x * (A^B/B^A) or further, (B/A)^x * A^(Δ+A) * B^(Δ-B) which I think is pretty elegant :)

Note that at 10:22 1/j! is strictly less than 1/2^(j-1) for j > 2, so on the next line a_n is _strictly_ less than 2 + 1/2 + 1/4 ... = 3. It's only a small point but it shows that a_n < 3, rather than a_n <= 3.