No one else has been able to explain the relationship between string theory’s 10- and 26- dimension affinities! Only you! THANK YOU!

Thank you for your channel and videos!!!

The missile knows where it is at all times. It knows this because it knows where it isn't. By subtracting where it is from where it isn't, or where it isn't from where it is (whichever is greater), it obtains a difference, or deviation. The guidance subsystem uses deviations to generate corrective commands to drive the missile from a position where it is to a position where it isn't, and arriving at a position where it wasn't, it now is. Consequently, the position where it is, is now the position that it wasn't, and it follows that the position that it was, is now the position that it isn't. In the event that the position that it is in is not the position that it wasn't, the system has acquired a variation, the variation being the difference between where the missile is, and where it wasn't. If variation is considered to be a significant factor, it too may be corrected by the GEA. However, the missile must also know where it was. The missile guidance computer scenario works as follows. Because a variation has modified some of the information the missile has obtained, it is not sure just where it is. However, it is sure where it isn't, within reason, and it knows where it was. It now subtracts where it should be from where it wasn't, or vice-versa, and by differentiating this from the algebraic sum of where it shouldn't be, and where it was, it is able to obtain the deviation and its variation, which is called error.

Is anyone preparing for imat for 2025????

Regards from india🇮🇳

@2:00 it is not a "joke played by the universe". It is a consequence of general relativity _with_ non-trivial local topology (so non-classical GR). The 8 and 24 arise naturally from the 4D Dirac spinors, via the Clifford algebra. You will not so easily see this if you use the matrix algebra rep. There are 8 disjoint 24-cells in the Standard Model of particle physics. An automorphism group of the whole lot is CPTt symmetry: CPT plus a triality "reflection".

9:00 Defining definability turns out to be a quite tricky problem.

I love your explanation of terms. I am currently in 7th grade and find your videos very interesting and easy to understand despite the wide range and difficulty of the information you convey to your audience.

Professor, there are a few luminaries for whom it is unequivocally true their candles were extinguished too soon. Terribly and sadly too soon. Among the short list imho would be Frank P. Ramsey, Srinivasa Ramanujan, Evariste Galois, and of course the potential Gauss successor, Gotthold Eisenstein. He had such a tremendously uncanny intuition for a man of his age. That we utilize Eisenstein's algorithm largely without credit while teaching our first year linear algebra students hand matrix multiplication is a shame

When you realize that he’s actually calculating these results as he’s teaching… the brain on this guy is incredible

what do the rectangle symmetry groups have to do with Z/2Z*Z/2Z?

23:24 How is this true? The bottom one should contain a much bigger set of primes since n doesn't need to be a power of 2.

Would you say infinity, 0 or i are prime numbers?🤔

i have a doubt at 20:58 how does sum of probabilities being infinite suggest that there are infinite mersenne primes? Also the probability 1/log(2n-1) is of any number being prime till 2n-1?

2:48 cos in complex analysis vs real number of Unit circle

Thank you, Richard! I have to admit, I discovered you through 3Blue1Brown, but your videos have completely captivated me. I'm even grateful for the technology that makes this possible-listening to a speech by one of the top mathematicians is truly a gift for math enthusiasts worldwide. Keep up the fantastic work!

gm. The comments at the end are extremely helpful

gm

around 4:40 at the end of the exact sequence, shouldn't we have A tensor B rather than just A?

16:10 - 22:01 is great discussion of Lang’s algebra, chapter I problem 41 (though with a very slightly different proof)

"If you go as far as "ubiquity of go"", I mean, that's only one letter away...

Thanks for the video ! Did someone get why the injection from Z to Q is an epimorphism ?

Wonderful!!!

5:19 here if such n does not exist, then the map can be extended freely, namely you can choose any element in I as image of b, without resulting in any contradiction.

Thank You, Professor.

Construction with a ruler and compass (2 instruments) corresponds to some Galois Group of order 2^n. The problem of trisecting an angle (in the general case) corresponds to a Galois Group of order 3. Does this mean that generally a angle should always be trisectable using 3 instruments?

"... and probably won't make a lot of sense unless you've seen most of the earlier talks. In fact, frankly, it may well not make much sense even if you have seen the other talks" lol love it

It's just great! With new claims from Google, revisiting this short talk feels really good

30:35 very Fermatian

why you choose those numbers for cover page of your channel ? why you select this diagram as your channel logo ?

The section about crackpots constantly sending proofs of Riemann hypothesis is the best.

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This information is very helpful and valuable. Thank you!

Physics have more math than math in pg level

Thanks for doing this!

This is the highest production video of yours I’ve seen so far, very nice

You are trying the Birch Swinnerton or RH I am sure. Had it been not RH, you might not have kept it a secret.

Thank you so much for your content, you have an incredibily captivating way of presenting things

4π^2 All prime products are 4π^2= [4π^2]^(8/8)= ⇔ [4π^2]^8 × [4π^2]^(1/8) E8 ÷ E8=1 (E8)^24 ×(E8)^(1/24) 3×8=24 The 3rd generation of quark・lepton And it gives a hint to the ABC problem❗️

That was awesome proof and very easy to understand thanks a lot

If 2x+1 is not divisible by three then 2x+1 is a prime number

Great lecture, thank you.

Do you have an explanation of the key claim at 14:10 - that the string of p elements can’t repeat in <p rotations.

e^π +ie^πi +je^πj +ke^πk +le^πl =MC ^2 e^πi-1=0 jkl=0 Quarternion Octonion Principle of the constancy of the speed of light Law of conservation of energy Law of conservation of momentum ζ(s),η(s),Γ(s) The infinite sum of natural numbers is ∞, -1/12 Differential calculus, integral calculus

π= 3.141592653589793…, π≒√2+√3= 3.146264369941972…, This gives some hints. SU(2) → √1√2 U(1) → √1 Next, if we write it in this form √1√[√2+√3]√4 → 2√1√[√2+√3]→2√π √2+√3≡√5 If there are some special spaces where this relationship holds, √2+√3=√5 √5= √2+√3 √7=2×√2+√3=√1+2×√3 √11=√1+2×√5=2×√2+√7 √13=2×√5+√3 √17=√13+2×√2 √19=2×√7+√5=√9+√10 ...=... -∞<...<-4<-3<-2<-1<0<1<2<3<4<5<7<8<9<10<11<12<13<14<15<16<17<18<19<...<+∞ These are true in a space with many one-dimensional lines, which at first glance appear to form triangles and intersect at three-dimensional angles. Each line has an origin 0, and the distance from the origin 0 is 1 unit. The axis of the one-dimensional real part that constitutes the Cartesian coordinate system and the oblique coordinate system has an origin 0 and a unit 1. When you look up at the night sky, the stars shine like the moon, galaxies, and constellations in the sky, and numbers are scattered like shooting stars that suddenly shine brightly.

Can't believe you slandered potatoes. Shameful.

6:40 all groups of order 6 are solvable, but not simple, so I assume it's S₃ 8:00 the eigenvalue gives the left column, not the bottom row

solved. P=peneka. NP=nau-peneka. next...

Absolute life saver!!

7:10 In the example of the construction of "a" splitting field of p(x) = x^3-2 over Q, the field of rational numbers, we factor p(x) = x^3-2 = (x-2^{1/3})(x^2 + 2^{1/3} x + 2^{2/3}) = p_1(x)p_2(x), so that p_2(x) is just irreducible over the field L = Q(2^{1/3}) not over the Q. this irreduciblity over L = Q(2^{1/3}) leads us to the new field M := L [x]/p_2(x) 8:12. 13:33 In the general construction of the fields, i.e., the proof of the existence theorem of the splittiing field of any polynomial p(x) over the field K 13:33, we factor p(x) = p_1(x)p_2(x) .... so that each p_i(x) is irreducible over K but ... as in the example 7:10 should we assume that each p_i(x) is irreducible over K_i, not just K, where K_i is defined by the recursive manner K_{i+1} = K_i[x]/(p_i(x)), and each K_i would be a field if we assumed that each p_i(x) was irreducible over K_i, not K. Or I am not sure but is each p_i(x) automatically irreducible over K_i, if we assume that each p_i(x) irreducible over K???