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  @AdarshKumar-wy8zw7 сағат бұрын

  Thanks bro for the explanation

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  @UzziWallendorfКүн бұрын

  F(a) b c d 0¹ 0² 0³ 0⁴ 1² 1³ 1⁴ ->0

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  @UzziWallendorfКүн бұрын

  X+1/×+6-×+6/×+1+×+1/×+6/×+1/×+3,׳+1+/׳->x⁶/×¹+×⁶+×¹-×¹/×⁶->׳,×¹->ײ->0

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  @UzziWallendorfКүн бұрын

  Fx² x⁰ ft² ft² dt 1990 ->f(×)²+f(t²l dþ/×+1990/1990ײ

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  @UzziWallendorfКүн бұрын

  4/2 /in9-×/in9-×+inx+3=x²->in+x⁶in⁶x ->inx⁹ in ⁹ x ->inx³ inx² in ² ->2

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  @UzziWallendorfКүн бұрын

  B²/a² ex-¹i/a² b² x²

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  @UzziWallendorfКүн бұрын

  A b /0 b ² x² a² b² ->e-i² ->x² /y²->e-×¹/x²->a²/b²×²->0-

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  @UzziWallendorfКүн бұрын

  L m =2pie/⁰ cosx cos²x cosmx d. M 1 m<10;lm 0 '-[cosx¹ cos2x+¹->vos³x cosmx² dx ->¼

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  @UzziWallendorfКүн бұрын

  ¹ ⁰c-1-y 1/y+1 +1/y+2+1+y/3+.+1/1992->1/y+×¹+×+1->y¹+y²+y³->1192+×¹->1993->1993-1992->y¹->1

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  @shipraarora5389Күн бұрын

  You are saviour 🤝

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  @user-fd9id9gl9s7 күн бұрын

  Whats the source?

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  @chrisyang042313 күн бұрын

  Finding the normal covering spaces is relatively easy, since we know that every normal subgroup of a group G is the kernel of some homomorphism from G to another group. So we can find the normal subgroups of index 3 by finding surjective homomorphisms from <a, b> to Z/3. By the universal property of free groups, specifying where the generators a and b go uniquely specifies a homomorphism <a, b> to Z/3. Then play around to get all surjective homomorphisms (there are 8), and identify the ones that give the same kernel (reducing the number of normal subgroups to 4). So there are 4 normal covers. This doesn't work with the non-normal covering spaces though. :(

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  @kingminato521914 күн бұрын

  Very good.

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  @robloxiangojocuzwhynot23 күн бұрын

  2nd lol

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  @LetsSolveMathProblemsАй бұрын

  Correction: For the conditional expectation at 4:35, to use the correct definition of a martingale, we should condition on all the preceding terms (so (q/p)^(S_0), ..., (q/p)^(S_(n-1))), not just the immediately preceding term. Despite this, the same argument goes through since X_n is independent of all such previous terms.

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  @replicaacliperАй бұрын

  I think at 7:00 its also crucial that the probability of ruin/winning is nonvanishing at T increases, i.e. you can put a lower bound on the probability that is independent of T. If the probability of escaping decreased quick enough with T then I'm sure you can have a nonzero probability of having an infinite sequence

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  @LetsSolveMathProblemsАй бұрын

  I'm guessing you meant T as in just the generic time step (not stopping time), and in that case, you are right that it is crucial that the probability of ruin/winning is bounded away from 0 at any time step (which is true here since the entire "game board" is finite).

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  @replicaacliperАй бұрын

  @@LetsSolveMathProblems indeed

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  @ccolombeАй бұрын

  Love your videos man! Glad to see another in the feed

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  @joeblogg9937Ай бұрын

  There is a limit as r->1. 2/5 ?

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  @LetsSolveMathProblemsАй бұрын

  Yes, that's correct! In the electric network argument, we do not even have to take the limit (and try to justify the limit we get is actual probability at r = 1): Just plug in r = 1 into expressions for x and y (see 21:55).

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  @circe3976Ай бұрын

  It was very enjoyable to watch, thank you.

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  @sajidrizvi4665Ай бұрын

  I'm sure I'm gonna enjoy this one!

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  @vatsalchaturvedi6602Ай бұрын

  Wow thanks a lot it really helped me

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  @EpicMethGamingАй бұрын

  when amc 8

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  @xdgolden4847Ай бұрын

  extremely helpful

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  @subasinipati9668Ай бұрын

  In India also it's not taught but it's in competition

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  @user-mv5sf1bv8cАй бұрын

  3-1 = 1 wow I learned something new

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  @vidyas.4531Ай бұрын

  haha I forgot that this was the amc 12 and that you could use trig, so I didn't think of using that triangle XD

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  @cameronspalding97922 ай бұрын

  @5:17 use angle addition identities

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  @karolkurek92012 ай бұрын

  I do not understand one thing: how |r|<1 is satisfied by e^(-x) if e^0=1 and this is not <1. Integral goes from 0, so r=1 occurs.

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  @booli85422 ай бұрын

  Great proof, but it requires the definition of e as the infinite sum of 1/k! which is a but unnatural (it isn't how you define e, in general)

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  @sudhirsirohi72022 ай бұрын

  Solution is too long.This can be done within 30 seconds without totient,without Chinese remainder theorem.

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  @mx36032 ай бұрын

  atleast he's fast at explaining it :>

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  @anonymousperson68662 ай бұрын

  why is 3b not matching calc bc's scoring guideline?

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  @matangranot1002 ай бұрын

  4:47 - Minecraft Villager Sound

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  @Sh4dowbanned2 ай бұрын

  Balancing chemical equations, only time I got a headache & fear of failing class

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  @matteomoioli52102 ай бұрын

  Hi, i do not quite understand why at minute 13:21 you don't consider exp(i\theta) and i no more, can anyone tell me why?

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  @masondear3 ай бұрын

  my brain froze when i saw this problem but you really helped me get through it

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  @pierceingram98313 ай бұрын

  This video goes in depth about the angles of a shape. Interior and exterior. I loved how much writing he did and how he explained himself. Very helpful.

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  @user-fi8ji8jx8b3 ай бұрын

  Wheeler? WHEELER?!?!?! euler please. Flammable maths will claim your soul xD

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  @akif76993 ай бұрын

  I searched for this thanks. Also i have a question: in our math lecture my teacher explain this curve for e^-x^2. Is there a difference?

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  @mohammadmoradi97483 ай бұрын

  It was really beautiful and elegant. Thank you for sharing it and thanks to the Great Joseph Fourier for establishing such a nice and easy-to-understand proof.

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  @emmanueljosegarcia19753 ай бұрын

  Any integral of the form ∫f(x, √(x^2+a^2)) dx can be transformed into an exponential function (typically simpler to integrate) using the following general formula: ∫f(x, √(x^2+a^2)) dx = ∫f((e^θ-e^(-θ))/2 * a, (e^θ+e^(-θ))/2 * a) * (e^θ+e^(-θ))/2 * a dθ, where θ = sinh^(-1)(x) and x ≥ 0. The previous formula is part of a broader approach that serves as an alternative to trigonometric or hyperbolic substitution. You can check out this new trick here: geometriadominicana.blogspot.com/2024/03/integration-using-some-euler-like.html

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  @221e3 ай бұрын

  Okay I'm 6 years late but at 12:18 could you use small angle formulae? sin(x/2^k) approaches x/2^k because the expression becomes arbitrarily small as k tends to infinity. That gives you the same answer, as the limit equals 1/((2^infinity)*(x/2^infinity)) = 1/x, giving the whole product equal to sinx/x.

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  @guycohen44033 ай бұрын

  How will the function change if instead x²sin(1/x) it will be x²sin(1/x²)?

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  @CyrusSaiyed3 ай бұрын

  Please make a video on (x - 3)(x - 1)|(x - 3)(x - 1)|

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  @theuttamrathore90773 ай бұрын

  Jee

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  @conradolacerda4 ай бұрын

  Sorry, but this is not a proof by contradiction. Explanation: "if x is a rational number, then there exists a natural number b such that b!x is an integer" is a true statement. By showing that b!e is not an integer for all natural b, you proved that e fails the above test and so the assertion follows by a mere contraposition.

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  @mishabs4 ай бұрын

  16y old me trying to understand 💀

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  @markgraham23124 ай бұрын

  a and b don't have to be positive integers, they can also be negative integers, and and b ≠ 0.

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  @HazbinBrainRot4 ай бұрын

  THANK YOU SO MUCH! I understand!

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  @stof29054 ай бұрын

  WaaaaaW❤❤❤❤❤