30 Comentários

1. _That was quite some fun, don't you agree? 🙂 Check deez out to support the channel btw <3 _
   Train your Calculus Expertise by trying out Brilliant! =D https://brilliant.org/FlammableMaths
   
   Check out my newest video over on @Flammy's Wood ! =D https://youtu.be/bRCV9ZaTeeY
   
   Support the channel by checking out Deez Nutz over on https://stemerch.eu/products/deez-nuts-premium-3d-cutting-board?_pos=1&_sid=3ebd4ef06&_ss=r ! :3
   
   Engi Watch: https://stemerch.com/products/the-incredibly-unrigorous-engineering-watch?variant=40377075728562

2. amazing………………………..https://youtu.be/v9cEwsNt5mw

3. would it be acceptable to just use trial and error to figure out the values of S? Or would it be a given that you have to use calculus

4. i don't know why i haven't been able to place your accent, when you spoke German i felt like more of an idiot than when trying to understand half of the maths
   other half helps though don't worry

5. Hello teacher, how are you? I have a question for you, please answer. What books do you recommend for me to prepare for the Mathematical Olympiads? Thanks in advance! ❤️😔

6. Good job! Correct!

7. Always the weird noises in the beginning of the video that i look forward everyday

8. I love your videos Flammy, but tbh this should not qualify as a hard problem as my two guess solutions would immediately give me the answer

9. I’m having the Δp to guess what your shirt means

10. wait i never noticed your german accent before

11. This is not an olympiad question (olympiads are harder). This problem is from a math contest called the AMC12, specifically the 2021 AMC12B (Part B test/exam) Problem 21. This is pre-olympiad level in the US, along with AIME.

12. Achsenbeschriftung fehlt, keine Punkte

13. I don’t understand 80% of the stuff you do but the 20% i understand is insanely valuable to me. Thank you ❤️

14. As S is the SUM of all +ve values of x between √2 – 4, shouldn't the answer b S<6 (as there's 2,3 etc between em), or am i missing something. It's bothering me 🥲

15. I can't believe you approximated sqrt(2) to be 1.5 when it is so close to 1.4.

    6:07

16. No log needed
    say f = x^2^sqrt(2) and g = sqrt (2)^2^x
    Sqrt(2) solves by inspection
    f is less than g @ x=0, overtakes g @ sqrt (2) (check by seeing f'>g' @ x =sqrt(2)) but must be overtaken by g at some point bc g is growing exponentially and f is a polynomial.
    Try x = 4, see that f<g, so already overtaken
    So I1=sqrt(2) & sqrt (2)<I2<4 2sqrt(2)<S<4+sqrt(2), which corresponds to sqrt(2)<S<6, or C
    Don't know how to prove these are the only intersections tho

17. Is sqrt(2) not a trivial solution by looking at the original equation?

18. Whoa YouTube decided to just block your channel notifs for some reason and didnt realise you posted so many vids 😐

    also nice biceps

19. Man roasted America and I love it since I am from America and hate that obesity rate

20. Always love and humble respect from BANGLADESH.

21. Ayo you in a sauna Papa?

22. Shirt says Don't be a third derivative of displacement with respect to time [JERK]

23. Love the joke at the beginning 👍👍👍😂😂😂

24. papa jens with da tan(gent) line😳

25. Can someone esplain me the shirt joke? Is it an obvious one and I'm just dumb hahaha?

    Edit I gonna try to think it out lol brb…

    Ohhh, I remember my physics teacher telling us those terms, and apparently the next are snap ,crackle and pop… idk if he was lying to make me look like and idiot tho… oh wikipedia agrees about the rice crispies mascots

26. Papa likes it tough!

27. How does one justify that there are only two intersection points ? I understand that the logarithm is "slower" than linear functions but what would be the formal proof ?

28. x^a = b^x | log( … )

    a * log(x) = x * log(b)

    log(x)/x = log(b)/a

    Substitute: α = – log(b)/a, y = 1/x

    y log(1/y) = -α | * (-1)

    y log(y) = α

    Substitute: y = e^z, z = log(y)

    z * e^z = α | W( … )

    z = W(α)

    Resubstitute and use property of Lambert function:

    y = e^z = e^(W(σ)) = α/W(α)

    Resubstitute:

    x = 1/y = W(α)/α

    —————–

    EDIT: Oops, I had read the task wrong.

    x^a = b^(c^x) | log( … )

    a * log(x) = c^x * log(b) | Log( … )

    Log(a) + Log(log(x)) = x * Log(c) + Log(log(b))

    Log(log(x)) = x * Log(c) + Log(log(b)) – Log(a)

    Define:

    f(x) = Log(log(x))

    g(x) = Log(c) * x + Log(log(b)) – Log(a)

    Zero Point of g(x):

    x0 = -(Log(log(b)) – Log(a))/Log(c)

    Choose:

    log( … ) = log[b]( … )

    Log( … ) = log[c]( … )

    Therefore:

    g(x) = x – Log(a)

    x0 = Log(a)

    Use condition: a = c^b

    g(x) = x – b

    x0 = b

    Consider: x = b^n

    F(n) = f(b^n) = Log(log(b^n)) = Log(n)

    G(n) = g(b^n) = b^n – b

    Find n, such that F(n) = G(n). We get:

    n = 1: F(1) = Log(1) = 0, G(1) = b – b = 0

    First solution: x1 = b.

    For n = c^k, we get:

    F(c^k) = Log(c^k) = k

    G(c) = b^(c^k) – b = R

    If R > k, then b^(c^k) is upper boundary:

    b^(c^k) – b > k <=> c^k > log(k + b)

    With b = sqrt(2) and c = 2, we get with k = 2:

    sqrt(2)^4 – sqrt(2) = 4 – sqrt(2) ≈ 2.68.. > 2

    Therefore b^(c^k) = 4 is upper boundary.

29. idk why youtube's algorithm brought me here, but i ain't complaining

30. 69% obesity rate math olimpiad, good one

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