Відео

very interesting example, love your teaching style!

You are great thanks guys so much 10Q a lot

tetration(x,2)=16 x^x=16 ln(x^x)=ln 16 x ln x = ln 16 W(x ln x) = W(ln 16) ln x = W(ln 16) x = e^(W(ln 16))

This is the best I have ever heard this concept explained. I was struggling with understanding how the proof worked for the past 2 days, and I finally understand now. Thank you!

But you didnt get the fish back

bro just use lhopitals rule before applying the limit

Do you have a video on Descartes rule of signs?, we haven't learnt one at school, where I live

2^(2^2) = 2^4 = 2 x 2 x 2 x 2 = 64

10 to the power of 2 to the power of 2 or, in other words, 10 to the power of 4, so 10,000

Is right hand side rationalization correct?? As denominator has a power x but numerator has not

Sir u are too much I learned a lot from you

The equation 3 + √(3 + √x) = x obviously does not involve an infinitely nested square root, and there really is no need to resort to infinitely nested square roots to solve this equation. If you look at the given equation you can see that there is some kind of repetition at the left hand side. Working from the inside out we take a number x, take the square root, add 3, and then _again_ take the square root and add 3. This means that if we define a function f(x) = 3 + √x on the domain [0, ∞) then the equation can be written as f(f(x)) = x Now let f(x) = y and substitute that in f(f(x)) = x and we have f(y) = x The function f(x) = 3 + √x is _strictly increasing_ on its domain [0, ∞) so, for any two real values x and y in this domain, if y > x then f(y) > f(x) but since f(y) = x and f(x) = y this would imply x > y which is a contradiction. Similarly, if y < x then f(y) < f(x) would imply x < y which is again a contradiction. Since y > x and y < x are excluded we must have y = x. Thus, it follows from f(f(x)) = x that f(x) = x and, conversely, f(x) = x evidently implies f(f(x)) = x. Therefore, the equations f(f(x)) = x and f(x) = x have _the same set of solutions_ on the domain [0, ∞) which means that we can solve the equation f(x) = x to obtain the solutions of f(f(x)) = x. So, we only need to solve 3 + √x = x Subtracting 3 from both sides and then squaring both sides gives x = (x − 3)² x² − 7x + 9 = 0 (x − ⁷⁄₂)² = ¹³⁄₄ x = ⁷⁄₂ + ¹⁄₂√13 ⋁ x = ⁷⁄₂ − ¹⁄₂√13 Since 3 + √x = x implies x > 3 and ⁷⁄₂ − ¹⁄₂√13 < 3 only x = ⁷⁄₂ + ¹⁄₂√13 is a solution of the original equation. Of course you can express the sole solution of the equation 3 + √(3 + √x) = x as an infinitely nested square root, but then you need to prove that your infinitely nested square root actually converges, which is not really trivial. Let us define an infinite sequence of finitely nested square roots by means of the recurrence relation u₀ = 3, uₙ₊₁ = 3 + √uₙ then proving that your inifinitely nested square root converges is equivalent to proving that the limit of uₙ for n → ∞ exists. We have proved this if we can prove that (1) the sequence (uₙ) is (strictly) increasing and (2) the sequence (uₙ) has an upper bound. To prove (1) we need to prove that uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺. This statement is evidently true for n = 0 since u₁ = 3 + √3 > 3 = u₀. Now suppose the statement is true for some nonnegative integer n = k. Then we have uₖ₊₂ − uₖ₊₁ = (3 + √uₖ₊₁) − (3 + √uₖ) = √uₖ₊₁ − √uₖ > 0 since uₖ₊₁ − uₖ > 0. So, the statement uₙ₊₁ > uₙ is true for n = 0 and also true for n = k + 1 if it is true for n = k, which implies that uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺. To prove (2) we note that since uₙ₊₁ > uₙ for any n ∈ ℤ₀⁺ we have 3 + √uₙ > uₙ for any n ∈ ℤ₀⁺ which implies √uₙ > uₙ − 3 and therefore uₙ > (uₙ − 3)² since uₙ − 3 ≥ 0 and therefore uₙ² − 7uₙ + 9 < 0 which implies ⁷⁄₂ − ¹⁄₂√13 < uₙ < ⁷⁄₂ + ¹⁄₂√13 for any n ∈ ℤ₀⁺. Of course, since ⁷⁄₂ − ¹⁄₂√13 < 3 and 3 ≤ uₙ this also means that 3 ≤ uₙ < ⁷⁄₂ + ¹⁄₂√13 for any n ∈ ℤ₀⁺. Since the sequence (uₙ) is (strictly) increasing with an upper bound it follows that lim uₙ for n → ∞ exists. If L is this limit, then it follows from uₙ₊₁ = 3 + √uₙ that L = 3 + √L and therefore L = ⁷⁄₂ + ¹⁄₂√13 since uₙ ≥ 0 for any n ∈ ℤ₀⁺. And, as has already been proved algebraically, x = ⁷⁄₂ + ¹⁄₂√13 is the sole solution of the equation 3 + √(3 + √x) = x which can therefore indeed be represented as the infinitely nested square root defined by the sequence (uₙ).

Vous êtes un homme

I got that wrong. The real answer is two billion zeros.

That's gonna be a number with 8 billion zeros, right?

Nice explanation. I have just a little problem with this subject. The idea of infinitely nested radicals (and infinitely nested fractures as well) is something that may be understood intuitively. The question is if these notions are also formally defined. After all, we are dealing with mathematics.

Please make a video on Newton's method

Thanks you so much

Put value of y=5-x in equation 1

the answer is 256😮😮😮

You’re my favorite math youtuber :).

Good job bro❤

Why u call square root as rad? I don’t get it

So it’s 2 to the power of 2 to the power of 2. 2 x 2 is 4, now it’s 2 to the power of 4. 2 x 2 x 2 x 2 is 16

2^2^2

I expected it to be like 1 followed by 15-50 zeros not one billion now I understand why schools don’t teach you this anymore

Try not to use L’Hôpital💀💀💀

Is it 16?

log n!/n^n=n*log n-n+o(n)-n*log n=-n+o(n) ->-infty. Thus n!/n^n -> 0. QED As usual, this guy is verbose and primitive.

This fuction is not continu how could it be derivable ???

Wpi;d it be incorrect to treat both sums as integrals and then equate? The you get n^4/4=(n^2/2)^2. This seems to prove it. What am I missing?

So. Who uses tetration? What's it for? I think you forgot to say.

why don t you ask if this fuction is derivable before anything

its 16 the answer.

God bless you 🙏🏿

16 cause 2^2^2 =2^4 = 16

My mind couldn't accept that -2 was an answer. But after cross-multiplying the two fractions to get a common denominator, it worked because the common denominator simplified to 1. That helped me to understand the point you were making about conjugates.

Sir please make a video on subspaces

nice one!

You're so underrated thank you

You have a good impression with amazing knowledge. You have Every element that a best teacher could own! Thank you!

this equation is a mosnter

It's 16 Thank you Sir for more Information like this You are really grate

I did the work on my own and I managed to get to the answer on my own, but it ended up with some supplemental guess and check after i got to (x-2)! = x+1. I see why the n’s were used, but it was easier without the extra step for me

Am failing for how you got 3<X<5,,,, Was it not supposed to be 2<X<4?

I found 1 and 3 by observation but my dumb ass didn't think of exploring the negative integer side of the number line 😭

Thank you.

Well i actully solved this in my mind with a different solution. (X+2)²=X²+4X+4 (X+2+X-2)(X+2-X+2)=(2X)(4) = 8X so we can say: f(a/b) = (a²)/(a+b)(a-b) -> f(X/1) = X²/(X+1)(X-1) -> f(X) = X²/X²-1 😊 Pls like until he see this😢

It’s 20000000000