An Infinite Tower
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- Опубліковано 5 січ 2024
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The chain rule was incorrectly calculated. Should be: 1/x^2 - (ln x)/x^2 = (1 - ln x )/x^2. So the result of extreme ocurring at x=e does not change.
The function y=x^(1/x) can be calculated for all x between 0+ and infinity. You stated that the original equation has only solutions when x is between (1/e) and e. How did you arrive at that?
For the sake of clarity, you should state the convergence conclusion is derived from the Banach fixed point theorem.
I regularly see your videos and learn many complex way of solution.
The sequence x_n+1= y^x_n has 2 stationary points for 1 < y < e^ (1/e) = 1.445 at x_l < e and x_u > e.
Derivative of the function (y^x)' = ln y ⋅ y^x = ln x (for y^x = x at the stationary point).
This is less than one for x_l and greater then one for x_u.The lower x_l is therefore a stabile stationary point and the upper x_u is a labile stationary point.
If we start our sequence at x_1 = y (which is our issue, y = 3^(1/3) = 1.442) it will end up in the attractive lower stationary point x_l = 2.478.
if we see intuitively the cube root of 3 at the very top exponent ( at infinity) then (cube root 3)^(cube root 3)>(cube root 3)...now coming down the exponent lane, each of the exponent will be larger than the previous one, hence finally we will have (cube root 3) raise to power infinite which will be infinite.. it's definitely a divergent series..
You always have the best way and cold problem❤
😊 thank you
Why then does the answer x = 3 work?
One step y^3 gives (3^(1/3))^3 = 3^(1/3*3) = 3, so you can repeat that infinitely.
Maybe y needs to be between 1/e and e for it to converge, but x doesn't?
If so then all y = a^(1/a) could work, and for any problem y^y^y^... = x the answer would be x = a.
What introduced the extraneous solution?
When you assume convergence, you may do some calculation that isn't justified. Just because you end up with a finite result, doesn't automatically imply that the quantity converged in the first place; this is why you should always check for convergence first before calculating an infinite sequence.
@@Debg91 appreciate you!!
Fun one. What's the minimum and maximum a among the reals where a^a^a^a^... = x and x is a defined value?
I remember video of sybermath about it. a^a^a^…=x, so a^x=x. so a=x^1/x.
@@user-asdlfhadfhal Yes, but we know 2^2^2^2^... is going to diverge. 2^x = x. Well 2^2 = 4, 2^1 = 2, 2^0 = 1, 2^0.5 = 1.414..., see they can never be equal. The exponent is always less than the result. So a = 2 diverges.
So what's the largest real a that won't diverge?
@@Qermaq hmm… I think you are smart than me
@@Qermaq there is no root about 2^x=x, so a’s value that make the equation a^x=x has root is a that y=a^x and x are contact in line, so let’s say the point that both encounter is (t,a^t). differential coefficient of a^x at t is ln a*a^t, it’s same with differential coefficient of x, which is 1. So a^t=t, ln a*a^t=1. by calculation, t*ln a=1, so a^t=e. and a^t was t, so t=e.
So, a^e=e, so a=e^1/e, if a is larger than e^1/e, both won’t be in contact, and if less, both will be in contact at two point. So largest a is e^1/e??
@@Qermaq To ask for an excuse, I’m bad at English
0,3?
Try this problem
Integral 1/(1+(sin(x))^2) dx
Hint multiple top and bottom sec(x)^2
3
Towards the end of the video, you recast the problem as finding the limit of a sequence: y, y^y, y^(y^y), .... You really need to show that this sequence converges before any of the previous analysis can be accepted. This was never done and there's no reason to think that the "correct" result isn't as spurious as 3. Just because it falls within the extreme values for the equation y^x=x isn't enough.
x=e^(-W(-a))=2,4777...a=(1/3)ln3
I derived x=W(-a)/(-a), equivalent to your result because W(-a)/(-a)=e^(-W(-a)) is implied by W(-a)*e^(W(-a))=(-a), which is implied by W(W(-a)*e^(W(-a)))=W(-a).