Is there a nice closed form for this tetration derivative?
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- Опубліковано 22 вер 2022
- We compute the derivative of the n-th tetration of x using mathematical induction to help support our claim of its closed form.
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Let’s define d/dx(ⁿx) = 비벤밥ⁿ(x)
That's got to be the most horrific pi/sigma formula I've ever seen.
Lucky you. That means you have not seen a lot...
Willian's formula for prime number
@anurupsil8216 that's just....boss level formula
@@anurupsil8216 But that has no pi
Ligma you Said?
The complexity of this formula reminds me of Willian's formula for prime number
I think I found a better way to describe it that ignores sigma and pi notation at the cost of being recursive. It's so much cleaner looking in its generalized form tho
How would it be?
Nvm, he literally writes it in 12:29
@@adolfojasso796fuck I'm dumb lmao
Usually, "recursive" is what "closed forms" are not, but sigmas and pis are often (as in this case) not that different from recursion in terms of amount of work if you can't get around them, and I guess you didn't say it was a closed form, but it's still interesting to the get a recurrence formula for the derivative of a sequence of functions (which themselves can be described by a recurrence formula).
11:11 To be be clear, you just need to assume that it's true for SOME integer (which you have just proven), not for all natural numbers, and then proving it's true for n+1 proves it's true for all integer greater than or equal to n. The way you wrote it, someone could imagine you meant FOR ALL n in N, which would mean you were starting by assuming what you wanted to prove. This sort of thing might be why some people apparently think induction sounds like circular reasoning for a long time.
Great video! I somehow missed this one when it came out. I can’t wait to integrate this proof into my son’s cycle of proof writing exercises…
amazing video!!! interesting to see powers of lnx show up. It suggests to me the best way to think about tetration is with base e as well...
I prefer logarithmizing both sides instead while differentiating equations involving hyperpowers.
This is excellent, but it’s got me wondering what would happen if you differentiated with respect to n if n wasn’t restricted to the naturals
Is tetration defined for non-natural numbers?
@@FranLegon there are formulas for it I think but idk if there’s a standardized one
Mathematicians are still exploring this extension for n a complex number, but there are now mainly 2 ways of doing it. They are listed on the wikipedia, but i don't understand much of it
It can be extended to rationals pretty easily although there are no clean formulas for directly computing the 1/2 tetration. You have to use some sort of regression algorithm to compute that @@FranLegon
Wow this formula is look so beautiful 😍😍
great video
Yes it helps a bit that the derivative contains the function so to speak. Nice work finding that recursive closed formula.
Great 👍👍😃
you can see this closed formula better if you calculate the logarithmic derivative... then you can even write the kth derivative of the tetration of "n" more simply... hint - use the formula for the kth derivative of the product
Now do integral of tetration
This derivative is undefined at x=1, since for (ln(x))^k, where k=0 is lower bound in the sum, 0^0 is an indeterminate form, and so ln(x) not = 0 and hence x not equal 1.
Nicely done derivation and proof! I just published an equivalent proof:
ua-cam.com/video/dJWUCEuripk/v-deo.html
and what I think is an equivalent closed form solution. Please feel free to critique!
Or you use category theory inspired definitions where empty sum is 0, empty product is 1 and thus 0^0 is 1
didn't BPRP recently prove the limit of 0^0 is 1?
@@Appri BPRP proved with an example that there are functions f(x) and g(x) so that both the limit of f(x) and the limit of g(x) for x approaching 0 is 0, but the limit of f(x)^g(x) for x approaching 0 is 1.
@@Djake3tooth yeah I know the limit isn't actually 1, haha
wish I'd never heard about tetration. UA-cam keeps recommending it to me 💀
Why?
@@awesomethegreatamazing2651 Because it's very interesting but reminds me how much I hate maths 😅
Man...college maths was a LONG time ago for me. Induction? I remember hating it :D
Now do the same for pentation.
Now do the integral
I’m here from blackpenredpen
Now, next challenge: define this is for non integer n?
Now time to extend this derivative for all Real Number ! Make n^^x becomes x^^x !!
I thought power series solutions in diff eq was nasty. 😂
I think it is better to define an operator D which sends ^nx to ^(n-1)x and then figure out what D(x) is 😂
I know the answer to this video but that contains a new type of mathematical notation.
Now do n^^x instead of x^^n
shouldn’t d/dx(x^^0)=1?
Now how do you derive this derivative? 😈
Wow
Now do pentation
I don't think the first derivative of nth pentation of x is possible.
Before you can take the derivative of a pentation, you have to define x^^x for real numbers x
Great, now integrate x^^n :)
How about n being Real (or even worse, the derivative of let's say 2^^x)
I want to know about the derivative of e^^x, where ^^ is tetration and x is a real number.
I'd hope it would be the same form when complex z and e^^z
Is there even a definition for non integer tetrations yet?
@@mathieuaurousseau100 I have a B.S. in mathematics, so I'm very unqualified and suck at researching and understanding papers.
Though I found one that shows plots of tetration for complex inputs. I struggle to understand how the extension works though. Its really neat to see how the graph changes for different bases. (see "Evaluation of holomorphic ackermanns" by
Dmitrii Kouznetsov)
Furthermore, multiple analytic extensions have been proposed. Different ones have different properties, which are subjectively preferential. This is because exponentiation is not commutative, so when repeating exponentiation, you have to carefully choose a way to evaluate your expressions. So I'm not aware of any analytic extension for tetration that has been widely agreed upon as the most useful by mathematicians, the way it has been done for repeated multiplication (aka exponention).
Not a canonical one
@@mathieuaurousseau100 From what I have heard e^^(1/2) for example means find the function f(x) such that f(f(x))=e^x, and to do f(1). It's impossible to find an exact function f(x) where f(f(x))=e^x, but you don't need an exact function, you only need an approximate function that is approximately true for values of x between 0 and 1, and to use the recursion formula from there, and to get an every so closer approximate function for f(x). It turns out that e^^(1/2) is approximately 1.6464... We can also do the same thing to find e^^0.01, e^^0.02, ... if instead of finding f(x) such that f(f(x))=e^x, we instead find f(f(...100 times(f(x))...))=e^x. It's impossible to be true for the whole number line, it only needs to be approximately true for values between 0 and 1, and we can use the recursion formula from there.
Easy... f'(x)=f(x)*f'(x-1)... Next step.. f'(x-1)=f(x-1)*f'(x-2)... See it?product To The infinity if 'x' in not natural...
Now find the derivative of x titration x
Add phenolphthalein to the solution
😂
The background music is suuuuper distracting
C U R S E D
Damn
wtf
I subscribed to this channel over a year ago, but unsubscribed due to the creator’s unwillingness to reply to comments.
He seems to want his subscriber count to increase.
He should write more and talk less.
What about x tetrated to the x, and also the integral of the same?