derivative of tetration of x (hyperpower)
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- čas přidán 18. 08. 2018
- Derivative of tetration of x, derivative of (x↑↑3), derivative of double up arrows, derivative of tetration of x, Knuth Arrow Notation. This is a hard calculus 1 problem when we differentiate this power tower of x
T-shirt 👉 teespring.com/derivatives-for...
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#blackpenredpen #calculus #math #tutorial #college
Solve x^x^x=2? Check out here: czcams.com/video/ef-TSTg-2sI/video.html
What if
d(3↑↑X)/dx?
Professor: "On exam, I will test whether you know mathematical induction"
Exam: "Find derivative of n-th tetration of x." 🤣🤣
sqrt(2)
@@MrAleksander59 The notation would be hard enough already
@@snekback. Yes, but I learned it years ago and well understood. I tried to do something interesting with it so I researched hyper roots and logs, tried to differentiate it. When this video came out I thought: what if the topic will be researched more?
Now integrate it
No, it's hard!
😂😂😂😂😂
X^x^x + c
If you integrate x^(x^x) (i.e. 3rd tetration of x), then there’s no answer since it’s non-elementary. If you integrate x^3, it would be x^4/4+C.
@@justabunga1 What does elementary mean?
“i don’t know how to integrate this so don’t ask me”
we found his kryptonite
Lmaooo
no one even knows what x^^0.5 is iirc, but derivitive of x^^n is.. well.. math.stackexchange.com/questions/617009/finding-the-derivative-of-x-uparrow-uparrow-n
@@incription scary
Wha...? Well now, I don't feel so dumb.
it would just be x^(x^(x))+1)/x^x +1) +c
"Im just gonna put this in the thumbnail to make a little clickbait"
Transparency
Sree Kommalapati lolllll and then i changed
IU am 66 yrs old. I earned a MS in Mathematical Physics in 1977. I never heard about Tetration till just now THANK YOU so much!!!!!!!!!!!!!!!!!!!!!!!!!
Thank you!!
Hm, seems like the education back then was pretty shitty if you didn't hear about that. Where did you earn your degree?
@@19midnightsun87 Grow up.
19midnightsun87 since titration wasn’t coined until the 1980s it would make sense that Cold Logic Crusader would not know it. Heck, no computer of that day could even try to perform the operation.
Ah, I see! Very interesting. Thank you for the info.
If you'd put d/dx (x³) for my final exam, you'd be my favourite teacher!
If that appears in your final exam i assume you dont need a teacher
at least solve it by base principles
Let me give you one@@user-vp7in7bk7z
get vaxxed! You need at least 3 or you won't pass the test
@suyunbek1399 no sir. I will never remember the limits definition of derivatives. Power rule all the way!
X: This isn't even my final form!
Is this a reference to Dragon Ball? Cell's final form or something.
@@blackpenredpen shaggy
@@blackpenredpen Freeza
For those who'd like to do more research on ⁿa, the notation is called tetration.
Also x↑↑↑y = x↑³y is called pentation and other notation to represent it is x [5] y
x↑ⁿ⁻² y = x [n] y, where n is the number of operation (ex: sum n=1, multiplication n=2, potentiation n=3, tetration n=4, etc).
Thanks! I didn't know nothing about this.
Hows the thing called you need for g_64?
ahh interesting, thanks :-)
@Connor Gaughan if thats actually the case.. how boring
I found a nice recursive formula for the derivative of x↑↑n by setting y = x↑↑n, taking the log on both sides and doing implicit differentiation:
d(x↑↑n) = x↑↑n * ( d(x↑↑(n-1)) * log(x) + x↑↑(n-1) / x )
I'm not sure this is that clean and nice.
@@hugoburton5222 well it's recursive, so it probably can't get cleaner than this
I think it is simpler ;-) -> d(x↑↑n) = x↑↑n * d(x↑↑[n-1] * ln[x])
@ you mean n -> to |R? the question would be, what it would mean......
Now do d/dx(x↑↑x)
Tetration, not to be confused with titration. I wonder how many chemistry students came here looking for curves and then subsequently ran away because of some light mathing. Love the video. Clear and concise. It's clear that your professor chops are strong.
none what so ever
@@amineaboutalib Prove it
@@amineaboutalib *whatsoever
Im a chemist and I came for the math :)
Lmao 🤣
Good use of the Chen Lu; 70/10
Yay!!!
"Chen Lu" The Goddess of Derivatives
Chain rule haha i died
Lmao
LMAO
Lol I always thought it was "chair rule" because that's how my professor pronounced it. Awesome teacher tho
just wait till you see the "prada lu" the mightiest of all!
I have never seen this notation before!
You might find this interesting. (Watch 175 and 176).
It starts off a little "hokey", but does get pretty interesting (and introduces that notation)
Silas Rodrigues Its very uncommon. Only really used to describe for extremely large numbers, like Graham's numbee
Doh, I didnt include a link! Here they are.
175- czcams.com/video/EUvFXd1y1Ho/video.html
176- czcams.com/video/Z8I68E7yZeY/video.html
Sorry about that.
By that you mean "I have never seen Numberphile before"?
Me too! Today I learned a brand new thing! >:3
For any function f(x)^g(x), the derivative can be found by adding the power rule to the exponent rule. That is to say d/dx (f(x)^g(x)) = f’(x)*g(x)*f(x)^(g(x)-1)+g’(x)*ln(f(x))*f(x)^g(x)
This is a great simplifying formula to show properties in elements in hypercubes :)
This is stolen. You clearly don't know how this is derived. You don't just magically add the formulas together and get this.
@@gregsouza7564 I actually proved this in my math class lol it’s not that difficult to derivate f(x)^g(x) and determine this result. In fact if you note that in cases c^a and a^c the constant parts have their half of the equation canceled out due to having a derivative of 0. Therefore the power rule and the exponent rule are both just simplified versions of this general rule
@@gregsouza7564 lmao whatt? What do you mean "this is stolen". goofy
If you dont derive all math yourself you basically stole it
You taught me so many things like double factorials, hyperpowers... I never thought such things exist. Well done!
It's so amazing to see a (mostly) friendly community of people who like math as much as me (:
My mans ADMITTED he was gonna put tetration of x in video for a clickbait. HahAHA
Love your videos: I haven't studied Maths for a long time, and neither do I teach it, but these make difficult problems so easy to follow.
: ))))
I'm so pleased I can follow these. You are so fun to watch. Thank you for sharing the awesome math
I didn't read through the comments so if someone has already posted the derivative then kudos to them.
The form of the derivative of x^^n for n >=4 is:
x^^n*x^^n-1*(1/x+x^^n-2*(lnx/x+x^^n-3*(ln^2x/x+x^^n-4*(ln^3x/x+...+x^^2*(ln^(n-3)x/x+ln^(n-2)x+ln^(n-1)x)...)
You can prove this by induction. The inductive step is shown by the recursion derivative of x^^n = x^^n*(x^^n-1/x+lnx*d/dx(x^^n-1)) and the base case is that the derivative of x^^4 is x^^4*x^^3*(1/x+x^^2*(lnx/x+ln^2x+ln^3x)).
I put the base case in the same form as my answer to show that it's true because yt comments are hard to format and anyways loads of people in the comments did the x^^4 case.
Moving on to the induction we have d/dx(x^^n+1)=x^^n+1*x^^n*(1/x+lnx/x^^n*d/dx(x^^n)) from the recursion. Then we plug in the same form from above.
d/dx(x^^n)=x^^n*x^^n-1*(1/x+x^^n-2*(lnx/x+x^^n-3*(ln^2x/x+x^^n-4*(ln^3x/x+...+x^^2*(ln^(n-3)x/x+ln^(n-2)x+ln^(n-1)x)...)
Therefore:
d/dx(x^^n+1)=x^^n+1*x^^n*(1/x+x^^n-1*(lnx/x+x^^n-2*(ln^2x/x+x^^n-3*(ln^3x/x+x^^n-4*(ln^4x/x+...+x^^2*(ln^(n-2)x/x+ln^(n-1)x+ln^(n)x)...))
Done.
Recursion? nah. Hold my beer: \frac{d}{dx}{^{n}x} = \frac{1}{x}\sum_{k=1}^n\left ( \ln^{k-1}(x) \prod_{i=0}^k {^{n-i}x}
ight )
Hi, I really like your videos and since you used tetration in this one, I would like to ask you a question that's been on my mind for quite some time now, but for which I could not find a solution yet.
What I realized was the following: If you try to complete the natural numbers with respect to subtraction, which is the inverse operation to multiplication, you get the integers.
If you complete these with respect to quotients, which are inverse to multiplication, i.e. form the quotient field, you obtain the rational numbers.
By completing these wrt. roots of polynomials, i.e. wrt. exponentiation, you obtain the algebraic numbers.
But what if you complete these using tetration, i.e. add superroots, superlogs and other solutions of "tetration equations"? E.g. the superroot of 2, i.e. the number x with x^x=2, seems to not be algebraic, so can you form a field extending the rational numbers that is closed wrt. these operations? It can't be a field extension of finite dimension since it is not algebraic, and it also has to be a field, I think, and it should still be countable so it isn't the real numbers, but I could not figure out much more.
Also, if you continue this process with pentation and higher order operations (see Knuth's up-arrow notation), you should get other fields extending each other. Since they are a mapping family, you can take the direct limit of those, and get a very big field - are these the whole real numbers? A lot of questions, I know, but I hope someone else already thought about it and figured it out, since it is far beyond my current scope. Anyway, I would be happy about any kind of help.
have you got an answer to this yet?? sounds insane i wanna know
They should be included in the real numbers, moreover if we keep going towards the higher orders, we could find the transcendental constants (Like π, e) midway through. Or, we might even try to approach infinitation (as in, the infinth order, writing with infinitely many up arrows) and
1) find certain numbers which can't be written without the inverse infinth order. Or,
2) already cover up every real number, thus having no real numbers left, furthermore having found a way to cover up the first uncountable infinity. Or,
3) just get to the transcendental numbers with a clear definition about them.
New math lore
In the ending result you also could simplify: x^(x^x)*x^x = x^(x^x+x), but this depends on which notation you prefer. ;-)
Hey BPRP. I'm about to graduate with a math degree. I want to tell you, your vids are awesome and I love your pronunciation. Thanks for what you do! You bring me snippets of math I can enjoy when I feel bogged down in technical math i have to learn for a grade (which I'm sure you know can suck the fun out of it). So thanks. Truly.
Thank you!! I am glad to hear this!! Best of luck to you in everything you do.
So I was just learning multivariable calculus and I realized how much simpler this is if you just use the multi variable chain rule on f(x, x^x) where f(y,z)=y^z
I love learning new notations! #YAY
Loved it!!!
New notations.
Thanks Steve.. 😊😊😊
Tetration is read as "the nth tetration of a"
x^^3, x^x^x, "the third tetration of x"
or 'x tetrated to 3'?
For pentation, you have to use the up arrows, presumably because you run out of upper corners to write in after tetration due to the number of upper corners being 2...
Your ball is getting smaller and smaller. 😁😁😁
@Venky Wank good one XD
Which one of the ~two~...three...?
balls
by ball u mean mic or real balls
@@trustnobody90....the mic
really enjoy your style 👍
Wow. I learned a new thing today. Thanks.
Thanks, I've always wondered this!
Awesome as always!
As always he made it waaaayy longer than it needed to be!
What is i^^i?
Flamingpaper e^(-pi/2)
bl00dwork No, this would be for i^i...
novidsonmychannel justcommenting but I thought that ¡^^¡ would be ¡^¡ since i^^^i would be i^i^i
bl00dwork OK, I read the "^^" as the notation for what bprp introduced in this video, so e.g. x^x^x would be x^^3. But I have no idea myself about what i^^i could mean... I'm not sure if it even makes sense anyway...
By replacing those "i"es with e^(i*pi/2), i^i^i would be e^((i*pi/2)*e^(i*pi/2)^i) :(
and simplifing this we get e^(i*pi/2 * e^(-pi/2)):-|
is's still a mess but that's it.
if you want to expand it with Euler's formula again, it would be much more "complex":-)
Awesome! Thank you!!
I think you can also do it with implicit differentiation where you take the natural log on both sides. You'll need to repeat for x^^3 so for x^^2 you'll have:
x^^2=y
XLn(X)=ln(y)
Ln(X)+1=Dy/DX •1/y
Therefore
Dy/DX=x^^2(ln(X)+1)
I think you forgot to divide by x in the second term on the left-hand side of line 3, but yeah I used implicit differentiation and got the right answer, too(as I tend to do first when I don't know how to take a derivative in general)
Edit: I don't know why I thought you made a mistake in line 3, but it all looks fine upon rereading it.
Very clear explanation... every time i learn something 😀
Glad to hear that!
Thanks!
I'm from Indonesia and i find this interesting!
Brother u are really a genius !!!💖👏👏👌👑
It’s so cool!
I am ... amazed
You can combine x^x with x^x^x by using the multiplication of exponential expressions adds the exponent rule where (x^x)*(x^x^x)=x^(x+x^x).
Left-upper index is usually used for bottom-order tetration, for right-order hyperoperators arrow notation is basic. Left hyperoperators are rarely used though so it's specified explicitly in that case.
I have never heard that
So, this notation that blackpenredpen just told, I had actually once accidently discovered it myself.
So, when I did a bit of research on values of x^^a for fractional values of 'a' and found out an elegant relation between x^^(1/2) and Lambert - W function. It is:
x^^(1/2)=e^(W(x))
or
W(x)=ln(x^^(1/2)).
I am also working on the derivative of x^^a, which is partially answered by Calyo Delphi in one another comment.
I did some research and thought I should share it here.
3 years late but I think you might be wrong because e^^x is (as far as I know) an injective function, so if e^^a = e^^b then a = b. But e^^1 by definition equals e, and if you plug in e into your formula {x^^(1/2)=e^(W(x)}, you get e^^(1/2) = e^W(e), which is still e. This however contradicts e^^x being an injective function, since 1/2 ≠ 1, therefore the formula is incorrect
@@fantiscious yup you're right. i prob miscalculated then or sth. the actual relation was
(e^x)^^1/2 = e^W(x)
Proof:
x = t^t => t = x^^1/2 -- 1.
x = te^t => t = W(x) -- 2.
=> e^(x) = e^(te^t)
=> e^x = (e^t)^(e^t)
=> e^t = (e^x)^^1/2 = e^(W(x)) (from 1.)
=> (e^x)^^1/2 = e^W(x)
@@vedantneema Wow thanks for replying early lol. However could there be a misconception from your first line of proof? You see,
log_x(x^^n) = x^^(n-1)
=> log_x(x^^1) = x^^0
x^^1 = x ∀ x (by definition)
=> log_x(x^^1) = log_x(x) = 1 = x^^0
∴ x^^0 = 1 ∀ x
But in your 1st line of proof, you imply that x = t^^n => t = x^^(1/n) for all n. This says that 2 = t^^n => t = 2^^(1/n).
However as n approaches to infinity, t = 2^^(1/n) approaches 2^^0, which from the lemma above shows it approaches 1. This cannot be true though since;
2 = lim_(n -->∞)(t^^n)
=> 2 = t^t^t^t^...
=> 2 = t^2 (since t^t^t^... appears in itself)
=> t = √2
This shows that t = √2 and t = 1, but √2 ≠ 1, making a contradiction
@@fantiscious "x = t^^n => t = x^^(1/n) for all n". No, I only meant it for when n = 2. Also in your last statement you conclude that t can have two solutions for the given equation, one each is obvious in two different but equivalent forms. That stems from the fact that unlike addition, multiplication or exponentiation; tetration is not monotonic. Refer to the graph of x^x within [0, 2](decreasing in [0, 1] and increasing within [1,2]). The graph of y^y=x (== y = x^^1/2) is equally weird.
Now the limits within which x^^n is monotonic is something that may be interesting to work out. I speculate it remains the same for n>=1. I'll report back if I find something.
For Expert:
Tetration n times:
Write it as T(x,n) so that T(x,1) = x, T(x,2) = x^x and so on, write its derivative as T'(x,n)
Construct a reduction formula, start from n=2, try to solve T'(x,n) for all integers n. (tedious like hell)
Well, I’m glad he at least admits that he makes clickbait thumbnails. Didn’t know about that notation, but already knew about tetration. Great vid
the 10 mins I spent here was worthier than my existence
I love this
I got the answer... yeah, I watched the video quite late, but I found the answer all on my own, and then watched the rest of the video, I also got the answer to that last question you asked 😄( it might not seem like a big deal, but I am in 10th grade 😅😅😅 ) it was fun doing it... keep posting such videos 👍🏻👍🏻👍🏻
어느순간부터 내 알고리즘에떠서 12math처럼 잼게 보고있슴다 ㅋㄴㅋㅋ
I did get the generalisation for the tetration of x to any natural index(excluding 1).Well,it goes like this,
Let x(n) be the tetration of x to a natural number n
* denotes the product of the tetrations(pi-product function)[*(k=r to k=n)
x(k)=x(r)x(r+1)x(r+2)........x(n)
# denotes summation(sigma function) [#(r=1 to r=n) x(r)=x(1)+x(2)+.........x(n)]
Then,
d[x(n)]/d(x)=(1/x)[*(k=1 to k=n)x(k)(ln(x))^n-1 + #(r=1 to r=n-1)[*(k=r to k=n)x(k)[(ln(x))]^n-r-1]]
Where #(r=1 to r=n-1)*(k=1 to k=n) denotes the summation of the products.
Put it on paper for easy understanding.Do note it is only valid for natural number greater than or equal to 2.
Just put a reply if I had gone wrong somewhere.
Differentiation made easier by taking logs of both sides, twice.
taking y = x^x^x ; logy = x^x logx
Taking logs of both sides again, log(logy) = log(x^x) + log(logx)
ie log(logy) = x logx + log(logx)
Now differaentiating both sides with respect to x,
(1/logy)(1/y) dy/dx = x(1/x) + logx + (1/logx)(1/x)
Hence dy/dx = y logy [ 1 + logx + (1/xlogx) ]
= x^x^x . x^x logx [ 1 + logx + (1/xlogx) ]
Do a vid about integrating e^((x^x)*ln(x)) next please
you can't integrate x^x, so pretty sure you cant integrate x^(x^x)(which is what you asked here)
dolev goaz That's no fun.
@dolev goaz:
Since we're talking mathematics here, I sincerely wonder: *"do you have proof for your two claims?"*
(The claims I mean: 1: "you can't integrate x^x" and 2: "so you can't integrate x^(x^x)" )
For x>0 both functions (i.e. exp(ln(x)*exp(ln(x))) and exp(ln(x)*exp(ln(x)*exp(ln(x)))) ) are continuous and differentiable, so why wouldn't you be able to integrate them?
More precisely, their indefinite integrals aren't expressible in terms of elementary functions.
YES! He can't tease us like that!!
It can further simplify by bringing the 2 from the natural log of x in front of it then add it to the other one
(I'm new in calculus so please don't judge me too badly)
A nice rule for the derivative of n^xcould be d/dx n^x=n^x*(n-1)^x...2^x(ln(x)^(n-1)+ln(x)^(n-2)+ 1/x (ln(x)^(n-3)+ln(x)^(n-4) +...+1)).
The approach have been to do the derivative for 3^x 4^x and 5^x and I noticed this trend.
Hello sir i am from india.
Awesome your maths
Thanks
I tried it via the log method and found it to be easier.
Great 👍
i can't believe that my friend who have never seen tetration or heard about thought of this concept and wrote it in the same annotation of this and chose to call it superpostion .then he sarted studyng it as a functon and he got some pretty cool stuff .bt he was stuck on a problem.while searching the net for a solution he fond the same concept in the name of tetration and it shook us how similar his invention is to it .
YAY I finally got something correct!
Also, for me it was easier to solve via implicit differentiation.
You could also write it in a more "finished" (?) form as:
d(x↑↑3)/dx = ln(x↑↑6) + ln(x↑↑4)ln(x↑↑3) + (x↑↑3)(x↑↑2)/x
there is a formula for diff(x^x),whatever!I really enjoyed your video
One thing my Maths Sir once told about differentiating x^x is that "first differentiate it as constant^variable then differentiate it as variable^constant and add them that is the derivative of x^x" idk if it works for any tetration
Do a video on solving tetration equations please as well as tetration with imaginary numbers and taking the super-root of those numbers.
Please make videos on mulrivariable calculus
I was also thinking of using substitution and logarithmic differentiation. So solve for x^u, where u= x^x. Use logarithmic differentiation for u first, then after finding u’, use it to find x^u. I think this is long and messy though.
It's awesome
Alternatively, you can rewrite the given equation as: ln(ln(y)) = ln(ln(x)) + x*ln(x), and then try differentiating the equation in this form.
Solve for this; Cow tied at the corner of a circular field is able to graze 3/4th of the field. What is ratio of length of its rope and radius of the field?
Man it would be cool to have this guy as a calculus teachers
Thanks , I didn't know .
I am very intrigue by tetration and pentation..... please do some videos on the topic. (also integral equation, when there are so many video on differential equation, integral equation is very rare)
very nice . i hope you can explain how to integrate this term .
I’ve taken 3 levels of calc and that notation never came up. Great video
Amazing !!!!!
You make calculas very easy for us.
Can you please teach Probability with fun.
Because it seems very confusing some times.😇😊😊😊😊😊
Tetration in general needs a complete follow up video
I think I have the general formula where I set x↑↑0=1 (Is that right?).
First a recursion formula:
d/dx(x↑↑n) = x↑↑n · (d/dx(x↑↑(n-1)) · lnx + x↑↑(n-1) · 1/x)
After applying that a few times I found a pattern, so here's the formula (I think):
d/dx(x↑↑n) = 1/x · sum_(k=0 to n-1) of [(lnx)^(n-1-k) · product_(m=k to n) of (x↑↑m)]
d/dx(x↑↑0) = d/dx(1) = 1/x · sum_(k=0 to -1) of [(lnx)^(-1-k) · product_(m=k to 0) of (x↑↑m)]
= 1/x · 0
= 0
d/dx(x↑↑1) = d/dx(x) = 1/x · sum_(k=0 to 0) of [(lnx)^(-k) · product_(m=k to 1) of (x↑↑m)]
= 1/x · [(lnx)^(0) · product_(m=0 to 1) of (x↑↑m)]
= 1/x · x↑↑0 · x↑↑1 = 1/x · 1 · x
= 1
d/dx(x↑↑2) = d/dx(x^x) = 1/x · sum_(k=0 to 1) of [(lnx)^(1-k) · product_(m=k to 2) of (x↑↑m)]
= 1/x · [(lnx)^(1) · product_(m=0 to 2) of (x↑↑m) + (lnx)^(0) · product_(m=1 to 2) of (x↑↑m)]
= 1/x · [lnx · x↑↑0· x↑↑1 · x↑↑2 + x↑↑1 · x↑↑2]
= 1/x · [lnx · 1· x· x^x + x · x^x]
= lnx· x^x + x^x
= x^x · (lnx+1)
d/dx(x↑↑3) = d/dx(x^x^x) = 1/x · sum_(k=0 to 2) of [(lnx)^(2-k) · product_(m=k to 3) of (x↑↑m)]
= 1/x · [(lnx)^(2) · product_(m=0 to 3) of (x↑↑m) + (lnx)^(1) · product_(m=1 to 3) of (x↑↑m) + (lnx)^(0) · product_(m=2 to 3) of (x↑↑m)]
= 1/x · [ln²x · x↑↑0 · x↑↑1 · x↑↑2 · x↑↑3 + lnx · x↑↑1 · x↑↑2 · x↑↑3 + x↑↑2 · x↑↑3]
= 1/x · [ln²x · 1 · x · x^x · x^x^x + lnx · x · x^x · x^x^x + x^x · x^x^x]
= ln²x · x^x · x^x^x + lnx · x^x · x^x^x + x^x · x^x^x · 1/x
= x^x^x · x^x · (ln²x+lnx+1/x)
Since multiplying a number by a non-integer number and taking a number to a non-integer power are both possible, is tetrating a number to an non-integer number possible? Is it also possible for the other infinitely many arithmetic operations?
Amazing
Instead of plugging in exp(xln(x)) each and every time to replace x^x,
why not just use the power rule for differentiating the function power of another function,
thus if y = u^v, then
dy = vu^(v-1)du + (u^v)ln(u)dv
So if y = x^x, then
dy = xx^(x-1)dx = x^xln(x)dx
= x^xdx = x^xln(x)dx
= x^x[1 + ln(x)]dx
I like the middle one best.
I found a quite complicated formula for general integration of tetration functions in wikipedia, here: en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
It is given as exactly 20th example from the top; the formula involves incomplete gamma function and some parameters (n, m, i, j), but it's still not quite clear to me. I have just subscribed your channel :-) Could you devote some time to this formula in one of your future lectures? I mean, how it has been found and how it can be used In practice? Anything that can make it more familiar and clear?
It's cool and all, but the page contains integrals. Can you find one which lists the formula for derivatives?
@@nishilsheth9076 You know, derivatives can be calculated using quite simple set of a few rules. They include: a formula for derivative of the product (1), derivative of the quotient (2), derivative of a composite function (3), derivative of a power (4), and derivative of an inverse function (5). So here you are:
(1) [f(x)*g(x)]' = f'(x)*g(x) + g'(x)*f(x), or - more simply, (fg)' = f'g + g'f
(2) [f(x)/g(x)]' = [f'(x)*g(x) - g'(x)*f(x)] / [g(x)]^2, more simply (f/g)' = (f'g - g'f)/g^2
(3) when u(x) = f(g(x)), u'(x) = g'(x) * f'(u), for example [ln(sinx)]' = (1/sinx)*(-cosx) etc.
(4) [f(x)^g(x)]' = [f(x)^g(x)] * {[g'(x) * ln f(x)] + [g(x)/f(x) * f'(x)]}, or more simply:
(f^g)' = (f^g) * (g'* lnf + g*f'/f ) - which is an applied formula for derivative of a composite function after making an identity transformation: f^g = e^(g*lnf)
(5) when y=f(x) and x=g(y), g'(y)=1/f'(x).
For example when x=y^y and y=ssrt(x), (y^y)* (1+lny) = 1/[ssrt(x)]', which yields [ssrt(x)]' = 1/[y^y * (1+ lny)], so:
when f(x) = ssrt(x), f'(x) = 1/ [x + ln(ssrt(x)], since when y= ssrt(x), y^y = x by definition
When using these five formulas, we can determine a derivative of any function expressed with a finite formula.
Now Dr. Peyam has to make a video about the generalization d/dx(x↑↑a), where a represents any rational number!
I would like to know if tetration has properties. like, if it was a x^2 * x^3 right there, you would be able to add the exponents, but it dosn't seem to be the case with tetration
Wow, that's really cool! Now I'm curious if it's possible to differentiate n↑↑x.
We would first have to figure out what x^^1.5 would even mean. I've done a lot of research on this and there seems to be no conclusive answer.
The value of d/dx(x^x^x) at x=3 is 5.4x10^14
I was unable to come up with an explicit formula for (d(x↑↑n)/dx), but I was able to come up with a recursive formula (I hope). If someone could tell me if the formula is correct or how to make it explicit, that would be fantastic.
Formula:
d(x↑↑n)/dx = (x↑↑n)(ln(x)(d(x↑↑(n-1))/dx)+((x↑↑n)/x))
#yay Chen lu!
Please make a video on TREE(x).
Very interesting. I have following question: how can I calculate the first derivative of the function "a tetrated x" (a is positive)?
We still have no clear way of defining how to even calculate non-integer raised tetration, so good luck with that! I mean what would 2^^1.5 even mean?! I've looked into this quite a bit and there seems to be no conclusive answer.
@@DoctorT144 Using super Logarithm (inverse of Tetration)
By definition sLog2 (2^^3) = 3
NOTE: "sLog" is a notation for super Logarithm. Like how Logarithm cancels the base leaving the exponent ex. Log2 (2^3) = 3 super Logarithm does the same with Tetration leaving the super power.
We can use super Logarithm to solve non integer super powers since super Logarithm is repeated Logarithm by definition.
Let's let sLog2 (16) = 3+x
Where 0 ≤ x < 1 (represents a 0 or decimal)
sLog2 (2^^3) = sLog2 (2^2^2) => Log2(2^2^2) = 2^2
=> Log2(2^2) = 2
=>Log2(2) = 1
At this point we've taken three logs representing our integer part of the solution (given by the fact that the answer is equal to 1). We just take log again for the decimal x (the remainder of 2's that we need.)
Log2 (1) = 0
Thus sLog2 (16) = 3+0 = 3
Well let's look at what happens when we go backwards through the same process to see what happens to the remainder.
Log2 (Log2 (Log2 (Log2 (16)))) = 0
Log2 (Log2 (Log2 (16))) = 2^0
Log2 (Log2 (16)) = 2^2^0
Log2 (16) = 2^2^2^0
16 = 2^2^2^2^0 = 2^2^2 = 2^^(3+0)
The remainder adds an extra '2' to the top of the power tower and the additional 2 is raised to the power of the remainder
For 0 ≤ x ≤ 1
By definition sLog a(a^^3+x) => a^a^a^a^x
By definition of Tetration a^^3+x = a^a^^2+x = a^a^a^^1+x = a^a^a^a^^x
a^a^a^a^^x = a^a^a^a^x
a^a^a^^x = a^a^a^x
a^a^^x = a^a^x
a^^x = a^x by definition for 0 ≤ x ≤ 1
2^^1.5 = 2^2^0.5 = 2^√2 ≈ 2.6651441427...
Can you set y=x^x^x
And then take natural log on both sides and use implicit differentiation like
dy/dx
The rule is x↑↑n = (x↑↑n) * (d/dx ( x↑↑(n-1) * ln(x) + x↑↑(n-1) *(1/x))
Thats the best you can do
its basically the original function multiplied by the product rule of x↑↑(n-1) and ln(x)
09:58 Using Knuth's arrow notation, "1/x" (another way to write this is of course "x^(-1)") could be written as "x & (-1)" (where the symbol "&" represents the "up arrow" symbol in the Knuth's arrow notation)?
Nice work :)
Now lets differentiate x(arrow)x
😆
This is what I got, though I’m not sure if correct! It’s technically an explicit function 😂
d(x || n)/dx = E(k=1 to n) [ {((ln(x))^(n-k))/x} * P(i=k-1 to n) [x || i] ]
Notation
E ~ the sigma summation function
P ~ the sigma pin function
|| ~ the double arrow function
Domain:
For n within positive integers and assuming that x || 0 = 1
Graham's number on the spot! :o
Check by integration, perhaps?