# $n^{th}$ derivative of a tetration function

I stumbled upon this very peculiar function last summer, namely: $f(x)=x^{x^{x^{...^{x}}}}$, where there is a number $n$ of $x$'s in the exponent, I tried to find the derivative for the function and I was successful, it turned out not to be the most elegant formula but it worked. (Firstly, I invented a new notation, namely, a function such as $f(x)$ we can write it as the following: $f(x) =x^{\langle x \vert n\rangle}$ where $x$ is the exponent that is getting "powered" up $n$ times.) The formula I obtained by pattern matching was: $$f^{\prime}(x)=x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -1}\left[1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j\right]\tag{n\geqslant 2}.$$ I know this looks like a mad mess and I am aware that people like this have done it more elegantely, but now for the question. This is only the first derivative of the function, is there a way, or rather is there a general derivative i.e a $n^{th}$ derivative of this function?

Update: December 23th

I have tried to approach the problem myself since I asked the question and I have not gotten to a stage to say if it is impossible or possible to do, however I think I am on the right track. At first, I thought of distributing the factor $x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ to all the terms in the parentheses, but I quickly realized I had to deal with at least derivatives of triple products. Now I have come to realize that the easiest way is to differentiate the function just as it is and get a normal product and thus I must use the following formula: $$(f \cdot g)^{(n)}=\sum_{k=0}^{n}{n\choose k}f^{(k)}\cdot g^{(n-k)}$$ where $f =x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ and $g=1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j$. Since $k$ and $n-k$ are arbitrary numbers this leads us to find the general derivative for $f$ and $g$, this is where I am right now. (I do realize that I am trying to find the $n^{th}$ derivative of the first derivative but that is easily fixed later). Please come with suggestions on how to tackle this problem.

Update: December 24th

I have made progress with the help of Maple 17, namely, I have found a repeating pattern in at least a part of the general derivative, but there is still a part of it I cannot yet explain. Nonetheless, I present to you the part of the general derivative I have found: $$D_x^{\xi}f(x) = x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -\xi} \Big[(-1)^{\xi}\cdot\xi! +O(x)\Big]$$

I renamed the degree of the derivative as $\xi$ since $n$ is taken for the number of $x$s. The $O(x)$ is the (perhaps) series which I am currently working on finding, I do think I am on the right track though. The approach above with the product rule turned out to be less successful.

I found a method of doing the first derivative but it's relatively messy and makes you go all the way down the "n" chain as it were. Hopefully my ideas can give you some inspiration or spark a secondary idea.

I am using $$^nx$$ as the nth tetration of x.

What I did to reach my answer was start taking the derivatives of $$^nx$$ with increasing values of $$n$$ using $$e^{lnx}$$. So for n=1 you obviously get $$\frac{d}{dx}(^1x) = \frac{d}{dx}(x) = 1$$

For $$n=2$$ you get $$\frac{d}{dx}(^2x) = \frac{d}{dx}(x^x)$$ $$=\frac{d}{dx}(e^{\ln x^x})$$ $$=\frac{d}{dx}(e^{x\ln x})$$ $$=e^{xlnx}\frac{d}{dx}(x\ln x)$$ $$=(^2x)\Bigl(\ln x\frac{d}{dx}(x)+x\frac{d}{dx}(\ln x)\Bigl)$$ We already know the value of $$\frac{d}{dx}(x)$$ from the last problem, so we can plug it right in. $$=(^2x)\Bigl(\ln x+\frac{x}{x}\Bigl)$$ $$=(^2x)(\ln x+1)$$

For n=3 you get $$\frac{d}{dx}(^3x) = \frac{d}{dx}(x^{x^x})$$ $$=\frac{d}{dx}(e^{\ln(x^{x^x})})$$ $$=\frac{d}{dx}(e^{(x^x)(\ln x)})$$ $$=\frac{d}{dx}(e^{(e^{\ln(x^x)})(\ln x)})$$ $$=\frac{d}{dx}(e^{(e^{x\ln(x)})(\ln x)})$$ $$=(^3x)\frac{d}{dx}(e^{x\ln(x)})(\ln x)$$ $$=(^3x)\Bigl((\ln x)\frac{d}{dx}(e^{x\ln(x)})+(e^{x\ln(x)})\frac{d}{dx}(\ln x)\Bigl)$$ Note here that $$e^{x\ln(x)}$$ equals $$^2x$$. This means that we can substitute in values we already know, just like in the last problem, and a pattern starts to emerge. $$=(^3x)\Bigl((\ln x)\frac{d}{dx}(^2x)+(^2x)\frac{d}{dx}(\ln x)\Bigl)$$ $$=(^3x)\Bigl((\ln x)\frac{d}{dx}(^2x)+\frac{^2x}{x}\Bigl)$$ We don't need to plug in the end result of $$\frac{d}{dx}(e^{x\ln(x)})$$ here, because the form the equation is in now will end up fitting our generalization later.

For now, let's check by plugging in $$n=4$$ $$\frac{d}{dx}(^4x) = \frac{d}{dx}(x^{x^{x^x}})$$ $$=\frac{d}{dx}(e^{\ln(x^{x^{x^x}})})$$ $$.$$ $$.$$ $$.$$ $$=\frac{d}{dx}(e^{(e^{(e^{x\ln(x)})(\ln x)}(\ln x)})$$ $$=(^4x)\frac{d}{dx}(e^{(e^{x\ln(x)})(\ln x)}(\ln x)$$ $$=(^4x)\Bigl((\ln x)\frac{d}{dx}(e^{(e^{x\ln(x)})(\ln x)})+(e^{(e^{x\ln(x)}(\ln x)})\frac{d}{dx}(\ln x)\Bigl)$$ Here again, $$e^{(e^{x\ln(x)})(\ln x)}$$ is the same as $$^3x$$. So if we rewrite our equation as $$(^4x)\Bigl((\ln x)\frac{d}{dx}(^3x)+\frac{^3x}{x}\Bigl)$$ we can see that a general form of the first derivative can be written as $$\frac{d}{dx}(^nx) = (^nx)\Bigl((\ln x)\frac{d}{dx}(^{n-1}x)+\frac{^{n-1}x}{x}\Bigl)$$

Obviously this has the problem of relying on the derivatives down the power tower, but I think this could have interesting applications. Hope these inane ramblings of a 17 year old help!

Excellent question, and a good result. I am also impressed that you have developed your own notation. That is often a very effective way of getting to grips with a problem, especially one that has not yet become popular. I think there is a trend, however.

The notation tends to evolve with use. The notation here involves a redundancy which one can ill-afford in a subject already pushing at against conceptual boundaries. if you study your remarkable formula for the derivative, you will see that all the references to tetration involve the incomplete symbol: $$x^{

In this usage, the initial exponent symbol $$x$$ is redundant, and complicates the expression. Thus the evolutionary pressure of being concise will force the rejection of this appendage, and one may use the symbol $$\langle x \vert n \rangle$$ by itself. this is conveniently defined by (if I have understood correctly): $$\langle x \vert 0 \rangle = 1 \\ \langle x \vert n+1 \rangle = x^{\langle x \vert n \rangle}$$ For the purpose of differentiation, the logarithm is useful i.e. since $$\ln \langle x \vert n+1 \rangle = \langle x \vert n \rangle \ln \;x$$ we obtain : $$\frac{\langle x \vert n+1 \rangle'} {\langle x \vert n+1 \rangle} = \frac{ \langle x \vert n \rangle }{x} \left( \frac{\langle x \vert n \rangle'}{\langle x \vert n \rangle}x\ln \;x + 1 \right)$$perhaps as might be expected, the logarithmic derivative $$\frac{f'}f$$ looms large here, and it is hardly surprising to see the "entropy" function also make an appearance.

We may abbreviate the form considerably if we define: $$T^n(x) = \frac{\langle x \vert n \rangle'}{\langle x \vert n \rangle}$$ So that we have a form fairly well-suited to recursive evaluation : $$T^{n+1}(x) = \frac{ \langle x \vert n \rangle }{x} \left( x \ln\ x\; T^n(x)+ 1 \right)$$ Congratulations on your achievement! I hope these casual remarks will be of some use or interest.

• My mentor told me almost the same thing and I have considered using the notation in that way. (And yes, you have understood the notation correctly). However when I first started I wanted it to be a general notation, i.e we can have a base $x$ and and exponent $\langle a \vert n \rangle$. If we only write $\langle x \vert n \rangle$ we cannot see which base it is, however as you've said, notation evolve and there might not be any good use for the "general case" notation of mine. Commented Dec 23, 2013 at 12:04
• you are right that tetration is only a very special case (albeit the most interesting and tractable one) of iterated exponentiation. I have been playing with this idea for a while, and it was very satisfying to be able to crystallize my intuitions only a few days ago. basically all the integers $\gt 1$ can be formed from the primes by two very simple rules, one of which is exponentiation. this leads to a very different view of the integers than the dominant (pragmatic) view. perhaps because of this, my attempts to expound it have been hampered by a sort of soft censorship. (ctd) Commented Dec 23, 2013 at 13:02
• a fuller view is given in a question that has been put on hold, but I don't know if such held questions are generally accessible. the link is math.stackexchange.com/questions/613807/… - I did the question no favors by baptizing it with such an obscure title, but that came from the fact that an earlier attempt was marred by a silly typo. (ctd) Commented Dec 23, 2013 at 13:10
• I presented this a third time in a much more concise manner, adding a very specific question math.stackexchange.com/questions/614698/… but have had no feedback. I mention this only because you have an obvious interest in something that is still widely regarded as very marginal, and also because you are evidently someone who enjoys exploring ideas for their own sake and I admire the honesty and fortitude with which you have pursued your investigation of tetration. Commented Dec 23, 2013 at 13:11
• Well I thank you for presenting me with this information, it truly is very interesting. However I lack the mathematical skill to understand your formal mathematical language (I am still in high school so I have yet to be learnt such notation). I can tell you something that I've had in my mind though. If I (or anyone else) find this general derivative it is possible to expand the function via a Taylor series and then... find the anti-derivative! Of course it's a long way there, and I'm not sure if it will work, but I know it's worth spending endless nights sitting with only pen and paper for. Commented Dec 23, 2013 at 13:29

$$\frac{\mathrm{d}}{\mathrm{d} x} x↑↑n = x^{x↑↑(n-1)+x↑↑(n-2)-1} \space \cdot {\left(1+\sum_{i=1}^{n-1}\left[ \prod_{j=1}^{i}x↑↑(n-(j+1)) \cdot (\operatorname{ln}x)^i \right] \right)}, \tag{n\geqslant 2}$$

Summary:

The derivative of $$x$$ to the tetration of $$n$$ is equal to

$$x$$ raised to the quantity of ($$x↑↑(n- 1)$$ + $$x ↑↑(n-2)$$ - $$1$$)

multiplied by the quantity $$(1 + \sum_{i=1}^{n-1}$$ $$\prod_{j=1}^{i}$$ ($$x ↑↑(n-(j+1)$$ $$\cdot \space (\operatorname{ln}x)^i)$$.

UPDATE:

If we were to use the proof for Negative Heights, we can prove that the formula works for n = 1 as well.

$$\frac{\mathrm{d}}{\mathrm{d} x} x↑↑1 = x^{(1+ 0-1)} \cdot (1 + 0) = 1$$

I added the proof to Desmos as well.

UPDATE 2: I'm not super informed about the Pi Product symbol, but from what I got from Desmos, $$\prod_{i=1}^{0}x = 1$$ for all real values of x. If this is the case, you can simplify the formula even further:

$$\frac{\mathrm{d}}{\mathrm{d} x} x↑↑n = x^{x↑↑(n-1)+x↑↑(n-2)-1} \space \cdot {\left(\sum_{i=0}^{n-1}\left[ \prod_{j=1}^{i}x↑↑(n-(j+1)) \cdot (\operatorname{ln}x)^i \right] \right)}, \tag{n\geqslant 1}$$

Since $$(\operatorname{ln}x)^0 = 1$$ & $$\prod_{j=1}^{0}x↑↑(n-(j+1)) = 1$$, you will get 1 as the first term.

Here is a link to Desmos where you can see test cases from n = 1 to n = 4

The solution is the one on towards the bottom.

– Community Bot
Commented Jan 1 at 8:15
• @ScipioAemilianus I am unsure how to notate using MathJax. Give me some time and I will rewrite the answer in the proper notation when I figure out how to. Commented Jan 1 at 8:48
• @ScipioAemilianus I think I did a decent job writing it out. Thoughts? Commented Jan 1 at 9:16
• @ScipioAemilianus I am using en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation to denote tetration. Commented Jan 1 at 9:24
– user1173615
Commented Jan 1 at 9:25

I tried finding the tetration of x to a natural index n and I got a similar result but since I could not use notations like sigma and pi(in my phone)therefore I used * and #.But now I can finally.

I did get the generalisation for the tetration of x to any natural index(excluding 1). Let x(n) be the tetration of x to a natural number n(n>1).For example,

$$x(3)=x^{x^{x}}$$ Then,

$$\frac{d(x(n))}{dx}=\frac{\prod_{k=1}^{n}x(k)(lnx)^{n-1} + \sum_{k=1}^{n-1}\prod_{r=k}^{n}x(r)(lnx)^{n-k-1}}{x}$$

Let’s test this!

$$\frac{d(x(3))}{dx}=\frac{\prod_{k=1}^{3}x(k)(lnx)^{2}+\sum_{k=1}^2\prod_{r=k}^{3}x(r)(lnx)^{2-k}}{x}$$ Then, $$\frac{d(x(3))}{dx}=\frac{x(1)x(2)x(3)(lnx)^{2}+\sum_{k=1}^{2}x(k)x(k+1)......x(3)(lnx)^{2-k}}{x}$$

> $$\frac{d(x(3))}{dx}=\frac{x(1)x(2)x(3)(lnx)^{2}+x(1)x(2)x(3)(lnx)+x(2)x(3)}{x}$$

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.

I actually had a similar question, and my friend and I found a general formula which in all honestly, is not too complex. For all n$$\le$$2, the $$\frac{d}{dx}$$[$$^n$$x]=

$$\displaystyle\sum_{k=1}^{n-2}$$ $$(_nx_{n-k}^{k-1}) * \frac{1}{x}$$ + $$_nx_2^{n-2}$$ + $$_nx_2^{n-1}$$

Take note of the notation im using here:

$$_ax_z$$ = $$^ax$$ * $$^bx$$* …$$^zx$$

$$x^{k-1}$$ = x * ln$$^{k-1}(x)$$

I think y'all are over complicating it a bit... Is it not the following?

This isn't based on hard scientific fact, this is just the pattern which seems to appear as it goes along.

Based on the following ( "X" is multiplication, as asterisk makes italics ) :

g( x ) = x^f ( x )

g'( x ) = x^f ( x ) X ( f ( x ) / x + ln( x ) X f'( x ) )

If applied to f ( x ) = x^^( k - 1 ), this pattern seems to follow.

(I'm not sure how to input the notation into the text thing, so here's a picture)

Additionally, in the actual picture I use tetration notation similar to an exponent, but the number of times the exponent is repeated in to the top-left of the base.

NOTE: X is multiplication, PI is product, SIGMA is summation, "^^" is tetration, and parenthesis following product and summation are first the bounds, then the function being evaluated ------------------------------------------------------------------------------------------------------------------------------------------------------------->>> d/dx( x^^k ) = PI(i = 2, k ) ( x^^i ) X SIGMA ( i = k - 2, k - 1 ) ( ln( x )^i ) + ( x^( -1 ) ) X ( SIGMA ( a = 2, k-1 ) ( ln( x )^( k - 1 - a ) X PI ( i = a, k )( x^^i ) ) )