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[✓] Tetration function in Wolfram?Alpha?

GROUPS:

Is there any way to enter tetration in WolframAlpha, in such a way that I can use it with the limit function? More specific, can I make Wolfram Alpha calculate the limit of sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^... You get it? And i know it will be 2, i just wanted to ask for the general case.

POSTED BY: Alexander Wolters
Answer
2 months ago

This can be seen as a FixedPoint[], after looking into what Mathematica makes out of it

In[27]:= Sqrt[2]^Sqrt[2]^Sqrt[2]^Sqrt[2]^Sqrt[2]^Sqrt[2] // FullForm
Out[27]=Power[2,Power[2,Plus[-1,Power[2,Plus[-1,Power[2,Plus[-1,Power[2,Plus[-1,Power[2,Rational[-1,2]]]]]]]]]]]

and keeping in mind that

In[18]:= ((((Sqrt[2]^Sqrt[2])^Sqrt[2])^Sqrt[2])^Sqrt[2])^Sqrt[2] // N
Out[18]= 7.10299

diverges, one finds the expression to reach the fixed point is

In[30]:= 2^FixedPoint[Power[2, Plus[-1, #]] &, Sqrt[2.]]
Out[30]= 2.

which can - to return to the question - be entered into W|A. Note

In[31]:= FixedPoint[Power[2, Plus[-1, #]] &, Sqrt[2.]]
Out[31]= 1.

to no surprize.

POSTED BY: Udo Krause
Answer
2 months ago

Slightly off-question but I think this can help. I actually did some research on tetration (and extending it to complex realms) some weeks ago and I learnt the following through this Wolfram MathWorld page.

The value of the infinite power tower power tower can be computed analytically by writing

tetration

Taking the logarithm of both sides and plugging back in to obtain

log procedure

Solving for h(z) gives

final.

Where W(z) is the Lambert W-function (Corless et al. 1996). h(z) converges iff e^(-e)<=x<=e^(1/e) (0.0659<=x<=1.4446; OEIS A073230 and A073229), as shown by Euler (1783) and Eisenstein (1844) (Le Lionnais 1983; Wells 1986, p. 35). This function is already in mathematica with the following name:

    ProductLog[x]

Just to prove it works:

proof it works

I don't know what things you're working on but probably this method will be more efficient than solving for the limit manually. Hope it helped :)

Answer
2 months ago

Observe - despite this argument has a weakness if repeated - that (because of $\infty-1=\infty$) one might single out one term $\sqrt2$ from a=Sqrt[2]^Sqrt[2]^Sqrt[2]^... coming to

In[1]:= Solve[Sqrt[2]^a == a, a]
Out[1]= {{a -> 2}, {a -> -((2 ProductLog[-1, -(Log[2]/2)])/Log[2])}}
POSTED BY: Udo Krause
Answer
2 months ago

One can enter solve sqrt(2)^x=x and obtain a result from W|A. It seems that W|A will not handle the more general case of a^x=x, though Mathematica is fine with it.

In[63]:= Solve[a^x == x, x]

During evaluation of In[63]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Out[63]= {{x -> -(ProductLog[-Log[a]]/Log[a])}}

This does not fully address the question since it is a multivalued result, so one must figure out which case is correct for the limit problem at hand.

POSTED BY: Daniel Lichtblau
Answer
2 months ago

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