What is the infinite hyper root of 1000? See details?

Question

{hyper root is an opposite of tetration]

So, unless I am misunderstanding what I read, there is a paradox. [x ^^infinity=c} then [x = c root of c]. but people have pointed out that the [cubed root of 3)^^infinity is more like 2.44 than  3. Why is this?

And I know that the infinite “regular root” of a positive number is always 1. But the infinity “hyper root” of a positive value greater than 1 should NOT be equal to 1.
For x^^infinite=c, the maximum x value is [e^1/e] or 1.4446

...so would the “ infinite hyperroot of 1000” be something close to 1.445? anyone calculate what it would be?

thanks

Captain Matticus, LandPiratesInc · Accepted Answer

x^(x^(x^(x^....) = 1000
x^(1000) = 1000
x = 1000^(1/1000)
x = 1.006931668851804169929660787266...

x^(x^(x^(x^(...) = n
x^n = n
x = n^(1/n)

x = 3^(1/3) = 1.4422495703074083823216383107801

Don't know where you got 2.44 from.

x = n^(1/n)
ln(x) = (1/n) * ln(n)
dx/x = (1/n) * dn/n + ln(n) * (-dn/n^2)
dx/x = (1/n^2) * (1 - ln(n)) * dn
dx/dn = x * (1 - ln(n)) / n^2

dx/dn = 0

0 = x * (1 - ln(n))

x = 0

1 - ln(n) = 0
1 = ln(n)
e = n

So your critical values occur when x = 0 and when n = e

What is the infinite hyper root of 1000? See details?

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