super square root of 256 = ?

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Solving x^2=2^x by using the super square root:

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Consider the function tet : R+ —> R, tet(x) = x^x, with x^x being understood as an abbreviation of exp[x·ln(x)]. If you find the critical points of tet, then you will notice that 1/e is where tet is equal to its lowest value, (1/e)^(1/e). In other words, its minimum is located at 1/e. With this, we know that there exists a function, ss, which is a domain-restriction of tet, that maps the interval [+1/e, +♾) to [+(1/e)^(1/e), +♾), with ss(x) = x^x still. Notice how, with this definition, ss is not an injective function and a surjective function. Therefore, we can define its compositional inverse, called ssrt : [+(1/e)^(1/e), +♾) —> [+1/e, +♾), such that ssrt(x^x) = x for every x in [+1/e, +♾), and ssrt(x)^ssrt(x) = x for every x in [+(1/e)^(1/e), +♾). We call it the super square root because it is analogous to the square root function: if we consider a function f : R —> [0, +♾), f(x) = x^2, then you can notice that sq : [0, +♾) —> [0, +♾), sq(x) = x^2 is a bijective domain-restriction of f, so sq has a compositional inverse sqrt : [0, +♾) —> [0, +♾) such that sqrt(x^2) = x and sqrt(x)^2 = x for every x in [0, +♾) (and not for every real x). Notice how there is no actual square-rooting being done when evaluating ssrt(x), but we call it the super square root because it is analogous to the square root in how it is defined.

For the record, an explicit expression for ssrt can be found in terms of other functions that BPRP has covered in the channel. Notice that x = y^y = exp[y·ln(y)]. Hence ln(x) = y·ln(y) = ln(y)·exp[ln(y)], so W[ln(x)] = ln(y), where W is the Lambert function, also called the product logarithm, and BPRP already has several videos on the main channel about this function, so check them out. Therefore, y = exp(W[ln(x)]). In other words, ssrt(x) = exp(W[ln(x)]). This also matches its domain definition perfectly, because W has domain [–1/e, +♾) as explained in the video, and ln(x) = –1/e occurs when x = (1/e)^(1/e), exactly what you would expect.

So ssrt(256) = exp(W[ln(256)]) = exp(W[4·ln(4)]) = exp[ln(4)] = 4.

angelmendez-rivera

I was curious about the Pokeball. Now I'm more interested in the math than ever.

varunika

Nice video! I wish there was a closed form to ssrt(x) that would be so cool

MathElite

Bruh I just learned some of this in class today what a weird coincidence

defaultyanny

I never heard it called "hyperpower"; "tetration", as I recall, is the usual term, originated by Goodstein.

WriteRightMathNation

x↑2 = x²
x↑↑2 = x↑x
So, ssrt(x↑x) = x

How about super cube root?
scrt(x↑x↑x) = x
x↑↑3 = x↑x↑x

r.a.

Everything in maths comes down to adding and it's counterpart, subtracting. The other basic operations, that is multiplying and dividing, are just glorified versions of the same thing. So for example, if you add 3 + 3 + 3 + 3 + 3, you can shorten the expression by just writing 3 * 5. Likewise, if you have 3 * 3 * 3 * 3 * 3, you can again shorten it, because it's just a 3 multiplied 5 times with itself, so you can write 3^5. If you have 3^3^3^3^3 you can shorten the expression with tetration. Next would be pentation, hexation and so on.

The standard form of a root is in the multiplicative sense.

When you take the square root of a number, you look for an answer that, when multiplied with itself, will give you your inital number. To get the answer, you divide the exponent of that number by 2.

n * n = m <=> n = m^(1/2)

If you take a step back and ask the same question in the additive sense, so which number do you have to add with itself to get the value, you can just divide your number by 2.

n + n = m <=> n = m/2

To get the next higher root, we have to consider the next higher operation to multiplying something, which is raising something to a power. The question now becomes: which number do you have to raise to itself to get the initial value? So what we have is a root in the exponential sense.

n^n = m <=> n = e^(W[ln(m)])

Sadly, you can't just divide the tetration by 2 here, so the pattern breaks. Instead, you have to use the lambert W function. This is the idea behind the "super square root".

weinsterle

OMG
I don't even know what's a Super Square Root 😳

juanpablozamoramoncada

These special functions just blow my mind. I find it amazing how the regular functions are extended to solve more complex equations.

Ninja

When are you changing your channel's name?

mathevengers

For those people who didn't understand super square root, it means square rooting the square root of something.

silvally

Did you make this notation up? What you call the super square root should be some sort of Lambert W.

godowskygodowsky

Only real ogs remember the first super square video

SuperRoj

@ blackpenredpen -- Regardless of the "channel name, " instead of going through the lessons like
sprints, which seems you want to do it as an ego thing, please do them them at closer to normal
rates so as not to lose some viewers. You remind me of certain bank tellers who are impressed
with themselves to count out the change in denominations of a large value bill in about two seconds.

robertveith