The Mystery Of The 0th Root

This video explores the tricky concept of the zeroth root. Looking at limits, we find the zeroth root seems to lead to contradictions and inconsistencies. As we approach zero in the denominator, the quantity goes to infinity or zero depending on the number we start with. With no clear intuitive definition, the zeroth root remains an elusive mathematical idea that is traditionally left undefined.

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

[BriTheMathGuy]

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[marcusscience]

an iconic trio: division by 0, 0th root, log base 1

[Rhovanion]

Zeroth also sounds like the final boss of a video game

[ianmathwiz]

We could just have the zeroth root be a multifunction, like with other roots or with the complex logarithm. The zeroth root would have two branches if |z| is not 1, evaluating to either zero or infinity on each branch, and infinitely many branches if |z|=1.

Granted, the resulting Riemann surface would be highly pathological, but I think that's to be expected in a case like this.

[MusicEngineeer]

Yes - as others pointed out, maybe a multifunction could come to the rescue. At 1:20, you ask, how can the 0-th root of 1 be 2 and 3 at the same time. That made me immediately think: well, the square-root of 4 is 2 and -2 at the same time - so having multiple solutions is not such a crazy thing actually. The somewhat weird thing is only that now our set of solutions would become infinite - I guess even uncountably so.

[EdKolis]

The zeroth root is basically the infinite power, or what you'd get by applying compound interest for an infinite amount of time. Either zero, infinity, or one.

[abhidababy]

There is another problem AFTER

negative numbers:
complex numbers:
___---___
I clarify
Zeroeth root of Negative/Complex Numbers

[denki]

The zeroth root actually appears on the "generalized mean". The formula is `(sum(x_i^p)/n) root p` where n is the number of data points, and p is a parameter to control what type of mean to compute. p=-1, 1, 0.5, 2 correspond to the harmonic mean, arithmetic mean, RMS, and SMR respectively.

The mean should be undefined when p=0 (zeroth root), but if we used the limit, the parameter surprisingly correspond to the geometric mean, whose traditional formula has no additions nor divisions.

[deltapi]

I really like your style of doing math. Reasoning about which approaches you could take. Following them through and reflecting the merit of each approach. And one of those is going to be more suitable in the context of the field. This is totally not what you do in schools, because in schools you don't reason or try dead end paths for enquire. In schools everything has a "right" and "wrong" drawer things need to put in as fast as possible. Neither thinking nor science works this way.

[lllll]

the zeroth root is not a function, but a relation. zeroth root of 1 is the set of all real numbers, and the zeroth root of anything else is the empty set

[MichaelRothwell]

Nice topic the zeroth root - I hadn't thought of this before!

Interesting that in the first part you point out that ⁰√1 can take any value, then in the second part you say we had better define ⁰√1 to be 1...

For me it's like 0/0 - it can take any value, so is undefined.

[ssaamil]

3:46 nice! I like this way of showing the "process" of taking limits. :D

[signisot]

keeping in mind that this is the definitional inverse function of x^0, it would make sense that 1 maps to every positive and negative value simultaneously, and that 0 itself would be as hard to define in this equation as 0^0. When taking the limit of the function from the right we're effectively re-charting x^infinity, and from the left we're taking x^(-infinity) which is to say (1/x)^(infinity).
The behavior of the zeroth root really makes sense for all numbers, with some special behavior at 1 and some undefined value at 0 as per the 0^0 reason.

[profdc]

Another way to look at it is to examine the limit for y = exp(x) = lim n->infinity (1+x/n)^n

You can get a parameterized version of the log function with n: lim n->infinity y^(1/n) = 1 + (1/n)log y

changing parameter a = 1/n

y^a = 1 + a log y for a->0

so for power a tends to zero, the result is linear with the log function.

[periodictable]

The zeroth root of 1 can be literally any number since any number^0 = 1 (Even 0^0=1 according to many sources). What this really means is that 1^infinity can equal anything, which becomes painfully obvious when you first learn limits.

[dmimiguo]

- Daddy, why they say it is impossible to divide by zero?
- C'mon son. Sit down. We need to talk about USS Yorktown.

[masonwheeler]

3:10: "So the zeroth root of 64 should be 64^1/0. And herein lies another problem with the zeroth root: we're dividing by zero."

Is it really another problem, or is it exactly the same problem stated more clearly?

[PanduPoluan]

I always immediately think of roots as the reciprocal exponentiation.

So square root = x ^ (1/2)
Cube root = x ^ (1/3)

Zeroth root = x ^ (1/0)

Define 1/0 first, _then_ you can calculate the 0th root.

[gblargg]

1:30 Square roots in general have two values, e.g. sqrt(4) is +/-2. So having more than one value isn't immediately a problem.

[alejrandom]

0:31 the arrow is in the wrong direction. Example: (-1)²=1 => -1=sqrt(1) which is nonsense.