The Mystery Of The 0th Root

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2023-07-27に共有
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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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コメント (21)
  • @Rhovanion85
    Zeroth also sounds like the final boss of a video game
  • @ianmathwiz7
    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.
  • @abhidababy6746
    There is another problem AFTER negative numbers: complex numbers: _---_ I clarify Zeroeth root of Negative/Complex Numbers
  • @denki2558
    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.
  • @ssaamil
    3:46 nice! I like this way of showing the "process" of taking limits. :D
  • @deltapi8859
    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.
  • @pressfinchat
    The problem with limits is that it describes how the function behaves at it gets closer and closer to 64^(1/0), but NOT at 64^(1/0) itself. For example, you can have a limit x-> 1 equal to 1 but f(1) can be something completely different, like f(1) = 2, even though it should be 1 if we trust what the limit is telling us. So, even with regular numbers, the argument that “for number a, the limit of this function approaches this value so the function must be this value” is already completely invalid, let alone when using it to explain a number like 0.
  • @7lllll
    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
  • 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.
  • @JuhaKona
    your videos are a beacon of clarity and inspiration!
  • @profdc9501
    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.
  • 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.
  • 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?
  • @RSLT
    Very good and unconventional explanation of limits. Great job!