Can you please tell me how Aleph-Null differs from Infinity? Like, why use Infinity as the limit? Could Aleph-Null be the limit because it's the first number larger than any finite number.
@@FlatTopRoband You don’t win the battle…you merely convince Zeroth that fighting is meaningless by dragging the fight out for several hours of trying to endure.
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.
@@sakesaurus It most certainly is a multifunction, because it's defined as f(z) = exp(1/2*log(z)), and log(z) is a multifunction and the 1/2 means you get two branches of the complex logarithm before you run into the periodicity of the exponential function.
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.
The square-root of 4 is only 2. You can have -2 if resolving an equation, for example, x^2=4. Then, x=2 or x=-2. (Since sqrt(x^2)=|x|). But the square root of a number returns the positive output only, and not the negative one. Take the square-root function f(x)=sqrt(x). If f(4)=sqrt(4) would be equal to 2 and -2, then f would not be a properly defined function. I hope this clear things out!
@@Guillau213 What you said is correct for the square root function as it's conventionally defined for real-number inputs. However, the complex square root is typically defined to be a multivalued function, or a function defined on a Riemann surface. In this case, both 2 and -2 are valid square roots of 4. It's also possible to only take the principal branch of the complex square root function, in which case only 2 is a valid square root. But IMO this is not only arbitrary, but it destroys the mathematical properties of the Riemann surface.
@@Guillau213the complex squareroot is always multivalued, having 2 values that are negatives of other. The real squareroot only has 1 positive output
@@ianmathwiz7 You are right for complex roots. I was referring to the problem using real values only, since I thought the question was to define the function for real values. But at this point, if we are using the complex root, we are no longer in the real analysis world, so the rules are different ;)
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.
That's because teaching maths in schools is circumscribed by the need to maximise the performance of as many pupils as possible in a terminal exam on an arbitrary curriculum. If you want to change that, then vote out the politicians who think that they are the best people to decide what (and how) our children should learn.
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.
That's because this isn't just the zeroth root, it's the zeroth root of something to the zeroth power. Anything to the zeroth power is 1 and the zeroth root of 1 is 1 so it checks out, the geometric mean is recovered in the infinitesimal neighborhood of 1
@@viliml2763 If we write the top of the exponent tower as A, and logarithms from that as A-1, A-2, A-3 etc, then it's not such a big deal to change A to 0 and add instead of negate. Stern-Brocot algorithm generates inverse fractions which are pairs of fractions which have versions of 1/1 as their mediants, and unitary operators manifest IIRC also in other ways.
@@viliml2763 Other fascinating unitary operator contained in Stern-Brocot tree is that if we modify the fractions a/b to "simplicities" 1/ab, then each new row of simplicities adds up to 1.
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.
That's because he was using different definitions of the zeroth root in different sections of the video. Also, he only made these statements to show that they lead to absurd conclusions, not because he actually believes both statements.
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.
Yes, but no. It’s +♾️ when going + to zero, but -♾️ going - to zero. Plot y=1/x to get a better idea. However, I’m an engineer, so for all intents and purposes 1/0 is +♾️ and -1/0 is -♾️
There are cases where it is none of those 3. ie (-1)^∞ does not converge, but is bounded. Its also not just x^∞, its technically x^(e^iθ*∞) for whatever thats worth
@@msq7041I think OP's point is a practical use of it. There are applications of mathematics that inherently place conditions around what's feasible. For instance, a real world optimization problem in which the optimal value has units that can't possibly be negative in the real world despite the math alone concluding so.
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.
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
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.
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.
This logics fails however for x^(1/x), as when x approaches 0+, x^(1/x) approaches 0, and when x approaches 0-, x^(1/x) approaches 0. Thus x at 0 is defined here.
@@Tom-vb6fk No. If x at 0 is defined, then you just put in x=0 in the formula... which you can't. Rather, the *_limit_* of the formula as x approaches 0 from both sides is defined. But a limit is just that: a limit. It's not the actual result of x at 0.
@@PanduPoluan thats wrong, this shows that x is continuos at 0, and it converges to 0. Matter of fact x^(1/0) is defined for x is an element from (-1,1). You should learn what makes 1/0 undefine first
For some reason, I can’t add this video to my watch later list. And I can’t add it to my playlist. It was hard to physically watch this video. WTF youtube? I don’t think I’ve ever had this problem. I like this video
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?
This COULD make an interesting GRAPH. Plot y = "xth root of k" by plotting y = k ^ (1/x), for some selected values of k. • With k=1, the plot is a horizontal line at y=1. • With k=1.25, the plot has y=1.25 at x=+∞, exploding to y=+∞ as x approaches 0 from the right, AND y=0.8 at x=-∞, collapsing to y=0 as x approaches 0 from the left. • With k=0.8, the plot is a mirror image of the previous graph. The plot has y=0.8 at x=+∞, collapsing to y=0 as x approaches 0 from the right, AND y=1.25 at x=-∞, exploding to y=+∞ as x approaches 0 from the left.
Square roots do not have two values, otherwise it would not be a well defined function. The solution to x^2 = y will have two solutions for x, but that's a quirk of ^2, not the square root.
@@AdamBomb5794 I looked it up more and it's a subtle linguistic distinction. The number 4 has two square roots, 2 and -2. However the symbol √ denotes only the non-negative (principal) square root.
@@AdamBomb5794 Rather, it DOES have two values, thus is NOT a well defined function. You may CHOOSE to define it so it is well defined (for example so that the square root of a positive number is a positive number), but that's a rather arbitrary choice in general. Powers are generally well defined only if the exponent is an integer.
IEEE-754 numbers can solve this: 1/+0 = +∞, 1/-0 = -∞ . This is because +0 and -0 aren't "exactly 0", *they represent the limit as we approach 0 from the corresponding side.* This explains most of the weird quirks of FP math
My understanding of calculus allows me to agree with your take: That a number (other than 0) "a" raised to the (1/0) power (infinity) (from the left or the right) should remain undefined, it is as if one is trying to define a number which has a 0 factor in the denominator, and while limits allow for this, each equation should be treated with the respect it deserves, every problem is different. I'll note that 0^(1/0) or 0^(infinity) is equal to 0 in all cases when using a limit, such that it can and has been defined properly, but any other number cannot be used in such a manner, it breaks.
In my opinion, the function at 5:07 is the best way to define 0th root of a real number. The very last argument you made about the limiting values interchanging when taking x
No it cannot be dismissed Even when written as 1/64^(1/x) , it also tends to 0 64^(1/x) CANNOT be assigned a value at x = 0 because it doesn't approch a single value when x approches 0
@@olixx1213i think you might have misunderstood my point. To put it in simpler terms, my point is, that the case {x < 0 and a > 1} can be rewritten as the case {x > 0 and 0 < a < 1}, and vice versa. Hence making the problem at hand a bit less complex and making the function at 5:07 the closest way we can represent the 0th root.
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.
The 0th root has been one of my greatest sources of entertainment in mathematics... and this video brilliantly highlights why. I obviously completely agree with the conclusions reached... and with the conclusion of the linked video. 🙂
With division by 0, approaching 1/x from the left and right results in negative infinity and infinity, and it makes sense to define a number system where those are the same. Makes less sense to try to make 0 and infinity equal I guess.
@@vindi167 it's not really a question of objective truth, it's a question of what you want to do. In some cases, having them be the same and the number line being circular is useful, so use that. (E.g. dividing by 0). In other cases, having separate positive and negative infinity may be more useful. They're different number systems, just be clear which one you're using and don't mix them up.
The zeroth power of any number is one for a very good reason. Exponents are subtracted in division, so if the numerator and denominator have the same exponent meaning the numbers are equal the result is 1. n^x/n^x = n^(x-x) = n^0 = 1. It is not defined to be one, it IS one by calculation.
@@wayneyadams Yes, but it only makes sense on the surface. Given a > 0, the very standard approach is to *define* a^0 as 1, then a^n inductively for positive n, then a^-n as 1/a^n. (The definition can be further extended to rational and real exponents, but for the purposes of this discussion it is not necessary.) Next the identities a^(x+y) = a^x * a^y and (a^x)^y = a^(xy) are *proved* from these definitions. You seem to be doing something completely different. You *assume* the identity a^(x+y) = a^x * a^y (or similar) without saying how a^x is defined, and then you prove a^0 = 1 from that identity. For this approach to make sense, you would need to define a^x in such a way that the identity a^(x+y) = a^x * a^y becomes a more direct consequence of the definition then a^0 = 1. It's technically possible, but rather exotic and nobody really does that, unless for fun or education. Moreover, your approach fails to show that 0^0 = 1, which is an equally useful convention as a^0 = 1 for a > 0. That's why it is better to have that as a definition, not a consequence. If you disagree, let me ask: how do you define a^x?
@@adayah2933 I am using standard mathematical operations for exponents. There are no assumptions, or unsupported definitions. Exponentiation is shorthand notation for repetitive multiplication. For example, nxn = n^2 and nxnxn=n^3 (nxn)x(nxnxn) = nxnxnxnxn = n^5 n^2 x n^3 = n^5 In the same way division is accomplished by subtracting exponents. Let's divide. (nxn)/(nxnxn) = 1/n n^2/n^3 = n^(2-3) = n^-1 n^-1 = 1/n There is no reason n cannot be negative. n0. The only place it breaks down is when n=0 which is the only place we have to define the value.
then when x < -1 and we’re talking the limit from the right, it explodes into infinity but half of the time it crosses into the complex world depending on if there’s an even-th root
Hey, just wanted to say I only recently found your channel and I love what you are doing here. I do mathematical modeling in biotech R&D for living, and believe it or not, for most of my life i struggled with math, wspwcially as a kid: the way math was taught to me was very dogmatic, with little room for creativity or debates. It was only in uni when I literally started to build my own math out of necessity that I started to ubderstand what math really is. Now I have a newborn and I am often thinking how would I teach math to him: your approach is what I am going to follow with my kid. You are demistifying math as an a-priori god-given universal truth, and showing instead what math really is all about: a set of logically coherent rules which have consequences and limits. These rules can be whatever you want them to be, as long as you build them in a way that makes sense, but of course the limits, consequences, range of applicability and usefulness of your mathematics will very much vary based on which rules you decide to use. Thanks gor putting your content out there and keep u lp the good work!
The reason why it does matter from which side is the limit is because the power is negative so the limit of n to the 0- of 64 is like saying the limit of n to 0+ of 1/64 and the opposite with the 0.5
you can show √4 As 4^1/2 Hence to define the root 0 You'll need to get 1/0 and since 1/0 is infinity and not a number everything ends here it's like you have one end of a infinite line but since you don't have the other side You can not measure it. (summary)
Man, the comments section is a paradise.. all dudes talking maths with dignity and respect.. "you can't derivate that you f..of..x!!", "transform that function up your integral, you nondetermined polymath!!".. none 😊 peace man, you make the world go round 🙏
I mean, it makes sense to leave it undefined. Anything to the 0th power is 1, so the 0th root of one could be anything, but you can't even take the 0th root of any number other than 1 because the 0th power of any real number could never be anything except for 1. It's kind of like the 0th root of 1 is undefined and the 0th root of any number other than 1 is a domain error.
The only case it where might work is the 0thRoot(1)={x, x∈R/{0}} because we are basically solving for x^0=1 and that’s defined for all real numbers except 0. And for non 1 numbers we have 0thRoot(x≠1)= ∅ basically no solution for anything else. So I would be implying that 1^(1/0)= x which doesn’t make any sense except if you look at it as x^0=1. So to go from that to the other expression I’ll need to take the 0th power then to evaluate 0/0=1!? Anyway, stuff is weird ignore this comment, it just breaks down.
The cool thing about Undefined, is it is “not defined”. So if you’re looking to make use of applications of the concept of 0-rt(x), then you can define what it should do… in a particular Axiomatic system. Ofcourse, the contradictions matter a lot. There is actually a field of math that focuses on having good axioms (avoiding paradoxes and stuff),, which I haven’t studied. Regardless. It’s kinda fair game. Especially if you’re analyzing 0-rt(x) within the context of some sort of other mathematical model: For example, it could be useful for some specific kind of geometry, who knows.. An example of this is “Fractional dimensions”. The idea is: “Fractional dimensions” don’t exist/ they’re made up (probably: this might be a bad example though, because someone probably analyzes fractional-dimension shapes),,, anyhow… Fractals are easy to explain ~as though they are shapes belonging to a fractional dimensional space. So, it is useful to define things in math,,, when the numbers work out okay.
The reason that this concept is so undefinable is that f(x) = a^(1/x) has what is known as an “essential singularity” at x=0. If you approach 0 along the imaginary axis, you get infinite oscillations over finite space. Approaching from an arbitrary curve can let you approach any number you want. Essential singularities are a neat topic.
When it comes to mathematical computations and their limits such as a/0, tan(PI/2 + PI*N), vertical slope, etc... I personally rather list them or determine them to be indeterminate as opposed to "undefined". This for me pertains more to the context of the language than anything else. Take vertical slope for example which is basically the same as a/0 where a != 0, and tan(PI/2). They are basically the same thing. Vertical slope approaches either +/- infinity. For me this isn't undefined. It is however indeterminate because this is a many to one solution. Undefined to me means something that doesn't have a definition. And for things such as division by 0, vertical slope, tan(PI/2), etc... they are actually well defined. It's just that their results don't pass the vertical line test as they have more than one output. Take the general equation of a circle: (x-h)^2 + (y-k)^2 = r^2. Where the point (x,y) lies on its circumference, the point (h,k) is its center and r is its radius. This also doesn't pass the vertical line test because there are two outputs for most of its inputs. Yet, this equation or expression although isn't a function is well defined and isn't considered undefined. In fact, the equation of the circle for all tense and purposes is the same thing as the Pythagorean Theorem: A^2 + B^2 = C^2. It's just that one is relative to circles, where the other is relative to right triangles. Also, the slope of a given line from within the slope-intercept form y = mx+b is defined as (y2-y1)/(x2-x1) = dy/dx = sin(t)/cost(t) = tan(t) where t, theta is the angle that is between the line y = mx+b and the +x-axis. So for me I don't care for this idea that we've been taught that division by 0, and other phenomenon within mathematics is undefined. To me that means it doesn't have a definition, that it's not defined. Now, I can completely agree with indeterminate or ambiguous. As there is a many to one solution, or its output jumps all over the place based on specific ranges within its domain. This is just my take though. What if I were to tell you that that tangent function is actually continuous... Consider the fact that tan(t) = sin(t)/cos(t). We know that both sin and cos are continuous for all of its domain and that its domain is at least All Real Values since they are continuous wave functions that are periodic, oscillatory, rotational and transcendental. And the tan function can be composed of their ratios! Thus, for me, the tangent function also shares those same properties. Yeah, many will argue that the limit of the tangent function isn't the same when taking the left hand from the right hand limits... and to this argument I still wouldn't claim that it is undefined. I would claim that it is either indeterminate or ambiguous. The reason for this, is that tan(PI/2) approaches both + and - infinity and from the left and right hand limits it also approaches 0 besides just +/-infinity. This leads to the conclusion that it isn't undefined as this is no different than vertical slope or division by 0, it's just that we can not determine its output based on that single input as it is a many to one solution. This is what makes it indeterminate or ambiguous.
A few questions: 1:07 You said that any positive real number raised to the exponent of zero equals 1, but negative reals raised to the exponent zero also equal 1, so you don’t need to make that distinction: any nonzero real number will work. In fact, more broadly, any nonzero complex number raised to the exponent zero equals 1. As for zero itself, that’s a bit more debatable. 😄 Those immersed in calculus would say zero to the exponent zero is undefined, but if you ask someone who deals mostly with combinatorics, they’d probably say it equals 1. (The Google calculator, for example, says it equals 1.) 5:16 I’m not sure it’s so cut-and-dried to define the zeroth root of 1 to be 1. The “note” says that 1 to any exponent equals 1, which is true, but any nonzero real raised to the exponent of 0 equals one, which is just as consistent of a pattern, so mathematicians could just as easily define it as ♾️, though technically you could define it as anything you want if you don’t want to call it undefined. (I suppose defining it as 1 gives it a nice symmetry.)
Pretty close to the infinite root, simplifying as x^1/infinity, 1/infinity is approaching zero if not already. X^0 is 1 and therefore the infinite root of x is 1. But this works for any equation no matter the number, so they’re all 1. 1^infinity is undefined as well this being one of those reasons. This root also states every number is every other number and is therefore undefined. 0 is the opposite of infinity and negative infinity and as it involves 1/0 which approaches those exact values, and numbers approach two seperate values depending on the way we approach zero.
// just note that in geometry, there are objects that square to 1, 0, or -1. // these are square roots of 1. directions in space are such a thing. // these are exactly as weird as i*i=-1. u*u=1. v*v=1. // note that when you square a number, there are multiple possible choices // it is not defined which one it is, without other context x^2=4 implies (x=2 or x=-2 or x=2u or x=2v or ...). which can be converted into: x^2=4 implies x=2 or x^2=4 implies x=-2 or x^2=4 implies x=2u or ... We are not bothered by this, where a multiply by zero loses information about a: 0 a = b if you were to divide by 0, then you are saying that a could be anything. b/0 implies (a=0 or a=1 or a=1/2 or a=-2 or a=i or ...) // note that this is not a problem if a=0. // but by convention, we won't write b=0a as b/a=0 when a=0. 0a = b implies (b=0 and b/a=0) There is something we are not doing right in our notation when we track side-conditions.
I’m an aspiring mathematician and I’m actually working on division by 0, on the exponential part I found that if x is positive then x^1/0 (basically the 0th root)=1/0 If x=1 then x^1/0=1 And if x is negative then x^1/0=+/-1/0, I can’t detail everything but if y’all really want maybe I’ll make a video on the subject
x^(-y) is the same as (1/x)^y; this is why scientific notation for small decimals is generally written as x*10^(-y), such as the pressure of space being notated as 1x10^(-17) torr rather than 0.00000000000000001 torr. So in your examples at 5:25, 64^-(1/x) would equal (1/64)^(1/x) which would be between 0 and 1, while 0.5^-(1/x) would be the same as *1/0.5)^(1/x), which would just be 2^(1/x) and thus be greater than 1...thus the values for 64 coalescing to 0 and the values for 0.5 coalescing to infinity holds true to the earlier definition.
You're statement about those 2 things, x^(-y) and (1/x)^y, being the same in the first sentence, is not why scientific notation for small decimals is generally written the way it is. It is written that way out of convenience. In the way you have stated it, you claim a fact about exponents is the reason why and technically it's not. It is indeed a fact that what you said is true for x for real numbers except 0, but yeah.
This also goes into the rabbit hole of x divided by 0. A log (or anything else) divided into 2 pieces is one of the ways people teach division to children. If you want to cut a log into 1 piece, you don't make a cut at all. If you want to divide a log into 0 pieces, you delete the log from existence. So x/0 might be 0.
In the third grade when we were exposed to multiplying and dividing for the first time, we learned that multiplication was just repeated addition and division was repeated subtraction. Therefore, division by zero means subtracting zero until you can't subtract anymore. Therefore division by zero equals infinity. That has nothing to do with this video, though, even though it was mentioned as being undefined.
With the fractional formula instead (exponent outside the root), we can get: (sqrt[0](x))^y = x^(y/0) Let y = 0¹ (while adding in other 0 exponents): (sqrt[0¹](x))^0 = x^(0¹/0¹) Simplify: (sqrt[0¹](x))^0 = x¹ = x Rewrite this with the expression you used (sqrt[0¹](x) = x^(1/0) Substitute infinity for my personal definition (-1)! (x^(-1)!)^0¹ = x^(0(-1)!) = x^(0!) = x¹ = x Calculate y = 0² (sqrt[0¹](x))^0² = x^(0²/0) Simplify: (sqrt[0¹](x))^0² = x⁰ Same as before: (x^(-1)!)^0² = x⁰ x^(0(0)(-1)!) = x⁰ x⁰ = x⁰ One more thing, 0²th root: (sqrt[0²](x))^y = x^(y/0²) Define: y = 0³ (sqrt[0²](x))^0³ = x^(0³/0²) = x^0¹ = 1 And we can say: sqrt[0²](x) = x^(-1)!² Finally: x^(0(-1)!(0)(-1)!(0)) = x⁰ = 1 y = 0² (sqrt[0²](x))^0² = x^(0²/0²) = x¹ = x Substitute: sqrt[0²](x) = x^(-1)!² (x^(-1)!²)^0² = x^(0(-1)!(0)(-1)!) = x¹ = x y = 0¹ (sqrt[0²](x))⁰ = x^(0¹/0²) = x^(1/0) Substitute: x^((-1)!(0)(-1)!) = x^(1/0) x^(-1)! = x^(1/0) x^(1(infinity)¹) = x^(1(infinity)¹) Using our first expression; (sqrt[0¹](x))^y = x^(y/0¹) Let's define y = 1, so we get x^(1/0¹) : (sqrt[0¹](x))^1 = x^(1/0¹) Tidy it up: sqrt[0¹](x) = x^(1/0) And now we also know: (sqrt[0²](x))^0¹ = sqrt[0¹](x) = x^(1/0) = x^(-1)!
Also one more definition since I did sqrt[0¹] and sqrt[0²], I wanna do sqrt[0⁰]: (sqrt[0⁰](x))^y = x^(y/0⁰)) = x^(y/1) = x^y Also (sqrt[0⁰](x))^y = (sqrt[0/0](x))^y = x^(y/(0/0)) = x^(0y/0) = x^(1y) = x^y So sqrt[0⁰](x)^y = sqrt[0/0](x)^y = x^y
If we raise both sides of the equation x=°√A (A is some consonant) by the power of 0 we get: x^0=A For A=1 we get x=any number. For any A other than 1, no x satisfies the equation.
Self evident I’m sure, and bit more work, but I’m confident that a simple graph could better illustrate the the limits and the resulting disconnects. Still a concise explanation.
You can rewrite zeroth root of x as (e^(1/0))^lnx. Rewrite the expression , now you take the limit as n approaches 0 on, and you can see that the positive limit expression is undefined (+inf) at 1, and 0 for all the negative values, and the negative limit expression is undefined at 1 (+inf as well) and 0 for all the positive values. That means that the function has no real value anywhere, because the left side contradicts the right side of the limit, and the one point where both limits agree is undefined.
This behavior will only become more ridiculous when considering negative number bases, complex number bases, or approaching 0 from a phase other than 0 or π.
You said the answer. x is the set of all nonnegative real numbers. It's just that you can only take the 0th root of the number 1. But x can be anything. Using undefined /0 to get something else - verboten. BTW what is (-1)^0? BTW what is (i)^0?
a number divided by 0 is not undefined, it is infinity. only 0/0 is undefined. so root 0 is same as raised to power infinity. negative power just means 1/number with positive power.. so its pretty straight forward and does not really need limit to figure it out.
another way of finding that is it undefinied is for example. x^1/2 = the sqrt of 2 and x^1/3 = sqrt of 3 so if we were to do the zeroth root we will have to have x^1/0 which is underfined therefore the zeroth root is undefined
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Great video! But I dont agree with your final conclusion (of it being too hard to explain). Negative roots are just 1/|the root|, so this makes total sense...
Of course, 0^0 IS defined and equal to 1, in spite of many maths text books pretending it is undefined. It is not just your personal opinion! There are many ways to show it, starting with the definition of arithmetic operations based on set theory : x^y is the number of applications from a set of y elements to a set of x elements; there are no application from a non-empty set to the empty set, but exactly one application from the empty set to iself. That is, 0^n = 0 for n>1 and 1 for n=0. Or, to use a less abstract but equivalent concept, x^y is the number of possible words of y letters taken from an alphabet of x letters. Again, you cannot write any word if the alphabet is empty, apart from the empty word, so that there no word of length > 0 but one word of length 0 with an empty alphabet. Another argument comes from algebra. Whenever we use a formula like (a+b)^n=sum(binomial(i,n)·a^i·b^(m-i) for i=0 to n, one has to assume 0^0 = 0 for the formula to remain valid for any value of a and b. Just set a=0 and b=1 and it becomes obvious. Even those who say that 0^0 is undefined use this kind of formula without noticing... The confusion comes from the fact that in calculus, the limit of x^y when both x and y tend to 0 is undefined (or can be any number, including 0 or infinity if x and y are not independant). It means that this function of two arguments is discontinuous at (0,0) - a point which is on the border of its domain) - not that it is undefined.
You can say that 64^1/0.5 is bigger than 64^1/0.6, but you cannot interpolate it to "64^1/0.5 is less than 64^1/0" because you have just zero with no sign. And for +0 or -0 (which are only symbols but not numbers) you will get a different solution. So a small answer: in ordinary math analysis x^1/0 is undefined, end of story
Take a hint from the IEEE-754 floating point math standard: NaN = Not-a-Number. This turns strange results like this into something at least vaguely computable. It has been a part of every FPU you have ever touched, but most people still don't know about it.
logic method: 0th root of any number is infinite except 1 because 0th root of 1 is equal to any number, because for example, n^0 is 1, it means that 0th root of 1 can be any number
having a zero on the bottom of a fraction defies axioms as it is undefined, when you write 0rootx as x^(1/0), the 0 is on the bottom of the fraction making it outside the bounds of our maths,
Zeroth root of -0.5 tends to 0 because you keep repeatedly multiplying itself by itself infinitely so it's the same result as +0.5 just oscillating + and - but still shrinking. Not infinity as the video says. i.e. -0.5 1/4 -1/8 1/16 -1/32 etc
Oh i see your confusion. It was kind of confusing on his part, but what you got wrong is that he meant the "x" in the exponent approaches zero. He wasn't talking about the base, which he also called "x" later in the piece "definition" of zeroth root. He should have called the base "n" or something
This is not always undefined though, for x an element from (-1,1) it is defined for x^(1/0) as when when approach from 0+, x^(1/0+) approaches 0, while approaching 0-, x^(1/0-) approaches 0, meaning x^(1/0) is defined for x in (-1,1), which is 0.
The eternal issue about about deviding by 0 is that the lim is not the final answer unless you accept it as the final answer. "Lim" meaning "infinitely close to but not quite there". The devil is in the details... Whenever infinity is involved the rules break down, hence 'undefined'. In fact infinity is undefined, it belongs to the realm of philosophy. There you will be scating an very thin ice and all bets are off.
A root is just an exponent in fraction form A square root is N^1/2 So if we do root 0 we would have N^1/0 1/0 is undefined, you cannot define the answer when you divide by 0
idk it seems internally consistent to me. you defined 2^1/0 to be infinity using the limit process. .5^1/0 can be written as 1/(2^1/0), which would be 1/infinity which we usually say is 0. this is exactly what you got. and when making the exponent negative, the order flips because a negative exponent flips a fraction making each thing be the opposite. it’s all technically internally consistent
When the number is 0 < X < 1; isn't that quantity in some way essentially 1/x? So it makes sense that 1/inf goes to zero. It makes sense to me that the inverse of of the set of whole numbers (found in 0
To define, say 0th root of n = NAN(n) and NAN(n)^0 =n and NAN(NAN(n))=NAN^2(n), where NAN^2(n))^0=NAN(n). Can generalize to get NAN^b(n)=n^(1/(0^b)). You just can’t replace 0^b by 0. fixed in this “axiom” : a not congruent to b implies a cannot be substituted for b, and b cannot be substituted for a, even if a is c*0 and b is d*0, unless c=d. 0 is congruent to 1*0, but not z*0 for all z not equal to 1. 0 is congruent to 1-1. This also resolves issues with 1/0 when you realize n/0 is not congruent to m/0 unless n=m, or otherwise math breaks, and how (-1)/0 is not congruent to (+1)/0 but 1/(-0)=1/(+0) by stating 1/(-0) is not congruent to 1/(+0). There is also a nice video on an extension to include log(0) by looking at properties + and * and e^x have, instead of repeated multiplication. Not so undefined anymore… 1/0 is the number of points of length 0 on a line of length 1. 1/0 is also (-1)! Or 0!/0=1/0 and is -1*-2*-3*… Binomial theorem without conditions, says 0!=0^0 so since 0!=1, 0^0=1 Log base 1 could also probably be defined in some kind of similar manner.
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Can you please tell me how Aleph-Null differs from Infinity? Like, why use Infinity as the limit? Could Aleph-Null be the limit because it's the first number larger than any finite number.
@@arcaltoby5772 aleph of null contains infinity but not infinity times infinity
in my opinion the ⁰√x is just simultaneously 0 and ∞ for any positive number that isn't 1
If x
@@dinh1cam36 the way i see it, for x=1 it's 1, for x=-1 it's the unit circle on the complex plane, otherwise it's simultaneously 0 and infinity
an iconic trio: division by 0, 0th root, log base 1
They are all division by 0 in disguise.
⁰√x=x^(1/0)
log_1(x)=ln(x)/ln(1)=ln(x)/0
I imagine this as one of those "tell us a little bit about yourself" segments but instead 1/0 keeps appearing with fake mustaches and wigs
More iconic than Lebron D Wade and Bosh
Div0, div0 in a trench coat, div0 with a fake mustache
Well log base 1 has complex solutions. Log base 0 on the other hand...
Schrödinger’s root
😂😂😂
Lmao
Shouldn't comment about something you know about 0.064% about.
@@UdayShankar0shouldnt be on the internet if you cant take simple jokes
@@OzOMega well said and destroyed
Zeroth also sounds like the final boss of a video game
And there's no strategy guide to beat it.
@@FlatTopRoband You don’t win the battle…you merely convince Zeroth that fighting is meaningless by dragging the fight out for several hours of trying to endure.
and now im reading “zeroth” not as zero-th but zer-oth
Zeroth hand of golden buddha
Sephizeroth?
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.
this idiot doesn't know about multifunctions
I just wanted to say. This only screams for an analogue to (a little bit more complex, no pun intended) complex numbers. I see no issue here.
no. Square root is not a multi function. Why should 0th root be?
@@sakesaurus It most certainly is a multifunction, because it's defined as
f(z) = exp(1/2*log(z)), and log(z) is a multifunction and the 1/2 means you get two branches of the complex logarithm before you run into the periodicity of the exponential function.
The only word I understood was "if"
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.
The square-root of 4 is only 2. You can have -2 if resolving an equation, for example, x^2=4. Then, x=2 or x=-2. (Since sqrt(x^2)=|x|). But the square root of a number returns the positive output only, and not the negative one. Take the square-root function f(x)=sqrt(x). If f(4)=sqrt(4) would be equal to 2 and -2, then f would not be a properly defined function. I hope this clear things out!
@@Guillau213 What you said is correct for the square root function as it's conventionally defined for real-number inputs. However, the complex square root is typically defined to be a multivalued function, or a function defined on a Riemann surface. In this case, both 2 and -2 are valid square roots of 4.
It's also possible to only take the principal branch of the complex square root function, in which case only 2 is a valid square root. But IMO this is not only arbitrary, but it destroys the mathematical properties of the Riemann surface.
@@Guillau213the complex squareroot is always multivalued, having 2 values that are negatives of other. The real squareroot only has 1 positive output
@@ianmathwiz7 You are right for complex roots. I was referring to the problem using real values only, since I thought the question was to define the function for real values. But at this point, if we are using the complex root, we are no longer in the real analysis world, so the rules are different ;)
yeah just like the complex log :)
3:46 nice! I like this way of showing the "process" of taking limits. :D
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.
Excellent point! Thinking many options through is such an important process in math!
That's because teaching maths in schools is circumscribed by the need to maximise the performance of as many pupils as possible in a terminal exam on an arbitrary curriculum. If you want to change that, then vote out the politicians who think that they are the best people to decide what (and how) our children should learn.
@@RexxSchneider Don't even know how to answer this comment other than: perfect.
There is another problem AFTER
negative numbers:
complex numbers:
___---___
I clarify
Zeroeth root of Negative/Complex Numbers
-1√X = y where y^-1=X
y^-1 = 1/y
-1√2 = 0.5
Xth root of y = y^(1÷x) and for I, that would be y^(-i) so i√2 = 0.7692389-0.6389613i
@@lsf_yunjin-fearnotI'm never reading math video comments ever again, my brain hurts
I love the image of the same attitude applied to complex numbers 'cause it limits to a circle at infinity. It's such a pretty spiral!
Fancy a McDonald's
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.
Youp, just explained in another comment how the thingie can be takens as the root seed of Stern-Brocot style concatenation of mediants.
That's because this isn't just the zeroth root, it's the zeroth root of something to the zeroth power.
Anything to the zeroth power is 1 and the zeroth root of 1 is 1 so it checks out, the geometric mean is recovered in the infinitesimal neighborhood of 1
@@viliml2763 If we write the top of the exponent tower as A, and logarithms from that as A-1, A-2, A-3 etc, then it's not such a big deal to change A to 0 and add instead of negate.
Stern-Brocot algorithm generates inverse fractions which are pairs of fractions which have versions of 1/1 as their mediants, and unitary operators manifest IIRC also in other ways.
@@viliml2763 Other fascinating unitary operator contained in Stern-Brocot tree is that if we modify the fractions a/b to "simplicities" 1/ab, then each new row of simplicities adds up to 1.
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.
That's because he was using different definitions of the zeroth root in different sections of the video. Also, he only made these statements to show that they lead to absurd conclusions, not because he actually believes both statements.
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.
That point is addressed in the video, the left and right limits to those values don't "meet up"
Yes, but no. It’s +♾️ when going + to zero, but -♾️ going - to zero. Plot y=1/x to get a better idea. However, I’m an engineer, so for all intents and purposes 1/0 is +♾️ and -1/0 is -♾️
There are cases where it is none of those 3. ie (-1)^∞ does not converge, but is bounded. Its also not just x^∞, its technically x^(e^iθ*∞) for whatever thats worth
@@ptrkmras a mathematician, you cannot imagine how cursed that is...
@@msq7041I think OP's point is a practical use of it. There are applications of mathematics that inherently place conditions around what's feasible. For instance, a real world optimization problem in which the optimal value has units that can't possibly be negative in the real world despite the math alone concluding so.
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.
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
Not real numbers, but something much better which includes also 1/0. ;)
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.
Your claim that 1^infinity can equal anything is not "painfully obvious". It's not even meaningful.
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.
This logics fails however for x^(1/x), as when x approaches 0+, x^(1/x) approaches 0, and when x approaches 0-, x^(1/x) approaches 0. Thus x at 0 is defined here.
@@Tom-vb6fk No. If x at 0 is defined, then you just put in x=0 in the formula... which you can't.
Rather, the *_limit_* of the formula as x approaches 0 from both sides is defined.
But a limit is just that: a limit. It's not the actual result of x at 0.
@@PanduPoluan thats wrong, this shows that x is continuos at 0, and it converges to 0. Matter of fact x^(1/0) is defined for x is an element from (-1,1). You should learn what makes 1/0 undefine first
For some reason, I can’t add this video to my watch later list. And I can’t add it to my playlist.
It was hard to physically watch this video. WTF youtube? I don’t think I’ve ever had this problem.
I like this video
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?
This COULD make an interesting GRAPH. Plot y = "xth root of k" by plotting y = k ^ (1/x), for some selected values of k.
• With k=1, the plot is a horizontal line at y=1.
• With k=1.25, the plot has y=1.25 at x=+∞, exploding to y=+∞ as x approaches 0 from the right, AND y=0.8 at x=-∞, collapsing to y=0 as x approaches 0 from the left.
• With k=0.8, the plot is a mirror image of the previous graph. The plot has y=0.8 at x=+∞, collapsing to y=0 as x approaches 0 from the right, AND y=1.25 at x=-∞, exploding to y=+∞ as x approaches 0 from the left.
your videos are a beacon of clarity and inspiration!
Very good and unconventional explanation of limits. Great job!
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.
what
Square roots do not have two values, otherwise it would not be a well defined function. The solution to x^2 = y will have two solutions for x, but that's a quirk of ^2, not the square root.
@@giabao576 SQUARE ROOTS IN GENERAL HAVE TWO VALUES. What you might be thinking of is the *principal* square root, the non-negative of the two.
@@AdamBomb5794 I looked it up more and it's a subtle linguistic distinction. The number 4 has two square roots, 2 and -2. However the symbol √ denotes only the non-negative (principal) square root.
@@AdamBomb5794 Rather, it DOES have two values, thus is NOT a well defined function. You may CHOOSE to define it so it is well defined (for example so that the square root of a positive number is a positive number), but that's a rather arbitrary choice in general. Powers are generally well defined only if the exponent is an integer.
IEEE-754 numbers can solve this: 1/+0 = +∞, 1/-0 = -∞ . This is because +0 and -0 aren't "exactly 0", *they represent the limit as we approach 0 from the corresponding side.* This explains most of the weird quirks of FP math
My understanding of calculus allows me to agree with your take: That a number (other than 0) "a" raised to the (1/0) power (infinity) (from the left or the right) should remain undefined, it is as if one is trying to define a number which has a 0 factor in the denominator, and while limits allow for this, each equation should be treated with the respect it deserves, every problem is different. I'll note that 0^(1/0) or 0^(infinity) is equal to 0 in all cases when using a limit, such that it can and has been defined properly, but any other number cannot be used in such a manner, it breaks.
You know stuff gets crazy when even Bri leaves the expression undefined
In my opinion, the function at 5:07 is the best way to define 0th root of a real number. The very last argument you made about the limiting values interchanging when taking x
No it cannot be dismissed
Even when written as 1/64^(1/x) , it also tends to 0
64^(1/x) CANNOT be assigned a value at x = 0 because it doesn't approch a single value when x approches 0
@@olixx1213i think you might have misunderstood my point. To put it in simpler terms, my point is, that the case {x < 0 and a > 1} can be rewritten as the case {x > 0 and 0 < a < 1}, and vice versa. Hence making the problem at hand a bit less complex and making the function at 5:07 the closest way we can represent the 0th root.
@@a_minor No you cannot , because we need x to be 0 since its the 0th root of a
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.
0:31 the arrow is in the wrong direction. Example: (-1)²=1 => -1=sqrt(1) which is nonsense.
The 0th root has been one of my greatest sources of entertainment in mathematics... and this video brilliantly highlights why. I obviously completely agree with the conclusions reached... and with the conclusion of the linked video. 🙂
With division by 0, approaching 1/x from the left and right results in negative infinity and infinity, and it makes sense to define a number system where those are the same.
Makes less sense to try to make 0 and infinity equal I guess.
Saying +/- inf is correct tho
i agree, i think +∞ and -∞ are equal. everything i've seen points to the idea of the numberline looping at ∞
@@vindi167 it's not really a question of objective truth, it's a question of what you want to do. In some cases, having them be the same and the number line being circular is useful, so use that. (E.g. dividing by 0). In other cases, having separate positive and negative infinity may be more useful. They're different number systems, just be clear which one you're using and don't mix them up.
The zeroth power of any number is one for a very good reason. Exponents are subtracted in division, so if the numerator and denominator have the same exponent meaning the numbers are equal the result is 1. n^x/n^x = n^(x-x) = n^0 = 1. It is not defined to be one, it IS one by calculation.
It is defined to be 1...
@@adayah2933 Did you read and understand my post about why it is one, not defined as one?
@@wayneyadams Yes, but it only makes sense on the surface. Given a > 0, the very standard approach is to *define* a^0 as 1, then a^n inductively for positive n, then a^-n as 1/a^n. (The definition can be further extended to rational and real exponents, but for the purposes of this discussion it is not necessary.) Next the identities a^(x+y) = a^x * a^y and (a^x)^y = a^(xy) are *proved* from these definitions.
You seem to be doing something completely different. You *assume* the identity a^(x+y) = a^x * a^y (or similar) without saying how a^x is defined, and then you prove a^0 = 1 from that identity. For this approach to make sense, you would need to define a^x in such a way that the identity a^(x+y) = a^x * a^y becomes a more direct consequence of the definition then a^0 = 1. It's technically possible, but rather exotic and nobody really does that, unless for fun or education.
Moreover, your approach fails to show that 0^0 = 1, which is an equally useful convention as a^0 = 1 for a > 0. That's why it is better to have that as a definition, not a consequence.
If you disagree, let me ask: how do you define a^x?
@@adayah2933 I am using standard mathematical operations for exponents. There are no assumptions, or unsupported definitions. Exponentiation is shorthand notation for repetitive multiplication.
For example, nxn = n^2 and nxnxn=n^3
(nxn)x(nxnxn) = nxnxnxnxn = n^5
n^2 x n^3 = n^5
In the same way division is accomplished by subtracting exponents.
Let's divide. (nxn)/(nxnxn) = 1/n
n^2/n^3 = n^(2-3) = n^-1
n^-1 = 1/n
There is no reason n cannot be negative. n0. The only place it breaks down is when n=0 which is the only place we have to define the value.
The zeroth root is n^1/0, so it would inherently be undefined. I’m interested to see how you made a video out of it though.
then when x < -1 and we’re talking the limit from the right, it explodes into infinity but half of the time it crosses into the complex world depending on if there’s an even-th root
- Daddy, why they say it is impossible to divide by zero?
- C'mon son. Sit down. We need to talk about USS Yorktown.
Hey, just wanted to say I only recently found your channel and I love what you are doing here. I do mathematical modeling in biotech R&D for living, and believe it or not, for most of my life i struggled with math, wspwcially as a kid: the way math was taught to me was very dogmatic, with little room for creativity or debates. It was only in uni when I literally started to build my own math out of necessity that I started to ubderstand what math really is. Now I have a newborn and I am often thinking how would I teach math to him: your approach is what I am going to follow with my kid. You are demistifying math as an a-priori god-given universal truth, and showing instead what math really is all about: a set of logically coherent rules which have consequences and limits. These rules can be whatever you want them to be, as long as you build them in a way that makes sense, but of course the limits, consequences, range of applicability and usefulness of your mathematics will very much vary based on which rules you decide to use. Thanks gor putting your content out there and keep u lp the good work!
The reason why it does matter from which side is the limit is because the power is negative so the limit of n to the 0- of 64 is like saying the limit of n to 0+ of 1/64 and the opposite with the 0.5
you can show
√4 As 4^1/2 Hence to define the root 0 You'll need to get 1/0
and since 1/0 is infinity and not a number everything ends here
it's like you have one end of a infinite line but since you don't have the other side You can not measure it.
(summary)
Man, the comments section is a paradise.. all dudes talking maths with dignity and respect.. "you can't derivate that you f..of..x!!", "transform that function up your integral, you nondetermined polymath!!".. none 😊 peace man, you make the world go round 🙏
x^½ = ²√x
x^⅓ = ³√x
Therefore ⁰√x = x^(1/0)
1/0 is undefined
A number to an undefined number is undefined
that's why i love math. Sometimes things so complicated that they make you laugh.
I mean, it makes sense to leave it undefined. Anything to the 0th power is 1, so the 0th root of one could be anything, but you can't even take the 0th root of any number other than 1 because the 0th power of any real number could never be anything except for 1. It's kind of like the 0th root of 1 is undefined and the 0th root of any number other than 1 is a domain error.
I like the idea of it being equal to multiple values. It looks similar to superposition.
What is if you try defining it with an "any of" prefix?
The only case it where might work is the 0thRoot(1)={x, x∈R/{0}} because we are basically solving for x^0=1 and that’s defined for all real numbers except 0. And for non 1 numbers we have 0thRoot(x≠1)= ∅ basically no solution for anything else. So I would be implying that 1^(1/0)= x which doesn’t make any sense except if you look at it as x^0=1. So to go from that to the other expression I’ll need to take the 0th power then to evaluate 0/0=1!? Anyway, stuff is weird ignore this comment, it just breaks down.
The cool thing about Undefined, is it is “not defined”. So if you’re looking to make use of applications of the concept of 0-rt(x), then you can define what it should do… in a particular Axiomatic system.
Ofcourse, the contradictions matter a lot. There is actually a field of math that focuses on having good axioms (avoiding paradoxes and stuff),, which I haven’t studied.
Regardless. It’s kinda fair game. Especially if you’re analyzing 0-rt(x) within the context of some sort of other mathematical model: For example, it could be useful for some specific kind of geometry, who knows..
An example of this is “Fractional dimensions”. The idea is: “Fractional dimensions” don’t exist/ they’re made up (probably: this might be a bad example though, because someone probably analyzes fractional-dimension shapes),,, anyhow… Fractals are easy to explain ~as though they are shapes belonging to a fractional dimensional space. So, it is useful to define things in math,,, when the numbers work out okay.
Glad to see you back buddy
The reason that this concept is so undefinable is that f(x) = a^(1/x) has what is known as an “essential singularity” at x=0. If you approach 0 along the imaginary axis, you get infinite oscillations over finite space. Approaching from an arbitrary curve can let you approach any number you want. Essential singularities are a neat topic.
brilliantly twisted and interesting.
When it comes to mathematical computations and their limits such as a/0, tan(PI/2 + PI*N), vertical slope, etc... I personally rather list them or determine them to be indeterminate as opposed to "undefined". This for me pertains more to the context of the language than anything else. Take vertical slope for example which is basically the same as a/0 where a != 0, and tan(PI/2). They are basically the same thing. Vertical slope approaches either +/- infinity. For me this isn't undefined. It is however indeterminate because this is a many to one solution. Undefined to me means something that doesn't have a definition. And for things such as division by 0, vertical slope, tan(PI/2), etc... they are actually well defined. It's just that their results don't pass the vertical line test as they have more than one output.
Take the general equation of a circle: (x-h)^2 + (y-k)^2 = r^2. Where the point (x,y) lies on its circumference, the point (h,k) is its center and r is its radius. This also doesn't pass the vertical line test because there are two outputs for most of its inputs. Yet, this equation or expression although isn't a function is well defined and isn't considered undefined. In fact, the equation of the circle for all tense and purposes is the same thing as the Pythagorean Theorem: A^2 + B^2 = C^2. It's just that one is relative to circles, where the other is relative to right triangles.
Also, the slope of a given line from within the slope-intercept form y = mx+b is defined as (y2-y1)/(x2-x1) = dy/dx = sin(t)/cost(t) = tan(t) where t, theta is the angle that is between the line y = mx+b and the +x-axis. So for me I don't care for this idea that we've been taught that division by 0, and other phenomenon within mathematics is undefined. To me that means it doesn't have a definition, that it's not defined. Now, I can completely agree with indeterminate or ambiguous. As there is a many to one solution, or its output jumps all over the place based on specific ranges within its domain.
This is just my take though. What if I were to tell you that that tangent function is actually continuous... Consider the fact that tan(t) = sin(t)/cos(t). We know that both sin and cos are continuous for all of its domain and that its domain is at least All Real Values since they are continuous wave functions that are periodic, oscillatory, rotational and transcendental. And the tan function can be composed of their ratios! Thus, for me, the tangent function also shares those same properties. Yeah, many will argue that the limit of the tangent function isn't the same when taking the left hand from the right hand limits... and to this argument I still wouldn't claim that it is undefined. I would claim that it is either indeterminate or ambiguous. The reason for this, is that tan(PI/2) approaches both + and - infinity and from the left and right hand limits it also approaches 0 besides just +/-infinity. This leads to the conclusion that it isn't undefined as this is no different than vertical slope or division by 0, it's just that we can not determine its output based on that single input as it is a many to one solution. This is what makes it indeterminate or ambiguous.
awesome intuitive approach
A few questions:
1:07 You said that any positive real number raised to the exponent of zero equals 1, but negative reals raised to the exponent zero also equal 1, so you don’t need to make that distinction: any nonzero real number will work. In fact, more broadly, any nonzero complex number raised to the exponent zero equals 1. As for zero itself, that’s a bit more debatable. 😄 Those immersed in calculus would say zero to the exponent zero is undefined, but if you ask someone who deals mostly with combinatorics, they’d probably say it equals 1. (The Google calculator, for example, says it equals 1.)
5:16 I’m not sure it’s so cut-and-dried to define the zeroth root of 1 to be 1. The “note” says that 1 to any exponent equals 1, which is true, but any nonzero real raised to the exponent of 0 equals one, which is just as consistent of a pattern, so mathematicians could just as easily define it as ♾️, though technically you could define it as anything you want if you don’t want to call it undefined. (I suppose defining it as 1 gives it a nice symmetry.)
Pretty close to the infinite root, simplifying as x^1/infinity, 1/infinity is approaching zero if not already. X^0 is 1 and therefore the infinite root of x is 1. But this works for any equation no matter the number, so they’re all 1. 1^infinity is undefined as well this being one of those reasons. This root also states every number is every other number and is therefore undefined. 0 is the opposite of infinity and negative infinity and as it involves 1/0 which approaches those exact values, and numbers approach two seperate values depending on the way we approach zero.
// just note that in geometry, there are objects that square to 1, 0, or -1.
// these are square roots of 1. directions in space are such a thing.
// these are exactly as weird as i*i=-1.
u*u=1. v*v=1.
// note that when you square a number, there are multiple possible choices
// it is not defined which one it is, without other context
x^2=4 implies (x=2 or x=-2 or x=2u or x=2v or ...).
which can be converted into:
x^2=4 implies x=2
or
x^2=4 implies x=-2
or
x^2=4 implies x=2u
or
...
We are not bothered by this, where a multiply by zero loses information about a:
0 a = b
if you were to divide by 0, then you are saying that a could be anything.
b/0 implies (a=0 or a=1 or a=1/2 or a=-2 or a=i or ...)
// note that this is not a problem if a=0.
// but by convention, we won't write b=0a as b/a=0 when a=0.
0a = b implies (b=0 and b/a=0)
There is something we are not doing right in our notation when we track side-conditions.
I’m an aspiring mathematician and I’m actually working on division by 0, on the exponential part I found that if x is positive then
x^1/0 (basically the 0th root)=1/0
If x=1 then
x^1/0=1
And if x is negative then
x^1/0=+/-1/0, I can’t detail everything but if y’all really want maybe I’ll make a video on the subject
x^(-y) is the same as (1/x)^y; this is why scientific notation for small decimals is generally written as x*10^(-y), such as the pressure of space being notated as 1x10^(-17) torr rather than 0.00000000000000001 torr.
So in your examples at 5:25, 64^-(1/x) would equal (1/64)^(1/x) which would be between 0 and 1, while 0.5^-(1/x) would be the same as *1/0.5)^(1/x), which would just be 2^(1/x) and thus be greater than 1...thus the values for 64 coalescing to 0 and the values for 0.5 coalescing to infinity holds true to the earlier definition.
The values don't coalesce, though. Coalesce means to come together.
You're statement about those 2 things, x^(-y) and (1/x)^y, being the same in the first sentence, is not why scientific notation for small decimals is generally written the way it is.
It is written that way out of convenience.
In the way you have stated it, you claim a fact about exponents is the reason why and technically it's not.
It is indeed a fact that what you said is true for x for real numbers except 0, but yeah.
This also goes into the rabbit hole of x divided by 0. A log (or anything else) divided into 2 pieces is one of the ways people teach division to children. If you want to cut a log into 1 piece, you don't make a cut at all. If you want to divide a log into 0 pieces, you delete the log from existence. So x/0 might be 0.
Because x^0 is 1 ... But in complex 1=exp(2*pi*k*i). (K is Natural number) Then roots zero deg. Is for example 2^(1/2*pi*i*k)
0:17
we already started badly
The root of 4 can also be -2; since (-2)²=4.
And cube root of 8 can also be -1±√3i; since that cubed is also 8.
you could also treat it as n^0=x, which makes n be anything if x=1 and undefined if x≠1
You could have added a graph showing the limit trends (0 or ∞) on either side of -1 and +1.
You should make a video with the graph.
In the third grade when we were exposed to multiplying and dividing for the first time, we learned that multiplication was just repeated addition and division was repeated subtraction. Therefore, division by zero means subtracting zero until you can't subtract anymore. Therefore division by zero equals infinity. That has nothing to do with this video, though, even though it was mentioned as being undefined.
With the fractional formula instead (exponent outside the root), we can get:
(sqrt[0](x))^y = x^(y/0)
Let y = 0¹ (while adding in other 0 exponents):
(sqrt[0¹](x))^0 = x^(0¹/0¹)
Simplify:
(sqrt[0¹](x))^0 = x¹ = x
Rewrite this with the expression you used
(sqrt[0¹](x) = x^(1/0)
Substitute infinity for my personal definition (-1)!
(x^(-1)!)^0¹ = x^(0(-1)!) = x^(0!) = x¹ = x
Calculate y = 0²
(sqrt[0¹](x))^0² = x^(0²/0)
Simplify:
(sqrt[0¹](x))^0² = x⁰
Same as before:
(x^(-1)!)^0² = x⁰
x^(0(0)(-1)!) = x⁰
x⁰ = x⁰
One more thing, 0²th root:
(sqrt[0²](x))^y = x^(y/0²)
Define: y = 0³
(sqrt[0²](x))^0³ = x^(0³/0²) = x^0¹ = 1
And we can say:
sqrt[0²](x) = x^(-1)!²
Finally:
x^(0(-1)!(0)(-1)!(0)) = x⁰ = 1
y = 0²
(sqrt[0²](x))^0² = x^(0²/0²) = x¹ = x
Substitute:
sqrt[0²](x) = x^(-1)!²
(x^(-1)!²)^0² = x^(0(-1)!(0)(-1)!) = x¹ = x
y = 0¹
(sqrt[0²](x))⁰ = x^(0¹/0²) = x^(1/0)
Substitute:
x^((-1)!(0)(-1)!) = x^(1/0)
x^(-1)! = x^(1/0)
x^(1(infinity)¹) = x^(1(infinity)¹)
Using our first expression;
(sqrt[0¹](x))^y = x^(y/0¹)
Let's define y = 1, so we get x^(1/0¹) :
(sqrt[0¹](x))^1 = x^(1/0¹)
Tidy it up:
sqrt[0¹](x) = x^(1/0)
And now we also know:
(sqrt[0²](x))^0¹ = sqrt[0¹](x) = x^(1/0) = x^(-1)!
Inverses quickly
Okay let's take:
(sqrt[0¹](x))^y = x^(y/0¹)
Define y = -1(0) or -0¹
sqrt[0¹](x))^(-1(0)) = x^(-0¹/0¹)
Let's take the second-half first:
x^(-0¹/0¹) = x^(-1) = 1/x
Substitute sqrt[0¹](x) for x^(-1)!:
(sqrt[0¹](x))^(-0) = x^((-1)(0)(-1)!) = x^(-1) = 1/x
Now one more: y = -0²
(sqrt[0¹](x))^(-0²) = x^(-0²/0¹)
Substitute sqrt[0¹](x) for x^(-1)!, simplify second-half:
(x^(-1)!)^(0²(-1)) = x^((-1)(0²/0¹)) = (1/x)⁰ = 1
Simplify first half now:
(x^(-1)!)^(0²(-1)) = x^((-1)(0)) = (1/x)⁰ = 1
Also one more definition since I did sqrt[0¹] and sqrt[0²], I wanna do sqrt[0⁰]:
(sqrt[0⁰](x))^y = x^(y/0⁰)) = x^(y/1) = x^y
Also
(sqrt[0⁰](x))^y = (sqrt[0/0](x))^y = x^(y/(0/0)) = x^(0y/0) = x^(1y) = x^y
So sqrt[0⁰](x)^y = sqrt[0/0](x)^y = x^y
Did anyone else start laughing when seeing the thumbnail? This is the funniest thing I've seen all week. The concept is hilarious impossible.
If we raise both sides of the equation x=°√A (A is some consonant) by the power of 0 we get:
x^0=A
For A=1 we get x=any number.
For any A other than 1, no x satisfies the equation.
Self evident I’m sure, and bit more work, but I’m confident that a simple graph could better illustrate the the limits and the resulting disconnects. Still a concise explanation.
You can rewrite zeroth root of x as (e^(1/0))^lnx. Rewrite the expression , now you take the limit as n approaches 0 on, and you can see that the positive limit expression is undefined (+inf) at 1, and 0 for all the negative values, and the negative limit expression is undefined at 1 (+inf as well) and 0 for all the positive values. That means that the function has no real value anywhere, because the left side contradicts the right side of the limit, and the one point where both limits agree is undefined.
TO SUMMERISE -
Zeroeth root is undefined
Positive number (>1)^(1÷(10^a very huge number)) =infinity
Positive number (0
what happens when you put it on a graph, will it be something like a quadratic where its big with in one set of rules and small in another?
This behavior will only become more ridiculous when considering negative number bases, complex number bases, or approaching 0 from a phase other than 0 or π.
You said the answer. x is the set of all nonnegative real numbers. It's just that you can only take the 0th root of the number 1. But x can be anything. Using undefined /0 to get something else - verboten.
BTW what is (-1)^0?
BTW what is (i)^0?
How to Scare Your Math Teacher and/or Math Class:
1. Write 0th root of any number
2. Write “Solve this” under it
3. Let chaos ensue
a number divided by 0 is not undefined, it is infinity. only 0/0 is undefined. so root 0 is same as raised to power infinity. negative power just means 1/number with positive power.. so its pretty straight forward and does not really need limit to figure it out.
Division by 0 is commonly undefined, though the limit of x/y as y approaches 0 is infinity for x>0 and -infinity for x
another way of finding that is it undefinied is for example. x^1/2 = the sqrt of 2 and x^1/3 = sqrt of 3 so if we were to do the zeroth root we will have to have x^1/0 which is underfined therefore the zeroth root is undefined
Hello dear professor
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Your Student from Algeria
On the complex plane, n^(1/z) has ESSENTIAL SINGULARITY at z = 0. So undefined is the only possible answer.
Great video! But I dont agree with your final conclusion (of it being too hard to explain). Negative roots are just 1/|the root|, so this makes total sense...
Of course, 0^0 IS defined and equal to 1, in spite of many maths text books pretending it is undefined. It is not just your personal opinion! There are many ways to show it, starting with the definition of arithmetic operations based on set theory : x^y is the number of applications from a set of y elements to a set of x elements; there are no application from a non-empty set to the empty set, but exactly one application from the empty set to iself. That is, 0^n = 0 for n>1 and 1 for n=0.
Or, to use a less abstract but equivalent concept, x^y is the number of possible words of y letters taken from an alphabet of x letters. Again, you cannot write any word if the alphabet is empty, apart from the empty word, so that there no word of length > 0 but one word of length 0 with an empty alphabet.
Another argument comes from algebra. Whenever we use a formula like (a+b)^n=sum(binomial(i,n)·a^i·b^(m-i) for i=0 to n, one has to assume 0^0 = 0 for the formula to remain valid for any value of a and b. Just set a=0 and b=1 and it becomes obvious. Even those who say that 0^0 is undefined use this kind of formula without noticing...
The confusion comes from the fact that in calculus, the limit of x^y when both x and y tend to 0 is undefined (or can be any number, including 0 or infinity if x and y are not independant). It means that this function of two arguments is discontinuous at (0,0) - a point which is on the border of its domain) - not that it is undefined.
You are absolutetly Blasphemous! I suscribe!
You can say that 64^1/0.5 is bigger than 64^1/0.6, but you cannot interpolate it to "64^1/0.5 is less than 64^1/0" because you have just zero with no sign. And for +0 or -0 (which are only symbols but not numbers) you will get a different solution. So a small answer: in ordinary math analysis x^1/0 is undefined, end of story
Take a hint from the IEEE-754 floating point math standard: NaN = Not-a-Number. This turns strange results like this into something at least vaguely computable. It has been a part of every FPU you have ever touched, but most people still don't know about it.
NaN happens because FPUs don't know the answer to something It isn't inherently related to mathematics
Really good exploration. I say 0^0 = 1.
Very interesting video and It was as interesting as it was underwhelming: a lot !
logic method:
0th root of any number is infinite except 1 because 0th root of 1 is equal to any number, because for example, n^0 is 1, it means that 0th root of 1 can be any number
I don’t get it, if you put a different input then you get a different output right?
Correct me if i’m missing something
having a zero on the bottom of a fraction defies axioms as it is undefined, when you write 0rootx as x^(1/0), the 0 is on the bottom of the fraction making it outside the bounds of our maths,
Let y=0th root of x. So y^0=x.
Multiplying both side by y, y=xy.
So x=1 or y=0. The 0th root of 1 can be any number and 0th root of x=0 if x is not 1.
Amazing!!!
And Reload. Those are good too. Not many know about them!
Zeroth root of -0.5 tends to 0 because you keep repeatedly multiplying itself by itself infinitely so it's the same result as +0.5 just oscillating + and - but still shrinking. Not infinity as the video says.
i.e.
-0.5
1/4
-1/8
1/16
-1/32
etc
When did he take the zeroth root of -0,5?
Oh i see your confusion. It was kind of confusing on his part, but what you got wrong is that he meant the "x" in the exponent approaches zero. He wasn't talking about the base, which he also called "x" later in the piece "definition" of zeroth root. He should have called the base "n" or something
ok :)
5:11 I was just starting to think about that...
But what about where x < 0?
This is not always undefined though, for x an element from (-1,1) it is defined for x^(1/0) as when when approach from 0+, x^(1/0+) approaches 0, while approaching 0-, x^(1/0-) approaches 0, meaning x^(1/0) is defined for x in (-1,1), which is 0.
Yeah I’ve seen the brilliant ad for like 5 years straight everyday but your the lucky soul who got the commission off of me 😂 finally folded
The eternal issue about about deviding by 0 is that the lim is not the final answer unless you accept it as the final answer. "Lim" meaning "infinitely close to but not quite there". The devil is in the details... Whenever infinity is involved the rules break down, hence 'undefined'. In fact infinity is undefined, it belongs to the realm of philosophy. There you will be scating an very thin ice and all bets are off.
A root is just an exponent in fraction form
A square root is N^1/2
So if we do root 0 we would have N^1/0
1/0 is undefined, you cannot define the answer when you divide by 0
It’s one of those situations where something is undefined, so redefine it (or use category theory to see why we can/ cant)
2:28 itd be infinity or negative infinity depending on the *numerators* value, positive is positive infinity, negative is negative infinity
idk it seems internally consistent to me. you defined 2^1/0 to be infinity using the limit process. .5^1/0 can be written as 1/(2^1/0), which would be 1/infinity which we usually say is 0. this is exactly what you got.
and when making the exponent negative, the order flips because a negative exponent flips a fraction making each thing be the opposite. it’s all technically internally consistent
When the number is 0 < X < 1; isn't that quantity in some way essentially 1/x? So it makes sense that 1/inf goes to zero.
It makes sense to me that the inverse of of the set of whole numbers (found in 0
To define, say 0th root of n = NAN(n) and NAN(n)^0 =n and NAN(NAN(n))=NAN^2(n), where NAN^2(n))^0=NAN(n). Can generalize to get NAN^b(n)=n^(1/(0^b)). You just can’t replace 0^b by 0. fixed in this “axiom” : a not congruent to b implies a cannot be substituted for b, and b cannot be substituted for a, even if a is c*0 and b is d*0, unless c=d. 0 is congruent to 1*0, but not z*0 for all z not equal to 1. 0 is congruent to 1-1. This also resolves issues with 1/0 when you realize n/0 is not congruent to m/0 unless n=m, or otherwise math breaks, and how (-1)/0 is not congruent to (+1)/0 but 1/(-0)=1/(+0) by stating 1/(-0) is not congruent to 1/(+0). There is also a nice video on an extension to include log(0) by looking at properties + and * and e^x have, instead of repeated multiplication. Not so undefined anymore…
1/0 is the number of points of length 0 on a line of length 1.
1/0 is also (-1)! Or 0!/0=1/0 and is -1*-2*-3*…
Binomial theorem without conditions, says 0!=0^0 so since 0!=1, 0^0=1
Log base 1 could also probably be defined in some kind of similar manner.
log base 0* not log(0)
Concluding the original math system was based 9.
1-9
11-21
22-31
32-41
42-51
52-61
62-71
(N)1 - (N)1 + 9
The old maths system was then numbers.