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 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.
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
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.
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.
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.
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.
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.
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
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
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.
1^infinity cannot be “equal” to anything since infinity isn’t even a number to begin with. You’re using the properties of 0/0 here and that just doesn’t work because division/rooting of zero causes mathematical contraction, not a lack of specific answers.
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.
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.
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
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
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 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. 🙂
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.
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
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.
I have no problem with square root of 4 having two values (2 and -2), and similarly I don't have a problem with 0th root of 1 being every possible number. It's the 0th root of other numbers that I have a problem with.
The thing is that expressions that can be equal to any number are conventionally undefined. 0/0, for example. Any number can be 0/0, but 0/0 is said to be undefined.
In that part of the video, he is defining the 0th root of b to be lim(x→0^+) b^(1/x). If we use b = 1 here, we get lim(x→0^+) 1^(1/x), which is unambiguously 1.
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)
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.
Finalmente alguém fez um vídeo sobre esse assunto que eu já passei dezenas de horas fazendo inúmeras abordagens usando álgebra, comparação com exponenciação e até limites no expoente e no índice de das raízes.
@@rudradeepchatterjee3640bassically if we make a graf y=x^x then we can identify y just one problem like y=1/x there no point when x=0 maybe that graf have the same in point 0?
My reason why it isn't is thus: The reason everything to the power of 0 is one is because of the alternate way to calculate exponents. For example, 2² is equal to 2⁴ * 2^(-2) (16 * ¼ = 4, therefore 2² = 4). If we try to do this in the case of x⁰ for any other number, you will get one because you will be multiplying a number by its reciprocal. Example: 2⁰ = 2¹ * 2^(-1) = 2 * ½ = 1. Do this for 0, and you get 0¹ * 0^(-1) which is undefined. Therefore, 0⁰ is undefined.
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.
and, a extra, the zeroth square root of 1 its probably e, one of definitions of e in limits is: lim (1+x)^1/x -> by definition is e, but if we pull 0 in x, will be 1^1/0, so, 1^1/0=e? x->0
The zeroth root of negative numbers is basically impossible to define for negative numbers less than -1. Say you tried to solve the zeroth root of -2; that's equivalent to (-2)^(1/0) = (-2)^infinity. The issue is whether (-2)^infinity is positive or negative infinity, since if you raise a negative number to a power, the sign of the result depends on whether the power is even or odd. On the other hand, (-0.5)^infinity is 0, which means that the sign doesn't matter.
@@jeffdege4786 I don't see how it doesn't make sense. (-2)^x can be either negative or positive depending on if x is even or odd; for example, (-2)^4 = 16, but (-2)^5 = -32. This is actually the reason why even roots of negative numbers don't exist in the real number system. This issue also exists for numbers that are less than 0 but greater than -1, but it's moot by the fact that such numbers tend towards 0 if they're raised to increasingly large powers. For example, (-0.5)^2 = 0.25, but (-0.5)^3 = -0.125; however, the result gets closer and closer to 0, which doesn't have a sign, so the zeroth root of negative numbers between -1 and 0 is 0.
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!
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.
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.
// 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.
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.
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.
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?
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.
Is it actually non defined or do not exist? When we say right hand limit and left hand limit is not equal, limit of x as approaches a does not exist. Then the case we disguise in the videos has one property that i mentioned. The left limit and right limits of the integers between 1 and 0 and bigger than one differs. And also 0th root of 1 is not just equal to one but also equal every real number. So, it seems "DNE" fits more to this analysis.
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.)
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.
you added the negative sign to the exponent. But your x was the 64 or 0.5 not the exponent :/ If x is negative, you get a complex value, which will results into cool spirals...
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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
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.
In general there are n nth complex roots of a number. So there are 0 0th roots. Now the problem shifts to the meaning of zth roots where z does not belong to I+.
So if the "first root of x would be x , like x to the power of 1 is x , why can't one step below partly be seen as the power of 2? in some sense the 0th root of 4 would be 16 , the "minus-1th" root be 64 and so on? or only below the zeroth root, to avoid that "point"? but i won't propose in which section of mathematics. Or is that variety idea too lame, too simple for you guys? I haven't seen this derivation in the comments section so far.
Well, I'm trying to understand your reasoning which seems to be: 1rt(x) = x cuz x^1 = x Therefore 0rt(4) = 16 cuz... 0rt(4) should be seen as 1rt(4)^2 = 4^2? I don't see why you would define 0rt(x) in relation to 1rt(x) in this way. And negative roots should be treated like adding 1 to the power, like (-1)rt(4) = 1rt(4)^(2+1) = 4^3 = 64? I also don't see why you'd add 1 to the power. If your reasoning is something different, please feel free to explain because I'm... just not understanding.
Would you say that replacing the zero in the maths system with a square and the square works just like 0, however the square represents space as a zero. Would be a easier way to represent zero being added to maths to represent space.
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.
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 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.
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.
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
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.
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.
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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
@@occho_movie 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...
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"
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
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 :)
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
@@lesserafim_fan_31724I'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 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.
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. ;)
3:46 nice! I like this way of showing the "process" of taking limits. :D
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.
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.
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.
0:31 the arrow is in the wrong direction. Example: (-1)²=1 => -1=sqrt(1) which is nonsense.
It's not nonsense. Idk much about algebraic proofs, axioms, etc. but I can with all certainty say that if ±a²=b then sqrt(b)=±a
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.
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.
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
Very good and unconventional explanation of limits. Great job!
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
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
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.
1^infinity cannot be “equal” to anything since infinity isn’t even a number to begin with. You’re using the properties of 0/0 here and that just doesn’t work because division/rooting of zero causes mathematical contraction, not a lack of specific answers.
You know stuff gets crazy when even Bri leaves the expression undefined
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.
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
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.
your videos are a beacon of clarity and inspiration!
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
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?
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
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 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. 🙂
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.
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
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.
i was taught by my teacher that xth root or n^x means x many solutions, i guess it checks out
I have no problem with square root of 4 having two values (2 and -2), and similarly I don't have a problem with 0th root of 1 being every possible number. It's the 0th root of other numbers that I have a problem with.
the square root of 4 have only 1 value (It's 2)
The thing is that expressions that can be equal to any number are conventionally undefined. 0/0, for example. Any number can be 0/0, but 0/0 is said to be undefined.
@@briogochill6450No. If you multiply -2 by -2, you also get four. Since, -2*-2 = -2/2 = 2*2
Edit: Just remembered about the ± symbol. So, ±2.
@@OmegaCat9999bro do you even know that square root opens with mod
@@OmegaCat9999 x^2 = 4 has two possible solutions, but the square root of 4 is by definition just 2
5:11 I was just starting to think about that...
But what about where x < 0?
- Daddy, why they say it is impossible to divide by zero?
- C'mon son. Sit down. We need to talk about USS Yorktown.
5:04 If x = 1 then zeroth root should equal to every real number since every real number to power 0 is 1
In that part of the video, he is defining the 0th root of b to be lim(x→0^+) b^(1/x). If we use b = 1 here, we get lim(x→0^+) 1^(1/x), which is unambiguously 1.
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)
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.
approacing a phase other than pi,
why pi?
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?
Finalmente alguém fez um vídeo sobre esse assunto que eu já passei dezenas de horas fazendo inúmeras abordagens usando álgebra, comparação com exponenciação e até limites no expoente e no índice de das raízes.
f(x) = x^(1/n) is a continuous function for all n. We could look at the behavior of this seqence of functions as n goes to infinity?
controveseal? 0⁰ is just 1. my favourite way of writing 1 infact.
0⁰ is undefined in reality but in the next video its shown how are we getting 1 as the answer
@@rudradeepchatterjee3640 in reality it is much more common for it to be 1 than for it to be undefined
@@rudradeepchatterjee3640bassically if we make a graf y=x^x then we can identify y
just one problem
like y=1/x there no point when x=0
maybe that graf have the same in point 0?
My reason why it isn't is thus:
The reason everything to the power of 0 is one is because of the alternate way to calculate exponents. For example, 2² is equal to 2⁴ * 2^(-2) (16 * ¼ = 4, therefore 2² = 4). If we try to do this in the case of x⁰ for any other number, you will get one because you will be multiplying a number by its reciprocal. Example: 2⁰ = 2¹ * 2^(-1) = 2 * ½ = 1. Do this for 0, and you get 0¹ * 0^(-1) which is undefined. Therefore, 0⁰ is undefined.
@@justyouraveragecorgi This is terrible reasoning. Your reasoning also proves that 0^x should be undefined for all x
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.
Sometimes I feel like people forget that 0 is a placeholder, and not a number in situations like this.
and, a extra, the zeroth square root of 1 its probably e, one of definitions of e in limits is:
lim (1+x)^1/x -> by definition is e, but if we pull 0 in x, will be 1^1/0, so, 1^1/0=e?
x->0
I can't believe you didn't consider the zeroth root of negative numbers...
fax, bad video
@@sogga_fan😂
The zeroth root of negative numbers is basically impossible to define for negative numbers less than -1. Say you tried to solve the zeroth root of -2; that's equivalent to (-2)^(1/0) = (-2)^infinity. The issue is whether (-2)^infinity is positive or negative infinity, since if you raise a negative number to a power, the sign of the result depends on whether the power is even or odd. On the other hand, (-0.5)^infinity is 0, which means that the sign doesn't matter.
@@SsvbxxYT So, what do you do when your math didn't make sense? You invent new math...
@@jeffdege4786 I don't see how it doesn't make sense. (-2)^x can be either negative or positive depending on if x is even or odd; for example, (-2)^4 = 16, but (-2)^5 = -32. This is actually the reason why even roots of negative numbers don't exist in the real number system. This issue also exists for numbers that are less than 0 but greater than -1, but it's moot by the fact that such numbers tend towards 0 if they're raised to increasingly large powers. For example, (-0.5)^2 = 0.25, but (-0.5)^3 = -0.125; however, the result gets closer and closer to 0, which doesn't have a sign, so the zeroth root of negative numbers between -1 and 0 is 0.
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.
So the zeroth root is undefined, huh? you wasted how many minutes of my life to tell me that?
Get lost
It wouldn’t be a mystery if we knew what it did idiot
Cry harder
Cry harder
What?, 5 mins.
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!
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.
1:36 what if x={R} ?
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.
The cards for you videos dont appear for me I have a lot of youtube addons so maybe that has something to do with it ?
that's why i love math. Sometimes things so complicated that they make you laugh.
// 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.
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.
Video idea if you up for, play with negative x in the complex plane. Try to show why it's not discontinuous if you specify a unique root.
So this video is saying that x/0 is essentially infinity, nice
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.
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?
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.
Glad to see you back buddy
Is it actually non defined or do not exist? When we say right hand limit and left hand limit is not equal, limit of x as approaches a does not exist. Then the case we disguise in the videos has one property that i mentioned. The left limit and right limits of the integers between 1 and 0 and bigger than one differs. And also 0th root of 1 is not just equal to one but also equal every real number. So, it seems "DNE" fits more to this analysis.
How about imaginary-th square root?
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.)
awesome intuitive approach
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.
you added the negative sign to the exponent. But your x was the 64 or 0.5 not the exponent :/
If x is negative, you get a complex value, which will results into cool spirals...
Hello dear professor
Your lessons are really interesting and crucial,i do appreciate your job,i wish you peace and happiness under the sky of prosperity,all the best. Take care and have a good time.
Your Student from Algeria
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
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.
X^0=a
X≠0√a
that's because it's raised to 1/0 , but then you accept that 1/0×0=1, which you can't!
Did anyone else start laughing when seeing the thumbnail? This is the funniest thing I've seen all week. The concept is hilarious impossible.
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
In general there are n nth complex roots of a number. So there are 0 0th roots. Now the problem shifts to the meaning of zth roots where z does not belong to I+.
brilliantly twisted and interesting.
So if the "first root of x would be x , like x to the power of 1 is x , why can't one step below partly be seen as the power of 2? in some sense the 0th root of 4 would be 16 , the "minus-1th" root be 64 and so on? or only below the zeroth root, to avoid that "point"? but i won't propose in which section of mathematics. Or is that variety idea too lame, too simple for you guys? I haven't seen this derivation in the comments section so far.
Well, I'm trying to understand your reasoning which seems to be:
1rt(x) = x cuz x^1 = x
Therefore 0rt(4) = 16 cuz... 0rt(4) should be seen as 1rt(4)^2 = 4^2? I don't see why you would define 0rt(x) in relation to 1rt(x) in this way.
And negative roots should be treated like adding 1 to the power, like (-1)rt(4) = 1rt(4)^(2+1) = 4^3 = 64? I also don't see why you'd add 1 to the power.
If your reasoning is something different, please feel free to explain because I'm... just not understanding.
you could also treat it as n^0=x, which makes n be anything if x=1 and undefined if x≠1
Would you say that replacing the zero in the maths system with a square and the square works just like 0, however the square represents space as a zero. Would be a easier way to represent zero being added to maths to represent space.
I'd say it's colloquially defined as "undefined" and more accurately defined as "will blow up to infinity".
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
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
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.
What is the problem with saying it means to the power of infinity according to the limit
2:28 itd be infinity or negative infinity depending on the *numerators* value, positive is positive infinity, negative is negative infinity
Try defining the inverse element of zero. That should do the trick.
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 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 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.
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.
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
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.
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.