Speaking for myself, I think we should stop using those notations for "indeterminate forms".

@farklegriffen2624

3 ай бұрын

What are you suggesting exactly?

@sporkstar1911

2 ай бұрын

0⁰ is 0 the reason is a special case for the number zero because it represents Nothing... no small part of anything such as 0.0000... its actually Zero and its nothing. The Definition of Zero is stronger than the undefined number of infinity because for the latter you don't actually know what you have, for Zero you absolutely do and there is no amount of anything in it. If there were some amount of something in it, it would be a Non-Zero number and become 1 when raised to the power of 0. Thus, to divide 0/0 you also get Zero because even if it would be divided by a number so strong you would get Zero as a result... well... you already have Zero to start with. It doesn't matter what the reciprocal is in this case. This problem is solved with logic rather than mathematical rules. Other forms of commutation do not apply.

@angelmendez-rivera351

2 ай бұрын

@@farklegriffen2624 He explained it in the video. Did you watch?

@angelmendez-rivera351

2 ай бұрын

@@sporkstar1911 *0^0 = 0.* It is not. *The reason is a special case for the number zero because it represents Nothing...* This is incoherent gibberish. 0 has a precise definition in mathematics: it is the additive identity. This is all it is. *...its actually Zero and its nothing.* Again, this is meaningless gibberish. The word "nothing" has no philosophical meaning, much less a mathematical meaning. *The Definition of Zero is stronger than the undefined number of infinity,...* The string of words "undefined number of infinity" is just word salad. It does not mean anything. An "undefined" thing cannot be a number, and "infinity" is not a number anyway. The word "infinity" is essentially a buzzowrd nonmathematicians use to refer to many distinct mathematical concepts as if they were all the same concept. *...because for the latter you don't actually know what you have, for Zero you absolutely do and there is no amount of anything in it.* What you are talking about is the empty set, which is irrelevant to the discussion we are having. *If there were some amount of something in it, it would be a Non-Zero number and become 1 when raised to the power of 0.* The definition of the arithmetic function we call "exponentiation" has nothing to do with amounts at all. It is merely a generalization of multiplication. *Thus, to divide 0/0 you also get Zero because even if it would be divided by a number so strong you would get Zero as a result...* This is more incoherent word salad. Division is defined as multiplication by a multiplicative inverse. The symbols x/y are a shorthand for x • inv(y), where inv(y) denotes the multiplicative inverse of y. 0/0 does not exist, because 0 has no multiplicative inverse. *It doesn't matter what the reciprocal is in this case.* No, of course it matters. A mathematical object cannot be equal to a nonexisting thing. *This problem is solved with logic rathed than mathematical rules.* With this, you have demonstrated you have no understanding of logic whatsoever. Let me burst your bubble here: mathematical rules _are_ logic. Mathematical analysis is a direct application of formal logic. *Other forms of commutation do not apply.* I have some advice for you: before using words, you should learn what those words mean.

@angelmendez-rivera351

2 ай бұрын

@@sporkstar1911 *0^0 = 0.* It is not. *The reason is a special case for the number zero because it represents Nothing...* This is incoherent gibberish. 0 has a precise definition in mathematics: it is the additive identity. This is all it is. *...its actually Zero and its nothing.* Again, this is meaningless gibberish. The word "nothing" has no philosophical meaning, much less a mathematical meaning. *The Definition of Zero is stronger than the undefined number of infinity,...* The string of words "undefined number of infinity" is just word salad. It does not mean anything. An "undefined" thing cannot be a number, and "infinity" is not a number anyway. The word "infinity" is essentially a buzzowrd nonmathematicians use to refer to many distinct mathematical concepts as if they were all the same concept. *...because for the latter you don't actually know what you have, for Zero you absolutely do and there is no amount of anything in it.* What you are talking about is the empty set, which is irrelevant to the discussion we are having. *If there were some amount of something in it, it would be a Non-Zero number and become 1 when raised to the power of 0.* The definition of the arithmetic function we call "exponentiation" has nothing to do with amounts at all. It is merely a generalization of multiplication. *Thus, to divide 0/0 you also get Zero because even if it would be divided by a number so strong you would get Zero as a result...* This is more incoherent word salad. Division is defined as multiplication by a multiplicative inverse. The symbols x/y are a shorthand for x • inv(y), where inv(y) denotes the multiplicative inverse of y. 0/0 does not exist, because 0 has no multiplicative inverse. *It doesn't matter what the reciprocal is in this case.* No, of course it matters. A mathematical object cannot be equal to a nonexisting thing. *This problem is solved with logic rathed than mathematical rules.* With this, you have demonstrated you have no understanding of logic whatsoever. Let me burst your bubble here: mathematical rules _are_ logic. Mathematical analysis is a direct application of formal logic. *Other forms of commutation do not apply.* Let me give you some advice: before using words, you should actually learn what those words mean.

fun fact: whether you define 0^0 as 1 or 0, you will get that 0^(0^0) = 0

@alexbennie

3 ай бұрын

This is awesome!

@daniellewilson8527

3 ай бұрын

Agreed, it is really cool

@goncalofreitas2094

3 ай бұрын

Really nice

@angelmendez-rivera351

3 ай бұрын

Indeed.

@BS-bd4xo

2 ай бұрын

0^indeterminate tho???

simple solution, 1 is actually undefined

@shruggzdastr8-facedclown

3 ай бұрын

​@Fire_Axus : one-zero-one-zero-zero-one-one-zero-one-zero

@jordanrodrigues1279

3 ай бұрын

I like it. {0} 0+0=0 0*0=0 0*(0+0)=0*0+0*0 The additive inverse exists for all elements. The multiplicative inverse exists for all elements that are not an additive identity. (I feel like this last statement will be problematic for people who don't agree that 0^0 should be 1.) It's almost a field. Almost! It's only missing a multiplicative identity.

@angelmendez-rivera351

3 ай бұрын

@@jordanrodrigues1279 Well, it does have a multiplicative identity: 0. The reason it is not a field is because fields are defined specifically so that the multiplicative identity and additive identity are not equal. However, you do have a commutative division ring here, which is as close to having a field as you can get without being a field.

@Tabu11211

2 ай бұрын

Lmfao

@adityamishra7711

2 ай бұрын

the very first thing you define in arithmetic is 1, either directly or indirectly (peano)

Zero is just a scribble that represents the unscribblable

I think the biggest problem with the language of indetermine forms (which is the main reason why any debate that 0^0 might not equal 1 ultimately exists, historically) is that it treats lim f(x)^g(x) (x -> p) and f(p)^g(p) as if the were the same thing, and this is just incorrect usage of notation. In fact, calculus textbooks often make it a point to emphasize that, because a function may not be continuous at x = p, lim f(x) (x -> p) and f(p) may not be equal; yet when it comes to binary operations, they suddenly forget about the very thing they emphasized, and conflate the two notations as if they were the same thing. It is no wonder so many people out there have such misunderstanding of how limits work, and it probably also fuels the debate surrounding 0.999... = 1 as well, since it also fundamentally revolves around misunderstanding limits.

@ron-math

3 ай бұрын

Well said.

@alexbennie

3 ай бұрын

Context is everything! Do you encounter 0^0 whilst doing some number theory (I.e Integers & Algebra), or do you encounter it doing some calculus (I.e. Continuity, Functions, etc)? In my opinion, the notation should be be standardised from the fundamental arithmetic level upwards, so if I see 0^0, then I know it's 1, because 0! = 1. If anyone wants to define it to be not 1, well then, use a different notation, like 0^(0),or something.

@ron-math

3 ай бұрын

Check this community post: kzread.infoUgkxxpywNpVowmNe0rbIEHLa7ps-oTJHO05L @@alexbennie

@angelmendez-rivera351

3 ай бұрын

@@alexbennie I agree.

@TheGrinningViking

2 ай бұрын

There's people who think .9 repeating isn't mathematically equal to one after seeing the proof? It's not even a long proof!

It's *2/3* !! Proof: Cauchy said 0, Knuth 1 and I trust Knuth exactly double as much as I trust Cauchy 🤣

@YouTube_username_not_found

2 ай бұрын

The usefulness of probability theory!!

@HugoHabicht12

2 ай бұрын

But ⅓ of 0 is 0 🤓

@harriehausenman8623

2 ай бұрын

@@HugoHabicht12 Nah. I don't believe it.

@thomaskaldahl196

2 ай бұрын

actually I take the geometric mean to get the cube root of 1 × 1 × 0, which is 0

So happy I came across this channel, love the videos, keep it up ❤

@ron-math

3 ай бұрын

Will do!

@angelmendez-rivera351

2 ай бұрын

@@Fire_Axus And? Emotions not being rational is entirely fine.

14:20 - 14:30 Wow, I am so happy to finally see a content creator who dislikes the "indeterminate form" language and notation. I have been fighting against this paradigm, because I have worked in education previously, and My Siesta, the language of indeterminate forms textbooks and teachers use make it so damn difficult to help students who are learning calculus, and I think we have far outlived the need for this language, and we need to rethink how we teach limits in an institutional setting. Frankly, the entire calculus is a hot mess in the U.S.A., but limits in particular are an egregious example.

@ccbgaming6994

3 ай бұрын

Agreed! I took “normal” calculus class involving limits my senior year of high school but took “honors calculus” (basically an intro to real analysis) my freshman year at university. Much more rigorous and accurate, but also much more rewarding.

@MattMcIrvin

3 ай бұрын

I've noticed that math texts tend to emphasize that 0^0 is an indeterminate form right up to the point where they start talking about general polynomials and power series, at which point they often implicitly assume that 0^0 = 1 without noting that there was a change.

@angelmendez-rivera351

2 ай бұрын

@@MattMcIrvin Yes, and I find it somewhat irritating.

Any non-zero number to the power of 0 is always 1, so consider the following: *e^0 = 1* e^x is equal to the sum of *x^n / n!* for all n >= 0. That is its taylor series If e is raised to the power of 0, that makes the first part *0^0 / 0!* 0 to any power greater than 0 is of course equal to 0, and 0! is equal to 1, so all that leaves is *e^0 = 0^0!* If we know that all non-zero numbers to the power of 0 equal 1, then for *e^0 = 1* to be true, *0^0 = 1* must also be true

@cameronbigley7483

2 ай бұрын

While I agree, and that's the main argument I'd use, I can imagine there's a special case for e^0's Taylor series being incorrect, since math is funky that way.

@keilafleischbein59

2 ай бұрын

What if x^0=/=1 when x=0?

@alephnull8377

2 ай бұрын

Powers of zero are always one… is this a fact?

@JJean64

20 күн бұрын

​@@keilafleischbein59 Then e^x = 0 when x = 0, which is obviously wrong

18:30 0^0 can also be defined to be 1 in the context of natural Numbers (0, 1, 2, …) exponentiation using PI notation (repeated product), and then be expanded to integer exponentiation. Let m, n be natural numbers; m^n = [product {m} from i = 1 to n]. Using the nullary product rule, when n = 0, m^n is defined to be 1. This can be expanded to all integers with the constraint that when m = 0, n must be non-negative; then if n This also means that the argument at 15:30 doesn’t make sense, as that exponentiation rule is explicitly defined for a =/= 0: 0^(b-b) here is implied to be equal to 0^b * 0 ^(-b), but either b or -b must be negative if se suppose b =/= 0, and that’s not allowed. This kind of proof is like the proofs you can find online that 1=2 using a disguised faulty argument in the middle of the proof. Anyway, very interesting video, keep up the good work Edit: grammar and added small details

@ron-math

3 ай бұрын

Thanks for the comment! Early day math proofs are indeed very messy and sometimes even childish from today's perspective.

@gabritex14

3 ай бұрын

@ron-math Yeah, sometimes you just wish you could reinvent old notations from the ground up with more consistency, but nowadays is basically impossible to do so since they're widely adopted basically everywhere. Things like renaming immaginary numbers to something better (like idk, "lateral numbers" or something like that), using the tau value instead of pi, geometric algebra instead of vector analisys, and gamma functions to not have that offset by one. Those would be my top priority

@dageustice

2 ай бұрын

If I can strengthen your latter argument: most times I see 0⁰ being stated as undefined, they use the 0⁰=0^(1-1)=0/0 argument, as you mentioned. But naturally, we can extend this for all integers a: 0^a = 0^(a+1-1) = 0^(a+1)/0. So if the argument for 0⁰=0/0 is true, one would have to conclude that 0^a is undefined for all integers.

0:44 - 1:17 I would strongly suggest slowing down here and rewinding this explanation. It goes very fast, and it suffers from a problem of circular reasoning, which in mathematics and formal logic, is unacceptable. The problem is, what is being said in this segment relies on the definition that a^(-1) = 1/a, but this definition cannot even be properly motivated without knowing beforehand that a^0 = 1. Therefore, you cannot prove a^0 = 1 by starting with a^(-1) = 1/a with your assumption. If we use the definition of exponentiation as I explained it in my previous comment, then we automatically obtain a^1 = a and a^0 = 1 as theorems, because if I apply the action "multiply by a" 1 time to the integer 1, then I get a • 1 = a, and if I apply the same action 0 times to the integer 1, then this is the same thing as not actually doing the action: or in other words, doing nothing. However, since doing nothing and "multiply by 1" are the same thing, we have a^0 = 1 • 1 = 1. Having said this, one can use the principle of mathematical induction to prove that a^(m + 1) = a^m • a for all nonnegative integers m and all integers a, and then it can be used again to prove that a^(m + n) = a^m • a^n for all nonnegative integers m, n. This property is the fundamental property of exponent operations, and any extension of the domain of definition ought to preserve this property: otherwise, the extended operation can no longer be called "exponentiation." Having said this, if we want to extend a^b to include negative integers b, then this extension ought to be such that a^(m + n) = a^m • a^n still holds when m, n are arbitrary integers. In particular, we ought to start by understanding what a^(-1) is, and if it can exist at all, based on the restrictions we have. By definition, 1 + (-1) = 0, so 1 = a^0 = a^(1 + (-1)) = a^1 • a^(-1) = a • a^(-1). Now, look the equation a • a^(-1) = 1. Since the product is equal to 1, this means that a^(-1) must be equal to the multiplicative inverse of a, which is only possible if a has any inverse at all. This excludes, for example, a = 0. This is important, because when b is a nonnegative integer, then a^b is always well-defined, given the definition I presented, including when a = 0, b = 0, which, if you test the definition, clearly gives 0^0 = 1 as a theorem. However when b is a negative integer, then a is no longer arbitrary. a must satisfy the restriction of being invertible. So, even if 0^0 = 1, 0^(-1) does not exist, and it cannot be defined as exponentiation. There is nothing wrong with this: it just means that you cannot write 0^(1 + (-1)) = 0^1 • 0^(-1), because in order to be able to write it, 0^(-1) has to exist. This is not exclusive to a = 0. If a is some singular matrix, then the same restriction applies. Why am I making a huge deal out of this? Because it is truly important to conceptually understand the underlying mathematics that the notation is hiding, and it is also important to understand the notation itself. This segment of the video uses division as part of the derivation, but remember: division is not a primary operation. It is a secondary operation, defined in terms of multiplication and multiplicative inverse (i.e., reciprocal), which are primary operations. The definition of x/y is x • inverse(y), or x • y^(-1), if you will. This is also why one cannot divide by 0: it would mean y = 0, so you would have x/0 = x • 0^(-1) = x • inverse(0), but inverse(0) does not exist, because 0 has no multiplicative inverse. There does not exist some q such that 0 • q = 1. As such, one cannot be using division willy-nilly when studying the properties of exponentiation without properly establishing how exponentiation works when limiting yourself to multiplication and multiplicative inverses. When a student or teacher fails to account for this, then they will make a mistake, often a conceptual one too, and they will come up with an invalid mathematical derivation, and not be able to realize what is wrong with it. I have seen it happen far too often. Ultimately, you can only establish a^(b - c) = a^b/a^c when you acknowledge that b - c is just shorter way of writing b + (-c), and assuming a is invertible, a^(b + (-c)) = a^b • a^(-c). Now, a^(-c) is the multiplicative inverse of a^c, so a^(-c) = 1/a^c, so a^b • 1/a^c = a^b/a^c. See how that works? But this all assumed a is invertible, which the video totally failed to take into account. On the other hand, a^0 = 1 is true for all a, invertible or not, because it directly follows from the original definition of a^b to begin with. You cannot use a^(b - c) = a^b/a^c to prove a^0 = 1.

@ron-math

3 ай бұрын

From 0:44 to 1:17, a is still a positive integer as assumed earlier so I wouldn't worry about a being uninvertible . But I do agree that the case b

The 0^x counterargument leaves me unconvinced. It is true that 0^x = 0 for all POSITIVE real numbers, but 0^x is undefined for all NEGATIVE real numbers, so why would it be surprising that the value on the boundary in between them (at x = 0) might be an entirely different value: 0^0 = 1? Also, the formula a^(b − c) = a^b/a^c can be used to show ANY power of zero is undefined, and is clearly untrue for a = 0.

@ron-math

2 ай бұрын

Yeah. Those are history

Consider it in the context of, say, 5×0³. This expands to 5×0×0×0=0. 5×0² -> 5×0×0=0. 5×0¹ -> 5×0=0. 5×0⁰ -> 5×... The 0 has vanished, leaving what looks like a void, but it isn't. After all, the mathematics you perform all the time are surrounded by voids. Even the number 5 alone is multiplied unseen by an infinite sequence of ×1s. The numerical void left by taking any number to its 0th power gets filled in by this endless ocean of 1s. To put it another way, if a variable is undefined such as might be when you write program code to calculate 5×num without declaring `num`, then the calculation can't actually transform the 5 in any way, and so it remains 5.

A primitive implementation of integer powers is repeated multiplication. Even multiplying all numbers in an array can work like that. Same with addition. Addition: Start with 0, add to it all numbers in array. Array is empty? Duh, output is 0. 0*0 = 0 ({+0} zero times) = 0. Multiplication: Start with 1, multiply it by all numbers in array. Array is empty? Duh, output 1. 0^0 = 1 ({*0} zero times) = 1. Easy, problem solved.

0:33 - 0:43 The more accurate phrasing is that a^b denotes a multiplied with itself b - 1 times. Remember, a appears in the product b times, but the multiplication symbol actually only appears b - 1 times. This is an important distinction, and when we fail to make this distinction, students get very confused. I know this from both first-hand and second-hand experience, and it happens in educational institutions, as well as on the Internet. By the way, this is also not the most useful approach to explaining the definition, because the explanation fails when considering a^0 and a^1. You may intuitively expect it to fail for a^0, but not being able to conclude a^1 = a through the definition is a fatal mistake, so this should make it clear that this is not how we ought to define exponent operations. Instead, I think we need to start be reframing the idea of multiplication by a as being an _action_ you perform on some object (in this case, on an integer). a^b can be defined as the result of applying the action "multiply by a" b times to the integer 1. In my experience, this analogy is the most helpful for conceptually explaining the definition to grade school students.

@ron-math

3 ай бұрын

Thank you so much for your input. It is impossible for me to know without your input. Really appreciate.

I havent seen this take from anyone else, but the way i interprete it is that 0^0 = 1, by many of the arguments presented on the video, and the reason why the limits are badly behaved is that the function x^y is discontinuous at (0,0) and we are just having a path that approaches (0,0)

@angelmendez-rivera351

2 ай бұрын

This is my take too

We can sort of "prove" that if 0⁰ were to be defined at the most basic level, for exponentiation with a natural exponent, where the property aᵐ⁺ⁿ = aᵐ·aⁿ should hold for all natural m, n ∈ ℕ₀, then it can be only 1 or 0: let 0⁰ = x, then 0⁰ = 0⁰⁺⁰ = 0⁰·0⁰, so x = x², x² − x = 0, x·(x − 1) = 0, and either x = 0 or x = 1. This is better than nothing, but still not satisfactory, so we eventually have to simply stick to the definition a⁰ = 1, aⁿ⁺¹ = aⁿ·a for any base a, so 0⁰ = 1 as an empty product, which by convention is equal to the neutral element of multiplication, that is 1.

@angelmendez-rivera351

3 ай бұрын

If we are thinking of exponentiation as the next hyperoperation in the hyperoperation sequence after multiplication, and if we are axiomatically defining these operations from the most fundamental mathematical grounding, then exponentiation ought to be defined set-theoretically (which has been done before), and with this, 0^0 = 1 is a theorem.

@omp199

3 ай бұрын

@@angelmendez-rivera351 How are you defining this "hyperoperation sequence"?

@angelmendez-rivera351

2 ай бұрын

@@omp199 The hyperoperation sequence is a well-known recursive sequence: the 0th operation is succession, the 1st is addition, the 2nd is multiplication, the 3rd is exponentiation, the 4th is tetration, and so on. Let H(k, m, n) denote the nth operation being applied to the natural number pair (k, m). We have that H(k, m, 0) = S(m), H(k, 0, 1) = k, H(k, 0, 2) = 0, H(k, 0, n) = 1 for all n >= 3, and H(k, m + 1, n + 1) = H(k, H(k, m, n + 1), n). This sequence is naturally motivated by applying recursion to the most basic set-theoretic operations you can perform. The video hints at this.

3:43 - 4:36 I appreciate that the formal definition of a function was given here, especially because when I see content creators talk about the empty set and the empty function, they rarely, if ever, clarify this, leaving their audience confused, in my experience. This is also related the set-theoretic definition of the factorial, explaining why 0! = 1, but again, when such explanations are invoked, the definitions being used are rarely, if ever, clarified, so this is very refreshing to see here. Well done! (Or medium-*rare*?)

@ron-math

3 ай бұрын

I see you are watching and commenting. The surprise of 0! = 1 is waiting for you. Great to have your comment here. Feels great :)

@shruggzdastr8-facedclown

3 ай бұрын

Now, we're cookin'! 🥩

Excellent video.

@ron-math

Ай бұрын

Thank you very much!

Classic Hewlett-Packard calculators played both sides of the street: the four-level RPN models like the HP32S and HP42S were Team Cauchy, their algebraic models tended to also be Team Cauchy, but their high-end RPL series (HP28, HP48, HP50g) was Team Euler.

@ron-math

3 ай бұрын

Yes. When I did research for this video, I found those interesting behaviors. However, I couldn't locate the reason behind the change for sure so I skipped that. Thanks for commenting!

@MattMcIrvin

2 ай бұрын

@@ron-math The RPL models were the first ones that had symbolic algebra and calculus support and the ability to manipulate series expansions, whereas the others only did numerical calculations. If I had to guess I'd say the switch was so that the calculator didn't have to worry about the x=0 case whenever it handled a general power series formula. But the RPL models differed from the others in many other ways as well--they were a world unto themselves.

@angelmendez-rivera351

2 ай бұрын

Very intriguing.

Think of it this way: 2² = 1×2×2, 2¹ = 1×2, 2⁰ = 1 1² = 1×1×1, 1¹ = 1×1, 1⁰ = 1 0² = 1×0×0, 0¹ = 1×0, 0⁰ = 1 Now it looks obvious

@juliannewman2ndchannelmusi475

Ай бұрын

Nice! 🙂

great video!

@ron-math

3 ай бұрын

Glad you enjoyed it :)

Amazing. The quality of this video is superb.

@ron-math

3 ай бұрын

Thank you very much!

@TheGildedMackerel0

3 ай бұрын

@@ron-math As are the rest of your library after inspection. 🙏

2:16 i think its useful to think about that by starting at 1, and then doing "×m" n times, to get m^n. that way its consistent in that your doing the same operation all those times, rather than starting with m and then doing "×m" (n-1) times which feels a bit arbitrary. the only thing this changes is that for n=0 you are just left with 1, which i think allows an intuition for why it can be the case even for m=0 because you just haven't done any multiplying by 0 so its not 0. of course im not saying this is proof, just that it can help you intuitively understand it (i think its useful to define 0^0 as 1 but i do agree that its not a fundamental necessity, and just proves useful, like how we can say 0! is 1 [though in that case we can always apply the working backwards rule but the starting at 1 also applies], or how we can say 1 is not a prime number not because we cant have it be one but because its not useful to be one, so we have our definition exclude 1)

Got you! You used Grays codes to create animation at 3:20. Can't hide from me

@ron-math

3 ай бұрын

🧐 Who is Gray(s)?

@alexyz9430

3 ай бұрын

Who's Gray???

@diribigal

3 ай бұрын

​@@ron-mathThe Gray Code, named after Frank Gray and findable on Wikipedia, is basically a standardized sequence of n-bit binary numbers where you change only one bit each time. Whether you knew it or not, your animation of the functions showed the Gray Code for three bits: 000, 001, 011, 010, 110, 111, 101, 100.

@ron-math

3 ай бұрын

Thanks! This is the first time I know it.@@diribigal

@peterfireflylund

2 ай бұрын

@@ron-mathit’s used to reduce wear when storing counters in EEPROMs. It is also used when we want to read several parallel bits as a number without an explicit synch mechanism. Take for example a device that measures people’s height. It’s just like an old sk00l version at the doctor’s but it has an electronic readout. The “ruler” has the numbers printed as black/white patterns using a Gray code and the electronics reads them horizontally. Or another example: when we want to use pointers into a FIFO buffer on a chip and we want to be able to compare them across clock domains.

thank you i didn't know you could define the notion of a function formally thank you now, after 35 years of existing i can accept "the function is defined in the real domain except at zero" it seemed arbitrary enough for any reasonable person to go insane

Beautiful excursion! 🤗 Very clear voice and the effort you put into speaking clearly is much appreciated 👍

I think of x^0 as zero factors of x. Having no factors is the same as having 1, because all powers, products, factorial, etc., start with 1 and then multiply by further factors. Having no further factors leaves 1, so 0^0 is 1. The argument that x^x approaches 1 as x approaches 0 is only valid if both x's (base and exponent) approach 0 at the same rate. Also, I remember learning in highschool that 0^0 is equal to 1 *by definition*! So, that neatly takes care of it not being defined! But seriously, that definition is the only one that seems to be consistent with the rest of mathematics.

6:02 - 11:20 I am somewhat conflicted here. The binomial theorem is essentially a special case of multiplication of formal power series, and so is the geometric series. However, the problem with these examples is that they also rely on circular reasoning. The truth is, if it were demonstrable that 0^0 ≠ 1, then nothing really would change, other than the notation we use for denoting these theorems about formal power series, as well as the notation for the series themselves. As far as the actual mathematical concepts go, nothing would change. It just so happens that because there is very strong reasoning to support 0^0 = 1, this kind of notation has become the norm for power series. On the other hand, I once again want to express my appreciation for how refreshing it is to see you going over the formal details of the definition of a ring, although as of now, it is not entirely clear to me what the relevance of the formal definition is to this video.

@ron-math

3 ай бұрын

I was hoping that the audience will check that those properties are preserved under the evaluation mapping to gain the motivation that 0^0 != 1 is not acceptable :)

I would say that 0^0=1 and f(x,y) is not continuous in (0,0), ao we can't use it in limits.

@angelmendez-rivera351

3 ай бұрын

@@Fire_Axus Who said the function has to be continuous?

Indetermined forms shouldn't be even used as an argument against. The function RxR->R defined by x^y is not continuous in standard topology no matter how ypu define it at 0 but that still doesn't mean we cannot define it.

@angelmendez-rivera351

2 ай бұрын

Exactly. I am so glad to encounter more people who understand this. Maybe there is hope for the future of the education system.

@__christopher__

2 ай бұрын

I think you mean: "defined by x^y". I'm pretty sure you didn't mean to completely ignore the second argument (and BTW, it is totally possible to make x^x continuous at 0, just set it to 1).

@sieni221

2 ай бұрын

@@__christopher__ I meant the function x^y sorry

1:56 - 2:20 This is why I said that this type of empirical argument base on patterns is inappropriate when doing mathematics. This problem is very easily solved if we rethink our approach for how we explain the definition. I proposed in another comment that, instead of saying a^b is equal to a multiplied with itself b times (which is inaccurate), we should say that a^b is equal to the result of applying to the integer the action "multiply by a" b times. This is how we learn that a^1 = a, a^0 = 1, and therefore, 0^0 = 1. This very same definition also tells us that 0^b = 0 when b > 0, and this is yet another example of why using pattern-recognition to study arithmetic is completely inappropriate. Just like with the sequence f from my previous comment, pattern-recognition will lead you to the wrong answers when it comes to arithmetic operations. This is why, today, mathematicians instead rely on axioms and deductive logic to study mathematics. Besides, with this definition, saying you are performing an action 0 times has a very intuitive meaning, and it generalizes to almost all other mathematical contexts. Feed 10 people with one meal.

Cauchy and Excel being allied is chef kiss.

Team Cauchy definitely, Just as the real numbers is only a smaller part of complex numbers, the result 0^0 = 1 is only one of the answers and Cauchy opened our eyes to the other answers but Euler kept his eyes shut /joking

@peterfireflylund

2 ай бұрын

Abelist!

@angelmendez-rivera351

2 ай бұрын

@@peterfireflylund ??

@Qreator06

2 ай бұрын

On a slightly more serious note, I like to think that every single case of integers Exponentiation causing 0^0=1 has an extended form that allow for real number inputs that just haven’t discovered yet and if we were to take the limiting process, it’s just the case of lim x->0 (x^x) which as we all know is one. If only there is a nice integer Exponentiation formula that has the form of a 0^0 limit that equal something other than one… This is a genuine question, is there such a thing?

@angelmendez-rivera351

2 ай бұрын

@@Qreator06 I do not believe such a thing can exist, but if someone thinks otherwise, they could provide an example.

Great video! I do think using the factorial formula as your DEFINITION of the binomial coefficients would be inappropriate in this context, as a more intrinsic definition would make it clear that nC0 = 1, thereby demonstrating the necessity of 0! = 1 in the process.

Consider the surface z=x^y -- what does it look like near the origin?

@ron-math

2 ай бұрын

It is not about the function z = x^y. It is about whether we should directly treat this symbol like a function.

@bilkishchowdhury8318

Ай бұрын

I like your Grosse Fugue

I cannot remember where I first learned this but I have "always" (i.e., since then) thought of 1 as being the product of 0 factors. To me this makes intuitive sense and I'm obviously on team 0^0 = 1.

@farklegriffen2624

3 ай бұрын

Then what is 2⁻¹ ? What does it mean to be the product of -1 factors?

@WhoisTheOtherVindAzz

3 ай бұрын

@@farklegriffen2624 An empty list/set/etc., makes sense, a list/set/etc. with -1 elements does not. Of course, if you can come up with a way for it to make sense then I'll be very intrigued!

@angelmendez-rivera351

2 ай бұрын

@@farklegriffen2624 -1 is not a natural number or a cardinal number, so asking about having -1 of a thing is not semantically meaningful. This is why, when we use definitions for the integers, as opposed to only the natural numbers, we use axioms for these definitions. Here, the defining axiom is that x^(m + n) = x^m • x^n for integers m, n.

@apteropith

2 ай бұрын

@@farklegriffen2624 negative-one factors is the amount of factors where, if you multiply by one factor more, you get zero factors you could take this to be either implicit or explicit depending on your interpretation of negative numbers, but you can work out the necessary value from there (spoiler: 2^-1 = 1/2) the same argument holds for zero factors: it's the amount of factors such that, if you multiply by one more factor, you have exactly one total factor

We can define infinity as 1/0 then if… 0^0=L, ln(L)=0ln(0)=0*-inf=0(1/0)=0/0. 0/0 can be seen as any number. Lets look at the equation 0x=0. 0/0x=0/0. Let 0/0=a, then ax=a, so a(x-1)=0. Ok, this is weird, but we can make x=1 so a is any number. Or a=0 and x is any number. Ok so 0/0 is 0 or any number, so it’s any number. So ln(L)=a, where a is any number (even complex). So L=e^a, so L is any number. 0^0 can be any number.

If you multiply zero by itself zero times, and you insert it into something else, the zero never gets expressed, so it is still logical for 0 to the 0 to be one.

Simple harmonic pendulum initial angle x approaching zero and first iteration n=0 function sin(x)^n=1, it makes sense it shows pendulum A bob of mass M at zero angle x deviation, but sin(pi/2)=1 it means is perpendicular to Earth ground that contains potential energy equal to 1. Pendulum contains potential even it is 0^0=1

@ron-math

3 ай бұрын

I feel this argument is very very interesting. Can you elaborate more or paste a link, etc.? Thanks!

@phyarth8082

3 ай бұрын

en.wikipedia.org/wiki/Pendulum_(mechanics)#Legendre_polynomial_solution_for_the_elliptic_integral. For very small angle approx. x=0 sin(x)^n is Legendre polynomial first iteration sequence value equal 1.

If I multiply something by zero no times then I have multiplied it by 1, therefore 0^0=1

@1ballad

2 ай бұрын

if you multiply 0 by 1 then you have 0

@benjaminojeda8094

2 ай бұрын

@@1ballad if you don't multiply by 0 you get 1, so 0⁰ = 1

@angelmendez-rivera351

2 ай бұрын

@@1ballad Yes, but 0 • 1 = 0^1 ≠ 0^0.

The thing is if you want x^0 you're best defining x^n as equalling 1*x_1*x_2...*x_n. So that when you have 5^3=1*5*5*5 and 5^0=1

Watch the video before rushing to argue with others. I promise it is different.

When it comes to mapping between sets what does 'mapping from the empty set to [anything else]' even mean?

@ron-math

3 ай бұрын

This is "cost" of formal definition: some intuition is lost. The question is great. In introductory calculus course, a mapping focuses on the "map" step but when making it more rigorous, it becomes a combination of (domain, codomain, relationship) and it is defined based on set theory axioms, etc. As long as this triple is unique, the mapping is unique. It is fine if some or even all of them are empty sets.

@MichaelRothwell1

Ай бұрын

Mapping from a set A to a set B means for every element of A assign an element in B. If A is empty (regardless of whether B is also empty) there is no work to do as there are no assignments to make. So there's just one way to do it.

0^0 = 1 on natural exponentiation. 0^0 is undefined on real exponentiation

@angelmendez-rivera351

3 ай бұрын

I think you are partially missing the point. "Exponentiation" is a word that refers to a specific operation, and using the word to also refer to a completely different operation is inappropriate. Besides, the undefinedness argument is grounded on a conceptual misunderstanding of real analytic concepts anyway.

@omp199

3 ай бұрын

@@angelmendez-rivera351 Would you mind explaining what conceptual misunderstanding of real analytic concepts you are referring to here?

@angelmendez-rivera351

3 ай бұрын

@@omp199 In the video, indeterminate forms were mentioned. They are a concept created by mathematician Augustin-Louis Cauchy, back when the application of the theory of functional analysis to set as the rigorous foundations for doing multivariable calculus had not been sufficiently developed. Here is the general idea: suppose I have a real-valued function h in R^2, and suppose that we also have real-valued functions f, g in R, and suppose we want to evaluate lim h(f(x), g(x)) (x -> p), with the information that lim f(x) (x -> p) = L and lim g(x) (x -> p) = M. Can it be done? If so, how would you do it? Most calculus teachers would say that the way to do it is to plug L for f(x) and M for g(x), to obtain h(L, M), and this is the limit. Now, suppose you had two other real-valued functions i and j in R, and you also knew lim i(x) (x -> p) = L and lim j(x) (x -> p) = M. Presumably, you would expect that if you replace lim h(f(x), g(x)) (x -> p) with lim h(i(x), j(x)) (x -> p), then you still get h(L, M). Therefore, lim h(f(x), g(x)) (x -> p) = lim h(i(x), j(x)) (x -> p). However, the problem is that it could be the case that if you evaluate lim h(f(x), g(x)) (x -> p) some other way, such as for example, with L'Hôpital's theorem or the squeeze theorem, then you may get some number N, and if you do the same with lim h(i(x), j(x)) (x -> p), then you may get some number P, and it could be the case that N ≠ P, even though lim h(f(x), g(x)) (x -> p) = lim h(i(x), j(x)) (x -> p) = h(L, M). This is a contradiction, and so, what Cauchy would have said during his time, and what calculus teachers would say today, is the expression h(L, M) is an indeterminate form. Cauchy listed 7 examples on his table, so in textbooks and on Wikipedia, they are referred to as "the seven indeterminate forms." However, there are infinitely many example one can come up with, using this type of mathematics. What is the issue with the above? The issue is... everything. All of the mathematics I showed you is completely incorrect from the very beginning. For starters, when you evaluate lim h(f(x), g(x)) (x -> p), you cannot replace f(x) with L and g(x) with M. This is completely invalid, and obviously absurd, since this is definitely not how limits work. In fact, the point (L, M) may not even be in the domain of the function h, so of course you cannot do this. The definition of the limit of a function is as follows: we say that lim k(x) (x -> p) = S if and only if there exists a unique real number S such that for all real ε > 0, there exists a real δ > 0, such that for all real x, if 0 p) = L, and μ is defined at L and continuous at L, then lim μ(k(x)) (x -> p) = μ(L). This follows directly from the definition of continuity: we say μ is continuous at L if and only if lim μ(y) (y -> L) = μ(L). The above theorem can then be understood as saying that the function μ and the limit operator can be interchanged: lim μ(k(x)) (x -> p) = μ(lim k(x) (x -> p)), and since lim k(x) (x -> p) = L, μ(lim k(x) (x -> p)) = μ(L). Of course, Cauchy knew all of this, but he failed to realize that these ideas do, in fact, extent to functions in R^2. He failed to realize that in order for lim h(f(x), g(x)) (x -> p) = h(L, M) to be true, h has to actually be both well-defined and continuous at the point (L, M). Otherwise, the equality above is false. And, if it _does_ turn out that h is continuous at (L, M), then you will never ever get a situation where lim h(f(x), g(x)) (x -> p) = N, lim h(i(x), j(x)) (x -> p) = P, yet N ≠ P, as long as lim f(x) (x -> p) = lim i(x) (x -> p) = L and lim g(x) (x -> p) = lim j(x) (x -> p) = M. In other words: the only situations where replacing f(x) with L and g(x) with M are precisely those situations where the continuity properties of h guarantee that N = P. Because Cauchy failed to realize this, he also failed to realize that in the cases where h is not continuous at (L, M), the fact that N ≠ P tells you absolutely nothing about the value of the expression h(L, M): in fact, the expression is simply irrelevant to the discussion, as you can never obtain by performing valid computational techniques in this case. Therefore, in all cases, calling h(L, M) an "indeterminate form" would be utterly nonsensical.

@angelmendez-rivera351

3 ай бұрын

I would try, but KZread is being extremely unpleasant and not letting me actually post my analysis without immediately deleting my comment. The useless algorithm is too fickle and unpredictable.

@omp199

3 ай бұрын

@@angelmendez-rivera351 I feel your pain. It does that to me so much: maybe half of my comments disappear, these days.

And the empty product? That is the most important!

0 is a strange number with rules entirely it's own. In any case, any number times 0 is 0 just as any number divided by it is both a positive and negative infinite value at the same time. Sometimes people can get smart and smug enough that they actually become quite stupid.

@angelmendez-rivera351

2 ай бұрын

*0 is a strange number with rules entirely its own.* No, not really. By definition, 0 is the additive identity, and this property naturally has consequences which lead to certain mathematical theorems being proven, but beyond this, 0 is just a mathematical object like any other. All mathematical objects have unique properties when you use sufficient axioms to specify them. *In any case, any number times 0 is 0...* I agree, but this has no relevance to what 0^0 is. 0^0 is an empty product, so 0 is not being multiplied by anything at all: in fact, no multiplication of any sort is taking place here. *...just as any number divided by it is both a positive and negative infinite value at the same time.* No, this is incorrect. Dividing by 0 is not a coherent idea to speak about, because what we call "division" is just a shortcut of notation. The symbols x/y are a shortcut for denoting x • inv(y), and inv(y) denotes the multiplicative inverse of 0. The issue here is that 0 has no multiplicative inverse.

@TheGrinningViking

2 ай бұрын

@@angelmendez-rivera351 Thank you for being a good example of what I was talking about.

@angelmendez-rivera351

2 ай бұрын

@@TheGrinningViking I was not an example of anything. What you said was genuinely asinine, you were called out on it, and now you are offended, like half of the people in the comments section 🤷🏼‍♂️🤦🏼‍♂️ I am not sorry for the fact that several of you cannot handle being corrected on a mistake. Besides, calling smart people "stupid" is ironically far more arrogant and smug than anything else I could ever say here.

@TheGrinningViking

2 ай бұрын

@@angelmendez-rivera351 Yeah, exactly this thing! Thank you for being a cautionary take. You probably are right though. I mean, you care enough to know about things. That's cool no matter how it comes across personally.

Pronunciation note: the vowel in “set” is NOT a diphthong. It is NOT pronounced the same as “site”, “sight”, or “side”.

@ron-math

2 ай бұрын

Thanks!

I use 0^0 as a little needy face, and this explains that much more

They said we couldn't divide by ZERO - then we discovered entangled particles 😂😂😂

@angelmendez-rivera351

2 ай бұрын

What does entanglement have to do with division by zero?

Wait, 1^infinity, is it the same as 0⁰? Because as a limiting process it is undefined, but it makes sense for it to be 1 right? Or saying that implies also that 0×infinity=0 and infinity⁰=1?

@ron-math

3 ай бұрын

Each of those "notations", if out of the context of limiting process, need to be investigated and interpreted differently, thus deserve their own videos. For example 1^infinity, using the set mapping analogy, would be equivalent to the number of functions mapping from an infinite-element set to a one-element set.

@LCDL6

3 ай бұрын

@@ron-math Those would be interesting videos, as for the mapping analogy it really seems like it is a 1, but with infinity not being a number and all I don't know, also, infinity⁰ also seems to be 1 by the mapping analogy...

@angelmendez-rivera351

2 ай бұрын

@@LCDL6 1^∞ is not a well-defined symbol, because the definitions of the operation ^, the real number 1, and the top object ∞ of the Dedekind-MacNeille completion of the real numbers, are not compatible with each other, and it can be proven that there is no way they can be made compatible with each other. The symbol ∞ here does not denote a transfinite cardinal or ordinal. It denotes the supremum of the affinely extended real line, when considered as a complete lattice.

@__christopher__

2 ай бұрын

@@LCDL6 While "infinity" by itself is not a number, in the context of cardinalities (that is, sizes of sets) there are indeed infinite cardinal numbers (yes, plural!), and 1 to the power of any of those numbers is 1.

It is 1, what the point?

We can just say that the answer is a superposition of one and zero.

@angelmendez-rivera351

2 ай бұрын

This would be nonsensical. The concept of superposition has little to do with number of arithmetic.

@hoagieland3716

2 ай бұрын

@@angelmendez-rivera351That’s not necessarily true. Take for example the expression: y^2 + x^2 = 1. At x=0, there are two different y values that satisfy the equation: y=1 and y=-1. So you could make the argument that y in the expression, y^2 + x^2 = 1, exists in a superposition of 1 and -1 when x=0. Is this practical for most situations, no probably not, but the idea at least seems plausible albeit with unconventional wording

@angelmendez-rivera351

2 ай бұрын

@@hoagieland3716 No, this would be totally incorrect. What you are doing is describing the solution set of an equation, which has absolutely nothing to do with the concept of superposition. I get very annoyed at how everyone these days throws around the word "superposition" despite having no clue of what the definition of the word is. Calling this a superposition of anything is not merely "unconventional wording," it is factually wrong.

@hoagieland3716

2 ай бұрын

@@angelmendez-rivera351 I mean, isn’t a solution set kind of like a superposition though? If a superposition is supposed to describe the state of a quantum system being in multiple states before it is measured, that is somewhat like an expression being two different values at the same time for a given x value. I acknowledge that it’s not really a 1:1 correlation and am stretching the definition a little. I guess a better example would be to talk about things like evaluating a limit for an expression that yields a result of 0/0 at a particular value. 0/0 is not defined in normal arithmetic since division by zero is not allowed. But the limit of an expression that evaluates to 0/0 at a particular x value can be determined. For example, lim x->0 for 2x/x is 2 whereas lim x->0 for x/x is 1. Somewhat backwards, but treating 0/0 as the “quantum system” and a limit expression as the “act of measuring” could be seen as analogous for saying 0/0 in the context of limits can be seen as a superposition of multiple numbers until a limit expression is applied to collapse the superposition into a singular result. Am I probably reaching a bit, yeah. But it’s still a fun thought experiment to try and apply terms from other fields to see if the logic surrounding them makes sense in any context in a different field and if there are any examples in said different field that can be analogous

@angelmendez-rivera351

2 ай бұрын

@@hoagieland3716 *I mean, isn't a solution set kind of like a superposition, though?* No, not at all, not even a little bit. *If a superposition is supposed to describe the state of a quantum system being in multiple states before it is measured, that is somewhat like an expression being two different values at the same time for a given x value.* No, it is not, not in the slightest. Once again, I feel compelled to remark how irritating it is for a person to use a word without actually trying to look up the definition of the word beforehand. I have no idea how communication is even supposed to work if this is how you approach conversations in your everyday life. A quantum system is characterized by a mathematical structure called a Hilbert space over the complex numbers, and specifically, it has to be something called a separable Hilbert space. I am not going to explain every single detail of what a separable Hilbert space is here, because it is not feasible for me to do it within a comment response, so I just explain the general gist: a separable Hilbert space is a vector space (the objects are vectors, and you can add them and scale them "in the usual way"), satisfying some additional properties, and having additional structure (such as an inner product). On a vector space, you can have some functions, called linear operators, and these are the object of study in linear algebra. In quantum mechanics, the physically observable quantities we are interested in when describing a system are the operators acting on this Hilbert space, and specifically, they must be self-adjoint (they are a concept analogous to symmetric matrices in linear algebra). The goal is to find the eigenstates and eigenvalues of these operators, and quantum states are normalized linear combinations of these eigenstates. In this case, a physicist would say the system is in a "superposition" of these eigenstates. _This_ is what the word means. Meawhile, what you are talking about has nothing to do with this. If x^2 + y^2 = 1, and x, y are real, and x = 0, then y = -1 or y = 1. This is nothing more than the disjunction of two logical propositions. Also, this is not even analogous to talking about 0^0 here, because 0^0 is an arithmetic expression not the solution set to an equation. The claim that 0^0 = 0 and 0^0 = 1 is a nonsensical claim, because equality is transitive, so therefore, 0 = 1, which is false. *I acknowledge that it's not 1:1 correlation and I'm stretching the definition a little.* Stretching the definition a little? Which definition? You have not used any definitions here. You are not using these words as if they had any definitions at all. This is not merely a little "stretch," it is completely wrong in multiple ways. Look, you can make the classic joke about how no one wants to talk me at parties if you want, or whatever, but people need to learn to get comfortable being told that there is a _huge_ difference between stretching things a little bit, and being completely wrong.

This would seem to imply that 0/0 also equals 1, since, for instance, 0^3/0^3 = 0^0 19:25

@MuffinsAPlenty

2 ай бұрын

This is a common line of reasoning, but it's flawed reasoning. The "rule" a^(b-c) = a^b/a^c does not hold when a^c = 0. If it did hold when a^c = 0, then you would be able to argue that _every_ power of 0 is equal to 0/0. For instance, 0^2 = 0^(4-2) = 0^4/0^2 = 0/0.

@johnrichardson7629

2 ай бұрын

@@MuffinsAPlenty 0 is annoying

it seems to me that if you don't define your singularities it will always be more difficult to define the results of their interactions (and calculus is all about defining those zeroes) ... but if you're sticking to the reals or a subset the result here is probably 1, you know?

@angelmendez-rivera351

2 ай бұрын

We don't perform arithmetic with singularities. This is the entire point of the concept of singularities. As such, a function being discontinuous should have no bearing on whether 0^0 = 1 or not.

Simple answer 0^0=🗿 End of discussion. Now give me my $1 million dollar ...

How are people arguing about this still? Just make a 3D plot and you can see that there's a vertical line at that value, meaning it can have multiple different values depending on which side you approach it.

@ron-math

3 ай бұрын

What you said is true and that's not the problem. The main problem is that people rush to treat it as a limiting process notation directly.

@bayleev7494

3 ай бұрын

there is no inherent relationship between an operation and the limit of the operation

@MuffinsAPlenty

2 ай бұрын

I have, from time to time, remarked that otherwise logically-minded people often discard logic when it comes to the topic of 0^0. Thank, Tim, for providing yet another example of this.

@angelmendez-rivera351

2 ай бұрын

You do know that calculators cannot prove things, right?

19:00 My question/comment is based on the heading on the Team Cauchy side... Why is team Cauchy using two different functions? In other words, why are they using Lim_(x->0) f(x) ^ g(x), for some f(0) = g(0) = 0 in the appropriate function space, to calculate Lim_(x->0) 0^0, in stead of using Lim_(x->0) f(x) ^ f(x), for some f(0) = 0 in the appropriate function space? Surely it's reasonable to argue that two different functions would introduce different behaviors at the limit? E.g. Assuming either Option A) f(x) = x and g(x) = e^x - 1 Option B) f(x) = g(x) = x It's been too long since my last ε-δ proof, but surely there's an intuitive/logical "difference" in the "speed" at which one approaches the limit? In other words: Saying 0^0 is not well defined without context... The "speed" of the limit of the base and exponent must be dealt with. Edit: Some formatting and I forgot to say that I loved the video! Immediate sub!

@ron-math

3 ай бұрын

Thanks for the sub! Right. Team Cauchy had the freedom to adjust the "speed" of the limit. People then already know that the speeds can be vastly different.

@budderman3rd

2 ай бұрын

Limits don't tell much when they approach but never actually gets there. Like you can still find where the hole of a function would be, but it's not actually there. Or where a line end it shows where the end would be but not it's actually there.

@angelmendez-rivera351

2 ай бұрын

Team Cauchy is relying on old mathematical ideas which have long been corrected. Cauchy himself could not be blamed for making those mistakes, since the full-machinery of set-theoretic topology, and the consequences this had for vector analysis (as an extension of real analysis) had not been discovered/invented during his time. However, the education system can absolutely be blamed, because we have the knowledge to do better than this. We have the privilege to be standing on the shoulders of giants, so for once, we should actually stop and appreciate the view from the top. Cauchy would be disappointed that the school system has refused to move from the ideas he proposed and has neglected the more modern approaches.

I've never seen this argument for 0^0=1 (doesn't mean nobody used it) a^(-b) = 1/(a^b) So a^(-0) = 1/(a^0) But a^(-0) = a^0 , it means a^0 = 1/(a^0) , so a^0 must equal 1 or -1 (we'll just keep the 1 because why not) This should also apply for a=0 , we get 0^0 = 1/(0^0) , this can be true only if 0^0 = 1 (or -1)

@MuffinsAPlenty

3 ай бұрын

If you want to continue those lines, you can invoke the rule (a^b)^2 = a^(2b) and replace b with 0 to show that (a^0)^2 = a^0. And the only two numbers which square to themselves are 0 and 1. Now, I'm pretty confident these combination of rule application does not hold in general. But it's a fun heuristic which does not contradict 0^0 = 1.

@akin0m

2 ай бұрын

It's not obvious that you can apply the exponential identity in the first step in the case that a (or b) could be 0.

@dreakexacios349

2 ай бұрын

@@akin0m Oh ok

@angelmendez-rivera351

2 ай бұрын

@@MuffinsAPlenty To be fair, if you do combine both arguments, then the deductive implication is that 0^0 = -1 or 0^0 = 1, and 0^0 = 0 or 0^0 = 1. However, this still implies 0^0 = 1. Ultimately, though, I agree that these arguments are not technically valid.

@__christopher__

2 ай бұрын

Well, your first line doesn't apply for a=0 and b≠0, as you are dividing by 0. Since the whole argumentation builds on that first line, the argument fails for a=0.

0^0=1^0. Both are true because (plus or minus 1)=(plus or minus 0)=(plus or minus i.) Proof. 1. Draw a unit circle as the Argand diagram 2. The length of the entire x axis is two. Half of that x axis is 1 which is the origin, thus 1=0. 3. The length of the entire y axis is 2i. Half of that y axis is i which is the origin, thus 1=0=i. 4. Since 1 and zero are equal (equivalent) then 0^0=0^1=1^0=1^1. QED

Zero equals one because it calls dibs on that spot

If f,g are holomorphic and f(a)=g(a)=0 then limit at a of f^g = 1

The logic with geometric series and power series do not work as they are based on the assumption that x^0 = 1. The actual series would be 1+x+x^2+... with 1 replaceble as x^0

@angelmendez-rivera351

2 ай бұрын

This is a fair objection.

I literally said this is stupid before he said you may think this argument is stupid.

Is there a proof for e^x Taylor Series is "(x^n)/n!" ? If so, then just plugging in 0 it will end being e^0=(0^0)/0! Which 0!=1. Then it has to equal 1. Maybe unrelated(?): Limits approachs, but doesn't actually get there. While series actually adds to it and actually gets there, or converges.

@ron-math

2 ай бұрын

Yes. The Taylor expansion of e^x is indeed another example.

@budderman3rd

2 ай бұрын

@@ron-math I honestly don't understand the debate about it tbh lmao

@ron-math

2 ай бұрын

The fact that you can plug in 0 to those series expansions to get 0^0 =1 can be treated as special cases of the homomorphism argument. The debate is only fruitful when people focus on the mathematical interpretation of the notation. 😄@@budderman3rd

@budderman3rd

2 ай бұрын

@@ron-math Can't you not put exactly that in mathematical notation it just takes alot? Like it's a proof being proven by a proof by a proof, etc. Till axioms which are needed for math to actually work. It took like 500 pages to have a rigorous proof of 1+1=2

@budderman3rd

2 ай бұрын

@@ron-math So basically if their is an actual proof that the series is exactly equal to e^x, even if its a special case of tayler series, it's still a proof. And you can't have a proof unless what's in the proof has been proven itself. Which leads a chain of proofs all the way down to axioms. So the actual question that actually matters is: Does it have a proof that the series is exactly equal to e^x? And if so, the debate is over and it makes no actual logical sense including mathimatical sense that it's still going on and 0⁰ is just 1.

Wolfram let me down here when I check it out.😢

I haven't watched the video yet, but ain't 0^0 just 0/0 like how x^0 is just x^1 × x^-1, hence x^0 =1 while x != 0?

@angelmendez-rivera351

2 ай бұрын

0^0 ≠ 0/0. Many claim otherwise, but their "proofs" are incorrect. Usually, the "proof" is that 0^0 = 0^(1 - 1) = 0^1/0^1 = 0/0. However, this is invalid, because 0^(1 - 1) ≠ 0^1/0^1.

@theswordslay3542

2 ай бұрын

i mean, if you put it like that, 0^2 is also becomes 0/0. 0^2 = 0^(3-2) = 0^3/0^2 = 0/0 am i right? that's the flaw of this argument, a^(m-n) = a^m/a^n only works if a is NOT 0. so unless you wanted 0 to the power of anything to be 0/0, yeah this argument isn't valid.

I love when mathematicians fight each other, like religious combatants 😂.

no, euler is right. cauchy can GFH

@ron-math

3 ай бұрын

😆

@doctorscoot

3 ай бұрын

@@ron-math I will die on this hill (and a couple of others)

@angelmendez-rivera351

3 ай бұрын

I mean, Cauchy really is in the wrong here. His argument is based entirely on a misunderstanding of how the limit concept and the definition of continuity apply when discussing multivariable functions. This, and when you realize that x^y is not appropriate notation for denoting exp(yln(x)), as both expressions are legitimately different and even have different domains (and calling the latter "exponentiation" is inappropriate), it completely kills all of the arguments trying to assert that 0^0 is undefined.

@sbares

3 ай бұрын

@@angelmendez-rivera351^^^^this guy gets it.

Why would it have to be one thing anyways? We could define it pragmatically until we face with an issue that requires a clear-cut definition, which I doubt we would.

@angelmendez-rivera351

2 ай бұрын

This video demonstrates exactly why it is has a clear-cut definition. Also, why would it have be one thing? Oh, I don't know, why would 2 + 2 have to equal 4?

@atlas_19

2 ай бұрын

@@angelmendez-rivera351 It doesn't have a clear cut definition tho? Also, did you even read my comment? Not that you need to read it to see that basic arithmetic that you use every single day not having clear cut definitions is way more detrimental than some edge case that a normal person never, and a mathematician only occasionally encounters, that is not yet critical for or at the heart of any theorem or field.

@atlas_19

2 ай бұрын

@@angelmendez-rivera351 Oh and btw, 2+2 does not have to equal 4 either, as long as you stay consistent with it.

@angelmendez-rivera351

2 ай бұрын

@@atlas_19 *It doesn't have a clear cut definition tho?* It does, which is why the vast majority of mathematicians today use it frequently across multiple disciplines of mathematical analysis. *Also, did you even read my comment?* I did, and there is nothing that you said that has not been addressed by me or the video. Did you watch the video? *Not that you need to read it to see that basic arithmetic that you use every single day not having clear cut definitions is way more detrimental than...* You say this now, but in your next comment, you pointed out how 2 + 2 does not actually have to be equal 4, which completely contradicts what you said here, and it does my job for me. *...some edge case that a normal person never, and a mathematician only occasionally encounters, that is not critical yet or at the heart of any field or theorem.* You are extremely, sorely mistaken about this being the case, which is exactly why I compared it to 2 + 2 = 4. It is not an edge case, and mathematicians do not encounter it "only occasionally," and it is also untrue that normal people do not encounter it. The video actually gave plenty of examples. Even those who do are not mathematicians encounter it frequently, which is why this topic is mentioned in grade school. The kinds of topics only mathematicians encounter are topics I guarantee you that you have never heard of, nor have even heard their names.

@atlas_19

2 ай бұрын

@@angelmendez-rivera351 You can take your smart ass somewhere else, learn some communication skills, than come back. I'm not dealing with your pedantics or hair-splitting.

I don't know if I should give this video 1 like or 0 like. On a side note: there are people who think that 0 means plural. Crazy, huh?

So if you want 0^x to be continuous, 0^0 is 0. If you want x^0 to be: 0^0=1. If you want polynomial evaluation to be consistent, it must also be 1, but if you care about f^g in general, then it should be any value. What a tricky conundrum indeed.

@angelmendez-rivera351

2 ай бұрын

This is *not* it works at all, though. This has nothing to do with "wanting" a function to be continuous. Whether a function is continuous at a point or not is not something you choose. The function either is continuous, or it is not, and if it is continuous at a point, then it can be formally proven to be continuous at that point using the definition. If it is not continuous, then it is not continuous. Mathematical facts (by which I mean theorems, specifically) are not something we get to decide: they are the inevitable consequences of the proof calculus being used.

@haniyasu8236

2 ай бұрын

@@angelmendez-rivera351 but that is exactly how it works though... you may not decide your results, but you can choose your axioms and definitions. The exponential function itself is a prime example. By your logic, x⁰, x^1/2, and x^-1 would all be undefined since you can't multiply x by itself 0, 1/2, or -1 times. It's only by deciding that we want to change the definition (of which there are about a dozen different options btw) do we actually get results for different kinds of numbers as exponents.

@angelmendez-rivera351

2 ай бұрын

@@haniyasu8236 *but that is exactly how it works, though...* No, it isn't. You don't get to choose whether the Pythagorean theorem follows from the Euclidean axioms of flat geometry or not. *...you may not decide your results, but you can choose your axioms and definitions.* So, you still agree with me that you cannot choose the results, which is sufficient to undermine what you originally said. Whether a function is continuous or not is something that is proven from the properties of the function. You don't get to choose whether a particular set of given, fixed properties implies continuity or not. *The exponential itself is a prime example.* It isn't. The fact that there exists an analytic function whose Maclaurin coefficients are 1/n! is not something we had a choice on. It was a fact that just so happened to be true and we were able to prove it. *By your logic, x^0, x^(1/2), and x^(-1) would all be undefined, since you can't multiply x by itself 0, 1/2, or -1 times.* What are you talking about? Nothing I said here remotely implies any of that. You're just changing the topic now. Also, it makes perfect sense to talk about doing a thing 0 times. We do this all the time even in colloquial English, and it still makes perfect sense in mathematics too. If you ask me "how many cars do you own?" and I say "zero," are you going to reply with "whaaaaat? That makes no sense! You can't have 0 cars!" C'mon now. Don't be obtuse.

@angelmendez-rivera351

2 ай бұрын

@@haniyasu8236 *but that is exactly how it works, though...* No, it isn't. You don't get to choose whether the Pythagorean theorem follows from the Euclidean axioms of flat geometry or not. *...you may not decide your results, but you can choose your axioms and definitions.* So, you still agree with me that you cannot choose the results, which is sufficient to undermine what you originally said. Whether a function is continuous or not is something that is proven from the properties of the function. You don't get to choose whether a particular set of given, fixed properties implies continuity or not. *The exponential itself is a prime example.* It isn't. The fact that there exists an analytic function whose Maclaurin coefficients are 1/n! is not something we had a choice on. It was a fact that just so happened to be true and we were able to prove it. *By your logic, x^0, x^(1/2), and x^(-1) would all be undefined, since you can't multiply x by itself 0, 1/2, or -1 times.* What are you talking about? Nothing I said here remotely implies any of that. You're just changing the topic now. Also, it makes perfect sense to talk about doing a thing 0 times. We do this all the time even in colloquial English, and it still makes perfect sense in mathematics too. If you ask me "how many cars do you own?" and I say "zero," are you going to reply with "whaaaaat? That makes no sense! You can't have 0 cars!" C'mon now.

@angelmendez-rivera351

2 ай бұрын

*but that is exactly how it works, though...* No, it isn't. You don't get to choose whether the Pythagorean theorem follows from the Euclidean axioms of flat geometry or not. *...you may not decide your results, but you can choose your axioms and definitions.* So, you still agree with me that you cannot choose the results, which is sufficient to undermine what you originally said. Whether a function is continuous or not is something that is proven from the properties of the function. You don't get to choose whether a particular set of given, fixed properties implies continuity or not. *The exponential itself is a prime example.* It isn't. The fact that there exists an analytic function whose Maclaurin coefficients are 1/n! is not something we had a choice on. It was a fact that just so happened to be true and we were able to prove it. *By your logic, x^0, x^(1/2), and x^(-1) would all be undefined, since you can't multiply x by itself 0, 1/2, or -1 times.* What are you talking about? Nothing I said here remotely implies any of that. You're just changing the topic now. Also, it makes perfect sense to talk about doing a thing 0 times. We do this all the time even in colloquial English, and it still makes perfect sense in mathematics too. If you ask me "how many cars do you own?" and I say "zero," are you going to reply with "whaaaaat? That makes no sense! You can't have 0 cars!" C'mon now. Don't be obtuse.

You have a new sub, very interesting video. If you don't mind I have a video I need help on... I just dropped a video where I try to tie Riemann Hypothesis to Yang-Mills 3D gauge theory. F(x)F(y) represent massless (zero mass) particles traveling position functions, and 2/|x-y|^n multiplied by a Perturbation of g^2 (the coupling constant). |x-y| goes to zero, and g^2 for to zero. When n=0 where n is dimension. You just get 2 since 2/(0)^0 = 2. The reciprocal of 1/2, the critical line to the Riemann Zeta function where "trivial zeroes" fall on alongside negative even numbers (-n/2 whole numbers). In my examples, I made huge rings that span a distance of 2 with 2 subset rings every time, and 1/2 and -1/2 seems to work as an empty set for the entire sequence, an additive inverse since 1 is 1/2 of 2 and 1+1 = 2 and 1x1= 1. 1/2 also works as a singleton to the first subset ring , with "e" and "π" being transcendentals that seem to tie the rings together into knots. Blah, blah. Long story short I'm glad to see a video teaching this to viewers like me more in detail. I dropped a video about rings and their subsets, and how (0)^0 seems to play a role in the identity of the mass gap in Yang-Mills 3D. Anyways thanks if you read this

@ron-math

3 ай бұрын

Hi! Thank you very much for commenting! I am not familiar with the tie you are trying to build, unfortunately.

@drawsolidart

3 ай бұрын

@@ron-math thank you so much for the response. i have a few videos about it, unfortunately i'm not much of a math student or pro.

its actually irrational....

@angelmendez-rivera351

2 ай бұрын

How is 0^0 irrational, when you do not even believe it is defined?

Team Knuth. Makes more sense.

1:19 - 1:34 This is a bad argument, and educators and content creators need to stop using it. This is not how mathematicians do mathematics. Patterns are not a substitute for definitions. If 0^0 has any value at all, then said value is established by defining a^b, and then letting a = 0 and b = 0 in the definition. This is how all of arithmetic is done. Computing 2 + 2 is no different: you define a + b in general, and let a = 2, b = 2. No one goes around looking for patterns in a + 2 using values of a ≠ 2 in order to learn what 2 + 2 is, and if they did, then it would be silly to do so, because this is not how mathematics is done. Besides, there are entire video series on KZread dedicated to explaining why this kind of empirical pattern-recognition is invalid in mathematics, and even dangerous. For example, consider the following sequence: f(0) = 1, f(1) = 2, f(2) = 4, f(3) = 8, and f(4) = 16. What is f(5)? Do you know what f(5) is? The answer is no. It is impossible to know the answer, because I have not actually defined what the sequence f is. If you say f(5) = 32, then you are incorrect. f(5) = 31, and this is because f(m), for all nonnegative integers m, is defined to be the number of maximum regions a disk can be partitioned into with m chords (a chord is line segment connecting two points in the circumference of the disk). See how pattern recognition probably led most of the readers astray? Asking you to guess what f(5) is by only telling you what f(0), f(1), .., f(4) are is completely inappropriate of me. By the way, 3blue1brown has an entire series of videos on this topic, and at least 1 of the videos discusses this exact sequence I just described. 1:35 - 1:55 Yes, but as it was said in the video: this was 300 years ago. How mathematicians did mathematics back then was completely different from how mathematics is done today, and because of the lack of rigor in their methodology, many of the discoveries were merely accidental, while many others were just incorrect. The history of mathematics is certainly important, but teaching the history of mathematics is not a substitute for actually teaching mathematics. This is a major pet peeve I have with not only content creators on KZread, but also with the education system in English-speaking countries. Schools pretend to teach mathematics, but really, all they are doing is teach history of mathematics, rather than any of the actual concepts, and they do it poorly too. However, content creators are not improving significantly on that, and repeat many of the same kind of mistakes the school system does. I still appreciate the work of course, but being honest and fair with the scrutiny is important if we want to improve even further.

@ron-math

3 ай бұрын

Agree. I am trying to "show" it, not to "teach" it. Need to make my attitude more explicit.

@angelmendez-rivera351

3 ай бұрын

@@ron-math This is fair. I think discussing the argument is fine, but it was unclear to me that you realized that the argument has its problems. This may just be a consequence of the fact that I am reacting as I am watching, which I normally tend to not do.

Wouldnt defining 0⁰ as 1 make 0/0 equal 1 let n be a non 0 integer 0⁰ = 0ⁿ⁻ⁿ = 0ⁿ/0ⁿ = 0/0

@MuffinsAPlenty

3 ай бұрын

You provide a very widespread wrong argument for 0⁰ = 0/0 by invoking an exponential rule aᵇ⁻ᶜ = aᵇ/aᶜ when aᶜ = 0, despite the fact that the rule does not hold for aᶜ = 0. If this rule held for a = 0, then the following slight tweak of your argument would be _completely valid._ 0² = 0³⁻¹ = 0³/0¹ = 0/0. Hooray! We have just proven that 0² = 0/0. Obviously, seeing such a result, and knowing that 0² has a completely unambiguous, well-defined meaning, something in this argument must be wrong. And if you check every step, the only place something could have gone wrong is the equality 0³⁻¹ = 0³/0¹ In general, how do you prove the rule aᵇ⁻ᶜ = aᵇ/aᶜ, provided that b and c are nonnegative integers? One way to do so would be to start with proving the related equation aⁿaᵐ = aⁿ⁺ᵐ, which can be done with the definition of exponentiation as repeated multiplication, the associativity of multiplication, and mathematical induction. From here, one can set m = c and n = b-c (in which case n+m = b), giving aᵇ⁻ᶜaᶜ = aᵇ. Then, one can divide both sides by aᶜ. At least, _provided that aᶜ is not zero,_ and obtains the familiar rule you invoked aᵇ⁻ᶜ = aᵇ/aᶜ. But again, we needed to divide by aᶜ to obtain this rule, and such is not valid when aᶜ = 0. In other words, your argument that 0⁰ = 0/0 starts with the hidden premise that division by 0 is totally legal. But we all know that it isn't.

@theswordslay3542

2 ай бұрын

ill make a tl;dr version of the above reply. 0^2 = 0^(3-1) = 0^2/0^1 = 0/0. but we know 0^2 is not 0/0. something's wrong, and that is your rule doesn't work when the base is set to 0.

Integer exponentiation and real exponentiation are inherently different operations, strictly speaking. Integer exponentiation is defined recursively, where the exponent counts how many times you multiply the base. In particular, the empty product is 1, so 0^0 must be 1; or rather, this argument is really about the empty product, and whether "multiplication over no elements" is a coherent concept. And in a precise sense, it is a very natural concept and can even be used to define 1 itself. Real exponentiation, very contrastingly, involves two transcendental functions, exp and ln, and the latter is only defined for positive numbers. So "0^0", when shorthand for exp(0ln(0)), is an ill formed operation-this, I think, is the crux of the confusion. People forget that "e multiplied by itself pi times" doesn't *truly* make sense; most of the time, we just have no reason not to pretend it does: It is merely the function exp turning addition into multiplication. Consequently, in places where we can well-tune it, so that "x^y":=exp(yln(x)), it can be made to act like integer exponentiation-it's just that x=0 is not one of those places.

@ron-math

3 ай бұрын

You may like this video: kzread.info/dash/bejne/e61ly9mknNW0nps.html

@angelmendez-rivera351

2 ай бұрын

I would argue that calling it "real exponentiation" is just abuse of language, and denoting it x^y is abuse of notation.

@denizgoksu9868

2 ай бұрын

"Real numbers" is a bizarre name for them from the get go tbf, but I haven't heard of or thought up any alternatives

@__christopher__

2 ай бұрын

That's not how we defined it in school.We defined rational exponents via roots (that is, inverse functions of integer exponents), and then irrational exponents as limits of rational exponents. Note that 0 is not an irrational number, and therefore the limit definition doesn't apply. The exp(y ln(x)) is then a derived formula, not a definition, and it obviously only applies where ln(x) is defined. In particular, 0^(1/2) = sqrt(0) = 0 is well defined, and so is 0^pi = 0.

@angelmendez-rivera351

2 ай бұрын

@@__christopher__ Yes, but unfortunately, schools teach many things which are incorrect. The indeterminate forms concept is just another example, and there are many other problems. The way decimal notation is taught is wrong (which is why 0.(9) = 1 is such a debated topic among non-mathematicians); the way we teach the concept of functions (which is so few individuals understand that sqrt(4) = 2 and sqrt(4) ≠ 2, even though, given real x, x^2 = 4 if and only if x = -2 or x = 2); the way we teach notation in arithmetic is wrong (which is why we have this debate among non-mathematicians concerning 0^0); the way we teach limits and continuity is wrong (which is why there are many so-called "paradoxes" involving basic concepts in real analysis); we even just teach division wrong (which is why few individuals have an understanding of the problems with division by 0).

0^0 is 1 and undefined

@angelmendez-rivera351

3 ай бұрын

This is a contradiction.

@v8torque932

3 ай бұрын

@angelmendez-rivera351 how so

@angelmendez-rivera351

3 ай бұрын

@@v8torque932 Equality is transitive. By definition, if a = b, and b = c, then a = c. If 0^0 = 1, then 0^0 is defined, because 1 is defined. If 0^0 = 1 AND is undefined, then not only are you implying that 1 is undefined (which is false), but that 0^0 is both defined and undefined, and this is the contradiction.

0^0 = 0/0 = undefined

@angelmendez-rivera351

2 ай бұрын

0^0 ≠ 0/0. By your logic, 0^2 is undefined, since 0^2 = 0^(3 - 1) = 0^3/0^1 = 0/0.

0/0 can be anything, so the debate is what anything has to be instead, so lets use 1 lol :D

@angelmendez-rivera351

2 ай бұрын

This video is about 0^0, not 0/0. Also, strictly speaking, there is no value possible for 0/0, because the multiplicative inverse of 0 _does not exist._

kzread.info/dash/bejne/pnWW3LSBqKXZZqw.html 0^0 = 1? Exploring 0^0 in 5 Levels from Exponents to Math Major 19 hours ago Seems like it's a popular theme this week 😀

@ron-math

3 ай бұрын

Interesting. It seems like a pure coincidence as he has a series of similar stuff.

@ron-math

3 ай бұрын

I watched his video, pretty good.

Lim x^x, (x->0) = 1

Team Cauchy for me :))

Ig you could say argument holds no power and is based on nothing

@angelmendez-rivera351

2 ай бұрын

Sometimes.

0^0 no answer because no operation of division. 0^0 not equal to 1, and not equal to any kind of numbers. Thank you. Taha M. Muhammad/USA Kurd Iraq

0^0 = 0^(1-1) = 0/0 = undefined

@AlbertTheGamer-gk7sn

2 ай бұрын

0! = (-1)! * 0 = infinity * 0 = undefined as well.

@angelmendez-rivera351

2 ай бұрын

@@AlbertTheGamer-gk7sn LOL

@angelmendez-rivera351

2 ай бұрын

Your mistake is that you said 0^(1 - 1) = 0/0. This is false.

@kashphlinktu

2 ай бұрын

@@angelmendez-rivera351 but x^(a-b) = (x^a)/(x^b)

@angelmendez-rivera351

2 ай бұрын

@@kashphlinktu This is not true in general. For example, let x = 0, a = 3, b = 1. When you substitute into the equation and simplify, you obtain 0^2 = 0/0. However, we know 0^2 = 0, and 0 ≠ 0/0, so it follows that 0^2 ≠ 0/0. Therefore, the formula given is not true for all x, a, b.

While it is true that 0^0==1 is true (in most contexts), there's still a part of me that resists. From a non-mathematical viewpoint, it seems to go against common sense. How can two nothings make a something? I feel the same way about 1^∞ being indeterminate. Common sense would tell you that no matter how many times you iterate 1x1, you'll never get anything besides 1. While there are ways to demonstrate its indeterminacy (for instance by graphing x^∞ as it approaches 1 from both sides), it can still feel "wrong". I guess what I'm saying is that sometimes math makes no dang sense. I patiently await Mathematics 2.0. Or was that calculus?

@angelmendez-rivera351

2 ай бұрын

*While it is true that 0^0 = 1 is true (in most contexts),...* In what context is 0^0 ≠ 1? *From a non-mathematical viewpoint, it seems to go against common sense.* Should it matter that it goes against common sense? Most mathematical theorems go against common sense. Should we throw out all of mathematics? *How can two nothings make a something?* The words "nothing" and "something" have no meaning in mathematics, so this not a question any mathematician can answer. *I feel the same way about 1^∞ being indeterminate.* There is nothing in mathematics that is truly "indeterminate" in the way that mathematics teachers will say there is. This whole idea of an "indeterminate form" is an obsolete pseudo-concept based on a misunderstanding of a relatively advanced mathematical theory. 1^∞ is not indeterminate, it is just undefined. *Common sense would tell you that no matter how many times you iterate 1 • 1, you'll never get anything besides 1.* Common sense would be mistaken. To begin with, the mathematical object ∞ does not represent an 'amount' of things or a 'number of times.' ∞ is not a real number, nor is it a cardinal number, nor is it any kind of thing you would recognize as a number. It has a formal mathematical definition, as the supremum of the affinely extended real line, but the affinely extended line lacks the ability to have the arithmetic operations we use when working with real numbers or complex numbers. Therefore, writing the symbol '∞' next to any arithmetic operation will automatically give you an undefined result. 1^∞ is not an exception.

@patcat8950

2 ай бұрын

@@angelmendez-rivera351 I never said anything about math being wrong, just that to the layperson, it *appears* wrong. There are facts of math (and of physics, and of many other scientific disciplines) that go against what one would naively expect the answer to be. Yet, *eppur si muove* -- belief doesn't trump observation. GPS devices have to compensate for the relativistic time differential between our "static" frame of reference and those of the satellites; sticking to common sense will literally get you lost. Factz iz factz; consensus is irrelevant. To be clear, **I am not arguing against you**. I'm just stating, as you did, that math doesn't always align with common sense, and it doesn't have to. As for your context question -- 0^0 is indeterminate/undrfined in calculus, as stated in the video. Doesn't mean it isn't equal to 1 -- just that it's not demonstrable using calculus. (Of course, we all know that calculus was just made up by Newton and Leibniz while they were high. "Hey. Hey, dude. You know that Zeno guy, with his distances and his footraces and his arrows? What if-- what if we came up with something that unparadoxed his paradoxes?" "Dude, that's an awesome idea; we just gotta come up with a name. ...I am sooo stoned. Wait, what's the Latin word for 'stone'?")

0^0 is only controversial because the commonly used definition of exponentiation sucks. Define x^y as 1 multiplied by x y number of times, and all is good.

Wolfram Alpha is Team Cauchy, but Sagemath is Team Euler. I find myself on Team Euler most of the time--the situations in which it's more useful to take 0^0 = 1 are far more common in my experience than the ones in which it's important to regard it as indeterminate.

@ClementinesmWTF

3 ай бұрын

“Proof by usefulness” unfortunately is not a valid method. Ultimately, I agree that 0^0=1 is the case that comes up the most in areas of more elementary mathematics, but that does not mean we can take it to be true always. There are places where/axioms that imply it is false and that is why people argue against people just straight up saying it like it’s a fact and instead giving it an ambiguous case in elementary algebra. Once you graduate to higher mathematics, it’s ok to begin talking about sqrts of negative numbers as existing, or 0^0 having a value, or 1/0 being able to be defined, but in the most general, elementary cases, it should all be avoided.

@MuffinsAPlenty

2 ай бұрын

@@ClementinesmWTF "There are places where/axioms that imply it is false" Could you give an example? Usually when people make a blanket statement like this, I've found their examples to be rather lacking. Things like "if you make a logical error and want to continue making that logical error, then 0^0 = 1 causes problems." But maybe you have a legitimate example!

However, in the L0 norm, 0^0=0.

@ron-math

2 ай бұрын

Mind be more specific?

@angelmendez-rivera351

2 ай бұрын

The L0 norm? There is no L0 norm. Lp norms only exist for p >= 1, and last time I checked, 0 >= 1 is false.

@MuffinsAPlenty

2 ай бұрын

Since NXTangl has not responded, I can explain to the best of my knowledge, since I've seen this sort of argument before. For real numbers p >= 1, the Lp norm of an n-vector is (|a₁|ᵖ + |a₂|ᵖ + ... + |aₙ|ᵖ)^(1/p). These are genuine norms. Some people have chosen to extend the notion of Lp norms below p = 1 by dropping the "1/p" power. So if 0 Lp "norm" of an n-vector is |a₁|ᵖ + |a₂|ᵖ + ... + |aₙ|ᵖ. These are not genuine norms since they are not homogeneous. (Keeping the 1/p power would keep them homogeneous, but would break subadditivity - i.e., it wouldn't have a triangle inequality. The triangle inequality wins the battle of which is the more important property to keep.) If one then takes the limit as p approaches 0 (from the right), one obtains a count of the number of nonzero entries of the vector, since if a is nonzero, lim(p→0⁺) |a|ᵖ = 1 and also lim(p→0⁺) |0|ᵖ = 0. In other words, if we let ||·||ₚ denote the Lp "norm", then for a given vector x, we have lim(p→0⁺) ||x||ₚ = the number of nonzero entries in x. And if one is not concerned with issues of continuity, it might seem reasonable to some people to just "replace all the p's with 0's" to get an L0 norm (although, again, it is not actually a norm since it is not homogeneous), and denote it ||·||₀, and to denote its formula as |a₁|⁰ + |a₂|⁰ + ... + |aₙ|⁰. Denoting the formula for the L0 "norm" in such a way requires us to use 0⁰ = 0, since lim(p→0⁺) |0|ᵖ = 0. Of course the justification for this is "just replace all p's with 0's" which amounts to "just assume aᵖ is continuous as p→0⁺ for all nonnegative a, even in the case where a = 0". Despite not being an actual norm, if one considers the distance function induced by the "norm", one gets the legitimate and useful Hamming distance. And I can understand the temptation to just replace all p's with 0's. It seems to make a nice compact way to represent this idea which is consistent with the notation for other Lp "norms" for 0 Of course, I'm am very much in favor of 0⁰ = 1 and I don't know much about this topic beyond what I just wrote, so others might have other things to say which spin this in a more positive light or give other insights not written here or given corrections to what I've said here.

@angelmendez-rivera351

2 ай бұрын

@@MuffinsAPlenty I appreciate the thorough exposition here. I must admit, though, that I fail to see how, even in this context, 0^0 = 1 would be false. This ultimately just seems like a rehash of the fallacious argument from continuity against 0^0 = 1. I know the Hamming distance is quite useful, but strictly speaking, it does not require letting 0^0 = 0, much in the same way that using the binomial theorem does not strictly require 0^0 = 1 either. It seems like another argument of the form "facts are inconvenient, so we should throw out the facts," while pretending to be something else.

@MuffinsAPlenty

2 ай бұрын

@@angelmendez-rivera351 Yes, I agree. I will admit, though, that I find it to be the most genuinely appealing case for 0^0 = 0 in any context. (I still do not find it particularly convincing, though.) And I think it was even you who mentioned to me before that using 0^0 = 1, we could similarly get an "indicator function" for any number r as f_r(x) = 0^|x−r|. In which case, one could also easily get a "nonzero counting function" for an n-vector by taking n−(0^|a₁| + 0^|a₂| + ... + 0^|aₙ|). Another rather unique example I've heard someone suggest was Münchhausen numbers was an argument. A positive integer in base ten is a Münchhausen number if the sum of its digits raised to the digit's power is the number itself. So, for example, 3435 is a Münchhausen number in base ten since 3435 = 3^3 + 4^4 + 3^3 + 5^5. If 0^0 is disallowed or has a value of 1, then there are only two Münchhausen numbers in base ten: 1 and 3435. However, if one allows 0^0 = 0, then 0 and 438579088 become Münchhausen numbers as well (leaving only four Münchhausen numbers in total in base ten). So apparently this was justification for why we should have 0^0 = 0. I didn't find this to be a convincing argument and went on to point out that Münchhausen numbers seem more of a novelty than anything meaningful, especially since they are so base-dependent. I went on to also show that in some bases, 0^0 = 1 produces more "additional" Münchhausen numbers than 0^0 = 0 does, reiterating the point that this was a silly justification for 0^0 = 0 since it was so incredibly base-dependent and didn't even perform as well as 0^0 = 1 in certain bases to boot. But it was a rather unique example I haven't seen before, so I thought I might as well share it here with you, too.

The answer is 180

0^0 x 1 = 1 x 1 I don't see any credible argument against this fact.

1

0⁰ = e⁰ˡⁿ⁰ = (e⁰)ˡⁿ⁰ = 1ˡⁿ⁰ = 1 despite ln0 approaching -∞. It boils down to 1/(1^∞), and I'm not sure why this is a problem. No matter how many times you multiply 1 by itself, it always will stay at 1.

@angelmendez-rivera351

2 ай бұрын

0^0 ≠ exp(0ln(0)). ln(0) does not exist, because there exists no real (or complex) number z such that exp(z) = 0.

Although I appreciate you not explicitly choosing a side at the end there, this video is just dripping with bias. You make it sound like people are leaving 0^0 undefined because there hasn't been a proof for the contrary. But you never talk about the consequences of defining 0^0. First and foremost, integer-exponentiation is now non-analytic, because you're introducing a puncture into the function. That ruins the lives of a LOT of analysts, because for them, leaving 0^0 undefined *is* the nicer option.

@ron-math

3 ай бұрын

Thank you for the comments! The bias you noticed is my "preference". I have no problem leaving 0^0 as undefined or 1 ( although I went over several arguments why it "should" be defined as 1). The main issue I tried to address is the damage that the "indeterminate forms" notations have done, blocking students to think this notation in a bigger picture rather than playing limit construction games. That being said, I also acknowledge the status quo of calculus teaching and it made me sad: we can do better.

@MrHerhor67

2 ай бұрын

Why would it be non-analytic? What does it even mean for a noncontinuous function?

@farklegriffen2624

2 ай бұрын

@@MrHerhor67 x^y is continuous if and only if you leave 0^0 undefined. If you define it, it's then discontinuous, and thus non-analytic.

@angelmendez-rivera351

2 ай бұрын

@@farklegriffen2624 *...this video is just dripping with bias.* Well, yes, but there is nothing wrong with it. There is nothing wrong with bias _existing._ Everyone has biases, including me, and yes, it includes you too, whether you want to believe otherwise. Besides, some biases are just factually correct. Saying "this is a bias" does not give you justification to outright dismiss what is being said. Also, it is very strange to bring up bias when discussing something as objectively studied as mathematics are. *You make it sound like people are leaving 0^0 undefined because there hasn't been proof for the contrary.* No, he did not. In fact, he did present the arguments which are usually made in favor of leaving 0^0 undefined. *But you never talk about the consequences of defining 0^0.* False. He clearly did. He explained how 0^0 = 1 enables us to consolidate the notation we use for expressing polynomials and power series in Σ-sum notation. He also explained how this has consequences in combinatorics and set theory as well. *First and foremost, integer-exponentiation is now non-analytic,...* Uh... yes? Trivially._All_ functions from the set of all pairs of natural numbers N^2 to set N are trivially non-analytic, since they are not even functions of real numbers. *...because you're introducing a puncture into the function.* What are you talking about? *That ruins the lives of a LOT of analysts, because for them, leaving 0^0 undefined is the nicer option.* Oh, you misspoke. You were talking about so-called "real exponentiation," not integer exponentiation. Okay, well, what you said is still completely incorrect. Regardless of whether or not you define 0^0 or not, the function g defined by g(x, y) = exp(xln(y)) is still non-analytic. It cannot be analytic, because ln is not an analytic function either. Thus, defining 0^0 = 1 has no bearing on the analyticity of this function. It does not ruin the lives of analysts at all. 0^0 = 1 does not change the fact that lim g(x, y) (x -> 0, y -> 0) does not exist. And I have never met any analysts who think this is a problem, because there is never a situation where you are required to evaluate g(0, 0). In fact, 0^0 ≠ g(0, 0), and even excluding this, it is still true that y^x ≠ g(x, y) in general. *x^y is continuous if and only if you leave 0^0 undefined.* No. x^y is an operation with integers. g(x, y) is continuous if and only if you leave g(0, 0) undefined, and we are not suggest that g(0, 0) = 1.

@ChrisDreher

2 ай бұрын

​@ron-math I came here to post something similar to the original commenter above. It would have been better to go deeper on the 0^0 = undefined argument during the limits section (even if you disagree and later present why). As it stands, it only gets a description which is preceded by emotional statements of the limit approach as "lazy" and makes you feel "sad". While the rest of the video is good, the limit section of the video itself ironically comes across as lazy and sad.

4:25

One thing I wonder: can `0^0=1` be used as an axiom for `0*inf=0`? Since: `0^0 = exp(ln(0^0)) = exp(0*ln(0)) = exp(0*-inf)`, and `0^0 = 1 = exp(0)`, therefore `0*-inf = 0`. You can use similar logic for: `0/0=0`, `inf/inf=0`, `inf-inf=-inf`, `1^inf=1`. It all stays pretty self-consistent. You can take other axioms too; if `0*inf=1` then `0^0` can end up being either `e^1` or `e^-1` depending on the sign of the zero power. I think it's awesome.

@yokinman5550

2 ай бұрын

One thing to note though - in order for "extrema" logic to stay truly consistent, I don't think you can use extrema to cancel constants, coefficients, exponents, etc. Like, none of `inf^2`, `2*inf`, `1+inf`, or `inf` can count as the same thing. Same with `0`, `-0`, etc.

@angelmendez-rivera351

2 ай бұрын

*can 0^0 = 1 be used as an axiom for 0 • ∞ = 0?* You cannot. The problem is that, by definition, ∞ is the supremum of the affinely extended real line, and the affinely extended real line is defined to be Dedekind-MacNeille completion of the real numbers when the real numbers are taken specifically as a lattice (as opposed to a field). This means the affinely extended real line is a complete lattice, but the problem is that it can be proven that a complete lattice cannot be endowed with field operations. Actually, the situation is much worse: they cannot even manage to be a semiring. Whether you define 0^0 = 1 fails to address the issue, because this fact I just mentioned has nothing to do with exponentiation. *Since 0^0 = exp(0ln(0)) = ...* 0^0 ≠ exp(0ln(0)). Think about it: ln(0) does not exist as a quantity. I know you are postulating here that ln(0) = -∞, but this is not true. You _can_ say lim ln(x) (x -> 0) = -∞, but this and ln(0) = -∞ are completely different statements in general.

Zero is not even a number. What's the debate about 😂

@angelmendez-rivera351

2 ай бұрын

0 is most definitely a number. It is a natural number, an ordinal number, a cardinal number, a rational number, a real number, a complex number, a hyperreal number, and a surreal number, among other things.

@YouTube_username_not_found

2 ай бұрын

@@angelmendez-rivera351 Is this the definition of "natural number"? a finite cardinal number.

@angelmendez-rivera351

2 ай бұрын

@@KZread_username_not_found Strictly speaking, I would define a natural number as a finite ordinal number, but there exists a theorem which states that the set of all finite ordinal numbers, and the set of all finite cardinal numbers, are equal, so the distinction happens to not be relevant here. By this definition, 0 is a natural number, since it is a finite ordinal number. It is the only limit ordinal which is finite. I know some authors exclude 0 from the definition of a limit ordinal, but I find this convention to be very silly.

@YouTube_username_not_found

2 ай бұрын

@@angelmendez-rivera351 >> "It is the only limit ordinal which is finite." What does that have to do with the definition of a natural number? >> "I know some authors exclude 0 from the definition of a limit ordinal, but I find this convention to be very silly." That's irrelevent to it being a natural number. Perhaps you mean to say _"definition of a finite ordinal"_ ?

@angelmendez-rivera351

2 ай бұрын

@@KZread_username_not_found It has nothing to do with the definition of a natural number. It is merely a bonus fact. If you wanted to define 0 specifically as a mathematical object, then defining it as the unique finite limit ordinal would be adequate, but this alone has no bearing on how the natural numbers are defined.

Your proof that there is one zapping from 0 to 0 is incomplete (thus wrong). Lets recale that 0 is by definition the emptyset and any integer n:={0,1,..,n-1} If you use the same argument with 0^1 you will obtain the same conclusion that is wrong. Indeed an element in 0^1 is a subset of 0×1=0 so if it existe it is 0. The fact is that it does not exist in this case : 0 is not a map from 1 to 0. There is NO such map What we can say is that 0^a or a^0 cannot contain more than 1 element, but it can contain no element at all. So your argument is incomplete : The reason why there is actually one fonction from 0 to 0 is grammatical : For any property p we have : For any x in 0 p(x) is amways true And Three existe x in 0 s.t. p(x) Is always wrong : So for all set a , a^0 has always exactly 1 element : that is 0 Indeed f in a^0 iff for any x in 0 blablabla : ALWAYS TRUE So any f satisfy this, by what you said f is a subset of 0 and then it is 0, and a^0={0} witch has cqrdinality 1 (it IS 1 btw) And if b is not 0 : 0^b cas NO element : Because any element of 0^b has to satisfy : For any element x in b there exists y in 0 s.t. blablabla And there is no such a Mao, the cardinality of the set of such maps is then 0

@angelmendez-rivera351

2 ай бұрын

*If you use the same argument with 0^1, you will obtain the same conclusion that is wrong.* No, this is false. Here, 0^1 is the cardinality of the set of all functions from the set 1 to the set 0, but there are no functions from the set 1 to the set 0, so 0^1 = 0. The argument in the video is not incomplete. It implies both 0^0 = 1 and 0^1 = 0, and there is no contradiction. *Indeed, an element in 0^1 is a subset of 0 × 1 = 0, so if it exists it is 0.* No, not quite. The graph of a function in the set 0^1 is equal to some subset of the set 0 × 1, so if it exists, then the graph is equal to the set 0, which means the domain of the function is the set 0, which contradicts the fact that the function is in the set 0^1. Therefore, 0^1 is equal to the set 0, because it is empty. *0 is not a map from 1 to 0.* The video never claims this. You are conflating the graph of a function with the function itself. *The reason why there is actually one function from 0 to 0 is grammatical.* No, it is not. Grammar has no relevance in mathematical analysis. The reason is entirely set-theoretic.

@YouTube_username_not_found

2 ай бұрын

​@@angelmendez-rivera351 He probably meant to say the reason is _syntactic_ . Going back to the definition of a function, when the domain A is ∅ , the definition will give a proposition of the form *∀x∈A ; P(x)* but because A is ∅ , P(x) is *"vacuously" true* , and when the domain A is not ∅ , P(x) , which is *∃y∈B=∅ ; (x,y)∈A×B* , is no longer true because now there is x∈A and there is no y∈∅ . All of this is contingent on these two properties: *∀x∈∅ ; P(x)* being always true and *∃x∈∅ ; P(x)* being always false .

@YouTube_username_not_found

2 ай бұрын

Edit: I should have added ! to the ∃ at the 4th line.

@YouTube_username_not_found

2 ай бұрын

The fact that he didn't say it explicitly does not mean that the definition of a mapping in the video suddenly disappeared. Indeed, a subset of A×B is potentially a graph of a function but one should later check whether it satisfy the property in the video. Edit: this reply is directed to @savonliquide7677

@savonliquide7677

2 ай бұрын

@@angelmendez-rivera351 read more carefully please