1 ^ â, It's Not What You Think
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- Äas pĆidĂĄn 29. 01. 2023
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In this video, we explore the concept of 1 raised to the power of infinity. At first glance, it may seem like the answer should be 1, but as we dive deeper into the topic, we quickly realize that it's not that simple. We discuss the limitations of using limits to try and determine the value of 1^infinity and how the traditional definition of exponentiation doesn't allow for infinity to be used as an exponent. We also explore an alternative approach, using logarithms, to try and determine the value of 1^infinity. We show how logarithms allow us to rewrite exponential expressions as a power of the base, and how in this case, 1^infinity can be rewritten as log(1^infinity) = infinity * log(1) = infinity * 0, which is undefined.
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THIS WAS UPLOADED 5 DAYS AGO HOW DOES IT SHOW 8 DAYS AGO????????????????????????
lol
This is all speculation. Infinity is not a number. It is like trying to divide infinity by number 2.
@@ericephemetherson3964 This is the correct answer, and I find that this entire discussion is an exercise in sophistry, a waste of time, and rooted in a fundamental misunderstanding of arithmetic concepts, originated from misunderstandings in calculus. Ultimately, the problem is that people's uncultivated intuition tells them that we should be able to define infinite numbers, and arithmetic with them, even though they have no idea as to how one would go about doing this. If an arithmetic operation with an infinite quantity is undefined, then we should be able to just define it, and so, they fail to understand why mathematiciand cannot do that. You have to explain how a mathematician would go about formalizing these intuitions into a rigorous theory, and then explain how the theory demonstrates that the above wishes are impossible to accomplish.
@@angelmendez-rivera351 You have put your post very eloquently and running parallelly with my thinking. I studied calculus which I did not finish due to my illness and my main subject was physics which I still love till today. But physicists are not mathematicians. Actually mathematicians are not able to see the physical world as well as phycists because mathematicians require numbers whereas phycists possess great imagination. And you are correct; some mathematicians really do not know what they are talking about. Take, for example the calculations of probabilities. If you put probabilities into numbers then logic must tell you that the calcualtion runs on THE probablity. Same with an electron in an atom; you are only certain of some probability that the electron is going to be in certain position. So, calculating probability automatically makes thta calculation probable and you will never know anything for certain. I hated that part of physics because why calculate a probability? It seemed futile to me. I would rather be certain. Also, mathematicians and physicists use the 'time'' in their equations without a definition of time. And that's another strange issue to me.
What we learn in class: 2ÂČ = 4
Whats on the homework: 2Âł + 3 = ?
The exam:
this deserves many likes
The answer for homework is 11
8
@@VoidGravitational I really hope youre joking
@@N.I.G.H.T.M.A.R.E its 9
As a physicist, I'm going to forget I ever saw this so I can still do... well a surprising amount of physics that relies on 1^infty = 1 lol.
Well, I have a question for you.
Could the practice of mathematics be reworked so three is no use of "infinity"?
It doesn't have a counterpart in reality, anyway. There is no such thing as an infinite amount of anything.
And, the point of the science of mathematics is to be applied to the real world.
Why not just throw out the concept of "infinity" altogether?
â@@TheNoiseySpectator In physics values aren't exact, we rely on experimental data: As long as a shortcut doesn't impact the final result any more than measurement uncertainties do, physicists will take them so that the calculations may be simplified and easier to understand since it won't matter anyways in the end.
@@TheNoiseySpectator That's a really interesting question. What you're saying on it's face feels like it should be correct and in some sense it is. You're right that everything we experience is finite. However, just because everything is finite in existence doesn't meen it is 100% describable using finite mathematics. I don't know your mathematical background so I'm gonna give some examples that might be a bit math heavy, but I'll try to explain everything explicitly so if you're not familiar with the jargon then it should still make sense.
One of the most useful things in physics is the Taylor series. Effectively what this is is a way of turning mathematical objects, such as sine functions and exponentials, into an infinite summation of terms and vise versa. If you only take some of the terms instead of the infinite number of terms, this is not equivalent to the original mathematical objects. In theory, this sounds like it should make things far worse, however it is extremely often that transforming something into a Taylor series allows you to simplify the function in some way before transforming it back out of a Taylor series. Personally, in my field of physics, I use this all the time. For example I used it for simplifying matrix exponentials in quantum mechanics which comes up all the time when looking at how a system evolves over time.
Going a bit back before that, infinity is used in the basis of calculus. Calculus is,in essence, the study of rates of change in mathematics. Infinity and infinitesimal (effectively the inverse of infinity [smallest number possible]) come up in 2 of the most important calculus concepts: the derivative and the integral. Without this, we wouldn't have the backing for the vast VAST majority of physics. Think about how when a car drives down a road, it has a certain speed. That speed is the rate of change in position of the car. The time evolution of thestate vector of a quantum system I was talking about before that gives rise to a matrix exponential? That's derived using differential equations (aka calculus).
Basicly what I'm saying is that while infinity itself isn't something we see in our day to day, it is a key concept that governs the mathematics that backs out understanding of the dynamics of our day to day. A world without considering infinity would be a world that is unrecognizable to us today.
â@@Eric-jh5mpsuperb answer
â@@Eric-jh5mp i appreciate your answer it was very easy to understand for a highschooler like me!
I rely on the âcome on manâ theorem on questions like this. The way it works is if thereâs a difficult question in math you assert your opinion without backing it up in any way. And if someone challenges it you just say âcome on man.â
That got me.
With what tone and expression? I wanna use this.
@@timeslice "come on MAN" read in a southern accent/Biden's accent
â@@timeslice There is no wrong tone. All tones are equally correct and valid here to say "Come on, man", they all work perfectly.
Proof by intimidation
The only sensible interpretation is that 1^oo = lim n->oo 1^n and that is 1.
That wouldn't help for a long enough video though and the advertiser would be unhappy.
@@azimuth4850 I have already shown why what you say is nonsense in another thread that you have seen. Repeating your nonsense doesn't make it become true.
Here is what I said (to Tigris Callidus):
"I have already stated that you cannot induce to infinity because there is no natural number n such that n+1 = oo. How isn't that obvious to you beggars belief. Whatever, it is a fact, whether or not you understand it or agree with it.
Here's an example of why induction to infinity cannot possibly be OK.
1 > 0.9
1 > 0.99
1 > 0.999
...
ergo
1 > 0.999...
You might think that's fine, however:
0.999... > 0.9
0.999... > 0.99
0.999... > 0.999
...
ergo
0.999... > 0.999...
Here's another: 1 is finite, 1+1 is finite, 1+1+1 is finite, ergo 1+1+1+... is finite.
You could restate that as, 1 is a natural number, 2 is a natural number, 3 is a natural number, ergo oo is a natural number.
Do some research, unless you like embarrassing yourself"
@@azimuth4850 Pay attention, it is not me that is saying that we should induce to infinity. I have never said that you can use induction to get to infinity. I am the one say that it is not possible to use induction to get to infinity. How can you get it so backwards?
You can prove, by induction, that for EVERY natural number that 1 > 0.999...9 (n 9s) and 0.999... > 0.999...9 (n 9s). Yet 1 = 0.999... and 0.999... = 0.999....
You cannot prove that 10 > n for every natural number using induction (or any math at all).
Stop saying that I have said things that I have not said.
It is my pleasure to insult you. You are a deserving cause.
@@Chris_5318 Ok, I see your example that certain types of naive induction do not work with the infinite, and I don't dispute that. I have never seen those example before, and I'll admit they made me think and I learned some math I didn't know. I think my confusion arose because the reason you gave initially that induction doesn't work to infinity because there is no n such that n + 1 = oo. I think that's not quite the exact reason, because as I was saying the same criticism could be leveled at the definition of a limit. The reason the limit works and induction doesn't (in some cases), is because the limit is taking the difference to be arbitrarily small. It's actually a form of boosted induction now that I think about it.
In any case, the induction does work with 1^oo, which was my original point. Proving 1^N = 1 for any arbitrary N proves that 1^oo = 1. Because nothing is changing. The examples you gave all have to do with infinite repeating decimals, or logical statements about a sum being "finite" vs "infinite." Well I agree that creates problems with simple induction. But I have to stand firm that poses no issue for 1^oo.
â@@azimuth4850 I have said that you CANNOT induce over the naturals to get to infinity. I also said that Induction has nothing whatsoever to do with the definition of limit. In case you haven't see it, here is the relevant epsilon-N definition:
Let S_n be a sequence of real numbers and let S be a real number.
Then S = lim n->oo S_n means that for every real Δ > 0
there is a natural N such that for every natural n > N that |S_n - S| < Δ.
That is what the definition is, regardless of whether you agree or not. NB The body of the definition does not refer to oo and the lim expression is essentially archaic, but far too convenient to ditch.
Next we have that expressions such as f(oo) are abuse of notation. The usual convention is that it means lim n->oo f(n). (I'm assuming a discrete function whose domain is the natural numbers. By that convention, 1^oo means lim n->oo 1^n and that is 1.
Proof: Choose an arbitrary Δ and then N = ceiling(1/Δ), in fact any natural number will do.
Then for all n > N we have |1^n - 1| = 0 < Δ. As Δ is arbitrary, that must be true for every Δ > 0, and so the limit is 1.
Based on my calculations, this video will be epic
đż
@@coolbasedgigachad6820 based
I saw you at Fireship!
@@radda_radda im famous, wanna take a photo with me?
@@draido-dev No, it's okay
1:46 The reason why you get different answers is because both the bases are different numbers. They are not equal to 1, just two different numbers that are really really close to 1.
An exact 1 to the power of infinity is equal to one.
If you substitute infinity with a very large number you will quickly see that the extra term added to one does not actually tends towards zero. You need to use binomial expansion which generate new terms the higher the exponent is and with exponent near infinity the number of additional terms being added is also near infinity.
No. 1^â is undefined. What you mean to say is lim 1^x (x -> â) = 1, which is true, but irrelevant, since it has nothing to do with the question.
@@kazedcat Consider lim f(n) (n -> â) = 0, and lim g(n) (n -> â) = â. (1 + f(n))^g(n) = 1 + f(n)âąg(n)/1 + f(n)^2âąg(n)âą(g(n) - 1)/2 + f(n)^3âąg(n)âą(g(n) - 1)âą(g(n) - 2)/6 + âąâąâą + f(n)^(g(n) - 1)âąg(n) + f(n)^g(n) = 1 + f(n)^g(n) + f(n)âąg(n)âąh(n). Hence, lim (1 + f(n))^g(n) (n -> â) = 1 + lim f(n)âąg(n)âąh(n) (n -> â), where h(n) = 1 + f(n)âą(g(n) - 1)/2 + f(n)^2âą(g(n) - 1)âą(g(n) - 2)/6 + âąâąâą + f(n)^(g(n) - 2).
@@angelmendez-rivera351itâs only undefined when you are talking about limits, which I was not.
This video is a bit misleading, since it failed to show the difference between limits and computations.
I was just adding extra information as to why you get different answers for what seems to be the same question. 1^infinity equals e while another case 1^infinity equals 1
@@sanauj15 *It's the only undefined when you are talking about limits, which I was not.*
Okay, you need to go back and read my comment again, because not only does this not address anything I said, I also actually already addressed this misconception here.
*This video is a bit misleading, since it failed to show the difference between limits and computations.*
False. He explicitly brings up this distinction during 2:43 - 3:00 in the video. It seems you need to watch the video again.
*I was just adding extra information as to why get different answers for what seems to be the same question.*
You did not. You just parroted the same misconception that hundreds of other comments in the comments section have already stated, all while failing to understand that lim 1^x (x -> â) is not the same as 1^â, which is ironic, since you complained that the video failed to make this distinction. You failed to make it yourself.
Rather than getting all defensive as you did, when someone corrects you, I recommend that you actually (a) read their comment, for one (b) understand why the comment is correcting you (c) acknowledge the correction.
"its not what you think"
proceeds to say it is exactly that
What I don't understand about analysis like this is the whole idea is to get mathematics consistent, correct? So we, like you said, can choose 1â to be whatever we want as long as it's mathematically rigorous and consistent. So let's say 1^â is 1, is there any other (potentially seemingly unrelated) mathematics that breaks with this definition? Because if there isn't, I don't see any issue with defining it this way.
The example used here where the limit as x -> â of (1 + 1/x)^x = e might be a good counter-example to this point, but the issue I have here is that it doesn't really tend to exactly 1^â, it's more tending to (1+)^â, but I guess an argument could be made here that this should diverge to infinity, while it clearly doesn't - this is another issue though.
he's not defining anything, 1^inf it's undefined. There are many function where if you take the limit x->inf those goes to a specific number or diverge. That's it
@@laxis3155 Well a definition is completely arbritrary, â-1 is "undefined" but then we defined it to be the imaginary unit 'i', why can't we do the same here and take 1^â, which is undefined, and then define it to be 1 if it is rigorous and consistent with the rest of Mathematics? This is all devil's advocate though, I'm 100% sure there is a very good reason we can't do such a thing because defining 1^â in this way messes something up in some context or another.
The thing is the symbol infty isn't really valid to exponentiate with. We can surely say something like lim as x--> infty f(x) = k and then talk about k, but with the understanding that k is not "f(infty)" because it doesn't actually mean anything to take a function's value at infinity. We use stuff like f(infty) as a shorthand but there's always some understood context for what that "really means". This is the source of supposed controversy of a slew of notation. Including 0^infty, and 1^infty, when these things are really only defined as limits, and writing it this way is somewhat an abuse of notation.
@@laxis3155 *There are many functions where, if you take the limit x -> â, it goes to a specific number, or diverges.*
This is completely irrelevant, though. The question is, "what is 1^â equal to?" The question is _not_ "what is lim (f -> 1)^(g -> â) equal to?" These are completely different questions, with different answers. If I ask you "what is floor(0) equal to?" your answer should not be "well, lim floor as x -> 0 does not exist, because..." No, your answer should be "floor(0) = 0, because floor(x) is defined as max({n in Z : n =< x}), and max({n in Z : n =< 0}) = 0."
@@squeezy8414 *So, we, like you said, can choose 1^â to be whatever we want, as long as it's mathematically rigorous, and consistent. So let's say 1^â is 1, is there any other (potentially seemingly unrelated) mathematics that breaks with this definition?*
1^â is undefined, but it actually has nothing to do with limits. It has nothing to do with real analysis. If you want an answer for why 1^â cannot be defined, then the best place to start is to learn about ordered groups, lattices, and the concept of a Dedekind-McNeille completion. I will not delve into any details in my comment, but all I will say is that there actually does exist a reason for why you cannot do arithmetic involving +, âą, and ^ with â (well, unless you restrict yourself to the natural numbers, specifically, for a strange reason). There are key theorems that imply this, so this is not a matter of "we have not found a way to make it work," but a matter of "formal logic itself demands that it be impossible." It is akin to the reason behind why we cannot divide by 0.
*The example used here, where the limit as x -> â of (1 + 1/x)^x = e, might be a good counterexample to this point, but the issue I have here is that it doesn't really tend to exactly to 1^â, it's more tending to (1+)^â,...*
Exactly! This is why I think this video's answer to the question is just completely wrong. In fact, I think _every_ video on CZcams is wrong about this, because the language of the so-called "indeterminate forms," which is very problematic on its own right and has many fundamental problems, just does not actually answer the question, it answers a completely different, unrelated question, which has nothing to do with the topic. Questions about arithmetic expressions are ultimately in the domain of abstract algebra, not real analysis. Limits do not belong in these discussions at all. This is true of the problem with 1^â, but also true of the expressions 0^â, 0^0, 0/0, 1/0, â^0, 0âąâ, â/â, â - â, etc.
âBut, is it?â
**cues the VSauce Intro Music*
Also you can approach this like the 0^0 value is approached: if we want a^m a^n = a^(m + n) to hold and we allow ourselves â + â = â, then we have a condition 1^â 1^â = 1^â, i. e. this should be a multiplicative idempotent. If we want it to be a real or complex number then we already have just two options left: 1 and 0. Now, we can make a couple of arguments to discredit 0 as the potential value, accurately enough to leave 1 as valid. VoilĂ !
The problem with this is that, unlike with 0, where arithmetic is well-defined, arithmetic with â is not well-defined, generally speaking.
@@angelmendez-rivera351 Well, in general yes but if we constrain ourselves to a particular structure like Briâs favorite wheel algebra (where thereâs just one signless â and also we need to add â„ as a kind of plug for holes) or a plain projective line over a ring (where â is also signless) then it works too but just not all operations get to be defined, as multiplication by â (a valid projective transformation, albeit degenerate) evaluated at 0 gives 0 and multiplication by 0 evaluated at â gives â so either we donât get to define 0 Ă â or we get noncommutative Ă, and so on for a couple of other cases like this one. IMO having all operations of multiplication/addition by a constant defined albeit binary operations end up a bit ill-defined is an okay situation.
@@05degrees *...if we constrain ourselves to a particular structure, like Briâs favorite wheel algebra (where thereâs just one signless â and also we need to add â„ as a kind of plug for holes), or a plain projective line over a ring (where â is also signless), then it works too,...*
It does not. Arithmetic operations are not well-defined on a projective space. The operations that we define on a projective space are projective transformations, which are more complicated than just arithmetic operations. Also, in both cases, using the symbols / and â is entirely an abuse of notation.
*...but just not all operations get to be defined, as multiplication by â (a valid projective transformation, albeit degenerate) evaluated at 0 gives 0 and multiplication by 0 evaluated at â gives â so either we donât get to define 0 Ă â or we get noncommutative Ă, and so on for a couple of other cases like this one.*
As stated, this is not how projective transformations work.
*IMO having all operations of multiplication/addition by a constant defined albeit binary operations end up a bit ill-defined is an okay situation.*
It is not, and if it truly were, then there would be no problem with leaving 1^â undefined.
@@angelmendez-rivera351 me waiting for someone to say ur a nerd and completely roast them (good argument, tho)
â@@angelmendez-rivera351
Ur a nerd and completely roast them
(good argument, tho)
Okay, so before I start giving mathematical explanations, I need to get this out of the way: mathematics educators and communicators seriously need to stop talking about "indeterminate forms" as a legitimate mathematical concept, because I think they are doing a great disservice to people who are trying to get a deeper understanding of the mathematics of calculus and arithmetic. *There is no such a thing as an 'indeterminate form.'* No, in mathematics, there are two, and only two classes of expressions: those which are well-defined, and those which are undefined, or ill-defined (I take "undefined" and "ill-defined" to mean the same thing, as there ultimately is no coherent way for the two to be meaningfully distinguished). An expression like 2^3 is well-defined. An expression like 1^â is not. The latter is in the exact same category as 1/0, ln(0), 0/0, 0âąâ, and so on. Please, let us start calling these things for what they are: undefined expressions, because that is all that they are.
An expression is just a string of symbols. We can take symbols to either be fundamental in signficance (i.e., the symbol by itself means something), or we can take a few symbols that, when combined into a string, that string has fundamental significance (e.g., the string "sin," which is comprised of the symbols 's,' 'i,' 'n,' symbols which have no significance of their own, but when combined into the string "sin," they form a string with some actual mathematical significance). When you combine multiple fundamental, 'atomic' strings into an expression, the significance of this composite expression depends, in some way, on the significance of the fundamental symbols, and how that significance emerges from the combination is dictated by rules of notation. The expression 1^â is an expression with three atoms: the atom "1," the atom "^," and the atom "â." We need to discuss how these individual symbols are defined, before we can say anything about 1^â. What is the symbol 1 defined as? Well, we are secretly working with some fundamental type of mathematical structure obeying certain axioms formulated in some type of logic. This structure is called the ordered field of real numbers, and these are ultimately just some abstract objects that do nothing except satisfy certain properties we call 'axioms,' properties that are meant to be understood rules for how these objects interact, and these objects we take as being primitive or fundamental in this context: they are not defined in terms of anything else, they kind of just "are" for the sake of being alone. These real numbers satisfy one axiom, that there exists some y, such that for all x, xâąy = yâąx = x. This axiom is one of the defining properties of âą, a binary operation, and when we use the other axioms, we can prove formally that this y object satisfying the above property is unique. Since it is unique, we give it a name. The symbol "1" is that name. So, 1 is defined by the property that it uniquely satisfies: that 1âąx = xâą1 = 1, for all x. That is one symbol out of the way. Now, what is ^ defined as? It is a function, specifically a function on the Cartesian product between R+ and R, with R as the codomain. R stands for the ordered field of real numbers, the fundamental structure we are interested in studying, and R+ refers to those real numbers which satisfy 0 < x, they are called positive real numbers. ^ is defined by the property that x^1 = x, that if x < y, then (1 + 1)^x < (1 + 1)^y, and that x^(y + z) = (x^y)âą(x^z). Notice how, in turn, this requires understanding the definitions of < and +, which are all defined by axioms themselves. Everyone here intuitively understands how these objects work, so I will not bother with every detail, but the point is, it matters to know where the foundations of the discussion lie.
The one symbol here that is troublesome here is the symbol â. This symbol is technically well-defined: it is the maximum of the Dedekind-McNeille completion of the real numbers, which is a mathematical structure that extends the real numbers in a very specific way. However, the problem is that the way the symbol is defined, it does not for any arithmetic. There is no mathematically consistent way to extend to the domains of + and âą as functions to include â in the domain, and continue having well-defined expressions. We have tried, but because of the axioms these structures are defined by, it can be proven that there is no way to make it work. Things like â - â and 0âąâ will always cause problems. So, arithemetic expressions with â are *undefined.* They are *undefined,* because the way mathematical definitions work makes them a very specific kind of thing. Mathematical definitions have to make reference to mathematical properties: to well-defined mathematical concepts, and we have to ensure that the definition does not secretly hide a contradiction within it (which makes it incoherent), and we have to make sure that whatever properties are being referenced are actually uniquely satisfied if the definition is meant to denote a particular object, and not an entire class of objects. Since the axioms make it impossible for â to be in the domain of any suitable extensions of +, âą, and ^, the objects that satisfy the properties that the expression 1^â would make a reference to simply do not exist. In other words, the expression does not refer to anything at all! It has no actual significance, and this what we mean when we say the expressions are undefined. Definitions in mathematics are not an arbitrary endeavor, like everyone likes to say. There is some degree of arbitrariety, yes, but only after you get past many walls of restrictions that must be satisfied.
Now, what is the funny business with indeterminate forms? Why do people insist to interpret 1^â as being a "shorthand expression" for lim f(x)^g(x) (x -> p), specifically in the cases that lim f(x) (x -> p) = 1 and lim g(x) (x -> p) = â (whatever this expression even is supposed to mean)? The problem here is that these two notations actually make reference to two completely unrelated mathematical topics that are different. We need to stop making an association and pretending these symbols are synonymous. Do we use the symbols floor(0) and lim floor (x -> 0) to mean the same thing? No, we do not, and the questions "what is floor(0)?" and "what is lim floor(-x^2) (x -> 0)?" have completely different answers. floor(0) = 0, while lim floor(-x^2) (x -> 0) = -1. Incidentally, lim floor(x^2) (x -> 0) = 0, which is not equal to -1. Does that mean floor(0) is "an indeterminate form"? Hmm??? I ask this in a completely serious fashion, just to highlight how, um, inaccurate (to put it politely) the notion of "indeterminate forms" is. No, floor(0) is not an indeterminate form. No one thinks of it as one. We all know that floor(0) = 0, and if I ask you "what is lim floo(x) (x -> 0)?," then you would, correctly say, that the expression is undefined/ill-defined. Alternatively, you would say that the limit being referenced by the expression does not actually exist: there is no mathematical object that we can call "the limit of floor(x), x -> 0," because the definition is not satisfied by any objects at all. So, despite this being the case, why are people still insisting on this indeterminate form concept? I do not understand it. The question "what is 1^â?" is a completely different question from "if lim f (x -> p) = 1 and lim g (x - p) = â, then what is lim f(x)^g(x) (x -> p) equal to?," and the two questions have different answers. For the former question, the expression given is undefined. *Why* is it undefined? To answer that, you have to get into a discussion of Dedekind-McNeille completions as mathematical structures, and why their properties do not allow for + and âą as operations. The latter question, though, has a simple answer: it depends on the properties of f and g as functions. Do you see how these are different questions? One question is about arithemetic, abstract algebra, and about mathematical structures. The other question is about functions of real numbers and their asymptotic behaviors. Completely unrelated topics. So, why does the education system insist in treating them as the same question and as the same topic? I have no idea why, but we need to stop doing it. And, yes, students do need to understand that knowing lim f (x -> p) = 1 and lim g (x -> p) = â is *not* sufficient information to determine what lim f^g (x -> p) is. But using the language of "indeterminate forms" to teach this is unhelpful, and very misleading. Firstly, it gives people the impression that being "indeterminate" is somehow fundamentally different from "undefined." Secondly, it reinforces the already too common misconception that to evaluate limits, you just "plug in" and see what happens (not how limits work at all). Thirdly, it leads to people developing a bunch of mystical, crackpot nonsense like "ah, so this number/expression is equal to all numbers at once, which proves we are all one [insert more mysticism]." Fourthly, it is just an unnecessarily obfsucating, convoluted way of telling the student "we do not have sufficent information to answer the question." Fifthly, whenever we bring up limits to answer questions like the value of 1^â, we do a disservice by actually failing to answer the question, and instead, answering a different, unrelated question.
Woah.
Thanks
I agree with you that all of these expressions involving \infty are undefined, not 'indeterminate'. I disagree, however, that the expressions of limits of functions and evaluations of functions at points are *unrelated* . They are related in the case of continuous functions and in fact that relationship is so important that I believe it is the source of these controversies and misunderstandings. The logical issue starts by saying something like "well, I'd sure like the function 1^x to be continuous everywhere and oh look, \infty is somewhere so 1^\infty = lim x--> \infty 1^x." Then, one looks at say (1 + 1/x)^x and says the same thing. So now you have 2 incorrect statements that are set equal to each other and thus making 1^\infty "indeterminate" by the incorrect assumptions that 1) \infty is a real number you can evaluate functions at and 2) you can assume continuity and use it to claim that an arithmetic expression is indeterminate. The same thing happens with 0^0 all of the time, only in this case the point of evaluation is a real number, 0. The issue arguably is in not stressing that we can not assume continuity of functions and then using this to make statements about arithmetic expressions. The issue exists, however, because for the overwhelming amount of practitioners of math, you *can* just assume your functions are continuous that and \infty is just a shorthand for some very large number. This is why the shorthand for writing limits as evaluations exists in the first place; people are lazy and most people who use math are not mathematicians. I don't think this point is ignorable as to why the issue exists and why this misconception is so common.
Edited for spelling and grammar. I hate dyslexia.
@@azursmile You're welcome!
*I disagree, however, that the expressions are **_unrelated._** They are related in the case of continuous functions, and in fact, that relationship is so important, that I believe it is the source of these controversies and misunderstandings.*
The relationship between lim(x -> p)(f) and f(p) is incidental. Some functions just so happen to satisfy lim(x -> p)(f) = f(p) for some p in dom(f), and among those functions, some functions just so happen to satisfy it for all p in dom(f). In reality, this is almost none of the functions from a nontrivial subset of R to R. Since the relationship is only incidental, I think I am justified in presenting the concepts as unrelated. The polynomial x^3 - 5âąx + 3, for real numbers x, is defined in a fashion that is conceptually unrelated to the definitions for the expression Î(x + 1), where Î is the well-known Gamma function. Yet, for those real numbers which satisfy the equation x^3 - 5âąx + 3 = Î(x + 1), could you not also say that the two are actually related? Yes, but I think you would agree this relationship is incidental. Just because both expressions are equal for some real x does not mean that that the concept of x^3 - 5âąx + 3 is related to the concept of Î(x + 1). The same is true of my comment. It is entirely incidental that some functions f satisfy lim(x -> p)(f) = f(p), but the concepts behind
the definitions of lim(x -> p)(f) and f(p) remain unrelated in the sense explained above.
*The logical issue starts by saying something "well, I'd sure like the function 1^x to be continuous everywhere,... oh look, â is somewhere so 1^â = lim (x -> â) 1^x." And then, looking at, say, (1 + 1/x)^x, and saying the same thing, so now you have 2 incorrect statements that are set equal to each other and thus making 1^infty "indeterminate" by the incorrect assumptions that 1) â is a real number you can evaluate functions at; and 2) you can assume continuity and use it to claim that an algebraic expression is indeterminate.*
I disagree that this is the source of the problem. Why? Because it assumes that people experiencing the confusion are aware that the continuity of a function f at a point p is defined by the equation lim(x -> p)(f) = f(p). In my experience, this is false. Most people who have completed a calculus course that I have interacted with have no idea that this is how continuity is defined. Ideally, this should be taught in calculus courses, but unfortunately, it seems that in most calculus courses, this is not being taught. But this is just a consequence of what I hinted at to be the real source of the problem: that the way we teach calculus, from beginning to end, is just completely wrong. We teach the concept of limits wrong, and because all the other concepts taught in calculus start with limits at their foundation, those concepts, too, are taught incorrectly. This is what has led to the development of nonsensical language used in calculus classrooms to justify theorems and ideas that, in actuality, if explained in the correct framework, would be very intuitive, without requiring so much of the terminological baggage. And it all starts with the nonsense that teachers tell their students of "when you compute the limit of a function, you have to start by plugging in." If this where your foundations lie, as it is the case for many people, then it is only natural that they will confuse things like 1^â with lim f(x)^g(x) (x -> p), even when they have no clue about the fact that, secretly, whether they know it or not, they are appealing to some sort of continuity argument. As far as they are concerned, they are just evaluating limits the way they were taught to do it.
Now, one point in favor of what you are saying could be that maybe the teachers themselves are making the choice to teach this thing I call "nonsense," using the logic that you outlied. However, the students would never know that. The students have no idea that they are being taught some type of fallacious appeal to the continuity of arithmetic operations. Also, I would hope that a first-year college professor, who presumably has a Ph.D in mathematics, is not actually using the logic you outlied here as justification for teaching students "when you take the limit, you have to plug in" and other such incorrect ideas. If that really is the case, though, then I will lose all faith in humanity.
*The issue, arguably, is in not stressing that we cannot assume continuity of functions, and then use this to make statements about arithmetic expressions, but the issue exists because for the overwhelming practitioners in math, you can just assume your functions are continuous, and â is just a shorthand for some very large number.*
I agree that not emphasizing that one cannot merely assume the continuity of arithmetic operations to then make arguments about arithmetic expressions. I disagree that this is the source of the problem in itself. My understanding is that this is just a by-product of limits themselves being poorly taught in general. Outside of this continuity topic, there is plenty of evidence that limits in themselves are taught poorly. For example, I have looked at what many people have to say on the topic of 0.(9), and all the controversies that emerge from it. I have seen people with degrees in S.T.E.M disciplines make bad, erroneous arguments about how limits work, and why 0.(9) is or is not equal to 1. This tells me that, apparently, it is possible to complete a calculus course with "straight As" despite having no frigging idea of what the limit concept _actually_ is, and how it is supposed to work. Seeing this, I have no reason to think that people are even aware that they are making an appeal to continuity when discussing these so-called "indeterminate forms."
*This is why the shorthand exists in the first place, people are lazy, and most people who use math are not mathematicians.*
I have no issue with shorthands existing, in and of themselves. It sometimes looks like I do, because I make it a point to tell people to not use shorthands. But the reason I tell people to not use shorthands is because they either suck at using them, or they are using them to communicate with others who evidently do not know that these are merely shorthands. Mathematics are so poorly taught at all levels, that it is impossible for most people to tell which pieces of notation are shorthands, and which are not. This includes students with degrees in mathematics, because that is what I have seen in my experience. And yes, shorthands are supposed to be used with the clarification that what is being used is a shorthand. I have no issue with that. But the problem is: in practice, no one clarifies the shorthands to begin with. So, that already tells me that we should not he hoping that shorthands will ever be explained in practice. If that is the case, then it is true to say that, at least from the standpoint of education, we are better off not having shorthands at all, even if that comes at the expense of the people who want to use the shorthands. I mean, this is their fault for not clarifying the shorthands to start with. But again, the shorthands in themselves are not my problem. I would be completely fine with all of this if the education system went out of its way to teach (a) which pieces of notation being used are shorthands, (b) why are they used, (c) when is it not appropriate to use them. But, I have zero hope that there ever will come a time when the education system will satisfy those three demands, or even only 1 of the 3. And for that reason, we should be against teaching students to use those shorthands. If the students believe that the shorthands are a problem, they will be less likely to use them. It will not stop the problem completely, no, because as you said, people are lazy: there will always be someone who will use shorthands, regardless of what they are told. But it will certainly mitigate the problem to an extent and reduce the scale of the "controversy." Teach them the correct notation and correct terminology. There are legitimately zero reasons not to. This can only lead to less confusion.
the thing is 1 times 1 will always result in 1,no matter how many times you repeat the operation
2:55 an interesting reference here could be mentioning that X^Y is set of all functions from Y to X, then say about Von Neumann's definition of natural numbers.
Empty set is 0, while 1 = {0}, thus if we think about â as the set Y with infinitely many elements, then 1^â can be interpret as {0}^Y, i.e. set of all functions from Y into {0}. However, it contains only one constant function that is always equal to 0 đ
The result is still not a number, but the same could be said about "1^â", so it's an interesting interpretation of it đ
This interpretation is not correct, because â is not the cardinality class of an infinite. The cardinality classes of the infinite sets are given by the Aleph numbers. Aleph(0) is the cardinality class of the set of natural numbers. 2^Aleph(0) is the cardinality class of the set of real numbers. Aleph(1) is the cardinality class of the set of all countable ordinal sets. Etc. In terms of cardinal arithmetic, you are correct that the cardinality class of the set Y^X is equal to |Y|^|X|, where here, we are dealing with exponentiation of cardinal numbers. In this context, it is indeed true that 1^Aleph(0) = 1, 1^Aleph(1) = 1, etc. 1^Îș = 1, for all cardinality classes Îș. However, as I said, â is not a cardinal number. This symbol does not denote "infinity" in the sense of "infinitely many elements." What it does denote is some object which, when ordered with the real numbers, is greater than/comes after all real numbers. Historically, such quantities have been called infinite, but strictly speaking, this is not the correct way to think about these quantities, and the name "unbounded" or "unreachable" may be more accurate, though still not a perfect name. â is defined by the property that for all real numbers x, x < â, and there does not exist any other objects y such that â < y. In other words, â is the supremum of the real numbers.
@@angelmendez-rivera351 Most of modern mathematics is based on set theory, So, â must be a set, if defined. Since for any set S, 1^S = 1, 1^â = 1, as well.
Btw, the Aleph stuff works only with the continuums hypotheses.
19th century points of view of analysis are obsolete.
@@geraldeichstaedt Nothing I have said requires the continuum hypothesis. Also, your statement is incorrect. It is not true that 1^S = 1 for all sets S. Set exponentiation is defined as follows: the set Y^X is defined as the set of all functions with domain X and codomain Y.
You are indeed correct that â is a set, but this clarifies nothing, since set theory is ultimately irrelevant here. â is defined in terms of order theory, as the top object of the Dedekind-Macneille of the ordered field of real numbers when taken as an order lattice. Lattices are algebraic structures built from sets, but the construction is by no means trivial. Also, 1^â being undefined has nothing to do with real analysis.
Thanks!
i will go straight to using this new definition without any context or explanation in my real analysis exam
I think this is more a question of mathematical language than mathematical reasoning. If 1^x were a standalone function like sin x or ln x, setting 1^â to 1 would be fine. However, we're dealing with a two-input operator x^y here. That means you can approach the (projective) point x=1, y=â from infinitely many directions. Tracing the value of x^y on those approaches 1 or 0 or â, or even any value between those (needs curved approach): let a_n, b_n > 0 infinite sequences with a_nâ 1, lim a_n = 1, lim b_n = b. Set x=a_n, y=(ln b_n)/(ln a_n) then x^y approaches b as n approaches infinity, while x approaches 1. If bâ 1, |y| approaches infinity. The sign of y can be controlled by putting a_n on the same side of 1 as b is.
2:32 - that is really interesting how we have gone from something obvious to something impossible
I remember hearing this once and it will always be stuck in my head: Infinity is not a number, itâs a concept
Even so, i feel like the answer would still be one. 1 times itself will *always* be one, so even if you repeat the multiplication an infinite amount of times, it should still be one.
That's a cop out because infinity is strongly related to numbers.
@@pooperdooper3576 The answer is one, but for set theoretical reasons. The limit is not a proof, since the ^operator isn't continuous.
@@pooperdooper3576 You keep on multiplying 1 infinitely, that is, you never stop. The operation never completes. You never get an answer. So it's indeterminate. If you stop at a number really really large, it will be one, but that number will then not be infinity. Infinity isn't a number in the first place. Does this help? :)
The limit of q ^ n when n goes to infinity depends on the value of q
q = -1 undefined since it's an oscillating function like sin and cos
-1 < q < 1 it's 0
q > 1 it's infinity
q = 1 it's actually equal to 1 but only because it's the variable n which is part of Z
Assuming the value is exactly 1 and not (1+) like in that example, it's undefined if the variable is a real number and 1 if the variable is from Z
I've seen so many videos about all these things lately, and what amazes me the most is that everybody in their channel talks like if they were the first ones ever trying to find an answer. All these things must be already defined by mathematicians long ago or they wouldn't be able to solve many problems.
What I learned is that my infinite bank account with $1 on it is never gonna increase.
Finance still hurts these 2-3 years
Is your interest rate 0%?
X^infinite= 1 if X is absolute 1.But if X is trending to 1 then,X^infinite is not equal to 1.
Should be, but it seems mathematics somehow say otherwise because "that's not how infinity is defined", even this would be the common sense answer.
@@melonenlord2723 There is no such a thing as "common sense" in mathematics. Mathematicians decided to throw it away, because it is unreliable: because it has been demonstrated that common sense is wrong, more often than not. For the example, people used common sense to believe the Earth is flat, and the Sun orbits around the Earth. Later, we learned that this is false: the Earth is spheroidal, and the Earth orbits the Sun. People used common sense to believe bloodletting would cure diseases. Later, modern medicine taught us that bloodletting does not work, which is why you have never had to do bloodletting to cure a disease. People used common sense to believe that lightning was caused by celestial beings unleashing their anger at humans. Later, we learned this is false: lightning is the result of the presence of an electromagnetic potential between the clouds and the ground. People used common sense to believe that humans flying would never be possible. Later, planes were invented, and now humans can fly quite easily. In fact, we have flown higher than any bird ever has, and have traveled to the Moon.
So, as you can see, "common sense" is very unreliable when it comes to establishing facts about the real world, and when it comes to establishing facts about abstract concepts that no one understood for millennia. Sure, common sense is perfectly fine when it comes to throwing out the trash, when cleaning the bathroom, and when making hamburgers on a barbecue, but when it comes to learning facts about the universe, you need to leave it to the experts, and their rigorous, non-common-sense based methodology. This methodology is the only reason why the Internet exists, and thus, the only reason you can go around on your electronic device spouting the nonsense that you do on these comments sections. By the way: you are welcome.
1^â = 1. There is exactly one map from any set to the set 1 = {0}. If â is defined in the modern context of set theory, it requires to be a set. Btw., it doesn't matter in which way â is defined, i.e., whether +â = -â or not. See 1-point compacitfication vs. 2-point-compactification of R.
A video from Bri a day keeps ignorance away
This is one of the best math videos I have seen in a while! Very concise and informative. You have earned yourself a subscriber :)
Nice video, as always.
I have a problem with the sentence "You take the base and multiply it by itself exponent times" (1:35).
There are clearly 3 multiplications in the decomposition of 2 to the 4th.
If you multiply the base by itself ONCE you get the base squared.
I think a better sentence would be "You take 1 and multiply it by the base exponent times".
This would also account for why any base to the 0th power is 1.
Yes, I always say this, but no one listens.
And it also explains why negative power gives you 1/x^n, since you divide 1 by x, n times.
Yeah, since 1 is the empty product or product of no factors. This even provides a rationale for treating 0^0 as 1, which is used in a lot of contexts.
I waited sooooo much for this video! Thanks a lot!
Hope you enjoyed it!
1 (as the numerical constant 1) elevated to the whatever is 1, unless that "1" isn't really 1 aka some function approaching 1
Yeah that should stop all the arguments now
clearly the answer is -1/12
wtf
â@@thedream6203 -1/12 is for some reson sum of all natural numbers
Wrong video bro
I like viewing things through complex/harmonic analysis, exponents are scaling/rotations, and exponents of 1 are just the null rotation around the origin. An infinite sequence of null rotations is just Ξ + 0 + 0 + 0 ... = Ξ
Well, strictly speaking, multiplication by â. What you described is an example of a series, which is not the same thing.
Great video and super interesting. Quick note 1^â is an indeterminate form meaning that we need to determine it.
I have a problem with this. I understand it is standard in calculus classrooms to use the language of indeterminate forms to teach students, but this is simply bad practice in education. In mathematics, there truly is no such a thing as an indeterminate form. A string of symbols, called an expression, is either well-defined, or not well-defined. If it is well-defined, then this means that the symbols are an unambiguous name for an object, or a class of objects, which satisfy some property, stated in the formal language of mathematics. To understand if 1^â is well-defined or not, you need to understand what the individual symbols mean, and the syntax for how they work together. Calling a string of symbols "indeterminate," as such, is nonsensical.
Besides, the language of indeterminate foms only causes confusion for students, and it reinforces basic misunderstandings of how the limit operator on functions works. Primarily, it teaches students that the correct way to compute limits is to "plug in into the expression," and if they "get one of these 'indeterminate forms,'" then this just means they have to compute the limit some other way. This is incorrect. This is not how limits are computed, and arithmetic operations are completely unrelated to the limit operator.
@@angelmendez-rivera351 Maybe Indeterminate is just an English translation that doesn't capture the true meaning of this form. Sometimes the wrong name stays forever. For example, Columbus thought that he was going to India, and when he reached America, he called the people Indian, and names didn't change ever since. So the meaning of Indeterminate doesn't mean what it represents in math. If we call it ambiguous forms like other countries, people will know there is a process to disambiguate them. So like Indians that have a different meanings, indeterminate means ambiguous in math.
If the limit definition goes against basic math( which it does in several cases), that shows that it is incomplete and needs to be polished.
@@RSLT Here is the issue: provided the condition that lim f(x) (x -> p) = 1, lim g(x) (x -> p) = â, there are multple expressions we can discuss. We can discuss 1^â, lim b^a (a -> â, b -> 1), lim 1^a (a -> â), lim b^â (b -> 1), lim f(x)^g(x) (x -> p), lim 1^g(x) (x -> p), and lim f(x)^â (x -> p). There are 7 distinct expressions I listed. Is it logical, is it rational, justifiable, to denote _all seven_ expressions, with a single symbol? No! Of course not. These expressions all represent a distinct mathematical object, and if those objects actually exist, they may be potentially different, after all. So, no, it does not make sense to insist that the symbols 1^â, lim b^a (a -> â, b -> 1), and lim f(x)^g(x) (x -> p), for example, should all be treated as the same symbol, since they are all different objects.
What does the symbol 1^â mean? It means that you take the exponentiation function ^, and you evaluate it at the point (1, â). So, what happens when do that? Well, the answer is that you _cannot_ do that. You probably knew that already, but here is the question you probably do not know the answer to: *why* is it that you cannot do that? You may be inclined to answer the question by saying "well, if you look at these different limits..." No no no no. We talked about this. 1^â is not a limit expression. It is different from lim b^a (a -> â, b -> 1). These are notated with different symbols because they are different expressions. I am not asking you what the limit of ^ is as (x, y) -> (1, â). I am asking you what ^ is evaluated at (1, â). Do you see the problem? This is why I consider that everyone answering the question "why is 1^â undefined?" is answering incorrectly. So, again, why can you not evaluate ^ at (1, â)? Well, you may respond by saying "â is not a real number." Okay, but who cares? We do arithmetic with objects that are not real numbers all the time. We do arithmetic with functions, with sets of objects, with vectors, spinors, games, etc. So, again, why can you not define 1^â? You may reply with "well, if you do arithmetic with â, then you have to accept some very unintuitive results, such as â + 1 = â, and even â^2 = â." Okay, but again, who cares? Counterintuitive results occur in mathematical theorems all the time. Just look at every theorem in set theory. Do you see? All of the traditional answers you have heard do not actually provide any mathematically correct and satisfactory explanation for why you cannot do arithmetic with â.
Now, do not get it twisted. A mathematically correct answer for why you cannot do arithmetic with â *does exist.* However, I have never seen anyone on CZcams, or anyone on Internet forums, being able to provide the answer, and in fact, the answer lies with a mathematical topic called lattice theory, which I have never seen anyone on CZcams cover. Here is the gist of it: a lattice is a mathematical structure with a concept of ordering, an order set, but this ordered set satisfies a special property: every pair of objects has an infimum (greatest lower bound), and a supremum (smallest upper bound), and these are unique, necessarily. For example, the rational numbers form a lattice, as do the natural numbers: for any two objects in these sets, I can tell you which object is bigger, so that object is the supremum, and analogously with the infimum. However, some lattices are complete lattices. A complete lattice not only has every pair of objects have a supremum and an infimum, but _all_ subsets of the lattice have a supremum and an infimum. Here is the important part: all lattices can be uniquely extended to a complete lattice. By the way, this is conceptually analogous to how you can uniquely extend the natural numbers to the whole numbers (integers), the whole numbers to the rational numbers, and the real numbers to the complex numbers. When you extend the integers to a complete lattice, you introduce two new symbols to do so, -â and â. This results in the affinely extended integers. If you do this with the rational numbers instead, you obtain the affinely extended real line. In these complete lattices, â is simply defined as the supremum of the entire lattice, and -â is the supremum of the empty set, also equal to infimum of the lattice. The problem is, though, that you can use proof by contradiction to demonstrate that it is not consistent to equip this structures with addition and multiplication operations that preserve the ordering. *This* is the reason why you cannot do arithmetic with â. Ultimately, it is not an issue of "mathematicians would rather not define it." Rather, there are theorems that tell us that we cannot do it, no matter how much we want to. As such, there is no mathematical object that the symbol 1^â can correspond to, so it must be undefined.
By the way, this tells us _absolutely nothing_ about all the other 6 expressions I listed earlier in my comment. 1^â is undefined, but lim 1^a (a -> â) = 1 is still a fact. You can prove this with basic calculus. Do you understand what I was getting at? We cannot treat these symbols as all having the same meaning, because they do not have the same meaning, and yet, calculus professors do that all the time. This is why students are always so confused. lim b^a (a -> â, b -> 1) does not exist, but proving this would require having some knowledge of vector calculus. What about lim f(x)^g(x) (x -> p)? This limit expression is a function of (f, g), in the sense that for different pairs of functions (f, g), you obtain a different object L as the result of the limit (when it exists). The point is, though, that this function is not a constant. Its output varies, depending on the input. The label "indeterminate form" is supposed to capture this idea, but it fails, because the label gets applied to the expression 1^â, which has nothing to do with limits! Honestly, the correct way of teaching this in a calculus classroom is to avoid using the label, and instead, simply explaining that knowing lim f(x) (x -> p) = 1, lim g(x) (x -> p) = â is not sufficient information to compute lim f(x)^g(x) (x -> p). You need additional information, such as lim f'(x) (x -> p), lim g'(x) (x -> p), lim f''(x) (x -> p), lim g''(x) (x -> p), etc. You can do this without ever mentioning the expression 1^â, and without ever using the label "indeterminate form." In fact, you do better than this: all of calculus can be taught without ever making any reference to the symbol â at all!
By the way, everything I have said here is not unique to 1^â. It applies equally to all the so-called 'indeterminate forms,' and it even applies to other arithmetic expressions, such as 1/0 and 0^â.
@@angelmendez-rivera351 I think the name is the source of confusion. It's not indeterminate in itself; _you_ haven't determined it.
@@pyrodynamic4144 It partially is a problem of naming, but also of notation.
Great idea for a math discussion!
In case anyone is wondering what does happen when you put one raised to infinity into a calculator, it goes like this:
Casio fx-CG500: 1
TI-89 Titanium: 1
HP Prime: undef
TI-nspire CX II: 1 with the message âWarning: 1^oo replaced by 1â
It's interesting that my calculator calculates 1^âŸïž as 1, but puts a note that "1^âŸïž is replaced by 1"
what kind of calculator do you use
@@dvoid4968 TI-nspire CX CAS calculator. I only use the version on a computer. It's quite good and for me a requirement for school as it's used in exams.
It sounds like instead of doing any calculations, the calculator is programmed at that point to just assign a value of "1" to the variable, and show that message.
Again, much like a piece of wood or wax being shaped along an assembly line just falling through a trap door if it gets molded into a specific shape, just assigning that value when asked to perform that specific calculation, it instead doesn't do any calculations.
Mine says indeterminate form
Sorry, but this seems about as useful as medieval bishops arguing over how many angels would fit on the head of a pin.
Well, one way to approach this would be to take a number system where you do have infinite numbers and power functions, take some reasonable choice on what infinite number the infinity symbol should stand for (or alternatively, find out that for the number system of choice it doesn't matter), and enjoy the well defined result here.
Two obvious choices here would be ordinal numbers and cardinal numbers. In ordinal numbers, one might choose the smallest infinite number, omega. Now by definition (well, I think there are several equivalent definitions, but this one is particularly convenient) a^omega = sup(a^n), n
I once read a sybolic logic book that took 137 pages to come to : "therefore we can assume that 1+1=2"
So anything is possible if you put enough mental gymnastics into it.
can you do a video about multivariable calculus and it's applications and cool properties?
idea: what about the 0 root of a number? I saw that thi is a number elevated at the 1/0 power, which tends to +/- infinity, but the calculator says 0. Could you explain it?
Zeroth root meaning power 1/0 since anything by zero is undefined we say 0th root of n ( any real number) is undefined
@@srividhyamoorthy761 oh ok
Why is the example at 1:05 used as what the written form for the limit would be, instead of the one at the end?
Also, why are you calling into question if the audience is sure of the definition at 3:13? That is what I imagine all exponents are, in reality. Am I misunderstanding something? Even in that example, it seems to me like it would just lead back to always being 1, instead of being undefined.
Thank you for the video.
When dealing with infinity, which is not a number, the result you get depends on specifically how you pose the question. Thus, it is indeterminant.
Your math example compelled me to think about the following problem.
Consider. y= (1+ (1/x^z))^x
As you show, when z = 1, the value of y when x approaches infinity is e. And when z=2, the value of y when x approaches infinity is 1. So, what is the limit for y for values of z between 1 and 2?
The interesting answer is that y=e only at x=1. For x1, y=1. If you graph this, you get a right angle, with the vertix at (1,1).
There is no exponent Infinity since the exponent has to be a number.
However Infinity is not a number.
Yes and no. Thatâs what variables are for. An infinite exponent can definitely exist. We just call it by a different name.
@@r4nd0mguy99 :
In mathematics there is:
no yes and no
tertium non datur
@@r4nd0mguy99 No. Variables are not the same as â. 1^â is undefined. lim 1^n (n -> â) = 1. They do not mean the same thing.
A^B is the set of all maps from B to A. For any set S, there is exactly one map to the set 1 of one element. Hence 1^S = 1 for any set S, including S = â. If you don't define â, then don't use it. Otherwise, it's a set.
@@geraldeichstaedt
Hence? No, I cannot follow.
Where is the implication?
Infinity is the most interesting number even though it's not a number
Indeterminate forms in mathematics are a treat to watch damn
Love your videos
My math training is limited (and decades-old), but a number to the power of infinity being undefined strikes me as a matter of convention, not fundamental mathematical "truth" (assuming the latter concept actualy makes sense...)
I can't know for sure, but would a conventional definition that defined it to be equal to the limit of the number to the power of x as x grows without bound ACTUALLY result in problematic mathematical outcomes?
Personally, I don't see how " X^ âŸïž" makes it clear that it represents the limit of what X could be.
Maybe we could have a new symbol for that, such as ^--> |
Literally meaning "up to the point where it cannot continue".
It's just because 1^inf doesn't give you context over what it means, that's why it's problematic.
The same way you can put the infinity in context by saying it's the lim 1^x as x tends to inf, you can use calculus in defining what 1 is, should you wish to. This is valid because the second you touch calculus you're talking about limits, and that could apply to both the infinity and the one
Usually that doesn't result in different answers, but it does here based on the nature of the 1
Setting 1^inf to 1 would actually be the convention rather than true math
Calculator: *Domain Error*
The answer to the problem I put in:
Yes in calculus we deal with such problems where f(x) and g(x) are two functions such that f(x) ^ g(x) such that f(x) tends to 1 and g(x) approaches to infinity when 'x' tends to infinity........ so here we can use the limit concepts such that limit 'x' approaches to infinity where f(x) tends to 1 and g(x) tends to infinity the we take " e^ lim x approaches to infinity ln (g(x). (f(x)-1) ) " .....and by applying limit we could find answer
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It also depends what u mean by the symbol infinity. Is it some aleph cardinality or for example compactyfying point in topology. In cardinal arithmetic in set theory u can well define exponents with cardinals.
Never heard of this but sounds super interesting. Do you have a nice reference for the definition of such exponents?
@@bio241 Elements of Set Theory by Enderton is standard for ZFC set theory.
@@sieni221 thanks alot! Never dug deep into set theory, only learned the 'basics' you get during the standard calculus classes (and some more during category theory).
@@bio241 I havent gone much outside of that book other than some extensions of ZFC.
Fun fact: if you put x^infinity+y^infinity=1 in desmos, it draws nothing, but if we put x^infinity+y^infinity
Nice! Thanks for sharing friend!
It is somewhat defined to be 1 by rule that every value can be multiplied by one and not change, in the e example we have limit that aproaches 1 but it is not equal 1.
I mean in some region of math infinity just should mean that something never ends for example anything divided by 0 excluding itself should be infinity becouse we can always substract 0 forever and not end up with 0.
I know It would mess up with math in general but it should have its own category
Look up projective geometry where they define 1/0=infinity. You can do nice things with them but it is not normal mathematics.
Yeah my teacher showed this to us in school. Cool
It will be either one or undefined I guess
The best way i can describe it is 1^inf is always 1 if the base is 1 however in the case like the limit definition of e which is limit as x approaches inf of (1+1/x)^x the base 1+1/x doesnt equal 1 it approaches 1+ its approaching 1 but while it approaches 1 its always a number larger than 1 thats why its indeterminate you are taking a function that approaches 1 and raising it to a function that approaches infinity
1^â = 1, 0^0 = 1, but the ^ operator isn't continous. That's all.
Very intetesting you said you'd done 0^* Have you done one on infinity^0 or is it an easy one?
(Note: inf means infinity)
There is a person saying you all are wrong but, to be honest, the core of the problem just lies in the different ways to interpret the expression '1^inf'
(1) Most of us regard it as lim n->inf (1^n), which should be 1. (From the CZcamsr at 3:50)
I can (TRY TO) prove it via the rigorous definition of limit. Let c > 0 be arbitrary. For every c, there exists a natural number N (Trivially choose N = 1 for any c) such that n >= N implies |1^n - 1| = |1 - 1| (n is a natural number) = 0 < c
Equivalently, lim n->inf (1^n) = 1.
(2) However, as the CZcamsr (at 0:55) unveils in his video, '1^inf' means lim f->1,g->inf (f^g), which indeed has different values depending what f and g are, so it is indeterminate.
(3) Lastly, as a commenter insists (and somewhat mentioned by the CZcamsr at 1:57), the '1^inf' literally means '1^inf'. As I spent like 1 and a half hour reading the comments, that 'inf' is probably the 'inf' in the 'extended real number set'. 'inf' is the supremum of the 'real number set' R. It is an object but doesn't have nice arithmetic properties so that we can 'find' a value for '1^(inf)', i.e. it is undefined.
Well. I probably get it since we should have 'real number set' R := (-inf,inf) and the 'n -> inf' in (1) probably lies in R and we can just preceive n as some real number and trivially say '1^(inf) = 1'
Yet, with 'inf' actually being 'inf' as mentioned in (3), the arithmetics are totally different since we are not talking about R =: (-inf,inf) anymore, thus '1^(inf)' is not defined. (Not in accordance with the rules in R especially)
So, I, as a high schooler, reckon the problem lies in how we interpret '1' (causes confusion between (1) and (2)) and 'inf' (causes confusion between (1) and (3)) in '1^inf'.
We can absolutely put the constraints you mention of the expression "1^inf" if we want to. But those constraints must be specified when the expression is brought up. If they are not, we must use the broadest possible meaning of the expression. That being your definition (2).
If we just have the expression "1^inf" with no other information, we must assume it to mean this. It can only mean (1) or (3) if someone explicitly states that that is what they are using it to mean (though (3) still doesn't make much sense. Infinity cannot be used as a numeric object).
are you real?
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Seems like an example of overthinking this, to infinity.
You can also look at the limit as x approaches 1 of x^infinity.
For any number greater than 1, the result is infinity.
For any number less than 1, the result is 1/infinity or 0.
So 1^infinity is a discontinuity in that plot.
1^infinity is indeterminate.
But there are contexts where one type of evaluation is appropriate, and in those situations, you can use a fixed value.
Ï is the variable that represents ordinal infinity. Ordinal just means there's a number before, Ï-1, and a number after, Ï+1. Ï and others like it belong to a class of numbers called transfinite. You can do math to Ï; Ï+2 does not equal Ï, instead it is two more than Ï. There are countably infinite numbers between 0 and Ï. In fact, there are precisely Ï-1 whole numbers between 0 and Ï. While â-â = NaN, Ï-Ï = 0.
I'd imagine that 1^Ï would indeed be 1, instead of NaN.
What is it with 0, â, and their copious undef and NaN results? Is math broken? Was something big overlooked? Is there another imaginary plane of numbers where 1/0 is another i?
Yes math is broken and there is no way to fix it. Any mathematical system complex enough to do addition and multiplication is either incomplete or inconsistent or both. Infinity is the boundary where this incompleteness usually shows up. And yes zero also has some incompleteness.
@@kazedcat Are you sure it could not be fixed?
What if we threw the idea of "infinity" altogether?
It has no basis in reality, anyway.
There is not an infinite amount of _anything,_ not even points in space in all of the Universe.
Would our current understanding of mathematics and how to use it be permanently damaged if we just cut it away?
@Noobly.
I am not too familiar with the use of "w" (and I don't have a button for it on my smartphone). What about just writing that limit as;
w --> |
Or, | --> w
@@TheNoiseySpectator Infinity is not the problem it is just a symptom. the mathematics itself is the problem. Any system complex enough to do addition and multiplication is incomplete and or inconsistent. You can see it immediately with zero. Zero has no multiplicative inverse so it is not close under division. If you want to fix mathematics you need to throw out the concept of multiplication. A mathematics without multiplication will be very useless.
There's actually no such concept as Ï-1 or Ï-Ï, because subtraction is not defined in terms of ordinals. To see why, we need to remember that the next ordinal number after ordinal number a is aâȘ{a} and that Ï actually is the set of natural numbers. So if we try to define Ï-1, we see that Ï-1 actually is set of natural numbers without the "last natural number", which doesn't exist. Furthermore, all infinite cardinal numbers don't have previous ordinals.
it feels to me kinda similar to 0^0. It is an indeterminate form, but we "define" it to be the limit as x goes to 0 (from above) of x^x, which is 1
I donât think any similarity youâre seeing is right.
An indeterminate form is a concept in calculus. Meaning if you have *functions* of said form, their *limit* fails to be determined (without further information). It is unrelated to what any actual numerical values of any actual numbers are.
1^x = 1 for every single number x is true and obvious, but â is not a number, instead it is a shorthand for the limit of a function. Meaning 1^â is both an indeterminate form and not anything else.
x^0 = 1 for every single number x is true and obvious, and 0 is a number. Meaning 0^0 is both an indeterminate form and the number 1.
@@Lucaazade well, the similarity is that they're both an indeterminate form with an "obvious" answer of 1.
They are, of course, completely different objects, but they have that in common
One of the fun things that you can do in math is just axiomatically state x^inf=1 as a rule.
In basically every interpretation of powers, 1^inf equals one. Including the most important section: area of an inf-dimensional cube
This guy is AI generated
Try to put an infinity symbol in any purely arithmetic expression and you're bound to mess up. As said in the video, infinity is not a number. So asking such questions, at least to me, feels weird anyways. I don't expect any arithmetic to work if I substitute a number with something "not-numberly". Like, what's one raised to the apple?
Tho, I'm surprised not to see an argument using the power series in this video. That would get out of hand decently quickly having to deal with â^(âĂ0). (Which again sounds to me like "Apple raised to the power of apple but for the case that you never had apples anyways")
Thanks for the video, keep up the good work ^~^
Yes, this.
"â" is a symbol without a well-defined meaning in this context. It shows up in calculus, but for example lim [n -> â] is defined separately from lim [x -> c] (for c \in \R). I'd say that given that we do use "â" as a symbol in some places, it's a little more like saying "What's 1 to the â?" Those are both symbols which I recognise from mathematics, but they don't fit together like that. I suppose the answer to "What's 1 to the â?" is "A syntactical error." đ
@@johnaldis9832 1^S = 1 for any set S, since there is always exactly one map from a set S to the set 1 = {0}. On a set theoretical basis of mathematics, â is a set, no matter which specific compactification of R is chosen.
It doesnt matter how many times you multiply 1 by itself, it will always be 1 even if you do it forever, and thats really all the expression means, its not raising something to no apple.
One raised to the apple is one, since there is exactly one map that assigns each element of the apple to the set with one element.
And, btw, the apple raised to the zero is also one since there is only one map that assigns the zero elements of the empty set to the elements of the apple.
Zero is also called the initial object, and one the terminal object due to just this property within the category of sets and maps.
Since we do most of mathematics on the basis of set theory, this question should be sufficiently answered in the most general way.
@@geraldeichstaedt Youâre assuming that âappleâ is the cardinality of a set here. 1^$ only makes sense if $ has a number-like meaning. Of course, you can define notation to mean whatever you like in a specific context (I defined a relation denoted ~^\infty_N in my PhD thesis, for reasons which made sense at the time) but if youâre asking for the âstandardâ definition then for example â1^=â is a string of symbols with no defined meaning, and so is â1^\inftyâ.
I was thinking that for sooo many days thx for vedeo
Interesting approach to define 1â as indefinite is looking at the formula aâ.
For all aâ(a^x)=0, for all a>1, lim x>â(a^x)=â, so there's a discontinuation at a=1 similar to 1/x for x=0.
I am not a mathematician but i cant believe that if we do 1*1 even until the end of time, or what ever infinity means, will somehow the result be changed. If you can tell me a reason, i am all ears..
You can also approach it with the limit of 1^x, with x-->inf.
Oh. You just did the straightforward thing after taking the totally BS path.
Well done.
Except 1^oo could be interpreted as e.g. lim x_>oo (1 + a/x)^x= e^a and that is only 1 if a = 0.
I think the limit as x->â (1^x)=1
But if you have anything in the function that approaches 0 you need to remember itâs 0^+ or 0^-. Meaning an arbitrarily small value, arbitrarily close to 0, making the limit indeterminate.
0^0=1, since 0^0 is the set of all maps of the empty set to the empty set, where there is exactly one. The assumption that the ^ function has to be continuous is the flaw.
@@geraldeichstaedt is it assumed to be continuous or is it defined to be continuous?
@@goopyzoomer5798 It is wrongly assumed to be continuous. It's actually discontinous at (0;0). Therefore, the lim argument doesn't work. It's like with the sgn function. the lim of sgn to -0 is -1; the lim of sgn to +0 is +1. But sgn(0) = 0. So, the function is defined at 0 but not the lim.
How my math teacher described how e was defined: "If you mutiply enough 1s together, eventually you get 3"
Indeterminate
quick info: this guy really hate undefined things
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1 to the power of infinity is only indeterminate if 1 is the approach, if you only apply the variable in the exponent it is 1 and even the log of it is only indeterminate when you use limits on the obviously static parts rather than infinity which is the only approaching number
Yes, many of the indeterminate forms are only "indeterminate" because we're using limits on both parts. If you take the *actual number* 1 and raise it to an infinite power, you will always get 1. If you take the *actual number* 1 and take powers tending to infinity, then you will always get 1. It's only when the base is *tending toward* 1 that other things can happen.
So what if I do something like this. I choose an abitrarily large number x. I define y as floor(x). z as x - floor(x). Then I define f(x)=1^x. If x is positive, then f(x)=1*f(x-1). I can do something similar for negative values. Then by induction, f(x) is 1, right? 1^(x+y)=1^x*1^y I think. And for all y > -1, and < 1, 1^y is also 1. Weak azz proof I know, but where am wrong on this?
What is the zeroth root of 1 please help
there is no zeroth root of a number
1^(infinity)=1^(1/0)= 1/(1^0)=1
The zeroth root of 1 would be a number x such that xâ°=1. Since this is true for all real x, we don't get a well defined answer, so the zeroth root of 1 is undefined. It is exactly the same problem as the one we get when we try to define 0/0, where we are looking for x satisfying xĂ0=0, again, satisfied by all real x, so undefined.
There is no such a thing as "zeroth root."
@@agiri891 0th root is the number raised to infinity. Try inputting y=sqrt0(x) in Desmos and you will get a defined function.
1 elevated to inf is 1, the problem was not saying something close to 1, but exactly 1, those other limits are a whole different problem which is the case when something approaches to 1
2:34 this is illegal here had me laughing lmao
A more interesting definition from set-theory is that a^b := |b -> a| where you let the numbers a and b be sets of a and b elements respectively in the second half.
That is, the power-function is defined by the total number of functions between sets of a given size.
Or reversed, the number of functions between two sets A and B, |A -> B| = |B|^|A|.
It gives you the normal values like:
x^0 = 1,
0^y = 0, if y > 0
and everything else when base and exponent are natural numbers.
But for the case of 1^infinity, you can use that (N -> {1}) only contains the constant function f(n) := 1 to say that 1^infinity = 1.
I believe using infinity as an exponent is like saying one to the power of an apple, a random object from real life used in mathematical calculation as not a number
We know its 1 White out any calculation your title of video is 100% wrong bc i expect it would be 1 and in end of video you show is 1
In the field of set-theory alpha^beta is defined for all ordinals (more or less an extension of the natural numbers) alpha and beta. From the definition of ordinal arithmetic it follows that 1^alpha = 1 for all ordinals alpha, even infinite ones. This approach is obviously very different from the analytical view presented in the video, however, in my opinion this is still an interesting approach.
(1+a)^N with a>0 goes to infinity, for -1
This video is mathematical misinformation, and I ask that you take it down or add a significant disclaimer to the description.
Around the one minute mark, you begin discussing how 1^inf can be described using limits, but don't actually look at the limit of 1^x as x --> inf. Instead, you swap in (1 + 1/x)^x and (1 + 1/x^2)^x in a bit of flimflam. The misinformation is most apparent at the 1:30 mark, in the sentence beginning, "Essentially..." The statement you make there is simply *false*; you use flawed logic and violate the definition of a limit and the rules we follow to evaluate them at infinity, and then use this to support why 1^inf is indeterminate.
Please be clear: the limit of 1^x as x --> inf is 1, period. For any sequence of x that diverges to infinity, the sequence 1^x converges to 1; in fact, every element of the sequence is 1.
It might be that 1^inf is considered indeterminate in mathematics for *other* reasons, but the interpretation of this statement as a matter of limits is clear. Please don't put up pseudo-educational videos like this.
I am tempted to agree... this video is like forgetting the value of 1, and pretending 1 could be defined as 1+1/x (when x->oo). There are obviously a lot of cases where you have SOMETHING^x, where x->oo. But we did not start with SOMETHING. We started with 1. Nothing else.
If it is true as the video says that ( 1 + 1/x^2)^x goes to 1 when x goes to oo...
...why would not ( 1 + 0 )^x go to 1 when x goes to oo?
1 raise to infinity is equal 1.this is simple math.
Hahahaha no
i feel like it makes sense because no matter how many times you multiply 1 by 1, its not going to get any bigger. the same way i know 2 multiplied by 2 infinitely will never have a prime factorial of 3 or 5.
1^â is 1 for most values of 1 used as input. For values that only approach 1 it can be any non-negative finite value.
1^â is undefined. lim 1^g(n) (n -> â) = 1. As for lim f(n)^g(n) (n -> â), it depends further on the relationship between f and g. 1^â is not synonymous with lim 1^g(n) (n -> â), nor with lim f(n)^g(n) (n -> â), which is why the latter two have their own notation.
@@angelmendez-rivera351 1^â is is undefined, so let's define it then, I choose 21. Now 1^â=21.
It was more than 40 years ago, the teacher wrote in on the board 1â one to the power of infinity. Then he asked how much is it? Some of us said one. He said that is what you think. He smiled and wiped off the board. Never mentioned again. I don't know why did he do that?
"One" was the correct answer. But I don't know whether your teacher knew it.
That's because we typically treat a number and limes being equal to that number as the same thing, while technically they aren't.
lim x->â 1/x = 0
â * 0 = â * x * 0 = â * x * lim x->â 1/x = â * lim x->â x/x = â * 1 = â, so 0 * â = â and by analogy â * 0 = anything (and by "anything" I mean literally anything, not just a real or complex number)
1 to the infinity is not indeterminate form. A number very close to one to the infinity is indeterminate. Absolutely one to the infinity is just one. 1.000000000000000000000001 to the infinity is also indeterminate but not absolute one which has infinite zeros after decimal point.
I just finished taking Calculus and I immediately recognized the definition of e at 1:41 ! I am officially flabbergasted
To accurately determine the value of 1^â, you would need more information about the specific function or expression in which it appears. It's crucial to consider the context and apply appropriate techniques, such as L'HĂŽpital's Rule, limit laws, or other methods from calculus, to evaluate the limit.
1^potential infinity is undefined (the type of infinity used in calculus), but in Cantor's paradise, where infinity is actually actual infinity, 1^â”0 = 1
what we take:1^5=1
the homework:0^0=
the test:1^n=1 find n
3:58 - "This is a defin tion" definition you mean?
If you have an expression in x, say f(x), and you plug in a value, say a, for x and get 1^infinity, I would express that expression as 1^(1/0) then take the log of this, in hopes of getting an expression that evaluates to 0/0, then apply L'Hopital's Rule.
I want to ask what's won't with taking infinity as 1/0. I think it works in every way. It we do the same for this problem, we get 1^infinity is any real no. because x^0=1.
"But viewers of this channel know that I hate to stop there"
Me: Well... Let me get my popcornsđđ»đż
What is the result in the surreals? I'm pretty sure that it would be defined?