The Hyperreal Number System (and simplifying limit calculations)

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BP Learning

BP Learning

Күн бұрын

Пікірлер: 41
@siulapwa
@siulapwa 2 жыл бұрын
Interesting stuff. Though infinite zeros then 1 is a bit hard to digest if the definition of infinite is endless
@YuanAurion
@YuanAurion 2 ай бұрын
11:30 Man, I totally would've called it 'dropping out of hyperspace'.
@DropMyBassBro
@DropMyBassBro 3 жыл бұрын
Great intuitive explanation! Easily the best video on KZbin. Is there a way to intuitively write omega like you wrote epsilon = 0.000...1 ?
@rufusneumann9703
@rufusneumann9703 3 жыл бұрын
1000...0
@好吧-h6k
@好吧-h6k 3 жыл бұрын
If you’re never gonna reach 1, infinitesimals really don’t exist or are just 0 because it takes an infinite amount of numbers to get to 1 and only used in limits
@alice_in_wonderland42
@alice_in_wonderland42 3 жыл бұрын
@@好吧-h6k no number actually exists
@TheBasikShow
@TheBasikShow 2 жыл бұрын
(Hopping on to a year old question lmao) Unfortunately, there is no such expression for ω. Not only that, the expression he gave for ε was wrong; the hyperreal number 0.000...1 is actually equal to 10^-ω (up to some multiplicative factor), and is much MUCH smaller than ε. What he doesn’t say in the video is how to actually define the hyperreals. (This is the actual reason why hyperreal calculus seems better than traditional limit-based calculus: he’s sweeping all the rigor under the rug.) In “practice” the hyperreals are defined as lists of infinitely many real numbers; for example = 3. Then ω is defined as and ε is defined as 1/ω = . Unfortunately neither of these have nice decimal representations... but we can make a hyperreal number that does! For example, 10^ω = , which could be written as 1000...0. It takes a bit more work to properly define the “dot dot dot” but this is the basic idea.
@DropMyBassBro
@DropMyBassBro 2 жыл бұрын
@@TheBasikShow that's good to know, thanks. If all of the rigor is in defining the hyperreals does that mean once that once that obstacle is overcome, problems can be made easier than using the typical epsilon-delta analysis approach?
@manicmath3557
@manicmath3557 3 жыл бұрын
You can also call omega a transfinite number. Coz infinity is pretty much a concept
@Frankie3sticks
@Frankie3sticks 5 жыл бұрын
Too cool! Thank you
@tricky778
@tricky778 2 жыл бұрын
16:50 - "the infinity in the real numbers" but the real numbers don't have an infinity in them - how can std(omega) map to infinity in the reals when the reals don't have infinity in them?
@BPLearningTV
@BPLearningTV 2 жыл бұрын
I mean that it maps to ∞, which is how most people working in the reals refer to infinity, not that ∞ is a definite object in the real numbers.
@tricky778
@tricky778 2 жыл бұрын
@@BPLearningTV I was under the impression that they say there is no mapping from omega to the reals and might say as the argument approaches omega the mapping to the reals approaches infinity
@alejrandom6592
@alejrandom6592 3 жыл бұрын
This is amazing!
@manicmath3557
@manicmath3557 3 жыл бұрын
Is epsilon rational? If I can see it as one divided by omega and a rational has to be A quotient of two integers, then if omega is an integer. Epsilon should be rational right
@NoNamer123456789
@NoNamer123456789 3 жыл бұрын
To my understanding rationals are a subset of real numbers, but hyperreal numbers contain real numbers and more. Omega is defined outside the real numbers. More specifically, it is bigger than any real number. That means that Epsilon (1/Omega) is smaller than any real number that is bigger than zero.
@manicmath3557
@manicmath3557 3 жыл бұрын
@@NoNamer123456789 hmm. But i saw somewhere If you made the hyper real number line only Have integers, omega would still be there after tge infinite integers. So does that mean one over a mega is technically a question of two integers making it Rational. Maybe a hyper real rational?
@NoNamer123456789
@NoNamer123456789 3 жыл бұрын
​@@manicmath3557 Was it Vsauce's video? I'm no expert, but that's what I think: Omega is basically just a concept not a number (it comes after all real numbers) and even if you expanded integers to "hyper" integers, I don't think it can be (dis)proven to be an "hyper" integer. I mean, maybe the whole concept of integers breaks down, simply because we're not dealing with numbers anymore.
@TheBasikShow
@TheBasikShow 2 жыл бұрын
@@manicmath3557 While I have heard of hyper-integers, which are basically “integers plus integer copies of integer functions of ω”, I’ve never heard of hyper-rational numbers. If they do exist, though, then they’re pretty strange, because any real number (including all irrational numbers) is infinitely close to infinitely many ratios of hyper-integers.
@manicmath3557
@manicmath3557 2 жыл бұрын
@@TheBasikShow ooo good point. Thank you
@giuseppepapari8870
@giuseppepapari8870 4 жыл бұрын
well, it just looks to me like computing limits by Taylor expansion, no need to increase the set of reals...
@DanielContrerasPino
@DanielContrerasPino 4 жыл бұрын
The importance of hyperreals are mostly historical. Leibniz developed his calculus using infinitesimals but those were far from rigorous, making more mathematicians (including Leibniz himself) very uneasy, even when the results looks right. Note that this was way before limits, series or even functions. In other cases, like "imaginary" (complex) numbers, sooner or later a formal construction was presented to validate the operations they did before, in a more "unrigorous" way. But infinitesimals resisted to being properly constructed. Then the idea of "limits" was presented, then formalized/proven using epsilon-delta construction purely in terms of real numbers, making infinitesimals unnecessary. Centuries after that, with a now more developed algebraic language and logic, infinitesimals were constructed, proven they are indeed consistent. Through this extension of the real field all the calculus can be formalized without the need of limits, in the now called "non-standard analysis". Does it worth? Difficult to say. It's very likely that it never goes beyond a "math curiosity" and the "hurra!" moment of proved them and know that it is possible to do it. Nonetheless, there are various pieces of calculus that are derived more "elegantly" from infinitesimals than the traditional style, including the dy/dx notation that is then a true ratio of (infinitesimal) numbers instead of "just a notation in steroids".
@ItsAllEnzynes
@ItsAllEnzynes 4 жыл бұрын
I’m sure that functionally limits can do everything the hyper reals can. It’s just more intuitive (in my opinion) to use the hyper reals as opposed to traditional limits. I’d say this is the equivalent of the Tau vs Pi or Decimal vs Duodecimal. These are unnecessary changes that may add some clarity or simplicity, but ultimately come down to preference.
@carlosmehicano8052
@carlosmehicano8052 3 жыл бұрын
its a lot simpler than that, its basically reduced to an algebra. And expanding on the reals isn't really that difficult, the tiny book 'infinitesimal calculus' by Henle develops them in a very simple manner, he also proves the transfer theorem.
@alejrandom6592
@alejrandom6592 3 жыл бұрын
No
@alejrandom6592
@alejrandom6592 3 жыл бұрын
No
@NoNamer123456789
@NoNamer123456789 3 жыл бұрын
Sorry for the stupid question, but if we have a function y = x^0 = x/x, is it possible to solve it anyhow (a different method)? The way I see it epsilon doesn't help here, because: lim x->0 (from the right side) is (x+epsilon)/(x+epsilon) and while epsilon/epsilon equals 1, we have to substitute 0 for epsilon which isn't defined.
@diribigal
@diribigal 2 жыл бұрын
What do you mean by "solve it"? In regular calculus we know the limit of x/x is 1. And with hyperreals, you might note that ε/ε equals 1 for any infinitesimal ε.
@NoNamer123456789
@NoNamer123456789 2 жыл бұрын
@@diribigal I guess I meant the limit as x approaches 0. I mean I intuitively assumed that would be 1, but after all those years I forgot it's like a rule or something.
@vuufke4327
@vuufke4327 3 жыл бұрын
can you divide epsilon by 2?
@BPLearningTV
@BPLearningTV 3 жыл бұрын
Yes. It just yields epsilon/2
@vuufke4327
@vuufke4327 3 жыл бұрын
@@BPLearningTV how does it make sense then that epsilon is the smallest number that isn't 0?
@BPLearningTV
@BPLearningTV 3 жыл бұрын
@@vuufke4327 Epsilon isn't the smallest number. It *is* smaller than all real numbers that aren't zero. But there are innumerable infinitesimals available. Epsilon is merely one of them.
@NeoiconMintNet
@NeoiconMintNet 3 жыл бұрын
@@BPLearningTV dividing means start at 0 and count how many times the denominator needs to be added to get the value of the numerator. 0/0=0; any other numerator except 0 results in infinite counts, with infinity not being a number because numbers change value when adding values, but infinity plus 1 is still infinity. limits simply means the value a function can't be equal, like not being able to walk into walls, the limit of movement or changes in value. infinitesimals are numbers like any other number representing a value. for example : dividing 1/3 gets you 0.3+0.1 with 1=10/10; 3×0.3=0.9; 0.9+0.1=10/10; the 0.1 is a decimal number and a remainder, a value, and no matter how far you expand, the remainder is a value and is always a decimal number, nothing special. so the difference between 0.3... and 1/3 is an infinitesimal value but still a decimal number like 0.3... ; in the form of 1 over 10^n. why make this so complicated?
@BPLearningTV
@BPLearningTV 3 жыл бұрын
@@NeoiconMintNet All my students think my presentation is very straightforward. I'm not sure what you are finding complicated about it. As far as your definition of infinity, you are conflating ordinal and cardinal infinities. In cardinal infinities, adding one has no effect, but, in ordinal infinities, adding one to an infinite value yields a distinct value.
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