Not quite, 0.999999... is equal to 1 in some sense, but 0.999999999... to the power of infinity is not necessarily 1. If you don't think 0.999999... is 1, please never call 0.33333... 1/3.
Yes precisely that! Notice that exactly one to the power of infinity is actually something that appears quite often in both statistics and number theory. So it's not something that should be regarded as sloppy to write at all but it should preferably be expressed as how he writes it at around 08:16 as infinity is a philosophical concept, not a number. What it means is that proper 1^k remains 1 even when k goes to infinity. So it's also about limits. If I throw a dice where all sides shows the value 6, the probability I will only throw sixes after k throws will always remain even if I do it forever. What is sloppy however is to, without having any context just ASSUME 1 is not 1, just because it usually wasn't in some calculus course. Of course 1 is 1. If it's not, specify! It's important to be comfortable to set such a value to 1 and not be afraid and think "can I really do that?". There is something wrong with how limits in calculus are being taught making many think things are indeterminate when they actually are not. This was a great video by blackpenredpen.
I think the reason indeterminate forms are so confusing is because the notation 0♾ or 1^♾ is simply completely misleading. These forms are not arithmetic operations. They are limits. The correct way it should be presented is that the limit of f(x)^g(x) as f(x) -> 1 and g(x) -> ♾ is indeterminate, as is the limit of f(x)g(x) as f(x) -> 0 and g(x) -> ♾ is indeterminate. The expressions 0♾ and 1^♾ have no meaning. Of course, in the special case that f(x) = 1, then the limit f(x)^g(x) is simply 1. However, this is not so if f(x) is increasing or decreasing monotonically. Similarly, if f(x) = 0 as a special case in f(x)g(x), then the limit is 0, but in the general case of arbitrary f, the limit is indeterminate.
Arteyyy "♾ isn't a real number" Most students (and most people in general) treat it as a number. So they will interpret it as a numerical expression, regardless of whether it is correct or not. Deal with it. "...and exponentiation isn't one of the four basic arithmetic operations..." Many, if not most mathematicians, treat it as n arithmetic operations in their research and papers. Deal with it. "A better criticism... misleading shorthands in mathematics..." Yes, my comment talked about that. In fact, I said that nigh-verbatim in my comment. Perhaps you're not sufficiently skilled at reading basic difficulty English sentences, but if that is the case, then I am telling you that is in fact in the comment. Ask anyone else reading. "...as long as we don't work with symbolic expressions, which in this case, we aren't requires to." No one actually cares aboit whether we are required to or not. What we cared about is whether we actually do it or not. And spoiler alert: we do it everytime we have the opportunity. Never met a professor who doesn't. And that's what matters. Thank you for proving my point. Deal with it. Oh, and for the record, in mathematics, technically nothing that is not a consequence of a set of axioms isn't required. So your point is moot. "Oh, and you must be on acid too." I'd rather be on acid and be correct than not be on acid and be say something awfully irrelevant like you did, and I'd also rather be on acid than being so immature and childlike as to be hopelessly petty and unable to hold a conversation where you go around insulting people because you couldn't simply handle the mere fact that I corrected you and schooled you in the previous conversations we had. Deal with it. Didn't know I'd have a little kid stalking me on UA-cam for all of my existence trying to act like a ghost.
mike burns No, it's understood by mathematicians. It's obviously NOT understood by students generally. That's why this video was made in the first place. That's precisely what the video is about in the first place. Maybe you should watch the video again.
Arteyyy Don't bother responding, because I'm not looking at this comment section again. After all, unlike you, I actually have a life and don't have time to be wasting with a childish one like you. It signifies you've already gotten lost. Farewell. So long.
I like how you call me immature and childish after you started insulting me for no reason. Or how you say you don't waste your time with me after doing just that.
This is brutally next level. I am a Chemistry grad student and I've always had a strong interest in mathematics. Can't believe I'm learning all these things from you. Thank you.
I know the notation might look a little wonky, but how about we use something like "(→1)^(→∞)" and "(→0)/(→0)" for limiting forms instead of just 1^∞? I think this might help students understand better that indeterminate limiting forms are about functions approaching values, rather than about literal computation.
@@aryanjoshi3342 The superscript +/- are used for _one-directional_ limits. However, (→1)^(→∞) and (→0)/(→0) are not one-directional indeterminate forms.
"Liquid Numbers": at the end of this video there is a link to the "BE CAREFUL" video (1 year ago) which has similar statements such as: Lim[n->inf] (1 + 1/n)^n is "not a solid e" and so... "liquid_e" - e 0 ... Also, in many videos about the use of Limits, including the "BE CAREFUL" one, there are lot of simplification of polynomial fractions based on the fact of: Lim[x->inf] x^2 > Lim[x->inf] (ax + b) or with other powers p>0: Lim[x->inf] x^p > Lim[x->inf] x^(p-1) Supposedly we all can live with that, no problem. As in "BE CAREFUL" video the fraction is "conveniently simplified" as: Lim[x->inf] -x^2 / (x^2 + ax + b) = -1 and the answer for that video becomes: -e/2 ... Now, is that a "solid -e/2"? Supposedly not... as the simplification of polynomial fraction is also a "non-solid -1" or it is more likely a "liquid -1" (or "fluidic"?). That is the concept of "non-solidity" of numbers and quantities which makes the support of Newtonian mathematical principles of Calculus since: Lim[dx->0] dy/dx while dx is "not a solid 0"; Paradoxically, the word "calculus" means "small stone", a "relatively solid" object. Personally, despite I don't have any alternative to offer, I am not a fan of such simplification of polynomial fractions. Acceptance doesn't mean "like" or "consider it brilliant". Instead of a "solid" equality to -e/2, it could be a continuous fraction as an answer? Is not unfair, in order to discriminate from "regular" Calculus, Ramanujan shall have his own (more) "fluidic" equality notation in: 1 + 2 + 3 +... {R}= -1/12 ? I leave to you all these questions. Regards.
A horizontal shaft rotates in bearings at its ends. At its midpoint is keyed a disk weighing 40 lbs, whose center of gravity is 0.1 inch from axis of rotation. If a static force of 200 lbs deflects the shaft and disk through 0.1 inch, determine the critical speed of rotation of the shaft. @Blackpenredpen
I think there is convention, that if you write a formula that represents a limit, you assume as many things as constant as possible. Of course you can read 1^inf as a double limit, but intuition would suggest to read it as lim (1^x) x to inf. Without this convention neutral operations really hurt you : 1^inf would also be 1*1^inf which can easily be used to break any meaning of the notation.
Wolf Ram Alpha I wanted to clarify that e being equal to that limit is not a coincidence. It is a definition. I'm putting emphasis on the word "definition" so that no one in the comments starts asking to "prove that it equals e": such a request is nonsensical since you cannot prove a definition.
@@misotanniold787 i don't think that this way of defining e is the best. First of all: why there even exist such function, that is the same as it's derivative. Secondly: even if exp(0)=1, why this function is unique, i mean you said exp(z)=(exp(1))^z, but you said it as a result of first definition, but not as a scond part of definition, if f(x)=C*e^x, then f'(x)=C*e^x=f(x) see? And even so, why then exp(z)=(exp(1))^z? Also with that definition, how you prove that lim(x tends to infinity) (1+1/x)^x=exp(1) I think that this definition of e as lim(1+1/x)^x as x going to infinity is more beatiful and show us a lot of its properties. When we investigate derivatives first time and doing (e^x)' with definition of derivatives, only relying on this definition we could see that (e^x)'=e^x. It really seems for me kinda illogical to define e like that if it would be harder to prove other properties of e, even if it shows other more beautiful properties.
Now, i would say the misconception arrives because the teacher often don't write the approaching arrow... coz its not exactly one but its tending towards one ... And there the mystery remains ... When it tends towards one it can go from negative side and also from positive side so the fractions changes from proper to improper fractions ... As if we suppose write limit x tends towards 1 negative and write 1^infinity , then the answer to it is zero ... But if we do the same with x tends towards positive 1 then the answer becomes infinity ... So here the contradictory factor arrives which makes it the boundary line(1) for the indeterminastic results ...
@@Revanth292000 I think that would diverge as n^0 for n->\inf equals to 1 and n^(p/(n+1)) goes to some other limit, depentant on p (except p=0). We have now two boundaries, the sin(n!) Causes it to bounce between those, ergo it diverges.
Back to basic, you may plot a curve 1^x. You'll find it a straight line and can see 1^(1) is 1, 1^100 is 1, 1^100,000 is 1, so 1^inf is also 1. There isn't any decimals behind the base 1. Your proof by lim(x->inf)(1+1/x)^x is the shift of meaning. We are talking about the integer 1 and 1^inf, instead of the rational number (1+1/inf) and (1+1/inf)^inf.
I noticed that in this limit the numerator results in the power of e number (consider n tends to infinity in limits below) so lim((1 + a/n)^n) = e^a lim((1 + 0/n)^n) = lim(1^n) = e^0 = 1
I usually keep it indeterminate because the infinite root of x is the same as x^1/infinity, this is one of 2 things. Either it approaches 1, or it is 1. Both would suggest the infinite root of x is 1. But I never defined x meaning this is true for any number and therefore 1^infinity is any number. Therefore if we can’t determine it it’s indeterminate.
Could you get any number you want for the limit, just putting ln(a) over the x? I mean e^n=lim(x-inf)(1+n/x)^x... If n is ln(a) you get a for the limit
Thank you! According to my university (The university of buenos aires) I should magically understand all of this without explanation. You saved me, and the explanation was very good, now everything makes sense :)
It'd just be 1/(ln(5) right? Any constant times infinite is still infinity, so as long as you get lim x -> infinity of ln(5) gives you e^ln(5) = 5 So to do that you'd need ln(x) / (1/(ln(5)) ln(x) Something like that Hopefully i wrote it right im on mobile So like ln(x) / 7ln(x) gives you 1/7, so ln(x) / (1/7)ln(x) gets you 7 So you just sub out the 7 in (1/7) for ln(5)
Greeaat matter to think about the matter of Maths, Exactitude and Approaching (and why not approach the higher, then approach the lower side, then compare, in French "encadrement").
I don't see why it would be a difficult concept. In 12th grade, my calculus teacher told me that infinity/infinity, 0/0, infinity-infinity, 0^0 and 1^infinity were the general indeterminate forms, and they all seemed obvious to me. 1^infinity, REALLY means (1+delta)^f(delta) where lim as x->0 of f(x) = infinity, and since you don't know HOW quickly f(x) goes to infinity as x goes to 0, it could be anything, it could be a power that is insufficient to get it far away from 1, it could blow up MUCH faster than the 1+delta approaches 1 and thus make it REALLY big, and it could have complex values to make the result negative too.
The problem is that if log_a(1^inf) = inf*log_a(1), where a could be any number as you presumed, then the following is true: a^log_a(1^inf) = a^inf*log_a(1) 1^inf=a^(inf*0) Now, here you already see that a^(inf*0) is an indeterminate form Because depending on how you arrange inf and 0 you get two different things, contradicting the commutative property of multiplication: (a^inf)^0 = inf^0 = ??? (a^0)^inf = 1^inf = 1
0:18 we’ll obv inf/inf is the mom because it’s top heavy and bottom heavy (you can guess what that means lol) that’s how my cal teacher taught us limits #/inf is bottom heavy so big O Inf/# is top heavy so big ∞
Could you do this in general by breaking aparant the 1 and ∞ using e^ln(x), then you get e^∞0, invert either the ∞ or 0 to get a 0/0 or ∞/∞ form, then use L'H to solve?
Леша Кузнецзов Use integration by parts, integrating xsqrt(1 + x^2) and differentiating x. To integrate xsqrt(1 + x^2), let y = 1 + x^2, dy/2 = xdx, so xsqrt(1 + x^2)dx = sqrt(y)dy/2. The integral of this is 1/(4sqrt(y)), and converting back gives 1/(4sqrt(1 + x^2)). Therefore, the integral of (x^2)sqrt(1 + x^2)dx is (x/4)/sqrt(1 + x^2) - (1/4)integral of dx/sqrt(1 + x^2). Integral of dx/sqrt(1 + x^2 can be found by letting x = sinh(u), dx = cosh(u)du, sqrt(1 + x^2) = cosh(u), so dx/sqrt(1 + x^2) = du. The integral of this is u = arsinh(x). Therefore, integral[(x^2)sqrt(1 + x^2)dx] = (1/4)[x/sqrt(1 + x^2) - arsinh(x)] + C
First, let me thank you for your videos and say that I like your approach, although I would modify it a little. My main modification would be not to include infinity in the table to avoid confusion. It is abundantly Lear that you know what you are doing, but students tend to get confused with the difference between limit and evaluate a function and include infinity in the table could feed this thought.
Infinity is the same value at either end of the number-line, connecting each end and forming a loop. It's not just in the positive direction, therefore if inf-inf = ind then inf+inf also = ind.
Infinity is not a number. It just indicates where the value tends increase or decrease without any bonds. Just because some large number minus itself is 0, infinity minus infinity is indeterminate. We can't tell what value is going towards to, so you have to do more work. Infinity plus infinity has the limit that tends to infinity.
@@justabunga1 I agree with your explanation of infinity but I'm kind of thinking about infinity - infinity as removing something infinitely large from something that is also infinitely large but has the same infinite mass, because the notation is exactly the same, so I'd just say infinity - infinity is 0. With this idea in mind, except for 0/0 and 0^0 there would be no indeterminate family, infinity over infinity would be 1, infinity^0 would be 1, 1^infinity = 1, 0*infinity=0, and infinity-infinity=0. Right what I'm going here would mean infinity is a number, which it's not.
Surely the use of 'infinity' as a power is undefined, as 'infinity' (I don't have the proper symbol so I'm just using 'infinity' as an equivalent) is not an integer - or any real number at all. So it makes no sense to attempt to raise an integer (here, 1) to a power that is not any kind of real number. As for 0.999999999999999999 ... where the string of 9s never ends, that is identical to 1. If the string of 9s is any finite integer long, than that string is always less than 1, tho approaching it ever more closely. We needn't use limits here. Just the value 1 - 0.999999999999 ... to see the shortfall at a particular number of 9s.
I like to see this as: x^1, then x^2, then x^4, and so on, you will sea that the higher the exponent, the more the hiperbola looks like a right angle at 1, and as it aproaches 1^inf it becomes a right angle with all values possible for x=1
If you have the number above 1 that's infinitely close to 1, then that raised to the infinity would be e if I'm not wrong, but idk what 1- 1/infty would be
I actually accidentally proved that 1^infinity could be any number, was messing around with x^y = 1+x formula in wolfram and figured out that solving for x, x=(1+x)^1/y and taking the limit of y showed that x = 1, and using the original equation it shows that 1^infinity = 1+1 or 2, then i figured out that you could replace 1 in the original formula with any number, or f in this case, and the limit still works out to be 1 which means you can make it equal any number by replacing f with a number like 5 and following through the proof and you would get 1^infinity = 5+1 or 6, therefore proving that 1^infinity can be any number and therefore it is indeterminate (only in a limiting sense like bprp said which is when its indeterminate)
Hi bprp i enjoy & learn from you, however your assumption of 1 in Eulers Formula based on my understanding is not correct. Here 1 is a continous growing magnitude of 1 which stays as constant 1 no matter how big is gonna get. it is like the game of " pack man " . you see 1 in the beginning of growth is not the same as 1 at ends. Due to Eulers this 1is aktually 3 times the 1 in the first place , but STILL represented as 1. Therefor a^n=1, if a=1, but 1^inf.= 3 and no more bc. we re talking about Limits. please let me know if you or fan´s think over it differently 😊 .
The problem with current mathematics is that we only understand zero as "nothing" or when we shrink down something until it disappears (zero dimension). And secondly that adding an infinite amount of zero is zero. We fail to recognized that everything is sorrounded by nothingness. Like if you remove everything in the universe including space. There only exists nothingness which neither big not small.
I beg your pardon but 1 is neutral to multiplication, hence 1^inf is 1 by definition of "neutral element of multiplication". Or I'm missing something fundamental
So it is (let me use Latex notation) \lim_{ε→0){ (1+ε)^∞ } Well, it wasn't clear at all! At first it seemed to be something like \lim_{n→∞){ 1^n } which I hope we all agree is 1.
Could you use transfinite numbers in limits? Especially the infinity we talk about tends to be the well ordered set of an infinite set, and its power sets
Nonstandard analysis use "infinite-valued" numbers to compute limits "at infinity". Nonstandard analysis is based on constructing the hyperreal number system with infinitesimals and infinite-valued numbers. But this is very different from using cardinal or ordinal numbers. Cardinal numbers don't really work for this sort of thing. _Possibly_ ordinal numbers can be used to describe it (particularly ω), but generally only in the context of sequences, not for functions.
What about lim x->infinity (1-1/x)^x? You mentioned that 1 is not exactly one, and instead it could be 1.00000...1 or 0.999...9, so surely it would only be fair to consider the answer from below one as well as above one?
Joël Ganesh The thing is that numbers are not very well behaved around 1 Thus 1^∞ is indeterminate. 1^a = 1 but values around it can take different values. .9 is away from 1 by .1 but (.9)² is .19 units away 1.1 is away from 1 by .1 units but (1.1)² is .21 units away .19 ≠ .21 Thus, .9 and 1.1 aren’t very similar even though they are in the neighborhood of 1.
In case 1^inf=1, and the 1 is exact, shouldn't infinity not be exact infinity (like transfinite), if we want to determine it and get 1? Just asking to get intuition
Also, isn't this only as you approach 1 from the positive side? Because .99999^10000 (10000 being a big number and .99999 approaching 1) does not equal e it's not even close to e
No, even if it is not exactly 0 (approaching it). Think why. When you multiply a really small number a lot of times with itself, the answer will become veeery small very fast. If you do it 'infinitely' many times, it will approach 0.
@@aasyjepale5210 Yeah, bprp explained it in the video. For a little less than one, it'd go toward zero, and for a little greater than one, it'd go toward infinity.
1/0 is not defined, lim_{x->0} (1/x) -> infinity but 1/0 itself is not defined. If it were defined, you could: 0 = 0 3 * 0 = 5 * 0 3 * 0 / 0 = 5 * 0 / 0 3 = 5
Can someone explain what is meant by "not a solid 1" or "not exactly 1"? I don't understand why the 1 in 1^inf would not be considered "actually 1" if it is numerically shown as 1. Am I missing something?
There is really only one indeterminate expression, which is infinity minus infinity. All the others are basically just exp(indeterminate) or exp(exp(indeterminate)) phrased in different ways, and you could probably continue that game, but the resulting expressions are rare to encounter.
I thought he was a student until he show his happiness when we have to do more work. Now I know he is the teacher.
lol
Basically: 1 to the infinity power is 1, but something _approaching_ 1 to the infinity power is not.
not really, because when we talk about the infinite, we talk about Limits, no about numbers
but there's a reason it's written as 1 and not a number approaching one. math formality makes no sense to me sometimes.
Not quite, 0.999999... is equal to 1 in some sense, but 0.999999999... to the power of infinity is not necessarily 1. If you don't think 0.999999... is 1, please never call 0.33333... 1/3.
Yes, maths was interested before Chinese people invented shitty things
Yes precisely that!
Notice that exactly one to the power of infinity is actually something that appears quite often in both statistics and number theory. So it's not something that should be regarded as sloppy to write at all but it should preferably be expressed as how he writes it at around 08:16 as infinity is a philosophical concept, not a number. What it means is that proper 1^k remains 1 even when k goes to infinity. So it's also about limits. If I throw a dice where all sides shows the value 6, the probability I will only throw sixes after k throws will always remain even if I do it forever. What is sloppy however is to, without having any context just ASSUME 1 is not 1, just because it usually wasn't in some calculus course. Of course 1 is 1. If it's not, specify!
It's important to be comfortable to set such a value to 1 and not be afraid and think "can I really do that?". There is something wrong with how limits in calculus are being taught making many think things are indeterminate when they actually are not. This was a great video by blackpenredpen.
9:19 this is the way you say 'no' to somebody
I think the reason indeterminate forms are so confusing is because the notation 0♾ or 1^♾ is simply completely misleading. These forms are not arithmetic operations. They are limits. The correct way it should be presented is that the limit of f(x)^g(x) as f(x) -> 1 and g(x) -> ♾ is indeterminate, as is the limit of f(x)g(x) as f(x) -> 0 and g(x) -> ♾ is indeterminate. The expressions 0♾ and 1^♾ have no meaning. Of course, in the special case that f(x) = 1, then the limit f(x)^g(x) is simply 1. However, this is not so if f(x) is increasing or decreasing monotonically. Similarly, if f(x) = 0 as a special case in f(x)g(x), then the limit is 0, but in the general case of arbitrary f, the limit is indeterminate.
1^inf is just a shorthand. It's understood that this is for the limit.
Arteyyy "♾ isn't a real number"
Most students (and most people in general) treat it as a number. So they will interpret it as a numerical expression, regardless of whether it is correct or not. Deal with it.
"...and exponentiation isn't one of the four basic arithmetic operations..."
Many, if not most mathematicians, treat it as n arithmetic operations in their research and papers. Deal with it.
"A better criticism... misleading shorthands in mathematics..."
Yes, my comment talked about that. In fact, I said that nigh-verbatim in my comment. Perhaps you're not sufficiently skilled at reading basic difficulty English sentences, but if that is the case, then I am telling you that is in fact in the comment. Ask anyone else reading.
"...as long as we don't work with symbolic expressions, which in this case, we aren't requires to."
No one actually cares aboit whether we are required to or not. What we cared about is whether we actually do it or not. And spoiler alert: we do it everytime we have the opportunity. Never met a professor who doesn't. And that's what matters. Thank you for proving my point. Deal with it.
Oh, and for the record, in mathematics, technically nothing that is not a consequence of a set of axioms isn't required. So your point is moot.
"Oh, and you must be on acid too."
I'd rather be on acid and be correct than not be on acid and be say something awfully irrelevant like you did, and I'd also rather be on acid than being so immature and childlike as to be hopelessly petty and unable to hold a conversation where you go around insulting people because you couldn't simply handle the mere fact that I corrected you and schooled you in the previous conversations we had. Deal with it. Didn't know I'd have a little kid stalking me on UA-cam for all of my existence trying to act like a ghost.
mike burns No, it's understood by mathematicians. It's obviously NOT understood by students generally. That's why this video was made in the first place. That's precisely what the video is about in the first place. Maybe you should watch the video again.
Arteyyy Don't bother responding, because I'm not looking at this comment section again. After all, unlike you, I actually have a life and don't have time to be wasting with a childish one like you. It signifies you've already gotten lost. Farewell. So long.
I like how you call me immature and childish after you started insulting me for no reason. Or how you say you don't waste your time with me after doing just that.
OMG thank you so much man, i never understood why 1 to the infinity was not 1, it s all clear now, keep up the good work, you're the best
This is brutally next level. I am a Chemistry grad student and I've always had a strong interest in mathematics. Can't believe I'm learning all these things from you. Thank you.
0^0 mom inf over inf dad
We are number one, y'all are only approaching
Could you do the limit of (sin x - cos x)^tanx when the limit is π/2 ? Love your channel, keep up the good work!
the answer is 2
1 power infinity is the ans
Using logarithms, and l'hopital's rule, u should get 1/e, why is e everywhere >:(
I know the notation might look a little wonky, but how about we use something like "(→1)^(→∞)" and "(→0)/(→0)" for limiting forms instead of just 1^∞? I think this might help students understand better that indeterminate limiting forms are about functions approaching values, rather than about literal computation.
we put + or - in the exponent for those
@@aryanjoshi3342 The superscript +/- are used for _one-directional_ limits. However, (→1)^(→∞) and (→0)/(→0) are not one-directional indeterminate forms.
"Liquid Numbers":
at the end of this video there is a link to the "BE CAREFUL" video (1 year ago) which has similar statements such as:
Lim[n->inf] (1 + 1/n)^n is "not a solid e" and so...
"liquid_e" - e 0 ...
Also, in many videos about the use of Limits, including the "BE CAREFUL" one, there are lot of simplification of polynomial fractions based on the fact of:
Lim[x->inf] x^2 > Lim[x->inf] (ax + b)
or with other powers p>0:
Lim[x->inf] x^p > Lim[x->inf] x^(p-1)
Supposedly we all can live with that, no problem.
As in "BE CAREFUL" video the fraction is "conveniently simplified" as:
Lim[x->inf] -x^2 / (x^2 + ax + b) = -1
and the answer for that video becomes:
-e/2 ...
Now, is that a "solid -e/2"? Supposedly not... as the simplification of polynomial fraction is also a "non-solid -1" or it is more likely a "liquid -1" (or "fluidic"?).
That is the concept of "non-solidity" of numbers and quantities which makes the support of Newtonian mathematical principles of Calculus since:
Lim[dx->0] dy/dx while dx is "not a solid 0";
Paradoxically, the word "calculus" means "small stone", a "relatively solid" object.
Personally, despite I don't have any alternative to offer, I am not a fan of such simplification of polynomial fractions. Acceptance doesn't mean "like" or "consider it brilliant". Instead of a "solid" equality to -e/2, it could be a continuous fraction as an answer?
Is not unfair, in order to discriminate from "regular" Calculus, Ramanujan shall have his own (more) "fluidic" equality notation in:
1 + 2 + 3 +... {R}= -1/12 ?
I leave to you all these questions.
Regards.
A horizontal shaft rotates in bearings at its ends. At its midpoint is keyed a disk weighing 40 lbs, whose center of gravity is 0.1 inch from axis of rotation. If a static force of 200 lbs deflects the shaft and disk through 0.1 inch, determine the critical speed of rotation of the shaft.
@Blackpenredpen
Gr8 video ,big fan of bprp, could you please make a video on TAYLOR & MACLURIN series.
Actually, he already has.
I love the awesome work you are doing!!! Can you please make a video on Complex Analysis...
I think there is convention, that if you write a formula that represents a limit, you assume as many things as constant as possible.
Of course you can read 1^inf as a double limit, but intuition would suggest to read it as lim (1^x) x to inf.
Without this convention neutral operations really hurt you : 1^inf would also be 1*1^inf which can easily be used to break any meaning of the notation.
Beautifully explained
Thank you so much
Actualy 1 ^i is also 1, so 1 raised to any complex number is always 1
Felipe Lorenzzon We typically only deal with real numbers in calculus
@@angelmendez-rivera351 I know, but I were referring to the first case he showed
@@angelmendez-rivera351 Typically? I bring up complex analysis then. Or is that not typical?
Fun fact: i^i is real
@@anabang1251 More specifically, e ^ (-π / 2)
I know the answer to some indeterminates, infinity/infinity = 1, 1^infinity = 1, infinity^0 = 1, 0^0 = 1, 0 * infinity = 0, and infinity - infinity = 0
e is e-rational!!! and I've always wondered what is 'e' and where does it come from!! Thanks BPRP!!
Wolf Ram Alpha e is defined as the limit as x -> ♾ (1 + 1/n)^n. It comes from theoretical investigation of compound interest rate.
@@angelmendez-rivera351 yea I saw, bprp explained it pretty well tho👍!!
Wolf Ram Alpha I wanted to clarify that e being equal to that limit is not a coincidence. It is a definition. I'm putting emphasis on the word "definition" so that no one in the comments starts asking to "prove that it equals e": such a request is nonsensical since you cannot prove a definition.
@@angelmendez-rivera351 okey i got it 🙌✌✌ so it's just a definition!!
@@misotanniold787 i don't think that this way of defining e is the best. First of all: why there even exist such function, that is the same as it's derivative. Secondly: even if exp(0)=1, why this function is unique, i mean you said exp(z)=(exp(1))^z, but you said it as a result of first definition, but not as a scond part of definition, if f(x)=C*e^x, then f'(x)=C*e^x=f(x) see? And even so, why then exp(z)=(exp(1))^z? Also with that definition, how you prove that lim(x tends to infinity) (1+1/x)^x=exp(1)
I think that this definition of e as lim(1+1/x)^x as x going to infinity is more beatiful and show us a lot of its properties. When we investigate derivatives first time and doing (e^x)' with definition of derivatives, only relying on this definition we could see that (e^x)'=e^x. It really seems for me kinda illogical to define e like that if it would be harder to prove other properties of e, even if it shows other more beautiful properties.
Yep. Kids can, at times, cause "missed conceptions". Usually by means of, uh, a certain kind of interruption.
1^infinity = e^( ln(1) * infinity) = e^( 0 * infinity) => 0*infinity, indeterminate form
Now, i would say the misconception arrives because the teacher often don't write the approaching arrow... coz its not exactly one but its tending towards one ... And there the mystery remains ... When it tends towards one it can go from negative side and also from positive side so the fractions changes from proper to improper fractions ... As if we suppose write limit x tends towards 1 negative and write 1^infinity , then the answer to it is zero ... But if we do the same with x tends towards positive 1 then the answer becomes infinity ... So here the contradictory factor arrives which makes it the boundary line(1) for the indeterminastic results ...
Watching this video at this time makes me very very happy.....thanks for your videos...
now i get why my mum always says im the 1^infinity out of the family
Because you are exponentially growing smarter with time
You have an amazing mom
@@yamenarhim9336 And exponential functions grow the fastest!
@@alexanderpoltzer8885 You got the Point :D
@@yamenarhim9336 but one stay the same
I was bad at math, hardly passed my last year of bach but I find this kind of things very interesting
I love your videos. Can you explore hyper-operation more maybe with complex numbers? Thanks :))
Can you please solve this
Lim n tends to infinity .... n raised to p into sin square (n!) Upon n+1
Given : 0
Anybody else ?
Your expression is absolutely impossible to understand. Can you write it using symbols?
Angel Mendez-Rivera I think she says, (n) ^ ( (p * sin^2(n!)) / (n+1) ) as n approaches inf
Rev K I see, thank you.
@@Revanth292000 I think that would diverge as n^0 for n->\inf equals to 1 and n^(p/(n+1)) goes to some other limit, depentant on p (except p=0). We have now two boundaries, the sin(n!) Causes it to bounce between those, ergo it diverges.
If we're talking about 1^∞ we should consider exact 1. But if you consider base to be approximately 1 you should use the limit denotation.
Back to basic, you may plot a curve 1^x. You'll find it a straight line and can see 1^(1) is 1, 1^100 is 1, 1^100,000 is 1, so 1^inf is also 1. There isn't any decimals behind the base 1. Your proof by lim(x->inf)(1+1/x)^x is the shift of meaning. We are talking about the integer 1 and 1^inf, instead of the rational number (1+1/inf) and (1+1/inf)^inf.
you have to turn off the math part of your brain and turn on the business part. It was clickbait!
I noticed that in this limit the numerator results in the power of e number (consider n tends to infinity in limits below)
so lim((1 + a/n)^n) = e^a
lim((1 + 0/n)^n) = lim(1^n) = e^0 = 1
lim x->0 (1+ax)^(b/x) = lim y->inf (1+a/y)^(by) = exp(ab). So at end of video the answers are (exp(ln(4))=4, exp(0)=1 and exp(inf) = inf.
I usually keep it indeterminate because the infinite root of x is the same as x^1/infinity, this is one of 2 things. Either it approaches 1, or it is 1. Both would suggest the infinite root of x is 1. But I never defined x meaning this is true for any number and therefore 1^infinity is any number. Therefore if we can’t determine it it’s indeterminate.
Could you get any number you want for the limit, just putting ln(a) over the x? I mean e^n=lim(x-inf)(1+n/x)^x... If n is ln(a) you get a for the limit
Thank you! According to my university (The university of buenos aires) I should magically understand all of this without explanation. You saved me, and the explanation was very good, now everything makes sense :)
Thank you so much...
So my math teacher told us that 1^infinity is not 1 and then some kid yelled "it's 2!"
they're partly correct...
3:05 why don't complex numbers work? 1^z=a^(log_a(1)*z)=a^(0*z)=a^0=1 for any arbitrary value of a (except a=0 and a=1), like e or 10.
Wow, I almost didn't notice a little note that you wrote with the white marker, which says that 1 is not actually one, but lim x->1)
Great!
You're enlighting my understanding!
Well done!
Please what is the boi?
It'd just be 1/(ln(5) right?
Any constant times infinite is still infinity, so as long as you get lim x -> infinity of ln(5) gives you e^ln(5) = 5
So to do that you'd need ln(x) / (1/(ln(5)) ln(x)
Something like that
Hopefully i wrote it right im on mobile
So like ln(x) / 7ln(x) gives you 1/7, so ln(x) / (1/7)ln(x) gets you 7
So you just sub out the 7 in (1/7) for ln(5)
Greeaat matter to think about the matter of Maths, Exactitude and Approaching (and why not approach the higher, then approach the lower side, then compare, in French "encadrement").
Heyy bprp, which calculus book can you recommend for students like me? Greetings from the Philippines
I have never in my life seen such a beatiful 1 1:43
I don't see why it would be a difficult concept. In 12th grade, my calculus teacher told me that infinity/infinity, 0/0, infinity-infinity, 0^0 and 1^infinity were the general indeterminate forms, and they all seemed obvious to me. 1^infinity, REALLY means (1+delta)^f(delta) where lim as x->0 of f(x) = infinity, and since you don't know HOW quickly f(x) goes to infinity as x goes to 0, it could be anything, it could be a power that is insufficient to get it far away from 1, it could blow up MUCH faster than the 1+delta approaches 1 and thus make it REALLY big, and it could have complex values to make the result negative too.
There are actually 7 indeterminate forms. You forgot 2 more of these, which are 0 times infinity and infinity^0.
Good job man
If a solid 1^inf=1
Then log(1^inf)=log1
Inf*0=0
Correct me if I am wrong
No because in the moment you take the log of 1^inf you change the whole limit.
Maybe.
I dunno
@@valeriobertoncello1809 but how
The problem is that if log_a(1^inf) = inf*log_a(1), where a could be any number as you presumed, then the following is true:
a^log_a(1^inf) = a^inf*log_a(1)
1^inf=a^(inf*0)
Now, here you already see that a^(inf*0) is an indeterminate form
Because depending on how you arrange inf and 0 you get two different things, contradicting the commutative property of multiplication:
(a^inf)^0 = inf^0 = ???
(a^0)^inf = 1^inf = 1
And you not also presume that inf^0 is determinate but that it is equal to 1^inf, i.e. that inf^0=1
Still a bit confused
But thanks
Great video!
0:18 we’ll obv inf/inf is the mom because it’s top heavy and bottom heavy (you can guess what that means lol) that’s how my cal teacher taught us limits
#/inf is bottom heavy so big O
Inf/# is top heavy so big ∞
e^(pi*i*2) = 1 = e^0
2*pi*i = 0😨
sin(pi) = sin(2*pi)
arcsin(sin(pi)) = arcsin(sin(2*pi))
pi = 2*pi
1 = 2 😨
1^2=1
1^3=1
2=3
I know this sucks😑
(-2)^2 = 2^2 = 4
-2 = 2
0*1 = 0*(10^100)
So 1 = 10^100
x^x^x^x^x^... = 2
x^2 = 2
x = √2
x^x^x^x^x^... = 4
x^4 = 4
x^2 = 2
x = √2
√2^√2^√2^√2^... = 2 = 4
Bro.. you looked way better without beard..
Could you do this in general by breaking aparant the 1 and ∞ using e^ln(x), then you get e^∞0, invert either the ∞ or 0 to get a 0/0 or ∞/∞ form, then use L'H to solve?
1 to a complex power is still 1.
1^(a+bi)=e^(ln(1)(a+bi))=e^(0(a+bi))=e^0=1
I wonder if the (x root of x)^^infinity is indeterminate, that is ^^ referring to tetration
Please, int x^2 sqrt(1+x^2)
Леша Кузнецзов Use integration by parts, integrating xsqrt(1 + x^2) and differentiating x. To integrate xsqrt(1 + x^2), let y = 1 + x^2, dy/2 = xdx, so xsqrt(1 + x^2)dx = sqrt(y)dy/2. The integral of this is 1/(4sqrt(y)), and converting back gives 1/(4sqrt(1 + x^2)). Therefore, the integral of (x^2)sqrt(1 + x^2)dx is (x/4)/sqrt(1 + x^2) - (1/4)integral of dx/sqrt(1 + x^2). Integral of dx/sqrt(1 + x^2 can be found by letting x = sinh(u), dx = cosh(u)du, sqrt(1 + x^2) = cosh(u), so dx/sqrt(1 + x^2) = du. The integral of this is u = arsinh(x). Therefore, integral[(x^2)sqrt(1 + x^2)dx] = (1/4)[x/sqrt(1 + x^2) - arsinh(x)] + C
@@angelmendez-rivera351 why, int sqrt(y) dy/2=1/4sqrt(y)
First, let me thank you for your videos and say that I like your approach, although I would modify it a little. My main modification would be not to include infinity in the table to avoid confusion. It is abundantly Lear that you know what you are doing, but students tend to get confused with the difference between limit and evaluate a function and include infinity in the table could feed this thought.
Infinity is the same value at either end of the number-line, connecting each end and forming a loop. It's not just in the positive direction, therefore if inf-inf = ind then inf+inf also = ind.
Infinity is not a number. It just indicates where the value tends increase or decrease without any bonds. Just because some large number minus itself is 0, infinity minus infinity is indeterminate. We can't tell what value is going towards to, so you have to do more work. Infinity plus infinity has the limit that tends to infinity.
@@justabunga1 I agree with your explanation of infinity but I'm kind of thinking about infinity - infinity as removing something infinitely large from something that is also infinitely large but has the same infinite mass, because the notation is exactly the same, so I'd just say infinity - infinity is 0. With this idea in mind, except for 0/0 and 0^0 there would be no indeterminate family, infinity over infinity would be 1, infinity^0 would be 1, 1^infinity = 1, 0*infinity=0, and infinity-infinity=0. Right what I'm going here would mean infinity is a number, which it's not.
0:52
Me who keeps running to those numbers in limits: *NOOOOO* :
Surely the use of 'infinity' as a power is undefined, as 'infinity' (I don't have the proper symbol so I'm just using 'infinity' as an equivalent) is not an integer - or any real number at all. So it makes no sense to attempt to raise an integer (here, 1) to a power that is not any kind of real number. As for 0.999999999999999999 ... where the string of 9s never ends, that is identical to 1. If the string of 9s is any finite integer long, than that string is always less than 1, tho approaching it ever more closely. We needn't use limits here. Just the value 1 - 0.999999999999 ... to see the shortfall at a particular number of 9s.
I love this guy sooooo muchh
If 1 is not exactly equal to 1, then why not take the limit of 1 approaching 1 (lim[1 -> 1] ...)?
I like to see this as: x^1, then x^2, then x^4, and so on, you will sea that the higher the exponent, the more the hiperbola looks like a right angle at 1, and as it aproaches 1^inf it becomes a right angle with all values possible for x=1
If you have the number above 1 that's infinitely close to 1, then that raised to the infinity would be e if I'm not wrong, but idk what 1- 1/infty would be
What about 0^inf?
If you have a number really close to 0, when you raise it to the ∞ power, it stays 0, so 0^∞ -> 0.
A big problem it is approaching 0 from the negative side and it is part of a bigger thing.
(1^inf)/(0^înf). 1 isn't 1 and 0 isn't 0 and inf isn't inf.
This is not an indeterminate form. The limit goes to 0. A good example would be y=x^(1/x) as x approaches 0 from the right.
He got out of the house and adopted by a determined family
Can u pls integrate 1/(x³+1)½ w t r x
Didn't get it.....
Thank you
I've always though of 1 to the infinity as Euler's number e. because 1 is an infinitesimal value larger than 1
I actually accidentally proved that 1^infinity could be any number, was messing around with x^y = 1+x formula in wolfram and figured out that solving for x, x=(1+x)^1/y and taking the limit of y showed that x = 1, and using the original equation it shows that 1^infinity = 1+1 or 2, then i figured out that you could replace 1 in the original formula with any number, or f in this case, and the limit still works out to be 1 which means you can make it equal any number by replacing f with a number like 5 and following through the proof and you would get 1^infinity = 5+1 or 6, therefore proving that 1^infinity can be any number and therefore it is indeterminate (only in a limiting sense like bprp said which is when its indeterminate)
The compound formula takes me to the "e" constant.
The first person who loves ideterminate forms!
Hi bprp
i enjoy & learn from you, however your assumption of 1 in Eulers Formula based on my understanding is not correct. Here 1 is a continous growing magnitude of 1 which stays as constant 1 no matter how big is gonna get. it is like the game of
" pack man " . you see 1 in the beginning of growth is not the same as 1 at ends. Due to Eulers this 1is aktually 3 times the 1 in the first place , but STILL represented as 1. Therefor a^n=1, if a=1, but 1^inf.= 3 and no more bc. we re talking about Limits. please let me know if you or fan´s think over it differently 😊 .
The problem with current mathematics is that we only understand zero as "nothing" or when we shrink down something until it disappears (zero dimension). And secondly that adding an infinite amount of zero is zero. We fail to recognized that everything is sorrounded by nothingness. Like if you remove everything in the universe including space. There only exists nothingness which neither big not small.
I beg your pardon but 1 is neutral to multiplication, hence 1^inf is 1 by definition of "neutral element of multiplication". Or I'm missing something fundamental
When you are dealing with a limit, the form " 1^inf " mostly means (something_approaching_1)^inf, which isn't aways one.
So it is (let me use Latex notation)
\lim_{ε→0){ (1+ε)^∞ }
Well, it wasn't clear at all! At first it seemed to be something like
\lim_{n→∞){ 1^n }
which I hope we all agree is 1.
Her: How come you don’t understand my signs, they’re so obvious
Her signs: The indeterminate family
Would be clearer if we used Richardson’s hyper reals notation for infinitesimals. But few people are familiar with that.
I love the way he says Dot Dot Dot xD
Congratulation, now you have to do more work.
I cried
a lot
Could you use transfinite numbers in limits? Especially the infinity we talk about tends to be the well ordered set of an infinite set, and its power sets
Nonstandard analysis use "infinite-valued" numbers to compute limits "at infinity". Nonstandard analysis is based on constructing the hyperreal number system with infinitesimals and infinite-valued numbers.
But this is very different from using cardinal or ordinal numbers. Cardinal numbers don't really work for this sort of thing. _Possibly_ ordinal numbers can be used to describe it (particularly ω), but generally only in the context of sequences, not for functions.
How I thought about it is taking 1^\inf = e^(ln(1^\inf)) = e^(\inf * ln(1) ) = e^(\inf * 0) = e^(something indeterminant) = something indeterminant
there is also lim_->inf (1-1/x)^(1/x) = 1/e
2020 raise to the power 2019 - 2020 divided by 2020 square + 2021=N
then find the sum of digits of n
bro plz solve this?? trying from last 5 weeks
Nice video thanks a lot
Can any of the three limits at the end of the video be rearranged into forms such that L’Hôpital’s Rule can apply?
What about lim x->infinity (1-1/x)^x? You mentioned that 1 is not exactly one, and instead it could be 1.00000...1 or 0.999...9, so surely it would only be fair to consider the answer from below one as well as above one?
9:28 Its 'E'rational
The numbers are not symmetrical around 1
Eg
.5^2 = .25
And
1.5^ = 2.25
No relation whatsoever (except ending in the same digits)
Joël Ganesh
The thing is that numbers are not very well behaved around 1
Thus 1^∞ is indeterminate. 1^a = 1 but values around it can take different values.
.9 is away from 1 by .1 but (.9)² is .19 units away
1.1 is away from 1 by .1 units but (1.1)² is .21 units away
.19 ≠ .21
Thus, .9 and 1.1 aren’t very similar even though they are in the neighborhood of 1.
Joël Ganesh
As for your question
15² = (10+5)² = 10^2 + 2(10)(5) + 5² = (10)(10+10) + 25 = 10(20) + 25
Which is why the pattern exists
lim x->1- (x^infinity)=0
In case 1^inf=1, and the 1 is exact, shouldn't infinity not be exact infinity (like transfinite), if we want to determine it and get 1? Just asking to get intuition
What is the limit of (1+(1/ix))^x as x approaches infinity?
Challenge
Convert this Consecutive fraction to proper fraction using G.P.
X = 0.0 123 123 123 ....
Well, if x=0.0123123•••
then 10x=0.123123••• and also 10000x=123.123123•••
so 10000x-10x=123
9990x=123 and x=123/9990=41/3330
@@DeathGryphon99 thanks bro.
But using geometric series
Just buy a lacavier please
Also, isn't this only as you approach 1 from the positive side? Because .99999^10000 (10000 being a big number and .99999 approaching 1) does not equal e it's not even close to e
What if there is minus in this formuła? (1-(1/x))^x
lim( (1+y/x)^x ) = e^y
your limit= lim ((1 - 1/x)^x)
= lim ((1 + (-1/x)^x)
we can see that y = -1
hence lim ((1-1/x)^x)= e^(-1)
What about 0^(infinity)? Wouldn't that be indeterminate?
No, even if it is not exactly 0 (approaching it). Think why. When you multiply a really small number a lot of times with itself, the answer will become veeery small very fast. If you do it 'infinitely' many times, it will approach 0.
@@futfan9092 Isn't 1 multiplied by itself (finitely) many times always 1? So why is 0^inf 0, but 1^inf indeterminate?
Never mind, bprp mentions it.
@@snbeast9545 as powers get higher they go away from 1 so for 1 there is no specific direction to go - its how i see it
@@aasyjepale5210 Yeah, bprp explained it in the video. For a little less than one, it'd go toward zero, and for a little greater than one, it'd go toward infinity.
It goes to zero. That’s not an indeterminate form. A good example of this would be y=x^(1/x) as x approaches 0 from the right.
1^infinity =x ? How to do calculations
(any number)^0=1
1=(any number)^0
1^(1/0)=any number
1^(inf)=any number, so it is very undefined even when we don't assume that 1=1.0000000000001
1/0 is not defined, lim_{x->0} (1/x) -> infinity but 1/0 itself is not defined. If it were defined, you could:
0 = 0
3 * 0 = 5 * 0
3 * 0 / 0 = 5 * 0 / 0
3 = 5
Can someone explain what is meant by "not a solid 1" or "not exactly 1"? I don't understand why the 1 in 1^inf would not be considered "actually 1" if it is numerically shown as 1. Am I missing something?
0.999 recurring = 1
@@JasperWilliams42 but 1^ifinity means 1x1x1... infinite times, so surely we can't get 0.999 from that?
but in this case it's exact one and not approaching 1@hybmnzz2658
please do sqrt(i^2)
xd
Does lim (1)^n when n=infinite converge or diverge
Can somebody help explain the 2nd expression at the end of the video? Why is it 1?
he is operating with pens like with food sticks lol
There is really only one indeterminate expression, which is infinity minus infinity. All the others are basically just exp(indeterminate) or exp(exp(indeterminate)) phrased in different ways, and you could probably continue that game, but the resulting expressions are rare to encounter.
1 raised to power iota is equal to?