1^∞ is equal to 1 as long as the base 1 is exact 1, I mean if the value of the base approaches 1 (with some limits like lim(x->0 sin(x)/x)) then only the limit approaches to some value else 1^x is exactly 1 if the base 1 is exact and the power x is a real value (for complex values it changes). Here I would like to give an example, we can cancel out the 2^∞ on Numerator and denominator because they are equal. ∞s can be non equal. Some infinitys are larger than others. for example, if ∞1 is the total number of real numbers between 0 and 1, and ∞2 is the total number of natural numbers existing, then ∞1 > ∞2. ie. ∞1/∞2= ∞ and ∞2/∞1 is 0. If someone read this comment completely then they are very patient. I appreciate.
@@wallaceOne5900 Hello, Thank you for your response, the first part just say that the quantity 1^x = 1 is always true if the bese 1 is exact (not any approx or limit) and the power is a real value. Thus, 1^∞=1. I stated that for complex values it changes, check in wolfram alpha for better results. 1) 3^(1/x)=1 which is same as 1^x=3 Search 3^(1/x)=1 in wolfram alpha. Have a great day ahead.
Can we all agree that when we define 1 if we order the natural numbers set it's the first element in the set? If we agree then 1^z equals 1 and z contains in complex numbers set Proof of that using the definition of holomorphic function or analytic function By knowing that 1^z is e^(z*ln(1)) and knowing that e^z is holomorphic function then e^z*ln(1) is also holomorphic function meaning that every complex number in the domain Z has its own defined complex number in the codomain then even if it's a big number the result will be defined and of course Z^0 is 1 except if Z is 0 But in our case e^z is "infinity" which is far from 0 But Now question to our brother what makes my proof false and yours right?
The fact that it takes just one example (that the guy used in this video) to make an entire demonstration worthless proving one example that stands by it
RANT !!! i hate the fact that youtube doesnt have some sort of latex mode so that math expressions look way easier to see because i wasnt use to divisions as having slashes and multiplications with asterisks. like if i js said $\frac{5}{x}=2$ and it would upload an image, thaf would be very cool.
One to the power x as x approaches infinity (the correct way of writing a mathematical expression with an infinity in it) - I would call 1 - else you break field theory over what defines a unit value in a number field. When one breaks the 2 over 2 apart removing the brackets and applies the infinity to each separate component is where you are playing with fire - as you have applied the rules of numbers "illegally" as it comes to dealing with the priniciples of manipulating with infinities (which aren't numbers) with algebraic integrity. Do that (which is not a correct way of manipulating and infinity in a numerical expression) and that is where you can arrive at any answer you desire.
@@gruntilda2 that is the mistake most commonly made by folk - thinking of it as a number - that you can do general numerical operations on it sensibly - rather than of it being a concept that numbers never end / aren't bounded. Folk ask if its about numbers why can't I do number type things to it and/or with it? They find it very unnatural to learn one absolutely must not do "number type things" with concepts about numbers that one can do to numbers! It is counter intuitive - like saying 1 appears as a first digit of all integers 30.2% of the time but 9 only 4.5% - you get a huh - that doesn't make sense reply - which is great - as its an excellent way for spotting fraud in business data :)
If it's exactly 1 raised to infinity (1 multiplied by itself an infinite number of times) then the answer is indeed 1 However, the reason we say 1^∞ is an indeterminate limit form is because we usually don't deal with an exact 1 For example lim x-->∞ of (1+1/x)^x=e We actually have a number ever so slightly larger than 1 raised to infinity which will give us approximately 2.7182818284590452353
1^inf is undefined because infinity is NOT a number. Its a concept, which is observed using the tool called limit. If you are familiar with limits you might find notation 1^inf familiar. Its used to categorize which indeterminate form is presented. These are two completely different entities.
It's a number and a concept, it's a transfinite number. You can do things like aleph-0 + aleph-1 = aleph-1. I'm not totally sure to be honest, I would just use limits though to avoid the unpleasantness. I'd say it was undefined if I had to give an answer.
there are some models that allow the use of infinites and infinitesimal as a number but there are restrictions on how youre allowed to use it in the same way than we cant divide by zero
In mathematics and in subsequent applications in physics or chemistry… etc. once you stumble over a divergent operation (like an infinite power here) you should take the limit just before applying the divergence, not the other way around. Simple rules that could save you from insane conclusions.
I think it was some kind of a cruel math joke from Prime, sort of "the proof that these reasonings are incorrect is left as an exercise to the viewer", but I don't think that this is a kind of a good quality math humor, since most people in the comments seem to be deeply confused with this "proof".
@@allozovsky it could very well be a joke but there are others out there who get confused like you said. and then you wonder how the hell there are people today who are convinced the earth is flat and they are in the illusion of possessing “proofs” !!
let f(x) be 1^x = 1 when x = 1, 1^1=1 so f(1) is true let y be some positive real number that 1^y = 1 when x = y+1 1^(x) = 1^(y+1) = (1^y)*1 = 1*1 = 1 so f(x+1) is also true by the principle of mathmetical induction, f(x) is true for all positive real numbers therefore f(infinity) is true, 1^inf = 1
It is only not defined as such when the result of a limit where both the exponent approaches infinity and the base approaches 1 at different rates whilst taking on different real values
I love your channel, but I have to disagree with this proof. In reality, lim(x->inf) 1^x = 1, and I'll explain why. First off, everyone needs to understand that infinity is not a real number. That flipped up eight is an expression that only makes sense in calculus, which expresses a hypothetical number that WON'T STOP going up. It's not a plain number, but a symbol used in calculus to explain the behavior of functions as x goes to higher and higher positive numbers or lower and lower negative numbers. Thus, in reality, infinity does not make sense in arithmetics. It can only make sense in LIMITS, which is why this video is already flawed. However, even if we put it as a limit (as I did in the beginning of my comment), it still computes as one. This is because 1^a = 1 regardless of the number used. This misconception comes thanks to two situations in limits that give you different numbers, but are completely different : - First, the 1^inf indeterminate, which computes to e : e = lim(x->inf) (1+1/x)^x This can look similar, but notice that the base of the exponent is not one, but 1+1/x This completely changes matters, since the base is approaching one and therefore computing erratic results in the process. Those results do not stop at 1, but rather e, since the base changing and therefore not one. - Second, the proof used in this video, inf/inf. This is another indeterminate form, but is not undefined. When solving this indetermination, it simply boils down to knowing which function grows faster, either the numerator or the denominator. Inf is a vague term, since it expresses a growing number, but not how fast it grows, and that will depend on the function we are working with. To find out, we can simply look at the functions and compare them on the "order of infinity", or we can use derivatives, but that's a step further in the calculus world. If the numerator grows faster, then it's a faster growing number divided by a slower growing number, so it's infinity. If the denominator grows faster, then it's the other way around, so it ends up computing zero. And, if numerator and denominator grow at the same speed, which is the case when they are both the same function, they end up as one. That's the case used in the video, which is wrongly crossed out as undefined, but really, if you got two functions that are exactly the same, no matter how big those numbers are, it's going to be one. Again, let me highlight that infinity/infinity is a vague expression, not a number, and that it can compute different things depending on the case.
It is right but let consider: a(n)=n (for example), than we have: b(n) = 1^a(n)=1 for any natural n. By definition of limit that will be: Limit b(n) = 1 while n -> + infinity.
Don't upload garbage on the media and confuse people. If the base is exact 1, 1 to any power is 1. You must make it crystal clear that the base is approaching 1 (1 in the limit sense) and yet it may be defined or undefined. Limit of floor(x+1)^(1/x) as x goes to 0 from above is 1. Limit of (1+1/x)^x as x goes to infinity is e. I'd recommend that you take off this bs.
5 місяців тому
Leíste los procedimientos de límites al revés: primero se simplifica la expresión y luego se aplica el límite. Lim n-> inf of (x/x)^n with x dif 0 Lim n-> inf of 1^n 1^(inf) 1
Undefined doesnt means that the limit doesnt exists. It means that you are not calculating it in a correct way. Base is exactly 1, then the limit is 1.
@@sherinah4812 that's too bad, because it's false. You can't treat the inifinity symbol as a number (the same mistake is being made in the video itself)
Indeterminate*. And btw if the base is exactly 1, then 1 to the infinity is 1, otherwise it's either 0 or infinity whether the base goes to 1 from the left or the right. This is the case if we decide to make sense of what you wrote, because infinity is not a number, therefore you cant use it like that
Sir I love your videos so much but I have a request from you if you are free then please 🙏 make video's on math's chapters of IIT JEE (it's a competitive exam in india) Please help Sir 🙏🙏🙏🙏😢😢😢
that’s wrong. 2^x/2^y is not reduced to x/y but to 2^(x-y) and if x=y then this is 2^0=1 as required. However, infinity is not a number and so any arithmetic with it is undefined
In this video, you just can't break 'infinity' like a number like you did (2/2)^(ꝏ) = (2^ꝏ)/(2^ꝏ) this is not possible algebraically. Because if it is then (2^(ꝏ))/(2^(ꝏ)) could be (2^ꝏ)/(2^ꝏ) = (1^ꝏ + 1^ꝏ)/(1^ꝏ+1^ꝏ) it will became like that if u treat infinity(ꝏ) as a algebraic number. So this video is just wrong way to put it and, like me i would a bit agree that 1^ꝏ=1 by some limits way.
It's alright if raising to the power of ∞ is a valid operation. The initial assumption of the video is that raising to the power of ∞ is a valid operation, as he takes 1 to the power of ∞.
@@xinpingdonohoe3978 that's a far stretch you're making 😄 but i see your point. I'd still rather the video would work with limits here. The infinity sign isn't even a number and if it were, what would, for example, inf - inf be? Lots and lots of problems incoming... 😄
Assume lim x approaches inf in (x+3/x+2)^x this gives us 1, you can calculate, without any operation of limit it's already impossible to write infinity as power of a number but in this case if we really take 1 as base it will give us 1 there is no wrong.
IF YOU LOVE MATH, 🟨 MUST READ: I'll say it clearly: stop messing with the term infinity as if it's a number!! All the problems, and all the unnecessary philosophies happen from one mistake that everyone makes: analyzing a problem with infinity as if it were a number. enough with that!!! The concept of infinity is not a number on the number's axis, infinity is an "idea" not a point on the axis. You can ask how much is 3x+3x and the answer will be 6x, but you can't ask how much is infinity plus infinity, the answer is not two infinity. Because it's not a number. I can say: I went 50 kilometers forward and went back 42 kilometers, therefore I am 8 kilometers from the starting point. But I can't say I walked an infinite number of miles (what is that???) and returned three-quarters of an infinity. Does that mean I'm a quarter of an infinity away from the start?? Is it ridiculous enough to understand what I'm saying? Well, no need to ask how much is 1 to the power of infinity. unnecessary, meaningless, and a little annoying. Tamir Erez Math teacher.
We can't just treat 1^∞ as a "number". It is not even defined. If you want to give it some sense then you have to define it. Say for example 1^∞=lim_n 1^n, in which case for sure it is 1. But as I said before it depends on your definition, for example if you define 1^∞ to be lim_n->∞ lim_x->1 x^n then it even depends on the order of the limit if we do first n then we get 1 again, if we do x first then it doesn't exist as for 0
@danielmunozgeorge2876 > What fais in your proof? Well you can't divide limits unless both exist, basic property of limits. Yeah, that "surprising" distribution of the exponent over the numerator and denominator was quite unexpected, even shocking to me. I guess Prime should understand such basic things. Unless it was some kind of a bad math joke and spotting the inaccuracy was left as an exercise to the reader.
@@allozovsky yeah, I bet Prime knows that, but of course these videos are not meant to math students but rather any audence. It is always easier to catch the audience with these kind of "surprising" math.
If you use this proof, you prove also that any other fraction is infinite, as infinite power of any number is infinite. That way it is just a misconception, not a proof. Good thing you wrote "proof".
The point here is to prove that 1^infinity is not 1. We are proving this because 1 to the power any real no. Is 1. But 1 to the power infinity is not 1. That's all that's done.
Right, infinity/infinity is indeterminate, not undefined. And you actually end up with infinity/infinity fairly frequently in calculus! Sometimes you can't determine what that ratio should turn out to be... other times you can. That's why it's a "proof" we've been given, not a proof.
Well, even i could've told you that 1^inf ≠ 1 because they're obviously in different forms. One is an integer with an exponent while the other is just an integer; therefore, they are not the same and not exactly equal since a different rule is being applied for the infinite exponential series. Even though, it can definitely be defined using logical rationale and the essential coherency of mathematical laws. We can easily extrapolate that 1^inf can never provide any other output besides 1 because any number selected from 1 -> inf will always be 1 since infinity in itself is 1 system, as well as all numbers themselves being the result of various sizes (amounts) of 1. Oh yeah, AND 1^n = 1!
Of course its undefined though, you cant raise to the power of infinity because infinity isnt a number.. If you choose to pretend that infinity is a number then the answer would be 1, because assuming it is a number, infinity over infinity would be 1 (but it's not a number)
this is so wrong on multiple levels. First of all, inf is not a real number, so essentially 1^inf has to be defined using limit and in that case, 1^inf is actually 1. Second, in the proof, you are so freely using inf like it's a real number and using various operations and identities that are valid only for real/complex numbers, but not for inf. And if you define every step in the proof using the concept of limit, then proof becomes invalid by default because it would come out to be 1, which is contradictory to what you are claiming
Is it not a mistake to say 1^inf = 2^inf x 2^-inf because you're saying that 1^inf x 2^inf = 2^inf, introducing an infinity where there may not have been one because you evaluate the infinity twice? 0 = 1 x 0. 1 = 0/0 = undefined. Just introduced an infinity to say that 1 is undefined. It may be undefined but I don't think this shows it.
The same reason 0/0 is not 1; they can represent different values depending on context, therefor without any context they are referred to as “indeterminate form.”
@@strangex1524 while technically true I’d call this an oversimplification. After all, every number is a concept. There are number systems wherein infinity can be treated as a number with special properties.
0:24 I have confused even in that, why (2/2) is to the power of infinite, if 1 is to the power of infinite is not equal to 1, 😂😂😂😂 , (1)^infinite =!= 1
This is one, to any number even if INFINITY DIVIDED BY INFINITY IS 1 I AM SURE THAHT IT IS 1, IT IS NOT UNDERFINED. 1 to infinity is (2/2) to infinity = 2 to inf/ 2 to inf = inf / inf is 1 WRONG proof
@@PrimeNewtonsOk, my comment was not constructive, sorry for that. And you are contributing making „the boring math“ subject more attractive to a wider audience, which is great! However, in this video, you are doing calculus with notations that have no (canonical) definition. Therefore, the rules that you are applying are also without any meaning in this context.
Thats a contradiction in your proof, since you are using inf symbol as a number leading to inf/inf not a number. Since it is a contradiction lim (n- inf) 1^n = 1.
1 is not exactly 1 it may be like either 1.0000000000001 or 0.999999999999 1.000000000001power Infinity = also infinity or 0.99999999999 power Infinity = 0 . Hence answer is undefined😊
Infinity is not a number. There's no such quantity as 1^infinity or any other quantity mentioned in this video. (Of course, there is a theory of transfinite ordinal and cardinal numbers in set theory, but this has nothing to do with the "infinity" symbol you see in this video or in a calculus course.)
To talk about infinite you need to talk about limits. Then the argument shifts to how close to 1 are you when taking it to the infinity power. If it is exactly 1 then 1 to the infinite is actually equal to 1, otherwise you get shinanigens
This is one, to any number even if INFINITY DIVIDED BY INFINITY IS 1 I AM SURE THAHT IT IS 1, IT IS NOT UNDERFINED. 1 to infinity is (2/2) to infinity = 2 to inf/ 2 to inf = inf / inf is 1 WRONG proof
1^∞ is equal to 1 as long as the base 1 is exact 1, I mean if the value of the base approaches 1 (with some limits like lim(x->0 sin(x)/x)) then only the limit approaches to some value else 1^x is exactly 1 if the base 1 is exact and the power x is a real value (for complex values it changes).
Here I would like to give an example, we can cancel out the 2^∞ on Numerator and denominator because they are equal. ∞s can be non equal. Some infinitys are larger than others. for example, if ∞1 is the total number of real numbers between 0 and 1, and ∞2 is the total number of natural numbers existing, then ∞1 > ∞2. ie. ∞1/∞2= ∞ and ∞2/∞1 is 0.
If someone read this comment completely then they are very patient. I appreciate.
Funny part at the end 😂
great explanation
I don't understand the first part too well, but I get the latter parts. Please could you explain it better to me?
@@wallaceOne5900 Hello,
Thank you for your response, the first part just say that the quantity 1^x = 1 is always true if the bese 1 is exact (not any approx or limit) and the power is a real value. Thus, 1^∞=1. I stated that for complex values it changes, check in wolfram alpha for better results.
1) 3^(1/x)=1 which is same as 1^x=3
Search 3^(1/x)=1 in wolfram alpha.
Have a great day ahead.
@@technicalpoint6268 I think I get your point. Thank you😊
Can we all agree that when we define 1 if we order the natural numbers set it's the first element in the set?
If we agree then
1^z equals 1 and z contains in complex numbers set
Proof of that using the definition of holomorphic function or analytic function
By knowing that 1^z is e^(z*ln(1)) and knowing that e^z is holomorphic function then e^z*ln(1) is also holomorphic function meaning that every complex number in the domain Z has its own defined complex number in the codomain then even if it's a big number the result will be defined and of course Z^0 is 1 except if Z is 0
But in our case e^z is "infinity" which is far from 0
But Now question to our brother what makes my proof false and yours right?
The fact that it takes just one example (that the guy used in this video) to make an entire demonstration worthless proving one example that stands by it
RANT !!!
i hate the fact that youtube doesnt have some sort of latex mode so that math expressions look way easier to see because i wasnt use to divisions as having slashes and multiplications with asterisks. like if i js said
$\frac{5}{x}=2$
and it would upload an image, thaf would be very cool.
One to the power x as x approaches infinity (the correct way of writing a mathematical expression with an infinity in it) - I would call 1 - else you break field theory over what defines a unit value in a number field.
When one breaks the 2 over 2 apart removing the brackets and applies the infinity to each separate component is where you are playing with fire - as you have applied the rules of numbers "illegally" as it comes to dealing with the priniciples of manipulating with infinities (which aren't numbers) with algebraic integrity. Do that (which is not a correct way of manipulating and infinity in a numerical expression) and that is where you can arrive at any answer you desire.
Haha if only we could use infinity as a number 😂
@@gruntilda2 how to interact infinity with numbers has been understood for a very, very long time - almost as long as its been mis-understood...
@@matthewkendall5235 I mean using it as a definable, numerical number using the symbol 😂
@@gruntilda2 that is the mistake most commonly made by folk - thinking of it as a number - that you can do general numerical operations on it sensibly - rather than of it being a concept that numbers never end / aren't bounded. Folk ask if its about numbers why can't I do number type things to it and/or with it? They find it very unnatural to learn one absolutely must not do "number type things" with concepts about numbers that one can do to numbers!
It is counter intuitive - like saying 1 appears as a first digit of all integers 30.2% of the time but 9 only 4.5% - you get a huh - that doesn't make sense reply - which is great - as its an excellent way for spotting fraud in business data :)
If it's exactly 1 raised to infinity (1 multiplied by itself an infinite number of times) then the answer is indeed 1
However, the reason we say 1^∞ is an indeterminate limit form is because we usually don't deal with an exact 1
For example lim x-->∞ of (1+1/x)^x=e
We actually have a number ever so slightly larger than 1 raised to infinity which will give us approximately 2.7182818284590452353
Wow
This is trying to say x^inf can be said to be "technically" indeterminate at x=1 though if we take the limit as x -> 1 from both sides, it's clearly 1
Now how about Lim[x->1] (2x/(1+x)) ^ inf??? NVM on the first comment. The limit from the right is inf and is 0 from the left.
@@DavyCDiamondback you can't just put infinity in the expression, you have to say limit as it approaches infinity
@@Samir-zb3xk I realized that. How about (x+1)^(1/x) as x approaches zero from the right?
1^inf is undefined because infinity is NOT a number. Its a concept, which is observed using the tool called limit.
If you are familiar with limits you might find notation 1^inf familiar. Its used to categorize which indeterminate form is presented. These are two completely different entities.
It's a number and a concept, it's a transfinite number.
You can do things like aleph-0 + aleph-1 = aleph-1.
I'm not totally sure to be honest, I would just use limits though to avoid the unpleasantness.
I'd say it was undefined if I had to give an answer.
there are some models that allow the use of infinites and infinitesimal as a number but there are restrictions on how youre allowed to use it in the same way than we cant divide by zero
In mathematics and in subsequent applications in physics or chemistry… etc. once you stumble over a divergent operation (like an infinite power here) you should take the limit just before applying the divergence, not the other way around. Simple rules that could save you from insane conclusions.
I think it was some kind of a cruel math joke from Prime, sort of "the proof that these reasonings are incorrect is left as an exercise to the viewer", but I don't think that this is a kind of a good quality math humor, since most people in the comments seem to be deeply confused with this "proof".
@@allozovsky it could very well be a joke but there are others out there who get confused like you said. and then you wonder how the hell there are people today who are convinced the earth is flat and they are in the illusion of possessing “proofs” !!
That is true only if the "1" raised to infinity is the result of a limit, so it's not really "1". If the "1" is exactly 1, the the result is 1.
Simple and clever! That is the Mathematics beauty!!
Brief and quick. I love that.
let f(x) be 1^x = 1
when x = 1, 1^1=1
so f(1) is true
let y be some positive real number that 1^y = 1
when x = y+1
1^(x) = 1^(y+1) = (1^y)*1 = 1*1 = 1
so f(x+1) is also true
by the principle of mathmetical induction, f(x) is true for all positive real numbers
therefore f(infinity) is true, 1^inf = 1
I have a doubt, can we actually consider infinity as a "real number"?
@@giovishow depends?
@@Taokyle on what?
Infinity is not a number kiddo 🤡
this is not aleph bro maybe you're not a kiddo but you're definitely not educated well
It is only not defined as such when the result of a limit where both the exponent approaches infinity and the base approaches 1 at different rates whilst taking on different real values
I love your channel, but I have to disagree with this proof.
In reality, lim(x->inf) 1^x = 1, and I'll explain why.
First off, everyone needs to understand that infinity is not a real number. That flipped up eight is an expression that only makes sense in calculus, which expresses a hypothetical number that WON'T STOP going up. It's not a plain number, but a symbol used in calculus to explain the behavior of functions as x goes to higher and higher positive numbers or lower and lower negative numbers.
Thus, in reality, infinity does not make sense in arithmetics. It can only make sense in LIMITS, which is why this video is already flawed.
However, even if we put it as a limit (as I did in the beginning of my comment), it still computes as one. This is because 1^a = 1 regardless of the number used.
This misconception comes thanks to two situations in limits that give you different numbers, but are completely different :
- First, the 1^inf indeterminate, which computes to e :
e = lim(x->inf) (1+1/x)^x
This can look similar, but notice that the base of the exponent is not one, but
1+1/x
This completely changes matters, since the base is approaching one and therefore computing erratic results in the process. Those results do not stop at 1, but rather e, since the base changing and therefore not one.
- Second, the proof used in this video, inf/inf.
This is another indeterminate form, but is not undefined. When solving this indetermination, it simply boils down to knowing which function grows faster, either the numerator or the denominator.
Inf is a vague term, since it expresses a growing number, but not how fast it grows, and that will depend on the function we are working with. To find out, we can simply look at the functions and compare them on the "order of infinity", or we can use derivatives, but that's a step further in the calculus world.
If the numerator grows faster, then it's a faster growing number divided by a slower growing number, so it's infinity.
If the denominator grows faster, then it's the other way around, so it ends up computing zero.
And, if numerator and denominator grow at the same speed, which is the case when they are both the same function, they end up as one.
That's the case used in the video, which is wrongly crossed out as undefined, but really, if you got two functions that are exactly the same, no matter how big those numbers are, it's going to be one.
Again, let me highlight that infinity/infinity is a vague expression, not a number, and that it can compute different things depending on the case.
Justified👏👏
雖然∞∕∞沒定義,但lim x->∞ (n^x/n^x) 還是等於1的
It is right but let consider: a(n)=n (for example), than we have: b(n) = 1^a(n)=1 for any natural n. By definition of limit that will be: Limit b(n) = 1 while n -> + infinity.
Don't upload garbage on the media and confuse people. If the base is exact 1, 1 to any power is 1. You must make it crystal clear that the base is approaching 1 (1 in the limit sense) and yet it may be defined or undefined. Limit of floor(x+1)^(1/x) as x goes to 0 from above is 1. Limit of (1+1/x)^x as x goes to infinity is e. I'd recommend that you take off this bs.
Leíste los procedimientos de límites al revés: primero se simplifica la expresión y luego se aplica el límite.
Lim n-> inf of (x/x)^n with x dif 0
Lim n-> inf of 1^n
1^(inf)
1
Undefined doesnt means that the limit doesnt exists. It means that you are not calculating it in a correct way. Base is exactly 1, then the limit is 1.
Of course it's undefined,but still the limit of 1 to the power of x as x approaches infinity is still equal to 1 so you're basically onto nothing
Another proof:
If 1^∞ = 1, ln(1^∞) = ln(1) = 0
But the properties of logs says : ln(x^y) = y*ln(x)
So,
ln(1^∞) = ∞*ln(1) = ∞*0 wich is indefined
One of the few comments I actually understood 😭
you need to prove that ∞×0 is undefined
Or you can write it as ln1/1/infinity so 1/infinity is 0 so ln1/1/infinity=0/0=indeterminate
You mentioned the log property but failed to state the arguments it works for.
@@sherinah4812 that's too bad, because it's false. You can't treat the inifinity symbol as a number (the same mistake is being made in the video itself)
Indeterminate*. And btw if the base is exactly 1, then 1 to the infinity is 1, otherwise it's either 0 or infinity whether the base goes to 1 from the left or the right. This is the case if we decide to make sense of what you wrote, because infinity is not a number, therefore you cant use it like that
i agree but also it can be any number like the definition of euler number e
Sir I love your videos so much but I have a request from you if you are free then please 🙏 make video's on math's chapters of IIT JEE (it's a competitive exam in india) Please help Sir 🙏🙏🙏🙏😢😢😢
I have received many similar requests. I have no material to work with. You should email me something to study or problem to solve.
that’s wrong.
2^x/2^y is not reduced to x/y but to 2^(x-y) and if x=y then this is 2^0=1 as required.
However, infinity is not a number and so any arithmetic with it is undefined
What part of the video claimed 2^x/2^y reduces to x/y? I believe that's the whole point of your comment.
what app are you using?
Samsung notes
Use L'Hopital's Rule: Take derivative if you get infinity over infinity.
In this video, you just can't break 'infinity' like a number like you did (2/2)^(ꝏ) = (2^ꝏ)/(2^ꝏ) this is not possible algebraically.
Because if it is then (2^(ꝏ))/(2^(ꝏ)) could be
(2^ꝏ)/(2^ꝏ) = (1^ꝏ + 1^ꝏ)/(1^ꝏ+1^ꝏ)
it will became like that if u treat infinity(ꝏ) as a algebraic number. So this video is just wrong way to put it and, like me i would a bit agree that 1^ꝏ=1 by some limits way.
Except 2^x =/= 1^x + 1^x for any value of x other than 1.
Beautiful
I don't think that (2/2)^inf = 2^inf/2^inf, it is right under the condition when the exponent is finite , I think
You're correct :)
It's alright if raising to the power of ∞ is a valid operation. The initial assumption of the video is that raising to the power of ∞ is a valid operation, as he takes 1 to the power of ∞.
@@xinpingdonohoe3978 that's a far stretch you're making 😄 but i see your point. I'd still rather the video would work with limits here. The infinity sign isn't even a number and if it were, what would, for example, inf - inf be? Lots and lots of problems incoming... 😄
@patrickfrei9322 > what would, for example, inf - inf be?
Michael Penn says (∞ + ∞) and (∞ − ∞) is a perp (⊥) - one more "number" :)
Assume lim x approaches inf in (x+3/x+2)^x this gives us 1, you can calculate, without any operation of limit it's already impossible to write infinity as power of a number but in this case if we really take 1 as base it will give us 1 there is no wrong.
IF YOU LOVE MATH,
🟨 MUST READ:
I'll say it clearly: stop messing with the term infinity as if it's a number!!
All the problems, and all the unnecessary philosophies happen from one mistake that everyone makes: analyzing a problem with infinity as if it were a number.
enough with that!!!
The concept of infinity is not a number on the number's axis, infinity is an "idea" not a point on the axis.
You can ask how much is 3x+3x and the answer will be 6x, but you can't ask how much is infinity plus infinity, the answer is not two infinity. Because it's not a number.
I can say: I went 50 kilometers forward and went back 42 kilometers, therefore I am 8 kilometers from the starting point.
But I can't say I walked an infinite number of miles (what is that???) and returned three-quarters of an infinity. Does that mean I'm a quarter of an infinity away from the start??
Is it ridiculous enough to understand what I'm saying?
Well, no need to ask how much is 1 to the power of infinity.
unnecessary, meaningless, and a little annoying.
Tamir Erez
Math teacher.
Well, we must define what '1' is before we carry on to other complex problems like 1+1= 2 or not.
1^♾️=♾️ for large values of 1.
no..
😂😂😂
Dude… Read your "proof" backwards and it will be the exact way how to eliminate that undefined infinity over infinity expression…🤦
Sir, infinity divide by infinity =1
We can't just treat 1^∞ as a "number". It is not even defined. If you want to give it some sense then you have to define it. Say for example 1^∞=lim_n 1^n, in which case for sure it is 1. But as I said before it depends on your definition, for example if you define 1^∞ to be lim_n->∞ lim_x->1 x^n then it even depends on the order of the limit if we do first n then we get 1 again, if we do x first then it doesn't exist as for 0
@danielmunozgeorge2876 > What fais in your proof? Well you can't divide limits unless both exist, basic property of limits.
Yeah, that "surprising" distribution of the exponent over the numerator and denominator was quite unexpected, even shocking to me. I guess Prime should understand such basic things. Unless it was some kind of a bad math joke and spotting the inaccuracy was left as an exercise to the reader.
@@allozovsky yeah, I bet Prime knows that, but of course these videos are not meant to math students but rather any audence. It is always easier to catch the audience with these kind of "surprising" math.
@@danielmunozgeorge2876 That was a cruel joke, though.
Bro has the LOTR handwriting
If you use this proof, you prove also that any other fraction is infinite, as infinite power of any number is infinite. That way it is just a misconception, not a proof.
Good thing you wrote "proof".
Sorry, I'm not following you. What do you mean.
The point here is to prove that 1^infinity is not 1. We are proving this because 1 to the power any real no. Is 1. But 1 to the power infinity is not 1. That's all that's done.
No, exact one to the power infinity is alllllllllllllllways infinity
How is infinity/infinity undefined ?
Indeterminate not undefined
Right, infinity/infinity is indeterminate, not undefined. And you actually end up with infinity/infinity fairly frequently in calculus! Sometimes you can't determine what that ratio should turn out to be... other times you can. That's why it's a "proof" we've been given, not a proof.
@@PrimeNewtons how is it indeterminate? I’m assuming by infinity here you mean a very large real number (Because you say it is not undefined)
You cannot raise a number to a representation. Infinity is not the same as the let say a VARIABLE "x"
Its an energy
Well, even i could've told you that 1^inf ≠ 1 because they're obviously in different forms. One is an integer with an exponent while the other is just an integer; therefore, they are not the same and not exactly equal since a different rule is being applied for the infinite exponential series. Even though, it can definitely be defined using logical rationale and the essential coherency of mathematical laws. We can easily extrapolate that 1^inf can never provide any other output besides 1 because any number selected from 1 -> inf will always be 1 since infinity in itself is 1 system, as well as all numbers themselves being the result of various sizes (amounts) of 1. Oh yeah, AND 1^n = 1!
Of course its undefined though, you cant raise to the power of infinity because infinity isnt a number..
If you choose to pretend that infinity is a number then the answer would be 1, because assuming it is a number, infinity over infinity would be 1 (but it's not a number)
It’s not true because the line where (2/2)^infty=2^infty/2^infty is not true. We don’t have this relation for when exponent is infinity
That's probably why it was a "proof" but not a proof.
The real intention was to say that it is 1 indeed, can it be so?
Just checked in Python - it say that
print(1**float('inf'))
print(1**(1/0))
print(1**(0/0))
is
1.0
1.0
1.0
That’s wrong!!!
If 1= a/a, then (a/a)^b, the operations’s order is:
1st parentheses
2nd exponential
Or 2 to the power of inf minus inf which still undef :)
this is so wrong on multiple levels. First of all, inf is not a real number, so essentially 1^inf has to be defined using limit and in that case, 1^inf is actually 1.
Second, in the proof, you are so freely using inf like it's a real number and using various operations and identities that are valid only for real/complex numbers, but not for inf.
And if you define every step in the proof using the concept of limit, then proof becomes invalid by default because it would come out to be 1, which is contradictory to what you are claiming
If 1^(infinity) is lim 1^x when x to infinity, then it is 1
Sure, what possibly else can it be.
Is it not a mistake to say 1^inf = 2^inf x 2^-inf because you're saying that 1^inf x 2^inf = 2^inf, introducing an infinity where there may not have been one because you evaluate the infinity twice? 0 = 1 x 0. 1 = 0/0 = undefined. Just introduced an infinity to say that 1 is undefined. It may be undefined but I don't think this shows it.
That is the meaning of undefined. You can make it whatever you want.
Just generalize this idea then everything to the infinity is just undefined.
if 2/2 = 1, then why inf/inf is not 1?
inf/inf is an indeterminate form
Both ♾️ have to be the same
The same reason 0/0 is not 1; they can represent different values depending on context, therefor without any context they are referred to as “indeterminate form.”
Because infinity is not a number it's a concept
@@strangex1524 while technically true I’d call this an oversimplification. After all, every number is a concept. There are number systems wherein infinity can be treated as a number with special properties.
not if they're countable infinities.
infinity divided by infinity in that problem equals 1
1= 1/1 = 2/2 = 3/3 = inf/inf
1 is undefined bro, u can't use it
1^∞ doesn't approach anything so it still 1
But 1 to the infinity is derfined . This is when u have somethint that converge to 1 at the power of something going to infinity
lim x->∞ (1+1/x)^x = e
lim x->∞ 1^x = 1
They don’t agree, and this is enough to conclude that 1^∞ is an indeterminate form of
0:24 I have confused even in that, why (2/2) is to the power of infinite, if 1 is to the power of infinite is not equal to 1, 😂😂😂😂 , (1)^infinite =!= 1
No 1^infinity is equal to 1
But (1+)^infinity is not
Well it IS equal to 1 if we talking exactly 1 and not the limit of 1
Can’t divide by 0
2 power infinity-infinity then take log on both sides. still infinuty so double cheked
Infinity isn't a number, is this a bad attempt at humour?
0.1^∞/0.1^∞
This is one, to any number even if INFINITY DIVIDED BY INFINITY IS 1
I AM SURE THAHT IT IS 1, IT IS NOT UNDERFINED. 1 to infinity is (2/2) to infinity = 2 to inf/ 2 to inf = inf / inf is 1
WRONG proof
just no.
This is wrong. It is equal to 1.
this is too obvious cause 1^∞ is not mathematically defined expression
im sorry, you don't understand the symbols you are writing.
I did not agree with this sol.but i mean the number divided by itself always 1 exept 0
Just another nonsense math video. But I enjoyed the music.
A wise man once said, "One man's sense is another man's nonsense".
@@PrimeNewtonsOk, my comment was not constructive, sorry for that. And you are contributing making „the boring math“ subject more attractive to a wider audience, which is great!
However, in this video, you are doing calculus with notations that have no (canonical) definition. Therefore, the rules that you are applying are also without any meaning in this context.
I understand. I am planning a video on this topic in the near future.
W😮hy? Is this?
Thats a contradiction in your proof, since you are using inf symbol as a number leading to inf/inf not a number.
Since it is a contradiction lim (n- inf) 1^n = 1.
1 is not exactly 1 it may be like either 1.0000000000001 or 0.999999999999 1.000000000001power Infinity = also infinity or 0.99999999999 power Infinity = 0 . Hence answer is undefined😊
But 2&/2& #(2/2)& .thanks
I think inf/inf is (-inf;inf)
How can 2 to the power infinity is infinity? 2 is literally 4/2
Infinity is not a number. There's no such quantity as 1^infinity or any other quantity mentioned in this video. (Of course, there is a theory of transfinite ordinal and cardinal numbers in set theory, but this has nothing to do with the "infinity" symbol you see in this video or in a calculus course.)
To talk about infinite you need to talk about limits.
Then the argument shifts to how close to 1 are you when taking it to the infinity power. If it is exactly 1 then 1 to the infinite is actually equal to 1, otherwise you get shinanigens
it is undefined.
While 1^infinity is not equal to 1.
It is equal to 1 under the condition that you are dealing with EXACTLY one. And not 1 in the limit form.
shut up, ∞ is not even a number
Wrong. Wrong. Wrong. Learn limits
nonsense!
Boooooo
I love your videos but i expected better from you @PrimeNewtons. 😢
잘못된 증명.
각각이 수렴 하지 않을때는 극한은 분리할수 없다
This is one, to any number even if INFINITY DIVIDED BY INFINITY IS 1
I AM SURE THAHT IT IS 1, IT IS NOT UNDERFINED. 1 to infinity is (2/2) to infinity = 2 to inf/ 2 to inf = inf / inf is 1
WRONG proof