1+0+0+...=?
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- Опубліковано 31 тра 2024
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0:00 A wrong way to evaluate the limit of the Riemann zeta function
4:35 An example that 0+0+0+... doesn't approach 0 with Riemann sum
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#calculus #zetafunction
You will get “0” if you do this on a test.
Nope🙅 I'll get 0+
☹️☹️☹️🤭
no, it is 1.
except for the first one, every term is approaching to 0, and the number of terms is fixed, because this is the infinite series. And each term is continuous whatever the 's' is.
so, the answer is 1. (and when I calculated this by calculator, the result was 1.)
@@user-donghun04 No, every term approaching to 0 may result in any sum
@@user-donghun04 The result is indeed 1, but the method for computing it is incorrect. The correct way to prove it is 1 is by noting that the series converges uniformly on both N and s. Hence the limits can be interchanged.
@@user-donghun04 We know it's 1 but your argument is incomplete. Consider:
s(1) = 1 + 1 + 0 + 0 + 0 + 0 +...
s(2) = 1 + 0 + 1 + 0 + 0 + 0 +...
s(3) = 1 + 0 + 0 + 1 + 0 + 0 +...
...
Except for the first one, every term is approaching to 0, and the number of terms is fixed, because this is the infinite series. so, the answer is 1.
Except it's actually 2.
It gets serious when this guy breaks out the blue pen.
Wow, I've never seen that joke before.
Order of seriousness:
black -> red -> blue -> green -> purple ?
@@WerewolfLord I’ve never seen purple . . .
@@Camp_RB it's very rare but I think I've seen it a couple times.
@@Camp_RB i think ive seen it in some of his calc vids
I love how he solves hard math while holding a pokeball lol
@@Gammalol it looks like just a plushie. I remember him using a sphere-shaped mic, but now he might have gotten used to holding something while recording, and the pokeball was the first thing in mind
@@igormatheus8698 It's the mic. You can clearly see the cord
The only animé with an Erdős Number!
@@igormatheus8698 you can see the wire
your feelings are irrational
The way he uses different colors itself gives you the clue that something is wrong with those zeroes...
That definition of e blew my mind. Intuitively the base is either 1 or slightly greater than 1. 1 raised to any power is 1 but anything greater than 1 raised to an infinite power should be infinity, so the fact that it somehow arrives at e, which is a relatively small number is crazy to me. Is there another video that goes more in depth about that? Edit. I just tested with wolfram alpha and (1+1/(10^10))^10^10 is a good approximation for e. This is incredible.
It can be expanded (carefully) to give the Taylor series expansion of e^x where x=1
Here's a nice explanation of where e comes from, by Eddie Woo: ua-cam.com/video/pg827uDPFqA/v-deo.html
But you never are raising it to an infinite power. You’re raising nearly 1 to a massive power. There’s a difference.
@@AngadSingh-bv7vn good luck with the tailor series.
The base is never 1. The base is 1 + 1/n, and 1/n is never 0. Also, the exponent is not infinite. The exponent is n. The exponent diverges to infinity, but is never infinite.
For those who want to know, inverting limit and series in this case is actually possible given the correct justifications. Considering a series of functions fn defined of (a,b) a or b can be respectively -inf and +inf, if the series of fn converges uniformly on (a,b) and all functions fn have a limit « ln » at b (for example) then the numerical series of ln converges and is equal to the limit as x approaches b of the numerical series of fn(x). In other terms you can invert the series and the limit given a uniform convergence. Zeta is the sum of the series of fn where fn(x) = 1/n^x which does converge uniformly on [2,inf). Since f1(x) converges to 1 and for all n>1 fn(x) converges to 0, we can use the theorem previously stated. Warning!!! the reciprocal is not necessarily true (for example zeta is defined on (1,inf) but the series of fn does not converge uniformly on this interval, eventhough the limit of the series and the series of the limit have the same value)
If you do this on a test, you'll be wrong, but you'll get 1/12 of a bonus mark.
Not -1/12?
@@blackpenredpen It's wrong. So you get docked -1/12 of a mark.That works out in your favour, doesn't it? I think. Ramanujan, where are you when I need you.
@@blackpenredpen well, you are great but not ramanujan
If he lost -1/12 marks that means he got a net +ve
Hey, blackpenredpen,
You raised a point that is very critical, and some people don't seem to get right!
WHEN the real 0 (and not the 0 as a limit of a function) and the real 1 (and not the 1 as a limit of a function) are involved, many people seem to confuse them with the limits, and throw the "Indeterminate form" just like that.
I really hope you make a video explaining the difference once and for all: 1 is not 1+, and 0 is not 0+. I also hope you give some examples.
I think it will help many people get that once and for all.
Have a wonderful day!
A good point.
I do somewhat hope the explanation will avoid appealing to indeterminate forms and will instead the idea of these limits on its own terms.
At the risk of no one reading, I guess I will lead by example and provide a preliminary explanation of my own.
In calculus, there exists the so-called algebraic limit theorem. This is the limit theorem that states that if lim f (x -> p) and lim g (x -> p) both exist and are equal to L and M, respectively, then lim [f + g] (x -> p) = [lim f (x -> p)] + [lim g (x -> p)], and lim [f·g] (x -> p) = [lim f (x -> p)]·[lim g (x -> p)]. This theorem is awesome, because in practice, it is the theorem that allows us to compute the limit of any function at any point symbolically, without needing to use the ε-δ definition of the limit of a function every time. However, the algebraic limit theorem does not hold if lim |f| (x -> p) = ♾ or lim |g| (x -> p) = ♾, because then the limit does not exist for f or g. lim |f| (x -> p) = ♾ is just shorthand for lim 1/|f| (x -> p) = 0. It does not actually mean that lim |f| (x -> p) is literally equal to ♾, it just means that lim |f| (x -> p) does not exist in a special way.
Nonetheless, even if the algebraic limit theorem does not hold, it is still possible to symbolically compute several limits. For example, consider lim 0(x)·1/x (x -> 0). 0(x) is the constant function with output 0, so lim 0(x) (x -> 0) = 0, but lim 1/x (x -> 0) does not exist, so it is nonsensical to say that lim 0(x)·1/x (x -> 0) = [lim 0(x) (x -> 0)]·[lim 1/x (x -> 0)]. You cannot multiply by something that does not exist. Yet, lim 0(x)·1/x (x -> 0) can still be computed. Please remember that 0·y = 0 for any real number y. 1/x is a real number, so 0(x)·1/x = 0(x) for any nonzero x. Therefore, lim 0(x)·1/x (x -> 0) = lim 0(x) (x -> 0) = 0. Notice that while the product [lim 0(x) (x -> 0)]·[lim 1/x (x -> 0)] is utter nonsense, lim 0(x)·1/x (x -> 0) = 0. Notwithstanding, if I replace 0(x) by x^1, I get that lim x^1·1/x (x -> 0) = 1, even though lim x^1 (x -> 0) = lim 0(x) (x -> 0) = 0. If lim x^1 (x -> 0) = 0, then evaluating lim x^1·1/x (x -> 0) is like multiplying by 0, and yet we are not getting 0.... except, this is false. Multiplying by x^1 and then letting x -> 0 is not the same as multiplying by lim x^1 (x -> 0), since the algebraic limit theorem does not hold. So multiplying by x^1 and then letting x -> 0 is not the same as multiplying by 0. This is why it makes perfect sense for lim x^1·1/x (x -> 0) to not be equal to 0: you are not actually multiplying by 0 at any point.
That being said, teachers have the misleading and inaccurate habit of noting that lim x^1 (x -> 0) = 0 and lim 1/|x| (x -> 0) = ♾, hence lim x^1·1/x (x -> 0) is an indeterminate form of the type 0·♾. The problem with doing this is that it pretends the algebraic limit theorem holds in this situation, even though it is not. One is never actually multiplying by 0, so writing this as 0·♾ is misleading. Similarly, ♾ is not a real number, and one is never actually multiplying by ♾: x^1·1/x is always a real number for nonzero x, and 1/x is always real for every nonzero x. So calling this an indeterminate form of type 0·♾ is inaccurate. What would be accurate is to note that the algebraic limit theorem does not hold, but that the limit may still exist, and that it may be easily computable symbolically.
This occurs similarly for lim f^g (x -> p). Remember that if lim ln(f)·g (x -> p) exists, then lim exp[ln(f)·g] (x -> p) exp[lim ln(f)·g (x -> p)]. If lim g (x -> p) = ♾ and lim ln(f) (x -> p) = 1, then we have the sitiation with 1^♾. Many teachers will claim this limit is an indeterminate form of type 1^♾. This is inaccurate, though: ln(f) need not equal 1, and since lim g (x -> p) does not exist, the algebraic limit theorem does not hold, hence cannot say exp[lim ln(f)·g (x -> p)] = exp{[lim ln(f) (x -> p)]·[lim g (x -> p)]} = exp{ln[lim f (x -> p)]·♾} = exp[ln(1)·♾] = 1^♾. Exponentiating g with base f and then letting x -> p is not the same as exponentiating lim g (x -> p) with base lim f (x -> p). Hence you are never exponentiating by base 1. This is clear in the example with lim (1 + 1/n)^n (n -> ♾) = e. Yes, lim (1 + 1/n) (n -> ♾) = 1, but 1 + 1/n is never equal to 1. Similarly, n is never equal to ♾. n is a natural number, so it always finite.
A similar analysis works for every indeterminate form, and what it demonstrates is that the indeterminate forms are misleading, which is to say, that whenever a teacher says that a limit is of the form 0/0, you are not actually dividing by 0. You are letting a function f with lim f (x -> p) = 0 be divided by g with lim g (x -> p) = 0, and then let x -> p, which is different than dividing lim f (x -> p) by lim g (x -> p). Only the latter constitutes division by 0.
I understand calculus is hard. But it is important that we understand limits and limit theorems on their own terms, and that we stop trying to reduce them to simple "arithmetic" involving the symbol ♾. You cannot do arithmetic with ♾ in the first place. Also, it is incorrect to talk about 0·g when you are actually multiplying by f, and then evaluating the limit, rather than multiplying by 0.
1+dx = 1+ ?
@@POPO-od8jb 1+ could an infinitely small bit (like dx) bigger than 1 if you want, such that it has properties like 1 but doesn't explicitly give indeterminate forms in situations such as 1/ln(x).
You can probably define:
n+ = inf(n,∞) and
n- = sup(-∞,n)
if you so feel it needs a definition.
The main issue with the Riemann zeta function limit is that a series is the limit of a sum as the number of terms increase. So when you're expanding it like 1+0+0+... you first assume that you can interchange these limits freely and evaluate the sum with the outer limit first before you take the limit of it as the number of terms approach infinity. In general this cannot be done freely, and requires some special conditions on the limits to hold first.
I wish that I understood your statement. Can you elaborate on "these"?
ong tho
breaking: wolf becomes chihuahua
@@jasonbroadway8027 The infinite sum implicitly has a limit in it: an infinite sum of f(n) from n=1 to infinity is technically the limit as N->infinity of the sum of f(n) from n=1 to N. So here you have lim_{s->infty}[lim_{N->infty} sum_{n=1}^N 1/n^s]
This is not necessarily the same as lim_{N->infty}[lim_{s->infty} sum_{n=1}^N 1/n^s] - you can't just swap the order of limits without further justification.
@@jasonbroadway8027 let me give it a shot.
When a sum goes from 1 to infinity, it's shorthand for the limit of that sum from 1 to n as n goes to infinity. So you can take the sum from 1 to 1 and write down the answer. Then write down the sum from 1 to 2, then from 1 to 3, and so on. Youll end up with an infinite sequence. The limit of that sequence is equal to the infinite sum you stated with.
This mean that the limit in the video was really equal to this:
Limit as s goes to infinity of (limit as m goes to infinity of (the sum from 1 to m of ())).
The thing that went wrong in this video is that he swapped the order of the limits. He took each term of the sum and sent s to infinity before he sent m to infinity. That's the wrong order! You need to take the limit in m first and then do the limit in s. This is kinda like the difference between f(g(x)) and g(f(x)) if where f and g are functions.
Now, there are some conditions where you can swap the order of the limits. That works when the function you're summing meets a certain "nice" criteria. The expression 1/n^s as a function of n is only "nice" for a few values of s -- certainly not while going to infinity.
When doing it wrongly you are switching the order of the limits which cannot be done generally.
However, we can switch the order of the limits if we note that the terms don't just approach 0 but decrease monotonically to 0.
Exactly what I was thinking. By switching the limits, the "wrong" approach is actually right. I don't think he ecplained it that well.
Ahh u guys probably don't know the rules, he is doing correctly
@@Calculus_is_power You can't always interchange integrals and limits. One counterexample is the sequence of functions
f_n(x) = n if 0 oo is identically 0, so the integral of the limit is 0. You can't interchange the integral and the limit in this case.
@@martinepstein9826 ooh yea thanks for clearing my misconcept
This is a fascinating video! Videos like yours inspire me to share my own maths content!
Use Taylor's remainder theorem to show that you get a epsilon-good approximation to zeta(s) by truncating the sum after the first N = epsilon^{-1/s} terms. So you have that many terms, and each is only as large as 2^{-s}. So the sum from n = 2 to N is at most N 2^{-s} = 1/(2^s eps^{1/s}) , and the limit s to infinity of this is zero.
The explanation with Riemann sum is brilliant!
Hello, Bprp. Thank you for the inspiring video of your teaching Math, especially the advanced Math. Speaking concerning Riemann sum, can you make a video showing us that how can the Riemann sum be the same as integral? (Even though I learned integrals a lot, I still cannot find the reason to this)
Now he made a double pun.
Both are 0 when the calculation ends and as a final grade for the exam.
The limit of a series is not necessarily the series of the limit.
How do I rotate a region around a like other than one parallel to the x or y axis? I’ve been working on it, and I think I’m pretty close, but I’m stuck. The region I’m working with is the area between y=x and y=(x-2)^2, and I’m trying to rotate around the line y=x. It has been eating away at my brain, please help.
I don't know what you are talking about, but it is fun to hear your maths while eating lunch.
I thought you will go for lim n->inf of sum from n to infinity of 1/n.
Although, your riemann sum of square root of x may be nicer 😊
I have a question sir, why the second sum at 4:35 isnt equat to infinite ? Like you have a sum of an infinity numbers who are not equal to 0 smth like 0.001*infinity right ?
Blackpenredpen... you are the one that keeps the original "Scream".... 😃
Thanks a lot. Now I have to resit my test.
Damn, no one has explained Riemann sums so well to me before
This is going to be awesome
Honestly, I watched this because of the pokéball. Worth it.
🤣🤣🤣
What is the answer
Is the solution to
(x + 1/x)^x = e
Infinity?
I like these long ones
1:26 he can't avoid a smirk on his face at the plot twist... :-)
From your 2/3 limit example we can have anice estimation for sqrt1 +sqrt2 +..sqrtn =2/3 n sqrtn when n is big
I have quite an interesting question for you: find the solution(s) of √y = y - 20. Here's a tricky part to this question: you'll arrive at √y = -4, can there be a complex solution to this equation?
If you use the standard definition of the radical symbol, then no, there cannot be a complex solution to sqrt(y) = -4, because by definition, Re[sqrt(y)] >= 0.
sqrt(y) = y - 20 y - sqrt(y) - 20 = 0 [sqrt(y)]^2 - sqrt(y) - 20 = 0 [sqrt(y) + 4]·[sqrt(y) - 5] = 0 sqrt(y) - 5 = 0 sqrt(y) = 5 y = 25.
sqrt(y) = -4 doesn't come from a complex solution, it comes from y = 16 and picking a non-standard square root function (which is admittedly usually conceptually tied to complex numbers)
Why can't y be 25
Woww thats actualy True sqrt25=25-20 can i learn how you did it
Without watching all of it, I'd say it equals to 1+ (converging towards 1 from above)
Well not so fast. 1=1^-s=>exp(ln(1)) =exp(-s(ln(1)) what is the limit s to infinity?
How many zeros are trivial?
Me : "That's definitely 1"
Me after 5 minutes : "oh."
Actually, it is still 1. BPRP is s both right and wrong here.
It was really great mathematician
You shell not pass
Where infinity is assumed a quantity versus a symbol to a function, you get errors.
Love it
I want to ask a question, can 0+ be understood as the next real number of 0?
if 0+ were a real number not equal to 0 then 0+ / 2 would have to be inbetween 0 and 0+
I wonder what could be sqrt(2) ⇈ sqrt(2)
@Siddhartha Chaganti He creator and ruler of the universe and source of all moral authority; the supreme being and uncreated doesnt depend on his creation or its law like death ,time ,maths etc..
why you need to talk about religion and god in a math video?
@@timetraveller2818 i want to know does mathematical geniuses believe in god and how
If the function series converges uniformly then the limits can be inverted!
Yes. Finally someone who knows.
You are indeed worthy of using the username Lelouch
Right, good point. And in this case, the series does converge uniformly on, say, [2,\infty), by the M test, with Mn=1/n^2. So we can switch the two limits
実におもしろいです😊
Cool vid
I thought the same idea One day But I didn't knew how to calculate it...... 🙁🙁
In a similar vein, d/dx [x^2] = d/dx [x + x + ... + x] = 1 + 1 + ... + 1 = x, what's wrong with this? ;)
x^2 is not equal to x + x + ••• + x. This is only true for natural numbers x, but the natural numbers correspond to isolated points of the topology of the real numbers, so it is nonsensical to talk about differentiability here.
No, the integral is bigger than the serie. Its known as integral criteria for abel series
Usually it cannot switch the order for the limit and series sum😉
Isn't it true, however, that the terms get infinitely smaller each n? The 0+ses aren't equal, they get smaller every term, so even though there are infinitrly many of them, they do seem to get infinitely smaller, which makes me wonder if the answer wouldn't end up being 1 after all (or rather, limited to 1).
I think you have the right idea. The sum does not collapse to zero, but neither does it blow up to infinity. Instead, it sums up to some finite value like 2/3, 1 or e.
The limit is indeed 1. However, BPRP is still correct in that the method used is invalid, although his explanation for why the method is invalid is incorrect. See my comment to the video for more details.
Why this problem is a little tricky is because there is a tug of war between the terms that are getting smaller and the fact that there is infinitely many of them. Your « explanation » could be reversed the other way around : You could say that eventhough the terms go to zero, they are still bigger than zero, there are infinitely many of them, so the series should not even converge for any s and diverge to infinity.
The key for solving these problems is to know how quickly does the terms of the series converge to zero in the neighbourhood of infinity.
@@CrazyPlayerFR That is not an accurate explanation either. There are not "infinitely many " terms. In fact, indeterminate forms have absolutely no relevance in this question. The problem here is that of uniform convergence: we intuitively assume we can just switch the order of the limits and call it a day, but we actually cannot do this in general (though in this particular example, we actually can, making this video a terrible example to illustrate the subject, *sigh*).
@@angelmendez-rivera351 yes i am aware of this, I just posted a comment explaining how we could have done this formally using uniform convergence. I wasn’t giving any « proof » in my comment, just trying to explain why using the same arguements that one could use with a finite amount of terms which all have a limit can not be generalised so easily to an infinite amount of terms. I might have done it wrong though, dont’t know! :).
Lesson: for large s, the zeta function converges, by not *necessarily* uniformly. Therefore, we cannot necessarily interchange the sum and the limit.
Actually, it does converge uniformly. This video is inaccurate. lim ζ(s) (s -> +♾) = 1 is indeed true. You can look this up. The only part about this video that is accurate is the conclusion that the method utilized was invalid (because uniform convergence had not been established), but the explanation behind the conclusion was wrong too.
@@angelmendez-rivera351 I added the word "necessarily" now so if the convergence is truly uniform, I won't contradict this truth.
@@angelmendez-rivera351 It is inaccurate to say that ζ converges uniformly, because it doesn't converge uniformly on its domain, (1,infinity).
But it does converge uniformly on [a,infinity) for any a>1, which is enough to justify switching the limits.
@@kyleneilson1457 I see. Now that you commented that, I know how to prove the uniform convergence.
I respectfully disagree with the explanation given in the video. I think appealing to the concept of indeterminate forms is incredibly misleading, as it has nothing to do with why the method being discussed for computing lim ζ(s) (s -> +♾) is invalid. There is something legitimately deeper going on here that does deserve to be mentioned.
Notice that ζ(s) = lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾). Hence lim ζ(s) (s -> +♾) = lim lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾) (s -> +♾). However, in the video, the method was to _first_ let s -> +♾, then evaluate the series, which is equivalent to computing lim lim 1/1^s + 1/2^s + ••• + 1/x^s (s -> +♾) (x -> +♾) instead. Notice how the order of the limits is different. You cannot switch the order of the limits without some proper justification, since, in general, lim lim f(x, s) (x -> +♾) (s -> +♾) is not equal to lim lim f(x, s) (s -> +♾) (s -> +♾). That is why the method is invalid.
That being said, I think this is a bad example to choose to prove that point, because in this case, it _does_ turn out that the two quantities are the same, and indeed, lim ζ(s) (s -> +♾) = 1 is true.
this guys is smart
"1^infinity is 1" No it's undefined because multiplying by 1 is the same than adding 0 so 1*1*1*1*1*...=0+0+0+0+0+...=0infinity=undefined or 1^inf=e^(ln(1^inf))=e^(infinity * ln(1))=e^(infinity*0)
1^∞ is undefined
lim x→∞ 1^x is defined and =1
Lol
I still am not so sure the first trivial method is incorrect. The counter example you showed for the area of sqrt(x) has the number of the terms increasing with the value of n. When the number of terms increases, I can clearly see the indeterminate form 0 times infinity, since at every step each term is getting smaller but there are more terms overall. For the Riemann series however, the number of terms is always constant and, even though it is an infinite amount, all of them are going to 0 at the same time, without adding more terms to the sequence, and can also get arbitrarily close to 0. I guess the only way I can clearly see things is if I go through the epsilon-delta definition, maybe it's a problem when interchanging the limits.
The problem is indeed when interchanging limits. lim lim f(n, m) (n -> ♾) (m -> ♾) is not necessarily the same as lim lim f(n, m) (m -> ♾) (n -> ♾). There is a theorem, called the Moore-Osgood theorem, that indicates the conditions under which the above two are equal.
Of course there are important differences between the zeta series and those Riemann sums, but since the first method doesn't cite any of those differences it can't be correct.
It doesn't obviously matter that each Riemann sum is a finite sum, since we can always just rewrite it with an infinite tail of zeros or replace the last term T with T/2 + T/4 + T/8 +... However, no matter how we do this we will never be able to say that for all n the n'th term monotonically decreases from one sum to the next. This turns out to be the key difference.
Interchanging limits can only be done under certain conditions.
Yeah, when function is continuous 🙄🙄, right 🙄🙄
I dropped out of calc because I personally hated it. And I still watch your videos for some reason. And actually learn the solutions
You know it's serious when BLACKpenREDpen uses a blue pen
I liked the video be cause i liked it
Dude, I gotta say... You have a most epic beard, and I would love to be in your tute!
I mean you have a finite numbers divided by infinity, but you have infinitely many of them - it kinda makes sense the infinities ”cancel each other”
1^inf is undefined or am i wrong
1^♾ is undefined because ♾ is undefined. ♾ does not denote a mathematical object, and it certainly does not denote an element of an algebraic structure, so talking about the symbol alongside +, ·, and ^ is nonsensical.
That being said, for every nonnegative real number L, there exist functions f and g with lim f (x -> p) = 1 and lim 1/|g| (x -> p) = 0, such that lim f^g (x -> p) = L. Hence, the properties lim f (x -> p) = 1 and lim 1/|g| (x -> p) = 0 alone do not allow you to make a conclusion about whether lim f^g (x -> p) exists or not, let alone what nonnegative real number it is equal to. This is why some people say 1^♾ is an indeterminate form. However, this conclusion is inaccurate. The correct conclusion is not that 1^♾ is an indeterminate form, but that the algebraic limit theorem does not hold, meaning that lim f^g (x -> p) does not equal [lim f (x -> p)]^[lim g (x -> p)]. This is because lim g (x -> p) does not exist.
@@angelmendez-rivera351 ty for your explanation 😊
5:05
Are you Singaporean?
The solution to the thumbnail and title "1+0+0+...=?" still is 1 - literal 0s are still zero.
Also i would have given a slightly simpler and more intuitive example:
Divide a cake to N people equally. If N goes to infinity everybody gets 1/N => 1/inf => ''0" of the cake, but we know that adding up all those "zeros" must give us back the whole cake.
i noticed that he was holding a pokeball a 6:20
Cool.
If that 0 is the actual zero that follows the definition of zero in IN or IR (neutral element with respect to addition and following the multiplication property that states that multiplying it by any number results in 0), then all these zeros are worth nothing, even if you do add them infinite times.
Obviously that is different from the fake 0 that everyone treats for a real zero (e.g., when 0 is the limit of some function of x as x approaches infinity).
So I am wondering what is the trick here?
Is it the same kind of trick that he used last time to say that the integral between -inf and +inf of x is undefined?
That's what I was wondering as well, not 100% sure
Well, I just want to clarify that that lim ζ(s) (s -> +♾) = 1 is indeed true. What BPRP is disputing is not the answer, but the method for getting to the answer. The issue here is that if you "plug in ♾" and then evaluate the series, you are actually computing lim lim 1/1^s + 1/2^s + ••• + 1/x^s (s -> +♾) (x -> +♾), which is not necessarily equal to lim lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾) (s -> +♾), which is what you _should_ be trying to compute, since lim ζ(s) (s -> +♾) := lim lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾) (s -> +♾). In this case, it just so happens that the two are equal, but switching the order of the limits actually requires proper justification, since in general, changing the order of the limits changes the value.
@@angelmendez-rivera351 well, I actually commented on the preview of the video as it was showing
If you say 1 + 0 + 0 + ... = 1, you'll get zero.
When I commented, the video was not released yet :)
But now that he premiered the video, it is obvious that he did exactly what I feared: he treats a limit that approaches 0 as zero, which is DIFFERENT from saying 1 + 0 + 0 + ... != 1.
So it is basically a clickbait :D
The 0 turned out to be NOT the zero that we know and love (at least I love it)!
@@mondherbouazizi4433 I agree with you. In my view, the concept of indeterminate forms is itself clickbait. People say that 0/0 is indeterminate, but this is not the case. 0/0 is undefined. ε/δ, for ε, δ very close to 0, is indeterminate. Similarly, they say 0^0 is indeterminate. It is not. 0^0 = 1. ε^δ, for ε, δ very close to 0, is indeterminate. The same can be said for every so-called "indeterminate form", and really, even the name "indeterminate form" is misleading, since these "forms" represent points in the domain of these functions where a discontinuity exists. This is why I dislike it when maths channels utilize the concept, and I dislike it even more when school teachers bring it up, though admittedly, this is mostly the curriculum's fault. Rather than intuitively illuminate the students, the concept just creates more confusion.
I am waiting for the day when I have the time, energy, and resources to start my own UA-cam channel, and dedicate myself to just bust the commonly believed mathematical myths learned within school systems, especially the ones associated with subjects within calculus. Calculus is probably the worst taught mathematics course that I know.
@@angelmendez-rivera351 I couldn't agree more.
You summarized exactly all my thoughts regarding this issue! I really really like the way you put it together.
And I hope you do create such a channel.. I will definitely subscribe. I actually like the "understand what's going on before diving in" approach.
Again, thanks for explaining things in such a nice way.
Have a great day!
Pls do integral of 1/(xe^x) from 1 to ∞ (I'm struggling lol🤦♂️)
That’s not possible with elementary function tho.
No wonder you're struggling, it's non-elementary
@@theimmux3034That's the point I was trying to do indefinite non elementary integrals
@@blackpenredpen Sir one youtuber solved Riemann hypothesis this is not a joke really this is link his video ua-cam.com/video/8y7fMmdAcb8/v-deo.html
is this joke video? it seems like joke but i cannot tell
Well but if you calculate the area of 1/n and so on you get closer and closer to the area of curve of the square root of x but you will NEVER get the exact value due to our "last" value for calculation being n/n or rather said infinity divided by infinity
I do not know what your point was, but there is no such a thing as "Infinity over Infinity". This is utter nonsense.
@@angelmendez-rivera351 its like what the ancient method for the calculation of pi was, you would get closer and closer to the real value but you can't get the exact value, n/n is non definable. So basically you could go up to infinity with it
@@romipog9337 I am not sure of that. lim n/n (n -> ♾) = 1.
@@angelmendez-rivera351 well im not very good at maths but i might add that you should first define n.
@@romipog9337 What? n is a natural number. This much was at least obvious from the video.
nice kirby plush
Isn’t it wrong to assume that infinitely many rectangles under the curve would equal the area under the curve? The same reason why a unit square’s perimeter doesn’t equal a unit circle’s circumference, because while it looks like a circle, it still has tiny parts which are out of the circle, and an infinite number of them.
Squeeze the orem
The entire video I was like,
„Why do I understand this? I’m not supposed to, I’m in high school, stop this madness“
所以原本那個作法過程是錯的，但結果是對的？
1/1^inf=1??? I forgot something
Wow the limit that’s 2/3 is cool
Ιt is 3/2.
What is your educational degree ?
I'm guessing a doctorate
black got red got 😄
why do u only do calculs u need to expand the topic
That's an alpha male beard right there
I’d like to respectfully disagree, the counterexample you provided , the number of terms constituting the sum were variable and dependent on the variable which is tending to infinity .In The Riemann Zeta example,irrespective of us taking any value of s(whether it be 1000,10000000 etc) the number of terms will stay the same(which is infinity).
an infinite sum is a limit of a partial sum. what you're actually looking at when you look at the riemann zeta limit, is a limit of a limit of a sum. in fact, the main issue of the "wrong" way to do it, is that you're switching the order of those two limits (which can be done sometimes, but not always).
basically, the numbers of terms is dependant on n in both examples.
@@tabeh- how will it be dependant on n(or s for first example)? the infinite sum ,which as you say can be defined as the limit of a partial sum, here let’s say the number of terms is b , why should b and n be dependent on each other?
@@manik1477 basically [lim(s->inf) Σ (from 1 to inf) 1/n^s] can be written as let's say [lim(s->inf) lim (N->inf) Σ (from 1 to N) 1/n^s]. hard to show this through text, but this is as clear as it gets.
So essentially you're taking a sum of N terms, then the limit of N -> inf and THEN you take the limit of s -> inf. There is a limit inside the sigma notation. The partial sum is not really dependant on n, but some other letter that I called N.
When he does the "wrong" way of calculating it, instead of
[lim(s->inf) lim (N->inf) Σ (from 1 to N) 1/n^s]
he calculates
[lim(N->inf) lim (s->inf) Σ (from 1 to N) 1/n^s]
that is, by using the s limit first he gets the sum 1+0+0+0 of N-1 zeroes, which he then continues to infinity by using the N limit.
@@tabeh- Oh okay, I get the picture now,Thank you so much for explaining this!
1
I thought the "wrong" thing was going to be something about Riemann being a complex-valued function and we didn't specify the real or imaginary part, but I also realized I have no idea what "complex infinity" would mean. 😅
Complex infinity, interesting 🤣🤣
What Pokémon is inside your Pokéball?
The only one to have an Erdős Number!
wait what,
2:28 HOWEVER
The ygo progression series has ruined that word for me
Can you prove 0!=1? 🥺
Bro gamma function proof 0! =1
Gamma(1) = 0! =1
well of course, 0 != 1 anyways
Well of course Since (n-1)! =n!/n
And if you put n=1 Then you will get
(1-1)!=1!/1
Therefore
0!=1
n! is the product of the n-tuple (1, 2, ..., n). Thus 0! is the product of the 0-tuple (). The product of () is 1. Thus 0! = 1. Q. E. D.
@@numberandfacts6174
Technically, your proof does not show that 0! = 1. Why?
Given that identity,
Gamma(alpha) = (alpha - 1)!.
If alpha = 1, then
Gamma(1) = (1 - 1)! = 0!
And my question is, why 0! = 1? 😁
I never heard his favorite pokemon
Approaching zero is not the same as here.... actually it will never be zero, that s why it is infinity
He didnt precise + or - infinity
Limits like this make everything so unintuitive 😂
Wait, if 1^+ and 1^- exist as valid numbers, I’d like to see a proof that 0.999 repeating is a strictly different number that 1^- please.
This is not a proof and is just informal intuition but may answer your question:
Say the notation for 0.9 with n repeating decimals is 0.9_n. So 0.9_1 is 9/10, 0.9_2 is 99/100, and so on. We will say that 1 minus this value is 10^-n which you can confirm for yourself by checking 0.9 + 0.1 = 1, 0.99 + 0.01 = 1, etc.
So we have that 1 - 0.9_n = 10^-n, or 1 = 0.9_n + 10^-n, and we will take the limit as n approaches infinity on both sides.
Lim n-->inf 1 is still exactly 1.
Lim n-->inf (0.9_n + 10^-n) = 0.999 repeating + 0.
You may feel inclined to argue that Lim n-->inf 10^-n is actually 0^+ as discussed in the video, but since it is not a term in an infinite sum, it equals exactly 0.
So, 0.999... equals exactly 1.
0+0+.......=0×infinite =indeterminate
Wrong. It is indeterminate when the 0 is not a 0, but rather a limit that approaches 0.
The definition of 0 says explicitly that it is the neutral element with respect to addition in IN or IR, thus no matter how many times you add it, the result won't change.
In other words, if your "zero" is the limit as x approaches infinity of 1/x, then yeah, that's indeterminate.
However, if it is the real 0 that cancels anything you multiply it by, and adds literally nothing (not even a tiny infinitesimal fraction), then your logic is wrong unfortunately.
There is something a bit deeper here. Not sure what trick Mr. Blackpenredpen hides, but it's not your simple transformation of addition into multiplication.
@@mondherbouazizi4433 Thank you for clarifying that. Appealing to indeterminate forms is so misleading here.
😕😒
Saun saun saun
BLUE PENNNN!!!!!!
WTFWTFWTFWTFWTFWTFWTFWTF
If 0 is not d, the answer will be 1. It would be e otherwise.
#GottaCatchEmAll' ^.^
Well... 0 is still better than -1/12.
conclusion : 1/∞ = 0+ but 1/∞ ≠ 0
1/∞=0 but sometimes 0+
No please after years of you helping me through my physics maths degree, I prefer you beardless...lol