1+0+0+...=?

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blackpenredpen

blackpenredpen

Күн бұрын

Пікірлер: 250
@blackpenredpen
@blackpenredpen 3 жыл бұрын
You will get “0” if you do this on a test.
@namantenguriya
@namantenguriya 3 жыл бұрын
Nope🙅 I'll get 0+ ☹️☹️☹️🤭
@user-donghun04
@user-donghun04 3 жыл бұрын
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.)
@moregirl4585
@moregirl4585 3 жыл бұрын
@@user-donghun04 No, every term approaching to 0 may result in any sum
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@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.
@martinepstein9826
@martinepstein9826 3 жыл бұрын
@@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.
@grady730
@grady730 3 жыл бұрын
It gets serious when this guy breaks out the blue pen.
@PubicGore
@PubicGore 3 жыл бұрын
Wow, I've never seen that joke before.
@WerewolfLord
@WerewolfLord 3 жыл бұрын
Order of seriousness: black -> red -> blue -> green -> purple ?
@Camp_RB
@Camp_RB 3 жыл бұрын
@@WerewolfLord I’ve never seen purple . . .
@mptness4389
@mptness4389 3 жыл бұрын
@@Camp_RB it's very rare but I think I've seen it a couple times.
@myfishcalledbobble6923
@myfishcalledbobble6923 6 ай бұрын
@@Camp_RB i think ive seen it in some of his calc vids
@APotatoWT
@APotatoWT 3 жыл бұрын
I love how he solves hard math while holding a pokeball lol
@igormatheus8698
@igormatheus8698 3 жыл бұрын
@@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
@lex224ification
@lex224ification 3 жыл бұрын
@@igormatheus8698 It's the mic. You can clearly see the cord
@Perririri
@Perririri 3 жыл бұрын
The only animé with an Erdős Number!
@pieTone
@pieTone 2 жыл бұрын
@@igormatheus8698 you can see the wire
@Fire_Axus
@Fire_Axus 11 ай бұрын
your feelings are irrational
@diedoktor
@diedoktor 3 жыл бұрын
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.
@AngadSingh-bv7vn
@AngadSingh-bv7vn 3 жыл бұрын
It can be expanded (carefully) to give the Taylor series expansion of e^x where x=1
@bjorneriksson2404
@bjorneriksson2404 3 жыл бұрын
Here's a nice explanation of where e comes from, by Eddie Woo: kzbin.info/www/bejne/ppibY2qrebV5p6M
@jessehammer123
@jessehammer123 3 жыл бұрын
But you never are raising it to an infinite power. You’re raising nearly 1 to a massive power. There’s a difference.
@moonshifter0
@moonshifter0 3 жыл бұрын
@@AngadSingh-bv7vn good luck with the tailor series.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@mayurchaudhari850
@mayurchaudhari850 3 жыл бұрын
The way he uses different colors itself gives you the clue that something is wrong with those zeroes...
@CrazyPlayerFR
@CrazyPlayerFR 3 жыл бұрын
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)
@viktorandersson4945
@viktorandersson4945 3 жыл бұрын
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.
@jasonbroadway8027
@jasonbroadway8027 3 жыл бұрын
I wish that I understood your statement. Can you elaborate on "these"?
@nolancary7675
@nolancary7675 3 жыл бұрын
ong tho
@awelotta
@awelotta 3 жыл бұрын
breaking: wolf becomes chihuahua
@Stirdix
@Stirdix 3 жыл бұрын
@@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.
@jacobtech7
@jacobtech7 2 жыл бұрын
@@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.
@iyziejane
@iyziejane 2 жыл бұрын
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.
@TheBodyOnPC
@TheBodyOnPC 3 жыл бұрын
When doing it wrongly you are switching the order of the limits which cannot be done generally.
@martinepstein9826
@martinepstein9826 3 жыл бұрын
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.
@Last_Resort991
@Last_Resort991 3 жыл бұрын
Exactly what I was thinking. By switching the limits, the "wrong" approach is actually right. I don't think he ecplained it that well.
@Calculus_is_power
@Calculus_is_power 2 жыл бұрын
Ahh u guys probably don't know the rules, he is doing correctly
@martinepstein9826
@martinepstein9826 2 жыл бұрын
@@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.
@Calculus_is_power
@Calculus_is_power 2 жыл бұрын
@@martinepstein9826 ooh yea thanks for clearing my misconcept
@MathAdam
@MathAdam 3 жыл бұрын
If you do this on a test, you'll be wrong, but you'll get 1/12 of a bonus mark.
@blackpenredpen
@blackpenredpen 3 жыл бұрын
Not -1/12?
@MathAdam
@MathAdam 3 жыл бұрын
@@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.
@janami-dharmam
@janami-dharmam 3 жыл бұрын
@@blackpenredpen well, you are great but not ramanujan
@lakshay-musicalscientist2144
@lakshay-musicalscientist2144 2 жыл бұрын
If he lost -1/12 marks that means he got a net +ve
@martind2520
@martind2520 3 жыл бұрын
The limit of a series is not necessarily the series of the limit.
@bm7502
@bm7502 3 жыл бұрын
I don't know what you are talking about, but it is fun to hear your maths while eating lunch.
@azizkerim8910
@azizkerim8910 3 жыл бұрын
What is the answer
@nikitakipriyanov7260
@nikitakipriyanov7260 2 жыл бұрын
The explanation with Riemann sum is brilliant!
@AliKhanMaths
@AliKhanMaths 3 жыл бұрын
This is a fascinating video! Videos like yours inspire me to share my own maths content!
@factsheet4930
@factsheet4930 3 жыл бұрын
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 😊
@yoav613
@yoav613 3 жыл бұрын
From your 2/3 limit example we can have anice estimation for sqrt1 +sqrt2 +..sqrtn =2/3 n sqrtn when n is big
@einsteingonzalez4336
@einsteingonzalez4336 3 жыл бұрын
Now he made a double pun. Both are 0 when the calculation ends and as a final grade for the exam.
@Math342010
@Math342010 3 жыл бұрын
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)
@pyrielt
@pyrielt 3 жыл бұрын
Without watching all of it, I'd say it equals to 1+ (converging towards 1 from above)
@DTN001.
@DTN001. 3 ай бұрын
In the college years, I would call then "positive zero" and "negative zero". Sounds fun and true at the same time.
@timothyaugustine7093
@timothyaugustine7093 3 жыл бұрын
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?
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@CauchyIntegralFormula
@CauchyIntegralFormula 3 жыл бұрын
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)
@simon_fox_youtube
@simon_fox_youtube 2 жыл бұрын
Why can't y be 25
@iloveevermore13
@iloveevermore13 2 жыл бұрын
Woww thats actualy True sqrt25=25-20 can i learn how you did it
@JavSusLar
@JavSusLar 2 жыл бұрын
1:26 he can't avoid a smirk on his face at the plot twist... :-)
@sh_lucius3426
@sh_lucius3426 3 жыл бұрын
Honestly, I watched this because of the pokéball. Worth it.
@RAJ-mn4nm
@RAJ-mn4nm 3 жыл бұрын
🤣🤣🤣
@hassanalihusseini1717
@hassanalihusseini1717 3 жыл бұрын
Blackpenredpen... you are the one that keeps the original "Scream".... 😃
@syvisaur7735
@syvisaur7735 3 жыл бұрын
Damn, no one has explained Riemann sums so well to me before
@Hanyamanusiabiasa
@Hanyamanusiabiasa 3 жыл бұрын
Me : "That's definitely 1" Me after 5 minutes : "oh."
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
Actually, it is still 1. BPRP is s both right and wrong here.
@lelouch1722
@lelouch1722 3 жыл бұрын
If the function series converges uniformly then the limits can be inverted!
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
Yes. Finally someone who knows.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
You are indeed worthy of using the username Lelouch
@kyleneilson1457
@kyleneilson1457 3 жыл бұрын
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
@Grassmpl
@Grassmpl 3 жыл бұрын
Lesson: for large s, the zeta function converges, by not *necessarily* uniformly. Therefore, we cannot necessarily interchange the sum and the limit.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@Grassmpl
@Grassmpl 3 жыл бұрын
@@angelmendez-rivera351 I added the word "necessarily" now so if the convergence is truly uniform, I won't contradict this truth.
@kyleneilson1457
@kyleneilson1457 3 жыл бұрын
@@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.
@Grassmpl
@Grassmpl 3 жыл бұрын
@@kyleneilson1457 I see. Now that you commented that, I know how to prove the uniform convergence.
@ianmathwiz7
@ianmathwiz7 3 жыл бұрын
Interchanging limits can only be done under certain conditions.
@neelparekh1759
@neelparekh1759 2 жыл бұрын
Yeah, when function is continuous 🙄🙄, right 🙄🙄
@reidflemingworldstoughestm1394
@reidflemingworldstoughestm1394 3 жыл бұрын
I like these long ones
@lacasadeacero
@lacasadeacero 3 жыл бұрын
No, the integral is bigger than the serie. Its known as integral criteria for abel series
@adamp9553
@adamp9553 3 жыл бұрын
Where infinity is assumed a quantity versus a symbol to a function, you get errors.
@isjosh8064
@isjosh8064 11 ай бұрын
Is the solution to (x + 1/x)^x = e Infinity?
@shapejustanormaltriangle7659
@shapejustanormaltriangle7659 3 жыл бұрын
This is going to be awesome
@ABaumstumpf
@ABaumstumpf 3 жыл бұрын
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.
@swimaaarxd50
@swimaaarxd50 3 жыл бұрын
Thanks a lot. Now I have to resit my test.
@69k_gold
@69k_gold 3 жыл бұрын
I dropped out of calc because I personally hated it. And I still watch your videos for some reason. And actually learn the solutions
@malabarmappilaanti-sudapin7581
@malabarmappilaanti-sudapin7581 3 жыл бұрын
@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..
@timetraveller2818
@timetraveller2818 3 жыл бұрын
why you need to talk about religion and god in a math video?
@malabarmappilaanti-sudapin7581
@malabarmappilaanti-sudapin7581 3 жыл бұрын
@@timetraveller2818 i want to know does mathematical geniuses believe in god and how
@RSLT
@RSLT 2 жыл бұрын
Well not so fast. 1=1^-s=>exp(ln(1)) =exp(-s(ln(1)) what is the limit s to infinity?
@terrynoah10
@terrynoah10 3 жыл бұрын
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.
@alexandresibert6589
@alexandresibert6589 3 жыл бұрын
I wonder what could be sqrt(2) ⇈ sqrt(2)
@Cycy98MinecraftFR-xh9lr
@Cycy98MinecraftFR-xh9lr 11 ай бұрын
"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)
@Ostup_Burtik
@Ostup_Burtik 8 ай бұрын
1^∞ is undefined lim x→∞ 1^x is defined and =1 Lol
@sudhakarmathsacademy
@sudhakarmathsacademy 3 жыл бұрын
It was really great mathematician
@janeza382
@janeza382 2 жыл бұрын
You shell not pass
@louisthebigcheese5833
@louisthebigcheese5833 11 ай бұрын
i noticed that he was holding a pokeball a 6:20
@MathSolvingChannel
@MathSolvingChannel 3 жыл бұрын
Usually it cannot switch the order for the limit and series sum😉
@TR_Arial
@TR_Arial 3 жыл бұрын
You know it's serious when BLACKpenREDpen uses a blue pen
@adityajha5500
@adityajha5500 3 жыл бұрын
I thought the same idea One day But I didn't knew how to calculate it...... 🙁🙁
@dqrksun
@dqrksun 3 жыл бұрын
Pls do integral of 1/(xe^x) from 1 to ∞ (I'm struggling lol🤦‍♂️)
@blackpenredpen
@blackpenredpen 3 жыл бұрын
That’s not possible with elementary function tho.
@theimmux3034
@theimmux3034 3 жыл бұрын
No wonder you're struggling, it's non-elementary
@dqrksun
@dqrksun 3 жыл бұрын
@@theimmux3034That's the point I was trying to do indefinite non elementary integrals
@numberandfacts6174
@numberandfacts6174 3 жыл бұрын
@@blackpenredpen Sir one youtuber solved Riemann hypothesis this is not a joke really this is link his video kzbin.info/www/bejne/bqqal4CjmaaWmJo
@degeimofer3272
@degeimofer3272 3 жыл бұрын
is this joke video? it seems like joke but i cannot tell
@dodododo7098
@dodododo7098 3 жыл бұрын
this guys is smart
@محمدالنجفي-ظ1ه
@محمدالنجفي-ظ1ه 2 ай бұрын
In 2:50 Why u raise (1+1/x) to the power of x ?? , shouldn't we just done the (1+1/x) that this might make slightly bigger than 1 right?? , i understand that (1+1/x)^x ≈e .
@catcatcatcatcatcatcatcatcatca
@catcatcatcatcatcatcatcatcatca 2 жыл бұрын
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”
@user-violets
@user-violets 11 ай бұрын
I want to ask a question, can 0+ be understood as the next real number of 0?
@hhhhhh0175
@hhhhhh0175 11 ай бұрын
if 0+ were a real number not equal to 0 then 0+ / 2 would have to be inbetween 0 and 0+
@aperinich
@aperinich 7 ай бұрын
Dude, I gotta say... You have a most epic beard, and I would love to be in your tute!
@purim_sakamoto
@purim_sakamoto 3 жыл бұрын
実におもしろいです😊
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@frendlyleaf6187
@frendlyleaf6187 6 ай бұрын
5:05
@shadowhunterevil8214
@shadowhunterevil8214 3 жыл бұрын
The entire video I was like, „Why do I understand this? I’m not supposed to, I’m in high school, stop this madness“
@abdellatifdz8748
@abdellatifdz8748 3 жыл бұрын
I liked the video be cause i liked it
@awesomeleozejia8098
@awesomeleozejia8098 3 жыл бұрын
Wow the limit that’s 2/3 is cool
@elen1ap
@elen1ap 2 жыл бұрын
Ιt is 3/2.
@johnchessant3012
@johnchessant3012 3 жыл бұрын
In a similar vein, d/dx [x^2] = d/dx [x + x + ... + x] = 1 + 1 + ... + 1 = x, what's wrong with this? ;)
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@romipog9337
@romipog9337 3 жыл бұрын
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
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
I do not know what your point was, but there is no such a thing as "Infinity over Infinity". This is utter nonsense.
@romipog9337
@romipog9337 3 жыл бұрын
@@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
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@romipog9337 I am not sure of that. lim n/n (n -> ♾) = 1.
@romipog9337
@romipog9337 3 жыл бұрын
@@angelmendez-rivera351 well im not very good at maths but i might add that you should first define n.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@romipog9337 What? n is a natural number. This much was at least obvious from the video.
@jeidbekoo1170
@jeidbekoo1170 2 жыл бұрын
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 ?
@filipve73
@filipve73 3 жыл бұрын
How many zeros are trivial?
@Schlaousilein67
@Schlaousilein67 3 жыл бұрын
Cool.
@jojohansi6592
@jojohansi6592 3 жыл бұрын
Love it
@Okyyy6666
@Okyyy6666 5 ай бұрын
1^(infinity) not equals one
@yatu9002
@yatu9002 2 жыл бұрын
1
@aneeshsrinivas9088
@aneeshsrinivas9088 3 жыл бұрын
nice kirby plush
@alphabeta1337
@alphabeta1337 3 жыл бұрын
Cool vid
@stapler942
@stapler942 3 жыл бұрын
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. 😅
@neelparekh1759
@neelparekh1759 2 жыл бұрын
Complex infinity, interesting 🤣🤣
@gkms6638
@gkms6638 3 жыл бұрын
What is your educational degree ?
@BrandyBrandalia
@BrandyBrandalia 3 жыл бұрын
I'm guessing a doctorate
@pranavchauhan2494
@pranavchauhan2494 3 жыл бұрын
wait what,
@britishpiperygo
@britishpiperygo 3 жыл бұрын
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.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@martinepstein9826
@martinepstein9826 3 жыл бұрын
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.
@nicolaslinhares2820
@nicolaslinhares2820 2 жыл бұрын
1/1^inf=1??? I forgot something
@njgamer7712
@njgamer7712 3 жыл бұрын
0+0+.......=0×infinite =indeterminate
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@mondherbouazizi4433 Thank you for clarifying that. Appealing to indeterminate forms is so misleading here.
@gibson2623
@gibson2623 3 жыл бұрын
Approaching zero is not the same as here.... actually it will never be zero, that s why it is infinity
@hellokitty7399
@hellokitty7399 2 жыл бұрын
所以原本那個作法過程是錯的,但結果是對的?
@intensiveadvancedmath5281
@intensiveadvancedmath5281 3 жыл бұрын
black got red got 😄
@jackwayne1626
@jackwayne1626 3 жыл бұрын
That's an alpha male beard right there
@hatembahri4314
@hatembahri4314 3 жыл бұрын
1^inf is undefined or am i wrong
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@hatembahri4314
@hatembahri4314 3 жыл бұрын
@@angelmendez-rivera351 ty for your explanation 😊
@Eichro
@Eichro 3 жыл бұрын
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).
@msittig
@msittig 3 жыл бұрын
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.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@CrazyPlayerFR
@CrazyPlayerFR 3 жыл бұрын
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.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@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*).
@CrazyPlayerFR
@CrazyPlayerFR 3 жыл бұрын
@@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! :).
@stickmanbattle997
@stickmanbattle997 3 жыл бұрын
I never heard his favorite pokemon
@omegapirat8623
@omegapirat8623 3 жыл бұрын
What Pokémon is inside your Pokéball?
@Perririri
@Perririri 3 жыл бұрын
The only one to have an Erdős Number!
@manik1477
@manik1477 3 жыл бұрын
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).
@tabeh-
@tabeh- 3 жыл бұрын
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.
@manik1477
@manik1477 3 жыл бұрын
@@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?
@tabeh-
@tabeh- 3 жыл бұрын
@@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.
@manik1477
@manik1477 3 жыл бұрын
@@tabeh- Oh okay, I get the picture now,Thank you so much for explaining this!
@Davidamp
@Davidamp 3 жыл бұрын
2:28 HOWEVER The ygo progression series has ruined that word for me
@ShaunJW1
@ShaunJW1 3 жыл бұрын
No please after years of you helping me through my physics maths degree, I prefer you beardless...lol
@peterwan816
@peterwan816 11 ай бұрын
If 0 is not d, the answer will be 1. It would be e otherwise.
@TheMemesofDestruction
@TheMemesofDestruction 3 жыл бұрын
#GottaCatchEmAll' ^.^
@saliryakouli1260
@saliryakouli1260 11 ай бұрын
He didnt precise + or - infinity
@htspencer9084
@htspencer9084 11 ай бұрын
Limits like this make everything so unintuitive 😂
@thisiswhoiam7249
@thisiswhoiam7249 3 жыл бұрын
Can you prove 0!=1? 🥺
@numberandfacts6174
@numberandfacts6174 3 жыл бұрын
Bro gamma function proof 0! =1 Gamma(1) = 0! =1
@1224chrisng
@1224chrisng 3 жыл бұрын
well of course, 0 != 1 anyways
@adityajha5500
@adityajha5500 3 жыл бұрын
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
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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.
@thisiswhoiam7249
@thisiswhoiam7249 3 жыл бұрын
@@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? 😁
@agame2545
@agame2545 3 жыл бұрын
why do u only do calculs u need to expand the topic
@migs6674
@migs6674 3 жыл бұрын
Are you Singaporean?
@NobodyYouKnow01
@NobodyYouKnow01 11 ай бұрын
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.
@kaisenramen4563
@kaisenramen4563 8 ай бұрын
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.
@xr_xharprazoraxtra5428
@xr_xharprazoraxtra5428 3 жыл бұрын
conclusion : 1/∞ = 0+ but 1/∞ ≠ 0
@Ostup_Burtik
@Ostup_Burtik 8 ай бұрын
1/∞=0 but sometimes 0+
@Vishw_1234
@Vishw_1234 3 жыл бұрын
😕😒
@MishTheMash
@MishTheMash 6 ай бұрын
BLUE PENNNN!!!!!! WTFWTFWTFWTFWTFWTFWTFWTF
@SuperMario-jd9zt
@SuperMario-jd9zt 3 жыл бұрын
Saun saun saun
@mu11668B
@mu11668B 3 жыл бұрын
Well... 0 is still better than -1/12.
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