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.)
@moregirl45853 жыл бұрын
@@user-donghun04 No, every term approaching to 0 may result in any sum
@angelmendez-rivera3513 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.
@martinepstein98263 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.
@grady7303 жыл бұрын
It gets serious when this guy breaks out the blue pen.
@PubicGore3 жыл бұрын
Wow, I've never seen that joke before.
@WerewolfLord3 жыл бұрын
Order of seriousness: black -> red -> blue -> green -> purple ?
@Camp_RB3 жыл бұрын
@@WerewolfLord I’ve never seen purple . . .
@mptness43893 жыл бұрын
@@Camp_RB it's very rare but I think I've seen it a couple times.
@myfishcalledbobble69236 ай бұрын
@@Camp_RB i think ive seen it in some of his calc vids
@APotatoWT3 жыл бұрын
I love how he solves hard math while holding a pokeball lol
@igormatheus86983 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
@lex224ification3 жыл бұрын
@@igormatheus8698 It's the mic. You can clearly see the cord
@Perririri3 жыл бұрын
The only animé with an Erdős Number!
@pieTone2 жыл бұрын
@@igormatheus8698 you can see the wire
@Fire_Axus11 ай бұрын
your feelings are irrational
@diedoktor3 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-bv7vn3 жыл бұрын
It can be expanded (carefully) to give the Taylor series expansion of e^x where x=1
@bjorneriksson24043 жыл бұрын
Here's a nice explanation of where e comes from, by Eddie Woo: kzbin.info/www/bejne/ppibY2qrebV5p6M
@jessehammer1233 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.
@moonshifter03 жыл бұрын
@@AngadSingh-bv7vn good luck with the tailor series.
@angelmendez-rivera3513 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.
@mayurchaudhari8503 жыл бұрын
The way he uses different colors itself gives you the clue that something is wrong with those zeroes...
@CrazyPlayerFR3 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)
@viktorandersson49453 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.
@jasonbroadway80273 жыл бұрын
I wish that I understood your statement. Can you elaborate on "these"?
@nolancary76753 жыл бұрын
ong tho
@awelotta3 жыл бұрын
breaking: wolf becomes chihuahua
@Stirdix3 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.
@jacobtech72 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.
@iyziejane2 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.
@TheBodyOnPC3 жыл бұрын
When doing it wrongly you are switching the order of the limits which cannot be done generally.
@martinepstein98263 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_Resort9913 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_power2 жыл бұрын
Ahh u guys probably don't know the rules, he is doing correctly
@martinepstein98262 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_power2 жыл бұрын
@@martinepstein9826 ooh yea thanks for clearing my misconcept
@MathAdam3 жыл бұрын
If you do this on a test, you'll be wrong, but you'll get 1/12 of a bonus mark.
@blackpenredpen3 жыл бұрын
Not -1/12?
@MathAdam3 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-dharmam3 жыл бұрын
@@blackpenredpen well, you are great but not ramanujan
@lakshay-musicalscientist21442 жыл бұрын
If he lost -1/12 marks that means he got a net +ve
@martind25203 жыл бұрын
The limit of a series is not necessarily the series of the limit.
@bm75023 жыл бұрын
I don't know what you are talking about, but it is fun to hear your maths while eating lunch.
@azizkerim89103 жыл бұрын
What is the answer
@nikitakipriyanov72602 жыл бұрын
The explanation with Riemann sum is brilliant!
@AliKhanMaths3 жыл бұрын
This is a fascinating video! Videos like yours inspire me to share my own maths content!
@factsheet49303 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 😊
@yoav6133 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
@einsteingonzalez43363 жыл бұрын
Now he made a double pun. Both are 0 when the calculation ends and as a final grade for the exam.
@Math3420103 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)
@pyrielt3 жыл бұрын
Without watching all of it, I'd say it equals to 1+ (converging towards 1 from above)
@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.
@timothyaugustine70933 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-rivera3513 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.
@CauchyIntegralFormula3 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_youtube2 жыл бұрын
Why can't y be 25
@iloveevermore132 жыл бұрын
Woww thats actualy True sqrt25=25-20 can i learn how you did it
@JavSusLar2 жыл бұрын
1:26 he can't avoid a smirk on his face at the plot twist... :-)
@sh_lucius34263 жыл бұрын
Honestly, I watched this because of the pokéball. Worth it.
@RAJ-mn4nm3 жыл бұрын
🤣🤣🤣
@hassanalihusseini17173 жыл бұрын
Blackpenredpen... you are the one that keeps the original "Scream".... 😃
@syvisaur77353 жыл бұрын
Damn, no one has explained Riemann sums so well to me before
@Hanyamanusiabiasa3 жыл бұрын
Me : "That's definitely 1" Me after 5 minutes : "oh."
@angelmendez-rivera3513 жыл бұрын
Actually, it is still 1. BPRP is s both right and wrong here.
@lelouch17223 жыл бұрын
If the function series converges uniformly then the limits can be inverted!
@angelmendez-rivera3513 жыл бұрын
Yes. Finally someone who knows.
@angelmendez-rivera3513 жыл бұрын
You are indeed worthy of using the username Lelouch
@kyleneilson14573 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
@Grassmpl3 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-rivera3513 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.
@Grassmpl3 жыл бұрын
@@angelmendez-rivera351 I added the word "necessarily" now so if the convergence is truly uniform, I won't contradict this truth.
@kyleneilson14573 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.
@Grassmpl3 жыл бұрын
@@kyleneilson1457 I see. Now that you commented that, I know how to prove the uniform convergence.
@ianmathwiz73 жыл бұрын
Interchanging limits can only be done under certain conditions.
@neelparekh17592 жыл бұрын
Yeah, when function is continuous 🙄🙄, right 🙄🙄
@reidflemingworldstoughestm13943 жыл бұрын
I like these long ones
@lacasadeacero3 жыл бұрын
No, the integral is bigger than the serie. Its known as integral criteria for abel series
@adamp95533 жыл бұрын
Where infinity is assumed a quantity versus a symbol to a function, you get errors.
@isjosh806411 ай бұрын
Is the solution to (x + 1/x)^x = e Infinity?
@shapejustanormaltriangle76593 жыл бұрын
This is going to be awesome
@ABaumstumpf3 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.
@swimaaarxd503 жыл бұрын
Thanks a lot. Now I have to resit my test.
@69k_gold3 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-sudapin75813 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..
@timetraveller28183 жыл бұрын
why you need to talk about religion and god in a math video?
@malabarmappilaanti-sudapin75813 жыл бұрын
@@timetraveller2818 i want to know does mathematical geniuses believe in god and how
@RSLT2 жыл бұрын
Well not so fast. 1=1^-s=>exp(ln(1)) =exp(-s(ln(1)) what is the limit s to infinity?
@terrynoah103 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.
@alexandresibert65893 жыл бұрын
I wonder what could be sqrt(2) ⇈ sqrt(2)
@Cycy98MinecraftFR-xh9lr11 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_Burtik8 ай бұрын
1^∞ is undefined lim x→∞ 1^x is defined and =1 Lol
@sudhakarmathsacademy3 жыл бұрын
It was really great mathematician
@janeza3822 жыл бұрын
You shell not pass
@louisthebigcheese583311 ай бұрын
i noticed that he was holding a pokeball a 6:20
@MathSolvingChannel3 жыл бұрын
Usually it cannot switch the order for the limit and series sum😉
@TR_Arial3 жыл бұрын
You know it's serious when BLACKpenREDpen uses a blue pen
@adityajha55003 жыл бұрын
I thought the same idea One day But I didn't knew how to calculate it...... 🙁🙁
@dqrksun3 жыл бұрын
Pls do integral of 1/(xe^x) from 1 to ∞ (I'm struggling lol🤦♂️)
@blackpenredpen3 жыл бұрын
That’s not possible with elementary function tho.
@theimmux30343 жыл бұрын
No wonder you're struggling, it's non-elementary
@dqrksun3 жыл бұрын
@@theimmux3034That's the point I was trying to do indefinite non elementary integrals
@numberandfacts61743 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
@degeimofer32723 жыл бұрын
is this joke video? it seems like joke but i cannot tell
@dodododo70983 жыл бұрын
this guys is smart
@محمدالنجفي-ظ1ه2 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 .
@catcatcatcatcatcatcatcatcatca2 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-violets11 ай бұрын
I want to ask a question, can 0+ be understood as the next real number of 0?
@hhhhhh017511 ай бұрын
if 0+ were a real number not equal to 0 then 0+ / 2 would have to be inbetween 0 and 0+
@aperinich7 ай бұрын
Dude, I gotta say... You have a most epic beard, and I would love to be in your tute!
@purim_sakamoto3 жыл бұрын
実におもしろいです😊
@angelmendez-rivera3513 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.
@frendlyleaf61876 ай бұрын
5:05
@shadowhunterevil82143 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“
@abdellatifdz87483 жыл бұрын
I liked the video be cause i liked it
@awesomeleozejia80983 жыл бұрын
Wow the limit that’s 2/3 is cool
@elen1ap2 жыл бұрын
Ιt is 3/2.
@johnchessant30123 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-rivera3513 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.
@romipog93373 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-rivera3513 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.
@romipog93373 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-rivera3513 жыл бұрын
@@romipog9337 I am not sure of that. lim n/n (n -> ♾) = 1.
@romipog93373 жыл бұрын
@@angelmendez-rivera351 well im not very good at maths but i might add that you should first define n.
@angelmendez-rivera3513 жыл бұрын
@@romipog9337 What? n is a natural number. This much was at least obvious from the video.
@jeidbekoo11702 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 ?
@filipve733 жыл бұрын
How many zeros are trivial?
@Schlaousilein673 жыл бұрын
Cool.
@jojohansi65923 жыл бұрын
Love it
@Okyyy66665 ай бұрын
1^(infinity) not equals one
@yatu90022 жыл бұрын
1
@aneeshsrinivas90883 жыл бұрын
nice kirby plush
@alphabeta13373 жыл бұрын
Cool vid
@stapler9423 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. 😅
@neelparekh17592 жыл бұрын
Complex infinity, interesting 🤣🤣
@gkms66383 жыл бұрын
What is your educational degree ?
@BrandyBrandalia3 жыл бұрын
I'm guessing a doctorate
@pranavchauhan24943 жыл бұрын
wait what,
@britishpiperygo3 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-rivera3513 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.
@martinepstein98263 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.
@nicolaslinhares28202 жыл бұрын
1/1^inf=1??? I forgot something
@njgamer77123 жыл бұрын
0+0+.......=0×infinite =indeterminate
@angelmendez-rivera3513 жыл бұрын
@@mondherbouazizi4433 Thank you for clarifying that. Appealing to indeterminate forms is so misleading here.
@gibson26233 жыл бұрын
Approaching zero is not the same as here.... actually it will never be zero, that s why it is infinity
@hellokitty73992 жыл бұрын
所以原本那個作法過程是錯的,但結果是對的?
@intensiveadvancedmath52813 жыл бұрын
black got red got 😄
@jackwayne16263 жыл бұрын
That's an alpha male beard right there
@hatembahri43143 жыл бұрын
1^inf is undefined or am i wrong
@angelmendez-rivera3513 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.
@hatembahri43143 жыл бұрын
@@angelmendez-rivera351 ty for your explanation 😊
@Eichro3 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).
@msittig3 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-rivera3513 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.
@CrazyPlayerFR3 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-rivera3513 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*).
@CrazyPlayerFR3 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! :).
@stickmanbattle9973 жыл бұрын
I never heard his favorite pokemon
@omegapirat86233 жыл бұрын
What Pokémon is inside your Pokéball?
@Perririri3 жыл бұрын
The only one to have an Erdős Number!
@manik14773 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-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.
@manik14773 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-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.
@manik14773 жыл бұрын
@@tabeh- Oh okay, I get the picture now,Thank you so much for explaining this!
@Davidamp3 жыл бұрын
2:28 HOWEVER The ygo progression series has ruined that word for me
@ShaunJW13 жыл бұрын
No please after years of you helping me through my physics maths degree, I prefer you beardless...lol
@peterwan81611 ай бұрын
If 0 is not d, the answer will be 1. It would be e otherwise.
@TheMemesofDestruction3 жыл бұрын
#GottaCatchEmAll' ^.^
@saliryakouli126011 ай бұрын
He didnt precise + or - infinity
@htspencer908411 ай бұрын
Limits like this make everything so unintuitive 😂
@thisiswhoiam72493 жыл бұрын
Can you prove 0!=1? 🥺
@numberandfacts61743 жыл бұрын
Bro gamma function proof 0! =1 Gamma(1) = 0! =1
@1224chrisng3 жыл бұрын
well of course, 0 != 1 anyways
@adityajha55003 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-rivera3513 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.
@thisiswhoiam72493 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? 😁
@agame25453 жыл бұрын
why do u only do calculs u need to expand the topic
@migs66743 жыл бұрын
Are you Singaporean?
@NobodyYouKnow0111 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.
@kaisenramen45638 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.