Thank you for all of your work ! You're an amazing teacher and you make me like and understand math even more !
@General12th6 жыл бұрын
I thought you were going to talk about the _product_ of all those functions, IE: f(x) = product(fn) = product(x^n). But just the limiting function is cool too.
@Koisheep6 жыл бұрын
I got functional analysis vibes from this, when you learn about L²([a,b]) and work with the metric
@rudboy95996 жыл бұрын
Konhat Lee Sakurai yes! Loved that class. Uniform and pointwise convergence. L^1,2,p spaces. Fun times
@volcanic31046 жыл бұрын
f(-1) is in a superposition!
@danielbenyair3006 жыл бұрын
VOLC4NICxSP4RK Lol...
@danielbenyair3006 жыл бұрын
VOLC4NICxSP4RK It is not both!... it is nitere!!
@waishingtseung69304 жыл бұрын
✓✓✓✓✓✓✓
@frankielai57536 жыл бұрын
Lol....colour consistency.....love it.
@ffggddss6 жыл бұрын
Nicely explained! ∞ can also ruin continuity when you add continuous functions. Ex.: a square wave (with averaged value assigned at each of its discontinuities) is a discontinuous function. Yet it is equal to its Fourier series, in which each term is a continuous function. Fred
@rudboy95996 жыл бұрын
Hence why uniform convergence is so nice
@thexoxob94486 ай бұрын
in general: f(x) = limn->inf x^n is ambiguous because it doesn't specify by integers or real numbers. You put x belongs to what set below the arrow. If you are curious if n is integer it's in the video but if real numbers the negative part is gone
@JayTemple2 жыл бұрын
When I took Functional Analysis, this was literally the first example they showed us of a sequence of continuous functions whose limit was a non-continuous function.
@splodinatekabloominate8466 жыл бұрын
math 🅱roke
@patrickgleason30536 жыл бұрын
Infinitely many infinites #YAY
@SteamPunkLV6 жыл бұрын
:P
@h3bb16 жыл бұрын
Would it be correct to say that this infinity function is continous at -1
@ffggddss6 жыл бұрын
Yes! Remember, the definition: f(x) is continuous at x=a iff lim[x→a⁻]f(x) = lim[x→a⁺]f(x) = f(a) This is true for all a in the open interval (-1,1). Fred
@h3bb16 жыл бұрын
ffggddss awesome and thank you for the reply
@assiddiq73603 жыл бұрын
Visually, the graph would converge to a reflected L-shape; a horizontal line from (-1,0) to (1,0), and a vertical line from (1,0) towards (1,∞). The beauty of the graph is that it visualize how 1^∞ is indeterminate, although only for positive number
@HeliosBeats6 жыл бұрын
it's the opposite for x^4, the higher n is n f(x)= x^n, (when it's even), the thinner/taller the graph
@ruhanbhandohal98766 жыл бұрын
ReDesign Beats He is referring to x^n with the domain from -1 to 1. In that case it gets wider.
@HeliosBeats6 жыл бұрын
oh true
@pablojulianjimenezcano43626 жыл бұрын
The t-shirt is incredible, best idea ever! #tshirt #Yay
@twwc9606 жыл бұрын
It's also the case that infinite sums of continuous functions need not be continuous, even if the infinite sums converge everywhere: consider the Fourier series for the square wave function, for example. Each term in the series is continuous, and the series converges everywhere on the real line, but the function it converges to is not continuous. In fact, before Fourier, most mathematicians believed that series of continuous functions, when they converged, converged to continuous functions. Fourier found the first counterexamples, but his work was not very rigorous. I think Dirichlet was the first to establish these results rigorously.
@Guil1186 жыл бұрын
#t-shirt Man! Never have I ever considered buying any merch from any youtuber. But yours made me change my mind
@Guil1186 жыл бұрын
Also Im tackling the 3n+1 problem and its driving me insane!
@jorgesponja30426 жыл бұрын
#t-shirt hype!! Your videos are preparing me for starting college in 2 weeks, keep doing it, you're the best #BPRP
@ogorangeduck6 жыл бұрын
不错! also that T-shirt is LIT
@blackpenredpen6 жыл бұрын
orange duck 謝啦!
@hopp21846 жыл бұрын
Do a video about tetrations!
@blackpenredpen6 жыл бұрын
I will, one day.
@mike4ty46 жыл бұрын
And what happens for negative values of n - i.e. "f(x) = x^{-\infty}"? You get the opposite effect - everything outside of [-1, 1] is okay, but the divergent bit is inside!
@Shsbzyqn123 Жыл бұрын
Good going sir
@SeriousApache6 жыл бұрын
Calculating around infinity it's like approach speed of light, normal law stop working here.
@General12th6 жыл бұрын
Laws work just fine as we approach light speed. Special relativity handles it all perfectly well.
@GarryDumblowski6 жыл бұрын
It does, but I think they were referring to Newtonian physics. The strange thing is, due to the boundlessness of infinity, ordinary numbers all have the same properties. Even huge numbers like TREE(3) or G(64) have all the same properties as other numbers; they have prime factorizations and you can take their square roots and graph functions involving them. Same goes with this; the graph of x raised to some ginormous power is still perfectly continuous. Putting it in the context of a limit almost makes this property seem to instantaneously jump to unruliness, so it's quite separated from the more gradual nature of relativity
@SeriousApache6 жыл бұрын
So it is no longer normal laws
@General12th6 жыл бұрын
RUSapache There's nothing "normal" about Newtonian physics.
@GarryDumblowski6 жыл бұрын
I mean, that's a fair assessment, but it's "normal" in the sense that at the pitiful speeds we humans are able to achieve, you have to know where to look to see the effects of relativity. It's what's considered "normal" to us.
@yrcmurthy83236 жыл бұрын
#YAY Hats off to you !
@chupetaparabose16 жыл бұрын
I have a question about the solid numbers , for example i have the limit of 0/(x*x) with x-->Infinity. By me it should be 0 , but i m not really sure about it...THX
@samokoribanic91846 жыл бұрын
#YAY
@muse06224 жыл бұрын
How 'bout differentiating the x^inf?
@estuardodiaz27206 жыл бұрын
Hey, but what about x being equal to 1+1/n. By def. of e, lim n-> infinity (1+1/n)^n = e, so, in the limiting function, you would have more than one point different from 1 converging to a number different than 0, right??
@thomashaas43816 жыл бұрын
No, because the limit function f(x) = lim {n->inf} f_n(x) is independent of n. By definition of f, you can't have the argument x be dependent on n. You basically pick a fixed value for x and then look at the limit of the sequence f_1(x), f_2(x), f_3(x), ... which you then define to be f(x). You see that x is fixed for the sequence.
@estuardodiaz27206 жыл бұрын
oh, okay, but what if we fix x to be lim {k->inf} (1+1/k) or perhaps x = 1+ epsilon, I know in this case we cannot use the definition of e, but what assures us that any x grater than 1 diverges, perhaps the boundry of 1 does not (thanks for answering by the way haha)
@thomashaas43816 жыл бұрын
Well if you fix x = lim {k -> inf} (1 + 1/k), then x = 1. Hence you again get f(x) = 1 as result. x = 1 + epsilon > 1 (assuming epsilon > 0), hence you get divergence. The only way to get what you want is to look at the sequence f_n(x_n) where x_n = 1 + 1/n. However, that is not how the limit function was defined because the argument x has to be fixed/constant (it can't be sequence that is dependent on n)
@estuardodiaz27206 жыл бұрын
oh, okay, I think I got it, thank you!! :)
@n0ita6 жыл бұрын
Pretty cool, i like to imagine this fuction as a inverse black hole, everything inside stay safe, but outside, it becomes bizarre...
@blue_blue-16 жыл бұрын
That‘s why greek philosophers didn‘t dare to think about infinity. Healthy philosophy.
@Bhamilton-ws4go6 жыл бұрын
Funny thing to point out, but at the beginning, x⁴ is a skinnier parabola, not a fatter one, than x² ;)
@blackpenredpen6 жыл бұрын
the bottom part... : )
@Bhamilton-ws4go6 жыл бұрын
blackpenredpen I guess with parabolas, it's all about perspective, since every parabola is similar :)
@KnakuanaRka6 жыл бұрын
Well, actually the part of the curve below 1 is fatter, but the rest above is skinnier. With the scale of his drawings, they stop around 1, so they’re perfectly right.
@factsheet49306 жыл бұрын
I feel like I need to point out that while the function is 0 between -1 and 1, it doesn't necessarily jumps to 1 at x=1... since... x is approaching 1 it's like saying: lim x->1 of x^infinity... which is undefined but personally ill just continue the curve upwards to 1 at very close to 1 instead of jumping because the values should map there :>
@stephenbeck72226 жыл бұрын
No think about the “r” example (like geometric series). The left hand limit as x goes to 1 is indeed 0. The right hand limit is infinity. And the value of f(1) itself is 1. A very unusual graph but that is what makes it a great example.
@ffggddss6 жыл бұрын
For each finite value of n, the curve goes smoothly up to 1 at x=1. In the limiting case, it actually *does* jump from 0 to 1; it's discontinuous there. Fred
@harrisidh6 жыл бұрын
okay but how about we look at it as lim (1-1/n)^n = 1/e n→∞ so it'd have to pass through 1/e when it's very close to 1
@factsheet49306 жыл бұрын
But that's a different limit question sadly... look at the graph of: y=(x+1)^(1/x) this graph does have a point which 1^infinity appears (when x=0) and yet it doesn't pass through 1/e! (in fact it never has a value of 1/e because as x->infinity, y->1). The reason I believe it does go up to 1 as lim x->1, is because for any other value of n (in x^n) it does go up to 1...
@thomashaas43816 жыл бұрын
Just to make it clear: the limiting process in Blackpenredpen's case is with respect to n. The limit function f(x) is the limit of the sequence f_1(x), f_2(x), .... In case of x = 1, the sequence is constantly 1 and hence the limit is 1 as well. At no point does he consider a limit with respect to x. But since lim x->1 f(x) = 0 != 1 = f(lim x ->1 x) = f(1) we see that the function is discontinous at x = 1.
@adamzeggai55064 жыл бұрын
the graph of even numbers doesn't get wider, it actually gets tighter the greater the value of n gets lol
@andresxj16 жыл бұрын
But the divergence in the negative part (x1), isn't it?
@sikandarbakht20766 жыл бұрын
What about t-shirts, i cannot wait much, what have u thought about it??....BTW great video... Thanks
@blackpenredpen6 жыл бұрын
Sikandar Bakht it should be coming soon.
@MrYadaization6 жыл бұрын
Could you do a video on quaternions?
@thexoxob94486 ай бұрын
In general: the limit of a sequence may not have the same properties as the sequence, like 1.4, 1.41, 1.414, 1.4142, .... These numbers are rational, but if you take the limit as the sequence goes to infinity, the limit is irrational. (Take a guess what number it is)
@jercki726 жыл бұрын
lol I saw the thumbnail and knew exactly what it was about
@weykasemolibdenodio71846 жыл бұрын
Functional Analysis!
@milos_radovanovic6 жыл бұрын
Great video! And this time finaly no contraversal statments! You are geting more consistant! Will you make any suplemental for that video you puled down on infinite nested square roots?
@milos_radovanovic6 жыл бұрын
I also think that videos on integral and derivative of infinite nested square roots deserve some added rigor on domain definitions!
@blackpenredpen6 жыл бұрын
This is actually a prep for that : )
@milos_radovanovic6 жыл бұрын
Just keep a clickbait at bay, and all will be well! ;D
@ReyhanMehta6 жыл бұрын
But is infinity an integer?
@revilo73896 жыл бұрын
What about at x = (1 - 1/n)?
@thomashaas43816 жыл бұрын
the limit function has no parameter n hence you can't really look at f(1-1/n) unless you assume n to just be some constant value in which case you have f(1-1/n) = 0 since 0
@JamalAhmadMalik6 жыл бұрын
#yay!
@EMorgensztern6 жыл бұрын
can you find the continuity (or not) of y=x^(1/x) from -inf to 0 ? I love your videos about complex # Thanks for what you do #yay from France !
@helloitsme75536 жыл бұрын
Eliott Morgensztern x cannot be 0 since 1/0 is not possible. Ehm x cannot be -2 either , cause roots of negative numbers. I think x cannot be in the form of -2a where a is a natural number, but idk how to prove there are no other incontinuities
@Vojtaniz016 жыл бұрын
HelloItsMe That is enough to prove that the function y=x^(1/x) is not continuous on given interval. No need to search for any other discontinuities.
@ahmadkalaoun34736 жыл бұрын
You said 1 to infinity's powee is equal to 1...but i don't think so... We have Lim(n->infinity) (1+1/n)^n=e If we remplace n by infinity we get (1+0)^inf=e 1^inf=e and - 1^inf=-e What do you think 🤔 🤔🤔
@ffggddss6 жыл бұрын
But you can't use that argument to define 1^∞, because lim[n→∞] (1 + a/n)ⁿ = eª which, if you "set" n=∞, says (1+0)^∞ = 1^∞ = eª So you can get any positive number you want out of that. Fred
@ahmadkalaoun34736 жыл бұрын
ffggddss you're right... So we can say 1^inf=0/0=inf/inf Al of those terms are indefinite... Aren't they?
@MiguelGonzalez-hy4sd6 жыл бұрын
7:51 that wouldn't be a indeterminate?
@icew0lf986 жыл бұрын
would arctan(f(x)) be pi/2?
@L1N3R1D3R6 жыл бұрын
Well, f(x) is a piecewise function, so any composition of that function would also have to be defined piecewise as well. For x 1, f(x) always tends to positive infinity, so arctan(f(x)) would indeed tend to π/2; however, I'm not confident enough with defining values for infinity as opposed to just saying the limiting value, so think of that how you will.
@blackpenredpen6 жыл бұрын
Depending on the domain
@icew0lf986 жыл бұрын
yeah I forgot to mention "for x larger than 1"
@xaxuser50336 жыл бұрын
#Yaaay
@alexdarcovich93496 жыл бұрын
Cheeky :) #yay
@helloitsme75536 жыл бұрын
I thought you were gonna talk about 1+x+x^2+x^3..... which is not continuous at 1 cause it's 1+1+1+1+1+1+1+1 and also 1/1-x = 1+x+x^2... If you plug in x you get 1/0
@blackpenredpen6 жыл бұрын
Oh yea, that too!
@NAMEhzj6 жыл бұрын
but that limit (1 + x + x² + x³ + ...) is actually (kinda) fine. It just diverges for |x| >= 1, but if you take all of the domain, namely the open interval (-1, 1) its just 1/(1-x) which is perfectly continuous there.
@helloitsme75536 жыл бұрын
NAMEhzj yeah I know , only discontinuous on |x|>1 or equal to 1
@NAMEhzj6 жыл бұрын
i just meant that the problem at 1 is not one of continuity, it just isnt defined there.
@hankseda6 жыл бұрын
Hard to be rigorous on pointwise convergence in less than 12 minutes. 👏
@blackpenredpen6 жыл бұрын
Yea... I just wanted to provide an example in this video
@willbishop13556 жыл бұрын
Doesn't really make sense to write down 1^inf = 1, as you've pointed out in other videos, e.g. lim[n->inf] (1 + 1/n)^n = e, but obviously we can write lim[n->inf] (1^n) = 1.
@shadowatom6 жыл бұрын
I don't believe you need the limit in this case because 1 is a constant. Your first example is two functions with one approaching 1 and the other approaching infinity which results in an indeterminate form. Since 1 is a constant in this case and 1 to any power is 1, you can safely say 1^inf is still 1.
@gopalrajpurohit8016 Жыл бұрын
Is it high school math?
@OonHan6 жыл бұрын
#t-shirt
@Cloud88Skywalker6 жыл бұрын
8:48, isn't 0.9999... = 1 ?
@tommero65846 жыл бұрын
He only said a finite amount of 9s. Not infinitely many.
@sugarfrosted20056 жыл бұрын
It's easy to make this rigorous just by taking limit supremum or limit of X^(T_n)
@animeshpradhan56836 жыл бұрын
integrate1/ ((cosx)^5+(sinx)^5) please
@gabob69926 жыл бұрын
Proof that : x^2-y^3=1
@loganallomes43056 жыл бұрын
Do you know the numbers have an end? -Mohamed Ababou-
@pramodgupta2998 Жыл бұрын
10:55 *Dirty mind activated* If you know you know
@danielbenyair3006 жыл бұрын
1st! Infinity is NOT a bumber!!! therefore CANNOT be positive or negative!!! 2nd! 1 does NOT equle to 0.99999999999999999999999... !!!! 3rd! This 0.000000000001^10000000000000000000 also does NOT equle to 0 !!!!!! 4th! the slope near 1 is INFINITY!!!!!!
@Alex-dn7jq6 жыл бұрын
Wait, 1^inf=1? I thought that this was indefinite.
@Alex-dn7jq6 жыл бұрын
misotanni got it! Thx!
@romanhredil37996 жыл бұрын
Actually, 1^inf is divergent.
@thomashaas43816 жыл бұрын
"If 1 is variable" is a very strange way to put it. The most natural interpretation of the term 1^inf is lim {n->inf} (1^n) which indeed is 1. Any other interpretation of that term is bogus. In any case, infinity is no real number and therefore shouldn't be treated as such. For convenvience reasons however, it is usually interpreted as a limit in case there is only one infinity symbol used. Terms like inf/inf or (1+1/inf)^inf containing 2 or more infinities are ambiguous since there is no reason to assume that both infinities are obtained as the limit of the same function.
@danielbenyair3006 жыл бұрын
alex ubuntu (-1)^infinity is indefinite... not +1
@KnakuanaRka6 жыл бұрын
That’s only in limits, where it represents the limit of a function in the form of one expression to the power of a second expression, where those expressions tend towards 1 and i at the limit, such as the (1+1/x)^x limit used to evaluate e. As the value x approaches i, the base of that formula approaches 1, but never truly hits it, so the value doesn’t get stuck at 1. In this case, there’s no limit, so the base is exactly 1, and all the exponentiation isn’t he World makes no difference.
@deenaaalkotb6 жыл бұрын
did you figure my problem out
@blackpenredpen6 жыл бұрын
I have a vid on rotation a polar curve kzbin.info/www/bejne/Z4jcomOnmdWMp7s For the parametric version, Sen Zen has done an awesome video on that already. kzbin.info/www/bejne/eIHKomV3qsmie9E
@deenaaalkotb6 жыл бұрын
thank you for that
@deenaaalkotb6 жыл бұрын
but I have another problem if f(x)=f(y) then x=y but sometimes it doesn't work like sin(3π/2)=sin(-π/4) what happened
@l3th4lv1p3r6 жыл бұрын
deena alkotb f(x) = f(y) implies x=y is only true if f is an injection function en.m.wikipedia.org/wiki/Injective_function
@deenaaalkotb6 жыл бұрын
@@l3th4lv1p3r thank you it was really nice 😎
@boda19485 жыл бұрын
I would like to call this "the binary function" because it only has 2 y values 1 and 0
@gg.38126 жыл бұрын
At -1 it s not defined, it does not diverge
@ffggddss6 жыл бұрын
Terminology to describe that behavior differs. In my university courses long ago, any failure to converge was said to diverge. Nowadays a distinction is sometimes made between escaping to ±∞ ("divergence"), and failure to converge that is bounded ("oscillation"). In both cases, the function is said to be undefined at such a point. Fred
@rudboy95996 жыл бұрын
Got my master's last year. It's still the same. If it doesn't converge, it is said to diverge.
@Ben.20006 жыл бұрын
inf broke math. #yay!
@romanhredil37996 жыл бұрын
but I can say 1^inf=e, or e^2, or e^3, or any positive number so 1^inf also diverges i'll show you the proof: You know this formula: lim_n→inf ((1+1/n)^n)=e so, if you plug infinity, you'll get (1+1/inf)^inf, 1/inf=0, so we just have 1^inf, but by the formula higher, we get this: 1^inf=e, but 1^inf=(1^n)^inf=(1^inf)^n=e^n, where n is any real number, and we get 1=n, which destroys all maths so, you must agree that 1^inf is divergent
@iabervon6 жыл бұрын
Roman Hredil' When he says "a solid 1", he means that the base is a constant. That is, it's flatter than any other function. In particular, ln (base)
@adityatripathi16486 жыл бұрын
Someone's watching Doraemon in background
@PhasmidTutorials6 жыл бұрын
@weinsim38566 жыл бұрын
2nd
@EchoHeo6 жыл бұрын
LOL i'm doing MATHS
@ogorangeduck6 жыл бұрын
2:41 hehe 太胖
@firstnamelastname79416 жыл бұрын
I’m pretty sure 1^infinity is an indeterminate form.
@thomashaas43816 жыл бұрын
There is no reason why it should be. No matter how you resolve the infinity as a limit of some function, you will end up with lim {n -> inf} 1^g(n) = 1 (for any g(n) which diverges towards infinity as n goes towards infinity). In the video you have g(n) = n, in which case you can directly see that the limit is 1 since it is the limit of the constant sequence 1^1, 1^2, 1^3, ...
@redsalmon99666 жыл бұрын
Anything outside of this picture, the function just, NOT GOOD ANYMORE