Limit as an integral (Riemann Sum)
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- เผยแพร่เมื่อ 2 ต.ค. 2024
- Riemann sum limit. In fact, we will interpret this limit as an integral. We will see the indeterminate form of 0+0+0+... doesn't necessary approach 1. This is actually the integral of sqrt(x) from 0 to 1 and we get 2/3 for the answer. If you would like more calculus tutorials like this, then check out my just calculus channel: / justcalculus
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blackpenredpen
Wasn't in chinese
7.8/10
Too easy
e/π
Sir Rahmed
You still loyal,
10/10
: )
Pokemon reference😂
@@blackpenredpen next, you need to do calculus 100 integrals challenge in pure mandarin 😂
From all I've watched of your vids, this one was the most amazing
Thanks!!!
Wait, we didn't do integration until Cal II.
Joshua Hillerup
R u on quarter system?
blackpenredpen two semesters per year (plus summer, but most don't do that). Cal I is limits and derivatives and whatnot. Cal II is integration, Cal III is multivariate.
Joshua Hillerup I see. We do a bit of integration at the end of the calc 1 semester. Mainly just on area and volume. And some u subs. Then calc 2 is all the other integration techniques and applications.
And inf series
So, do you guys do this stuff in high school or university? Because “cal 2” sounds like the name of a subject in university but we do it in high school so I am a bit confused
Hi Black pen and Red pen,
I'm Robin from France, and I wanted to tell you that I love your videos ! Thanks to you I got better in analysis and had very good grades for my exams, and I wondered if you'd consider doing more probabilities / vecotrial, linear/ bilienar algebra problems ?
I wanted to share with you and everyone one probability exercice that I find pretty fun :
p balls ranging from 1 to N are drawn with replacement. Let X be the random variable that is the maximum (biggest number) of the p balls.
1)Describe (omega, P)
2) Find P(X = k), k ranging from 1 to N (hint : consider the event (X
I like your explanation of what intuiton one should have. But i also want to submit a perhaps more formal answer:
- infinitely 0's added is indeed 0. But this is a series, and each term approaching zero doesnt imply convergence, and specifically not convergence to 0, (in this case it is convergent though) :)
#YAY could you make video about complex derivatives?
Guys this is why he takes upper limit as 1. If you know riemann sum, your width of each rectangle is what? Well it's whatever length you have divided by how many rectangles right? Well 1/n is saying you have a width of 1 of the total function (so upper limit 1 lower limit 0) and divided by n rectangles
0:03 Camera shy
#YAY! I was able to figure this one out quickly because of another video you did a while back in your "Math for Fun" playlist. Now, when I see a limit of the form "0+0+0+...+0," I start by looking to see if I can turn the limit into an integral. So, with this limit, once I realized it was an integral, I just wrote down the definition of the integral as a limit, then found the Δx, the x_k, and the bounds of integration. It was easy from that point; I did the rest in my head!
I love your sense of humor (e.g. the "awkward" beginning of the video). I bet your math classes are loads of fun!
I think you can imagine my face at 5:35 when I realised it's basically the upper sum of the function sqrt(x), aka the integral on [0,1]. I really need to review my calc definitions...
Thanks to you, I now _hate_ math. Took 25 years, but you did it!
Not sure why you left such a comment...
Give us some reasons, we will listen : )
He's being sarcastic :)
Another awesome video/explanation.
Black pen, red pen yaaaayy!!! ☺
Thanks for you 💕💕
شكرا جزيلا
4 seconds and i got it, nice try my friend
Did i hear more math videos in the summer? Now that's worth a #YAY!
Yay!!!
So today I was called up by my mathematics teacher in school to solve a math problem on the board.
I realised how hard it was to write on it. My writing turned out to be in shape of a curve.... Which you just integrated!!
#YAY!
Bro💀💀💀
Too easy
Since you forgot, let me do this for you #YAY
My previous comment is wrong.Sorry,the answer is pi/4.
Dont wanna spam you bad will ya explain how to tackle x = ln x?
x=ln(x) has no solutions, ln(x) is always less than x
phunmaster2000 it has solution. Just not real.
oh yeah you're entirely right, I was under the assumption they didn't want a complex answer
Dank Kush
Here is an intuitive way of thing about it:
Raise e to the power of each side; the equation becomes x=e^x. It's hard to compare the functions, so let's use the power series for e^x. (The first step is because the power series for e^x is simpler than for ln(x).
e^x=1+x+1/2*x^2... and so on. We only need to worry about comparing values within the radius of convergence, since the radius is the entire domain when there are infinitely many terms. Anything true within the radius is true for the entire function. Looking at the first two terms of the power series, we get e^x≈x+1. x+1 is never equal to x, and so within the radius of convergence, e^x is never equal to x. As we add more terms to the series, the radius will expand, showing that x is never equal to e^x, and therefore x never equals ln(x).
Yes, this is a little bit like "killing a fly with a bazooka," but it is more rigorous and can be used to solve more complicated equations and *prove* statements such as "ln(x) is always less than x."
Dank Kush
You used x, not z, so I assumed real values. The explanation for complex numbers is quite lengthy, so I will link to Wikipedia. If you need the explanation for a class, there is enough information to solve in the page.
en.m.wikipedia.org/wiki/Lambert_W_function
Easiest way to calculate ths integral is the reverse of power rule and use Newton - Leibniz theorem
My teacher asked us to find the integral by taking an infinite reimann sum. Is that a thing?
actually we can use squeeze theorem to do this problem as well
Ok, but why is the integral of √x from 0 to 1 equal to 2/3? What is the antidetivative of √x?
Write sqrt(x) as an exponent and use reverse power rule
i dont looked answer. My answer is:2/3
infinity times 0 should be 0^infinity
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Can you do a Tutorial for JEE ADVANCED MATHS (Indian premiere Engineering college entrances)
- can you help me with my limits homework please?
+ well, that's actually an integral...
- whaaat....
Well is it 2/3+ or 2/3- ?
Could somebody please explain to me why does he take as an upper limit 1? Is it by definition or some property?
Julián Felipe Herrera Mejía Because the last term has √(n/n)... if it had had something like √(5n/n), we would've taken the upper bound to be 5. More generally, if it ended with √(kn/n), we'd take the upper bound as k. Try doing it in reverse; taking the integral from 0 to an arbitrary constant k, but instead of directly applying the power rule to find the integral, break the area up into 'n' pieces and take the sum of all pieces, with n going to infinity. You'll see for yourself why 1 is the upper bound in this case.
Thank you very much!
Julián Felipe Herrera Mejía You're welcome :)
This is of format 0* infinity
This is worst than 3=0
Tried to apply this for 1/x from the last video: (1/n)*(n/1)+(1/n)*(n/2)+...+(1/n)*[n/(n-1)]+(1/n)*(n/n) => 1+1/2+1/3+...1/n. It does look divergent.
Yeeeeeeeeeeeeh!! #yay
#YAY
this is amazing
Thanks for you
#YAAY#YAY#YAYAYA#AYY#YAY
Anofhena Hemtinabek ❤
Sum from k=1 to n of 1/n f(k/n) converges to integral from 0 to 1 of f(x)dx
That's calculus bending on itself #YAY#limtointegral
#Yay
Ur great buddy!
This is just the Riemann-integral method, isn't it?
Yeah, I suspected that those were not exact zeros that got added up. A number and a limit (a number too!) are two different things, and is source of confusion for many.
This is the first video of yours I was able to do before watching
Holy cow
Hello
I am in the intro!! #Yay
#yay!!!!!!!
0:10 best song from the best anime (fight me)
: )
Is this a joke or are you talking about the background being an actual anime song?
The show's called doraemon. It's one of the best shows from my childhood.
Ah okay. I didn't know the song was from that. It's one of my childhood favourites too.
nice example of the 0+0+0+0... madness
You can also do that with sandwich theorem without using integration
Do you think it would be reasonable to attempt a Delta epsilon proof here?
Also could you do a dummy variable substitution letting n=x where x is a real number and then factor out 1/n, Then use L'Hopital?
I don't think I've ever seen a limit that expands with more terms. Maybe with a limit of a series I guess.
A series IS a limit that expands with more terms
lol that satisfying ending music
This reminds me of another problem with limits that sound counter intuitive. Each interior angle of a regular polygon is 180-360/n if n is the number of sides. One can think of a circle as a regular polygon with infinitely number of sides. Thus, each interior angle would be 180 degrees. So therefore, a circle would be a straight line.
Contradiction!
Well it is true that a circle is differentiable at every point while finite polygons are not differentiable at each vertex. One way of stating differentiability is that a tangent line can be defined at the point. In simpler terms, we sometimes say that each differentiable point has "local linearity," i.e., behaves like a straight line if you zoom in far enough.
There is no contracdiction here. When n goes to infinity you get aleph0 points where there's an angle of 180, as many as the rational numbers on the number line, which have a measure of 0.
So in total there is a 0 length out of 2pi where the circle is a straight line. I'm fine with that.
You make see things in different ways! So awesome! :-)
Very smart to determine the identity of this limit sum to an integral of a square root so it becomes a no brainer in the end! Very admirable!
noice
I guess I'm first Yay! #yay
no
i don't know if you've made a video about how to calculate the infinite sum of 1/n * (-1)^n using calculus but i just saw the proof of it and it's really neat and if you haven't made a video about it i think you should
Oh yes, I did! I also used complex numbers.
Thanks for your comment.
bprp
*awful intro*
How old are you?
There's an exact formula for Σ(x), Σ(x²), etc. Is there a formula for Σ(sqrt(x))?
as far as i know no, but in this video you can notice that for large n is asymptotic to 2/3*sqrt(n^3)
damn, this channel never ceases to amaze me
I have a lot of trouble in relating the sum with integration.
Integrals can be modelled as being the same as an infinite sum of tiny sliced parts of a function, the width of the slices being modelled to tend towards zero giving the most accurate result for an area under a function as possible- here the sum is slices of sqrt(x), 1/n is the size of the slices being summed, and the slices are fractions of the total n in the radical- as n tends to infinity, this limits towards the area under the graph of y=sqrt(x) from 0 to 1
MGtv The video uses the formal definition of an integral, sometimes called a 'Riemann sum' I think. You would probably find it in any introductory calculus textbook.
You might have also heard the idea introduced in a topic known as the 'limit of sum'.
read into Riemann summation.
Thank you guys
why you didn't solve x=e^x i will give you a hint it is a complex solution
whooooo such a mathmaster, i think he'll probably know that it is a complex number XD
Very cool, isn’t it?
How can you proove that the limit of this sum is the area under the curve (the integral)?
Just say “trust me bro, the math checks out” 😅
@@navjotsingh2251 It's actually not so hard to prove if you use well the uniform continuity of the function. But's that's actually a theorem.
impressive👍🏻👍🏻😃
Happy birthday! :)
Hi! Are you going to do some geometry problems, because the video contains "areas"? Leak? : )PS: If you add infinitely many zeros, that's 0*infinity, so we don't really know what the result is. So, I can't see any other method of doing this.
if you add ap exatly 0 infinite time you get 0. n*0 is equal to 0 for any n. The difference come when you try to sum up an infinite number of element that each tends to 0
Yeah, you're right, I was thinking of it, but wrote it in the wrong way. If we could change 0*infinity into the form infinity / infinity, we would be able to use L'H. But it seems we can't do it here.
What I did was build a square at the points 0,0 1,0 0,1 and 1,1. The function intersects it at 0,0 and 1,1, splitting it into 2 parts (the part we need to integrate and a part above the function but still in the square), nevertheless, the total area of the square must equal 1, since 1*1 = 1. If you rotate the graph switching the x-and-y-axes, you now have the function x^2 (one side, at least since the other if unrotated would be the negative root but that's not the point...(i.e. turning sqrt(x) into x^2 by rotation)). The other area which was formerly above sqrt(x) is now below x^2, and its area is of course integral(0 to 1) of x^2 which is 1/3 (1/3* 1^3). Then, the area above it is 1-1/3, or 2/3......
#Yay!
I wish I could show a drawing, but I can't......
Integral of sqrt(x) is, according to the power rule, x^(3/2) over 3/2, which is 2/3 * x^(3/2). So from 0 to 1 we have 2/3 - 0 = 2/3. But I get your solution, it's definitely more creative and shows some interesting properties. Look, that if you have such square, functions sqrt(x) and x^2 divide the square to the parts of the same areas = 1/3.
I think what is missing from this video is an explanation, why "other" ways of solving this integral are not valid. Those terms on their own would converge to zero, but the sum doesn't - if we get a sum, how can we know? We now have one specific example, but no general intuition.
Silvus when i have problem with this intuition i think to the sum of 1/n n times
If the sum boundaries do not depend on your variable then you take the result of summation (as if it's finite), then take its lim it (e.g. as you do with geometric series - sum x^n = lim of 1-x^(n+1) / 1 - x = 1 / 1 - x).
If the sum boundaries DO depend on your variable then do something clever :-P
I still dont get why the interval is from 0-1 and not 2-3, etc.
Because we integrate from 1/n to n/n which for n->bignumber is from 0 to 1
If the interval was 2-3, we would have inside the square roots (2+1/n), (2+2/n), (2+3/n), …, (2+n/n)
How is this not in conflict with all the limes theorems such as:
If a_n -> a, b_n -> b, then a_n + b_n -> a+b.
Der Justus The number of terms is proportional to n in this case, not a finite constant. As each term gets closer to zero, you add another term to the sum, which is not something those theorems permit.
As an easy example, what's the limit as n goes to infinity of 1/n+1/n+... with n terms? For any value of n, the sum is 1, so the limit is 1, despite each term going to 0.
iabervon oh yeah of course. Didn't think of that, thanks!
because you're adding finite terms in that theorem, things mess up with infinity numbers
Wow, nice ~
Is there any way to solve this without turning it into an integral?
Not without a bunch of convoluted advanced math. The "textbook" way to evaluate the given area with Riemann sums (before they start solving all areas with Fund. Theorem of Calc and using antiderivatives) uses a bit of a trick to avoid having to sum square roots. I think Ramanujan helped develop a formula for summing square roots but there are pieces of it way beyond regular calculus.
Very interesting! #Yay!
Tricky
Thank you! #yay
This is brilliant !
Infinite sums = Integral
MINDBLOWN
#yay
Its more like integral = specific infinite sums
not all infinite sums can be described with integrals. Probably.
I know that. I didn't want a long comment
Each term would need to approach zero, or we could just define it to be an integral between 0 and infinity
tu chipotes #french
It’s a Riemann integral!
Mi Les ikr?!
yeees but where is all that confusing notation?? This video is impresise i don't like it :(
why? it seems clear to me
I mean it is not mathematically rigurous. It is made with intuition, he did not write the riemann sum theory. The division, the function etc.. I would have like to see those too. I don't know the proper therms in english.
But doesnt the part of the rectangles above the curve make the area slightly bigger?
Expanding on misotanni's point, when n = 1, Σ = 1. When n = 10, Σ ≈ 0.71051. When n = 100, Σ ≈ 0.67146.
When n = 1000, Σ ≈ 0.66716. When n = 10000, Σ ≈ 0.66672.
Yes but that's why we take the limit (when there are infinitely many rectangles the extra parts tend to zero)
JGLP doesn't*
Yeeehh
#yayyyyyyy
Excellent job