this channel is amazing. The first video i watched was the 101 integrals. I sat up 5 hours doing the 101 integrals with him, and ever since ive been amazing at calculus. I might do it again soon.
I like it, very well explained. Alternatively, in the second series we could add the "finite" 1 in both sides of inequality, since: 1/(1^3) = 1 , so Sum[n=1 to inf] 1/(n^3) < 1.5 , it converges.
Now, a bit further, if we use same inequality of first series, we could get: Sum[n=1 to inf] 1/(n^3) > 0.5 The convergence value (Cv) of second series would be in a interval such as: 0.5 < Cv < 1.5 Since the second series already has 1 on the first term, and all other terms are positive, we could reduce the interval such as: 1 < Cv < 1.5 The arithmetic mean would lead us to: Cv ~ 1.25 Geometric mean: Cv ~ 1.225 and Harmonic mean: Cv ~ 1.2 The Harmonic mean involves the Harmonic series of values, maybe we can have faith on this one, using the calculator for the first ten terms (n=1 to 10) we got 1.19753... with a slow increment rate (
Yes, we can try Weighted mean for the sake of curiosity, using the differences of area of the first rectangle [x=1 to 2] from Integral ("exact") area. Proposal: use the differences of areas in a Weighted mean in such a way that the area with less difference has more weight. Integral area (Ai): -1/(2.x^2) ]1 to 2 = (-1/8) - (-1/2) = 3/8 = 0.375 "Under curve" area (Au): 1 × 1/(2^3) = 1/8 = 0.125 ... Ai - Au = 0.25 "Over curve" area (Ao): 1 × 1/(1^3) = 1/1 = 1 ... Ao - Ai = 0.625 Meaning the "under curve" area Au is more approximate to the Integral area Ai than "over curve" area Ao. Then, in the case of the second series, the convergence value Cv would be better approximated by a lower value in the interval, the weighted average would be: Cv ~ Wm = (0.625×1 + 0.25×1.5) / (0.625 + 0.25) ~ 1.143 Hmmm.. not quite as we could expect maybe because just the first rectangle was used as weight despite the fact it has the most evident difference. Can we assume that for series involving 1/(n^p) the Harmonic mean of two Integral tests is the best way to find an approximated convergence value?
Well explained! Cool I'm graduating by the end of this month and I'm still learning new things with you guy. My teacher's (calculus I) explanation didn't even get close to yours. Thanks to make it easier
BPRP I AM STUDYING IN IIT BOMBAY THIS CHANNEL REALLY HELPFUL FOR MY SEMI FINALS AND MY SEMI FINALS WERE COMPLETED NOW I AM PREPARING FOR FINALS PLEASE TELL GOOD LUCK TO ME IHAVE WRITTEN VERY WELL THAT I GOT 3 RD RANK ALL OVER INDIA IN MY PREVIOUS EXAMS THANK YOU BPRP
My method for the first one: comparing to the harmonic series, concluding every term is bigger or equal to the term in the harmonic series, thus the series is bigger than the harmonic series and it diverges
The f(x) does not have to be a decreasing one. You can show that S(n) = 1+2+3.. diverges while the corresponding f(x) increases. Take f(x) = x-1. Int f(x) = (x^2/2)-x = x(x/2 -1). This gives positive infinity. S(n) is certainly not -1/12.
as a high school senior that is done with school, part of my brain wants to keep learning so I don't forget everything when I get to college in the fall, and the other part of my brain makes me wanna run away from anything having to do with math.
Oh my god, I just used integral test (Riemann Sum) to solve the challenging problem 93 posted by @LetsSolveMathProblems , and I see this video right after that!
How do i know whether the rectangles will be under the curve or below the curve. The drawing of the curve was random then the rectangles seem to be either over them or below them, what am I getting wrong?
The short answer is: You DON'T need it. I think the reason why many textbooks include it is to make sure that f is actually integrable on bounded intervals. However, this is not necessary since decreasing (not necessarily continuous) functions are already integrable on bounded intervals.
hey bprp! can you do a video explaining the limit x-> -inf of ((sqrt(x+1)-sqrt(1-x))/x there’s a complex argument involved, and i dont know how to conceptually understand i times infinity
The question does not make sense. If you are taking a calculus II course, and the professor assigns the question, then the professor is making a fundamental mistake, because the expression does not mean anything. You need complex analysis to be able to work with a question that uses functions from the set of reals to the set of complex numbers, and then notate it properly. You also need topology, since you need to specify a topology for the complex plane. Are we using Alexandroff compactification to give the complex plane a projective topology? Are we giving it a torus topology? Or are we using affine topology? The answer to this matters, because as you said in your comment, you do not know how to work with i♾, and the reason you do not how to do it is because there genuinely is no unambiguous way to do it without first specifying the topology, since again, the topology determines the answer. The question, on a fundamental level, makes no sense for that reason.
@@angelmendez-rivera351 well, nice. now i know more. i would have much preferred if you answered with that to begin with instead of condescendingly saying my question was meaningless. i think it's important to ask questions even if they seem stupid, because now i can go research more about what you mean and deepen my understanding of my question
trace williams You're putting words in my mouth. I was never condescending, and I never said your question was stupid. I don't appreciate that you're pulling these false accusations out of your ass for no reason. If you want to have an excuse to be mad at me, I can give it to you now, because now I'm disappointed, but you had genuinely no reason to assume I was being condescending. "This question is meaningless" is just another phrasing of "this question makes no sense", except the former is a bit more precise. You asked for an elaboration, I elaborated. What is your problem?
My finals r in a few days, this channel has been really useful,
Wish me luck everyone
SUBSCRIBE TO PEWDIEPIE best of luck!!!
@@blackpenredpen - I hope T Series bombs his finals.
😍❤️
All the best!
Good luck 🍀👍 ✨
It’s an integral part to our understanding 🙂
I'm paying attention, but the pen switching algorithm at 5:53 is mesmerizing!
You have to be the coolest calc teacher ever (seriously). Keep it up steve we love you
Brilliant explanation with useful figures that aid visualisation. Thank you.
Thank you!
Thank you so much. Clear and concise teaching!
I've thought for this for a while, it's exactly the explanation I want!
Thank you!
I got my IB calc paper 3. Your series on series has been crazy helpful:D
Yay!! I am glad to hear!
No way, I was looking up last night for a video on this topic, and then my favorite channel uploads one. Fantastic
: ))) Thanks!
And I guess you'd need this one too: ruclips.net/video/Gfnfmut6zP0/видео.html
this channel is amazing. The first video i watched was the 101 integrals. I sat up 5 hours doing the 101 integrals with him, and ever since ive been amazing at calculus. I might do it again soon.
Steve biju john nice!! I m glad to hear!
who would win?
p-series vs t-series
The infinite series xD
I like it, very well explained. Alternatively, in the second series we could add the "finite" 1 in both sides of inequality, since:
1/(1^3) = 1 , so
Sum[n=1 to inf] 1/(n^3) < 1.5 , it converges.
Now, a bit further, if we use same inequality of first series, we could get:
Sum[n=1 to inf] 1/(n^3) > 0.5
The convergence value (Cv) of second series would be in a interval such as:
0.5 < Cv < 1.5
Since the second series already has 1 on the first term, and all other terms are positive, we could reduce the interval such as:
1 < Cv < 1.5
The arithmetic mean would lead us to:
Cv ~ 1.25
Geometric mean: Cv ~ 1.225
and Harmonic mean: Cv ~ 1.2
The Harmonic mean involves the Harmonic series of values, maybe we can have faith on this one, using the calculator for the first ten terms (n=1 to 10) we got 1.19753... with a slow increment rate (
Yes, we can try Weighted mean for the sake of curiosity, using the differences of area of the first rectangle [x=1 to 2] from Integral ("exact") area. Proposal: use the differences of areas in a Weighted mean in such a way that the area with less difference has more weight.
Integral area (Ai): -1/(2.x^2) ]1 to 2 = (-1/8) - (-1/2) = 3/8 = 0.375
"Under curve" area (Au): 1 × 1/(2^3) = 1/8 = 0.125 ... Ai - Au = 0.25
"Over curve" area (Ao): 1 × 1/(1^3) = 1/1 = 1 ... Ao - Ai = 0.625
Meaning the "under curve" area Au is more approximate to the Integral area Ai than "over curve" area Ao.
Then, in the case of the second series, the convergence value Cv would be better approximated by a lower value in the interval, the weighted average would be:
Cv ~ Wm = (0.625×1 + 0.25×1.5) / (0.625 + 0.25) ~ 1.143
Hmmm.. not quite as we could expect maybe because just the first rectangle was used as weight despite the fact it has the most evident difference. Can we assume that for series involving 1/(n^p) the Harmonic mean of two Integral tests is the best way to find an approximated convergence value?
Concise and informative. The best resource on the why AND the how I've come across. Thank you!
this guy is great...love his videos
P thank you!
Always The best explanations! We all really appreciate you 🙌
Beautiful! Thank you for doing this.
Well explained! Cool
I'm graduating by the end of this month and I'm still learning new things with you guy. My teacher's (calculus I) explanation didn't even get close to yours. Thanks to make it easier
Thanks a lot.
You are my inspiration.
Keep rocking.
Thank u bprp I wish u were my high school teacher
Huge Thanks Blackpennnnnn
Very nice explaination
You can use this integral test to determine how fast a series either goes off to infinity or how fast it converges
聽完學校的再聽曹老師的解說更懂了~
葉子禎 💪💪好欸！
technically if the first term is infinite then you can't conclude just from its corresponding integral that the series is convergent
I actually understood this video! I can't believe it!
Great ideas and brilliant.please keep uploading
BPRP I AM STUDYING IN IIT BOMBAY THIS CHANNEL REALLY HELPFUL FOR MY SEMI FINALS AND MY SEMI FINALS WERE COMPLETED NOW I AM PREPARING FOR FINALS PLEASE TELL GOOD LUCK TO ME IHAVE WRITTEN VERY WELL THAT I GOT 3 RD RANK ALL OVER INDIA IN MY PREVIOUS EXAMS THANK YOU BPRP
WOW!! Best of luck to you!!!!
@@blackpenredpen tq bprp
Nice work 😉👍
Great video! :D
Thanks you ur maths is easy to do thanks u my dear
fantastic, thank you
My method for the first one: comparing to the harmonic series, concluding every term is bigger or equal to the term in the harmonic series, thus the series is bigger than the harmonic series and it diverges
HelloItsMe
Yea. Comparison test!
And you usually shows that the harmonic series diverges thanks to an integral so its the same thing in fact :p
@@lelouch1722 I do it with 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8....
HelloItsMe But this video is about why the integral test works, not about why 1/sqrt(n) diverges.
@@angelmendez-rivera351 that is true indeed
The f(x) does not have to be a decreasing one. You can show that S(n) = 1+2+3.. diverges while the corresponding f(x) increases. Take f(x) = x-1. Int f(x) = (x^2/2)-x = x(x/2 -1). This gives positive infinity. S(n) is certainly not -1/12.
as a high school senior that is done with school, part of my brain wants to keep learning so I don't forget everything when I get to college in the fall, and the other part of my brain makes me wanna run away from anything having to do with math.
Amazing!
Can you evaluate more infinite sums
Oh my god, I just used integral test (Riemann Sum) to solve the challenging problem 93 posted by @LetsSolveMathProblems , and I see this video right after that!
Wow nice!
wow learned something new today
Say, couldn't you use the same argument in the first case (1/(sqrt n)) such that n=2 to infinity?
Really good . Thank you
First comment: Clear and concise!
Thank you!
why must it be decreasing? Can't you apply the same idea to an increasing function (that maybe asymptotes to a finite value)
In that case it would be divergent anyway. Remember we need a_n to approach 0 in order to even have a chance of Sum(a_n) to be convergent.
@@blackpenredpen uh I see, thanks!
*blackpenredpen* What if it increases to 0?
How do i know whether the rectangles will be under the curve or below the curve. The drawing of the curve was random then the rectangles seem to be either over them or below them, what am I getting wrong?
I have a major crush on him from now on
Nice video!
could you also prove 1/n^3 converges even with left end point rectangles if you first proved 1/n^2 converges?
Hey! I'm starting math videos on my channel? Any tips? Should I use a camera and a whiteboard at the uni? Or do I use my surface and microphone?
Hey, if F(x) is integral of f(x), then why does area under the curve, f(x), from a to b equals F(b) - F(a)? Can you please make a video on that?
Fundamental theorem of calculus
Hey will you please tell why the continuity is always needed?
Btw great video😀
The short answer is: You DON'T need it.
I think the reason why many textbooks include it is to make sure that f is actually integrable on bounded intervals. However, this is not necessary since decreasing (not necessarily continuous) functions are already integrable on bounded intervals.
I got the continuous and decreasing part but why should it be positive? Can't we draw the rectangles under the x-axis?
For example f(x)=-1/x^2
That’s because in that case the result would be a negative area both in the integral and in the sum
great keep the videos up
Thank you
You are adorable 🌿
woulda been really useful for that one calc test two weeks ago
Now you're in this you could make a video about Stirling's approximation.
hey bprp! can you do a video explaining
the limit x-> -inf of ((sqrt(x+1)-sqrt(1-x))/x
there’s a complex argument involved, and i dont know how to conceptually understand i times infinity
trace williams This is not a meaningful question.
@@angelmendez-rivera351 what do you mean by meaningful?
The question does not make sense. If you are taking a calculus II course, and the professor assigns the question, then the professor is making a fundamental mistake, because the expression does not mean anything. You need complex analysis to be able to work with a question that uses functions from the set of reals to the set of complex numbers, and then notate it properly. You also need topology, since you need to specify a topology for the complex plane. Are we using Alexandroff compactification to give the complex plane a projective topology? Are we giving it a torus topology? Or are we using affine topology? The answer to this matters, because as you said in your comment, you do not know how to work with i♾, and the reason you do not how to do it is because there genuinely is no unambiguous way to do it without first specifying the topology, since again, the topology determines the answer. The question, on a fundamental level, makes no sense for that reason.
@@angelmendez-rivera351 well, nice. now i know more. i would have much preferred if you answered with that to begin with instead of condescendingly saying my question was meaningless. i think it's important to ask questions even if they seem stupid, because now i can go research more about what you mean and deepen my understanding of my question
trace williams You're putting words in my mouth. I was never condescending, and I never said your question was stupid. I don't appreciate that you're pulling these false accusations out of your ass for no reason. If you want to have an excuse to be mad at me, I can give it to you now, because now I'm disappointed, but you had genuinely no reason to assume I was being condescending. "This question is meaningless" is just another phrasing of "this question makes no sense", except the former is a bit more precise. You asked for an elaboration, I elaborated. What is your problem?
谢谢!
Hey, can you try to Integrate f(x) = tanx/cos(lnx)
You can't.
It may be possible for tan x/ln(cos x), perhaps...
It cannot be done
"Isn't it" is back. I missed it :)
Ok they equal to infinity. What about partial sum of this kind of series?
It was helpful
You forgot to put " [ " on the primitive line such as at 6:15 with " [2 sqrt(x)] 1 -> positive infinit
Raygun Nito No, actually, one does not write it. That is the convention. We are not bracketing the expression
@@angelmendez-rivera351 ok i see, because i learnt with bracketting the expression.
Good
Is it zeta function?)
Thanks. Where are you from? China?
Born in Taiwan if I remember well
But xan you integral test a multiplication of two sums?
The only problem :
Teespring delivery charges are high for countries like India.
I really liked the best friend one.
calc 2 final soon :((((
Also you know that the Zeta function converges only if the absolute value of s is greater than one,however don't put that in a test
Nacho That only works if the function you are adding is n^(-s). It does not work for any other type of function.
Aren't you making this 10x harder then needed. Wouldn't the P series test work on both?
I am watching your video for more than 1 year and till now I don't know your name because you have never told it. Is it a secret???😁😁😁
His name is Steve
Uhhh, you could've simplified 1/sqrt(1) to 1.
Jared Kaiser No need to
D' alembert vs Leibniz test...
You say "take a look of" all the time, but grammatically correct would be "take a look at" in English...
Timur Sultanov
Omg, thank you!!
@@blackpenredpen you are welcome
is it possible to put more than 1 like !
Second !
First!!
me first
your ascent is so heavy
Anything else you wanna say?