When I was at high school in Iran in 1972-1973, one of my dreams was to go to a US University such as MIT. It didn't happen due to many reasons including lack of financial resources. Now at 55 as I watch these courses, I feel this dream has finally come true and I feel very young and enthousiastics; I will watch them all ! Thanks to MIT, to its professors and all its staff.
This is a very difficult way to proove that limits! In France I've never seen that before, I did it with equivalents like sin(x)~x when x goes to 0 and that's muxh easier I think! Anyway this is a good video
@IAmBenyamin Fantastic that you feel that way! However, when you get to college, you might encounter a professor teaching this that isn't anywhere near as good as this guy is .Don't let that discourage you, though! This guy alone got me through calculus. :)
@IAmBenyamin If you follow this through and add a bit of digital signal processing, you'll understand how your MP3 player works. Understanding the mathematics is vital at the conception phase of an idea, which then gets handed off to the engineering people who make these wonders real.
I concur. The Squeeze Theorem makes a cleaner and easier case in proving sinx/x. I was fine the first 2 videos, but I have spent MY WHOLE day meditating and pondering on his geometrical proof to sin/x and cos-1/x. I don't see yet geometrically how sinx/x approaches 1. Is it because the ratio between the curve and the line tends to 1?
It's because the length of the curve gets closer and closer to the length of the line (infinitely close gives the same length and therefor the ratio between the curve and the line tends to 1) where the linelength = sin x and arclength = x.
My confusion: As θ ->0 for sinθ/θ, sinθ goes to 0 faster than θ but not by much. I thought to myself that since 1-cosθ/θ=0 because the numerator is going to 0 faster than the denominator, then sinθ/θ must also be going to 0 since the same is true. But the rate at which 1-cosθ is going to 0 faster than θ is grows infinitely larger, while the rate sinθ is going to 0 faster than θ is growing infinitely smaller, which is why the ratios tend toward their corresponding 1 or 0. I'm better now.
i found a similar proof for the derivative of trigonometric functions. to prove sinx/x=1 we can imagine that as x----->0, the ratio of the opposite side to the hypotenuse approaches zero, but so does x. since they both because infintesimally small, but not exactly zero. their quotient should equal 1 right?
Interesting how watching 3-50 min videos from this guy has taught me and made me actually UNDERSTAND limits, derivatives, continuity, and differentiation MORE than 2 previous semesters of calculus at my local community college where its $20+ a unit (5 unit course) PLUS cost of books PLUS endless hrs at math tutoring PLUS endless supply of caffeine PLUS full cancellation of my social calendar. THANKS MIT !!!!
@brayosaintdead note that the geometric proof is conditioned for theta = curve length in radians. that's why limit of sin(θ)/θ tends to 1 as θ --> 0, for the ratio between the vertical length of the cord (sin θ) and the curve lenght (θ radian) = 1. However for (1 - cos θ) / θ, where the horizontal distance from origin to the cord is cos θ. The ratio between the gap (radius - cosθ) and curve length (in θ) ---> 0, for the gap is much shorter than the the curve length, and thus goes to zero faster
Thank you for making these vids available, MIT! I'm taking calc II at my university in the Spring after a semester away from calculus, so I really appreciate being able to refresh my memory by "taking" a calculus course to prep! This professor is a lot more into proofs than the professor I had, so I'm even seeing familiar things in a new light. Thanks again!
He has good teaching style, but sometimes he use some uglish ways to prove. For example at 43:38 - realy strange way, I? offer? better look to that picture:
He's an amazing teacher. He is actually one of the few teachers I know of who can DELIVER his train of thoughts to the students. I wish I had him as my teacher x] He makes calculus so fun lol.
He has good teaching style, but sometimes he use some uglish ways to prove. For example at 43:38 - realy strange way, I? offer? better look to that picture:
I laughed so hard at this epic explanation form Prof. Jerison at 27:40
"...ah, so the question is what about this business of arc length. So the word arc length, that, that shape, that orange shape is an arc, and we're just talking about the length of that arc, and so we're calling it arc length, that's what the word arc length means, it just means the length of the arc."
43:40 the idea is very hard to comprehend based on his explanation, he could have just drawn the two triangles and shown theta and the third angle of the first triangleb are complementary. Then considering that if you extend the veritcle line that is PR down to the unit length radius 1, then the third angle of the first triangle added with the 90 degrees from the secant line and angle QPR would be supplementary, so than we can conclude that the third angle of the first triangle added with angle
@kitman I don't think he means that the value of the radius will increase, I think he means that to represent the angle in a visible method, he would have to magnify the drawing, hence a longer "looking" distance of the radius
In italian calculus courses there are better proofs for limits sinx/x=1 and (cosx-1)/x=0... here the explanation is very confused, and in particular how can I say that the limits would be 1 and 0 and not, for example, 0.99 and 0.01? We use instead the theorem of comparison for the first limit (sin x < x < tan x) and a trick for the second limit (1-cos x = 2sin^2(x/2)). Very simpler and clearer.
@sanjaigupta Here there is an elegant demonstration:
It's not the only one: you can infer that x>sin(x) banally if you consider that an arc of circle is always longer than the associated chord (arc=2x and chord=2sin(x))
@MatteoPascal also, most Italian calculus teachers (not all, of course) are smug bastards who couldn't care less about their students actually learning something. I quit attending my calculus course because every time I would ask a question it would be simply be labeled as elementary and left unanswered by the professor.
@drsiegfriediseman You are right: in italian calculus courses there are better proofs, NOT better professors. Big difference here! :) We have a long math tradition but I think our teachers are now too proud and far from students. It isn't a coincidence that there aren't italian open courses on YouTube! Anyway, I'm very sorry for your bad experience, I hope you will have better experiences with italian colleagues.
Why doesn't he just say the denominator is decreasing slower than the numerator as theta approaches zero, therefore the quantity approaches zero, as the denominator becomes much larger than the numerator. That is precisely what is happening. I feel if he said that, there would not be so many questions.
i have another idea for the proof of sin(x)/x=1 as x tends to 0
imagine the graph of sin(x)
as x tends to 0 the slope approaches 1 (f'(sin(x))=cos(x) and cos(0)=1). This would mean that x is the x value and sinx is the y. Since the slope approaches 1 then this means that the rise over the run (y/x)=1 or sinx/x=1
@BarbierNicholas That doesn't really work/ make sense. But there is something called L'Hospitals rule, that will be learned later. It makes that kind of problem very routine.
@BarbierNicholas you have interesting idea. But your x never goes to zero. If we take y=sin(x) and (y/x)=1, well we'll get a graph of y=x which is a diagonal line of 45 degree; not x-->0.
oh come on... he's not so good at explaining... what he should say is that the (1-cosine) approaches zero at a much faster rate than theta approaches zero
Very enjoyable and time-saving watching this as a refresher course at 2.7x the speed with a program such as Enounce's MySpeed. I recommend to anyone who likes MIT OCW to search for a program that increases the rate of play of Flash movies such as these. The text for this course is outstanding as well, George Simmon's Calculus with Analytic Geometry, 2nd ed.
@hypeblast he makes the students go beyond being calculus robots. Calculus robots are people who just have all the formulas memorized and can compute crazy derivatives, integrals etc. However, calculus robots, like people who waste their time memorizing pie, do not have the geometric and conceptual understanding of calculus, which is essential to solving the more complicated, real life problems. In Real life problems you aren't just given functions and asked to find derivatives.
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
@hypeblast Calculus is a prerequisite for all physics and engineering courses, which is where the real applications come in. Also many economics programs require calculus. MIT offers a course called calculus with applications. For people who major in math like myself, the real application and understanding of advanced mathematics doesn't start until you learn analysis. In a basic calculus course like 18.01 it is not the professors job to teach advanced applications.
@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
Of course it's going to be easy for a high schooler to understand this. It's the third lecture... People in that classroom only have high school math behind them.
That is bs proofs of the sinx/x geometrical proofs aren't that good. What you should do is the use maclaurins theorem and make sin(x) into a polynomial function. Which in zero acctually equals sinx. Then it's real easy to see that sinx/x is equal to 1
But how would you construct the sine series in the first place? It is usually done by using the nth-order derivatives which makes this a circular proof not something you want. Just defining the infinite series to be sin and cos is really unsatisfying, since they at first glance don't seem to have anything to do with trigonometric ratios in triangles. That's why it's really reasonable to use a geometric proof.
Yea but if you use an errorfunction constricted to a higher power function then that function will go to zero when x goes to zero. Use ordo.
So in that case you just need to use the second order maclaurin term to solve the problem and ofcourse the ordo term constricted to x^4. Then you get the a really easy problem.
What I meant was that to construct the series in the first place, you need to know the derivatives of sin and cos, making the argument circular.
I agree the limit is easy to evaluate with a maclaurin series, but that is only a verification of the limit, not a proof since you have already needed the liimit to construct the series.
@bgreeson I don't see why you mention recursion, my only point was that it is reasonable, though not rigorous, to use a geometric proof of the sin(x)/x limit, since a power series is just a power series and one can't see immediately that those power series are the same as the functions with triangle definitions, at least I still can't, I only know. On a related note, power series come a while further in the course, so it would not make sense anyway to use them.
i think the proof on lim x=0, sinx/x is not mathematically rigorous. The genuine proof is by using squeeze theorem, which restricts sinx/x between 0 and cosx.
agreed 100% thesearcher1, and yes, ftisland1, calc may be difficult as a sophmore in high school, but these videos will help tremendously for an ap calc exam
Yes and no. Watch these videos and then check-out a calc book from a local library or just go to your local community college and read the book in there without checking it out/make copies. Honestly though, you should concentrate on algebra and applied geometry for the SAT. Then, after making algebra and geometry your bitch, you should start on calc. These videos dont seem that hard (he goes pretty slow), but I guess they can be to a new person.
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Well,without any ambiguity, i claim that the A and B expression could have been proved somewhat easily and more plausible beyond the ground of probabilities by simply using L'HOSPITAL'S RULE!
The proofs he used were totally solid. They're also nice because they're very intuitive. L'Hospital's would be premature on the third day (assuming you want to be very rigorous).
It might help if he explained that, as theta goes to zero (in the limit), the lines of the angle get closer and closer to parallel. Then he could draw the picture with two parallel lines, in which case the 1 - Cos(theta) segment would be gone (zero).
This comment has received too many negative votesshow
parent angle of the base triangle must add with 90 - theta to equal 90. Evaluating this as an expression we see (theta x) + (90 - theta) = 90
this form leads to theta x - theta = 0 or theta x = theta. This gives a simple quantitative proof of the two angle's equivalence without the qualitative dribble the teacher attempted to use.
I think the lecturer did quite a bad job in clarifying why exactly the thetas in the two triangles he used for the derivative of sin(theta) were equal. If you notice, as delta(theta) approaches zero the angle between the triangle PQR and the triangle formed by extending a line from R to the base of the unit straight line approaches 90. Also, the angle opposite the theta obviously has the value 90 - theta. The other angle on this perpindicular which the teacher claims to be congruent to the
Wow, don't use this class as a litmus for evaluating the difficulty of MIT in general. My friend is attending MIT this year and she took the class "Calculus with Theory". In fact, many of their students don't take single variable calculus first year. Most of the people taking this class are doing it for the general institute reuirement
MIT will not assume that 4 yrs of HIGH SCHOOL Calculus has sufficienty prepared an ENGINEERING student in the subject. These kids are obviously top of their class, so to speak, but they still start at the beginning like all freshman.
I understand his argument for A and B, but I wouldn't really call it a proof. More like a persuasive description. There seems to be a magic moment where he says use your imagination as things go to the limit and you see that it does not blow up it just continues its trend. I can see that but it does not seem to pbe proven, just plausible. The other proofs he has shown cancellations that leave only real definite quantities behind. I know what he says is true, but I would not call his proof tight.
Well of course - I think the only way to make the proof tight would be to introduce the concept of the delta-epsilon proof and dabble a bit in Real Analysis. These are introduced quite rigorously though in the MIT course Calculus with Theory, which is for well prepared first year students.
"oh these must be kids who didn't have calculus or these classes are introductory classes for kids who never had them in high school"
is that it CLEARLY says on the MIT admissions website that a student MUST have 4 years of math education in high school INCLUDING calculus. And I don't expect MIT recruiting students who SUCK in math either.
Maybe these kids just want an easy A. I don't know.
Should any of you be able to find the general theorem for integrating a system of n variables/n dimensions by yourselves then yes, MIT is too easy for your level.
funnily enough I was at secondary school in UK in 1968
and I completely agree with free-from-influence
well done Gabriella for badgering her husband to make this available
warnford 2 days ago
When I was at high school in Iran in 1972-1973, one of my dreams was to go to a US University such as MIT. It didn't happen due to many reasons including lack of financial resources. Now at 55 as I watch these courses, I feel this dream has finally come true and I feel very young and enthousiastics; I will watch them all ! Thanks to MIT, to its professors and all its staff.
freefrominfluence 1 week ago 4
I am very happy to see the vidoe after you give this Derivatives of products, quotients, sine, cosine
NganaJHone 2 weeks ago
Steady I Really Like This Video Derivatives of products, quotients, sine, cosine
Ondelendo 2 weeks ago
Good, I like that you share this video, I wish success always Derivatives of products, quotients, sine, cosine
bebeheuy 2 weeks ago
Nice Video That You Share , So Very Nice Thanks You Derivatives of products, quotients, sine, cosine
willamricard 2 weeks ago
I Really Like The Video From Your Derivatives of products, quotients, sine, cosine
imegatrone 2 weeks ago
Your Video Derivatives of products, quotients, sine, cosine Is Very Useful Sharing
bundawartini 2 weeks ago
after i watched this video, my insight is very open because the video is very good to give information Derivatives of products, quotients
anakmudajaman 2 weeks ago 3
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anakmudajaman 2 weeks ago
I think multiplying by (1+cosx) up and down, would lead to sin²x/x(1+cosx)
We assume sinx/x is 1 and we are left with sinx/1+cosx which is 0/2.
That's a better proof.
And sinx/x is 1 due to sinx<x<tgx and sinx tends to tgx as x goes to 0
Sheikarthur 3 weeks ago
great lecture thanks again mit
stavrospod 1 month ago
Cool Video!!
Flusercom 1 month ago
Yes. Very - Very Infomative. Thanks MIT
ngeliatduit 1 month ago
Very Infomative. Good Video. Thanks
Spasatcom 1 month ago
Good!! very good lecture!!
kerupukrase 1 month ago
Good Video!!
penyidikanca 1 month ago
I like this video. I think it's great that I don't have to pay to learn calculus
keithl28 2 months ago
i see how the students are so quite in college is that real?
alimunth 2 months ago
@alimunth yes, that is real.
qwertyuiopertyuiopty 1 month ago
That awkward moment when you realize the teacher doesn't see your hand and isn't going to any time soon.
SmileySmillie 2 months ago
man we do these at high school (no im not trying to be an asss but we ACTUALLY do these on high school) i mean im 17 and im doin em right now
4JohN7 2 months ago
No jodas!!
CHmn93 2 months ago
NERD DETECTED
LaryFallcoN 2 months ago
This is a very difficult way to proove that limits! In France I've never seen that before, I did it with equivalents like sin(x)~x when x goes to 0 and that's muxh easier I think! Anyway this is a good video
jojolapin8847 3 months ago
This kind of calculus have aplications in linguistic???
kytypot 3 months ago
wow i envy ppl who can understand this stuff!
Saxxx1991 3 months ago
thabk you subtitles T_T
stevekerr0221 3 months ago
I'm 15, live in england and don't have a clue why im watching this. but ill tell you what, this seems so interesting.
IAmBenyamin 3 months ago 18
@IAmBenyamin dont worry u will get a hang of it wen ur 16!
MyChatwin 2 months ago
@IAmBenyamin Fantastic that you feel that way! However, when you get to college, you might encounter a professor teaching this that isn't anywhere near as good as this guy is .Don't let that discourage you, though! This guy alone got me through calculus. :)
qwertyuiopertyuiopty 1 month ago
@IAmBenyamin If you follow this through and add a bit of digital signal processing, you'll understand how your MP3 player works. Understanding the mathematics is vital at the conception phase of an idea, which then gets handed off to the engineering people who make these wonders real.
Nyquistic 3 weeks ago
lovely blackboard :)
PatinyaHan 4 months ago
i wish he wouldnt do calculus proofs on trig. makes me feel insecure about the result
jennigerr 4 months ago in playlist MIT 18.01 Single Variable Calculus, Fall 2006
I concur. The Squeeze Theorem makes a cleaner and easier case in proving sinx/x. I was fine the first 2 videos, but I have spent MY WHOLE day meditating and pondering on his geometrical proof to sin/x and cos-1/x. I don't see yet geometrically how sinx/x approaches 1. Is it because the ratio between the curve and the line tends to 1?
DNYAP 5 months ago 2
@DNYAP Exactly, thats the geometric way of seeing it.
In fact for very small numbers you can aproximate with low error sin(x)=x
Thats part of the Taylor Polinom
LordWargus 4 months ago
@DNYAP
It's because the length of the curve gets closer and closer to the length of the line (infinitely close gives the same length and therefor the ratio between the curve and the line tends to 1) where the linelength = sin x and arclength = x.
prantare 2 months ago
Wow. I wish I had him when I was taking this stuff 10 yrs ago.Thank you Dr. whoever you are.Thanks MIT!
f1mercury 5 months ago
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My confusion: As θ ->0 for sinθ/θ, sinθ goes to 0 faster than θ but not by much. I thought to myself that since 1-cosθ/θ=0 because the numerator is going to 0 faster than the denominator, then sinθ/θ must also be going to 0 since the same is true. But the rate at which 1-cosθ is going to 0 faster than θ is grows infinitely larger, while the rate sinθ is going to 0 faster than θ is growing infinitely smaller, which is why the ratios tend toward their corresponding 1 or 0. I'm better now.
theshapeoftime 7 months ago
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theshapeoftime 7 months ago
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theshapeoftime 7 months ago
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theshapeoftime 7 months ago
4:20 Minha terra tem palmeiras onde canta o sabiá.
sen(a)cos(b)+sen(b)cos(a)
My land has palmtrees, where sings the thrush bird
sin(a)cos(b)+sin(b)cos(a)
(it doesn't rhime in English...)
caracaes 8 months ago 4
Excellent stuff.
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jenita39 8 months ago
1-cosx just changes faster than x, when x -> 0, for example, how is the limit of n^2/n when n->0. Think about it.
wzhang7 11 months ago
well he said in lecture 1 that his job is to make it more complicated lol...i wish professor strang taught this
SpoiledLogic 11 months ago
i found a similar proof for the derivative of trigonometric functions. to prove sinx/x=1 we can imagine that as x----->0, the ratio of the opposite side to the hypotenuse approaches zero, but so does x. since they both because infintesimally small, but not exactly zero. their quotient should equal 1 right?
amapmwt 11 months ago
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adidasguy87 11 months ago
6 people dont have brains
TheKarate10 11 months ago
This guy must drink a lot of coffee before class lol
xscarlet7 11 months ago
Interesting how watching 3-50 min videos from this guy has taught me and made me actually UNDERSTAND limits, derivatives, continuity, and differentiation MORE than 2 previous semesters of calculus at my local community college where its $20+ a unit (5 unit course) PLUS cost of books PLUS endless hrs at math tutoring PLUS endless supply of caffeine PLUS full cancellation of my social calendar. THANKS MIT !!!!
chesteronline2010 11 months ago 47
@chesteronline2010 really :O ?? then i wont study
MadCowsGaming 3 months ago
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Jameswb98 1 year ago
hOW DID YOU GET LIM > 1
AND THE OTHER PART IS LIM > 0 ?? 7:30
brayosaintdead 1 year ago
@brayosaintdead note that the geometric proof is conditioned for theta = curve length in radians. that's why limit of sin(θ)/θ tends to 1 as θ --> 0, for the ratio between the vertical length of the cord (sin θ) and the curve lenght (θ radian) = 1. However for (1 - cos θ) / θ, where the horizontal distance from origin to the cord is cos θ. The ratio between the gap (radius - cosθ) and curve length (in θ) ---> 0, for the gap is much shorter than the the curve length, and thus goes to zero faster
msirichit 1 year ago
Thank you for making these vids available, MIT! I'm taking calc II at my university in the Spring after a semester away from calculus, so I really appreciate being able to refresh my memory by "taking" a calculus course to prep! This professor is a lot more into proofs than the professor I had, so I'm even seeing familiar things in a new light. Thanks again!
slavito90 1 year ago
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He has good teaching style, but sometimes he use some uglish ways to prove. For example at 43:38 - realy strange way, I? offer? better look to that picture:
tinyurl [dot] com [slash] mit3prove
PikoJn 1 year ago
love how the teacher tried to play it off like he knew the sine sum formula all along when it was clear he forgot it.
adidasguy87 1 year ago
@adidasguy87 lol tryied to play it off? he immedietely listened to his students' opinion.
pithikoulis 1 year ago
@pithikoulis not the impression i got
adidasguy87 1 year ago
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@pithikoulis not the impression i got
adidasguy87 1 year ago
He's an amazing teacher. He is actually one of the few teachers I know of who can DELIVER his train of thoughts to the students. I wish I had him as my teacher x] He makes calculus so fun lol.
cpu194 1 year ago
What a scary lecture....
RobertFromKilikia 1 year ago
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He has good teaching style, but sometimes he use some uglish ways to prove. For example at 43:38 - realy strange way, I? offer? better look to that picture:
tinyurl [dot] com [slash] mit3prove
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lavoniaharris 1 year ago
yeah , it gets complicate when derivated formulus are thrwon in the mix.
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jamyacollins 1 year ago
CAN SOMEONE HELP ME UNDESTAND THE FUNCTION OF cos and sine???when are they used?
junior1984able 1 year ago
@junior1984able do you clearly know what cos and sin are?
VELOCITYangus 1 year ago
@junior1984able there angles of triangles
junior1984able 1 year ago
GOD PLEASE HELP ME UNDERSTAND CALCULUS...
junior1984able 1 year ago
My Brain Hurts!
Slayerfan84 1 year ago
that man is great
strangercandy 1 year ago
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nadeeraist 1 year ago
hehe ... I'm in math class at MIT ... hehe =)
New987user 1 year ago
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New987user 1 year ago
lol awh, teachers who still use chalk.
belathequeen 1 year ago
I laughed so hard at this epic explanation form Prof. Jerison at 27:40
"...ah, so the question is what about this business of arc length. So the word arc length, that, that shape, that orange shape is an arc, and we're just talking about the length of that arc, and so we're calling it arc length, that's what the word arc length means, it just means the length of the arc."
LOL
EgoSumIndigentia 1 year ago 2
Geez...I like how his lecture is so slow
onetimesometime 1 year ago
i like it how he said zero was his favorite number. My calc teacher said zero is boring. They should get into a math teacher debate.
BYMYSYD 1 year ago
Where does he prove that sin(x)/x= 1?
michalchik 1 year ago
@michalchik it's the math dragon theorem
theopakalypse 1 year ago
@michalchik
He talks about it in lec 2. He doesnt prove it, though.
farlyso 1 year ago
Some impressively dumb questions from MIT students, huh...?
Raizdecimal 1 year ago
@Raizdecimal Many of them are there because they work very hgard and make few mistakes, not because if the depth of their insight,
michalchik 1 year ago
QPR would also be complementary, therefore angle QPR must equate to theta.
biksengrity 1 year ago
43:40 the idea is very hard to comprehend based on his explanation, he could have just drawn the two triangles and shown theta and the third angle of the first triangleb are complementary. Then considering that if you extend the veritcle line that is PR down to the unit length radius 1, then the third angle of the first triangle added with the 90 degrees from the secant line and angle QPR would be supplementary, so than we can conclude that the third angle of the first triangle added with angle
biksengrity 1 year ago
33:50 don't understand why theta get smaller and smaller, the origin go farther away?
kitman 1 year ago
@kitman I don't think he means that the value of the radius will increase, I think he means that to represent the angle in a visible method, he would have to magnify the drawing, hence a longer "looking" distance of the radius
biksengrity 1 year ago
lol took long enough on the sum identity
yodilingturkey 1 year ago
In italian calculus courses there are better proofs for limits sinx/x=1 and (cosx-1)/x=0... here the explanation is very confused, and in particular how can I say that the limits would be 1 and 0 and not, for example, 0.99 and 0.01? We use instead the theorem of comparison for the first limit (sin x < x < tan x) and a trick for the second limit (1-cos x = 2sin^2(x/2)). Very simpler and clearer.
MatteoPascal 1 year ago 6
@MatteoPascal for completeness, how would you prove that sin x < x < tan x in a small interval containing 0?
sanjaigupta 10 months ago
@sanjaigupta Here there is an elegant demonstration:
It's not the only one: you can infer that x>sin(x) banally if you consider that an arc of circle is always longer than the associated chord (arc=2x and chord=2sin(x))
MatteoPascal 10 months ago
@sanjaigupta
I couldn't write the link, so I will send you by e-mail.
MatteoPascal 10 months ago
Some of the questions made were very naive.
caracaes 8 months ago 2
@MatteoPascal also, most Italian calculus teachers (not all, of course) are smug bastards who couldn't care less about their students actually learning something. I quit attending my calculus course because every time I would ask a question it would be simply be labeled as elementary and left unanswered by the professor.
I WISH I had had this professor in class.
drsiegfriediseman 7 months ago 2
@drsiegfriediseman You are right: in italian calculus courses there are better proofs, NOT better professors. Big difference here! :) We have a long math tradition but I think our teachers are now too proud and far from students. It isn't a coincidence that there aren't italian open courses on YouTube! Anyway, I'm very sorry for your bad experience, I hope you will have better experiences with italian colleagues.
MatteoPascal 7 months ago
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mathforphysics 5 months ago
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mathforphysics 5 months ago
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mathforphysics 5 months ago
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mathforphysics 5 months ago
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Why doesn't he just say the denominator is decreasing slower than the numerator as theta approaches zero, therefore the quantity approaches zero, as the denominator becomes much larger than the numerator. That is precisely what is happening. I feel if he said that, there would not be so many questions.
bgreeson 1 year ago
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bgreeson 1 year ago
He never got to showing lim x --> 0 of ((cos(x) - 1)/x) or of sinx/x last lecture, but in this lecture he asks if we remember from last time???
bgreeson 1 year ago
kevin spacey
IceCork90 1 year ago
i have another idea for the proof of sin(x)/x=1 as x tends to 0
imagine the graph of sin(x)
as x tends to 0 the slope approaches 1 (f'(sin(x))=cos(x) and cos(0)=1). This would mean that x is the x value and sinx is the y. Since the slope approaches 1 then this means that the rise over the run (y/x)=1 or sinx/x=1
BarbierNicholas 1 year ago
@BarbierNicholas That doesn't really work/ make sense. But there is something called L'Hospitals rule, that will be learned later. It makes that kind of problem very routine.
bgreeson 1 year ago
@BarbierNicholas you have interesting idea. But your x never goes to zero. If we take y=sin(x) and (y/x)=1, well we'll get a graph of y=x which is a diagonal line of 45 degree; not x-->0.
msirichit 1 year ago
whats 18.02?
MyHugestFan 1 year ago
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Raizdecimal 1 year ago
@MyHugestFan Multivariable Calculus
Raizdecimal 1 year ago
oh come on... he's not so good at explaining... what he should say is that the (1-cosine) approaches zero at a much faster rate than theta approaches zero
alquiora 1 year ago
Very enjoyable and time-saving watching this as a refresher course at 2.7x the speed with a program such as Enounce's MySpeed. I recommend to anyone who likes MIT OCW to search for a program that increases the rate of play of Flash movies such as these. The text for this course is outstanding as well, George Simmon's Calculus with Analytic Geometry, 2nd ed.
caramonspace 1 year ago
calc is easy this guy makes this subject complicated?
hypeblast 1 year ago
@hypeblast he makes the students go beyond being calculus robots. Calculus robots are people who just have all the formulas memorized and can compute crazy derivatives, integrals etc. However, calculus robots, like people who waste their time memorizing pie, do not have the geometric and conceptual understanding of calculus, which is essential to solving the more complicated, real life problems. In Real life problems you aren't just given functions and asked to find derivatives.
hot4math 1 year ago
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
hypeblast 1 year ago
@hypeblast Calculus is a prerequisite for all physics and engineering courses, which is where the real applications come in. Also many economics programs require calculus. MIT offers a course called calculus with applications. For people who major in math like myself, the real application and understanding of advanced mathematics doesn't start until you learn analysis. In a basic calculus course like 18.01 it is not the professors job to teach advanced applications.
hot4math 1 year ago
@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
hypeblast 1 year ago
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@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
hypeblast 1 year ago
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@hot4math well If ts not the professor job to teach, then why will you go to school? Why would you waste thousands and thousands for school? what are you doing in school? Not everyone is George Green.
hypeblast 1 year ago
This has been flagged as spam show
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
hypeblast 1 year ago
This has been flagged as spam show
@hot4math its funny how you mentioned robots, ive seen many Asians just solved so much problems but an no knowledge in applications but this guy here sucks for any culture and he will make studying difficult ; this guy at MIT is a waste of time..
If you have taken calculus you will know this guy is weak.. He never explains the definition trigonometric d/dx functions which is Lim-> 0 ((f( x+h)-f(x) / h))) = ?
he makes something else up.. he dont teach you any applications? what concept?? =)
hypeblast 1 year ago
@hypeblast That defn of derivative was in the first lecture...
bgreeson 1 year ago
@hypeblast What parts do you think he's making complicated? I'm at 26:30 and so far he's been quite clear for me.
5w33ti3 1 year ago
Of course it's going to be easy for a high schooler to understand this. It's the third lecture... People in that classroom only have high school math behind them.
ELkuken 2 years ago
yea, a high schooler could do as well as these students at MIT!
siver1992 2 years ago
its a highschool course, taught in case there are people who missed it(ie art students and such, who just need it to pass)
zantrua 2 years ago
If you take an example by replacing theta or x with a number, it becomes instantly clear why SinX/X = 0 and why (1-CosX)/X = 1
use your calculator and enter X as 0.000000000000000000000000009 and find the cos and sin.
cos of that is 1
and sin is 0.
the number is still not zero. Imagine what if it were a decimal point followed by 100000 zeros and then a nine.
Dividing sin and (1-cos) of that number with the number does not result in an indeterminate.
gunghojoeno 2 years ago 2
@gunghojoeno
Thats not what he's proofing though. He's showing the sin(x)/x = 1, not 0.
schlynn 2 years ago
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gunghojoeno 2 years ago
@schlynn
Sorry guys, I got that ass backwards
Sinx/x = 1 and (1-cosx)/x = 0. Thanks to schlynn for pointing that out.
gunghojoeno 2 years ago
That is bs proofs of the sinx/x geometrical proofs aren't that good. What you should do is the use maclaurins theorem and make sin(x) into a polynomial function. Which in zero acctually equals sinx. Then it's real easy to see that sinx/x is equal to 1
liwa2k7 2 years ago
But how would you construct the sine series in the first place? It is usually done by using the nth-order derivatives which makes this a circular proof not something you want. Just defining the infinite series to be sin and cos is really unsatisfying, since they at first glance don't seem to have anything to do with trigonometric ratios in triangles. That's why it's really reasonable to use a geometric proof.
sikory 2 years ago
Yea but if you use an errorfunction constricted to a higher power function then that function will go to zero when x goes to zero. Use ordo.
So in that case you just need to use the second order maclaurin term to solve the problem and ofcourse the ordo term constricted to x^4. Then you get the a really easy problem.
liwa2k7 2 years ago
What I meant was that to construct the series in the first place, you need to know the derivatives of sin and cos, making the argument circular.
I agree the limit is easy to evaluate with a maclaurin series, but that is only a verification of the limit, not a proof since you have already needed the liimit to construct the series.
sikory 2 years ago
@sikory And recursion doesn't exist.
bgreeson 1 year ago
@bgreeson I don't see why you mention recursion, my only point was that it is reasonable, though not rigorous, to use a geometric proof of the sin(x)/x limit, since a power series is just a power series and one can't see immediately that those power series are the same as the functions with triangle definitions, at least I still can't, I only know. On a related note, power series come a while further in the course, so it would not make sense anyway to use them.
sikory 1 year ago
@sikory OK, got it.
bgreeson 1 year ago
i think the proof on lim x=0, sinx/x is not mathematically rigorous. The genuine proof is by using squeeze theorem, which restricts sinx/x between 0 and cosx.
oliverzya 2 years ago
I think you ment restricting f(x) = sin(x)/x between f(x) = cos(x) and f(x) = 1.
Arkantosi 2 years ago
agreed 100% thesearcher1, and yes, ftisland1, calc may be difficult as a sophmore in high school, but these videos will help tremendously for an ap calc exam
stuckintraffic123 2 years ago
so i'm a sophmore in high school n i was wondering if i took all these lectures, will it give me enough information to take the ap calc ab exam??
and i'm finding these lectures really hard....is tht cause i'm only a sophmore??
ftisland1 2 years ago
Yes but you should do Calc problems as well to get in down.
jasonmathias 2 years ago
Yes and no. Watch these videos and then check-out a calc book from a local library or just go to your local community college and read the book in there without checking it out/make copies. Honestly though, you should concentrate on algebra and applied geometry for the SAT. Then, after making algebra and geometry your bitch, you should start on calc. These videos dont seem that hard (he goes pretty slow), but I guess they can be to a new person.
DomingoLuis 2 years ago
It should help you on the exam, but I'm a freshman in high school and i think these lessons are easy as hell...
stupidfleshmonkies 2 years ago
I like his style,
I think he is very good for obsessive students,
I like his (concise/big picture) verbal use
(I claim, visualize, kinds of interpretations,etc...)
He is also funny: "I will use this to make things
more confusing, I reversed it to make things more complicated, calculus made hard"
I hope these students know his value.
Thank you MIT for your videos.
TheSearcher1 2 years ago 33
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PikoJn 1 year ago
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Well,without any ambiguity, i claim that the A and B expression could have been proved somewhat easily and more plausible beyond the ground of probabilities by simply using L'HOSPITAL'S RULE!
furthermaths 2 years ago
The proofs he used were totally solid. They're also nice because they're very intuitive. L'Hospital's would be premature on the third day (assuming you want to be very rigorous).
It might help if he explained that, as theta goes to zero (in the limit), the lines of the angle get closer and closer to parallel. Then he could draw the picture with two parallel lines, in which case the 1 - Cos(theta) segment would be gone (zero).
1000101er 2 years ago 3
And they call this the basic course. They have an "advanced" version of this...
wornslipper 2 years ago 4
These lectures are soo helpful!! I watch these before going to my calc class and I dominate :D
fghijkaylee 2 years ago
in which College are you and what is ur class's number? I mean MAT115? MAT114????
thanks
gulloogtg 2 years ago
This comment has received too many negative votes show
parent angle of the base triangle must add with 90 - theta to equal 90. Evaluating this as an expression we see (theta x) + (90 - theta) = 90
this form leads to theta x - theta = 0 or theta x = theta. This gives a simple quantitative proof of the two angle's equivalence without the qualitative dribble the teacher attempted to use.
iridethewave 2 years ago
I think the lecturer did quite a bad job in clarifying why exactly the thetas in the two triangles he used for the derivative of sin(theta) were equal. If you notice, as delta(theta) approaches zero the angle between the triangle PQR and the triangle formed by extending a line from R to the base of the unit straight line approaches 90. Also, the angle opposite the theta obviously has the value 90 - theta. The other angle on this perpindicular which the teacher claims to be congruent to the
iridethewave 2 years ago
Wow, don't use this class as a litmus for evaluating the difficulty of MIT in general. My friend is attending MIT this year and she took the class "Calculus with Theory". In fact, many of their students don't take single variable calculus first year. Most of the people taking this class are doing it for the general institute reuirement
iridethewave 2 years ago
MIT will not assume that 4 yrs of HIGH SCHOOL Calculus has sufficienty prepared an ENGINEERING student in the subject. These kids are obviously top of their class, so to speak, but they still start at the beginning like all freshman.
LV07TSK 2 years ago
I understand his argument for A and B, but I wouldn't really call it a proof. More like a persuasive description. There seems to be a magic moment where he says use your imagination as things go to the limit and you see that it does not blow up it just continues its trend. I can see that but it does not seem to pbe proven, just plausible. The other proofs he has shown cancellations that leave only real definite quantities behind. I know what he says is true, but I would not call his proof tight.
michalchik 2 years ago
Well of course - I think the only way to make the proof tight would be to introduce the concept of the delta-epsilon proof and dabble a bit in Real Analysis. These are introduced quite rigorously though in the MIT course Calculus with Theory, which is for well prepared first year students.
iridethewave 2 years ago
This A course is quite simple compared to linear algebra in every school. Its the same in Sweden
arnoldbertil 2 years ago 2
Es fácil, no se porque se complican tanto. Gracias por el material, gracias al MIT.
IVANLUISLG 2 years ago
my only beef with people here saying things like
"oh these must be kids who didn't have calculus or these classes are introductory classes for kids who never had them in high school"
is that it CLEARLY says on the MIT admissions website that a student MUST have 4 years of math education in high school INCLUDING calculus. And I don't expect MIT recruiting students who SUCK in math either.
Maybe these kids just want an easy A. I don't know.
lakers4shoPSD 2 years ago
these lectures should have background music ;)
kid29a 2 years ago
Oh, so you kids find calculus to be too easy ?
Should any of you be able to find the general theorem for integrating a system of n variables/n dimensions by yourselves then yes, MIT is too easy for your level.
igoronline 2 years ago 2
YouTube comments are a joke.
NellyMath 2 years ago 28
@NellyMath your grammar makes me laugh too
BladeOfDawn 1 year ago
lol at all you guys criticizing the school. do you really think its this easy in the upper year classes?
freedomtoprosperity 2 years ago 2
Be nice guys. They're freshman just starting, after a long summer.
JustSemantics 2 years ago 2
where is Cal 2. I only see Cal 1 & 3.
Kinghercules 2 years ago