This is what I call a "valuable" math lecture. Time is spent to explain to students "why" things are true. While in institutions I've been to, time is spent to read slides for students and show them how to solve problems without even having a slightest idea about the procedure there making to solve a problem.
So do I! Whenever I asked why, the reply was either it is the way it is or that was a stupid question. But until today, almost 30 years after academic years, I am still into why then what then how.
This attitude of MIT is incredible. Offer the best engineering course in the world to all people. Now, anyone can learn with the highest quality. Thanks, MIT.
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.
For 13 minutes (21:00-34:30), Professor Jerison tries to justify (1-cos x)/x goes to zero, without, I think, making the case. While he convinces me that the "1-cosx gap" goes to zero faster than the arc length, he doesn't show that it goes to zero fast enough. For example, if it went to zero 10 times faster than the arc length, the limit would approach 1/10, not zero. I'm comfortable with his argument about (sinx)/x going to 1 and have in fact used that argument to defend the "small angle" approximation in Physics. (Larson 3rd edition, pg. 81 however, gives a rigorous explanation using limiting cases of triangular areas bounding the section of the circular area swept out by the angle, and then applying the squeeze theorem, (1-cosx)/x is left as an exercise. But once you have the limit of (sinx)/x going to 1, just multiply top and bottom of (1-cosx)/x by (1+cos), the numerator becomes (sinx)^2 and the expression becomes: [(sinx)/x] [(sinx)/(1+cosx)], first factor goes to 1 and the second to zero. It's good to review properties of limits for this. Where fault can be found, the good is often ignored. MIT and the professors involved are doing a GREAT public service by constructing these online courses. I'm amazed at the breadth and complexity of the courses offered. I deeply thank all of the MIT people involved in this. The examples and problems done in the class are excellent. I find many of the proofs easier to follow than in some texts. Thanks Professor Jerison!.
I think that geometric proof is not rigorous or even misleading. The correct proof should be this one --www.csun.edu/~ama5348/calculus/presentations/math150a_2_7video.pdf
I always found it interesting how the US system of teaching calculus has evolved into two subjects - calculus and real analysis. I guess the motivation for calculus is to introduce some of the concepts and build the practical skill of calculating derivatives, integrals and apply them to analysis of functions and generally solving practical problems. The rigorous introduction of the subject is left to real analysis which I guess not all students learn. For me, analysis became clear once I looked into how the subject developed over time, how Newton/Leibniz saw the connection between the derivative and the integral. I always wondered why all those people were so busy with calculating areas and why the tangent was something they were interested in. Also, the limit process always looked as a cheat until the moment someone suggested it was a best approximation, the way to do away with infinity and replace it with the nearest best approximation. Just wondering if calculus presented in such a form rises more questions than it answers.. I guess the joy of applying the method to different problems overshadows the gaps here and there..
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!
this teacher is to be appreciated . no matter the question , he still answers every single one with equale seriousness . my teachers never did this , they even tell me that my question is stupid sometimes
You're paying 80k for attend that school, that's the minimum thing they can do. And also, you're one of the most brilliant kids of the country (If not the world), they know that even though your question may sound stupid, if it's something easy to answer then you'll have no problems to understand the explanation
You don't have to memorize precisely how to start the addition rule for sin. You just have to remember to include the two possible pairings of cos and sin with a and b, and add them up. The order does not matter because of the symmetry.
The proofs that the limit of sin x/x and (1-cos x)/x tend to 1 and 0, respectively, as x approaches 0 could have been rigorously shown using the squeeze theorem. But, at this point, they haven't been introduced to the squeeze theorem yet.
This guy is a good teacher, but he leaves some points out, in which case you really have to just sit down, pause the video, and go through the steps he's doing very carefully. Especially for the geometric proof. But it's very understandable.
As x tends to 0, sinx/x tends to 1. At x=0.1, Sinx/x = sin(0.1)/0.1 = 0.99833416646.... At x = 0.01, Sin(0.01)/0.01 = 0.99998333341.... At x = 0.0001 Sin(0.0001)/0.0001 = 0.99999999833.... At x = 0.00000001 sin(0.00000001)/0.00000001 = 0.99999999999....
+The Skooma Cat It's precisely the opposite for me. Like the pictures make intuitive, real-world sense to me, the algebra feels just like symbol pushing.
I really have learnt all of thesein high school but my teachers never explain it clearly but rather force me to remember the equations and do the problems. That ways of study left me lots of knowledge gaps. I should have realized it earlier. 😭😭😭
Good observation, In Portugal we learned it using the proof ot the squezze theorem, and that is the correct way. I am actually a mathematician. the intuitive aproach used in the lesson only had the benefit of spare the students the intrincacy of point set topology in its basic sense.
I'm not even here for learning. I'm an Engineer. I learnt all this in class 12. I just love watching these Math lectures. And I'm trying to revise everything. The professor here is like the American guy CZcams, for me, an Indian guy. Irony.
congratulations is a great class of calculus, I admire you so much; and some day I want visit to MIT. See you soon Teacher Jose Luis Benavides Cepeda Colombia
the formula for cos(x-y) is identical to the dot product of two vectors written in trigonometric form. the formula for sin(x+y) is equal to a determinate with cosx, cosy in the first row and sinx, siny in the second row. - using the properties of even and odd functions, one can then get back and forth between sin(x-y) and cos(x+y);
In the case where x approaches 0 in the expression of (cosx-1)/x, one can find the value of the expression using geometric intuition. When "angle x" approaches 0 then the arc length "x" becomes equal to sinx, just as Professor explained in the video. But (cosx-1)/x is another nut to crack. Let there be a *single* triangle made by the radius of the unit circle and the x-axis with hypotenuse "H", adjacent side "A" and opposite side "P". Let the angle made by the radius to the x-axis be "x". Now let's take a look at (cosx-1)/x We know from Professor's explanation that Arc length *"x"= sinx* (when x approaches 0), so put the value of x=sinx in (cosx-1)/x, and we can write (cosx-1)/x as: *(cosx-1)/sinx* Rearrange it as: *(Cosx/sinx) - (1/sinx)* Note that (Cosx/sinx)=A/P and (1/sinx)=H/P {basic definitions} (cosx-1)/x= (A/P)-(H/P)=(A-H)/P Now take a look at Hypotenuse and the Adjacent side. As angle x gets smaller and smaller, Hypotenuse moves towards the adjacent and when x becomes infinitesimally small, *Hypotenuse "H" will nearly become equal to Adjacent "A"* ( or you can say that "H" will nearly superimpose on "A") Therefore *H=A* for a very small angle x. So : (cosx-1)/x =(A-A)/P=0 P.S》I thought about sharing this because there were a lot comments regarding this problem. I'm not trying to act like a nerd or genius, as I have been learning calculus for only some time and and it is pretty obvious that my skill is infinitesimally weak with respect to that of the Professors of MIT. This is what I've leart from his geometrical approach.
Hello all. I had to watch Khan Academy's Trigonometry videos to get back up to speed with this lecture. The law of sines and cosines didn't stick with me from high school.
There are a lot of comments that some of the Prof.'s explanations are not rigorous enough. MIT has 2 or 3 levels of introductory calculus. This one is for freshmen with little or no high school calculus, and who will probably major in eng. where the key is to get the right answer rather than to be rigorously correct. If you go to the OCW website, you can find the one aimed at students who want to major in math or physics. It's totally different and much harder.
If you want an excellent and straighforward proof that lim sin(x)/x = 1 , Kahn academy has a nine minute video that is perfectly explained using the squeeze theorem. That's about the only thing I would have included in this video - but he was pressed for time.
I wish he could have explained better how it is that the angle QPR is theta. He was just like "you flip it over 90 degrees and there you go." There was barely any explanation of how it is that those triangles are similar! I had to pause the video, draw my own picture, and stare at it for half an hour to figure it out.
Lol i guess it's a bit too late but if you consider the line perpendicular to the two lines that form an angle theta then you can see why angle QPR would also be theta
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
In the last geometric proof of y' where y=sin theta, teacher could easily show us that angle QPR=theta if he draw a perpendicular from P to horizontal radius from O and suppose it meets that horizontal radius at S then RPS will be a straight line and angle QPR+ angleOPQ +angle OPS =180• angle QPR + (90- theta)+90 =180 Angle QPR=theta
After three courses I can say this is definitely a challenging course. I can tell some remarkable MIT students back then were confused by the geometry interpretation.
the proof at 15:00 is the best geometric proof for sin(theta)/theta -> 1 as theta ->0 that i have seen....the bow thing will help me never forget....thanx
You could say that about a lot of math, the trick isn't usually in doing the algebra. It's understanding the relationships between numbers and equations.
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).
@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))
Those who are seeking a more rigorous point of view may refer to Appendix A.2 of the 18.01 (Fall 2006) course textbook: Simmons, George F. Calculus with Analytic Geometry.
Really really great explanation for trig diff especially the geometric proofs which gives the illusion of hard ( the last one ) but they aren't just thier visualizations try to draw it on a paper it makes sense and you can proof that QPR = theta in a different ways
X value oscillated between differentiation and integration and becomes a vertical line devouring the horizontal line and shifted by 90 degrees for the notice of the professor please.
@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
for A and B it can be used another geometrical proof using the f(theta) function's graphic, taking its tangent at theta equals zero and comparing to the tangent of y=theta . The graphic of y=1-cos(theta) shows a minimun in theta=zero and the graphic of Y=sin(theta) shows the same tangent than y=theta (the relation between arclenght and angle)
and if you accept than y=sin(theta) tends to equal theta when theta tends to zero then 1-cos(theta) must tend to zero faster than theta to satisfied pythagoras theorem
If you aren't convinced that the derivative sinx = cosx only works if the angle is in radians, you may want to google "derivative of sin measured in degrees". You will see that they are not the same. As for the 0/0, the proofs are done geometrically here, so you must look at the side lengths and the ratios between them rather than from an algebraic perspective. There are also algebraic proofs that can give the same answer, but these are intuitive and easy to visualize.
If angle is measured in radians then the formula for arc length=theta*radius, as radius=1, arc length=theta, Now as angle theta gets closer to 0, arc length and that straight line, i.e. sin(theta), sits on top of each other making sin(theta) = arc length, so that is why sin(theta)/theta=1, as value of both sin(theta) & theta becomes equal. and NOW as those two quantities are equal, it makes cos(theta)= radius= 1. and when calculated (1-cos(theta)/theta)=(1-1/theta)=(0/theta). and 0/anything=0. Correct me if I m wrong.
While explaining the geometric part of sin theta/theta as theta tends to zero vertical sin theta is also squeezed to zero but you give the answer as 1 that really cofuse as where as 1- cos theta /theta geometry is not confused we are able to understand. So they take the conjugate to prove this enigmatic result of sin theta/theta. Sankaravelayudhan Nandakumar. the audience may please write to me in this regard. How can a vertical line becomes a horizontal line which is only cos xero as one.
I always found it hard to grasp the concepts of limits, especially when it looked like 0/0 but I learned that you should never think of it as 0/0 because really, that is an undefined number and what is important is that the quantity in the limit tends towards 0, infinity or some other number for that matter. The professor should maybe have emphasised this when addressing the students question about sin(dx)/dx -> 1 because sin(0)/0 is really an undefined number.
@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.
The classes you are talking about are introductory classes. They are for those who never had these subjects in high school before and thus have to take it in their first year of college. Many students at MIT place out of these classes. The difficult courses come later.
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
No. Because here you have a fixed ratio of 1:2 between the numerator & denominator. But in case of (1-cosx)/x; the ratio will keep going upward as x approaches 0.
Limit of x/2x as x approaches 0 is 1/2. If you add an infinitely small amount to x (the limit from the right) then it is essentially 1/2. If you subtract an infinitely small amount from x ( limit from the left) you end up with 1/2. So the limit as x-> 0 is 1/2. You could also simply cancel the terms and have x/2x 1/2 limit x-> 0 = 1/2
In the last geometric proof to show angle QPR is equal to theta, he could have drawn a line from R perpendicular to the horizontal radius, instead of making the "rotation argument".
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.
There is no issue of rotification of the formula, it depends on understanding to explain it to those who are learners. If a person rotifies all formulae but unable to understand to explain it to the learners would not bring forward any results to educate them.
Various people, in commenting, are dissatisfied with the presentation here. First off, the professor is clearly in command of the subject matter. This is MIT and the pace in tough. In this third lecture he is up to a weeks-ahead point of a srandard calculus course. And these students, if they put the standard 3 hours of study per hour of lecture, will master the lecture material. One may search for a presentation that is superior, but if you find it, it will not be found at the third hour lecture. Or one may put in the work to create a satisfactory ptrsentation, to see how you yourself would teach it.
+DonLicuala The students with strong mathematical background can do the honors calculus class. The level of the students is normal for a class that has everything from physics majors to biology or management majors. You cant judge the whole class by the students with the weakest background, which will be the ones doing most of the questions.
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...)
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.
Wait...why is a topic that requires L'Hôpital's Rule to prove... ...being used as a "foundational" principle?! I'm all for --integrating-- incorporating transcendental functions early, but this is a bit absurd. WHY?!?!?!? *edit:* "Integrating" was a poor choice of descriptor (obviously)
@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
He could have answered the students question better by explaining that theta was tending to zero by the center moving out rather than the lines closing in. He tried to say that but didn't explain it right. If you view it this way, the top is tending towards zero while the bottom is a constant.
This is what I call a "valuable" math lecture. Time is spent to explain to students "why" things are true. While in institutions I've been to, time is spent to read slides for students and show them how to solve problems without even having a slightest idea about the procedure there making to solve a problem.
Accurate...
So do I! Whenever I asked why, the reply was either it is the way it is or that was a stupid question. But until today, almost 30 years after academic years, I am still into why then what then how.
Are u kidding me, i think u never have visited institute, i mean these proof i have learnt while preparing for IIT .😅
9
Agreed
This attitude of MIT is incredible. Offer the best engineering course in the world to all people. Now, anyone can learn with the highest quality. Thanks, MIT.
At last. my dream. I'm studying at MIT! Thanks OCW!
Pls tell me this lecture is helpful for undergraduate mathematics
@@tasruph9690 yes obviously calculus is the soul of mathematics even for undergrad math
We're so impressed about the fact that you all learned this in high school. Shut up about it.
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.
For 13 minutes (21:00-34:30), Professor Jerison tries to justify (1-cos x)/x goes to zero, without, I think, making the case. While he convinces me that the "1-cosx gap" goes to zero faster than the arc length, he doesn't show that it goes to zero fast enough. For example, if it went to zero 10 times faster than the arc length, the limit would approach 1/10, not zero.
I'm comfortable with his argument about (sinx)/x going to 1 and have in fact used that argument to defend the "small angle" approximation in Physics. (Larson 3rd edition, pg. 81 however, gives a rigorous explanation using limiting cases of triangular areas bounding the section of the circular area swept out by the angle, and then applying the squeeze theorem, (1-cosx)/x is left as an exercise.
But once you have the limit of (sinx)/x going to 1, just multiply top and bottom of (1-cosx)/x by (1+cos), the numerator becomes (sinx)^2 and the expression becomes:
[(sinx)/x] [(sinx)/(1+cosx)], first factor goes to 1 and the second to zero. It's good to review properties of limits for this.
Where fault can be found, the good is often ignored. MIT and the professors involved are doing a GREAT public service by constructing these online courses. I'm amazed at the breadth and complexity of the courses offered. I deeply thank all of the MIT people involved in this. The examples and problems done in the class are excellent. I find many of the proofs easier to follow than in some texts. Thanks Professor Jerison!.
Thanks, dude! I appreciate this!
I think that geometric proof is not rigorous or even misleading.
The correct proof should be this one --www.csun.edu/~ama5348/calculus/presentations/math150a_2_7video.pdf
Shiro Emiya
Thanks， dude!
@@Dragonfly-jk7wo Thanks for the link. Now I understand what's going on
Very well written comment.
I always found it interesting how the US system of teaching calculus has evolved into two subjects - calculus and real analysis. I guess the motivation for calculus is to introduce some of the concepts and build the practical skill of calculating derivatives, integrals and apply them to analysis of functions and generally solving practical problems. The rigorous introduction of the subject is left to real analysis which I guess not all students learn. For me, analysis became clear once I looked into how the subject developed over time, how Newton/Leibniz saw the connection between the derivative and the integral. I always wondered why all those people were so busy with calculating areas and why the tangent was something they were interested in. Also, the limit process always looked as a cheat until the moment someone suggested it was a best approximation, the way to do away with infinity and replace it with the nearest best approximation. Just wondering if calculus presented in such a form rises more questions than it answers.. I guess the joy of applying the method to different problems overshadows the gaps here and there..
This is a gem of a lecture. Wish every teacher spent more time on the why and the how.
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!
this teacher is to be appreciated . no matter the question , he still answers every single one with equale seriousness . my teachers never did this , they even tell me that my question is stupid sometimes
You're paying 80k for attend that school, that's the minimum thing they can do.
And also, you're one of the most brilliant kids of the country (If not the world), they know that even though your question may sound stupid, if it's something easy to answer then you'll have no problems to understand the explanation
This guy makes more sense in 50 minutes than my lecturer does in 2 hours.
TheLithiumTerror Than my lecturer in whole semester,
Than my lecturers in a whole business career(?)
a student from those who were present in this lecture would have said the same
problem is with the students
@@tahataherali8616 So not true.
You don't have to memorize precisely how to start the addition rule for sin. You just have to remember to include the two possible pairings of cos and sin with a and b, and add them up. The order does not matter because of the symmetry.
So what if he forgot it? My teacher forgets a lot of formulae, he's still a beast at integrating though.
5:45 group the terms to get 0/0
11:50 remarks about sin & cos's derivatives
15:00 geometric proofs
It's pleasure to learn maths from an MIT professor
4:14: Okay let's take a vote, is it sin-sin or sin-cosine? Proof by popularity!
That was hysterical...especially when he jokingly reproved them that they have to know that stuff off the top of their heads.
He was testing the audience knowledge, just that.
maybe that's how they do it at mit
The proofs that the limit of sin x/x and (1-cos x)/x tend to 1 and 0, respectively, as x approaches 0 could have been rigorously shown using the squeeze theorem. But, at this point, they haven't been introduced to the squeeze theorem yet.
This guy is a good teacher, but he leaves some points out, in which case you really have to just sit down, pause the video, and go through the steps he's doing very carefully. Especially for the geometric proof. But it's very understandable.
You are supposed to 1) have an IQ over 125 to even be there 2) contact tutors 3) use the recitations 4) study and do the problems
He's at MIT, not high school.
As x tends to 0,
sinx/x tends to 1.
At x=0.1,
Sinx/x = sin(0.1)/0.1
= 0.99833416646....
At x = 0.01,
Sin(0.01)/0.01
= 0.99998333341....
At x = 0.0001
Sin(0.0001)/0.0001
= 0.99999999833....
At x = 0.00000001
sin(0.00000001)/0.00000001
= 0.99999999999....
Holy ****. The algebra explanation is so much more undersandable for me than the geometric one.
+The Skooma Cat It's precisely the opposite for me. Like the pictures make intuitive, real-world sense to me, the algebra feels just like symbol pushing.
use corresponding angels between two parall lines, that tiny delta-y is parallel to y axis
This is great. This guy is soo much better than my teacher and much more enthusiastic.
I really have learnt all of thesein high school but my teachers never explain it clearly but rather force me to remember the equations and do the problems. That ways of study left me lots of knowledge gaps. I should have realized it earlier. 😭😭😭
Good observation, In Portugal we learned it using the proof ot the squezze theorem, and that is the correct way. I am actually a mathematician. the intuitive aproach used in the lesson only had the benefit of spare the students the intrincacy of point set topology in its basic sense.
I'm not even here for learning. I'm an Engineer. I learnt all this in class 12. I just love watching these Math lectures. And I'm trying to revise everything.
The professor here is like the American guy CZcams, for me, an Indian guy. Irony.
congratulations is a great class of calculus, I admire you so much; and some day I want visit to MIT.
See you soon
Teacher Jose Luis Benavides Cepeda
Colombia
the formula for cos(x-y) is identical to the dot product of two vectors written in trigonometric form. the formula for sin(x+y) is equal to a determinate with cosx, cosy in the first row and sinx, siny in the second row.
-
using the properties of even and odd functions, one can then get back and forth between sin(x-y) and cos(x+y);
In the case where x approaches 0 in the expression of (cosx-1)/x, one can find the value of the expression using geometric intuition.
When "angle x" approaches 0 then the arc length "x" becomes equal to sinx, just as Professor explained in the video. But (cosx-1)/x is another nut to crack.
Let there be a *single* triangle made by the radius of the unit circle and the x-axis with hypotenuse "H", adjacent side "A" and opposite side "P". Let the angle made by the radius to the x-axis be "x". Now let's take a look at (cosx-1)/x
We know from Professor's explanation that Arc length *"x"= sinx* (when x approaches 0), so put the value of x=sinx in (cosx-1)/x, and we can write (cosx-1)/x as:
*(cosx-1)/sinx*
Rearrange it as:
*(Cosx/sinx) - (1/sinx)*
Note that (Cosx/sinx)=A/P and (1/sinx)=H/P {basic definitions}
(cosx-1)/x=
(A/P)-(H/P)=(A-H)/P
Now take a look at Hypotenuse and the Adjacent side.
As angle x gets smaller and smaller, Hypotenuse moves towards the adjacent and when x becomes infinitesimally small, *Hypotenuse "H" will nearly become equal to Adjacent "A"* ( or you can say that "H" will nearly superimpose on "A")
Therefore *H=A* for a very small angle x.
So :
(cosx-1)/x =(A-A)/P=0
P.S》I thought about sharing this because there were a lot comments regarding this problem. I'm not trying to act like a nerd or genius, as I have been learning calculus for only some time and and it is pretty obvious that my skill is infinitesimally weak with respect to that of the Professors of MIT. This is what I've leart from his geometrical approach.
thanks alot
Hello all. I had to watch Khan Academy's Trigonometry videos to get back up to speed with this lecture. The law of sines and cosines didn't stick with me from high school.
There are a lot of comments that some of the Prof.'s explanations are not rigorous enough. MIT has 2 or 3 levels of introductory calculus. This one is for freshmen with little or no high school calculus, and who will probably major in eng. where the key is to get the right answer rather than to be rigorously correct. If you go to the OCW website, you can find the one aimed at students who want to major in math or physics. It's totally different and much harder.
Can you tell me which One is it please? Who is it taught by?
Such a brilliant teacher.
If you want an excellent and straighforward proof that lim sin(x)/x = 1 , Kahn academy has a nine minute video that is perfectly explained using the squeeze theorem. That's about the only thing I would have included in this video - but he was pressed for time.
I wish he could have explained better how it is that the angle QPR is theta. He was just like "you flip it over 90 degrees and there you go." There was barely any explanation of how it is that those triangles are similar! I had to pause the video, draw my own picture, and stare at it for half an hour to figure it out.
Lol i guess it's a bit too late but if you consider the line perpendicular to the two lines that form an angle theta then you can see why angle QPR would also be theta
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
In the last geometric proof of y' where y=sin theta, teacher could easily show us that angle QPR=theta if he draw a perpendicular from P to horizontal radius from O and suppose it meets that horizontal radius at S then RPS will be a straight line and angle QPR+ angleOPQ +angle OPS =180•
angle QPR + (90- theta)+90 =180
Angle QPR=theta
After three courses I can say this is definitely a challenging course. I can tell some remarkable MIT students back then were confused by the geometry interpretation.
Wow. I wish I had him when I was taking this stuff 10 yrs ago.Thank you Dr. whoever you are.Thanks MIT!
Beauty of mathematics lies in the proofs, the arguments made for proof rather than mugging the result and finding the answer :)
Thank you so much.
the proof at 15:00 is the best geometric proof for sin(theta)/theta -> 1 as theta ->0 that i have seen....the bow thing will help me never forget....thanx
You could say that about a lot of math, the trick isn't usually in doing the algebra. It's understanding the relationships between numbers and equations.
Never thought calculus could be so beautiful.
Props to Saurav Bastola, czcams.com/channels/7hkeE-LRyzietNCecWnHLg.html for the listed topics.
Lecture 1: Rate of Change
Lecture 2: Limits
Lecture 3: Derivatives
Lecture 4: Chain Rule
Lecture 5: Implicit Differentiation
Lecture 6: Exponential and Log
Lecture 7: Exam 1 Review
Lecture 9: Linear and Quadratic Approximations
Lecture 10: Curve Sketching
Lecture 11: Max-min
Lecture 12: Related Rates
Lecture 13: Newton's Method
Lecture 14: Mean Value Theorem
Lecture 15: Antiderivative
Lecture 16: Differential Equations
Lecture 18: Definite Integrals
Lecture 19: First Fundamental Theorem
Lecture 20: Second Fundamental Theorem
Lecture 21: Applications to Logarithms
Lecture 22: Volumes
Lecture 23: Work, Probability
Lecture 24: Numerical Integration
Lecture 25: Exam 3 Review
Lecture 27: Trig Integrals
Lecture 28: Inverse Substitution
Lecture 29: Partial Fractions
Lecture 30: Integration by Parts
Lecture 31: Parametric Equations
Lecture 32: Polar Coordinates
Lecture 33: Exam 4 Review
Lecture 35: Indeterminate Forms
Lecture 36: Improper Integrals
Lecture 37: Infinite Series
Lecture 38: Taylor's Series
Lecture 39: Final Review
Prof. David Jerison, i'm glad to have access to his class, thank you from Peru
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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).
Thank you for this comment even though you might not see, this comment helped me immensely
Thanks so much for putting this online! Helps me to refresh my mind in preparation for Analysis class in university.
CAN SOMEONE HELP ME UNDESTAND THE FUNCTION OF cos and sine???when are they used?
Sinx/X when integrared as 1becomes x as it reflectx something more. and the first part is a differentiated and seco d one is Integrated to produce X..
The information shows that 0/0 =1 as sinxdelx/sin x =1 as del x goes to zero.
7:37 isnt (cos(deltax) - 1)/(deltax) in an indeterminate form? if you substitute deltax = 0, (cos(0) - 1) / (0) => 0/0
Awesome teacher...
The geometric proof was great.
The trouble is deltheta goes to zero in cosx sindel x/delx it does not become zero either by geometry or by substitute of a function.
Stanford also posts its lectures on CZcams
@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))
Those who are seeking a more rigorous point of view may refer to Appendix A.2 of the 18.01 (Fall 2006) course textbook: Simmons, George F. Calculus with Analytic Geometry.
Really really great explanation for trig diff especially the geometric proofs which gives the illusion of hard ( the last one ) but they aren't just thier visualizations try to draw it on a paper it makes sense and you can proof that QPR = theta in a different ways
X value oscillated between differentiation and integration and becomes a vertical line devouring the horizontal line and shifted by 90 degrees for the notice of the professor please.
I love this.
@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
for A and B it can be used another geometrical proof using the f(theta) function's graphic, taking its tangent at theta equals zero and comparing to the tangent of y=theta . The graphic of y=1-cos(theta) shows a minimun in theta=zero and the graphic of Y=sin(theta) shows the same tangent than y=theta (the relation between arclenght and angle)
and if you accept than y=sin(theta) tends to equal theta when theta tends to zero then 1-cos(theta) must tend to zero faster than theta to satisfied pythagoras theorem
If you aren't convinced that the derivative sinx = cosx only works if the angle is in radians, you may want to google "derivative of sin measured in degrees". You will see that they are not the same.
As for the 0/0, the proofs are done geometrically here, so you must look at the side lengths and the ratios between them rather than from an algebraic perspective. There are also algebraic proofs that can give the same answer, but these are intuitive and easy to visualize.
If angle is measured in radians then the formula for arc length=theta*radius, as radius=1, arc length=theta, Now as angle theta gets closer to 0, arc length and that straight line, i.e. sin(theta), sits on top of each other making sin(theta) = arc length, so that is why sin(theta)/theta=1, as value of both sin(theta) & theta becomes equal. and NOW as those two quantities are equal, it makes cos(theta)= radius= 1. and when calculated (1-cos(theta)/theta)=(1-1/theta)=(0/theta). and 0/anything=0. Correct me if I m wrong.
Thanks 3brown2blue to make this easier.
18:59. why there is 2 theta in the denominator. Can someone explain it for me, please
While explaining the geometric part of sin theta/theta as theta tends to zero vertical sin theta is also squeezed to zero but you give the answer as 1 that really cofuse as where as 1- cos theta /theta geometry is not confused we are able to understand. So they take the conjugate to prove this enigmatic result of sin theta/theta.
Sankaravelayudhan Nandakumar. the audience may please write to me in this regard. How can a vertical line becomes a horizontal line which is only cos xero as one.
this is so elementary to be taught on 1st year UNIVERSITY , but it's well taught GREAT
We are not able to understand when Delta x goes to zero cosx sin deltax/Delta x goes to 1.
I always found it hard to grasp the concepts of limits, especially when it looked like 0/0 but I learned that you should never think of it as 0/0 because really, that is an undefined number and what is important is that the quantity in the limit tends towards 0, infinity or some other number for that matter.
The professor should maybe have emphasised this when addressing the students question about sin(dx)/dx -> 1 because sin(0)/0 is really an undefined number.
Eso se explicó en la lección 2.
Thanks MIT
thanks for giving these video series online and free
financially week students can also study by these lectures.thanks a lot
@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.
The classes you are talking about are introductory classes. They are for those who never had these subjects in high school before and thus have to take it in their first year of college. Many students at MIT place out of these classes. The difficult courses come later.
What an amazing teacher
Omg when you shifted that 90° I just realised you just made a Pi/2 phase shift and turned it into a Cos from Sin
I think😂
beautiful just beautiful
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
so, if his explanation is correct, then the limit of x/2x when x->0 = 0 because x aproches faster to 0 than 2x. It sounds ridiculous
No. Because here you have a fixed ratio of 1:2 between the numerator & denominator. But in case of (1-cosx)/x; the ratio will keep going upward as x approaches 0.
Limit of x/2x as x approaches 0 is 1/2. If you add an infinitely small amount to x (the limit from the right) then it is essentially 1/2. If you subtract an infinitely small amount from x ( limit from the left) you end up with 1/2. So the limit as x-> 0 is 1/2.
You could also simply cancel the terms and have
x/2x
1/2
limit x-> 0 = 1/2
There are questions which let me doubt that this course was really thought at MIT for MIT students. Nevertheless prof is super.
1-cos(theta) tends to zero faster than arc length (= theta), so it becomes like -- 0 divided my any number so its zero
This aldo if analysed gives information that when zero is inegrated gives 1 considering the basic proof.
i really don't get the rotation bit at around 45:00. ive been staring at it for like 5m at this point.
Its good they still use blackboards, significantly better and easier to follow than the bland powerpoints we seem to get at my uni..
In the last geometric proof to show angle QPR is equal to theta, he could have drawn a line from R perpendicular to the horizontal radius, instead of making the "rotation argument".
When differentiating the sine and cosine functions, he mapped delta x to 0 on the first term, and 1 on the second term. How does this work?
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.
watch these videos and pickup openstax calculus 1 book,you have a solid combination to understand calculus.
14:34 - if these formulae are explained graphically it would be much easier to understand and interesting.
does anyone know a good textbook to buy for this course?
I Really Like The Video From Your Derivatives of products, quotients, sine, cosine
There is no issue of rotification of the formula, it depends on understanding to explain it to those who are learners. If a person rotifies all formulae but unable to understand to explain it to the learners would not bring forward any results to educate them.
Various people, in commenting, are dissatisfied with the presentation here. First off, the professor is clearly in command of the subject matter. This is MIT and the pace in tough. In this third lecture he is up to a weeks-ahead point of a srandard calculus course. And these students, if they put the standard 3 hours of study per hour of lecture, will master the lecture material.
One may search for a presentation that is superior, but if you find it, it will not be found at the third hour lecture. Or one may put in the work to create a satisfactory ptrsentation, to see how you yourself would teach it.
+DonLicuala The students with strong mathematical background can do the honors calculus class. The level of the students is normal for a class that has everything from physics majors to biology or management majors. You cant judge the whole class by the students with the weakest background, which will be the ones doing most of the questions.
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...)
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.
Wait...why is a topic that requires L'Hôpital's Rule to prove...
...being used as a "foundational" principle?!
I'm all for --integrating-- incorporating transcendental functions early, but this is a bit absurd.
WHY?!?!?!?
*edit:* "Integrating" was a poor choice of descriptor (obviously)
Thank you MIT. #Grateful.
Please explain cosx sindeltax/deltax becomes 1.
@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
He could have answered the students question better by explaining that theta was tending to zero by the center moving out rather than the lines closing in. He tried to say that but didn't explain it right.
If you view it this way, the top is tending towards zero while the bottom is a constant.
Don't give up. ( 加油！)
I don't get it. So it's okay if we have a limit like 0/0, but a limit like 0/1 is wrong, is that right?
1/0*
You can't make anything out of a 1/0 form.