For anyone who may be confused in the first part of this video, it helps to keep in mind that he is talking specifically about a function with a tangent plane z=0 at the origin. Also keep in mind what the graphs of x^2 - y^2 and x^2 + y^2 look like!
In a traditional multivariable calculus class, the concept of Hessian matrix will be introduced at this point. I wonder why he doesn't. Anyway, a Hessian matrix of 2 variables is just a matrix of [ fxx fxy; fxy fyy ]. Let D = det(H), then: 1) If D < 0, saddle point 2) If D > 0, either relative min or max. Furthermore, if fxx > 0, then it's relative min. If fxx < 0, it's a relative max.
In the 2x2 case yes, but for an nxn case, there is the concept of positive definite, and it's not about determinate only anymore - one needs to start talking about eigenvalues.
@@GabrielCsillaz can you tell me the proof of this Hessian matrix is not a formula it is just an another way to look into the things when you go in higher dimension then this Matrix is very helpful because this is some kind of matrix which represent the simple way to look into the problem ..if you go through this AC-B^2 its very complicated in higher dimension.
At 27:30 it wasn't clear to me why having 2 roots yields a saddle point, then I came to this: Remembering how a saddle point looks like in a Contow representation, there are two directions for which we can stay at the level curve with value 0, those directions are given by the ratios we obtain solving the quadratic formula for (x/y), also using this, the fact that we have a degenerate min/max when we have b²-4ac=0 (when we only have one root with multiplicity 2) comes easily (the root indicates the direction for which we stay at the min/max=0). I hope this is actually accurate and helps someone.
I think this could act as a proof of min 18:15 claims: * for the degenerate case, we have a line ( x + b / (2a) * y ) = 0 for wich z is allways 0 . It acts like the case of z = x^2 but rotated ( because the term affecting y^2 is zero in the deg. case ). * for the opposite signs (saddle): if we fix y=0 and let x increase we get possitive values but if we move for the line ( x + b / (2a) * y ) = 0 (the same as for the deg case) we get a decrease in the z values. (because y^2 is positive and the term affecting it is negative for saddle point case). For the conditions of the exercise we start moving from (0,0) and the signs from above are in the case ( 1 / 4a ) is possitive (so change if negative). You can plot some functions on Google: e.g copy paste 6*x**2+5*x*y+y**2 and x**2+4*x*y+4*y**2
There are many reasons for the applause and amusement, so it's not a simple answer. First you have to go back to the mind of a 20 year old. Everything is amusing, even a contest race of a professor's eraser skills vs. auto-board drop down. Second, Denis Auroux is not just any professor, he's immensely popular with MIT students. This popularity includes not just one, but two fan clubs. The students lucky enough to attend his lectures actually understand that it is indeed a privilege to be there. If you follow through his lectures beginning to end, he's quite famous for taking partially understood concepts, mixing new material into a craftily tied together composite lecture, and by the end, you've learned something new in multiple dimensions. No pun intended of course.
uh, I misunderstood. After seeing the justification from Taylor expansion or quadratic approximation, I understand better why Prof. Auroux spent so much time on exploring the quadratic function earlier. This is a relatively harder topic, as reflected by those "dislikes". Thanks for the lecture.
So far, this was the only difficult lecture. This lecture, I think, I'll need to work out a lot of what he's saying. It was ok in the beginning. It got tough until 30 min, then it gets back to being easy.
16:30 He says that the function "depend only on one direction of things". I guess he ment it only depends on one of the variables. However, when 4ac-b^2=0, we are still left with the square on the left side, which contains both x and y. Can someone explain what i'm not seeing?
First lecture of the series that I've slightly struggled with :( I understand the second derivative test, and how to carry it out, but in the specific quadratic function example, why does a "difference of two squares" imply a saddle point? I see how a sum of two squares implies a local min or max depending on the value of a, but confused about how the algebra leads to the conclusion that we have a saddle. Also not quite clear on the degenerate case and why 4ac - b^2 = 0 implies degeneracy. Need to do some further reading on this before moving on methinks...
David Robins You might have found the answer in 3 years but anyway... When a function is written as a difference of 2 squares.... U understand that it can be both negative or positive depending upon which of these 2 squares gets a greater value on plugging in various values of x and y...now for a few values of (x,y) one square maybe greater while for other values of (x,y) the other square maybe greater...so u realise that around the critical point there are going to be points taking values of z both greater or lesser than the value of z at the critical point....which means that the surface can go uphill in some directions and downhill in others....which is the very definition of a saddle point
I think the saddle point relates back to the canonical example of saddle point that we had before. Its form was x^2 - y^2, so if your function looks like that - it's going to have a saddle point at x=0, y=0. The trick is that what we got is the canonical x^2 - y^2, but we could transform the variables to get it, as he mentions it here: ua-cam.com/video/3_goGnJm5sA/v-deo.html For the degeneracy - again, with the change of variables, it turns out that the shape is the same for any slice along one of the axes, which is sort of degenerate (2d shape replicated many times, instead of a really interesting 3d function
An excellent lecturer. I'm growing less fond of the students in the audience, who seem easily amused by random occurrences, are obsessed with whether a concept is going to be on the exam rather than being interested in learning for its own sake, and they repeatedly joke/complain about the difficulty of Taylor's series, even though for someone who got into MIT that should be elementary.
@@domenickeller2189 They're of adult age. They're old enough to drive a car, or hold down a job. They've been accepted into an elite institution with an illustrious history. To top it all off, they're being recorded. Why should I accept that they should be allowed to act like elementary school children? It's embarrassing.
@@wlockhart Sure, all of your points are valid. But they still don't matter because regardless of everything, they are college students. A lot of college students act like that regardless of the institution. They're mostly freshmen meaning they were recently in high school, so many of them still act like high schoolers anyway. I believe that their laughter adds a casual sort of feel that allows the class to relax a bit. They most likely aren't interested in learning for its own sake; many just want good grades and a degree. I personally don't have problems with Taylor series but I can see how one could. So yes, your points are valid but quickly go out the window.
If anyone is struggling with this lecture, Wikipedia does a wonderful job of explaining the second derivative test intuitively: en.wikipedia.org/wiki/Second_partial_derivative_test#Geometric_interpretation_in_the_two-variable_case
Wow this stuff is hard. Definitely takes a lot of pondering and practice to get it down. Im sure this was explained masterfully but now I need to address my skill issue. Thanks!
Around 34:00 : what about the case there AC - B^2 > 0 but A = 0? I take it that is also a case where we have a local max, since -B^2 is always negative; i'm just sorta surprised no-one noticed the omission?
maybe this works: for the conditions we start at (0,0), w=ax^2+bxy+cy^2 is cont. and a is not 0. If we proove (0,0) is the only crit point we proove is a saddle point; ( if it would be a min-max point we would need another crit. to get the neg-positive behaviour because w(0,0)=0 ) . So partial derivatives give as a 2x2 system. You can check the posibilities with the conditions as boundaries, including a not 0 and b^2 - 4ac > 0 to proove it's the only crit. point.
+Supreeth Ravish Remember that: (1) There is a single critical point for w, at x = y = 0 (you can check by finding the partial derivatives for w wrt x and y and manipulating to find (b^2-4ac)x = 0 and (b^2-4ac)y = 0 (2) w = 0 at the critical point (when x = y = 0) When rearranged, the quadratic form w = ax^2 + bxy + cy^2 is re-expressed as a scaled quadratic function w = y^2 [ a(x/y)^2 + b(x/y) + c ] (the y^2 factor is always non-negative). Looking only at the quadratic function in the square bracket (which is just w/y^2) and using the standard mechanisms for quadratic functions, note that: (1) if the discriminant b^2 - 4ac < 0 there are no real-valued roots, meaning that the function w/y^2 is either negative everywhere or positive everywhere (it never crosses the (x/y) axis). Since w is just that function multiplied by a non-negative factor y^2, the surface of w *will never cross the xy plane*. Consequently, since the critical point is at the origin, it must be a maximum or a minimum (depending on the sign of a). (2) if b^2 - 4ac > 0, there are two real roots and w/y^2 crosses the x/y axis. So w/y^2 has both positive and negative values, which means w (which is the same function scaled by a non-negative value y^2) *also* crosses the xy plane and has both positive and negative values. Since the sole critical point of w is at the origin, it *must* be a saddle point (in some directions the function will drop below the xy plane and in other directions it will rise above). So the discriminant b^2-4ac for the quadratic function has corresponding meanings with the 4ac-b^2 discriminant found for the quadratic form (one just swaps the < and > signs since one is the negated version of the other). Incidentally, one can put into context the quadratic function w/y^2 (for the given a, b and c) by realising that its range is just the set of heights of the surface w relative to the circumference of a circle segment found by rotating the unit vector ai (for any nonzero a) in the xy plane at angles 0 < theta < pi (pi < theta < 2pi is just a reflection), all divided by y^2. In other words, the range of a(x/y)^2 + b(x/y) + c is w(a, theta)/y^2 (in polar coordinates) for all 0 < theta < pi and any nonzero a. The corresponding rectangular coordinates yield the ratio values in the domain of w(x,y)/y^2. Since the x/y ratio is a function of direction from the origin and not distance, w(a, theta)/y^2 is invariant to (nonzero) values of a.
At 16:21 Why does function depends only on one direction if 4ac - b^2 = 0 ? Yes the term with y^2 is 0 but don't we have y in first term as well? (x + b/2a * y)^2
imagine a line 'x + b/2a * y = some_constant' in the xy-plane: every dot in this line will have the same z value (i.e., the same height; a* some_constant). So it produces a horizontal line with only one direction.
He had mentioned in a previous lecture that we'll be only dealing with differentiable functions , and don't bother about continuity (because it won't be discussed)
In India it is a compulsory question for board exam and my teacher just said the conditions as formulas and then made us practise a lot of problems of this kind. That's it!
Linear approximation is the first order fit - i.e. tangent plane approximation. See previous lecture. Let z ~ z0 + fsubxat(x0) *(x-x0) + fsubyat(y0) * (y-y0). Pardon terrible notation.
+MegaDespotic Why do you think so? I cannot see any mistake... It means you first do a derivative of f just with respect to x (and other variables behave like constants) and then with respect to y ( - || - ). It is actually the same as (df/dx)*(df/dy)... but of course with partials
+MegaDespotic Well, I accidently swapped the letters, but still you know :-D The letters can be written in any order as you first do the partial derivetive of f with respect to y and you multiply it by the partial derivative of f with respect to x... And multiplication is commutative Or am I getting you wrong? :-D
@@selcuk2649 From 16:10 he talks a bit about it, which is basically saying that it's the inconclusive case, which I think is why he just ignored going over it again at 26:50.
I like this professsor a lot but this lecture in particular was not very clarifying. I have studied this concept before and for me was easier to understand.
@@proghostbusters1627 I was able to follow it too, but I think it depends on your previous knowledge, this might be very difficult for beginners to these conscepts because he's not giving a lot of intuitive understanding.
I mean I know this is MIT but if at 9:40 he would just do factoring without skipping steps, it would be much clearer for everyone and he wouldn't have to go over everything to make sure people understood.
i have learned a lot from these lectures--but i just which prof Auroux would just step throgh the algebra--he always skips steps--then says "everyone see that"--no one does--and then back tracks--tedious. Well--i am sure having to do the steps at all is tedious for him. Overall--he manages to pack a lot of insight into 45 minutes each and every time
I am stuck at this point where he finds out the minima and maxima of ax^2+bxy+cx^2 through sum of squares. As per my understanding, to find out the critical point, we need to find if the slope at that point is 0. But, through the quadratic equation, we did not actually find the slope. How was the the critical point determined in that case?
(0, 0) is the critical point, which was mentioned at the beginning of this lecture. The topic of the course is to determine what type of critical point (min, max, or saddle) for this point (0,0)
Suppose f(x, y) = x^3, then f_x = 0 gives x = 0, and f_y = 0 is always true. So we have a bunch of critical points that form the y axis, what should we call these points?
U were certainly not following all the lectures. This great man is great at rubbing the 2nd board before the upper one comes down. This wow stufd has been done in previous lectures as well
sl6006095 It's called Clairaut's Theorem (or Schwarz' Theorem); it states that if the second partial derivatives are continuous, then it doesn't matter which order you do them in. You can probably find a proof online. I wouldn't call it an easy proof if you're at the level for this class, though. And almost every function that you encounter in the sciences that you might want to differentiate will have continuous partials, so it isn't much of a concern (unless you are in a pure math class).
In the function w=a.x^2 + b.xy + c.y^2 x and y play the same role, there is a form of symmetry. That means the sign of c would matter the same as the sign of a, right? But how he ends up discussing only the sign of, and not that of c?!! Anyone?
for the determinant to be negative (which is the case we care about the sign of a in) we need both a and c to have the same signs, otherwise, the determinant can never be negative.
If 4ac - b^2 = 0, then the y^2 component disappears. However, you still have the x and y inside the first term, which you square. I don't get how it's degenerate. What if W = x^2 + 2xy + y^2? 4ac = 4, and b^2 = 4, so 4ac - b^2 = 0. However, it doesn't have a degenerate critical point.
Think about it like this. When 4ac-b^2 = 0, the second square disappears, and you're left with w = a(x+by/2a)^2. You do indeed have both the x and y terms, so you can still pick any arbitrary x and y and come up with unique values. However, there EXISTS a direction in the x-y plane in which the value of w does not change. You can pretty easily see this if you just look at the inner x and y terms, (x + ky), where k = b/2a. If you set x + ky = h, where h represents a fixed (i.e., non-changing) value of the function w, then you can always find a line of degeneracy in the form of y = (h - x)/k. The important point is that, by following this line, (x+ky) never changes, and therefore neither does a(x+ky)^2 = w. Using your example, W = x^2 + 2xy + y^2, we can look at it in the sum of squares form, W = (x+y)^2. It has no second square because 4ac-b^2 = 0. You can set (x+y) = h, and get that y = h-x. So for instance, if you wanted to go off into infinity and hang out at the value of W = 0 forever, you can do that by following the line y = -x. If you instead prefer the value W = 4, you can set x+y = 2 and follow the line y=2-x. When 4ac-b^2 is NOT 0, and you have something like W = (x+y)^2 + cy^2, you can still find a line in which the first term does not change, however the second term will "screw it up" along the line and change the value of W. Hope that helps.
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But (with some help from Taylor expansion) he used that specific case to justify what he claimed about general case. So it's only irrelevant if you only need to know how to compute stuff and don't strive for deeper understanding of the subject.
@@alyasker2194Taylor's series gives a decent approximation for the behavior of the function around the critical point; namely it produces a quadratic polynomial based on the second derivatives at the point of interest. You can use facts about this approximation to deduce facts about the function itself; regardless of how complicated it may be. Hope that helps tie things together.
Well, what i mean is that the students that were taking that class all of them were studying engineering or there were some other students from other careers?
All MIT students take this course (or the more advanced, theoretical 18.022). It is a General Institute Requirement at MIT (it's required for graduation).
sad. the formal definition of the saddle point isnt given...and where is the proof that the difference of two linear functions squared always yields a saddle?
he constantly says yesterday and tomorrow, do these guy have lessons all days ? when do they do the homework ?! however if they only have 1hs of calculus thats not such a pain. having 4hs x 2 days is the same pain or worse, but you can at least do the homework between these two days !
For anyone who may be confused in the first part of this video, it helps to keep in mind that he is talking specifically about a function with a tangent plane z=0 at the origin. Also keep in mind what the graphs of x^2 - y^2 and x^2 + y^2 look like!
One of the best and youngest professor ever...thanks very much Auroux and MIT
In a traditional multivariable calculus class, the concept of Hessian matrix will be introduced at this point. I wonder why he doesn't.
Anyway, a Hessian matrix of 2 variables is just a matrix of [ fxx fxy; fxy fyy ]. Let D = det(H), then:
1) If D < 0, saddle point
2) If D > 0, either relative min or max. Furthermore, if fxx > 0, then it's relative min. If fxx < 0, it's a relative max.
Did u realize that AC-B² = D ?
he just doesnt want his students memorizing formulaes, he wants that they actually know what they are doing
In the 2x2 case yes, but for an nxn case, there is the concept of positive definite, and it's not about determinate only anymore - one needs to start talking about eigenvalues.
@@GabrielCsillaz can you tell me the proof of this
Hessian matrix is not a formula it is just an another way to look into the things when you go in higher dimension then this Matrix is very helpful because this is some kind of matrix which represent the simple way to look into the problem ..if you go through this AC-B^2 its very complicated in higher dimension.
At 27:30 it wasn't clear to me why having 2 roots yields a saddle point, then I came to this: Remembering how a saddle point looks like in a Contow representation, there are two directions for which we can stay at the level curve with value 0, those directions are given by the ratios we obtain solving the quadratic formula for (x/y), also using this, the fact that we have a degenerate min/max when we have b²-4ac=0 (when we only have one root with multiplicity 2) comes easily (the root indicates the direction for which we stay at the min/max=0). I hope this is actually accurate and helps someone.
Wonderful explanation that I haven't seen before! You just don't see this level of teaching in usual calculus courses
These people are a stand-up comedian's wet dream.
Why in the world does this video have so many dislikes. Absolutely amazing lecture.
I think this could act as a proof of min 18:15 claims:
* for the degenerate case, we have a line ( x + b / (2a) * y ) = 0 for wich z is allways 0 . It acts like the case of z = x^2 but rotated ( because the term affecting y^2 is zero in the deg. case ).
* for the opposite signs (saddle): if we fix y=0 and let x increase we get possitive values but if we move for the line ( x + b / (2a) * y ) = 0 (the same as for the deg case) we get a decrease in the z values. (because y^2 is positive and the term affecting it is negative for saddle point case).
For the conditions of the exercise we start moving from (0,0) and the signs from above are in the case ( 1 / 4a ) is possitive (so change if negative). You can plot some functions on Google: e.g copy paste 6*x**2+5*x*y+y**2 and x**2+4*x*y+4*y**2
Thank you very much for the proof. You're the real MVP!
I'm actually confused how easily amused the crowd was lol
Same here.
Looked like he wrote generate clit pt.
There are many reasons for the applause and amusement, so it's not a simple answer. First you have to go back to the mind of a 20 year old. Everything is amusing, even a contest race of a professor's eraser skills vs. auto-board drop down. Second, Denis Auroux is not just any professor, he's immensely popular with MIT students. This popularity includes not just one, but two fan clubs. The students lucky enough to attend his lectures actually understand that it is indeed a privilege to be there. If you follow through his lectures beginning to end, he's quite famous for taking partially understood concepts, mixing new material into a craftily tied together composite lecture, and by the end, you've learned something new in multiple dimensions. No pun intended of course.
Toan Ngo 35:42 lol
+Toan Ngo it's scripted
uh, I misunderstood. After seeing the justification from Taylor expansion or quadratic approximation, I understand better why Prof. Auroux spent so much time on exploring the quadratic function earlier. This is a relatively harder topic, as reflected by those "dislikes". Thanks for the lecture.
So far, this was the only difficult lecture. This lecture, I think, I'll need to work out a lot of what he's saying. It was ok in the beginning. It got tough until 30 min, then it gets back to being easy.
16:30 He says that the function "depend only on one direction of things". I guess he ment it only depends on one of the variables. However, when 4ac-b^2=0, we are still left with the square on the left side, which contains both x and y. Can someone explain what i'm not seeing?
Yes, good point, I don't understand that either.
Someone asks the same question at 21:20, and the answer follows that.
Second derivative test starts at about 30:00
:)
*They are enjoying it more then netflix*
First lecture of the series that I've slightly struggled with :( I understand the second derivative test, and how to carry it out, but in the specific quadratic function example, why does a "difference of two squares" imply a saddle point? I see how a sum of two squares implies a local min or max depending on the value of a, but confused about how the algebra leads to the conclusion that we have a saddle. Also not quite clear on the degenerate case and why 4ac - b^2 = 0 implies degeneracy. Need to do some further reading on this before moving on methinks...
David Robins You might have found the answer in 3 years but anyway...
When a function is written as a difference of 2 squares.... U understand that it can be both negative or positive depending upon which of these 2 squares gets a greater value on plugging in various values of x and y...now for a few values of (x,y) one square maybe greater while for other values of (x,y) the other square maybe greater...so u realise that around the critical point there are going to be points taking values of z both greater or lesser than the value of z at the critical point....which means that the surface can go uphill in some directions and downhill in others....which is the very definition of a saddle point
I think the saddle point relates back to the canonical example of saddle point that we had before. Its form was x^2 - y^2, so if your function looks like that - it's going to have a saddle point at x=0, y=0.
The trick is that what we got is the canonical x^2 - y^2, but we could transform the variables to get it, as he mentions it here: ua-cam.com/video/3_goGnJm5sA/v-deo.html
For the degeneracy - again, with the change of variables, it turns out that the shape is the same for any slice along one of the axes, which is sort of degenerate (2d shape replicated many times, instead of a really interesting 3d function
An excellent lecturer. I'm growing less fond of the students in the audience, who seem easily amused by random occurrences, are obsessed with whether a concept is going to be on the exam rather than being interested in learning for its own sake, and they repeatedly joke/complain about the difficulty of Taylor's series, even though for someone who got into MIT that should be elementary.
bro chill they are college students lol
@@domenickeller2189 They're of adult age. They're old enough to drive a car, or hold down a job. They've been accepted into an elite institution with an illustrious history. To top it all off, they're being recorded. Why should I accept that they should be allowed to act like elementary school children? It's embarrassing.
@@wlockhart Sure, all of your points are valid. But they still don't matter because regardless of everything, they are college students. A lot of college students act like that regardless of the institution. They're mostly freshmen meaning they were recently in high school, so many of them still act like high schoolers anyway. I believe that their laughter adds a casual sort of feel that allows the class to relax a bit. They most likely aren't interested in learning for its own sake; many just want good grades and a degree. I personally don't have problems with Taylor series but I can see how one could. So yes, your points are valid but quickly go out the window.
had no idea what was going on till 30:00
same
You need to watch last two videos of the series, he explains saddle points well and parabola
If anyone is struggling with this lecture, Wikipedia does a wonderful job of explaining the second derivative test intuitively: en.wikipedia.org/wiki/Second_partial_derivative_test#Geometric_interpretation_in_the_two-variable_case
Such an incredible lecture! Great Professor Auroux.
Thank you for these videos, they are helping a lot in my class :).
This man is a great teacher!
31:05 female function male function
I thought why he was spending so much time on this particular case at the beginning, then at 38th minute... wow!!
Wow this stuff is hard. Definitely takes a lot of pondering and practice to get it down. Im sure this was explained masterfully but now I need to address my skill issue. Thanks!
Around 34:00 : what about the case there AC - B^2 > 0 but A = 0? I take it that is also a case where we have a local max, since -B^2 is always negative; i'm just sorta surprised no-one noticed the omission?
If A = 0, then AC = 0
AC - B² = -B² is always non positive, so the case AC - B² > 0 and A = 0 simply does not exist.
Because this is all constructed for an equation of the form where a is preciseley not 0, at 8:58.
26:59 I'm sorry, I don't get this. w takes both positive and negative values. How does that lead to saddle point?
maybe this works: for the conditions we start at (0,0), w=ax^2+bxy+cy^2 is cont. and a is not 0. If we proove (0,0) is the only crit point we proove is a saddle point; ( if it would be a min-max point we would need another crit. to get the neg-positive behaviour because w(0,0)=0 ) . So partial derivatives give as a 2x2 system. You can check the posibilities with the conditions as boundaries, including a not 0 and b^2 - 4ac > 0 to proove it's the only crit. point.
+Supreeth Ravish
Remember that:
(1) There is a single critical point for w, at x = y = 0 (you can check by finding the partial derivatives for w wrt x and y and manipulating to find (b^2-4ac)x = 0 and (b^2-4ac)y = 0
(2) w = 0 at the critical point (when x = y = 0)
When rearranged, the quadratic form w = ax^2 + bxy + cy^2 is re-expressed as a scaled quadratic function w = y^2 [ a(x/y)^2 + b(x/y) + c ] (the y^2 factor is always non-negative). Looking only at the quadratic function in the square bracket (which is just w/y^2) and using the standard mechanisms for quadratic functions, note that:
(1) if the discriminant b^2 - 4ac < 0 there are no real-valued roots, meaning that the function w/y^2 is either negative everywhere or positive everywhere (it never crosses the (x/y) axis). Since w is just that function multiplied by a non-negative factor y^2, the surface of w *will never cross the xy plane*. Consequently, since the critical point is at the origin, it must be a maximum or a minimum (depending on the sign of a).
(2) if b^2 - 4ac > 0, there are two real roots and w/y^2 crosses the x/y axis. So w/y^2 has both positive and negative values, which means w (which is the same function scaled by a non-negative value y^2) *also* crosses the xy plane and has both positive and negative values. Since the sole critical point of w is at the origin, it *must* be a saddle point (in some directions the function will drop below the xy plane and in other directions it will rise above).
So the discriminant b^2-4ac for the quadratic function has corresponding meanings with the 4ac-b^2 discriminant found for the quadratic form (one just swaps the < and > signs since one is the negated version of the other).
Incidentally, one can put into context the quadratic function w/y^2 (for the given a, b and c) by realising that its range is just the set of heights of the surface w relative to the circumference of a circle segment found by rotating the unit vector ai (for any nonzero a) in the xy plane at angles 0 < theta < pi (pi < theta < 2pi is just a reflection), all divided by y^2. In other words, the range of a(x/y)^2 + b(x/y) + c is w(a, theta)/y^2 (in polar coordinates) for all 0 < theta < pi and any nonzero a. The corresponding rectangular coordinates yield the ratio values in the domain of w(x,y)/y^2. Since the x/y ratio is a function of direction from the origin and not distance, w(a, theta)/y^2 is invariant to (nonzero) values of a.
Thank you very much for your answer sir.
At 16:21 Why does function depends only on one direction if 4ac - b^2 = 0 ? Yes the term with y^2 is 0 but don't we have y in first term as well? (x + b/2a * y)^2
imagine a line 'x + b/2a * y = some_constant' in the xy-plane: every dot in this line will have the same z value (i.e., the same height; a* some_constant). So it produces a horizontal line with only one direction.
I Really Like The Video From Your Second derivative test boundaries and infinity.
U glitch
at 07:28, you might want to make clear that w denotes one component involved involved in the second derivative. Then we care about its sign.
Fxy = Fyx if and only if F is continous.
Denis, why you lyin?
He had mentioned in a previous lecture that we'll be only dealing with differentiable functions , and don't bother about continuity (because it won't be discussed)
Awesome stuff, thanks MIT!
This is MIT freshman math - all students in all departments are taking it and the next level of treatment starting 2nd year gets into the real meat.
At 31:18. The professor says that Fxy ALWAYS = Fyx. But Fxy doesn't ALWAYS = Fyx according to Khan Academy. Isn't Khan Academy right ?
It's always true for continuous partial derivatives. Very rarely will someone be dealing with stuff other than that
he said that we will assume for now throughout the course that the functions that will be dealing with are continuous
In India it is a compulsory question for board exam and my teacher just said the conditions as formulas and then made us practise a lot of problems of this kind. That's it!
bruh, which board of education has multivariable calculus in India for highschool?
@@cantfindagoodchannelname7359 same question
Bro which board has multivariable calculus?
@@cantfindagoodchannelname7359 maybe he meant exam on a black board haha
Linear approximation is the first order fit - i.e. tangent plane approximation. See previous lecture.
Let z ~ z0 + fsubxat(x0) *(x-x0) + fsubyat(y0) * (y-y0). Pardon terrible notation.
That notation is so confusing!
the notation with curly d's at 31:05 is wrong
+MegaDespotic Why do you think so? I cannot see any mistake... It means you first do a derivative of f just with respect to x (and other variables behave like constants) and then with respect to y ( - || - ). It is actually the same as (df/dx)*(df/dy)... but of course with partials
it should be fxy=(((d^2)f)/(dydx))
+MegaDespotic fxy=((d^2)f/(dydx))=(df/dx)*(df/dy) with curly ds
that is the same...
+MegaDespotic Well, I accidently swapped the letters, but still you know :-D The letters can be written in any order as you first do the partial derivetive of f with respect to y and you multiply it by the partial derivative of f with respect to x... And multiplication is commutative
Or am I getting you wrong? :-D
you can not write them in any order
At around 26:50, why does the professor not mention anything about the case where the discriminant is zero and there is only one root?
I am searching for the same thing did u figure out the problem
@@selcuk2649 From 16:10 he talks a bit about it, which is basically saying that it's the inconclusive case, which I think is why he just ignored going over it again at 26:50.
this lecture is so helpful
What does he mean at @7.45 'these guys are much smaller than x and y when x and y are zero' ?
"I'm sure that 300 different students means 300 different ringtones, but I'm not eager to hear all of them." and ON QUE my phone rings.
nice lecture, thanks mr auroux
Why does this have so many dislikes?
Probably some unsmart people who just want to put down anything academic.
But then why didn't they quit in lesson 1?
Also, this guy is a great teacher that even unsmart people such as myself can learn from!
I like this professsor a lot but this lecture in particular was not very clarifying. I have studied this concept before and for me was easier to understand.
@Diego Alonso I am and I was able to follow it like every other lecture. So what gives?
@@proghostbusters1627 I was able to follow it too, but I think it depends on your previous knowledge, this might be very difficult for beginners to these conscepts because he's not giving a lot of intuitive understanding.
Beautiful lecture!
21:00 wow, I finally understood it!
"Ok, you are easily amused"
I know right?
I mean I know this is MIT but if at 9:40 he would just do factoring without skipping steps, it would be much clearer for everyone and he wouldn't have to go over everything to make sure people understood.
i have learned a lot from these lectures--but i just which prof Auroux would just step throgh the algebra--he always skips steps--then says "everyone see that"--no one does--and then back tracks--tedious. Well--i am sure having to do the steps at all is tedious for him. Overall--he manages to pack a lot of insight into 45 minutes each and every time
Prof: "you are easily amused" hahahahaha @35:47
I am stuck at this point where he finds out the minima and maxima of ax^2+bxy+cx^2 through sum of squares. As per my understanding, to find out the critical point, we need to find if the slope at that point is 0. But, through the quadratic equation, we did not actually find the slope. How was the the critical point determined in that case?
(0, 0) is the critical point, which was mentioned at the beginning of this lecture. The topic of the course is to determine what type of critical point (min, max, or saddle) for this point (0,0)
I love the moment when students said wow when prof verified 4ac-b^2 by second derivative test.
Dont have money for book on amazon, how can i download it to solve the part 1 problem sets?
Suppose f(x, y) = x^3, then f_x = 0 gives x = 0, and f_y = 0 is always true. So we have a bunch of critical points that form the y axis, what should we call these points?
35:32 "ok you are easily amused" 😂 lolz
35:36 I found it disrespectful, why on earth is erasing blackboard worth laughing at?????????
U were certainly not following all the lectures. This great man is great at rubbing the 2nd board before the upper one comes down. This wow stufd has been done in previous lectures as well
sir, f sub xy is not always equal to f sub yx
really, but how to check?
sl6006095 It's called Clairaut's Theorem (or Schwarz' Theorem); it states that if the second partial derivatives are continuous, then it doesn't matter which order you do them in. You can probably find a proof online. I wouldn't call it an easy proof if you're at the level for this class, though.
And almost every function that you encounter in the sciences that you might want to differentiate will have continuous partials, so it isn't much of a concern (unless you are in a pure math class).
@@Bignic2008 thanks
38:06 Professor spitting a fire line haha
In the function
w=a.x^2 + b.xy + c.y^2
x and y play the same role, there is a form of symmetry.
That means the sign of c would matter the same as the sign of a, right?
But how he ends up discussing only the sign of, and not that of c?!!
Anyone?
for the determinant to be negative (which is the case we care about the sign of a in) we need both a and c to have the same signs, otherwise, the determinant can never be negative.
@siggascation There is, and it's the same as f sub xy.
Great lecture series!!
How is the second derivative of x⁵ = 0? Isn't it the constant function y= 5?
how is 4ac-b^2 a square term? and how is that in the third case if a0 it is min
If 4ac - b^2 = 0, then the y^2 component disappears. However, you still have the x and y inside the first term, which you square. I don't get how it's degenerate. What if W = x^2 + 2xy + y^2? 4ac = 4, and b^2 = 4, so 4ac - b^2 = 0. However, it doesn't have a degenerate critical point.
Empire State well x^2 +2xy + y^2 indeed has degenerate critical points...whole line x=-y is a line of degenerate points...think about it.
Think about it like this.
When 4ac-b^2 = 0, the second square disappears, and you're left with w = a(x+by/2a)^2. You do indeed have both the x and y terms, so you can still pick any arbitrary x and y and come up with unique values.
However, there EXISTS a direction in the x-y plane in which the value of w does not change. You can pretty easily see this if you just look at the inner x and y terms, (x + ky), where k = b/2a. If you set x + ky = h, where h represents a fixed (i.e., non-changing) value of the function w, then you can always find a line of degeneracy in the form of y = (h - x)/k. The important point is that, by following this line, (x+ky) never changes, and therefore neither does a(x+ky)^2 = w.
Using your example, W = x^2 + 2xy + y^2, we can look at it in the sum of squares form, W = (x+y)^2. It has no second square because 4ac-b^2 = 0. You can set (x+y) = h, and get that y = h-x. So for instance, if you wanted to go off into infinity and hang out at the value of W = 0 forever, you can do that by following the line y = -x. If you instead prefer the value W = 4, you can set x+y = 2 and follow the line y=2-x.
When 4ac-b^2 is NOT 0, and you have something like W = (x+y)^2 + cy^2, you can still find a line in which the first term does not change, however the second term will "screw it up" along the line and change the value of W.
Hope that helps.
To remind myself, I'm putting the list here:
Lecture 1: Dot Product
Lecture 2: Determinants
Lecture 3: Matrices
Lecture 4: Square Systems
Lecture 5: Parametric Equations
Lecture 6: Kepler's Second Law
Lecture 7: Exam Review (goes over practice exam 1a at 24 min 40 seconds)
Lecture 8: Partial Derivatives
Lecture 9: Max-Min and Least Squares
Lecture 10: Second Derivative Test
Lecture 11: Chain Rule
Lecture 12: Gradient
Lecture 13: Lagrange Multipliers
Lecture 14: Non-Independent Variables
Lecture 15: Partial Differential Equations
Lecture 16: Double Integrals
Lecture 17: Polar Coordinates
Lecture 18: Change of Variables
Lecture 19: Vector Fields
Lecture 20: Path Independence
Lecture 21: Gradient Fields
Lecture 22: Green's Theorem
Lecture 23: Flux
Lecture 24: Simply Connected Regions
Lecture 25: Triple Integrals
Lecture 26: Spherical Coordinates
Lecture 27: Vector Fields in 3D
Lecture 28: Divergence Theorem
Lecture 29: Divergence Theorem (cont.)
Lecture 30: Line Integrals
Lecture 31: Stokes' Theorem
Lecture 32: Stokes' Theorem (cont.)
Lecture 33: Maxwell's Equations
Lecture 34: Final Review
Lecture 35: Final Review (cont.)
credit: dadgum it
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@Rrumiiii take the partial derivative with respect to x, then to y....
This is helpful 🤍❤️
36:59 he changed the way whe wrote x
Excellent.
@zyflair dude that wasnt unexpected, it was an extreme moment, the board could have cut his hand off !
What if the second derivative in the denominator evaluated at the critical point equals zero...?
But (with some help from Taylor expansion) he used that specific case to justify what he claimed about general case. So it's only irrelevant if you only need to know how to compute stuff and don't strive for deeper understanding of the subject.
i honestly cant see the connection between the taylor expansion and the second derivative test how does this justify that
@@alyasker2194Taylor's series gives a decent approximation for the behavior of the function around the critical point; namely it produces a quadratic polynomial based on the second derivatives at the point of interest. You can use facts about this approximation to deduce facts about the function itself; regardless of how complicated it may be. Hope that helps tie things together.
@@xnonqme3716 hey thanks a lot!
He says fxy = fyx , that's assuming both fxy and fyx are continuous
@20.50 .why even if w is positive if a is positive we have minima
anyone know if this course is for engineers only?
This not just for engineers. This course is for anyone who wants to learn multivariable calculus.
Well, what i mean is that the students that were taking that class all of them were studying engineering or there were some other students from other careers?
All MIT students take this course (or the more advanced, theoretical 18.022). It is a General Institute Requirement at MIT (it's required for graduation).
Thanks 🤍❤️
sad. the formal definition of the saddle point isnt given...and where is the proof that the difference of two linear functions squared always yields a saddle?
VIVA LA FRANCE! This man knows his stuff.
I'VE GOT A DEVIL'S HAIRCUT IN MY MIND
LOOKING FOR MATLAB'S GREATEST HITS, 10 HRS STRAIGHT OF EXTREME COMPUTATIONAL MATH WITH BACKGROUND VISUALIZER ON UA-cam
No idea what he was doing between 24:43 to 30:00
Why do they clap sometimes when hes moving the board?
This is great!
@35:25 top 10 anime comebacks
Why is that
dw/dx=2ax+by
and
d²w/dxdy=b
and not
d²w/dxdy=2a+b
What's so funny @17:55?
yackawaytube A phone rang :/
what is the definition of saddle point?
Why were there so many dislikes?
35:54: Like a boss.
he constantly says yesterday and tomorrow, do these guy have lessons all days ?
when do they do the homework ?!
however if they only have 1hs of calculus thats not such a pain. having 4hs x 2 days is the same pain or worse, but you can at least do the homework between these two days !
the first 30 mins were ;/ ;/
and the next 20 mins were :)
The black board trick never gets old...XD
the class is so rambunctious.. it must be that sexy french accent
But that was a critical moment in the lecture... another one of prof. Auroux's famous "speed erases". :)
83 dislikes from dumb people who hate math.
410 likes from the future innovators of this world.
Lecture had them so shook they had to resort to comedy
What happene'd at 18：00？
skipping this lecture for now...too much messy algebra for me.
Hello Dr. Adler's class hope y'all can see my comment
the prof said "u are easily amused"
wtaf is going on in this lecture 🤣🤣 definitely going to need to rewatch
haha the second derivative test is just the determinant for a 2x2 hessian matrix, linear algebra FTW
Wow, just wow.