Second partial derivative test intuition
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- เผยแพร่เมื่อ 29 มิ.ย. 2024
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The second partial derivative test is based on a formula which seems to come out of nowhere. Here, you can see a little more intuition for why it looks the way it does. - แนวปฏิบัติและการใช้ชีวิต
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This guy is a gift to humanity
It should be illegal to make calculus this simple.
Justin Ward absolutely right
good visuals for intuition and lack of proofs go a long way
Hari Sahan not, and no such thing as simple or not simple, any can be ok
Quite amazing, innit?
hnnn
I'll never look at pringles in the same way again..
Genius 😂
Another way to think about the Second partial derivative test is using Linear Algebra:
Consider a 2x2 Matrix H with elements as [[fxx, fxy], [fxy, fyy]], with derivatives evaluated at (x0,y0)
Now, think about how the point (x0, y0) will move once it is exposed to the matrix H. This is a Linear transformation case.
If the linear transformation stretches the area onto the same plane, then that point (x0,y0) represents either a maxima or minima.
On the other hand, if the linear transformation flips the plane ==> (x0,y0) is a saddle point.
Mathematically, if
det(H) > 0 ==> maxima/ minima
det(H) saddle point
det(H) = 0 ==> cant decide.
Ankit Gupta never notices that, thanks!
The Hessian matrix :)
Cool, but I do not understand how the flipping of space is equivalent to a saddle point
@@robmarks6800 he's just talking because the formula looks like determinant of a 2x2 matrice, nothing related between these 2 topics
@@robmarks6800 same question here
It's also sort of helpful to plot a vector field of the gradient function. If d^2f/dx^2 and d^2f/dy^2 are both positive, we'd have divergence around the center that'd make the point a minimum, however, if you plot some vector fields you can see that for saddle points the vectors sort of fly off in top right and bottom left corners leaving the diagonal as a saddle point(if you plot the example in the previous video). That measure of how much they fly off is df/dydx; you could imagine moving in x direction and looking at the change in y component of the vectors, or move along y axis it doesn't really matter. So there's a certain threshold *(d^2f/dx^2)*(d^2f/dy^2) = the mixed partial derivative^2* after which we get a saddle point.
Dude??? Why is Khan Academy so freaking amazing!!!!!!
This should be called the Pringle Test.
Can you please provide a link to your paper?
i owe my whole tuition fee for your videos man... U guys are rocking.. God bless you
how can you be this good?
omg!! the series of multivariable function is incredibly clear!!! OH my god!! YOU are a talented teacher when doing a job like this!
I thought I was too dumb for any form of teaching and got my mind ready for not understanding anything. But your video headshotted all of my doubts to my pleasant surprise. Layman term yet fundamentally solid explanations like yours are what I need
where can i fine the article that he mentioned?
Where can we find that article you are talking about? It would be nice to put it in the description
I wish my teachers were more like you :)
Hats off to your teaching skills
Why do we only consider xy term? There are several terms (such as xy^2) which have non-zero mixed partial derivatve. Why this theorem holds for them too?
Can u tell me the app u use fro graph in 3d
As Grant said, there is a more rigorous way to show why the test has that form. From the formula of total derivative df, we can construct a formula for d^2(f) which is a quadratic form of dx and dy. Then, we can classify the critical point based on the sign of d^2(f) (it is a saddle point if d^2(f) can be positive or negative; it is a maximum or a minimum if d^2(f) is always positive or always negative). And, determining the sign posibilities of d^2(f) by the discriminant of the quadratic form- the minus H.
Very cool. Actually, if there is disagreement between fxx and fyy, that is sufficient to conclude it is a saddle point. Only if there is disagreement do we need to deduct fxy to see whether the gradients that are not strictly along the x- or y-axis identify the point as a saddle.
Yes, critical points exist even if they are not any of extrema both in 2D & 3D.
Amazing video that clarifies a big ? I had when learning this
You are just amazing man!
Can u just say it’s the determinant of the hessian for future reference? Wish I had known sooner
where can i find the article
could you tell us please where is the Article you have mentioned .
Also worth pointing out that H is just the determinant of the Hessian matrix, which is, I'm assuming, why he labeled that variable as H. Kind of surprised he didn't identify that explicitly in the video, because it makes it easier to remember the test, knowing that we're just looking at the sign of the determinant of the Hessian. It also enforces the parallel with the single-variable case, where we're looking at the sign of the second derivative, because we can imagine the Hessian as analogous to a multivariable second derivative.
I'm surprised Grant didn't say something like "the mixed second partial derivative is a kind of diagonal saddlification of the surface" :D
can't find the related link about the formula, is it removed?
Bruh i wish i saw these while i took multivariabel calc
Why we always check the sign of partial double derivative of x only and not y to determine max or min when that h is positive
Because we know that if H is positive it's either a max or a min. The (fxy)^2 term is always positive so we're always subtracting a positive number (adding a negative number if you like) and the only way for H to be positive in this condition is if the first term is positive so either both of them (fxx and fyy) are positive or both of them are negative therefore if we know the sign for either of them it'll be enough.
Hope that helped.
@8:10, the height alone x axis is a constant, which corresponds to df/dx=0, not d2f/dx2=0, like the tutor said. Anyboday any comments?
df/dx = 0 also implies that d²f/dx² is also 0
Where is the article??? :( i want it xd
did you find it?
What if your 2nd partial derivatives equal 0? Not H, but fxx. How can you absolutely have a max/min but not have concavity? What does that mean? Or did I just make a mistake?
Rockyzach88 if fxx and/or fyy are equal to cero, then the first term of H is going to be cero, making H less than cero (assuming fxy exists) and, therefore, indicating that you are in a saddle point
You can only conclude you have a saddle point if H < 0. If H=0, the 2nd derivative test is inconclusive.
If both direct 2nd partial derivatives equal zero, then this means you simply don't have curvature in the x and y directions, and all your curvature is on a direction that is rotated from them. The mixed derivative would tell you the information about the function's curvature.
If both fxx and fyy are equal to zero, but fxy and fyx are not, then you have a saddle point, where your most-positive and most-negative curvatures will occur in rotated axes from the x and y directions. You can construct an equivalent of Mohr's circle, to find those directions, and find the principal second partial derivatives where curvature is maximum.
If the entire 2nd derivative matrix has a determinant of zero, then this means you have a situation where the 2nd partial derivatives aren't good enough to tell you the nature of the stationary point. You could have an inflection point, or you could have the nature of the stationary point buried in a higher order derivative, like it is in the single variable case of y=x^4.
Which playlist is this in???? @khanacademy
i think multivariable calculus
anybody got the link to the article mentioned at 9:42?
xy 성분이 커질수록, 대각선 방향에서 포물선을 만드는 경향이 커진다는 건가..?
Why is the 2nd partial derivative with respect to x twice the one that determines if it is a maximum or a minimum ?
It’s not, doing it with y would work too. It’s just if one is positive so would the other be because this is in the scenario where the point is either a minima or a maxima meaning both the x and y planes have the same concavity so only testing one is enough
Makes absolutely no mention that the second derivative test is the determinant of the Hessian matrix, which he introduced 7 videos ago in talking about the much less important quadratic approximation!
Yea this stuff makes a lot more sense if you know that the determinant of a matrix is the product of eigenvalues and the trace is the sum of eigenvalues.
but why double?
Grant Sanderson of 3Blue1Brown
is this the 3blue1brown guy(cant remember his name)? their voices sound sooo similar
Yes there the same person
Yes, and his name is Grant Sanderson.
Where are the articles you mention?
I think the articles are in Khan Academy site.
Oh thanks!
I think it's this www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/second-partial-derivative-test
Here's directly the actual full reasoning article, which is linked on that one: www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/reasoning-behind-the-second-partial-derivative-test
@@Ropbastos thanks a lot...!
Bro i love you
Why are you assuming fxy at x0, y0 is the same as fyx at x0, y0? Isn't that something that requires explanation because it is not obvious.
Look up Clairaut's theorem on mixed partials
take a look at video 18 in the series: th-cam.com/video/J08-L2buigM/w-d-xo.html
He talks about mixed partials in another video
i LOVE you.
I love you.
Why fxy not fyx
they are the same so long as the second partial derivative is continuous