You are very good at explaining - I find your talking speed fantastic for grappling with these concepts myself, giving me time to ponder the meaning of implications. Thanks for making these videos!
Thanks! I'm not sure if this would answer your question, but let's see. Since the differential D_{c}f of f at a point c tells us what happens to tangent vectors at c, if the determinant is non-zero, then D_{c}f has no kernel, i.e., there are no tangent vectors that get sent to the zero tangent vector on the codomain. If there was such a vector, then there would be a direction in which the image points of f get "bunched up" together along that direction. The function f(x)=x^{3} gives a good example of this at the point 0. A simple example with 2 variables is f(x,y)=y+x^{3}. Imagine traffic near a stop sign and how cars still have a positive velocity as they approach the stop sign, eventually stop, and then continue forward. Now, if points get bunched up like this, then there would be no smooth way to "unbunch" these points (via composition). In the example of the traffic at the stop sign, we would be looking for a function h such that, in particular, g(f(x))=x and g is differentiable. This means that we would be looking for a function (g) such that when I apply it to the position function (f) of the car as a function of time (x), then I would get a resulting function that had constant unit velocity as if there was no stop sign there. But since f'(x)=0 at the stop sign, g would have to compensate for that by increasing its derivative to infinity at that point. This same idea holds in higher dimensions by looking at any particular direction in which the derivative vanishes.
@@ArthurParzygnat Thank you so much for giving a detailed answer! I promise you that the first thing I will do when I wake up tomorrow is to go through your answer properly. It's midnight here now so I will reply as to how much it helped tomorrow 👍
@@NoNTr1v1aL No worries. Does the response make sense right now for a function of just one variable, such as f(x)=x^{3}? The inverse of this function is given by g(y)=y^{1/3}, which is continuous, but if you compute its derivative at y=0, what do you get?
You are very good at explaining - I find your talking speed fantastic for grappling with these concepts myself, giving me time to ponder the meaning of implications. Thanks for making these videos!
thanks! 👏🏻👏🏻❤️
Is there a geometric intuition for the inverse function theorem? Great video btw!
Thanks! I'm not sure if this would answer your question, but let's see. Since the differential D_{c}f of f at a point c tells us what happens to tangent vectors at c, if the determinant is non-zero, then D_{c}f has no kernel, i.e., there are no tangent vectors that get sent to the zero tangent vector on the codomain. If there was such a vector, then there would be a direction in which the image points of f get "bunched up" together along that direction. The function f(x)=x^{3} gives a good example of this at the point 0. A simple example with 2 variables is f(x,y)=y+x^{3}. Imagine traffic near a stop sign and how cars still have a positive velocity as they approach the stop sign, eventually stop, and then continue forward. Now, if points get bunched up like this, then there would be no smooth way to "unbunch" these points (via composition). In the example of the traffic at the stop sign, we would be looking for a function h such that, in particular, g(f(x))=x and g is differentiable. This means that we would be looking for a function (g) such that when I apply it to the position function (f) of the car as a function of time (x), then I would get a resulting function that had constant unit velocity as if there was no stop sign there. But since f'(x)=0 at the stop sign, g would have to compensate for that by increasing its derivative to infinity at that point. This same idea holds in higher dimensions by looking at any particular direction in which the derivative vanishes.
@@ArthurParzygnat Thank you so much for giving a detailed answer! I promise you that the first thing I will do when I wake up tomorrow is to go through your answer properly. It's midnight here now so I will reply as to how much it helped tomorrow 👍
It looks like I have to learn a bit more to understand the answer. One day I will get it for sure.
@@NoNTr1v1aL No worries. Does the response make sense right now for a function of just one variable, such as f(x)=x^{3}? The inverse of this function is given by g(y)=y^{1/3}, which is continuous, but if you compute its derivative at y=0, what do you get?
@@ArthurParzygnat The slope is infinite at the origin.