User:IssaRice/Chain rule proofs
a differentiable function looks locally like a linear transformation. if you compose two differentiable functions, then it seems pretty obvious that locally, that composed map also looks like a linear transformation. but which linear transformation? well, you first apply the linear transformation that locally approximates the inner function. then you land in some target space, and you find what the outer function locally looks like at the place you land. and that's all the chain rule says.
So Folland actually says this on page 109: "Since the product of two matrices gives the composition of the linear transformations defined by those matrices, the chain rule just says that the linear approximation of a composition is the composition of the linear approximations." But... putting this short paragraph away after a proof, and only after you've already discussed two versions of the chain rule in previous chapters, is in my opinion not trying hard enough to communicate the central message. This should be in bold flashing letters at the TOP of the FIRST discussion of the chain rule.
Using Newton's approximation
The main idea of using Newton's approximation to prove the chain rule is that since f is differentiable at we have the approximation when is near . Similarly since g is differentiable at we have the approximation when is near . Since f is differentiable at , it is continuous there also, so we know that is near whenever is near . This allows us to substitute into whenever is near . So we get
Thus we get , which is what the chain rule says.
Slightly more formally:
when is near
when is near
is continuous at so is near whenever is near .
Thus if is near
to complete the proof, we just need to show that the error is . The former term clearly is. The latter term is so it is as well.
We want to show is differentiable at with derivative . By Newton's approximation, this is equivalent to showing that for every there exists such that
whenever . So let .
Now we do some algebraic manipulation. Write
where . This holds for every . Since we thus have
Substituting the expression for in the expression for we get
we can rewrite this as
Thus our goal now is to show .
But by the triangle inequality it suffices to show .
where we are free to choose .
To get the bound for (using Newton's approximation), we need to make sure is small. But by continuity of at we can do this.
where again we are free to choose .
TODO: can we do this same proof but without using the error term notation?
TODO: somehow Folland does this without explicitly using continuity of f; i need to understand if he's using it implicitly somehow or he's actually proving it when bounding using
Since is differentiable at , we know is a real number, and we can write
(there is no magic: the terms just cancel out)
If we define we can write
Newton's approximation says that as long as .
Since is differentiable at , we know that it must be continuous at . This means we can keep as long as we keep .
Since and , this means we can substitute and get
Now we use the differentiability of . We can write
Again, we can define and write this as
Now we can substitute this into the expression for to get
where we have canceled out two terms using .
Thus we have
We can write this as
where . Now the left hand side looks like the expression in Newton's approximation. This means to show is differentiable at , we just need to show that .
The stuff in square brackets is our "error term" for . Now we just need to make sure it is small, even after dividing by .
But f is differentiable at , so by Newton's approximation,
we also have
We can bound this from above using the triangle inequality:
Now we can just choose small enough.
Limits of sequences
Let be a sequence taking values in that converges to . Then we want to write
Now use the limit laws to conclude that the limit is . The problem is that can be zero even when .
Let be a sequence taking values in that converges to .
Define a function by
The idea is that we want to say is going to , so we just define it at the undefined points to already be at that limit.
Now we have
for all . (Why? Consider the cases and separately.)
Differentiability of at says that if is a sequence taking values in that converges to , then as . What if is instead a sequence taking values in ? Then we can say as . To show this, let .
Now we can find such that for all , if , then . (TODO: I think here we need to break off into two cases: one where there's an infinite number of such that , and one where there's only a finite number so that eventually the sequence is all just . Only in the former case can we find , by considering the subsequence that isn't equal to , but this is not a problem because in the latter case the sequence's tail is already at the place where we need it to be, so we don't even need to find . The question is, is there some more elegant way to do this that doesn't break off into cases?)
But this means if , then we have two cases: either and , in which case as above, or else , in which case so .
Differentiability of at implies continuity of at , so this means that as . Since for each , we can use as our sequence in to conclude that as we have .
Now by the limit laws
Since the sequence was arbitrary, we can conclude that .
TODO: Tao says that division by zero occurs when , which seems strange to me.