# User:IssaRice/Chain rule proofs

## Contents

## Using Newton's approximation

### Main idea

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.

### Proof

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

Similarly write

where .

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

### old proof

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

### Main idea

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 .

### Proof

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.