We had a homework assignment a couple of weeks back. It was looking at mathematics in a very different way from how many had seen it before, and it caused a lot of confusion. I would like to try and add some clarity to what we were doing. My thought was, rather than going through the questions themselves, I would like to add annotations to the proof itself. Let’s see how this works. The proof that you were given is in black, the annotations are in blue, and after I’ve been through the proof, I will expand on it in a simplified form.


Theorem: The function f is differentiable at x=a if and only if there is a constant m and a function E of x, defined for all x \not = a, such that


f(x)=f(a)+m(x-a)+E(x)(x-a) for all x \not = a      – (eq 1)




\lim\limits_{x\to a}E(x)=0.


(If both these conditions are satisfied, then f'(a)=m.)


What we are doing here is giving another definition of differentiability (at a particular point, a). You already have a definition of differentiability at a point which is related to the formal definition of the derivative of the function. ie. if:


lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}


exists, then the function is differentiable at a.


Sometimes this isn’t the most convenient definition of differentiability at a point, and indeed to prove the chain rule, it’s not the most convenient way to go either. Now we have two ways to show that a function is differentiable. What we need to prove is that our original definition of differentiability at a point implies that you can write the function in the form above, and vice versa.


Proof. If f is differentiable at x=a, then f'(a)=\lim\limits_{x\to a}\dfrac{f(x)-f(a)}{x-a} exists.


ok, that was our original definition.


For x\neq a, put E(x)=\dfrac{f(x)-f(a)}{x-a}-f'(a).


Here we’re simply defining a function E(x) in a particular way, defined in terms of the derivative of f. This is ok, because at the beginning we said “if f is differentiable at x=a”, and so we are allowed to use f'(a) here. You might ask the question why we define E(x) like this, but if you look at the thing that we’re trying to prove (in particular equation (1)), then it makes sense.




\lim\limits_{x\to a}E(x)=\lim\limits_{x\to a}\,\left[\dfrac{f(x)-f(a)}{x-a}-f'(a)\right]=f'(a)-f'(a)=0


Here we are looking at a property of E(x) given how we’ve defined it.


We’ve now shown that if f(x) is differentiable at a, then we can write down a function E(x), as above, and indeed its limit as x tends to a is 0, as required by the theorem.






If we put m=f'(a), we have




This proves the first half of the theorem.


OK, so what we’ve seen is that f(x) can be written in the form above (involving m and E(x)) if f is differentiable. Now we need to show that if we can write it in that form, then f is necessarily differentiable.


For the second half, suppose there is a constant m and a function E of x such that:


f(x)=f(a)+m(x-a)+E(x)(x-a) and \lim\limits_{x\to a}E(x)=0.


Now we are assuming that you can write f(x) in this form and seeing what the consequences are. We now rearrange this equation and see that:




\lim\limits_{x\to a}\dfrac{f(x)-f(a)}{x-a}=\lim\limits_{x\to a}\,[m+E(x)]=m.


Hence f is differentiable at x=a.


What we have shown is that writing f(x) in this form can be rearranged and the limit taken which is precisely the definition of the derivative. Because of the property of E(x) (its limit), this shows that the limit definition of the derivative exists and thus the function is differentiable.


Let’s now go through what the order of logic is. I’m going to change things around completely and use a different example so that we don’t have to think about the maths, but we can think about the logical implications:


Theorem: Tumi will be going to the party on Saturday night if and only if Esti is going to the party.


In the original theorem “Tumi going to the party” was “f(x) is differentiable at x=a”, and Esti going to the party was “f(x) can be written in the form…”.


What we are saying here is that Tumi going to the party is equivalent to Esti going to the party (one implies the other).


Proof of the “only if” party:


If Tumi goes to the party then we can be sure that he will let Esti know, and whenever Esti knows about a party, you can be sure that she will be there.


Here we have to be careful about what we’ve just proven. It could still be that Esti goes to the party but Tumi is not going. But so long as Tumi is going, Esti will definitely be there. In the original theorem this is the “only if” part of the statement:


Tumi will be going to the party on Saturday night only if Esti is going.


We haven’t proved the whole theorem. We still need to show that If Esti is going then Tumi will definitely be going. We do this by using some other knowledge we have:


Proof of the “if” part:


Whenever Esti knows about a party, she gives Tumi a call and he will definitely turn up if he knows about it.


This means that Tumi will go if Esti goes.


The fact that we’ve proved that Tumi will go if Esti goes and Tumi will go only if Esti goes means that one implies the other.


Make sure that you understand the order of the implications (ie. what implies what when you have an if and only if statement). In even simpler language:


a is true if and only if b is true:


The only if part is the same as “if a is true, then b is true”


The if part is the same as “if b is true, then a is true”.

How clear is this post?