Arriving at the conclusion that zero multiplied by undefined is zero

Question

I was working on a question which asked whether $f(x).g(x)$ is differentiable at $x=a$, given that $g(x)$ is not differentiable at $x=a$. My solution is as following:

I deduced that $f(x).g(x)$ is not differentiable at $a$.

But then I checked my solution using an example, and it seems that my solution was wrong.

In the above example, $f(x)=x-1$ and $g(x)= (mod(x-1)+1)$. Although $g(x)$ is not differentiable at $x=1$, $f(x).g(x)$ IS differentiable at $x=1$.

And this is how I am arriving at the conclusion that zero multiplied by something undefined is zero (at least in this case).

So my questions are:

1. Can zero multiplied by something undefined be zero?
2. If the answer to the above question is no, then what is the mistake in my argument?
3. Is $f(x).g(x)$ differentiable at $x=a$, when $g(x)$ is not differentiable at $x=a$? What are the necessary conditions for $f(x).g(x)$ to be differentiable at x=a?

Note: $f(x)$ is supposed to be differentiable at x=a.

PS: I have just begun studying calculus, and I am not really a genius in Maths. So please excuse me for any grave msitakes that I may have committed above. Thanks.

Answer 1

There is a key issue in your post: you can't use the product rule the way you're using it. The product rule is correctly states as follows:

Let $f,g:\mathbb{R} \rightarrow \mathbb{R}$ be functions that are both differentiable at $a \in \mathbb{R}$. Then the function $h(x) = f(x)g(x)$ is differentiable at $a$ and this derivative is given by: $$h'(a) = f'(a)g(a) + g'(a)f(a)$$ where $f'$ and $g'$ are the derivatives of $f$ and $g$ respectively.

Now it may be that even in the case where $g$ is not differentiable at $a$, the function $h$ is differentiable at $a$. However the derivative cannot properly be written using the product rule.

This observation is the root of your problem. In general, you can't define quantities like $0 * [\text{undefined}]$. That's exactly what it means to be undefined! You can't do arithmetic with [undefined] in the traditional sense!

So, to go through your questions:

1. You can't define such a quantity in general. If you're doing arithmetic with [undefined]s, you've probably made a mistake earlier in your thought process
2. The error in your reasoning is applying the product rule in a situation where the necessary conditions to apply the product rule haven't been met
3. In general, there isn't going to be a nice set of necessary and sufficient conditions to show when $f(x)g(x)$ is differentiable at a point in terms of the differentiability of $f(x)$ and $g(x)$. See the following pathology:

\begin{align*} f(x) &= \begin{cases} 0 & \quad \text{if $x$ is rational} \\ 1 & \quad \text{if $x$ is irrational} \end{cases} \\ g(x) &= 1 - f(x) \end{align*} Notice that both $f$ and $g$ are nowhere differentiable, but we have $h(x) = f(x)g(x) = 0$ for all $x$. $h$ is differentiable everywhere.

Answer 2

1. Yes, you showed it yourself! Another fun example would be to look at $$ x\cdot\sin(1/x) $$ at $x=0$.
2. Your mistake lies in the fact that you have just written down $f(a) g'(a)$. Now, this does not exist but it might be the case that $$ \lim_{x\to a} f(x) g'(x) $$ exists.
3. Really depends on the situation, but if $f(a)$ is non-zero, it will definitely be not the case. In the other case, you can find examples at both ends.

I hope this helps!

Answer 3

1. Zero multiplied by something undefined is undefined. So the answer is no.

2. Your mistake lies in a wrong application of the theorem which says: If $h=fg$ and $g$ and $f$ are differentiable at $a$, then $h$ is differentiable at $a$ and $h'(a)=g'(a)f(a)+g(a)f'(a)$. In your case, the hypotheses of the theorem are not fulfilled.

3. Depends on the situation. You show one example in which none are differentiable or continuous and the product is everywhere differentiable. What is not possible is that $f$ and $g$ are differentiable and $h$ is not. The product rule theorem gives you sufficient conditions, not necessary. Both differentiable, then the product is also. If any one is not differentiable, $h$ could either be or not.