Derivative of floor function

Question

I thought the derivative of function $\text{floor}(x)$ should be $\infty$ for integer values of $x$ and 0 elsewhere. But wolframalpha plot showed something different. Is there any explanation?

Answer 1

The Alpha plot is badly wrong. The derivative of $\lfloor x \rfloor$ is 0 at non integers and not defined at integers. You would have to ask the people at Wolfram why this happens.

Update: Alpha has changed its behavior in response to the input. It clearly now does not understand the request at all.

Answer 2

If you try asking Wolfram Alpha to differentiate the floor function, it will just output "Floor'(x)". If you force Wolfram Alpha to plot the derivative of the floor function, I think what Wolfram Alpha does is it as an infinite sum of dirac deltas, so that when you integrate, you can still get back the floor function. See http://reference.wolfram.com/mathematica/ref/DiracDelta.html and http://mathworld.wolfram.com/DeltaFunction.html

Clearly that's not what you had in mind when you asked Wolfram Alpha to plot the graph, but well...

Answer 3

It's probably calculating the derivative numerically instead of symbolically. An approach that gives a good approximation for functions that actually are differentiable, but behaves weird when the function is discontinuous.

For example, using a simple centered difference approximation $f'(x) \approx \frac{f(x + h) - f(x - h)}{2h}$, with $h = 0.05$, gives $f'(1) \approx \frac{f(1.05) - f(0.95)}{0.1} = \frac{1 - 0}{0.1} = 10$. This is nonsense, but an artifact of the computation.

Answer 4

the derivative of a step function $$ \sum_{n\le x}a(n) $$ is always defined in the sense of distribution as $$ \sum_{n=0}^{\infty}a(n)\delta (x-n) $$