On Tue, Feb 3, 2015 at 11:14 PM, Sturla Molden sturla.molden@gmail.com wrote:

Warren Weckesser warren.weckesser@gmail.com wrote:

...

                0    if x < 0
heaviside(x) =  0.5  if x == 0
                1    if x > 0

This is not correct. The discrete form of the Heaviside step function has the value 1 for x == 0.

heaviside = lambda x : 1 - (x < 0).astype(int)

By "discrete form", do you mean discrete time (i.e. a function defined on the integers)? Then I agree, the discrete time unit step function is defined as

u(k) = 0 k < 0 1 k >= 0

for integer k.

The domain of the proposed Heaviside function is not discrete; it is defined for arbitrary floating point (real) arguments. In this case, the choice heaviside(0) = 0.5 is a common convention. See for example,

* http://mathworld.wolfram.com/HeavisideStepFunction.html * http://www.mathworks.com/help/symbolic/heaviside.html * http://en.wikipedia.org/wiki/Heaviside_step_function, in particular http://en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument

Other common conventions are the right-continuous version that you prefer (heavisde(0) = 1), or the left-continuous version (heaviside(0) = 0).

We can accommodate the alternatives with an additional argument that sets the value at 0:

heaviside(x, zero_value=0.5)

Warren

Sturla

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On 04/02/15 06:18, Warren Weckesser wrote:

By "discrete form", do you mean discrete time (i.e. a function defined on the integers)? Then I agree, the discrete time unit step function is defined as

It is the cumulative integral of the delta function, and thus it can never obtain the value 0.5. The delta function is defined to have an integral of 0 or 1.

Sturla

On 04.02.2015 11:45, Daπid wrote:

There are several definitions. Abramowitz and Stegun (http://people.math.sfu.ca/~cbm/aands/page_1020.htm) assign the value 0.5 at x=0.

The NIST handbook uses the value 0 at x=0.

Perhaps a Heaviside with an optional argument that defines the value at x=0 would be good. I'd love to see that in NumPy.

Actually, Heaviside function is better seen as a distribution instead of a function, and then there is no problem with the value at 0, as long as it is finite.

Understanding a distribution as the limit of a sequence of functions, the value at x=0 then depended on the choice of function in the sequence, I guess. Using something symmetrical such a Gaussian or a centred box then makes the value of 0.5 plausible.

Alex

Le 04/02/2015 06:58, Jaime Fernández del Río a écrit :

I have an implementation of the Heaviside function as numpy
ufunc.  Is there any interest in adding this to numpy?  The
function is simply:

                    0    if x < 0
    heaviside(x) =  0.5  if x == 0
                    1    if x > 0

I don't think there's anything like it in numpy. Wouldn't scipy.special be a better home for it?

scipy.signal could also host it, since it already contains functions for linear systems (e.g. step response, which are closely related), and also some waveform generators like square()

http://docs.scipy.org/doc/scipy-0.14.0/reference/signal.html

However, I agree with Joseph when he says that this function is a bit thin.

best, Pierre