# Is there a way to write the convolution form of a derivative?

In image filtering and computer simulations, derivatives can be applied to a image by performing a convolution of the image/data with an appropriate kernel matrix. For example, in 1D this could be

$$k = [-1,0,1]/2$$

This form makes sense for discrete, evenly spaced data. How would you express this in a general way?

I'm thinking of something of the form:

$$\frac{df}{dx} = f * k$$

$$k = ?$$

Does this make sense?

• That should be k=[1,0,-1]/2. – Cris Luengo Dec 16 '18 at 16:09
• @CrisLuengo, yes you are right. – user668074 Dec 17 '18 at 5:50

To answer your question directly: No there is no such function There is not even a function k such that $$f = f*k$$.

BUT: If one is not very mathematically precise, and one is familiar with the Dirac-„Function“ one could say „$$f = f*\delta_0$$“. Then one could look at one of the differential quatients (vor example the symmetric one, since you already mentioned [-1,0,1]) which would then look like this $$k:=\lim_{h\rightarrow 0} \frac{\delta_{h}-\delta_{-h}}{2*h}$$ And use this as a kind of pseudo definition for such a „function“

If you then assume, that limit exists, and you can pull out the limit from the integral you get $$f*k(x)=f*(\lim_{h\rightarrow 0} \frac{\delta_{h}-\delta_{-h}}{2h})(x) = \lim_{h\rightarrow 0} \frac{1}{2h}(f*\delta_{h}-f*\delta_{-h})(x) =\lim_{h\rightarrow 0} \frac{1}{2h}(f(x+h)-f(x-h)) = f‘(x)$$

If you are interested in the real workings behind that, you would have to learn something about (tempered) Distributions.

I guess this answer is “not very precise”. I also use “image” and “function” interchangeably.

In the continuous domain,

$$\frac{\partial}{\partial x} f = \frac{\partial}{\partial x} (f * \delta_0) = f * \frac{\partial}{\partial x} \delta_0\;,$$

easily provable using Fourier analysis.

In the discrete domain, this kernel can be approximated by [1,0,-1]/2 or [1,-1], or in many other ways (e.g. using different samplings of the derivative of the sinc function). With a non-regular grid this becomes more complicated to approximate.

One very good approach to estimating derivatives in images uses Gaussian regularization (images hace noise, regularization helps there; a Gaussian is a good filter because it is perfectly isotropic yet separable). In fact, one can compute the exact derivative of a function smoothed by a Gaussian by convolving with the derivative of a Gaussian,

$$\frac{\partial}{\partial x} (f*g) = f * \frac{\partial}{\partial x} g\;,$$

which can be sampled with very little loss.

For irregular sampling, this regularization helps too: normalized convolution [1] allows to apply a Gaussian smoothing to a function with missing or uncertain samples, as well as an irregularly sampled function.

The normalized differential convolution is as follows, with $$m$$ an indicator function that is 1 where there is a sample in $$f$$, and $$g$$ the Gaussian kernel.

$$\frac{\partial}{\partial x} \frac{(f \, m) \ast g}{m \ast g} = \frac{(f \, m) \ast \frac{\partial}{\partial x} g}{m \ast g} - \frac{(f \, m) \ast g}{m \ast g} \frac{m \ast \frac{\partial}{\partial x} g}{m \ast g} \;.$$

[1] H. Knutsson and C. F. Westin, "Normalized and differential convolution," Proceedings of IEEE Conference on Computer Vision and Pattern Recognition, New York, NY, 1993, pp. 515-523.