Maple versus Mathlab

How is the convolution of a function calculated with itself in MATLAB and WolframAlpha?

I don't know much about Mathematica, so I can only tell you (partially) about the Matlab part.

Doing the convolution with Matlab conv functions means it's numeric. What do you mean by the integral definition is that you want to do it symbolically. For this you need the Matlab Symbolic Toolbox. This is essentially a maple plugin for Matlab. So what you want to know is how it works in maple.

You might find this link helpful (wikibooks) for an introduction to the MuPad in Matlab.

I was trying to implement your box function as a function in Matlab as

However, this does not work with the t-ü symbolic type.

Could you explain it as a piecewise function see (matlab-online-help), this didn't work in my matlab though. The examples are all Maple syntax, so you would be working in Maple right away.

I have to work around it

Unfortunately this did not work


and not to an explicit result. You can still use the evaluation of this expression, e.g.

But Matlab / MuPad doesn't give you and is an explicit expression in the sense of t. This is not really surprising as the function is discontinuous. This is not so easy for symbolic calculations.

I would now go and help the computer, luckily for the example that you asked, the answer is very simple. The fold in your example is just that. The ring rolled products is again a box function simply not one that goes out from [-1,1], but a smaller interval (that depends on t). Then you just need to integrate the function over this smaller interval.

Some of these steps have to be done by hand first, then you might try going back to Matlab. You don't even need a computer algebra program to get the answer.

Maybe Matematica - or Maple - actually could solve that problem, just remember that the MuPad included with Matlab is just a stripped-down version of Maple. Perhaps the above could still help you; it should give you the idea of ​​how things work together. Try looking in a nicer function for f, e.g. a polynomial, and it should work.

Source of informationAuthor H. Brandsmeier