Defining a counting function in mathematica

  • Mathematica
  • Thread starter michael71828
  • Start date
  • #1
michael71828
4
0
Hi and thank you for reading this.

I'm learning to use mathematica and among those things I'm trying to do, is to define a function that can count for me, say, the number of positive zeros less than a given number Z of a familly of function.

For exemple, let f_n(x) = sin(x/n) for any natural number n. I want mathematica (or any other program you may suggest) to compute the cardinality of {x \in ]0,Z[ = (0,Z) : \exists n \in N for which f_n(x)=0}. For Z = \pi + 1, I would expect that cardinality to be 1, in that case. For Z=2\pi +1, it'd be 2, and so on.

I want a general method, so I can apply it to my specific problem.

Just let you know that I've google'd it, I looked wolfram's forums, wolfram documentation, asked friends, and couldn't find it !

I hope I was clear, and that my english wasn't too bad.

Thank you !
 

Answers and Replies

  • #2
michael71828
4
0
Maybe the 2\pi +1 and so on wasn't so clear.

I really meant ''x over n'' in sin(x/n), but I used LaTeX terminology for ''pi'' 2\pi is really 2*pi.

I don't think I fooled many of you, but just in case... :)
 
  • #3
michael71828
4
0
Or maybe matlab would be better ?
 
  • #4
Dale
Mentor
Insights Author
2021 Award
33,291
10,675
Hi michael71828, welcome to PF,

For future reference you can display LaTeX using tex tags. Also, you will get faster answers if you don't respond to your own OP since some people look for unanswered posts and respond to those first.

I would recommend the FindInstance function for your purpose. Look at the documentation and see if you have any questions on how to use it. Be sure to use the option which allows you to specify the maximum number of instances to return.
 

Suggested for: Defining a counting function in mathematica

  • Last Post
Replies
2
Views
3K
Replies
5
Views
557
Replies
4
Views
923
Replies
1
Views
193
Replies
6
Views
434
Replies
0
Views
154
Replies
4
Views
189
Replies
1
Views
217
Replies
1
Views
638
  • Last Post
Replies
1
Views
540
Top