# [Maxima] Discontinuity Points

Stavros Macrakis macrakis at alum.mit.edu
Thu Jun 28 14:00:44 CDT 2012

```There are two ways I can think of to introduce discontinuities into a
function:

* Functions with discontinuities, e.g. tan(x), 1/x, mod(x,2), etc.  That
reduces to the problem of solving for the appropriate values of the
argument.
* Explicit conditionals, e.g. if x>0 then 1 else 0.

Everything else is a composition of these two.

It is easy enough to recursively search the expression tree for these
cases.  There are some refinements to the basic idea that are harder to
implement:

* Dealing with removable or spurious discontinuities (if you want to), e.g.
what is the desired result for sin(x)/x or mod(x,%pi)*sin(x)?
* Combining discontinuities nicely, e.g. discontinuities(1/sin(x)) is
%z1*%pi and discontinuities(1/sin(2*x)) is %z2*%pi/2; but the second case
subsumes the first case, so discontinuities(1/(sin(x)*sin(2*x)) is simply
%z3*%pi/2.

-s

On Wed, Jun 27, 2012 at 1:34 AM, Aleksas Domarkas <aleksasd873 at gmail.com>wrote:

>   How to find the points of discontinuity af a function on interval?
>
>  Application
>  Theorem(Generalized Newton-Leibnitz formula).
> Let F = integrate(f,x) and
> x[1], x[2], ..., x[n]  discontinuity  points of  F  from interval [a, b],
> p is list:  p=[a, x[1], x[2], ..., x[n], b]. Then
> integrate(f,x,a,b)=sum(F(p[k+1]-0)-F(p[k]+0),k,1,n+1)
>
>  Example.  integate(cos(3*x)/(5-4*cos(x)),x, 0, 10*%pi)
> (%i1) f:cos(3*x)/(5-4*cos(x))\$
> (%i2) F:integrate(f,x)\$
>  discontinuity  points of  F is    %pi, 3*%pi, 5*%pi, 7*%pi, 9*%pi
> (%i3) p:[0,%pi,3*%pi,5*%pi,7*%pi,9*%pi,10*%pi];
> (%o3) [0,%pi,3*%pi,5*%pi,7*%pi,9*%pi,10*%pi]
> (%i4)
> 'integrate(f,x,0,10*%pi)=sum(limit(F,x,p[k+1],minus)-limit(F,x,p[k],plus),k,1,6);
> (%o4) integrate(cos(3*x)/(5-4*cos(x)),x,0,10*%pi)=(5*%pi)/12
>
>  Wrong:
> (%i5) integrate(f,x,0,10*%pi);
> (%o5) 0
>
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