On 6/26/2012 10:34 PM, Aleksas Domarkas wrote:
> How to find the points of discontinuity af a function on interval?
You could help if you can describe how you would do this algorithmically
without the computer.
Then someone could write a program to do it.
There are some special cases that can be detected; this is done by the
definite integration
program, for example. Apparently not your example though.
RJF
>
> 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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