find_root return values

edit

def find_all_roots(func,a,b,nsteps):    
    roots=[]                    
    n=0
    x0=a                #initializations
    step=(b-a)/(nsteps)     #size of a sub-interval
                    #the rest is quite self-explanatory
    while (n<nsteps):
        try:
            roots.append(find_root(func,x0,x0+step))
            root_exist = True
        except RuntimeError:       
            root_exist = False
        x0+=step
        n+=1
    roots.sort()
    return roots

I just wanted to write a piece of code to find ALL roots in given interval by dividing it into small subintervals. I found two difficulties: 1. pay attention if your function makes sense on boarders of the interval you plan to scan (solution -- just pay attention to it); 2. handling the situation when there is no root (solved by the comment of tmonteil, thanks!). The code is primitive and surely not optimal, but it does the job. Find it attached. Good luck! Archie

def find_all_roots(func,a,b,nsteps):
roots=[]
n=0 x0=a #initializations step=(b-a)/(nsteps) #size of a sub-interval #the rest is quite self-explanatory while (n<nsteps): try:="" roots.append(find_root(func,x0,x0+step))="" root_exist="True" except="" runtimeerror:="" <br=""> root_exist = False x0+=step n+=1 roots.sort() return roots