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find_root return values

asked 2013-07-03 01:59:34 -0500

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Hi there!

What are return values of find_root in the case when there is no root in the interval?

Thanks!

Archie

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answered 2013-07-03 02:17:57 -0500

tmonteil gravatar image

When there is no root in the given interval, the function find_root() does not return anything but raises a RuntimeError exception:

sage: f(x) = 2 ; f
x |--> 2
sage: find_root(f,-10,10)
...
RuntimeError: f appears to have no zero on the interval

If you want to deal with that, you have to catch the error (otherwise your program will stop here):

try:
    root = find_root(f,-10,10)
    root_exist = True
except RuntimeError:
    root_exist = False

You can read more here.

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Great answer! Exactly what I wanted, thank you very much! I will post my naive piece of code for finding all roots in given interval. Just in case anyone likes it. Best...

Archie gravatar imageArchie ( 2013-07-03 02:22:36 -0500 )edit
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answered 2013-07-03 07:31:47 -0500

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answered 2013-07-03 03:38:03 -0500

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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
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Comments

Just in case you are interested; there is a root function which does this already with a lot higher precision and more options. def find_all_roots(func,a,b): assume( x>=a , x<=b) try: roots = [cp for cp in (func==0).roots(x, ring=RR, multiplicities=False)] except: roots = [ ] return roots find_all_roots((x+x^2-10), -10,10) returns: [-3.70156211871642, 2.70156211871642] find_all_roots((x^2+1), -10, 10) returns: [ ] (sry about the bad spacing, not sure how to format the text here yet)

Pavel Yartsev gravatar imagePavel Yartsev ( 2013-07-27 15:44:49 -0500 )edit
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answered 2013-07-03 02:17:39 -0500

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I think it raises an error, but surely you can check this very quickly by trying to find a root of x^2 + 1 .

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answered 2013-07-03 03:36:25 -0500

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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

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Asked: 2013-07-03 01:59:34 -0500

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Last updated: Jul 03 '13