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obtaining all numerical roots of a function in an interval

asked 2012-04-14 18:17:01 -0500

Nil gravatar image

Hello, thanks for reading.

I'm working on single variable calculus here: Basically what I need is what "find_root" does, but I need a list of ALL roots in a given interval, not just one.

So I've been playing with "solve". I found this piece of code which works in most cases:

sage: roots = solve(f(x),x,solution_dict=True)

sage: roots = [s[x] for s in roots]

sage: num_roots = map(n, roots)

but it gives an error if the function is periodic and has inifinite roots, becuase the symbolic expression that "solve" gets has infinite solutions too.

Defining a desired interval should solve this issue, but I have no idea how to implement such thing!

Thanks you and have a good day.

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answered 2012-04-17 22:41:35 -0500

ndomes gravatar image

updated 2012-05-02 11:48:15 -0500

def find_root_recursive(func,a,b,tol=0.000000000001):
    L = []
        x0 = find_root(func,a,b)
        L += find_root_recursive(func,a,x0-tol,tol)       
        L += find_root_recursive(func,x0+tol,b,tol)       
    return L

L = find_root_recursive(sin,-50,50)
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answered 2012-04-14 18:58:13 -0500

Shashank gravatar image

If you run the following code you get [-0.973248989467730, 0.973248989467730, -0.229752920547361, 0.229752920547361], but if you uncomment assume(x>0) you get [0.973248989467730, 0.229752920547361].

f(x) = -x*x+x**4+0.05
roots = solve(f(x),x,solution_dict=True)
roots = [s[x] for s in roots]
num_roots = map(n, roots)
print num_roots
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Thanks for the reply, but this does not fix the problem with periodic infinte roots. I run into the same error: Traceback (most recent call last): MT = maxim(f, a, b) File "", line 1, in <module> File "/tmp/tmpGBdQkB/", line 8, in <module> MT = maxim(f, a, b) File "/tmp/tmpbtKw0f/", line 20, in maxim roots = [s[x] for s in roots]; KeyError: x

Nil gravatar imageNil ( 2012-04-14 19:33:40 -0500 )edit

Can you tell me what is the f you are using?

Shashank gravatar imageShashank ( 2012-04-15 12:58:41 -0500 )edit

yes. it is: f(x) = 8 x^3 sin(x^2)-12 x cos(x^2) , or simply (cos(x^2))'''. But I believe the problem happens with any function with inifinite roots.

Nil gravatar imageNil ( 2012-04-16 09:21:31 -0500 )edit

For cos(x*x) I get two roots [-1/2*sqrt(pi)*sqrt(2), 1/2*sqrt(pi)*sqrt(2)], but I don't get any error.

Shashank gravatar imageShashank ( 2012-04-16 10:16:14 -0500 )edit

answered 2019-11-19 13:53:29 -0500

Emmanuel Charpentier gravatar image

Formally, it can't be done. For example, $\sin\displaystyle\frac{1}{x}$ has an infinity of roots between -1 and 1...

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Asked: 2012-04-14 18:17:01 -0500

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Last updated: Nov 19 '19