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August 10th, 2010, 09:00 AM   #1
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Equilibrium

I have got a continuous and n-times differentiable function f(s) defined on the interval (0,1/2) in which f(1/2)=0. f(0) and f’(1/2), which depend on the parameters’ sign, might be either sign, whilst f’’(1/2)=0.
Questions:
-Is there any way to know if the function displays other zeros in the specified interval?
-If yes, is there any way to know how many those zeros are?
Thanks to everybody.
Best Regards,
Francesco
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August 10th, 2010, 08:36 PM   #2
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Re: Equilibrium

Let f(s)=cos(?ns), with n odd. Then the number of zeros depends on n - you can have as many as you want.
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August 11th, 2010, 04:47 AM   #3
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Re: Equilibrium

Dear Mathman,
Thanks a lot for your answer. Actually I am not looking for a class of function with many zeros, on the contrary I would like to have a sufficient and necessary condition such that there are no many zeros, or at most 1.
Thanks
Francesco
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August 11th, 2010, 01:26 PM   #4
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Re: Equilibrium

As you can see from the example I gave, you need more conditions than can be prescribed on the end points alone.
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September 27th, 2010, 02:49 PM   #5
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Re: Equilibrium

Use the Nth derivative to compute the error term for a Taylor series of order (N-1?). If the magnitude of this error term over the sub-interval is smaller then your distance from zero at the point which you computed the Taylor series in this sub-interval then you know their can't be a root in that sub-interval. If you wish to consider the complex plane then use a two dimensional Taylor series (one for the real and imaginary part and rule out the possibility of roots in given sub neighborhoods by bounding the magnate of the Nth derivative over that region.

To search quickly for places near where roots might be their is an algorithm MATLAB uses which finds all roots of a polynomial by computing the eigenvalues of a characteristic matrix.

Edit: Following considerations for complex roots:

Also note that we can compute the magnitude of a polynomial by multiplying it by the complex conjugate. The complex conjugate part is given by replacing z with x-iy and expanding. The non conjugate part can be computed by replace z with x+iy. The product of these the product of two resulting 2D polynomial which gives the square of the magnitude.

Subtracting the magnitude by the largest value of the error term in this region will give points where their may be enough error for their to be a root at or beyond this point. However, this gives a point and not a region. To construct a region consider the subset of two domains. The first domain is the smallest enclosing polynomial curve where the imaginary part is equal to zero. The second region is the smallest region where the real part is equal to zero. The roots are vertexes where these two curves cross. The region can be drawn by drawing piece wise polynomial curves between these vertexes.
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