[Numpy-discussion] polynomial fromroots
Charles R Harris
charlesr.harris@gmail....
Sat Oct 9 22:09:24 CDT 2010
On Sat, Oct 9, 2010 at 9:01 PM, Charles R Harris
<charlesr.harris@gmail.com>wrote:
>
>
> On Sat, Oct 9, 2010 at 8:36 PM, <josef.pktd@gmail.com> wrote:
>
>> On Sat, Oct 9, 2010 at 10:01 PM, Charles R Harris
>> <charlesr.harris@gmail.com> wrote:
>> >
>> >
>> > On Sat, Oct 9, 2010 at 7:47 PM, <josef.pktd@gmail.com> wrote:
>> >>
>> >> I'm trying to see whether I can do this without reading the full
>> manual.
>> >>
>> >> Is it intended that fromroots normalizes the highest order term
>> >> instead of the lowest?
>> >>
>> >>
>> >> >>> import numpy.polynomial as poly
>> >>
>> >> >>> p = poly.Polynomial([1, -1.88494037, 0.0178126 ])
>> >> >>> p
>> >> Polynomial([ 1. , -1.88494037, 0.0178126 ], [-1., 1.])
>> >> >>> pr = p.roots()
>> >> >>> pr
>> >> array([ 0.53320748, 105.28741219])
>> >> >>> poly.Polynomial.fromroots(pr)
>> >> Polynomial([ 56.14003571, -105.82061967, 1. ], [-1., 1.])
>> >> >>>
>> >>
>> >> renormalizing
>> >>
>> >> >>> p2 = poly.Polynomial.fromroots(pr)
>> >> >>> p2/p2.coef[0]
>> >> Polynomial([ 1. , -1.88494037, 0.0178126 ], [-1., 1.])
>> >>
>> >>
>> >> this is, I think what I want to do, invert roots that are
>> >> inside/outside the unit circle (whatever that means
>> >>
>> >> >>> pr[np.abs(pr)<1] = 1./pr[np.abs(pr)<1]
>> >> >>> p3 = poly.Polynomial.fromroots(pr)
>> >> >>> p3/p3.coef[0]
>> >> Polynomial([ 1. , -0.54270529, 0.0050643 ], [-1., 1.])
>> >>
>> >
>> > Wrong function ;) You defined the polynomial by its coefficients. What
>> you
>> > want to do is
>>
>> My coefficients are from a lag-polynomial in time series analysis
>> (ARMA), and they really are the (estimated) coefficients. It is
>> essentially the same as the model for scipy.signal.lfilter.
>> I just need to check the roots to see whether the process is
>> stationary and invertible. If one of the two lag-polynomials (moving
>> average) has roots on the wrong side of the unit circle, then I can
>> invert them.
>>
>> I'm coding from memory of how this is supposed to work, so maybe I'm
>> back to RTFM and RTFTB (TB=text book).
>>
>> (I think what I really would need is a z-transform, but I don't have
>> much of an idea how to do this on a computer)
>>
>> Thanks, the main thing I need to do is check the convention or
>> definition for the normalization. And as btw, I like that the coef are
>> in increasing order
>> e.g. seasonal differencing multiplied with 1 lag autoregressive
>> poly.Polynomial([1.,0,0,-1])*poly.Polynomial([1,0.8])
>>
>>
> The polynomial is obtained from the product of the terms (x - r_i), so it
> is monic. In a different basis it may not appear that way:
>
> In [23]: p = poly.Legendre.fromroots([1, -1.88494037, 0.0178126 ])
>
> In [24]: p
> Out[24]: Legendre([ 0.32261828, -1.30070346, 0.57808518, 0.4 ],
> [-1., 1.])
>
> The z transform of a sequence is simply the polynomial with 1/z in place of
> x, so you just need to invert all the roots.
>
More precisely, a unit lag is 1/z. You can turn it into the Fourier
transform by substituting exp(j*2*pi*f) for z.
Chuck
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