[Numpy-discussion] [ANN]: Taylorpoly, an implementation of vectorized Taylor polynomial operations and request for opinions

Sebastian Walter sebastian.walter@gmail....
Sat Feb 27 16:39:03 CST 2010


On Sat, Feb 27, 2010 at 10:02 PM, Friedrich Romstedt
<friedrichromstedt@gmail.com> wrote:
> To the core developers (of numpy.polynomial e.g.): Skip the mess and
> read the last paragraph.
>
> The other things I will post back to the list, where they belong to.
> I just didn't want to have off-topic discussion there.
>
>> I wanted to stress that one can do arithmetic on Taylor polynomials in
>> a very similar was as with complex numbers.
>
> I do not understand completely.  What is the analogy with complex
> numbers which one cannot draw to real numbers?  Or, more precise: What
> /is/ actually the analogy besides that there are operations?  With i,
> i * i = -1, thus one has not to discard terms, contrary to the
> polynomial product as you defined it, no?

I'm sorry this comment turns out to be confusing.
It has apparently quite the contrary effect of what I wanted to achieve:
Since there is already a polynomial module  in numpy I wanted to
highlight their difference
and stress that they are used to do arithmetic, e.g. compute

f([x],[y]) = [x] * (sin([x])**2 + [y])

in Taylor arithmetic.

>
>> I guess there are also situations when you have polynomials   z =
>> \sum_{d=0}^{D-1} z_d t^d, where z_d are complex numbers.
>
>>> I like it more to implement operators as overloads of the __mul__
>>
>> I thought about something like
>> def __mul__(self, other):
>>    return mul(self,other)
>
> Yes, I know!  And in fact, it may symmetrise the thing a bit.
>
>>> etc., but this is a matter of taste.
>>>  In fact, you /have/ to provide
>>> external binary operators, because I guess you also want to have
>>> numpy.ndarrays as left operand.  In that case, the undarray will have
>>> higher precedence, and will treat your data structure as a scalar,
>>> applying it to all the undarray's elements.
>>
>> well, actually it should treat it as a scalar since the Taylor
>> polynomial is something like a real or complex number.
>
> Maybe I misunderstood you completely, but didn't you want to implement
> arrays of polynomials using numpy?  So I guess you want to apply a
> vector from numpy pairwise to the polynomials in the P-object?

no, the vectorization is something different. It's purpose becomes
only clear when applied in Algorithmic Differentiation.
 E.g. if you have a function
f: R^N -> R
x -> y=f(x)
where x = [x1,...,xN]

and you want to compute the gradient g(x) of f(x), then you can compute
df(x)/dxn by propagating  the following array of Taylor polynomials:

x = numpy.array( UTPS([x1_0, 0]), ..., UTPS([xn_0, 1]), ...,
UTPS([xN_0,0]), dtype=object)
y = f(x)

if you want to have the complete gradient, you will have to repeat N
times. Each time for the same zero'th coefficients [x1,...,xN].

Using the vecorized version, you would do only one propagation
x = numpy.array( UTPS([x1_0, 1,0,...,0]), ..., UTPS([xn_0,
0,...,1,...0]), ..., UTPS([xN_0,0,....,1]), dtype=object)
y = f(x)

i.e. you save the overhead of calling the same function N times.

>
>> [z]_{D+1} &=_{D+1}& [x]_{D+1} y_{D+1} \\
>> \sum_{d=0}^D z_d t^d & =_{D+1} & \left( \sum_{d=0}^D x_d t^d \right)
>> \left( \sum_{c=0}^D y_c t^c \right)
>
> Did your forget the [] around the y in line 1, or is this intentional?
>  Actually, I had to compile the things before I could imagine what you
> could mean.
Yes, I'm sorry, this is a typo.
>
> Why don't you use multidimensional arrays?  Has it reasons in the C
> accessibility?  Now, as I see it, you implement your strides manually.
>  With a multidimensional array, you could even create arrays of shape
> (10, 12) of D-polynomials by storing an ndarray of shape (10, 12, D)
> or (D, 10, 12).

the goal is to have several Taylor polynomials evaluated in the same
base point, e.g.
[x_0, x_{11}, x_{21}, x_{31}]
[x_0, x_{12}, x_{22}, x_{32}]
[x_0, x_{13}, x_{23}, x_{33}]

i.e. P=3, D=3
One could use an (P,D) array. However, one would do unnecessary
computations since x_0 is the same for all P polynomials.
I.e. one implements the data structure as
[x]_{D,P} := [x_0, x_{1,1},...,x_{1,D-1},x_{2,1},...,x_{P,D-1}].

This results in a non-const stride access.

>
> Just because of curiosity:  Why do you set X_{0, 1} = ... X_{0, P} ?
>
> Furthermore, I believe there is some elegant way formulating the
> product for a single polynomial.  Let me think a bit ...
>
> For one entry of [z]_E, you want to sum up all pairs:
>    x_{0} y{E} + ... + x{D} y{E - D} ,     (1)
> right?  In a matrix, containing all mutual products x_{i} y{j}, this
> are diagonals.  Rotating the matrix by 45° counterclockwise, they are
> sums in columns.  Hmmm... how to rotate a matrix by 45°?
>
> Another fresh look: (1) looks pretty much like the discrete
> convolution of x_{i} and y_{i} at argument E.  Going into Fourier
> space, this becomes a product.  x_i and y_i have finite support,
> because one sets x_{i} and y_{i} = 0 outside 0 <= i <= D.  The support
> of (1) in the running index is at most [0, D].  The support in E is at
> most [0, 2 D].  Thus you don't make mistakes by using DFT, when you
> pad x and y by D zeros on the right.  My strategy is: Push the padded
> versions of x and y into Fourier space, calculate their product,
> transform back, cut away the D last entries, and you're done!
>
> I guess you mainly want to parallelise the calculation instead of
> improving the singleton calculation itself?  So then, even non-FFT
> would incorporate 2 D explicit python sums for Fourier transform, and
> again 2 D sums for transforming back, is this an improvement over the
> method you are using currently?  And you could code it in pure Python
> :-)
>
> I will investigate this tonight.  I'm curious.  Irrespective of that I
> have also other fish to fry ... ;-)
>
> iirc, in fact DFT is used for efficient polynomial multiplication.
> Maybe Chuck or another core developer of numpy can tell whether
> numpy.polynomial does it actually that way?

I believe the degree D is typically much to small (i.e. D <= 4) to
justify the additional overhead of using FFT,
though there may be cases when really high order polynomials are used.

>
> Friedrich
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