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First Order Derivatives in the Forward Mode of Algorithmic Differentiation

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Univariate Taylor Series ExpansionsΒΆ

As an easy example we want to compute the Taylor series expansion of

\[y = f(x) = \sin(\cos(x) + \sin(x))\]

about \(x_0 = 0.3\). The first thing to notice is that we can as well compute the Taylor series expansion of

\[y = f(x_0 + t) = \sin(\cos(x_0 + t) + \sin(x_0 + t))\]

about \(t = 0\). Taylor’s theorem yields

\[\begin{split}f(x_0 + t) &= \sum_{d=0}^{D-1} y_d t^d + R_{D}(t) \\ \mbox{where } \quad y_d &= \left. \frac{1}{d!} \frac{d^d }{d t^d}f(x_0 + t) \right|_{t = 0} \;.\end{split}\]

and \(R_D(x)\) is the remainder term.

Slightly rewritten one has

\[y(t) = f(x(t)) + \mathcal O(t^D)\]

i.e., one has a polynomial \(x(t) = \sum_{d=0}^{D-1} x_d t^d\) as input and computes a polynomial \(y(t) = \sum_{d=0}^{D-1} y_d t^d + \mathcal O(t^d)\) as output.

This is now formulated in a way that can be used with ALGOPY.

import numpy; from numpy import sin,cos
from algopy import UTPM

def f(x):
    return sin(cos(x) + sin(x))

D = 100; P = 1
x = UTPM(numpy.zeros((D,P)))[0,0] = 0.3[1,0] = 1

y = f(x)
print('coefficients of y =',[:,0])

Don’t be confused by the P. It can be used to evaluate several Taylor series expansions at once. The important point to notice is that the D in the code is the same D as in the formula above. I.e., it is the number of coefficients in the polynomials. The important point is


The coefficients of the univariate Taylor polynomial (UTP) are stored in the attribute It is a x.ndarray with shape (D,P) + shape of the coefficient. In this example, the coefficients \(x_d\) are scalars and thus = (D,P). However, if the the coefficients were vectors of size N, then would be (D,P,N), and if the coefficients were matrices with shape (M,N), then would be (D,P,M,N).

To see that ALGOPY indeed computes the correct Taylor series expansion we plot the original function and the Taylor polynomials evaluated at different orders.

import matplotlib.pyplot as pyplot; import os
zs = numpy.linspace(-1,2,100)
ts = zs -0.3
fzs = f(zs)

for d in [2,4,10,50,100]:
    yzs = numpy.polyval([:d,0][::-1],ts)
    pyplot.plot(zs,yzs, label='%d\'th order approx.'%(d-1))
pyplot.plot([0.3], f([0.3]), 'ro')
pyplot.plot(zs,fzs, 'k.')

This plot shows Taylor approximations of different orders. The point \(x_0 = 0.3\) is plotted as a red dot and the original function is plotted as black dots. One can see that the higher the order, the better the approximation.