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I. Python Object Browser.
II. Python to R Communicator.
III. Manipulation of piecewise polynomial functions.
1. History of changes (poly).
2. Installation of the poly module.
3. Introduction into the poly module.
4. Calculus behind the poly module.
IV. Building C++ projects.
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Calculus behind the poly module.


o understand the code of the functions S(),T(), review the formulas ( Property of scale and transport 1 )-( Property of scale and transport 7 ).

The following propositions are reflected in the classes Poly and PiecewisePoly.

Proposition

(Transport of polynomial) Suppose the polynomial $p\left( x\right) $ is given by the sequence MATH : MATH then $T_{k}^{\ast}p$ is given by MATH , MATH

Proof

Recall that MATH According to the definition ( Scale and transport operators 2 ), MATH MATH

Proposition

(Scale of polynomial) Suppose the polynomial $p\left( x\right) $ is given by the sequence MATH : MATH then $S_{d}^{\ast}p$ is given by MATH , MATH

Proof

According to the definition ( Scale and transport operators 2 ), MATH

Proposition

(Multiplication of polynomials) Suppose the polynomials $a\left( x\right) $ and $b\left( x\right) $ are given by the sequences MATH and MATH MATH then the product MATH is given by the sequence MATH MATH where MATH

Proof

We calculate MATH Make the change MATH , $p=i+j$ : MATH We now change the order of summation, see the picture ( Order of summation 0 ).
Order of summation 0
Order of summation 0

MATH

Proposition

(Integration of polynomials) Suppose the polynomial $c\left( x\right) $ is given by the sequence MATH : MATH then MATH

Proposition

(Convolution of polynomials 1) Let MATH for some polynomials $a,b$ and intervals $A,B$ . Then MATH MATH where MATH and we use the convention MATH

Proof

We calculate MATH MATH MATH There are several mutually exclusive cases, see the picture ( Piecewise convolution ).


Piecewise convolution figure
Piecewise convolution.

Let MATH then the integral $\left( \#\right) $ might have a non empty interval of integration in the following mutually exclusive cases: MATH MATH MATH MATH We transform the above inequalities: MATH MATH MATH MATH Therefore the integral $\left( \#\right) $ takes the form MATH

Proposition

(Convolution of polynomials 2) Let MATH then MATH

Proof

We calculate MATH We change the order of summation, see the figure ( Order of summation 1 ).


Order of summation 1
Order of summation 1

MATH MATH MATH We introduce the notation MATH then MATH We permute the order of $k$ and $q$ -summations, see the figure ( Order of summation 2 ).


Order of summation 2
Order of summation 2

MATH We permute the order of $q$ and $p$ -summations, see the figure ( Order of summation 3 ).


Order of summation 3
Order of summation 3

MATH MATH We make the change MATH , $r=m+q$ . MATH We permute the order of $r$ and $m$ -summations, see the figure ( Order of summation 4 ).


Order of summation 4
Order of summation 4

MATH





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