MathAction SandBoxBugLaurentSeries

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    - SandBoxBugLaurentSeries <-- You are here.

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(1) -> Z==>Integer; Q==>Fraction Z

Type: Void

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z: Symbol := 'z; P==>UnivariatePolynomial(z,Q)

Type: Void

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t:P := monomial(1,1)

$\label{eq1}z$

(1)

Type: UnivariatePolynomial(z,Fraction(Integer))

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p:P := (1-t)*(1-t^2)*(1-t^3)

$\label{eq2}-{{z}^{6}}+{{z}^{5}}+{{z}^{4}}-{{z}^{2}}- z + 1$

(2)

Type: UnivariatePolynomial(z,Fraction(Integer))

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L==>UnivariateLaurentSeries(Q,z,0)

Type: Void

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R ==> Record(k: Z, c: Q)

Type: Void

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l: List R := reverse [[degree m, leadingCoefficient m]$R for m in monomials p]

$\label{eq3}\begin{array}{@{}l} \displaystyle \left[{\left[{k = 0}, \:{c = 1}\right]}, \:{\left[{k = 1}, \:{c = - 1}\right]}, \:{\left[{k = 2}, \:{c = - 1}\right]}, \: \right. \ \ \displaystyle \left.{\left[{k = 4}, \:{c = 1}\right]}, \:{\left[{k = 5}, \:{c = 1}\right]}, \:{\left[{k = 6}, \:{c = - 1}\right]}\right]$

(3)

Type: List(Record(k: Integer,c: Fraction(Integer)))

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series(l)$L

$\label{eq4}1 - z -{{z}^{2}}+{{z}^{4}}+{{z}^{5}}-{{z}^{6}}$

(4)

Type: UnivariateLaurentSeries?(Fraction(Integer),z,0)

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l1: List R := [[n, if n=5 then 0 else n/1]$R for n in 1..7]

$\label{eq5}\begin{array}{@{}l} \displaystyle \left[{\left[{k = 1}, \:{c = 1}\right]}, \:{\left[{k = 2}, \:{c = 2}\right]}, \:{\left[{k = 3}, \:{c = 3}\right]}, \:{\left[{k = 4}, \:{c = 4}\right]}, \: \right. \ \ \displaystyle \left.{\left[{k = 5}, \:{c = 0}\right]}, \:{\left[{k = 6}, \:{c = 6}\right]}, \:{\left[{k = 7}, \:{c = 7}\right]}\right]$

(5)

Type: List(Record(k: Integer,c: Fraction(Integer)))

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series(l1)$L

$\label{eq6}z +{2 \ {{z}^{2}}}+{3 \ {{z}^{3}}}+{4 \ {{z}^{4}}}+{6 \ {{z}^{6}}}+{7 \ {{z}^{7}}}$

(6)

Type: UnivariateLaurentSeries?(Fraction(Integer),z,0)

It seems weird that the resulting series is aborted at a non-existing coefficient in the input list.

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l2: List R := [[n, n/1]$R for n in 1..7|n~=5]

$\label{eq7}\begin{array}{@{}l} \displaystyle \left[{\left[{k = 1}, \:{c = 1}\right]}, \:{\left[{k = 2}, \:{c = 2}\right]}, \:{\left[{k = 3}, \:{c = 3}\right]}, \:{\left[{k = 4}, \:{c = 4}\right]}, \: \right. \ \ \displaystyle \left.{\left[{k = 6}, \:{c = 6}\right]}, \:{\left[{k = 7}, \:{c = 7}\right]}\right]$

(7)

Type: List(Record(k: Integer,c: Fraction(Integer)))

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series(l2)$L

$\label{eq8}z +{2 \ {{z}^{2}}}+{3 \ {{z}^{3}}}+{4 \ {{z}^{4}}}+{6 \ {{z}^{6}}}+{7 \ {{z}^{7}}}$

(8)

Type: UnivariateLaurentSeries?(Fraction(Integer),z,0)

There is obviously also a bug here, because the zero coefficient should have been removed. I would, however, accept such a result if the specification of series were made precise as to rely on the input stream not to contain zero coefficients.

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S ==> SparseUnivariateLaurentSeries(Q,z,0)

Type: Void

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series(l)$S

$\label{eq9}1 - z -{{z}^{2}}+{{z}^{4}}+{{z}^{5}}-{{z}^{6}}$

(9)

Type: SparseUnivariateLaurentSeries?(Fraction(Integer),z,0)

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series(l1)$S

$\label{eq10}z +{2 \ {{z}^{2}}}+{3 \ {{z}^{3}}}+{4 \ {{z}^{4}}}+{0 \ {{z}^{5}}}+{6 \ {{z}^{6}}}+{7 \ {{z}^{7}}}$

(10)

Type: SparseUnivariateLaurentSeries?(Fraction(Integer),z,0)

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series(l2)$S

$\label{eq11}z +{2 \ {{z}^{2}}}+{3 \ {{z}^{3}}}+{4 \ {{z}^{4}}}+{6 \ {{z}^{6}}}+{7 \ {{z}^{7}}}$

(11)

Type: SparseUnivariateLaurentSeries?(Fraction(Integer),z,0)

Subject: Be Bold !!

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