Taking elements in solve and multiplying to a polynomial

whatupmatt
175 ●8 ●11 ●20

slelievre
17789 ●22 ●163 ●352 http://carva.org/samue...

So I want a function that solves a system of two equations and the move those numerical results to polynomials f and g.

def PolynomialCoefficient(fcn1, fcn2):
    var ('a b')
    eqn1 = 2*a + 2*b == 0
    eqn2 = 3*a - b == 1
    List = solve([eqn1, eqn2], a, b)

Now I want to move this to a*f(x)+b*g(x) without hitting solve each time and manually typing what a and b are. For example, PolynomialCoefficient(x, y) should output 1/4 x - 1/4 y.

The issue is len(List) outputs 1. If it was length 2 with a in first slot and b in 2nd slot, I could do

    Poly = List[0]*fcn1 + List[1]*fcn2
    return Poly

Any ideas on how to move a and b?

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answered 4 years ago

slelievre
17789 ●22 ●163 ●352 http://carva.org/samue...

updated 4 years ago

The solve function has an optional argument solution_dict.

Using it, solutions are returned as dictionaries, which makes it easier to access the solved values of each unknown.

Here is a way to use that in the function from the question.

def polynomial_coefficient(fcn1, fcn2):
    a, b = SR.var('a, b')
    eqn1 = 2*a + 2*b == 0
    eqn2 = 3*a - b == 1
    sol = solve([eqn1, eqn2], a, b, solution_dict=True)[0]
    return sol[a]*fcn1 + sol[b]*fcn2

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Great thanks

whatupmatt ( 4 years ago )

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answered 4 years ago

Max Alekseyev

6932 ●9 ●52 ●162

updated 4 years ago

Your List is a list of solutions with respect to a,b. If you want just one solution you can re-define it as

List=solve([eqn1,eqn2],a,b)[0]

and then proceed as in your code; or if you want to get Poly as a list of polynomials (one for each solution), you can define it as

Poly=[ List[0]*fcn1+List[1]*fcn2 for List in solve([eqn1,eqn2],a,b) ]

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Thanks for your help.

whatupmatt ( 4 years ago )

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