Concatenation of symbolic vectors.

vectors

Cyrille
1409 ●38 ●82 ●144

This

NB_0=vector(var('x', n=3, latex_name='x'))
B_0=vector(var('y', n=4, latex_name='ϵ'))
show(LatexExpr(r"NB_0 =" + latex(NB_0)),LatexExpr(r"\, \text{ et }\, B_0 =" + latex(B_0)))

gives me two vectors.

But I need to concatenate them -- i.e. to create a vector $V1= [x_1, x_2, x_3, x_4, \epsilon_1, \epsilon_2, \epsilon_3, b]$ (I need the $b$ ) and $V2= [\epsilon_1, \epsilon_2, \epsilon_3, z]$ (I need the $z$ ). And when I call them in a table I want to create I need the Latex names not the sagemath one. How could I do ?

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answered 4 years ago

dan_fulea
5650 ●5 ●44 ●97

It is hard to understand what and why should be done, but maybe the following will do the job. The presented solution is just blind plain work to avoid follow up questions. (It is hard to understand why vectors should be used, lists are better suited. And lists can be easily concatenated. But ok... )

x1 = var('x1', latex_name='x_1')                                                                                                                        
x2 = var('x2', latex_name='x_2')                                                                                                                        
x3 = var('x3', latex_name='x_3')                                                                                                                        
x4 = var('x4', latex_name='x_4')                                                                                                                        

e1 = var('e1', latex_name=r'\epsilon_1')                                                                                                                
e2 = var('e2', latex_name=r'\epsilon_2')                                                                                                                
e3 = var('e3', latex_name=r'\epsilon_3')                                                                                                                

b  = var('b' , latex_name='b')                                                                                                                          

v = [x1, x2, x3, x4, e1, e2, e3, b]

Then in the sage interpreter:

sage: vector(v)                                                                                                                                               
(x1, x2, x3, x4, e1, e2, e3, b)
sage: show(vector(v))                                                                                                                                         
\newcommand{\Bold}[1]{\mathbf{#1}}\left({x_1},\,{x_2},\,{x_3},\,{x_4},\,{\epsilon_1},\,{\epsilon_2},\,{\epsilon_3},\,{b}\right)
sage: latex(v)                                                                                                                                                
\left[{x_1}, {x_2}, {x_3}, {x_4}, {\epsilon_1}, {\epsilon_2}, {\epsilon_3}, {b}\right]

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answered 4 years ago

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

Not sure I understand the question. Maybe this helps a little.

Names are slightly changed with respect to the question.

Define two vectors and two symbolic variables:

sage: A = vector(var('x', n=3, latex_name='x'))
sage: B = vector(var('y', n=4, latex_name=r'\epsilon'))
sage: b = SR.var('b', latex_name='b')
sage: z = SR.var('z', latex_name='z')

Produce two new vectors combining the above:

sage: U = vector(SR, len(A) + len(B) + 1, A.list() + B.list() + [b])
sage: U
(x0, x1, x2, y0, y1, y2, y3, b)

sage: V = vector(SR, len(B) + 1, B.list() + [z])
sage: V
(y0, y1, y2, y3, z)

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Comments

It' was nearly what I was expecting less we lost the $\epsilon$

Cyrille ( 4 years ago )

we can see $\epsilon$ with show()

A = vector(var('x', n=3, latex_name='x'))
B = vector(var('y', n=4, latex_name=r'\epsilon'))
b = SR.var('b', latex_name='b')
z = SR.var('z', latex_name='z')
U = vector(SR, len(A) + len(B) + 1, A.list() + B.list() + [b])
V = vector(SR, len(B) + 1, B.list() + [z])
show('U : ',U)
show('V : ',V)

ortollj ( 4 years ago )

Thanks for your kind answer but I have a problem I will explain it in another question. It will be easier.

Cyrille ( 4 years ago )

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