symbolic | R Documentation |
Returns a character string representing \mjseqnk-tensor and
\mjseqnk-form objects in symbolic form. Used by the print method if
either option kform_symbolic_print
or
ktensor_symbolic_print
is non-null.
as.symbolic(M,symbols=letters,d="")
M |
Object of class |
symbols |
A character vector giving the names of the symbols |
d |
String specifying the appearance of the differential operator |
Spivak (p89), in archetypically terse writing, states:
A function \mjseqnf is considered to be a 0-form and \mjeqnf\cdot\omegaomitted is also written \mjeqnf\wedge\omegaomitted. If \mjeqnf\colon\mathcalR^n\longrightarrow\mathcalRf: R^n -> R is differentiable, then \mjeqnDf(p)\in\Lambda^1\left(\mathcalR^n\right)omitted; see PDF. By a minor modification we therefore obtain a 1-form \mjseqndf, defined by
\mjdeqndf(p)\left(v_p\right)=Df(p)(v)df(p)(v_p)=Df(p)(v).
Let us consider in particular the 1-forms \mjeqnd\pi^iomitted; see PDF. It is customary to let \mjseqnx^i denote the function \mjeqn\pi^iomitted; see PDF (On \mjeqn\mathcalR^3R^3 we often denote \mjseqnx^1, \mjseqnx^2, and \mjseqnx^3 by \mjseqnx, \mjseqny, and \mjseqnz). This standard notation has obvious disadvantages but it allows many classical results to be expressed by formulas of equally classical appearance. Since \mjeqndx^i(p)(v_p)=d\pi^i(p)(v_p)=D\pi^i(p)(v)=v^i(omitted; see PDF), we see that \mjeqndx^1(p),...,dx^n(p)dx^1(p),...,dx^n(p) is just the dual basis to \mjeqn(e_1)_p,...,(e_n)_p(e_1)_n,...,(e_n)_p. Thus every k-form \mjeqn\omegaomitted can be written
\mjdeqn \omega=\sum_i_1 < \cdots < i_k\omega_i_1,...,i_k dx^i_1\wedge\cdots\wedge dx^i_k.omitted.
Function as.symbolic()
uses this format. For completeness, we
add (p77) that k-tensors may be expressed in the form
_i_1,..., i_k=1^n a_i_1,...,i_k\cdot \phi_i_1\otimes\cdots\otimes\phi_i_k.omitted.
and this form is used for k-tensors.
Returns a “noquote” character string.
Robin K. S. Hankin
print.stokes
,dx
(o <- kform_general(3,2,1:3))
as.symbolic(o,d="d",symbols=letters[23:26])
(a <- rform(n=50))
as.symbolic(a,symbols=state.abb)
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