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DEtools[remove_RootOf] - rewrite a given expression removing RootOf.
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Calling Sequence
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remove_RootOf(expr, R)
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Parameters
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expr
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any Maple algebraic expression or equation
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R
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(optional) identifies which specific RootOf is to be removed
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Description
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The remove_RootOf command receives an algebraic expression containing a RootOf and removes it by rewriting the expression. In general, this command is used to directly and naturally convey the mathematical meaning of algebraic expressions that contain RootOfs.
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Note that remove_RootOf is different from allvalues; remove_RootOf attempts to remove the RootOf -- typically resulting in implicit algebraic expressions -- while allvalues attempts to return all the values implied by the RootOf.
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remove_RootOf is of particular value for understanding or further manipulating ODE solutions, and especially when allvalues fails in evaluating the RootOf.
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When the given expression has no RootOf inside it, remove_RootOf returns it as given; when the expression has many RootOfs, only one RootOf (the one with greatest length) is removed. It is also possible to direct remove_RootOf to remove a specific RootOf, say R, by passing R to remove_RootOf as its second argument.
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When the removal of a RootOf itself requires the introduction of another RootOf, the removal is not performed and the expression received is returned as given. Also remove_RootOf only removes RootOfs of one argument (so RootOfs with indices or labels will remain in the output).
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This function is part of the DEtools package, and so it can be used in the form remove_RootOf(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[remove_RootOf](..).
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Examples
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Here is an example that shows the difference between remove_RootOf and allvalues.
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is rewritten by remove_RootOf as:
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](/support/helpjp/helpview.aspx?si=8673/file01371/math172.png)
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An example with nested RootOfs
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If we now evaluate the integral, the resulting solution is
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One RootOf can be removed in the expression above (and the solution can be further evaluated using allvalues).
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Here, the sum over the Roots of the third degree polynomial is equal to zero.
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A second order ODE example in which three solutions with RootOf are returned by dsolve
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![[dsolve(diff(y(x), `$`(x, 2))-h(y(x)/x^(1/2))/x^(3/2) = 0)]](/support/helpjp/helpview.aspx?si=8673/file01371/math247.png)
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![[y(x) = RootOf(_Z*x^(3/2)+4*h(_Z/x^(1/2))*x^2), y(x) = RootOf(-ln(x)+2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. _Z))+2*_C2)*x^(1/2), y(x) = RootOf(-ln(x)-2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. _Z))+2*_C2)*x^(1/2)]](/support/helpjp/helpview.aspx?si=8673/file01371/math250.png)
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![map(remove_RootOf, [y(x) = RootOf(_Z*x^(3/2)+4*h(_Z/x^(1/2))*x^2), y(x) = RootOf(-ln(x)+2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. _Z))+2*_C2)*x^(1/2), y(x) = RootOf(-ln(x)-2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. _Z))+2*_C2)*x^(1/2)])](/support/helpjp/helpview.aspx?si=8673/file01371/math254.png)
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![[y(x)*x^(3/2)+4*h(y(x)/x^(1/2))*x^2 = 0, -ln(x)+2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. y(x)/x^(1/2)))+2*_C2 = 0, -ln(x)-2*(Int(1/(_g^2+8*(Int(h(_g), _g))+_C1)^(1/2), _g = `` .. y(x)/x^(1/2)))+2*_C2 = 0]](/support/helpjp/helpview.aspx?si=8673/file01371/math257.png)
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Although remove_RootOf is mainly useful for manipulating nonlinear ODE solutions, it can also be of help with solutions to linear ODEs, for instance when allvalues fails in evaluating the RootOfs.
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