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So I'm working on a physics problem. I have two functions:

$\rho = \frac{F}{(\beta F^\alpha+1)^\frac{1}{\alpha}}$, $p = - \frac{F}{(\beta F^\alpha+1)^\frac{1}{\alpha}} + \frac{4}{3}\frac{F}{(\beta F^\alpha+1)^{1+\frac{1}{\alpha}}}$

Just by looking at $p$, I can tell it can be algebraically manipulated in order to write it, explicitly, in terms of $\rho$. Indeed:

$p= \frac{\rho}{3}(1-4\beta\rho^\alpha)$

Every math operation I do by hand on this project, I check it on Mathematica just to make sure I don't screw it up. Sometimes I use functions like Simplify or FullSimplify to make my life easier. For example, if I use Simplify[$\rho$], I get a nice reduced expression, and the same for Simplify[$p$]. Nevertheless, I'd love for my program to write $p$ in terms of $\rho$ so I don't have to do it manually as I did above. I'm assuming a way to do this is to tell Mathematica to factor the explicit form of $\rho$ out of $p$, but I'm not sure. Either way, I haven't been able to do this (the factorizing). So far, I've read about the function FactorTerms and even an specific one shown in this answer, but none of this work as I want them to.

I'm assuming a function like this already exists or one may be defined simply, but I just can't figure it out. Any help is appreciated.

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    $\begingroup$ Please edit your question to include a concrete example of what you are trying to do. $\endgroup$
    – Bob Hanlon
    Jan 24, 2023 at 2:41
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    $\begingroup$ Hi @Vicente, and welcome to MSE! In addition to what Bob suggested, could you please include Mathematica's code for your functions, so that people don't need to retype it? $\endgroup$
    – Victor K.
    Jan 24, 2023 at 4:01
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    $\begingroup$ I can tell it can be algebraically manipulated are you sure? I just checked,. Your desired $d$ does not match. When I do it by hand, this is what I get !Mathematica graphics $$\rho \left(\frac{4}{3} \left(\frac{\rho }{F}\right)^{\alpha }-1\right)$$ $\endgroup$
    – Nasser
    Jan 24, 2023 at 4:31
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    $\begingroup$ Copy-pasteable input? Without which this is not a viable question for this forum. $\endgroup$ Jan 24, 2023 at 5:13

2 Answers 2

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ClearAll["Global`*"]
ρ = F/(β*F^α + 1)^(1/α)
d = -(F/(β*F^α + 1)^((1/α))) + 4/3 F/(β*F^α + 1)^(1 + 1/α)
d /. (1 + F^α β) :> (F/HoldForm[ρ])^α
PowerExpand[%] // Simplify

Mathematica graphics


Screen shot

Mathematica graphics

V 13.2

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Solution

A more hands off approach, that still requires an guessed substitution, in this case

(β*F^α + 1) -> q

and then we eliminate q.

Also, I choose a somehow more verbose code to avoid lingering definitions and keep a clean kernel.

Surprisingly, not all combinations of Reduce, Eliminate and Solve work as effectively. These two options do work for this case:

Eliminate (β*F^α + 1) by hand

Block[
    {
        q,
        subs = ( (β*F^α + 1) -> q ), (* Put your intuition here *)
        expre1 = ρ == F/(β*F^α + 1)^(1/α),
        expre2 = d == (F/(β*F^α + 1)^((1/α))) + 4/3 F/(β*F^α + 1)^(1 + 1/α)
    },
    Assuming[
        {F, β, α, ρ, q} ∈ PositiveReals,
        FullSimplify[
            expre2 /. subs /. Solve[expre1 /. subs, q]
        ]
    ]
]

enter image description here

Eliminate (β*F^α + 1) using the third argument of Solve

Or also alternatively

Block[
    {
        q,
        subs = ( (β*F^α + 1) -> q ), (* Put your intuition here *)
        expre1 = ρ == F/(β*F^α + 1)^(1/α),
        expre2 = d ==(F/(β*F^α + 1)^((1/α))) + 4/3 F/(β*F^α + 1)^(1 + 1/α)
    },
    Assuming[
        {F, β, α, ρ, q} ∈ PositiveReals,
        Solve[
            Reduce[{expre2, expre1}/.subs]
            , {d}
            , {q} (* Undocumented "Eliminate" feature*)
            , InverseFunctions->True
        ]
    ]
]
Simplify@%
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  • $\begingroup$ I'm asking further clarification about this undocumented way of evaluating Solve in this other question. $\endgroup$
    – rhermans
    Jan 24, 2023 at 12:09

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