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Mathematica Algebra

How to factor out known multipliers in algebraic expressions in mathematica

Alissa Bans

Posted 12 years ago

Hi There, I have some very complicated expressions in the form of F[x,y,z,A,B,C]/(x^2 - y^2 - z^2) . Here F[ ] is a very long function of x,y,z as well as some other terms, A,B,C. There are places in F where (x^2 - y^2 - z^2) could be completely canceled out, however, no matter what algebraic manipulation I try in mathematica (Simplify,Cancel,Factor, etc), I cannot get these terms to cancel! What I want is a way to say: rewrite F in terms of groupings of (x^2 - y^2 - z^2) wherever you can (even if it makes F look overall messier). Note I've tried "Collect" and doesn't seem to work in this sense. Can anybody suggest a good thing to try? I can send the expression for F in a notebook, but I warn you it's very VERY long . Thanks, Alissa

POSTED BY: Alissa Bans

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 12 years ago

You might try mapping `Factor` over the sum of terms obtained from using `Expand` on the expression.

POSTED BY: Daniel Lichtblau

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 12 years ago

By "map Factor over..." I mean something like this (which maybe is what you did): Map[Factor, Expand[F[x,y,z,A,B,C]/(x^2 - y^2 - z^2] This way Factor is used on each term separately.

POSTED BY: Daniel Lichtblau

Alissa Bans

Posted 12 years ago

Okay, also I have no idea of the best way to express mathematica notebooks in this forum, my paste job up there seems pretty messy. Can I just attach the notebook somehow?

POSTED BY: Alissa Bans

Alissa Bans

Posted 12 years ago

Thanks for the suggestion. Applying 'Factor' to the 'Expand'ed form seems to just mix up the terms of F more, i.e. it sort of multiplies everything out. But I may be mistaking what you mean by mapping Factor? Thanks either way for the reply!

POSTED BY: Alissa Bans

Maarten van de Meent

Posted 12 years ago

You might want to try: Simplify[F,R==(x^2 - y^2 - z^2)]

POSTED BY: Maarten van de Meent

Vitaliy Kaurov

Vitaliy Kaurov, WOLFRAM Research

Posted 12 years ago

Alissa, even thought your expressions are large, do you think it is possible to find a compact sub-expression that illustrates the nature of the problem?

POSTED BY: Vitaliy Kaurov

Alissa Bans

Posted 12 years ago

Thanks! Simplify[ F, R==(x^2 - y^2 - z^2)] helps a lot, I believe it's grouping all the factors of (x^2 - y^2 - z^2) into 'R' which makes it very easy for me to cancel it out. Wow, this forum is really a great tool. Thanks everybody!

POSTED BY: Alissa Bans

Alissa Bans

Posted 12 years ago

Like I said Simplify[ F, R==(x^2 - y^2 - z^2)] seems to mostly do the job, but here a notebook to illustrate the problem in case people have other suggestions. Basically, I know that part of "Denom" can be factored out of "F" , the output I'm really interested in is F/Denom, but I'm aiming cancel out Denom every place I can. Does that make sense? Again, thanks to everyone for replying! F := (m r (K^2)[r] \[Sigma][r] \[CapitalSigma][ r]^2 (bz^2 r^(2 nz) + c[r]^2 \[CapitalSigma][r]) (-bz^2 r^(2 nz) - c[r]^2 \[CapitalSigma][r] + r^2 \[Sigma][r]^2 \[CapitalSigma][r]) + 2 \[CapitalOmega][ r] (-I bz^5 b\[Theta]g r^(1 + 5 nz) Derivative[1][\[CapitalSigma]][r] + bz^3 r^(3 nz) \[CapitalSigma][ r] (I bz^2 b\[Theta]g (1 + nz) r^(2 nz) - 3 I b\[Theta]g r c[r]^2 Derivative[1][\[CapitalSigma]][r] + r^2 (2 I b\[Theta]g r + bz m r^nz) \[Sigma][r]^2 Derivative[ 1][\[CapitalSigma]][r]) - m r c[r] \[Sigma][r] \[CapitalSigma][ r]^4 (2 r^3 \[Sigma][r]^3 Derivative[1][c][r] + r^2 c[r] \[Sigma][ r]^2 (\[Sigma][r] + 2 \[CapitalOmega][r] + r Derivative[1][\[CapitalOmega]][r]) - c[r]^3 (3 \[Sigma][r] + 2 \[CapitalOmega][r] + 3 r Derivative[1][\[CapitalOmega]][r])) + bz r^nz \[CapitalSigma][ r]^2 (-2 I b\[Theta]g r c[r]^4 Derivative[1][\[CapitalSigma]][ r] + c[r]^2 (I bz^2 b\[Theta]g (2 + 3 nz) r^(2 nz) + 2 r^2 (I b\[Theta]g r + bz m r^nz) \[Sigma][ r]^2 Derivative[1][\[CapitalSigma]][r]) + bz^2 r^(1 + 2 nz) \[Sigma][ r] ((-I b\[Theta]g (-1 + 3 nz) r + 3 bz m r^nz) \[Sigma][ r] + 2 (I b\[Theta]g r + bz m r^nz) \[CapitalOmega][r] + r (2 I b\[Theta]g r + 3 bz m r^nz) Derivative[ 1][\[CapitalOmega]][r])) + \[CapitalSigma][ r]^3 (-2 I bz b\[Theta]g r^(3 + nz) c[r] \[Sigma][r]^2 Derivative[1][c][r] + c[r]^4 (I bz b\[Theta]g (1 + 2 nz) r^nz + m r^2 \[Sigma][r]^2 Derivative[1][\[CapitalSigma]][r]) - bz r^(3 + nz) \[Sigma][ r]^3 (bz m (1 + 2 nz) r^nz \[Sigma][r] + 2 I b\[Theta]g r \[CapitalOmega][r] + 2 bz m r^nz \[CapitalOmega][r] + bz m r^(1 + nz) Derivative[1][\[CapitalOmega]][r]) + r c[r]^2 \[Sigma][ r] (bz r^ nz (-I b\[Theta]g (-1 + 2 nz) r + 6 bz m r^nz) \[Sigma][ r] - m r^3 \[Sigma][r]^3 Derivative[1][\[CapitalSigma]][ r] + 2 bz r^ nz ((I b\[Theta]g r + 2 bz m r^nz) \[CapitalOmega][r] + r (I b\[Theta]g r + 3 bz m r^nz) Derivative[ 1][\[CapitalOmega]][r]))))) Denom := (2 m c[r]^2 \[Sigma][r]^2 \[CapitalSigma][ r]^3 (bz^2 r^(2 nz) + c[r]^2 \[CapitalSigma][r] - r^2 \[Sigma][r]^2 \[CapitalSigma][r]) \[CapitalOmega][r]) (here the (x^2-y^2-z^2) -> (bz^2 r^(2 nz)+c[r]^2 \ \[CapitalSigma][r]-r^2 \[Sigma][r]^2 \[CapitalSigma][r]) ) (I am looking for a for sure way to cancel (bz^2 r^(2 nz)+c[r]^2 \ \[CapitalSigma][r]-r^2 \[Sigma][r]^2 \[CapitalSigma][r]) from parts \ of F, so I guess I want every term that doesn't have to be over that \ denomator to not be over that denomenator )

POSTED BY: Alissa Bans

Vitaliy Kaurov

Vitaliy Kaurov, WOLFRAM Research

Posted 12 years ago

We are considering this functionality, but at the moment you could use any free upload services and provide a link for large notebooks. But anyway - your paste job looks good to me ;)

POSTED BY: Vitaliy Kaurov

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