       Re: Factoring

• To: mathgroup at smc.vnet.net
• Subject: [mg65111] Re: Factoring
• From: dh <dh at metrohm.ch>
• Date: Wed, 15 Mar 2006 06:28:23 -0500 (EST)
• References: <dv6847\$nq4\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Matt,
the following may be too simple for you, but try it.
expr is the expression you want to factor and pat is the factor you want
to extract:

MyFactor[expr_, pat_] := pat((#/pat) & /@ expr);

E.g. with 4*x*I*Sin[t] + 28*x^3*I*Cos[t] you would write:
expr = 4*x*I*Sin[t] + 28*x^3*I*Cos[t];
MyFactor[expr, 2 x I]

and will get:
2*I*x*(14*x^2*Cos[t] + 2*Sin[t])

Daniel

Matt wrote:
> Hello Mathgroup,
>   I'm sure that I've overlooked something obvious, but for the past two
> and a half hours, I've been trying to figure out how to use built-in
> Mathematica functions to just factor the imaginary number 'I' out of
> this:
>
> 4*I*r^2*Sin[2*Î¸] + I*r^5*Sin[5*Î¸]
>
> using Factor[] gives me I*r^2*(4*Sin[2*Î¸] + r^3*Sin[5*Î¸])
>
> using FactorTerms[] gives me I*(4*r^2*Sin[2*Î¸] + r^5*Sin[5*Î¸]) which
> is what I want, but as soon as I add in a common numerical factor, it
> also factors that out as well,
>
> e.g. 4*I*r^2*Sin[2*Î¸] + 8*I*r^5*Sin[5*Î¸]
>
> I finally just decided to divide the whole thing by 'I', and do some
> reconstruction of my expression, but I have the distinct feeling I've
> totally missed something very simple.  As an additional example, what
> if I just wanted to extract 2*x*I out of the following:
>
> 4*x*I*Sin[t] + 28*x^3*I*Cos[t]
>
> how would I do that?  In general, I'm looking for a function that says
> "Given an expression 'expr' and another expression 'sub' that is common
> to all additive terms of 'expr', give a result that is the product of
> 'sub' and the result of factoring 'sub' out of 'expr'".
>
> Thanks,
>
> Matt
>

```

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