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Re: factor out a term to cancel in a fraction
*To*: mathgroup at smc.vnet.net
*Subject*: [mg98887] Re: factor out a term to cancel in a fraction
*From*: sean_incali at yahoo.com
*Date*: Mon, 20 Apr 2009 19:14:31 -0400 (EDT)
*References*: <200904200531.BAA00876@smc.vnet.net> <gshfsf$bvm$1@smc.vnet.net>
Initially, I thought the reason it didn't cancel was because maybe
mathematica thought k1 could be = 0. So i figured, if I told it that
k1>0, it may work. But now I see that Assuming probably has some other
application...
On Apr 20, 2:39 am, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> Factor[(e0*s[t]*k1*k2)/(s[t]*k1 + k1*Km)]
> (e0*k2*s[t])/(Km + s[t])
>
> or
>
> Cancel[(e0*s[t]*k1*k2)/(s[t]*k1 + k1*Km)]
> (e0*k2*s[t])/(Km + s[t])
>
> Difficult ? Give an example of "easy".
>
> And by the way, where did you get the idea that Assuming does anything =
> at all with Collect?
>
> Andrzej Kozlowski
>
> On 20 Apr 2009, at 14:31, sean_inc... at yahoo.com wrote:
>
> > Hello group.
>
> > I'm trying to do a little algebra which i can do in my head using
> > Mathematica. Why? Well, it's neater.
>
> > Consider the following.
>
> > p'[t] == (e0 s[t] k1 k2 )/ (s[t] k1 + k1 Km)
>
> > Factor out k1 and cancel it to simplify the expression. Why is this so
> > difficult to accomplish in mathematica? Following does nothing.
>
> > p'[t] == (e0 s[t] k1 k2 )/ (s[t] k1 + k1 Km) //
> > Assuming[k1 > 0, {Collect[#, k1]}] &
>
> > Thanks for any info.
>
> > Sean
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