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Groups > comp.soft-sys.math.mathematica > #1797 > unrolled thread
| Started by | Murray Eisenberg <murray@math.umass.edu> |
|---|---|
| First post | 2011-04-20 08:29 +0000 |
| Last post | 2011-04-20 08:29 +0000 |
| Articles | 1 — 1 participant |
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Re: Simplify results Murray Eisenberg <murray@math.umass.edu> - 2011-04-20 08:29 +0000
| From | Murray Eisenberg <murray@math.umass.edu> |
|---|---|
| Date | 2011-04-20 08:29 +0000 |
| Subject | Re: Simplify results |
| Message-ID | <iom5gr$v$1@smc.vnet.net> |
First, a "silly" answer: Why do you think Mathematica ought to give the
same form as you get by hand calculation?
Second, a more serious answer: look at the way Mathematica represents
the original expression:
FullForm[(8 a^2 b^3 (c^2 + 1)^4 - 6 a^3 b^2 (c^2 + 1)^3 +
14 a^4 b > (c^2 + 1)^2)/(4 a^3 b^2 (c^2 + 1)^2 -
10 a^4 b^3 (c^2 + 1)^3)]
On 4/19/2011 6:56 AM, Berthold Hamburger wrote:
> Hi,
>
> This might be a silly question, so please bear with me, but I have been
> scratching my head about it for some time now.
>
> Is there a particular reason why Mathematica (8.01) simplifies the following
> fraction reversing the signs in the result:
>
> IN:
>
> Simplify[(8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b
> (c^2+1)^2)/(4a^3b^2(c^2+1)^2-10a^4b^3(c^2+1)^3)]
>
> OUT:
>
> (-7 a^2+3 a b (1+c^2)-4 b^2 (1+c^2)^2)/(a b (-2+5 a b (1+c^2)))
>
> Reducing the fraction by hand gives me:
>
> (7 a^2-3 a b (1+c^2)+4 b^2 (1+c^2)^2)/(a b (2-5 a b (1+c^2)))
>
> Thanks
>
> Berthold
>
--
Murray Eisenberg murray@math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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