Re: Why can't Mathematica find this root?
- To: mathgroup at smc.vnet.net
- Subject: [mg38431] Re: Why can't Mathematica find this root?
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Sat, 14 Dec 2002 03:21:03 -0500 (EST)
- References: <at4dgq$f1c$1@smc.vnet.net> <at9b6n$pqk$1@smc.vnet.net> <atc81n$34a$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Konrad Den Ende" <konrad at voxway.com> wrote:
> > No, you don't. There are infinitely many roots, of which Mathematica
> > gives you only three: -Pi/2, 0, and Pi/2.
>
> Actually, you get five different roots, not three.
Oops! Sorry, I mistakenly had thought you were solving just f[x]==0, rather
than f'[x]==0, which indeed gives five roots.
> Nevertheless - point taken.
> I ment of course that i'd like Mathematica to give me ALL the roots on
> form: angle + (period * n). I'll try to be more exact the next time.
Well, here's something that might interest you, although it's not really
what you wanted. Let's be a bit more general, now asking Mathematica to
Solve[f'[x]==a, x]. This gives eight expressions:
{{x -> -ArcCos[-Sqrt[5/8 - Sqrt[9 - 16*a]/8]]},
{x -> ArcCos[-Sqrt[5/8 - Sqrt[9 - 16*a]/8]]},
{x -> -ArcCos[Sqrt[5/8 - Sqrt[9 - 16*a]/8]]},
{x -> ArcCos[Sqrt[5/8 - Sqrt[9 - 16*a]/8]]},
{x -> -ArcCos[-Sqrt[5/8 + Sqrt[9 - 16*a]/8]]},
{x -> ArcCos[-Sqrt[5/8 + Sqrt[9 - 16*a]/8]]},
{x -> -ArcCos[Sqrt[5/8 + Sqrt[9 - 16*a]/8]]},
{x -> ArcCos[Sqrt[5/8 + Sqrt[9 - 16*a]/8]]}}
Merely substituting 0 for a in the above gives one double and six single
roots:
{{x -> (-2*Pi)/3}, {x -> (2*Pi)/3}, {x -> -Pi/3},
{x -> Pi/3}, {x -> -Pi}, {x -> Pi}, {x -> 0}, {x -> 0}}
That's still not what you wanted. But to get, at least in some sense, what
you want, you simply need to interpret all of the inverse cosines above as
denoting not just the principal value (which is what Mathematica does, of
course), but rather all values of the inverse cosine relation. Then each
of the eight expressions above would represent infinitely many values.
Finally, substituting 0 for a, you would have all solutions represented in
terms of the multivalued inverse cosine relation:
-arccos(-1/2), arccos(-1/2), -arccos(1/2), arccos(1/2),
-arccos(-1), arccos(-1), -arccos(1), arccos(1)
or equivalently, for integer N,
(-2*Pi)/3+2*Pi*N, (2*Pi)/3+2*Pi*N, -Pi/3+2*Pi*N, Pi/3+2*Pi*N,
-Pi+2*Pi*N, Pi+2*Pi*N, 0+2*Pi*N, 0+2*Pi*N.
To get this final form -- at least the way I did it -- required human
intervention, which I'm sure you had hoped to avoid. Can anyone suggest
a simple way, avoiding human intervention, to have Mathematica give an
equivalent form of the result?
David Cantrell
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