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MathGroup Archive 1998

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Re: A very difficult equation ...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14077] Re: A very difficult equation ...
  • From: bertronj at indra.com (Jean-Denis S Bertron)
  • Date: Tue, 22 Sep 1998 03:25:08 -0400
  • Organization: Indra's Net - Public Internet Access
  • References: <6tvn9u$1r6@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Julia Kamke (tobi.kamke at t-online.de) wrote:
: Hi!!!


: f[x_] := -2 + 2 * Cos[x] + x * Sin[x]

: How can I solve the equation f[x] == 0 in a symbolic way? Solve[f[x] ==
: 0, x] says, that it isn't possible to solve this equation! Furthermore
: I can't simplify this equation. But I think, that x = z * 2 * Pi (z =
: .. -2, -1, 0, 1, 2 ...) are some of the possible solutions. Is there a
: trick or a possibility to find out this solutions and the rest of all
: solutions with Mathematica??

: Please help me!!!


: Thanks,

:   Tobi


: P. S. Is f[x] == Sin[x](Sin[x](x^2 + 4) - 4x)??? How can I find this out
: with Mathematica??

I don't know much about mathematica, but I know that if you provide it
with some extra algebraic rules to work with, it may be able to
simplify  the equation partially. Look up the 'book' under
AlgrebraicRules. They have a good example where cos^2 + sin^2 = 1 is
added as a rule and it helps get a solution.
J.D.


--
Jean-Denis Bertron	jd.bertron at pobox.com 
http://rainbow.rmi.net/~bertronj  
Disclaimer(message):- Offending(message)!.


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