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Re: Why is Newton's method failing to "find sufficient increase in function"?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112070] Re: Why is Newton's method failing to "find sufficient increase in function"?
  • From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
  • Date: Sat, 28 Aug 2010 07:03:54 -0400 (EDT)
  • References: <i55gre$2e6$1@smc.vnet.net>

Setting WorkingPrecision to some number (WorkingPrecision -> 20) works
for me.

Cheers -- Sjoerd

 On Aug 26, 12:49 pm, Yaroslav Bulatov <yarosla... at gmail.com> wrote:
> I'm getting FindMaximum::lstol warning in the code below...why? How
> can I get rid of it?
>
> For this particular function I can fix it by changing Method to
> Automatic, but this breaks optimization for other functions in my
> optimization task where Newton's method works fine.
>
> o = 1/5 Log[E^(-(h/Sqrt[3]))/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
>    1/5 Log[E^(h/Sqrt[3])/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
>    1/10 Log[E^(-(h/Sqrt[3]) - Sqrt[2] j)/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
>    3/10 Log[E^(h/Sqrt[3] - Sqrt[2] j)/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
>    1/10 Log[E^(-Sqrt[3] h + Sqrt[2] j)/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
>    1/10 Log[E^(Sqrt[3] h + Sqrt[2] j)/(
>      2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
>       E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
>       E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))];
> ContourPlot @@ {o, {j, -1, 1}, {h, -1, 1}}
> FindMaximum @@ {o, {{j, -0.008983550852535105`}, {h,
>     0.06931364191023386`}}, Method -> "Newton"}



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