Re: Something that FullSimplify knows but Reduce does not ?
- To: mathgroup at smc.vnet.net
- Subject: [mg118052] Re: Something that FullSimplify knows but Reduce does not ?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 9 Apr 2011 18:00:48 -0400 (EDT)
In case anyone is interested in the answer to the question I posted below, here it is.
FullSimplify is indeed able to prove by means of exact methods that ArcTan[Sqrt[3 - 2*Sqrt[2]]==Pi/8. Reduce does not know this because it does not use FullSimplify (since it can be too slow). Neither Reduce nor FullSimplify will use approximate numerical methods (or at least approximate numerical methods that have not been symbolically validated) to decide equality between exact expressions. However it is sometimes possible that some functions called up by Reduce of FullSimplify may use such methods without producing a warning message. Such situations should be rare and relatively innocuous.
Andrzej Kozlowski
On 8 Apr 2011, at 10:16, Andrzej Kozlowski wrote:
> Consider the following (parametric) function of t:
>
> g[t_] := (1/2)*Sin[2*t]*(((a - b)*(a + b))/
> Sqrt[a^2*Sin[t]^2 + b^2*Cos[t]^2] +
> (b^2 - a^2)/Sqrt[a^2*Cos[t]^2 + b^2*Sin[t]^2])
>
> We try to find the roots of this for 0<=t<=Pi/4 using Reduce:
>
> Reduce[
> g[t] == 0 && a > 0 && b > a && 0 <= t <= Pi/4, t, Reals]
>
> Reduce[
> g[t] == 0 && a > 0 && b > a && 0 <= t <= Pi/4, t, Reals]
>
> During evaluation of In[474]:= Reduce::ztest:Unable to decide whether numeric quantities HoldForm[{Pi - 8*ArcTan[Sqrt[3 - 2*Sqrt[2]]], -Pi + 8*ArcTan[Sqrt[3 - 2*Sqrt[2]]]}] are equal to zero. Assuming they are. >>
>
>
> a > 0 && b > a && (t == 2*ArcTan[Sqrt[3 - 2*Sqrt[2]]] || t == 0)
>
> So, Reduce claims that it was unable to decide if Pi - 8*ArcTan[Sqrt[3 - 2*Sqrt[2]]] is zero or not. But
>
> FullSimplify[Pi - 8*ArcTan[Sqrt[3 - 2*Sqrt[2]]]]
>
> 0
>
> Is everything here really as it should be?
>
>
> Andrzej Kozlowski
>