       Re: Numerical expression

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• Subject: [mg128825] Re: Numerical expression
• From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>
• Date: Thu, 29 Nov 2012 06:03:26 -0500 (EST)
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```Hy Massimo,

let us define the 3-order unity-roots

w0 = 1
w1 = Exp[I 2 \[Pi]/3]
w2 = Exp[I 4 \[Pi]/3]

and the following subexpressions

h1=(9 (3)^(1/2) + 4 (23)^(1/2))^(1/3)
h2=(9 (3)^(1/2) - 4 (23)^(1/2))^(1/3)

ee=(h1+h2)/3^(1/2)
ee//N= 2.32666 + 0.765945 I

Using unity-roots this correspond to the choice of unity-roots

ee=(w0*h1+w0*h2)/3^(1/2)
ee//N= 2.32666 + 0.765945 I

but if you use another (of the 9 possible) choice of unity roots  you can obtain 1. By trials

ee=(w1*h1+w0*h2)/3^(1/2)
ee//N=1+(3.30384*10^-16 I)

Other combinations give other values.

Bye Rob

-----Messaggio originale-----
Da: Dana DeLouis [mailto:dana01 at me.com]
Inviato: mercoled=EC 28 novembre 2012 9.17
A: mathgroup at smc.vnet.net
Oggetto: Re: Numerical expression

> ( ( 9(3)^1/2+4(23)^1/2)^1/3 + (9(3)^1/2+4(23)^1/2)^1/3  )  /  (3)^1/2)

Hi.   Just to add to the others, if we "Assume" the following

k = (9*Sqrt + 4*Sqrt)^ (1 / 3)  ;

// Or...
k=k//FullSimplify
Sqrt[1/2 (13+Sqrt)]

Then what you have is:

(k+k) / Sqrt[3.]
3.76887

// Which matches what others have mentioned:

For this to equal 1,   then k would have to equal

Sqrt / 2

Perhaps your equation for the k part is off a little. ??

= = = = = = = = = =
HTH  :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =

On Sunday, November 25, 2012 11:29:24 PM UTC-5, Massimo wrote:
> How could I handle in Mathematica this expression?
>
>
>
> ( ( 9(3)^1/2+4(23)^1/2)^1/3 + (9(3)^1/2+4(23)^1/2)^1/3  )  /  (3)^1/2)
>
>
>
> With a lot of trouble I have found out that is equal 1,
>
> but how to get it with Mathematica?
>
>
>
> Thanks very much.
>
>
>
>
>
>
>
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