Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2012

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Numerical expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128813] Re: Numerical expression
  • From: "Massimo" <linusx++ at mail.com>
  • Date: Wed, 28 Nov 2012 03:15:54 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: l-mathgroup@wolfram.com
  • Delivered-to: mathgroup-newout@smc.vnet.net
  • Delivered-to: mathgroup-newsend@smc.vnet.net
  • References: <20121126042617.44760687B@smc.vnet.net> <k91u1k$5k8$1@smc.vnet.net>

"Dave Snead" <dsnead6 at charter.net> ha scritto nel messaggio 
news:k91u1k$5k8$1 at smc.vnet.net...
> Adding the necessary parentheses:
> ((9 (3)^(1/2) + 4 (23)^(1/2))^(1/3) + (9 (3)^(1/2) + 4 (23)^(1/2))^(1/
>      3))/(3)^(1/2)
> this comes out to
> (2*(9*Sqrt[3] + 4*Sqrt[23])^(1/3))/Sqrt[3]
>
> (2*(9*Sqrt[3] + 4*Sqrt[23])^(1/3))/Sqrt[3]//N
> gives
> 3.76887
> So it's not 1.

Thank you all  for your anwers.

I was wrong with a minus sign. The correct expression is:

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

With the RealOnly package I get the Real result that I was searching for.

FullSimplify[((9 (3)^(1/2) + 4 (23)^(1/2))^(1/3) + (9 (3)^(1/2) -
4 (23)^(1/2))^(1/3))/(3)^(1/2)] == 1

Now I ask you how to get the same result using online WolframAlpha.

Thanks again.



__________ Informazioni da ESET Smart Security, versione del database delle firme digitali 7736 (20121127) __________

Il messaggio è stato controllato da ESET Smart Security.

www.nod32.it







  • Prev by Date: Re: Bessel integral - Strange Hypergeometric Function
  • Next by Date: how to orient Polar Plot 0 Degrees at the North Pole
  • Previous by thread: Re: Numerical expression
  • Next by thread: Re: Numerical expression