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How do I get a numerical cube root?

Posted 11 years ago
Can't paste code on my machine.


\.b2r = CubeRoot[((¹V)/Pi ) (3/4)];           (* need cube root of \.b2r 3Sqrt[\.b2r] *)

Out put

"\.b2r =" Surd[Quantity[5.46439*10^-69, ("Meters")^3], 3]

I require a numerical value like 1.8 x 10^-23 meters

PS 1V is a calculated volume, but any cube root will do.
Many Thanks
Andrew Pepes
POSTED BY: Andrew Pepes
3 Replies
Posted 10 years ago

Hi, I'm very new to Mathematica and I can't figure out how power simplifications are performed. If I try with

N[Power[Quantity[4, ("Centimeters")^3]*3/4/Pi, 1/3], 3]

I get 0.985 cm which is what I expect.

But, if I try with something like

Clear[r1, r2, v1, v2];
r1 = Quantity[2 , "Centimeters"]
v1 = r1^3*3/4*Pi
v2 = v1*5
r2 = N[CubeRoot[v2*3/4/Pi], 2]

I get

Surd[Quantity[45/2, "Centimeters"^3], 3]

where I would expect CubeRoot["Centimeters"^3] to be simplified to "Centimeters".

What am I doing wrong?

Thank you for your help, Luca-

POSTED BY: Luca Ivaldi
Would something like this work?
Power[Quantity[5.46439*10^-69, ("Meters")^3], 1/3]

It gives
Quantity[1.76136*10^-23, "Meters"]
(* 1.76136*10^(-23) m *)

POSTED BY: Marco Thiel
Posted 11 years ago
Hi Marco

Brilliant, I used it with my variables
\.b2r = Power[Quantity[((¹V)/Pi ) (3/4)], 1/3];
and it gives
"\.b2r =" (Quantity[1.76136*10^-23, "Meters"])
Many Thanks
POSTED BY: Andrew Pepes
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