| View previous topic :: View next topic |
| Author |
Message |
midan
Newbie
Joined: 11 Feb 2009
Posts: 2
|
 Posted: Wed Feb 11, 2009 2:59 am Post subject: Calc rise to power |
 |
|
I need to resolve this formula S=(J3+SQRT(I3^3+J3^2))^(1/3) where
i3=-0.1111111 and j3=-0.9629630 i get an "#VALUE!" result
for other values like i3=0.5 and j3=0.4 is ok
---
Sorry for my bad english |
|
| Back to top |
|
 |
Villeroy
Super User
Joined: 04 Oct 2004
Posts: 10065
Location: Germany
|
| Posted: Wed Feb 11, 2009 5:32 am Post subject: |
|
|
You try to get the root of a negative number.
_________________
Rest in peace, oooforum.org
Get help on http://forum.openoffice.org |
|
| Back to top |
|
|
midan
Newbie
Joined: 11 Feb 2009
Posts: 2
|
| Posted: Wed Feb 11, 2009 6:30 am Post subject: |
|
|
Calculated with OOCalc I3^3+J3^2=0.9259259 , is greather then 0
(J3+SQRT(I3^3+J3^2))=-0.0007125
the problem is in this step -0.0007125^(1/3) |
|
| Back to top |
|
|
keme
Moderator
Joined: 30 Aug 2004
Posts: 2744
Location: Egersund, Norway
|
| Posted: Wed Feb 11, 2009 1:32 pm Post subject: |
|
|
To find the cubic root of a negative base, you can use IMABS(IMPOWER(base;1/3))*SIGN(base). It's useful to find the base first (a cell with an intermediate calculation). Explanation follows...
With integer powers, Calc will use the "natural" procedure, and all is well. If the exponent is not an integer, Calc will use a different way to calculate the power. This involves using logarithms, which do not work well with negative arguments.
Some math knowledge is necessary:
When the base is negative and the exponent is a fraction, the power will be a number somewhere in the complex plane. Generally we do not know that the power has an answer in the real numbers (i.e. no imaginary part). However, with the exponent of 1/3 we know that for every real base there is a real answer.
Still, in this situation we need to use the tools for working with complex numbers, the IM...() functions.
Say we have the number to find the cube root of in cell A1. =IMPOWER(A1;1/3) will give a correct answer in the complex plane, but never the real number we want.
However, we know that all answers will have the same absolute value, so we'll extract that: =IMABS(IMPOWER(A1;1/3)).
The absolute value is always a positive number, but the root will have the same sign as the base, so to keep the sign: =IMABS(IMPOWER(A1;1/3))*SIGN(A1) |
|
| Back to top |
|
|
Ed
Super User
Joined: 28 May 2003
Posts: 1040
|
| Posted: Wed Feb 11, 2009 2:07 pm Post subject: |
|
|
| keme wrote: | | To find the cubic root of a negative base, you can use IMABS(IMPOWER(base;1/3))*SIGN(base). It's useful to find the base first (a cell with an intermediate calculation). |
For a cube root, you could use SIGN(base)*ABS(base)^(1/3). Of course if you know that base is a negative number then you could just use -(-base)^(1/3)
Avoiding the use of complex number functions should make the calculation more accurate. |
|
| Back to top |
|
|
keme
Moderator
Joined: 30 Aug 2004
Posts: 2744
Location: Egersund, Norway
|
| Posted: Wed Feb 11, 2009 2:09 pm Post subject: |
|
|
Of course, but where's the fun in that... 
 <hiding face in hands and will not be able to face other people for a week at least> |
|
| Back to top |
|
|
bblackmoor
Newbie
Joined: 11 Jan 2007
Posts: 4
|
| Posted: Wed Jul 29, 2009 2:16 pm Post subject: Re: Calc rise to power |
|
|
| midan wrote: | I need to resolve this formula S=(J3+SQRT(I3^3+J3^2))^(1/3) where
i3=-0.1111111 and j3=-0.9629630 i get an "#VALUE!" result
for other values like i3=0.5 and j3=0.4 is ok |
The problem is that OpenOffice Calc returns an error when it attempts to find the root of a negative number, because it uses flawed logic to determine the value of roots (i.e., not taking into account the sign of the base) -- a bug it has had for at least four years, and probably since its creation.
Yes, you can do it manually: you should not need to.
Microsoft Excel does not have this bug. Every time I have a conversation with someone and I tell them they have no need to ever use Microsoft Office again, this years-old trival-to-fix bug sits on my shoulder and laughs at me. |
|
| Back to top |
|
|
|
