Message Boards Message Boards

0
|
4178 Views
|
4 Replies
|
3 Total Likes
View groups...
Share
Share this post:
GROUPS:

If I really need to find the abbreviated formula for a given result

Posted 11 years ago
Hi. If I really need to find the abbreviated formula for the given numeric result,

What other computer program can do this job other than Mathematica..?

Could you please recommend any of them?

I really do appreciate your help! 
POSTED BY: Woo Young Kang
4 Replies
Posted 11 years ago
Thank you!

It seems like the FindSequenceFunctin works with the list of numbers, not with the one type of long variable.;

The type of term that I want to abbreviate is like:

(Sqrt[3] *F* (u[1, 0] - 2 u[1, 1] + u[1, 2] + 2 (u[2, 0] - 2 u[2, 1] + u[2, 2])))/(2 \(u[1, 0]^2 + 3 u[1, 1]^2 + 3 u[1, 2]^2 - 3 u[1, 2] u[1, 3] + u[1, 3]^2 + 2 u[1, 3] u[2, 0] + u[2, 0]^2 - 6 u[1, 2] u[2, 1] - 3 u[2, 0] u[2, 1] + 3 u[2, 1]^2 + 12 u[1, 2] u[2, 2] - 6 u[1, 3] u[2, 2] - 3 u[2, 1] u[2, 2] + 3 u[2, 2]^2 - 3 u[1, 1] (u[1, 2] + 2 (u[2, 0] - 2 u[2, 1] + u[2, 2])) + (-6 u[1, 2] + 4 u[1, 3] + u[2, 0] - 3 u[2, 2]) u[2, 3] + u[2, 3]^2 + u[1, 0] (-3 u[1, 1] + u[1, 3] + 4 u[2, 0] - 6 u[2, 1] + 2 u[2, 3])))


-> which is like a very long one number, not the list of numbers.

If this is so, is there any other function that may work with this sort of problem?

If not, is there any alternative software that can do this sort of work..?


Thank you very very much again!
POSTED BY: Woo Young Kang
In a one dimensional, linear problem, FindSequenceFunction would be the function to use.

I don't know where to start with a two-dimensional, square-root problem.
You might have the simplest form already.
POSTED BY: Bruce Miller
Posted 11 years ago
Thank you!

I am trying to use the result of the following code:
a2 = Variance[Table[u[i, t] - u[i, t - 1], {t, 1, 3}, {i, 1, 2}]];
aa2 = Total[Flatten[FullSimplify[a2, Element[u[_, _], Reals]]]];
c2 = Covariance[Table[u[i, t] - u[i, t - 1], {t, 1, 3}, {i, 1, 2}]];
m2 = FullSimplify[c2, Element[u[_, _], Reals]];
b2 = Total[Flatten[FullSimplify[m2 - IdentityMatrix[2]*Diagonal[m2]]]];
FullSimplify[D[Refine[((aa2 + 2*b2)^(1/2))*F, Element[u[_, _], Reals]], u[1, 1]]]
--> Here, the variable at the right end of the last line which is "u[1,1]" will vary according to my needs,

such as u[1,2], u[2,2], u[2,3] and so on.

I am trying to find a pattern here and try to find a general expression for the resulting equation.

Is there a way to do so..?

Thank you so much!
POSTED BY: Woo Young Kang
I am not sure what sort of result and what sort of formula.

Do you have a list of numbers you want to fit a simple polynomial to?
(See FindFit and related functions.)

Do you have one number you for which want a formula in exact numbers and roots?
(Example 1.618 is approximately  (Sqrt[5]+1)/2.)

Do you have an example?
POSTED BY: Bruce Miller
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract