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 Solving Higher Order Equations
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LewisLabAZ

USA
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Posted - 07/20/2004 :  2:01:10 PM  Show Profile  Edit Topic  Reply with Quote  View user's IP address  Delete Topic
Is the root solving function on Origin 7.5 capable of handling higher order equations such as a quartic or a quintic? We want to solve these equations for x and do not want the coefficients as fitting parameters in our user defined NLSF fit.
Thank you very much
Lewis Lab

easwar

USA
1964 Posts

Posted - 07/20/2004 :  4:42:15 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Hi Lewis Lab,

The objective of that "tool" in the File Exchange area was to provide a tool for finding roots of simple equations using bisection method. Solving quintics seems to be a rather complex problem as ca be seem from this page:
http://mathworld.wolfram.com/QuinticEquation.html
and this tool on the File Exchange area is not capable of doing that.

Easwar
OriginLab




Edited by - easwar on 07/20/2004 4:42:41 PM
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easwar

USA
1964 Posts

Posted - 07/21/2004 :  2:39:06 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Hi Lewis Lab,

Perhaps I did not understand your question correctly. What this tool does is just a numerical bisectional search for roots, with some preset accurarcy (how close is the y value to zero). So as long as a numerical solution is possible, it should find the root as long as the equation can be computed - no matter whether it is a cubic or quintic etc. The tool does not do any algerabic root finding, if that is what you are looking for.

Are you having trouble with a particular equation/expression? Did you first try generating values with your expression in a column, then plot that colummn to see if any roots would exist in your X range?

Easwar
OriginLab




Edited by - easwar on 07/21/2004 2:39:44 PM
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rdremov

USA
28 Posts

Posted - 07/21/2004 :  2:57:04 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Quick try of the tool:
find_roots "(x-1)*(x-2)*(x-3)*(x-4)*(x-5)" 0 10;
will correctly find all five roots: x = 1, 2, 3, 4 and 5

Roman
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LewisLabAZ

USA
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Posted - 07/23/2004 :  7:13:34 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Thank you easwar and rdremov for the replies. We really appreciate the attention.
Lewis Lab
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