Re: Finding only the real roots
- To: mathgroup at smc.vnet.net
- Subject: [mg113635] Re: Finding only the real roots
- From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
- Date: Fri, 5 Nov 2010 05:13:29 -0500 (EST)
Hi, Peter,
All your questions have simple answers, but first your should write down correct Mathematica expressions.
NSolve, rather than Nsolve, Sqrt, rather than root and [] at Sqrt, rather than () at root.
You will hardly move further without learning few basic things about Mathematica, so do it.
In particular,
NSolve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p]
{{p -> (0.125 (-1. (-8. - 8. c - 8. g) g -
11.3137 Sqrt[-0.125 g^2 + 1. c g^2 + 0.5 g^3]))/g^2}, {p -> (
0.125 (-1. (-8. - 8. c - 8. g) g +
11.3137 Sqrt[-0.125 g^2 + 1. c g^2 + 0.5 g^3]))/g^2}}
but I would do
Solve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p]
{{p -> (2 g + 2 c g + 2 g^2 - Sqrt[-g^2 + 8 c g^2 + 4 g^3])/(
2 g^2)}, {p -> (
2 g + 2 c g + 2 g^2 + Sqrt[-g^2 + 8 c g^2 + 4 g^3])/(2 g^2)}}
instead. Also Manipulate works. Try this for instance:
Manipulate[
Solve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p], {c, 0,
1}, {g, 0, 1}]
Have fun, Alexei
Hi all,
I'm trying to numerically evaluate a function using NSolve, for
example
Nsolve[1/2+c+g(1-p)+root(c+p*g-1)==0,p] where my two variables g and c
are both in the interval [0,1]. But, since the root is not always
positive I do not get a solution to NSolve. The idea was therefore to
first define the positive roots and then to do NSolve.
My questions are therefore;
1) Is this the best approach, or are there more direct ways?
2) What should I do to only get the real roots?
3) If I first solve the root, and then use NSolve, will I still be
able to use functions such as Manipulate on the NSolve solution?
Best,
Petter
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Alexei Boulbitch, Dr. habil.
Senior Scientist
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