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Topic: Simple non-linear equation solving question
Replies: 7   Last Post: Feb 13, 2012 12:31 PM

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 Patrick Posts: 4 Registered: 2/12/12
Re: Simple non-linear equation solving question
Posted: Feb 13, 2012 12:32 AM
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Thanks for the response. My Matlab version is 7.10.0 (R2010a).

The full code:

clear all
clc
close all

x=[-.5:.01:1];

m=-x.^3 + (sin(x)).^2;

plot(x,m)

clear x
x = sym('x');
solve(-x.^3 + (sin(x)).^2)

Which provides a plot, and:

ans =

matrix([[0]])

From the plot, I can see that there is also a solution at about 0.803.

Thanks again...

"Nasser M. Abbasi" <nma@12000.org> wrote in message <jha64h\$kls\$1@speranza.aioe.org>...
> On 2/12/2012 10:42 PM, Patrick wrote:
> > Hi,
> >
> > I'm trying to solve the following non-linear equation: 0=-x^3 + (sin(x))^2
> >
> > EDU>> solve(-x^3+sin(x)^2)
> >
> > ans =
> >
> > matrix([[0]])
> >
> > By graphing this equation, it appears that there are solutions at 0 and 0.803. What am I doing wrong?
> >
> > Thanks...

>
> Which verssion do you have?
>
> on mine:
>
> MATLAB Version 7.12.0.635 (R2011a)
>
> I get
>
> --------------------
> EDU>> syms x
> solve(-x^3+sin(x)^2)
>
> ans =
>
> 0
> ----------------------
>
> matrix([[0]]) means zero solution also? i..e a vector, which one
> entry, and that entry is zero. so same answer I have.
>
> So, I do not know how you got 0.803 there. What command you used
> to plot? Please show complete code you did, not one line.
>
> --Nasser

Date Subject Author
2/12/12 Patrick
2/13/12 Nasser Abbasi
2/13/12 Patrick
2/13/12 Roger Stafford
2/13/12 Nasser Abbasi
2/13/12 Alan Weiss
2/13/12 Alan Weiss
2/13/12 Patrick

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