Re: Plotting
- To: mathgroup at smc.vnet.net
- Subject: [mg72996] Re: Plotting
- From: Bill Rowe <readnewsciv at sbcglobal.net>
- Date: Sun, 28 Jan 2007 00:51:51 -0500 (EST)
On 1/27/07 at 6:37 AM, ghcgwgwg at singnet.com.sg (Rayne) wrote:
>Hi all, I want to plot the cubic equation f(x) == x^3 - 16x^2 + 3x +
>1 == 0. I had:
>f==x^3-16 x^2+3 x+1 ==== 0 Plot[f,{x,-1,20}]
>but I was unable to get the plot.
If you have accurately posted what you tried, it should not be
surprising Plot generates error messages.
x^3-16 x^2+3 x+1 === 0 will always evaluate to False since the
expression x^3-16 x^2+3 x+1 is clearly not identical to 0. So,
the expression
f==x^3-16 x^2+3 x+1 === 0
Evaluates to f==False.
Now this is an equation and does not define a value for f. So,
when you try
Plot[f, {x, -1, 20}]
you are asking Mathematica to plot an undefined variable with no
dependence on x. And the result is error message stating f has
no numerical values.
Presumably you meant to define f as follows:
f = x^3-16 x^2+3 x+1 == 0
But while this does assign f a value, this is not something Plot
will work with since
In[3]:=
f = x^3 - 16*x^2 + 3*x + 1 == 0
Out[3]=
x^3 - 16*x^2 + 3*x + 1 == 0
In[4]:=
f /. x -> 2
Out[4]=
False
That is f will evaluate to either True or False for any numeric
value assigned to x.
You can of course do
Plot[x^3 - 16*x^2 + 3*x + 1, {x, -1, 20}];
Is this what you want?
Note:
In[5]:=
roots=NSolve[f, x]
Out[5]=
{{x -> -0.1726471308825491}, {x -> 0.3664487167738349},
{x -> 15.806198414108714}}
That is your cubic has three real roots. So, perhaps you want a
plot of the value of the cubic at these roots, i.e.,
ListPlot[{x /. #, x^3 - 16 x^2 + 3 x + 1 /. #}&/@roots,
Frame->True, Axes->None, PlotStyle ->PointSize[.02]];
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