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]]; -- To reply via email subtract one hundred and four