       Re: Not all points plot on my graph...

• To: mathgroup at smc.vnet.net
• Subject: [mg105192] Re: [mg105158] Not all points plot on my graph...
• From: Tomas Garza <tgarza10 at msn.com>
• Date: Tue, 24 Nov 2009 05:46:53 -0500 (EST)
• References: <200911231149.GAA21721@smc.vnet.net>

```Well, f is complex-valued in the range {-2, -3}.
E.g.,

In:= Table[f[j],{j, -2, -3, -0.1}]

Out= {0.,-0.107722+0.18658 I,-0.170998+0.296177 I,-0.22407+0.388101 I,-0.271442+0.470151 I,-0.31498+0.545562 I,-0.355689+0.616072 I,-0.394187+0.682751 I,-0.430887+0.746318 I,-0.466085+0.807283 I,-0.5+0.866025 I}

No way to graph it. Also, probably the point (-3, 1) is too small and almost invisible.

BTW, I would use SetDelayed (:=) instead of Set (=) in the definition of f.

Tomas

> Date: Mon, 23 Nov 2009 06:49:43 -0500
> From: davidfrick2003 at yahoo.com
> Subject: [mg105158] Not all points plot on my graph...
> To: mathgroup at smc.vnet.net
>
> Why is it when I write this...
>
> Clear[f, x]
> f[x_] = (x + 2)^(2/3)
> Plot[f[x], {x, -3, 0}, PlotRange -> {{-3, 0}, {-4, 4}}]
> f[-3]
>
> .. I get no values show in the graph for x<-2 even though f[-3]
> evaluates to (-1)^(2/3).
>
> (-1)^(2/3) evaluates as f[-3]=1 so why wouldn't the point (-3,1) plot
> on my graph?
>
>

```

• Prev by Date: Re: Re: Re: I broke the sum into pieces
• Next by Date: Mathematica and LabelME
• Previous by thread: Re: Not all points plot on my graph...
• Next by thread: Re: Not all points plot on my graph...