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Re: Clarification [Overlay graphs ]

On 6/29/04 at 4:50 AM, J_o_s_e_f at (JosefG) wrote:

>You have hit the nail on the head. I would like the domains to

>If I added Pi/4 to x, the proof I am working on would
>automatically be solved. So the first Plot you provided will not
>work for me. I did experiment with the first Plot and I am pretty
>sure that I would get the same results overlaping the domains, but
>I have to be sure about results.

I still don't really understand what you are trying to do. The way you've specified the problem, you are plotting Cos[x] over Pi/4 to Pi/2 and Sin[x] over 0 to Pi/4. That is you've specified a domain for one function of (0, Pi/4) and for the other (Pi/4, Pi/2). These clearly do not overlap. In fact, there is no way to get these to overlap.

The trick I used of adding Pi/4 to the argument of one function in effect changes the one domain from (0, Pi/4) to (Pi/4, Pi/2). It is as if I defined a new variable, y = x + Pi/4 and now plotted Sin[y] with y having a domain of (0, Pi/4).

>I keep getting the error "Show::shx" when I use your second Plot.
>I don't know if it works.

>> Show[Block[{$DisplayFunction = Identity}, 
>>     {Plot[Cos[x], {x, Pi/4, Pi/2}, PlotStyle -> Hue[0]], 
>>      Plot[Sin[x], {x, 0, Pi/4}], PlotStyle -> Hue[0.6]}]
>>    PlotRange -> All];

Somehow the code above got garbled. I thought I pasted it, but perhaps I actually typed it with a typo.

The "]" character that comes after the range for Sin[x] should be placed after the PlotStyle->Hue[0.6] instead of where it appears. That is it should read

    {$DisplayFunction = Identity}, 
    {Plot[Cos[x], {x, Pi/4, Pi/2}, PlotStyle ->Hue[0]], 
     Plot[Sin[x], {x, 0, Pi/4}, PlotStyle -> Hue[0.6]]}], 
   PlotRange -> All]; 
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