Re: Clarification [Overlay graphs ]

*To*: mathgroup at smc.vnet.net*Subject*: [mg49083] Re: Clarification [Overlay graphs ]*From*: Bill Rowe <readnewsciv at earthlink.net>*Date*: Wed, 30 Jun 2004 05:34:29 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

On 6/29/04 at 4:50 AM, J_o_s_e_f at hotmail.com (JosefG) wrote: >You have hit the nail on the head. I would like the domains to >overlap. >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[{$DisplayFunction = Identity}, >> {Plot[Cos[x], {x, Pi/4, Pi/2}, PlotStyle -> Hue[0]], >> Plot[Sin[x], {x, 0, Pi/4}], PlotStyle -> Hue[0.6]}] >> PlotRange -> All]; Somehow the code above got garbled. I thought I pasted it, but perhaps I actually typed it with a typo. The "]" character that comes after the range for Sin[x] should be placed after the PlotStyle->Hue[0.6] instead of where it appears. That is it should read Show[Block[ {$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]; -- To reply via email subtract one hundred and four