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Re: Nullcline and getting "2" values for y

  • To: mathgroup at smc.vnet.net
  • Subject: [mg81503] Re: [mg81468] Nullcline and getting "2" values for y
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Wed, 26 Sep 2007 06:38:49 -0400 (EDT)
  • References: <28556801.1190690438836.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

A function of y has ONE value for each y. (By definition of the term  
"function".)

What you mean to say, I think, is that there are two solutions to a  
certain equation. You haven't mentioned any such equation specifically, =

but... if I have to guess, I'd say you mean something like

fun = (b c y - a y^n + y^(1 + n))/(b + y^n);
f == fun

f == (10 y - 11 y^5 + y^6)/(10 + y^5)

Solving for y in terms of f gives up to SIX solutions, none of which are 
expressible in radicals:

roots = y /. Solve[f == fun, y] // ToRadicals

{Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 1],
  Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 2],
  Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 3],
  Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 4],
  Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 5],
  Root[-10 f + 10 #1 + (-11 - f) #1^5 + #1^6 &, 6]}

Plot[roots, {f, -5, 1}]

Plot[roots, {f, -50, 10}]

Plot omits complex results, so those give an incomplete picture of the  
situation.

Bobby

On Mon, 24 Sep 2007 03:21:14 -0500, sean_incali <sean_incali at yahoo.com> 
wrote:

> I have this equation I can plot using the following
>
> ClearAll["Global`*"]
>
>
> fun = (b c y - a y^ n + y^(1 + n))/(b + y^n);
> d = 0.1;
> c = 1;
> b = 10;
> a = 11;
> n = 5;
> ParametricPlot[{fun, (y)}, {y, 0, 1}, PlotRange -> All]
>
> I am plotting the function "fun" against values of y.
>
> When I do that I get a curvew that clearly shows that, for a given
> value of y, I should get two value for the function.
>
> How do I get what they are?
>
> If I just go..
>
> d = 0.1;
> c = 1;
> b = 10;
> a = 11;
> n = 2;
> y = 0.5;
>
> (b c y - a y^ n + y^(1 + n))/(b + y^n)
>
> I get
>
> 0.231707
>
> but starting at y=0 all the way to y= 0.6 or so, there are two values!
> (according to the graph)
>
> How do I get what those values are?
>
>
> Thanks for any input.
>
> sean
>
>
>



-- 

DrMajorBob at bigfoot.com


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