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Re: Root again
*To*: mathgroup at smc.vnet.net
*Subject*: [mg109129] Re: Root again
*From*: Scott Hemphill <hemphill at hemphills.net>
*Date*: Mon, 19 Apr 2010 04:08:21 -0400 (EDT)
*References*: <hps1an$6fv$1@smc.vnet.net> <hq0mn0$jtd$1@smc.vnet.net>
*Reply-to*: hemphill at alumni.caltech.edu
Scott Hemphill <hemphill at hemphills.net> writes:
[snip]
> Anyway, the roots aren't all real for any choice of t.
>
> If y = x^6 + t*x + 1, then D[y,x] == 6x^5 + t and D[y,x,x] == 30x^4.
> Since D[y,x,x] is non-negative, D[y,x] will be monotonically increasing.
> Since D[y,x] always has exactly one real zero, y will have exactly one
> minimum. It can therefore have zero, one, or two real roots.
>
>
> In[1]:= y=x^6+t*x+1
>
> 6
> Out[1]= 1 + t x + x
>
>
> We can find the minimum by solving for the location where the derivative
> is zero.
>
>
> In[2]:= Reduce[D[y,x]==0,x,Reals]
>
> Out[2]= x == Root[t + 6 #1 & , 1]
My cut and paste missed a line here. This should read:
5
Out[2]= x == Root[t + 6 #1 & , 1]
> We can find the critical values for t for which the minimum occurs at y==0.
>
>
> In[3]:= Reduce[(y/.Rule@@%)==0,t]
>
> -6 6
> Out[3]= t == ---- || t == ----
> 5/6 5/6
> 5 5
>
> In[4]:= N[%]
>
> Out[4]= t == -1.56919 || t == 1.56919
>
> So at these two values of t, y will have just one real root. In between
> them there will be no real roots (e.g. t==0 => y==x^6+1 has no real
> roots) and outside these two values of t there will be two real roots.
Scott
--
Scott Hemphill hemphill at alumni.caltech.edu
"This isn't flying. This is falling, with style." -- Buzz Lightyear
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