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