       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>

```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:= y=x^6+t*x+1
>
>                      6
>   Out= 1 + t x + x
>
>
> We can find the minimum by solving for the location where the derivative
> is zero.
>
>
>   In:= Reduce[D[y,x]==0,x,Reals]
>
>   Out= x == Root[t + 6 #1  & , 1]

My cut and paste missed a line here.  This should read:

5
Out= x == Root[t + 6 #1  & , 1]

> We can find the critical values for t for which the minimum occurs at y==0.
>
>
>   In:= Reduce[(y/.Rule@@%)==0,t]
>
>                 -6           6
>   Out= t == ---- || t == ----
>                 5/6          5/6
>                5            5
>
>   In:= N[%]
>
>   Out= 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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