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#2 2012-11-22 19:50:58
- bobbym
- Administrator
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Re: real roots
Hi;
The first thing you can do is use a Cauchy bound to determine that all real roots would have to be in the closed interval [-4,4].
Another improved bound proves the real roots if they exist are between [-2,2].
Interesting but non essential for the following method.
Form a Sturm chain:
Substitute two endpoints of of -∞ and ∞ into x.
You will get the 7 x 2 matrix
Count the number of sign changes in the first column and subtract the number of sign changes in the second column.
3 - 3 = 0
There are no real roots!
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#3 2012-11-22 19:53:32
- bob bundy
- Moderator
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Re: real roots
hi jacks,
I thought I'd start by looking at the graph y = F(x)
It has a single minimum as you can see and y > 0 for all x.
That's enough to show it has no real roots.
But a graph alone doesn't constitute a proof as it relies on a tool and doesn't show all values of x.
So one approach would be to prove that the graph truely has those properties.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#4 2012-11-22 20:07:56
- zetafunc.
- Guest
Re: real roots
bobbym wrote:
Hi;
The first thing you can do is use a Cauchy bound to determine that all real roots would have to be in the closed interval [-4,4].
How did you get that? With Cauchy's bound I am getting it in [-7,7]... since
where c is a root.
#6 2012-11-22 20:18:55
- bobbym
- Administrator
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Re: real roots
Hi;
There are many variants of the Cauchy bound. I am using 2 of them:
http://en.wikipedia.org/wiki/Sturm%27s_theorem
and
http://www.mathsisfun.com/algebra/polyn … zeros.html
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#8 2012-11-25 14:39:56
- bobbym
- Administrator
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Re: real roots
Hi jacks;
I forgot that idea. That is best.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#9 2012-11-25 15:52:17
- noelevans
- Full Member
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Re: real roots
Hi! 
Or if we evaluate f(-1) by the remainder theorem the signs in the quotient followed by the
remainder strictly alternate in sign: + - + - + - + so there are no negative roots less than or
equal to -1. (Lower bound theorem)
Similarly if we evaluate f(1) by the remainder theorem the signs in the quotient followed by
the remainder are all positive. So there are no positive roots greater than or equal to 1.
(Upper bound theorem)
If |x|<=1 we have x+3>=2. Also the even powered terms cannot be negative. Hence the
complete sum is >= 2 for all x in [-1,1].
Hence f(x)=0 has no real roots, and in fact is always positive since if there were a value a outside
of [-1,1] for which f(a)<0 then the intermediate value theorem would guarantee a root between
either 1 and a if a>1 or between -1 and a if a<-1.
Or more elegantly along the lines that Jack pointed out: Completing the square on x^2+x+3 we get
x^2+x+3 = (x+1/2)^2 - 1/4 + 3 = (x+1/2)^2 + 11/4 which has minimum 11/4.
Also x^6 + x^4 is always >= zero. Hence f(x) >= 11/4 (actually strictly >) for all x in the reals.
f(-1/2) = 181/65 = 2.828125.
Have a very blessed day!
Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.