roll theorem prove question..

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March 9th 2009, 11:11 AM
transgalactic

roll theorem prove question..

question :
suppose that to the quadratic equation x^2+px+q=0
has two roots $x_1<x_2$

prove that for
$<br /> \frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0<br />$
there are n different roots on the interval (x1,x2)
??

the prove:
by rolls theorem we have a point "c" on the interval of (x1,x2) for which

f'(c)=0

then they say
"that we can get another solution "g" on the interval of (x1,c) using rolls theorem"

but its not rolls theorem
we dont have two points for which there is a point "j" for which f'(j)=0

we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )

??
March 9th 2009, 11:11 PM
transgalactic

is it ok to use rolls theorem here??
March 10th 2009, 01:53 AM
transgalactic

ok
my problem is very simple

if i have a function f(x)
and i have a point "a" for which f(a)=0
and i have point "b" for which f'(b)=0
is the a point "c" between [a,b] for which f''(c)=0
??

(i got one oint of f(x) and the other on f'(x) they are not on the same level)
March 10th 2009, 03:20 AM
HallsofIvy

Quote:

Originally Posted by transgalactic

question :
suppose that to the quadratic equation x^2+px+q=0
has two roots $x_1<x_2$

so we can write $x^2+ px+ q= (x- x_1)(x- x_2)$ .

Quote:

prove that for
$<br /> \frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0<br />$
there are n different roots on the interval (x1,x2)
??

Write that as $\frac{d^n (x-x_1)^n(x-x_2)^n}{dx^n}$

Quote:

the prove:
by rolls theorem we have a point "c" on the interval of (x1,x2) for which

f'(c)=0

then they say
"that we can get another solution "g" on the interval of (x1,c) using rolls theorem"

but its not rolls theorem
we dont have two points for which there is a point "j" for which f'(j)=0we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )

??

They aren't referring to f now, they are applying Rolle's theorem to f'.
It should be obvious that for $f(x)= (x^2+ px+ q)^n= (x-x_1)^n(x-x_2)^n$ , f' is also 0 at $x_1$ and $x_2$ . In fact that is true for all deritives up to the n-1 derivative.
March 10th 2009, 03:36 AM
HallsofIvy

Quote:

Originally Posted by transgalactic

ok
my problem is very simple

if i have a function f(x)
and i have a point "a" for which f(a)=0
and i have point "b" for which f'(b)=0
is the a point "c" between [a,b] for which f''(c)=0
??

(i got one oint of f(x) and the other on f'(x) they are not on the same level)

Do you mean "is there a point "c" between a and b for which f"(c)= 0"?
Not necessarily. It depends on whether f'(a)= 0 also which is true for the polynomial in your first post.

For example if f(x)= (x- 1)(x- 2)= $x^2- 3x+ 2$ , f(1)= 0 and f(2)= 0 so there must exist "b" between 1 and 2 such that f'(b)= 2x- 3= 0. It is obvious that that point is b= 3/2 but f"(x)= 2 which is never 0.

But if f(x)= $(x^2- 3x+ 2)^2= (x-1)^2(x-2)^2$ , then $f'(x)= 2(x-1)(x-2)^2+ 2(x-1)^2(x-2)$ which is 0 at x= 1, x= 2 and at x= 3/2. Now applying Rolle's theorem to f', on the intervals from 1 to 3/2 and from 3/2 to 2 there must be a number b between 1 and 3/2 such that f"(b)= 0 and and a number c between 3/2 and 2 such that f"(c)= 0. That is, there are two values between 1 and 2 that make the second derivative 0 as the theorem says.
March 10th 2009, 04:49 AM
transgalactic

when you say
"on the intervals from 1 to 3/2 and from 3/2 to 1"
this is the same solution
you cant take the same interval but on the opposite direction
and say that its a different solution (its the same solution for the second derivative)

and when you say:
"there must be a number b between 0 and 3/2 such that f"(b)= 0 "

b need to be between the interval(between 1 to 3/2) not
"between 0 and 3/2"
??
March 10th 2009, 04:59 AM
transgalactic

if you want to apply rolls theorem on f'(x)
then we got a one solution for f''(x) between 1 and 3/2 and onther solution for f''(x) between 3/2 and 2

??
March 10th 2009, 05:37 AM
transgalactic

i cant understand what is the general rule.

in my solution they say "we have one point on f(a)=0
the other on f'(b) so there is a point c where f''(c)=0

that is not rolls theorem
??
March 10th 2009, 11:11 AM
HallsofIvy

Quote:

Originally Posted by transgalactic

when you say
"on the intervals from 1 to 3/2 and from 3/2 to 1"

Sorry, that was a typo. I meant "1 to 3/2 and 3/2 to 2".
I have edited my previous post.

Quote:

this is the same solution
you cant take the same interval but on the opposite direction
and say that its a different solution (its the same solution for the second derivative)

and when you say:
"there must be a number b between 0 and 3/2 such that f"(b)= 0 "

b need to be between the interval(between 1 to 3/2) not
"between 0 and 3/2"
??
March 10th 2009, 11:16 AM
HallsofIvy

Quote:

Originally Posted by transgalactic

i cant understand what is the general rule.

in my solution they say "we have one point on f(a)=0
the other on f'(b) so there is a point c where f''(c)=0

that is not rolls theorem
??

Are you certain that was what they said? it should be "we have one point on f'(a)=0 the other on f'(b) so there is a point c where f''(c)=0" which is Rolle's theorem applied to f' instead of f.
March 10th 2009, 11:21 AM
transgalactic

i am given that f(x)=x^2 + px + q
so i can present f(x) as f(x)=(x-x1)(x-x2)
so if we look at n=2 we get
f(x)=(x-x1)^2(x-x2)^2

f'(x)=2(x-x1)(x-x2)^2 + 2(x-x1)^2(x-x2)=2(x-x1)(x-x2)[x-x2+x-x1]=
=2(x-x1)(x-x2)[2x-x1-x2]

so x1 and x2 are solution on f'(x)
but this is a practical observation.
what do i need to say so for g(x)=[f(x)]^n
then x1,x2 are the roots till the n'th derivative??
March 10th 2009, 01:22 PM
HallsofIvy

Quote:

Originally Posted by transgalactic

i am given that f(x)=x^2 + px + q
so i can present f(x) as f(x)=(x-x1)(x-x2)
so if we look at n=2 we get
f(x)=(x-x1)^2(x-x2)^2

f'(x)=2(x-x1)(x-x2)^2 + 2(x-x1)^2(x-x2)=2(x-x1)(x-x2)[x-x2+x-x1]=
=2(x-x1)(x-x2)[2x-x1-x2]

so x1 and x2 are solution on f'(x)
but this is a practical observation.
what do i need to say so for g(x)=[f(x)]^n
then x1,x2 are the roots till the n'th derivative??

If f(x)= (x-x1)^n(x-x2)^n then f'(x= n(x-x1)^(n-1)(x-x2)^n+ n(x-x1)^n(x-x2)^(n-1). f"= n(n-1)(x-x1)^(n-2)(x-x2)^n+ 2n^2(x-x1)^(n-1)(x-x2)^(n-1)+ n(n-1)(x-x1)^n(x-x2)^(n-2), etc. That is, the ith derivative is a sum of terms each including factors of (x-x1) and (x-x2) to at least the n-i power. As long as i< n, each term will include at least one factor of each of (x-x1) and (x-x2) and so will be 0 at x= x1 and x= x2.
March 10th 2009, 10:45 PM
transgalactic

so in our example
x1 and x2 are solutions till the n-1
int the nth derivative x1 and x2 are not a solution