Proving Multiple Roots of a Function Using Roll's Theorem: A Scientific Analysis

[transgalactic]

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

prove that for
$$\frac{\mathrm{d^n} }{\mathrm{d} x^n}[(x^2+px+q)^n]=0$$
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 don't 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 )

??

 

[Mark44]

I'm guessing that no one has responded because you haven't given enough information.

the prove:
by rolls theorem we have a point "c" on the interval of (x1,x2) for which
f'(c)=0

That's "proof." A proof is the mathematical statements you write to show that some conclusion is reached when you start with a certain hypothesis. To prove something, you come up with those statements. A proof is a thing; proving it is an action.

Anyway, Rolle's Theorem does guarantee that there is a number c in (x1, x2) for which f'(c) = 0. I'm assuming that f(x) = x^2 + px + q.

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

Another solution to what? If f(x) = x^2 + px + q, there is only one solution to f'(x) = 0. You need to be specific about what equation g is a solution to.

Many of your posts come from questions you are asking as you read someone else's proof. If you don't include all of the preceding steps, it is difficult or impossible for us to explain how the author of the proof can make some statement.

Since the statement you're trying to prove (or whose proof you're trying to understand - this is not clear) involves the n-th derivative of a function raised to the n-th power, the most natural approach would seem to be a proof by induction, but there is no mention in your post that this is what is happening.

 

[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)

 

[Mark44]

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
??

No, not in general. Here's a counterexample.
f(x) = x^2 - 1
f'(x) = 2x
f''(x) = 2
f(-1) = 0 and f'(0) = 0 so a = -1 and b = 0 and the interval is [-1, 0].
There is no number in the interval [-1, 0] for which f''(x) = 0. In fact, for this function, there is no number anywhere for which f''(x) = 0.

 

[transgalactic]

they say it in the proof .
they say that we got a point f'(c)=0 so we check [x1,c]
and between them there is another point "g" for which f''(g)=0
and they say that we keep doing then till the n'th derivative

you said that its not working for all functions
what conditions do i need for this to work??
i thought that for this stuff to work
for example:
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
and if f'(a)=0
is the a point "c" between [a,b] for which f''(c)=0

but i can't see that they mention that each new solution is a solution to its anti derivative too??

 

[Mark44]

they say it in the proof .
they say that we got a point f'(c)=0 so we check [x1,c]
and between them there is another point "g" for which f''(g)=0
and they say that we keep doing then till the n'th derivative

you said that its not working for all functions
what conditions do i need for this to work??

Probably the stated conditions of the problem.

i thought that for this stuff to work
for example:
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
and if f'(a)=0
is the a point "c" between [a,b] for which f''(c)=0

My counterexample shows that this is not true for all functions.

but i can't see that they mention that each new solution is a solution to its anti derivative too??

"each new solution is a solution to its antiderivate" - this doesn't make any sense. Also, I asked in post 2 what you meant by "each new solution." Whenever you talk about a solution, it's always relative to some equation or inequality or a system of equations or inequalities. If I tell you that x = 6 is a solution, I'm not telling you anything meaningful if I don't tell you what this is a solution to.

You said in your first post that they say "we can get another solution "g" on the interval of (x1,c) using rolls theorem." What equation is g a solution to?

 

[transgalactic]

"You said in your first post that they say "we can get another solution "g" on the interval of (x1,c) using rolls theorem." What equation is g a solution to?"

i ment f''(g)=0

g is solution for the second derivative

 

[transgalactic]

the trick that they are doing is taking a solution of a function on n'th derivative
and a solution of the same function on the n+1 derivative

and they say that using rolls theorem we have a solution on n+2 derivative of this function on the interval of the n'th and n+1 x value solution

i can't understand how it could happen
??

 

[Mark44]

Here's what I would do:
You're given that x^2 + px + q = 0 has two roots, x1 and x2, where x1 < x2.

Show that the equation d/dx[(x^2 + px + q)^1] = 0 has 1 root in the interval (x1, x2).
Show that the equation d^2/dx^2[(x^2 + px + q)^1] = 0 has 2 different roots in the interval (x1, x2).

If you can follow the logic of the proof you're reading for n = 1 and n = 2, that would be a good start at following what they're doing for an arbitrary n.

 

[transgalactic]

ok if i am given that x^2 + px + q = 0
and it has two roots f(x1)=f(x2)=0

so between them we have point 'c' for which f'(c)=0
of f'(x) i have only point c
there is no way to get a solution on f''(x)

??

for what condition x1, and x2 are solution on every derivative
??

 

[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??

 

[Mark44]

ok if i am given that x^2 + px + q = 0
and it has two roots f(x1)=f(x2)=0

so between them we have point 'c' for which f'(c)=0

OK, so you have verified that the original statement is true for n = 1, namely that the equation d/dx[(x^2 + px + q)^1] = 0 has 1 root in the interval (x1, x2). You're done with that part.

of f'(x) i have only point c
there is no way to get a solution on f''(x)

??

for what condition x1, and x2 are solution on every derivative
??

 

[Mark44]

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

Don't use f again, since you have already defined it above as f(x) = x^2 + px + q. You will confuse any readers, and yourself if you come along and redefine it as some different function. Use another letter, say g. (But don't also use g as a number in an interval.)

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)

So x1 and x2 are solutions of g'(x) = 0. (I changed the name of the function.)
Did you not notice that there is another solution to the equation g'(x) = 0? It's also possible to see exactly where this solution is in relation to x1 and x2.

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??