Proving f(x) > 0 for All x: Analyzing f(x)

[ritwik06]

Homework Statement

$$f(x)=x^{12}-x^{9}+x^{4}-x+1$$

Find th minimum possible value of the function, and show that its an always increasing function.
Actually I want to prove that f(x)>0 for all x.

The Attempt at a Solution

Its difficult for me to factorize this.
$$dy/dx=12 x^{11} - 9 x^{8} + 4 x^{3} -1$$
How shall I find the required information from the derivative. Is there any other way?

 

[kreil]

Graphing the fcn with a calculator is the easiest way to solve this.

- f'(x) = 0 at x = 0.67460257...
- f'(x) monotonically increases, but is not always positive.
- f'(x) changes from - to + through the zero listed above (so f(x) achieves an absolute min at x = 0.6746...)
- This minimum is y = 0.5124...

It follows from the statements above that since f(x) has 1 extremum, a minimum, and since this minimum is positive that f(x) > 0 for all x.

 

[foxjwill]

I did this a little differently. We start by plugging two values of x, a and -a, where a>0. From this we have

$$f(-a)=a^{12}+a^9+a^4+a+1>0$$ and
$$f(a)=a^{12}-a^9+a^4-a+1$$​

Now, since $$a^{12}>a^9$$ and $$a^4>a$$, $$f(a)>0$$. Since f(-a) and f(0) are both obviously always going to be positive, we can conclude that $$\forall x \in \mathbb{R}, f(x)>0$$. Q.E.D.

 

[ritwik06]

Graphing the fcn with a calculator is the easiest way to solve this.

- f'(x) = 0 at x = 0.67460257...
- f'(x) monotonically increases, but is not always positive.
- f'(x) changes from - to + through the zero listed above (so f(x) achieves an absolute min at x = 0.6746...)
- This minimum is y = 0.5124...

It follows from the statements above that since f(x) has 1 extremum, a minimum, and since this minimum is positive that f(x) > 0 for all x.

You are right exactly! But the problem is friend that we are not allowed to use the calculator.

 

[ritwik06]

I did this a little differently. We start by plugging two values of x, a and -a, where a>0. From this we have

$$f(-a)=a^{12}+a^9+a^4+a+1>0$$ and
$$f(a)=a^{12}-a^9+a^4-a+1$$​

Now, since $$a^{12}>a^9$$ and $$a^4>a$$, $$f(a)>0$$. Since f(-a) and f(0) are both obviously always going to be positive, we can conclude that $$\forall x \in \mathbb{R}, f(x)>0$$. Q.E.D.

Thanks a lot foxjwill. But what u said is not true for 0<x<1
$$a^{12}>a^9$$ and $$a^4>a$$
These inequalities are not true for the interval x$$\in$$(0,1)

 

[matt grime]

Could you clarify what you mean by 'an always increasing function', since with my reasonable assumption of what that means, that polynomial is not an "always increasing function" on the real line, since the derivative at 0 is negative.

 

[Gib Z]

First of all, let us factor the function is this manner - $$f(x) = x^{12} - x^9 + x^4 - x +1 = x^9( x^3 -1) + x( x^3-1) + 1 = (x^9 + x) (x^3 -1) + 1 = x(x^8 + 1) (x-1)(x^2 +x +1) +1$$.

Up to that point, it should all be obvious steps. Now, let us take those 2 linear terms to make one quadratic, and also take the other quadratic, and complete the square on both of them:

$$f(x) =(x^2 - x)(x^2+x+1)(x^8 +1) + 1 = \left( \left(x- \frac{1}{2}\right)^2 -\frac{1}{4} \right) \left( \left(x+\frac{1}{2} \right)^2 +\frac{3}{4}\right) (x^8 +1) + 1$$

From there, expand the first two brackets as such:

$$f(x) = \left[ ( x-\frac{1}{2})^2 \left( (x+\frac{1}{2})^2 + \frac{3}{4} \right) - \frac{1}{4} \left( (x+\frac{1}{2})^2 + \frac{3}{4} \right) \right] (x^8+1) + 1$$
To get the minimum value of the function, we first must find the minimum value of the terms in the square brackets; this means we want the first term to be as small as possible, and our second completed square to be as large as possible. For the positive term, the minimum value of the perfect square in [0,1] is 1/4, whilst the minimum value of the completed square is 1. Hence, their product can never be less than 1/4.

For the negative term, we maximize the completed square in [0,1] to to see our negative term has at most a magnitude of 3/4. Hence, the minimum value of the square brackets in -1/2. Since we want to subtract as much as possible, maximize the degree 8 term in (0,1). That term will always be less than 2. Hence we will always subtract less than 1 from the remaining constant term if x (0,1), and if we let x=0 and x=1 into f(x) we see that f(x) =1. And so f(x) will always be greater than zero in [0,1]


EDIT: Not only did I express that horribly, I'm certain there has got to be a more elegant way than that.

 

[m_s_a]

Reference in rotation among the transactions
+,-,+,-,+

if f(r) =0 then r>0
r=root in f

 

[matt grime]

Are you trying to say that the roots of, say, x^3-x must all be strictly positive, m_s_a? Or perhaps you should explain your cryptic comment?

 

[m_s_a]

Yes .. Yes
This is what I want to say

 

[matt grime]

So, if x^3=x, then x must be strictly positive? Even though (-1)^3=-1 and 0^3=0? And indeed for any polynomial with zero constant term 0 is always a root irrespective of signs.

 

[m_s_a]

So, if x^3=x, then x must be strictly positive? Even though (-1)^3=-1 and 0^3=0? And indeed for any polynomial with zero constant term 0 is always a root irrespective of signs.

Challenge
write Polynomail
More than double the rotation
an=/=0
r=0 or A negative number

 

[matt grime]

I'm afraid I cannot understand what you've written. That's why I asked you to clarify what you mean. You really need to define 'rotation' and then '(more than) double rotation'. One can only guess that you're referring to the signs of the coefficients changing, and as I pointed out to you something like

x^6 - x^5 + x^4 - x^3 + x^2 - x

has a root at x=0, which seems to contradict what you *might* be saying.

 

[m_s_a]

I'm afraid I cannot understand what you've written. That's why I asked you to clarify what you mean. You really need to define 'rotation' and then '(more than) double rotation'. One can only guess that you're referring to the signs of the coefficients changing, and as I pointed out to you something like

x^6 - x^5 + x^4 - x^3 + x^2 - x

has a root at x=0, which seems to contradict what you *might* be saying.

an=/=0
Because
Rotation does not appear in a double -

 

[m_s_a]

Try
Term is not defined
Time is Open

 

[ritwik06]

First of all, let us factor the function is this manner - $$f(x) = x^{12} - x^9 + x^4 - x +1 = x^9( x^3 -1) + x( x^3-1) + 1 = (x^9 + x) (x^3 -1) + 1 = x(x^8 + 1) (x-1)(x^2 +x +1) +1$$.

Up to that point, it should all be obvious steps. Now, let us take those 2 linear terms to make one quadratic, and also take the other quadratic, and complete the square on both of them:

$$f(x) =(x^2 - x)(x^2+x+1)(x^8 +1) + 1 = \left( \left(x- \frac{1}{2}\right)^2 -\frac{1}{4} \right) \left( \left(x+\frac{1}{2} \right)^2 +\frac{3}{4}\right) (x^8 +1) + 1$$

From there, expand the first two brackets as such:

$$f(x) = \left[ ( x-\frac{1}{2})^2 \left( (x+\frac{1}{2})^2 + \frac{3}{4} \right) - \frac{1}{4} \left( (x+\frac{1}{2})^2 + \frac{3}{4} \right) \right] (x^8+1) + 1$$
To get the minimum value of the function, we first must find the minimum value of the terms in the square brackets; this means we want the first term to be as small as possible, and our second completed square to be as large as possible. For the positive term, the minimum value of the perfect square in [0,1] is 1/4, whilst the minimum value of the completed square is 1. Hence, their product can never be less than 1/4.

For the negative term, we maximize the completed square in [0,1] to to see our negative term has at most a magnitude of 3/4. Hence, the minimum value of the square brackets in -1/2. Since we want to subtract as much as possible, maximize the degree 8 term in (0,1). That term will always be less than 2. Hence we will always subtract less than 1 from the remaining constant term if x (0,1), and if we let x=0 and x=1 into f(x) we see that f(x) =1. And so f(x) will always be greater than zero in [0,1]


EDIT: Not only did I express that horribly, I'm certain there has got to be a more elegant way than that.

I think U r wrong there. The terms inside the square bracket:
the minimum of first is 0
the maximum of the term to be subtracted is 3/4
and the max value of $$x^{8}+1$$ is two
which on multiplication gives< -1.
SO I am still having a problem with this.

 

[matt grime]

Does anyone know what m_s_a is on about?

Since you (m_s_a) can't be explain yourself I'm going to stop trying to understand.

 

[ritwik06]

Try
Term is not defined
Time is Open

an=/=0
Because
Rotation does not appear in a double -

I'm afraid I cannot understand what you've written. That's why I asked you to clarify what you mean. You really need to define 'rotation' and then '(more than) double rotation'. One can only guess that you're referring to the signs of the coefficients changing, and as I pointed out to you something like

x^6 - x^5 + x^4 - x^3 + x^2 - x

has a root at x=0, which seems to contradict what you *might* be saying.

Challenge
write Polynomail
More than double the rotation
an=/=0
r=0 or A negative number

So, if x^3=x, then x must be strictly positive? Even though (-1)^3=-1 and 0^3=0? And indeed for any polynomial with zero constant term 0 is always a root irrespective of signs.

Yes .. Yes
This is what I want to say

Are you trying to say that the roots of, say, x^3-x must all be strictly positive, m_s_a? Or perhaps you should explain your cryptic comment?

Reference in rotation among the transactions
+,-,+,-,+

if f(r) =0 then r>0
r=root in f

Hey guys! I am unable to understand a word u say. Please be more explicit, and don't deviate from the original question :)
regards,
Ritwik

 

[matt grime]

I believe that m_s_a was attempting to shed some light onto something... I have no idea what though.

 

[Gib Z]

Ahh God damn it.

$$f(x) = (x^4 - x^9) + (-x + x^{12} ) + 1$$.