Proof that an even degree polynomial has a minimum

[B3NR4Y]

Homework Statement

Let $$p(x) = a_{2n} x^{2n} + ... + a_{1} x + a_{0} $$ be any polynomial of even degree.
If $$ a_{2n} > 0 $$ then p has a minimum value on R.

Homework Equations

We say f has a minimum value "m" on D, provided there exists an $$x_m \in D$$ such that
$$ f(x) \geq f(x_m) = m $$

for all x in D.

The Attempt at a Solution

I know I should prove that p(x) goes to infinity on both sides, but I'm not sure how to start doing that.
I can rewrite $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^{2n}}) $$

But I'm not sure how to prove that it goes to infinity. If I can do that I can use the definition of the minimum and the intermediate value theorem to prove that the minimum exists.

 

[axmls]

Unless I'm missing something here--as in, if this isn't an assumption you can make--it's readily apparent that every term as you've written the polynomial will drop out except $a_{2n}$ as $x \to \infty$.

 

[B3NR4Y]

Unless I'm missing something here--as in, if this isn't an assumption you can make--it's readily apparent that every term as you've written the polynomial will drop out except $a_{2n}$ as $x \to \infty$.

Unfortunately that's not an assumption I can make, I have to prove everything.

 

[SteamKing]

Homework Statement

Let $$p(x) = a_{2n} x^{2n} + ... + a_{1} x + a_{0} $$ be any polynomial of even degree.
If $$ a_{2n} > 0 $$ then p has a minimum value on R.

Homework Equations

We say f has a minimum value "m" on D, provided there exists an $$x_m \in D$$ such that
$$ f(x) \geq f(x_m) = m $$

for all x in D.

The Attempt at a Solution

I know I should prove that p(x) goes to infinity on both sides, but I'm not sure how to start doing that.
I can rewrite $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^{2n}}) $$

But I'm not sure how to prove that it goes to infinity. If I can do that I can use the definition of the minimum and the intermediate value theorem to prove that the minimum exists.

Remember, the limit of a sum is the sum of the limits of each individual term.

Can you show that a_2nx²ⁿ → ∞ as x → ∞ ? Then work your way thru the terms of p(x) until you reach a₀ .

 

[axmls]

Remember, the limit of a sum is the sum of the limits of each individual term.

Can you show that a_2nx²ⁿ → ∞ as x → ∞ ? Then work your way thru the terms of p(x) until you reach a₀ .

Here would be my issues with this:

1. can the OP assume said property of limits? (I presume he can--but you never know)

2. Are you sure this will work for negative infinity? After all, who's to say the odd-integer exponents don't drag down the even-integer exponent terms as $x \to -\infty$? That'd be another thing to prove for the $x < 0$ case.

 

[Mark44]

Homework Statement

Let $$p(x) = a_{2n} x^{2n} + ... + a_{1} x + a_{0} $$ be any polynomial of even degree.
If $$ a_{2n} > 0 $$ then p has a minimum value on R.

Homework Equations

We say f has a minimum value "m" on D, provided there exists an $$x_m \in D$$ such that
$$ f(x) \geq f(x_m) = m $$

for all x in D.

The Attempt at a Solution

I know I should prove that p(x) goes to infinity on both sides, but I'm not sure how to start doing that.
I can rewrite $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^{2n}}) $$

But I'm not sure how to prove that it goes to infinity. If I can do that I can use the definition of the minimum and the intermediate value theorem to prove that the minimum exists.

p'(x) is an odd-degree polynomial with a positive leading coefficient. It will have to have at least one x-intercept where p' changes sign (hence p changes from decreasing to increasing or vice versa). Do you have any theorems you can invoke about odd-degree polynomials?

 

[SteamKing]

Here would be my issues with this:

1. can the OP assume said property of limits? (I presume he can--but you never know)

2. Are you sure this will work for negative infinity? After all, who's to say the odd-integer exponents don't drag down the even-integer exponent terms as $x \to -\infty$? That'd be another thing to prove for the $x < 0$ case.

Beats me. Just trying to give the OP a nudge to do something. I usually write my name at the top of the paper when I get stumped.

 

[B3NR4Y]

Okay I think I am able to prove it using what SteamKing said, but I see what axmls means by some of the lower power terms may overcome the upper power terms. That's why I preferred writing it as $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^2n} ) $$ it seems to show that $$a_{2n}$$ is the prevalent term. The only thing that sucks is I can't go limit by limit.

 

[axmls]

Perhaps you could show as well that $\lim_{x \to \infty} \frac{1}{x^k} = 0$ first for the case $k=1$, then the rest of the terms follow easily from the squeeze theorem.

The only drawback would be if you can't assume, for instance, that the limit of the sum is the sum of the limits, or if you can't assume the squeeze theorem.

Those assumptions you can and can't make is what makes these difficult to work with.