Prove that the following polynomial has a minimum in R p(x)=x^4+a_3 x^3+a_2 x^2+a_1x+a_0.

Cale Terrell 2022-10-24 Answered
Proof that a polynomial has a minimum in R
I have to prove to following statement and I am having a really hard time here.
There it is:
Prove that the following polynomial has a minimum in R
p ( x ) = x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0
I tried to make the following proof, but I am stuck:
The polynomial can be shown like this:
x 4 ( 1 + a 3 x + a 2 x 2 + a 1 x 3 + a 0 x 4 )
Let
g ( x ) = ( 1 + a 3 x + a 2 x 2 + a 1 x 3 + a 0 x 4 )
So we can now write the polynomial like
p ( x ) = x 4 g ( x )
Since x , a R we can say that lim x = lim x = 1 (from limit arithmetic).
We know that it is always the case that x 4 > 0, so the sign of
p ( x ) = x 4 g ( x )
is determined by g(x). For certain values, g ( x ) < 0 so we can say that there is some a , b R such that g ( a ) < 0 and g ( b ) > 0.
p(x) is continouos everywhere, certainly in [a,b], so from the Extreme Value Theorem, the function has a minimum in [a,b].
There are two problems in my proof:
I can't find any value in which g(x) is negative, only if there is some a which is negative. What if a is always positive?
If the first problem is solved, I managed to prove the statement for some [a,b]. Is it just enough? It seems to me that it isn't.
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Answers (2)

pararevisarii
Answered 2022-10-25 Author has 9 answers
Step 1
The claim itself is true more generally for any polynomial of even degree with positive leading coefficient.
Step 2
We do not need that g ( x ) < 0 for some x and in fact such x may not exist. What we need is that lim x ± g ( x ) = 1 implies that there exists M R such that | x | > M implies | g ( x ) 1 | < 1 2 (so that g ( x ) > 1 2 ). Then for | x | > M we conclude f ( x ) > 1 2 x 4 > 1 2 M 4 . We may assume wlog. that 1 2 M 4 > a 0 = f ( 0 ). Then the minimum the continuous function has on [-M,M] is in fact a global minimum.
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Chaim Ferguson
Answered 2022-10-26 Author has 4 answers
Step 1
You can't just write f ( x ) = x 4 g ( x ) because f(0) is defined and g(0) isn't. Instead, you can simply prove
lim x ± f ( x ) =
(because the x 4 term dominates all others in the limit)
With that settled, by the definition of these limits you can find some R > 0 such that
f ( x ) > a 0 x : | x | > R
(the > a 0 is arbitrary here, any constant works as long as f attains it inside [-R,R]; also, R need not be the optimal choice) Now look at f | [ R , R ] . This is a continuous function on a closed and bounded set. You should have a theorem giving you that f | [ R , R ] attains a minimum.
Since 0 [ R , R ], we know that this minimum is at most f ( 0 ) = a 0 , so f doesn't attain smaller values outside of [-R,R] by construction of R.
Step 2
This proves
min x R f ( x ) = min x [ R , R ] f ( x ) R
exists.
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