Prove that there exists an integer greater than x such that any polynomial f(x) will be strictly non-negative and get large?I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as followsSuppose f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 0 is a polynomial of degree n &gt; 0. Then there exists an integer k such that if x &gt; k then f ( x ) &gt; 0.I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

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Prove that there exists an integer greater than x such that any polynomial f(x) will be strictly non-negative and get large?I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as followsSuppose   f  (  x  )  =      a    n        x    n    +      a          n      −      1            x          n      −      1        +  ⋯  +      a    0   is a polynomial of degree   n  &amp;gt;  0. Then there exists an integer k such that if   x  &amp;gt;  k then   f  (  x  )  &amp;gt;  0.I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

metodikkf6z · Accepted Answer

Step 1One should note that the field of interest here (presumably) is R, so that f(x) is a real polynomial. Notice first that for       a    n    &amp;lt;  0, this fails completely, because we can just choose   f  (  x  )  =  −  x and it does not have the property.Step 2If we assume that       a    n    &amp;gt;  0, then write            f      (      x      )              x      n        =      a    n    +      a          n      −      1            x          −      1        +  ⋯  +      a    1        x          1      −      n        +      a    0        x          −      n      so that       lim          x      →      ∞                  f      (      x      )              x      n        =      a    n    &amp;gt;  0. This implies that       lim          x      →      ∞        f  (  x  )  =  ∞, so that   f  (  x  )  &amp;gt;  0 for   x  ≥  N for some   N  ∈      R

Owen Mathis · Accepted Answer

Step 1It is obvious that the polynomial cannot be written as a product of more than n linear polynomials, possibly with complex coefficients (for otherwise, it should have a degree greater than n). Thus, f(x) does not vanish at more than n points on the real line. Hence, there is some real x′ so that for all   x  &amp;gt;      x    ′    , we must have   f  (  x  )  ≠  0. By continuity (and the fact that   f  (  x  )  →  +  ∞ as   x  →  +  ∞ - factor out       x    n   to see this, recalling that       a    n    &amp;gt;  0), it follows that   f  (  x  )  &amp;lt;  0 just before the last zero, so that just after this   f  (  x  )  &amp;gt;  0..Step 2Taking x′ to be sufficiently close to, but after, this last zero, and recalling after x′ the polynomial retains constant sign, it follows that it must be positive for all   x  &amp;gt;      x    ′    . Finally, by the principle of Archimedes you can find some positive integer greater than this x′.

Prove that there exists an integer greater than x such that any polynomial f(x) will be strictly non-negative and get large?

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