Does a polynomial exist such that $|P(x)| < a$ for some real $a$ and all real $x$?

Question

I'm pretty sure the answer is in the negative.

Can someone show me the proof?

$-x^2$ is never positive for all $x\in\Bbb R$. – Andrea Mori Aug 11 '12 at 12:17 — Andrea Mori, Aug 11 '12 at 12:17
I think he is talking about positive definite P(x) – Seyhmus Güngören Aug 11 '12 at 12:18 — Seyhmus Güngören, Aug 11 '12 at 12:18
yes that's what I meant, sorry for the earlier error. – hollow7 Aug 11 '12 at 12:18 — hollow7, Aug 11 '12 at 12:18

score 5 · Answer 1 · answered Aug 11 '12 at 12:21

5

Only constant polynomial will do the job. If $P$ has degree $d\geq 1$, assuming it WLOG monic, we have $|P(x)|\geq \frac{|x|^d}2$ for $|x|$ large enough, as $\lim_{x\to +\infty}\frac{P(x)}{x^d}=1$.

answered Aug 11 '12 at 12:21

Davide Giraudo

172,925

Andrea Mori · Accepted Answer · 2012-08-11T12:35:57.300

3

The answer is no. Write $$ P(x)=ax^n+\text{terms of lower degree} $$ with $a\neq0$ and $n\geq1$. Then $$ \lim_{x\to\infty}P(x)=\pm\infty $$ according to the sign of $a$. This shows that $|P(x)|$ is unbounded.

edited Aug 11 '12 at 12:35

answered Aug 11 '12 at 12:21

Andrea Mori

26,969

Except constant polynomials. – Michael Hardy Aug 11 '12 at 19:49
@MichaelHardy : of course. This is somewhat implicit in my answer since I specified $n\geq1$. Obviously constant polymonials are bounded. – Andrea Mori Aug 11 '12 at 20:20

Gerry Myerson · Answer 3 · 2012-08-11T12:23:43.830

1

(The question has been changed since the answer below was posted, so it no longer applies)

$P(x)=-x^2$ is less than 17 for all real $x$.

edited Aug 11 '12 at 12:23

answered Aug 11 '12 at 12:17

Gerry Myerson

179,216

Seyhmus Güngören · Answer 4 · 2012-08-11T12:38:11.087

0

I think there exist such if $n\rightarrow\infty$ where $n$ is the order of the polinomial. For example when you have a cosine and use the taylor series expansion then you will get a polinomial and this polinomial will be bounded by the absolute value. Since the polinomials are of not infinite degree then $|P(x)|$ can not be bounded.

edited Aug 11 '12 at 12:38

answered Aug 11 '12 at 12:23

Seyhmus Güngören

7,848

Taylor polynomials only give an approximation in some interval. – Andrea Mori Aug 11 '12 at 12:25
There are exact polinomial approximations for some bounded functions. And I think if we let the polinomial to be of infinite degree it is possible to claim that there exist a bound. http://en.wikipedia.org/wiki/Taylor_series – Seyhmus Güngören Aug 11 '12 at 12:28
A polynomial has finite degree. If it has infinite degree is not a polynomial. Your link reads "Taylor series". – Andrea Mori Aug 11 '12 at 12:31
Ok you are right so I must correct my post as when the degree goes to $\infty$ and else not. – Seyhmus Güngören Aug 11 '12 at 12:35
How does it look like now? I hope I wont get more downvotes since I am trying to be constructive. – Seyhmus Güngören Aug 11 '12 at 12:38

Does a polynomial exist such that $|P(x)| < a$ for some real $a$ and all real $x$?

4 Answers4