Let $f:\mathbb{R}\rightarrow \mathbb{R}$ a polynomial of a even $n$ degree, such that $0\leq f(x)$
let $g=f+f'+f''+\cdots+f^{(k)}$, prove $g$ has a global minimum in $\mathbb{R}$ when $k$ is the $k$-th derivative
How should I approach this?
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What is $k$? Any integer? – Clement C. Jul 26 '16 at 17:36
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@ClementC. edited – gbox Jul 26 '16 at 17:38
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My question is -- $k$ can be anything (the statement is to be proven for all $k\geq 0$)? – Clement C. Jul 26 '16 at 17:39
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there is no restriction on $k$ – gbox Jul 26 '16 at 17:40
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A start could be : if $f \geq 0$ for all $x$ then what can you say about coefficient of highest exponent monomial in f? – mathreadler Jul 26 '16 at 17:51
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1Also please try and be more careful when you write questions. It can be very off-putting for some users to answer and then have the question changed afterwards. – mathreadler Jul 26 '16 at 18:06
2 Answers
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when $f$ is polynomial, derivatives of $f$ are polynomial as well. Further $g$ is polynomial.
Because $f \gt 0$ and $f$ is even, $\lim_{x \to \pm \infty} f(x)= + \infty$. And because $f$ is the highest order polynomial among $f', f^{(2)},f^{(3)}$ etc. we can conclude $\lim_{x \to \pm \infty} g(x)= + \infty$. Thus, g has global minimum.
user115350
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Since the coefficient of the highest-order term is positive and the degree of $g$ is even we can choose $C$ big enough such that $g(z)>f(0)$ for all $|z| > C $ . Then since $g$ is continuous and $[-C,C]$ is compact, there is $ x\in [-C,C]$ such that $f(x) \leq f(y)$ for all $y\in [-C,C]$. Then $f(x)$ is a global minimum.
math635
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Sorry, you are right, I forgot to write that $g$ starts with the function itself – gbox Jul 26 '16 at 17:43
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g is even because f is even: the monomial with the highest power of g will be the monomial with highest power of f because the derivatives of f all have lower degrees... – math635 Jul 26 '16 at 18:46
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1no its not. I meant that the degree of the highest power is even. I edited the answer. – math635 Jul 26 '16 at 19:30