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hbghlyj
Posted 2023-3-29 02:53
Last edited by hbghlyj 2023-3-29 09:452.
Let $\mathcal{F}$ be the set of continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that $$e^{f(x)}+f(x) \geq x+1, \: \forall x \in \mathbb{R}$$For $f \in \mathcal{F},$ let $$I(f)=\int_0^ef(x) dx$$Determine $\min_{f \in \mathcal{F}}I(f).$
AOPS
Note that $e^y + y$ is increasing and because of that for $x\in[0,e]$ the equation $e^y+y=x+1$ has unique solution with respect to $y$, thus identifying a function $y(x)\,,\, x\in [0,e]$. It means for any $f\in \mathcal{F}$ it holds $f(x)\leq y(x),x\in[0,e]$.
Hence, $I(f)$ takes its maximal value when $f=y(x)$. It remains to calculate $\int_0^e y(x)\,dx$.
Now, $y(x)$ is an increasing function for $x\in[0,e]$ taking values in $[0,1]$ and let $x(y)$ be its inverse when $x\in[0,1]$. We have:
$$\int_0^1 (e^y+y)\,dy = \int_0^1 (x(y)+1)\,dy \implies$$$$e-1+\frac{1}{2} = \int_0^1 x(y)\,dy +1 \implies e-1+\frac{1}{2} = e\cdot 1 -\int_0^e y(x)\,dx +1$$Hence $\int_0^e y(x)\,dx = \frac{3}{2}$.
3.
Let $f:[a,b] \to \mathbb{R}$ be an integrable function and $(a_n) \subset \mathbb{R}$ such that $a_n \to 0.$
$\textbf{a) }$ If $A= \{m \cdot a_n \mid m,n \in \mathbb{N}^* \},$ prove that every open interval of strictly positive real numbers contains elements from $A$.
$\textbf{b) }$ If, for any $n \in \mathbb{N}^*$ and for any $x,y \in [a,b]$ with $|x-y|=a_n,$ the inequality $\left| \int_x^yf(t)dt \right| \leq |x-y|$ is true, prove that $$\left| \int_x^y f(t)dt \right| \leq |x-y|, \: \forall x,y \in [a,b]$$
AOPS
The sequence must consists of infinitely many positive numbers.
(a)Let $(x,y)$ be an open, with $0< x<y$.Since the sequence converges to 0 with infinitely many positive numbers, hence there is $n$ such that,
$$2\le \frac{y-x}{a_n}$$Hence, there is some integer $m$ between $\frac{x}{a_n}< m<\frac{y}{a_n}$.We are done.$\blacksquare$
(b)Let $x,y\in [a,b]$ with $x<y$.Pick any $\varepsilon$ such that $y-x>\varepsilon>0$.By (a) we can find integer $i$ and some $a_n$ such that
$$y-x-\varepsilon <ia_n\le y-x$$Hence $x+ia_n\in(y-\varepsilon,y)$.Since $f$ is integrable there is $M\ge |f(x)|,\forall x\in [a,b]$.Now observe that,
$$|\int_x^y f(t) dt|= |\int_x^{x+ia_n} f(t) dt+\int_{i+a_n}^y f(t) dt|\le |\int_x^{x+ia_n} f(t) dt|+|\int_{i+a_n}^y f(t) dt|$$Now partitioning the interval $[x,x+ia_n]$ into $i$ intervals of length $a_n$ we get
$$|\int_x^{x+ia_n} f(t) dt|\le ia_n\le y-x$$Again, $|\int_{i+a_n}^y f(t) dt|\le M(y-ia_n)\le M\varepsilon$
Since $\varepsilon $ is arbitrary,we are done.$\blacksquare$
4.
For any $k \in \mathbb{Z},$ define $$F_k=X^4+2(1-k)X^2+(1+k)^2.$$Find all values $k \in \mathbb{Z}$ such that $F_k$ is irreducible over $\mathbb{Z}$ and reducible over $\mathbb{Z}_p,$ for any prime $p$.
AOPS
We first claim that $F_k$ is reducible over each $\mathbb{Z}_p$. Indeed, let $K$ be any such field. If $-1$ is a square in $K$, then $F_k = (X^2 - (k+1))^2 + 4X^2$ is clearly reducible over $K$. If $k$ is a square in $K$, then $F_k = (X^2 + (k+1))^2 - 4kX^2$ is clearly reducible over $K$. So now assume that $-1$ and $k$ are not squares in $K$. Then $-k$ is a square in $K$, say $-k = a^2$ for $a\in K$. Then $F_k = (X^2 + (1-k+2a))(X^2 + (1-k-2a))$ is reducible. This proves the claim.
So the part "such that $F_k$ is reducible over all $\mathbb{Z}_p$" is actually totally redundant... The question is merely "find all $k$ such that $F_k$ is irreducible over $\mathbb{Z}$". As noted above, if $k$ or $-k$ is a square in $\mathbb{Z}$, $F_k$ is reducible, so the answer is negative. Now assume that $k$ is not a perfect square or minus a perfect square in $\mathbb{Z}$. We claim that $F_k$ is irreducible over $\mathbb{Z}$ in that case.
To prove this point, first note that $F_k = (X^2 - (k+1))^2 + 4X^2$ can not have a real root, hence certainly not a linear factor over $\mathbb{Z}$. So if $F_k$ is reducible, it must be the product of two quadratic factors, say $F_k = (X^2+aX+b)(X^2-aX+c)$ with $a,b,c\in\mathbb{Z}$. Then $0 = a(b-c)$, $bc = (k+1)^2$ and $b+c-a^2 = 2(1-k)$. The first equation shows that there are two cases. If $a = 0$ then $b+c = 2(1-k)$ and $bc = (k+1)^2$, so the equation $T^2 + 2(k-1)T + (k+1)^2$ would have a solution in $\mathbb{Z}$. However, its discriminant $4(k-1)^2-4(k+1)^2 = -16k$ is not a square because $-k$ is (supposed to) not (be) a perfect square. The second case is the one in which $b=c$; then $b = c = \varepsilon (k+1)$ for some $\varepsilon\in\{-1,1\}$ and $a^2 = 2(b+k-1) \in \{-4,4k\}$, but neither $-4$ or $4k$ is a perfect square, so this is impossible as well. We conclude that $F_k$ is irreducible over $\mathbb{Z}$. |
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