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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
ISI 2019 : Problem #2
integrated_JRC   40
N a few seconds ago by kamatadu
Source: I.S.I. 2019
Let $f:(0,\infty)\to\mathbb{R}$ be defined by $$f(x)=\lim_{n\to\infty}\cos^n\bigg(\frac{1}{n^x}\bigg)$$(a) Show that $f$ has exactly one point of discontinuity.
(b) Evaluate $f$ at its point of discontinuity.
40 replies
1 viewing
integrated_JRC
May 5, 2019
kamatadu
a few seconds ago
Putnam 1958 November A7
sqrtX   1
N an hour ago by centslordm
Source: Putnam 1958 November
Let $a$ and $b$ be relatively prime positive integers, $b$ even. For each positive integer $q$, let $p=p(q)$ be chosen so that
$$ \left| \frac{p}{q} - \frac{a}{b}  \right|$$is a minimum. Prove that
$$ \lim_{n \to \infty} \sum_{q=1 }^{n} \frac{ q\left| \frac{p}{q} - \frac{a}{b}  \right|}{n} = \frac{1}{4}.$$
1 reply
sqrtX
Jul 19, 2022
centslordm
an hour ago
Putnam 1958 November B7
sqrtX   5
N 2 hours ago by centslordm
Source: Putnam 1958 November
Let $a_1 ,a_2 ,\ldots, a_n$ be a permutation of the integers $1,2,\ldots, n.$ Call $a_i$ a big integer if $a_i >a_j$ for all $i<j.$ Find the mean number of big integers over all permutations on the first $n$ postive integers.
5 replies
sqrtX
Jul 19, 2022
centslordm
2 hours ago
System of two matrices of the same rank
Assassino9931   3
N 2 hours ago by RobertRogo
Source: Vojtech Jarnik IMC 2025, Category II, P2
Let $A,B$ be two $n\times n$ complex matrices of the same rank, and let $k$ be a positive integer. Prove that $A^{k+1}B^k = A$ if and only if $B^{k+1}A^k = B$.
3 replies
Assassino9931
Today at 1:02 AM
RobertRogo
2 hours ago
No more topics!
polynomial with real coefficients
Peter   9
N Apr 15, 2025 by Rohit-2006
Source: IMC 1998 day 1 problem 5
Let $P$ be a polynomial of degree $n$ with only real zeros and real coefficients.
Prove that for every real $x$ we have $(n-1)(P'(x))^2\ge nP(x)P''(x)$. When does equality occur?
9 replies
Peter
Nov 1, 2005
Rohit-2006
Apr 15, 2025
polynomial with real coefficients
G H J
Source: IMC 1998 day 1 problem 5
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Peter
3615 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $P$ be a polynomial of degree $n$ with only real zeros and real coefficients.
Prove that for every real $x$ we have $(n-1)(P'(x))^2\ge nP(x)P''(x)$. When does equality occur?
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Diogene
808 posts
#2 • 3 Y
Y by dinamometre1123, Adventure10, Mango247
I will proceed by induction. For $n=1$ that's manifestly true. Let $P$ be polynomial of degree $n+1$ , so
$P(x)=u(x)Q(x)$ where $u(x)$ is a monic real polynomial of degree $1$ and $Q(x)$ is a polynomial of degree $n$ with only real roots and real coefficients.
$P(x)=u(x)Q(x) \Longrightarrow P'= Q+uQ'\Longrightarrow P'' = 2Q' + uQ''$ , where $(n-1)(Q')^2\geq nQQ''$ , so :

$n(P')^2-(n+1)PP'' = nQ^2-2uQQ'+nu^2(Q')^2-(n+1)u^2QQ''\geq$$nQ^2-2uQQ' + \frac{u^2(Q')^2}n = (Q\sqrt n - \frac{uQ'}{\sqrt n})^2 \geq 0$... (q.e.d)
:cool:
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Diogene
808 posts
#3 • 3 Y
Y by Adventure10, Mango247, Rohit-2006
Sorry, I forgot the equality case ! Remember : $u(x)$ is a monic real polynomial , $u(x) = x-a$
$n(P')^2-(n+1)PP'' = 0 \Longrightarrow Q\sqrt n - \frac{uQ'}{\sqrt n} = 0 \Longrightarrow \frac{Q'}{Q} = \frac nu$$\Longrightarrow Q(x)= cu^n_{(x)}\Longrightarrow P(x)=c u^{n+1}_{(x)}$
Also, It's easy to verify that the equality holds for $P(x)=c u^{n+1}_{(x)}$
:cool:
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bandrak
17 posts
#4 • 2 Y
Y by Davrbek, Adventure10
without induction, by the condition we have that
\[ P = b\prod^n_{i=1} (x-a_i)\]
Suppose $ x$ is not a root of $ P$ and Let $ c_i = \frac{1}{x-a_i}$, then
\[ P' = P (c_1 + c_2 + \ldots + c_n), P'' = P (\sum_{i \neq j} 2c_i c_j)\]
The problem thus becomes
\[ (n-1)(c_1 + c_2 + \ldots + c_n)^2 \ge 2n (\sum_{i \neq j}c_i c_j)\]
factoring gives
\[ \sum_{i \neq j} (c_i - c_j)^2 \ge 0\]
Which is trivially true. Also the equality holds iff all the $ c_i$ are equal, which means all the $ a_i$ are equal and $ P=b(x-a)^n$
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chakrabortyahan
380 posts
#5
Y by
Another solution
Let $\alpha_1,\alpha_2,...\alpha_n$ be the zeroes (non-necessarily distinct) of $P$. The given inequality holds trivially for $\alpha_i's$. So we assume that $x\neq \alpha_i$ in the latter part of the solution.
Now note that $P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$
Now by C-S inequality ,
$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$
Now note that $(i)$ can be re-written as $-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$
Hence proved as $P(x)^2> 0 $ for all $x\neq \alpha_i$
And the equality holds iff all $\alpha_i$ s are equal (equality case of Cauchy -Schewarz)
$\blacksquare\smiley$
This post has been edited 1 time. Last edited by chakrabortyahan, May 5, 2024, 6:22 PM
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Moubinool
5567 posts
#6
Y by
This problem was used for oral examination ENS PSLR in 2023
problem 23 oral PSLR
https://www.rms-math.com/exos-etoiles-2023-site.pdf
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mqoi_KOLA
85 posts
#7
Y by
chakrabortyahan wrote:
Another solution
Let $\alpha_1,\alpha_2,...\alpha_n$ be the zeroes (non-necessarily distinct) of $P$. The given inequality holds trivially for $\alpha_i's$. So we assume that $x\neq \alpha_i$ in the latter part of the solution.
Now note that $P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$
Now by C-S inequality ,
$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$
Now note that $(i)$ can be re-written as $-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$
Hence proved as $P(x)^2> 0 $ for all $x\neq \alpha_i$
And the equality holds iff all $\alpha_i$ s are equal (equality case of Cauchy -Schewarz)
$\blacksquare\smiley$


$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$
why did you have modulus here in the thing between? CS inequality is true for all reals right
This post has been edited 1 time. Last edited by mqoi_KOLA, Mar 24, 2025, 11:39 AM
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quasar_lord
241 posts
#8
Y by
@above $\alpha_i$ can be complex ig?
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mqoi_KOLA
85 posts
#9
Y by
quasar_lord wrote:
@above $\alpha_i$ can be complex ig?

its given in q P has all real zeroes
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Rohit-2006
238 posts
#10
Y by
chakrabortyahan wrote:
Another solution
Let $\alpha_1,\alpha_2,...\alpha_n$ be the zeroes (non-necessarily distinct) of $P$. The given inequality holds trivially for $\alpha_i's$. So we assume that $x\neq \alpha_i$ in the latter part of the solution.
Now note that $P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$
Now by C-S inequality ,
$$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$$
Now note that $(i)$ can be re-written as $-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$
Hence proved as $P(x)^2> 0 $ for all $x\neq \alpha_i$
And the equality holds iff all $\alpha_i$ s are equal (equality case of Cauchy -Schewarz)
$\blacksquare\smiley$

Dada nvm....I don't think it's a complete solution because in the question said that for all $x$ but you are excluding the roots.....this is the first problem
The second problem is you are only told that degree is $n$ so I can consider that $n=0$ but in that case if it's a constant polynomial then it has infinite roots...that is basically for all real $x$....then your statement is totally wrong....I don't think the $(\frac{u}{v})'$ idea is good for admitting all values of $x$ in real.....so if possible please post another solution....we are waiting for that
This post has been edited 1 time. Last edited by Rohit-2006, Apr 15, 2025, 5:50 PM
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