UFM Additional Further Pure

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We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Demonstrate that, for all $x \in ] - 1,1 [$ and all $q \in \mathbb { N } ^ { * }$, we have $$( 1 - x ) ^ { - q / 2 } = \sum _ { p = 0 } ^ { + \infty } H _ { p } \left( \frac { q } { 2 } + p - 1 \right) x ^ { p }$$
We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Deduce $$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = 1 + \sum _ { i = 0 } ^ { + \infty } \left( \sum _ { j = 0 } ^ { + \infty } a _ { i , j } ( x ) \right) \right.$$ where we have set $$\forall ( i , j ) \in \mathbb { N } ^ { 2 } , \quad a _ { i , j } ( x ) = \frac { ( - 1 ) ^ { i + 1 } } { ( i + 1 ) ! } H _ { j } \left( \frac { i - 1 } { 2 } + j \right) x ^ { i + j + 1 }$$
We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Demonstrate that, for all $x \in ] - 1,1 \left[ , \sum _ { i = 0 } ^ { + \infty } \left( \sum _ { j = 0 } ^ { + \infty } \left| a _ { i , j } ( x ) \right| \right) = \exp \left( \frac { | x | } { \sqrt { 1 - | x | } } \right) - 1 \right.$.
We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Use the results admitted in the preamble to establish the equality $$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } \right.$$ where $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Using the Cauchy product of power series, deduce that, for all integers $n$ and all real numbers $\alpha$ and $\beta$, $$L_{n}(\alpha + \beta) = \sum_{k=0}^{n} \binom{n}{k} L_{k}(\alpha) L_{n-k}(\beta).$$
grandes-ecoles 2019 Q5 Evaluation of a Finite or Infinite Sum View
For $x \in ]-1,1[$, give the value of the sum of the power series $\sum_{p=1}^{+\infty} x^{p}$ as well as that of its derivative.
Prove by induction that, for all integers $n \in \mathbb{N}^{*}$, there exists a unique polynomial $R_{n} \in \mathbb{R}_{n}[X]$ such that, for all $x \in ]-1,1[$, $$\sum_{p=1}^{+\infty} p^{n} x^{p} = \frac{R_{n}(x)}{(1-x)^{n+1}}$$
In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have $\operatorname{card}(\Omega_{n}) = n!$. For $0 < u < v$ and $|x|$ sufficiently small, $$H(x,u,v) = \sum_{n=0}^{+\infty} \frac{x^{n}}{n!} \left( \sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p} \right)$$
Using question 6, justify that, for all integers $n$ and all $u$ and $v$ such that $0 < u < v$, the sum $$\sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p}$$ is a polynomial function of $u$ and $v$.
In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have for all integers $n$ and all $t \in ]0,1[$, $$g_{n}(t) = \frac{1}{n!} \sum_{p=1}^{+\infty} p^{n} t^{p} (1-t)^{n+1}.$$ Fix an integer $n \geqslant 2$.
Show that $\sum_{p=n+1}^{+\infty} p^{n} t^{p} (1-t)^{n+1} \underset{t \rightarrow 0^{+}}{=} O(t^{n+1})$.
In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have for all integers $n$ and all $t \in ]0,1[$, $$g_{n}(t) = \frac{1}{n!} \sum_{p=1}^{+\infty} p^{n} t^{p} (1-t)^{n+1}.$$ Fix an integer $n \geqslant 2$.
Using the result of question 30 and by expanding $(1-t)^{n+1}$, determine the Taylor expansion of $g_{n}$ to order $n$ at 0.
In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), fix an integer $n \geqslant 2$.
Using the Taylor expansion of $g_{n}$ to order $n$ at 0, deduce that, for all $m$ in $\llbracket 1, n \rrbracket$, $$P(X_{n} = m) = \frac{1}{n!} \sum_{k=0}^{m-1} (-1)^{k} \binom{n+1}{k} (m-k)^{n}.$$
Recall Stirling's formula, then determine a real number $c > 0$ such that $$\binom { 2 n } { n } \underset { n \rightarrow + \infty } { \sim } c \frac { 4 ^ { n } } { \sqrt { n } }$$
If $\alpha$ is an element of $]0,1[$, show, for example by using a series-integral comparison, that $$\sum _ { k = 1 } ^ { n } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { n ^ { 1 - \alpha } } { 1 - \alpha }$$ If $\alpha$ is an element of $]1 , + \infty[$, show similarly that $$\sum _ { k = n + 1 } ^ { + \infty } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { ( \alpha - 1 ) n ^ { \alpha - 1 } }$$
grandes-ecoles 2020 Q4 Prove an Integral Inequality or Bound View
For $x \in [ 2 , + \infty[$, we set $$I ( x ) = \int _ { 2 } ^ { x } \frac { \mathrm{dt} } { \ln ( t ) }$$ Justify, for $x \in [ 2 , + \infty[$, the relation $$I ( x ) = \frac { x } { \ln ( x ) } - \frac { 2 } { \ln ( 2 ) } + \int _ { 2 } ^ { x } \frac { \mathrm{dt} } { ( \ln ( t ) ) ^ { 2 } }$$ Establish moreover the relation $$\int _ { 2 } ^ { x } \frac { \mathrm{dt} } { ( \ln ( t ) ) ^ { 2 } } \underset { x \rightarrow + \infty } { = } o ( I ( x ) )$$ Deduce finally an equivalent of $I ( x )$ as $x$ tends to $+ \infty$.
We consider the functions $F$ and $G$ defined by the formulas $$\begin{aligned} & \forall x \in ]-1,1[ , \quad F ( x ) = \sum _ { n = 0 } ^ { + \infty } P \left( S _ { n } = 0 _ { d } \right) x ^ { n } \\ & \forall x \in [ - 1,1 ] , \quad G ( x ) = \sum _ { n = 1 } ^ { + \infty } P ( R = n ) x ^ { n } \end{aligned}$$ Show that the power series defining $F$ and $G$ have radius of convergence greater than or equal to 1. Justify then that the functions $F$ and $G$ are defined and of class $C^{\infty}$ on $]-1,1[$.
Show that $G$ is defined and continuous on $[-1,1]$ and that $$G ( 1 ) = P ( R \neq + \infty ) .$$
We consider the functions $F$ and $G$ defined by the formulas $$\begin{aligned} & \forall x \in ]-1,1[ , \quad F ( x ) = \sum _ { n = 0 } ^ { + \infty } P \left( S _ { n } = 0 _ { d } \right) x ^ { n } \\ & \forall x \in [ - 1,1 ] , \quad G ( x ) = \sum _ { n = 1 } ^ { + \infty } P ( R = n ) x ^ { n } \end{aligned}$$ If $k$ and $n$ are positive integers such that $k \leq n$, show that $$P \left( \left( S _ { n } = 0 _ { d } \right) \cap ( R = k ) \right) = P ( R = k ) P \left( S _ { n - k } = 0 _ { d } \right) .$$ Deduce that $$\forall n \in \mathbb{N}^{*} , \quad P \left( S _ { n } = 0 _ { d } \right) = \sum _ { k = 1 } ^ { n } P ( R = k ) P \left( S _ { n - k } = 0 _ { d } \right) .$$
grandes-ecoles 2020 Q8 Compound or random-sum PGF View
We consider the functions $F$ and $G$ defined by the formulas $$\begin{aligned} & \forall x \in ]-1,1[ , \quad F ( x ) = \sum _ { n = 0 } ^ { + \infty } P \left( S _ { n } = 0 _ { d } \right) x ^ { n } \\ & \forall x \in [ - 1,1 ] , \quad G ( x ) = \sum _ { n = 1 } ^ { + \infty } P ( R = n ) x ^ { n } \end{aligned}$$ Show that $$\forall x \in ]-1,1[ , \quad F ( x ) = 1 + F ( x ) G ( x ) .$$ Determine the limit of $F ( x )$ as $x$ tends to $1^{-}$, discussing according to the value of $P ( R \neq + \infty )$.
grandes-ecoles 2020 Q9 Limit Evaluation Involving Sequences View
Let $\left( c _ { k } \right) _ { k \in \mathbb{N} }$ be a sequence of elements of $\mathbb{R}^{+}$ such that the power series $\sum c _ { k } x ^ { k }$ has radius of convergence 1 and the series $\sum c _ { k }$ diverges. Show that $$\sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } \underset { x \rightarrow 1 ^ { - } } { \longrightarrow } + \infty$$ With the element $A$ of $\mathbb{R}^{+*}$ fixed, one will show that there exists $\alpha \in ]0,1[$ such that $$\forall x \in ]1 - \alpha , 1[ , \quad \sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } > A$$
Let $p \in \mathbb{N}$. Show that the sequence with general term $u_n = \binom{n}{p}$ is hypergeometric.
Prove that the set of sequences satisfying relation $$P(n) u_n = Q(n) u_{n+1}$$ with $$P(X) = X(X-1)(X-2) \quad \text{and} \quad Q(X) = X(X-2)$$ is a vector space for which we will specify a basis and the dimension.
grandes-ecoles 2021 Q3.24 Evaluation of a Finite or Infinite Sum View
We set $C = \exp\left(\frac{I}{2\pi}\right)$ with: $$I = \int_0^{2\pi} \ln\left(\max\left\{\left|e^{i\theta} - 1\right|, \left|e^{i\theta} + 1\right|\right\}\right) d\theta$$ Show that: $$I = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}$$ You may use the result from question 2.13.
grandes-ecoles 2021 Q3.25 Estimation or Bounding of a Sum View
We set $C = \exp\left(\frac{I}{2\pi}\right)$ with $I = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}$. The calculator gives: $$\begin{aligned} \exp\left(\frac{2}{\pi}\sum_{k=0}^5 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}78774486868 \\ \exp\left(\frac{2}{\pi}\sum_{k=0}^6 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}79449196958 \end{aligned}$$ Can we deduce the rounding of $C$ to $10^{-2}$ precision? If yes, give the value of this rounding. In any case, justify the answer properly.
grandes-ecoles 2021 Q4 Convergence proof and limit determination View
Let $\left(u_n\right)_{n \in \mathbb{N}}$ be a hypergeometric sequence with associated polynomials $P$ and $Q$. Suppose that there exists a natural integer $n_0$ such that $P\left(n_0\right) = 0$ and, $\forall n \geqslant n_0, Q(n) \neq 0$. Justify that the sequence $\left(u_n\right)_{n \in \mathbb{N}}$ is zero from a certain rank onwards.
Using question 4, show $$\forall n \in \mathbb { N } , \quad C _ { n + 1 } = \sum _ { r = 0 } ^ { n } C _ { r } C _ { n - r } .$$
Using the random variable $T$, show that the series $\sum _ { n \geqslant 0 } \frac { C _ { n } } { 4 ^ { n } }$ converges.