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Tutor profile: Frank G.

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Frank G.
Tutor for six years
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Questions

Subject: Linear Algebra

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Question:

Let $$ E $$ be a finite dimensional Euclidean space. If $$ S $$ is a finite vector system of $$ E $$, we say that a vector $$ v \in E $$ admits a $$ S-$$ Fourier sum if $(\lVert v \rVert^{2} = \sum_ {s \in S} \langle v, s \rangle^{2} $) Show that if $$ S $$ is a finite orthonormal system such that every vector $$ v \in E $$ admits a $$S-$$Fourier sum, then $$ S $$ is an orthonormal basis of $$ E $$.

Inactive
Frank G.
Answer:

Let's first see that the $$S $$ is linearly independent, $$S=\{ v_{1},v_{2},..., v_{m}\}$$. Let's assume that: $( w = a_ {1} v_ {1} + a_ {2} v_ {2} + \ldots + a_ {m} v_ {m} = 0 $) By the condition of the problem it is then satisfied that $( 0 = \lVert w \rVert^{2} = \sum_ {i = 1}^{m} \left \langle w, v_ {i} \right \rangle = \sum_ {i = 1}^{m} \left \langle \sum_ {j = 1}^{m} a_ {j} v_ {j}, v_ {i} \right \rangle = \sum_ {i = 1}^{m} \sum_ {j = 1}^{m} a_ {j} \langle v_ {j}, v_ {i} \rangle = \sum_ {i = 1}^{m} a_ {i }^{2} $) from which it is concluded that $$ a_ {i} = 0 $$ for all $$ i = 1,2, \ldots, m $$, that is $$ S $$ is linearly independent. To see that $$ S $$ is a generator, it is enough, for all $$ v \in E $$, to find scalars $$a_ {1}, a_ {2}, \ldots, a_ {m} $$ such that $( v = a_ {1} v_ {1} + a_ {2} v_ {2} + \ldots + a_ {m} v_ {m} $) Then: $( \langle v, v_ {i} \rangle = \left \langle \sum_ {j = 1} ^ {m} a_ {j} v_ {j}, v_ {i} \right \rangle = \sum_ {j = 1} ^ {m} a_ {j} \langle v_ {j}, v_ {i} \rangle = a_ {i} $) and the scalars are determined. We conclude that $$ S $$ is the basis of $$ E $$.

Subject: Number Theory

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Question:

Let $$ n $$ be a positive integer for which there exists a positive integer $$ N $$ such that $$ x ^ {n} -2x + 1 $$ is a perfect square for every integer $$ x> N $$ . Prove that $$ n = 2 $$.

Inactive
Frank G.
Answer:

Let's assume that $$ n> 2 $$. Consider a prime $$ q \geqslant N $$ that does not divide $$ n-2 $$ (it exists because $$ n-2 \neq 0 $$). Let's take $$ x = q + 1> N $$. Then: $( x^{n}-2x+1=(x-1)\left( \sum_{j=1}^{n-1} (x^{j}-1)+n-2 \right) =qA $) To obtain a contradiction, just prove that $$ q $$ does not divide $$ A $$, since $$ q $$ would appear only once in the prime decomposition of $$ x^{n} -2x + 1 $$ implying which is not a perfect square. Notice that $$ x-1 $$ divides a $$ x ^ {m} -1 $$ for all $$ m $$, that is $$ q $$ divides a $$ x ^ {m} -1 $$ for all $$ m $$. In particular, $$ q $$ divides $$ A- (n-2) $$, and since $$ q $$ does not divide $$ n-2 $$, neither does it divide $$ A $$.

Subject: Statistics

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Question:

Suppose $$X_{1}, X_{2},...,X_{n}$$ are iid Poisson$$(\lambda)$$. Find a function $$ g $$ such that $$ \sqrt {n} (g (\overline {X_ {n}}) -g (\lambda)) $$ converges in law to a variable $$ N(0,1)$$.

Inactive
Frank G.
Answer:

According to the Central Limit Theorem, it is true that: $$\sqrt{n} (\overline {X_{n}} - \lambda) \longrightarrow^{D} N(0,\lambda)$$ because the variance of the Poisson$$(\lambda)$$ is equal to the mean, and its value is $$ \lambda$$. By the $$\Delta-$$method, if $$ g $$ is continuous at $$ \lambda $$, it follows that: $(\sqrt {n} (g (\overline {X_ {n}}) -g (\lambda))\longrightarrow^{D} g'(\lambda)N(0,\lambda)=N\left( 0,\lambda [g'(\lambda)]^{2} \right)$) Just look for $$ g $$ among those functions such that $$w [g'(w)]^{2}=1$$, that is $$g'(w)=w^{-1/2}$$. Integrating we obtain that $$g(w)=2\sqrt{w}$$.

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