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# Tutor profile: Frank G.

Frank G.
Tutor for six years

## 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 $$E$$.

Frank G.

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 $$n = 2$$.

Frank G.

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 $$N(0,1)$$.

Frank G.

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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