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by dr.tasos
Sat Jul 29, 2017 1:06 pm
Forum: Real Analysis
Topic: No rational function
Replies: 1
Views: 978

Re: No rational function

Suppose there exists a rational function ( i assume you mean the function is a quotient of two polynomials ) . Let $ f(x)=\frac{P(x)}{Q(x)} $ Clearly since the harmonic series diverges $ lim_{ n \to \infty} f(n)=+\infty $ that means that $ deg(P(x)) > deg(Q(x)) $ Since $ lim_{ n \to \infty} \frac{H_...
by dr.tasos
Sat Jul 29, 2017 12:08 am
Forum: Linear Algebra
Topic: Inequality on Hermitian matrices
Replies: 2
Views: 1703

Re: Inequality on Hermitian matrices

I think it's direct from cauchy schwartz . Let me explain myself : because A is Hermitian from spectral theorem it is diagonizable so one may deduce that $ ker(A)=ker(T) $ with $T$ some diagonal matrix . If all eigenvalues are different from zero then $ dimker(A)=0 \Rightarrow rank(A)=n $ and the in...
by dr.tasos
Fri Jul 28, 2017 11:43 pm
Forum: Linear Algebra
Topic: The determinant is zero
Replies: 1
Views: 1891

Re: The determinant is zero

I think i got a solution : First suppose $ A $ is invertible so $ \exists A^{-1} $ I will prove by induction that $ A^nB-BA^n=nA^n $ $ \forall n \in \mathbb{N} $ The $ n=1 $ follows trivially from the hypotheses . Suppose it holds for all $ k \leq n $ i will prove it for $ k=n+1 $ $ A^nB-BA^n=nA^n $...
by dr.tasos
Wed Jul 26, 2017 12:33 pm
Forum: Real Analysis
Topic: A limit
Replies: 4
Views: 1911

Re: A limit

I think you need to give an initial value though.
by dr.tasos
Wed Jul 26, 2017 4:58 am
Forum: Real Analysis
Topic: Convergence of a series
Replies: 1
Views: 1124

Re: Convergence of a series

First $ a_{n+1}=sin(a_n) \leq a_{n} $ Since $ a_1 \in (0, \frac{\pi}{2} ) $ then using induction one can prove $ a_n \in (0, \frac{\pi}{2}) $ Hence $ a_n $ being bounded from below $ \lim_{n \to \infty} a_n=0 $ To estimate what the sequence feels like we compute the following limit $$ \lim_{n \to \i...
by dr.tasos
Wed Jul 26, 2017 4:32 am
Forum: Real Analysis
Topic: On measure theory
Replies: 1
Views: 1027

Re: On measure theory

I will change the notation slightly denoting $ m $ the Lebesgue measure . Suppose $ f=g $ on $ \mathbb{R} \setminus E $ with $ m(E)=0 $ so it's safe to assume that $ E^{\mathrm{o}}= \varnothing $ because if $ E^{\mathrm{o}} \ne \varnothing $ Then we get $ m(E)>0 $ Since $ \overline{\mathbb{R} \setmi...
by dr.tasos
Sun Apr 23, 2017 3:05 am
Forum: Functional Analysis
Topic: Closed subspace of finite dimension
Replies: 2
Views: 1456

Re: Closed subspace of finite dimension

It is enough to prove that the $B_{X} $ ( unit ball) is compact . From arzela ascoli we know that if we have a family $ F $ of uniformly bounded and equicontinuous functions ( in $C[0,1]$) then every sequence $ \left \{ f_{n} \right \}^{\infty}_{n=1} \subseteq F $ has a convergent subsequence . Give...
by dr.tasos
Sun Mar 20, 2016 1:29 pm
Forum: Number theory
Topic: ${\rm{gcd}}(\varphi(n), n)=1$
Replies: 6
Views: 2781

Re: ${\rm{gcd}}(\varphi(n), n)=1$

If $ n=p_{1}^{k_{1}}...p_{n}^{k_{n}} $ isnt square free then $ \exists \quad 1 \leq i \leq n $ Such that $ k_{i}\geq 2 $ $ \phi(n)=(p_{1}-1)p_{1}^{k_{1}-1}... (p_{i}-1)p_{i}^{k_{i}-1}..(p_{n}-1)p_{n}^{k_{n}-1} $ So $ p_{i} | n \quad \wedge \quad p_{i} | \phi(n) $ Contradiction to $ (n,\phi(n))=1 $
by dr.tasos
Sun Mar 20, 2016 12:58 pm
Forum: Number theory
Topic: ${\rm{gcd}}(\varphi(n), n)=1$
Replies: 6
Views: 2781

Re: ${\rm{gcd}}(\varphi(n), n)=1$

21 is free of square because $ 21=3 \times 7 $

But $ (21,φ(3) *φ(7))= 3 $
by dr.tasos
Sun Mar 20, 2016 11:47 am
Forum: Number theory
Topic: ${\rm{gcd}}(\varphi(n), n)=1$
Replies: 6
Views: 2781

Re: ${\rm{gcd}}(\varphi(n), n)=1$

15 isnt prime but $ (15,\phi(3) \phi(5))=1 $