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

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 $lim_{ n \to \infty} \frac{H_... Sat Jul 29, 2017 12:08 am Forum: Linear Algebra Topic: Inequality on Hermitian matrices Replies: 2 Views: 1703 ### Re: Inequality on Hermitian matrices I think it's direct from cauchy schwartz . Let me explain myself : because A is Hermitian from spectral theorem it is diagonizable so one may deduce that$ ker(A)=ker(T) $with$T$some diagonal matrix . If all eigenvalues are different from zero then$ dimker(A)=0 \Rightarrow rank(A)=n $and the in... Fri Jul 28, 2017 11:43 pm Forum: Linear Algebra Topic: The determinant is zero Replies: 1 Views: 1891 ### Re: The determinant is zero I think i got a solution : First suppose$ 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 $... Wed Jul 26, 2017 12:33 pm Forum: Real Analysis Topic: A limit Replies: 4 Views: 1911 ### Re: A limit I think you need to give an initial value though. Wed Jul 26, 2017 4:58 am Forum: Real Analysis Topic: Convergence of a series Replies: 1 Views: 1124 ### Re: Convergence of a series First$ 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... 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...
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...
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 $(n,\phi(n))=1$
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 $(21,φ(3) *φ(7))= 3$
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 $(15,\phi(3) \phi(5))=1$