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You can use Dirichlet criteria because $\sum_{k=1}^{n}\ \sin{kx}$ is bounded. To see this, notice that $\sum_{k=1}^{n}\ \sin{kx}$ = $Im(\sum_{k=1}^{n}\ e^{ikx})$.

Write $A=2I+N$ where $$N=\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$ Then $\exp(A)=\exp(2I)\exp(N)$ since $I$ and $N$ commute. $\exp(N)$ is not too difficult to compute as $N^2=0$...

Hint Write $f(x)=(x-2)g(x)$ Then $f'(x)=g(x)+g'(x)(x-2)$.

The two not obvious points needed to prove are : $a,b \in E \implies ab \in E$ Every $a \in E$ such that $a \neq 0$ is invertible. Let's start with a useful lemma: Lemma: Let $P \in K[X]$. Then $P(... View answer 4 votes$f(x) = \frac{u(x)}{v(x)}$so$f(x) v(x) = u(x)$Differentiating gives:$f'(x) v(x) + v'(x) f(x) = u'(x)$Differentiating again:$f''(x) v(x) + f'(x) v'(x) + v''(x)f(x)+f'(x)v'(x) = u''(x)$... View answer 4 votes It is much faster to use the fact that$Cov$is a bi-linear map. Hence$Cov(aX+b,Y+Z) = a Cov(X,Y+Z) + Cov(b,Y+Z)$(linearity w.r.t first variable)$b$is constant so$Cov(b,Y+Z)=0$. Then$Cov(aX+b,...

Yes because of the linearity of integration.

Take $(a_n)$ defined by : $a_n = 0$ if n is even. $a_n = \frac{1}{2}$ if n is odd. This sequence satisfies (3) but it does not converge.

For (2) by the same reasoning by contradiction, there exist $L>0$ and $x_0$ such that for all $x>x_0$, $|f(x)| \geq L$. Because $f$ is continuous, either $f(x) \geq L$ for all $x>x_0$ or $f(... View answer 3 votes If is$f$is the pdf of$X$then$\mathbb{E}[e^{\lambda X}] = \int e^{\lambda x}f(x) dx$. View answer 3 votes Let's show that$\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $By Cauchy-Schwarz inequality:$\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{...

Hint: Try the change of variable $t=3x.$

Hint : You can see this easily by triangulating $V$. (which is always possible in $M_n(\mathbb{C})$) You would get that $\det(I+\epsilon V) = \textstyle \prod_{i=1}^{n} (1+\epsilon \lambda_i)$ where $... View answer 2 votes From $$\alpha{\begin{pmatrix} 1\\ 1\\ 0\\ \end{pmatrix}} +\beta{\begin{pmatrix} 1\\ 0\\ 1\\ \end{pmatrix}}= {\begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix}}$$ If you consider the third ... View answer Accepted answer 2 votes Hint$f(x) \leq f(y) \Leftrightarrow -f(x)\geq -f(y)$View answer 2 votes Take$f(x)=\sin(x)$if$x\in[0,2\pi]$and$f(x)=0$otherwise. Then$f$is continuous and$|\int_0^\infty f(x)dx|=0$but$f \neq 0$. View answer 2 votes Hint Use induction along with the product rule. View answer Accepted answer 2 votes Let$(h_n)$a sequence in$V$converging in$\mathbb{C}$. Let$u_n = \frac{f(z_0+h_n)-f(z_0)}{h_n}$The bounded$\sup$condition implies$||u_p - u_q|| \leq M |h_p-h_q|$Hence$(u_n)$is a Cauchy ... View answer 2 votes Because$N=1$, the equation is:$300S + 3X = 2020 + 170I +2E$Let's write it$R(S,X) = L(I,E)$. We notice that$L(5,0)=2870$and$R(9,9)=2727 < L(5,0)$so$I<5$. Let's check the remaining 5 ... View answer 2 votes I think the fastest way would be to do it by contradiction. Let$X=(3,-1,1,1)$and$Y=(1,3,-1,1)$Assume$X$and$Y$are linearly dependent. Then there is$a$such that$X= aY$. Using this on the ... View answer 2 votes Hint: Let$g$=$f'$. You have shown that there exists$x_1$and$x_2$such that$g(x_1) = g(x_2)$. Apply Rolle's theorem. View answer 2 votes 1) Let$\lambda_1, \lambda_2$the eigenvalues of$A$. We know that$\lambda_1=2$. We have that$\det A = \lambda_1 \lambda_2 $By assumption,$\det A = 1$so$1= 2\lambda_2$and hence$\lambda_2 = ...

Yes. More generally, if $E$ and $F$ are both $\mathbb{K}$-vector spaces of dimension $n$ then $E \simeq F$.

Work by induction. For $k=1$ it is true as $e >1$ Assume it is true for $k$. Then $(k+1)! = (k+1) k! > k k!> k \frac{k^k}{e^k} = e \frac{k^{k+1}}{e^{k+1}} > \frac{k^{k+1}}{e^{k+1}}$ ...

Hint : $A^{n}=(P^{-1} D P)^{n}=P^{-1} D^{n} P$

Let $a_n = ( \frac{1}{n} - 1)^{n^2}$ Then $|a_n| = (1- \frac{1}{n})^{n^2} =\exp(n^2 \ln (1-\frac{1}{n})) \sim e^{-n+\frac{1}{2}}$ $\sum e^{-n+\frac{1}{2}}$ converges so $\sum |a_n|$ converges ...

$T^3 = T$ implies that T is diagonalizable. Hence T is diagonalizable and 0 is its only eigenvalue. What can you conclude ?

You are right that it is straightforward that $X+Y$ is bounded. In fact, $X+Y$ is also closed: Let $(u_n)$ a sequence of $X+Y$ elements such that $\lim(u_n) = u$. We want to show that $u \in X+Y$. ...
The last implication directly stems from : $\forall n\geq N : \forall x\in X, x\neq0,\frac{\|(\hat T-T_n)x\|}{\|x\|}\leq \varepsilon$ Taking the supremum for all $x\neq0$, we get by definition : $\|\... View answer 1 votes You can use Cauchy-Schwarz inequality with$u_i = \sqrt{x_i}$and$v_i=\frac{1}{\sqrt{x_i}}\$.