TITLE: How to find evaluate this infinite series
QUESTION [0 upvotes]: I obtained this series as the one which describes an integral I'm trying to calculate from the definition
$$\lim_{n\to \infty} \sum_{k=0}^{n-1} \frac{4n}{(2n+2k+1)^2}$$
From plugging in values of $n$ it looks like this converges to $\frac 12$.
Can you guys give me a hint as to how to show this?

REPLY [1 votes]: Notice, $$\lim_{n\to \infty}\sum_{k=1}^{n-1}\frac{4n}{(2n+2k+1)^2}=\lim_{n\to \infty}\sum_{k=1}^{n-1}\frac{4n}{4n^2\left(1+\frac{k}{n}+\frac{1}{2n}\right)^2}$$
$$=\lim_{n\to \infty}\sum_{k=1}^{n-1}\frac{\frac{1}{n}}{\left(\frac{k}{n}+1+\frac{1}{2n}\right)^2}$$
let $\frac{k}{n}=u\implies \frac{1}{n}=du\to  0$
Upper limit at $k=n-1$ $\implies u=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{n-1}{n}=1$
Lower limit at $k=1$ $\implies u=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{1}{n}=0$
using integration, with proper limits
$$=\lim_{n\to \infty}\sum_{k=1}^{n-1}\frac{\frac{1}{n}}{\left(\frac{k}{n}+1+\frac{1}{2n}\right)^2}=\frac{1}{4}\int_{0}^{1}\frac{du}{\left(u+1\right)^2}$$
$$=\left(\frac{-1}{u+1}\right)_{0}^{1}$$
$$=\left(\frac{-1}{1+1}+\frac{1}{0+1}\right)$$
$$=\color{red}{\frac{1}{2}}$$


TITLE: An inequality involving operator and trace norms
QUESTION [4 upvotes]: Consider two square matrices $A, B \in \mathbb{R}^{n \times n}$ and let $\| \cdot\|_1$ and $\|\cdot\|$ be, respectively, the trace norm (the sum of singular values) and the usual operator norm (the maximum of singular values).
Is there a known bound for the following quantity?
$$
\sup\{\alpha > 0: \; \alpha \, \text{tr}(A^TB) \le \| A+B\|_1, \, \forall B \; \text{s.t.} \;\|B\| \le 1\}
$$

REPLY [4 votes]: Since I cant comment, I will leave this thought here. Since $||\cdot||_1$ and $||\cdot||$ (as you defined them) are dual norms, it must be that tr$((A+B)^TX)\leq||A+B||_1$ for any $X$ such that $||X||\leq 1$. Therefore, tr$(A^TB)\leq ||A+B||_1 - ||B||^2<||A+B||_1$ (since tr$(B^TB)\geq||B||^2$).
(edit: fixed typo and replaced $||B||$ with $||B||^2$ in the final step)


TITLE: $\lim_{n\rightarrow \infty} (n+1)\cdot x^{n\cdot n!} < 1$
QUESTION [1 upvotes]: If $\vert x \vert < 1$ then I want to show $\lim_{n\rightarrow \infty}
 (n+1)\cdot x^{n\cdot n!} < 1$

It makes perfectly sense in my world, because the factor $x^{n\cdot n!}$ is smaller than the factor $(n+1)$ when n goes to infinity. I have tried to use L'Hoptial but it doesn't work. Then I tried to find an example of an expression which is greater than $(n+1)\cdot x^{n\cdot n!}$ but still smaller than 1, when n goes to infinity. But all the examples I have found diverges. Now I'm stuck - any ideas?

REPLY [0 votes]: The case ${x}=0$ is clear. For the rest of the values we use the fact that $\vert x \vert < 1$, and we may replace ${x}$ by $\frac{1}{t}$ where $\vert t \vert > 1$. Lastly, we get that: $$\lim_{n\rightarrow \infty} \frac{n+1}{t^{{n} \cdot {n!}}} =\frac{polynomial \space growth \space behaviour}{exponential \space growth \space behaviour} =0$$
The proof is complete.


TITLE: Prove that $(S \cap T = \varnothing) \land (S \cup T = T) \rightarrow S = \varnothing$.
QUESTION [2 upvotes]: Logically, the following proposition makes sense:
$(S \cap T = \varnothing) \land (S \cup T = T) \rightarrow S = \varnothing$
Or, in english, if sets $S$ and $T$ share no elements, and the union of the sets is equal to one of the sets, then the other set must be empty.
However, what I'm struggling with is how to translate this english into mathematical symbols.
Would anyone know how to solve this with mathematical symbols?

REPLY [0 votes]: Here is an element-level calculational solution, which only uses the set theory definitions and predicate logic.
Expanding the definitions in the first part, and doing some simplification, results in
\begin{align}
& S \cap T = \varnothing \\
\equiv & \;\;\;\;\;\text{"extensionality; definitions of $\;\varnothing,\cap\;$"} \\
& \langle \forall x :: \lnot(x \in S \land x \in T) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: DeMorgan"} \\
& \langle \forall x :: x \not\in S \lor x \not\in T \rangle \\
\end{align}
For the second part, we get
\begin{align}
& S \cup T = T \\
\equiv & \;\;\;\;\;\text{"extensionality; definition of $\;\cup\;$"} \\
& \langle \forall x :: x \in S \lor x \in T \equiv x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: $\;p \lor q \equiv q\;$ is one way to write $\;p \Rightarrow q\;$"} \\
& \langle \forall x :: x \in S \Rightarrow x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: $\;\lnot p \lor q\;$ is another way to write $\;p \Rightarrow q\;$"} \\
& \langle \forall x :: x \not\in S \lor x \in T \rangle \\
\end{align}
Combining these, we can simplify even further:
\begin{align}
& S \cap T = \varnothing \;\land\; S \cup T = T \\
\equiv & \;\;\;\;\;\text{"the above two calculations"} \\
& \langle \forall x :: x \not\in S \lor x \not\in T \rangle \;\land\; \langle \forall x :: x \not\in S \lor x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"$\;\land\;$ distributes over $\;\forall\;$"} \\
& \langle \forall x :: (x \not\in S \lor x \not\in T) \;\land\; (x \not\in S \lor x \in T) \rangle \\
\equiv & \;\;\;\;\;\text{"$\;\lor\;$ distributes over $\;\land\;$"} \\
& \langle \forall x :: x \not\in S \lor (x \not\in T \land x \in T) \rangle \\
\equiv & \;\;\;\;\;\text{"contradiction; simplify $\;p \lor \text{false}\;$ to $\;p\;$"} \\
& \langle \forall x :: x \not\in S \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\varnothing\;$"} \\
& S = \varnothing \\
\end{align}
This proves (a stronger version of) the original statement.


TITLE: Convert a linear difference equation into a controllable state-space model
QUESTION [2 upvotes]: I have the following linear difference equation (which is an discrete-time SISO ARX model):
$y(k)+\sum_{i=1}^{n}a_iy(k-i)=\sum_{i=1}^{n}b_iu(k-i)$
and I need to transform it in an equivalent state-space model in controllable canonical form, that is:
$\begin{pmatrix}x_1(k+1)\\ x_2(k+1)\\ \vdots\\ x_{n-1}(k+1)\\ x_n(k+1)\end{pmatrix} = \begin{pmatrix}0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & 0 & \ldots & 1 \\ -a_n & -a_{n-1} & -a_{n-2} & \ldots & -a_1\end{pmatrix}\begin{pmatrix}x_1(k)\\ x_2(k)\\ \vdots\\ x_{n-1}(k)\\ x_n(k)\end{pmatrix}+\begin{pmatrix}0\\ 0\\ \vdots\\ 0\\ 1\end{pmatrix}u(k),$
$y(k) = \begin{pmatrix}b_n & b_{n-1} & \ldots & b_2 & b_1\end{pmatrix}\begin{pmatrix}x_1(k)\\ x_2(k)\\ \vdots\\ x_{n-1}(k)\\ x_n(k)\end{pmatrix}$
I am unable to find a connection between $y(\cdot)$ and $x_i(\cdot)$ variables, so that if I have data representing $y$ values I can find the related $x_i$ values.
I need this because I want to use state-feedback control (e.g., LQR).
Can someone help me?
Any hint for the MIMO case?
Thank you very much!!

REPLY [2 votes]: Let's start by moving the summation terms all over to the right hand side.
$y(k) = -\sum_{i=1}^n a_i y(k-i) + \sum_{i=1}^n b_i u(k-i)$
Now, let's define $\mathbf{x_k} = \left(\begin{array}{c} y(k-1) \\ y(k-2) \\ \vdots \\ y(k-n) \end{array}\right)$, and $\mathbf{u_k} = \left(\begin{array}{c} u(k-1) \\ u(k-2) \\ \vdots \\ u(k-n) \end{array}\right)$.
Then, $\mathbf{x_{k+1}} = \mathbf{Ax_k}+\mathbf{Bu_k}$, where $\mathbf{A} = \rm{diag}(-a_1,-a_2,\ldots,-a_n)$, and $\mathbf{B} = \rm{diag}(b_1,b_2,\ldots,b_n)$.
Finally, $y(k) = \left(1\ 1\ \cdots\ 1\right)\mathbf{x_k}$.


TITLE: $X$ is a random variable. $f_X(x)=2xdx$ is the density function of $X$ such that $x\in[0,1]$.$Y=\lfloor NX\rfloor$ such that $N\in \mathbb {N}$.
QUESTION [1 upvotes]: $X$ is a random variable. $f_X(x)=2xdx$ is the density function of $X$ such that $x\in[0,1]$.
$Y=\lfloor NX\rfloor$ such that $N\in \mathbb {N}$.
For $N=2L$ , find $P(Y=L) , L \in \mathbb {N}.$
My solution :
$P(Y=L) \implies $$\lfloor 2LX\rfloor=L \implies \frac{1}{2}\leq x < 1$.
Then $P(Y=L)=\int_\frac{1}{2}^12xdx=\frac{3}{4}.$
The correct answer is $\frac{2L+1}{4L^2}$ , where am I wrong ? It looks very simple to me.
Thanks !

REPLY [1 votes]: I believe there is an error in your treatment of the floor function.
$\lfloor 2LX \rfloor \neq L$ for all $X > \frac{1}{2}$
Consider $X = \frac{3}{4}$ and $L = 8$:
$$\lfloor 2 \cdot 8 \cdot \frac{3}{4} \rfloor = \lfloor 12 \rfloor \neq 8$$
Does this help you work through the rest of the solution?