Monday, January 31, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Laplace transform of 3, simple but what happens to $e^{st}$ term?

Posted: 31 Jan 2022 09:08 AM PST

I want to find the Laplace transform of f(t)=3, simply it is calculated by the integration procedure:

\begin{equation} \int_{-\infty}^{\infty}f(t)e^{-st}dt=3\int_{-\infty}^{\infty}e^{-st}dt=\bigg[\frac{3}{s}e^{-st}\bigg]_{\infty}^\infty=\bigg(\frac{3}{s}e^{-s\infty}+\frac{3}{s}e^{s\infty}\bigg)=\frac{3}{s}\bigg(e^{-s\infty}+e^{s\infty}\bigg) \end{equation}

But here the last term in the brackets is certainly not 1, in fact it evolves towards infinity, as given here.

But wolfram suggests on $f(s)=\frac{3}{s}$. What happened with $e^{-st}$ terms then?

Thanks

Very basic trig question: how to calculate an angle given a slope (an angle from a ratio of sides)

Posted: 31 Jan 2022 09:07 AM PST

I am trying to write a very basic trig primer, from scratch. I asked a basic question on Mathematics Educators, but they suggest to ask here. Here it is.

Say I discuss slopes (say, the slope of a line through the origin), and wish to give the 'right' hints on how the slope of a line is related to the angle of the line with the x-axis.

It would be nice to show how to calculate the angle, given the slope. I am looking for easy examples, say for lines with slopes 1,1/2,1/3,1/4,...; or any suitable family of lines.

I am not completely sure, but my question seems to be something like: how do I design an algorithm (and/or a geometric construction) to determine the angle from the slope (i.e, from the ratio of sides).

Sorry to be quite vague.

Convergence in the operator norm

Posted: 31 Jan 2022 08:58 AM PST

Given the following operator: $T:\ell^2\to\ell^2$, which acts in the following way on the standard basis vectors: $$Te_{2k-1}=\frac{1}{k}(e_{2k-1}-ie_{2k})$$ $$Te_{2k}=\frac{1}{k}(ie_{2k-1}+e_{2k})$$ I need to prove that this operator is compact by showing that it is the limit of finite rank operators. My attempt:

Define $(T_nx)_j=(Tx)_j$, for $j\le 2n$, and for $j\gt 2n: (T_nx)_j = 0$. Intuitively this should converge to $T$ In the operator norm and all these operators $(T_n)_{n\ge1}$ are of finite rank thus $T$ is compact!

Let us try and show that $\|T-T_n\|\to0$, for $n\to \infty$. $$\|T-T_n\|=\sup_{\|x\|=1}\|(T-T_n)x\|$$, but from here I am lost I don't see how this expression goes to $0$. Can anyone show me how it works?

Boolean Algebra; comprehending a given solution

Posted: 31 Jan 2022 08:52 AM PST

I am reading a text on Boolean Algebra. Right after the definition of the disjunctive normal form and the conjunctive normal form the author gives an example. Unfortunately I am not able to understand a transformation, that is needed to find the CNF of the given Boolean expression using the four Axioms. (The DNF was easy to find, so I know that there are probably no mistakes in the text.)

Let $z(x'y'+xy)'+x'y$ be a Boolean expression. Using De Morgan's laws we have $$z(x+y)(x'+y')+xy'.$$ That was easy. The author then goes on stating:

It follows immediately that $$z(x+y)(x'+y')+xy'=(z+xy')(x+y+xy')(x'+y'+xy')$$ $$=(x+z)(y'+z)(x+y)(x'+y').$$

I do not understand what is happening in these two lines, so I would be grateful if someone could tell me in detail what is going on.

If you are interested, the "final" CNF supposedly is $$(x+y'+z)(x+y+z)(x'+y'+z)(x+y+z')(x'+y'+z')$$

Area of Triangle and Diophantine Equations

Posted: 31 Jan 2022 08:56 AM PST

We are given some arbitrary solution $(x_0, y_0)$ to $ax-by=1$ and another point $(b,a)$. We are asked to prove that the area of the triangle with vertices at $(0,0)$, $(b,a)$, and $(x_0, y_0)$ is $\frac{1}{2}$. I figured that I should use the equation for a right triangle, and apply some substitution as follows:

$\frac{1}{2}\sqrt{x_0^2+y_0^2}\sqrt{(b-x_0)^2+(a-y_0)^2}$

plugging in $x_0$ and $y_0$ in terms of the diophantine equation doesn't seem to cancel out any terms and the algebra is messy.

Derivative of a matrix w.r.t to another?

Posted: 31 Jan 2022 08:46 AM PST

What's the general formula for taking the derivative of a matrix w.r.t to another matrix? For example, what's $\frac{\partial A}{X}$, $\frac{\partial A}{X^\top}$, $\frac{\partial A}{\tilde{X}}$, $\frac{\partial A}{\tilde{X}^\top}$ for $A = X\tilde{X}^\top$, $A = X\tilde{X}$, $A = X^\top\tilde{X}$, and $A = X^\top\tilde{X}^\top$.

Or how about $\frac{\partial A}{X}$, $\frac{\partial A}{X^\top}$ for $A = XX^\top$, $A = X^2$, $A = X^\top X$

I struggle to derive these generally, are there known formulas for these? How do I in general go about deriving these? There are also more complicated cases of $\frac{\partial X\tilde{X}}{X^\top \tilde{X}}$ and so on.

Confusion about the necessity of the axiom of choice for infinite sets [duplicate]

Posted: 31 Jan 2022 08:46 AM PST

Given a family $(X_i)_{i \in I}$ of non-empty sets, why cannot one introduce a function $f : I \to \bigcup_{i \in I} X_i$ as follows :

for all $i \in I$, there exists $x \in X_i$ — let $f(i) := x$,

thus producing a "choice function" ?

Why isn't the above construction valid in $\sf {ZF}$ ?

Solve the complex number equation $|z|-\bar{z}=i$

Posted: 31 Jan 2022 08:39 AM PST

Solve the complex number equation $|z|-\bar{z}=i$.

The following was my thought process:

$$|z|=i+\bar{z}$$ Given that $Im(|z|)=0$, $Im(\bar{z})=-i$. Hence, $\bar{z}=a-i$ and $z=a+i$.

$$|z|=i+(a-i)$$

$$|z|=a$$

But since $z=a+i$, $|z|=\sqrt{a^2+1}$ and hence I've reached a contradiction, as $\sqrt{a^2+1}\neq a$.

Hence, there is no solution for $|z|-\bar{z}=i$, for $z\in\mathbb{C}$.

How do we calculate the inverse Laplace transform of $F(s)=\frac{s^2+1}{(s+1)(s-1)}$?

Posted: 31 Jan 2022 08:43 AM PST

We have

\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)} \end{equation}

which I want to use Heavisde method to find the fractions.

We start

\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)}=\frac{A}{(s+1)}+\frac{B}{(s-1)} \end{equation}

By Heaviside method, we multiply both sides by the denumerator of A, ($s+1$)

\begin{equation} (s+1)\frac{s^2+1}{(s-1)}=A+\frac{B(s+1)}{(s-1)} \end{equation}

We insert for the pole of the A-fraction, for s, that is $s=-1$ and we obtain

\begin{equation} (-1+1)\frac{s^2+1}{(s-1)}=A+\frac{B(0)}{(-2)} \end{equation}

which gives the incorrect results, that $A=0$

What should I do here to prepare the fractions correctly for the Heaviside method to be applied?

Thanks

Update, I devise the method suggested below in the accepted solution, and try the Heaviside method on that.

\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)}=\frac{(s^2-1)+2}{(s+1)(s-1)} \end{equation}

\begin{equation} F(s)=\frac{(s^2-1)+2}{(s+1)(s-1)}=\frac{(s+1)(s-1)+2}{(s+1)(s-1)} \end{equation}

\begin{equation} F(s)=\frac{(s+1)(s-1)+2}{(s+1)(s-1)}=\frac{(s+1)(s-1)}{(s+1)(s-1)}+\frac{2}{{(s+1)(s-1)}} \end{equation}

\begin{equation} F(s)=1+\frac{2}{{(s+1)(s-1)}} \end{equation}

So from here we treat the last fraction with the Heaviside method:

\begin{equation} F(s)=\frac{2}{{(s+1)(s-1)}}=\frac{A}{(s+1)}+\frac{B}{(s-1)} \end{equation}

Find A by multiplying with $(s+1)$ on both sides and obtain

\begin{equation} (s-1)=A+\frac{B(s+1)}{(s-1)} \end{equation}

Insert for the pole of the A-fraction, $s=-1$

\begin{equation} (-1-1)=A+\frac{B(-1+1)}{(s-1)} \end{equation}

This yields that $A=-2$. We do the same for B,

\begin{equation} (s+1)=\frac{A(s-1)}{(s+1)}+B \end{equation}

Insert for $s=1$, and get that $B=2$, hence:

\begin{equation} F(s)=1+\frac{2}{(s-1)}-\frac{2}{(s+1)} \end{equation}

Inverse transform of F(s):

\begin{equation} \mathscr{L}^{-1}\{1+\frac{2}{(s-1)}-\frac{2}{(s+1)}\}=\delta(t)+2e^{t}-2e^{-t} \end{equation}

Hard Integral without Partial Fraction Decomposition

Posted: 31 Jan 2022 08:48 AM PST

So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.

$\int{\frac{3x+4}{(x^2+5)^2}dx}=$

Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$

$(x^2+5)^2[\frac{3x+4}{(x^2+5)^2}]=(x^2+5)^2[\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}]$

$3x+4=(Ax+B)(x^2+5)+(Cx+D)$

$3x+4=Ax^3+5Ax+Bx^2+5B+Cx+D$

$3x+4=(A)x^3+(B)x^2+(5A+C)x+(5B+D)$

So, $A=0$, $B=0$, $5A+C=3$, and $5B+D=4$

So, $A=0$, $B=0$, $C=3$, and $D=4$

$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{(0)x+(0)}{x^2+5}+\frac{(3)x+(4)}{(x^2+5)^2}}dx$

$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{3x+4}{(x^2+5)^2}}dx.$

As this clearly doesn't work, I am wondering what the integral would be. WolframAlpha tells me that the integral is as follows:

$\int{\frac{3x+4}{(x^2+5)^2}dx}=\frac{1}{50}(\frac{5(4x-15)}{x^2+5}+4\sqrt{5}\tan^{-1}(\frac{x}{\sqrt{5}}))+c$

I am not sure how to get here. Any help would be greatly appreciated.

Prove Linearity

Posted: 31 Jan 2022 08:54 AM PST

Given the following time varying model: $Z(t) = \alpha*t*z(t-1)$, how do I go about proving linearity? The issue I keep running into with a number of approaches is that because Z(t) is dependent on Z(t-1), I always end up with some inscrutable function within the function.

For transparency this is homework, some I'm happy for general guidance over a full solution.

Edit: The question explicitly says "Prove linearity" not "prove or disprove linearity." And the professor has used these homework's for years. So... take that for what it's worth. This is in the context of time series.

If $x\in X$ and $r > 0$, then for every $y\in Y\cap B^{X}_{r}(x)$ there is $s > 0$ such that $y\in B^{Y}_{s}(y)\subset Y\cap B^{X}_{r}(x)$.

Posted: 31 Jan 2022 08:47 AM PST

Exercise

If $x\in X$ and $r > 0$, then for every $y\in Y\cap B^{X}_{r}(x)$ there is $s > 0$ such that $y\in B^{Y}_{s}(y)\subset Y\cap B^{X}_{r}(x)$.

My attempt

So far it has not been discussed the characterization of open sets as sets that each point is contained in an open ball contained in the set. Having said, could anyone give me some hint to solve such exercise? It is not homework. I am really interested in understanding the theory properly.

Notation

The notation $B^{X}_{r}(x)$ represents the open ball centered in $x$ with radius $r > 0$ within the set $X$.

Verifying continuity by topology

Posted: 31 Jan 2022 08:46 AM PST

I want to show that sum of two continuous functions is continuous using the topological definition only .

I know the question could be answered by composing two continuous functions $(.,.):X\to\mathbb{K}^2$ and $+:\mathbb{K}^2\to\mathbb{K}$ over some infinite field $\mathbb{K}$ , and that has been answered several times in this site as well . But if I would want to use the basic open set definition , what would be the correct approach ?

Here's what I tried . Consider two continuous maps $f:X\to\mathbb{K}$ and $g:X\to\mathbb{K}$ where $X$ is some topological space . Then for all open sets $U$ in $\mathbb{K}$ , we have $f^{-1}(U)=\{x\in X:f(x)\in U\}$ and $g^{-1}(U)=\{x\in X:g(x)\in U\}$ are open in $X$ . We need to show $(f+g)^{-1}(U)=\{x\in X:(f+g)(x)\in U\}=\{x\in X:f(x)+g(x)\in U\}$ is open in $X$ . But how to relate the $f^{-1}(U),g^{-1}(U)$ with $(f+g)^{-1}(U)$ I don't understand . I have a feeling that the fact union of two open sets is open might be useful here .

Any help is appreciated .

Given a flow network and a max flow f on it, Determine whether there are at least 4 different max flows.

Posted: 31 Jan 2022 09:02 AM PST

I'm having trouble solving this one and would really appreciate any help. thank you in advance!

so, the problem is:

given a flow network with integer capacities on the edges and a max flow f on that network, I need to write an algorithm that determine whether there are at least 4 more different max flows on that given network.

thanks again!

Solving second order differential equations for forced oscillation

Posted: 31 Jan 2022 09:05 AM PST

please refer to this image

Please refer to this image

I have several doubts regarding the stapes involved in solving this Differential equations

  1. Why the $-\delta$ term is interduced in $x_{c}(t)=A e^{i(\omega t-\delta)}$? what is the intuition behind this ? is it still mathematicaly currect all though we have altered the Equation ?

2.How we reached $A(\omega)=\frac{1}{\sqrt{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+\gamma^{2} \omega^{2}}}$ (Please give some detailed Solution)

Please someone help

Show that two random variables have equal distribution

Posted: 31 Jan 2022 08:49 AM PST

Let $(X, \{\mathbb{P}_\vartheta; \vartheta \in (0,1)\})$ be a statistical model whith the density of $X$ given as $$ f_\vartheta(x) = \frac{\vartheta^{-1}}{2}\mathbb{1}_{[0,\vartheta]}(x) + \frac{(1 - \vartheta)^{-1}}{2}\mathbb{1}_{(\vartheta,1]}(x). $$

  1. Let $U,V$ ~ $U([0,1])$ independent. Show that $X$ is equally distributed as the random variable $\vartheta V\mathbb{1}_{U \leq 1/2}+(1-(1-\vartheta)V)\mathbb{1}_{U>1/2}$.
  2. Show that $\mathbb{E}_\vartheta[X] = \frac{1}{2}\vartheta + \frac{1}{4}$
  3. Let $((X_1,\dots,X_n), (\mathbb{P}_{\vartheta)\in(0,1)})$ be such that $X_1,\dots,X_n$ are independent and identically distributed with density $f_\vartheta$. Find an unbiased and consistent estimator for $\vartheta$ and proof unbias and consistent.

I tried this exercise for a very long time and didn't get any further so some help would be nice.

What is the largest $n\times n$ square lattice that can be $k$-colored such that no rectangles with same-colored vertices exist?

Posted: 31 Jan 2022 09:06 AM PST

It is fairly well known that for any $k$-coloring of $\mathbb{Z}^2$, it is impossible to color the vertices such that you never get a rectangle with axes parallel to the x and y-axis with vertices of the same color. In fact, it is known that you cannot color an $(k+1) \times (\frac{k^2(k+1)}{2}+1)$ lattice of points that meets this criterion. For example, you cannot 2-color the $3\times 7$ lattice in this manner. You can, however, color the $(k+1) \times (\frac{k^2(k+1)}{2})$ lattice in this way. One such successful 2-coloring of the $3\times 6$ lattice is below.

![enter image description here

So, if the $3\times 6$ lattice is 2-colorable in this fashion, it is a natural question to ask, "what about the $6\times 6$ lattice?". As it turns out, the answer to this is "no". We can prove this by first showing you cannot 2-color the $5\times 6$ lattice.

After some experimentation, one sees that the only successful 2-coloring of the $4\times 6$ lattice is some permutation of the rows and columns of the below.

enter image description here

If one adds a column to the right, it becomes apparent that the top-right vertex can neither be red nor blue: in the case of the former, we eventually see that the fourth-row vertex cannot be red or blue, and in the case of the latter, we see the same for the third-row vertex. Therefore, the $6\times 6$ lattice cannot be colored like this - QED.

However, what about the $5\times 5$ lattice? It becomes clear that we essentially have six unique ways to color in the first three rows, seen below (again, barring permutations of the rows and columns). I know that out of these six, you cannot 2-color the top-left one successfully. I have yet to try the others, but would not be surprised to find you cannot achieve this for any configuration.

enter image description here

For what it is worth, it can be shown that yes, the $4\times 4$ lattice can be successfully colored like this, which will be shown at the end. However, this is only for 2-coloring. What about 3-coloring, or 4-coloring? What about $n$-coloring in general? If I had to make a conjecture, the most I would be willing to say is that you can never make a $(\frac{k^2(k+1)}{2})\times (\frac{k^2(k+1)}{2})$ lattice that can be colored like this, but even for that I don't know where to start proving it, and I don't know how to move that upper bound any lower. Any thoughts?

enter image description here

Prove that $X_t = W_t \mathbb{1}(\tau > t) + (2W_{\tau} - W_t) \mathbb{1}(\tau \le t)$ is a martingale.

Posted: 31 Jan 2022 09:04 AM PST

Consider $W_t$ is a standard Wiener process. Assume $\tau = \inf\{t : W_t = a\}$. Consider $X_t = W_t \mathbb{1}(\tau > t) + (2W_{\tau} - W_t) \mathbb{1}(\tau \le t)$. We want to show that $(X_t, \mathcal{F}_t)$ is a martingale, where $\{\mathcal{F}_t\}$ is natural filtration of the Wiener process.

I know that it's possible to show that the process is also Wiener, but it's much more complicated (since we use strong Markov property). Here I guess the situation is much easier.

It's not hard to show that $X_t$ is $\mathcal{F}_t$ measurable and $L_1$. The major problem is to show $\mathbb{E}(X_t | F_s) = X_s$ precisely.

I tried so: consider different cases

  1. $s < t < \tau$: $\mathbb{E}(W_t|F_s) = W_s$

  2. $s < \tau < t$: $\mathbb{E}(2W_\tau - W_t | F_s) = \mathbb{E}(2W_\tau - 2W_s + W_s - W_t + W_s | F_s) = W_s$

  3. $\tau < s < t$: $\mathbb{E}(2W_\tau - W_t| F_s) = \mathbb{E}(2W_\tau - W_s - (W_t - W_s)|F_s) = 2W_\tau - W_s$

But I guess it's not so fair to write so, since $\tau$ is random variable and for some $\omega \in \Omega$ it might be true and for others not. Any hints?

Equation of a cubic Bezier curve

Posted: 31 Jan 2022 09:08 AM PST

quadratic bezier curve

For a quadratic Bezier Curve defined by points $A, B, C$, with point $M$ on the curve interpolated by $i$,

When points $A, M, C$ and angle $\alpha$ are given, $i$ and $B$ are:

$$i^2 = \frac{(y_A-y_M)\sin\alpha-(x_A-x_M)\cos\alpha}{(y_A-y_C)\sin\alpha-(x_A-x_C)\cos\alpha}$$ $$x_B=\frac{x_A(1-i)^2+x_Ci^2-x_M}{2i(i-1)}$$ $$y_B=\frac{y_A(1-i)^2+y_Ci^2-y_M}{2i(i-1)}$$

for a full working out: $$x_D = x_A+i(x_B-x_A)$$ $$x_E = x_B+i(x_C-x_B)$$ $$x_M = x_D+i(x_E-x_D)$$

$$x_M = x_A(1-i)^2+2ix_B(1-i)+i^2x_C$$ $$x_B=\frac{x_A(1-i)^2+i^2x_C-x_M}{-2i(1-i)} = AB\sin\alpha+x_A$$ $$y_B=\frac{y_A(1-i)^2+i^2y_C-y_M}{-2i(1-i)} = AB\cos\alpha+y_A$$ $$AB = \frac{i^2(x_A-x_C)+x_A-x_M}{2i\sin\alpha(1-i)}= \frac{i^2(y_A-y_C)+y_A-y_M}{2i\cos\alpha(1-i)}$$

hence resulting the above 3 equations.

cubic bezier curve

Now consider a cubic Bezier curve defined by points $A, B, C, D$

$1)$With given points $A, D$, $M$ interpolated by $i$,angles $\alpha, \beta$, Can $B, C$ and $i$ be derived?

$2)$ if not, With given points $A, D$, $M$ (interpolated by $i$), $N$ (interpolated by $j$), angles $\alpha, \beta$, What are the equations for $B, C$, $i$ and $j$?

Why subtracting each data point from the mean and then squaring it and divided it by n-1 is a good measure of spread?

Posted: 31 Jan 2022 08:45 AM PST

So, we are trying to find the distance of all of the data points about the mean. But, isn't it the other way around? When we use things like variance and MAD, aren't we actually finding the average distance of mean to all of the data points and not the distance of all of the data points from the mean? Isn't a more convincing way of finding a good number for a spread would be like this:

Sum of "average distance" of each data point to the other divided by n(sample pop.). Now, think of this thing for a second. Because it has been over a week that I am trying to make sense of why do are they finding just the average distance of one number, i.e. mean, in variance?

Formula is:

Sum of the "average distances" of each point divided by n.

By the sum of average distances I meant is: Let's say you have 4 data points: 2,4,8,12

  • I will ask how far is 2 from all of the data points, in absolute terms;

  • I get these numbers: 2,6,10 (4 is 2 units away, 8 is 6 units away, and 12 is 10 units away);

  • Then, I will ask a statistical question how are the data points from 2. I need one number.

  • Thus, I will take the average of these differences and the number I'll get is the distance of all of the data points from 2 on average.

  • Then, I will repeat the process for all of the data points.

  • And, left with 4 averages, which is the average distance of each point to all of the data points.

  • at last, I will again ask a statistical question, how far are the data points from each other.

  • Thus, I will sum all of these averages and divided them by n, which is 4 in this case.

  • And, the number I'll get is the number representing the spread of the data. Not, only the number is greater than what you could've found otherwise in variance.

I agree that variance and SD are the conventions. But, they give mean a special position. They say let's find the distance of all of the data points from the mean, but the way I see it they actually measure how far is mean to all of the data points. The analysis is incomplete. The way I proposed the formula, I know it's a little bit lengthy, but I take into account each number in the data set, find how far is each of them from each other, and then take an average to finally get to the point. I didn't give any special position to any number.

Note: I understand that it will take time to find the measure for spread relative to what we are used to, i.e., variance and standard deviation. But, I think the important thing to realize is that we will, at last, use all of the data and measure it through programs. And, I can make a code in R that can easily calculate the spread with the formula I just proposed.

Existence elimination in Lean 3

Posted: 31 Jan 2022 08:58 AM PST

Lean 3 is a theorem prover that implements the calculus of inductive constructions. Differently than Coq, Lean 3s kernel works proof irrelevant. This means that in the kernel of Lean all proofs of the same theorem are judgmentally (definitionally) equal and can be substituted for each other.

Does that mean that if I have A : Type, P : A -> Prop, and p : (∃ x : A, P x), then there is no way to extract an actual w : A from p which satisfies P w?

I thought being able to extract witnesses from existence proofs is one of the main arguments for Martin-Löf style type theories.

In how many ways one can arrange the numbers $0,1,2,\dots,n$ using each number exactly once so that no two adjacent numbers sum greater than $n+1$?

Posted: 31 Jan 2022 08:51 AM PST

Given the numbers $0,1,2,\dots,n$. In how many ways one can arrange the numbers using each number exactly once so that no two adjacent numbers sum greater than $n+1$?

I found the answers for small values of $n$. For $n=1$, the answer is $2$. For $n=2$, the answer is $6$. For $n=3$, the answer is $12$. But how to solve this problem for any $n$?

Equivalent condition to boundedness of a linear operator

Posted: 31 Jan 2022 08:44 AM PST

I have to prove the following statement:

Let $X$ and $Y$ be normed spaces over $\mathbb{K}$ and $T:X\to Y$ a linear operator. Prove that $T$ is bounded iff $\sum_{n=1}^\infty T(x_n)$ converges for all sequences $\{x_n\}$ for which $\sum_{n=1}^\infty x_n$ converges absolutely.

I have been stuck on this problem for days, yet I feel I'm really close to solving it. Any hint, idea or advice would be appreciated. I haven't been able to prove any of the implications, but this is what I have right now:

For the $(\implies)$ direction, assume that $T$ is bounded, and let $\{x_n\}$ be a sequence such that $\sum_{n=1}^\infty x_n$ converges absolutely. I'm trying to prove that $\sum_{n=1}^\infty T(x_n)$ converges on two steps. First I prove that the sequence $\{S_k\}_{k=1}^\infty$ where

$$ S_k=\sum_{n=1}^kT(x_n) $$

is Cauchy. This is easy to show using the boundedness and linearity of $T$, and the fact that $\{\sum_{n=1}^k ||x_n||\}_{k=1}^\infty$ is Cauchy, because it converges. Then I would like to prove that the sequence $\{S_k\}$ has a convergent subsequence, as that would imply the result. However, I don't know how to show this. The only idea I had was to prove and use this statement:

If $\{x_n\}$ is a sequence on a normed space $X$ such that $\sum_{n=1}^\infty x_n$ converges absolutely, then there exists a sequence $\{n_k\}$ of positive integers such that the sequence $\{\sum_{n=1}^{n_k} x_n\}_{k=1}^\infty$ converges.

as the continuity and linearity of $T$ would then imply what I want. However, I don't think that statement is true, as it would then imply that every normed space is Banach, which is obviously not true. So I have no idea on how to proceed now. In general I am having trouble to prove the convergence of almost anything, any advice for that would be really helpful.

For the ($\impliedby$) direction, I am trying to use the sequential criterion for continuity to prove that $T$ is continuous, and since $T$ is linear that would prove that $T$ is bounded. Let $a\in X$ and $\{x_n\}$ a sequence in $X$ that converges to $a$. I want to prove that $T(x_n)\to T(a)$. I trying to use the same steps as in the other implication, i.e. I need to show that $\{T(x_n)\}$ is Cauchy and has a subsequence that converges to $T(a)$. Here however, I managed to show the existence of a convergent subsequence, but I don't see how to show that the sequence is Cauchy. To see that it has a subsequence convergent to $T(a)$, notice the following:

As $x_n\to a$, for each $k\in\mathbb{Z}^+$ there exists an $n_k\in\mathbb{Z}^+$ such that

$$ ||x_{n_k}-a||<\dfrac{1}{2^k} $$

And since $\sum_{k=1}^\infty 2^{-k}$ converges, we have that $\sum_{k=1}^\infty||x_{n_k}-a||$ converges. By hypothesis, this implies that $\sum_{k=1}^\infty T(x_{n_k}-a)$ converges, and from here we get that $||T(x_{n_k})-T(a)||=||T(x_{n_k}-a)||\to 0$ as $k\to\infty$, and as such $T(x_{n_k})\to T(a)$.

If now I manage to prove that $\{T(x_n)\}$ is Cauchy, I would get the result. However I have no idea on how to do this.

Thanks in advance for any help you could give me!

Local minima of $f(x) = \begin{cases} 5 x^2 (2 - \sin \tfrac{1}{x}) & \text{ if } x \neq 0 \\ 0 & \text{ if } x = 0 \end{cases}$ dense near $0$

Posted: 31 Jan 2022 08:48 AM PST

Consider $$f(x) = \begin{cases} 5 x^2 (2 - \sin \tfrac{1}{x}) & \text{ if } x \neq 0 \\ 0 & \text{ if } x = 0. \end{cases} $$ For any given $\delta > 0$ it is very clear from the graph of $f$ that there is at least one (in fact infinitely many) point $x_\delta$ in $(-\delta, \delta)$ at which $f'(x_\delta) = 0$. Somehow, I am not getting how to mathematically prove this. Can anyone help me out?

enter image description here

Is this derivation of $\int{\frac{1}{x^2-a^2}}dx$ correct?

Posted: 31 Jan 2022 08:41 AM PST

Let $a>0.$ Using substitution, determine $$\int{\frac{1}{x^2-a^2}}\,\mathbb{d}x.$$

My book's attempt:

$$\int{\frac{1}{x^2-a^2}}\mathbb{d}x\\ [\text{let}\ x=a\sec\theta,\ \therefore dx=a\sec\theta\tan\theta d\theta]\\ =\int{\frac{1}{a^2\sec^2\theta-a^2}}a\sec^2\theta d\theta\\ =\int{\frac{1}{a^2(\sec^2\theta-1)}}a\sec\theta\tan\theta d\theta\\ =\int{\frac{1}{a^2\tan^2\theta}}a\sec\theta\tan\theta d\theta\\ =\frac{1}{a}\int{\frac{\sec\theta}{\tan\theta}}d\theta\\ =\frac{1}{a}\int{\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}}d\theta\\ =\frac{1}{a}\int{\csc\theta d\theta}\\ =\frac{1}{a}\int{\frac{\csc\theta(\csc\theta+\cot\theta)}{(\csc\theta+\cot\theta)} d\theta}\\ =-\frac{1}{a}\int{-\frac{\csc^2\theta+\csc\theta\cot\theta}{(\csc\theta+\cot\theta)} d\theta}\\ =-\frac{1}{a}\ln|\csc\theta+\cot\theta|+C$$

enter image description here

$$=-\frac{1}{a}\ln\left|\frac{x}{\sqrt{x^2-a^2}}+\frac{a}{\sqrt{x^2-a^2}}\right|+C\\ =-\frac{1}{a}\ln\left|\frac{x+a}{\sqrt{x-a}\sqrt{x+a}}\right|+C\\ =-\frac{1}{a}\ln\left|\frac{\sqrt{x+a}}{\sqrt{x-a}}\right|+C\\ =-\frac{1}{a}{\ln\left|\frac{x+a}{x-a}\right|}^{\frac{1}{2}}+C\\ =-\frac{1}{2}.\frac{1}{a}\ln\left|\frac{x+a}{x-a}\right|+C\\ =\frac{1}{2a}{\ln(\left|\frac{x+a}{x-a}\right|)}^{-1}+C\\ =\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C.$$

Someone told me that this work contains many errors. The second line clearly has a typo ($\sec^2\theta$ should be replaced with $\sec\theta\tan\theta$ there); what are the other errors?

Related: 1, 2

finding derivatives of variables in multivariable taylor polynomial

Posted: 31 Jan 2022 08:51 AM PST

Given: $$F(x,y) = -6 -4(x-4) +6(y-6) +8(x-4)^2 +9(x-4)(y-6) -4(y-6)^2 + R_2$$ and that $F(4,6)=-6$
find y'(4), y''(4), make sure that the function is applicable for the implicit function theorem.
my attempt: so I've basically made sure that the conditions applied, implicitly derived F and found out y'. It was a lot of hard work and while I got a result my method didn't at all utilize the fact it's taylor -I think there must be an easier way but I just can't put my finger on it. I think it'll utilize the fact that $-4 = F_x(x,y)$ and that $6 = F_y(x,y)$ - and maybe the chain rule? since $$-4 = \frac{\partial F(4,6)}{\partial{x}}=\frac{\partial{y}}{\partial{x}}* \frac{\partial F(4,6)}{\partial{y}} = 6*y'(x) \Rightarrow y'(x) = \frac{-2}3$$ but both my calculation and this calculator https://www.emathhelp.net/en/calculators/calculus-1/implicit-differentiation-calculator/?f=-4%28x-4%29%2B6%28y-6%29%2B8%28x-4%29%5E2+%2B9%28x-4%29%28y-6%29+-4%28y-6%29%5E2%3D0&type=x&px=4&py=6 output $2/3$
To sum up - where is my logic faulty? is there a similar easier way? how do I extend this to y''(4)?

An 8th grade contest-math puzzle

Posted: 31 Jan 2022 08:56 AM PST

In a competitive sort of exam, the following question was aimed at 8th graders and above-

Sophie had written the numbers 1 to 22 in the 22 discs in the figure, but Adelaide, her big annoying sister, erased fourteen.

enter image description here

Find the positions of the numbers erased knowing that

  1. Each number written in the centre of a hexagon represents the sum numbers placed at the vertices of this hexagon
  2. Two discs directly connected by a line never contain consecutive numbers.

On the answer sheet, write the values of the numbers $a, b, c, d$ and $e$.

On top of this, the question may have multiple answers (and the student is expected to report all of them) as well.

I don't have any idea about how to attack this problem. Of course, brute forcing is not a solution since in that case you have to check through $\sim 22^{14}\approx 6.2\times 10^{18}$ cases which is an impossibility even for a computer programme.

After doing some Google search recently, I found out something similar called Hexagonal Tortoise Problem and a paper about it. But, these couldn't take me anywhere.

One random observation is that if we indeed need to brute force the solution with some educated guesses, then maybe the disc connecting $4,7,12$ is the one to start with (since that's the one which is most restricted). But, even in that case, there are too many choices to deal with.

All ideas are welcome.

Prove |Re z| less than or equal to |z| and |Im z| less than or |z|

Posted: 31 Jan 2022 09:07 AM PST

|Re z| less than or equal to |z| ; z = x + yi

|Re z|= |x| --> sqrt(x^2) |z| = sqrt(x^2 + y^2)

sqrt(x^2) less than or equal to sqrt(x^2 + y^2) (therefore y^2 greater than or equal to 0)

then the answer is sqrt(x^2) less than or equal to sqrt(x^2 + y^2) is this correct?

and do the same thing for Im z?

is this correct? Thanks!

Good approximation to $\ln(x)$ for $x$ in $1 < x < e$

Posted: 31 Jan 2022 09:05 AM PST

I'm looking for a simple function that gives a good approximation to $\ln(x) $ within $1 < x < e$.

Do you have anything in mind? I'm not looking for an infinite function, but a short and finite version of it might be good, if it's also a simple solution as well.

RYB and RGB Color Space Conversion

Posted: 31 Jan 2022 09:06 AM PST

I am working on a project where I need to convert colors defined in RGB (Red, Green, Blue) color space to RYB (Red Yellow Blue).

I managed to solve converting a color from RYB to RGB space based on the article - Paint Inspired Color Mixing and Compositing for Visualization.

I convert a color from RYB to RGB with this "algorithm":

So the values of r (red), y (yellow), and b (blue) are known, also these arrays/sets are constants:

white: [1, 1, 1]  red: [1, 0, 0]  yellow: [1, 1, 0]  blue: [0.163, 0.373, 0.6]  violet: [0.5, 0, 0.5]  green: [0, 0.66, 0.2]  orange: [1, 0.5, 0]  black: [0.2, 0.094, 0.0]

here is how I calculate the value of red for the RGB space based on the parameters above:

i = 1;  rgb_red = white[i] * (1 - r) * (1 - b) * (1 - y) +   red[i] * r * (1 - b) * (1 - y) +   blue[i] * (1 - r) * b * (1 - y) +   violet[i] * r * b * (1 - y) +   yellow[i] * (1 - r) * (1 - b) * y +   orange[i] * r * (1 - b) * y +   green[i] * (1 - r) * b * y +   black[i] * r * b * y);

for rgb_green exactly the same thing but for i=2, and i=3 for rgb_blue.

My problem is that now I want to convert from RGB to RYB back. In other words, knowing the values of rgb_red, rgb_green and rgb_blue I want to calculate the values of r, y, and b. So I need a kind of inverse function for this, but I don't know how to get it.

Any help is appreciated.
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