Saturday, July 3, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


What is the graph of$|x|+2|y|+3|z|\leq 1$?

Posted: 03 Jul 2021 08:37 PM PDT

What is the graph of$|x|+2|y|+3|z|\leq 1$ ? And determine its geometrical figure. I know that the graph will be 3D. If I assume that $x \geq 0$,$y \geq 0$,$z \geq 0$. So let $x'^2 = x \geq 0$,$y'^2 = y \geq 0$,$y'^2 = z \geq 0$. Then above inequality can be written as $x+2y+3z\leq 1$ and $x'^2+2y'^2+3z'^2\leq 1 \implies \frac{x'^2}{1}+\frac{y'^2}{1\over2}+\frac{z'^2}{1\over3}\leq 1$ which is ellipsoid and its interior. Please help me.

Ask Riemann integrable function inequality

Posted: 03 Jul 2021 08:35 PM PDT

Let f and g be Riemann integrable functions on the interval [a,b]. Prove that

$$\int_a^b |f(x)| \cdot |g(x)| \leq (\int_a^b |f(x)|^2)^{1/2} \cdot (\int_a^b |g(x)|^2)^{1/2}$$

I don't know how to do this proof. Any hint will be helpful.

Determine the sequence of functions is uniformly convergent or not.

Posted: 03 Jul 2021 08:32 PM PDT

Let $\{f_n\}^{+∞}_{n=1}$ be a sequence of functions where $f_{n}(x)=\frac{nx}{1+n+x}$ where x belongs to [0,1] and n belongs to N (all natural numbers).

Determine whether the sequences of functions $\{f_n\}$ is uniformly convergent or not.

I'm having trouble with the problem and looking for help. Thanks in advance!

Plot the points in polar form (p,p(n+1)) and examine the spirals.

Posted: 03 Jul 2021 08:27 PM PDT

Referring to https://www.youtube.com/watch?v=EK32jo7i5LQ how would the spiral arms change if instead of plotting in polar form (p(n),p(n)) one plotted points (p(n),p(n+1))? Or plotted the points (p(n),p(n+a)) for a=2,3,4...? Would the result be pure chaos?

Coordinate free description of a map induced by a base-point-free divisor on a compact Riemann surface

Posted: 03 Jul 2021 08:25 PM PDT

Let $X$ be a compact Rieman surface, and $D$ be a divisor on $X$ with $|D|$ base-point-free. Note that the vector space $L(D)$ is finite-dimensional, and the complete linear system $|D|$ is naturally identified with the projectivization $\Bbb P(L(D))$. If we choose a basis $f_0,\dots,f_n$ for $L(D)$, then these meromorphic functions induce a map $\phi_D:X\to \Bbb P^n$ by $\phi_D(x)=[f_0(x):\cdots:f_n(x)]$ (and this map is unique up to coordinate change in $\Bbb P^n$). On the other hand, we have a map $\psi_D:X\to |D|^*=(\Bbb P(L(D)))^* \cong \Bbb P^n$ defined by sending $x\in X$ to the hyperplane $\{E\in |D|:E\geq x\} $ of the dual projective space $(\Bbb P(L(D)))^*$. In p.166 of Miranda's book Algebraic Curves and Riemann Surfaces, it is asserted that with suitable coordinates of $(\Bbb P(L(D)))^*$ (i.e. with a suitable identification of $(\Bbb P(L(D)))^*$ and $\Bbb P^n$), the map $\phi_D$ coincides with $\psi_D$, but I can't see why. What coordinates of $(\Bbb P(L(D)))^*$ should I choose?

How to solve this integral containing floor function?

Posted: 03 Jul 2021 08:22 PM PDT

Let $\alpha(t)=2t+ \lfloor{t} \rfloor $. Calculate $$\int_0^3 t^2 d\alpha(t)$$.

My idea is since $\alpha(t)$ is the sum of two increasing function. So I can split into two integrals. But I don't know how to solve $d\alpha(t)$. I hope someone can give me the complete solution.

$X_n\xrightarrow{d} X$ and $X_n/Y_n\xrightarrow{P} 1$ implies $Y_n\xrightarrow{d} X$?

Posted: 03 Jul 2021 08:22 PM PDT

Let $(X_n)_{n\ge 1}$ and $(Y_n)_{n\ge 1}$ be sequences of random variables in a probability space such that $Y_n(\omega)\neq 0\ \forall \omega$ and $$X_n\xrightarrow{d} X,\ \frac{X_n}{Y_n}\xrightarrow{P} 1$$ Is it true that $Y_n\xrightarrow{d} X$? I know it holds in the case $X_n-Y_n\xrightarrow{P} 0$ then if $\{Y_n\}_{n\ge 1}$ is uniformly bounded $$\frac{X_n}{Y_n}\xrightarrow{P} 1\Rightarrow X_n-Y_n\xrightarrow{P} 0$$ and the property is valid. In general, I think intuitively that it is true as well but cannot prove it, thus I would appreciate some hint. Thanks in advance!

Summation over roots of a polynomial

Posted: 03 Jul 2021 08:17 PM PDT

Take the polynomial $P(x)=x^6+3x^5-x^4+11x^3-x^2+3x+1$. Suppose $r_1, r_2,...,r_6$ are complex numbers such that $P(r_i)=0\forall i$. Compute $$\sum_{cyc}\frac{r}{r^2+r+1}$$

For this problem I have tried a few things. First of all, $r^2+r+1=\frac{r^3-1}{r-1}$. Putting this in, I get nowhere. Secondly, $r^2+r+1=r(r+1)+1$. I tried to deal with this by constructing a polynomial with roots $r_i+1$. After I had done this, I got nowhere. Finally, I tried actually finding the roots. Wolfram gives messy solutions.

It would be great if someone could help me solve this (no calculus, only elementary methods please).

Multivariate normal density converging to Dirac delta function

Posted: 03 Jul 2021 07:58 PM PDT

Say $X_1,\ldots,X_n$ is a random sample from a multivariate normal distribution $N(\boldsymbol{\beta},\Sigma)$ where $\Sigma$ is known, and $\widehat{\boldsymbol{\beta}}_n = (1/n)\sum_{i=1}^n X_i$. Consider the following integral: $$ I(\widehat{\boldsymbol{\beta}}_n) := \int f(\boldsymbol{a}) \phi(\boldsymbol{a}\mid \widehat{\boldsymbol{\beta}}_n,\Sigma/n)\,\mathrm{d}\boldsymbol{a}, $$ where $\phi(x\mid \mu,\Sigma)$ is the density of the multivariate normal distribution with mean vector $\mu$ and covariance matrix $\Sigma$.

My claim/conjecture is that as long as $f$ is a compactly supported continuous function, $I(\widehat{\boldsymbol{\beta}}_n) \xrightarrow{p} f(\boldsymbol{\beta})$, implying that $\phi(\boldsymbol{a}\mid \widehat{\boldsymbol{\beta}}_n,\Sigma/n) \to \delta(\boldsymbol{a}-\boldsymbol{\beta})$ in some mode of convergence. To prove this, I need to replace $\widehat{\boldsymbol{\beta}}_n$ with $\boldsymbol{\beta}$, but consistency doesn't help here. How should I go about proving or disproving this?

The way to configure n points that results in the maximum number of lines that contain more than or equals to m points

Posted: 03 Jul 2021 07:53 PM PDT

I was thinking of the question described below when designing a program that can find the collinear lines in a plane. https://www.cs.princeton.edu/courses/archive/fall12/cos226/checklist/collinear.html

If provided with n points, what is the maximum number of lines that contains more than or equals to m collinear points?

For instance, if 20 points are provided, what is the maximum possible number of lines(and the corresponding configuration of these points that results in the maximum number of lines) that contains more than or equals 4 points.

Please give me a combinatorial answer if possible. Thanks.

Euclidean algorithm and its formal proof

Posted: 03 Jul 2021 08:37 PM PDT

Suppose that $a,b\in \mathbb{N}$ and our goal to find $\text{gcd}(a,b)$ and the most effective way to do that is the Euclidean algorithm. I know how the algorithm works but I'd like to understand thoroughly and that is why I created this topic because some moments are not crystal clear to me.

Suppose we are given two natural numbers $a,b\in \mathbb{N}$ and assume that $a>b$ with $a\nmid b$.

Step #0. Then there is the unique pair of integers $q_0, r_2$ such that $a=bq_0+r_2,$ where $0<r_2<b$. Let's introduce the new notation $r_0:=a$ and $r_1:=b$. Then

$$r_0=r_1q_0+r_2, \text{where} \ 0<r_2<r_1.$$

Step #1. Then we apply the same procedure to the pair $\{r_1,r_2\}$ and obtain that

$$r_1=r_2q_1+r_3, \text{where} \ 0\leq r_3<r_2.$$

If $r_3=0$ then $(a,b)=r_2$ and I know how to prove it.

If $r_3>0$ then we move on to the Step #2 where $r_2=r_3q_2+r_4$ with $0\leq r_4<r_3$.

Claim: On some step #m with $m\geq 1$ the remainder $r_{m+2}$ will vanish.

Proof: Suppose this is false. Then on any step #m the remained is nonzero, i.e. $r_{m+2}>0$. Then it is easy to show that $r_{n+2}\leq b-n-1$. Hence $0<r_{b+2}\leq -1<0$. But this is a contradiction.

Hence $\exists m\geq 1$ such that $r_{m+2}=0$. Then it implies that $r_{m+1}\neq 0$ (otherwise if $r_{m+1}=0$ then $r_m=\dots=r_1=0$, right? Can anyone show the rigorous proof of that? Probably the proof relies on induction).

Hence $r_{m+1}\mid r_m$ and $(a,b)=(b,r_2)=(r_1,r_2)=\dots=(r_m,r_{m+1})=r_{m+1}.$

Is my reasoning correct? If yes, can anyone explain the question which I've asked?

Thanks in advance!

Evaluating $\int_0^{1/\sqrt{3}}\dfrac{\arctan(x)\ln(1-3x^2)}{1+x^2}\,\mathrm{d}x$

Posted: 03 Jul 2021 07:42 PM PDT

Recently while working on an interesting problem, I'm stuck on evaluating the following daunting but interesting integral:

$$\int_0^{1/\sqrt{3}}\dfrac{\arctan(x)\ln(1-3x^2)}{1+x^2}\,\mathrm{d}x$$

I have been working on it since quite some time, I've tried some trivial substitutions and integration by parts but it didn't get me anywhere. Any help would be highly appreciated.

Thanks for reading.

Is the path of a brownian motion (for some fixed time interval) a well defined random variable?

Posted: 03 Jul 2021 07:22 PM PDT

I might not have stated the question as clearly as I could have in the title.

Basically my thought process is the following; If I have a brownian process that I'm going to let run from, say, $0$ to $1$ minute, then this brownian process will trace out some non-differentiable continuous path in $\mathbb{R}^2$. If I let the process run again, then I will get a different path. This suggests to me that there is a random variable on the set of paths a brownian process can trace out.

My question is; is the set of all these paths a well defined probability space, and if so, do we know what that distribution looks like (does it have a mean? or variance? etc)? Or, does this set of paths exhibit certain pathologies that prevent it from being a probability space?

Related (perhaps easier): if we just restrict our attention to the set of values that a brownian process hits (so, ignoring the actual paths), do we have a way of answering something like "What is the probability that this brownian process hits a value of $1000$ in the first $30$ seconds?", or similar questions?

Thanks in advance.

Why is the number of elements in a group called "order"? [duplicate]

Posted: 03 Jul 2021 07:52 PM PDT

This is a question that I have for a long time, Maybe it is something silly, but I really want to know. Why is the number of elements in a group called "order"? I mean, the word "order" in Spanish (which is my language) has a very strong meaning in terms of "ordering", but it does not refer to quantities. What was the motivation for this?

Hint for $\int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{\left(\sqrt{a^2+b^2}-a\right)^v}{b^v\sqrt{a^2+b^2}}$

Posted: 03 Jul 2021 07:28 PM PDT

I'm looking for a hint of how to show that: \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{\left(\sqrt{a^2+b^2}-a\right)^v}{b^v\sqrt{a^2+b^2}},\qquad \Re v>-1,\quad a>0,\quad b>0, \end{align*} where $J_v$ is the Bessel function of the first kind of order $v$.

Two of my approaches, which didn't lead me to the result, are:

  • (The motivation for this approach is this) Use the integral representation: \begin{align*} J_v(z)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i\left(v\theta-z\sin\theta\right)}d\theta-\frac{\sin v\pi}{\pi}\int_0^\infty e^{-z\sinh t-vt}dt,\qquad \Re z>0, \end{align*} apply Fubini's Theorem, and after a substitution I arrived at: \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{1}{\pi i}\oint_{|z|=1}\frac{z^v}{bz^2+2az-b}dz-\frac{2\sin v\pi}{\pi}\int_1^\infty\frac{u^{-v}}{bu^2+2au-b}du, \end{align*} which I couldn't show that has de desired closed-form.
  • Use the integral representation: \begin{align*} J_v\left(z\right)=\frac{1}{2\pi i}\int_{C} e^{z(t-t^{-1})/2}t^{-v-1}dt,\qquad |\arg z|<\frac{\pi}{2}, \end{align*} where $C$ is a path from $(-\infty,\delta)$, around the origin counterclockwise and back to $(-\infty,-\delta)$; that is, a reflection of a Hankel contour about the line $\Re z=0$. Using this integral representation I arrived at \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{-1}{\pi i}\oint_D\frac{z^v}{bz^2+2az-b}du, \end{align*} where $D$ is the inversion of $C$ over the circle $|z|=1$; that is, the path resulting from $C$ under the mapping $z\to\frac{1}{z}$. I couldn't evaluate this last integral either.

Describing Locus of Complex region

Posted: 03 Jul 2021 07:27 PM PDT

I have an equation $|1-z_1z_2|<1$ How do I find the range of values z1 that satisfies this inequality? $z_2$ is also complex.

On the number of $Aut(G)$-orbits in the set of $k[G]$-modules

Posted: 03 Jul 2021 07:19 PM PDT

Let $k$ be any field, and $G$ a finite group. For an arbitrary $k[G]$-module $V$ and an automorphism $\sigma $ of the group $G$ we determine a $k[G]$-module $V^{\sigma }$ by considering that $V^{\sigma }=V$ as the $k$-module and the group $G$ act in $V^{\sigma }$ according to the rule: $g.v=\sigma(g)v$ $(v\in V)$?. Let $R(k[G],n)$ denote the set of isomorphism classes of all $k[G]$-modules of rank $n$. Define an action of $Aut(G)$ on $R(k[G],n)$ by $\overline{V}\varphi=\overline{V^{\varphi}}$ for all $\overline{V} \in R(k[G],n)$, $\varphi\in Aut(G)$. Let $G^{V}$ be the set of elements in $G$ which act as the identity on $V$. Applying the third part of (Theorem 2.4.5, p. 19) to the group $G/G^{V}$.

Does the number $N_{G}(R)$ of $Aut(G)$-orbits in $R(k[G],n)$ equals to the number of conjugacy classes of subgroups of $\mathrm{GL}_{n}(k)$ of order dividing $|G|$?.

Assume that $Aut(G)=Inn(G)$, what can we say about the number $N_{G}(R)$?. Are there other special cases which simplify the actions and the number $N_{G}(R)$?.

Thank you in advance.

Finding the resultant and direction of the resultant

Posted: 03 Jul 2021 07:27 PM PDT

I need to resolve the forces into components the horizontal and vertical then compute for the resultant and the direction of the resultant enter image description here

Find all ring homomorphisms from $\mathbb{Z} \rightarrow \mathbb{Z}_m$

Posted: 03 Jul 2021 07:20 PM PDT

I want to find all rings homomorphisms from:

i) $$\mathbb{Z} \rightarrow \mathbb{Z}_m$$

ii) $$\mathbb{Z}_m\rightarrow \mathbb{Z}$$

iii) $$\mathbb{Z}_n \rightarrow \mathbb{Z}_m$$


I don't know how to work this exercise, can someone explain to me how we think about these types of questions?

For the first one I only found $2$ the trivial and $\varphi(m)=m \pmod{n} $

for ii) I think of the trivial and $\varphi(m)=m $

and for iii) I found the trivial and $\varphi(m)=am, a\in \mathbb{Z}_m$ such that $na=0 \pmod{m}$

Proving a function $g(z)$ non zero inside the unit disc.

Posted: 03 Jul 2021 08:19 PM PDT

Denote the unit disc as $\mathbb{D}=\{z\in\mathbb{C}\mid |z|<1\}$. Define a Blaschke product in $\mathbb{D}$ as, $$B(z)=\prod_{n=1}^\infty \left(\frac{\alpha_n-z}{1-\overline{\alpha_n}z}\right)\frac{|\alpha_n|}{\alpha_n}$$ where $\alpha_n $ are the zeros of $f(z)$. Define a function $$g(z)=\frac{B(z)}{f(z)}$$ $$g(z)=\frac{1}{f(z)} \prod_{n=1}^\infty \left(\frac{\alpha_n-z}{1-\overline{\alpha_n}z}\right)\frac{|\alpha_n|}{\alpha_n} $$ Question Prove that $g(z)\neq 0$, $\forall z\in \mathbb{D}$

My try- Assume on the contrary that there exists $\alpha_0\in \mathbb{D}$ such that $g(\alpha_0)=0$ $$\frac{1}{f(\alpha_0)} \prod_{n=1}^\infty \left(\frac{\alpha_n-\alpha_0}{1-\overline{\alpha_n}\alpha_0}\right)\frac{|\alpha_n|}{\alpha_n} =0 $$ Since $f(\alpha_n)=0$ for all $n\geq 1$ , so all the zeros of $B(z)$ are cancelled by the zeros of $f(z)$. Hence we should have $g(z)\neq 0$ $\forall z\in \mathbb{D}$. Please give a proof of this question. An answer will be appreciated. Thanks for your time.

I'm trying to learn how to prove by induction

Posted: 03 Jul 2021 08:27 PM PDT

Please, I need some help to prove this $$1+\sum_{i=1}^n 5^i = \frac{5^{n+1}-1}{4}$$ I got this but I'm not sure how to proceed

Base n=1 $$=\frac{(5^2)-1}{4}$$ $$=\frac{25-1}{4}$$ $$=\frac{24}{4}$$ $$={6}$$ It works for n=1

Induction hypothesis

Let n=k $$1+5+25+125+...+k=\frac{5^{k+1}-1}{4}$$

Induction step

Let n=k+1

Prove $$1+5+25+125+...+k+(k+1)=\frac{5^{k+2}-1}{4}$$

By induction hypothesis $$1+5+25+125+...+k=\frac{5^{k+1}-1}{4}$$

And I tried this way using laws of exponents $$1+5+25+125+...+k+(k+1)=\frac{5^{k+1}-1}{4}+(k+1)$$ $$=\frac{5^{k+1}-1}{4}+\frac{4(k+1)}{4}$$ $$=\frac{5^k*5+4k+3}{4}$$

I think that I'm doing it wrong but would be so grateful if you could help me

EDIT: I just realized how bad my summation was, but now I understand the proof, ty all

Differentiate $\sqrt{x} e^x sec x$

Posted: 03 Jul 2021 08:36 PM PDT

Differentiate the following equation : $$\sqrt{x} e^x \sec x$$

Here I can see 3 values.

  1. $\sqrt{x}$
  2. $e^x$
  3. $\sec x$

This time, how can I use product rule?

My book said

$$\sqrt{x} \frac{d}{dx}(e^x \sec (x)) + e^x \sec {x} \frac{d}{dx} (\sqrt{x})$$

But, I found three values. So, according to product rule if I differentiate than, I would have 3 variable, wouldn't I? What am I missing?

$$\frac{d}{dx}ab = a\frac{d}{dx}b + b\frac{d}{dx}a$$

It's for two values (a,b). So, It would have 3 variables for 3 values(a,b,c).

Exercise 7 page 93 Functional Analysis book of Conway

Posted: 03 Jul 2021 08:32 PM PDT

The following is Exercise 7 page 93 in Functional Analysis book of Conway :

Let $1 \le p \le \infty$ and suppose $(a_{ij})$ is a matrix such that $(Af)(i) = \sum_{j=1}^{\infty} a_{ij} f(j)$ defines an element $Af$ of $\ell^p$ for every $f$ in $\ell^p$. Show that $A \in \mathcal{B}(\ell^p)$.

Case $p=\infty$ : If $f(j)$ is such that $||f(j)||_{\infty} < \infty$ and $\sum_{j=1}^{\infty} a_{ij} f(j) < \infty $ how to show that $(Af)(i)$ is bounded so that A maps $\ell^{\infty}$ to $\ell^{\infty}$?

Case $1 \le p < \infty$ : For this case same approach I think but this time we use Holder's inequality I suppose?

There is an answer here that uses Closed Graph Theorem, but I needed an approach that uses Uniform Boundedness Principle.

Alternative proofs of convergence of geometric series

Posted: 03 Jul 2021 07:54 PM PDT

The usual proof for the convergence of a geometric series of ratio $C: |C|\in [0,1)$ makes use of the formula $$\sum_{0\leq k \leq n} C^k = \frac{1-C^{n+1}}{1-C}.$$

I'm looking for alternative ways to prove it. The motivation for this is that, if someone who never saw this formula tried to prove the geometric series converges might have a hard time, unless maybe there are other, perhaps more insightful ways to prove it.

Where are the imaginary components in a moment generating function (MGF) of a distribution?

Posted: 03 Jul 2021 07:27 PM PDT

An MGF is

$$M_X(t)=E(e^{tX})=\int_{-\infty}^\infty e^{tx}f_X(x) dx$$

whereas a Laplace transform is

$$\mathcal L\{f_X(x)\}(s)= \int_0^\infty e^{-sx}f_X(x)dx$$

I am not referring to the sign difference, or the limits of integration.

My question is about $s\in \mathbb C$ being a complex number in the Laplace transform, whereas I only see real numbers in the idea of the MGF (i.e. series expansion of $e^{tx}$). Indeed, $M_X:\mathbb R \to [0,\infty]$ with the domain of $M_X$ defined as the set $D_X=\{t \mid M_X(t)<\infty\}.$ In characteristic functions complex numbers are introduced, but they are Fourier transforms, not Laplace. Obviously nobody really cares, but I don't understand why: the LT captures oscillations and exponential decay (in standard engineering uses), whereas the FT only captures sinusoidal components.

The imaginary numbers do surface, but in the characteristic function of a distribution.

THE QUESTION: So is there really an imaginary component in the MGF (like in the Laplace), and if there is not, why is it taught that the MGF is the Laplace transform of the pdf?

My intuition is that in probability the Laplace transform with domain in the complex numbers has been somewhat limited to $t\in \mathbb R,$ corresponding to a segment of the real line around zero in which the $E[e^{tX}]$ for different reasons:

  1. To be able to extract moments through the Taylor series.
  2. To classify distributions in relation to the exponential distribution, which is very useful in extreme value theory - analysis of the tails. This is best exemplified in the Chernoff inequality:

$$\Pr(X \geq a) = \Pr\left(e^{tX} \geq e^{ta}\right)\leq \frac{E[e^{tX}]}{e^{ta}}= \frac{M_X(t)}{e^{ta}}=e^{-ta}M_X(t)$$

which indicates that a finite MGF will result in exponentially tapering tails.

The third reason (to uniquely determine the distribution) would not really be a reason to keep the MGF: The characteristic function does have this role with the additional advantage that it exists for all distributions.

So it is as if the Laplace transform was somehow limited to the real numbers (needless to say the usual random variables are real, but the domain of the MGF is not the random variable), so as to capture the exponential nature of the distribution, while the characteristic function deals with the complex part.

If $f\cdot g$ is decomposable and $g$ is decomposable, is $f$ decomposable?

Posted: 03 Jul 2021 07:50 PM PDT

A decomposition of a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a sum of products of single-value functions: $f(x,y)=\sum_{i=1}^n u_i(x)\cdot v_i(y)$. If a function has at least one decomposition, it is called decomposable; otherwise, it is undecomposable.

It is straightforward to show that sums and products of decomposable functions are likewise decomposable. I do not know of specific undecomposable functions, but my intuition is that they exist and that most functions are undecomposable.

Question. I'd like to know: given $f\cdot g$ is decomposable and $g$ is decomposable, [when] is $f$ decomposable?

If that's hard, I'd appreciate any additional results about when products and sums involving undecomposable functions can result in decomposable functions.

  • So far, my intuition is that decomposable functions and undecomposable functions will have similar arithmetic closure properties to rational and irrational numbers.

  • I have been looking for a provably undecompsable function—besides functions like $x^y$, I've thought about using an infinite sequence of linearly-independent functions, because I've shown that if a function has a decomposition, it has a minimal decomposition where all the $u_i$ are linearly independent and all the $v_i$ are linearly independent. If a function were somehow an irreducibly infinite sum of pairs $u_iv_i$ where the $u_i$ are all linearly independent as are the $v_i$, it would not be decomposable.

  • I've also been looking into function spaces where these functions and their decompositions live. Based on what I've learned, I believe the space of decomposable functions is the tensor space $\mathscr{F}_X \otimes \mathscr{F}_Y$, consisting of linear combinations of pairs of functions of $x$ and of $y$, and this can be embedded in the vector space $\mathscr{F}_{X\times Y}$ of two-variable functions. But I have not yet found a way to prove my desired result by arguing generally about vector/tensor spaces.

Tensor Product of Quaternion Algebras is a Real Matrix Algebra

Posted: 03 Jul 2021 07:41 PM PDT

I found following isomorphisms in Cor. 13.24 in the book "Topological Geometry" (by Porteous): $$H \otimes H(2) \cong R(8)$$ $$H(2) \otimes H(2) \cong R(16)$$

($K(n)$ is the algebra of all real n x n matrices with entries from the algebra $K$. By 'algebra' I mean a unital associative ring.)

I cant find proofs for them. I am guessing that in general, $H(k) \otimes H(l) \cong R(2^{k+l})$ and that this can be proved inductively by first proving that $H \otimes H = R(4)$. Are my claims true? Can you provide some proofs?

I am aware of the representaion of $H$ as a subalgebra of $R(4)$ (https://en.wikipedia.org/wiki/Quaternion#Matrix_representations). But I am not sure if that can help here.

$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $

Posted: 03 Jul 2021 07:56 PM PDT

I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.

What do we call a "function" which is not defined on part of its domain?

Posted: 03 Jul 2021 08:18 PM PDT

Before the immediate responses come in, I realize that a properly defined function means that it is defined for every value in its domain.

My question is this: if $f:A\to B$ has the property $f(a)=b_1$ and $f(a)=b_2$, then it is often still called a function, but one which is "not well-defined".

If there is $b$ in $B$ such that there is no pre-image under $f$ then we say $f$ is "not surjective".

So what would we call a "function" which has the property that $f(a)$ is not defined for some $a$ in $A$? It seems like there should be a word for this, other than just saying $f$ is not a function.


Edit: I realize that a function which is not well defined is not actually a function. I'm talking about informal speak, for example in class how we say "let's check if this function is well defined" as though it were a function even if it weren't well defined. I'm wondering if there is an analogous phrase for maps which aren't defined on their whole domain. This is all informal, which is why I tagged it a soft question.

How to find one side of a rectangle, if I have other side and slope

Posted: 03 Jul 2021 08:02 PM PDT

So suppose I have the width of a rectangle that is 1000 cms and its slope that is at 20 degrees. How can I calculate the other length required for the height of this rectangle....

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