Thursday, December 16, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

An identity on binomial coefficients without using explicit formulas

Posted: 16 Dec 2021 02:53 AM PST

I need to prove that

$\ {n-1 \choose k-1} \ {n \choose k+1} \ {n+1 \choose k} = \ {n-1 \choose k} \ {n+1 \choose k+1}\ {n \choose k-1}$

without using $\ {n \choose k} = \frac{n!}{k!(n-k)!}$

so I am supposed to either use the idea similar to proving that $ (1-1)^n = 2^n = \sum \ {n \choose k} $ or find something that both RHS and LHS counts.

Unfortunately, I can't show any attempt here because all I could think of is to apply recurrence relations several times to both sides hoping to be able to see what is going on.

Limit of a sum becomes an integral

Posted: 16 Dec 2021 02:51 AM PST

Suppose I have a number $N>1$ and a torus $\mathbb{T}_{N} = \mathbb{Z}^{n}/N\mathbb{Z}^{n}$. Suppose $f$ is continuous on this torus. Is it true that: $$\frac{1}{N^{n}}\sum_{x\in \mathbb{T}_{N}}f(x) \to \int_{\mathbb{R}^{n}} f(x) dx$$ and $N \to \infty$?

Entire function bounded on real line and growth bounded by function of imaginary component

Posted: 16 Dec 2021 02:47 AM PST

Let $f\colon \mathbb{C} \to \mathbb{C}$ be an entire function taking reals to reals and that is bounded on $\mathbb{R}$. Suppose there exists a continuous function $g\colon \mathbb{R} \to \mathbb{R}$ satisfying $$ |f(x)| \le g(\text{Im}(x)) \quad\quad (*) $$ for every $x\in \mathbb{C}$.

One example is $$ f(x) = e^{-x^2}\,. $$ Of course, if $h\colon \mathbb{C}\to \mathbb{C}$ is an entire function taking reals to reals and satisfying $$ |h(x)|\le k(|x|)\,, $$ and $f$ satisfies ($*$), then the composition $h\circ f$ satisfies ($*$).

Can we say anything else about real entire functions satisfying ($*$)? I'm interested in a Paley–Wiener type theorem for this class of functions but broadly interested in simple examples of this class of entire functions.

Extension of the Gershgorin circle theorem for symmetric matrices and localization of positive eigenvalues

Posted: 16 Dec 2021 02:46 AM PST

In mathematics, the Gershgorin circle theorem can be used to localize eigenvalues of a matrix (including symmetric). Let $A$ be a real symmetry $n × n$ matrix, with entries $a_{ij}$. For $i∈{1,…,n}$ let $R_i$ be the sum of the absolute values of the non-diagonal entries in the $i$-th row:

$R_{i}=\sum _{j\neq {i}}\left|a_{ij}\right|$

Let $D(a_{ii},R_{i})\subseteq \mathbb {C}$ be a closed disc centered at $a_{ii}$ with radius $R_{i}$. Such a disc is called a Gershgorin disc.

Theorem: Every eigenvalue of $A$ lies within at least one of the Gershgorin discs $D(a_{ii},R_{i}).$

I was interested in the case when the center of the circle closest to the imaginary axis is in the right half-plane, but part of the circle also intersects the negative half-plane. In this case, the eigenvalue of the matrix may be negative, but satisfies the Gershgorin circle theorem.

enter image description here

Problem: What condition can be added to the Gershgorin theorem so that the eigenvalues localized with its help always lie only in the right half-plane, i.e. were only positive? This condition should also make it work in cases where the circle may slightly touch the left half-plane.

Meaning of the symbol ꓡꓶ

Posted: 16 Dec 2021 02:49 AM PST

I came across this expression in an algorithm that I am working on. p=M/2*pi m=ꓡpꓶ Could anyone help me with what this symbol means 'ꓡꓶ'? Thank you.

Can a discrete valuation ring be finitely generated over a field?

Posted: 16 Dec 2021 02:40 AM PST

In my homework of schemes, the professor proposed the following exercise:

"Let $ X $ be a scheme of finite type over a field $ k $ and $ f \in \mathcal{O}_X (X) $ a global section. Show that $ f $ is nilpotent $\iff f (x) = 0 $ for every $ x $ closed point in $ X $. Show by a counterexample that this assertion is false in general, that is, if we no longer assume $ X $ to be a scheme of finite type over a field."

Before proving the main assertion, I will exhibit a contre-example based of the spectrum of a Discrete Valuation Ring:

(Counterexample when $ X $ is not a scheme of finite type over a field):

Let $ X = Spec A $ with $ (A, m) $ a discrete valuation ring. Topologically $ X = \{x, \eta \}$ where the open sets are $ \{\emptyset, X, \{\eta \} \} $ and the closed ones are $ \{\emptyset, X, \{x \} \} $. Let $ \pi $ be a uniformizing element of $ A $, that is, a generator of the maximal ideal $ m$. We know that $ \pi $ is not nilpotent, but $ \pi (x) = 0 $ where $ x $ is the unique closed point of $X$.

We can even find examples where $A$ a $k$-algebra. This is the case, for example of $A = k [[T]] $. The problem here will be precisely the hypothesis of finite type which is not respected:

Indeed if $ X = Spec k [[T]] $ and $ f = \pi $, Proposition 2 does not apply. More precisely, let $ U = D (f) \cong Spec \; \kappa (\eta) $, then: $ \eta $ is a closed point on $ U $, whose residual field is $\kappa (\eta) $ is a transcendent extension of $k$, and which is not closed on $X$. The problem is that $ \mathcal{O}_U (U) $ and $\mathcal{O}_X (X) $ are not finitely generated $k$-algebras.

(My question): "Assuming that the main assertion of the exercise is true and that my contre-example based the spectrum of a DVR is right, I have the impression that this shows indirectly that a discrete valuation ring cannot be a finitely generated $k$-algebra.

It is evident for the case when $A = k [[T]]$, but is there a more direct proof of this apparently true fact?

If so, does this result have any relation with the fact that DVR over a field $k$ are usually stalks of the sheafs of regular functions in non-singular points of algebraic curves over $k$?

(Proof of the main assertion)

We start by recalling some properties of the nilpotent global sections of quasi-compact schemes:

(Proposition 1): Let $Z$ be a quasi-compact scheme. So for all $f \in \mathcal{O}_Z (Z) $, we have: $f$ vanishes on all points of $Z \iff f $ is nilpotent.

(Proof - Prop 1):

  1. $ (\Rightarrow) $ If $ Z $ is affine, it is straightforward from: $ \bigcap \limits_ {p \subset B} \, p = \sqrt {( 0)} $.

If $ Z $ is not affine, we take a finite cover $ Z = \cup_i U_i $ by affine open subschemes. Since the proposition is true for affine schemes, $ f_ {| U_i} $ is nilpotent for all $ i $.

By hypothesis, we only have a finite number of $ i $, so there exists a $ N $ such that: $ f_ {| U_i} ^ N = 0 $, $ \forall i $. Since $\mathcal{O}_Z$ is a sheaf, the $ f_ {| U_i} ^ N $ glue to $ f ^ N $, which is therefore, equal to $ 0 $.

  1. $(\Leftarrow)$ It's trivial.

In order to study the nilpotent global sections of a scheme of finite type over $k$. We first recall the following property:

(Proposition 2) Let $ X $ be a scheme of finite type over a field $ k $. For any point $ x \in X $, the following assertions are equivalent:

  1. $ x \in X $ is a closed point.
  2. $ k \hookrightarrow \kappa (x) $ is a finite extension.
  3. $ k \hookrightarrow \kappa (x) $ is an algebraic extension.

(Proof - Prop 2) Let $ U = Spec \, A $ be an affine open subscheme containing $ x $. In $ U $, the point $ x $ corresponds to a prime ideal $ p \subset A $. We then consider the compositions $ k \hookrightarrow A \to \frac{A}{p} \hookrightarrow Frac (\frac {A}{p}) = \kappa (x) $.

  • $(1 \Rightarrow 2) $: In this case, $x$ is also closed in $ U $ and $p $ is therefore a maximal ideal of $ A $. This implies that $ \kappa (x) = \frac {A} {p} $ and, it is both a field and a $ k $-algebra of finite type . According to the Nullstellesatz, $ k \hookrightarrow \kappa (x) $ is therefore a finite extension.

  • $ (2 \Rightarrow 3) $: Any finite field extension is algebraic.

  • $ (3 \Rightarrow 1) $: As $\kappa (x) / k $ is assumed to be algebraic, the composition $ k \ to \frac {A} {p} $ is an integral morphism. Since it is injective, it can be identified with an integral extension of integral domains. In this case: $ k $ is a field $\Rightarrow \frac {A} {p} $ is a field and, this shows that $ x $ is a closed point of $ U $ .

    We have just shown that if $ \kappa (x) / k $ is algebraic, then: for every affine open subset $ U \ni x $, $ x $ is a closed point of $ U $. It is a standard fact that a point is closed in a scheme if and only if it's closed in every affine open subscheme, hence $ x $ is a closed point of $ X $.

(Solution to the exercise):

Since $ X $ is quasi-compact, Proposition 1 shows that $ f \in \mathcal{O}_X (X) $ is nilpotent $ \iff $ $ D (f) = \emptyset $. In general, $ D (f) $ is a scheme of finite type over $ k $. Since $ D (f) $ is in particular Noetherian, $ D (f) = \emptyset \iff D (f) $ has no closed point.

By Proposition 2: $ D (f) $ has no closed point $ \iff $ No point in $D(f)$ is closed in $X \iff f (x) = 0 $, $ \forall x $ closed point in $ X $.

4 dimensional complex vector space | orthogonal vectors

Posted: 16 Dec 2021 02:33 AM PST

I never worked with complex numbers on this topic, so maybe someone can help me. Unfortunately it is german but I hope it isn't to complicated to translate. If so, ask me.

https://i.stack.imgur.com/txUFY.png

Thanks a lot.

How to find constants which are applicable to maths like e,π

Posted: 16 Dec 2021 02:43 AM PST

How to find constants like e which are applicable to maths.like e ,π π= circumference/ diameter=3.142... Nearest fractional form =22/7 I heard that some convergence equations Give this type of results. Any other constants or expressions Present in maths which give such constants.

Existence of a filtration with respect to which a sequence becomes a martingale difference.

Posted: 16 Dec 2021 02:40 AM PST

Suppose $(X_t,\mathcal F_t)_{t \in \mathbb Z}$ is such that $\mathbb E(X_t|\mathcal F_{t-1}) = 0$ for every $t$, where $(\mathcal F_t)$ is a filtration on a probability space $(\Omega,\mathcal F,\mathbb P)$. But $(X_t)$ is not necessarily adapted to $(\mathcal F_t)$.

Then is there any well-known situation or fact by which we can ensure the existence of some filtration $(\mathcal G_t)$ with respect to which $(X_t)$ is adapted and still $\mathbb E(X_t|\mathcal G_{t-1}) = 0$ for every $t$?

That is, is there any special example or condition under which we can make a sequence or an array of random variables a martingale difference by changing(enlarging) the original filtration? My naive guess is that one could realize this by using the natural filtration of $(X_t)$ and the original filtration $(\mathcal F_t)$

In a finite-dimentional Hausdorff locally convex vector space, how to prove there exists a seminorm which is a norm?

Posted: 16 Dec 2021 02:36 AM PST

Let E be a finite-dimensional Hausdorff locally convex vector space, and $e_1,\ldots, e_n$ is its basis. I know that from the Hausdorff and locally convexity, there exists a seminorm $p$ satisfying that $p(e_1)\geq 1,\ldots$, and $p(e_n)\geq 1$. I want to prove that for any $x\neq 0$, $p(x)>0$. This immediately means that $p$ is a continous norm on this tvs.

By the way, I know that a finite-dimensional locally convex vector space can be induced by a norm. This has been discussed in this post. @Stephen Montgomery-Smith has posted a detailed solution. But I really want to prove that:

If $p(e_1)\geq 1,\ldots$, and $p(e_n)\geq 1$, then for any $x\neq 0$, $p(x)>0$.

In fact, if the above is true, then the fact that a finite-dimensional locally convex vector space can be induced by a norm would be proved more directly!

Conditional expectation with proof of linearity

Posted: 16 Dec 2021 02:35 AM PST

I'm working on proving that

$E(aY+bZ | X) = aE(Y|X) + bE(Z | X)$

Where X, Y and Z are discrete random variables. Where we assume that all their (joint/marginal) probability mass functions and expectations exist

Approach:

So I figured I have to use linearity of expectations. What bugs me, is what to do with the conditional statement and the constants.

My thought is to use independence, so $p(y|x)= p(y)$

And the constants I thought I could move outside, since $E(a) = a$

Then I could proceed with the proof:

$E(X+Y) = \sum_x \sum_y (x + y)P_{XY}(x,y)$

$=\sum_x \sum_y x \cdot P_{XY}(x,y) + \sum_x \sum_y y \cdot P_{XY}(x,y) $

$= \sum_x x \cdot \sum_y P_{XY} (x,y) + \sum_y y \cdot \sum_x P_{XY} (x,y)$

$= \sum_x x \cdot P_{X}(x) + \sum_y y \cdot P_{Y}(y) $

$= E(X) + E(Y)$

My question is, I don't really know if I just can apply the use of independence like that and then proceed with the proof of $E(X + Y) = E(X) + E(Y)$?

Hope you can help!

Examine the uniform convergence of $\sum_{n=1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)}$

Posted: 16 Dec 2021 02:38 AM PST

The problem is stated as:

Examine the uniform convergence of $\sum_{n=1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)}$ on the interval $[0,\infty)$ and $[a,\infty)$ for $a>0$.

My attempt:

Case I: $[0,\infty)$

We form our partial sum $S_m(x) := \sum_{n=1}^{m} \frac{x}{(1+nx)(1+(n+1)x)}$, hence we have that:

$$ |S(x)-S_m(x)| = \sum_{n=m+1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)} $$

Picking $x_n = 1/n \in [0,1] \subset [0,\infty) \forall n \geq 1 $, we have that the series can be rewritten as:

$$ | S(x)-S_m(x)| = \sum_{n=m+1}^{\infty} \frac{1}{(2n)(2+1/n)} \geq \sum_{n=m+1}^{\infty} \frac{1}{4n+2}$$ which clearly diverges.

Therefore, we have found a counter example of uniform convergence within this interval, since if we would have uniform convergence, no matter what x we pick, the series must converge uniformly.

Case II: $[a,\infty)$

Notice that $$S(x) = \sum_{n=1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)} = \sum_{n=1}^{\infty} \left (\frac{1}{(1+2n)(1+nx)} - \frac{1}{(1+2n)(1+(n+1)x)} \right )$$

We can form our Majorant term by calculating the upper bound of the term in the series above and we have that:

$$\left |\frac{1}{(1+2n)(1+nx)} - \frac{1}{(1+2n)(1+(n+1)x)} \right | \leq \left (\frac{1}{(1+2n)(1+na)} + \frac{1}{(1+2n)(1+(n+1)a)} \right )$$

Let $M_n := \left (\frac{1}{(1+2n)(1+na)} + \frac{1}{(1+2n)(1+(n+1)a)} \right )$, and since both terms act asymptotically as $1/n^2$, it becomes evident that the series of terms given by $M_n$ does converge. By Weierstrass M - test, we have uniform convergence within this interval.

I hope you can give me some feedback on my solution. Maybe some tips on what to improve, and what steps went wrong. I'm really trying to get good at problems with uniform convergence, so any help would be appreciated!

Thanks.

Isometric axis problem : prove that $OS\perp PQ$

Posted: 16 Dec 2021 02:48 AM PST

Let quadrilateral $ABCD$ be inscribed in $(O)$. Let $I_1, I_2, I_3, I_4$ be the center of the circle inscribed in triangles $ABD, ADC, DBC, ABC$ respectively. $I_1I_3$ cuts $AC$ at$ P$, $I_2I_4$ cuts $BD$ at $Q.$ $I_1I_3 $cuts $I_2I_4$ at $S$.

  • Prove that $OS\perp PQ$.
  • Prove that $I_1I_2I_3I_4$ is a rectangle.

enter image description here

We have $\widehat{I_2CI_3} = \frac{\widehat{BCD}-\widehat{ACD}}{2} = \frac{\widehat{BCA}}{2}$

$\widehat{I_2DI_3} = \frac{\widehat{ADB}}{2} = \widehat{I_2CI_3} $

So we have $I_2I_3DC$ as inscribed quadrilateral. Similarly, we also have $I_1I_4BA,I_4I_3CB,I_1I_2DA$ as inscribed quadrilaterals.

$\Rightarrow \widehat{I_4I_3I_2} = 360^\circ-\widehat{I_2I_3C} - \widehat{I_4I_3C} = 90^\circ$

Similarly, we have $I_1I_2I_3I_4$ is a rectangle.

That's all I can infer, hope to get help from everyone. Thanks very much !

Difference between Theorems 2.1.4 and 2.1.5 in Tao, Analysis II (2015, 3e)

Posted: 16 Dec 2021 02:48 AM PST

To my untrained eye, (a) and (b) in Theorems 2.1.4 and 2.1.5 below seem the same.

What's the difference? What am I missing or why's he repeating himself? From pp. 28-29:

We now generalize much of the discussion in Chapter 9 . We first observe that continuous functions preserve convergence:

Theorem 2.1.4 (Continuity preserves convergence). Suppose that $\left(X, d_{X}\right)$ and $\left(Y, d_{Y}\right)$ are metric spaces. Let $f: X \rightarrow Y$ be a function, and let $x_{0} \in X$ be a point in $X$. Then the following three statements are logically equivalent:
(a) $f$ is continuous at $x_{0}$.
(b) Whenever $\left(x^{(n)}\right)_{n=1}^{\infty}$ is a sequence in $X$ which converges to $x_{0}$ with respect to the metric $d_{X}$, the sequence $\left(f\left(x^{(n)}\right)\right)_{n=1}^{\infty}$ converges to $f\left(x_{0}\right)$ with respect to the metric $d_{Y}$. (c) For every open set $V \subset Y$ that contains $f\left(x_{0}\right)$, there exists an open set $U \subset X$ containing $x_{0}$ such that $f(U) \subseteq V$.
Proof. See Exercise 2.1.1. Another important characterization of continuous functions involves open sets.

Another important characterization of continuous functions involves open sets.

Theorem 2.1.5. Let $\left(X, d_{X}\right)$ be a metric space, and let $\left(Y, d_{Y}\right)$ be another metric space. Let $f: X \rightarrow Y$ be a function. Then the following four statements are equivalent:
(a) $f$ is continuous.
(b) Whenever $\left(x^{(n)}\right)_{n=1}^{\infty}$ is a sequence in $X$ which converges to some point $x_{0} \in X$ with respect to the metric $d_{X}$, the sequence $\left(f\left(x^{(n)}\right)\right)_{n=1}^{\infty}$ converges to $f\left(x_{0}\right)$ with respect to the metric $d_{Y}$.
(c) Whenever $V$ is an open set in $Y$, the set $f^{-1}(V):=\{x \in X:$ $f(x) \in V\}$ is an open set in $X$.
(d) Whenever $F$ is a closed set in $Y$, the set $f^{-1}(F):=\{x \in X:$ $f(x) \in F\}$ is a closed set in $X$.
Proof. See Exercise 2.1.2.

Find degree of arc with four points on a circle

Posted: 16 Dec 2021 02:26 AM PST

As shown in the figure below, $\angle CAP = \angle CBP = 13^\circ$

$\overset{\huge\frown}{MA}=38^\circ$, find the degree measure for the arc $\overset{\LARGE\frown}{BN}$

Figure

Due to the fact that $\angle{CAP}$ and $\angle{CBP}$ are the same, we can make another circle passing over points $A,C,P,B$.

Now, my initial thought is that we can name one of the angles we have to be $\theta$, and try to make up an equation that can find $\theta$,

I tried a few, for example in the new circle we made:

$$\angle{CAB}+\angle{CPB}=180^\circ$$

where I named $\angle{APB}=\theta$

But After subsituting I get:

$$... + 2\theta + ... - 2\theta = 180 $$

Indicating I cannot find $\theta$ with the equation.

Note: I am bad at LaTeX, so if there's anything weird in my LaTex code feel free to improve it.

Is this a valid or invalid argument?

Posted: 16 Dec 2021 02:48 AM PST

I was trying to solve this problem:

Wearing a mask $(M)$ is known to prevent other people from getting infected $(I)$, e.g. $$(M) \to (\neg I),$$ if people get infected $(I)$ then the virus spreads $(S)$, e.g. $$I \to S$$; also is known that a virus spreads if and only if people do not wear a mask and travel e.g. $$(\neg M) ∧ (T) \leftrightarrow (S).$$

Do the previous premises entail that if virus spreads, then people are not wearing masks? e.g. $$(S)\to (\neg M).$$

Is that a valid argument? Is it invalid but satisfiable? Is it unsatisfiable?

Prove it and justify the method used.

Understand the mistake in solving the continuity problem

Posted: 16 Dec 2021 02:23 AM PST

I have seen more than one answer about this question but I would like to understand why my first idea of solution is not right!
Let $$ \displaystyle f(x) = \begin{cases} x^2 \qquad x \in \mathbb{Q} \\ -x^2 \qquad x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} $$ I want to prove that $f$ is continuous only in $x=0$. I know that in order to prove the function is not continuous in $x_0\neq 0$ I have to find two sequence $x_n\to x_0\in\mathbb{Q}$ with $f(x_n)\to l\neq x_0^2$ and $y_n\to x_0\in\mathbb{R} \setminus \mathbb{Q}$ with $f(y_n)\to t\neq -x_0^2$.

But my first idea was to consider that: $$\lim_{x\to x_0}f(x)=x_0^2\,\,\, \text{if}\,\, x_0\in\mathbb{Q}$$ $$\lim_{x\to x_0}f(x)=-x_0^2\,\,\, \text{if}\,\, x_0\in\mathbb{R} \setminus \mathbb{Q}$$ Then $x_0=-x_0^2\iff x_0=0$...but I think something is wrong...maybe this depends on the fact that I am considering an equality between a rational quantity ($x_0^2$) and a possible irrational one ($-x_0^2$).
Can you help me in understanding the mistake in this idea?

Are numbers of form $n^2-n+41$ prime numbers? [duplicate]

Posted: 16 Dec 2021 02:29 AM PST

Today searching about how to find prime numbers, I read on one website that numbers of form $6n\pm1$ are prime numbers and $n^2-n+41$ are prime numbers.

I tested first formula and found some discrepancy (pseudo prime). However, other formula looked correct, I even checked for small $n$ values and did not find discrepancy (pseudo prime numbers).

Is there any pseudo prime number with 2nd formula?

How to prove this Bessel equality?

Posted: 16 Dec 2021 02:35 AM PST

I've got a problem with proving this Bessel's equality: $$x^2 = 2\sum_{n=1}^{\infty} (2n)^2 J_{2n}(x)$$

The Bessel generating function is $\exp(\frac{x}{2}(t-t^{-1})) = \sum_{n=-\infty}^{\infty}J_{n}(x)t^n$.

I think the solution should is replace $t$ with some other forms and proving the equality by comparing the coefficient, but I can't find a way out.

Permutation of 5 girls and 5 boys

Posted: 16 Dec 2021 02:51 AM PST

  1. When no 2 girls wants to sit together. => Now we can use gap method here, $5! * {6 \choose 5} * 5! $

  2. When boy or girls want to seat alternatively. => There can 2 cases: $b_1g_1b_2g_2b_3g_3b_4g_4b_5g_5$ or $g_1b_1b_2g_2b_3g_3b_4g_4b_5g_5$

$\therefore$ $2*5!*5!$

  1. When all girls and all boys want to seat together: we can use string method here $2 * 5! * 5!$

  2. When all girls do not want to seat together: $10! - 6!5!$// Total - All girls are together using string method. Is it correct? Answer given is $10! - 5!*5!*2$

Riesz representation theorem in one dimension

Posted: 16 Dec 2021 02:53 AM PST

Let E be the bounded measurable set in R,1<p<$\infty$.
Show that
$\forall f\in L^p(E)^*,\exists y=y(t)\in L^q(E)$，p and q are conjugate exponents.
s.t.$\Vert f \Vert=\Vert y \Vert, f(x)=\int_Ex(t)y(t)dt,\forall x=x(t)\in L^p(E).$
I have proved the situation when E is any closed interval, and I want to generalize from the closed interval situation to any bounded measurable set in R.
My attempt is as follows
Obtaining the existence of a and b from the boundedness of E makes $E\subset[a,b]$.Considering Hahn Banach's theorem,since f is a bounded linear functional on E, then f can be extended to $L^P([a, b])$ with the norm invariant to obtain F s.t.$F|_Z=f$.Define a new function$$\widetilde{x}(t)=\begin{cases} x(t)&\ t\in E,\\ 0&\ t\in[a,b]-E. \end{cases}$$ Let the space formed by all such $\widetilde{x}(t)$be Z.
$$ Z\subset L^p([a,b]),f(\widetilde{x})=f(x),\exists y_0\in L^q([a,b])$$ s.t.$$\int_{[a,b]}\widetilde{x}(t)y_0(t)dt=F(\widetilde{x})=f(\widetilde{x})=f(x).$$ $$\Vert F \Vert=\Vert y_0 \Vert=\Vert f \Vert,f(x)=\int_E\widetilde{x}(t)y_0(t)dt=\int_Ex(t)y_0(t)dt.$$
I have no way to get y(t) on$L^q(E)$.Any hints would be greatly appreciated！

Beginner feedback on real analysis proof

Posted: 16 Dec 2021 02:24 AM PST

I am a bio student self-studying Abbott's Understanding Analysis and would love some feedback on one of my answers to an exercise. I have no experience writing proofs, and I'm used to plug-n-chug math taught by school, but I'm determined to get through this book as I find it fascinating. Thanks!

Q: If $x ∈ (A ∩ B)^c$, explain why $x ∈ A^c ∪ B^c$. This shows that $(A ∩ B)^c ⊆ A^c ∪ B^c$

Pf: If $x \in A\cap B$, then $x \in A,B$ and $x \in A\cup B$. We can think of $A\cap B$ as the collection of elements in both $A$ and $B$. The complement $(A\cap B)^c$ is therefore the set of elements not in $A$ and $B$, elements can still originate from $A$ or $B$, just not those in both.

The set $(A\cap B)^c$ is equal to $A^c\cup B^c$ because $A^c$ is the set of elements not in A (but, again, can contain elements in $B$). But, if an element is in A and also in B, neither $A^c$ nor $B^c$ will contain that elements. Thus, $A^c\cup B^c$ can be thought of as the set of elements not in $A$ and $B$, just as with $(A ∩ B)^c$.

The deformation as a functor

Posted: 16 Dec 2021 02:45 AM PST

I'm studying some deformation theory of associative algebras and my professor told me that the "set of deformations" of an associative algebra isn't in fact a set but more precisely a functor, but i'm having some troubles understanding how, so:

Let's consider $\mathbf{C}$ the category of some commutative algebras, and let's consider $\text{Def}(A)$ the deformation space of an element $A$ in our category.

How $\text{Def}:\mathbf{C}\to\mathbf{Set}$ is a functor?

If the question is too vage could you at least give me some more details?

Hope someone can help me.

Thanks!

Continuity and inequality

Posted: 16 Dec 2021 02:24 AM PST

Let $f,g : [0,T] \rightarrow \mathbb R$ non-negative functions s.t. $f$ is differentiable and $g$ integrable and satisfy $$f(t)f'(t) \le f(t) g(t), \qquad \forall t \in (0,T).$$ Does that imply $$f(t) \le f(a) + \int_a^t g(s) ds, \qquad \forall 0\le a \le t \le T?$$ Note that the first inequality only implies that $$\forall t \in (0,T), \quad (f(t)=0 \quad \text{or} \quad f'(t) \le g(t)),$$ while the second needs a stronger inequality. Maybe the continuity fixes this problem. Any hint?

Edit: The question is motivated by the following in "The mathematical theory of viscous incompressible flow by Ladyzhenskaia, page 88" enter image description here

Limit of an almost periodic function

Posted: 16 Dec 2021 02:46 AM PST

Consider the function: $$ f(t)=\sum_{\nu\in F} c_\nu e^{i \nu t} $$ where $F\subset \mathbb{R}$ is a countably finite or infinite set (but $F\neq\{0\}$) and $c_\nu\in \mathbb{C}$.

The question is to give a simple argument which shows that the limit $\lim_{t\to\infty}f(t)$ does not exist. A way to proceed would be to prove that $f(t)$ is almost-periodic: $$ |f(t)-f(t+\tau_\epsilon)|<\epsilon $$ for any $\epsilon$ and $t$. However I do not see a simple way to obtain this latter result avoiding some theorems and lemmas. Any ideas?

If we can't change the order of the terms of a conditionally convergent series, but can change the sign of terms, can we get whatever number we want?

Posted: 16 Dec 2021 02:35 AM PST

Let $\ \displaystyle\sum_{n\in\mathbb{N}} x_n\ $ be a conditionally convergent series, and we can choose $\ a_j=-1\ $ or $\ a_j=1\ $ for each $j\in\mathbb{N}.$

Let $\ \beta\in\mathbb{R}.$

Can we choose the $\ a_j\ $ so that

$$\sum_{n\in\mathbb{N}} a_n x_n = \beta\quad ? $$

If so, we can solve this question I came across this morning by letting $\ x_n = \left(-1\right)^n\frac{1}{p_n},\ $ where $\ p_n\ $ is the $\ n-$th prime number, because this is a conditionally convergent series. Edit: this is not what the question was asking.

Note that my question is different to the Riemann Series Theorem, which says that we can keep the terms, but rearrange the order. In my question we are changing the terms.

Solving $uu_{x_1}+u_{x_2}=u$ with boundary conditions

Posted: 16 Dec 2021 02:42 AM PST

I'm trying to solve the problem \begin{cases} uu_{x_1}+u_{x_2}=u & (x_1,x_2)\in U\\ u(x_1,x_2)=2x_1, &(x_1,x_2)\in\Gamma \end{cases}

where $U:=1<x_1<2,x_2>0$ and $\Gamma=1<x_1<2,x_2=0$. I think the relevant characteristic equations are $$\frac{dx_1}{u}=\frac{dx_2}{1}=\frac{du}{u}$$

I used these relations to solve two separable differential equations and came up with the solution $u=2e^{x_2}(x_1-ux_2)$.

EDIT for bounty : The solution is not correct. I would be very much interested in seeing a model solution method. I have myself used the above differential relations as well as the initial data $$x_1 =s, \enspace x_2=0,\enspace z=2s$$

and proceeding afterwards with the characteristic method.

Showing $\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\left(\frac{e}{2}-\frac{1}{e}\right)<1$ without a calculator

Posted: 16 Dec 2021 02:53 AM PST

Show that:

$$\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\left(\frac{e}{2}-\frac{1}{e}\right)<1$$

I have tried power series of exponential around $0$ wich is :

$$e^x=1+x+\frac{x^2}{2}+O(x^3)$$

We can pursue it with another order .

Edit :

An inequality due to Nanjundiah states for $n\geq 1$ a natural number:

$$\left(1+\frac{1}{n}\right)^{\left(n+\frac{1}{2}\right)}>e$$

Edit 2 :

An inequality due to Bennett states for $x$ a real number $m,n$ natural numbers and $m,n>x$ then :

$$\left(1+\frac{x}{m}\right)^{m}\left(1-\frac{x}{n}\right)^{n}<1$$

There is a little mistake in the statement of the edit 2 we need $x\neq 0$

For $\pi$ I have tried the continued fraction see wikipedia

Question: How to show it by hand without a calculator?

Constraint qualification for linear constraints [closed]

Posted: 16 Dec 2021 02:25 AM PST

How to formulate conditions under which the LI CQ is fulfilled at arbitrary feasible point of the set of feasible solutions given by linear inequalities $S=(x\in\mathbb{R}^{n}:Ax\le b, x\ge 0)$? I know I need to show that gradients of constraints are linear independent, however, it depends on the number of active inequalities. I do not know how to count with this fact. Any idea, thank you?

Equation of a plane, given two points and a perpendicular plane

Posted: 16 Dec 2021 02:51 AM PST

The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.

What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
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