Saturday, May 29, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Finding the values of the constants a and b

Posted: 29 May 2021 07:20 PM PDT

      { ax+3,    if x ≤ 5   f(x)={    8 ,    if x=5           to be continuous for all x        { x^2+bx+1, if x>5

I'm looking at a blank paper for minutes now, and I still don't understand how I am supposed to find the value of the constants. This is only the first question of the 10, I only need an example to have an answer to the rest. Can you help me with this?

Perfect Hash Function that detects invalid inputs?

Posted: 29 May 2021 07:14 PM PDT

I work in hardware/electronics where everything is power of 2.

To avoid storing the key along with the value, I can find a perfect hash function for the known-in-advance keys.

Nevertheless, the PHF doesn't tell me if a key is invalid (not part of the keys the PHF has been generated with).

Example:

I have 4 values: 2,7,12,61 They are all mapped uniquely to the output space via the PHF. Inputting 1,3..6,8... into the PHF obviously creates collisions.

  • Is there a way to detect these values?

If not, I need to store the key along with the value. If I can detect them, I don't need to store the key with the value and save memory.

I could for example reserve the key 0 (or any key the algorithm used to generate the PHF gives me), to detect such invalid key.

Limit of the given function

Posted: 29 May 2021 07:10 PM PDT

Please help me out with this limit given below-

\begin{equation} \operatorname{Lim}_{x \rightarrow 0^+}(\sin x)^{x} \end{equation}


I tried the following substitution-

\begin{equation} x \rightarrow \frac{\pi}{2}-x \end{equation} Then \begin{equation} \operatorname{Lim}_{x \rightarrow \frac{\pi}{2}^-} \frac{(\cos x)^{\pi / 2}}{(\cos x)^{x}} \end{equation} Which is of the form zero/zero so tried to apply L'Hospital's Rule but didn't reach to the answer.

Dedekind-infinite set stays infinite even if losing one element

Posted: 29 May 2021 06:57 PM PDT

I learn set theory and took notes from a proof in a book, so that it makes sense to me. Can someone look if my notes/thinking is (formally and mathematically but most importantly conceptually) correct? Where does it look fishy or underexplained?

Theorem: Be $A$ a (dedekind-) infinite set. Then $A\setminus{a}$ is an infinite set too.

Proof: Be $A' \subset A$ and $f: A \to A'$ bijective (since A is dedekind-infinite).

Because of $A' \subset A$ there is some $a \in A\setminus A'$.

We assume $g: f$ with $dom(f) = A\setminus {a}$. $g$ is injective because $f$ is bijective and therefore also injective which isn't changed by the take away from $a$ from $f$'s domain (but bijectivity might).

Now, $f(a) \notin rng(g)$ because $f(a) \in rng(f)$ and its preimage is $a \in A$, but such an element is impossible in $rng(g)$ because $g$ forbids a preimage $a \in A$.

Furthermore, $f(a) \neq a$ because of $rng(f) = A'$.

Furthermore $a \notin rng(g)$ because $rng(g) \subseteq A'$.

Furthermore and trivially $a \notin A \setminus a$, but $f(a) \in A \setminus a$, because $f(a) \neq a$ and else $rng(f) = A' \subset A$.

Because $rng(g) \subseteq A' \subset A$ and because $rng(g)$ misses $a, f(a)$ while $A \setminus a$ only misses $a$, we can conclude $rng(g) \subset A \setminus a$.

So is $g: f$ with $dom(f) = (A \setminus a \to A') = (A \setminus a \to rng(g))$ bijective?

Yes, because $g$ is injective (see above) and it's also surjective because of the very meaning of rng(g) to have at least a pre-image in its domain.

So we have $rng(g) \subset A \setminus a$ and $g: A \setminus a \to rng(g)$ bijective which makes $A\setminus a$ dedekind-infinite by Dedekind's definition. $q.e.d.$

This theorem even holds for any finite amount of n elements taken away from $A$ because you can just apply the proof technique iteratively: $A\setminus a$ can be proved to be dedekind infinite and so also $(A\setminus a)\setminus a$ and so on for $((A\setminus a) \setminus b)... \setminus n$.

Corrollary: Is $B$ (dedekind-) finite then so is $B \cup a$ because if $B \cup a$ was dedekind-infinite then our above theorem would make $B\setminus a$ (= $B$) infinite as well contrary to the assumption.

a closed form for the following product.

Posted: 29 May 2021 06:55 PM PDT

while solving some infinite products I recently encountered a terribly hard infinite product which is (product (1-e^(-(4n+1)pi)) ,n=0 to infinity) I tried many methods theorems but still can't obtain a closed-form or a perfect answer, but what I did obtain is an approximation which is as follows ≈ (4)^(1/16)/(e^(pi))^(1/24). Any help would be appreciated.

Criteria for the representability of a functor

Posted: 29 May 2021 06:54 PM PDT

Let $F: (Sch/S)^{op} \to Sets$ be a functor that is both a sheaf in the Zariski topology and has an open covering $(f_i: F_i \to F)_{i \in I}$, where each of the $F_i$ is representable.

In theorem 8.9 of Görtz-Wedhorn's book on algebraic geometry, it is proven that the functors $F_i$ glue to a scheme $X$. Then, because $F$ is a sheaf, the morphisms $f_i: F_i \to F$ glue to a morphism $f: X \to F$ - which the authors claim to be an isomorphism. Is there an elementary way to see that $f$ is an isomorphism? The book doesn't use universal objects, so an answer that doesn't address them would be great.

Finding unknown values given cumulative distribution for set of data.

Posted: 29 May 2021 06:55 PM PDT

enter image description here

I am confused on the correct answers to this problem.

The data is given by the age of the first 44 presidents upon inauguration -$(57,61,57,57,58,57,61,54,68,51,49,64,50,48,65,52,56,46,54,49,51,47,55,55,54,42,51,56,55,51,54,51,60,61,43,55,56,61,52,69,64,46,54,47)$

1.For 1. would the answer be $.2$ or $.22$? Where should I read the value off on the graph? I know the cumulative distribution function is right continuous and if $F$ is the cumulative distribution function then $F_{44}(50)=\frac{1}{44}\{\text{number of observations less than or equal to 50}\}$. I'm guessing it should be around $.22$.

enter image description here

Any suggestions?

  1. Again I am confused because of the right continuity of the graph. Would the answer be $.2$ or $.22$? I am guessing it should be exactly $.2$.

enter image description here

  1. I suppose the answer would be about $51$ based on the graph. Is this correct?

enter image description here

Inverse function theorem /local inverse

Posted: 29 May 2021 06:39 PM PDT

Let $T: {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ be a continuous, linear and bijective mapping and let $f(\vec{x})$ be a continuously differentiable function, such that $$ \exists C > 0, \forall \vec{x} \in {\mathbb{R}}^{n}, ||f(\vec{x})|| \leq C|| \vec{x} ||^{2}$$ show that $$g(\vec{x}) : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$$

with

$$ g(\vec{x}) = T\vec{x} + f(\vec{x})$$ has a continuously differentiable inverse mapping in a neighborhood of $\vec{0}$.

———

I think I need to show that the Jacobian of g(x) is regular, but how can I show that exactly and how to gon from there?

I have difficulty solving with $xt$ on the right hand side.

Posted: 29 May 2021 06:46 PM PDT

$$\tag{1}u_t - u_{xx} = xt; ~0<x<\pi $$

$$ \tag{2}u(x,0)=1 ; ~ 0\le x \le\pi$$

$$\tag{3} u_x(0,t)=0; ~ u_x(\pi,t) =0 $$

How can I find the solution u using separation of variable?

Conway exercise 5

Posted: 29 May 2021 06:56 PM PDT

Lef $f$ be analytic on $\overline{B}(0,R)$ with $|f(z)|\leq M$ for $|z|\leq R$ y $|f(0)|=a>0$. Show that the number of zeros of $f$ in $B(0;R/3)$ is less than or equal to $$\frac{\log(M/a)}{\log2}.$$

(Hint: If $z_1,\ldots,z_n$ are the zeros of $f$ in $B(0;R/3)$, consider the function $$g(z)=f(z)\left[\prod_{k=1}^n \left(1-\frac{z}{z_k}\right)\right]^{-1}$$ and note that $g(0)=f(0)$) It is in the section of The maximum modulus theorem (mmt), so I must use that theorem.

My idea: I think that since $g$ is holomorphic and doesnt have zeros in $B(0;R/3)$ then $$g(z)=e^{s(z)}$$ with $s(z)$ holomorphic in $B(0;R/3)$. Then I don't know how to relate it with the zeros of $f$. Any suggestions? I think the theorem of Rouche may help, but I don't know.

When the radius of the circle tends to infinity, does the circle tend to be a straight line?

Posted: 29 May 2021 06:42 PM PDT

When the radius of the circle tends to infinity, does the circle tend to be a straight line? Because the curvature of a circle is 1/R, when R tends to infinity, 1/R tends to zero.

Could You Please Check My Answers to These Exercises?

Posted: 29 May 2021 06:43 PM PDT

I have just started reading Elements of Set Theory by Herbert Enderton, and his book includes exercises for readers but has no answer key. I'm just getting started and trying to make sure I understand. Could you be so kind as to validate these answers for me?

  1. {∅} is both an element and a subset of {∅,{∅}}
  2. {∅} is a subset of {∅,{{∅}}}
  3. {{∅}} is a subset of {∅, {∅}}
  4. {{∅}} is an element of {∅,{{∅}}}
  5. {{∅}} is neither an element nor a subset of {∅,{∅,{∅}}}

If I have erred in any of these, could you please let me know and identify the error? Thank you!

Constant distance ratio from all points of a circumference to a point inside and outside of it

Posted: 29 May 2021 06:44 PM PDT

given a circle of radius $r$, centered at the origin, and a point $P_1 = (x_1,y_1)$ inside the circle ($x_1^2 + y_1^2 < r^2$), find a point $P_2 = (x_2,y_2)$ outside of it ($x_2^2 + y_2^2 > r^2$) such that, the ratio of the distances from any point $P = (x,y)$ of the circumference (x^2 + y^2 = r^2) is constant: \begin{equation*} \frac{d(P,P_1)}{d(P,P_2)} = k, \;\;\;\; k \in \mathbb{R}^+ \end{equation*} what i want to know is if $P_2$ exists, if its unique and if a formula can be found to determine it from the coordinates of $P_1$ and the value of $r$.

regarding the motivation for this problem, i was thinking about this as it relates to poles and zeros of Z transfer functions in all pass filters, i know and have confirmed that in the case of the unit circle, with $P_1 = re^{i\theta}$ (0 < r < 1) then choosing $P_2 = \frac{1}{P_1^*} = \frac{1}{r}e^{i\theta}$ satisfies the condition and k = r. with that i have the existence and formula part of my question answered but i wish to know if there is more than one possible choice of $P_2$ and also how to come up with the solution in the first place.

How to prove than $a+b+c = 2^n-1$ and $a^2+b^2+c^2 = (4^n-1)/3$ have integer solutions only with Mersenne exponent or exponents of Mersenne exponent?

Posted: 29 May 2021 06:48 PM PDT

I noticed something with Mersenne numbers : you can write it with the form $a+b+c = 2^n-1$ and $a²+b²+c² = (4^n-1)/3$ when $n$ is a odd Mersenne exponent (3, 5, 7, 13 ...) or an exponent of a odd Mersenne exponent ($3^2, 5^4, 7^3 ...$)

For example with Mersenne exponent :

  • $4+2+1 = 7 = 2^3-1$ and $4^2+2^2+1^2 = 21 = (4^3-1)/3$
  • $14+9+8 = 31 = 2^5-1$ and $14^2+9^2+8^2 = 341 = (4^5-1)/3$
  • $2760+2761+2670 = 8191 = 2^{13}-1$ and $2760^2+2761^2+2670^2 = 22369621 = (4^{13}-1)/3$

And exponent of Mersenne exponent :

  • $44732914+44738572+44746241 = 134217727 = 2^{27}-1$ and $44732914^2+44738572^2+44746241^2 = 6004799503160661 = (4^{27}-1)/3$ with $27 = 3^3$

You can notice there are each time two even numbers and one odd number for $a, b$ and $c$.

Another observation : it seems there are no integer solution for $a, b, c$ when $2^n-1$ a composite Mersenne numbers like $2^{11}-1$. Apparently, you can't write $a+b+c = 2^{11}-1$ and $a^2+b^2+c^2 = (4^{11}-1)/3$ for this case.

Is there a way to explain that ? I don't know to start for proving it.

Limit of the series $\sum_{k=1}^\infty \frac{n}{n^2+k^2}.$

Posted: 29 May 2021 06:43 PM PDT

I am trying to evaluate $$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$ Now I am aware that clearly $$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$ but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as $$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$ But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as $\frac{\pi}{2}$ but I am unaware of the steps and logic to get there.

Locus where $v: \mathscr{E} \to \mathscr{F}$ is surjective

Posted: 29 May 2021 06:53 PM PDT

Let $\mathscr{E}, \mathscr{F}$ be quasicoherent modules over a scheme $S$, $\mathscr{F}$ of finite type, and $v: \mathscr{E} \to \mathscr{F}$ a homomorphism. Consider the functor $F: (Sch/S)^{op} \to Sets$ defined by $F(T) = \{f \in \operatorname{Hom}_S(T, S): f^*(v) \mbox{ is surjective} \}$, for $S$-schemes $T$. I want to show that $F$ is represented by an open subscheme of $S$ (this is proposition 8.4 in Görtz-Wedhorn).

To do so, it is sufficient to find an open subschemes $U \subseteq S$ such that $f: T \to S$ factors through $U$ iff $f^*(v)$ is surjective. As $\mathscr{F}$ is of finite type, $\operatorname{Supp}(\operatorname{coker}(v))$ is closed. Let $U = S \setminus \operatorname{Supp}(\operatorname{coker}(v))$. If $f$ factors through $U$, then, for $t \in T$, $\operatorname{coker}(f^*(v))_t = f^*(\operatorname{coker}(v))_t = coker(v)_{f(t)} \otimes_{\mathscr{O}_{S, f(t)}} \mathscr{O}_{T, t} = 0$ - which implies that $f^*(v)$ is surjective.

I am having trouble proving the reciprocal: if $f^*(v)$ is surjective, then $f$ factors through $U$. The main reason of my struggle is because $\operatorname{coker}(f^*(v))_t =coker(v)_{f(t)} \otimes_{\mathscr{O}_{S, f(t)}} \mathscr{O}_{T, t} = 0$ doesn't seem to imply that $coker(v)_{f(t)} = 0$. Can anyone help me?

Solving $|\frac{x+1}{x-2}|<1$ and $\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$.

Posted: 29 May 2021 07:15 PM PDT

I want to find the real number such that $|\frac{x+1}{x-2}|<1$ and find the values $w$ such that :

$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$$.

By playing a little bit with this first inequality I have found out that the real numbers $(-\infty,0]$ satisfy the inequality. But when Im trying to prove this I got te following:

$|\frac{x+1}{x-2}|<1$ means that $-1< \frac{x+1}{x-2}<1$ from the left inequality I got $-x+2<x+1$. Then I got $\frac{1}{2}<x$. From the right inequality I got that $x+1 < x-2$, but this gives me no information for $x$ since its cancelled. So what can I conclude from the right inequality? Does all the real numbers satisfy this inequality are $x \in (- \infty, \frac{1}{2})$?

For the second equality I got that:

$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}= \frac{6w+7-6w-1}{(2w+1)(2w)}=\frac{6}{4w^{2}+2w}$$

Then I got $0=4w^{2}+2(w)-6$ which can be solved by the quadratic formula but this gives me values $\frac{-6}{4}$ and $1$. But clearly when $w=1$ the equality doesnt hold.

But Im stuck there. Thanks!

How to prove limits by using the formal definition (epsilon-delta)??

Posted: 29 May 2021 07:11 PM PDT

I was told to prove the limit of (10-2x)=16, as x=-3.

It says to use the formal definition (epsilon-delta).

This is what I tried so far, but I keep getting stuck. Can you help me solve this?

if |x-(-3)<delta then |(10-2x)-16<epsilon  Let delta= epsilon/2  if |x+3|< epsilon/2 then -epsilon/2 <x+3 <epsilon/2  So, -epsilon < 2x-6 <epsilon

Quasilinear PDE with Tricky Characteristcs [closed]

Posted: 29 May 2021 06:46 PM PDT

I have been working through solving quasilinear PDEs, and I am stuck solving the problem:

$$ (x+y) u_x+u_y=\left(u^{2}+x\right)\left(u^{2}+y\right), $$

with the parameterization,

$$ x = s+\kappa, \quad y = -s, \quad u=s^2. $$

where $s$ is the parameter and $\kappa\in\mathbb{R}$.

Normally I would compute the characteristics

$$ \frac{dy}{dx} = 1/(x+y), \quad \frac{du}{dy} = 1/\left(u^{2}+x\right)\left(u^{2}+y\right) $$ however, as the first ODE $$y'=1/(x+y)$$ cannot be solved in terms of elementary functions, and the second ODE is even less obliging, I'm wondering whether this problem is tractable using the method of characteristics, or whether there is some kind of 'trick' which I am missing.

If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following would be true

Posted: 29 May 2021 07:07 PM PDT

If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following is true:
A) $\frac{\sin ^{-1} x}{1+\cos ^{-1} y}=\frac{\pi}{2}$
B) $\frac{x+y}{\sin 1}=-2$
C) $\sin ^{-1} x+\sin ^{-1} y=1$
D) $\frac{x+y}{\cos 1}=1$

My Approach: Well I couldn't do really much. I tried to consider approaches related to factorising the given expression but that doesn't work in this case. How should I try to approach this problem? Any hints/solutions are appreciated. Thanks!

Computing the total variation for a multivariable function

Posted: 29 May 2021 06:37 PM PDT

I am trying to write an example computation with multivariable total variation to include in my functional analysis notes using the following definition from Wikipedia:

Let $\Omega$ be an open subset of $\mathbb{R}^n$. Given a function $f$ belonging to $L^1(\Omega)$, the total variation of $f$ in $\Omega$ is defined as $$V(f,\Omega):=\sup \left \{ \int_\Omega f(x) \text{div} \phi(x) dx : \phi \in C_c^1(\Omega, \mathbb{R}^n), ||\phi||_{L^\infty(\Omega)} \leq 1\right\}$$ $C_c^1(\Omega, \mathbb{R}^n)$ is the set of continuously differentiable vector functions of compact support contained in $\Omega$, $||\cdot||_{L^\infty(\Omega)}$ is the essential supremum norm, and $\text{div}$ is the divergence operator.

I would like to use this formula directly and demonstrate the process of taking the supremum. I understand that if $f$ is $C^1$ on $\overline \Omega$, then the formula for total variation is simplified to the computation $V(f, \Omega) = \int_\Omega |\nabla f(x)| dx$, which I am not trying to use here.

So, the problem is computing the total variation of $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x,y) =\frac{xy}{x^2 + y^2}$ on $\Omega$, where $\Omega$ is the open unit disk in $\mathbb{R}^2$. So, $\Omega = \{x : x \in \mathbb{R}^2 \text{ and }||x|| < 1\}.$ Here is a visual of this situation:

My Attempted Solution.

$f$ is a classic example of a function discontinuous at 0, so $f \notin C^1(\overline \Omega)$. We first show that $f \in L^1(\Omega)$. Recall $f \in L^1(\Omega) \iff \int_\Omega |f| < \infty$. So, $$\begin{align} \int_{\Omega} |f| &= \iint_D |f(x,y)| dA \\ &= \int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \left \lvert \frac{xy}{x^2 + y^2}\right \rvert dy dx \\ &= 0 < \infty \qquad \text{(C.A.S)}\end{align}.$$ Hence, $f \in L^1(\Omega)$. It remains to compute $$V(f,\Omega):=\sup \left \{ \int_\Omega f(x,y) \text{div} \phi(x,y) dx : \phi \in C_c^1(\Omega, \mathbb{R}^2), ||\phi||_{L^\infty(\Omega)} \leq 1\right\}$$

Before taking the supremum over $\phi \in C_c^1(\Omega, \mathbb{R}^2)$, we attempt the following simplification, $$\begin{align*} \int_\Omega f(x,y) \text{div} \phi(x,y) dx &= \iint_D f(x,y) \left(\nabla \cdot \left(\phi_x, \phi_y\right)\right) dA \\ &= \iint_D f(x,y) \left(\frac{\partial \phi_x}{\partial x} + \frac{\partial \phi_y}{\partial y}\right) dA \\ &= \iint_D f(x,y) \frac{\partial \phi_x}{\partial x} dA + \iint_D f(x,y) \frac{\partial \phi_y}{\partial y} dA \end{align*} $$

This is where I am stuck in terms of working with $\phi(x,y)$. It boils down to two main questions:

  1. Given what we know about $\phi$, can the above be simplified any further to an expression that does not depend on $\phi$? (So that we don't have to take a supremum) Otherwise,

  2. If we do have to take a supremum over the vector fields $\phi$, how would one go about this? I understand in the single variable case of total variation, one is simply constructing a family of partitions on an interval that enable you to take the supremum over all partitions. But here, how would I go about constructing a family of vector fields that enable me to take the supremum required?

Significant Proofs Understandable to Middle School Students with Pre-Algebra Background

Posted: 29 May 2021 07:16 PM PDT

I'm interested in introducing my middle school children to proofs using significant examples (i.e., not just basic geometry proofs in an intro to trig textbook). They have a pre-algebra background (so, polynomials, powers, greatest common factor, etc.). I can think of several proofs off the top of my head:

Proof of the irrationality of $\sqrt{2}$

Cantor's diagonalization proof

Pythagorean's rearrangement proof (and other proofs of the Pythagorean theorem).

What other significant proofs are there that would be explainable to someone with a pre-algebra background?

The space $\mathbb{R}^\mathbb{N}$ is Separable with respect to product topology

Posted: 29 May 2021 06:54 PM PDT

I know this is metrizable. How can I prove that it is separable? And the topology doesn't come from any norm.

On two-dimensional maximal commutative subalgebra of the complexification of the Heisenberg algebra

Posted: 29 May 2021 07:08 PM PDT

I have considered the Heisenberg algebra $\mathfrak{h}$, of $3 \times 3$ real upper triangular matrices with zero diagonal, and its complexification $\mathfrak{h}_{\mathbb{C}}$, which turns out to be the space of $3 \times 3$ complex upper triangular matrices with zero diagonal.

I now consider a two-dimensional commutative subalgebra $\mathfrak{g}$ of $\mathfrak{h}_{\mathbb{C}}$. I have proved that this is a maximal commutative subalgebra. Now, I want to prove that there is some $Z \in \mathfrak{g}$ such that the linear transformation $ad_Z : \mathfrak{h}_{\mathbb{C}} \rightarrow \mathfrak{h}_{\mathbb{C}}$ is not diagonalizable.

To see this, I first observed that the center of $\mathfrak{h}_{\mathbb{C}}$, which I will denote by $\mathfrak{z}$ is contained in $\mathfrak{g}$. Now, we know that the center $\mathfrak{z}$ is spanned b an element $H = \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right]$. Let $X \in \mathfrak{g} \setminus \mathfrak{z}$. Then, $\mathfrak{g} = \text{span} \left\lbrace H, X \right\rbrace$. Clearly, there is some $Y \in \mathfrak{h}_{\mathbb{C}} \setminus \mathfrak{g}$ such that $\mathfrak{h}_{\mathbb{C}} = \text{span} \left\lbrace H, X, Y \right\rbrace$.

Now, we notice that if we are to find an element $Z \in \mathfrak{g}$ such that $ad_Z$ is not diagonalizable, then it would have to be a linear combination of $H$ and $X$. But since $H$ is in the center of $\mathfrak{h}_{\mathbb{C}}$, we have $ad_H = 0$. Therefore, $ad_Z = \alpha ad_X$, for some $\alpha \neq 0$. Hence, my guess is that $ad_X$ is not diagonalizable (because that is what we want to prove).

However, when we compute the action of $ad_X$ on the basis $\left\lbrace H, X, Y \right\rbrace$ of $\mathfrak{h}_{\mathbb{C}}$, we get

$$ad_X \left( H \right) = 0, ad_X \left( X \right) = 0, ad_X \left( Y \right) = \alpha H + \beta X + \gamma Y,$$

where at least one of $\alpha, \beta, \gamma$ is non-zero. That is, the matrix of $ad_X$ in this basis is

$$ad_X = \left[ \begin{matrix} 0 & 0 & \alpha \\ 0 & 0 & \beta \\ 0 & 0 & \gamma \end{matrix} \right].$$

Hence, the eigenvlues of $ad_X$ are $0, 0, \gamma$ with the corresponding eigenvectors $H, X$ and $\left( \dfrac{\alpha}{\gamma} \right) H + \left( \dfrac{\alpha}{\gamma} \right) X + Y$, provided $\gamma \neq 0$.

That is, if $\gamma \neq 0$, then clearly $ad_X$ is diagonalizable and so is $ad_Z$ for every $Z \in \mathfrak{g}$. Thus, I now want to prove that $\gamma = 0$. However, I am unable to do so. Any hints about this will be appreciated!

Find $\int_{0}^{\infty} \frac{\sin x}{x+x\cos^2 x }\,\mathrm{d}x$

Posted: 29 May 2021 07:06 PM PDT

I have just done these, but I don't know what to do next......

\begin{align} \int \frac{\sin x}{x+x\cos^2 x}\,\mathrm{d}x & = \int \frac{1}{x}\cdot\frac{\sin x}{1+\cos^2 x}\,\mathrm{d}x\\ &=\int \frac{1}{x}\,\mathrm{d}(\arctan(-\cos x))\\\ &=\frac{1}{x} \arctan(-\cos x)-\int -\frac{1}{x^2} \arctan(-\cos x)\,\mathrm{d}x \\ &=\frac{1}{x} \arctan(-\cos x)-\int \frac{\arctan(\cos x)}{x^2}\,\mathrm{d}x \end{align}

Compound interest with % withdrawal

Posted: 29 May 2021 07:14 PM PDT

One has $P$ cash to invest in the market. They put 1/6 of $P$ in and set 5/6 aside. They get an $r$% return. The next day they put 1/6 of their new total cash into the market and get an $r$% return. This repeats, compounding daily for a year.

How much money do they have at the end of the year, including the amount they set aside?

How to calculate $\binom{852467439}{426} \mod{289}$?

Posted: 29 May 2021 07:10 PM PDT

How to calculate $\binom{852467439}{426} \mod{289}$?
My answer is 102, but the correct answer is 238.
I used Theorem 1. from Granville's paper but there is something wrong in my solution. http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/node2.html#SECTION00020000000000000000

This is the table of $(n!)_{17}$

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288
1 2 6 24 120 142 127 149 185 116 120 284 224 246 222 84 84 67 117 28 10 220 147 60 55 274 173 220 22 82 230 135 120 120 154 53 227 245 18 142 42 30 134 116 18 250 190 161 86 254 254 203 66 96 78 33 147 145 174 36 173 33 56 116 26 271 239 239 18 104 159 177 205 142 246 200 83 116 205 216 156 76 239 135 135 50 15 164 146 135 147 230 4 87 173 135 90 150 111 118 69 69 171 155 91 109 103 142 161 81 32 116 103 182 122 280 103 16 16 186 253 232 214 237 147 26 123 138 173 237 124 184 196 254 188 188 35 206 23 41 1 142 76 251 270 116 1 148 88 195 256 186 186 33 202 11 282 50 147 111 242 189 173 50 158 218 281 101 18 18 188 257 244 262 188 142 280 132 219 116 188 114 54 110 120 67 67 169 151 79 61 152 147 196 72 240 173 152 192 252 77 237 137 137 52 19 176 194 86 142 195 13 168 116 86 80 20 25 273 237 237 16 100 147 129 254 147 281 191 2 173 254 226 286 162 84 256 256 205 70 108 126 273 142 110 183 117 116 273 46 275 229 137 118 118 152 49 215 197 67 147 77 21 53 173 67 260 31 247 220 86 86 69 121 40 58 171 142 25 64 66 116 171 12 241 144 1 288

And here $n=852467439, m=426, r=852467013\,(n=m+r)$ in base $17$, respectively

.. 0 1 2 3 4 5 6 7 8
$n_i$ 8 5 10 10 6 5 1 2 0
$m_i$ 1 8 1 0 0 0 0 0 0
$r_i$ 7 14 8 10 6 5 1 2 0

$N_i = n_i + 17.n_{i+1}$
Also make the corresponding definitions for $M_i, m_i, R_i, r_i \,(n = m+r)$

And here is the table of $\frac{(N_i!)_p}{(M_i!)_p.(R_i!)_p}(i=0,...,8)$

i 1 2 3 4 5 6 7 8
$\frac{(N_i!)_p}{(M_i!)_p.(R_i!)_p}$ 1 42 141 1 1 1 1 1

My answer is $102$, but the correct answer is $238$

rectangular paddock, dimensions, maximise area it encloses

Posted: 29 May 2021 07:01 PM PDT

Having trouble trying to work out a question which involves finding a function to graph evidence of the correct answer, any advice would be greatly appreciated. I am struggling with part 'b' a lot, please help in any way. Ques. 1 a)A farmer has 100 metres of fencing materials and wishes to make a rectangular paddock. Find the dimensions of the paddock which will give him the maximum area. Use some form of electronic technology to justify why your answer is the maximum. b) The farmer wants to make another rectangular paddock along a straight stretch of river. If he uses the river as one side of the paddock and again uses 100m of fencing materials, what are the dimensions of the paddock which will maximise the area it encloses?

mean value inequality proof

Posted: 29 May 2021 06:40 PM PDT

I am currently working through a book on differential topology and Lie groups on my own.This features in the appendix on multi-variable calculus prerequisites. I am trying to go through an outline of the proof given and reason out every step. This question is a bit long, so please bear with me. The statement is as follows:

Given $ f: (a,b) \rightarrow E $ where $ E $ is a Banach space and $ f $ is differentiable, we have ,

$$ ||f(y)-f(x)|| \leq |y-x| \sup_{0\leq t \leq 1} ||f'(x+t(y-x))|| $$

$ \forall (x,y) \in (a,b) $

Now the proof runs as follows:

The author takes an $ M\gt M_0 =\sup_{0\leq t \leq 1} ||f'(x+t(y-x))|| $ and the set

$$ S = \{ t \in [0,1]:||f(x+t(y-x))-f(x)|| \leq Mt|x-y| \} $$

This construction didnt seem natural to me.Next the author claims that S is closed. I presume that if we have a limit point $ t' $ of $ S $ and consequently a sequence $ (t_n) $ in $ S $ , then by continuity of $ f $ and the right side of the inequality , we have $ t' $ in $ S $. Is this right?? Then as $ f $ is differentiable on $ (x,y) $, given $ \epsilon \lneq $ $ M-M_0 $, for all $ t $ near $ s $ and $ t \gt s $, we have:

$$ ||f(x + t(y-x))-f(x+s(y-x))-f'(x+s(y-x))(t-s)(y-x)|| \leq \epsilon |t-s||y-x| $$

I get that this inequality is due the Frechet derivative definition. But why $ t \gt s $? Its then shown that $ t \in S $ and hence $ s=1 $. How's this??

Thanks in advance.

Edit: Sorry, I have left out a glaring detail:that $ s = \sup S $ which exists in $ S $ obviously as $ S $ is closed and bounded.

$\mathbb{A}^{2}$ not isomorphic to affine space minus the origin

Posted: 29 May 2021 07:08 PM PDT

Why is the affine space $\mathbb{A}^{2}$ not isomorphic to $\mathbb{A}^{2}$ minus the origin?
Posted by at

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)