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Spectral norm inequality Suppose I have two matrices A, B
$$
A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{m \times n}
$$
Then on what conditions on A and B will the followig ineqiality hold:
$$
||A+B||_2 \geq ||A||_2
$$
I for some reason feel that this would hold when the spaces spanned by the columns these matrices are different. But maybe I am wrong, or I cannot mathematically articulate it.
|
You are wrong. In particular, if we take
$$
A = \pmatrix{1&100\\0&0}\\
B = \pmatrix{-1&-100\\1/100&1}
$$
Verify that the column space of the two matrices is different, but $\|A + B\| < \|A\|$.
|
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|
sums and product of density functions are density?
Given $f,g$ to be density functions. If the function $\psi = \lambda f +(1-\lambda)g$, for $\lambda \in [0,1]$, a density function? If the product $fg$ is also a density function?
I know that a random variable $X$ is called continuous if its distribution function can be expressed as
$$F(x) = \int_{-\infty}^xf(u) du,$$
for some $f: \mathbb{R} \to [0,\infty)$ called the (probability) density function of $X$.
To show that $\psi$ is a density function, I only need to show $\psi$ is integrable, and that the integral is equal to $1$, that is,
$$\int_{-\infty}^{\infty} \psi =1.$$
We have
$$\int_{-\infty}^{\infty} \psi = \lambda\int_{-\infty}^{\infty} f + (1-\lambda)\int_{-\infty}^{\infty} g.$$
Now, We have that since $f,g$ are density functions, then
$$\int_{-\infty}^{\infty} f(u)du = \lim_{x \to \infty} F(x) = 1,$$
and
$$\int_{-\infty}^{\infty} g(u)du = \lim_{x \to \infty} G(x) = 1,$$
thus $$\int_{-\infty}^{\infty} \psi = \lambda + 1 - \lambda = 1,$$ as required. Also, $\psi \geq 0$, because the image of $f,g$ are non-negative and $\lambda \in [0,1]$. For the other part, I was given a hint that the product $fg$ is not necessarily a density function. For example, let $f=g=1$ for $x \in [0,1]$, then $fg=1$ for $x \in [0,1]$. I just cannot see clearly, why this example is not a density function? Can someone show me why is it not a density function?
Thanks in advance.
|
I think the example $f=g=1$ on $[0,1]$ is not a counterexample— it's a positive example of two density functions whose product is also a density function. A counterexample might be something like:
Let $f(x) = 1$ if $x \in [0, 1]$, and zero elsewhere. Let $g(x) = 1$ if $x \in [2,3]$ and zero elsewhere. The product $fg$ is zero everywhere, and is not a density function. (The trick here is that the density functions don't "overlap"— they have nonoverlapping support.)
A side note: in order for $\psi$ to be a density function, its total integral must be equal to one (not just finite):
$$\int_{-\infty}^\infty \psi(x)dx = 1.$$
And $\psi$ must also have $\psi(x) \geq 0$ everywhere.
When $\lambda \in [0,1]$, the convex sum $\lambda f + (1-\lambda)g$ is a density function in this way, as you point out.
|
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|
Tangent planes of a surface parallel to a fixed direction I'm learning differential geometry, specifically the theory of surfaces, and need help with the following problem:
Consider the surface $S$ given by $x_1 = x_2 - f(x_2 - x_3)$, where $f \in C^2$.
Show that every tangent plane of $S$ are parallel to a fixed direction.
Here's my work so far:
We first need to find the parametric representation of $S$. Setting $x_2 = u$ and $x_3 = v$ give $x_1 = u - f(u - v)$. Therefore, the parametric representation of $S$ is given by
$$S : \vec x(u, v) = (u - f(u - v)) \vec e_1 + u \vec e_2 + v \vec e_3$$
or in a more conviennent/compact notation
$$S : \vec x(u, v) = (u - f, u ,v).$$
The equation of the tangent plane at $\vec x(u_P, v_P)$ in parametric form with parameters $\mu, \lambda$ is given by
$$\vec T_P(S) : \vec y = \vec x(u_P, v_P) + \lambda \vec x_u(u_P, v_P) + \mu \vec x_v(u_P, v_P) \tag{*}$$
where
$$\vec x_u(u_P, v_P) = (1 - f', 1, 0)$$
and
$$\vec x_v(u_P, v_P) = (f', 0, 1).$$
Substituting in $(*)$ yields
$$\vec T_P(S) : \vec y = (u - f, u ,v) + \lambda (1 - f', 1, 0) + \mu (f', 0, 1). \tag{**}$$
From $(**)$ I don't know how to prove that the tangent planes are parallel to a fixed direction. I'm also confused by the meaning of the phrase "parallel to a fixed direction". Does this means that the tangent planes of the surface are all parallel to some plane of $\mathbb R^3$?
|
Given
$$
x_{\,1} = x_{\,2} + f(x_{\,2} - x_{\,3} )\quad i.e.\quad F(x_{\,1} ,x_{\,2} ,x_{\,3} ) = 0
$$
we can rewrite it ,as
$$
x_{\,1} - x_{\,3} = x_{\,2} - x_{\,3} + f(x_{\,2} - x_{\,3} )\quad \Rightarrow \quad y_{\,1} = y_{\,2} + f(y_{\,2} )\quad i.e.\quad G(y_{\,1} ,y_{\,2} ) = 0
$$
or, otherwise, as
$$
x_{\,1} - x_{\,2} = f(x_{\,2} - x_{\,3} )\quad \Rightarrow \quad z_{\,1} = f(z_{\,2} )\quad i.e.\quad H(z_{\,1} ,z_{\,2} ) = 0
$$
Now, the tranformations
$$
\left( {x_{\,1} ,x_{\,2} ,x_{\,3} } \right) \Leftrightarrow \left( {y_{\,1} ,y_{\,2} ,y_{\,3} } \right)\quad \left( {x_{\,1} ,x_{\,2} ,x_{\,3} } \right) \Leftrightarrow \left( {z_{\,1} ,z_{\,2} ,z_{\,3} } \right)
$$
with the understanding that $y_3$ and $z_3$ are vectors independent from other two, so as to complete the respective basis,
are linear transformations, by which planes transform into planes, and "tangent to $F$" into "tangent to $G$" / "tangent to $H$".
$G$ and $H$ are clearly cylinders, with axes $y_3$ and $z_3$ respectively, so their tangent planes will all be parallel to those directions.
In the Euclidean space such a direction would be unique, that of $(1,1,1)$.
|
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|
The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$ What is $$\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy?$$ I split it as $\frac{y^{2}}{(y^2+1)^{0.5}} + \sqrt{y^2+1}.$ Now I substituted $y^{2}=u $ thus $2y\,dy=du$ so we get $0.5 \sqrt{\frac{u}{u + 1}} + 0.5 \sqrt{\frac{1 + u}{u}}$ but now what to do? Another idea was doing $+1-1$ in original question but that too doesn't lead anywhere. Now $y=\tan{x} $ as suggested below is an easy way but I am seeking for a purely algebraic way. Thanks.
|
$\displaystyle\int\frac{4y^2+2}{\sqrt{y^2+1}}dy=\int\frac{2y^2+2}{\sqrt{y^2+1}}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy=2\int\sqrt{y^2+1}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy.$
Now use $\displaystyle u=2y,\;dv=\frac{y}{\sqrt{y^2+1}}dy\;$ so $\;du=2dy,\;v=\sqrt{y^2+1}$ in the 2nd integral to obtain
$\displaystyle2\int\sqrt{y^2+1}dy+2y\sqrt{y^2+1}-2\int\sqrt{y^2+1}dy=\color{red}{2y\sqrt{y^2+1}+C}$
|
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|
Does topology apply to the integers? What is the natural topology (or topologies) on the integers. Can we define a metric on the integers?
|
We could add (since I don't see it or any equivalent when skimming the comments) the symmetric set topology on $\mathbb{Z}$: $U \subseteq \mathbb{Z}$ is open if $$n \in U \iff -n \in U$$ holds. This topology disconnects the integers and makes them non-compact, but does admit a countable basis (correct me if I'm messing something up here).
|
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|
Using Green's theorem find line integral $\oint_C (-x^2+x) dy $ enclosed by $x=2y^2$ and $y=2x$
Using Green's theorem find line integral $\oint_C (-x^2+x)\, dy $ enclosed by $x=2y^2$ and $y=2x$
The intersection points between the line and the parabola are
$$P_1 = \left( 0,0 \right) \quad P_2 = \left( \frac{1}{8} , \frac{1}{4} \right)$$
That said, knowing that
$$\frac{\partial Q}{\partial x} = -2x+1 \quad \frac{\partial P}{\partial y} = 0 \qquad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2x+1$$
So
$$\oint_C (-x^2+x)\, dy = \int_{0}^{\frac{1}{4}} \int_{2y^2}^{\frac{y}{2}} \left( -2x + 1 \right) \, dxdy = \cdots = \frac{3}{640}$$
The answer of the textbook is different.
Textbook's answer: $\frac{1}{5}$
Did I make a mistake somewhere?
|
You changed $-2x-1$ to $-2x+1$ in your last integral.
|
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|
Can we specifically prove that if $n\in\mathbb{Z^+}$ is composite and square-free then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ is not a UFD? Let $\mathcal{O}_K$ be the ring of integers of a field $K$. I have learnt about the Baker-Heegner-Stark theorem, which implies that if $K=\mathbb{Q}(\sqrt{-n})$ with $n\in\mathbb{Z}^+$ square-free, then $\mathcal{O}_K$ is a UFD if and only if $n$ is a Heegner number. However, I am interested in the possibility of obtaining an easier result: is it possible to prove that all Heegner numbers greater than $1$ are primes, with undergraduate mathematics?
|
This is not quite enough. It could be that the two factorizations can each be refined into the same factorization.
|
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|
Revisit "How can I visualize the nuclear norm ball" Revisiting How can I visualize the nuclear norm ball?
Two eigenvalues are reproduced as following:
$$ s_{1,2}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2\pm|x+z|\sqrt{(x-z)^2+4y^2}}. $$
According to the following (from a paper)
If a symmetric matrix: $$ A=\left( \begin{array}{cc} x & y\\ y & z\end{array} \right)$$
is rank $1$, then $y=\sqrt{xz}$, which comes from the fact that $vv^T$ is rank $1$ and any rank $1$ matrix can be represented in this form.
$$\left[\begin{array}{cc} v_1\\ v_2\end{array}\right]\left[\begin{array}{cc} v_1 & v_2\end{array}\right]=\left( \begin{array}{cc} v_1^2 & v_1v_2\\ v_1v_2 & v_2^2\end{array} \right)$$
My question: how to explain the red circle in figure (b) is the $2\times 2$ symmetric unit-Euclidean-norm rank $1$ matrix?
This is a circle in $3$-D, how to get the equation of this circle through the rank $1$ matrix provided above?
I believe just replace $y=\pm\sqrt{xz}$ in $s_{1,2}$ and can get the answer.
So I choose the larger one of $s_{1,2}$:
$$ s_{\max}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2 + |x+z|\sqrt{(x-z)^2+4y^2}}= \sqrt{2(x+z)^2}=1 (\text{unit norm})$$
|
The answer to the linked question shows that the curved side of the "cylinder" satisfies the equation
$$(x-z)^2+4y^2=1$$
and its planar caps satisfy
$$(x+z)^2=1.$$
The red curves are the intersection of the two surfaces, so they satisfy both equations. Therefore, they must also satisfy their sum,
$$x^2+4y^2+z^2=2.$$
Thus the curves lie on the intersection of the ellipsoid $x^2+4y^2+z^2=2$ and the planes $x+z=\pm1$, so they are ellipses (but not necessarily circles).
|
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|
Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.
One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.
|
The only thing I would add to T. Bongers' answer is that the key to these kinds of questions is generally to ask:
*
*How can I transform the limit to evaluate it using standard techniques?
*Especially with limits at infinity, what is 'dying off' quicker?
|
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|
What is meant by the fact that $k\langle x_1, \ldots, x_n\rangle$ denotes a free associative algebra in indeterminate $x_1, \ldots, x_n$? As the question title suggests, what is meant by the fact that $k\langle x_1, \ldots, x_n\rangle$ denotes a free associative algebra in indeterminates $x_1, \ldots, x_n$? Could anybody help me unpack this statement? Could anyone give me their intuition for working with such an object?
Thank you!
|
Just like a polynomial algebra is spanned by monomials, the free associative algebra with indeterminates $x_1, \dots, x_n$ has a basis given by monomials of the type
$$x_{i_1} \dots x_{i_k}$$
where $k \ge 0$ and $i_1, \dots, i_k \in \{ 1, \dots, n \}$, the case $k=0$ corresponding to the empty monomial, i.e. the unit of the algebra. The product is "concatenation" of such monomials extended linearly, i.e.
$$(x_{i_1} \dots x_{i_k}) \cdot (x_{i_{k+1}} \dots x_{i_{k+l}}) = x_{i_1} \dots x_{i_{k+l}}$$
The main difference with a polynomial algebra is that $x_i x_j \neq x_j x_i$ when $i \neq j$, so the order of the indeterminates matters in a monomial. For example $x_1 x_2 - x_2 x_1$ is not the zero element. This is also sometimes called the "tensor algebra".
The name comes from the fact that given any associative algebra $A$ and any elements $a_1, \dots, a_n \in A$, there exists an algebra morphism $f : k \langle x_1, \dots, x_n \rangle \to A$ such that $f(x_i) = a_i$, and that morphism is unique. If you know category theory, this means that the functor $k \langle - \rangle$ is left adjoint to the forgetful functor from associative algebras to sets.
|
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Two total orders Let $\mathcal{R}$ and $\mathcal{S}$ be two total orders on $E$ such that $\forall(x,y) \in E\times{E}, x\mathcal{R}y \Rightarrow x\mathcal{S}y$. Show that $\forall(x,y) \in E\times{E}, x\mathcal{R}y\Leftrightarrow x\mathcal{S}y$.
I need some help with this problem. I know that a total order is a set where every pair of elements in a relation such that anti-symmetry and transitivity holds. However I am having a problem from transferring from a relation $\mathcal{R}$ to a relation $\mathcal{S}$. Could you please give me some hints?
|
We are going to use the converse of
$$x\mathcal Ry \Rightarrow x\mathcal S y$$
which is
$$\neg (x\mathcal Ry) \Rightarrow \neg(x\mathcal S y).$$
The order is total, so if we have $\neg (x\mathcal Ry)$ we have $y\mathcal R x$.
By the first property it gives us
$$y\mathcal S x.$$
By anti-symetry we can conclude that $x=y$ or $\neg (x\mathcal Sy)$.
But $x=y$ can not be because $\neg(x\mathcal Ry)$ (and you always have $x\mathcal R x$).
So $\neg (x\mathcal Sy)$.
Finally,
$$x\mathcal Ry \iff x\mathcal S y.$$
|
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Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please? Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please?
I've been looking for a while now, and I just don't find it, the basics of the hypothesis !
If anyone can help !
|
The paper is listed in the Journal Comptes Rendus de l'Académie des Sciences and the full title of the article is Sur les zéros de la fonction $\zeta(s)$ de Riemann. The full reference is
Hardy G H. Sur les zéros de la fonction $\zeta(s)$ de Riemann. Comptes Rendus de l'Académie des Sciences,
1914, 158: 1012-1014
and you can access this paper on
*
*www.academie-sciences.fr with free access.
*EDIT (added another source): thanks to @Raymond Manzoni, who found this resource (also free access) archive.org
|
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Show that $\int_{-\pi}^{\pi} f(t) \sin (nt) dt \rightarrow 0 , \int_{-\pi}^{\pi} f(t) \cos (nt) dt \rightarrow 0$ For any integrable function $f: S^1 \rightarrow \mathbb{C}$ , I need to show that $$ \int_{-\pi}^{\pi} f(t) \sin (nt) dt \rightarrow 0$$ $$ \int_{-\pi}^{\pi} f(t) \cos (nt) dt \rightarrow 0$$ as $n \rightarrow \infty$
Now, here is what I did -
Using the Riemann-Lebesgue Lemma , I got $$\hat{f}(n) = \int_{-\pi}^{\pi} f(x) e^{- inx} dx \rightarrow 0 $$ as $n \rightarrow \infty$
Now breaking $e^{- inx}$ as $\cos(nx) - i\sin(nx)$ , I get
$$\int_{-\pi}^{\pi} f(x) (\cos(nx) - i\sin(nx)) dx \rightarrow 0 $$
If $f(x)$ is a real function then the proving is trivial. But I am not able to find a proof if $f$ is complex.
I have seen lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos nt\,dt$ but can someone suggest a different methodology from Fourier analysis point of view.
|
Um, I think you can just break $f$ into real and imaginary parts, and apply Riemann Lebesgue Lemma to each, just as you did for real valued $f$ here?
|
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no $k$-sparse vector in null space of $A$ where $y=Ax$ Here is a part of my notes on compressed sensing. Suppose we are given an underdetermined linear system of equations $Y=AX$. For a unique solution,
What we need:
subtraction of no two $k$-sparse vectors ($x_1$, $x_2$) must be in nullspace of $A$.
What is required for the following to hold:
since the subtraction of two $k$-sparse vectors is at most
$2k$-sparse, $2k$ columns of the matrix must be linearly independent
then no $k$-sparse vector will be in null space of $A$.
I don't understand the part where a connection between sparsity and the columns of the matrix being linearly independent is made. An explanation would be great.
|
I presume the formulation should be that every 2k columns of $A$ should be linearly independent, as this implies that there are no 2k sparse vector in the null space:
To see this, denote $(\vec{a}_1,...,\vec{a}_N)$ the column vectors of $A$ and let $X$ be a 2k-sparse vector, i.e. there are at most $2k$ nonvanishing components of $X$. This means that there is a set of at most 2k indices $i_1,\ldots,i_m$, with $m\leq 2k$ so that every component of $X$ outside this set vanishes.
In order for $X$ to be in the kernel of $A$ we must have:
$$ AX = \sum_{j=1}^m \vec{a}_{i_j} x_{i_j}=0 $$
But as every collecting of at most $2k$ vectors of $A$ are assumed linearly independent we must have every $x_{i_1}=...=x_{i_m}=0$. So every component of $X$ vanishes and $X$ was the zero vector.
|
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Cauchy subsequence problem Consider the set $S_0$ of all continuous functions $f : [0,1] \to [0,1]$.
Define a metric on the set $S_0$ by setting
$$
\rho(f,g)=\sup\limits_{x\in{[0,1]}}|f(x)-g(x)|.
$$
(a) Give an example of a sequence in $S_0$ that does not contain a Cauchy subsequence with respect to the metric $\rho$.
(b) For a given $\epsilon> 0$ consider $S_\epsilon \subset S_0$ consisting of the set of functions $f \in S_0$
such that
$$
|f(x)-f(y)|\leq|x-y|^{\epsilon}\qquad \rm{for~ all} ~x,y\in[0,1]
$$
Show from first principles that any sequence in $S_\epsilon$ contains a subsequence that is
Cauchy with respect to the metric $\rho$.
I cannot find an example in (a). And in (b), if $f : [0,1] \to [0,1]$ is finite dimentional with respect to metric $\rho$. Then by proving f is bounded and closed, we can get f is compact. But f is infinite dimensional, so how to tackle this problem?
|
Hint for (a): Try a sequence of functions $f_n$ such that, e.g., $f_n(1/n) = 1$ but $f_n(1/m) = 0$ for all other $m$.
Hint for (b): if you have a sequence $f_n$ in $S_\epsilon$ such that $f_n(r)$ is Cauchy for each rational $r$, show that $f_n$ is Cauchy in the metric $\rho$.
|
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Are there integers $a,b,c$ such that $a$ divides $bc$, but $a$ does not divide $b$ and $a$ does not divide $c$?
Are there integers $a,b,c$ such that $a$ divides $bc$, but $a$ does not divide $b$ and $a$ does not divide $c$?
I am not quite sure what to do with the given information. I know I could easily find an example.
We know that $a$ divides $bc$ so,
$$bc=aq \text{ for some integer } q.$$
And that $a$ does not divide $b$ or $c$ so,
how is that represented?
What would be my first step?
|
The general result on this question is this generalisation of Euclid's lemma:
Gauß's lemma: If a number divides a product of two factors, a,d is coprime with one of them, it divides the other.
|
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}
|
Find $f^{(100)}(0) $ and $f^{(101)}(0) $ if $f(x)=xe^{\arctan{x}}$ $$f(x)=xe^{\arctan{x}}$$
Part of my solution
$$f^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}x^{(k)}(e^{\arctan{x}})^{(n-k)}=x(e^{\arctan{x}})^{(n)}+(e^{\arctan{x}})^{(n-1)}$$
First term probably disappears because $x=0$ but i don't know what to do with second term.
|
Partial answer:
As you pointed out,
$$\begin{align}f^{[n]}(x)&=\sum_{i=0}^n{n\choose i}x^{[i]}\left(e^{\tan^{-1}(x)}\right)^{[n-i]}
\\
&=x\left(e^{\tan^{-1}(x)}\right)^{[n]}+n\left(e^{\tan^{-1}(x)}\right)^{[n-1]}
\end{align}$$
We set $g(x)=\left(e^{\tan^{-1}(x)}\right)$, $h(x)=\tan^{-1}(x)$ and consider $g^{[n]}(x)$.
$\begin{align}g'(x)&=\left(e^{h(x)}\right)'
\\
&=h'(x)g(x)
\\
g''&=h''g+h'g'
\\
&=h''g+h'h'g
\\
&=(h''+(h')^2)g
\\
g'''&=(h'''+2h'h'')g+(h''+(h')^2)h'g
\\
&=(h'''+3h'h''+(h')^3)g
\\
g^{[4]}&=(h^{[4]}+3(h''h''+h'h''')+3(h')^2h'')g+(h'''h'+3(h')^2h''+(h')^4)g
\\
&=(h^{[4]}+4(h')^2h''+3(h'')^2+4h'h'''+(h')^4)g
\\
g^{[5]}&=(h^{[5]}+4(2(h')h''h''+(h')^2h''')+6(h'')h'''+4(h''h'''+h'h^{[4]})+4(h')^3h'')g\dots
\\
&\quad\dots+(h'h^{[4]}+4(h')^3h''+3h'(h'')^2+4(h')^2h'''+(h')^5)g
\\
&=(h^{[5]}+8(h')^3h''+11h'(h'')^2+8(h')^2h'''+10h''h'''+5h'h^{[4]}+(h')^5)g
\end{align}$
Note that each bracketed term is a sum of the form:
$$\sum_{i=0}^{n}C_i\prod\left(h^{[m]}\right)^p$$
Where $C_i$ are integer constants and for each $i$, $\sum{pm}=n$. In fact, all sums of $pm$ equal to $n$ are represented as terms. For example, $3=(3\cdot1)=(1+2)=(1\cdot3)$ so there are terms with $(h')^3$, $h'h''$ and $h'''$. This is strikingly similar to the binomial series but I've yet to determine the constants. We would probably be able to use Faa di Bruno's formula for this.
We know that $h'(x)=\frac{1}{x^2+1}$, which we equate to a geometric series in $-x^2$ and generalise.
$$\begin{align}
h'(x)&=\sum_{i=0}^\infty(-x^2)^i
\\
h''(x)&=\sum_{i=1}^\infty(-1)^i\cdot2ix^{2i-1}
\\
h^{[n]}(x)&=\sum_{i=n-1}^\infty(-1)^i(2i)_{n-1}x^{2i-(n-1)}
\end{align}$$
Where $(x)_n=\prod_{i=0}^{n-1}(x-i)$ and $x^0=1$ when $x=0$.
Hence, $h'(0)=1, h''(0)=0, h'''(0)=-2, h^{[5]}=4!,h^{[7]}=-6!\dots$ and in general:
$$h^{[n]}(0)=\begin{cases}0&n\:\mathrm{even}
\\
(-1)^{\frac{n-1}{2}}(n-1)!&n\:\mathrm{odd}
\end{cases}$$
We may substitute this into the multi-chain-rule expansion of $g^{[n]}$(x) at $x=0$ using that $g(0)=1$.
$\begin{align}g'(0)&=1
\\
g''(0)&=1
\\
g'''(0)&=-2+1
\\
g^{[4]}(0)&=4(-2)+1
\\
g^{[5]}(0)&=4!+8(-2)+1
\dots
\end{align}$
The problem becomes finding the general coefficients $C_i$ of the odd terms in the expansion of $g^{[n]}$ but I'm unsure how to proceed.
|
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|
Prove each of the following conditions is sufficient to ensure that $f(x+y)≤f(x)+f(y)$ $f$ is increasing and $f$ satisfies $f(x)=0$ and $f(x)>0$ for all $x>0$.
Show that each of the following conditions is sufficient to ensure that $f(x+y)≤f(x)+f(y)$ for all $x,y≥0$.
*
*(a) $f$ has a second derivative satisfying $f''＜0$;
*(b) $f$ has a decreasing first derivative;
*(c) $\frac{f(x)}{x}$ is decreasing for $x>0$;
I know $a → b → c$.
I don't know how to prove $f(x+y)≤f(x)+f(y)$ using (c).
|
Hint: Assume (c). Then
$$x\frac{f(x+y)}{x+y}\le x\frac{f(x)}{x} $$
and
$$y\frac{f(x+y)}{x+y}\le y\frac{f(y)}{y}.$$
|
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|
Why is a reflection followed by another reflection is a rotation? I just started abstract algebra and we are working with dihedral groups. I've made Cayley tables for D3 and D4 but I can't explain why two reflections are the same as a rotation
|
Consider the dihedral group $D_5$, and consider its action on the pentagon. In particular, every element of the group can be thought of as some combination of rotations and reflections of a pentagon whose corners are labeled $1,2,3,4,5$ going clockwise.
First, notice that no matter what we do, the numbers will be in the order $1,2,3,4,5$ in either the clockwise (cw) or counterclockwise (ccw) direction.
If our change switches the order from ccw to cw (or vice versa), then we must have reflected the image. On the other hand, if no such change occurs, then we must have rotated the image.
Note that reflecting twice results in switching from ccw to cw, then to ccw. So, the numbers still go $1,2,3,4,5$ in the ccw direction. So, we must have rotated the image.
|
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|
What is the smallest and largest possible values for the variance? Suppose
$P( X \in \{1,2,3\}) = 1$ and $E(X) =2.5.$
What is the smallest and largest possible values for the variance?
My understand: So what I understand is variance finds the distance between each element and the mean. So the closer 2 is to E(X) the farther 1 and 3 can be from E(X). But I have no clue how to obtain this..
|
Although the two things are equal, I think it is easier to use $$\operatorname{Var}(X) = \mathbb{E}\Big[(X-\mathbb{E}X)^2\Big]$$
rather than $\operatorname{Var}(X)=\mathbb{E}X^2-(\mathbb{E}X)^2$.
Let $\mathbb{P}[X=i]=p_i$, then
\begin{align}
\operatorname{Var}(X) &= \mathbb{E}\Big[(X-\mathbb{E}X)^2\Big] \\
&= \sum_{i \in \{1,2,3\}}(i-2.5)^2 \cdot p_i \\
&= (-1.5)^2\cdot p_1 + (-0.5)^2 \cdot p_2 + (0.5)^2 p_3 \\
&= 2.25\cdot p_1 + 0.25\cdot p_2 + 0.25\cdot p_3. \tag{$\spadesuit$}
\end{align}
Maximizing and minimizing this under constraints $p_1+p_2+p_3=1$ and $1\cdot p_1 + 2\cdot p_2 + 3\cdot p_3 = 2.5$ is simple, because the weights of $p_2$ and $p_3$ are equal and these are both $0.5$ away from the mean:
*
*To minimize variance move all the probability mass away from $p_1$ –distribute everything between $p_2$ and $p_3$ in a way that gives you the correct expected value.
*To maximize variance move the probability mass as much as possible to $p_1$. To achieve the correct expected value you have to have some counterbalance in $p_3$, but $p_2$ should be equal to zero.
Taking into account the two bullet points, the formula marked by $(\spadesuit)$ has a nice intuition: to minimize variance, move the probability mass as much as possible in the direction of the mean. To maximize the variance do the opposite – move the probability mass away from the mean.
I hope this helps $\ddot\smile$
|
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|
Question about condtion of MVT for integral. For $f \in C[a,b], g\ge0$ (or $g\le0$) on $[a,b]$, then there exists $c\in(a,b)$ $\int _a^b f(x)g(x)dx = f(c)\int_a^bg(x)dx$.
It is MVT for integral.
Why $c \in (a,b)$ ? How about $c\in[a,b]$? Is it wrong?
I am wondering this because, in proving progress,
by IVT, there exist $m,M$ s.t $m\le f(x)\le M$
so $m\int_a^bg(x)dx$ $\le$ $\int_a^b f(x)g(x)dx$ $\le$ $M\int_a^bg(x)dx$.
when $\int_a^bg(x)dx>0$ then $m\le$ $\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$ $\le$ $M$ and
$\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$ becomes $f(c)$, so it seems $c\in [a,b]$ might work.
So, I want to know why $c\in(a,b)$, or $c\in[a,b] $ would be okay.
|
In the book Mathematical Analysis I by Vladimir A. Zorich (p. 352) I found:
As far as I can tell from the proof in the book, the proof depends on the intermediate value theorem, which states
But $c \in (a,b)$, since if either $f(a) = 0$ or $f(b) = 0$ would imply $f(a)f(b) = 0$ and so the function would not fulfill the hypothesis of the intermediate value theorem. Thus in fact it is $c \in (a,b)$, but $(a,b) \subseteq [a,b]$. Hence we can use $[a,b]$.
|
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|
Different seven-digit numbers could Sid have meant to type
Sid intended to type a seven-digit number, but the two 3's he meant to
type did not appear. What appeared instead was the five-digit number
$52115$. How many different seven-digit numbers could Sid have meant to type?
First time, I attempted this problem as follows-
I considered, there are 12 places available for two 3's (as they can appear together)
_ _ 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 _ _
So, we can select 2 places from 12 places in ${12\choose 2}$ = $66$ ways.
But, I found the correct approach is simply choosing 2 places from available 7 place in ${7\choose 2}$ = 21 ways. However, I can't find my mistake in previous one. Can anyone explain what am I missing?
|
Consider the following possibilities:
_ 3 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 _ 3
_ 3 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 3 _
3 _ 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 _ 3
3 _ 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 3 _
All of these would give the same result of 3521153, but are counted separately by your first method.
|
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|
Number of ways of distributing $N$ balls into $M$ bins such that at least one bin has at least $n$ balls in it?
What is the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that at least one bin has at least $n$ balls in it?
My attempt:
The number of ways of of placing $N$ balls into $M$ bins is $\binom{N+M-1}{N}$.
I tried, by stars and bars, to calculate the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it:
If the first or last bin contains the $n$ balls, we have used one partition, so there are $M-2$ left. This gives $2\binom{N-n+M-2}{N-n}$ ways.
If the second to $(M-1)$th bin contains the $n$ balls, we have used two partitions, which gives $(M-2)\binom{N-n+M-3}{N-n}$ ways.
So the total number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it is
$$2\binom{N-n+M-2}{N-n}+(M-2)\binom{N-n+M-3}{N-n}.$$
Is this correct? It seems like a strange result.
Then I thought about simply summing this expression for $n$ going from $n$ to $N$, i.e.,
$$\sum_{k=n}^N 2\binom{N-k+M-2}{N-k}+(M-2)\binom{N-k+M-3}{N-k},$$
in order to obtain the expression for "at least $n$ balls," but I feel like this would be over-counting somehow.
And then there is the issue of "at least one bin," which I am rather blunted by.
Any help is much appreciated!
Note that I'm looking for a closed form solution to the problem in the yellow box. Thanks.
|
Seems that it's not correct, since you going to have double counting for the case where cell 1 and cell 2 both have $n$ balls.
Try use inclusion-exclusion principle.
It's easy using the properties $p_i = \mathrm{The\space ith\space cell\space has\space at\space least\space n\space balls}$ to calculate $E(0)$ the case when no property holds.
For $r$ different properties to hold, we want that $r$ specific cells will have at least $n$ balls we first put $n$ balls in each of them and then the rest $N - rn$ we split however we want, so we get: $$ w(p_{i_1},...,p_{i_r}) = {{N + M - rn - 1}\choose N - rn}$$ and with the number of ways to pick $r$ different properties and the fact that the properties are symmetric we get $$ w(r) = {M \choose r} \cdot {{N + M - rn - 1}\choose N - rn} $$
Now we have that $$E(0) = {{N+M-1}\choose{N}} - \sum_{r=1}^M (-1)^r\cdot w(r)$$
After calculating $E(0)$ use the fact that whats youre loking for is the complement of $E(0)$ which gives you that the desired number is: $${{N+M-1}\choose{N}} - E(0) = \sum_{r=1}^M (-1)^r\cdot w(r)$$
|
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|
Does every theorem have a short proof? This question is somehow based on my belief that every theorem has a short and simple proof. By "proof" I mean:
*
*Proving an statement
*Disproving a statement
*Proving that a statement is undecidable
Once we have formalized what we understand for a "step" in a proof, could it be proven that every theorem has a proof consisting of less than $n$ steps? If so:
*
*What would be the (minimal) value of $n$?
*Such a proof would be about all proofs so what would it say about itself?
*Could there be (in some sense) proofs with a non-integer number of steps?
|
surely a theorem doesn't need to conform to any real quality requirements, so it could contain an huge number of steps that need to be proved, greater than any n you think of.
|
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|
Why do parabolas' arms eventually become parallel? Here is what this site states
All parabolas are the same shape, no matter how big they are. Although they are infinite, meaning that the arms will never close up, the arms will eventually become parallel.
Now, I have an argument against it. Let $f(x) = ax^2 + bx + c$ be a quadratic polynomial with $a , b$ and $c$ being real numbers and $a \ne 0$. So, its graph will give us a parabola ($\because$ graph of a quadratic polynomial is a parabola).
Now,
$$\dfrac{d (ax^2 + bx + c)}{dx}$$$$ = 2ax + b$$
i.e. the slope of a quadratic polynomial is given by $g(x) = 2ax + b$. Now, differentiating the equation for slope of the quadratic
$$\dfrac {d (2ax + b)}{d x}$$
$$ = 2a$$
So, if $a \gt 0$ then the slope of $g(x)$ will be increasing. This means that the slope of $f(x)$ will also be increasing. Similarly, if $a \lt 0$ then the slope of $f(x)$ will be decreasing.
This means that for all $x$ the slope of $f(x)$ will be different. So, this contradicts the fact (according to me) that the arms of a parabola will eventually be parallel. Where am I going wrong?
|
There is no contradiction. As you said, the slope keeps increasing, this means that as you go towards $x\rightarrow\pm\infty$, the slope also goes to $\pm\infty$, i.e the graph of the function gets more and more "vertical".
|
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|
Translating $\forall$ and $\exists$ in a statement The following statement has two versions – one where $d$ is quantified by $\forall$ and the second where it's quantified by $\exists$. The task here is to find a counterexample where the statements below are false. The domain is all integers.
*
*$\forall a \forall b \forall c \forall d(a^d + b^d = c^d$)
*$\forall a \forall b \forall c \exists d(a^d + b^d = c^d$)
The first statement is false for at least some values of the variables. When $a=1, b=2, c=3, d=4$, the statement does not hold for all variables $a$, $b$, $c$, and $d$. For instance:
$$
1^4 + 2^4 \ne 3^4
$$
The second statement is false when $a = 1, b = 2, c = 10$ because there doesn't exist a $d$ where the statement would be true. For instance, if $d$ was $5$:
$$
1^5 + 2^5 \ne 10^5
$$
Am I interpreting this correctly and do my counterxamples make sense?
|
Turning a comment into an answer:
For the first statement, you can remove the for all variables a, b, c, and d part, as the counterexample of $[a=1,b=2,c=3,d=4]$ already states this fact.
For the second statement, you need to prove that your counterexample of $[a=1,b=2,c=10]$ is true for ALL values of $d$.
I'm not sure whether or not it is indeed true for all values of $d$, but it is certainly not so trivial.
Alternatively, you can use the counterexample of $[a=0,b=0,c=1]$ which is a lot more trivial.
|
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|
Understanding why $\det(A) = \det(A^T)$ via the 3D Paralleliped I am trying to understand why, geometrically, we have that
$$
\det(A) = \det(A^T).
$$
To build intuition, I am thinking in 3 dimensions. So let $A$ be a $3 \times 3$ matrix of real numbers.
First, I know that if
$$
\det(A) = \left|
\begin{array}
xx_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{array}
\right|
$$
then $|\det(A)|$ can be thought of as the area of a parallelepiped formed by $(x_1, x_2, x_3)$, $(y_1, y_2, y_3)$, and $(z_1, z_2, z_3)$.
Moreover, the fact that $\det(A) = \det(A^T)$ implies that $|\det(A)|$ can also be thought of as the area of the parallelepiped formed by
$(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$.
Question: Does this also mean that the area of a parallelepiped formed by any permutation of the $x_i$, $y_i$, and $z_i$ is equal to $|\det(A)|$? For example, does the area of a parallelepiped formed by the coordinates $(x_1, y_2, z_3)$, $(x_2, y_1, z_2)$, and $(x_3, y_3, z_1)$ equal $|\det(A)$|?
|
Here is a 2D example.
$$\begin{pmatrix}
2&3\\
1&4\\
\end{pmatrix} \mapsto_{transpose}
\begin{pmatrix}
2&1\\ 3&4\\
\end{pmatrix}$$
I'm still thinking of a satisfying (geometric) explanation as to why this always works, but pictures are always good.
|
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|
How to compute this limit involving complementary error functions I am trying to take the following limit
$$\lim_{x\to \infty } \, \frac{2 x \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}{t \operatorname{erfc}\left[-\frac{x}{\sqrt{2}}\right]}$$
my first thoughts were to use LHospital's rule after making the top complementary error function as the denominator of the denominator. Upon differentiating I got the following
$$\lim_{x\to \infty } \, \frac{\sqrt{2 \pi } \exp\left[\frac{\left(t^2+1\right) x^2}{2 t^2}\right] \operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]^2}{\exp\left[\frac{x^2}{2}\right] \left(\operatorname{erf}\left[\frac{x}{\sqrt{2}}\right]+1\right)+t \exp\left[\frac{x^2}{2 t^2} \right]\operatorname{erfc}\left[\frac{x}{\sqrt{2} t}\right]}$$
which seems even more complicated.
Maybe using an identity or some form of expansion for the complementary error function may lead to a simpler expression.
Any help will be greatly appreciated.
Thank you.
|
By exploiting the continued fraction representation for the complementary error function
$$\frac{2x\int_{\frac{x}{t\sqrt{2}}}^{+\infty}e^{-z^2}\,dz}{t\int_{-\frac{x}{\sqrt{2}}}^{+\infty}e^{-z^2}\,dz}\approx \sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2t^2}}$$
hence the wanted limit is simply zero. You may also use concentration (tail) inequalities for the normal distribution (thanks to M.Spivey) or Hoeffding's inequality.
|
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Proof Proving that aj=aj−1+aj−2 cannot converge to a finite limit writing question. I was trying this problem and was hoping for some feedback as I am not sure if I proved it or not... I am trying to figure out where I went wrong as I feel I didn't run a good argument.
For $\epsilon>0,\exists N:|a_j-a_{j+1}|<\epsilon$ for$j>N$.
$a_j=a_{j-1}+a_{j-2}$
$|a_j|=|a_{N+1}+a_{j-1}+a_{j-2}-a_{N+1}|$ I added zero in a unique way.
$|a_j|\leq|a_{N+1}|+|a_{j-1}+a_{j-2}-a_{N+1}|$ Triangle inequality
$|a_j|\leq|a_{N+1}|+|a_j-a_{N+1}|$ Substitute in $a_j$
for $j>N$, $|a_j-a_{N+1}| < \epsilon$
$|a_j|\leq|a_{N+1}|+\epsilon$
$|a_j|-|a_{N+1}|\leq\epsilon$ Since epsilon can't be negative and we know $|a_j|-|a_{N+1}|<0$, is there a contradiction?
|
No, you do not arrive at a contradiction. For example, it is true that $|17|-|42|\le 0.001$ even though the left hand side is negative.
Instead, let $\phi$ be the larger solution of $\phi^2=\phi+1$. Show that $\phi>1$. Show by induction that $a_n>c\phi^n$ provided you pick $c>0$ such that $a_1>c\phi$ and $a_2>c\phi^2$.
|
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Are there infinitely many $\alpha \times \beta$ Chomp boards where player 2 wins? Let $\alpha$ and $\beta$ be nonzero ordinals.
Infinite chomp (called ordinal chomp by Wikipedia) on an $\alpha \times \beta$ board is played as follows.
We consider the set $\alpha \times \beta$, partially ordered under the natural ordering: $(x,y) \le (z,t)$ iff $x \le z$ and $y \le t$.
On your turn, you choose an uneaten square of chocolate $(x,y) \in \alpha \times \beta$ and "eat" all squares $(x',y') \ge (x,y)$. The square $(0,0)$ is poisoned, so the person who eats it loses.
This game can only progress a finite number of moves, so one player must have a winning strategy.
For finite ordinals $\alpha$ and $\beta$, not both $1$, it is well-known that player 1 wins.
On the other hand, player 2 wins for the $1 \times 1$ and $2 \times \omega$ boards.
Question: Are there infinitely many pairs $\alpha, \beta$ where player 2 wins on an $\alpha \times \beta$ board?
Bonus: Classify as many boards as possible where player 2 wins.
Additional observations:
*
*If $\alpha$ and $\beta$ are both successor ordinals, the well-known strategy-stealing argument applies: If player 2 had a winning strategy, player 1 could eat the largest chocolate and then copy player 2's supposed strategy. So player 1 must win.
References:
*
*My question was motivated by this answer, in which a winning strategy is presented for player 2 in $2 \times \omega$ chomp, and for player 1 in $\alpha \times \omega$ chomp which works for any ordinal $\alpha > 2$.
*This article by Ian Stewart covers finite Chomp and the $2 \times \omega$ case.
*$\alpha \times \beta$ chomp is a special case of a poset game, namely the poset game on the poset $(\alpha \times \beta) \setminus \{(0,0)\}$.
*This Wikipedia reference could be helpful:
p. 482 in: Games of No Chance (R. J. Nowakowski, ed.), Cambridge University Press, 1998
A google books link is here, but I cannot find electronic access to the book's contents.
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Yes, for every nonzero ordinal $\alpha$ there is an ordinal $\beta$ such that $\alpha \times \beta$ is a second player win (I can't find the source sorry). So far the only pairs that are known are $1 \times 1$ (trivial), $2 \times \omega$, $3 \times \omega^\omega$, and $4 \times \omega^2$.
The situation with five or more rows gets incredibly complicated, and it seems likely that the $\beta$ for 5 rows is a lot bigger than $\omega^\omega$, although bounds could be found by constructively applying the proof of existence.
It seems (from a very rough first guess) that even numbers will have a much smaller $\beta$ than odd numbers.
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Sketching Functions
Sketch the region enclosed by $x+y^2=30$ and $x+y=0$. Decide whether to integrate with respect to $x$ or $y$. Then find the area of the region.
How do I start off sketching? Very first step?
I'm in Calc 2 and it's sad that I don't know what to do but I am trying to learn.
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Can you sketch $y=30-x^2$? This is fairly standard highschool stuff. Once you do that, flip it over the line $y=x$ to interchange the $x$ and $y$ coordinates, to get $x=30-y^2$. That’s your first graph. The other one is easier, it’s $y=-x$ (you can also think of it as $x=-y$), still standard highschool stuff. With your two pictures, you are ready to go. You’ll need to know the points of intersection, those are where $x=30-y^2$ and $x=-y$, in other words where $30-y^2=-y$. Solve for $y$ by putting them all on one side of the equals sign, $0$ on the other side, and then factoring your quadratic in $y$.
|
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|
Naming scheme for the acceleration vector and acceleration when working with parametric space curves? The formula for the acceleration vector on a space curve is $$a=\kappa|v|^2N+\frac{d^2s}{dt^2}T$$
I understand the formula above and how to calculate the components. What I don't understand is the naming scheme between $a$ and $\frac{d^2s}{dt^2}$
The first is called the acceleration vector, and textbooks simply refer to it as the acceleration when working with parametric space curves. The second describes the "linear acceleration" as we move along the curve which is a scalar and not a vector. I do not want to refer to it as "acceleration" because that word is already being used. So what should I call it?
I've been using the term "linear acceleration" but is there a better term?
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Perhaps try a name like "intrinsic acceleration." The motivation for this schema is that if you have an ant skating along a curve embedded in $\Bbb{R}^3$, its acceleration will come from two sources:
*
*the curvature of the embedding in ambient space (i.e., how hard the ant has to grip the curve in order to not fly off into space), and
*the acceleration of the ant itself on the curve (i.e., how hard it's pushing the skateboard forward).
The first term describes the acceleration induced by the geometry of the embedding, so maybe we should call it "extrinsic acceleration." The second describes the effect of how fast the curve is being traversed by the observer, hence "intrinsic acceleration."
(Note that if you just want to study geometry, you can reparametrize the curve with arc-length. This renders $s'' = 0$ identically and isolates the interaction between the curve and its environment. The tradeoff is that this is computationally more difficult because it requires inverting a function defined implicitly by integration.)
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Extreme point of the unit ball in H(U) An extreme point of a convex subset $C$ of a vector space $X$ is a point which can not be expressed in the form $\lambda a+(1-\lambda)b$, with $a,b\in C$ and $0<\lambda<1$.
For an open subset $U$ of the complex plan, $\mathbb C$, we denote by $H(U)$ the set of all holomorhphic functions on $U$. The norm on $H(U)$ is the usual $\|.\|_\infty$ norm.
What are the extreme points of the unit ball in $H(U)$?
The unit ball in $H(U)$ is the set $\{f\in H(U): \|f\|_\infty\leq 1\}$
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Partial answer: If $A={\Bbb C}\setminus U$ contains at most two points then $B_1(H(U))$ consists of constants only. So assume $|A|\geq 3$ and let $\phi:U\rightarrow {\Bbb D}$ be a conformal bijection with the unit disk. On the unit disk a set of extremal points are $e_{n,\theta}=\theta z^n$, $n\geq 0$, $|\theta|=1$. And $e_{n,\theta} \circ \phi$ are then extremal points in $B_1(H(U))$. There are however functions in $B_1$ that are not generated by these extremal points.
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$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, for all integers $n\ge 1$. Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that
$$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$.
(Given Hint: For all positive integers $a$ and $b$, $\frac{a}{b}+\frac{b}{a} \ge 2$.)
Can anyone help? Thank you.
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This answer $\ds{\underline{is\ not}}$ based in 'Induction' as the OP requested. @RobertZ just called this point to my attention. Indeed, it's a straightforward point of view.
\begin{align}
\color{#f00}{\pars{\sum_{i = 1}^{n}x_{i}}\pars{\sum_{i = 1}^{n}{1 \over xi}}} & =
\sum_{i = 1}^{n}\sum_{k = 1}^{n}{x_{i} \over x_{k}} =
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}
\pars{{x_{i} \over x_{k}} + {x_{k} \over x_{i}}}
\\[5mm] & =
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}
\bracks{\pars{\root{x_{i} \over x_{k}} - \root{x_{k} \over x_{i}}}^{2} + 2} \geq
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}2 = \color{#f00}{n^{2}}
\end{align}
|
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Statement provable for all parameters, but unprovable when quantified I've been reading a book on Gödel's incompleteness theorems and it makes the following claim regarding provability of statements in Peano arithmetic (paraphrased):
There exists a formula $A(x)$ such that the statements $A(0), A(1), A(2), \dots$ are all provable, but $\forall x\, A(x)$ is not provable.
It goes on to say that while Gödel's first incompleteness theorem guarantees its existence, it is not easy to find such a property for a theory as strong as PA.
Is there a specific example of such a formula or has none been found yet?
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If you simulate a Turing Machine using a Universal Turing Machine and you run the simulation for exactly N steps you will see either that the machine Halts in N steps or that the machine does not Halts in N steps.
However in general you cannot proove that a Turing Machine Halts by mere simulation, because if it does not halt your simulation would run forever. I think that's the same but in different words.
You know the answer to the question:
*
*Does the machine halt in N steps? (Decideable)
You don't know the answer to the question:
*
*Does the machine halts in any number of steps? (Undecideable)
You could use as formula a machine that require the turing machine to reach a determined state. You know if the state can be reached in a limited amount of steps, but not if you let it run forever with unlimited amount of memory.
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Variable chord of hyperbola If a variable chord of hyperbola $x^2$$/4$ - $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
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Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
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How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
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Notice here,
$$x^2 - y^2 - x + y$$
$$(x+y)(x-y)-1(x-y)$$
$$(x-y) (x+y-1)$$
That's done.
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Injection - Bijection function Question Suppose $A, B$ are two non-empty sets and let $$f:A\to B$$ and $$g:B\to A$$ be two injective (one to one) functions.
QUESTIONS:
1. Is there exist a bijective function from $A$ to $B$?
2. If so, how can we prove it?
I think, there should be a such function. But I can not come up with an argument to prove it?
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I have seen 2 proofs, one complicated, and this one: For $x\in A$ the possible sequence $$x,\; f^{-1}(x),\; g^{-1}f^{-1}(x),\; f^{-1}g^{-1}f^{-1}(x),...$$ may have just one term (if $f^{-1}(x)$ doesn't exist), or 2 terms (if $f^{-1}(x)$ exists but $g^{-1}f^{-1}(x)$ does not), et cetera.
Let $x\in E$ if the sequence stops after an even number of terms. Let $x\in O$ if the sequence stops after an odd number of terms. Let $x\in I$ if the sequence has no end.
Let $h(x)=f(x)$ if $x\in O.$ Let $h(x)=f^{-1}(x)$ if $x\in E\cup I.$ I will leave it to you to verify that $h:A\to B$ is a bijection.
I have seen this referred to in books as the Schroeder-Berstein Theorem, the Cantor-Bernstein theorem, and the C-S-B theorem.
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Supremum of the product of two sets Here is a GRE math subject problem:
63. For any nonempty sets $A,B\subseteq \Bbb R$, let $A\cdot B$ be the set defined by
$$A\cdot B=\{xy\,:\, x\in A\wedge y\in B\}$$
If $A$ and $B$ are nonempty bounded sets and if $\sup A>\sup B$, then $\sup(A\cdot B)=$
(A) $\quad\sup(A)\sup( B)$
(B) $\quad\sup(A)\inf(B)$
(C) $\quad\max\{\sup(A)\sup(B),\ \inf(A)\inf(B)\}$
(D) $\quad\max\{\sup(A)\sup(B),\ \sup(A)\inf(B)\}$
(E) $\quad\max\{\sup(A)\sup(B),\ \inf(A)\sup(B),\ \inf(A)\inf(B)\}$
The answer is E while I thought it was C.
When could $\inf(A)\sup(B)=\sup(A\cdot B)$ under the condition $\sup(A)>\sup(B)$?
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When $\inf B<0$, $|\inf B|\ge \sup B$, $\sup A>0$ and $\sup A\ge |\inf A|$.
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Basis of the topology of a Stone space I'm trying to understand the general idea of Stone Representation Theorem (or, at least, the existence of a functor from the category of boolean algebras to the category of compact totally disconnected spaces) with the example $B=\{0,1,2,3\}$.
The corresponding Stone space contains all ultrafilters on $B$ (which are all subsets of $B$ that contains $1$ and exactly $a$ or $¬a$, $a \in B$), so $S(B)=\{ \{1,2\},\{1,3\} \}$.
Also, the basis of the space is $\beta_{S(B)}= \{ U \in S(B) \mid b \in U\} = \{ \emptyset,\{1,2\},\{1,3\} \}$. But that makes no sense, since the union of all elements of $\beta_{S(B)}$ is not an element of $S(B)$.
What am I missing, or misunderstanding?
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In that case $S(B)$ is simply the discrete two-point space. The points are $\{1,2\}$ and $\{1,3\}$, and the open sets are $\varnothing,\big\{\{1,2\}\big\},\big\{\{1,3\}\big\}$, and $S(B)$ itself. The base is
$$\beta_{S(B)}=\big\{\{U\color{red}{\subseteq}S(B):b\in U\}:b\in B\big\}\;.$$
Here
$$\begin{align*}
\{U\subseteq S(B):0\in U\}&=\varnothing\\
\{U\subseteq S(B):1\in U\}&=S(B)\\
\{U\subseteq S(B):2\in U\}&=\big\{\{1,2\}\big\}\;,\text{ and}\\
\{U\subseteq S(B):3\in U\}&=\big\{\{1,3\}\big\}\;.
\end{align*}$$
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Proving the mean value property for harmonic functions using Green's second identity Suppose that $\phi:\mathbb R^3\to\mathbb R$ is a harmonic function. I am asked to show that for any sphere centered at the origin, the average value of $\phi$ over the sphere is equal to $\phi(0)$. I am also directed to use Green's second identity: for any smooth functions $f,g:\mathbb R^3\to \mathbb R$, and any sphere $S$ enclosing a volume $V$, $$\int_S (f\nabla g-g\nabla f)\cdot dS=\int_V (f\nabla^2g-g\nabla^2 f)dV.$$
Here is what I have tried: left $f=\phi$ and $g(\mathbf r)=|\mathbf r|$ (distance from the origin). Then $\nabla g=\mathbf{\hat r}$, $\nabla^2 g=\frac1r$, and $\nabla^2 f=0$. Note also that $\int_Sg\nabla f\cdot dS=r\int_S\nabla f\cdot dS=0.$ Then $$\frac{1}{4\pi r^2}\int_S f(x)\,dA=\frac{1}{4\pi r^2}\int_S f(x)\nabla g(x)\cdot dS=\frac{1}{4\pi r^2}\int_S (f\nabla g-g\nabla f)\cdot dS.$$ Using Green's identity, this is equal to $$\frac{1}{4\pi r^2}\int_V (f\nabla^2g-g\nabla^2 f)\,dV=\frac{1}{4\pi r^2}\int_V\frac{f}{r}\,dV.$$ This reminds me of the Cauchy integral formula. If there is indeed some sort of identity that I can use to show that the last integral is equal to $f(0)$? Or is there another way to solve this problem?
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I appreciate the other answers, but I came up with my own answer which, in my humble opinion, is a bit simpler.
For posterity's sake, here is the answer: Let $g=1/|\mathbf r|$ and $f=\phi$. Then the following hold: $$\int_S g(\mathbf r)\nabla f(\mathbf r)\cdot d\mathbf S=-\frac{1}{r}\int_S \nabla f\cdot d\mathbf S=0,$$ $$\nabla^2 f(\mathbf r)=0\quad\quad\text{for } \mathbf r\in V,$$ $$\nabla g(\mathbf r)=-\frac{\mathbf{\hat r}}{r^2},$$ $$\nabla^2 g(\mathbf r)=-4\pi \delta(\mathbf r).$$ Thus, the average value of $\phi$ on $S$ is
\begin{align*}
\frac{1}{4\pi R^2}\int_S \phi(\mathbf r)\,dS&=\frac{1}{4\pi}\int_S f(\mathbf r)\left(\frac{\hat r}{R^2}\right)\cdot d\mathbf S\\
&=-\frac{1}{4\pi}\int_S f(\mathbf r)\nabla g(\mathbf r)\cdot d\mathbf S\\
&=-\frac{1}{4\pi}\int_S (f(\mathbf r)\nabla g(\mathbf r)-g(\mathbf r)\nabla f(\mathbf r)\cdot d\mathbf S\\
&=-\frac{1}{4\pi}\int_V(f(\mathbf r)\nabla^2 g(\mathbf r)-g(\mathbf r)\nabla^2 f(\mathbf r))\,dV\\
&=\frac{4\pi}{4\pi}\int_V\phi(\mathbf r) \delta(\mathbf r)\,dV\\
&=\phi(\mathbf 0).
\end{align*}
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Is a function a special kind of relation? Is a function a "special kind of relation", or, does it "describe a specific relation"?
My text on discrete mathematics explains:
A relation is a subset of a Cartesian product and a function is a
special kind of relation.
But it would make more sense to me if a function described a relation as a subset of the Cartesian product.
My thoughts being:
Given a function, f(x) = y, we can compute a set of (x,y) coordinates within the Cartesian plain. And this set of coordinates would be the relation that is the subset of the Cartesian product.
Am I confused? Could anyone help explain how a function IS a relation?
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Think of the relation "is less than" in $\{1,2,3\}^2$. That relation is $\mathrm{Less}=\{(1,2),(1,3),(2,3)\}$, however you normally will not write $(x,y)\in\mathrm{Less}$, but you will write $x<y$. You certainly have no problem with saying "$x<y$ is a relation". And of course that is true not only for $x<y$, but also for other relations like $x\le y$, $x<y^2$, $x=y$, or, for that matter, $f(x)=y$.
So we see, $f(x)=y$ is a relation. But as you correctly said, $f(x)=y$ is a function. So a function quite obviously is a relation.
It is, however, a very special relation; which is why the notation $f(x)$ makes sense: For any $x\in A$ there's exactly one $y\in B$ so that $(x,y)\in f$. That is, you can read "$f(x)$ as meaning "The unique $y$ such that $(x,y)\in f$", just as you can read $xRy$ as "$(x,y)\in R$". Note that this also means that in principle, $xfy$ would also be a meaningful way to write $f(x)=y$, although that notation is, well, extremely unusual.
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Integral using a parameter $k\in \mathbb Z$. I am trying to solve an exercise that says:
Compute the following integral for $k\in \mathbb Z$: $$\int _{\vert z \vert =1} \! z^k \sin {\frac{1}{z}} \, \mathrm d z$$
The only thing that I can think is making $z=\mathrm e ^{\mathrm i t}$, so the integral would be
$$\mathrm i \int _0 ^{2\pi} \mathrm e ^{\mathrm i kt} \sin ({\mathrm e ^{-\mathrm i t}}) \, \mathrm d t$$
Then, expanding the exponential ($\mathrm e ^{\mathrm i k t}=\cos(kt)+\mathrm i \sin (kt)$):
$$\mathrm i \int _0 ^{2\pi} \cos(kt) \sin ({\mathrm e ^{-\mathrm i t}}) \, \mathrm d t-\int _0 ^{2\pi} \sin(kt) \sin ({\mathrm e ^{-\mathrm i t}}) \, \mathrm d t$$
But I don't know what to do next. I'm afraid that maybe this is not the way to do it.
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\begin{align}
&\color{#f00}{\left.\oint_{\verts{z}\ =\ 1}
\,\,z^{k}\sin\pars{1 \over z}\,\dd z\,\right\vert_{\ k\ \in\ \mathbb{Z}}}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\left.\oint_{\verts{z}\ =\ 1}
\,\,{\sin\pars{z} \over z^{k + 2}}\,\dd z\,\right\vert_{\ k\ \in\ \mathbb{Z}}
\\[5mm] & =
2\pi\ic\bracks{k \geq -1}
\braces{\bracks{z^{k + 1}}\sin\pars{z}\vphantom{\Large A}} =
2\pi\ic\bracks{k \geq -1}
\braces{\ic^{k}\,\,{1 + \pars{-1}^{k} \over 2\pars{k + 1}!}}
\\[5mm] & =
\color{#f00}{\left\{\begin{array}{rcl}
\ds{2\pi\ic\,{\pars{-1}^{k/2} \over \pars{k + 1}!}} & \mbox{if} &
\ds{k}\ \mbox{is $\ul{even}$ and}\ \ds{k \geq 0}
\\[3mm]
\ds{0} & \mbox{if} &
\ds{k}\ \mbox{is odd or}\ \ds{k < 0}
\end{array}\right.}
\end{align}
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How can we parametrise this matricial hypersphere? What I call a matricial hypersphere for lack of a recognised name is the set in $\mathbb{R}^{p\times k}$ defined by
$$\mathfrak{H}=\left\{
a_1,\ldots,a_k\in \mathbb{R}^{p};\ \sum_{i=1}^k a_i a_i^\text{T} = \mathbf{A}
\right\}$$
where $\mathbf{A}$ is a $p\times p$ symmetric positive semi-definite matrix of rank $k$ $(k\le p)$. My questions are
*
*Is this a well-known object?
*Given the matrix $\mathbf{A}$ is there a completion of $\mathbf{A}$ into an object in bijection with $\{a_1,\ldots,a_k\}$, which is my meaning of parameterisation?
*what is the size or dimension of $\mathfrak{H}$?
Note: This object does not stem out of nowhere. It appears in linear
regression, where the $a_i$ vector is a collection of regression
coefficients, and in connection with Wishart distributions, where the
$a_i$'s are Normal variates. I actually need to find a
reparameterisation of the $a_i$'s given $A$ to proceed a research
problem.
|
Point 1) Let $B$ be the matrix with columns $a_i$: your description is equivalent to
$$\tag{1}BB^T=A$$
Thus, being given a symmetrical semi-definite positive $n \times n$ matrix $A$ with rank $k$, $\frak{H}$ can be identified with the set of $n \times k$ matrices $B$ such that $A$ can be written under the form (1).
Remark: formula (1) is "up to the multiplication by a $k \times k$ orthogonal matrix $\Omega$" (with property $\Omega\Omega^T=I_k$). More precisely, any decomposition of the form (1) generates a family of decompositions:
$$\tag{2}B\Omega\Omega^TB^T=A \ \ \Leftrightarrow \ \ B'B'^T=A \ \ \text{with} \ \ B':=B\Omega$$
Point 2): Concerning parameterization, couldn't you use the more or less classical parametrizations of the (grassmannian) manifold of $k$-dimensional subspaces in $\mathbb{R^n}$ ? A reference (http://www.macs.hw.ac.uk/~simonm/schubertcalculusreview.pdf). Let us take an example with $n=3$ and $k=2$ :
$$B^T=\begin{pmatrix}1&0&x\\0&1&y\end{pmatrix}$$
(I have taken $B^T$ because the "landscape" form is easier to work with).
The idea behind this parameterization which places into evidence a first block $I_k$ is this one :
Consider $B^T$, which is rank-$k$ matrix with $k$ rows and $n$ columns.
We can write it under the block form $B^T=(C|D)$ where $C$ is square.
By multiplying it (in the same spirit as in (2) by $C^{-1}$, one obtains $(I_k|E)=(I_k|C^{-1}D)$ ;
As a partial conclusion, matrices $B$ such that $B^TB=A$ correspond in a bijective way to k-dimensional subspaces in $\mathbb{R}^n$, thus can be parameterized in the same way as them, using $k \times (n-k)$ parameters.
Point 3) Consequently, $\frak{H}$ considered as a manifold (it is evidently not a vector space), has dimension $k(n-k)$. See for example Stack Exchange question (What is the dimension of this Grassmannian?).
Another reference linked to statistical applications: (http://www.cis.upenn.edu/~cis515/Turaga_Stiefel_2011.pdf).
|
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|
Generalization of Kronecker–Weber theorem: is an abelian extension always cyclotomic? It is well-known that any abelian extension of $\Bbb Q$ is contained within some cyclotomic field $\Bbb Q(\zeta_n)$. My question is about the more general statement:
If $L/K$ is an abelian extension, do we have $L \subset K(\zeta_n)$, where $\zeta_n \in \overline K$ is a root of $X^n-1$ (for some $n \in \Bbb N$) ?
For instance, any abelian extension of $\Bbb Q_p$ is contained within some cyclotomic field $\Bbb Q_p(\zeta_n)$. Morover, it holds when $K$ is a finite field (the Galois groups are cyclic, and any non zero element is a root of unity).
I believe that the statement is wrong, but I wasn't able to come up with a counterexample.
Thank you very much!
|
No.
Consider the extension $\mathbb{Q}(t^{1/2})$ of the field $\mathbb{Q}(t)$. It is an extension of degree $2$, and is thus abelian. However, adding any root of unity $\zeta$ to $\mathbb{Q}(t)$ will give you the field $(\mathbb{Q}(\zeta))(t)$, in which $t$ has no square root.
|
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|
Does the linear transformation that a matrix encodes depend on a choice of basis? Context
Let $M$ be an $m \times n$ matrix of real numbers. Let $\mathbf{x}$ be column vector of length $n$ with elements $x_1, \ldots , x_n \in \mathbb{R}$. Let $\vec{x} = (x_1, \ldots , x_n) \in \mathbb{R}^n$ be its analogue in $\mathbb{R}^n$.
Let $\vec{\mathbf{e}}^t$ denote the standard basis in $\mathbb{R}^n$ so that
$$
\vec{\mathbf{e}} =
\left[
\begin{array}
.\vec{e_1} \\
\vec{e_2} \\
\vdots \\
\vec{e_n}
\end{array}
\right]
$$
and let $\vec{\mathbf{b}}^t$ denote a non-standard basis in $\mathbb{R}^n$ with
$$
\vec{\mathbf{b}} =
\left[
\begin{array}
.\vec{b_1} \\
\vec{b_2} \\
\vdots \\
\vec{b_n}
\end{array}
\right]
$$
Fact from Linear Algebra: $M$ encodes a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that
$$
M \mathbf{x} = \mathbf{y}
$$
where $\mathbf{\vec{e}}^t M \mathbf{x} = T(\vec{x}) = \vec{y}$ is some vector in $\mathbf{R}^m$.
Question
Does the linear transformation $T$ that $M$ corresponds to depend on our choice of basis? That is, if instead of working with $\mathbf{\vec{e}}^t$ we instead worked with $\mathbf{\vec{b}}^t$, would $M$ encode a different linear transformation? For example, if
$$
\mathbf{\vec{e}}^t \mathbf{x} = \vec{x} = \mathbf{\vec{b}}^t \mathbf{x'}
$$
then do we have that
$$
\mathbf{\vec{e}}^t M \mathbf{x} = T(\vec{x}) = \mathbf{\vec{b}}^t M \mathbf{x'}?
$$
EDIT
Ok. Upon reflection, it is obvious that our choice of basis matters significantly. For example, let us work in $\mathbb{R}^2$ and let our non-standard basis be $(1,1)$ and $(1, -1)$.
Then the identity matrix
$$
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
will now send $(1,0)$ to $(1,1)$ since
$$
\begin{bmatrix} (1, 1) & (1, -1) \end{bmatrix} \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \left[
\begin{array}
& 1 \\
0
\end{array}
\right] = (1,1)
$$
which is clearly a different mapping than the identity mapping that would result were we to use the standard basis (where $(1, 0) \mapsto (1,0)$).
|
Yes, the choice of basis makes a difference. For this reason, some books will write something like the following: If $T$ is a linear transformation, then the matrix representation of $T$ relative to a basis $\beta = \{\beta_1, \beta_2,...,\beta_n\}$ will be written
$$[T]_\beta$$
For example, let the matrix
$$A = \begin{pmatrix} 6&4&-6\\4&6&-6\\ 6&6&-7 \end{pmatrix}$$
Typically when we write $A$ in this way without any subscripts, we are implying that $A$ is being represented relative to the standard basis,
$$\{ \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix} , \begin{pmatrix} 0\\0\\1 \end{pmatrix} \}$$
We can see that the columns of $A$ are just the action of the matrix $A$ on the usual basis vectors. In other words,
$$A \vec{e}_1 = (6,4,6)$$
$$A \vec{e}_2 = (4,6,6)$$
$$A \vec{e}_3 = (-6,-6,-7)$$
The matrix seems to "scramble up" the usual basis vectors. However, with some work (namely, the eigenvalue problem), we can find a different basis
$$\beta = \{\beta_1, \beta_2, \beta_3\} = \{ \begin{pmatrix} -1\\-2\\-2 \end{pmatrix}, \begin{pmatrix} -2\\-1\\-2 \end{pmatrix} , \begin{pmatrix} 2\\2\\3 \end{pmatrix} \}$$
where
$$A \beta_1 = \begin{pmatrix} 6&4&-6\\4&6&-6\\ 6&6&-7 \end{pmatrix} \begin{pmatrix} -1\\-2\\-2 \end{pmatrix} = \begin{pmatrix} -2\\-2\\-4 \end{pmatrix} = 2 \beta_1$$
Similarly,
$$A \beta_2 = 2 \beta_2 \quad \text{and} \quad A \beta_3 = \beta_3$$
So, the matrix $A$ (which represents some linear transformation) has the representation relative to the basis $\beta$ above
$$[A]_\beta = \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}$$
A nice, diagonal matrix! In this case, we can see that $A$ acts by simply lengthening the $\beta$-basis vectors. We may also say that $A$ dilates its eigenspaces.
Hope this helps.
|
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|
Rank of $vv^\top + ww^\top$ Given two non-colinear real unit vectors $v,w$, I believe the rank of $M=vv^\top + ww^\top$ is 2 and I'd like to prove it. $vv^\top$ and $ww^\top$ are obviously of rank one, $v$ is not in the kernel of $M$ because $vv^\top v=v$ and $\|ww^\top v\|<1$ (because $v$ and $w$ are not colinear), same with $w$. So the rank of $M$ is at least 2 [edit just realised this argument is wrong; the non-colinearity of the images is required]. Also, by subadditivity of the rank, $\text{rank}(vv^\top + ww^\top )\leq \text{rank}(vv^\top) +\text{rank}(ww^\top )$. Do you see any mistakes/do you have another proof?
|
Yet another proof: note that
$$
vv^T + ww^T = \pmatrix{v&w} \pmatrix{v & w}^T
$$
and for any matrix $M$, $M$ has the same rank as $MM^T$.
|
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|
Verify Alternative Formula for Expected Value I am studying for the first actuarial exam (Exam P) and came across a formula in my ACTEX prep manual that I had never seen before:
$$E[X] = a + \int_{a}^{b}{[1-F(x)]dx}$$
And the text said this was true as long as $x$ was continuously defined on the interval, and as long as $b\lt \infty$. True for continuous and discrete! I tried it out with a few different, very straightforward functions and could not get it to equal an expected value answer I found in the typical manner. Am I missing something in the application? Has anyone seen this before?
Any help is appreciated!
|
For a continuous random variable
$$
E[X]=\int_a^b dx\ x\ f_X(x)\ ,
$$
where $f_X(x)$ is the probability density function. But
$$
f_X(x)=F'(x)\ ,
$$
where $F$ is the cumulative distribution function.
Substituting in the integral above, and integrating by parts we have
$$
E[X]=\int_a^b dx\ x\ F'(x)=x F(x)\Big|_a^b -\underbrace{\int_a^b F(x)dx}_{\star}\ .
$$
Now, note that
$$
\int_a^b [1-F(x)]dx=(b-a)-\underbrace{\int_a^b F(x)dx}_{\star}\ .
$$
Hence
$$
E[X]=x F(x)\Big|_a^b -\int_a^b dx\ F(x)=b F(b)-a F(a)+\int_a^b [1-F(x)]dx-(b-a)
$$
$$
=a+\int_a^b [1-F(x)]dx\ ,
$$
using the fact that $F(a)=0$ and $F(b)=1$ (by definition of cumulative distribution function for a density supported on [a,b]).
|
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|
Is a Rotation Matrix Still a Rotation Matrix under a Non-Standard Basis? Let us work in $\mathbb{R}^2$. Consider the following rotation matrix:
$$
R = \begin{bmatrix}
\cos \theta & - \sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}
$$
I agree that this represents a rotation of angle $\theta$ in $\mathbb{R}^2$ under the standard basis of $\beta = \{(1,0), (0,1)\}$. But would it still represent a rotation if we changed the basis to something non-standard (in both the domain and the codomain), say $\beta_1 = \{1, 1), (1, -1) \}$?
That is, would we have that
$$
\mathbf{\vec{b}}^t R \mathbf{c}
$$
represents a rotation if $\mathbf{\vec{b}}$ represents our non-standard basis and $\mathbf{c}$ represents an arbitrary element in $\mathbb{R}^2$ with respect to the non-standard basis?
|
Yes, it's possible. In particular, $R=\pm I_2$ is invariant under every change of basis.
If you need a nontrivial example, note that when $R\ne\pm I_2$, the product $P^{-1}RP$ is a $2\times2$ rotation matrix if and only if $P$ is a nonzero multiple of any $2\times2$ real orthogonal matrix. The "if" part can be easily verified if you multiply out $P^{-1}RP$ directly; to prove the "only if" part, consider $(P^{-1}RP)^T(P^{-1}RP)$.
So, in your case, when the change-of-basis matrix is $P=\pmatrix{1&1\\ 1&-1}$, a rotation will always remain a rotation in the new basis, because $P$ is a nonzero multiple of the real orthogonal matrix $\frac1{\sqrt{2}}\pmatrix{1&1\\ 1&-1}$.
|
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|
How to write equation for half-life of caffeine consumed over time? If you ingest 100mg of caffeine instantly (say, as a pill), then given the six hour half-life of caffeine in the body, you'd calculate the milligrams of caffeine left in your system with $100(\frac{1}{2})^{\frac{t}{6}}$ (where $t$ is the time since ingestion in hours).
What is the equation if you consume the 100mg at a constant rate over the course of an hour (savoring your coffee)?
|
A solution based on the Laplace transform: you have $dy/dt=-ky+100(1-u(t-1)),y(0)=0$ where $u$ is the Heaviside step function and $k=\ln(2)/6$. Taking Laplace transforms gives $sY=-kY+\frac{100-100e^{-s}}{s}$, so that $Y=\frac{100-100e^{-s}}{s(s+k)}$. The inverse Laplace transform is then $y(t)=\frac{100}{k} \left ( 1-e^{-kt} + u(t-1) \left ( e^{k-kt} - 1 \right ) \right )$. This could also be written as $y(t)=\begin{cases}\frac{100}{k} \left ( 1-e^{-kt} \right ) & t \leq 1 \\ \frac{100}{k} e^{-kt} \left ( e^k-1 \right ) & t \geq 1 \end{cases}.$
|
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|
Determine if $y = x^2$ is injective I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($\sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
|
I think that the syntax of the definition from your class is the point of confusion. The class definition depends on functions being defined as
$$
f(x)=(\text{expression of }x)
$$ rather than as
$$
y=(\text{expression of }x).
$$ Consequently, the "$y$" in "$f(y)$" is just some dummy variable that gets input into $f$ and is unrelated to $x$. So, for your example, $f(x)=x^2$, $f(x)= 4$ for $x=2$ and $f(y)=f(4)=16$ for $y=4$. Again, note that $y$ is unrelated to $x$.
Finally, in order to show that $f(x)=x^2$ isn't injective, you can start with the definition or with its contrapositive, as you stated:
*
*Starting with the definition,
$$
f(-2)=f(2)\:\text{ but }-2\:\text{ isn't equal to }2.$$
*Starting with the contrapositive, let's consider $x=2$ and $y=-2$: then
$$
x\neq y\:\text{ but }\:f(x)=f(y)=4.
$$
|
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|
Integral with modulus in the denominator. I want to solve this integral:
Let $f$ be a holomorphic function in $\Omega$ such that $\bar D (0,1) \subset \Omega$ and let $a\in D(0,1)$. Compute $$\int_{\vert z\vert =1}\! \frac{f(z)}{\vert z-a \vert ^2}\, \mathrm d z$$
If the modulus wasn't in the denominator, I know this integral would be $2\pi \mathrm{i}\,f'(a)$ because of Cauchy Integral Formula for derivatives. But I don't know how to solve this actual problem.
|
In the unit circle, $\bar{z} = z^{-1}$, and thus $\lvert z - a\rvert^2 = (z - a)(\bar{z}-\bar{a}) = (z - a)(z^{-1} - \bar{a})$ in the unit circle. Hence
$$\int_{\lvert z\rvert = 1} \frac{f(z)}{\lvert z - a\rvert^2}\, dz = \int_{\lvert z\rvert = 1} \frac{f(z)}{(z - a)(z^{-1}-\bar{a})}\, dz = \int_{\lvert z\rvert = 1} \frac{zf(z)}{(z - a)(1 - \bar{a}z)}\, dz$$
We can decompose
$$\frac{z}{(z - a)(1 - \bar{a}z)} = \frac{1}{1 - \lvert a\rvert^2}\left(\frac{1}{z-a} + \frac{\bar{a}}{1 - \bar{a}z}\right)$$
This allows us to write
$$\int_{\lvert z \rvert = 1} \frac{zf(z)}{(z-a)(1-\bar{a}z)}\, dz = \frac{1}{1-\lvert a\rvert^2}\left(\int_{\lvert z\rvert = 1} \frac{f(z)}{z-a}\, dz + \bar{a}\int_{\lvert z \rvert = 1} \frac{f(z)}{1 - \bar{a}z}\, dz\right)$$
You'll be able to take it from here.
|
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|
Is the function ring $C^{\infty}(M)$ noetherian? Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?
|
I think that this is not true in general.
Consider the ring $C^{\infty}(\mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.
Assume by contradiction that $I=(f_{1},\ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},\ldots,f_{n})$ vanishes on $V:=\cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{\infty}(\mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $I\neq (f_{1},\ldots,f_{n})$. Hence $I$ is not finitely generated.
|
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|
Write the plane $y + 2z = 1$ in a parametric form I'm asked to write the plane
$$y + 2z = 1$$
in a parametric form.
I know how to do this with variables like
$$ax + by + cz = d$$
but the lack of an $x$-variable stumps me. How should I go about doing it?
|
Since $x$ does not appear in the equation $y+2z=1$, $x$ is simply independent of the values of $y$ and the corresponding $z$. We can thus use any other variables to parametrize $x$. For instance, $t:=x$.
Then we first parametrize $y$ or $z$. Here, we let $u=y$. Since $y$ and $z$ are dependent of each other, $z$ can be parametrize in terms of $u$, which is $\displaystyle z=\frac{1-u}2$.
Thus the possible parametrization is
$$(t,u,\frac{1-u}2), t,u\in\Bbb{R}$$
|
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|
Find a subring of $\Bbb Z \oplus \Bbb Z$ that is not an ideal of $\Bbb Z \oplus \Bbb Z$. Find a subring of $\Bbb Z \oplus \Bbb Z$ that is not an ideal of $\Bbb Z \oplus \Bbb Z$.
I can't see any way a subring of $\Bbb Z \oplus \Bbb Z$ can NOT be an ideal. Subrings of $\Bbb Z \oplus \Bbb Z$ are of the form $n\Bbb Z \oplus k\Bbb Z$ where $n$ and $k$ are integers. So for something to not be an ideal in $\Bbb Z \oplus \Bbb Z$ there must be an $x = (j_1, j_2)$ in $\Bbb Z \oplus \Bbb Z$ and an $y = (b_1n, b_2k)$ in $n\Bbb Z \oplus k\Bbb Z$ such that $xy$ or $yx \notin$ $n\Bbb Z \oplus k\Bbb Z$. But this doesn't make sense because it seems any integer pair can be reached.
Anyone have any ideas?
|
Any subring would I presume contain $(1,1)$, the ring's unity (this is not the only possible interpretation).
Then any proper subring will do.
The proper subrings are the $R_n=\{(a,b)\in\Bbb Z^2\mid a\equiv b\pmod n\},$ for $n\gt1.$ See @Qiaochu's proof here.
$R_\infty$ is @Arthur's example.
|
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|
Does 3-partite graph with at least n+1 edges per vertex have a triangle? I need help for one problem.
In every of 3 schools there are n students (in total = 3n vertex, n per school).
Every student knows at least n+1 students from the other two schools.
Prove that there are 3 students, one from each school, who know each
other.
Do you have some idea for this proving?
|
Let $G$ be a counterexample, i.e. a 3-partite graph with $n$ vertices in each partite set
and minimum degree $n+1$ that has no triangle.
Let $v$ be a vertex that maximizes the number of neighbors it has in one partite set, say $k$.
We may label the partite sets $A,B,C$, such that $v\in A$ and $v$ has $k$ neighbors in $B$.
Since $k\leq n$, $v$ has at least one neighbor in $C$, say $w$.
$w$ has at most $n-k$ neighbors in $B$ (otherwise we find a triangle involving $v,w$ and a vertex of $B$),
so $w$ has at least $(n+1)-(n-k)=k+1$ neighbors in $A$. This contradicts the choice of $v$.
Done.
(This problem awoke my interest in the sharpness question: is it true that for a given $n$ and even $k<3n(n+1)$
there always is a 3-partite graph with $n$ vertices in each partite set and total degree at most $k$
(and minimum degree at least $n$?) that has no triangle?)
|
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|
Repeating or clarifying proofs in cited papers Is it bad form to repeat proofs or arguments that exist already in cited papers (while making it clear this is not my own argument but from the cited paper)? Instead of just citing the argument... I find it more 'readable' if I rewrite the argument within my paper in my own words. Plus the original argument may be slightly unclear, and can be clarified.
Plus it makes the paper more self-contained.
|
It depends on how long the proof is (longer is worse, for your question) and how much of an improvement you are making (more is better).
|
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|
Vandermonde's identity check in probability The book on probability I'm reading states Vandermonde's identity as: $\binom{m+n}{k} = \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i}$, but further in the book I'm seeing it being used as: $\binom{m+n}{k} = \sum_{i=0}^n \binom{m}{k-i} \binom{n}{i}$ and I can't seem to show that these are the same. Is this identity wrong or am I missing something?
|
The identity is correct both ways. All that matters is that the index run over all non-zero products of the two binomials. In the first version we know that $\binom{n}{k-i}$ is $0$ if $i<0$ or $i>k$, so having $i$ run from $0$ to $k$, inclusive, ensures that we include all of the non-zero terms. We may also include some $0$ terms, but that does no harm. (This happens if $k>m$.) In the second we know that $\binom{n}i$ is non-zero precisely when $0\le i\le n$, so here again the range of summation ensures that we get all of the non-zero terms. The two versions could simply be written
$$\binom{m+n}k=\sum_i\binom{m}i\binom{n}{k-i}$$
and
$$\binom{m+n}k=\sum_i\binom{m}{k-i}\binom{n}i\;,$$
with $i$ ranging over all integers, since only finitely many terms are non-zero. This is why it’s convenient to adopt the convention that $\binom{n}k=0$ if $k<0$ or $k>n$.
|
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Why Two's Complement works About to read computer science, I have just stumbled accross the concept of "Two's complement". I understand how to apply the "algorithm" to calculate these on paper, but I have not yet obtained an understanding of why it works. I think this site: https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html provides an explanaition why "flipping the digits" and adding one produces the compliment. What I do not understand is why adding the complement is equivalent to substracting the original number. Could somebody please give an explanation (maybe with a decimal example of the same concept as well?)?
Many thanks!
|
With the help of the other answers on this post (especially Ben Grossmann's), I managed to figure out Two's Complement and why it works for myself, but I wanted to add another complete barebones answer for anyone who still can't understand. This is my first post, so thank you for reading in advance. Also, much of my mathematic notation is likely to be false, so please refer to other answers for more mathematically accurate explanations.
As Ben Grossmann pointed out, understanding that binary addition is modulo is the key to understanding how Two's Complement works. What that means in a binary sense is that the last carry doesn't get used, so:
1111 1111 + 1 = 0000 0000,
not 1 0000 0000.
In decimal, this looks like $(255+1)\bmod{2^8}$. A similar example that you may find easier to wrap your head around is $(a+b)\bmod{12}$, which should look familiar.
This works for addition, but how about modulo subtraction? Well, continuing with the clock example, if we want to subtract using modulo addition, there is an easy solution: $a+(12-b) \pmod{12}$, or in binary: $a+(2^8-b)\bmod{2^8}$. The $12$ and $2^8$ are canceled out by their corresponding modulo, leaving us with $a-b$. The trick is now getting $2^8-b$, and that trick lies in Two's Complement.
To derive $2^8-b$, or $1\ 0000\ 0000 - b$ using only 8 bits, we first have to convert that into a familiar format:
$1\ 0000\ 0000-b=1111\ 1111-b+1$
This is the equivalent of $11-b+1$ using 12 as our modulo. Subtracting a number from $1111\ 1111$ is the same as inverting it. If this is confusing, then consider the equation
$1111\ 1111-1101\ 1001$
As you may have noticed, no borrows occur, because there are no $0$s in $2^8$. This effectively means that every bit of b is inverted. And so $INV(b)$ can be substituted for $1111\ 1111 - b$.
Plug that into our previous equation, and we now have $a-b=a+INV(b)+1$. There you have Two's Complement.
What this means for negative numbers is that $1111/ 1111=-1$ (as explained above in more detail by Ben Grossmann) and $1000/ 0000=-128$, while $0000\ 0000=0$ and $0111\ 1111=127. The system cycles through first the positives, and once 128 is reached ($1000\ 0000$) it flips the sign and cycles the other way through the negatives. The seventh bit acts like that sign, and no actual sign flip is needed. Without any additional provisions needing to be made, we can now add both negative and positive numbers.
|
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|
Find the centre of volume, defined by $\frac 1V\int_V {\bf x}$ $\mathrm d\mkern1muV$ of the tetrahedron. Anyone who has seen any of the questions I have posted recently will be getting sick of this introduction, but; I'm currently teaching myself some Vector Calculus in preparation for university and I'm finding it significantly more difficult than previous topics I have studied for the course. As a result my attempt below will likely be comical, but I'm just trying to get to grips with things so forgive my ignorance.
A tetrahedron $V$ has vertices $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,
1)$. Find the centre of volume, defined by $\frac 1V\int_V {\bf x}$
$\mathrm d\mkern1muV$.
The only example similar to this that the notes I'm 'learning' from is about a hemisphere, for which it parametrises in $\theta$ and $\phi$ and takes each of $x,y,z$ in terms of these and integrates over all appropriate $\theta,\phi$ and $r$. My issue is that I can not find such a parametrisation for a tetrahedron, I first tried finding the $y$ coordinate of the centre of volume (for whatever reason) and let $x=t$ and varyied it over $[0,1]$ and then found all possible values $z$ could take for a given $t$ and then similarly all value $y$ could take for a given $z$. As a result I had $y$ in terms of $v$ which varied in terms of $u$ which varied in terms of $t$ if that makes sense (I doubt it does), I won't go further because even I'm not 100% sure why I did what I did and I'm convinced we're already completely wrong here.
Of course an answer to the question would be great and will likely
answer the other questions I have, but if not then here are some other
questions: Is my issue the approach? Is it that my 'parameters' are
not independent? Have I fundamentally misunderstood everything?
I realise this will probably be incredibly frustrating to read for a lot of you, so apologies. Any input is obviously greatly appreciated.
Thank you
|
The integral of function $f(x,y,z)$ on that tetrahedron is given by
$$\int_V f dV=\int_{x=0}^1\int_{y=0}^{1-x}\int_{z=0}^{1-x-y} f(x,y,z) dz dy dx.$$
Note that the volume is given by $\int_V 1 dV=1/6$ and, by symmetry
$$\int_V x dV=\int_V y dV=\int_V z dV.$$
By using the above parametrization, the easiest one is
$$\int_V x dV=\int_{x=0}^1x\int_{y=0}^{1-x}\int_{z=0}^{1-x-y} dz dy dx=\int_{x=0}^1x\cdot\frac{(x-1)^2}{2} dx=\frac{1}{24}.$$
|
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Prove that $\mathbb{Q}$ and $\mathbb{Q} \cup \{\pi, e\}$ have the same cardinality
Prove that $\mathbb{Q}$ and $\mathbb{Q} \cup \{\pi, e\}$ have the same cardinality.
I know I must show that there exists a bijection between these two sets but I'm having a difficult time trying to come up with a function that relates them. Any suggestions? Thanks.
|
Here's an outline of a proof:
*
*Show that the sets $\{0, 1, 2, \dots\}$ and $\{-2, -1, 0, 1, 2, \dots\}$ have the same cardinality.
*Show that one of these sets has the same cardinality as $\Bbb{Q}$, and the other has the same cardinality as $\Bbb{Q} \cup \{\pi, e\}$.
|
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|
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
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use that $$\cos(120^{\circ}+x)=-1/2\,\cos \left( x \right) -1/2\,\sqrt {3}\sin \left( x \right) $$
and $$\cos(240^{\circ}+x)=-1/2\,\cos \left( x \right) +1/2\,\sqrt {3}\sin \left( x \right) $$
|
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How can I improve the proof of this statement with convexity? $A\in\mathbb{R}^{m\times n}$. How do I prove that if $S\subseteq \mathbb{R}^m$ is
convex then so is $A^{-1}(S) =\{ x \in \mathbb{R}^n : Ax
\in S\}$ ?
My proof:
$\forall t \in [0,1], ta+(1-t)b \in S$
Since $Ax\in S$, define $Aa'\in S, Ab'\in S$,
$\forall t \in [0,1], tAa'+(1-t)Ab' \in S$
$A(ta'+(1-t)b') \in S$
$A^{+}A(ta'+(1-t)b')= ta'+(1-t)b'\in A^{+}S$ (1), where $A^{+}$ is the pseudo-inverse of matrix $A$
Since $a',b' \in \{x \in \mathbb{R}^n,Ax\in S\}$ and $x \in A^{+}S$,
we have $a',b' \in A^{+}S$ (2).
I have my proof by timing $A^+$ (the pseudo-inverse of A) before $Ax$, however that's not a common method, moreover, $A^{+}A$ is not guaranteed to be an identity matrix, which may be problematic. Is there any more common technique in proving this statement?
|
I think you're overcomplicating a touch,
let $x,y\in A^{-1}(S)$ then if
$$tx + (1-t)y \in A^{-1}(S)$$ for all $t\in [0,1]$, the set is convex.
Now, let $z = tx+(1-t)y$, then $z\in A^{-1}(S)$ iff $Az \in S$. You can now write:
$$Az = ta + (1-t)b$$
letting $a=Ax, b=Ay$, both $a,b\in S$ since $x,y\in A^{-1}(S)$, by convexity $ta+(1-t)b\in S$ and therefore $z\in A^{-1}(S)$.
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|
Laplace transform of $\sin (t) / t$ I need to find the Laplace transform of $$\frac{\sin(t)}{t}$$ without using the following rule $\mathcal{L}\left(\frac{f(t)}{t}\right)=\int_s^{\infty}F(u)du$.
We aren't allowed to use this rule unless we can prove it, and I'm assuming our lecturer does not want us to use it.
However, we were given the following hint "For this question, you may need to evaluate $ \int^\infty_0 \frac{\sin(t)}{t}dt$. Try substituting $\frac{1}{t} = \int^\infty_0 e^{-st} ds$."
I'm not sure what exactly this is supposed to mean and would appreciate it if someone could show me what this would look like. Does it mean evaluating $ \int^\infty_0 e^{-st} \int^\infty_0 e^{-st}\sin(t) dsdt$? I'm really confused.
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I think you should write
$$ \int_0^\infty e^{-ut} \frac{\sin{t}}{t} = \int_0^\infty \int_0^\infty e^{-(u+s)t} \frac1{2i}( e^{it}-e^{-it}) \;ds \; dt$$
and change the order of integration. Details depends on the expected level of math rigor.
|
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How to express sum of areas of triangular elements meeting a criteria I have a metric for rating the results of a finite element simulation.
I am summing the area of the triangular elements which have a single value associated with them.
If the value associated with each triangle meets my criteria, I add the area of the triangle to the total, and the total for the simulation forms my metric.
How can/should I express this using a capital sigma notation?
I think it should be quite simple - e.g. just summing the elements of a set $S$, where $S$ is the area of triangles for which the value $v$ is within range?
If this is the case, how can I express the set nicely?
I don't want/need to show the actual area calculation.
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You can use the Indicator function. Let $A_i$ be the area of the $i$th triangle with associated value $x_i$.
Then your sum is
$$\sum_i A_i \mathbf{1}_C(x_i),$$
where $C$ is set of values for $x_i$ that meets your criteria.
|
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Can every number be represented as a sum of different reciprocal numbers? Can every rational number be represented as a finite sum of reciprocal numbers? You are only allowed to use each reciprocal number one time per expression (So for example 3/2 cannot be 1/2+1/2+1/2).
You could then express these numbers in a kind of "binary-reciprocal" where a 1 in the nth place from the right denotes adding 1/n. E.g 17/10 could be 10011 as it is 1+1/2+1/5.
Thanks in advance!
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Yes, and there is a simple algorithm to achieve this. First, we subtract successive reciprocal numbers $(1/1,\ 1/2,\ 1/3,\ \ldots)$ until the remainder is less than the next reciprocal number. This is always possible because the harmonic series is divergent.
If the remainder is zero then we are done, so let us assume that
$$0 < \frac{a}{b} < 1.$$
There exists a unique positive integer $m$, greater than any of the preceding denominators, so that
$$\frac{1}{m} \le \frac{a}{b} < \frac{1}{m-1}.$$
Therefore,
$$\frac{a}{b} - \frac{1}{m} = \frac{am-b}{bm} \ge 0.$$
But
$$\frac{a}{b} < \frac{1}{m-1} \implies am-a < b \implies am-b < a,$$
so the numerator has decreased.
If the process is repeated, then the remainder will eventually be zero, since the numerators cannot decrease forever. Therefore, every positive rational number can be represented as a sum of distinct reciprocal numbers.
Example: To write 11/4 as a sum of distinct unit fractions, we subtract as many successive unit fractions as possible, starting with 1.
$$\frac{11}{4} - \frac11 - \frac12 - \frac13 - \frac14 - \frac15 - \frac16 - \frac17 - \frac18 = \frac9{280}.$$
The remainder lies between $\frac1{32}$ and $\frac1{31}$, so we subtract $\frac1{32}$.
$$\frac{9}{280} - \frac{1}{32} = \frac{1}{1120}.$$
Since the result is a unit fraction, we are done.
$$\frac{11}4 = \frac11 + \frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \frac1{32} + \frac1{1120}.$$
Thanks to Barry Cipra for pointing out an error in a previous version of this post.
|
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Powers of Möbius transformations equal to identity? I'm looking at "Mobius transformations" where $a,b,c,d\in\mathbb R$. I want to know for which $n$ there exists $a,b,c,d$ such that for $f(x) = \dfrac{ax+b}{cx+d}$,
$$f^n(x) = f(f(...(f(x)))) = x$$
and what relationships between $a,b,c,d$ are required. Or if it is for all $n$, if there is a pattern to these relationships.
For example,
$$f^1(x)=x \iff a-d=0, c= 0, b=0, a\neq 0$$
$$f^2(x)=x \iff a+d=0, a^2+bc\neq 0$$
I see that for $f^{2k}(x)$, we can get an iterative relationship from the above. With the same conditions as the $n=2$ case.
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As said by @Did, you need to find coefficients a,b,c,d such that
$$M(f)^n=\begin{pmatrix}a&b\\c&d\end{pmatrix}^n=k\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
Thinking to rotation matrices, there is an evident solution :
$$M=\begin{pmatrix}\cos(\frac{\pi}{n})&-\sin(\frac{\pi}{n})\\ \sin(\frac{\pi}{n})&\cos(\frac{\pi}{n})\end{pmatrix}$$
Otherwise said, a possible Möbius (or homographic) transformation is :
$$f_n(x)=\dfrac{\cos(\frac{\pi}{n})x-\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})x+\cos(\frac{\pi}{n})}$$
Edit : This rotation is not unique in general. Let us take an example: if $n=12$, you can take any constant $K=1,2,\cdots 11$ in the following matrix
$$M=\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}$$
and have $M^n=\pm I_2.$
(following a very judicious remark of "studiosus") a very general type of non trivial matrices $M$ such that $M^n=\pm I_2$, at least among diagonalizable matrices is obtained by thinking to the conjugation operation, that doesn't change the eigenvalues that will still be $e^{iK\pi/n}$ and $e^{-iK\pi/n}$:
$$M=P\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}P^{-1}$$
for any invertible matrix $P.$
|
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Evaluating an indefinite integral with exponents and logarithms I was taking a GRE practice exam and came across
$$ \int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})} dx $$
I noted that this can be expressed as
$$ \int_0^{\infty} \frac{1}{(1 + e^{bx})} - \frac{1}{(1 + e^{ax})} dx $$
And
$$ \int \frac{1}{(1 + e^{cx})} dx , u = e^{cx}, du = ce^{cx} dx \rightarrow$$
$$ \frac{1}{c} \int \frac{1}{u(1+u)} du = \frac{1}{c} \left(\ln(u) - \ln(1 + u) \right) = \frac{1}{c} \ln \left( \frac{e^{cx}}{1 + e^{cx}} \right)$$
So then
$$ \int_0^{\infty} \frac{1}{(1 + e^{bx})} - \frac{1}{(1 + e^{ax})} dx = \frac{1}{b} \ln \left( \frac{e^{bx}}{1 + e^{bx}} \right) - \frac{1}{a} \ln \left( \frac{e^{ax}}{1 + e^{ax}} \right) | _0^{\infty} $$
Which yields that we are attempting to evaluate
$$ \ln \left( \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}} \right) |_0^{\infty} $$
Which reduces to evaluating:
$$ \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}} |_0^{\infty} $$
Which is:
$$ \lim_{x \rightarrow \infty} \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}} - 1 $$
But at this point I can't seem to crack this with L'hopital's rule. And in general, this problem should be completed between 30-seconds to 1 minute so I think this entire approach is invalid since its taking longer than that.
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Hint:
$$
\ln \left(\frac{e^{cx}}{1+e^{cx}}\right)\Big|_{x=0}^{x=\infty} = \ln 1 - \ln \frac{1}{2}
$$
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|
Trouble Identifying Counting Problem Formula? Would this be the correct formula for the counting problem
Partition with Identical Items
$$
\begin{pmatrix}
n \\
r \\
\end{pmatrix}= \begin{pmatrix}
n-1 \\
r \\
\end{pmatrix} + \begin{pmatrix}
n-1 \\
r-1 \\
\end{pmatrix} \
$$
Moreover would anyone know a good example that might demonstrate a partition with Identical items?
|
I am a member of a group of $n$ people in a room, of whom $r \le n$ will be awarded
one of $r$ identical prizes. Number of ways: ${n \choose r}.$
But I am interested in whether I will get a prize. There are two
cases:
I do not win a one of the prizes: Number of ways: ${n-1 \choose r}.$
Send me out of the room. Award the $r$ prizes among the $n-1$ people remaining.
I win one of the prizes: Number of ways: ${n-1 \choose r-1}.$
Start by giving me my prize. There are $n-1$ remaining people and $r-1$
remaining prizes.
Add the number of ways for the two cases to get the number of ways overall.
Note: This identity is the basis of 'Pascal's triangle'. (With no
disrespect to Pascal, there are reports of Indian, Arab, and Chinese manuscripts
showing this triangular configuration of numbers from many hundreds of
years before Pascal, some with no obvious indication of purpose.)
|
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How many functions are there from $\mathbb Z$ to $\mathbb Z$? Repeating the question,
How many functions are there from $\mathbb Z$ to $\mathbb Z$?
A function $f \colon A \to B$ is a subset of $A \times B$ satisfying
$$(a,b) = (a,c) \qquad \Rightarrow \qquad b = c,$$
so it's enough (maybe) to look at subsets of $\mathbb Z \times \mathbb Z$. We know $|\mathbb Z \times \mathbb Z| = |\mathbb Z|$, and that the number of subsets of $\mathbb Z$ is $2^{|\mathbb Z|}$, but this counts finite subsets as well. Unsure of where to proceed from here.
|
You’ve made a start. As you say, each function from $\Bbb Z$ to $\Bbb Z$ is a subset of $\Bbb Z\times\Bbb Z$, so there are at most $2^{|\Bbb Z|}$ of them. To finish the argument you could prove that there are also at least $2^{|\Bbb Z|}$ such functions by actually finding that many that you can clearly identify as distinct.
HINT: For each $A\subseteq\Bbb Z$ let
$$f_A:\Bbb Z\to\Bbb Z:n\mapsto\begin{cases}
1,&\text{if }n\in A\\
0,&\text{otherwise}\;.
\end{cases}$$
|
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Axioms for category theory I'm confused of background logic on category theory.
In ZFC set theory, we can construct new sets from existing sets by axioms, such as power set axiom, axiom of pairing etx.
I read first few pages of MacLane's category theory text and now I'm reading Tom Leinster's category theory text. Neither of these texts say whether we need some axioms or not, however, they are using some axioms in some sense without saying. I want to know what are the standard axioms for category theory.
Here are examples:
Firstly, how do we construct $A\times B$ where $A,B$ are categories? It is written in texts that "if we define $\operatorname{Obj}(A\times B)=\operatorname{Obj}(A)\times \operatorname{Obj}(B)$ and $\operatorname{Mor}((A_1,B_1),(A_2,B_2))=(\operatorname{Mor}(A_1,A_2),\operatorname{Mor}(B_1,B_2))$, then $A\times B$ forms a category". What kind of axiom would make this collecting possible?
Secondly, how do we construct a functor category $[A,B]$? How do we make "Collecting functors" process possible?
Thirdly, it is a theorem in text that "fully faith and essenially surjective functors are equivalences". However, to prove this, we need some kind of axiom of choice for category theory.
What would be the standard axioms?
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The standard axioms vary: they're either ZFC with an axiom of choice for proper classes, some set theory such as NBG that axiomatizes classes more thoroughly, or ZFC with Grothendieck universes, so that "large" categories are interpreted as still being small, but relative to a larger "universe" of sets. There have been efforts to axiomatize category theory without set theory, most notably ETCC, the elementary theory of the category of categories, but these have not proven to be sufficient as a foundation.
|
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Difference between f(x) and f(x,y)? I recently started doing calculus and came across terms such as f(x) and f(x,y). What is the difference between them? Are they the same thing?
|
No, they are not the same thing. $f(x,y)$ is a function of two variables $x$ and $y$, e.g., $f(x,y) = 3x + \sin(y)$. But $f(x)$ is a function of only one variable, e.g., $f(x) = x^3$.
|
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Proof of quadratic inequality using AM-GM Proof of quadratic inequality using AM-GM
|
we have to show that $$x^2y^2+x^2+y^2+4-6xy\geq 0$$ dividing by $x^2+1$ gives
$$y^2-\frac{6xy}{x^2+1}+\frac{4+x^2}{x^2+1}\geq 0$$ this is equivalent to
$$\left(y-\frac{3x}{x^2+1}\right)^2+\frac{(4+x^2)(x^2+1)-9x^2}{(x^2+1)^2}\geq 0$$
and this is equivalent to $$\left(y-\frac{3x}{x^2+1}\right)^{ 2 }+\frac{(x^2-2)^2}{(x^2+1)^2}\geq 0$$
which is true.
|
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|
why is probability not considered part of pure mathematics? wikipedia considers probability a part of applied mathematics, and it doesn't seem to fall under one of the four areas of mathematics (algebra, number theory, topology/geometry,analysis).
Nevertheless it seems to me to be a very fundamental mathematical concept, so why is it not considered a part of pure mathematics?
|
The way I interpret the applied and pure dichotomy in math is:
Applied math is taking established mathematical results and utilizing them in describing, understanding and solving real-world problems.
Pure math is a collection of mathematical results that serve the purpose describing, understanding and solving abstract problems. That is, problems which aren't a real-world problems.
They are not absolutely disjoint but rather are important parts of the whole of mathematics. Moreover, despite my above "definitions", I would say there is no strongly distinct line between the two. A research paper could have both pure math results and applied math results within it.
Probability falls under measure theory which is part of analysis, but also has the ability to be applied to real-world problems. I would say it is both.
|
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Prime field of a characteristic zero field is isomorphic to $\mathbb{Q}$ I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic to $\mathbb{Q}$. Here, I will use the notation
$nx=\left\{\begin{array}{cc}\underbrace{x+\cdots+x}_{n \text{ times}} & \text{if }n>0\\ 0 & \text{if } n=0 \\ \underbrace{(-x)+\cdots+(-x)}_{n \text{ times}} & \text{if }n<0 \end{array}\right.$
Firstly, I showed that $P=\{(m1)\cdot(n1)^{-1}; m,n\in\mathbb{Z},n\neq 0\}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $\phi:\mathbb{Q}\to P$ such that $\dfrac{m}{n}\mapsto(m1)\cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $\phi$ is well-defined, i.e., given $\dfrac{a}{b}=\dfrac{c}{d}$, we have that $\phi\left(\dfrac{a}{b}\right)=\phi\left(\dfrac{c}{d}\right)$. Why is it necessary to show that $\phi$ is well-defined?
|
If you define a map $\phi : S\to T$, you must always check that $\phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $\phi$ based on a presentation of an element $x\in S$, you must check that if $x'$ is a different presentation of $x$, then $\phi(x) = \phi(x')$. If $\phi$ is to be a function, then $\phi(x)$ must have one and only one value. So, since $x = x'$, $\phi(x)$ must equal $\phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = \dots$. So, if you define a map $\phi : \Bbb Q\to S$, and you define $\phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $\phi : \Bbb Q\to\Bbb Z$ given by $\phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = \phi(8/4) = \phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $\phi : P\to\Bbb Q$ gives the same value $\phi(x)$ no matter which way you write $x$.
|
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Ways in which 38 can be divided into 3 positive parts such that the first is divisible by 8, the second by 7 and the third part by 3? I am stuck with the question, i have tried a couple of random approaches but none of them is correct. The answer is 2. Please help if you know how to solve this question.
|
Assuming that each part must be an integer, this is equivalent to the number of non negative integral solutions of
$$8x+7y+3z=38$$
We have $x,y,z\ge1$. So, let $a=x-1,b=y-1,c=z-1$ such that $a,b,c\ge0$.
Then,
$$8a-8+7b-7+3c-3=38$$
$$8a+7b+3c=56$$
Let $X=8a,Y=7b,Z=3c$.
$$X+Y+Z=56$$
Allowed values of $X=0,8,16,24,32,40,48,56$.
Allowed values of $Y=0,7,14,21,28,35,42,49,56$.
Allowed values of $Z=0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54$.
Number of solutions=Coefficient of $x^{56}$ in the expansion of
$$(1+x^8+x^{16}+\cdots+x^{56})(1+x^7+x^{14}+\cdots+x^{56})(1+x^{3}+x^6+\cdots+x^{54})$$
that is coefficient of $x^{56}$ in
$$\left(\frac{1-x^{64}}{1-x^8}\right)\left(\frac{1-x^{63}}{1-x^7}\right)\left(\frac{1-x^{57}}{1-x^3}\right)$$
which can be further simplified using binomial expansions.
Using this, you can find all possible solutions in general. In this case, however, trial and error may be more convenient.
|
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Is there a linear transformation $T$ from $ R^3$ into $R^2$ such that $T(1,-1,1) = (1,0)$ and $T(1,1,1)= (0,1)$ . Is there a linear transformation $T$ from $ R^3$ into $R^2$ such that
$T(1,-1,1) = (1,0)$ and $T(1,1,1)= (0,1)$ .
|
A linear transformation is uniquely specified by its action on a basis. We can extend the set of linearly independent vectors $\{(1, -1, 1), (1,1,1) \}$ to a basis for $\mathbb{R}^3$ by adjoining some vector $v\in \mathbb{R}^3$ to the set. The required linear transformation can then be specified by setting $T(v)$ to be any vector in $\mathbb{R}^2$.
|
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Induction proof $\sum_{k=1}^{n} k^{2} = \frac{1}{6} n(n+1)(2n+1)$ I'm asked to prove $$\sum_{k=1}^{n} k^{2} = \frac{1}{6} n(n+1)(2n+1)$$
using proof by induction.
Now, I know how to do induction proofs and I end up at this step, needing to prove that:
$$\frac{1}{6} n(n+1)(2n+1) + (n+1)^2 = \frac{1}{6} (n+1)(n+2)(2n+3)$$
So $$n(2n+1) + n+1 = (n+2)(2n+3)$$
$$2n^2 + 2n + 1 = 2n^2 + 7n + 6$$
which is obviously not equal.
What am I doing wrong?
|
Dividing by $(n+1)$ was good, but when you got rid of the $\frac16$ factor, you forgot to multiply the $(n+1)^2$ term by $6$.
|
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|
Can 2 + 2 ≈ 5 be true? I was wondering a little, about how to proof that 2+2 =5
And here I am:
2.4 + 2.4 = 4.8
If we approximated numbers in each side individually then :
2 + 2 ≈ 5
I know this may not be right, but I don't know why it's wrong.
|
When you are talking about approximations you are usually doing so in a real world context. And so whether or not $4 \approx 5$ will depend on what you are measuring.
For example, the distance from the earth to the moon is $239,000$ miles. This is approximately equal to $240,000$ miles. So there $1,000$ is basically equal to zero. (That is $9,000$ is approximately $10,000$.)
But if I am on a road trip then $1,000$ miles is definitely not approximately equal to zero.
|
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|
Showing isomorphism of field extensions I need to show that if $\alpha$ has minimal polynomial $t^2-2$ over $\mathbb{Q}$ and $\beta$ has minimal polynomial $t^2-4t+2$ over $\mathbb{Q}$, then the extensions $\mathbb{Q}(\alpha):\mathbb{Q}$ and $\mathbb{Q}(\beta):\mathbb{Q}$ are isomorphic.
I want to say that somehow $t^2-2 \equiv t^2-4t+2$ modulo something and this will be of help in the proof? I see that $t^2-4t+2-(t^2-2)=-4t+4$, so maybe I should work modulo $1-t$? Does this make any sense and am I on the right track?
Thanks.
|
Write $\Bbb Q(\alpha)\cong\Bbb Q[t]/(t^2 - 2)$ and $\Bbb Q(\beta)\cong\Bbb Q[x]/(x^2 -4x + 2)$. The roots of $t^2 - 2$ are $\pm\sqrt{2}$ and the roots of $t^2 - 4t + 2$ are $2\pm\sqrt{2}$. Since $t$ in the first quotient represents $\sqrt{2}$ or $-\sqrt{2}$, and $x$ in the second represents either $2 + \sqrt{2}$ or $2 - \sqrt{2}$, does this tell you where to send $t$ to define a map $\Bbb Q[t]/(t^2 - 2)\to\Bbb Q[x]/(x^2 - 4x + 2)$?
|
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Question about proof that $A\subset B \implies \overline{A}\subset \overline{B}$ Look at this simple proof that $A\subset B \implies \overline{A}\subset \overline{B}$:
$\overline{B}$ is a set that contains $B$, therefore it contains $A$. Since it's closed and contains $A$, it must contain $\overline{A}$.
This made sense for me intuitively, but how do I prove the last sentence?
Since $\overline{B}$ it's closed and contains $A$, it must contain $\overline{A}$.
|
$\bar C$ is the intersection of the set of all closed sets that have $C$ as a subset. Let $A^*$ be the set of all closed sets that have $A$ as a subset. Let $B^*$ be the set of all closed sets that have $B$ as a subset. If $A\subset B$ then $A^*\supset B^*,$ which implies $\bar A=\cap A^*\subset \cap B^*=\bar B.$
|
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|
How to determine level curves/contours of a 3D function? I have the function $f(x,y) = e^{y/x^2}$ and I need to draw a contour map for levels $e^{-2}$, $e^{-1}$, $1$, $e$, and $e^2$.
I set $e^{-2}$ equal to the function, and solved for $y$ so that $y = -2x^2$.
Isn't this curve impossible? What am I doing wrong?
Thanks for your help!
|
You are lucky in the sense that $f(t)=e^{t}$ is an invertible function. That means that you can just set the argument equal to each other just like you have done. But you would not have been allowed to do so if it wasn't invertible!
For example if you set $(x)^2=(-1)^2$ you can't just assume $x=-1$, because $f(t)=t^2$ is not invertible. You won't catch all solutions then.
|
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A function in which addition and multiplication behave the same way Exponents have a well-known property:
$$x^ax^b = x^{a+b}$$
but
$$x^{a} + x^{b} \neq x^{a+b}$$
Similarly,
$$\log(a) + \log(b) = \log(ab) $$
But
$$\log(a)\log(b) \neq \log(ab)$$
So my question is this:
Is there a function $f$ on $\mathbb{R}$ or some infinite subset of $\mathbb{R}$ with the following properties
$$(1)\quad f(x)f(y) = f(x+y)$$
$$(2)\quad f(x)+f(y) = f(x+y)$$
ie
$$(3)\quad f(x)+f(y) = f(x)f(y)$$
It seems that $(2)$ requires the function to be linear...
|
Your title expresses interest in "a function in which addition and multiplication behave the same way". That's condition (3) alone. Conditions (1) and (2) are unnecessarily-strong requirements that artificially restrict the possible solutions. Be that as it may ...
Let's invoke condition (3) with three arbitrary values, $x$, $y$, $z$.
$$\begin{align}
f(x) + f(y) = f(x)\cdot f(y) \\
f(x) + f(z) = f(x)\cdot f(z)
\end{align}$$
Subtracting, we get
$$f(y) - f(z) = f(x)\cdot(\;f(y)-f(z)\;)\quad\to\quad\left(f(x)-1\right)\cdot\left(f(y)-f(z)\right) = 0$$
For all choices of $x$, $y$, $z$, at least one factor must vanish. We conclude that $f$ must be some constant; say, $k$. (The vanishing of the first factor requires specifically that $k=1$, but we'll go ahead and absorb this into the more-general statement.)
Then condition (3) reduces to
$$k + k = k\cdot k \quad\to\quad k(k-2) = 0$$
so that $k = 0$ or $k = 2$. That is, we have two ways to satisfy condition (3):
$$f(x) \equiv 0 \qquad\text{or}\qquad f(x) \equiv 2$$
Imposing conditions (1) and (2) limits the solutions to just the first.
|
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|
Show that this random variable is uniformly distributed in $\left(0,1\right)$ Let $X$ be a continuous random variable with distribution function $F_{X}\left(x\right)$ and let $Y=F_{X}\left(x\right)$, show that $Y\sim U\left(0,1\right)$, where $U$ is the uniform distribution.
Since the density function $f_{X}\left(x\right)$ can be seen as the derivative of $F_{X}\left(x\right)$, this problem looks like an obvious application of the Change of Variable Theorem. Nevertheless, the statement of the theorem requires that the function in which the random variable is evaluated (composed) to be strictly monotonic (strictly increasing or strictly decreasing) and, as it is well known, distribution functions are non-decreasing only.
So my question is: can the Change of Variable Theorem somehow be applied in this case?
Thanks in advanced for any help provided.
Note: Sorry if there are any gramatical errors in the redaction, English is not my first language.
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As Did wrote, let $Y=F_X(X)$ and note that $F_X$ need not be strictly increasing for the following arguments to holds. If we define the generalized inverse of $F_X$ by $F_X^{\leftarrow}(y) =\inf \{ x \in \mathbb{R}: F_X(x)\geq y \}$, then try to prove the following steps
*
*If $F_X$ is continuous then (iff actually holds) $F_X^{\leftarrow}$ is strictly increasing on $[\inf\{\text{Range}(F_X)\},\sup\{\text{Range}(F_X)\}]=[0,1]$.
*Use 1. to argue that $P(F_X(X)\leq x)=P(F^{\leftarrow}_X(F_X(X))\leq F^{\leftarrow}_X(x))$, for any $x\in[0,1]$.
*Show that $F^{\leftarrow}_X(F_X(y))\leq y$ and determine for what values of $y$ one has $F^{\leftarrow}_X(F_X(y))<y$. Use this to show that $P(F_X^{\leftarrow}\circ F_X(X)=X)=1$ (or look at this).
*Show that if $F_X$ is continuous, then $F_X(F^{\leftarrow}_X(x))=x$ for any $x\in[0,1]$.
*Use 2.,3. and 4. together to prove that $F_X(X)\sim \text{Unif}(0,1)\iff P(F_X(X)\leq x)=x, \, \, \forall x\in[0,1]$.
|
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Each integer appears once in difference sequence Does there exist an increasing sequence $a_1,a_2,\dots$ of positive integers such that both of the following are fulfilled:
*
*Each positive integer appears exactly once among $a_2-a_1,a_3-a_2,\dots$
*For some $n$, each positive integer at least $n$ appears exactly once among $a_3-a_1,a_4-a_2,a_5-a_3,\dots$?
If we only require the first condition, we can take the sequence $1,2,4,7,11,\dots$, while if we only require the second condition, we can construct the sequence by induction, making sure that each difference appears once.
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To fulfil the first condition, the differences $a_i-a_{i-1}$ must be distinct positive integers. Thus, to avoid repeating a difference, $$a_n\ge a_0+\frac{n(n+1)}2.$$ The second condition requires that all but finitely many positive integers appear among the $a_i-a_{i-2}$. But the above growth is too fast to enable that. Thus the second condition cannot be fulfilled.
|
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How to solve $\int \sqrt{a+x^2} dx $, $a>0$? By setting $a+x^2 = t^2$ I can get $2xdx=2dt$ so $dx=dt/\sqrt{t^2-a}$. But such substitutions only lead to iterate integrals of the form
$$
\int \frac{t^2}{\sqrt{t^2-a}}dt,
$$
substituting again $t^2-a=u^2$ we get
$$
\int udu +\int\frac{2}{\sqrt{u^2+a}}du.
$$
How can I solve the second integral?
|
Hint: substitute $$x=\sqrt{a}\sinh(t)$$
|
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Avoiding circularity in explaining the meaning of real exponents In an answer to 'What does $2^x$ really mean when $x$ is not an integer?' Álvaro Lozano-Robledo explains that we can understand real number exponents in terms of the definition of $\log(x)$:
$$\log(x) := \int_1^x \frac{1}{t} dt$$
I understand that one can then define $e^x$ as the inverse function of $\log(x)$. In his answer he then shows that what $a^x$ really means for any real number base $a > 0$ and exponent $x$ can be defined in terms of $\log(x)$ and its inverse $e^x$ in the following way:
$$a^x = e^{\log(a^x)} = e^{x\log(a)}$$
But this definition seems to rely on the logarithm power property $\log(a^x)=x\log(a)$. My question is, how can this property be proven without already knowing what $a^x$ means for real numbers $a > 0$ and $x$?
My attempt at answering the question before posting it here yielded the following:
ProofWiki has a proof of the logarithm power property that depends on a proof of the power rule for derivatives with real number index. However, this last proof also seems to assume that the meaning of $a^x$ for real numbers $a > 0$ and $x$ is known.
P.S. I have not yet had the time to study calculus/analysis formally. Which means that this might all become obvious after reading some rigorous analysis book.
|
There are many approaches to define $a^{x}$ when $x$ is irrational. Note that if we are restricted to real numbers then we must have $a > 0$ for $a^{x}$ to make sense. The simplest (but non-intuitive) approach is to first develop a theory of logarithmic and exponential functions namely $\log x$ and $\exp(x)$. Since these are inverses to each other we only need to define one of them and treat the other as inverse. Once these functions are available we define $$a^{x} = \exp(x\log a)\tag{1}$$ This is a definition and it does not rely on any other result. However it is easy to see the motivation behind this definition.
Suppose initially that $x$ is a positive integer then $$\log(a^{x}) = \log(a\cdot a\cdots a\text{ upto }x\text{ times}) = \log a + \log a + \cdots + \log a\text{ upto }x\text{ times} = x\log a$$ If $x$ is a positive rational number say $p/q$ then $\{a^{x}\}^{q} = a^{p}$ and from what we have proved above this leads to $$q\log a^{x} = p\log a$$ so that $$\log a^{x} = (p/q)\log a = x\log a$$ Next let $x$ be a negative rational number so that $x = -y$ where $y \in \mathbb{Q}^{+}$. Then $$\log a^{x} = \log 1/a^{y} = -\log a^{y} = -y\log a = x\log a$$ Further note that $\log a^{x} = x\log a$ holds trivially when $x = 0$. It proves that the identity $$\log a^{x} = x\log a$$ is true for all rationals $x$ and we want this identity to hold for irrational values of $x$ also. Therefore it is essential to define $a^{x}$ as $\exp(x\log a)$. Thus the motivation for the definition $a^{x} = \exp(x\log a)$ is to extend the identity $\log a^{x} = x\log a$ for irrational values of $x$ also.
Another approach (which is more intuitive) to define $a^{x}$ is by taking a sequence $x_{n}$ of rationals tending $x$ and then define $$a^{x} = \lim_{n \to \infty}a^{x_{n}}\tag{2}$$ When we use this approach the definition of $\log$ comes later via the limit $$\log a = \lim_{x \to 0}\frac{a^{x} - 1}{x}\tag{3}$$ and $\exp$ is defined by $\exp(x) = y$ if $x = \log y$. Under this approach the equation $$a^{x} = \exp(x\log a)$$ becomes a result based on the definitions of $a^{x}, \log, \exp$.
|
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Correct method of integration involving two exponential terms I have an Integrand involving two exponential terms:
$$
\int_{0}^{\infty} \frac{\exp(x^2)}{(1+\exp(x^2))^2} dx
$$
I would like to know what is the best way to integrate such a function without blowing it up?
What if $x^2$ is replaced by two variables $(x^2 + y^2)$ and we have a double integral? Will the method of integration remain the same?
I use python and matlab for calculations.
Thanks
Notation fixed.
|
$$
\begin{align}
\int_0^\infty\frac{e^{x^2}}{\left(1+e^{x^2}\right)^2}\,\mathrm{d}x
&=\int_0^\infty\frac{e^{-x^2}}{\left(1+e^{-x^2}\right)^2}\,\mathrm{d}x\\
&=\frac12\int_0^\infty\frac{e^{-x}}{(1+e^{-x})^2}\frac{\mathrm{d}x}{\sqrt{x}}\tag{1}
\end{align}
$$
Consider
$$
\begin{align}
&\int_0^\infty\frac{e^{-x}}{(1+e^{-x})^2}x^{\alpha-1}\mathrm{d}x\tag{2}\\
&=\int_0^\infty\left(e^{-x}-2e^{-2x}+3e^{-3x}-\dots\right)x^{\alpha-1}\,\mathrm{d}x\tag{3}\\
&=\Gamma(\alpha)\left(1-2^{1-\alpha}+3^{1-\alpha}-\dots\right)\tag{4}\\[6pt]
&=\Gamma(\alpha)\zeta(\alpha-1)\left(1-2^{2-\alpha}\right)\tag{5}
\end{align}
$$
The integral in $(2)$ is analytic for $\mathrm{Re}(\alpha)\gt0$. For $\mathrm{Re}(\alpha)\gt1$ the sum in $(4)$ converges and takes the value in $(5)$. Since the functions in $(2)$ and $(5)$ are analytic and equal for $\mathrm{Re}(\alpha)\gt1$, they must be equal for $\mathrm{Re}(\alpha)\gt0$.
Therefore, with $\alpha=\frac12$, we get
$$
\begin{align}
\int_0^\infty\frac{e^{x^2}}{\left(1+e^{x^2}\right)^2}\,\mathrm{d}x
&=\frac{\sqrt\pi}2\zeta\left(-\tfrac12\right)\left(1-\sqrt8\right)\\
&\doteq0.336859119428876991346\tag{6}
\end{align}
$$
A note on computing $\boldsymbol{\zeta\!\left(-\frac12\right)}$
We can't use the standard series
$$
\zeta(s)=\sum_{n=1}^\infty n^{-s}\tag{7}
$$
to compute $\zeta\!\left(-\tfrac12\right)$ because the series in $(7)$ only converges for $s\gt1$.
In this answer, analytic continuation is used to show that
$$
\zeta\!\left(-\tfrac12\right)=\lim_{n\to\infty}\left(\sum_{k=1}^n\sqrt{k}\,-\tfrac23n^{3/2}-\tfrac12n^{1/2}\right)\tag{8}
$$
The convergence of $(8)$ is very slow; the error is approximately $\frac1{24\sqrt{n}}$. However, by using $8$ terms from the Euler-Maclaurin Sum Formula, the error is reduced to $\frac{52003}{100663296}n^{-25/2}$.
Thus, using $n=1000$, we get
$$
\zeta\!\left(-\tfrac12\right)=-0.2078862249773545660173067253970493022262\dots\tag{9}
$$
|
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|
Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}n_{22} + n_{11} + n_{22} + 1 - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{11} + n_{22} + 1 = n_{11} + n_{22} + 1$
Which shows that it's true for a 2*2 matrix.
Looking at other 2*2 matrices
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = n_{11}n_{22} + n_{22} - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{22} = n_{22}$
Similarly:
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}$
$\begin{vmatrix}
n_{11} & n_{12} + 1 \\
n_{21} & n_{22}
\end{vmatrix} = -n_{21}$
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} + 1 & n_{22}
\end{vmatrix} = -n_{12}$
Positive if on the main diagonal, negative if on the side diagonal.
Using that, we can show that
$\begin{vmatrix}
n_{11} + 1 & n_{12} & n_{13} \\
n_{21} & n_{22} + 1 & n_{23} \\
n_{31} & n_{32} & n_{33} + 1
\end{vmatrix} = (n_{11} + 1)(n_{22} + n_{33} + 1) - n_{12}n_{21} - n_{13}n_{31} = n_{11}n_{22} - n_{12}n_{21} + n_{11}n_{33} - n_{13}n_{31} + n_{11} + n_{22} + n_{33} + 1 = n_{11} + n_{22} + n_{33} + 1$
And in a similar fashion we can apply this to higher order matrices.
But I haven't the slightest clue how to word or show this "similar fashion".
|
The value of a determinant as well of a trace is independent of the choice of basis. So suppose that the image of $A$ is generated by a vector $v_1$. Complement this vector with $v_2,...,v_n$ to form a base. In this base the matrix of $A$ takes the form:
$$ \underline{A} =
\left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n}
\\ 0 & 0 & ... & 0 \\
\\ . & . & ... & . \\
\\ 0 & 0 & ... & 0
\end{matrix}
\right)
$$
and that of $E+A$:
$$ \underline{1+A} =
\left( \begin{matrix} 1+a_{11} & a_{12} & ... & a_{1n}
\\ 0 & 1 & ... & 0 \\
\\ . & . & ... & . \\
\\ 0 & 0 & ... & 1
\end{matrix}
\right)
$$
Then clearly ${\rm tr} \ A=a_{11}\ $ and $\ \det (1+A)=1+a_{11}$.
In less abstract terms, as $A$ has rank 1, we may write $A= u v^T$ where $u$ and $v$ are column vectors. Suppose that $e_1^T u\neq 0$. Then $u,e_2,...,e_n$ forms a basis. Carrying out the products one verifies:
$$ (1+ A)
\left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right] =
\left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right]
\left[ \begin{matrix} 1+v^T u & v_2 & ... & v_n\\
0 & 1 & ... & 0 \\
. & . & ... & . \\
0 & 0 & ... & 1 \end{matrix} \right]$$
so $1+A$ is conjugated to the matrix on the right which verifies the claimed identity. And determinant and trace is invariant under conjugation.
|
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|
show sequence $n^2+1$ diverges
Question: Show $S_n = n^2+1$ diverges.
My understanding of this question is proof by contradiction.
We first assume that $n^2+1$ converges to $s$. This implies that fix $\epsilon >0$, there exists $N$ s.t. $|n^2+1-s|< \epsilon$. I'm stuck here. anyhelp?
|
Hint: show that for any natural number $N$ there is some $k$ such that $S_{n} > N$ for all $n > k$.
|
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|
Did I solve $z^{n-1}=\overline{z}$ correctly?
Please could someone tell me if my solution of $z^{n-1}=\overline{z}$
is correct?
My solution:
We have $z^{n-1} =\overline{z}$ if and only if $z^{n-1}z =\overline{z}z$, so
$z^n = |z|^2$
In polar coordinates,
$$ r^n e^{i n \varphi} = r^2$$
or, equivalently,
$$ r^{n-2} e^{i n \varphi} = 1$$
This equation implies that $r=1$. Hence we are taking $n$-th roots of $1$:
$$e^{i n \varphi} = 1$$
So the resulting set is a discrete and finite subset of the unit circle (consisting of the $n$-th roots of $1$).
|
I assume $n$ stands for a positive integer.
If $n=1$, the equation is $1=\bar{z}$.
If $n=2$, the equation is $z=\bar{z}$, so any real number is a solution.
Suppose $n>2$. First of all, $z=0$ is a solution. Assume now $z\ne0$ and write it as $z=re^{i\varphi}$. The equation becomes
$$
r^{n-1}e^{i(n-1)\varphi}=re^{-i\varphi}
$$
Since $r>0$, we can write it as
$$
r^{n-2}e^{in\varphi}=1
$$
Since $n>2$, we get $r=1$ and $e^{in\varphi}=1$, so the $n$-th roots of unity are the solutions.
If $n$ is allowed to be any integer, the case $n=0$ becomes $z\bar{z}=1$, so any complex number of modulus $1$ is a solution.
If $n<0$, write $w=1/z$ and $m=-n$, so the equation becomes
$$
\frac{1}{w^{m+1}}=\frac{1}{\bar{w}}
$$
or
$$
w^{m+1}=\bar{w}
$$
and we're in the same situation as before with $n>2$ (but, of course, the solution $0$ cannot be considered).
|
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|
How to factorize $5$ in $\mathbb{Z}[\root 3 \of 2]$? Since $5$ has a norm of $125$ in this domain, and $N(1 + (\root 3 \of 2)^2) = 5$, it seems like a sensible proposition that $5 = (1 + (\root 3 \of 2)^2) \pi_2 \pi_3$, where $\pi_2, \pi_3$ are two other numbers in this domain having norms of $5$ or $-5$. This is supposed to be a unique factorization domain, right?
I am encouraged by the fact that $$N\left(\frac{5}{1 + (\root 3 \of 2)^2}\right) = N(1 + 2 \root 3 \of 2 - (\root 3 \of 2)^2) = 25.$$
But I am discouraged by the fact that $$\frac{5}{(-1 - (\root 3 \of 2)^2)(1 + (\root 3 \of 2)^2)} = \frac{7 - 6 \root 3 \of 2 - 2 (\root 3 \of 2)^2}{5}$$ is an algebraic number but not an algebraic integer. I have found a couple of other numbers with norms of $5$ or $-5$ but can't get them to multiply to $5$ in any of the combinations of three of them that I have tried. I feel like I'm going around in circles.
|
We have
$$ \mathbf Z[\sqrt[3]{2}]/(5) \cong \mathbf Z[x]/(x^3 - 2, 5) \cong \mathbf Z_5[x]/(x^3 - 2) \cong \mathbf Z_5[x]/(x+2) \times \mathbf Z_5[x]/(x^2 + 3x + 4) $$
so that the ideal $ (5) $ factors as $ (5) = \mathfrak p_1 \mathfrak p_2 $. To find the ideals $ \mathfrak p_1 $ and $ \mathfrak p_2 $, note that they correspond to the maximal ideals in the ring $ \mathbf Z_5[x]/(x^3 - 2) $. We therefore have the following factorization:
$$ (5) = (\sqrt[3]{2} + 2, 5)(\sqrt[3]{4} + 3 \sqrt[3]{2} + 4, 5) $$
Since the ring $ \mathbf Z[\sqrt[3]{2}] $ is a principal ideal domain, these ideals are principally generated by prime elements of norm $ 5 $ and $ 25 $, respectively. Some easy trial and error by hand with the norm form yields that the first ideal is generated by $ 2\sqrt[3]{4} - 3 $, and dividing $ 5 $ by this number finally gives the factorization
$$ 5 = (2\sqrt[3]{4} - 3)(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9) $$
Here's some more detail on the trial and error part: we know that $ \mathfrak p_1 $ is a prime ideal of norm $ 5 $, and we know that $ (5) = \mathfrak p_1 \mathfrak p_2 $. Since any number with norm $ 5 $ generates a prime ideal, we can conclude by unique factorization of ideals that any such element would have to generate $ \mathfrak p_1 $. After that, we just look for solutions to the equation $ x^3 + 2y^3 + 4z^3 - 6xyz = 5 $ (the norm form), and it is easily seen that $ x = -3 $ and $ z = 2 $ do the trick.
Note that the factorization you are asking for is impossible: the prime $ 5 $ does not factor as the product of three prime ideals, but as two.
|
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|
Let $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $\sqrt{pq}$ is also irrational. Progress:
Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cdot\sqrt{k}=\sqrt{pq}\cdot\sqrt{pq}=k$$
Which implies that $\sqrt{pq}$ is irrational, since, $k$ is irrational and that $p^2\cdot{q^2}$ cannot be a proper square being $p^2$ and $q^2$ relatively prime and that concludes the proof.
The above lines are my attempt to prove the assertion. Is the proof correct? If not then how can I improve or disprove it.
Regrads
|
Except that you haven't proved $p^2 * q^2 $ cannot be perfect square, rest of the proof is correct.
Since $p^2 $ and $q^2$ are relatively prime, they do not have common prime factors. And they are not perfect squares. Else p,q would be rational numbers. This is necessary since 25, 4 are relatively prime but their product is perfect square.
Hence atleast one of the prime factors of p^2 and q^2 will be raised to an odd power in prime factorisation and those primes will be distinct. So $p^2 * q^2 $ cannot be perfect square
|
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|
Derivative of scalar function $\rm v^\top A^n v$ w.r.t. matrix $\rm A$ I would like to derivative of a scalar function with respect to matrix.
In pariticular,
For given vector $v \in \mathbb{R}^n$, let $f(A)=v^\top A^n v$ for any integer $n\in\mathbb{N}$. I want to find $\nabla_A (v^\top A^n v)$
When $n=1$, $$\nabla_A (v^\top A v) = v^\top v$$ and for $n=2$,
$$\nabla_A(v^\top A^2 v) = Avv^\top + vv^\top A^\top$$
But I don't know how to obtain derivative for general $n$.
Is there general form of derivative?
|
Let
$$f (\mathrm X) := \mathrm a^{\top} \mathrm X^n \, \mathrm a = \mbox{tr} (\mathrm a^{\top} \mathrm X^n \, \mathrm a) = \mbox{tr} (\mathrm a \mathrm a^{\top} \mathrm X^n)$$
Since
$$(\mathrm X + h \mathrm M)^n = \mathrm X^n + h \left( \sum_{k=0}^{n-1} \mathrm X^k \mathrm M \mathrm X^{n-1-k} \right) + O (h^2)$$
then the directional derivative of $f$ in the direction of $\mathrm M$ at $\mathrm X$ is
$$D_{\mathrm M} \, f (\mathrm X) = \mbox{tr} \left( \sum_{k=0}^{n-1} \mathrm a \mathrm a^{\top} \mathrm X^k \mathrm M \mathrm X^{n-1-k} \right) = \mbox{tr} \left( \left( \sum_{k=0}^{n-1} \mathrm X^{n-1-k} \mathrm a \mathrm a^{\top} \mathrm X^k \right) \mathrm M \right)$$
Thus, the gradient is
$$\nabla_{\mathrm X} f (\mathrm X) = \left( \sum_{k=0}^{n-1} \mathrm X^{n-1-k} \mathrm a \mathrm a^{\top} \mathrm X^k \right)^{\top}$$
|
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|
Confusion on Summation Notation in Lagrange's Identity I'm working on a proof (the proof of Lagrange's Identity), but it includes a sum notation I'm not familiar with:
$$\sum_{1\le k\lt j\le n} (a_kb_j-a_jb_k)^2$$
I would appreciate any explanations of what this is saying, specifically in regards to the inequalities below the sigma. Thanks!
|
${1\leq j<k\leq n}$ is the domain of the operator. We sum the terms for all integer values of the bound variables, $(j,k)$, where this domain holds true.
This is sometimes more convenient than the double sum notation with which you might be more familiar.
$$\sum_{1\leq j<k\leq n} (a_k b_j −a_j b_k )^2
\\ ~=~ \sum_{j=1}^{n-1}\sum_{k=j+1}^{n}(a_k b_j −a_j b_k )^2
\\ ~=~ \sum_{k=2}^{n}\sum_{j=1}^{k-1}(a_k b_j −a_j b_k )^2 $$
|
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|
How can we express the statement "$f$ is a bijection from $A$ to $B$" in predicate logic? How can we express the statement
$f$ is a bijection from $A$ to $B$
in predicate logic? Can it be expressed in first-order logic?
A problem seems to be that the sets $A$ and $B$ are different sets, while predicate logic applies to a single set. Is that correct?
|
Establish Domain and Codomain:
$$\forall y~ \forall x~ F(x) = y \implies (x \in A \land y \in B)$$
Surjection:
$$\forall y \in B ~ \exists x ~F(x)=y$$
Injection:
$$\forall y ~ \forall x_1 ~ \forall x_2 ~ (F(x_1) = y \land F(x_2) = y) \implies (x_1 = x_2)$$
|
{
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|
Proof that columns of an invertible matrix are linearly independent
Explain why the columns of an $n \times n$ matrix $A$ are linearly
independent when $A$ is invertible.
The proof that I thought of was:
If $A$ is invertible, then $A \sim I$ ($A$ is row equivalent to the identity matrix). Therefore, $A$ has $n$ pivots, one in each column, which means that the columns of $A$ are linearly independent.
The proof that was provided was:
Suppose $A$ is invertible. Therefore the equation $Ax = 0$ has only one solution, namely, the zero solution. This means that the columns of $A$ are linearly independent.
I am not sure whether or not my proof is correct. If it is, would there be a reason to prefer one proof over the other?
As seen in the Wikipedia article and in Linear Algebra and Its Applications, $\sim$ indicates row equivalence between matrices.
|
I’ll use rows instead of columns, but the argument for columns holds similarly. Suppose a square matrix A is invertible that means it is row equivalent to the identity matrix. Now, observing that the rows of the identity are linearly independent, you can reapply the reverse operations on the rows of the identity to get the rows of A, this shows that the rows of A are linearly independent.
Suppose the rows of A are linearly independent. Applying row operations on the rows of A until you have a reduced echelon matrix, will give you a matrix with no zero rows and hence it must be the identity. Therefore A is row equivalent to the identity and so A is invertible.
|
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|
Selling price of coffee mixture
A trader buys $15$ Kg of Arabian coffee powder at $x$ dollars per kg and $25$ kg of Brazilian coffee at $y$ dollars per kg. He mixes the two types of coffee powder in a ratio $3 : 5$ and packs the mixture into packets each of which contains $100$ grams of the mixture. He sells the packs for $ 40x + 48 y / 640 $ dollars. Find an expression for the selling price of all the packets of coffee powder in terms of $x$ and $y$.
I clearly knows that I need to find the cost price of the coffee powder .
I done up a ratio of the price of the coffee powder = $ 3x : 5y $
Why is the cost price of the coffee powder not $ 3x + 5y $ ?
Can I get a hint on how to carry on? Thanks in advance !
|
"I done up a ratio of the price of the coffee powder = 3x:5y.
Why is the cost price of the coffee powder not 3x+5y? "
It is. What makes you think it isn't? Of course, that is not what the question asked so would not be the "answer" to the question. The question asked for the selling price.
|
{
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|
Extremality of $*$-automorphisms of $C^*$-algebras In a paper, I came across the notion of the extremality of a $*$-automorphism of a $C^*$-algebra. What is the definition of this property?
Edit: Basically, I have a unital completely positive map $\varphi: A \rightarrow A$, where $A$ is a $C^*$-algebra, such that $\psi = (\text{id}_A + \varphi)/2$ is a $*$-automorphism. Then the author states that from this property together with the extremality of a $*$-automorphism, $\varphi = \text{id}_A$, but I don't see how.
|
The author is saying that automorphisms are extreme among the ucp maps.
Write $\psi=(\phi_1+\phi_2)/2$, with $\phi_1,\phi_2$ ucp. Let $a\in A$ selfadjoint. Then, as $\phi_1(a)$ and $\phi_2(a)$ are selfadjoint, $(\phi_1(a)-\phi_2(a))^2\geq0$. So
\begin{align}
\psi(a)^2&=\frac{(\phi_1(a)+\phi_2(a))^2}4
=\frac{\phi_1(a)^2+\phi_2(a)^2}2-\frac{(\phi_1(a)-\phi_2(a))^2}4\\ \ \\
&\leq\frac{\phi_1(a)^2+\phi_2(a)^2}2
\leq\frac{\phi_1(a^2)+\phi_2(a^2)}2=\psi(a^2).
\end{align}
Because $\psi$ is multiplicative, $\psi(a)^2=\psi(a^2)$. So the inequalities above become equalities. In particular, we obtain
$$
(\phi_1(a)-\phi_2(a))^2=0.
$$
Then $\phi_1(a)-\phi_2(a)=0$, since it is selfadjoint. As this works for any selfadjoint $a\in A$, it follows that $\phi_1=\phi_2$.
|
{
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|
Cayley graph interpretation D3 I am trying to understand the Cayley graph for the group $D_3$, which from Mathematica, I got:
I tried to get the multiplication table in Mathematica:
$\left(
\begin{array}{cccccc}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 1 & 4 & 3 & 6 & 5 \\
3 & 5 & 1 & 6 & 2 & 4 \\
4 & 6 & 2 & 5 & 1 & 3 \\
5 & 3 & 6 & 1 & 4 & 2 \\
6 & 4 & 5 & 2 & 3 & 1 \\
\end{array}
\right)$
where the first row and column are the table headings.
How does this multiplication table relate to the graph?
Thanks!
|
The table and graph are related as follows. Let $X=\{1,2,3,4,5,6\}$ and let $S=\{2,4\}$. The table is missing its "headings" so that the actual table looks like this
$$
\begin{array}{c|cccccc}
& 1 & 2& 3& 4& 5&6\\
\hline
1 & 1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 1 & 4 & 3 & 6 & 5 \\
3 & 3 & 5 & 1 & 6 & 2 & 4 \\
4 & 4 & 6 & 2 & 5 & 1 & 3 \\
5 & 5 & 3 & 6 & 1 & 4 & 2 \\
6 & 6 & 4 & 5 & 2 & 3 & 1 \\
\end{array}
$$
where $i\cdot j$ is equal to the number in the $i$th row and $j$th column (so for example $4\cdot 2 =6$ and $2\cdot 4=3$).
Next notice that $S$ generates $X$ under the above multiplication. To see this note that $2^2=1,2,4,4^2=5,2\cdot 4=3,2\cdot 4^2=6$ are all the elements of $X$. Now let us turn to the graph, the vertices are the elements of $X$ and there is a red edge from $x$ to $y$ if $x\cdot 2 =y$. Similarly there is purple edge from $x$ to $y$ if $x\cdot 4=y$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
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