Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Using elementary row operations to solve intersection of two planes The question I am struggling with is the following:
Solve the following using elementary row operations, and interpret each system of equations geometrically:
\begin{align*}
x - 3y + 2z &= 8\\
3x - 9y + 2z &= 4
\end{align*}
The answer given in the book is $x = -2+3t, y = t, z = 5$, and the planes meet in a line.
I put this in matrix form, so
$$\left(\begin{array}{c c c|c}
1 & -3 & 2 & 8\\
3 & -9 & 2 & 4
\end{array}\right)
$$
Then I subtracted R1 from R2
$$\left(\begin{array}{c c c|c}
1 & -3 & 2 & 8\\
2 & -6 & 0 & -4
\end{array}\right)
$$
Then R1 - $\frac{1}{2}$R2
$$\left(\begin{array}{c c c|c}
0 & 0 & 2 & 10\\
-2 & -6 & 0 & -4
\end{array}\right)
$$
So I got that $2z =10$, so $z = 5$, but I am stumped on how to continue. How can I get $x$ and $y$ and interpret the results in terms of plane intersection?
| $2z = 10$, so $z = 5$.
$2x - 6y = -4 \Rightarrow2x = -4 + 6y \Rightarrow x = -2 + 3y$. Let $y = t$, we get $x = -2 + 3t$, $y = t$, $z = 5$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confusing description of torsion of a curve? In my textbook (by do Carmo) and both in wikipedia.
There are descriptions of what a torsion is, and both of them says it is a measure of "how fast a curve twists out of the plane of curvature"
I am aware of the definition of torsion which is the magnitude of the derivative of the binomial vector, but I fail to see how this describes "how fast the curve is twisting out of the plane" or "pulling out of the plane".
If we are talking about how fast the curve is traveling outside of its osculating plane, then this makes absolutely no sense to me at all, since the tangent vector is on the osculating plane.
| When a curve is planar, all osculating planes are equal.
When it is non planar, i.e. has some torsion, the osculating planes stop staying parallel when you move along the curve, and this change of direction is reflected by the binormal.
"Infinitesimal" insight:
Imagine the curve discretized with a fixed step.
Two successive points define a line, which is the tangent.
Three successive points define a plane, which is the osculating plane.
A fourth point can deviate from the plane and show the torsion. In other words, two triples of successive points will define two distinct planes and the angle between them corresponds to the torsion.
| {
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"url": "https://math.stackexchange.com/questions/2615811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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f is a continuous function from (X,$\tau$) to {0,1} with discrete topology, if f non constant then (X,$\tau$) disconnected Let $f$ be a continuous function such that $f : (X,\tau) \rightarrow (\{0,1\},\tau_1\}$. Where $(X,\tau)$ is a generic topological space and $\tau_1$ is the discrete topology. I want to prove that if f is non-constant then $(X,\tau)$ is disconnected.
I started by describing $(\{0,1\},\tau_1\}$. This topological space is compact, totally disconnected and Hausdorff. However,from here I do not know how to continue. Any tips?
| If $X$ is connected and $f:X\to Y$ is continuous then $f(X)$ is connected.
So if moreover $f$ is surjective then $Y=f(X)$ is connected.
In your case non-constant comes to the same as surjective.
Draw conclusions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that a limit does not exist with absolute values I'm to prove that the following limit does not exist
$\lim_{x\to -2} \frac{\vert 2x +4\vert -\vert x^3 +8 \vert}{x+2}$
From here, I have taken the method of finding $\lim_{x\to -2^+}$ and $\lim_{x\to -2^-}$ to show that they're not equal
However my problem is that both are simplified to become $\frac{x^3 - 2x +4}{x+2}$
Is there something I'm doing wrong or did i make a mistake when opening the absolutes?
| Because for $x\rightarrow-2+$ we have
$$\frac{|2x+4|-|x^3+8|}{x+2}\rightarrow2-4-4-4=-10$$ and
for $x\rightarrow-2-$ we have
$$\frac{|2x+4|-|x^3+8|}{x+2}\rightarrow-2+4+4+4=10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Uniformly minimum variance unbiased estimator of theta. Let $X_1,X_2,....X_n$ be a random sample from a population with probability density function
$$f(x|\theta)=\dfrac{\theta}{2}e^{-\theta|x|} ,-\infty < x < \infty,\theta>0$$
Then a UMVE of $\theta$ is ?
Can someone tell me if i did everything right or not in the following steps. I dont have answer written for this problem in my solution manual.
Steps :
1.Its an even function that's why pdf changes as $$f(x|\theta)=\theta e^{-\theta x} ,0 < x < \infty,\theta>0$$
2.Using Rao blackwell theorem i found MVUE $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$
Now MVUE is $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$ or $\dfrac{n-1}{\sum_{i=1}^{n} |X_i|}$ ? Please give your thoughts and tell me where did i follow wrong track?
| Write
$$f(x)=\dfrac{\theta}{2}e^{-\theta|x|}=h(x)g(\theta)\exp\left(\eta(\theta)\cdot T(x)\right)$$
with $h(x) = 1$, $g(\theta) = \theta/2$, $\eta(\theta)=-\theta$, and $T(x) = |x|$.
It follows that $f$ is of the exponential family.
Furthermore, note that the parameter space
$$\Theta=(0, \infty)\supset(0, 1)$$
so $\Theta$ contains an open set. Therefore, it follows that $T(\mathbf{X}) = \sum_{i=1}^{n}|X_i|$ is sufficient and complete.
Next, we need to find the CDF of $|X_1|=|X|$. Observe that due to that $f$ is even,
$$F_{|X|}(x)=\mathbb{P}(|X| \leq x)=\mathbb{P}(-x\leq X \leq x)=\int_{-x}^{x}f(t \mid \theta)\text{ d}t=2\int_{0}^{x}f(t \mid \theta)\text{ d}t$$
and we can see that
$$\int_{0}^{x}\dfrac{\theta}{2}e^{-\theta|t|}\text{ d}t=\dfrac{\theta}{2}\int_{0}^{x}e^{-\theta t}\text{ d}t=\dfrac{\theta}{2}\cdot\dfrac{1}{-\theta}(e^{-\theta x}-1)=\dfrac{1-e^{-\theta x}}{2}$$
so it follows that $$F_{|X|}(x)=1-e^{-\theta x}$$
for $x > 0$, with probability density function
$$f_{|X|}(x) = \theta e^{-\theta x}$$
for $x > 0$ and $f_{|X|}(x) = 0$ otherwise,
hence $|X|$ follows an exponential distribution with mean $1/\theta$. It follows that $T(\mathbf{X})$ has a Gamma distribution with $\alpha = n$ and $\beta = \theta$.
With some work, it can be shown that $\dfrac{1}{T(\mathbf{X})}$ follows an Inverse-Gamma distribution with mean $\dfrac{\theta}{n-1}$ as long as $n > 1$; hence, $$\mathbb{E}\left[\dfrac{n-1}{T(\mathbf{X})}\right]=\mathbb{E}\left[\dfrac{n-1}{\sum_{i=1}^{n}|X_i|}\right]=\theta$$
so by Lehmann-Scheffe, $\dfrac{n-1}{\sum_{i=1}^{n}|X_i|}$ is the UMVUE of $\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why this integral diverge? For $D=[0,+\infty)×[0,+\infty)$ and $f(x,y)=(1+x+y)^{-1}$, why does the integral over $D$ of $f(x,y)$ not converge?
It's not like $\rho/(1+\rho|\sin\theta+\cos\theta|)$ which to infinite is like $1/(|\sin\theta+\cos\theta|)$ and, close to $0$ is like $0$?
| Intuitively, the area of the domain grows like $xy$, while the integrand decreases like $\dfrac1{1+x+y}$, which is insufficient to compensate.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The energy method for $u_{tt}-du_t-u_{xx}=0, (0,1)\times(0,T) $ It is a problem in Evan's PDE. I want to prove the smooth solution of the following PDE is zero:
$$u_{tt}-du_t-u_{xx}=0, (0,1)\times(0,T) $$$u|_{x=0}=u|_{x=1}=u|_{t=0}=u_t|_{t=0}=0$.
The hint is to use the energy $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}+||\partial_xu||_{L^2[0,1]})$, but when I differentiate the energy, I can only get the first and third term of PDE
| Let us assume that the solution is sufficiently smooth and $d<0$. The time-derivative of the energy
$E(t) = \frac{1}{2}\left( \|u_t\|_{L^2[0,1]}^2 + \|u_x\|_{L^2[0,1]}^2\right)$ writes
$$
\begin{aligned}
\frac{\text{d}}{\text{d}t}E(t) &= \int_0^1 \left( u_{tt}\, u_{t} + u_{xt}\, u_{x} \right) \text{d}x \\
&= \int_0^1 \left( u_{tt}\, u_{t} + u_{tx}\, u_{x} \right) \text{d}x \\
&= \int_0^1 u_{tt}\, u_{t}\, \text{d}x + \left[ u_{t}\, u_{x} \right]_0^1 - \int_0^1 u_{t}\, u_{xx}\, \text{d}x \\
&= \int_0^1 \left(u_{tt} - u_{xx}\right) u_{t}\, \text{d}x \\
&= d\int_0^1 (u_{t})^2\, \text{d}x \\
&\leq 0 \, .
\end{aligned}
$$
To show that the energy is decreasing, we have used successively the equality of mixed derivatives, integration by parts, the fact that $u$ is constant-in-time at the boundaries $x=0$ and $x=1$ of the domain, and the PDE itself. Since at $t=0$, the energy is $E(0) = 0$ and the energy is always positive, we have shown that the energy is equal to zero for all $t$, which means that $u$ is constant in time and space. Now, since it equals zero at the boundaries of the domain, it is necessary that $u$ is identically zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the functional equation $f(x + f(x +y ) ) = f(2x) + y$? Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following equation:
$$ f(x + f(x +y ) ) = f(2x) + y,\quad \forall x,y\in\mathbb{R}$$
The only function I have found is $f(x) = x$, but I think there are more.
| Given $z$, let $x=f(z)$. and $y=z-x.$ Then you get:
$$f(x+f(x+y))=f(2f(z))$$ and $$f(2x)+y=f(2f(z))+z-f(z)$$
From this you get $z=f(z).$
A cute variation of Christian's very nice answer:
$$\begin{align}
2z+f(0)&=f(f(2z))&[x=0,y=2z]\\
&=f(f(z+f(z)))&[x=z,y=0]\\
&=z+f(z)+f(0)&[x=0,y=z+f(z)]
\end{align}$$
So $f(z)=z.$
This is avoiding the reference to being an injection, by implicitly using the right inverse $g(z)=f(z)-f(0).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Examples for $\max\{f(x)+g(x)\}\leq \max{f(x)}+ \max{g(x)}$ and $\min\{f(x)+g(x)\}\geq\min{f(x)}$+min{g(x)}. Can someone give me an example with concrete functions for the following relations $\max\{f(x)+g(x)\}\leq \max{f(x)}+ \max{g(x)}$ and $\min\{f(x)+g(x)\}\geq \min{f(x)}+\min{g(x)}$? I suppose one example would be enough to verify both.
| Another solution: $f(x)=g(x)=0$ (or equal to any other constants)
| {
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Parametrization of an arbitrary conic/ellipse I have the coefficients for a conic (I actually know that it is an ellipse) in the form $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$$
Is there an efficient algorithm which returns the parametrization of the eclipse, i.e., $\langle x(t), y(t) \rangle?$
| Based on the comments to your question, it looks like the underlying problem that you’re trying to solve is to determine efficiently whether or not a point lies within the ellipse. A straightforward way to do this is to plug the coordinates of the point into the left-hand side of the general equation. The sign of the result will discriminate among the three possibilities. If you arrange for the leading coefficient $A$ to be positive (as will $C$ in that case), then the point is outside of the ellipse if the result is positive, inside if negative and of course on the ellipse if zero. Points very near the boundary might run into machine resolution limitations, so beware of that.
The reason this works is that, if you start from the unit circle $x^2+y^2=1$, for which the appropriate inequalities are obvious, any ellipse can be obtained from it via a combination of scaling, rotation and translation. This transformation preserves the interior and exterior, so the inequalities involving the transformed equation have the same senses. With positive scale factors, the coefficients $A$ and $C$ in the general equation will also be positive.
If you have a lot of points to test against a fixed ellipse and most of them are outside of the ellipse’s bounding box, it could be more efficient overall to do some preliminary range checks before evaluating the general conic expression. Finding the points on the ellipse where the partial derivative w/r $y$ vanishes will give you the $x$-coordinate bounds, while the points where the $x$-derivative vanishes will give you the $y$ bounds.
Another approach to finding the bounding box is to find the horizontal and vertical tangents to the ellipse. Working in homogeneous coordinates, the tangent lines to an ellipse given by the matrix $Q$ that go through a point $\mathbf p$ are captured by the degenerate conic $T=\mathcal M_{\mathbf p}^TQ^{-1}\mathcal M_{\mathbf p}$. Here, $\mathcal M_{\mathbf p}$ is the “cross-product matrix” of $\mathbf p$, i.e., $\mathcal M_{\mathbf p}\mathbf x=\mathbf p\times\mathbf x$. Instead of $Q^{-1}$, we can use the adjugate of $Q$ or any other convenient non-zero multiple of it instead.
Taking $\mathbf p = [0:1:0]$ will produce the vertical tangents to $Q$. For the general conic equation, then, we have $$Q = \begin{bmatrix}A&\frac B2&\frac D2\\\frac B2&C&\frac E2\\\frac D2&\frac E2&F \end{bmatrix}$$ and so $$T=\begin{bmatrix}0&0&-1\\0&0&0\\1&0&0\end{bmatrix} \begin{bmatrix} 4CF-E^2 & DE-4BF & 2BE-2CD \\ DE-4BF & 4AF-D^2 & 2BD-2AE \\ 2BE-2CD & 2BD-2AE & 4AC-4B^2 \end{bmatrix} \begin{bmatrix}0&0&1\\0&0&0\\-1&0&0\end{bmatrix} = \begin{bmatrix} 4AC-4B^2 & 0 & 2CD-2BE \\ 0&0&0 \\ 2CD-2BE & 0 & 4CF-E^2 \end{bmatrix}.$$ Now, you could use this directly for the range check by checking the sign of $(4AC-4B^2)\,x^2+(4CD-4BE)\,x+(4CF-E^2)$, but that’s not a whole lot better than simply using the original conic equation. However, we can tease out the individual lines by “splitting” this degenerate conic: Find an $\alpha$ for which $T+\alpha\mathcal M_{\mathbf p}$ has rank one. The two lines are then any row and column of the resulting matrix that have a common nonzero diagonal element. All of the $2\times2$ minors of $T+\alpha\mathcal M_{\mathbf p}$ are identically zero except for $$\alpha^2-4C^2(D^2-4AF)-4C(AE^2+4B^2F-2BDE).$$ Taking either root results in the equations $$2(AC-B^2)\,x = (BE-CD)\pm\sqrt{C^2(D^2-4AF)+C(AE^2+4B^2F-2BDE)}$$ for the tangents, which give you the left and right edges of the bounding box. The calculation for the top and bottom is similar, but with $\mathbf p=[1:0:0]$ instead, and again there’ll only be one $2\times2$ minor of $T+\alpha\mathcal M_{\mathbf p}$ that doesn’t vanish identically. The resulting equations for the horizontal tangents are $$2(AC-B^2)\,y=(BD-AE)\pm\sqrt{A^2(E^2-4CF)+A(CD^2+4B^2F-2BDE)}$$ from which we can read the max. and min. $y$-coordinates of the ellipse.
| {
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"timestamp": "2023-03-29T00:00:00",
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Monotone convergence theorem for series (basic proof) My question is how to prove monotone convergence theorem for infinite series without more advanced technique like counting measure. I see this used a lot. But looking through books like Rudin, the theorem for series or elementary proof is not to be found.
The theorem is:
For a sequence $x_{mn} \geq 0$ if $\lim_{m \to \infty} x_{mn} = y_n$ (monotonically increasing for $m$) then
$$\lim_{m \to \infty} \sum_{n=1}^\infty x_{mn} = \sum_{n=1}^\infty
\lim_{m \to \infty} x _{mn}$$
| Here's a constructive $\epsilon$ proof.
Let $\epsilon > 0$ be arbitrary. Consider the the smallest value of $N$ such that the partial sum $$\sum_{n=1}^N
\lim_{m \to \infty} x _{mn}$$ is within $\epsilon/2$ of the actual sum.
Denote $\lim_{m \to \infty} x _{mn}$ by $x _{\infty,n}$ from now on.
Now let $M$ be such that whenever $m \geq M$ and $n \leq N$: $x_{\infty,n} - x_{mn} < \frac\epsilon{2N}$.
It follows that we are done, because for any $m\geq M$, $$\sum_{n=1}^\infty x_{\infty,n} \geq \sum_{n=1}^\infty x_{mn} \geq \sum_{n=1}^\infty x_{\infty,n} - \epsilon,$$
which can be proved in the following way:
$$\begin{aligned}
x_{\infty,n} &> x_{mn}\\
\therefore \sum_{n=1}^\infty x_{\infty,n} &\geq \sum_{n=1}^\infty x_{mn}\\
&\geq \sum_{n=1}^N (x_{mn} )\\
&\geq \sum_{n=1}^N (x_{\infty,n} - \frac\epsilon{2N})\\
&\geq \sum_{n=1}^N (x_{\infty,n}) - \frac\epsilon{2}\\
&\geq \sum_{n=1}^\infty (x_{\infty,n}) - \frac\epsilon{2} - \frac\epsilon{2}\\
&= \sum_{n=1}^\infty (x_{\infty,n}) - \epsilon.
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Group action on vector space Let $G$ be a group acting on (complex) vector spaces $V, W$ and let $G$ act trivially on the vector space $U$.
Let $Hom^G(V, W)$ denote the linear transformations $T:V \to W$ that respect the group action, that is, $g \cdot T(v) = T(g \cdot v)$ for all $v \in V$.
$W \otimes U$ has a natural $G$-action given by $g \cdot (w \otimes u) = gw \otimes gu = gw \otimes u$.
How do I see $Hom^G(V, W \otimes U) \cong Hom^G(V, W) \otimes U$?
There is a natural map going backwards given by sending $(T: V \to W) \otimes u$ to the map $v \mapsto T(v) \otimes u$.
This second map respects the $G$-action because $T$ respects the $G$-action, and $U$ has trivial $G$-action.
| I don't believe this is true in full generality; one needs $U$ to be finite-dimensional.
In the case where $G$ is the trivial group, this amounts to
$$\text{Hom}(V,W\otimes U)\cong\text{Hom}(V,W)\otimes U.$$
Specialising further to $W=\Bbb C$ (one-dimensional) we get
$$\text{Hom}(V,U)\cong\text{Hom}(V,\Bbb C)\otimes U.\tag{*}$$
The natural map from the right of (*) to the left has its image
the homomorphisms from $V$ to $U$ with finite-dimensional image,
so is not surjective.
But if $U$ is finite-dimensional one can write $U$ as finite direct
sum of trivial one-dimensional modules. As both functors
$U\mapsto \text{Hom}^G(V,W\otimes U)$ and $U\mapsto \text{Hom}^G(V,W)\otimes U$ preserve finite direct sums, we reduce to the case $U=\Bbb C$ with trivial $G$-action, in which case the result is immediate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum Area of inscribed rectangle how can we compute the maximum area of a rectangle which can be inscribed in a triangle of area 'M'
I have taken a special case here in the image file to calculate the inscribed rectangle area but how can we calculate it for general case??
| Let $EF=x$.
Thus, since $h_a=\frac{2M}{a}$ and $\Delta ABC\sim \Delta AEF$, we obtain:
$$\frac{x}{a}=\frac{\frac{2M}{a}-EG}{\frac{2M}{a}},$$
which gives $$EG=\frac{2M}{a}\left(1-\frac{x}{a}\right).$$
Id est, by AM-GM $$S_{EFHG}=\frac{2M}{a}\left(1-\frac{x}{a}\right)x=2M\left(1-\frac{x}{a}\right)\frac{x}{a}\leq2M\left(\frac{1-\frac{x}{a}+\frac{x}{a}}{2}\right)^2=\frac{M}{2}.$$
The equality occurs for $1-\frac{x}{a}=\frac{x}{a},$ which says that we got a maximal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number theory: find $a, b$ such that $\frac{a}{b} = b.a$ in a general base $\mathcal{B}\neq 10$ I was playing with numbers and thinking about this "coincidence"
$$\frac{5}{2} = 2.5$$
That is, for positive $a$ and $b$ we have
$$\frac{a}{b} = b.a$$
And those questions came into my mind:
1. Could we find all such integers pair $a, b$? (clearly in base $10$)
And due to the fact that we work in base $10$, a more general problem popped up, that is, to write our numbers in a base $\mathcal{B}\neq 10$ and thence look for triplets $(\mathcal{B}, a, b)$ such that
$$\frac{a}{b} = b.a$$
When $a, b$ are written in base $\mathcal{B}$.
2. Could we find a general formula that will produce infinitely many such integer triples?
I am not really into number theory, except for few little questions, so this is more a sort of "I am asking to you experts in the field" question.
If for some reason this problem is unclear or wrong or impossible, just tell me!
Thank you!
| My idea would be to do $\frac{a}{b}=b+\frac{a}{10}$ and this leads to $a=\frac{10b^2}{(10-b)}$. I think this may be a possible solution for case 1. Actually I do not have an idea for $\mathcal{B}\neq 10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$A+A\subseteq A\times A$ Define $A+A=\{a+b\colon a,b\in A\}$,$A\times A =\{ab\colon a,b\in A\}$.
Does there exist a finite integer set $A\subseteq \mathbb{Z}^+$, such that $|A|>1$ and $A+A\subseteq A\times A$ ?
| We assume such a set exists and derive a contradiction. First, if $\{1,2\} \subset A$ then $3 \in A$ as $1+2=3$ so we must have $1 \cdot 3 \in A \times A$. Also, since $|A| > 1$ then if $1$ or $2$ is not in $A$ then there exist an element of $A$ greater than $2$, so in either case such an element exists.
Edit: the rest is wrong so I struck it out.
Let $x$ be the smallest such element. Now as 2x \in A + A we must have 2 \in A because we need 2x \in A \times A and the minimality of x. This in turn implies implies x + 2 \in A \times A.
So 1,2,4,x,2x and x^2 are the smallest possible elements of A \times A and because of our choice of x we notes 2+x < 2x < x^2 so 2+x must be on that list. However proceeding by cases we see that x+2=1, x+2 = 2, x+2=4, x+2=x and x+2=2x all contradict our choice of x so no such set A can exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Embeddings between Hölder spaces $ C^{0,\beta} \hookrightarrow C^{0, \alpha} .$ Let $ \Omega \subset \mathbb R^n $ be an open subset and let $ 0 < \alpha < \beta \leq 1.$ We consider the space of Hölder continuous functions $C^{0, \alpha}$ which is a Banach space endowed with the norm
$$ \| f\|_{C^{0, \alpha}} := \| f \|_{\infty} + \sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}. $$
My questions has to do with the embedding $ C^{0,\beta} \hookrightarrow C^{0, \alpha} .$
If $ \Omega $ is bounded, then I can prove the estimate $ \| f\|_{C^{0, \alpha}} \leq \text{diam}(\Omega)^{\beta -\alpha} \| f\|_{C^{0, \beta}} ,$ which in turn implies that the embedding is bounded, i.e. continuous.
Question: How can I show that the embedding is still continuous in the case where $ \Omega $ is unbounded ?
Any help would be really appreciated.
|
Note that Embeddings between Holder space do not care about the boundedness of the domain
Patently we have $$\sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}+\sup_{ x,y \in \Omega \\ |x -y|\ge1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}$$
But since $|x-y|^\alpha\ge 1$ for $|x-y|\ge 1$. we obtain $$\sup_{ x,y \in \Omega \\ |x -y|\ge1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha} \le \sup_{ x,y \in \Omega \\ |x -y|\ge1} |f(x) -f(y)| \le 2\|f\|_\infty$$
whereas if $|x-y|\le 1$ and $0 < \alpha < \beta \leq 1.$ then
$$|x-y|^{\beta-\alpha}\le1\implies |x-y|^{\beta}\le|x-y|^{\alpha}$$
and hence, $$\sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\beta}\le \sup_{ x,y \in \Omega \\x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\beta}$$
It plainly follows that
$$\color{red}{\sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\beta}+2\|f\|_\infty\le 2\|f\|_{C^{0,\beta}}}$$
| {
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"url": "https://math.stackexchange.com/questions/2617513",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim \limits_{n \to \infty} n \int_{-1}^{0}(x+e^x)^n dx$. Evaluate $\lim \limits_{n \to \infty} n $$\int_{-1}^{0}(x+e^x)^n dx$. The answer should be $\frac{1}{2}$. I tried the substitution $x+e^x=u$ and then using the property that $\lim_{n \to \infty } n \int_{-a}^{1} x^n f(x)dx=f(1)$ but I don't know what to do further. If you solve this please do it so that a highschooler like me can understand it, thank you.
| Your property says that the limit is
$$
\frac{1}{1+e^{X(1)}},
$$
where $X(u)$ is the solution of $X(u)+e^{X(u)}=u$. Observe that $e^{X(1)}=1-X(1)$, so your limit is
$$
\frac{1}{2-X(1)}.
$$
It suffices to show that $X(1)=0$. But this is clear, since $1=X(1)+e^{X(1)}\ge e^{X(1)}\ge 1$, with equality if and only if $X(1)=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability problem involving geometric distribution and expected value We're given the following problem:
"An experiment is conducted until it results in success: the first step has probability $\frac{1}{2}$ to be successful, the second step (only conducted if the first step had no success) has probability $\frac{1}{3}$ to be successful, the third step (only conducted if the first two steps had no success) has probability $\frac{1}{4}$ to be successful : if none of the steps were successful, we repeat the experiment until success is achieved. Assuming the first step has cost $2$, and the second step as well as the third step have cost $1$, what is the overall cost until success ? "
Here's my approach. Let $X$ be the random variable the counts the number of steps required to achieve success. Thus giving me: $$P(X=1) = \frac{1}{2} \\ P(X=2)=\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(X=3) = \frac{1}{2} \cdot \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{48} $$
Now, to compute the expected cost, I did the following: $$E(X) = 2 \cdot 1 \cdot P(X=1) + 1 \cdot 2 \cdot P(X=2) + 1 \cdot 3 \ P(X=3)$$ (where the first coeffeicient of each term of the sum is the cost of each step). This gave me $\frac{79}{48}$ , which is far from the correct answer.
So, I do not understand what I did wrong (I guess there must be a problem with my reasoning), and I do not know how to find the correct result (which is $\frac{34}{9}$).
| Let $\mu_{0}$ denote the expectation of the cost of the steps yet
to be done if no steps have been made.
Let $\mu_{1}$ denote the expectation of the cost of the steps yet
to be done if step 1 has been made without success.
Let $\mu_{2}$ denote the expectation of the cost of the steps yet
to be done if step 1 and step 2 have been made without success.
Then we have the following equalities.
$\mu_{0}=\frac{1}{2}2+\frac{1}{2}\left[2+\mu_{1}\right]=2+\frac{1}{2}\mu_{1}$
$\mu_{1}=\frac{1}{3}1+\frac{2}{3}\left[1+\mu_{2}\right]=1+\frac{2}{3}\mu_{2}$
$\mu_{2}=\frac{1}{4}1+\frac{3}{4}\left[1+\mu_{0}\right]=1+\frac{3}{4}\mu_{0}$
Then:$$\mu_0=2+\frac12\left[1+\frac23\left[1+\frac34\mu_0\right]\right]$$
The solution of this equality is: $$\mu_0=\frac{34}9$$
| {
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Invertible Linear Map Suppose $V$ is finite-dimensional and $S,T\in \mathcal{L}(V)$. Prove that if $ST$ is invertible, then both $S$ and $T$ are invertible.
This is the partial question retrieved from Linear Algebra Done Right, and I have come out with different solution from the solution guide.
My solution:
Since $ST$ is invertible, then the only possibility for $(ST)u=0$ is when $u=0$. Therefore I have $(ST)(0)=0=S(T(0))=0$. Since $T(0)=0$ (as proven in book), I must have $S(0)=0$. Since $T(0)=0$ and $S(0)=0$ comes directly from $(ST)(0)=0$, both $S$ and $T$ are injective. Since both of them are linear operator, we can deduce they are both invertible (by equivalece statement.)
Is my proof valid?
| As José Carlos Santos stated, your proof is not correct. You have to show two things: injectivity and surjectivity of both $S$ and $T$. If you want to use zero element in your proof then injectivity can be shown by proving the following implication $$ f(u) = 0 \Rightarrow u = 0 \ ,$$
that is, you do not assume in the beginning that $u$ is zero.
Other hint that might be useful in the proof: If $ST$ is injective then $T$ must be injective. If $ST$ is surjective then $S$ must be surjective.
| {
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Why is the inverse of the derivative of f not the actual derivative of the inverse of f? So, I've explored this a little, but it is still confusing. When you calculate the inverse of a function, f, that is one-to-one, the points switch: a point (2,8) on f would be (8,2) on the inverse. So, one would assume that the derivatives of the functions would also constitute the reversal of points. However, that is not the case. For example, you have:
$f (x) = 5x^2 \phantom{=}\text{ for $x\geq0$}$
$f '(x) = 10x$
and
$(f^{-1}) (x) = \sqrt{\frac{x}{5}}$
Here is my question: Why is finding the inverse of the derivative of $f$, $f '(x)$, and taking its inverse not the real derivative of the inverse? I would think $(f^{-1}) '(x) = \frac{x}{10}$, but that is not the case. The real inverse would be taking the derivative of $(f^{-1}) (x)$ and finding $(f^{-1}) '(x) = (\frac{1}{10\sqrt{x/5}})$. In my mind, both of these seem like they could be the derivatives of the inverse, yet only the latter is true. Why is this?
Also, maybe I missed out in class, but is there some sort of quick relationship between (besides the formula) $f '(x)$ and $(f^{-1}) '(x)$ similar to how points switch between $f (x)$ and $(f^{-1}) (x)$.
Thanks.
| Intuitive thoughts to reflect on: draw the graph of $f$ and mark a point on it (say $(a,f(a))$). Draw the tangent at that point. It will have slope $f'(a)$.
Now flip the entire plane around the line $y=x$. The graph of $f$ has now become the graph of $f^{-1}$, the marked point has become $$(f(a),a)= (f(a),f^{-1}(f(a)))=(b,f^{-1}(b))$$ where $b=f(a)$. The tangent line is still the tangent line, but its slope is inverted ($\Delta y$ and $\Delta x$ have swapped roles for the line, so their ratio is inverted).
Putting this together, we get $f'(a)=\frac{1}{f^{-1}(b)}$. The fact that $a$ and $b$ both appear here is what makes the expressions for $f'(x)$ and $(f^{-1})'(x)$ look less related than they are. Geometrically, the derivative of $f$ and of $f'$ at the same point in the plane (allowing for flipping the plane, of course) are very related. Algebraically, the derivative of $f$ and of $f'$ at the same input value are less so.
| {
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Showing that $g(x) = \begin{cases} f(x), & \ x \ne 0 \\ c, & \ x = 0 \end{cases}$ is Riemann integrable where $f$ is Riemann integrable Suppose that $f: [-1,1] \to R$ Riemann integrable and let $g:[-1,1]\to R$ be defined
by $g(x) = \begin{cases} f(x), & \ x \ne 0 \\ c, & \ x = 0 \end{cases}$
Show that $g$ is integrable.
Attempt.
Since $f$ is integrable, it follows that $\forall \epsilon >0$, $\exists$ some partition $p \in P_{[-1,1]}$ such that $U(f,p) - L(f,p) < \epsilon$. Now, suppose there is at least one $x_i \in p$ such that $\sup_{x \in[x_{i-1}, x_{i}]} g(x) = c$. Then, it follows that $g(0) = c \ge f(0)$. Hence, we have that
$$U(g,p) \le U(f,p) + 2c \implies U(g,p) - L(g,p) \le U(f,p) - L(f,p) + 2c < \epsilon + 2c = \epsilon_{g}.$$
Now, suppose there is some $x_{i} \in p$ such that $\inf_{x\in [x_{i-1}, x_{i}]} g(x) = c$. Hence $g(0) = c < f(0)$. So, we have that $L(g,p) < L(f,p)$, which means that
$$U(g,p) - L(g,p) - 2f(0) < U(f,p) - L(f,p) < \epsilon.$$
If there is no $x_{i} \in p$ such that $\sup_{x \in[x_{i-1}, x_{i}]} g(x) = c$ or $\inf_{x\in [x_{i-1}, x_{i}]} g(x) = c$ then it follows that
$$U(g, p) - L(g,p) = U(f,p) - L(f,p) < \epsilon.$$
| hint
You just need to prove that $f-g $ is integrable.
put $d=|f (0)-c|$,
For a given $\epsilon>0,$ take the partition $$p=\{-1,-\frac {\epsilon}{4d},\frac {\epsilon}{4d},1\} $$
then
$$U (f-g,p)-L (f-g,p)=d\frac {\epsilon}{2d}<\epsilon.$$
| {
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Calculate segment distance to cover before turning via a pivot I'm building a visualisation where I have a body that is moving along-a-path, which is comprised of multiple segments, each with an arbitrary angle.
The body is moving along the path and:
*
*When the body's centre reaches the end of each segment it stops.
*It then rotates around it's centre to align itself with the next segment .
*It starts moving again along the next segment.
I can get this working just fine when the rotation point is the body's centre.
Here's an animation of the body rotating by it's centre (blue dot):
However now I'd like to rotate the body from a pivot point.
How can I calculate the distance I should cover in each segment before I stop and start turning around my pivot, so when the rotation ends my body's centre lies exactly in the centre of the next segment?
In short, when the body is moving it's centre must always lie on the segment line it moves on.
Here's an animation of the body rotating by it's pivot point (red dots):
In the above example the body overshoots the position on each segment where it should stop and start rotating, thus when it starts moving again - it's centre doesn't lie on the path.
FWIW I've got some code for this working in a browser sandbox, available here
| You should turn when pivot point $N$ reaches the angle bisector of $\angle ABC$, with a rotation of $2\angle BNO$. But the rotated body is turned by $180°-2\angle BON$ with respect to the direction of the path.
To make this work, then, you must choose your pivot so that $\angle BON=90°$.
| {
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Easy way to prove $|\text{curl}\ \mathbf n|^2=(\mathbf n\cdot \text{curl}\ \mathbf n)^2+|\mathbf n \wedge\text{curl}\ \mathbf n|^2$? Let $\mathbf n$ be a unit vector field. I would like to show that
$$|\text{curl}\ \mathbf n|^2=(\mathbf n\cdot \text{curl}\ \mathbf n)^2+|\mathbf n \wedge\text{curl}\ \mathbf n|^2$$
holds.
Expanding in coordinate is straightforward but looks ugly and doesn't provide much insight. Could anyone provide a better proof of this based on some properties of curl?
| You don't need any properties of curl - for any vectors $u,v$ we have $|u|^2|v|^2 = (u \cdot v)^2 + |u \wedge v|^2,$ so your formula follows from the fact that $|n|=1.$
| {
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2 different results using associative property in boolean expression I have the following expression that i'm trying to simplify: $Q \vee P \vee Q \vee T$
I simplified this like so: $Q \wedge P \vee Q \vee T \equiv Q \wedge P \vee (Q\vee T) \equiv Q\wedge P \vee T \equiv Q \wedge (P\vee T) \equiv Q\wedge T \equiv Q. $
But then I noticed there might also be another way to proceed, which would be like so: $Q \wedge P \vee Q \vee T \equiv (Q \wedge P \vee Q)\vee T \equiv T $
Both of these clearly get different results, what am I missing here?
| Order of operations matters. $Q\wedge(P\vee Q\vee T)$ and $(Q\wedge P)\vee Q\vee T$ are quite different expressions.
I simplified this like so: $Q \wedge P \vee Q \vee T \equiv Q \wedge P \vee (Q\vee T) \equiv Q\wedge P \vee T \equiv Q \wedge (P\vee T) \equiv Q\wedge T \equiv Q. $
You are treating that as ${Q\wedge (P\vee Q\vee T)\\Q\wedge((P\vee Q)\vee T)\\Q\wedge T\\ Q}$
But then I noticed there might also be another way to proceed, which would be like so: $Q \wedge P \vee Q \vee T \equiv (Q \wedge P \vee Q)\vee T \equiv T $
Here you treat it as: $(Q\wedge P)\vee Q\vee T\\ ((Q\wedge P)\vee Q)\vee T\\ T$
Which is the usual convention: that $\land$ has precedence over $\lor$ in the same way $\times$ has precedence over $+$.
Not everyone uses the same convention, so the best practice is to always use explicit bracketing to ensure the epression is read as intended.
Compare with: $q\times p+q+1 = (q\times p)+q+1$
| {
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Determine the critical points and identify them as asymptotically stable or unstable? Drawing phase lines? Here's the question:
Determine the critical (equilibrium) points, and classify each one as
asymptotically stable or unstable. Draw the phase line, and sketch
several graphs of solutions in the $ty$-plane
$$dy/dt = 1 − e^y,\; −∞ < y_0 < ∞.$$
This section of the textbook is all about population growth and is supposed to be in the form $dy/dt=r(1-y/k)y$, so I don't understand how to find the critical pts. in this case. Do I have to actually solve the DE to graph the solutions? Is the phase line just the $y$-axis?
| The only equilibrium point of $$dy/dt = 1 − e^y,\; −∞ < y_0 < ∞.$$ is at $dy/dt =0.$
Thus equilibrium happens at $$ e^y =1 $$ that is $y=0$
This equilibrium is asymptotically stable because $y>0 \implies dy/dx <0$ and $ y<0 \implies dy/dx>0.$
Therefore the solutions approach the equilibrium solution $y=0$ asymptotically.
| {
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Compare $2^{2016}$ and $10^{605}$ without a calculator So, I am supposed to compare $2^{2016}$ and $10^{605}$ without using a calculator, I have tried division by $2$ on both sides and then comparing $2^{1411}$ and $5^{605}$, and then substituting with $8,16,10$ and then raising to powers and trying to prove that but that did not go anywhere, I have also tried taking $\log$ of both sides, but did not help either. Also is there a more general approach to these kind of problems?
| $\log_{10}2^{2016}=2016\log_{10}2\approx 606.88>605$
If calculators are not allowed, we have
\begin{align*}
2^{2016}&=64(1000+24)^{201}\\
&>64\left[1000^{201}+\binom{201}{1}1000^{200}(24)\right]\\
&=64(10^{603})(5.824)\\
&>100(10^{603})\\
&=10^{605}
\end{align*}
| {
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How to Fourier Transform $\frac{\sin(x)^2}{x^2}$ without Contour Integration. In our lecture we need to Fourier transform $\frac{\sin(x)^2}{x^2}$, i.e. compute the integral:
$$\int_{-\infty}^{\infty} \mathrm e^{-iy x}\frac{\sin(x)^2}{x^2} \mathrm dx$$
Since it's a lecture on partial differential equations and not complex analysis, I don't think contour integration can be the solution here. I already tried to rewrite $\sin x$ getting
$$\int_{-\infty}^{\infty}\frac{-1}{4x^2} (\mathrm e^{ix(2-y)}-2\mathrm e^{-ixy}+\mathrm e^{-ix(y+2)})\mathrm dx$$
but now I'm still stuck with the computation of something like $$\int_{-\infty}^{\infty}\frac{\mathrm e^{-ixy}}{x^2}\mathrm dx$$
Any ideas or hints are greatly appreciated. Thanks in advance!
| Due to parity it is enough to compute
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx \stackrel{\text{def}}{=}\lim_{M\to +\infty}\int_{-M}^{M}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx $$
and by integration by parts the RHS equals
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{1}{2}\cos(xs)-\frac{1}{4}\cos(x(s+2))-\frac{1}{4}\cos(x(s-2))}{x^2}\,dx $$
or
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{s+2}{4}\sin(x(s+2))+\frac{s-2}{4}\sin(x(s-2))+\frac{s}{2}\sin(xs)}{x}\,dx $$
where we may exploit the standard result
$$ \forall \alpha\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx = \pi\,\text{Sign}(\alpha) $$
to get:
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx =\frac{\pi}{4}\left[|s-2|+|s+2|-2|s|\right]. $$
The RHS is a piecewise-linear function, supported on $[-2,2]$, going from $0$ to $\pi$ on $[-2,0]$ and from $\pi$ to $0$ on $[0,2]$. Not by chance, it is a multiple of the convolution between $\mathbb{1}_{(-1,1)}$ and itself.
| {
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Meaning of notation $f(x)$ in set theory In my book, a function $f$ is defined as a binary relation such that if $(x,y),(x,z)\in f$ then $y=z$. Moreover, as it is usual, author denotes $(x,y)\in f$ by
$$
y=f(x) .
\tag{1}
$$
So, from this notation, I understand $f(x)$ as the second component of the ordered pair $(x,y)\in f$, i.e $(x,f(x))$. Nevertheless, below, the author says that $f(x)=\bigcap\{y:(x,y)\in f\}$. But I think this notation is different from (1), since
$$
\bigcap\{y:(x,y)\in f\} = \{z: \forall y:(x,y)\in f \Longrightarrow z\in y\} ,
$$
I mean, $\bigcap\{y:(x,y)\in f\}$ is a set of elements of the class $y$, not such a $y$.
And, moreover, some other authors, as Herbert Enderton, define also $f(x)$ as the class
$$
f(x)=\bigcup\{y:(x,y)\in f\} .
$$
How can be all these notations/definitions compatibles?
| If $A$ is a singleton, $\{a\}$ then $\bigcap A=\bigcup A=a$. Since $f$ is a function, the set $\{y : (x,y)\in f\}$ is a singleton, for a fixed $x$, so the result follows.
The key point to remember is that everything is a set, including $x$ and $y$.
| {
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Variance of sum of $10$ random variables. Find the variance of sum of $10$ random variables if each has variance $5$ and if each pair has a correlation coefficient $.5$
Let $Y=X_1+X_2+X_3+\ldots+X_{10}$
I tried this problem by calculating variance of first $10$ random variables.
$V(Y)=50$. Then there will be $45$ pairs of covariance terms.
Correlation coefficient $\rho=.5$
$Cov(X_i,X_j)=2.5 $
Which gives variance $V(Y)=50+2.5=52.5$
Did i do everything right, please someone tell me.
| $$Y=\sum_{n=1}^{10}X_n\to var(Y)=E(Y^2)-E^2(Y)\\=E(\sum_{n=1}^{10}X^2_n+\sum_{m,n=1\\m\ne n}^{10}X_nX_m)-(\sum_{n=1}^{10}E(X_n))^2\\=50+90\times 2.5-(\sum_{n=1}^{10}E(X_n))^2$$if we take $E(X_n)=0$ we will have:$$var(Y)=275$$
| {
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Complete graphs in the plane with colored edges where an edge don't cross edges with same color The maximal number of nodes in a complete planar graph is $4$.
Suppose that the edges of the graph can be chosen with $m$ different colors and that edges with different colors are allowed to cross each other. What would be the maximal number of nodes for a complete graph like this occurring in the plane with this rule?
Down the case with $2$ colors and $7$ nodes.
I found this: layered graphs, and the results there doesn't contradict a conjecture
A complete graph with $4n$ vertices need n colors for the edges.
| This isn't a complete answer, but the quantity you're asking for is closely related to the thickness of a a graph, which is the minimum number of planar subgraphs which jointly cover the edges of the graph. It is known that the thickness of the complete graph $K_n$ is $\lfloor (n+7)/6 \rfloor$ except at $K_9$ and $K_{10}$.
I believe that the concepts aren't exactly equivalent, because your framing of the question requires all the planar subgraphs to use the same positions for the vertices, while thickness does not require the same restriction. However, the thickness still gives a lower bound on the number of colors required, and it's possible that diving into the papers in the MathWorld article would show that the constructions involved do use the same positions for all the subgraphs (or could be modified in that way).
| {
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"url": "https://math.stackexchange.com/questions/2619214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Squared Summation(Intermediate Step) I am studying Economics and are trying to get a firm grasp of summation rules and applications. Looking into the following relation,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$
The following "trick" is given below, to understand the above.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^n[(k+1)^3-k^3]=n^3 + 3n^2+3n$
As I expand the left-handside of the equation for a given sequence $S=[1^2, 2^2, 3^3,..., n^3]$, the following leads to the right-hand side of the equation,
$(2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + .... + (n^3 - (n - 1)^3) + ((n+1)^3 - n^3) = (n + 1)^3 - 1^3 = n^3 + 3n^2 + 3n$
Understanding the intermediate steps, ie. the above expanding, Im struggling with the intuition so to speak. Which kind of mentality should I have had applied on the second equation, in order get to the right hand side, without expanding it?
Any help, with some mathematical explanation is highly appreciated.
Thank you!
| I think this may be a question about how telescoping sums work. If so, then
$$\sum_{k=1}^n\left(f(k+1)-f(k)\right)=\sum_{k=1}^nf(k+1)-\sum_{k=1}^nf(k)=\sum_{j=2}^{n+1}f(j)-\sum_{j=1}^nf(j)$$
Where e have made the substitution $k=j-1$ in the first sum and $j=k$ in the second. When $j-1=k=1$, $j=2$ and when $j-1=k=n$, $j=n+1$. Then
$$\sum_{k=1}^n\left(f(k+1)-f(k)\right)=f(n+1)+\sum_{j=2}^nf(j)-f(1)-\sum_{j=2}^nf(j)=f(n+1)-f(1)$$
Where we have extracted the terms not common to both sums and canceled the common parts. In the instant case,
$$\sum_{k=1}^n\left((k+1)^3-k^3\right)=(n+1)^3-1^3=n^3+3n^2+3n$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is fraction has the same meaning with rational in number theory? I'm unable to get the difference between fraction and rational, we say $\frac{a}{b}$ is rational number if a and b are two integer with $b\neq 0$, and we can say also $\frac{a}{b}$ is a fraction but i don't know any reason for that, my question here is :Is fraction has the same meaning with rational in number theory ?
Note: one other thing wich is mixed me is that wolfram alpha considered rational number as fraction as shown here
Edit: I have edited the question just for specification and clarification
according to the gaven answers without any change in the meaning of question
| Any number written in the form $\frac{a}{b}$ is a fraction provided $b \neq 0$, because division by zero is not defined.
Thus $\frac{2}{3}$, $\frac{221}{5}$ and $\frac{\pi}{2}$ are all fractions but only $\frac{2}{3}$ are $\frac{221}{5}$ rational since the decimal representation either repeats $\frac{2}{3} \approx 0.666666666666666666666$ with the 6's repeating for ever or terminates $\frac{221}{5} = 44.2$.
$\frac{\pi}{2}\approx 1.5707963...$ Is not a rational however as the decimal expansion neither ends or repeats.
If we add the requirement that $a$ and $b$ are both integers then you have the fractional representation of a rational number. I would not call $0.5$ a fraction however as it is not written in the correct format, though it is certainly a rational number. I would call $\frac{1}{2}$ a fraction though and they both have the same value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many years the world's oil reserves will be enough? World oil reserves are estimated at around 240 billion tonnes. Its world production is 4.36 billion tons annually. Calculate how many years the world's oil reserves will be enough: a) if the current level of its production is maintained; b) taking into account the growth of extraction by 2% per year.
My try is:
a) $\frac{240}{4.36}≈55.05$ enough for 55 years.
b) $4.36+4.36\cdot 1.02+4.36\cdot (1.02)^2+⋯+4.36*(1.02)^n=4.36\cdot \frac{1\cdot(1.02^n-1)}{1.02-1}=4.36\cdot \frac{1.02^n-1}{0.02}
=218\cdot(1.02^n-1)>240$
$$1.02^n-1 >\frac{240}{218}$$
$$1.02^n>2.101$$
$$n > \frac{\ln (2.101)}{\ln(1.02)} ≈ 37.5$$
enough for 37 years.
But the answer in test is a)55 and b)26. Can you help please.
| I used Excel to calculate this, and found that after 37 years, 235 B will be extracted, so it appears your calculation is correct. And doing a rough estimate, we can use an approximation 2%*26*e = 141%, so in 26 years, production should be about 41% higher than today. But 26 is about half of 55, which would mean that production would have to be, on average, about twice current levels. So even without a calculator, 26 is clearly wrong.
As far as speculating as to where 26 came from, if you divide the current reserves by the current rate plus the reserves times 2%, you get 240/(4.36+240*.02) = 26.2, which rounds to 26. So this could be a coincidence, or perhaps your teacher did something along those lines.
| {
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False Dudeney : triangle to quiet a square. ABC is an equilateral triangle.
The same color polygons are isometric.
I can prove that MPQR is a rectangle, but not that it's not a square.
MI = IP and RF' = F'Q = FN.
As MPQR is a rectangle MI = IP = RF' = F'Q = FN.
This equalities are independent of the choice of E and F.
PQ = QR is equivalent to : EM + EN = 2FN
Why in this case : PQ is not equal to RQ ?
| Here is a solution to this problem.
| {
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Maximum possible reputation? (NOT a meta question) I had the following idea about the reputation system of MSE, that led to a math question:
Suppose a certain user on MSE gains an average of $+200$ reputation per day - the daily maximum. Suppose that this reputation comes from one vote from each of $20$ randomly selected (not necessarily distinct) users on the site each day. Then it seems that, at some point, this "superuser" would reach a sort of equilibrium by losing $200$ reputation each day just from other users who had previously upvoted her being deleted. Would this happen, or would the superuser grow in reputation unboundedly? If she is expected to top off, at what reputation would she stagnate?
To answer this question, one would need to use some of the statistics from this site, such as the average user deletion rate, average number of users at any time, and so on. I don't expect to find a perfect answer, but a pretty good estimate should be obtainable.
| I believe that your reputation will tend to infinity.
By the way, 200 is not the maximum reputation gain in a day; points from acceptances and bounties don't count toward the cap, and are unlimited. But that doesn't matter because I'm not assuming that you're maxing out your daily reputation gains. I only assume that your net reputation change (not counting deletions) is positive, i.e., you're gaining more points from upvotes etc. than you're losing to downvotes.
It is my understanding that death is not a cause for deleting a user's account. Therefore, anyone who manages to die (or become permanently inactive for any other reason) in a "state of grace" with Stack Exchange, will never be deleted.
Some of your votes come from "mortal" users, who are fated to be deleted someday; and some of them come from "immortal" users, who will never be deleted. Your accumulated reputation score at any time will be the sum of your "immortal" votes (which clearly grows to infinity) plus the current population of "mortal" votes (which fluctuates unpredictably). Q.E.D.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a representation of $D_n$ on $\mathbb{R}^2$? I am trying to write down a representation of $D_n = \langle \sigma, \tau \mid \sigma^n = \tau^2 =e, \tau \sigma = \sigma^{-1} \tau \rangle$ over $\mathbb{R}^2 \cong \mathbb{C}$ (as an $\mathbb{R}$-vector space).
What I Know:
My representation has to be a group homomorphism $\rho: D_n \rightarrow \text{GL}(2, \mathbb{R})$. I know that the homomorphism will be fully specified by specifying where I send the generators of $D_n$. Intuitively I know that I should be sending $\sigma$ to the $2 \times 2 $ rotation matrix for an angle of $2 \pi /n$ and $\tau$ to a horizontal reflection.
This makes me think I should be choosing
\begin{array}{c c c}
\sigma^k \rightarrow
\begin{pmatrix}
\cos{(2\pi k /n)} & -\sin{(2\pi k /n)} \\
\sin{(2\pi k /n)} & \cos{(2\pi k /n)}
\end{pmatrix}
&
\text{and
}&
\tau \rightarrow
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} \\
\end{array}
My Problem:
I am having issues checking that this map respects the group operation. I know I want to check that:
$$
\rho((\sigma^a \tau^b)(\sigma^c\tau^d)) = \rho(\sigma^a \tau^b) \rho(\sigma^c\tau^d)
$$
since an arbitrary member of $D_n$ can be written as $\sigma^a \tau^b$. I think it would suffice to show that $\rho(\sigma \tau)= \rho(\sigma) \rho(\tau)$. I can easily write down the matrices on the RHS as I have just defined them above. However, I don't know how to check to see if this product matrix equals the matrix for $\rho(\sigma \tau)$, since I don't know what that is! How should I go about verifying my homomorphism is in fact a homomorphism?
| All you have to do is set
$$\rho(\sigma)=\pmatrix{\cos (2\pi/n)&-\sin(2\pi/n)\\\sin(2\pi/n)&\cos(2\pi/n)}$$
and
$$\rho(\tau)=\pmatrix{1&0\\0&-1}$$
and then verify these preserve the relations, that is $\rho(\sigma)^n=I$
(standard property of rotation matrices), $\rho(\tau)^2=I$ (obvious)
and $\rho(\tau)\rho(\sigma)=\rho(\sigma)^{-1}\rho(\tau)$
(a couple of lines of calculation).
Anyway these are just the matrices representing the usual action of
$D_n$ as the symmetry group of the regular $n$-gon with centre at the origin.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Smallest prime $p$ which every integer $< n$ is a primitive root $\mod p$ Another interesting question related to primitive roots is what is the smallest prime $p$ for which the primes less than $n$ are primitive roots $\mod p$. The sequence of primes would be $[p, p_2, p_3,...]$ for every additional prime $k > n$. For instance, the first prime is $3$ because $3$ is the smallest prime $p$ such that $2$ is a primitive root $\mod p$. The next prime in this sequence is $5$ because $5$ is the smallest prime $p$ such that $2$ and $3$ are primitive roots $\mod p$.It follows that $53$ is the third prime in the sequence because it is the smallest prime $p$ such that $2, 3,$ and $5$ are primitive roots $\mod p$. The fourth prime in the sequence is $173$ the smallest prime $p$ such that $2, 3, 5,$ and $7$ are primitive roots $\mod p$. The fifth would also be $173$ the smallest prime $p$ such that $2, 3, 5, 7$ and $11$ are primitive roots $\mod p$, and so on. Duplicates in the sequence are allowed. Is anyone able to provide the list of primes in this sequence (up to say, the first $100$ of them) or a program that could do the work? Thanks in advance.
| The following PARI/GP-program does the job :
? k=3;x=primes(k);p=prime(k);gef=0;while(gef==0,p=nextprime(p+1);s=select(n->zno
rder(Mod(n,p))==p-1,x);if(s==x,gef=1));print(p)
53
?
Just change $k$ to get the smallest prime for another $k$.
The smallest primes for $k\le 18$ are :
? for(k=1,20,x=primes(k);p=prime(k);gef=0;while(gef==0,p=nextprime(p+1);s=select
(n->znorder(Mod(n,p))==p-1,x);if(s==x,gef=1));print(k," ",p))
1 3
2 5
3 53
4 173
5 173
6 293
7 2477
8 9173
9 9173
10 61613
11 74093
12 74093
13 74093
14 170957
15 360293
16 679733
17 2004917
18 2004917
The calculation for $k=19$ is currently running.
| {
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Question about proof of Stein's Lemma by Casella and Berger I am looking at the following proof from Casella and Berger's Statistical Inference. However, I don't understand the final statement. The only condition on $g'$ here is that $E|g'(X)|<\infty$. But how does this ensure that $g(x)e^{-(x-\theta)^2/(2\sigma^2)} \to 0$ as $x \to \pm \infty$?
| The hypotheses are that, for some nonnegative functions $f$ and $h$, the function $fh$ is integrable on $(0,+\infty)$, that the function $h$ is decreasing on some $(\theta,+\infty)$ and that $h(x)\to0$ when $x\to+\infty$, and the question is whether all this implies that $g(x)h(x)\to0$ when $x\to+\infty$, where $$g(x)=c+\int_0^xf(y)dy$$
To prove this is true, fix some $z>\theta$ and note that, for every $x>z$,
$$g(x)h(x)=g(z)h(x)+h(x)\int_z^xf(y)dy\leqslant g(z)h(x)+\int_z^xh(y)f(y)dy$$ Since $h(x)\to0$ when $x\to+\infty$, this yields $$\limsup_{x\to\infty}g(x)h(x)\leqslant\int_z^\infty h(y)f(y)dy$$ When $z\to\infty$, the RHS can be made as small as desired hence $$\limsup_{x\to\infty}g(x)h(x)=0$$ which proves the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
| As you were told in the comments, you must take some $\varepsilon>0$ and then prove that $\left|\frac{3n+5}{2m+6}-\frac{3n+5}{2n+6}\right|<\varepsilon$ if $m$ and $n$ are large enough.
Note that$$(\forall n\in\mathbb{N}):\frac{3n+5}{2n+6}=\frac{3n+9}{2n+6}-\frac4{2n+6}=\frac32-\frac2{n+3}.$$So, take $p\in\mathbb N$ such that $\frac2{p+3}<\frac\varepsilon2$. Then$$m,n\geqslant p\implies\left|\frac{3m+5}{2m+6}-\frac{3n+5}{2n+6}\right|=\left|\left(\frac32-\frac2{m+3}\right)-\left(\frac32-\frac2{n+3}\right)\right|<\varepsilon.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the difference between $x$ and $\{x\}$ when $x$ itself is a set? I've already asked this a part of another question, but thought it'd be easier to elaborate a bit more on my concern.
Let $x$ be a set. What is the difference between $x$ and $\{x\}$? I get that the latter is a set consisting of a single element - namely $x$, but what is the difference?
For example, we can have $x$ to be the set $\{1\}$, then $\{x\}=\{\{1\}\}$. Aren't those $2$ expressions the same?
Another problem are the brackets - when we have a set, do we always have to surround him with brackets, for instance, can we have $x$ to be the set $2$?
Thanks a lot
| $$\{1\} $$ is a set whose the unique element is the integer $1$
$$\{\{1\}\} $$ is a set whose the unique element is the set $\{1\} $.
| {
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Each vertex of the square has a value which is randomly chosen from a set. I don't need so much help computing this question (yet). However, I am still at the point of understanding what the question is really asking. If anyone understands it, and can explain it, that would be greatly appreciated!
I understand what pairwise and mutually independent are. I just don't understand how I would be able to show that. From what I understand, each vertex is randomly selected from {1, 2, ..., 6}. So say edge 1 is connected to a vertex that randomly selected 2, and another vertex that randomly selected 3 from the set, then the pdf would be X1 = 0 (because they are not the same value). Now say edge 2 connects two vertices that have randomly selected the same value, say 4, then its pdf would be X2 = 1 (since the vertices have the same value). I understand that they are pairwise independent (each set of pairs are independent), because they are randomly selected from the set {1, 2, ..., 6} but how can I show that? And how are X1, X2, X3, and X4 not mutually independent?
|
So say edge 1 is connected to a vertex that randomly selected 2, and another vertex that randomly selected 3 from the set, then the pdf would be $X_1 = 0$ (because they are not the same value).
Instead of fixing values for the vertices, look at the probability that $X_i=1$; let's call this $P(X_i)$. What is $P(X_1)$? What is $P(X_1 \cap X_2)$? How about $P(X_1 \cap X_2 \cap X_3)$ and $P(X_1 \cap X_2 \cap X_3 \cap X_4)$? Now use the definitions of pairwise independence and mutual independence.
Note also that for pairwise independence you should check the two kinds of pairs of edges: those that share a vertex, and those that are disjoint.
| {
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Induction on summation inequality stuck on Induction step working on a fairly simple induction problem, but stuck when decomposing the m+1^st summation and bringing in the I.H to show it holds.
I am bad with formatting so bare with me.
$$\sum_{k=1}^n \frac1{k^2} \leq 2 - \frac1n$$
Base Case:
Set $n = 1$ and show that the inequality holds
IH:
Assume for all $n = m$, $\sum_{k=1}^m \frac1{k^2} \leq 2 - \frac1m$ holds
I.S:
$$\sum_{k=1}^{m+1} \frac1{k^2} \leq 2 - \frac1{m+1}$$
since we can decompose m+1 to m we get
$$\sum_{k=1}^m \frac1{k^2} + \frac1{m^2} \le 2 - \frac1m$$
then we can use I.H since we have $\sum_{k=1}^m \frac1{k^2} \le 2 - \frac1{m+1}$
so I am trying to show $2 - 1/(m+1) + 1/(m^2)$, but I am stuck here since I cannot manipulate the equation to come to the m+1st equation on the RHS
| i think it must be
$$\sum_{k=1}^m\frac{1}{k^2}+\frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$ and it must be
$$\le 2-\frac{1}{m}+ \frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what does mean a zero eigenvalue in a PDE? I understand that in a PDE, the eigenvalues are some kind of speed of information propagation. For a hyperbolic system of PDE's, with three eigenvalues and one of them being zero, what does this mean? Is it the same than having a system of two equations?
This is the system of equations (SWE):
\begin{align}
\frac{\partial h}{\partial t} + \frac{\partial}{\partial x_i}(h \overline{u_i}) &= 0 \nonumber \\
%Depth averaged and filtered momentum equation
\frac{\partial (h{\overline{u_{i}}})}{\partial t}+\frac{\partial}{\partial x_{j}} \left(h {\overline{u_{i}}}\ {\overline{u_j}}\right)&=-\frac{1}{2} \frac{\partial g h^2}{\partial x_i}- gh\frac{\partial z}{\partial x_{i}}+ \frac{1}{\rho}(\overline{\tau_{\eta_i}}-\overline{\tau_{b_i}})
\label{swesimpl}
\end{align}
where $h,u_i$ and $u_j$ are the unknowns.
I'm using and augmented version of Roe solver to solve numerically this system of PDE's. So the augmented version of the Roe solver provides estimates for the three wave speeds (the eigenvalues), and depending of the sign of such waves, there is an expression for the fluxes to compute the solution at the next time step, via finite volume method. I'm working in a two dimensional case, that means there are three unknowns (three eigenvalues). But for certain numerical simulation, sometimes I get a zero eigenvalue, the solver does not take into account that possibility. So I'm wondering if I could use the traditional solver (which was developed for a one dimensional case, i.e. two eigenvalues).
However, I'm not sure about the physical meaning of having one of the three eigenvalues being zero and if this can be considered to be equal to have only two eigenvalues...
| If the eigenvalues are the wave propagation speeds, then a zero eigenvalue just says that one wave is stationary
| {
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French playing cards, probability to have a four (of a kind) We have a "French playing cards" composed of $52$ cards. Each kind of cards (Ace, King, Queen, ...) is composed of $4$ cards : There are $4$ Ace, $4$ Jack, etc. We pull $5$ cards among the $52$ cards.
We are a four (of a kind) if among the $5$ cards pulled, there are the $4$ of a same value ($4$ cards Ace, or King, or 2, or 3......).
What is the probability to have a four (of a kind) if the first card pulled is an Ace, and the fifth card pulled is a King ?
I hope you understand, it's very difficult for me to translate that. Thank you for help in advance.
| Here is a hint:
Given that you have drawn one ace and one king, you will need either to draw the remaining three aces, or draw the remaining three kings, from the remaining $50$ cards.
Can you continue from this point?
(Spoiler answer)
There are two winning ways to draw the remaining three cards, and ${50 \choose 3}$ total ways to draw the remaining three cards, so the probability is $2/{50 \choose 3}$.
| {
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Ramification in cyclotomic field I was trying to prove that:
$p$ is a odd prime and $\zeta$ is a primitive $p$-th root of unity and let $p^{*} = (-1)^{\frac{p-1}{2}}p$. Show that $\mathbb{Q}(\sqrt{p^{*}})$ is a quadratic subfield of $\mathbb{Q}(\zeta)$.
This is what I got so far:
The only prime ramified in $\mathbb{Q}(\zeta)$ is $p$, and the prime factorization in $\mathbb{Q}(\zeta)$ is: $(p) = (\mathcal{P_{1}}...\mathcal{P}_{n})^{\phi(p)=p-1}$ (Since it is Galois, so all primes have the same ramification degree). $p-1$is even, so I looked at $(\mathcal{P_{1}}...\mathcal{P}_{n})^{\frac{p-1}{2}}$. But I have trouble to determine what is this ideal exactly? Is this $(\sqrt{p^{*}})$? If it is, where does $(-1)^{\frac{p-1}{2}}$ comes from?
Any help is appreciated!
| Let $p$ be an odd prime, and consider the cyclotomic field $K=\mathbf Q(\zeta_p)$, which is a cyclic extension of degree $(p-1)$. Since $p$ is odd, by the Galois correspondence, $K$ contains a unique quadratic field $k=\mathbf Q (\sqrt d)$, and we want to show that actually $k=\mathbf Q (\sqrt {p^*})$ (in your notation).
To this end, we can study the ramification of primes in $k$ and $K$ as you started to do. The quickest way is to use the discriminants of the rings of integers $O_k$ and $O_K$. It is classically known that $disc(O_K)=\pm p^{p-2}$ and $disc(O_k)= 4d$ if $d\equiv 2$ or $3$ mod $4$ , $d$ if $d\equiv 1$ mod $4$. It follows that $p$ is the only prime which ramifies in $K$ (hence the same is true for $k$), and the previous reminder on ramification in $k$ and the unicity of $k$ show that $k=\mathbf Q (\sqrt {p^*})$ .
| {
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Evaluating the liminf and limsup So I am a little bit stuck on finding the liminf and limsup of the following interval:
$$[(-1/n)^{n} , 2]$$
I know that
$$\liminf_{n\to\infty} A_n = \cup^{\infty}_{n=1}(\cap^{\infty}_{j=n}A_{j})$$
$$\limsup_{n\to\infty} A_n = \cap^{\infty}_{n=1}(\cup^{\infty}_{j=n}A_{j})$$
but I do not know how to use this information to solve the question.
-Thanks!
| HINT
Note that $ |a_n|= |(-1/n)^n|\to 0$ is strictly decreasing thus
$$-1\le a_n\le \frac14$$
| {
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how to geometrically explain why pell numbers close to sqrt 2
How to graphically explain that the limit of yn/xn = $\sqrt 2$, as n approaches to infinity?
Like, I know how to prove it algebraically.
| Perhaps not quite what you want, but an explanation how the heron-method works.
We want to determine the length of a square with area $2$. We start with the rectangular with sides $1$ and $2$. To get closer to a square, we take the arithmetic mean of the sides (which is $\frac{3}{2}$ and determine the corresponding side (which is $\frac{4}{3}$). This can be repeated and leads to the iteration $$x_{n+1}=\frac{x_n+\frac{2}{x_n}}{2}=\frac{x_n^2+2}{2x_n}$$ which is exactly the heron-method. If we start with $x_0=1$, the fractions are among the partial fractions resulting from the continued fraction.
| {
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Prove that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac1x):x\in(0,1]\}$
The goal is to show that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac{1}{x}):x\in(0,1]\}$.
Proof.
Suppose there is path $f:[a,c]\to\overline S$ beginning at the origin and ending at a point of $S$.
The set of those $t$ for which $f(t)\in0\times[-1,1]$ is closed, so it has a largest element $b$.
Then $f:[b,c]\to\overline S$ is a path that maps $b$ into the vertical interval $0\times [-1,1]$ and maps the other points of $[b,c]$ to points of $S$.
Replace $[b,c]$ by $[0,1]$ for convenience; let $f(t)=(x(t),y(t)).$
Then $x(0)=0,$ while $x(t)>0$ and $y(t)=\sin(1/x(t))$ for $t>0$.
We show there is a sequence of points $t_n\to0$ such that $y(t_n)=(-1)^n.$ Then the sequence $y(t_n)$ diverges, contradicting the continuity of $f.$
To find $t_n$ we proceed as follows: Given $n,$ choose $u$ with $0<u<x(1/n)$ such that $\sin(1/u)=(-1)^n$. Then use the intermediate value theorem to find $t_n$ with $0<t_n<1/n$ such that $x(t_n)=u.$
*
*I don't understand how this function $f(t)=(x(t),y(t))$ is defined. $f(0)=(x(0),y(0))=(0,?)$
*About the sequence of points $t_n$, I don't understand the process of construction.
Given $n\in\mathbb N,$ let $u\in(0,x/n)\dots$ Is $x$ is a function right?
Using IVT, why the interval for $t_n$ is $(0,1/n)$?
| Let $H=\{(x, sin(1/x)) \;:\; x\in(0,1)\}$ and $T = \{0\} \times [-1, 1]$. Let
$\tag 1 S = H \cup T$
It is easy to show that $S$ is path-connected if and only if $S \cup \{\left(1,sin(1)\right)\}$ is path-connected. This is only mentioned because we modified (ever so slightly) the OP's question to fit the machinery found below.
The space $S$ is not path-connected.
Assume, to get a contradiction that we have a path $\gamma$ in $S$ connecting the point $(0,0)$ to $\left(\frac{1}{2}, sin(\frac{1}{2})\right)$. The inverse image of $T$ under $\gamma$ is a closed subset of $[0,1]$ and has a maximum value not equal to $1$. By modifying this path, we can define another path
$\tag 2 \omega: [0,1] \to S$
satisfying $\omega(0) \in T$, $\,\omega(1) = \left(\frac{1}{2}, sin(\frac{1}{2})\right)$ and $\omega\left( (0,1] \right) \cap T = \emptyset$.
But this path can also be regarded as connecting the two points inside the space
$\tag 3 G = H \cup \{ (0, \pi_y(\omega(0)) \}$ with $\pi_y$ the projection onto the $y\text{-axis}$
The concluding remark found here means that this is impossible.
| {
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$E = \mathbb{R}\times e$ is a n-dimensional measure zero if $e \subset \mathbb{R}^{n-1}$ of (n-1) dimensional measure zero In the book of Mathematical Analysis II by Zurich, at page 116 it is asked that
Show that if a set $E \subset \mathbb{R}^n$ is the direct product of
$\mathbb{R}\times e$ of the line $\mathbb{R}$, and a set $e \subset
\mathbb{R}^{n-1}$ of (n-1) dimensional measure zero, then $E$ is a set
of n-dimensional measure zero.
But, I cannot see how can this be, and do not know how to prove it, so I would appreciate any help for proving this statement.
Edit:
I do not understand the given answer, so please feel free to post another answers.
| From the construction of the product measure $\lambda_n$ (Lebesgue measure of $\mathbb{R}^n$),
if $A\in \mathcal{B}(\mathbb{R}^k)$ and $B\in \mathcal{B}(\mathbb{R}^{n-k})$, ($0<k<n$), then
$$ \lambda_n(A\times B) =\lambda_k(A)\lambda_{n-k}(B)$$
So here (using also that if $E_m \uparrow E$, $\lim_{m\rightarrow +\infty}\lambda_n(E_m)=\lambda_n(E)$)
$$\begin{align*}
\lambda_n(E)&= \lambda_n(\mathbb{R}\times e) \\
&=\lim_{m\rightarrow +\infty}\lambda_n((-m,m)\times e)\\
&=\lim_{m\rightarrow +\infty}\lambda_1((-m,m))\times \lambda_{n-1}(e)\\
&=\lim_{m\rightarrow +\infty}\lambda_1((-m,m)) \times 0 \\
&= 0
\end{align*}
$$
| {
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Proof by contrapositive: if $a$ and $b$ are consecutive integers then the sum $a + b$ is odd if $a$ and $b$ are consecutive integers then the sum $a + b$ is odd
Proof by contrapositive
Contrapositive form:
if the sum of $a$ and $b$ is not odd then $a$ and $b$ are not consecutive integers
I am struck here, so if $a + b$ is not odd means $a + b$ are even
$a + b = 2p$, where $p\in\mathbb Z$.
What are the next steps to show $a$ and $b$ are not consecutive?
| A direct proof is so much clearer:
$b=a\pm1$ implies $a+b= 2a\pm1$, which is odd.
But if you must use contrapositive:
Let $b=a+d$. Then $a+b=2a+d$ is even iff $d$ is even. Therefore, $|a-b|=|d|$ is even and so is never $1$.
| {
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Number of edges in the simple graph with the given conditions Let $G$ be a simple graph on $n$ vertices $v_1,v_2...v_n$. Let $G-v_i$ is having $m_i$ edges for $1\leq i \leq n$. Then prove that $m=\frac{1}{n-2}\sum_{i=1}^{n}m_i$.
| By handshake lemma we have $$\sum _{i=1}^n d_i = 2m$$
For each $i$ we have $$m_i = m-d_i$$
where $d_i$ is degree of $i$-th node. Now if we sum this over all $i$ we get
$$ \sum _{i=1}^n m_i = n\cdot m -\sum _{i=1}^n d_i = nm -2m = m(n-2)$$
and we are done.
| {
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Determine all ring homomorphisms of $\varphi:\mathbb{Z}_{10} \to \mathbb{Z}_{10}$ This is what I came up with as a solution and I wanted some suggestions on how to proceed with the last part or perhaps a new direction of thinking.
Given that $m=n=10$ we must have $0=m\cdot a=10a$ which is in $\mathbb{Z}_{10}$ $\Longleftrightarrow$ $10$ divides $10a$ $\Longleftrightarrow$ 1 divides $a$. Clearly this is trivial so $\varphi(1)\in \mathbb{Z}_{10}$. But the elements which are idempotent in $\mathbb{Z}_{10}$ are $0,1,5,6$. According to Gallian and Van Buskirk (1984) there should be $2^{\omega(n)-(\omega(n/(m,n))}$ ring homomorphisms. In other words there should be two. I'm leaning toward $0,5$ but I am unsure how to demonstrate that $1,6$ cannot be included. Any suggestions?
I'm not looking for a response where people can stroke their ego, I am genuinely not sure how to do this other than looking at the group homomorphisms, which I didn't fully understand either.
| I was incorrect in assuming there were only two ring homomorphisms. When I calculated $2^{\omega(n)-\omega(n/(m,n))}$ I neglected to realize that it yields 4 instead of 2. The four ring homomorphisms for $\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10}$ are in fact:
$$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto0$$
$$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto x$$
$$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto 5x$$
$$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto 6x$$
| {
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Median estimated from grouped data with a single class Given the formula for grouped median:
$Median = L_m + \left [ \frac { \frac{n}{2} - F_{m-1} }{f_m} \right ] \times c$
Where:
*
*$L_m$: lower boundary of median class
*$c$ : size of the median class
*$F_{m-1}$ : cumulative frequency of the class before median class
*$f_m$ : frequency of the median class
*$n$ : size data
Example: What should the median be for the following:
- 100, 100, 100, 100, 100, 100, 100, 100, 100, 100 (a repeat of 100 ten times)?
Calculation:
Using a bin/class size of 0.5:
$L_m$ = 100
$c$ = 0.5
$F_{m-1}$ = 0*
$f_m$ = 10
$n$ = 10
100 + [(5-0)/10]*0.5
= 100.25
| When you group data into intervals, information is lost.
So assumptions are made in order to make reasonable estimates of
the sample mean, median, etc.
The assumption of this formula for estimating the median from grouped data is that the data are spread roughly uniformly throughout the interval. Clearly, this assumption
is not met in your situation because all ten of the $100$'s lie at the lower
endpoint of the interval.
The idea of the formula is to estimate the median by interpolation, putting
the estimate somewhere within the interval. In your case the estimated
value $100.25$ is in the middle of the 'median interval' (the interval
known to contain the median).
If you were trying to contrive a situation in which the estimate is even
farther from the truth, you could put your ten $100$'s at the left end
of an interval $[100, 120).$ With no other data, your estimate of the
median would then be $110.$
There is nothing wrong with the formula, provided the assumption of data
spread evenly throughout the interval is close to the truth. But any
formula for estimating the median from grouped data will have to depend
on assumptions. All that can be said for sure is the the median lies
somewhere in the median interval. You have to recognize that the
information lost in grouping data into intervals cannot be precisely
recovered (unless the original data are saved and used).
Note: By contrast, the assumption usually made when trying to estimate the
sample mean from grouped data is that each observation lies precisely at the
midpoint of the interval that contains it. This idea gives rise to the
formula $\bar X \approx \frac 1 n \sum_{i=1}^k f_jm_j,$ where there are
$k$ intervals (usually of equal width), with midpoints $m_j$ and frequencies $f_j.$
| {
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A set of integers Assume that there is a set of ordered integers initially containing number $1$ to a given $n$. At each step, the lowest number in the set is removed, if the number was odd, then we go to the next step and if it was even, half of that number is inserted into the set and the cycle repeats until the set is empty.
My question is, how many steps does it take for a given number $n$ to finish the whole set?
I think it should be something the form of $2k - x$ where $x$ itself is likely a complex expression but I just can't seem to figure it out. Any help would be appreciated. (I got that from trial and error, by the way, no logic or proof behind it, I know that the answer is $\lfloor \frac{n}{2} \rfloor$+"The number of times each even number can be divided by 2", but I just can't find a closed formula)
| As in this algorithm, the actual order of the set isn't important, we can assign every $n\in\mathbb{N}$ the number of steps it takes till it's sorted out, which is namely how many times you'll have to divide by two till it's odd.
If we use prime factorization, that means if $z$ takes k steps till it's sorted out, it's prime factorization is $z = 2^{k-1} \cdot\, ...$
Now, we know the following:
Every second number has $2^0$ in its prime factorization (every odd number).
Every forth number has $2^1$ in its prime factorization.
Every eigth number has $2^2$ in its prime factorization.
...
So, for a given set $\{1,...,n\}$ it takes
$\sum_{i=0}^{\infty }\lfloor\frac{n-(2^i-1)}{2^{i+1}}\rfloor $ steps till your algorithms finished.
Here, $2^i-1$ is here how long we'll have to count from 1 till we reach the first number that has $2^i$ in its prime factorization (e.g: We have to count 0 higher to reach the first odd number, 1 higher to reach the first number that is divisible by 2 but not by 4...)
You can cut off the sum as soon as the divisor gets greater than the divident, so as rough approximation $log_2(n)$ works out. With that we get:
$$\sum_{i=0}^{log_2(n)}\lfloor\frac{n-(2^i-1)}{2^{i+1}}\rfloor $$
Another way to look at the problem is by keeping it a set. If $M$ is a set and $k\in\mathbb{N}$, let us define
$$M\cdot k := \{k\cdot m \mid m\in M\}$$
E.g. $\{1,2,3\}\cdot 2 = \{2,4,6\}$
Now we look at how our set looks like if we process it $i$ times, with our procss being:
For every number of the set, if the number is even, half it, if it is odd, remove it.
$i=0$ - The algorithm hasn't run yet, our set is the input set $\{1,..,n\}$.
$i=1$ - Only the even numbers are left $\{2,4,6,..,2\cdot\lfloor\frac{n}{2}\rfloor\} = 2\cdot \{1,2,3,..,\lfloor\frac{n}{2}\rfloor\}$
$i=2$ - Only the numbers divisble by four are left
$\{4,8,12,..,4\cdot\lfloor\frac{n}{4}\rfloor\} =
4\cdot \{1,2,3,..,\lfloor\frac{n}{4}\rfloor\}$
...
$i=k$ - Only the numbers divisble by $2^k$ are left
$\{2^k,2\cdot 2^k, 3\cdot 2^k, ..., 2^k\cdot\lfloor\frac{n}{2^k}\rfloor\} =
2^k\cdot \{1,2,3,..,\lfloor\frac{n}{2^k}\rfloor\}$
The number of operations our algorithm takes is now simply the sum of the numbers of each set in this chain. So, we get:
$$\sum_{i=0}^{\infty} \lfloor\frac{n}{2^i}\rfloor = \sum_{i=0}^{log_2(n)} \lfloor\frac{n}{2^i}\rfloor$$
Finally, we can let this sum look a little more refined:
Let $\{1,..,n\}$ be our set and $n = c_0\cdot 2^0 + c_1\cdot 2^1 +c_2\cdot 2^2+... + c_k\cdot 2^k$ its binary representation. Then the steps the algorithm needs for $\{1,..,n\}$ are equal to running our algorithm on the following sets:
$\{1,2,...,c_i\cdot 2^i\} \text{ where } i\in\mathbb{N}, i\leq k, c_i = 1 $
This let's us erase the $\lfloor \rfloor$-s in our sum, as for every step of the algorithm $ \lfloor\frac{c_i\cdot 2^i}{2^j}\rfloor\}$ is a whole number for $j\leq i$, and for $j>i$, the set is empty.
So, if $c_0\cdot 2^0 + c_1\cdot 2^1 +c_2\cdot 2^2+... + c_k\cdot 2^k$ is the binary represantation of our number, the steps our algorithm needs are:
$$\sum_{j=0}^k c_j\cdot\sum_{i=0}^j \frac{2^j}{2^i} = \sum_{j=0}^k c_j\cdot\sum_{i=0}^j 2^{j-i} = \sum_{j=0}^k c_j\cdot (2^{j+1}-1) $$
| {
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Show that $2n\choose n$ is divisible by $2n-1$ I have found many questions asking to prove $2n\choose n$ is divisible $2$, but I also observed by trying the first few "$n$" that $2n\choose n$ is divisible by $2n-1$.
It sure seems true when $2n-1$ is prime, but is it true in general for all $n$?
| Note that $\dbinom{2n}{n}$ is an integer. Note also that $1$ divides $\binom 21 $ and assume $n>1$ hereafter.
$\begin{align}
\dbinom{2n}{n} &= \dfrac{2n!}{n!\cdot n!} \\ &= \dfrac{2n\cdot (2n-1)\cdots (n+1)}{n\cdot (n-1)\cdots 1}\\
&= \dfrac{(2n)(2n-1)}{n\cdot n}\dbinom{2(n-1)}{n-1}\\
&= 2\dfrac{2n-1}{n}\dbinom{2(n-1)}{n-1}\\
\end{align}$
Now $\gcd(2n-1,n)=1$ so $n$ must divide $2\dbinom{2(n-1)}{n-1}$, that is, $2\dbinom{2(n-1)}{n-1}= kn$ for some integer $k$.
Then $\dbinom{2n}{n} = k(2n-1)$ as required.
| {
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Finding parametrized solutions to a trigonometric equation I want to find what pairs $(x,y)$ satisfy $\cos(x-y) = \cos(x) - \cos(y)$. To start, I defined a function $f(x,y) = \cos(x-y) - \cos(x) + \cos(y)$. I want to find functions $x(t), y(t)$ s.t. $f(x(t), y(t)) = 0$ I didn't really know where to go, so I tried putting the cosines in exponential form:
$$
\begin{align}
0 &= \cos(x-y) - \cos(x) - \cos(y)\\
0 &= \frac{1}{2}\big[e^{i(x-y)}+e^{i(y-x)}-e^{ix}-e^{-ix}+e^{iy}+e^{-iy}\big]\\
\end{align}
$$
From here I wasn't sure what to do. I tried multiplying through to get rid of negative exponents, rearranging into a neater format, factoring, etc. to no avail. I'm not even sure that's a useful first step, it just was the only one I could think of.
Any tips or ideas?
Also feel free to add or edit tags, wasn't quite sure what to put this under.
| I'm not sure if your method with exponentials allows you to conclude, but here's an overview of how I'd do it.
Although this isn't linear algebra, you can still get an idea of how to proceed by looking at your number of equations and unknowns. You have two unknowns for one equation, so if you add in one more unknown (parameter $t$) without any additional constraints, it'll only make things more complicated.
Instead you can just find an expression of $y$ as a function of $x$, that will give you a parametric expression of the form $x(t)=t$ and $y(t)$.
\begin{align*}
0=\cos(x-y)-\cos x+\cos y
&~\iff~ \cos x=\cos x\cos y+\sin x\sin y+\cos y \\
&~\iff~ \cos x=(\cos x+1) \cos y+\sin x\sin y
\end{align*}
Now there's somewhat of a trick to solve trigonometric equations of the form
$a\cos y+b\sin y=c$. Assuming that $a^2+b^2>0$, you can rewrite that as
$$
\frac a{\sqrt{a^2+b^2}}\cos y+\frac b{\sqrt{a^2+b^2}}\sin y=\frac c{\sqrt{a^2+b^2}}
~\iff~ A\cos y+B\sin y=C
$$
Notice that $A^2+B^2=1$, that means there is $\theta\in\mathbb R$ (unique if taken in $[0,2\pi)$) such that
$A\cos y+B\sin y=\cos\theta\cos y+\sin\theta\sin y=\cos(\theta-y)$.
From there, the equation admits solution(s) if and only if $\lvert C\rvert\le 1$.
This method can be applied to your equation
$\cos x= (\cos x+1)\cos y+\sin x\sin y$.
Just as a heads-up, if you apply the method I outlined in that exact manner,
you should obtain
$\sqrt{a^2+b^2}=2\lvert\cos\left(\frac x2\right)\rvert$, which is somewhat not nice. If you use the coefficient
$\sqrt{a^2+b^2}=2\cos\left(\frac x2\right)$ instead, it'll simplify things a little. In particular you should be able to prove that $\theta=\frac x2+2k\pi$, $k\in\mathbb Z$.
Last remark, but the cases $a^2+b^2=0$ and $\lvert C\rvert>1$ must be handled as exceptions if you want a complete proof of every possible solutions.
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Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$
My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive numbers to cancel the negative. One integer would have to cancel out the second, and the third wouldn't make the sum zero. Am I right?
| Hint: if $x+y+z=0$, then
$$
\begin{align}
\frac1x+\frac1y+\frac1z
&=\frac1x+\frac1y-\frac1{x+y}\\
&=\frac{(x+y)^2+x^2+y^2}{2xy(x+y)}\\
&=-\frac{x^2+y^2+z^2}{2xyz}
\end{align}
$$
| {
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Compute the determinant. If $a$, $b$, and $c$ are roots of unity. Then what is the value of
\begin{vmatrix}
e^a & e^{2a} & (e^{3a}-1) \\
e^b & e^{2b} & (e^{3b}-1) \\
e^c & e^{2c} & (e^{3c}-1)
\end{vmatrix}
I tried expanding it but the expression becomes unmanageable, is there some kind of simplification I can do?
| This is
$$\det\pmatrix{x&x^2&x^3-1\\y&y^2&y^3-1\\z&z^2&z^3-1}
=\det\pmatrix{x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3}
-\det\pmatrix{x&x^2&1\\y&y^2&1\\z&z^2&1}
$$
for $x=e^a$ etc. Both of these are essentially Vandermonde determinants.
| {
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Assumption of proof with contrapositive I want to prove by proof with contrapositive that
$\left \| u(0) \right \|=0 $ then $\left \| u(t) \right \|=0$ for all t belongs to $\left [ 0,T \right ]$.
Then I write
$\left \| u(t) \right \|\neq 0$ then $\left \| u(0) \right \|\neq 0$ for all t belongs to $\left [ 0,T \right ]$.
Is it correct ?
| The contrapositive of
If $∥u(0))∥=0$, then $‖u(t))‖=0$ for all $t$ belonging to $[0,T]$.
is the following:
If $‖u(t))‖\ne 0$ for some $t$ belonging to $[0,T]$, then $∥u(0))∥\ne 0$.
| {
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How to prove that $87!<16! \left(52^{71}\right)$ Prove that
$$87!<16! \left(52^{71}\right)$$
I do not how can i compare between the factorials or what the procedure to solve such questions?
| AM-GM gives:
$$\frac1{71}\sum_{i=1}^{71}(i+16)>\sqrt[71]{\prod_{i=1}^{71}(i+16)}$$
So:
$$\frac1{71}\left(\frac{71(71+1)}{2}+71\cdot 16\right)>\sqrt[71]{\frac{87!}{16!}}$$
or:
$$52^{71}>\frac{87!}{16!}\implies 16!(52^{71})>87!$$
| {
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How do you derive an equation if you know the asymptotes? I am trying to curve fit and let's say you have a data set in which you know the asymptotes are:
$y = 0$
$y = -x + 683$
The data always has a positive $x$ value, and the data converges to the line $y = -x + 683$ and has a parabola looking shape. The data is on the right side of the line $y = -x + 683$.
How do I back into an equation with these asymptotes? Is there a pattern or math trick/relationship that will allow me to derive curves in the future if I know the asymptotes?
Obviously I don't know how curved the curvature of the line is, but is there a general equation that I have and then I can use R to find the curvature? This question is similar to this one: Finding family of curve for given asymptotes
| A simple function that shoul make the work is the following
$$f(x)=-x+683+\frac{k}{x^a}$$
with $k,a>0$ parameters to fit the data.
| {
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Image of unit ball under twice differentiable function Let $f: \mathbb{R^n} \to \mathbb{R^n}$ be a twice differentiable function such that $f(x) = 0$ outside of the unit ball $B \in \mathbb{R^n}$. I have to show that (the measure is Lebesgue): $$\int_B \det Df = 0$$
I tried to prove that $|f(B)| = 0$ (which should be equivalent to the above) but I've had no luck. I'm quite stuck since I have no further theorem that I can think of, and honestly the equality doesn't even intuitively seem to hold.
| Let $f=(f_1,\ldots,f_n)$. Consider the form
$$
w=f_1df_2\wedge\ldots \wedge df_n-f_2df_1\wedge df_3\ldots \wedge df_n+\ldots+(-1)^nf_ndf_1\wedge \ldots \wedge df_{n-1}
$$
and apply the Stokes theorem:
$$
\int_{\partial B}w=\int_{B}dw.
$$
| {
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Prove that $A \smallsetminus (A \smallsetminus B) = A \cap B$ $A$ and $B$ are any sets, prove that $A \smallsetminus (A \smallsetminus B) = A \cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.
I have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways.
Could anyone please point me in the right direction?
Many thanks.
| Let $x \in A \setminus (A \setminus B).$
Then $x$ is an element such that $x \in A$ and $x \notin A \setminus B$. But if $x \notin A\setminus B$, with some additional work, we realize this implies that $x \in A$ and $ x \in B$. So $x \in A \cap B$.
Vice-versa: let $x \in A \cap B$ so $x \in A$ and $x \in B$. This implies that $x \notin A \setminus B$. But given that $x \in A$ and $x \notin A \setminus B$ this implies that $x \in A \setminus (A \setminus B)$.
| {
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How to prove that $|\ln(2+\sin(x)) - \ln(2+\sin(y))| <= |x-y| \space \forall \space x,y \in \mathbb R$ Question states
Prove that for all $x$ and $y$ $\in R$, the following inequality is true:
$\lvert \ln(2+\sin(x)) - \ln(2+\sin(y))\rvert \le \lvert x-y\rvert$
i've gotten to the point that
$\frac{y-x}{2+\sin(c)} = \ln\frac{2+\sin(y)}{2+\sin(x)}$ (y-x divided by 2+sin(c) is my f dash c from the mean value theorem
I asked my teacher that i should use mean value theorem here so please don't use anything other than this, but i have no idea how to push this problem further.
Also this is my first post so i'm sorry for the f dash (c) thing, mathjax is hard
|
I thought it might be instructive to present a way forward that relies on only elementary, pre-calculus tools. To that end we proceed.
First, note that $\log(2+\sin(x))-\log(2+\sin(y))=\log\left(\frac{2+\sin(x)}{2+\sin(y)}\right)$.
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the iequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1}$$
for all $t>0$. Hence, with $t=\frac{2+\sin(x)}{2+\sin(y)}$, we have
$$\frac{\sin(x)-\sin(y)}{2+\sin(x)}\le \log(2+\sin(x))-\log(2+\sin(y))\le \frac{\sin(x)-\sin(y)}{2+\sin(y)}\tag1$$
It is evident from $(1)$ that
$$\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |\sin(x)-\sin(y)|\tag2$$
Next, using the Prosthaphaeresis Identity, $\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$ along with $|\cos(\theta)|\le 1$ and $|\sin(\theta)|\le \theta$ reveals
$$|\sin(x)-\sin(y)|\le |x-y|\tag 3$$
Finally, using $(3)$ in $(2)$, we obtain the coveted inequality
$$\bbox[5px,border:2px solid #C0A000]{\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |x-y|}$$
| {
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"Continuity" of volume function on hyperbolic tetrahedra
Consider a sequence $T_i$ of tetrahedra in $\mathbb H^3$ whose
vertices tend to the vertices of a regular ideal tetrahedron $T$ in
$\partial \mathbb H^3$. Then $$Vol(T_i)\to v_3.$$
This should follow from Lebesgue dominated convergence if $T_i\subseteq T_{i+1}$ for (almost) all $i$, since, calling $\nu$ the volume form on $\mathbb H^3$, $$|\nu\chi_{T_i}|\leq|\nu\chi_T|$$ so the integrals converge.
I think one can always suppose to be in this case by moving the $T_i$ by isometries: is this true? Is there a formally satisfying way to see it?
Thank you in advance.
| There are some explicit formulae for volumes of hyperbolic tetrahedra in terms of dihedral angles which are not just continuous but real-analytic functions, say, one by Ushijima (Theorem 1.1):
A volume formula for generalized hyperbolic tetrahedra.
(See also references to earlier works that he gives in the paper.) Dihedral angles, in turn, depend continuously (actually, real-analytically) on the vertices. Hence, volume is a real-analytic function on the vertices.
| {
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Finding integer solutions to $3^y = x^2 + 56$ The original question is asking for the following:
Find all ordered pairs of positive integers
$(x, y)$ for which $3^y = x^2 + 56$.
I've found a solution I believe to be the only solution, but I'm struggling to prove it's the only solution. $(5,4)$
When I see a problem like this, I initially think to use modular arithmetic to find the pattern of the answer. For example, taking each side of the equation $\mod 4$ shows that $x$ must be odd and $y$ must be even. $\mod 3$ shows that $x$ is not divisible by $3$.
Using a spreadsheet and running all $\mod \ $ from $2$ to $81$ and looking at the remainders for $x = 1$ to $250$ shows that $x = 5$ is the only possible solution, but obviously this is not a proof. I'm struggling to find the steps necessary to prove this answer.
I've also simplified the equation to have $y=\frac{\ln(x^2+56)}{\ln(3)}$, but I don't know how to show the only time $y$ is an integer is when $x=5$. (Perhaps I am wrong and more solutions exist). Any help would be appreciated.
| Since $y>0$, $x$ must be odd. Also, as you said, since $3^y\equiv_4(-1)^y\equiv_4 x^2+56\equiv_4 0$, $y$ must be even. Notice that $$3^{y}-x^2=(3^{\frac{y}{2}}-x)(3^{\frac{y}{2}}+x)=56=2^3\cdot7$$
So there are only finitely many solutions, since $3^{\frac{y}{2}}+x>0$ divides $56$ (and take up atmost 8 values, since 56 has 8 divisors).
Generally, if $a\cdot b=c$ and $a\geq b$ the smallest value for $a$ is $\sqrt{c}$ (one can prove this by contradiction). Since $3^{\frac{y}{2}}+x$ is the larger factor, it is atleast $\sqrt{56}\approx 7.48>7$. So you only have to check the cases: $$3^{\frac{y}{2}}+x=14=2\cdot 7$$
$$3^{\frac{y}{2}}+x=28=2^2\cdot 7$$
$$3^{\frac{y}{2}}+x=56=2^3\cdot7$$
The first case is your solution $(5,4)$. I'm not sure how one would write the rest in a formal proof, but I'll guess using a computer (and refering to the calculations) is OK, since there are only finitely many solutions.
| {
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Joining the centers of the edges of Platonic solids We know that taking the centers of the faces of any 3d polyhedron (say, the Platonic solids) produces the dual solid. And repeating this operation gives us back the original solid. Another possible thing we can do is take the centers of the edges. This will produce other solids as well. If you do this to a tetrahedron, you get an octahedron. Do this to an octahedron and you get a cuboctahedron. My question is, why does taking the face centers preserve the properties of the Platonic solids and have this nice dual-solid property while taking the edge centers doesn't. What makes the operation of taking the face centers "superior" to taking the edge centers? And is there a name for the process of forming a new solid by taking the edge centers?
| The general setup here is dealing with regular polytopes. Then you are asking about taking the centers of k-faces (k-dimensional sub-elements). This process is usually called the k-rectification, cf. https://en.wikipedia.org/wiki/Rectification_(geometry).
When describing regular polytopes by Dynkin diagrams, those are of type xPoQoRoSo..., that is have a special node ("ringed", here demarked by x) at one end, and all other nodes are "unringed" (demarked by o), whereas the links bear several number marks (here represented by P, Q, R, etc.). The rectified polytope then gets described by oPxQoRoSo... and describes your edge-center figure, the birectified will be oPoQxRoSo... and describes the hull of the set of 2-boundary centers, etc.
What is special for the birectification, when applied to polyhedra, is that those would be oPoQx, that is the Special node happens to lye at the opposite end of the graph. And this - quite generaly - is just the dual of the starting regular polytope.
$$$$
Edit:
Whilst rectification (or multi-rectification like bi-rctified = oPoQxRoSo… etc.) is only defined on base of regular polytopes and which thus ensures all equal edges throughout (even when dealing with non-convex regulars), there is a closely related operation called ambification, which simply is defined by the convex hull of the edge centers of any convex polytope. That latter one has been used in the other answer. Applied to the convex regular ones those clearly coincide. But ambification generally will not ensure to produce equal edges in the new polytope.
--- rk
| {
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$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, with $a, b, c > 0$; $2b^2 > 4ac > b^2$; $z_1\in$ third quadrant; $z_2 \in$ second quadrant in the argand's plane then, show that $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$
| $z_2 = \overline{z_1}\,$ since the quadratic has real coefficients, so $\,\arg\left(\dfrac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)=2 \arg(z_1)\,$.
Since $\,a,b,c \gt 0\,$ the roots are in the left half-plane $\,\operatorname{Re}(z_1)=\operatorname{Re}(z_2) = -\,\dfrac{b}{2a} \lt 0\,$, and given the condition that "$z_1\in$ third quadrant" $\,\operatorname{Im}(z_1) \lt 0\,$, so $\,z_1\,$ is the root with negative imaginary part:
$$z_1 = \dfrac{-b - i \sqrt{4ac - b^2}}{2a} = \sqrt{\dfrac{c}{a}}\left(-\dfrac{b}{2\sqrt{ac}} - i \sqrt{1 - \dfrac{b^2}{4ac}}\right)$$
The latter can be written as $\,z_1=\sqrt{\dfrac{c}{a}}\big(\cos(\varphi)+ i \sin(\varphi)\big)\,$ where $\,\varphi=\arg(z_1)\,$ and:
$$
\begin{cases}
\begin{align}
\cos(\varphi) &= -\dfrac{b}{2\sqrt{ac}} \\[5px]
\sin(\varphi) &= -\sqrt{1 - \dfrac{b^2}{4ac}}
\end{align}
\end{cases}
$$
Therefore $\,\varphi=2k\pi \pm \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right)\,$, and the $\,\sin(\varphi) \lt 0\,$ condition trims the solution set down to $\,\varphi=2k\pi \color{red}{-} \arccos\left(-\dfrac{b}{2\sqrt{ac}}\right) = (2k-1)\pi + \arccos\left(\dfrac{b}{2\sqrt{ac}}\right)\,$, so in the end:
$$\,\arg\left(\dfrac{z_1}{z_2}\right)= 2 \varphi \;\;\equiv\;\; 2 \arccos\left(\dfrac{b}{2\sqrt{ac}}\right) = 2 \arccos\left(\sqrt{\dfrac{b^2}{4ac}}\right) \pmod{2 \pi} \,$$
| {
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For independent, symmetric random variables $(\xi_n)$, $E\left(\left(\sum\limits_n\xi_n\right)^2\land1\right)\le\sum\limits_nE(\xi_n^2\land1)$ The following problem appears as an exercise in the Russian version of Probability, Shiryaev, 2003 edition(it seems that no English version containing this problem is available yet).
Let $\xi_1,\xi_2,\ldots$ be independent, symmetrically distributed random variables. Then
$$\mathsf E\left(\left(\sum_n\xi_n\right)^2\land1 \right)\le\sum_n\mathsf E(\xi_n^2\land1)$$
To avoid discussing convergence issue, I would like to assume that the summation is finite. It's not quite clear from context what "symmetrically distributed" means, but it's reasonable to guess this means $\xi_n$ and $-\xi_n$ has the same distribution. In this sense, I tried to write,
$$\mathsf E\left(\left(\sum_{n=1}^N\xi_n\right)^2\land1 \right)=\int(x^2\land1)dF_{S_N}(x)=2\int_0^\infty(x^2\land1)dF_{S_N}(x)$$
and
$$\mathsf E(\xi_n^2\land1)=\int(x^2\land1)dF_{\xi_n}(x)=2\int_0^\infty(x^2\land1)dF_{\xi_n}(x)$$
and, after integrating by parts, reduce the problem to proving
$$\int_0^1x\left(\mathsf P(S_N\ge x)-\sum_{n=1}^N\mathsf P(\xi_n\ge x)\right)dx\le0$$
The problem would be settled if
$$\mathsf P\left(\sum_{n=1}^N\xi_n\ge x\right)\le\sum_{n=1}^N\mathsf P(\xi_n\ge x),\quad 0\le x\le1$$
which is, unfortunately, not true in general.
My question: Is my approach above completely nonsense? Is it possible to turn it into a proof? If not, how to prove this inequality?
Edit. By induction, it suffices to prove for $N=2$, i.e. $\mathsf E((\xi_1+\xi_2)^2\land1)\le\mathsf E(\xi_1^2\land1)+\mathsf E(\xi_2^2\land1)$. This should be easier, but still not quite obvious for me.
| What is wrong with the following much simpler argument: $$E\left\{\left(\sum {\xi_i}\right)^{2} \wedge 1 \right\} \leq \left\{E\left(\sum \xi_i\right)^{2}\right\}\wedge 1\ = \left\{\sum E(\xi_i)^{2}\right\} \wedge1 \leq \sum \left\{E(\xi_i)^{2} \wedge 1\right\}$$ where we have used the fact that $E\xi_i \xi_j =0$ for $i \neq j$?
Incidentally, if one of the variables has infinite expectation the inequality becomes trivial.
I have also used the fact that $$\left\{\sum a_i^{2}\right\} \wedge1 \leq \sum \{a_i^{2} \wedge 1\}$$ for all non-negative numbers $a_i$.
| {
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There is a natrual connection on the tangent bundle? I come up with a (maybe stupid) question: let $M$ be a smooth manifold, then the exterior differential $d$ is a natural connection on $\Omega^k(M)$, hence by dualizing we get a natural connection on $TM$? How can it be true (without a metric)?
| Since you say "dualizing", I assume you're talking about the case $k=1$. If we view differential forms as alternating tensors, the exterior derivative $\Omega^1(M) \to \Omega^2(M)$ can be viewed as a map $\Gamma(T^*M) \to \Gamma(T^*M \otimes T^*M),$ which has the right "type signature" to be a connection; but it does not satisfy the (correct kind of) Leibniz rule: for a smooth function $f$ and a one-form $\theta$ we have $$d(f \theta) = f d \theta + df \wedge \theta \ne f d \theta + df \otimes \theta.$$
| {
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What does Morley rank of a quotient group mean? $ \DeclareMathOperator{\RM}{RM} $
I have a problem with understanding part b) of Exercise 6.6.23 in David Marker's "Model Theory: An Introduction": ($ \mathbb M $ is the monster model.)
a) Suppose that $ \mathbb M $ is $ \omega $-stable, $ A , B \subseteq \mathbb M ^ n $ are definable, $ \RM ( A ) $ is finite and $ f : A \to B $ is a definable surjective map such that $ \RM \big( f ^ { - 1 } ( b ) \big) = k $ for all $ b \in B $. Show that $ \RM ( A ) \ge \RM ( B ) + k $. [Hint: Prove by induction on rank that $ \RM \big( f ^ { - 1 } ( X ) \big) \ge \RM ( X ) + k $ for all definable $ X \subseteq B $.]
b) Suppose that $ G $ is an $ \omega $-stable group of finite Morley rank and $ H \le G $ is an infinite definable subgroup. Show that $ \RM ( G ) \ge \RM ( H ) + \RM ( G / H ) $. In particular, $ \RM ( G ) > \RM ( G / H ) $.
c ) Show that b) is not true for all $ \omega $-stable groups. [Hint: Let $ K $ be a differentially closed field and consider the derivation $ \delta : K \to K $.]
First of all, doesn't $ H $ have to be a normal subgroup of $ G $ so that $ G / H $ can be defined? Or does $ G / H $ mean something other than the quotient group? And lastly, as far as I know, Morley rank is defined on formulas and definable subsets of the [monster] model. So, how is the question about the Morley rank of $ G / H $, which consists of cosets?
| The algebra question: When $H$ is an arbitrary subgroup of $G$, $G/H$ often denotes the set of left cosets of $H$, $\{gH\mid g\in G\}$. When $H$ fails to be normal, we can't equip $G/H$ with a natural group structure, but we can equip it with the structure of a $G$-set: the action of $G$ on $G/H$ is the obvious one $g'\bullet (gH)= g'gH$.
The model theory question: While $G/H$ is not a definable set in $G$, it is an imaginary definable set, i.e. a definable set in $G^{\text{eq}}$. Precisely, suppose $\varphi_H(x)$ is the formula defining $H$. Then there is a definable equivalence relation $E(x,y)$ on $G$, given by $\exists z\, (\varphi_H(z)\land xz = y$), and the equivalence classes mod $E$ are exactly the left cosets of $H$. Since Morley rank makes sense in $G^{\text{eq}}$, we can define $\mathrm{RM}(G/H)$ to be the Morley rank of the set of $E$-classes in $G^{\text{eq}}$.
| {
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Confusing permutation? Or combination? Is this problem a permutation or combination problem? My idea is that this is permutation but Im not that sure... any idea to solve this problem? I tried putting the first $3$ company in $3^3$ possible ways that is $27$, but i got boggled on the remaining $2$ organizations because of the given condition...
There are 5 student organizations. How many ways can 3 students join these organizations if no 2 students can join the same org? Thank you
| Since the students are distinguishable, it is a permutaiton problem.
It is just $\frac{5!}{(5-3)!}$.
You can also think of it as first out of the $5$ companies, choose $3$ companies and then rearrange the students.
$$\binom{5}{3} \times 3!.$$
Remark: $3^3$ doesn't seem to avoid the constraint that no $2$ students can join the same org.
| {
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Proving a limit exists using the definition of a limit $ \lim_{x\to\infty} {\sqrt{f(x)+1}}) = {\sqrt{L+1}} $ Let $f(x)$ be a function such that $f(x)\geq -1$ for every $x$. Suppose, $\lim_{x\to\infty} f(x) = L$ and that $L\leq-1$. Prove by using the definition of a limit that:
$ \lim_{x\to\infty} {\sqrt{f(x)+1}}) = {\sqrt{L+1}} $
Hi I would very much appreciate an explanation on how to solve this problem using the definition. I know the definitions well but don't succeed on solving accurately, maybe some tips in general of how to solve these kind of problems. Thank you!
| Supposing $L$ is finite and $L\geq -1$: $\displaystyle\lim_{x\to\infty}f(x)=L$ means that (by definition) for every given $\epsilon>0$, there exists a real number $N_1$ (depending on $\epsilon)$ such that the inequality $|f(x)-L|<\epsilon$ is satisfied for all $x\geq N_1$. Also for the same epsilon there is $N_2$ such that $|f(x)-L|<\sqrt \epsilon$ for all $x\geq N_2$. Now, look at $$\bigg|\sqrt{f(x)+1}-\sqrt{L+1}\big|^2=\big|\sqrt{f(x)+1}-\sqrt{L+1}\big|\big|\sqrt{f(x)+1}-\sqrt{L+1}\big|$$
$$\leq \big|\sqrt{f(x)+1}-\sqrt{L+1}\big|\big|\sqrt{f(x)+1}+\sqrt{L+1}\big|=|f(x)-L|^2<\epsilon\quad (\text{by difference of squares}).$$
Thus if we let $N=\max\{N_1, N_2\}$, then we see that for the given $\epsilon>0$
the inequality $$\big|\sqrt{f(x)+1}+\sqrt{L+1}\big|<\sqrt\epsilon$$
is satisfied for all $x\geq N$,
where we have also used the fact that $|a-b|\leq |a+b|$ for $a,\,b\geq 0.$ This shows that $\displaystyle\lim_{ x\to \infty}\sqrt{f(x)+1}=\sqrt{L+1}$.
If $L$ were $\infty$, we would use this definition:
$\displaystyle\lim_{x\to\infty}f(x)=\infty\Longrightarrow$ for every $\epsilon>0$, there exists a real number $N$ (depending on $\epsilon)$ such that the inequality $f(x)\geq \epsilon$ is satisfied for all $x\geq N.$ In this case using almost similar idea you will show that $\displaystyle\lim_{x\to\infty}\sqrt{f(x)+1}=\infty.$
| {
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Show that if $A,B$ are $2\times2$ matrices, then $(AB-BA)^2$ commutes with all $2\times2$ matrices. I tried to write it all out, but it becomes really messy... Is there a more elegant way to do it?
Note that I don't know about dimensions, vector spaces & bases yet
| $AB-BA$ has trace zero, so has the form $\pmatrix{u&v\\w&-u}$.
What happens when you square such a matrix?
| {
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Difference between minus one and plus one induction? I recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, however my professor has insisted I solve it by using $P(n-1)$ as opposed to $P(n+1)$, which I've done below.
Basis
$$\frac{1(1+1)(2*1+1)}{6} = 1$$
Inductive Step $(n-1)$
$$1^2+2^2+\ldots + (n-1)^2 = \frac{(n-1)(n)(2(n-1)+1)}{6}$$
Which Simplifies to
$$\frac{(n-1)(n)(2n-1)}{6} \rightarrow \frac{2n^3-3n^2+n}{6}$$
Add $6\frac{n^2}{6}$ to both sides and we've proven by induction.
My question is do there exists any mathematical proofs for which solving by Induction with $n+1$ and $n-1$ are not interchangeable and should I petition my professor to be able to use them interchangeably. I am aware that solving using $n-1$ and $n+1$ is identical, at least for every scenario I've come across (we're working with positive integers so I'm not expecting any variance from that), however given the overwhelming amount of resources, I can't for the life of me figure out why I am being instructed to use a method opposite what seems to be the norm for any other reason besides my teacher's personal preference.
| I would argue that, if anything, there are reasons to prefer $P(n) \Rightarrow P(n+1)$.
Natural numbers can be defined in many ways, but the usual inductive definition is the following:
*
*$0$ is a natural number;
*If $n$ is a natural number, then $s(n)$ is a natural number.
Here $s(n)$ denotes the successor of $n$.
These two rules define a set $\mathbb N$ together with an induction principle (which allows us to prove properties of all elements of $\mathbb N$ and is in fact the usual mathematical induction) and a recursion principle (which allows us to construct new objects from the elements of $\mathbb N$).
Then $\mathbb N$ can be endowed with the usual operations satisfying all the well-known properties. In particular, it is customary to write the successor $s(n)$ of $n$ as the sum $n+1$, although it is the sum between two natural numbers that is actually defined by recursion using the successor.
There are of course many inductive structures other than the set of natural numbers. For example, binary trees are defined by:
*
*$v$ is a binary tree (a single vertex, which is also the root);
*If $t_1$ and $t_2$ are binary trees, then $t_1 \bullet t_2$ is a binary tree (the graph formed by taking $t_1$ and $t_2$, adding a new vertex as a root and joining the roots of $t_1$ and $t_2$ to the new root).
How does the induction principle look like for binary trees? If you want to prove that $P(t)$ holds for any binary tree $t$, you have to prove:
*
*(Basis) $P(v)$ holds;
*(Inductive step) If $P(t_1)$ and $P(t_2)$ hold, then $P(t_1 \bullet t_2)$ holds.
In this case there is no equivalent to the predecessor of a natural number.
| {
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Where is the error in this proof for showing that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant? Show that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant.
$$1=1^x=(e^{2\pi i})^x=e^{2\pi ix}=\cos{(2\pi x)}+i\sin{(2\pi x)}$$
Where is the error in this "proof"? Does it not hold for all $x\in\mathbb{R}$, if yes which ones?
| Notice that $(-2)^2 = 4$. So $[(-2)^2]^{\frac 12} = 4^{\frac 12} = 2$.
But $[(-2)^2]^{\frac 12} = (-2)^{2\frac 12} = (-2)^1 = -2 \ne 2$.
What went wrong?
The assumption that $(a^m)^n = a^{mn}$ only holds if either $m, n \in \mathbb Z$ or if $a > 0; a \in \mathbb R; m,n \in \mathbb R$.
If we define $a^m = a*.... *a; m$ times then $(a^m)^n = a^{mn}$ follows by associativity.
If we define $b^{\frac mn} = \sqrt[n]{b^m}; b > 0; \frac mn \in \mathbb Q$ then $(b^q)^r = b^{qr}$ follows, by putting the fractions under a common denominator. This reduces to a radical base and integer powers.
If we define $b^x = \lim_{q\to x} b^q; b > 0$ then $(b^x)^y = b^{xy}$ can be deduced by the continuity of limits.
But when we define a fundimentally different definite for complex powers as $e^{iy} = \cos y + i \sin y$ then $(e^{iy})^z = (\cos y + i \sin y)^z \ne \cos yz + i \sin zy= e^{iyz}$ just plain doesn't follow.
| {
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Modular arithmetic and some applications Show that if $p> 2$ is a prime number, then $p$ divides $(p-2)! - 1$. I have tried using Fermat's Theorem, but I could not solve it.
| This follows from Wilson's theorem, which states that
$$(p-1)!\equiv_p -1$$ if and only if $p$ is a prime
If you accept this, then the rest follows easily since $(p-1)!\equiv_p (p-2)!(-1)\equiv_p-1$ implies: $$(p-2)!\equiv_p 1$$
This is the same as saying that $p$ divides $(p-2)!-1$.
Now for the proof of this theorem. In $\mathbb{Z}_p$ every non-zero element has an inverse, in particular $-1$ and $1$ are their own inverses, which means that if you multiply all these numbers you'll get: $$1\cdot 2\cdot\ldots\cdot (p-1)=(p-1)!\equiv_p 1\cdot 1\cdot\ldots\cdot 1\cdot (-1)\equiv_p-1$$
| {
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Prove the equation $\ln(x) = \frac1 {x-1}$ has exactly 2 real solutions.
Prove the equation $\ln(x) = \frac1 {x-1}$ has exactly 2 real solutions.
Hello all. I thought of defining the function $f(x)=x-e^{\frac 1 {x-1}}$ and showing it has only 2 single roots, though I am not sure on how to show it and I understand it is better to prove using Lagrange's theorem.
Would be happy to get some help on that question, thanks in advance :)
| $\ln x$ does not exist for $x \le 0$ and $\frac 1{x-1}$ does not exist for $x = 1$.
So if there are solution they will exist on the intervals $(0,1)$ and $(1, \infty)$.
On these intervals the function $f(x) = \ln x - \frac 1{x-1}$ is continuous.
$f'(x) = \frac 1x - (-1)\frac 1{(x-1)^2} = \frac 1x + \frac 1{(x-1)^2} > 0$ so $f'(x)$ is increasing on the intervals $(0,1)$ and $(1,0)$. So if $f(x)$ ever "crosses the $x$-axis". It can only do so at most once in each interval.
Taking the limits of $f(x)$ as $x \to 0^+; x \to 1^-; x\to 1^+$ and $x\to \infty$ we will see that $f(x)$ will have negative and positive values on each of these intervals. So by the intermediate value theorem there will be values in these intervals where $f(x) = 0$.
And as $f'(x) > 0$ there can only at most one on each interval, so there will be exactly one on each interval.
And, obviously, if $f(x) = 0$ then $\ln x = \frac 1{x-1}$.
| {
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An entire function whose integral is bounded is identically zero Suppose $f$ has a power series at $0$ that converges in all of $\mathbb{C}$ and $$\int_{\mathbb{C}} |f(x+iy)|dxdy$$
Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.
A hint is given: “Use polar coordinates to show $f(0)=0$”
Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc
| Let $f(z)=\sum_0 ^{\infty} a_n z^{n}$ be the power series expansion. Write $z=re^{i\theta}$ and integrate with respect to $\theta$ from 0 to $2\pi$. Integrating term by term is permitted because of uniform convergence. You get $2\pi a_0= \int_0 ^{2\pi} f(re^{i\theta}) d\theta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr d\theta =dxdy$ you will see that $|(\int_0^R rdr) 2\pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.
| {
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$\lim\limits_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$ Can anyone help me solve this?
I know the answer is 4, but I don't really know how do I find the biggest power of $x$ when there's a square root.
$$\lim_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$$
| HINT
Take $x^2$ as a common factor and simplify the quotient.
The limit will be $4$
| {
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Finding the 10th root of a matrix I want to find a $2 \times 2$ matrix, named $A$ in this situation, such that:
$$A^{10}=\begin {bmatrix} 1 & 1 \\ 0 & 1 \end {bmatrix} $$
How can I get started? I was thinking about filling $A$ with arbitrary values $a, b, c, d$ and then multiplying it by itself ten times, then setting those values equal to the given values but I quickly realized that would take too long. Is there a more efficient way?
| Hint: Another approach is to note that
$$
\exp\left(\begin{bmatrix}0&x\\0&0\end{bmatrix}\right)=\begin{bmatrix}1&x\\0&1\end{bmatrix}
$$
| {
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Meijer G-function Can you please help me devise a series for the Meijer's G-function (i) with inceces m=3, n=0, p=1 and q=3, for a general real variable?
The first difficulty that I am facing is the proper choice of an integration path, to use the residues' theorem. It seems to me that I may choose between two possibilities, one of which encloses no pole, whereas the other contour does. That is a nonsense, of course, but I am failing to see where my errors lies...
(i) The single a-coefficient is 0, and the three b-coefficients are z, 0, 0 (z is a general complex number).
| We have
$$G_{1,3}^{3,0}\left( x \middle| {0 \atop z, 0, 0} \right) =
\frac 1 {2 \pi i} \int_{\mathcal L}
\frac {\Gamma(z-y) \Gamma(-y)^2} {\Gamma(-y)} x^y dy.$$
Notice that a pair of the gamma functions cancels out.
To determine the integration contour, you need to analyze the asymptotic behavior of the gamma function and choose a contour (separating the poles) over which the integral converges, as this is implied in the definition of the G-function. In this case only the right loop will do.
Next, writing the integral as a sum of residues, you get a series expansion for small $x$. The integrand has residues at the points $y=k$ and $y=z+k$, thus the leading term of the expansion may be of order zero, or of order $z$, or there may be a logarithmic case.
| {
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Random point inside an equilateral triangle Take any equilateral triangle and pick a random point inside the triangle.
Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$.
If the three line segments from each vertex to the random point were removed out of the original triangle to form a new triangle , what would the new triangle's angles be?
Video about the problem
|
As in the attached diagram, let $ABC$ be the original equilateral triangle and let $D$ be a point in $\triangle ABC$.
We let point $E$ be on the opposite side of $BC$ as $D$ such that $\triangle BDE$ is equilateral. Then $BD=BE$, $BA=BC$ and $\angle DBA=\angle EBC=60^{\circ}-\angle DBC$. And therefore $\triangle DBA$ and $\triangle EBC$ are congruent. This implies that $EC=DA$ and since $DE=BD$, we now have $\triangle CDE$ as the triangle we want.
Let $\angle ADB=x$ and $\angle BDC=y$. Then $\angle EDC=y-60^{\circ}$, $\angle DEC=x-60^{\circ}$ and $\angle DCE=300^{\circ}-x-y$ are our desired angles.
| {
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If $a,b,c$ be in Arithmetic Progression, If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.
My Attempt:
$a,b,c$ are in AP so
$$b=\dfrac {a+c}{2}$$
$b,c,a$ are in HP so
$$c=\dfrac {2ab}{a+b}$$
Multiplying these relations:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$
| Hint:
Eliminate $c$
$$2ab=(a+b)c=(a+b)(2b-a)$$
Simplify to find $$0=a^2+ab-2b^2=(a+2b)(a-b)$$
| {
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Proposed definition of a countable set Consider the following proposed definition:
A set $X$ is countable iff $X=\emptyset$ or there exists $x_0\in X$ and $f:X\to X$ such that:
$\forall P\subset X: [x_0\in P \land \forall x\in P: [f(x) \in P] \implies P=X]$
In other words, a set $X$ is countable iff $X$ is empty or induction holds on $X$.
Note that, by this definition, if $X=\{x_0\}$ then $X$ is countable. The identity function on $X$ would be the required "successor function."
By this definition, the set of natural numbers is trivially countable.
Comments? Is this definition anything new?
| We can show that if $X$ satisfies this property, then there is a surjection $g:\mathbb N\rightarrow X$. Thus $X$ is countable in the usual sense. Of course the converse holds, as you already know.
Define $g$ as $g(0)=x_0$ and $g(n+1)=f(g(n)),\forall n\in\mathbb N$. Then by the property in question, the image of $g$ is equal to $X$, so $g$ is surjective.
Hope this helps.
| {
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Number of answers : $f(x)=f^{-1}(x)$
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$f(x)=f^{-1}(x) \ \ \ \ :f^{-1}(f(x))=x$$
MY Try :
$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$
now what do i do ?
| Hint:
Point of intersection of $f(x)$ and $f^{-1}(x)$ while same as that of $f(x)$ and the line $y=x$.
| {
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Does this entangled PDE capture the derivative? In a game-theory textbook, I encountered the following. Suppose we have player $1$ and and $2$ optimizing by playing strategies $x_1$ and $x_2$ $\in \mathbb{R}$. The first order conditions for player $1$ and $2$ are given respectively by:
$$f(x_1,x_2,a,b)=1$$
$$g(x_1,x_2,a,b)=1$$
What is $\frac{\partial x_1}{\partial b}$? Naively, we may apply the implicit function theorem:
$$\frac{\partial x_1}{\partial b}=-\frac{\frac{\partial f(..)}{\partial a}}{\frac{\partial f(..)}{\partial x_1}}$$
But given that $a$ is also a function of the first order condition for the second player, we must take into account the effect of $a$ through its effect on $x_2$. Hence, the PDE is given by
$$\frac{\partial x_1}{\partial a}=-\frac{\frac{\partial f(..)}{\partial a}+\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}}{\frac{\partial f(..)}{\partial x_1}+\frac{\partial x_2}{\partial x_1}*\frac{\partial f(..)}{\partial x_2}}$$
Where $x_2$ satisfies player $2$'s FOC. In the numerator, the term $\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}$ captures the effect of $a$ through $x_2$ on $f(..)$, and the $\frac{\partial x_2}{\partial x_1}*\frac{\partial f(..)}{\partial x_2}$ the effect of $x_1$ through $x_2$ on $f(..)$. Is this a correct application of the inverse function theorem that produces a good approximation of $\frac{\partial x_1}{\partial b}$?
One factor that makes me hesitant about this PDE is in the numerator, the term $\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}$ captures the effect of $a$ on $x_2$. However, this is exactly what we're trying to define all along, albeit for $x_1$. At any rate, this part will not completely characterize the effect of $a$ on $x_2$.
| If you differentiate both equations in $a$, you get
$$
f_a+x^1_af_1+x^2_af_2=0,\\
g_a+x^1_ag_1+x^2_ag_2=0,
$$
where I'm denoting $x^j_a=\partial x^j/\partial a$ and $f_j=\partial f/\partial x^j$. If we solve this system for $x^1_a$, we get
$$
x^1_a=\frac{f_2g_a-g_2f_a}{g_2f_1-g_1f_2}.
$$
Using Jacobians, you can express this as
$$
\frac{\partial x^1}{\partial a}=-\frac{\partial(f,g)/\partial(a,x^2)}{\partial (f,g)/\partial(x^1,x^2)}.
$$
| {
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How to solve $a$ in $\int_a^xf\left(t\right)dt=x^2-2x-3$ How do you solve for $a$ in $\int_a^xf\left(t\right)dt=x^2-2x-3$?
I have differentiated both sides with respect to $x$, but do not know how to continue after this.
| Put $F(x) = \displaystyle \int_{a}^x f(t)dt=x^2-2x-3$. Since $F(a) = 0$, $a^2-2a-3 = 0 \implies (a-1)^2 = 4\implies a - 1 = \pm 2 \implies a = -1, 3$. Thus there are $2$ answers: $a = -1,3$.
| {
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Why does x raised to an odd power result in a unique solution? I don't think I'm phrasing this correctly but anyway, say we are looking for solutions to x for $x^9=1/2$,
why is there only one unique solution as given by $x=(1/2)^{1/9}$
when by raising x to an even value gives two solutions as $\pm$ solution
I think I'm just missing a key concept so any help would be appreciated.
| The simple answer is there is not only one solution, but only one real solution. For example, the equation $x^3=1$ has the solutions $x=1,e^{\frac{2i\pi}{3}},$ and $e^{\frac{4i\pi}{3}}$.
| {
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Given sequence of nested intervals $I_n[a_n,b_n]$. Show that if $y=\inf\{b_{1},b_{2},b_{3},\ldots\}$ then $y\in[a_{n},b_{n}]$ $\forall n$ Let $I_{1}=[a_{1},b_{1}]$, $I_{2}=[a_{2},b_{2}]$, $I_{3}=[a_{3},b_{3}]$,
$\ldots$ be a sequence of closed bounded nested nonempty intervals
$I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq\cdots$. Show that if
$y=\inf\{b_{1},b_{2},b_{3},\ldots\}$ then $y\in[a_{n},b_{n}]$ $\forall n$
Since the set $\{b_{1},b_{2},b_{3},\ldots\}$ is bounded below by $a_{1}$, by the axiom of completeness the $\inf\{b_{1},b_{2},b_{3},\ldots\}$ exists let then let $y=\inf\{b_{1},b_{2},b_{3},\ldots\}$. Fix any $n$ we will show that $y\in[a_{n},b_{n}]$ $\forall n$. By the definition of $y$ being the infimum we know that $y\leq b_{n}$, $\forall b_{n}$. Thus if we can show that $a_{n}$ is a lowerbound for $\{b_{1},b_{2},b_{3},\ldots\}$ then $a_{n} \leq y$ where $y$ is the greatest lower bound. We will show that $a_{n} \leq b_{k}$, $\forall k$.
Case $1$: If $k \geq n$ then $a_{k} \leq a_{n} \leq b_{k} \leq b_{n}$ by the nested interval property.
Case $2$: if $k<n$ then we have that $a_{n} \leq a_{k} \leq b_{n} \leq b_{k}$ by the nested interval property.
Hence we have shown that $a_{n} \leq b_{k}$, $\forall k$ thus $y\in[a_{n},b_{n}]$ $\forall n$.
| You have made slight errors in the inequalities in both the cases.
If $k\geq n$, then the nested property gives $[a_n, b_n]\supseteq [a_k, b_k]$ so $a_n\leq a_k\leq b_k$ and therefore $a_n\leq b_k$.
If $k<n$, then $[a_k, b_k]\supseteq [a_n, b_n]$ so $a_n\leq b_n\leq b_k$ and so $a_n\leq b_k$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Value of $\sec^2 a+2\sec^2 b$
If $a,b$ are $2$ real number such that $2\sin a \sin b +3\cos b+6\cos a\sin b=7,$ Then $\sec^2 a+2\sec^2 b$ is
Try: i am trying to sve it using cauchy schwarz inequality
$$\bigg[(\sin b)(2\sin a+6\cos a)+(\cos b)(3)\bigg]^2\leq (\sin^2b+\cos^2b)\bigg((2\sin a+6\cos a)^2+3^2\bigg)$$
Could some help me to solve it , thanks
| Hint:
$2\sin a+6\cos a=\sqrt{40}\sin (a+x)$ where $x=\sin^{-1}\frac{6}{\sqrt{40}}$.
\begin{align*}
2\sin a \sin b +3\cos b+6\cos a\sin b&=\sqrt{40}\sin (a+x)\sin b+3\cos b\\
&=\sqrt{40\sin^2(a+x)+9}\cdot \sin(b+y)
\end{align*}
where $\sqrt{40}\sin(a+x)=\sqrt{40\sin^2(a+x)+9}\cdot\cos y$ and $3=\sqrt{40\sin^2(a+x)+9}\cdot\sin y$.
Since the maximum value of $\sqrt{40\sin^2(a+x)+9}\cdot\sin(b+y)$ is $7$, we need $\sqrt{40\sin^2(a+x)+9}=7$ and $\sin (b+y)=1$.
The answer should be $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute the period of this function $f(x)=7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}$ I'm starting my journey into Fourier Series. I am given this function: $$f(x)=7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}$$
Following my book, this function has a period of $T=2$ (this is the book I'm reading).
However from what I know:
*
*A function $f(x)$ is periodic with period $T$, if and only if each of its summands is periodic with period $T$.
*A function $g(x)$ is said to be periodic with period $T$ when it satisfies: $g(x+T)=g(x)$
So, for the function above, I can't seem to understand why the author says its period is 2, since: (please correct the following statements if I'm wrong)
*
*$\cos{(\pi x+T)}=\cos{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
*$\sin{(\pi x+T)}=\sin{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
*$\cos{(2\pi x+T)}=\cos{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
*$\sin{(2\pi x+T)}=\sin{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$
I derived the above periods using the formulas for the sine and cosine of a sum of 2 angles. For example:
$$\begin{align} \cos{(\pi x)} & = \cos{(\pi x+T)} \\ & = \cos{(\pi x)}\underbrace{\cos{(T)}}_{=1}-\sin{(\pi x)}\underbrace{\sin{(T)}}_{=0} \end{align} $$
Which is satisfied only when $T=\boxed{2\pi}, 4\pi,... = 2\pi k$
So... why did the book say $f(x)$ has a period equal to $T=2$? Where am I going wrong?
| You need $$7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}=$$
$$=7+3\cos{(\pi (x+T))}-8\sin{(\pi (x+T))}+4\cos{(2\pi( x+T))}-6\sin{(2\pi (x+T))}$$ for all real $x$, which indeed gives $T=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$f(x) = x^6, g(x) = x^{10}$ endomorphisms $\implies G$ is abelian
Let $(G, \cdot)$ be a group in which the functions $f: G \to G, f(x) = x^6$ and $g : G \to G, g(x) = x^{10}$ are endomorphisms and $f$ is injective. Prove that $G$ is an abelian group.
We need to prove that $f(xy) = f(yx), \forall x,y \in G$.
Because $f$ is an endomorphism, then $(xy)^6 = x^6y^6 \iff (yx)^5 = x^5y^5, \forall x,y \in G$. So, $x^6y^6 = (xy)^6 = (xy)^5(xy) = y^5x^5xy \implies x^6y^5 = y^5x^6, \forall x,y \in G$. Swaping $x$ and $y$, we get that $x^5y^6 = y^6x^5, \forall x,y \in G (*)$.
In the same way we obtain that $x^{10}y^9 = y^9x^{10}, \forall x,y \in G (**)$. Therefore, $x^5y^3 = y^3x^5$ (using $(*)$ and $(**)$), but I don't know how to continue from here.
| This is an extended comment rather than an answer.
The assumption that $f$ is injective is necessary.
Indeed, the article Abelian Forcing Sets by Joseph A. Gallian and Michael Reid contains the following result:
Definition: a set of integers $T$ is called abelian-forcing if for
any group $G$, if the map $x\mapsto x^t$ is an endomorphism for every
$t\in T$ then $G$ has to be abelian.
Theorem: a set of integers $T$ is abelian-forcing if and only if the
gcd of the numbers $t(t-1)$ where $t$ runs over $T$ is equal to $2$.
In the present case, the $\gcd (5\times 6,\ 9\times 10)=30\neq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Inverse Transform Method for a pdf Given a pdf:
$$f(x)=\tau x \exp\left(\frac{-\tau x^2}{2}\right); \quad x, \tau > 0$$
So I found the corresponsing cdf:
$$F(x)=1 - \exp\left(\frac{-\tau x^2}{2}\right)$$
Then I got given a value for tau: $\tau=0.2$ and I derived the inverse function $F^{-1}_X(u).$:
$$F^{-1}_X(u)=\sqrt{\color{red}{\frac{2}{\tau}}10 \log\left(\frac{1}{1-u}\right)}; \quad u\sim U[0,1]$$
Now from what I have been told, $F^{-1} \sim F$ but I just can't see it. I may be drawing wrong conclusions but $$0< F<1$$
and $$-\infty<F^{-1}<\infty$$
Is what I did correct?
EDIT: My lecture notes:
| The statement is
If $u \sim U[0,1]$ then $F_X^{-1}(u)\sim f$
To show you this I sample a uniform distribution and calculate
$$
x = F^{-1}_X(u) = \sqrt{-\frac{2}{\tau}\ln(1 - u)}
$$
Then I make a histogram and overplot the distribution
$$
f_X(x) = \tau x e^{-\tau x^2/2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear homogenous second order ODE without constant coefficients I am having trouble finding the general solution of the following second order ODE for $y = y(x)$ without constant coefficients:
$3x^2y'' = 6y$
$x>0$
I realise that it may be possible to simply guess the form of the solution and substitute it back into the the equation but i do not wish to use that approach here.
I would appreciate any help, thanks.
| Hint
Simple way
$$y''-2\dfrac {y}{x^2} = 0$$
$$x^2y''+2xy'-2xy'-2y= 0$$
$$(x^2y')'-2(xy)' = 0$$
Integrate
$$(x^2y')-2(xy) = K_1$$
Divide by $x^4$
$$\dfrac {x^2y'-2xy}{x^4} = \dfrac {K_1}{x^4}$$
$$(\dfrac {y}{x^2})' = \dfrac {K_1}{x^4}$$
integrate again
$$\dfrac {y}{x^2} = \int \dfrac {K_1}{x^4}dx +K_2$$
$$\boxed{y(x)=\frac {K_1} x +K_2 x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the largest eigenvalue of the following matrix?
Find the largest eigenvalue of the following matrix
$$\begin{bmatrix}
1 & 4 & 16\\
4 & 16 & 1\\
16 & 1 & 4
\end{bmatrix}$$
This matrix is symmetric and, thus, the eigenvalues are real. I solved for the possible eigenvalues and, fortunately, I found that the answer is $21$.
My approach:
The determinant on simplification leads to the following third degree polynomial.
$$\begin{vmatrix}
1-\lambda & 4 &16\\
4 &16-\lambda&1\\
16&1&4-\lambda
\end{vmatrix}
= \lambda^3-21\lambda^2-189\lambda+3969.$$
At a first glance seen how many people find the roots of this polynomial with pen and paper using elementary algebra. I managed to find the roots and they are $21$, $\sqrt{189}$, and $-\sqrt{189}$ and the largest value is $21$.
Now the problem is that my professor stared at this matrix for a few seconds and said that the largest eigenvalue is $21$. Obviously, he hadn't gone through all these steps to find that answer. So what enabled him answer this in a few seconds? Please don't say that he already knew the answer.
Is there any easy way to find the answer in a few seconds? What property of this matrix makes it easy to compute that answer?
Thanks in advance.
| The trick is that $\frac1{21}$ of your matrix is a doubly stochastic matrix with positive entries, hence the bound of 21 for the largest eigenvalue is a straightforward consequence of the Perron-Frobenius theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 11,
"answer_id": 3
} |
Basic algebra exercise I'm stuck with this problem. I think that my difficulties are more with dealing with complex numbers then with groups, but still.
Could you please help me?
Let $\mathbb{C}^{*}$ be $\mathbb{C} \setminus \{0\}$, the multiplicative group of the complex numbers without zero.
Let $\rho$ be the equivalence relation defined so that $a\rho b$ if $\frac{a^2}{b^2} \in \mathbb{R}$.
Describe the equivalence classes of $\rho$ as subsets of the Argand-Gauss plane.
Is $\rho$ a congruence relation compatible with the multiplication in $\mathbb{C}^{*}$?
What is the normal subgroup of $\mathbb{C}^{*}$ which corresponds to $\rho$?
Hints would be appreciated too (maybe even more than full solutions).
Thank you.
| If we write $a=re^{ix}$ and $b=se^{iy}$, then $a \sim b$ if and only if $e^{2i(x-y)} \in \mathbb R$ For what "angles" does this occur?
Step 1: if $a \sim b$ is it also true that $ac \sim bc$ for any $c \in \mathbb R$?
Step 2: note that $1 \in H$ for any subgroup of $\mathbb C$. We need at least all of the $z \sim 1$. What are these?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Strategy for finding integral roots for polynomials with large coefficients I'm trying to find the integer roots for $f(x) = x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420 = 0$. The techniques I'm expected to have at my disposal are:
*
*For a polynomial with integer coefficients $p(x) = a_nx^n + a_{n-1}x^{n-1}+ \dots + a_1x + a_0$, if $p(x)$ has a rational root $\frac{s}{t}$, then $s|a_0$ and $t|a_n$.
*"Einstein's Irreducibility Criterion": If there exists a prime $p$ such that $a_{n-1}\equiv_pa_{n-2}\equiv_p\dots\equiv_pa_0\equiv_p0$, $a_n\not\equiv_p0$ and $a_0\not\equiv_{p^2}0$ then $p(x)$ is irreducible over the rationals.
The book gives the answers $-12$ and $-35$. Using the first strategy, I could have solved for every divisor of $-420$ to find one solution $c$, then divide $(x-c)$ into $f(x)$ and perform the same process with the quotient. But is there a less time-consuming way to find these roots? I feel like I'm supposed to simplify $f(x)$ by substituting $x$ with something but I wouldn't know what that is.
| Try
\begin{align}
x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420
& \equiv (x^2+ax+b)(x^3+cx^2+dx+e) \\
& \equiv x^5+(a+c)x^4+(ac+b+d)x^3 \\
& \quad +(ad+bc+e)x^2+(ae+bd)x+be
\end{align}
with different combinations of $b$ and $e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Absorbing Markov Chain Probabilities Can someone explain to me why the answer to the following question is not 1/2? (The given answer is 1/4, but, after trying various methods, including both argument by symmetry (each absorbing state must get 1/2, as the probability of being absorbed is 1, and the process is symmetric with respect to either state) and a more fulsome listing of the transition probabilities, I still get 1/2).
It's possible I don't understand the meaning of "probability that the process reaches the state AA and AA", but on the meaning "eventually" it should be 1/2, and on the meaning, in the next transition, it would be 1/16. So I am unable to find a meaning that gives us 1/4.
"Biologists can model a population's genotype distribution across generations with a Markov chain. Suppose there are two individuals with genotype Aa, and in each successive generation, two individuals are selected from the (numerous) offspring of the previous generation. These pairs of individuals form the possible states: AA and AA; AA and Aa; Aa and Aa; AA and aa; Aa and aa; aa and aa. Note that the first and last states listed are absorbing states. If the process reaches the state Aa and Aa, determine the probability that the process reaches the state AA and AA.
Recall that a child receives one allele (letter) from each parent, with a 50% probability of either allele being selected; for instance, the parents with genotype Aa will have a child with genotype AA with probability 25%, Aa with probability 50% and aa with probability 25%."
| I, too, read the question as “What is the probability of reaching state $(\text{AA},\text{AA})$ given that the process starts in state $(\text{Aa},\text{Aa})$?” Your symmetry argument works and is borne out by an explicit calculation. There are some ways to get a probability of $\frac14$, by requiring that the process pass through state $(\text{Aa},\text{Aa})$ on its way to $(\text{AA},\text{AA})$, but that seems a bit of a stretch to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Measurable subset of Vitaly set has measure zero. Proof. $E_x = \{y \in [0,1]: x-y \in \Bbb{Q}\}$,
$ \varepsilon=\{A \subset [0,1]: \exists x \quad A=E_x\} $ .We chose one element from each set of family $\varepsilon$. This is a Vitaly set $V$.
Prove that if $E$ is measurable and $E \subset V$ then $E$ has measure $0$.
$E_q = [0,1] \bigcap \Bbb{Q} $, $q \in \Bbb{Q} $
I don't know how $E$ looks. I know for example that every singleton is measurable and has measure zero. But I don't know how to explain that every measurable set of $V$ has measure zero.
| Consider
$$E_{\mathbb Q} = \bigcup_{\substack{r \in \mathbb Q \\ -1 \le r \le 1}} (E+r) \subseteq [-1,2]$$
This is a countable infinite union of disjoints subsets, each of those having the measure of $E$. If the measure of $E$ would be strictly positive, $E_{\mathbb Q}$ would have an infinite measure, in contradiction with $E_{\mathbb Q} \subseteq [-1,2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
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