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Question in John Conway's book on Functional Analysis. | What you require is Cauchy -Schwarz inequality : $|u(x,y)| \leq \sqrt {|u(x,x)|} \sqrt {|u(y,y)|}$. To prove this inequality start with $u(x+ay,x+ay) \geq 0$, expand the inner product and choose $a$ suitably. Since this is a standard argument used in all versions of Cauchy -Schwarz inequality I am omitting the details, but feel free to aks me for details if needed. |
solve quadratic equation in two variable | General Case:
These equations belong to the conic sections:
$$
Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0
$$
If $x$ and $y$ are real values you end up with curves like hyperbolas and circles.
If you restrict the unknowns to integer values, these equations are called Diophantine equations.
An example is Pell's equation:
$$
x^2-ny^2=1
$$
which is equivalent to the type of your first equation.
You might restrict its general solutions to positive ones.
For the second type of equations one ends up with Pythgorean triples:
$$
a^2 + b^2 = c^2
$$
Your Case:
For your given equations there seem to be no solutions according to WA (link and link).
First Equation:
Looking for a solution of
$$
x^2 - y^2 = 6 \\
$$
Case $x \le y$: For $x$ and $y = x + k$ for $k \ge 0$ we have
$$
6 = x^2 - (x + k)^2 = -k^2 - 2kx \le 0
$$
so no solution for $x \le y$.
Case $x > y$: For $x = y + k$ and $k > 0$ we have
$$
6 = (y + k)^2 - y^2 = k^2 + 2ky = f(k,y)
$$
So we simply try out the first few positive integer values for $k$:
$$
f(1, y) = 1 + 2y \ne 6
$$
because the left hand side is odd and the right hand side even.
$$
f(2, y) = 4 + 4 y \ge 8 > 6
$$
As $f$ is increasing with $k$, we have $f(k,y) > 6$ for $k \ge 2$, so no solution here either.
Second Equation:
Looking for a solution of
$$
x^2 + y^2 = 6 \\
$$
We can assume $x \ge y$, otherwise we rename $x$ into $y$ and vice versa.
So we have $x = y + k$ and $k \ge 0$ and get
$$
6 = (y + k)^2 + y^2 = 2y^2 + k^2 + 2k = g(k, y)
$$
We try out the first few non-negative integer values for $k$:
$$
g(0, y) = 2y^2 = 6 \iff y = \pm \sqrt{3} \\
$$
which is no integer.
$$
g(1, y) = 2y^2 + 3 = 6 \iff y^2 = 3/2
$$
which is no integer as well.
$$
g(2, y) = 2 y^2 + 8 \ge 10 > 6
$$
as $g$ is increasing with $k$, it will not get better, so no solution here either. |
When will shapes like this be closed? | This may not be a full answer.
I take this to be a question about differential geometry of curves (see my comment to the question).
I assume we can think of this "shape" as a curve in $\mathbb{R}^2$. The curve is specified by your function $f: \left] a,b \right[ \to \mathbb{R}$ where the independent variable $x\in \left] a,b \right[$ can be thought of as the arc length as measured along the curve, and the function value $f(x)$ is a measure, in radians, of the "head" direction of the tangent vector to the curve, at $x$.
$f$ cannot be too pathological, you should probably assume that it is continuous or something like that.
If $\gamma: \left] a,b \right[ \to \mathbb{R}^2$ is a parameterization, also by arc length $x$, of the curve we are interested in, then the derivative $\gamma'$ with respect to $x$ gives the tangent vector. Since we parameterize by arc length, we can also consider $x$ as the time if we say the speed is the constant $1$. So $|\gamma'| \equiv 1$.
Then I think:
$$\gamma'(x) = \begin{pmatrix} \cos f(x) \\ \sin f(x) \end{pmatrix}$$
for all $x\in \left] a,b \right[$, since this is just the unit vector in the direction of $f(x)$.
Then the curve should just be the integral of this:
$$\gamma(x) = \begin{pmatrix} \int\cos f(x)\;\mathrm{d}x \\
\int\sin f(x)\;\mathrm{d}x \end{pmatrix}$$
where the indefinite integrals involve arbitrarily choosing integration constants. For example if $0\in\left] a,b \right[$, we can just set $\gamma(0)=\begin{pmatrix} 0\\ 0 \end{pmatrix}$, without loss of generality.
Then, if we know $\gamma(x)$ for all $x$, it still remains to determine whether the curve is "closed". So what do you mean by closed?
Clearly if $\gamma(x_1)=\gamma(x_2)$ for $x_1 \ne x_2$ (i.e. $\gamma$ is not injective), it will mean that the curve self-intersects (crosses over itself). If furthermore $f(x_1)=f(x_2)$ for the same $x_1,x_2$, it could appear that the curve repeats itself (but who knows if it will keep doing so)? If furthermore you can prove that $f$ is periodic with period $(x_2-x_1)$, then the curve is "closed" and periodic.
Maybe if the interval of definition is closed, $[a,b]$, you can check if $\gamma(a)=\gamma(b)$, and if needed also check if the restriction $\left. \gamma \right| _{\left[ a,b \right[ }$ is injective.
Doing this in practice will require specific knowledge of $f$.
Found with Google: Conditions for a parametric curve to avoid self-intersection? |
Lines projective space | Let's imagine this for $\mathbb P^2$. Your setup is something like this: the affine point $(x, y)$ corresponds to the projective point with homogenous coordinates $(x, y, 1)$. This assumes the standard embedding with last coordinate $z=1$; other choices are of course possible. For homogenous coordinates, any multiple of the coordinate vector denotes the same point, so $(\lambda x, \lambda y, \lambda)$ is the same, as long as $\lambda\neq0$. To reverse the process, you dehomogenize a vector $(x, y, z)$ by first dividing by the last component to obtain $(x/z, y/z, 1)$ and the omitting the last component to end up with $(x/z, y/z)$.
Now let us imagine all of this in $\mathbb R^3$. A single homogenous coordinate vector is a point in space, but as you are not really interested in that particular vector, but instead in all non-zero multiples of that vector, you beter imagine this as a line through the origin of $\mathbb R^3$. So a point in $\mathbb P^2$ corresponds to a 1-dimensional linear subspace (i.e. a line through the origin) of $\mathbb R^3$. To turn this back into an affine point, you intersect that line with the plane at $z=1$. That point of intersection is exactly the multiple of your coordinate vector which has $1$ as it's last coordinate, i.e. it is $(x/z, y/z, 1)$.
You can perform the same considerations for objects of one dimension higher: a line in $\mathbb P^2$ is a 2-dimensional subspace (i.e. a plane through the origin) of $\mathbb R^3$. The affine interpretation of that plane can be obtained by intersecting said plane with the plane $z=1$, which yields de desired line.
So now in the $\mathbb R^3$ image, you are given two lines through the origin, and are interested in finding the plane through the origin which contains them both. This is simply the span of the directions of the lines. Intersecting that plane with $z=1$ will therefore give the line connecting the two projective points. This carries over to higher dimensions, although things become harder to visualize.
This page has a Java applet which will visualize the line connecting two points, both in the $z=1$ plane and in the enclosing $\mathbb R^3$ space. It also draws the normal of the plane, which can be computed using the cross product. The text is all in German, but the applet itself should be useful enough. |
System of equations with integration | Hint: if $\int_{-1}^1 f(t)\ dt = c$, you have $g(x) = x(x-1)+c$. What then is
$f(x)$? |
Standard parametric equation of a parabola | It's good to be attentive to some coincidences that sometimes lead to discoveries. Is this case, as G. Sassatelli has said, it looks a mere coincidence.
But in the same vein, I remember having made a similar remark about the fact that the perimeter $p(r):=2 \pi r$ of a circle look as the derivative of the formula for its area $a(r):=\pi r^2$, and it is only some years later that I understood that it was by no means a coincidence.
Explanation: Taylor expansion amounts to the description of what happens in terms of augmentation of area:
$$\tag{1}a(r+h)=a(r)+a'(r)*h+ \text{terms in} \ h^2 \cdots \text{that are neglected}.$$
to be compared to:
$$\tag{2}\pi*(r+h)^2=\pi*r^2 + 2*\pi*r*h (+ \pi h^2 \ \text{that we neglect}$$
Identification between (1) and (2) give $a'(r)*h=2*\pi*r*h$ (giving $a'(r)=2*\pi*r$) and geometricaly interpreted as the augmentation of area that, up to an error of order $h^2$, is equivalent to the area of a rectangle with length $2 \pi r$ and width $h$. See for example (http://www.askamathematician.com/2013/02/q-is-it-a-coincidence-that-a-circles-circumference-is-the-derivative-of-its-area-as-well-as-the-volume-of-a-sphere-being-the-antiderivative-of-its-surface-area-what-is-the-explanation-for-this/) (what a title !) or(Why is the derivative of a circle's area its perimeter (and similarly for spheres)?) |
Condition number on perturbations cannot cancel $||x||$ in inequality. | \begin{align*}
||x+ \delta x || \leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
\leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
\leq & \frac{1}{||(I-A^{-1}\gamma ||A||)||} (||x|| + \gamma||A^{-1}|| \cdot ||A|| \cdot ||x||) \\
\leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\
\leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\
\leq & \frac{1}{1-\gamma \mu(A)} (||x|| + \gamma \mu(A) \cdot ||x||) \\
\leq & \frac{||x|| + \gamma \mu(A) \cdot ||x||}{1-\gamma \mu(A)} \\
\leq &\frac{||x||(1 + \gamma \mu(A)) }{1-\gamma \mu(A)} \\
\frac{||x+ \delta x ||}{||x||}\leq &\frac{1 + \gamma \mu(A) }{1-\gamma \mu(A)}
\end{align*} |
Solve the equation $\tan \theta = 2\sin \theta$. | HINT:
$$\tan\theta = 2\sin\theta \iff \sin \theta = 2 \sin \theta \cos \theta \iff \sin\theta(1 - 2\cos\theta)=0$$
ADDED: I am rewriting this in a form where you can "read off" solutions. $$ab = 0\; \iff \;a = 0 \;\text{ or }\; b = 0$$ So in the case at hand, we have that
$$\sin\theta(1 - 2\cos\theta)=0 $$ $$\iff\sin \theta = 0,\;\text{ or } \; 1 - 2\cos \theta = 0 $$ $$\iff \sin \theta = 0 \;\text{ or }\; \cos\theta = \frac 12$$ |
Finite number of subgroups $\Rightarrow$ finite group | Yes, it's correct, though perhaps somewhat disorganized.
The fact that a group is a union of its cyclic subgroups holds regardless of any hypothesis on the group. So you can state and prove that first.
Then: an infinite cyclic group has infinitely many (cyclic) subgroups. So, a group is infinite if and only if it has infinitely many cyclic subgroups: If the group is finite, then it has only finitely many subsets, so finitely many subgroups, so finitely many cyclic subgroups. If the group is infinite, then being a union of its cyclic subgroups, either one of the cyclic subgroups is infinite (and hence there will be infinitely many cyclic subgroups, since that subgroup has infinitely many cyclic subgroups), or else there are infinitely many finite cyclic subgroups in the union. Either way you are done.
That is your argument, just organized a bit more. |
Understand proof involving convolution and integration | It’s actually rather easy: just notice that the integral is over $y$ and the whole space, so shifting it by $x$ won’t make any difference. (In other words, you can replace $x-y$ by $y$).
Once you have done that, it’s just a matter of definition. |
Are $\mathbb{CP}^{n}$ and $\mathbb{RP}^{2n}$ diffeomorphic? | From the comment above by @RolfHoyer.
I understand that they are homeomorphic but couldn't find a proof that they are diffeomorphic.
This is incorrect: $\Bbb{RP}^{2n}$ and $\Bbb{CP}^n$ are not homeomorphic because $\Bbb{CP}^n$ is simply connected for all $n$ whereas $\Bbb{RP}^n$ is not simply connected for any $n$. So they don't even have the same homotopy type.
Hence, $\Bbb{RP}^{2n}$ and $\Bbb{CP}^n$ are not homeomorphic and hence not diffeomorphic. |
Laplace transform of the square of first derivative | The ODE is not a linear one. The Laplace transform method is not convenient. It is simpler to change of function first :
$${\operatorname{d}\!x(t)\over\operatorname{d}\!t}=Y(t)$$
$$4 -0.1\: Y(t) - 0.01 \big(Y(t)\big)^2 = 0.9{\operatorname{d}\!Y(t)\over\operatorname{d}\!t}$$
Then, continue with the usual method to solve the ODE of separable kind.
METHOT TO SOLVE YOUR PROBLEM :
1- Solve the above equation for $Y(t)$. So, you obtain explicitly $\frac{dx(t)}{dt}$.
2- Put $\quad Y(t)\quad $ into your second equation $\quad 0.9\frac{d^2y}{dt^2}=0.13\left(Y(t)\right)^2-9\quad $.
So, you obtain $\quad \frac{d^2y}{dt^2}=$ an explicit fonction of $t$.
3- Integrate two times this function of $t$ which leads to $y(t)$.
NOTE : Since the initial system of two equations includes numerical values which look like coming from a physical problem, the analytical solving might be not the simplest method. The above result on the form of functions is complicated for further use in numerical applications. Why not solving the ODEs directly with numerical calculus ? |
Best book on Sequences and Series | I agree with Ted Shifrin on Spivak's Calculus as a strong book on sequences and series (and he should know,he wrote most of the exercises for it from the second edition onwards.......lol) .
Most elementary analysis texts contains chapters on sequences and series, but most of the best sources on the subject are the older pre-Rudin analysis texts. One of my favorites is sadly long out of print-it's W. Ferrar's A Textbook of Convergence. If you can locate a copy of this forgotten classic, you definitely should do so. Another very good source along these lines is Titchmarsh's Theory of Functions,which also contains excellent classical treatments of measure theory and complex variables. |
Probability that a < b < c | [Completely basic probability point of view]
Given a set of $3$ numbers from the bag, what is the probability that the numbers are arranged the way you want? (Conditional)
Now use the theorem of total probability.
Same way, you can actually answer the question for drawing $k$ numbers out of the bag. |
Showing $U(f,P) - L(f,P) \le K||P||(b-a)$ | Note that by the mean value theorem, for any $x,y \in [x_{r-1},x_r]$ there exists $\xi$ between $x$ and $y$ such that
$$|f(x) -f(y)| = |f'(\xi)||x-y|\leqslant|f'(\xi)|(x_r - x_{r-1}) \leqslant K(x_r - x_{r-1}) \leqslant K\|P\|,$$
and
$$M_r - m_r = \sup_{x,y \in [x_{r-1},x_r]}|f(x) - f(y)| \leqslant K \|P\|$$
Thus,
$$U(f,P)-L(f,P) = \sum_{r=1}^n(M_r-m_r)(x_r-x_{r-1}) \leqslant \sum_{r=1}^nK\|P\|(x_r-x_{r-1}) = K\|P\|(b-a)$$ |
Can someone help me find the sum of the following series? | It is a geometric series. Pull out a factor $\frac19$ from the sum and you should see that the general term under the sum is $\frac{4^i}{9^i}$. |
Interpreting estimated percentage as a distribution | While this seems like a natural thing to do, I cannot wrap my head around what it means. In particular, $p$ is not actually a distribution, but an unknown constant.
Your thought process is exactly correct and leads to discussing Bayesian statistics.
Note that $p$, even though you think of it as having its own distribution, is actually conditional on the data $X_1, \dots, X_n$, where $X_1, \dots, X_n$ are Bernoulli distributed with probability $p$.
Thus, by Bayes' Theorem, we obtain (and I'm going to use $p_0$ for a value in the support of $p$):
$$f_{p \mid X_1, \dots, X_n}(p_0 \mid x_1, \dots, x_n)=\dfrac{f_{X_1, \dots, X_n \mid p}(x_1, \dots, x_n \mid p_0) \cdot f_{p}(p_0)}{f_{X_1, \dots, X_n}(x_1, \dots, x_n)}$$
Assuming that $X_1, \dots, X_n$, when conditioned on $p$, are independent, we may write
$$\begin{align}
f_{p \mid X_1, \dots, X_n}(p_0 \mid x_1, \dots, x_n)&=\dfrac{f_{X_1\mid p}(x_1 \mid p_0) f_{X_2 \mid p}(x_2 \mid p_0) \cdots f_{X_n \mid p}(p_0) \cdot f_{p}(p_0)}{f_{X_1, \dots, X_n}(x_1, \dots, x_n)} \\
&= \dfrac{p_0^{t}(1-p_0)^{n-t}f_p(p_0)}{c} \\
&\propto p_0^{t}(1-p_0)^{n-t}f_p(p_0)
\end{align}$$
where $c$ is a constant independent of $p_0$, and $t$ is the number of random variables, out of $X_1, \dots, X_n$, which result in the "success" with probability $p$. I use $\propto$ to mean "proportional to;" we don't need to worry about constants with respect to $p_0$ for now.
The constant $c$ isn't really that important. However, what remains is the important problem of assigning $f_p$: one popular model is to assume the Beta distribution for $p$ (known as a "prior" for $p$), for which
$$f_p(p_0) \propto p_0^{\alpha - 1}(1-p_0)^{\beta - 1}$$
so thus
$$f_{p \mid X_1, \dots, X_n}(p_0 \mid x_1, \dots, x_n) \propto p_0^{t+\alpha - 1}(1-p_0)^{n-t+\beta - 1}\text{.}$$
Since we know that $p_0 \in (0, 1)$, this is proportional to a Beta distribution! Thus, the result is as follows:
Suppose $X_1, \dots, X_n$ are Bernoulli distributed with success probability $p$, and are independent given $p$. Suppose also that $p$ follows a Beta distribution with parameters $\alpha, \beta$, known as the prior of $p$. Let $t < n$ be the number of $X_1, \dots, X_n$ which result in a success. Then
$$p \mid X_1, \dots, X_n \sim \text{Beta}(t+\alpha, n-t+\beta)\text{.}$$
This is known as the posterior distribution of $p$ given $X_1, \dots, X_n$, assuming a Bernoulli likelihood for $X_1, \dots, X_n$ and a Beta prior for $p$. In particular, because the Bernoulli likelihood and the Beta prior follow the same form (when ignoring constants with respect to $p$), this formulation is also known as the Bernoulli-Beta conjugate prior. I strongly recommend you read more into this interesting subject. |
Number of vectors in a span over a certain field | I would say the answer is $2^n$, but your reasoning is a bit off. First off, vector spaces are commutative under addition, so the elements you say are different, actually aren't. There is no need to consider rearrangements.
The reason why I say $2^n$ is because take $s \in \text{span }S$. Then
$$s = \sum_{i=1}^n a_i u_i.$$
So you have two choices for each $a_i$, namely $1$ or $0$. Since there are $n$ elements, with two choices each, you have $2^n$ elements. |
How many eigenvalues does $f$ have if ${f}^{m}$= Id for some $m$ | The mth power of $f$ is associated to the composition of the application with itself, repeated m times, whose associated matrix is $A^m$ if $A$ is the matrix for $f$.
Therefore if $m = 2$ and $\vec{v}$ is an eigenvector and $\lambda$ its eigenvalue it will happen that $f \circ f$ has as eigenvalue $\lambda^2$ as $f(f(\vec{v})) = f(\lambda\vec{v}) = \lambda(f(\vec{v})) = \lambda(\lambda\vec{v})= \lambda^2\vec{v}$. This can be generalized. If $m$ is generic then $f(f(f(....f(\vec{v}) ... ))) = \lambda^m\vec{v}$.
If the composition results in the identity matrix then $f(f( .... f(\vec{v}) ... )) = 1\vec{v} = \lambda^m\vec{v}$
All the complex solutions for $\lambda^m = 1$ are possible eigenvalues.
I could be wrong cause I did this long time ago, so double check everything. |
for poisson distribution, show that $P(0<X<2(\lambda +1)) \ge \frac \lambda{\lambda +1}$ | So it turned out I was doing some calculation mistakes before. As suggested in the question, I got the answer this way:
$$P(0<X<2(\lambda +1))=1-P(X=0)+P(X\ge 2\lambda +2)$$
$$\implies P(X=0)+P(X\ge 2\lambda +2)\le 1-\frac{\lambda}{1+\lambda}$$
$$\implies e^{-\lambda}+e^{-\lambda}\sum_{k=2\lambda +2}^{\infty}\frac{\lambda^{k}}{k!}\le \frac 1{\lambda +1 }$$
multiplying with $e^{\lambda}$ both sides and writing taylor series of $e^{\lambda}$ on the RHS gives the required answer.
{also use $\lambda \ge 1$ as this is trivially satisfied at $\lambda=0$.} |
Geometry problem: Line intersecting a semicircle | In this kind of problem, it is inevitable that plain old analytic geometry will work. A precise version of this assertion is an important theorem, due to Tarski. If "elementary geometry" is suitably defined, then there is an algorithm that will determine, given any sentence of elementary geometry, whether that sentence is true in $\mathbb{R}^n$.
So we might as well see what routine computation buys us. We can take the equation of the circle to be $(x+1)^2+y^2=1$, and the equation of the line to be (what else?) $y=mx+b$. Let our semicircle be the upper half of the circle.
Substitute $mx+b$ for $y$ in the equation of the circle. We get
$$(1+m^2)x^2+2(1+mb)x +b^2=0. \qquad\qquad(\ast)$$
Let the root nearest the origin be $r_1$, and the next one $r_2$. Note that the line meets the $x$-axis at $x=-b/m$.
From the geometry we can deduce that $-r_2=-2r_1$ and $b/m=-3r_1$, and therefore
$$r_1=-\frac{b}{3m} \qquad\text{and} \qquad r_2=-\frac{2b}{3m}.$$
By looking at $(\ast)$ we conclude that
$$-\frac{b}{m}=-\frac{2(1+mb)}{1+m^2} \qquad\text{and} \qquad \frac{2b^2}{9m^2}=\frac{b^2}{1+m^2}.$$
Thus the algebra gives us the candidates $m=\pm\sqrt{\frac{2}{7}}$. (Of course, the first equation was not needed.)
Sadly, we should not always believe what algebraic manipulation seems to tell us. I have checked out the details for the positive candidate for the slope, and everything is fine. Our line has equation $y=\sqrt{\frac{2}{7}}x+ \frac{2\sqrt{14}}{5}$. Pleasantly, the points $r_1$ and $r_2$ turn out to have rational coordinates.
However, the negative candidate is not fine. That can be checked by looking at the geometry. But it is also clear from the algebra, which has been symmetrical about the $x$-axis. The algebra was not told that we are dealing with a semicircle, not a circle. So naturally it offered us a mirror symmetric list of configurations.
We conclude that the slope is $\sqrt{\dfrac{2}{7}}$. |
Catalan Numbers: lattice paths with n+1 steps | The two paths enclose an array of unit squares forming a convex polyomino. (Note that in this context convex has a special meaning.) Say that the region has $m$ columns, with $c_k$ squares in column $k$. For $k=1,\ldots,m-1$ let $r_k$ be the number of rows shared by columns $k$ and $k+1$. The upper righthand corner is at the point $\langle m,n+1-m\rangle$.
The idea is to use these numbers to construct a Dyck path (mountain range) of length $2n$. It will have $m$ peaks, of heights $c_1,\ldots,c_m$ from left to right. For $k=1,\ldots,m-1$ the descent after peak $k$ will have length $c_k-r_k+1$, while the descent after peak $m$ will of course have length $c_m$.
The valley between peak $k$ and peak $k+1$ will have height $c_k-(c_k-r_k+1)=r_k-1$, so the ascent from it to peak $k+1$ will have length $c_{k+1}-r_k+1$.
Now $c_{k+1}-r_k$ is the number of squares by which the top of column $k+1$ rises above the top of column $k$, so
$$c_1+\sum_{k=1}^{m-1}(c_{k+1}-r_k)+c_m$$
is the height above the baseline of the top of column $m$, which is $n+1-m$, and
$$c_1+\sum_{k=1}^{m-1}(c_{k+1}-r_k+1)+c_m=n+1-m+(m-1)=n\;,$$
as desired.
I leave it to you to check that the total descents is also $n$, that the path never drops below the baseline, and that the convex polyomino and hence the original two paths can be recovered from the Dyck path. |
Determine $X-Y$ offset for rotated element given the $Y$-offset without rotation | Let $y$ be the initial vertical offset from the center of rotation. Let $x'$ and $y'$ be the offsets after rotation. Let $\theta$ be the angle through which the text is rotated clockwise.
$$x' = -y\cdot \sin \theta$$
$$y' = y\cdot \cos \theta$$ |
Simplifying $9^{3/4}$, I get $3\sqrt[4]{9}$, but that's not the answer. Why? | We know that $9=3^2$ .So, $$\sqrt [4]{9^3} =\sqrt [4]{(3^2)^3} =\sqrt [4]{3^2*3^2*3^2} =\sqrt {3*3*3*3*3*3} $$ After this you have proceeded correctly. You can simplify the last step as: $$ 3\sqrt [4]{3^2} =3\times 3^{2/4} =3\times 3^{1/2} =3\sqrt {3} $$ Hope it helps. |
Getting at least $k$ heads in a row $l$ times over after at most $m$ flips. | Let's take a look at a minimal example:
$$P = \frac 1 2\left[\begin{array}{ccc|ccc}
2&1&0&0&0&0\\
0&0&1&0&0&0\\
0&1&1&\color{red} 2&0&0\\\hline
0&0&0&\color{red} \uparrow&1&0\\
0&0&0&0&0&1\\
0&0&0&0&1&1
\end{array}\right]$$
Basic "building blocks" as in the answer here on the "diagonal" we can easily implement with Kronecker product with $\bf I$, the "storage states" are the 2s.
Now let but the upper leftmost block have storage state displaced 1 row to the block above.
This way each displacement will slow down (1 lower exponent than the previous part of the chain).
Any elegant way to avoid the displacement latency will be welcome!
Crazy checker (first time we get non-zero prob for 1 and 2 resp):
$${\bf v} = \left[\begin{array}{cccccc}0&0&0&0&0&1\end{array}\right]^T$$
$${\bf P}^2{\bf v} = \left[\begin{array}{cccccc}
0&0&0&0.25&0.25&0.5\end{array}\right]^T\\{\bf P}^5{\bf v} = \left[\begin{array}{cccccc}0.0625&0.125&0.3125&0.09375&0.15625&0.25\end{array}\right]^T$$
The chance for 2 heads in a row is $1/4 = 0.25$
The chance for 4 heads in a row is $1/16 = 0.0625$
So assuming $H{\bf H}H{\bf H}$ counts as two heads in a row twice then the solution works!
We can also calculate the probability of not having any two H in a row in a string of 2 and 5 respectively, and if we do, we do indeed get the sum of the two last states: $$1-\frac 1 4\approx 0.25+0.5 = 0.75 \\ 1-\frac {19}{32} \approx 0.15625+0.25= 0.40625$$
The systematic construction of matrices for arbitrary $k,l,m$ should now be obvious. |
How can I find the limit using L'Hospital? | The ratio of the derivatives is $$\dfrac{\dfrac{2^x \ln 2}{2^x + 1}}{\dfrac{3^x \ln 3}{3^x + 1}}.$$ This simplifies to $$\frac{2^x (3^x + 1)}{3^x (2^x + 1)} \cdot \frac{\ln 2}{\ln 3}.$$
You can evaluate the limit of the first fraction by first multiplying by $\dfrac{6^{-x}}{6^{-x}}$. |
In the following answer, since we assumed $a<b$, then how can we write $a \leq b$ in $\gcd(a,b)\leq a\leq b\leq\operatorname{lcm}(a,b)$ statement? | $a<b$ means that $a \le b$.
This is kind of intuitive, but if you want a sort of more formal proof, here (it's still pretty intuitive anyways).
Rearrange $a<b$ gets $0<b-a$, so $b-a$ is positive. Rearranging $a\le b$ gets us $0\le b-a$. But we know that $b-a$ is positive, so then $0\le b-a$ is true.
For example, as @fleablood stated in the comments, $2<5$ and $2\le 5$ are both true. |
Step in the proof of the Fundamental Theorem of Calculus | Since $f$ is continuous at $x$, there exists $r\gt 0$ such that for all $t$, if $t\in I$ and $|t-x|\lt r$, then $|f(t)-f(x)|\leq 1$.
Therefore $|f(t)|-|f(x)| \leq |f(t)-f(x)|\leq 1$, so $|f(t)|\leq 1+|f(x)|$. Set $K=1+|f(x)|$.
Intuitively: if $t$ is sufficiently close to $r$ (and in $I$), then $f(t)$ is close enough to $f(r)$. So all you need to to is pick a $K$ so that things "sufficiently close to $f(r)$" have absolute value less than $K$. |
Where to start with the Inverse Galois Problem | Great! The inverse Galois problem has always fascinated me.
A list of references is given in this answer. I particularly recommend Völklein's book, which is meant as an introduction. As he mentions, the basic techniques require only knowledge of introductory Galois theory, (finite) group theory, some algebra, and complex analysis since the theory begins by having groups realized via covering spaces of Riemann surfaces. The basic approach, dating back to Hilbert, who was the first to work seriously on the problem, is the irreducibility theorem, that ensures that groups realized over $\mathbb Q(x)$ are realized over $\mathbb Q$.
Deeper results require more background: An understanding of the theory of simple groups, for example, and techniques from algebraic geometry. There is the hope that the Classification of finite simple groups will allow an "inductive" solution of the inverse Galois problem, and this has greatly influenced modern research on the question. Some results, such as Shafarevich's realization of all solvable groups over $\mathbb Q$, are more ad hoc. Shafarevich's result is essentially number theoretic, for example. |
Modulo arithmetic: finding “c” when result of right-hand side expression is a float | \begin{array}{rr|rrl}
& 43 & 1 & 0 \\
-2 & 19 & 0 & 1 & (43 - 38 = 5, \, 1-0=1, \, 0-2=-2)\\
\hline
-4 & 5 & 1 & -2 & (19 - 20 = -1, \,0-4 = -4, \, 1+8 = 9) \\
& -1 & -4 & 9 \\
\hline
\end{array}
So \begin{align}
-1 = -4(43) + 9(19)
& \implies 1 = 4(43) - 9(19) \\
& \implies 1 \equiv -9(19) \pmod{43} \\
& \implies 19^{-1} \equiv -9 \pmod{43} \\
\end{align}
Given $a = 27$ and $b=19$, then
$c \equiv a^5 \cdot b^{-3} \equiv 27^5 \cdot 19^{-3} \equiv 1\pmod{43}$ |
pattern in decimal representation of powers of 5 | Write $a\%n$ for the least non-negative integer congruent to $a$ modulo $n$, i.e. $a\%n=r$ iff $a\equiv r\pmod{n}$ and $0\le r<n$.
In particular, $a\% 10^k$ is the number represented by the last $k$ digits of $a$.
Note that the cycle of the last four digits contains $625,3125,5625,8125$, which are $1,5,9,13$ times $5^4$. Similarly the cycle of the last five digits contains
$$
\begin{array}{ll}
3125 = 1\times 5^5 & 53125 = 17\times 5^5 \\
15625 = 5\times 5^5 & 65625 = 21\times 5^5 \\
28125 = 9\times 5^5 & 78125 = 25\times 5^5 \\
40625 = 13\times 5^5 & 90625 = 29\times 5^5
\end{array}
$$
In general, these cycles have the following properties for $k>1$:
P1: $5^n\%10^k$ repeats in a cycle of length $2^{k-2}$ for $n\ge k$.
P2: These sets, which we will denote by $S_k$, are the same, formalizing the pattern alluded to above:
$$S_k = \left\{5^n\% 10^k\right\}_{n=k}^{k+2^{k-2}-1}
= \left\{(4n+1)5^k\right\}_{n=0}^{2^{k-2}-1}$$
Let $$A_k = \sum_{x\in S_k} x \\
B_k = \sum_{x\in S_k} \left\lfloor \frac{x}{10^{k-1}}\right\rfloor = \sum_{x\in S_k} \frac{x - (x\%10^{k-1})}{10^{k-1}}
$$
Then $B_k$ is the sum of the digits in the number called the "period" in the question, which is congruent to the number modulo $9$.
From P1, for $k>2$ one cycle of the last $k$ digits contains two cycles of the last $k-1$ digits, so
$$
A_k = 10^{k-1} B_k + 2 A_{k-1} \\
B_k \equiv A_k-2A_{k-1} \pmod {9}
$$
that is, we can break down $A_k$ into two sums of the last $k-1$ digits and a sum of the first digits times $10^{k-1}$.
From P2 we can evaluate $A_k$:
$$
\begin{align}
A_k & = \sum_{n=0}^{2^{k-2}-1}(4n+1)5^k \\
& = 5^k\left[ \left(4\sum_{n=1}^{2^{k-2}-1} n\right) + \sum_{n=0}^{2^{k-2}-1} 1 \right] \\
& = 5^k\left(4 \frac{2^{k-2}(2^{k-2}-1)}{2} + 2^{k-2}\right) \\
& = 25\cdot 10^{k-2} \left(2^{k-1}-1\right) \\
& \equiv 7 \left(2^{k-1}-1\right) \pmod{9}
\end{align}
$$
from which it follows that
$$
\begin{align}
B_k & \equiv A_k-2A_{k-1} \\
& \equiv 7 (2^{k-1}-1) - 2\cdot 7(2^{k-2}-1) \\
& \equiv 7 \pmod {9}
\end{align}
$$
establishing the desired result.
Now to fill in the blanks we'll prove P1 and P2.
First $2^n\not\mid n!$. The power of $2$ in $n!$ is $v_2(n!)=\lfloor n/2\rfloor + \lfloor n/4 \rfloor + \lfloor n/8 \rfloor + \cdots<n$ (see e.g. this).
Hence $$2^n \left\vert \binom{2^n}{i} 2^i \right. = \frac{2^n C}{i!} 2^{i} $$
for some $C\in \mathbb{Z}$.
Then using the binomial theorem
$$
5^{2^{k-3}} = (1+4)^{2^{k-3}} = 1+2^{k-3}\cdot 4 + \sum_{i=2}^{2^{k-3}}\binom{2^{k-3}}{i} 4^i \equiv 2^{k-1}+1 \pmod {2^k} \\
5^{2^{k-2}} = (1+4)^{2^{k-2}} = 1 + \sum_{i=1}^{2^{k-2}}\binom{2^{k-2}}{i} 4^i \equiv 1 \pmod {2^k}
$$
which establishes that the order of $5$ in $\mathbb{Z}/2^k\mathbb{Z}^{\times}$ is $2^{k-2}$ (since it must also divide $2^{k-1}$).
Hence for $1<k\le n < k+2^{k-2}$, we always have $5^n\%5^k=0$, and $5^n\%2^k$ takes on exactly $2^{k-2}$ distinct values; by the Chinese Remainder Theorem $5^n\%10^k$ also takes on exactly $2^{k-2}$ distinct values. This establishes P1.
Furthermore for $1<k\le n < k+2^{k-2}$ every $5^n\equiv (4j+1)5^k\pmod{10^k}$ for some $j$, and there are exactly $2^{k-2}$ such possibilities, so every one must occur for exactly one choice of $n$ in this range. This establishes P2. |
What does the letter U mean in math? | It represents the union of all the sets $A_\alpha$. By the "union", what it basically means that:
$$ a \in \bigcup_{\alpha} A_\alpha \iff \exists \alpha :a \in A_\alpha $$
or in words - it is the set of elements which are contained in any of the $A_\alpha$. |
Closed Intervals of $\omega_1$ Are Compact | If you want to use a more general approach: for ordered spaces, compactness = order completeness (every subset has a supremum and an infimum), and for closed intervals of $\omega_1$ (where the induced topology = the order topology) this follows easily: we have a min for every non-empty subset, so certainly an infimum, and being bounded above gives us a minimum for the set of upper bounds hence a supremum as well, which by being a closed interval must be in the closed interval too. |
Functional Analysis: when is addition/multiplication of a closure not the same as the closure of the addition/multiplication? | First of all, we must have equality between $\lambda \overline{A}$ and $\overline{\lambda A}$. Continuity of scalar multiplication implies that
$$\lambda \overline{A} \subseteq \overline{\lambda A}.$$
If $\lambda \neq 0$, then multiplying both sides by $\lambda^{-1}$,
$$\overline{A} \subseteq \lambda^{-1}\overline{\lambda A} \subseteq \overline{\lambda^{-1}\lambda A} = \overline{A},$$
hence
$$\overline{A} = \lambda^{-1}\overline{\lambda A} \implies \lambda \overline{A} = \overline{\lambda A}.$$
When $\lambda = 0$, the equality clearly holds, as both sides are equal to $\{0\}$.
Now, the corresponding equality for summation need not hold. In order to figure out an example, it helps to know that if $\overline{A}$ or $\overline{B}$ are compact, then equality does indeed hold. To prove this, suppose $x_n \in A + B$ converges to some $x$. Then there exist sequences $a_n$ in $A$ and $b_n$ in $B$ such that $x_n = a_n + b_n$. If $\overline{A}$ or $\overline{B}$ is compact, then we can simply take a subsequence $n_k$ so that $a_{n_k}$ or $b_{n_k}$ converges. If $a_{n_k} \to a \in \overline{A}$, then
$$b_{n_k} = x_{n_k} - a_{n_k} \to x - a \in \overline{B},$$
hence $x = a + (x - a) \in \overline{A} + \overline{B}$. Similar logic works when $b_{n_k}$ converges to some $b \in \overline{B}$.
So, in order to get a counterexample, you'll need to consider the sum of two sequences, in $A$ and $B$ respectively, each without any convergent subsequence. Here's the classic example:
Consider the space $\Bbb{R}^2$ and let
\begin{align*}
A &= \{(x, y) : x > 0 \text{ and }y \ge 1/x\} \\
B &= \{(x, y) : x < 0 \text{ and }y \ge -1/x\}.
\end{align*}
Note that $A$ and $B$ are closed. Also, we have $a_n = (n, 1/n) \in A$ and $b_n = (-n, 1/n) \in B$, but
$$a_n + b_n = (n, 1/n) + (-n, 1/n) = (0, 2/n) \to (0, 0) \in \overline{A + B}.$$
However, $(0, 0) \notin \overline{A} + \overline{B} = A + B$, as every point in both $A$ and $B$ has a strictly positive $y$ coordinate. So, in this case, we have a strict subset, rather than equality. |
What are "set-theoretic maps"? | "Set-theoretic map" just means "function". The term "set-theoretic" is used to stress that any function between the two sets is allowed, without any further restrictions such as continuity or differentiability. This usage comes from category theory, where the term "map" can generically refer to morphisms in an arbitrary category, so "set-theoretic map" is referring to a morphism in the category of sets (as opposed to the category of topological spaces, or the category of smooth manifolds, for instance), which is just any old function.
(Other variants on the term "set-theoretic map" used in this way that you might encounter include "map of sets" or "set map".) |
Gelfand duality restricted to the category of Stone spaces | Here is a clean and short way to prove Stone duality:
show that the category of Stone spaces is the category of pro-objects in finite sets (profinite sets),
show that the category of Boolean rings is the category of ind-objects in finite Boolean rings,
show that the category of finite Boolean rings is equivalent to the opposite of the category of finite sets.
Now, under Gelfand duality, the opposite of the category of finite sets is also equivalent to the category of finite dimensional unital commutative C*-algebras. So it follows that the opposite of the category of profinite sets is also equivalent to the category of ind-finite dimensional unital commutative C*-algebras.
According to Wikipedia this condition is equivalent to being a (unital, commutative) AF C*-algebra. (A priori this condition is slightly more restrictive but Wikipedia claims it's equivalent to the spectrum being totally disconnected, so...) |
For standard deviation of grouped data shouldn't you have to convert frequencies given in decimals (in millions) to whole numbers. | I think I answered my own question now - the instructor is incorrect. Dividing by ( n-1 ) in the first photo is dividing by 9.4 million. You don't want to do that. You do not have one million degrees of freedom. Also, who has ever taken a sample size of 10.4 million? Even, if you somehow did that is such a large sample that the standard deviation is essentially the same as the population standard deviation, as shown in the second photo. |
Proving intersection of dense subsets of a metric space X is the isolated points of X. | If $x$ is an isolated point of the space $X$ then any dense subset of $X$ contains $x$, so
$ iso(X)\subset\bigcap\mathscr C$. From the other side, if $x$ is not an isolated point of the space $X$, then the set $X\setminus \{x\}$ is dense in $X$, so $x\not\in\bigcap\mathscr C$. |
Number of ordered triples with given condition | Given that $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ and $ 3\,|\,2^m+2^n+2^p$ where $m,n,p$ are integers.
Now if $2^m+2^n+2^p$ is divisible by $3$, then each of $2^m$, $2^n$, $2^p$ is divided by $3$.
Remainder of ${2^m}$ when it is divisible by $3$ is $(-1)^m$.
So remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}$. Now if ${(-1)^m+(-1)^n+(-1)^p}$ is divisible by $3$, the we are done.
Case I: If $m$ is odd and $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1+1+1}={1}$ which is not divisible by $3$.
Case II: If $m$ is even and $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1-1-1}={-1}$ which is not divisible by $3$.
Case III: If $m$, $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1+1+1}={3}$ which is divisible by $3$.
Case IV: If $m$, $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1-1-1}={-3}$ which is divisible by $3$.
So we are interested in the last two cases.
Now $1≤m≤100,\;1≤n≤50,\;1≤p≤25$
In the given range, the number of possible choice that all $m, n, p$ are even is $50, 25, 12$ respectively and the number of possible choice that all $m, n, p$ are odd is $50, 25, 13$ respectively.
So the total number of ordered triples $(m,n,p)$ where $m,n,p$ are integers that $ 3\,|\,2^m+2^n+2^p$ is $(50 \times 25 \times 12)+(50 \times 25 \times 13)=50 \times 25 \times 25= 31250 = 5\times 6250$.
So the possible value of $k=5$ . |
Counting the number of ways to get a sum of $2012$, given that any number of cards can be chosen | Recolour the cards of value $1$ blue and green. Let $a, b,c$ be the sum of the numbers on the blue, green and red cards respectively. Making use of the unique representation of numbers using binary digits:-
The problem is equivalent to finding the number of ordered triples of non-negative integers $(a,b,c)$ such that $a+b+c=2012$, where $c$ must be even.
Considering all the possibilities for $c$, gives $$1+3+5+ ... +2013=1007^2.$$ |
Relationship between a solution in $\mathbb{C}$ to a finite set of polynomials over $\mathbb{Z}$ and solutions in finite fields | It is probably possible to give a purely algebraic proof of this statement, but it also follows easily using standard facts from mathematical logic. So I hope you don't mind if I use this answer as an advertisement for the field of model theory.
First, note that a system of polynomial equations in $\mathbb{Z}[x_1,\dots,x_n]$ has a solution in a finite field of characteristic $p$ if and only if it has a solution in $\overline{\mathbb{F}_p}$, the algebraic closure of the prime field of characteristic $p$. Indeed, every finite field $K$ of characteristic $p$ embeds in $\overline{\mathbb{F}_p}$, and the image of a solution in $K$ under such an embedding is a solution in $\overline{\mathbb{F}_p}$. And conversely, given a solution $(a_1,\dots,a_n)\in (\overline{\mathbb{F}_p})^n$, this is also a solution in the finite subfield $\mathbb{F}_p[a_1,\dots,a_n]\subseteq \overline{\mathbb{F}_p}$.
So we have reduced to showing that a system of polynomial equations in $\mathbb{Z}[x_1,\dots,x_n]$ has a solution in $\mathbb{C}$ if and only if it has a solution in $\overline{\mathbb{F}_p}$ for infinitely many values of $p$. In fact, it turns out that this is also equivalent to having a solution in $\overline{\mathbb{F}_p}$ for all but finitely many values of $p$.
Now the reason this three-way equivalence is true is that given a system $\{p_1,\dots,p_k\}$ of polynomials with coefficients in $\mathbb{Z}$ (which we can assume to be finite, by Noetherianity of $\mathbb{Z}[x_1,\dots,x_n]$, as noted in the comments), we can express the existence of a solution by a sentence of first-order logic in the language of fields: $$\exists x_1\dots\exists x_n\,(p_1(\overline{x}) = 0\land \dots \land p_k(\overline{x}) = 0).$$ Now we can apply:
Theorem: Any sentence of first-order logic in the language of fields is true in $\mathbb{C}$ if and only if it is true in $\overline{\mathbb{F}_p}$ for infinitely many primes $p$ if and only if it is true in $\overline{\mathbb{F}_p}$ for all but finitely many primes $p$.
A closely related application of the Theorem is the Ax-Grothendieck theorem, the model-theoretic proof of which (due to Ax) lifts the fact that every injective function from a finite set to itself is surjective up to polynomial maps between algebraic varieties over $\mathbb{C}$.
The proof of the above Theorem is usually phrased using the compactness theorem. But here's another way to think of it, which may seem more intuitive if you're less familar with mathematical logic. By $\text{ACF}_0$, I mean the axioms of algebraically closed fields of characteristic $0$. These are the field axioms, together with an axiom for each degree $d$ saying that all non-constant polynomials of degree $d$ have roots, and finally an infinite list of axioms saying the characteristic is not any prime: $2\neq 0$, $3\neq 0$, $5\neq 0$, etc. Now it's a fact that $\text{ACF}_0$ is a complete theory, meaning that we can prove or disprove every sentence in the language of fields using the axioms of $\text{ACF}_0$. [This is a nontrivial fact, which is where all the algebra in the proof is hiding. You can also understand this as saying that if a sentence holds of one algebraically closed field of characteristic $0$, then it holds of all such fields, i.e. a manifestation of the Lefschetz principle.] In particular, if a sentence $\varphi$ is true in $\mathbb{C}$, then it must be provable from the axioms of $\text{ACF}_0$. But the proof of $\varphi$ will only use finitely many of the characteristic axioms (proofs being finite). So it follows that $\varphi$ is provable from the axioms of algebraically closed fields, together with finitely many axioms of the form $p\neq 0$ for finitely many primes $p$. Hence $\varphi$ is true in all algebraically closed fields except possibly for those whose characteristics are among these finitely many exceptional primes.
Of course, if $\varphi$ is true in the $\overline{\mathbb{F}_p}$ for all but finitely many primes $p$, then it is true in the $\overline{\mathbb{F}_p}$ for infinitely many primes $p$ (by Euclid's theorem). To complete the chain of equivalences, note that by the argument above, if $\varphi$ is true in infinitely many of the $\overline{\mathbb{F}_p}$, then it can't be that $\lnot \varphi$ is true in $\mathbb{C}$, otherwise we could run the argument in the last paragraph for the sentence $\lnot \varphi$, and find that $\lnot\varphi$ is true in all but finitely many of the $\overline{\mathbb{F}_p}$. Thus $\varphi$ must be true in $\mathbb{C}$. |
Optimization Calculus.. a box/shelter with sides missing.. | This looks about right to me. You would need to set the 1st derivative you found to zero, and find the critical value. To make sure it is a minimum, you would then calculate the 2nd derivative. If the 2nd derivative with the critical value put in is positive, then you have a minimum. |
Explicitly write down $g\in GL(n,\mathbb{C})$ so that $gAg^{-1}$ is upper triangular, where $A\in M_n(\mathbb{C})$ | An algorithm for finding $g$, with the added condition of taking $g$ to be unitary, is given in Hogben's Handbook of linear algebra. |
Language concatenation | Good question. The problem is that $L'$ isn't the concatenation $L\circ L$, since it's made up to the concatenation of each word in $L$ with itself. It would be if we had defined it as
$$
\{xy\mid x\in L, y\in L\}
$$
but a string is in $L'$ if it's of the form $ww$: two copies of a word in $L$. That's clearly not a regular language in general, since you could take as an example $L=\{\mathtt{0}^n\mathtt{1}\mid n\ge 0\}$ and then
$$
L'=\{\mathtt{0}^n\mathtt{1}\mathtt{0}^n\mathtt{1}\mid n\ge 0\}
$$
which a simple Pumping Lemma proof would show that isn't regular. Intuitively, the language above would require a finite automaton that could count the number of zeros in a string, which we know is impossible. |
Simple question about derivative under Integral Sign | this is called "integration by subsitution" (google it, or look in your lecture notes).
In this case you would write:
\begin{align}
\frac{dy}{dt} &= y' \Rightarrow dy = y' dt \\
\int\frac{y'(t) y(t)}{y^2(t)+1} dt &= \int\frac{y}{y^2+1}dy
\end{align}
Note that on the right side there is no $t$ anymore. The integration is directly over $y$ as a new variable. |
What is the name of this theorem of lemma? | The theorem that equal angles at the circumference of a circle stand on equal arcs is Euclid III,26. Hence the bisector of $\angle BAC$ bisects arc $BC$. It will pass through the circumcenter, however, only if $AB=AC$, since then it will be the perpendicular bisector of $BC$ (see Euclid III,1 Porism; VI,3). |
Maximizing Area of two squares in a circle | We consider the general case shown in $\bf{Fig. 1}$. Use the triangle $OQC$ to obtain $OC$ and $OA$.
$$OC=\frac{\sqrt{4 r^2-h^2}}{2},\space\space\space\space\therefore\space\space OA=AC-OC=s+h-\frac{\sqrt{4 r^2-h^2}}{2}.$$
Apply Pythagorean theorem to the triangle $OPA$ to express $s$ in terms of $r$ and $h$.
$$r^2=\left(\frac{s}{2}\right)^2+\left(s+h-\frac{\sqrt{4r^2-h^2}}{2}\right)\space\to\space s=\frac{4\sqrt{4 r^2-h^2}-3 h}{5} $$
The required area is given by,
$$A(S_1+S_2)=h^2+s^2=\frac{2}{25}\left(32 r^2+9 h^2-12 h\sqrt{4 r^2-h^2}\right)\tag{1}$$.
Now, we go on to find the minima ans maxima of (1) by equating its first derivative to zero.
$$\frac{d}{dh}\left(A\left(S_1+S_2\right)\right)=3 h\sqrt{4 r^2-h^2}-2\left(4 r^2-h^2 \right)+2 h^2=0$$
To make thing easy, we let $y=\sqrt{4 r^2-h^2}$ and obtain,
$$3 h y-2 y^2+2 h^2=\left(2 h-y\right)\left(h+2 y\right)=0.$$
From this, it is obvious that $y$ has only one acceptable value, i.e.
$$h=\frac{y}{2}=\frac{\sqrt{4 r^2-h^2}}{2}.$$
This can be solved to find the $y$ value, for which the $A\left(S_1+S_2\right)$ has an extremum.
$$h=\frac{2}{\sqrt{5}}r$$
Now we need the second derivative of (1) to determine what kind of extremum this is. The 2nd derivative is,
$$8h\sqrt{4 r^2-h^2}+12 r^2-6h^2.\tag{2}$$
When we substitute $\frac{2}{\sqrt{5}}r$ for $h$ in (2), we get,
$$\frac{64}{5}r^2+12 r^2-\frac{24}{5}r^2=20 r^2\ge 0.$$
Therefore, the value of $h$ we obtained above corresponds to a minimum of $A\left(S_1+S_2\right)$ and not to a maximum. This is most unfortunate, but then that is how life works. Now we have to show that (1) is monotonic. We do this by more arguing than calculating. But first, we need to know the minimum and maximum possible values of $h$. According to $\bf{Fig. 3}$ and $\bf{Fig. 2}$, maximum and minimum possible values of $h$ are $\sqrt{2}r$ and $\frac{2}{\sqrt{5}}r$ respectively.
We have seen that the minimum value of $A\left(S_1+S_2\right)$ corresponds to the minimum value of $h$. As per this and in the absence of any other local extrema, the value of $A\left(S_1+S_2\right)$, as a function, increases monotonically without an upper bound as the value of $h$ increases. However, $h$ has an upper bound equal to $\sqrt{2}r$. This also means that the maximum value of $A\left(S_1+S_2\right)$, as an area, occurs when $h$ has its maximum possible value.
$$\therefore\space\space\space A\left(S_1+S_2\right)_{\rm max}=\frac{52}{25}r^2\approx 2.08 r^2.$$ |
Check the pointwise and the uniform convergence of $(f_n)_n:D\rightarrow \mathbb{R}$ | What you did is fine, except that, in the case of the third sequence, you forgot that $0$ dos not belong to $(0,1)$. On this set $(f_n)_{n\in\Bbb N}$ converges pointwise to $1$.
In the case of the second sequence, the convergence is uniform. Differentiating, you can check that $\sqrt{\frac1{n^2}+x^2}-x$ is decreasing on $[0,\infty)$. So, its maximum is attained at $0$, where it takes the value $\frac1n$. By a similar argument, the maximum of $\sqrt{\frac1{n^2}+x^2}+x$ on $(-\infty,0]$ is also $\frac1n$. So,$$\sup\left|\sqrt{\frac1{n^2}+x^2}-|x|\right|=\frac1n$$and therefore the convergence is uniform
The third sequence doesn't converge uniformly to $1$, because$$\lim_{n\to\infty}f_n\left(\frac1{n^n}\right)=\lim_{n\to\infty}\frac1n=0.$$ |
Manipulating an infinite series | $$\sum_{k=1}^{+\infty}\frac{k^2}{k!}=\sum_{k=1}^{+\infty}\frac{k}{(k-1)!}=\sum_{k=1}^{+\infty}\frac{k-1}{(k-1)!}+\sum_{k\geq 1}\frac{1}{(k-1)!}=2\sum_{k\geq 0}\frac{1}{k!}=\color{red}{2e}.$$
An alternative approach is the following one. Since:
$$\sum_{k\geq 1}\frac{e^{kx}}{k!} = e^{e^x}-1, \tag{1} $$
we have:
$$ \frac{d^2}{dx^2}\sum_{k\geq 1}\frac{e^{kx}}{k!}=\sum_{k\geq 1}\frac{k^2 e^{kx}}{k!}=e^{x+e^x}+e^{2x+e^x}\tag{2}$$
and it is enough to replace $x$ with $0$. |
Linear programming modelling question | If the resource e.g. is a software engineer with five days shift and two days off :–), it makes sense:
Let $x_k$ be the amount which is put in use starting at day $k$.
The objective is to minimize $\sum_k x_k$.
At day $k$ we have $u_k$ of the resource in use where
$$
u_k = \sum_{i=0}^4 x_{((k-1-i) \bmod 7)+1}
$$
e.g
$$
u_3 = x_3 + x_2 + x_1 + x_7 + x_6
$$
The requirements give the constraints
$$
u_k \ge r_k
$$
e.g.
$$
x_3 + x_2 + x_1 + x_7 + x_6 \ge 15
$$ |
Transforming of Chi Square Distribution and F Distribution | If $X$ has chi-square($m$) distribution and $Y$ has chi-square($n$) distribution, and the two are independent, then it is either a theorem or a definition that
$(X/m)\big/(Y/n)$ has $F(m,n)$ distribution. So defining $W:=X/Y$, we have $$Z=\frac{\sqrt n}2\left(\sqrt W-\frac1{\sqrt W}\right).$$ |
Surface area of sphere derivation in polar co-ordinates query. | Since $r^2dx^2+dr^2=R^2(\cos^2x+\sin^2x)dx^2$, with appropriate limits $A=2\pi R\int ydx$. Now we get to two mistakes. First, since you rotated a semicircle, not a quarter-circle the integration range should be from $0$ to $\pi$. Secondly, since $y^2+r^2=R^2$ we have $y=R\sin x$, not $y=r\sin x$. Thus $A=2\pi R^2\int_0^\pi\sin xdx=4\pi R^2$. |
Proof of dense and closed sets | $S=\{x:f(x)=g(x)\}$ .to show that $S^c$ is open .
Let $a\in S^c\implies f(a)\neq g(a)$ .Let $d(f(a),g(a))=r$.
Since $Y$ is Hausdorff so $\exists r>0$ such that $B(f(a),\frac{r}{4})\cap B(g(a),\frac{r}{4})=\emptyset$.
Since $f,g$ are continuous then $\exists \delta_1,\delta _2>0$ such that $f(B(a,\delta _1))\subseteq B(f(a),\frac{r}{4})$ and $f(B(a,\delta _2))\subseteq(B(g(a),\frac{r}{4})$.
Take $\delta =\min\{\delta_1,\delta _2\}>0$
Then you can readily check that $a\in B(a,\delta )\subseteq S^c$.
Regarding the second part: $T\subseteq S\implies \overline{T}\subseteq \overline {S}\implies X\subseteq S$ since $S$ is closed and $T$ is dense . |
inverse of a tridiagonal matrix | As Henning points out, checking that two matrices are inverses is much easier than computing an inverse; all we need to do is find the product of the matrices. Also, note that your definition of $B$ is incomplete, but we can deduce the rest since $B$ must be symmetric.
By definition, the $i,j$ of the matrix product $AB$ is given by
$$
[AB](i,j) = \sum_{k=1}^n a_{ik}b_{kj}
$$
in the case of $i=1$, we have
$$
[AB](1,j) = \sum_{k=1}^n a_{1k}b_{kj} = \\
\begin{cases}
-2\cdot\frac{1(n+1 - j)}{n+1}
+ 1 \cdot \frac{2(n + 1 - j)}{n+1} & j \geq 2\\
-2 \cdot \frac{1(n+1-j)}{n+1}
+ 1 \cdot \frac{j(n + 1 - 2)}{n+1} & j=1
\end{cases} = \delta_{ij}
$$
We can repeat a similar computation for the cases $2 \leq i \leq n-1$ and $i = n$.
Alternatively: we can find the characteristic polynomial of $A$ using induction (Pavel's method). From there, we could find the inverse using the Cayley Hamilton theorem. |
Example for $|gN|<|g|$ | Consider $\mathbb{Z}_4$ (with addition) and set $g=1$. Obviously $|g|=4$. Now take $N=\{0, 2\}$. It is a subgroup of order $2$ and $2g\in N$.
More generally: let $G$ be a finite cyclic group of order $n$ and pick a generator $g\in G$, i.e. $|g|=n$. Now take any nontrivial subgroup $1\neq N\leq G$ (which is normal since $G$ is abelian). Then $|G/N|<|G|$ because $N$ is nontrivial. In particular $|gN|$ divides $|G/N|$ (by Lagrange's theorem) and hence is strictly smaller then $|g|$. |
Understand Picard-Lindelöf Proof | The big picture:
A certain subset $X\subset C[a,b]$ is introduced; we'll be looking for a solution of the integral equation in $X$.
$X$ is shown to be closed in $C[a,b]$. Hence, $X$ is a complete metric space with the induced metric.
The restriction of the integral operator to $X$ is shown to be a contraction of $X$ into $X$.
From 2 and 3 we conclude that the equation has a unique solution in $X$.
We rule out the possibility of solutions in $C[a,b]\setminus X$. This part is needed for the uniqueness part of the theorem. (We already know there is exactly one solution in $X$, but what if there are solutions outside of $X$?)
Part 5 is what you are asking about. |
Help finding value of N that minimizes a sum | The central limit theorem says that $X$ binomial $(1000,.5)$ is $X=500+5\sqrt{10}Z$ with $Z$ approximately standard normal. Using the erffunction, this suggests that $N$ is around $536.7$.
WA indicates $0.01046$ for $N=536$ and $0.00833$ for $N=537$ hence the solution is $N=\color{red}{537}$.
sum of binom(1000,k)/2^(1000) from k=537 to k=1000
Decimal approximation:
0.0104635553030424713317175500708088450031910719508853
sum of binom(1000,k)/2^(1000) from k=538 to k=1000
Decimal approximation:
0.0088311156677493199532392344343609404372379787296116 |
Automorphism of rational numbers | Can you guess a function $f:\mathbb Q^*\to Aut(\mathbb Q)$? If you have a rational number $q\ne 0$, it is very natural to consider the automorphism $g_q:\mathbb Q \to \mathbb Q$ given by $x\mapsto qx$. It is easily shown that $f(q)=g_q$ is then an injective group homomorphism. The last thing to do then is show that $f$ is surjective. So, if you start with an automorphism $g:\mathbb Q \to \mathbb Q$, you really want to show that $g=g_q$ for some $q\in \mathbb q$. If that were the case, then $g(1)=g_q(1)=q\cdot 1=q$, so that tells you that the only candidate is $q=g(1)$. Now think a bit about rational numbers and figure out how to show that $g=g_{g(1)}$. |
Let $R$ be the binary relation on $\mathbb{N}$ defined by $xRy$ ($x$ is in relation to $y$) defined by: $xRy$ if $xy=49$ | For a relation $R$ to be reflexive means that $xRx$ for ALL $x$ in the domain of the relation. Yes, it's true that in this example $x=7$ satisfies $xRx$ because $7\cdot7=49$. But is it true that $xRx$ for ALL $x\in\mathbb{N}$? I.e., is it true that for all natural numbers $x^2=49$? |
Solving a wave equation: $a^2\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$ | You've got it so far.
For the $T$ equation, you have that $T'' = \alpha^{2} \lambda T$ (you should really put the $\alpha^{2}$ in this equation to make your life easier). You need a function that, when differentiated twice, gives you back the same function times a constant. You seem like you know what that is.
For the $X$ equation, you have $X'' = \lambda X$ - and you can guess a similar solution function because you have the same sort of setup as the T equation.
The issue comes in dealing with the constraints. Depending on those, there are different ways of dealing with the problem.
You can guess a solution of the form $$ \sum_{n} a_{n}x^{n}$$ and go from there, differentiating term by term and combining sums with similar lowest power. Have you learned about Power Series? Frobenius Series? |
is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$? | Since $(X-1)(X^4 + X^3 + X^2 + X + 1) = X^5 - 1$, if $f$ wasn’t irreducible, you’d have a degree two factor $(X - α)(X - β) = X^2 - (α+β) X + αβ~$ of $X^5 - 1$ of $f$ in $ℤ/7ℤ[X]$, where $α$, $β$ and therefore $αβ$ ($∈ ℤ/7ℤ$) had to be $5$-th roots of unity. What $5$-th roots of unity are there in $ℤ/7ℤ$? What possibilities are there for $α$ and $β$? (Note that $α$ and $β$ don’t have to be elements of $ℤ/7ℤ$ a priori, but elements of some extension of $ℤ/7ℤ$.) |
To prove $f=0$ a.e under certain condition | Hint: Open sets in $\mathbb{R}^2$ are generated by products of open sets in $\mathbb{R}$. These, in turn, are generated by open intervals.
It might be that this is more complicated than it needs be, but try the following route:
$\quad (1)$: Using the hint above, show that $\int_A f=0$ on any open set $A\subset\mathbb{R}^2$.
$\quad (2)$: Argue by contradiction; suppose $\{x\,|\,f(x)\neq 0\}=\{x\,|\,f(x)> 0\}\cup\{x\,|\,f(x)<0\}$ has positive measure, so that without loss of generality $B=\{x\,|\,f(x)> 0\}$ has positive measure.
$\quad (3)$: Write $B=\bigcup B_n$, where $B_n=\{x\,|\,f(x)\geq\frac1n\}$.
There must be some $k$ with $\mu(B_k)>0$.
$\quad (4)$: Show that any measurable set $E$ can be approximated in measure by closed sets and by open sets.
In other words, for any $\epsilon >0$, there is
a closed set $F\subset E$ with $\mu(E\setminus F)<\epsilon$; and
an open set $\mathcal{O}\supset E$ with $\mu(\mathcal{O}\setminus E)<\epsilon$.
For each $\epsilon>0$, in the obvious notation let $F_\epsilon$ and $\mathcal{O}_\epsilon$ be such approximations of $B_k$.
$\quad (5)$: Consider $\mathcal{O}_\epsilon\setminus F_\epsilon$ and observe that it is open. We have that
$$\int_{B_k}f=\int_{\mathcal{O}_\epsilon}f-\int_{\mathcal{O}_\epsilon\setminus B_k}f$$
and
$$\int_{\mathcal{O}_\epsilon\setminus B_k}f=
\int_{\mathcal{O}_\epsilon\setminus F_\epsilon}f
- \int_{B_k\setminus F_\epsilon}f.$$
It follows that
$$\int_{B_k}f=\underbrace{\int_{\mathcal{O}_\epsilon}f}_{(*)}
-\underbrace{\int_{\mathcal{O}_\epsilon\setminus F_\epsilon}f}_{(**)}
+ \int_{B_k\setminus F_\epsilon}f$$
Now, by $(1)$, integrals $(*)$ and $(**)$ are both $0$.
Hence, $\int_{B_k}f=\int_{B_k\setminus F_\epsilon}f$, which implies $\int_{F_\epsilon}f=0$.
$\quad (6)$: On the other hand, notice that for $\epsilon$ small enough it must be that $\mu(F_\epsilon)>0$ and then the simple bound
$$\int_{F_\epsilon}f\geq \mu(F_\epsilon)\frac1k>0$$
contradicts our previous point. |
How to Find The outward Normal Vector in this case | The outward normal vector for the sphere is
$$\vec n =\frac1{\sqrt{x^2+y^2+z^2}}\left(x,y,z\right)$$
Note that the sphere and the cylinder have in common the circle in the $x-y$ plane therefore $S$ should be the hemisphere over the $x-y$ plane. |
Recostruct a function from the coefficients of its Chebyshev expansion | Sure. The projection coefficient is $\frac{c_k}{(T_k,T_k)}=\frac{2}{\pi} c_k$ and so a Fourier-like approximation of $f$ is
$$\frac{2}{\pi} \sum_{k=0}^n c_k T_k(x)$$
which will converge to $f$ in the $L^2((1-x^2)^{-1/2} \, dx)$ norm as $n \to \infty$. |
Does this function have closed form? | I am assuming that $p$ is prime and $r_i$ is relative prime to $i$, so $s_i$ is equal to $\nu_p(i)$, the $p$-adic valuation at $i$. Therefore
$$f(p,n)=\sum_{i=1}^n\nu_p(i)=\nu_p(n!)=\sum_{k=1}^\infty\biggl\lfloor\frac n{p^k}\biggr\rfloor\,,$$
by De Polignac-Legendre formula. |
Validation of a result (logic) | For given $l \in \Bbb N$ you want to show that $\lim_{x\to\infty}f(x)x^l=0$. To this end, you use the given hypothesis with “a suitable” $v$ such that the desired conclusion follows.
As you noticed, any $v > l$ works. So you can argue as follows:
For $v = l+1$ there exist $C, L > 0$ such that $f(x)\leq Cx^{-l-1}$ for $x > L$. Then
$$
0 \le f(x)x^l \le \frac{C}{x}
$$
for $x > L$, which implies $\lim_{x\to\infty}f(x)x^l=0$.
That is a complete proof. It does not matter that other choices of $v$ do not imply the desired conclusion. |
The dimensions of $V$ and $V^\perp$ are complementary | This doesn't have much to do with the particular field $\mathbf R$, number $4$, or form $f$—we get a similar formula for any non-degenerate bilinear form on a finite dimensional vector space. We know that $f$ is non-degenerate because the matrix of $f$ with respect to the standard basis has determinant $-1$.
Thus $f$ induces an isomorphism $\mathbf R^4 \to (\mathbf R^4)^*$ sending $x$ to the functional $y \mapsto f(x, y)$. Now compose with the restriction map $(\mathbf R^4)^* \to V^*$, which is surjective. What is the kernel of this composition?
[The map $(\mathbf R^4)^* \to V^*$ is dual to the inclusion $V \to \mathbf R^4$. It is surjective because we can find a decomposition $\mathbf R^4 = V \oplus W$ and use the universal property of direct sums to extend any functional on $V$.] |
Breaking down what the definition of an affine connection says | This is defining what ways to differentiate vector fields (called the covariant derivative), which we'll denote $\nabla_X Y$ for the covariant derivative of $Y$ in the direction $X$, we'll consider valid.
Property (1) is saying this is linear in $X$ over $\mathbb{R}$, and moreover that multiplication of $X$ by any function can be pulled out (you could call this linear over $C^\infty(M)$).
This implies that value at a point $p$, namely $(\nabla_X Y) (p)$, only depends on the value of $X$ at $p$, not any derivatives of $X$. (Intuitively, $X(p)$ is the direction we're differentiating in, so the derivatives of $X$ are not important, only those of $Y$.)
Property (2) plus property (3) for constant $f$ is saying this is linear in $Y$ over $\mathbb{R}$.
Property (3) is the product rule, aka Leibniz rule, and tells you that $\nabla_X (fY)$ is a sum of terms, one where you differentiate $Y$ (this is $f \nabla_X Y$) and one where you differentiate $f$ (this is $X(f) Y$).
If we work in coordinates, then what this definition works out to is that we can specify $\nabla_{\partial_i} \partial_j$ freely; let's call this $a_{ij}$; and once we specify these, $\nabla_X Y$ is specified for any $X$ and $Y$.
It would be a good exercise to work out $\nabla_{\sum_i b_i \partial_i} (\sum_j c_j \partial_j)$ in terms of $a_{ij}$, from the definition, and confirm that your formula defines an affine connection (in coordinates). |
Operator $T$ with $T^2=T$ which is not a projection | I suppose that your projections are orthogonal projections. You can take, for instance,$$\begin{array}{rccc}T\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&(x+y,0).\end{array}$$It is a projection from $\mathbb{R}^2$ onto $\{(x,0)\,|\,x\in\mathbb{R}\}$. But it is not orthogonal with respect to the usual inner product. |
Elementary way to extend properties of $+$, $\times$, and exponentiation to $\mathbb{R}$. | Let me offer some thoughts, in the hope that others will do better.
It all depends, doesn’t it, on what your students’ understanding is of the behavior and nature of real numbers? Let’s suppose you make a good correspondence between real numbers and lengths of segments. Then you can define multiplication by means of ratios: $ab$ is defined to be the thing that fits into the proportionality $a:1 = ab:b$, and this can be gotten computationally and defined by using similar triangles. Then if they believe the basic facts about similarity, they should be able to see (or prove) that you get the same thing with the proportionality $a:1=ab:b$.
For exponentiation, I think you’ll have to be satisfied with integer (or, conceivably, rational) exponents. The definition of $3^{\sqrt2}$ must certainly wait for analytic sophistication on the part of the student. |
Can A matrix A have different cofactor matrices? If So then, can you have different inverses for one matrix? | Elementary row operations lead, in general, to a different matrix $\bar A\neq A$.
There is not reason that, if it exists, $\bar A^{-1}=A^{-1}$ |
How to prove $x^2-34y^2=17$ has no integer solutions? | We must have $17\mid x$, hence the equation boils down to:
$$ 17 x^2 = 1 + 2y^2 = (1+y\sqrt{-2})(1-y\sqrt{-2}).\tag{1}$$
Since $\mathbb{Z}[\sqrt{-2}]$ is an euclidean domain and $17=3^2+2\cdot 2^2$, we just have to show that no square in $\mathbb{Z}[\sqrt{-2}]$ can have the form:
$$(1+y\sqrt{-2})(3+2\sqrt{-2})=(3-4y)+(2+3y)\sqrt{-2},$$
that is the same as saying that
$$ a^2-2b^2 = 3-4y,\qquad 2ab = 2+3y\tag{2}$$
is impossible. $(2)$ implies:
$$ 3a^2 + 8ab - 6b^2 = 17, $$
but $a$ and $b$ cannot be both even, so $3a^2+8ab-6b^2\in\{2,3,5,6\}\pmod{8}$, while $17\equiv 1\pmod{8}$, hence $(2)$, then $(1)$, have no integer solutions. |
Assume $E[X^2]=20$. Find the mean and variance of $X$ if the mean and variance are equal. | Hint
Use the fact that
$$Var(X)=E(X^2)-E(X)^2$$ |
A mirror focusing beams at one point | Suppose $y=f(x)$ is the equation of the mirror. Let $\alpha$ be the angle between the tangent line at $(x_0,f(x_0))$ and incident ray. Clearly, $\tan\alpha=\frac1{f'(x_0)}$. Also, the angle between reflected ray and $x$-axis is $\frac{\pi}{2}+2\left(\frac\pi 2-\alpha\right)$. The equation of the reflected ray is thus
$$\frac{y-f(x_0)}{x-x_0}=\tan\left(\frac{3\pi}{2}-2\alpha\right)=\frac{f'(x_0)^2-1}{2f'(x_0)}.$$
All of these rays (parameterized by $x_0$) should pass by focus $(0,a)$, hence the differential equation
$$\frac{f(x_0)-a}{x_0}=\frac{f'(x_0)^2-1}{2f'(x_0)}.$$
This has as the general solution the $1$-parameter family of symmetric parabolas
$$f(x_0)=\frac{x_0^2}{4C}+a-C,$$
corresponding to different focal distances. |
Number of ways to gain $10$ points while gaining conditional $1$, $2$, or $3$ points at a time? | I would just count by hand. Start with all wins and discover that you can't get there because three gives $8$ points and four gives $11$, so you can' start with more than three wins. Now backtrack. You find $WWWLL$ gives $10$ points, which is the only way starting from $WWW$. If you start with $WWL$ you have six so far, so can have four $L$s or two $L$s and a $W$ in some order. Keep going.
If we change the target to some larger number like $50$ so that hand counting is impractical I would make two coupled recurrences. Define $A(n)$ as the number of ways to score $n$ points ending with a loss and $B(n)$ as the number of ways to score $n$ points ending with a win. You have $A(n)=A(n-1)+B(n-1)$ because a loss always adds one point. What is the recurrence for $B(n)$? Now build a spreadsheet with a column for $n$, a column for $A(n)$, and a column for $B(n)$, write the recurrence formulas in one row and copy down. |
Prym variety associated to an étale cover of degree 2 of an hyperelliptic curve. | I know this question is old (came across it on a google search for something else) but the answer you want is in the (mostly expository) paper: http://arxiv.org/abs/alg-geom/9206008
In short, a genus 2 curve has six branch points for the hyperelliptic map, and a choice of two of them (unordered) is equivalent to a choice of two Weierstrass points, which is the same as the data of a point of order two/double cover of the curve.
The elliptic curve that is the Prym of that double cover is the one that is branched over $\mathbb{P}^1$ at the remaining 4 branch points.
As for a concrete way to picture $f$, the point of order two on the Jacobian is a line bundle $\mu$ that squares to the trivial line bundle $\mathscr{O}_C$, the isomorphism $\mu^{\otimes 2}\to \mathscr{O}_C$ gives an $\mathscr{O}_C$-algebra structure to the sheaf $\mathscr{O}_C\oplus\mu$, and $C'$ in your notation is the relative Spec of this algebra, and the map $f$ is projection.
Another way to construct $C'$ is to look locally at solutions to $z^2=1$ inside the total space of the line bundle $\mu$. |
Prove that the sum $a_1+a_2+....+a_n+b_1+b_2+...+b_n$ cannot equal to $0$ | The product of the $a_i$ is equal to the product of the $b_i$. This is because each of these products is the product of all the elements in the $n\times n$ array.
So the numbers of $-1$'s among the $a_i$, and among the $b_i$, have the same parity (both numbers are even or both are odd).
It follows that the combined number of $-1$'s among the $a_i$ and $b_i$ is even.
This ensures that the sum of all the $a_i$ and all the $b_i$ cannot be $0$. For if the sum is $0$, the total number of $-1$'s among the $a_i$ and $b_i$ must be $n$. And $n$ is odd. |
Show that there exists a partition $-\infty=t_0<t_1<...<t_k=\infty$ such that $\lim_{t\rightarrow t_j^{-}} F(t)-F(t_{j-1})<\epsilon$ | Since $F$ is a bounded, non decreasing function, and $F(t) \in [0,1]$ for all $t$. You need to choose $t_1,...,t_{k-1}$.
Pick $n$ such that ${1 \over n} < {1 \over 2}\epsilon$. Partition the range $[0,1]$ as $(0,{1 \over n},{2 \over n},..., 1-{1 \over n}, 1)$.
Let $J_i = [{i-1 \over n}, {i \over n})$, for $i=1,...,n-1$, $J_n = [1-{1 \over n},1]$ and $H_i = F^{-1}(J_i)$. Note that the
$H_i$ form a partition of $\mathbb{R}$. If $H_i$ is non empty, let $t_i = \sup H_i$.
Note that $t_i \notin H_i$, since if $t_i \in H_i$, then
$F(t_i) \in J_i = [{i-1 \over n}, {i \over n})$ and so
$F(t_i ) < {i \over n}$. Since $F$ is right continuous, there is some $t' > t_i$
such that $F(t_i ) \le F(t') < {i \over n}$, which is a contradiction.
Note that $H_1, H_k$ are non empty since $\lim_{t \to -\infty}F(t) = 0$ and $\lim_{t \to \infty}F(t) = 1$.
Let $i_j$, $j=1,...,m$ be the indices for which $H_{i_j}$ is non empty, note
that $i_1 = 1,i_m =k $. Let $i_0 = 0$. Then I claim that $t_{i_j}$, $j=0,...,m$ form the desired partition.
Note that $F(t_{i_1}) < {1 \over n}$, hence $\lim_{t \uparrow t_{i_1}} (F(t)-F(t_{i_0})) = 0$.
Now choose $j\ge 2$. If the interval $H_{i_j -1}$ is non empty, then we
have $t_{i_j} - t_{i_{j-1}} \le {2 \over n} < \epsilon$. If not, then
the intervals $H_{i_{j -1}+1},...,H_{i_{j }-1} $ are empty. Since
$t_{i_{j-1}} \notin H_{i_{j-1}}$, we must have
$t_{i_{j-1}} \in H_{i_{j }}$
in which case we have
$F(t_{i_j}) - F(t_{i_{j-1}}) \le {1 \over n} < \epsilon$. |
Plotting 2 parameters function in MATLAB | If I understood correctly your problem it should be really easy. For the first plot you do:
figure
X=[0.1 0.2 0.3];
Y=[5 4 1];
plot(X,Y);
hold on % This holds the lines you have already plot
For the second plot I suggest you to explicit the function $x^{2}y+3x+4y^{2}=1$ with respect to $y$ (i.e. $y = f(x)$) then you do (after the previous commands):
plot(X,f(X))
That should solve your problem.
Of course you can put labels and titles to the plot by doing:
xlabel('the label for the x axis')
ylabel('the label for the y axis')
title('The title you want for the plot')
I hope this helped you. |
Geometrically describe these Cosets and form a bijection with the Orbit-Stabilizer relation. | Regarding part (a)
You are close, but there are some issues. First of all, the left cosets will be of the form
$$
hJ = \begin{pmatrix} c & d \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} ac & d \\ 0 & 1 \end{pmatrix}
$$
Then to the conclusions:
$hJ$ constitutes a half line $(x,d)$, where $d$ is some fixed $y$-coordinate and $x=ac\in(0,\infty)$
For $Jh$ we have half lines of the form $(ac,ad)$ so in effect they can be described via $x=ac\in(0,\infty)$ and $y=ad$ which can be combined to $y=\frac{d}{c}\cdot x$ where $x>0$. For $d>0$ this is increasing, for $d=0$ it is constant and is one half of the x axis as you already stated, and for $d<0$ it is decreasing.
Regarding part (b)
You are close again, but again there are issues. First of all we have the stabilizer of $v=\begin{pmatrix} 0\\1\end{pmatrix}$ as
$$
H_v=\{\begin{pmatrix}a&0\\0&1\end{pmatrix}|\ a\in\mathbb R, a>0\}=J
$$
So in fact the stabilizer equals $J$ from part (a). There we already identified each left coset to be defined by $d$ being a line of the form $\{(x,d)|\ x>0\}$. Thus each left coset $hJ$ depends only on $d$ as does each element $\begin{pmatrix} d \\ 1 \end{pmatrix}$ in the orbit. For each fixed pair $c,d$ and the corresponding element $h=\begin{pmatrix}c&d\\0&1\end{pmatrix}\in H$ we have the bijections
$$
hJ\longmapsto\{(x,d)|\ x>0\}\longmapsto d\longmapsto\begin{pmatrix}d\\1\end{pmatrix}
$$ |
Is any continuous curve in $\mathbb{R}^n$ a 1-D manifold? | There are different answers, depending what exactly you mean, but all are no in last consequence.
If with manifold, you mean SUBmanifold, the answer is no. Just think of a line with corners (ok, that one could still be a topological submanifold) or a line that intersects itself.
If you ask if every image of a curve is a manifold when equipped with the subspace topology, a plane filling curve provides a counterexample. This is because the neighborhood of any points has infinitely many connected components and can therefore not be homeomorphic to any Rn. An even simpler counterexample is yet again a self-intersecting curve.
However, all these curves can be immersed submanifolds, but that is rather trivial and mostly useless. |
Proof of Pick's theorem for triangular case | The linear map with matrix $A=\pmatrix{a&b\\c&d}$ transforms the integer lattice $\Bbb Z^2$ into itself and maps $(0,0)$, $(1,0)$ and $(0,1)$ into $(0,0)$, $(a,b)$ and $(c,d)$. So the inverse of $A$ maps $S$ to $T$ with vertices
$(0,0)$, $(1,0)$ and $(0,1)$ and it is apparent this has no lattice points
in the interior. As $A$ maps $\Bbb Z^2$ bijectively to itself, then $S$ has
no interior lattice points either. |
If G is an abelian group with subgroups H,K of orders 3 and 5, prove there must exist an element of order 15. | If $G$ has subgroups $H_1=\langle h_1\rangle$ and $H_2=\langle h_2\rangle$ of orders $3$ and $5$ respectively, what can you say about the group $\langle h_1,h_2\rangle$?
Also, a note that your edit is not true. There are infinite finitely generated groups in which every element has finite order. The Grigorchuk group is an example. |
Express the range of a Domain Subset [Notation] | $$f(S)=\{f(s)\in B\ :\ s\in S\}$$ helps. |
Getting a Circular Crown's area and perimeter | Community wiki answer to allow the question to be resolved:
As stated in the comments, your results are correct and to be preferred over the rounded results in the book. The area is $\frac{95}{16}\pi\text{ cm}^2$, not $\frac{95}{16}\pi^2\text{ cm}^2$. |
Weak solution operator of Elliptic PDE is lower bounded or not? | No, this operator is not lower bounded in your sense. If it were, we would have
$$
\gamma \int_\Omega |\nabla u|^2\leq c\gamma\int_\Omega A\nabla u\cdot\nabla u=c\gamma\int_\Omega vu\leq c\gamma\|u\|_2\|v\|_2\leq c\|u\|_2^2
$$
for $u\in C_c^\infty(\Omega)$. To see that this is not true, take any $u\in L^2(\Omega)\setminus H^1(\Omega)$ and a sequence $(u_n)$ in $C_c^\infty(\Omega)$ such that $u_n\to u$ in $L^2$. |
I need a proof that a line cannot intersect a circle at three distinct points | Or a more geometric proof: If a circle intersects a line in $A$ and $B$, the center of the circle lies on the center normal of the line segment $AB$. If there is a third intersection point $C$, the center of the circle must also lie on the center normal of $BC$. But these two center normals are distinct parallel lines, and cannot have point in common. |
Implicit Function Theorem with $x^2+y^2+z^2=\psi(ax+by+cz)$ | We can express $F$ as $F(x,y,f(x,y))=(x^2+y^2+(f(x,y))^2-\psi(ax+by+cf(x,y)))$.
It's not immediately obvious to me what this means, but even if it meaningful and correct, I don't think it is going in the right direction.
The implicit function theorem tells you that there exist neighborhoods $V$ of $(0,0)$, $W$ of $0$ and a class $C^1$ function $f\colon V\to W$ such that $f(0,0)=(0,0)$ and $$\forall (x,y) \in V\left(F(x,y, f(x,y))=0\right) \tag 1$$
(I'm using the formulation of the IFT used on Wikipedia, which is very similar to the one I wrote here).
Now take a new function $H\colon V\to W$, $(x,y)\to F(x,y,f(x,y))$ and differentiate with respect to the first and second coordinates. Coupling this with $(1)$ should give you the answer. |
How is the radian measure of angles derived/defined? | (2):
Since $C=2\pi r$, if you double $r$, you double $C$ and you also double "a third of $C$", etc. This works with not just doubling, but whatever scaling you need to get from one circle to another. $C/r$ is always $2\pi$, for any circle, so
$(1/3)C/r$ is always $2\pi/3$, etc. Therefore, if you have an arc of length $s$ that's a certain ratio of the whole circumference (like 1/3), then $s/r$ is the same for all circles (in this case it's $2\pi/3$).
(1):
An angle from the center of the circle always covers a ratio of the whole circle that doesn't change, so if you care about angles, you care about fractions of the circumference. To see this, it might help to look at a picture like:
(Picture taken from http://ck022.k12.sd.us/) |
Menelaus' Theorem & Concurrency | Let us define $P$ as $BC\cap EF$: we just have to show that $MN$ goes through $P$.
The circle having diameter $AD$ and centre $O$ is the circumcircle of $AEDF$, hence $PD^2=PE\cdot PF$.
$ANDM$ is a parallelogram, hence the triangles $BND$ and $DMC$ are similar. If we set $H=PF\cap DM$ and prove that $CH\perp DM$ we are done, since we get that $P$ is a homotetic center for $BND$ and $DMC$. On the other hand $\widehat{HFD}=\widehat{EFD}=\widehat{EAD}=\widehat{EDB}$, so $DHFC$ is a cyclic quadrilateral and $CH\perp DM$ as wanted. |
Is the closed interval $[0,1]$ always needed to do homotopy? | The idea behind a homotopy is to allow an object to "live" for a finite period of time. The interval $[0,1]$ (and its topology) corresponds very naturally to the classically accepted idea of what a bounded time interval is.
If your idea of time is different, or if the space where your object "lives" has an unusual topology, it may be relevant to use another definition. But I have never seen that done. |
Proof that $(u+v)^p = u^p + v^p$ in vector space over finite field of characteristic $p$. | The characteristic of the field is $p$, so we have $a^p=0$ for every element $a$ with the given definition. This immediately implies the given equation. |
Computing Probabilities | $P(Y<1|X<1)=\frac {P(X<1,Y<1)} {P(X<1)}=\frac {\int_0^{1} \int_x^{1}2e^{-x-y} dydx} {\int_0^{1} \int_x^{\infty}2e^{-x-y} dydx}$. I will let you evaluate the integrals. |
Why is the probability of a continuous variable taking a particular value zero? Explain only logically | The distribution of the weights of people in the world is of course discrete: there is only a finite number of distinct weights associated with the finite number of people that exist. However, you can certainly consider a continuous probability distribution, like the bell curve, which approximates the probability that a randomly chosen person has a given weight. In this idealized world, a random person's weight may be an arbitrary real number, and this number can be known to arbitrary accuracy.
Now there is some finite probability that our random weight lies between 70 kg and 80 kg. Then the probability that it lies between 70 and 71 kg is roughly a tenth of that, because we're considering only one-tenth of the interval, right? And the probability between 70.0 and 70.1 kg is a tenth further, and between 70.00 and 70.01 kg a tenth still. What do you suppose that leaves the probability that the weight is exactly 70.000... kg? |
A particular DE as first order system | You can still put this system in state space form.
$$\begin{bmatrix}\dot{x}_1\\\dot{x}_2\end{bmatrix} = \begin{bmatrix}-a ~~~ 1 \\-b ~~~ 0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} + \begin{bmatrix}d\\c\end{bmatrix} u$$
You can check that $x_1=y$. This is called observable canonical form. Also this equation always has a unique solution for all initial conditions $x_0\in\mathbb{R}^n$. |
$M\simeq\ker(\beta)\oplus\frac{M}{\ker(\beta)}$ | As already remarked in the comments, your first claim is not true. What you need is a map $\rho \colon M \to \text{ker}(\beta)$ with $\rho|_{\text{ker}(\beta)} = 1_{\text{ker}(\beta)}$. With this additional assumption, there is hardly any difference between the first claim (with $\text{ker}(\beta)$) and the second claim (with the short exact sequences).
For the proof, just write down the obvious map $M \to \text{ker}(\beta) \oplus M/\text{ker}(\beta)$ (there now is one, because you have $\rho$) and prove that it is injective and surjective.
Edit (for completeness' sake). The obvious map $\phi \colon M \to \text{ker}(\beta) \oplus M/\text{ker}(\beta)$ is $\phi(m) = (\rho(m), \bar m)$.
Injectivity: Suppose $\phi(m) = 0$, i.e., $\rho(m) = 0$ and $\bar m = 0$. Then $m \in \text{ker}(\beta)$ and because $\rho|_{\text{ker}(\beta)} = 1_{\text{ker}(\beta)}$, $m = \rho(m) = 0$.
Surjectivity: Take $(l, \bar n) \in \text{ker}(\beta) \oplus M/\text{ker}(\beta)$. Now note that $\phi(n) = (\rho(n), \bar n)$. We can now 'fix' the left component by adding $l - \rho(n)$. Using $l - \rho(n) \in \text{ker}(\beta)$, we get $\phi(n + l - \rho(n)) = (l, \bar n)$.
It might be instructive to cast this argument purely in terms of the short exact sequences. |
Representations of knot groups | You can start by reading this paper by Eric Klassen. If you want to read about more recent topics, just take a look at Hans Boden's
homepage: Every other paper will be about representations of knot groups. Or you can dive into this paper by Aschenbrenner, Friedl and Wilton. (Warning: It is 149 pages long!) All in all, the literature on this subject is just enormous. |
2018 January Challenge: Prove inequality in geometry problem | We need to prove that
$$\sqrt{1-\frac{1}{\sqrt2}}-\sqrt{\frac{3}{2}}(\sqrt2-1)>\left(1-\frac{1}{\sqrt2}\right)^3,$$ which is true.
Indeed, let $DFMK$ be the given rectangle,
$\Delta FBM$ be given triangle ($FM=BM=1$,$\measuredangle FMB=90^{\circ}$) and
$\Delta EFC$ be another given triangle ($EF=FC=1$, $\measuredangle EFC=90^{\circ}$).
Thus, $$\frac{AC}{EF}=\frac{BC}{BF}$$ or
$$\frac{AC}{1}=\frac{\sqrt2-1}{\sqrt2},$$ which gives
$$AC=1-\frac{1}{\sqrt2}.$$
Hence,
$$AB=\sqrt{AC^2+BC^2}=\sqrt{\left(1-\frac{1}{\sqrt2}\right)^2+(\sqrt2-1)^2}=\sqrt{\frac{3}{2}}(\sqrt2-1).$$
Now, since $\measuredangle DCA=45^{\circ}$, by law of cosines for $\Delta DAC$ we obtain:
$$\begin{align}AD&=\sqrt{\left(\frac{1}{\sqrt2}\right)^2+\left(1-\frac{1}{\sqrt2}\right)^2-2\cdot\frac{1}{\sqrt2}\cdot\left(1-\frac{1}{\sqrt2}\right)\cdot\frac{1}{\sqrt2}}\\&=\sqrt{\frac{1}{2}+\frac{3}{2}-\sqrt2-1+\frac{1}{\sqrt2}}=\sqrt{1-\frac{1}{\sqrt2}}.\end{align}$$ |
How is formula for p value derived? | Although it looks intuitively alright, I am just wondering if there is a mathematical rigorous sequential way of obtaining the above formula?
Yes, there is!
First way, analytical
Using the Generalized Likelihood Ratio you get that
$$\sup_{\theta \in \Theta_0}L\propto\exp\left\{\frac{1}{2}\Sigma_i(x_i-\mu_0)^2 \right\} $$
$$\sup_{\theta \in \Theta}L\propto\exp\left\{\frac{1}{2}\Sigma_i(x_i-\overline{x}_n)^2 \right\} $$
Do the ratio, manipulate is algebraically (try because it is not difficult) and get the desired solution
Second way, using a property of Gaussian model
If the model is gaussian, in the two tails hypothesis test there is a well known method known as "Confidence interval method" by which you calculate the confidence interval of $\mu\in [a;b]$ with the desired confidence level $(1-\alpha)100\%$
You use the following decision rule: Reject $H_0$ if and only if $\mu_0 \notin [a;b]$
Following this method you get exactly the solution you want to prove
Finally observe that this test IS NOT a UMP test. |
Is there an algebraic formula that gives this weird multiplication?: $(-x)\circ(-x)=(-x)\circ(x)$ | You can avoid using the absolute value and sign functions if you change the representation of the domain! Let's use a sign-and-magnitude construction. An integer is a pair $(\sigma, n)$ where $\sigma$ is either the symbol $+$ or the symbol $-$, $n$ is an natural number, and we identify $(+,0)=(-,0)$. For a natural number $n$, we may use the shorthand $n=(+,n)$ and $-n=(-,n)$. But whenever we want to define any operation on integers, we may choose to define the operation on the sign and the magnitude independently; and the operation will be well-defined as long as we respect $(+,0)=(-,0)$.
With that setup, the definition is just:
$$(\sigma,x)\circ(\tau,y)=(\min(\sigma,\tau),xy)$$
Then observe that if either $x=0$ or $y=0$, then $xy=0$, so the sign doesn't matter, and we're done.
If you're unhappy with the $\min$ function, we can do away with that too, by representing the sign symbol as a sign bit. The usual computer representation would be $0=+$ and $1=-$. But it's slightly better for us to choose the opposite, $0=-$ and $1=+$. Then we have:
$$(\sigma,x)\circ(\tau,y)=(\sigma\tau,xy)$$
which is as simple as you could hope for! |
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