Find CDF of random variable which depends on other variable Let $X$ be a random variable with uniform distribution on $[-1, 1]$. Find the CDF of random variable Y given by the following formula:

  $Y = \left\{\begin{matrix}
 -\frac{1}{2},& X < - \frac{1}{2}\\
 X,& -\frac{1}{2} \leq X \leq \frac{1}{4}\\
 \frac{1}{4}, & X > \frac{1}{4}
\end{matrix}\right.$

So I've found PDF and CDF of $X$:

$f_X(x) = \begin{cases}
 \frac{1}{2}, & x \in [-1, 1]\\
 0, & \text{otherwise}
\end{cases}$

$F_X(a) = \int_{-\infty}^{a} f_X(x) dx = \left\{\begin{matrix}
 0, & a \leq -1\\
 \frac{a+1}{2}, & a \in (-1, 1) \\
 1, & a \geq 1
\end{matrix}\right.$

I tried to find Y's CDF by:

$F_Y(a) = P(Y \leq a)$

$= P(-\frac{1}{2} \leq a, X < - \frac{1}{2}) + P(X \leq a, - \frac{1}{2} \leq X \leq \frac{1}{4}) + P(\frac{1}{4} \leq a, X > -\frac{1}{4})$

But what should I do next? I'm finding such CDF for the first time and my notes say I need to consider a few different cases, but I have no clue what they should look like and how to do it. Any tips would be helpful.

Tip: You can find a CDF, but not a pdf, for all points on the support.

Notice that $Y$ will have a support of the interval $[-\tfrac 12;\tfrac 14]$, but will have a probability mass at the two end points, with a probability density in the interval between.   It is a mixed distribution.

That is $~\mathsf P(Y{=}-\tfrac 12)~=~\mathsf P(-1{\leq}X{<}-\tfrac 12)~$ and $~\mathsf P(Y{=}\tfrac 14)~=~\mathsf P(\tfrac 14{<}X{\leq}1)~$.

So then

   $$F_Y(a)~=~\mathsf P(Y\leq a) ~=~ \begin{cases} 0 &:& a<-\tfrac 12\\ \tfrac {2a+3}4 & : & -\tfrac 12\leq a < \tfrac 14 \\ 1 &:& \tfrac 14 \leq a \end{cases}$$

If you've peeked, notice the step discontinuities.

Note that $P(Y\leq y) = \int_\mathbb{R} P(Y \leq Y | X = x) f_X(x) dx$ by the law of total probability.

You can split the integral up by $x< -1/2$, $-1/2 \leq x \leq 1/4$ and $x> 1/4$ and do it.


How to show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd? Let G be a finite group G. Then How can I show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd ?
I read this question in an Algebra book. Since $e^3=e$, e must be one of those elements. But how to find for non trivial elements ?

Let $T$ be the set of all elements $x$ in $G$ such that $x^3=e$. Since $e\in T$, we will prove that $S=T\setminus\{e\}$ has an even cardinal.

Details are left to you:

If $x$ belongs to $S$, then its inverse $x^{-1}$ is also in $S$.
If $x$ is in $S$, then $x\neq x^{-1}$.
Every element of $S$ can be paired with another element of $S$, so $S$ has an even number of elements.

Disclaimer: this is a really worse solution, but I thought it was worth pointing out.

We claim that the number of elements in $T = \{g\in G: g^3 = e, g\neq e\}$ is an even number $N$. The claim follows from recalling that $e^3 = e$, hence the number of elements with trivial cubes is $N+1$, an odd number.

In fact, if $g\in T$, then $\{g,g^2,g^4...\}\subset T$, so, since $T$ is finite, there is a minimal number $N = |T|$ such that $g^{2N} \in T$, and moreover $g^{2N} = g$ since $N$ is minimal. Suppose by contradiction that
$$
2^N-1 = 1 \bmod{3} \text{ or } 2^N-1 = 2 \bmod{3}.
$$
Then, respectively, $g^{2N} = g^2$ and $g^{2N} = e$, both which cannot be since $g \neq g^{-1} = g^2$ and $e\notin T$. Hence
$$
2^N-1 = 0\bmod{3}.
$$

By induction, this is true if $N$ is a non-zero even integer, since $2^2 -1 = 3$ and
$$
2^N = 1+3k \implies 2^{N+2} = 1 + 3(k+1).
$$
Moreover, it follows that $2^{2N+1}-1 = 2 \bmod{3}$, hence $N$ is even. This completes the proof.


Fourier Series - Periodicity I don't know if I'm doing something wrong in this exercise.

$f(t)=\pi-t$, if $0<t<\pi$

$f(t)=0$, if $\pi<t<2\pi$

I have to find the Fourier Series of $f(t)$

I define the Fourier Series like this:

T: period

$$f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_n.\cos{\frac{2\pi nt}{T}}+b_n.\sin{\frac{2\pi nt}{T}}]$$

where ($\alpha$ is a real number)

$$a_0=\frac{2}{T}.\int_{\alpha}^{\alpha+T}f(t)dt$$

$$a_n=\frac{2}{T}.\int_{\alpha}^{\alpha+T}f(t).\cos{\frac{2\pi nt}{T}}dt$$

$$b_n=\frac{2}{T}.\int_{\alpha}^{\alpha+T}f(t).\sin{\frac{2\pi nt}{T}}dt$$

Here's the problem: I think that $T=4\pi$, because I extend the function $f(t)$ from $-2\pi$ to $2\pi$, so I have this:

$$a_0=\frac{1}{2\pi}.\int_{-2\pi}^{2\pi}f(t)dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t)dt$$

$$a_n=\frac{1}{2\pi}.\int_{-2\pi}^{2\pi}f(t).\cos{\frac{nt}{2}}dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t).\cos{\frac{nt}{2}}dt$$

$$b_n=\frac{1}{2\pi}.\int_{-2\pi}^{2\pi}f(t).\sin{\frac{nt}{2}}dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t).\sin{\frac{nt}{2}}dt$$

But other people took $T=2\pi$ and they have:

$$a_0=\frac{2}{2\pi}.\int_{0}^{2\pi}f(t)dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t)dt$$

$$a_n=\frac{2}{2\pi}.\int_{0}^{2\pi}f(t).\cos{nt} dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t).\cos{nt} \mathrm{d}t$$

$$a_n=\frac{2}{2\pi}.\int_{0}^{2\pi}f(t).\sin{nt} dt=\frac{1}{\pi}.\int_{0}^{2\pi}f(t).\sin{nt} \mathrm{d}t$$

Look that he took the integral between 0 and $2\pi$ and I took it from $-2\pi$ to $2\pi$

Did I do something wrong? Why this he took the period from 0 to $2\pi$? $f(t)$ is not a periodic function?

Thanks!

It does not matter whether we choose to extend $f$ so that the period is $2 \pi$ or $4 \pi$. Your problem is a simplification you made when determining the coefficients.

$$a_n = \frac{2}{4 \pi} \int_{-2 \pi}^{2\pi} f(t) \cos{\frac{n t}{2}} \ dt \neq \frac{1}{\pi} \int_{0}^{2\pi} f(t) \cos{\frac{n t}{2}} \ dt,$$ because when you extend your function $f(t)$ to the interval $[-2 \pi, 2 \pi]$ it is NOT an even function. So, you must compute the integral over the whole interval and not make this incorrect simplification.

What you will find is that when $n$ is odd, you will get $a_n = b_n = 0$ and that when $n$ is even you will get the same answer as you would get by thinking of the function as only having a period of $2 \pi$

$f$ is only defined in an interval, so that $f$ can be made periodic by extending it to the entire real line (or just to $[-2\pi,2\pi]$ if you prefer). In the end, the period $T$ of $f$ will be $2\pi$ because that's the length of the original interval ($[0,2\pi]$) in which $f$ was defined.

Your calculations are correct, because $4\pi$ is also a period of $f$, altough it is not the fundamental period (the smallest positive number that can be called "a period" of $f$). So both of the following are correct?

$$f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_n.\cos{\frac{2\pi nt}{T}}+b_n.\sin{\frac{2\pi nt}{T}}]$$
$$f(t)=\frac{\tilde a_0}{2}+\sum_{n=1}^{\infty}[\tilde a_n.\cos{\frac{2\pi nt}{\tilde T}}+\tilde b_n.\sin{\frac{2\pi nt}{\tilde T}}]$$

Where $a_n$ and $\tilde a_n$ (etc) are the coefficients from both formulas you posted (one considering $T=2\pi$, and the other $\tilde T=4\pi$).

Yes, they are the same. What's happening is that $\tilde a_n=\tilde b_n=0$ whenever $n$ is odd. You can check that this makes the formulas equal.

EDIT: the calculations are not entirely correct, because, as mentioned in Kevin Discroll's answer, you cannot make the simplifications for reducing the integral over $[-2\pi,2\pi]$ to twice the integral over $[0,2\pi]$ (you could make this simplification if the integrand was just $f(t)$, but you can't for the integrand $f(t)\cos nt/2$; see the comments in his answer). My answer above is only true if you define: $$\tilde a_n = \frac1{2\pi}\int_{-2\pi}^{2\pi}f(t)\cos (nt/2) dt$$ and likewise for $\tilde b_n$.


Intuitive interpretation of the Laplacian Just as the gradient is "the direction of steepest ascent", and the divergence is "amount of stuff created at a point", is there a nice interpretation of the Laplacian (a.k.a. divergence of gradient)?

Assume the function $f$ is $C^2$ in a neighbourhood of ${\bf 0}\in{\mathbb R}^n$. Using the Taylor expansion of $f$ at ${\bf 0}$ one can prove the following:
$$\Delta f({\bf 0}) =\lim_{r\to 0+}\ {2n\over r^2}{1\over \omega(S_r)} \int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})\ .\qquad (*)$$
This formula says that $\Delta f({\bf 0})$ is "essentially" (i.e., up to the scaling factor ${2n \over r^2}$) equal to the average difference $f({\bf x})-f({\bf 0})$ over small spheres around ${\bf 0}$.

Using this interpretation one gets, e.g., an intuitive understanding of the heat equation
$${\partial u\over\partial t}=a^2\ \Delta u\ ,$$
namely: If averaged over small spheres around a point ${\bf p}$ it is hotter than at ${\bf p}$ itself, then in the next second the temperature at ${\bf p}$ will rise.

Given the interest in the above formula $(*)$, here are some hints for the proof: By Taylor's theorem one has
$$f({\bf x})-f({\bf 0})= \sum_{i=1}^n f_{.i} x_i +{1\over2}\sum_{i,k} f_{.ik} x_i x_k + o(|{\bf x}|^2)\qquad({\bf x}\to{\bf 0}) .$$
Here the $f_{.i}$ and the $f_{.ik}$ are the partial derivatives of $f$ evaluated at ${\bf 0}$, whence constants. Now we integrate this over $S_r$ with respect to the surface measure ${\rm d}\omega$  and  obtain
$$\int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})={1\over2}\sum_{i}f_{.ii}\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x}) +o\bigl(r^{2+(n-1)}\bigr) \qquad(r\to0+)\ ,$$
because all other terms are odd in at least one variable. The integrals $\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x})$ are all equal; therefore we have
$$\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x})={1\over n}\int\nolimits_{S_r}\sum_k x_k^2\ {\rm d}\omega({\bf x})={r^2\over n}\omega(S_r)\qquad(1\leq i\leq n)\ .$$
Putting it all together we obtain
$$\int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})={r^2\over 2n}\omega(S_r)\Delta f({\bf 0}) +o(r^{n+1})\qquad(r\to 0+)\ ,$$
and solving for $\Delta f({\bf 0})$ we get the stated formula.

For a proof using Gauss' theorem see here:

Nice way of thinking about the Laplace operator... but what's the proof?

In addition to other answers, you also should pay attention to the fact that there is a connection between the Laplacian and the Gaussian kernel of the form
$$
\frac{1}{\sqrt{2\pi t}}e^{-\|x\|^2/2t}
$$
which in fact one of solutions of the equation $u_t = \Delta u$ mentioned by Chris. This kernel has a lot of applications to the averaging since it has a well-known "Bell shape" and is a density function for a normal (Gaussian) distribution.


Integer points on a surface I would like to understand ${\bf nonnegative \  integral}$ solutions $(x,y,z)$ on the surface
$$xyz-ax-ay-bz=d.$$
where $a,b,d$ are positive integers.

I can certainly prove that for a fixed $z$ there are finitely many non-negative solutions. But I'm having difficult time proving that it has finitely many solutions, even for some special triples $a,b,d$.

If $x=0$, then $-ay-bz=d$ where LHS is non-positive while RHS is positive. So, we have $x\not=0$. We have $y\not=0$ and $z\not=0$ similarly.

Now, we may suppose that $y\ge x\gt 0$ and $z\gt 0$.

If $\min(x,y,z)=z$, then
$$1=\frac{a}{yz}+\frac{a}{xz}+\frac{b}{xy}+\frac{d}{xyz}\le \frac{a}{z^2}+\frac{a}{z^2}+\frac{b}{z^2}+\frac{d}{z^3}$$
$$\implies z^3-(2a+b)z-d\le 0$$
The number of $z$ satisfying this inequality is finite.

Consindering
$$(zx-a)(zy-a)=bz^2+dz+a^2$$
for each $z$, we see that the number of $(x,y)$ is finite.

If $\min(x,y,z)=x$, then, similarly as above, the number of solutions is finite.

Example :

For your example where $a=9,b=4,d=37$, if $\min(x,y,z)=z$, we have
$$z^3-22z-37\le 0\implies z\le 5$$

For each $z=1,2,\cdots, 5$, consider
$$(zx-9)(zy-9)=4z^2+37z+81$$

For example, for $z=1$, we have
$$(x-9)(y-9)=2\times 61$$
$$\implies (x-9,y-9)=(1,2\times 61),(2,61)$$
$$\implies (x,y)=(10,131),(11,70)$$
since we already supposed that $y\ge x$.

There will always be finitely many (positive) solutions.

Supposing $a, b, c, d$ are all positive integers, then the positive integer solutions to

$$xyz = ax + by + cz + d \tag{1}$$

are bounded, for the following reason:

Observe that $(1)$ is equivalent to

$$\begin{align} z & = {ax + by + d \over xy + c} \\ & \le {ax + by + d \over xy} \tag2 \\ &  = \frac ay + \frac bx + \frac d{xy} \\ & \le a + b + d \tag3 \end{align}$$

$(2)$ because the new denominator is smaller (since $c \gt 0$), and $(3)$ similarly because $x, y \ge 1$.

By an identical argument you will find $1 \le x \le b + c + d$ and $1 \le y \le a + c + d$

Let $M = 3 \cdot \max \{a, b, c, d\}$ just for simplicity's sake; then all positive integer solutions to $(1)$ are bounded within the region $[1, M] \times [1, M] \times [1, M]$, within which there are only finitely many ($M^3$) integral points


Left and Right Cosets of a Subgroup of Index 2 Show that if $H$ is a subgroup of index $2$ in a finite group $G$, then every left coset of $H$ is also a right coset of $H$.

$H$ is a coset, and the only other possible coset is $G \setminus H$ by order considerations.


Let $A$ is nonsingular and each eigenvalue of $A$ is either $+1$ or $-1$.Why $A$ is similar to ${A^{ - 1}}$? Let $A \in {M_n}$ is nonsingular and each eigenvalue of $A$ is either $+1$ or $-1$.Why $A$ is similar to ${A^{ - 1}}$?

Characteristic equation of the matrix $A$ is of the form $(x-1)^a(x+1)^b$.
Since it is product of linear factors, it is similar to a matrix in Jordan form. You can continue from here

The eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$. By definition, similair matrices have the same eigenvalues.

The proof that $A^{-1}$ and $A$ have the same eigenvalues is simple:
$$Ax=\lambda x \iff A^{-1}Ax = A^{-1} \lambda x \iff \frac{1}{\lambda} x = A^{-1}x$$


What's new in higher dimensions? This is a very speculative/soft question; please keep this in mind when reading it. Here "higher" means "greater than 3".

What I am wondering about is what new geometrical phenomena are there in higher dimensions. When I say new I mean phenomena which are counterintuitive or not analogous to their lower dimensional counterparts. A good example could be hypersphere packing.

My main (and sad) impression is that almost all phenomena in higher dimensions could be thought intuitively by dimensional analogy. See for example, this link:

What this implies (for me) is the boring consequence that there is no new conceptual richness in higher dimensional geometry beyond the fact than the  numbers are larger (for example my field of study is string compactifications and though, at first sight, it could sound spectacular to use orientifolding which set a loci of fixed points which are O3 and O7 planes; the reasoning is pretty much the same as in lower dimensions...)

However the question of higher dimensional geometry is very related (for me) to the idea of beauty and complexity: these projections to 2-D of higher dimensional objects totally amazes me (for example this orthonormal projection of a 12-cube) and makes me think there must be interesting higher dimensional phenomena...

I would thank anyone who could give me examples of beautiful ideas implying “visualization” of higher dimensional geometry…

In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have
$$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$
Algebraically that's obvious, but geometrically I consider it highly surprising.

Edit:

Another surprising fact: In 4D and above, you can have a flat torus, that is, a torus without any intrinsic curvature (like a cylinder in 3D). Even more: You can draw such a torus (not an image of it, the flat torus itself) on the surface of a hyperball (that is, a hypersphere). Indeed, the three-dimensional hypersphere (surface of the four-dimensional hyperball) can be almost completely partitioned into such tori, with two circles remaining in two completely orthogonal planes (thanks to anon in the comments for reminding me of those two leftover circles). Note that the circles could be considered degenerate tori, as the flat tori continuously transform into them (in much the same way as the circles of latitude on the 2-sphere transform into a point at the poles).

If a 2-D plane is perpendicular to the diagonal of a 3-D cube and passes through the midpoint of this diagonal, its intersection with the cube is a regular hexagon.

If a 3-D hyperplane (i.e. given by an equation of the form ax + by + cz + dw = e) is perpendicular to the diagonal of a 4-D hypercube (or tesseract) and passes through the diagonal's midpont, its intersection with the hypercube is a regular octahedron.

I don't think this is very intuitive, but it makes sense after a little thought.
For the 3-D cube has six 2-D faces (squares) and that plane intersects all of them. So, it makes sense that the intersection is a hexagon.
The hypercube has eight 3-D hyperfaces (cubes), so that a hyperplane that intersects them all should do so in an octahedron.
And symmetry suggests the intersections should be regular (although that needs to be proved).


Equality of functions How a function which is not defined for some value can be equal to a function which is defined for the same value? How is $f(x) = \frac{(x-2)(x-3)}{(x-2)(x-4)}$ equal to $g(x) = \frac{(x-3)}{(x-4)}$ when their domains are different?

The actual problem is that what you wrote there are not the functions, but just two terms varying by a certain parameter.

In order to call those a function, you have to define the domain, which would look like this:

$$f: \mathbb{R}\setminus\{2, 4\}\rightarrow \mathbb{R}, x \mapsto f(x) := \frac{(x-2)(x-3)}{(x-2)(x-4)}\\
g: \mathbb{R}\rightarrow \mathbb{R}, x \mapsto g(x) := \frac{(x-3)}{(x-4)}\\$$
Now, $f\ne g$, but it equals the restriction of g to the domain of f: $f = g|_{\mathbb{R}\setminus \{2, 4\}}$

EDIT: Let me explain a bit more about how to compare terms and functions regarding equality.

Functions are defined as relations with certain properties (uniqueness & definedness of the argument):
$f: A\rightarrow B, x\mapsto f(x) :\Leftrightarrow f := \{(x, y)\in A\times B|y = f(x)\}$, which has to hold $\forall x\in A\exists!y\in B: (x, y)\in f$ as the defining property of a function.
If two functions are equal is therefore just a question of comparing two sets. It holds if both contain exactly the same elements $(x, f(x))$. Thus, they must have the same domain.
When comparing two expressions regarding equality, one has to be careful as well: Let $f(x), g(x)$ be expressions. $f(x) = g(x)$ would not be a logical statement, because We do not know what x is — therefore, we have to give an explicit x or quantorize it: $$\forall x: f(x) = g(x)$$ Because a quantorization without limiting x to a certain set (let's say $\mathbb{R}$) would only make sense if we would concern f(x) = g(x) = x, which would allow x to be something else than a number, you have to give a domain in some way as well.
For instance, let $f(x) := \frac{x-1}{x-1}, g(x) := 1$: $\forall x\in M: f(x)=g(x)$ would hold true for M = $\mathbb{R}\setminus \{1\}$, but not for $\mathbb{R}$, because there is no $f(x)$ at $x=1$.

The reason we can't check functions for equality is because:
 1. It reduces to The Halting Problem
 2. Functions tend to have an infinite set of inputs, and as others have mentioned, checking functions for equality is a question of comparing two sets.

But we could try to force both expressions to have the same format, then check symbol by symbol, I suppose.


PDE with homogeneous boundary conditions I'm trying to solve the following pde

$u_{xx}+u_{yy}=\sin(x)-\sin^3(x)$

where $x\in(0,\frac{\pi}{2}),y\in(0,2)$ with the conditions

$u(0,y)=u_{x}(\frac{\pi}{2},y)=u_{y}(x,0)=u_{y}(x,2)=0$.

I've tried solving through separation of variables but I get confused since the PDE is non-homogeneous. I also got advice that I should use Fourier transform. If anyone could help I would be grateful.

First, solve the homogeneous equation and its subsequent eigenvalue problems

\begin{cases} X''(x) + \lambda X(x) = 0, & X(0) = X'(\pi/2) = 0 \\
Y''(x) + \mu Y(x) = 0, & Y'(0) = Y'(2) = 0 \end{cases}

Since each set of eigenfunctions form a complete function space over $[0,\pi/2] \times [0,2]$, so we can write

$$ u(x,y) = \sum_{n=0}^\infty \sum_{m=0}^\infty c_{nm} \sin\big((2n+1)x\big)\cos\left(\frac{m\pi}{2}y\right) $$

Plugging this into the equation

$$ \sum_{n=0}^\infty \sum_{m=0}^\infty c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] \sin\big((2n+1)x\big)\cos\left(\frac{m\pi}{2}y\right) = -(\sin x - \sin^3 x) $$

Using the triple angle formula
$$ \sin x - \sin^3 x = \frac14\sin x + \frac14 \sin 3x $$

This is already a Fourier series in $x$ and $y$, which implies
\begin{cases}
c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] = -\dfrac14, & n=0, \ m=0 \\
c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] = -\dfrac14 , & n=1, \ m=0 \\
c_{nm} = 0, & \text{ otherwise} \end{cases}

or $c_{00} = -\frac14$, $c_{10} = -\frac{1}{36}$. Hence, the final solution is

$$ u(x,y) = -\frac14 \sin x - \frac{1}{36}\sin 3x $$

Seek a solution of the form

$$u(x,y)=\sum_{\lambda}X_{\lambda}(x)Y_{\lambda}(y)$$

where $X$ satisfies the following ODE and boundary conditions

$$X''(x)+\lambda X=0$$

$$X(0)=X\left(\frac{\pi}{2}\right)=0$$

Since we have Dirichlet boundary conditions for $X$, we know that $\lambda > 0$ and we have

$$X_{\lambda}(x)=\sin(\lambda^{\frac{1}{2}}x)$$

$$\lambda^{\frac{1}{2}}=2n, n \in \mathbb{N}$$

Substituting $u$ into the PDE gives

$$\sum_{n=1}^{\infty}\left \{ \sin\left( 2nx \right) \left[ Y''(y)-4n^2Y(y)\right] \right \}=\sin(x)-\sin^3(x)$$

By orthogonality,

$$Y''(y)-4n^2Y(y)=\frac{\left<\sin\left( 2nx \right), \sin(x)-\sin^3(x)\right>}{\left<\sin\left( 2nx \right), \sin\left( 2nx \right) \right>} = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(2nx)\sin(x)-\sin(2nx)\sin^3(x)dx$$

$$=\frac{16n(-1)^n}{\pi(2n-3)(2n-1)(2n+3)(2n+1)}$$

Let $G(y,\xi)$ be the Green function for this ODE satisfying $G(0,\xi)=G(2,\xi)=0$. Then,

$$Y(y)=\frac{16n(-1)^n}{\pi(2n-3)(2n-1)(2n+3)(2n+1)} \int_{0}^{2}G(y,\xi)d\xi$$

So the solution to the BVP is

$$u(x,y)=\sum_{n=1}^{\infty}\left \{\frac{16n(-1)^n\sin(2nx)}{\pi(2n-3)(2n-1)(2n+3)(2n+1)} \int_{0}^{2}G(y,\xi)d\xi \right \}$$


Find the distance between a function and a subspace in $L^2(0,\pi)$ Find a distance from vector $h(t)=t$ to the subspace
  $$M=\left\{x(t)\in L^2(0,\pi):\int_o^{\pi}x(t)\cos t \, dt=0\right\}.$$

So, it's a basic exercise, but I'm a bit confused solving that. Can someone explain how to do it? I know that by definition
 $$\operatorname{dist}(h,M)=\inf\|t-x(t)\|,$$
but this is a conditional extremum problem. Can I solve it using Fourier coefficients, we know that any function in a Hilbert space we can approximate of linear combination of basic elements?

I hope that someone explain how to find a distance using  the last way or may be there are exist more faster ways.

Start by writing
$$
                  x = \left[x-\frac{\langle x,\cos\rangle}{\langle \cos,\cos\rangle}\cos\right]+\frac{\langle x,\cos\rangle}{\langle \cos,\cos\rangle}\cos.
$$
$M$ consists of everything orthogonal to $\cos$, which includes the first term on the right. And the second term is orthogonal to $M$. So the orthogonal projection of $x$ onto $M$ is the first term, which must also be the closest point in $M$ to $x$. So, the distance from $x$ to $M$ is
\begin{align}
      \left\|x-\left[x-\frac{\langle x,\cos\rangle}{\langle\cos,\cos\rangle}\cos\right]\right\| & =\left\|\frac{\langle x,\cos\rangle}{\langle\cos,\cos\rangle}\cos\right\| \\
    & = \frac{|\langle x,\cos\rangle|}{\|\cos\|} \\
    & =\frac{\left|\int_{0}^{\pi}x\cos(x)dx\right|}{\left(\int_{0}^{\pi}\cos^2xdx\right)^{1/2}}.
\end{align}

Suppose you represent y(t) as x(t)+f(t), where x(t) is in M. Then the distance from y(t) to x(t) is f * f (note: I’m ignoring the square root because it’s irrelevant for optimization purposes). Setting the derivative to zero, 2f * df = 0. If y(t) is fixed and x(t) can vary, then x(t) can vary only within M, and thus dx is in M. Since f = y-x, and y is fixed, df = -dx, and thus df is in M as well. So f is a vector such that f*(elements of M) = 0. Intuitively, the distance from a point to a surface is the length of the normal vector from the surface to the point (a normal vector from the surface to the point is not in general unique, but with a hyperplane it is).

You are already given that cos(t) * (every element of M) = 0. With a little hand waving, it follows that f = a cos(t) for some scalar a, and so the distance from y to M is a*length of cos(t). a  is y * cos/(cos*cos), so the distance is y * cos/||cos||.


Linear programming with non-convex quadratic constraint Could anyone let me know if the following linear programming problem can be solved in polynomial time or should be NP-hard?

$\min c^Tx$

s.t. $x^TQx\geq C^2, x\in [0,1]^n,c\in \mathbb{R}_+^n,Q\in\mathbb{R}_+^{n\times n}$ is positive semi-definite matrix.

I am afraid you can't.

This is a not convex problem due to the quadratic constraint. If your $Q$ matrix were negative semidefinite, then it would have been solvable in polynomial time.

My guess is that this problem can be solved in polynomial time.

It is well known that $\min_x c'x$, such that $x'Qx \geq C$ can be solved in polynomial time (since SDP relaxation is tight). Your problem has extra bound constraints $x_i$ in $[0,1]$, which may or may not change the answer. But I think it is quite possible.


How to compute the quadratic variation of a compound poisson process? The jump diffusion model is defined as
$$dS_t = \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\;\;\;\;\;\;\;(1)$$
, where ${V_i}$ is a sequence of iid non-negative random variables and it is independent of $W_t$. In the Merton's jump diffusion model ,
$log(V) \sim N(\mu_J, \sigma^2_J)$ and $ N_t$ is a poisson process with rate $\lambda$.

I was asked to apply Ito lemma to $(d \;logS_t)$ to obtain the following:

$$S_t = S_0 exp \left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t\right) \prod^{N_t}_{j=1}V_j \;\;\;\;\;\;\;(2)$$

I literally do not know how to solve this problem because of the term $d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)$. This is how far I got to:

$$dlogS_t = \frac{1}{S_t} \left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\right) - \frac{1}{2S_t^2} d[S,S]_t$$

What exactly is $d[S,S]_t$ ? I know that
$$d[S,S]_t = \sigma^2 S^2_t dt + ...$$
But what is that "..." ?

I suspect that your SDE defining the $S_t$ process is problematic in the term "$S_t\text d\Big(\sum\limits_{i=1}^{N_t}(V_i-1)\Big)$". Intuitively, this term should capture instantaneous random jumps when they occur in any given realisation of the process. That is, if a jump occurs at time $t$, then the process jumps from $S_{t-}$ to $S_{t-}V_t$, where $S_{t-}$ is the left limit of the process at time $t$. So, it makes more sense for $S_t$ to be the jump-diffusion Levy process (so right-continuous, left-limit) defined by
$$
\text dS_t = \mu S_{t-}\text dt + \sigma S_{t-}\text dW_t + \text d\Big(\sum\limits_{0\leqslant r\leqslant t}S_{r-}(V_r - 1){\bf 1}_{\{N_r-N_{r-}=1\}}\Big),\tag{1}
$$

with jumps determined by a Compound Poisson process independent of $\{W_t\}_{t>0}$. That is, a Poisson process $\{N_t\}_{t>0}$ determines when jumps occur, and the size of such a jump at time $t$ is determined as $S_{t-}(V_t - 1)$ (where $\log V_t\sim\mathcal N(\mu_{J},\sigma^2_{J})\ $ for i.i.d. $V_t$).

Assuming $(1)$ for your SDE and $\mathbb E[\int_{0}^{T}S_{u-}^2\text du] < \infty$, Ito's lemma for jump-diffusion processes is applicable here. Note that, for a $\ C^{1,2}\ $ function $\ f:[0,T]\times\mathbb R\to\mathbb R$, the process $f(t,S_t)$ satisfies

  $$
\begin{eqnarray*}
f(t,S_t) - f(0,S_0) &=& \int_{0}^{t}(\partial_uf + \mu S_{u-}\partial_sf + \frac{1}{2}\sigma^2S_{u-}^2\partial_{ss}f)\text du \\
&& + \sigma \int_{0}^{t}S_{u-}\partial_sf~\text dW_u + \sum_{0\leqslant r\leqslant t}(f(r,S_r)-f(r-,S_{r-})){\bf 1}_{\{N_{r}-N_{r-}=1\}}\,.
\end{eqnarray*}
$$

This is simply Ito's lemma for a diffusion process but, in addition, the final term adds the change in the process due to the jumps determined by the compound Poisson process. So, choosing $f(S_t):=\log S_t$ gives us

$$
\begin{eqnarray*}
\log\Big(\frac{S_t}{S_0}\Big) &=& (\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log\Big(\frac{S_{r-}V_r}{S_{r-}}\Big) \\
&=&(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log V_r\,.
\end{eqnarray*}
$$
Therefore, in your notation,

  $$
S_t = S_0 e^{(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t}\prod_{i=1}^{N_t}V_i\,.
$$

In your case

the quadratic variation can be obtained by formally squaring the SDE
$
d[S,S]_t = dS_t \cdot dS_t =\left(  \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right) \right)^2
$
using the fact that
$
(\mu S_t)^2 dt^2 = 0
$
and
$
\mu S_t dt \cdot \sigma S_t dW_t =0
$
we get

$
d[S,S]_t = \sigma^2 S_t^2 dt +  2 [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), \mu S_t dt ] +
2 [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), dW_t ] +   [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), d\left(\sum^{N_t}_{i=1}(V_i - 1)\right) ]_t
$.

Since $V_i $ are i.i.d. $ [dV_i, dt ]=0$ and $ [dV_i, dW_t ]=0$ and  $ [dV_i, dV_j ]=\delta_{i\,j}$.
For the last term we also have $[dV_i  , dV_i ] = dV_i$.
Then finally

$
dlogS_t = \frac{1}{S_t} \left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\right) - \frac{1}{2S_t^2}  \sigma^2 S^2_t dt =
$

$
dlogS_t = \left( \mu - \frac{1}{2 }  \sigma^2 \right) dt + \sigma \;  dW_t + d\left(\sum^{N_t}_{i=1}(V_i - 1)\right)
$.


How can I solve this system of linear different equations? Here's the system $$\frac d{dt} \begin{bmatrix} x \\ y \\ z \\ p_1 \\ p_2 \\ p_3\end{bmatrix} = \begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ p_1 \\ p_2 \\ p_3\end{bmatrix}$$
where $\begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix}$ is a block matrix and $A = \frac 1{mR^2}I_3$ and $B = mR^2\omega^2\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{bmatrix}$.  $m, R,$ and $\omega$ are all just constants.  $x,y,z,p_1, p_2$, and $p_3$ are all functions of $t$.

I have a feeling I need to use the matrix exponential, but my brain isn't working right now.

Let $E = \left[ \matrix{1 & 1 & 1\cr 1 & 1 & 1\cr 1 & 1 & 1\cr} \right]$.
With Maple's help, I get (if $C$ is your $6 \times 6$ matrix)

$$ \exp(tC) = \dfrac{1}{3} \left[\matrix{E & (t/mR^2) E\cr 0 & E\cr} \right] +
\cos(\sqrt{3}\omega t) \left[ \matrix{I_3- E/3 & 0\cr 0 & I_3 - E/3\cr}\right] + \dfrac{\sin(\sqrt{3}\omega t)}{\sqrt{3}} \left[ \matrix{0 & \dfrac{I_3 - E/3}{mR^2\omega}\cr mR^2\omega(E-3I_3) & 0\cr} \right]$$


Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means.

The problem stated is

  Does a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ exist?

(Here P is the perimeter of the triangle. Sum of the sides. And A is the area of the triangle. a*b/2 in a triangle with a b c and c is the hypotenuse)

Obviously for the simple cases such triangles exists. For an example
when

$ A=2P \qquad $ the triangle $12,16,20$ works

$ A=P  \qquad \; \; $ the triangle $6,8,10$ works

$2A=P  \qquad $ the triangle $3,4,5$ works

I tried solving this by hand, first for the special case where $2P=A$. This ended up giving me

$$ \frac{a \cdot b}{2} = A $$

and

$$ P = \frac{A}{2} = \frac{a \cdot b}{4} \qquad \text{also} \qquad P = a + b + c  $$

So

$$ \frac{ab}{4} = a + b + c $$

By knowing that this is a right angle this leads to the equation (Using the Pythagorean theorem)

$$ a^2 + b^2 = c^2 $$

Now we have two equations and three unknowns, which also needs to be integers! Sadly I was not able to continue from here. I have only learned how to solve linear Diophantine equations. Not a system of nonlinear Diophantine equations.

Just to restate my question below =)

  Is there a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ ?

Regards, Werner

We have the following relations:
$$P=a+b+c,\quad 2A=ab,\quad a^2+b^2=c^2$$

In fact, there is one more that is implicit: the triangle inequality $a+b>c$ (which follows from $a^2+b^2=c^2$). We wish to relate the first and second relations, and so after staring at the problem for a while we realize the correct course of action is to multiply $P$ by $a+b-c$ (the latter guaranteed to be positive by the triangle inequality). This gives:

$$\begin{align}P(a+b-c)&=(a+b+c)(a+b-c)=(a+b)^2-c^2\\
&=a^2+b^2+2ab-c^2=2ab=4A
\end{align}$$

Hence in general for a right triangle we have $k=A/P=(a+b-c)/4$, so the question boils down to when the difference between the sum of the lengths of the legs and the length of the hypothenuse divides or is divisible by $4$, for a right triangle with integer side-lengths.

Here we actually need to generate pythagorean triples $(a,b,c)$ such that $a,b,c$ are integers and $a^2+b^2=c^2$; the wikipedia page here gives a good picture of Euclid's formula that I recommend thinking through. Euclid's formula is the following:

$$a=m^2-n^2,\quad b=2mn\quad c=m^2+n^2$$ for integers $m>n$.

Then the quantity $a+b-c$ is $m^2-n^2+2mn-m^2-n^2=2n(m-n)$, so $A/P=k=n(m-n)/2$. The options are now that either $2$ divides $n$ or $2$ divides $m-n$, so we either have $n=2d$ and $m=k/d+2d$ or $n=d$ and $m=2(k/d)+d$ for divisors $d$ of $k$.

(to get $P/A=k$ we need $2n(m-n)$ to divide $4$, so $n(m-n)$ must divide $2$, thus $n=1$, $m=2$, $n=1$, $m=3$, and $n=2$, $m=3$ are the only solutions, giving us $(3,4,5)$, $(8,6,10)$, $(5,12,13)$ with $k=2,1,2$ being the only possibilities).


Limit of an integral over a ball as the radius of the ball goes to zero 2 Let $D(a,r)$ be an open ball in $\mathbb{R}^{k}$ ($ k\geq1 $), and  $f$ locally integrable function in $\mathbb{R}^{k}$. Do we have:
$$\lim_{r\to 0}\int_{D(a,r)}f(t)dt=0?$$

The answer is YES. This is a classic result called "absolute continuity of The Lebesgue integral", for example see
rmh52 (https://math.stackexchange.com/users/35961/rmh52), Absolute continuity of the Lebesgue integral, URL (version: 2017-02-03): Absolute continuity of the Lebesgue integral


Using an impulse to bring an oscillator to rest The Problem: Consider the system 2y''+y'+2y=δ(t-5), in which an oscillation is excited by a unit impulse at t=5. Suppose that it is desired to bring the system to rest again after exactly one cycle--that is, when the response first returns to equilibrium moving in the positive direction.

(a) Determine the impulse kδ(t-t0) that should be applied to the system in order to accomplish this objective. Note that k is the magnitude of the impulse and t0 is the time of its application.

(b) Solve the resulting IVP.

The book's answer for (a): -e-T/4δ(t-5-T);
The book's answer for (b): T=(8*pi)/(sqrt(15))

So far: I found part (a) on my own and it is correct. I am having trouble properly setting up the resulting IVP. Thanks for your help!

You have the differential equation for the IVP.  After $t=5$ the right side is zero.  Just after $t=5$, the oscillator can't have moved instantaneously, so $y(5)=0$.  It will have a velocity of $\frac 12$ from the impulse, so after $t=5$ you have $2y''+y'+2y=0, y(5)=0, y'(5)=\frac 12$


AP statistics question A bag contains 2 black and 2 white marbles, and there is a supply of extra marbles of each color. A move consists of randomly drawing 2 marbles from the bag. If the marbles are the same color, they are replaced by two black marbles. If they are different collors they are replaced by two white marbles.

Find the probability that after 4 moves, all the marbles are black in the bag. Give Answer as fraction, reduced to Lowest terms.

Please help. I’m not doing well in statistics and need to answer this problem!

Hint:  There are five states of the bag.  From each state you can go to a maximum of three others.  So there are (at most) 15 transition probabilities to figure-in fact a few less.  For example, from (2B,2W) the chance to go to (4B,0W) is $\frac 16$ as you have to draw the two white balls.  What is the probability distribution after one draw?  Then keep going.


Three isosceles triangles △ABC', △AB'C, and △A'BC were constructed on the sides of triangle ABC. Three isosceles triangles $△ABC'$, $△AB'C$, and $△A'BC$ were constructed on the sides of triangle $ABC$. The vertex angles have the following measurements; $∠A'=α, ∠B'=β, ∠C'=γ$. Prove that if $α+β+γ=360°$ then the measurements of the angles of $△A'B'C'$ do not depend on the measurements of the angles of $△ABC$.

$D$, $E$, and $F$ are the midpoints of the sides of $△ABC$, making $A'E$, $B'F$, and $C'D$ the perpendicular bisectors of their respective triangles.

Let us consider the circles $\Gamma_A,\Gamma_B,\Gamma_C$ centered at $A',B',C'$ with radii $A'B,B'C,C'A$.
The constraint on the angles ensures that these circles meet at a point $P$.

We also have that $PB'AC', PC'BA'$ and $PA'CB'$ are kites, therefore $\widehat{B'A' C'}=\frac{1}{2}\alpha$ and so on.

Comment: These figures helps you the solution: In figure (b) the triangle is equilateral and $A'=B'=C'=120^o$ so angles of triangle A'B'C' are all $60^o=\frac{A'}2=\frac{B'}2=\frac{C'}2$
In figure (a) The sum of interior angle of triangle A'B'C' is $180^o=\frac{\alpha +\beta+\gamma}2$. Noting the fact that every triangle can be considered as a transformed equilateral triangle, keeping the point O which is intersection of perpendicular bisectors fixed, one may conclude that angles in triangle A'B'C' keep their particularities and have these values: $B'A'C'=\frac{\alpha}2$, $A'B'C'=\frac{\beta}2$ and $B'C'A'=\frac{\gamma}2$ which is not a function of the angles of triangle ABC.
PS: Figure d is transformed of figure d. Point O , measure of side AC and its distance to O is kept constant. Vertex B is moved to be co-linear with perpendicular bisector of AC. In all positions of B the conditions hold including when AB=BC=AC which is shown in figure d.Reverse action converts figure d to c.


Use Fubini's Theorem to verify the integration by parts formula, and justify that the hypotheses of Fubini's Theorem are satisfied. Let $f$ and $g$ be Lebesgue integrable real-valued functions on $[0,1]$, and define
$$F(x) = \int_0^x f(t)dt, \quad \text{and} \quad G(x) = \int_0^x g(t) dt.$$
Use Fubini's Theorem to verify the integration by parts formula
$$\int_0^1 F(x)g(x)dx = F(1)G(1) - \int_0^1 f(x)G(x)dx.$$
Justify that the hypotheses of Fubini's Theorem are satisfied.
This one seems straightforward to us, but for some reason we can't get anything nice to appear.
Here are our attempts for the left-hand side and the right-hand side respectively after the hint:
$$\begin{aligned} \int_0^1 F(x)g(x)dx &= \int_0^1 \int_0^x f(t)dtg(x)dx\\
&= \int_0^1\int_0^1 1_{[0,x]}f(t)g(x)dtdx\\
&= \int_0^1 f(t) \int_0^1 1_{[0,x]}g(x)dxdt. \end{aligned}$$
$$\begin{aligned} F(1)G(1) - \int_0^1 f(x)G(x)dx &= F(1)G(1) - \int_0^1 f(x)\int_0^x g(t)dtdx\\
&= F(1)G(1) - \int_0^1\int_0^1 1_{[0,x]}f(x)g(t)dtdx\\
&= F(1)G(1) - \int_0^1g(t)\int_0^1 1_{[0,x]}f(x)dxdt.\end{aligned}$$

The first few steps just involves rewriting the terms as they are defined and moving constants inside or outside of an integral as needed.
$$
\begin{align*}F(1)G(1)&=\Big(\int_0^1f(t)dt\Big)\Big(\int_0^1 g(x)dx\Big)= \int_0^1g(x)\Big(\int_0^1 f(t)dt\Big)dx\\
&= \int_0^1g(x)\Big(\int_0^x f(t)dt\Big)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\\
&= \int_0^1 g(x)F(x)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\end{align*}$$
Now we use Fubini's Theorem to show that the second integral takes the appropriate form. Well
$$\begin{align*}\int_0^1 g(x)\Big(\int_x^1 f(t)dt\Big)dx &= \int_0^1g(x)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)f(t)dt dx\\
&= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dt dx\\
&= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dxdt\\
&= \int_0^1f(t)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)dx dt \\
&= \int_0^1f(t)\Big(\int_0^1 1_{[0\leq x\leq t]}(x)g(x)dx\Big) dt \\
&= \int_0^1f(t)\Big(\int_0^t g(x)dx\Big) dt \\
&= \int_0^1 f(t)G(t)dt\end{align*}$$
Hence after a change of variables we have that
$$F(1)G(1) = \int_0^1 g(x)F(x)dx + \int_0^1 f(x)G(x)dx $$

$$f(x)=u(x)v(x)$$
$$\frac{df}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u$$
$$df=v.du+u.dv$$
$$\int df=\int v.du+\int u.dv$$
$$\int u.dv=f-\int v.du$$
$$\int u.dv=uv-\int v.du$$
If you don't have to use Fubini's theorem


Presentations: fewer relations than generators implies $G$ infinite. Take $r=0$. Let $G=\langle a_1\rangle \times ...\times \langle a_n\rangle, \langle a_i\rangle$ infinite cyclic. $G$ is generated by $n$ elements and all relations in $A$ are relations in $G$. Therefor by von  Dyck's theorem there is an epimorphism from $A$ to $G$. But $G$ is infinite and so, $A$ is infinite too.

For the case $r>0$ I have been unable to find a proof. I don't want a solution. I only ask for a hint which allows me to begin working in the problem.

Hint: $A$ is an abelian group, so think of it as a $\Bbb Z$-module.  $A$ can be presented as the quotient of $\Bbb Z^n$ by the (free) submodule generated by the $r$ relations.

I get it. I have $G=\langle m_1,...,m_n\mid s_i(m)\rangle$ where the $s_i$ are the same relations as in $A$. For instance, if $2x_1+x_2=0$ is a relation in $A$ then $2m_1+m_2=0$ is a relation in $G$. So $G$ is a subgroup of $Z$ and, as such, it is infinite. Now von Dyck gives an epimorphism $\beta:A\to G$ and $A$ is infinite.

However, how do I define the homomorphism $\alpha:A\to \langle t\rangle$?


Probability of getting more Heads between two gamblers I'm trying to solve this probability question I thought of. If two gamblers are playing a coin toss game and Gambler A has $(n+2)$ and B has $n$ fair coins. What is the probability that A will have more heads than B if both flip all their coins?

I tried to solve it like this, using symmetry. If we compare the first $n$ coins of of $A$ and $B$:

$E_{1}$: Event that A and B have the same number of Heads

$E_{2}$: Event that A has 2 Heads less than B

$E_{3}$: Event that A has 1 Head less than B

$E_{4}$: Event that A has 3 Heads less than B

$E_{5}$: Event that A has 2 Heads more than B

$E_{6}$: Event that A has 1 Head more than B

Let $P(E_{1}) = y$, $P(E_{2}) = P(E_{5}) = x$, $P(E_{3}) = P(E_{6}) = z$, $P(E_{4}) = k$ which implies $y + x + x + z + z + k = 1$ and probability of A having more heads = $y*0.75 + x + z + z(0.5^{2})$. However, I don't know how to solve the question from here.

Thank You

The probability that $A$ wins when it has two extra throws is related to the probabilities involving them both having $n$ throws ...

$$P( A \text{ wins in }n+2 \text{ throws})
 \\=P( A \text{ wins in }n \text{ throws})\\
 +\frac 34 P( A \text{ ties in }n\text{ throws})\\
 +\frac 14P( A \text{ down by one  in }n \text{ throws})$$

Consider the following problems ...

Assume $A$ and $B$ both throw $n$ fair coins

(a) What is the probability they toss the same number of heads ?

$$P(\text{ same }) = \sum_{k=0}^n  \binom nk^2 (0.5)^{2n}\\  = (0.5)^{2n}\sum_{k=0}^n  \binom nk^2 \\
= (0.5)^{2n}\binom {2n}{n} $$
Where the last equality comes from the normalization of the hypergeometric distribution

$$\sum_{k=0}^{X}\binom nk \binom {N-n}{X-k}=\binom N X$$
so
$$\sum_{k=0}^n  \binom nk^2 =\sum_{k=0}^n  \binom nk \binom {n}{n-k}\\ =\sum_{k=0}^{n}\binom nk \binom {2n-n}{n-k}=\binom {2n}{n}$$

So
$$ P( A \text{ ties in }n\text{ throws}) = (0.5)^{2n} \binom {2n}{n}$$
and by symmetry
$$ P( A \text{ wins in }n\text{ throws}) = \frac 12 \left( 1 - (0.5)^{2n} \binom {2n}{n} \right )$$

(b) What is the probability that $B$ tosses one more head than $A$  when they both throw $n$ fair coins ?

$$P(\text{ down by one  }) = \sum_{k=1}^{n}  \binom nk \binom{n}{k-1} (0.5)^{2n}
 \\  = \sum_{k=1}^{n}\binom nk \binom{2n-n}{n-k+1} (0.5)^{2n}
 \\=  \binom{2n}{n+1} (0.5)^{2n}
$$

So putting it all together

$$ P( A \text{ wins in } n+2 \text{ throws})
\\ = \frac 12 \left( 1 - (0.5)^{2n} \binom {2n}{n} \right ) + \frac 34 (0.5)^{2n} \binom {2n}{n}+\frac 14 (0.5)^{2n} \binom {2n}{n-1}
\\=\frac 12 +(0.5)^{2n+2}\left[ \binom {2n}{n} +\binom {2n}{n-1} \right ]
$$

You were right for using symmetry. Suppose A and B both have n coins. The probability of A winning would be 0.5. Now suppose A has $(n+2)$ coins and flips $n$ coins first then flips the other 2 coins. The probability is 0.5 for A to win with $n$ coins, and adding the 2 coins, the probability of at least one of them being heads is 3/4. So the probability of A winning is $3/4$.


Zero probability and impossibility I read a comment under this question:

  There are plenty of events that can occur that have zero probability.

This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering

if zero probability and impossibility mean the same?
if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?

Thanks and regards!

Two Schools

I think the crux of the matter is what probability actually is:

The Bayesian view - probabilities are measures of (personal) confidence or belief, so it's quite obvious why an event with probability zero is not the same thing as an impossible event. But perhaps this isn't such a satisfactory answer.
The frequentist view - probabilities are the asymptotic frequency of events as the number of independent trials tends to infinity. Here again wee see that something that happens with probability zero is not the same as something impossible; it's just something that happens so infrequently that the numerator in $\dfrac{\text{occurences}}{\text{trials}}$ is dominated by the denominator.

Technically Speaking

Putting aside such philosophical matters, there's also a technical matter to be discussed here. Under the usual measure-theoretic formulation of probability theory, we have a sample space $\Omega$ and a family $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ of events (measurable subsets of $\Omega$), and the probability of an event $A \in \mathcal{F}$ is its measure $\mathbb{P}(A)$. There is nothing in the axioms of measure theory which say that a non-empty set must have a non-zero measure; and if we interpret $\mathcal{F}$ as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability.

Example

To give a concrete example, consider a random variable $X$ which is uniformly distributed on the interval $[0, 1]$. Although $\mathbb{P}[X \in (a, b)] = b - a$ for all $(a, b) \subset [0, 1]$, the axioms of probability force us to conclude that $\mathbb{P}[X = x] = 0$ for any individual $x \in [0, 1]$: for if $\mathbb{P}[X = x] = \varepsilon > 0$, because $X$ is uniformly distributed, by additivity of the probabilities of disjoint events, we'd be forced to conclude that $[0, 1]$ contains at most $\frac{1}{\varepsilon}$ (a finite number!) points, which is absurd.

I don't know if this is correct, but this is how I made sense of it, I think this a very intuitive explanation.

Imagine a car moving (from left to right) in a straight line with constant speed of 100MPH, and at x=0 the breaks of the car are engaged. Naturally the car will start to decelerate until it comes to rest (speed = 0 MPH).

The car will certainly come to rest, but the question is:

At what distance D [meters] (with respect to x=0) will the car come to rest? (This value D is not known apriori and can vary from one experiment to the next by a myriad of factors, so we can ask probability questions about the variable D).

Since distance is modeled by real numbers, the variable D is continuous. If we ask:
what is P(D=6.354), the probability that it comes to rest at specifically D = 6.354 meters?

then, since the probability distribution of D is continuous, the answer is P(D=6.354)=0. This probability is zero.

Notice that there is nothing special about the number "6.354". That is, for any real number r the probability that D is exactly r is zero, i.e. P(D=r) = 0. This is simply a consequence of how continuous probability distributions are defined.

Regardless, we know that the car will certainly come to rest and that this happens at some specific distance d (where d is a real number). So even though P(D=d) = 0, it does not mean that coming to rest at D=d was impossible; clearly it was not because the car does come to rest at d.


Writing function as infinite Fourier sum with sine kernel Let $f\in L^2(\mathbb{R})$ be such that $\hat{f}$ is supported on $[-\pi,\pi]$. Also, $f$ is continuous and goes to $0$ at $\pm \infty$. Let $K(y)=\dfrac{1}{\pi y}\sin(\pi y)$. Show that $f(x)=\sum_{n=-\infty}^\infty f(n)K(x-n)$ for every $x$.

This looks like a Fourier sum of some sort, but it's rather strange that the index $n$ goes from $-\infty$ to $\infty$. How do we approach it?

Disclaimer: I do all the computation below without any care for the factors $2\pi$. No one shall ever do that, it's bad. But the ideas are there and I don't have a lot of time.

We set $BL^2 := \{f \in L^2 (\mathbb R) \mid \hat f = 0 \text{ almost everywhere on }\mathbb R \setminus [-\pi,\pi]\}$

First observe that $f\in BL^2 \Rightarrow f\in L^2(\mathbb R) \Rightarrow \hat f \in L^2(\mathbb R)$

{$\hat f(\mathbb R)$ has bounded support and $\hat f \in L^2(\mathbb R)$} $\Rightarrow \hat f \in L^1(\mathbb R)$

For $ f\in BL^2$, taking the inverse Fourier transform $\mathcal F^{-1}:L^2(\mathbb R)\to L^2(\mathbb R)$ of $\hat f$ then gives a continuous function, hence $f$ has a unique continuous version, which we still denote by $f$.

(In fact this version is even an holomorphic function.)

It thus makes sense to consider $f(n)$. (This was not obvious as $f$ was a priori only considered to be an element of $L^2(\mathbb R)$.)

Now we write $\hat f |_{[-\pi,\pi]}\in L^2([-\pi,\pi])$ as a Fourier series:

$$\hat f (\xi)=\sum_{n\in\mathbb Z} c_n(\hat f) e^{in\xi}$$
with
$$c_n(\hat f) = \frac{1}{2\pi}\int_{[-\pi,\pi]}\hat f (\xi) e^{-in\xi}d\xi= \frac{1}{2\pi}\int_{\mathbb R}\hat f (\xi) e^{-in\xi}d\xi=\mathcal F^{-1} \hat f (n) = f(n)$$
with $(c_n(\hat f))_{n\in \mathbb Z}=(f(n))_{n\in \mathbb Z} \in \ell^2(\mathbb Z)$.

We thus get the formula
$$\hat f (\xi)=\sum_{n\in\mathbb Z} f(n) e^{in\xi}$$
as an equality in $L^2([-\pi,\pi])$, but then
$$\hat f (\xi)=\sum_{n\in\mathbb Z} f(n) e^{in\xi}1_{[-\pi,\pi]}(\xi)$$
where the series converges in $L^2(\mathbb R)$. We can thus take the inverse Fourier transform of both sides and obtain the equality
$$f (x)=\sum_{n\in\mathbb Z} f(n) \mathcal F^{-1}\big(e^{in\xi}\big|_{[-\pi,\pi]}\big)(x)=\sum_{n\in\mathbb Z} f(n) K(x-n)\qquad (*)$$
in $L^2(\mathbb R)$.

Now as a last remark we observe that for $u\in BL^2$, $\|u\|_\infty\leq \frac{1}{\sqrt{2\pi}}\|u\|_2$ as
\begin{align}
|u(x)|&=|\frac{1}{2\pi}\int_{\mathbb R}\hat u (\xi)e^{-i\pi x\xi}d\xi|\\
& = |\frac{1}{2\pi}\int_{[-\pi,\pi]}\hat u (\xi)e^{-i\pi x\xi}d\xi|\\
(Cauchy-Schwarz)\qquad & \leq \frac{1}{\sqrt{2\pi}} (\int_{[-\pi,\pi]} |\hat u (\xi)|^2d\xi)^{1/2}\\
& = \frac{1}{\sqrt{2\pi}}(\int_{\mathbb R} |\hat u (\xi)|^2d\xi)^{1/2}\\
(Plancherel)\qquad& = \frac{1}{\sqrt{2\pi}}\|u\|_2 \,.
\end{align}

Thus in $BL^2$ the convergence in $L^2$ implies the convergence in $\|\|_\infty$ and the formula $(*)$ is not only valid almost everywhere but for all $x$ in $\mathbb R$.

For more details see Willem, Analyse harmonique réelle, 1995. Section 6.3.

Theoram 2.6 in
http://math.uchicago.edu/~may/REU2012/REUPapers/Cuddy.pdf
gives proof.
index going from −∞ to +∞ is equalent to convolution


Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I have been wondering whether the following limit is being used somehow, as a variation of the derivative:

$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} .$$

Edit:
I know that this limit is defined in some places where the derivative is not defined, but it gives us some useful information.

The question is not whether this limit is similar to the derivative, but whether it is useful somehow.

Thanks.

The "symmetric difference" form of the derivative is quite convenient for the purposes of numerical computation; to wit, note that the symmetric difference can be expanded in this way:

$$D_h f(x)=\frac{f(x+h)-f(x-h)}{2h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}h^2+\frac{f^{(5)}(x)}{5!}h^4+\dots$$

and one thing that should be noted here is that in this series expansion, only even powers of $h$ show up.

Consider the corresponding expansion when $h$ is halved:

$$D_{h/2} f(x)=\frac{f(x+h/2)-f(x-h/2)}{h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}\left(\frac{h}{2}\right)^2+\frac{f^{(5)}(x)}{5!}\left(\frac{h}{2}\right)^4+\dots$$

One could take a particular linear combination of this half-$h$ expansion and the previous expansion in $h$ such that the term with $h^2$ zeroes out:

$$4D_{h/2} f(x)-D_h f(x)=3f^\prime(x)-\frac{f^{(5)}(x)}{160}h^4+\dots$$

and we have after a division by $3$:

$$\frac{4D_{h/2} f(x)-D_h f(x)}{3}=f^\prime(x)-\frac{f^{(5)}(x)}{480}h^4+\dots$$

Note that the surviving terms after $f^\prime(x)$ are (supposed to be) much smaller than either of the terms after $f^\prime(x)$ in the expansions for $D_h f(x)$ and $D_{h/2} f(x)$. Numerically speaking, one could obtain a slightly more accurate estimate of the derivative by evaluating the symmetric difference at a certain (well-chosen) step size $h$ and at half of the given $h$, and computing the linear combination $\dfrac{4D_{h/2} f(x)-D_h f(x)}{3}$. (This is akin to deriving Simpson's rule from the trapezoidal rule). The procedure generalizes, as one keeps taking appropriate linear combinations of a symmetric difference for some $h$ and the symmetric difference at half $h$ to zero out successive powers of $h^2$; this is the famous Richardson extrapolation.

This is a wrong way of thinking, the comment by Jesse Madnick explains why.

This is symmetric differentiation and is useful when both left and right hand limits exists, as oppose to usual derivative definition when only it is sufficient that right hand limit should exist.

The only place that I have seen that to be of use to make things nicer was in Fractional Calculus, providing some nice formulas to generalise differentiation.

To see why, try using the above limit definition of differentiation to get 2nd , or 3rd order definitions of a derivative, you should be able to recognise the resulting pascal/binomial like looking formulas, from there one might just try to generalise that to get a derivative definition that would say something about $\tfrac{1}{2}^{\text {th}}$ derivative.

In short : there is nothing more to this definition than the other usual one, except that this one shows an extra symbol (2), why use one more extra symbol when there is no need? maybe just to reap the benefits of it being symmetric.

Edit : Is somebody going to point the mistake? What am I missing here?


Does the $k$th forward difference of Radó's $\Sigma$ eventually dominate every computable function? Let $\Sigma$ be Radó's Busy Beaver function, and let $\Delta[\Sigma]$ denote the forward difference of $\Sigma$, such that $\Delta[\Sigma] \ (n) = \Sigma(n+1) - \Sigma(n)$ for all $n \in \mathbb{N}$. Define $\Delta^0[\Sigma] = \Sigma$, and $\Delta^{k+1}[\Sigma] = \Delta[\Delta^{k}[\Sigma]]$, for all $k \in \mathbb{N}$.

Definition: Given functions $f,g$ with the common domain $\mathbb{N}$ and codomains that are subsets of  $\mathbb{Z}$, $g$ is said to eventually dominate $f$ iff $g(n) > f(n)$ for all sufficiently large $n\in\mathbb{N}$.

Question: Is it the case that for all $k \in \mathbb{N}$, the function
$\Delta^k [\Sigma]$ eventually dominates every computable function $f: \mathbb{N} \rightarrow \mathbb{N}$?

(I believe this is the case, but do not know how to prove it. Any pointers or sources on proving this -- or disproving it, if I'm wrong?)

Here's a difference table showing all the known values for these sequences:

     n:  0      1      2      3      4      5     ...
     ------------------------------------------------
     Σ:  0      1      4      6      13    ≥4098  ...
    ΔΣ:  1      3      2      7     ≥4085  ...
   ΔΔΣ:  2     -1      5     ≥4078  ...
  ΔΔΔΣ: -3      6     ≥4073  ...
 ΔΔΔΔΣ:  9     ≥4067  ...
ΔΔΔΔΔΣ: ≥4058  ...

Known sources: "A note on Busy Beavers and other creatures" (1996), by Ben-Amram, Julstrom, and Zwick, contains the conjecture that for any computable function $f$, there exists a constant $N_f$ such that $\forall n \gt N_f, \Sigma(n+1) > f(\Sigma(n))$. In support of this they prove the weaker result that for any computable function $f$, $\Sigma(n+1) > f(\Sigma(n))$ for infinitely many values of $n$. (Considering $f(x) = 2x$, for example, the conjecture implies that $\Delta[\Sigma]$ eventually dominates every computable function.)

Motivation: In the "Busy Beaver game" class of Turing machines, which eventually halt after  starting with a blank tape, the "score" of a machine is the number of $1$s remaining on the tape after halting. A variant of Kolmogorov complexity $C: \mathbb{N} \rightarrow \mathbb{N}$ defines $C(n)$ to be the least $k$ such that $n$ is the score of some $k$-state machine. (cf "Computability and Logic" by Boolos, Burgess & Jeffrey, 2007, p229.) I believe I can show that if the first difference $\Delta[\Sigma]$ eventually dominates every computable function, then this $C$ is not monotonically nondecreasing, i.e., there exist $m \lt n$ such that $C(m) \gt C(n)$.

NB: Due to receiving no satisfactory answer here, I've posted the simplest case of this question on cstheory.SE.

For the eventual conclusion "there exist $m<n$ such that $C(m)>C(n)$" it seems to be easier just to prove that for some $k$ there exists a $k$-state machine with a score higher than the number of possible $k$-state machines. Then by the pigeonhole principle there has to be a number less than this that is reached only by a machine with $>k$ states.

Depending on the details, there are at most about $(6k)^{2k}$ different $k$-state machines, and this is easily a computable function. Therefore if add some initial states that preload a large enough $k$ (which can be done in $O(\log k)$ states) we will eventually reach a machine with a score that is high enough relative to its size.

Use http://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.html
to get $D^k [z] (n)>f(k,n)z(n+k)$ for some computable $f(k,n)$ and every strictly increasing positive function $z$.
Thus if $g(n)>D^k [z](n)$ for some computable $g$, then there is a computable bound on any $z(n+k)$. Contradiction.


Proving bijectivity of $f:[0,\infty)\to[0,2)$, where $f(x)=\frac{1}{x}+1$ for $x>1$, and $f(x)=x$ for $x\le 1$ Let $f: [0, \infty) \rightarrow [0,2)$ be defined by $$f(x) = \left\{
\begin{array}{ll}
      \frac{1}{x} + 1, & x > 1 \\
      x, & x\le 1 \\
\end{array}
\right.$$

  The question asks to prove that the function is bijective.

I have to prove it is injective and surjective.

Injectivity:
if $x,y<1$, then $x=y$
if $x,y>1$ then, $\frac{1}{x}+1=\frac{1}{y}+1$, then $x=y$

  I get this bit, but I don't get how to prove the surjective part of the proof.

Please could you help?

Let $$f_1(x)=x$$ for $$x\in [0,1]$$
and
$$f_2(x)=\frac 1x +1$$ for $$x>1$$

$f_1$ is a bijection from $[0,1] $ to $[0,1]$ and $ f_2$ is a bijection from $(1,+\infty)$ to $ (1,2)$. thus $f$ is a bijection from $[0,1]\cup (1,\infty)$ to $[0,1]\cup (1,2)$.

Other proof.

Let $ y\in[0,2)$. We will prove it has only one antecedent.

If $y\in[0,1]$ then $f(y)=y$.
the antecedent is unique in $[0,1]$

if $y\in(1,2)$ then $f(\frac{1}{y-1})=y$ with $\frac{1}{y-1}>1$.
the antecedent is unique in $(1,\infty)$

Injectivity:  Suppose $f(x) = f(y)$.

There are four cases:  $0 \le x \le 1$ and $0\le y \le 1$ and $f(x) = x = f(y) = y$.  Thus $x = y$.

$0 \le x \le 1$ and $y > 1$ and $f(x) = x = f(y) = \frac 1y +1$.  So $x = \frac 1y + 1$. And $y= \frac 1{x-1}$.  But if $y > 1$ then $0 < \frac 1y < 1$ and $1 < \frac 1y + 1< 2$ which contradicts $0\le x \le 1$.  (Likewise $0\le x=\frac 1y + 1 \le 1\implies -1 \le \frac 1y \le 0\implies y < 0$ also a contradiction).

$x> 1$ and $0 \le y \le 1$.  .... also gives a contradiction...  Too similar to the above to bear discussing.

[Note: we could have avoided these contradictory options by noting that $0\le w \le 1 \implies 0\le w=f(w) \le 1$ and $w > 1 \implies 0<\frac 1w < 1\implies 1 < \frac 1w + 1 = f(w) < 2$.  So $f(w) \in [0,1] \iff w \in [0,1]$ and $f(w) \in (1,2) \iff w\in (0,1)$.]

Fourth and final case $x > 1$ and $y > 1$.  Then $\frac 1x + 1 = \frac 1y + 1$ so $\frac 1x = \frac 1y$ so $x=y$.

So $f(x)=f(y)\implies x=y$.

Surjectivity:

Let $k \in [0, 2)$.  If $1 < k < 2$ and if $2>\frac 1x + 1 = k >1$ then $1> \frac 1x = k- 1 > 0$ and $x =\frac 1{k-1} > 1$.

So if $1 < k < 2$ then $\frac 1{k-1} >1$ so $f(\frac 1{k-1}) = k$.

If if $0\le k \le 1$ then $f(k) = k$.

Either way for all $k\in [0,\infty)$ there is $x$ either equal to $k$ if $k \le 1$ or equal to $\frac 1{k-1}$ if $k > 1$  so that $f(x) = k$.


Split a set of numbers into 2 sets, where the sum of each set is as close to one another as possible Given a finite set $P = \{ p_1, \dots, p_n \} $ of integers, I'd like to split it into two subsets $A = \{ a_1, \dots, a_m \} \subseteq P$ and $B = \{ b_1, \dots, b_r \} \subseteq P$, where $m + r = n$, and, for each subset, the sum of the numbers is as close as possible to half the total of the sum of the numbers in $P$. More precisely, we have the following constraints
$$\sum_i^m a_i \leq \left \lceil \frac{\sum_j^n p_j}{2} \right \rceil $$
and
$$\sum_i^r b_i \leq \left \lceil \frac{\sum_j^n p_j}{2} \right \rceil $$
and
$$\sum_i^m a_i + \sum_i^r b_i \leq \sum_j^n p_j $$

How would I go about doing this in a programmatic way?

Sort your set DESC, like if you've [1,3,2,5,6], after DESC sort it will look like this: [6,5,3,2,1].
Now you'll have 'SET1' and 'SET2' as your sub sets.
if (SET1<=SET2){The element in the given set should come into the 'SET1';}
else{The element in the given set should come into the 'SET2';}
And of-course you'll eliminate the element you enter into the subsets 'SET1' OR 'SET2' from the given super set.


There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned? In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.

I thought to label the four cups $\alpha, \beta, \gamma, \delta$ with events

$A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$
$B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$
$C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$
$D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$

If she were to drink in order, then I would calculate $P(a) = {1}/{4}$.
Next $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \cap B)$, which I'm not completely sure how to calculate.

My doubt is that I shouldn't order the cups because that assumes $\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.

The probability of not being poisoned is exactly the same as the following problem:

You choose one cup and drink from the other three.  What is the probability of choosing the poisoned cup (and not being poisoned)?  That probability is 1/4.

Therefore, the probability of being poisoned is 3/4.

I thought it would be instructive to try a different approach.

If you drink three cups, that tells me for sure that two of the cups you drank (the first two) were not poisoned - otherwise you would not have gotten to the third cup.

So the poison must be in either cup 3 or cup 4, and since it is equally likely to be in any of the cups, there is a 50% chance of that happening.

Having survived the first two cups, you now have a 50-50 chance that cup 3 is poisoned (because it's either cup 3 or cup 4). To survive drinking cup 3, you need to beat those odds as well.

Multiplying these probabilities, you once again get 0.25 for the chance of surviving.

Of course it's the same answer - but I thought this would give additional insight.


Is there any three valued logic with this truth tables? T AND T = T
T AND U = T
T AND F = F
F AND T = F
F AND U = F
F AND F = F
U AND T = T
U AND U = U
U AND F = F

T OR T = T
T OR U = T
T OR F = T
F OR T = T
F OR U = F
F OR F = F
U OR T = T
U OR U = U
U OR F = F

NOT T = F
NOT U = U
NOT F = T

T => T = T
F => F = U
T => F = F
F => T = U

Add the remaining for implication:

U => U = U
T => U = U
F => U = U
U => T = U
U => F = U

I'm not very sure about the last two. But if I miss them before I guess it is because they don't add "knowledge".

The idea of U is to meant NOTHING or VOID a thing that doesn't provide any knowledge and from where we cannot obtain any information.

What is the worst paradox this logic could generate?

Edit just to add another concrete interpretation I have when deriving the truth tables. Imaging an expression that runs forever. This espression is VOID because I'm not able to evaluate its truthness.

But probably this is total garbage. Maybe I'm just mixing inconsistents intuitions in my head. If this thing doesn't have a name I have little expectations...

Edit: After some time, my current me thinks the ultimate reason of the question is I don't like LEM standard intepretation. To divide the world in true and false and nothing else.
Giving formal system limited powers to figure out thruth I find LEM inapropiate as an axiom.

Logically speaking, we should have an equivalence between
$$P \implies Q $$
and
$$not  Q \implies not  P$$
but in your three valued one,

$$T \implies T ( =T)$$ is not equivalent to
$$F \implies F  (=U)$$


Solving recurrence $X_n = 4X_{n-1}+5$ $X_n=4X_{n-1}+5$

How come the solution of this recurrence is this?

$X_n=\frac834^n+\frac53$

I also have that $X_0=1$.

I am using telescoping method and I am trying to solve it like this:

$X_n= 5 + 4X_{n-1}$

$X_n= 5 + 4(5+4X_{n-2})$

$X_n= 5 + 4\times5 + 4\times4\times X_{n-2}$

But this leads to me getting $5\times4^{n-1}\times4^n$.

Can some please explain this to me?

If you calculate the first few terms explicitly, you will find that the $n$th term is the sum of an exponential and a geometric series. For example,

$$
X_3 = 4^3 + 5(4^2+4+1).
$$
So in general,
$$
X_n = 4^n + 5\sum_{k=0}^{n-1}4^k = 4^n +5\frac{1-4^n}{1-4},
$$

which should simplify to the answer you gave.

This is a difference equation and it can be solved using Z-Transform. Take Z-transform of both sides of the equation and then use the initial condition. Ultimately, you will get Z-transform of X and then take its inverse z-transform to get the solution.


Why does the determinant $D$, have to be $0$ for equation to have a solution? Suppose $2\times2$ equation:

$$
  \begin{cases}
   a_1x + b_1y = c_1  \\
   a_2x + b_2y = c_2
  \end{cases} $$

We can make determinants:

$$D=\begin{vmatrix}{a_1}&{b_1}\\{a_2}&{b_2}\end{vmatrix}$$

$$A=\begin{vmatrix}{a_1}&{c_1}\\{a_2}&{c_2}\end{vmatrix}$$

$$B=\begin{vmatrix}{c_1}&{b_1}\\{c_2}&{b_2}\end{vmatrix}$$

Solution to the $2\times2$ equation exists if $D = 0$

Why?

Some intuitions.

From some point of view, your equation is $DX=B$, hence, if it is possible, $D^{-1}DX=X=D^{-1}B$ and  Cauchy's theorem about determinants  implies, that $\det D\neq 0$, because $D^{-1}D=I$.

Consider this linear system:
\begin{equation}
\begin{matrix}
a_1x+b_1y&={c_1}\\ a_2x + b_2y&= {c_2}\end{matrix}
\end{equation}
which, in matrix format is
\begin{equation}
\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} {c_1} \\ {c_2} \end{bmatrix}.
\end{equation}
Solutions to which can be found with Cramer's rule, as
\begin{equation}
x = \begin{vmatrix} {{c_1}} & b_1 \\ c_2 & b_2 \end{vmatrix}/\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}  = { c_1b_2 - b_1c_2 \over a_1b_2 - b_1a_2}
\end{equation}
Which shows that if the determinant is zero, there are no solutions. So in general this is not true.


Why are derivatives specified as d/dx? Is the purpose of the derivative notation d/dx strictly for symbolic manipulation purposes?

I remember being confused when I first saw the notation for derivatives - it looks vaguely like there's some division going on and there are some fancy 'd' characters that are added in... I recall thinking that it was a lot of characters to represent an action with respect to one variable. Of course, once you start moving the dx around it makes a little more sense as to why they exist - but is this the only reason?

Any history lesson or examples  where this notation is helpful or unhelpful is appreciated.

Because of their definition:

Start with a function, calculate the difference in value between two points and divide by the size of the interval between the two. You can represent this as such:

$$\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$$

or

$$\frac{\Delta f\left(x\right)}{\Delta x}$$

Where ∆, delta, is the Greek capital D and indicates an interval. Now, take the limit as $\Delta x$ goes to zero, and you have the differential. This is indicated by using a lower case $d$ instead of the $\Delta$.

$$\frac{df\left(x\right)}{dx}$$

Now, if this operation is treated as an operator applied to a function, it is usually represented as

$$\frac{d}{dx}f\left(x\right)$$

Note that (typically in physics), you can also use the letter $\delta$ to indicate very small intervals and in general you would use the symbol $\partial $ to represent partial differentials. They are all variations of the letter $D$.

This page, (which quotes from "Yearning for the Impossible, The Surprising Truths of Mathematics") has:

...It has to be explained that dy/dx is not the ratio of infinitesimal differences dy and dx -- since infinitesimals do not exist -- but is rather a symbol for the limit of the ratio y/x as x tends to zero...


solving given linear equation So before you guys judge me, I honestly am so clueless with this so please bare with my dumb questions.

I have been at this equation that I've been going for an hour now,

$\frac{(3x-1)}{2} -2 = 5-\frac{4x}{3+1}$ and I'm trying to solve for $x$.
So I go with
$6\cdot(3x-1)/2-2=(5-4x)/3 +1$ to cancel out the denominators, and I get
$9x+3-6=10-8x+2$
then I move all the like terms,
$9x+8x=12+9$
then my final answer is
$x=21/17$

that didn't seem correct with me so then I tried,
cancelling out the denominators by multiplying each side on their own number, 3 and 2.
then I got $3x-1-2=5-4x+1$
and my answer is
$x=9/7$

How do I go about solving this?.

Your Equation looks like:
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{3+1}$$
Adding in the denominator on the right.
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{4}$$
Which simplifies to
$$\frac {3x+1} {2} - 2 = 5-x$$
We want to eliminate the denominator so we multiply by $2$.
$$ 2\frac {3x+1} {2} - (2)2 = (2)5-(2)x$$
Simplifying
$$3x+1 - 4 = 10-2x$$
Adding on the right.
$$ 3x- 3 = 10-2x$$
Add $2x$ to both sides and add $3$ to both sides.

$$ 5x = 13$$
$$x=\frac{13}{5}$$

When you have minus or plus don't forget them. You must multiplaying all the side !
Example :
(2x-1)/3+2=x
3*(2x-1)/3+3*2=3*x
(2x-1)+6=3x
x=5


Is the sum of a Wishart matrix and a deterministic psd matrix "almost Wishart"? Let $XX^T$ be a Wishart matrix, generated by taking the columns of $X$ to be i.i.d. standard $p$-variate normal vectors. Let $AA^T$ be a non-random positive definite matrix.

Though it is not possible to get an eigendecomposition for the sum $XX^T+AA^T$ merely from information about the eigenstructure of the summands, I have a more basic question:

  Is it true that $XX^T+AA^T$ can be expressed as $YDY^T$, where the entries of $Y$ are i.i.d. variables (not necessarily normal?) and $D$
  is a self-adjoint non-negative matrix?  (or perhaps $D$ is even
  diagonal?)

This would be quite interesting, since one might be able to say something about the spacing of the eigenvectors of the sum without knowing its precise eigendecomposition.

Edit: My intuition is that the sum should be Wishart, since as I recall the sum of independent Wishart matrices is Wishart. The matrices are independent, and though the second matrix is not random, it seems that "from the perspective of $XX^T$, there is no difference between a Wishart matrix with expectation $AA^T$ and $AA^T$ itself. Can this be made rigorous?

EDIT 2 Please note that I am primarily asking for an understanding of the independence of the entries of $Y$.

You can do this with $D$ as the identity matrix. You have $X\in\mathbb R^{p\times n}$ and $AA^T\in\mathbb R^{p\times p}$. The matrix $XX^T + AA^T$ is a symmetric matrix with entries that are real, and so can be diagonalized by some orthogonal matrix $W\in\mathbb R^{p\times p}$:
$$
XX^T + AA^T = W \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \\ & & & \ddots \\ & & & & \lambda_p \end{bmatrix} W^T
$$
where $\Lambda$ and $W$ depend on $X$ and $A$, and the $\lambda$s can be taken $\ge 0$ because $XX^T+AA^T$ is non-negative definite.  So now let
$$
\Lambda^{1/2} = \begin{bmatrix} \sqrt{\lambda_1} \\ & \sqrt{\lambda_2} \\ & & \sqrt{\lambda_3} \\ & & & \ddots \\ & & & & \sqrt{\lambda_p} \end{bmatrix}.
$$
Then we can let
$$
Y = W\Lambda^{1/2}
$$
and let $D$ be the $p\times p$ identity matrix, and then we have
$$
XX^T + AA^T = Y D Y^T.
$$

If you let $X$ vary while $A$ is fixed, you get a function $X\mapsto Y$.

All of the above can be said without saying anything at all about randomness or probability distributions.  But now if you impose some probability distribution on the random matrix $X$ while saying $\Pr(A=\text{one specified fixed matrix})$, then that determines a probability distribution of the matrix $Y$.

But $Y D Y^T$ has neither a Wishart distribution nor a noncentral Wishart distribution unless $A=0$.  The reason for that is as follows.  The set of all non-negative-definite $p\times p$ symmetric matrices with real entries has a natural partial order defined as follows: $P\le Q$ precisely if $Q-P$ is non-negative definite.  It is not hard to show that that satisfies the definition of a partial order relation.

Now observe that your random matrix $YY^T$ is supported entirely on the set of matrices $\ge AA^T$ in this partial order.  That does not happen with Wishart matrices.  With a Wishart matrix $V$ one would have $0 < \Pr(V \le AA^T) \le \Pr(V \not\ge AA^T)$.  Consider in particular what happens when $p=1$.  The support is $[a^2,\infty)$.  The support of either a central or a non-central chi-square distribution is $[0,\infty)$.


Trouble with finding volume of a solid for $y=2-2x^2, y=0$ Here is my problem:

$$y=2-2x^2, y=0$$

I first found points of intersection by solving $y=2-2x^2, y=0$ and found that my points are $-1, 1$

Then I found the area by doing the following:

$$A= \pi r^2$$
$$A=\pi(2-2x^2)^2$$
$$A=64\pi(x^4+1)$$

I then anti-differentiated:

$$64\pi\int_{-1}^{1}(x^4+1)dx$$
$$64\pi(\frac{x^5}{5}+x)|_{-1}^{1}$$

Which I found to be:

$$\frac{768\pi}{5}$$

I exceeded my maximum amount of submission for this problem, but I was wondering if this attempt at the answer is correct or if I made another mistake?

I assume you are rotating about the $x$-axis. Then the volume of the solid of revolution is
$$\int_{-1}^1\pi(2-2x^2)^2\,dx.$$
Expand. We want
$$4\pi\int_{-1}^1 (1-2x^2+x^4)\,dx.$$
By symmetry, this is
$$8\pi\int_0^1 (1-2x^2+x^4)\,dx.$$
An antiderivative is $x-\frac{2}{3}x^3+\frac{1}{5}x^5$.
When we do the calculation, we end up with $\frac{64\pi}{15}$.

Remark: There were errors of algebra in expanding $(2-2x^2)^2$. I do not know where the $64$ came from. The other error came from forgetting the "middle term" when you square $a+b$.

@T. Bongers

$$
\infty
\quad\mbox{since}\quad
\int_{-\infty}^{\infty}{\rm d}z \to \infty
$$

$\mathbf{\mbox{ADDENDUM}}$
$$
{\displaystyle{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}}
 \atop
 y < 2 - 2x^{2}\,,\ y>0}\!\!\!\!
\displaystyle{{\rm d}x\,{\rm d}y\,{\rm d}z \to \infty\atop }
$$


Region of convergence for the series of functions Find out the region of convergence for the following series of function.

$$\sum_{m=1}^\infty x^{\log (m)}$$

Here $x \in \mathbb{R}$. This is a series of function. I was trying to find out the radius of convergence by Root Test and Ratio test. But no suitable solution I am getting.

What will be the region of convergence if we consider $x \in \mathbb{C}$?

Thank you for your help.

$x \in \mathbb{R}$ ??. How do we define $(-2)^{log(2)}$ ? And in my opinion, this is not really a power-series .
 As what I solved, this series converges if and only if $ x \in [0; \frac{1}{e})$
Hint:
Rewrite the series as :
 $ \sum_{k=0}^{\infty}  \sum_{m=[e^k]}^{[e^{k+1}]-1} x^{log(m)} $ (where $[.]$ is floor function)
And we see that:
$( [e^{k+1}]-1-[e^k]).x^{k-1} \le \sum_{m=[e^k]}^{[e^{k+1}]-1} x^{log(m)} \le ( [e^{k+1}]-1-[e^k])x^{k+1}$(*)
Which leads to the interval of convergence.

One idea would be to use the fact that $\:x^{\log(m)}=e^{\log(m)\log(x)}=e^{\log(m)\Large(\normalsize\log(x)/\log(m)\Large)}\underset{m\to\infty}{\sim}e^{\log(m)}.$


Sum of all numbers formed by the given digits The formula for sum of all numbers formed with all the given digits is:

(Sum of digits) (n-1)!(1111....ntimes)

n stands for number of digits.

For ex:
Sum of all numbers formed with 1,2,3 and 4 is
10 × 3! × 1111 = 66660.

Can anyone figure out how this was derived?

I figured the answer out.

If you fix 1 in the thousands place, all other numbers can be arranged in 3! Ways.

Therefore 1 occurs in thousands place 3! Times. Same for all the numbers. Therefore for thousands place, total sum= 1000(sum) 3!.

But same thing occurs for all other positions. Therefore sum= 1111...×(sum) ×(n-1)!

As total arrangements will be n factorial and there are n numbers.So each no. can occupy a place (n factorial/n) times then to get sum at a particular place value you have to multiply it with that place value and of course the digit. So doing it for all place values will give you a term (1111...n times) and also (sum of all digits in process) plus this all is occuring (n factorial/n) times or n-1 factorial times so take the product and there you get the result.


Does $\mathbb{F}_p((X))$ has only finitely many extension of a given degree? We know that $\mathbb{Q}_p$ has only finitely many extensions of a given degree in its algebraic closure. Is it the same for $\mathbb{F}_p((X))$?

No. There are infinitely many extensions of degree $p$.  Let $K = {\mathbf F}_p((X))$ and $\wp(x) = x^p - x$ (Artin--Schreier operator, which is additive in characteristic $p$). For $c \in K - \wp(K)$, the polynomial $T^p - T - c$ is irreducible in $K[T]$ and its splitting field over $K$ has degree $p$.  For two $c$ and $c'$ in $K - \wp(K)$, the splitting fields of $T^p - T - c$ and $T^p - T - c'$ in an alg. closure of $K$ are equal iff $c = c'$ in $K/\wp(K)$.  This quotient group is infinite. In particular, the reciprocal powers $1/X^n$ for $n$ positive and not divisible by $p$ are distinct in $K/\wp(K)$.  This is discussed in section 1 of http://math.stanford.edu/~conrad/248APage/handouts/weirdfield.pdf.

There are only finitely many extensions of each degree relatively prime to $p$ over ${\mathbf F}_p((X))$ in an algebraic closure, by the same proof as over ${\mathbf Q}_p$. When you deal with extensions of degree divisible by $p$ in characteristic $p$, watch out.

Yes.  The key properties are (1) the field is complete with respect to a nonarchimedean valuation, and (2) the residue field is finite.  With those two properties in hand, then the same proof as for $\mathbb{Q}_p$ goes through.  See (for example) this proof (Theorem 3.19).

EDIT: As others have pointed out, this isn't quite right; the proof fails for extensions of degree divisible by $p$.  See other answers.


Why should we prove obvious things? Obviously, there are obvious things in mathematics. Why we should prove them?

Prove that $\lim\limits_{n\to\infty}\dfrac{1}{n}=0$?
Prove that $f(x)=x$ is continuous on $\mathbb{R}$?
$\dotsc$

Just to list few examples.

Because sometimes, things that should be "obvious" turn out to be completely false.  Here are some examples:

Switching doors in the Monty Hall problem "obviously" should not affect the outcome.
Since Gabriel's horn has finite volume, then it "obviously" has finite surface area.
"Obviously" we cannot decompose sphere into a finite number of disjoint subsets and reconstruct them into two copies of the original sphere.
Since the Weierstrass function is everywhere continuous, then "obviously" it must have at least a few differentiable points.

Of course, mathematics has shown that switching doors is to the player's advantage, that Gabriel's horn actually has infinite surface area, that you can indeed get two copies of the original sphere (see Banach-Tarski paradox), and that the Weierstrass function is everywhere continuous but nowhere differentiable.  The point being, there are many things out there which are "obvious" but actually turn out to be entirely counterintuitive and opposite what we would otherwise expect.  This is the point of rigor: to double check and make sure our intuition is indeed correct, because it isn't always.

In my opinion, only axioms should be treated as obvious, above all while a theory is being explained to others. I think it is simply immoral for a mathematician who is writing a proof of a proposition in a book not to give every smallest detail in the chain of logical inferences, skipping the task of making the subject perfectly clear through a lazy abuse of the magic words trivial, obvious, exercise. If one wants to write mathematics for others, he must be as clear and rigorous as possible, at least while developing theory and examples. If instead he's too lazy to bear such a sacred task, then he should not even begin to write mathematics, because the essence of mathematics is in its logical clarity, and not in a pseudomystic vagueness - alas, many important books of great mathematicians are "written" in this terrible style! -.


Closed subspace of $l_2$ Let $$X=\{ a \in l_2 : \sum_{n=1}^{\infty}{\frac{1}{\sqrt{n}}a_n = 0} \}$$

Show that X is not a closed subset of $l_2$

My general approach is to find a converging sequence $a_n \in X$, whose limit $a$ does not belong to $X$. Any hints to this problem as I fail to see what sequence I can construct here

Here is a proof, which assumes that the sum in the definition of $X$ has to converge absolutely.
The sequence $x\in l^2$ with $x_n = \frac1{\sqrt n\log n}$ has the property
that $\sum_n x_n \frac1{\sqrt n}=+\infty$.
Now let me 'double' this sequence, construct a sequence $y$ such that
$$
y_{2n+1}, \ y_{2n+2} \sim \frac1{\sqrt n\log n}
$$
and
$$
\frac{y_{2n+1}}{\sqrt{2n+1}} = -\frac{y_{2n+2}}{\sqrt{2n+2}},
$$
i.e.
$$
y_{2n+1}:= \frac1{\sqrt n\log n}, \ y_{2n+2}:=-\frac1{\sqrt n\log n} \frac{\sqrt{2n+2}}{\sqrt{2n+1}}.
$$
Then by construction
$$
\sum_{n=1}^{2m} y_n \frac1{\sqrt n}=0
$$
for all $m$. Hence
$$
y^m:=(y_1, \dots, y_{2m},0,\dots)
$$
is a sequence in $X$ such that $y^m \to y$ in $l^2$. But $y$ is not in $X$,
as $\sum_{n=1}^\infty y_n\frac1{\sqrt n}$ does not converge absolutely. It does converge conditionally.

There's a problem in the very definition of $X$. The sum $\sum n^{-1/2} a_n$ doesn't have to converge, for example it's infinite for $a_n = n^{-1/2} (\ln n)^{-1}$ (which still lies in $\ell_2$).
Once you fix the definition of $X$ (in one way or another), you should be able to construct a counterexample to closedness, based on the sequence $(a_n)$ given above.


Disprove Homeomorphism I have a problem that puzzles me. I need to show that the two sets
$A = \{(x,y)  \in \mathbb{R}^2 \, \, \vert \, \, |x| \leq 1  \}$
and
$B = \{(x,y)  \in \mathbb{R}^2 \, \, \vert \, \, x \geq 0 \}$
are not homeomorphic; but I'm not able to figure out how start or what I need to arrive at.

The one-point compactification of $A$ is homeomorphic to the space $X$ obtained by identifying the points $\langle 0,1\rangle$ and $\langle 0,-1\rangle$ of the disk $D=\{\langle x,y\rangle:|x|+|y|\le 1\}$. The one-point compactification of $B$ is homeomorphic to $D$ itself. $X$ and $D$ are not homeomorphic, so $A$ and $B$ cannot be homeomorphic. However, I don’t immediately see a way to prove that $X$ and $D$ are not homeomorphic without using homotopy.

What about this: if A and B are not homeomorphic, then there exists a continuous bijection $ f: \, A \rightarrow B $ such that $ f^{-1} $ is not continuous. Then look at the function $f(x,y) = \left( \tan\left( \frac{(1+x) \pi}{4} \right), y  \right)$. Problem at $ x = 1 $ ?


Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$ For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$

My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$
Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$
Which is clearly wrong. :"(
Thank you very much.

Suppose the inequality is true,then,
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}$$
$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}\geq \frac{12}{a+b+c}$$
$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \frac {9}{a+b+c} $$$$\Rightarrow \frac {a+b+c}{3}\geq\frac{3}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$Which is obviously true ($AM\geq HM $).


Solve for $\theta$: $2 \sin \theta = 2 - \cos \theta$ Actually , I'm new to trigonometry .. So i want help in this question

  $$2 \sin \theta = 2 - \cos \theta $$

My attempt ->

$$\begin{align}
2 \sin \theta &= 2 - \cos \theta \\
2 \sin \theta &= 2 - {(1-\sin^{2}\theta)}^{1/2}
\end{align}$$

Beware that substituting $\cos\theta$ with $\sqrt{1-\sin^2\theta}$ is not correct. Beware also that squaring will add spurious solutions.

Here's a method: set $x=\cos\theta$ and $y=\sin\theta$, so you have
$$
\begin{cases}
2y=2-x \\[4px]
x^2+y^2=1
\end{cases}
$$
This basically transforms the problem into finding the intersection of a line with a circle.

Solving the first with respect to $x$ gives $x=2-2y$, so
$$
4-8y+4y^2+y^2=1
$$
becomes
$$
5y^2-8y+3=0
$$
yielding
$$
y=\frac{3}{5}
\qquad\text{or}\qquad
y=1
$$
The first solution gives $x=4/5$ and the second solution gives $x=0$. Finally you have
$$
\theta=\arcsin\frac{3}{5}+2k\pi \quad(k\text{ an integer})
$$
or
$$
\theta=\frac{\pi}{2}+2k\pi \quad(k\text{ an integer})
$$

Another way to solve this problem is to divide both sides by $\cos \theta$.
$$2\sin\theta = 2 - \cos\theta$$
$$2 = 2\sin\theta + \cos\theta$$
$$\frac{2}{\cos\theta} = \frac{2\sin\theta + \cos\theta}{\cos\theta}$$
$$2\sec\theta = 2\tan\theta + 1$$
$$\sec\theta = \tan\theta + \frac{1}{2}$$
$$\sec^2\theta = (\tan\theta + \frac{1}{2})^2$$
$$\sec^2\theta = \tan^2\theta +\tan\theta + \frac{1}{4}$$
$$\sec^2\theta = (\tan^2\theta +1)-1+\tan\theta + \frac{1}{4}$$
$$\sec^2\theta = \sec^2\theta-1+\tan\theta + \frac{1}{4}$$
$$\tan\theta = \frac{3}{4}$$


Coefficients for Laurent series expansion are unique Suppose $$f(z)=\sum_{n=-\infty}^\infty a_nz^n$$ for all $1<|z|<2$. Show that the coefficients $\{a_n\}$ are unique.

I want to use the fact that the Laurent series of any function $f$ is unique. But to use that, I have to know that $f$ is holomorphic in $1<|z|<2$. So how can I do it then?

Suppose there's another expansion
$$f(z)=\sum_{-\infty}^\infty b_n z^n.$$
But then we would have
$$0=f(z)-f(z)=\sum_{-\infty}^\infty a_n z^n-\sum_{-\infty}^\infty b_n z^n=\sum_{-\infty}^\infty (a_n-b_n) z^n.$$
But for this to be true $\forall z$ in the domain, we must have $a_n-b_n=0, \forall n$. So, the coefficients, are unique.


Show that the stalk of the kernel is the kernel of the stalks. This is Exercise 2.5.A. in Vakil's notes of Algebraic Geometry. This section is about sheaves of abelian groups on a topological space $X$.

  Show that the stalk of the kernel is the kernel of the stalks: for all $p\in X$, there is a natural isomorphism $$(\ker(\mathscr{F}\rightarrow\mathscr{G}))_p\cong \ker(\mathscr{F}_p\rightarrow \mathscr{G}_p).$$

My attempt:

We consider the following diagram:

where the downward arrow is the induced morphism on the stalks, as defined in Exercise 2.4.D. Since the three are all sheaves, the downward arrows are all injective. We want to show that $(\ker\phi)_p$ is the kernel of $\phi_p$ for all $p$. Since the second row is induced from the first row, we have $\phi_p\circ i_p=0$ for all $p$. Since all kernels are monomorphisms, by Exercise 2.4.N., the map $i_p$ is also a monomorphism. Hence it is a kernel of its cokernel.

My question:
I am not sure how to show that $i_p$ is the kernel of $\phi_p$. I tried the universal property, but cannot connect it with the upper arrows, so cannot find a unique map.

Thank you for your help!

If $p \in V \subset U$, then $\ker \varphi(U) \to \ker (\varphi_p)$ is the same as the composition $\ker \varphi(U) \to \ker\varphi(V) \to \ker (\varphi_p)$.
By the universal property of colimits, there exists a unique map $\alpha: (\ker \varphi)_p \to \ker (\varphi_p)$ where $[s,U] \mapsto [s,U]$.
It is straightforward to show that $\alpha$ is bijective.


necessary condition for Archimedean valuation To fix the terminology, a valuation of a field F will be a map from F to the real numbers such that, on writing $|x|$ for the image of $x$,
$|x|\geq 0$ for all $x$, $|x|=0$ iff $x=0$, $|xy|=|x||y|$ for all $x$,$y$ and $|x+y|\leq|x|+|y|$ for all $x,y$. This last inequality is called the triangle inequality. If it is replaced by the stronger inequality $|x+y|\leq max(|x|,|y|)$ for all $x,y$, which is called the ultrametric inequality, the valuation is called non-Archimedean. An Archimedean valuation is a valuation which is not non-Archimedean.

In Ribenboim’s book The Theory of Classical Valuations, he gives an easy proof on page 11 that a valuation is non-Archimedean iff $|n.1|\leq 1$ for all natural numbers $n$ where $n.1$ is the sum of $n$ copies of the identity element $1$ of F. On page 473 of Artin and Whaples’  paper “Axiomatic Characterization of Fields By the Product Formula for Valuations” Bull AMS 51 (1945) pp. 469-492 , they say that $|1+1|>1$ for every Archimedean valuation of F.

  It follows from Ribenboim’s theorem that $|1+1|>1$ is sufficient for a valuation to be Archimedean, but why is it necessary?

I tried showing the contrapositive, that a valuation such that $|1+1|≤1$ must be non-Archimedean, and using the fact that a valuation is non-Archimedean iff the set $$S=\{|n.1| : n \text{ is a natural number}\}$$ is bounded (p. 118 of Algebraic  Number Theory by Jurgen Neukirch) but could only obtain bounds on particular $|n.1|$, not on the whole set $S$.

Assume that $|2\cdot1_F|\leq1$. Then for each $n\geq1$ and its binary expansion $n=\sum_{k=0}^{m}a_k2^k$, we have

$$|n\cdot1_F|
\leq\sum_{k=0}^{m} \left| a_k 2^k\cdot1_F \right|
\leq\sum_{k=0}^{m} \left| a_k\cdot1_F \right|
\leq m+1
\leq\lceil\log_2 n\rceil+1.$$

Now utilize the tensor power trick:

$$|n\cdot1_F|=|n^k\cdot1_F|^{1/k}\leq(1+\lceil k\log_2 n\rceil)^{1/k} \xrightarrow[k\to\infty]{} 1. $$

In view of Ribenboim's theorem, this is enough to guarantee that $|\cdot|$ is non-Archimedean.

In fact, Ribenboim's theorem can also be proved by the tensor power trick. Assume that $|n\cdot1_F|\leq1$ for all $n\in\mathbb{N}$. Then for each $x, y \in F$,

$$ |x+y| = |(x+y)^n|^{1/n}
\leq \left( \sum_{k=0}^{n} \left|\binom{n}{k}x^k y^{n-k}\right| \right)^{1/n}
\leq (n+1)^{1/n}\max\{|x|,|y|\}. $$

Letting $n\to\infty$ shows that the ultrametric inequality holds.

Theorem: A valuation on a field $F$ is non-Archimedean if and only if $|1_F+1_F|\leq1$.

Proof:
See Lemma 8.2 (p. 18) of "Ultrametric Calculus: An Introduction to p-adic Analysis" - [W.H. Schikhof] - 2006


Iverson bracket help How can I show that if $G$ is a group and $x,y,m$ are in $G$ then,
$$[xy=m]=\sum_{r\in G}[x=r][y=r^{-1}m]$$

Where $[P]$ is the Iverson bracket.

Let $S$ be a set and suppose $x\in S$.
Then
$$\sum_{r\in S}[x=r]F(r) = F(x),$$
where $F(x)$ is any expression involving $x$.
(Roughly, the Iverson bracket acts here like a delta function.)

We have immediately that
$$\begin{eqnarray*}
[xy=m] &=& [y=x^{-1}m] \\
&=& \sum_{r\in G}[x=r][y=r^{-1}m].
\end{eqnarray*}$$

Assuming that the bracket on the left has an implied summation over all $x, y \in G$, then you are trying to show that for any fixed $m \in G$,
$$
\sum_{x,y\in G} [xy = m] = \sum_{r\in G} [x = r][y = r^{-1}m].
$$

In other words, you need to convince yourself that there's a bijection between the sets
$$
\{(x, y) \in G \times G \;|\; xy = m \}
\quad \longleftrightarrow \quad
\{r \in G \} = G.
$$

Once you identify $x = r$, then the condition on the left forces $y = x^{-1}m = r^{-1}m$.


Why don't we indicate the variable to summed as we do for integrals? When integrating over a certain variable $x$, we make sure to end the integral with $dx$, like so:

$$\int_{1}^{\infty}\frac{1}{x^2}dx$$ The reason for this of course becomes more clear as one goes deeper into single- and especially multivariable calculus, where one discovers that it does't just signify which variable to integrate.

But is there no valid reason to write, for example, the sum $1+1/4+1/9+\dots$ in this fashion:

$$\sum_{1}^{\infty} \frac{dn}{n^2}$$

Instead of the usual:

$$\sum_{n=1}^{\infty} \frac{1}{n^2}$$

Has it ever been done?

The "d$x$" is best regarded as a mnemonic symbol, and it reminds the reader (though somewhat misleading for novices) how an integral is carried out. In Russell's term, "d$x$" is called an incomplete symbol, which has no meaning itself when escaping from a given context.

I do not see apparent reasons keeping an author away from writing $\sum_{1}^{\infty}\frac{1}{n^{2}}dn,$ and indeed I write so when I wish to emphasize the similarity between the discrete and the continuous. Nonetheless, I think in discrete case d$n$ is simply = 1.

The $dx$ or $dn$ parts are used in integrals to show which variable we're integration over and so we can abuse the notation.

For sums and products the variable we're summing over is written below the sum or product sign, there's no reason to write which variable we're summing over twice.


Solve $y^2 = x^3 - 4$ where $x,y \in \mathbb{Z}$ Solve $y^2 = x^3 - 4$

So far i've tried to look at this:

$x^3 = (y-2i)(y+2i)$.

I think we need to prove that $y+-2i$ are coprime, so suppose not then:

$d = u+iv$ divides both numbers. So also d divides the sum and difference: $d|2y$ and $d|4i$.

If one applies the euclidean norm function we get that $u^2 + v^2|4y^2$ and $u^2 + v^2|16$.

And here i get stuck, any tips?

Kees

Your computation are done in the ring of Gaussian integers, which is a PID. Let $p$ be a prime in this ring dividing both $y+2i$ and $y-2i$. Then $p$ divides $2y$ and $4i$, so it is $1+i$ (up to invertible elements), because this is the only prime dividing $2$.

From $(1+i)(h+ki)=y+2i$ we get $h-k=y$, $h+k=2$, so $h=(2+y)/2$, $k=(2-y)/2$ and $y$ must be even.

So the two elements are indeed coprime if $y$ is odd. In this case $y+2i$ must be a cube times an invertible element, so
$$
y+2i=u(a+bi)^3
$$
In the case $u=1$, we get $y=a(a^2-3b^2)$, $2=b(3a^2-b^2)$. Since $y$ is odd, $a$ must be odd and $b$ even. Thus $b=\pm2$ and $3a^2-4=\pm1$. The only case is $b=-2$, $a=\pm1$. So $y=\pm11$.

A solution has been found: $x=5,y=\pm11$.

In the case $u=i$, we get $y=-b(3a^2-b^2)$, $2=-a(a^2-3b^2)$. So $b$ must be odd and $a$ even. Therefore $a=\pm2$. If $a=2$ we get $4-3b^2=-1$ (no solution). If $a=-2$ we get $4-3b^2=1$, so $b=\pm1$. This corresponds to $y=\pm11$, the same as before.

The cases $u=-1$ and $u=-i$ are the same, because the minus sign can be merged with the cube.

Suppose $y$ is even. Then also $x$ must be even and so we can set $x=2X$ and $y=2Y$, obtaining
$$
2X^3=Y^2+1=(Y+i)(Y-i)
$$
Since $1+i$ is prime and divides the left-hand side, it must divide $Y+i$ or $Y-i$. From $(1+i)(h+ki)=Y+i$ we get $h-k=Y$ and $h+k=1$. Thus $h=(Y+1)/2$ and $Y$ must be odd. Similarly, if $1+i$ divides $Y-i$. If we set $Y=2Z+1$, we get
$$
\frac{2Z+1+i}{1+i}=Z+1-Zi,
\qquad
\frac{2Z+1-i}{1-i}=Z+1+Zi.
$$
and the equation can be written as
$$
X^3=(Z+1-Zi)(Z+1+Zi)
$$
A prime dividing both factors must divide $2(Z+1)$ and $2Zi$. If it divides $Z$, it cannot divide $Z+1$, so it divides $2$. So the only prime that can divide both numbers is $1+i$. However, from
$$
(1+i)(h+ki)=Z+1+Zi
$$
we get $2h=2Z+1$, a contradiction. So the two factors are coprime and therefore they are cubes times an invertible.

Case $Z+1+Zi=(a+bi)^3$. We get $Z+1=a(a^2-3b^2)$ and $Z=b(3a^2-b^2)$.
Therefore $a^3-3ab^2-3a^2b+b^3=1$, or $(a+b)(a^2-4ab+b^2)=1$.

From $a+b=1$ we get $(a+b)^2-6ab=1$, hence or $6ab=0$. Thus $a=0$ or $b=0$ and we get the solutions $a=0$, $b=1$, that is $Z=-1$; $a=1$, $b=0$, that is $Z=0$. This corresponds to $Y=-1$ and $Y=1$.

From $a+b=-1$ we get $-6ab=-2$, that has no solution.

Case $Z+1+Zi=i(a+bi)^3$. $Z+1=-b(3a^2-b^2)$ and $Z=a(a^2-3b^2)$. Here we get $-a^3-3a^2b+3ab^2+b^3=1$, so $(b-a)(a^2+4ab+b^2)=1$ The case $b-a=1$ forces $6ab=0$. The case $b-a=-1$ forces $6ab=2$. Thus we don't find new solutions.

Another solution has been found $x=2$, $y=\pm2$

No other solutions exist