Mathematical Proof By Induction of $1 + 2 +\dots + n = (n + 1)n/2$ Trying to prove that $1 + 2 +\dots + n = (n + 1)n/2$.

I have let $n = 1$ in the basis step, which lead to $1=1$, so it is true for $n = 1$.

For the induction step, I have assumed that $k\geq 1$ and let $n = k$ leading to:

$$1 + 2 + . . . + k = (k + 1)k/2$$

For which we then have to use to prove that the statement is valid for $n = k + 1$, which is where I get confused because this the statement is supposed to become:

$$1 + 2 + . . . + k + (k + 1) = (k + 2)(k + 1)/2$$

While I thought it would have just been

$$1 + 2 + . . . + (k + 1) = (k + 2)(k + 1)/2  $$

Via the substitution of $k = k + 1$. Why is there still a singular $k$ in the $k + 1$ statement?

1 + 2 + . . . + k + (k + 1)

and

1 + 2 + . . . + (k + 1)

are exactly the same:  in both cases it is the sum of all numbers from $1$ to $k+1$ ... which includes $k$. It's just that the first representation makes the fact that this includes $k$ explicit, while the second does not.

It is because we assume that the statement is true for 'K' and 'K+1'.We have checked that the statement is true for k=1. Therefore, by what we have just seen, it must be true for k=2 and hence for the next integer 3. proceeding Inductively, we see that formula has to be true for all integers.


A countability problem of $\mathbb{R}$ I am having a really hard time coming up with a proof for this problem. For every finite set $F\subseteq \mathbb{R}$, let $\Sigma(F)$ denote the sum of the numbers in $F.(\Sigma(\emptyset)=0)$. Show that if $S\subseteq (0,+\infty)$ and $\Sigma(F)\leq100$ for every finite set $F \subseteq S$, then $S$ is finite or countable. I can't see why $S$ could be countable but not uncountable. Any suggestions are much appreciated.

Hint: Let $S_n=\{s\in S\mid s>\frac{1}{n}\}$. How many elements can be in $S_n$ if $\sum(F)<100$ for all finite $F\subseteq S$?

$S=\{1,1/2,1/4,1/8,1/16,\ldots\}$ is an example of a countably infinite set that meets the requirements.


How to prove that $J(M_n(R))=M_n(J(R))$? How to prove that $J(M_n(R))=M_n(J(R))$?
Here $M_n(R)$ is the ring of matrices of size $n^2$ over the ring $R$. And $J(M_n(R))$ is a two-sided ideal of the ring $M_n(R)$.

Take any matrix $A$ in $J$. Multiply it on the left by matrix that has $1$ on $i$-th place on the diagonal and zeroes elsewhere. The result will be a matrix $A'$ that has $i$-th row of $A$ and zeroes elsewhere. Since $J$ was double sided ideal, it's also in $J$. Now multiply $A'$ on the right by the matrix that has $1$ on $j$-th place on the diagonal and zeroes elsewhere. You'll get matrix which has $j$-th column of the matrix $A'$, so it will precisely be a matrix that has the same element as $A$ on the $i, j$ position, and zeroes elsewhere. Since $J$ was doublesided ideal, it will also be in $J$.

Do you see how to continue?


A finite set is closed Question:

Prove that a finite subset in a metric space is closed.

My proof-sketch: Let $A$ be finite set. Then $A=\{x_1, x_2,\dots, x_n\}.$ We know that $A$ has no limits points. What's next?
Definition: Set $E$ is called closed set if $E$ contains all his limits points.

Context: Principles of Mathematical Analysis, Rudin

Thinking $A$ as a subset of a metric space $M$. An easy approach will be to use that the single points $\{x_j\}\subset M$ are closed (you know why?), then of course
$$
A=\bigcup_{j=1}^n \{ x_j \}
$$
Since $A$ is a finite union of closed sets, it is itself closed.

If some open set $G_i$ containing $x_1$ does not contain $x_i$, for $i=1,\ldots,n$, then $\bigcap_{i=1}^n G_i$ is an open set containing $x_1$ and not containing any of $x_2,\ldots,x_n$.  Therefore $x_1$ is not a limit point of the set.  The same argument applies to show that $x_2,\ldots,x_n$ are not limit points.

If the space is a Hausdorff space, then certainly some open set $G_i$ containing $x_1$ does not contain $x_i$ if $i\ne1$.

If it's a metric space, you can take $G_i$ to be the open ball of radius $d(x_1,x_i)$ centered at $x_1$.


Continuous linear functionals Let L be a continuous  linear functional on a metric linear space X. Prove:  L(S) is a bounded set for any bounded subset S of X. The metric is translation invariant.

Note that this is true in general for $L:X\rightarrow Y$ linear continuous between two metric linear spaces.

Hints:

fix an open ball centered at $0$ (the one of radius $1$ for instance)
use continuity at $0$.
prove that $L$ is bounded on some open ball $B$ centered at $0$.
use that for every bounded set $S$, there is $\rho>0$ such that $S\subset \rho B$.

Consider the inverse image of any bounded open neighborhood $W$ of $0$ in the target space (let it be $\Bbb R$ or $\Bbb C$ or anything). Its preimage, $L^{-1}(W)$ is an open neighborhood of $0\in X$, so it contains a ball $B(\varepsilon)$ around $0$ with radius some $\ \varepsilon >0$. Let us fix this $\varepsilon$.

Now, if we have any bounded set $S\subseteq X$, then there exists a ball that contains it, say with radius $r_0>0$ and midpoint $Q$, then  $\ S\subseteq B(r)\ $ where $\ r=r_0+d(0,Q)$.

Now imagine the continuous mappings $f_\lambda:X\to X,\ \ v\mapsto \lambda\cdot v$. If $\lambda\neq 0$,  it is a homeomorphism, mapping any ball $B(\rho)$ (around $0$) to an open set, which also contains a ball around $0$, say with radius $f(\lambda,\rho)>0$. So that $B(\lambda\cdot\rho)=f_\lambda(B(\rho))\supseteq B(f(\lambda,\rho))$...

Then,  $L(S)\subseteq L(B(r))= L(f_{\frac r\varepsilon}( B(\varepsilon))) \subseteq f(\displaystyle\frac r\varepsilon)\cdot  W $, bounded.

Update: Hmm, in this generality it is not true. Let $X=\Bbb R$ with a bounded metric, for example $d'(x,y):=\displaystyle\frac{d(x,y)}{1+d(x,y)} <1$, and let $L$ be the identity $X\to\Bbb R$. Then the whole $X$ is bounded, but its image is not bounded.


Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$ Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$,
$$x^2 y + y^2 z + z^2 x \leq \frac {4}{27}
$$

I'm having trouble with this question. I suspect it may have a fairly simple proof using the AM-GM inequality and certain substitutions, however, I have been unable to complete such a proof.

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}$$

WLOG, assume $x\geq y\geq z$. You wish to maximize the function: $f(x,y,z)=x^2y+y^2z+z^2x$ given that $x,y,z \in \Re^{+} \cup\{0\}$ and $x+y+z=1$

Now, note that(we're trying to start of by obtaining a trivial inequality based on our assumption: $$\begin{split}f(x+z,y,0)-f(x,y,z)= & (x+z)^2y-(x^2y+y^2z+z^2x)=\\ & z^2y + yz(x - y) + xz(y - z) \geq 0\end{split}$$
This means that we can assume value of $z$ to be $0$. Letting $z=0 (\implies x+y=1)$ in our original function and applying AM-M inequality with the terms $x,x,2y$: $$f(x,y,0)=x^2y=2\frac{x^2y}{2}\leq\left(\frac{x+x+2y}{3}\right)^3=\left(\frac{2(x+y)}{3}\right)^3=\frac{4}{27}$$


How to solve a linear PDE like $\partial_t u(x,t) - \partial_x u(x,t) = a(x)\cdot b(t)$? Given a linear PDE like $\partial_t u(x,t) - \partial_x u(x,t) = a(x)\cdot b(t)$, how to solve it? A separation ansatz doesn't look promising due to the mixed term on the right hand side.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

Approach $1$:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=-1$ , letting $x(0)=x_0$ , we have $x=-s+x_0=-t+x_0$

$\dfrac{du}{ds}=a(x)b(t)=a(x_0-s)b(s)$ , we have $u(x,t)=f(x_0)+\int_k^sa(x_0-r)b(r)~dr=f(x+t)+\int_k^ta(x+t-s)b(s)~ds$

Approach $2$:

$\dfrac{dx}{ds}=1$ , letting $x(0)=0$ , we have $x=s$

$\dfrac{dt}{ds}=-1$ , letting $t(0)=t_0$ , we have $t=-s+t_0=-x+t_0$

$\dfrac{du}{ds}=a(x)b(t)=-a(s)b(t_0-s)$ , we have $u(x,t)=f(t_0)-\int_k^sa(r)b(t_0-r)~dr=f(x+t)-\int_k^xa(s)b(x+t-s)~ds$

Use a Fourier Transform:$\newcommand{\sgn}{\operatorname{sgn}}$

$$\begin{align}
  \partial_t u(x,t) - \partial_x u(x,t) &= a(x)\cdot b(t) \quad\Big|\quad \frac1{4\pi^2}\iint_{\mathbb R^2}dx\,dt\,e^{-i(\omega t + kx)}
\\ i\omega u_p(k,\omega) - ik u_p(k,\omega) &= -a(k)\cdot b(\omega)
\\\Rightarrow u_p(k,\omega) &= -i\frac{a(k)b(\omega)}{k-\omega}
\\ u_p(x,t) &= -i\iint_{\mathbb R^2}\frac{a(k)b(\omega)}{k-\omega}e^{i\omega t + ikx}\, dk\,d\omega
\\ &= -\frac{i}{4\pi^2}\iint_{\mathbb R^2}\iint_{\mathbb R^2}\frac{a(x')b(t')}{k-\omega}\exp\left[i\left(k(x-x')+\omega(t-t')\right)\right]\,dk\,d\omega\,dx'\,dt'
\\ &= -\frac{i}{4\pi^2}\iint_{\mathbb R^2}\int_{\mathbb R}a(x')b(t') \underbrace{\int_{\mathbb R}\frac{e^{ik(x-x')}}{k-\omega}dk}_{=-i\pi e^{-i\omega(x-x')}\sgn(x-x')}\, e^{i\omega(t-t')}\,d\omega\,dx'\,dt'
\\ &= \frac12\iint_{\mathbb R^2}a(x-h)b(t')\underbrace{\frac1{2\pi}\int_{\mathbb R}e^{i\omega(t-t'-h)}d\omega}_{=\delta(t-t'-h)}\sgn(h)\,dh\,dt' \quad\Big|\quad h\equiv x-x'
\\ &= \frac12\int_{\mathbb R}\sgn(h)a(x-h)b(t-h)dh
\\\Rightarrow u(x,t) &= u_h(x+t) + \frac12\int_0^\infty[a(x-h)b(t-h)-a(x+h)b(t+h)]dh
\end{align}$$

Note that this requires $\lim\limits_{h\to\infty}[a(h)b(h)-a(-h)b(-h)]=0$ in order to work, which is actually a weaker requirement than the existence of Fourier transforms of $a(x)$ and $b(t)$.


Two-point Taylor expansion with one assymptotic point? According to this paper, a two-point Taylor expansion can be definied like this:

$$\text{Let }f\left(z\right)\text{ be an analytic function and }z_1 \text{and }z_2\in \mathbb{C}, z_1\neq z_2\text{.}\\
\text{The two-point Taylor expansion is defined as:}\\
P_n\left(z_1, z_2;z\right)=\sum_{k=0}^n{\left[a_k\left(z_1,z_2\right)\left(z-z_1\right)+a_k\left(z_2,z_1\right)\left(z-z_2\right)\right]\left(z-z_1\right)^k\left(z-z_2\right)^k}\\
\text{with coefficients}\\
a_0\left(z_1,z_2\right)=\frac{f\left(z_2\right)}{z_2-z_1}\\
\forall_{n>0}{: a_n\left(z_1,z_2\right)=\sum_{k=0}^n{\frac{\left(n+k-1\right)!}{k!\left(n-k\right)!}\frac{(-1)^{n+1}n f^\left(n-k\right)\left(z_2\right)+(-1)^{k}k f^\left(n-k\right)\left(z_1\right)}{n!\left(z_1-z_2\right)^{n+k+1}}}}
$$
This is actually pretty nice. However, seeing that there are things like $z_1-z_2$, I'm not sure how you'd do an assymptotic expansion, where one of the two points is actually at infinity.

Is this possible? If so, how would you do it?

One thing I have done is use two sets of expansions: one at, say, zero and one at infinity.  So I would derive separate expansions and get two sets of coefficients; then save an extra parameter for matching the solutions to make a global approximation.


Showing continuity of $\dfrac{xy(x-2y)}{x^2+y^2}$ at the origin. Show that the function $f(x, y)=\dfrac{xy(x-2y)}{x^2+y^2}$ is continuous on $\mathbb{R}^2$.

It is easy to show continuity of $f(x, y)$ at points other than the origin but how is it continuous at the origin?

The function $f(x,y)$ is undefined at the origin, hence, it is not continuous at $(0,0)$.

Note that it does not have a limit as $(x,y)\to (0,0)$ because the limit of $f$  is different depending on the path that $(x,y)\to (0,0)$. For example,

$$ \lim_{y\to0} f(0,y) = -2$$
$$ \lim_{x\to0} f(x,0) = 1$$


How many ways are there of rolling 6 different coloured 6-sided dice so that exactly 3 different values are showing? Any help with this problem would be greatly appreciated. I have tried solving this problem like this:

1st die: 6 choices,
2nd die: 5 choices,
3rd die: 4 choices,
4th die: 3 choices,
5th die: 3 choices,
6th die: 3 choices

But this isn't getting me anywhere! Thanks.

Basically, you need to choose 3 distinct #s from 6,
choose patterns of 4-1-1 of a kind, 3-2-1 of a kind and 2-2-2 of a kind and permute them.

$4-1-1$ of a kind: $\binom31\binom63\cdot\frac{6!}{4!} = 1800$

$3-2-1$ of a kind: $\binom31\binom21\binom63\cdot\frac{6!}{3!2!} =7200$

$2-2-2$ of a kind: $\binom63\cdot\frac{6!}{2!2!2!} = 1800$

Add up to get a total of $10800$ ways

There are $\binom633^6$ ways with at most three distinct values ($\binom 63$ to pick the three values and $3^6$ ways of assigning values to the dice). Similarly, there are $\binom622^6$ ways with at most 2 values. So the answer is
$$
\binom633^6-\binom622^6=13620.
$$


simple question on division Does any natural number q divide 0 because there exists a natural number m such that qm=0 (m=0)?

Is it true that 0 does not divide any natural number >0, because there does not exist a number q such that 0*q=m where m is any natural number >0.

What about when m is 0. does 0 divide 0, is it true that the definition above holds but it is also undefined? thanks

Welcome to MSE!
As a general hint, let $R$ be ring. Then for each ring element $r$, $r0 = 0 = 0r$, since $r0 = r(r+(-r)) = r^2-r^2 = 0$; similarly, $0r=0$. So $0$ is absorbing in every ring.

If you look at the relation of divisiblity on the monoid of natural numbers, $a\mid b$ ($a$ divides $b$) means that $a\cdot c=d$ for some number $c$. Here $0$ is the largest element, since $a\mid 0$ by taking $c=0$ and $1$ is the smallest element, since $1\mid b$ by taking $c=b$.


What exponent law have I violated? Given the expression: $b^8(2b)^4$, simplify it.

I multiply $b^8$ and $2b$, which gives $(2b^9)^4.$

$(2b^9)^4 = 16b^{36}$. However this is incorrect.

The correct answer is to distribute the exponent first like so:
$b^8(2^4b^4) = 16b^{12}$. This makes sense, but I'd like to know why my original approach was incorrect.

You have violated the order of operations.  $b^8 (2b)^4$ means that $2b$ is raised to the fourth power, then multiplied by $b^8$.  If you were to perform multiplication of $b^8$ by $2b$ before you raise to the fourth power, you would also have raised $b^8$ to the fourth power, resulting in an additional factor of $b^{24}$.  As you can see, $36 = 24 + 12$.

To illustrate how one might have written the expression to obtain your erroneous result, we would write instead $(b^8 (2b))^4 = 16b^{36}$.


A function is said to be continuous. Can it still have vertical asymptotes? This is a general question. A function is said to be continuous. Can it still have vertical asymptotes? Looking at the definition of continuity, I would say no. Because near a vertical asymptote x-delta might have an y of close to minus infinity, while x+delta might have a value of near +infinity, for example.

It doesn't make sense to ask whether a function is continuous at a point where it's not defined. So if it has a vertical asymptote, then that's not a point of discontinuity, but rather a point that is not part of the domain.

To make it clear what I mean by "doesn't make sense", the definition of continuity at a point $x$ involves the expression $|f(x) - f(y)|$. If $x$ is such that $f(x)$ doesn't make sense, then neither does that expression, so asking about continuity is meaningless.

How about $f(x) = x^{1/3}$ (real branch)?  The asymptote in $(0,0)$ is vertical and the function is continuous on the reals.


Maximum incidence angle in optic fiber. A circular glass cylinder is a simplified model for fiber optics. Calculate the maximum incidence angle $\theta$ at the entrance face at which the light will be guided into the fiber by successive total reflections as a function of the refractive index $n$ of the fiber.

I started studying optics a short time ago and I'm having a difficulty in some exercises in the book. I appreciate any hints.

I know that I just need to calculate the maximum incidence angle for total internal reflection, since for larger angles, the refracted ray will leak out of the fiber. Also, angle of incidence equals angle of reflection. But, I don't know how to use this informations to get an explicit equation.

There is an easy tutorial on Thorlabs web page explaining this.  It's just a single application of Snell's law assuming you know the index of the fiber and the cladding.  The numerical aperture is defined by the the maximum half-acceptance angle $\theta_a$ associated with total internal reflection.

Let $n_i$ be the index of the outside material (usually 1) and $\theta_c$ be the critical angle for total internal reflection.  Then

$$\sin \theta_c = n_c/n_f = \cos \theta_t = \sqrt{1- \sin^2 \theta _t}$$
So the numerical aperture $NA$, is
$$ NA = n_i \sin \theta_a = \sqrt{n_f ^2 - n_c ^2}$$

If $n_i$ is 1 and $\theta_a$ is small, then you just the NA is approximately $\theta_a$.

Here is the link to Thorlabs:  https://www.thorlabs.com/tutorials.cfm?tabID=17485628-68dd-4d22-ad17-6dd4520974c7

You know that the maximum angle of incidence that can be used in an optical fiber is called the critical angle (and this angle produces and angle of refraction equal to $90$ degrees).
To solve for maximum angle of incidence $i$, use the formula
$\sin (i)/\sin (r) =n_2/n_1=C_1/C_2$

Where $i$ is the maximum incidence angle $u$ want to find,
$r$ is $90$ degrees giving $\sin(90)=1$,
$n_2$ is the index of the fiber core and $n_1$ is the index of air which is equal to $1$. Then $C_1$ is the speed of light in air which is equal to $3 \cdot 10^8$ m/s, then $C_2$ is the speed of light in the glass core.


Given two equations $(xax)^3 = bx$ and $x^2a = (xa)^{-1}$ in a nonabelian group, solve for $x$. This is for a basic non-commutative group.

My steps:

$(xax)^3 = bx$

$xaxxaxxax = bx$

$xax^2ax^2ax = bx$

$xax^2ax^2a = b$

$x^2a = (xax^2a)^{-1}b$

Now, substituting $x^2a$ into the second equation:

$(xax^2a)^{-1}b = (xa)^{-1}$

$b = (xax^2a)(xa)^{-1} $

$b = (xax^2a)(a^{-1}x^{-1})$

$b = xax$

Now back to the first equation:

$b^3 = bx$

$b^2 = x$

Apparently, this is not the correct answer according to my answer sheet.

Am I making a mistake somewhere?

The solution in the answer sheet is equivalent; your solution is fine.
Marking your work line by line . . .

My steps:
$(xax)^3 = bx$ $\color{red}{\quad\checkmark\text{Def}}$
$xaxxaxxax = bx$ $\color{red}{\quad\checkmark\text{Def}}$
$xax^2ax^2ax = bx$$\color{red}{\quad\checkmark\text{Rewrite}}$
$xax^2ax^2a = b$ $\color{red}{\quad\checkmark (\times x^{-1})}$
$x^2a = (xax^2a)^{-1}b$ $\color{red}{\quad\checkmark (\times (xax^2a)^{-1})}$
Now, substituting $x^2a$ into the second equation:
$(xax^2a)^{-1}b = (xa)^{-1}$ $\color{red}{\quad\checkmark (\text{Sub})}$
$b = (xax^2a)(xa)^{-1} $ $\color{red}{\quad\checkmark (\times (xax^2a))}$
$b = (xax^2a)(a^{-1}x^{-1})$ $\color{red}{\quad\checkmark (\text{Use of 'inverse of product' lemma})}$
$b = xax$ $\color{red}{\quad\checkmark (\text{Use of inverses})}$
Now back to the first equation:
$b^3 = bx$ $\color{red}{\quad\checkmark (\text{Sub})}$
$b^2 = x$ $\color{red}{\quad\checkmark (\times b^{-1})}$

The following is  solution in the image attached from the link.
solution Image on paper

Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.


Why do partial derivatives need to be continuous in order for the function to be differentiable Why it's not enough for the partial derivatives to exist for implying differentiability of the function?
Why is the continuity of the partial derivatives needed?

To answer your first question: because they're partial, i.e., derivatives in very special directions.  Imagine, for example, a surface which is generated by a ray parallel to the $x$-$y$-axis that is rotating in a way that for the first quarter of its rotation it raises by $1$, for the next quarter it lowers by $1$, then lowers $1$ again and in its last quarter it raises again by $1$.


question about two different definitions of limit point From the definition of a limit point it follows  :

A limit point $x$ of a subset $S$ of a topological space $X$ is a point such that every neighborhood of $y$ has a point in common with $S$.

also based on Wikipedia :

it doesn't make a difference if we restrict the condition to open neighborhoods only.

I know that if every neighborhood of $y$ intersects $S$, then also open neighborhoods of $y$ has a point in common with $H$
but why we can claim that if every open neighborhood of $y$ intersects $S$ then, so does every neighborhood of $y$?

First, by your question, you seem to have understood this direction but I will write it anyways: open neighborhoods are a special case of neighborhoods. So, if every neighborhood of $y$ intersects $S$, the same can be said for every open neighborhood.

Now, for the other direction, assume every open neighborhood of $y$ intersects $S$ and let $V$ be a neighborhood of $y$. By definition, $V$ contains an open set $U$ which contains $y$. $U$ is an open neighborhood of $y$, so $U$ intersects $S$. But $U\subset V$, so $V$ also intersects $S$. $V$ is an arbitrary neighborhood, so every neighborhood intersects $S$.

Mathematicians often shorten the statement

  $U$ is an open set containing $x$

to the phrase

  $U$ is a neighborhood of $x$.

Borrowing from Munkres:

  If $A$ is a subset of the topological space $X$ and if $x$ is a point
  of $X$, we say that $x$ is a limit point (or “cluster point”, or
  “point of accumulation”) of $A$ if every neighborhood of $x$
  intersects $A$ in some point other than $x$ itself. Said
  differently, $x$ is a limit point of $A$ if it belongs to the closure
  of $A − \{x\}$. The point $x$ may lie in $A$ or not; for this
  definition it does not matter.


A problem in mechanics This is an apparently simple problem that I could not solve. The source is a colleague who interviewed for the European space agency.
You are standing in a field with a chronometer and a camera. Someone throws a ball (a point mass) in the air, with an unknown speed and velocity. You can take pictures with the camera and record the time between pictures with the chronometer, but you can’t directly measure the distance between you and the ball. How many pictures do you need to take to determine where the ball will hit the ground?
Assume that the earth is flat, that $g$ is known to arbitrary precision, that air has no effect and that the trajectory is generic. Assume moreover that the camera is at a known position and at a known height above the ground (may assume that the height is $0$ if useful for the solution).
Attempts at solution
The state of the ball is determined by six parameters, but we only want to know where the ball hits the ground, so we may need only five or less parameters.
The first picture gives two parameters and the second one three, so two pictures may be sufficient. My gut feeling is that they are not, but I wouldn’t know how to prove it.
The third picture gives three new parameters, but they cannot all be independent. Indeed at most one is, so the answer should be three, but again I could not find a way to prove it.

I refuse to assume the earth is flat...
But I would guess four, because we are talking about a parabolic curve, which is represented by y = ax²+bx+c. We need to determine the three coefficients a, b and c, but we only obtain the information of the difference between two photos. So to get three differences we need four points.

I think the answer would be that the problem is unsolvable. No matter how many pictures you take, range cannot be predicted. Let me illustrate with a more intuitive situation. Suppose you live in a high-rise building, and you have one wall made of glass. You cannot look down on the ground, and don't know how high up you are. You can only look straight ahead. Now, someone from the ground throws a ball up in front of you. You can see the ball go up from a certain point on the floor of your room, reach a maximum height, and go down again from some other point. Suppose you start taking pictures using the same set-up you had before. Would you ever be able to know where the ball will hit the ground? Of course, you can assign origin to be that point where the ball first comes into your sight, have an $x$-axis along the floor of the room and a vertical $y$-axis, and let the time be $t=0$ at that instant. If you then take one other picture, that would be enough for you to draw the trajectory of the ball with respect to the coordinate system I described. However, since you do not know what the equation of the horizontal "ground line" is, that is, don't know how far below you the ground lies, you can never figure out where exactly the point of projection of the ball was, or where it will end up.
Hope this makes it clear.


Problem: Sum of absolute values of polynomial roots Can you please give me some hints as to how I might approach this problem? Thanks!

  Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$

HINT

$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$

Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equality holds implies $|x_1| = |x_2| = |x_3| = 1$

We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and $x_1 x_2 x_3$ to be real,  and further $x=-1$ satisfies the equation.

Hence $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And $|x_1| = 1$ and $|x_2| = 1$. Hence, $x_1 = e^{i \theta}$ and $x_2 = e^{i \phi}$.

$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$

$x_1 x_2 = 1 \implies \phi = -\theta$

$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$

Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.

Hence, $a = 4 \cos(\theta) - 2$ and the roots are $-1,e^{i \theta},e^{-i \theta}$

The way to solve this is actually quite simple, apparently:

$$|x_2| + |x_3| =2 \implies -2 \le x_1 + x_1 \le 2$$


How to know when estimated coefficients of a quadratic function are zero by looking at graph? I have a quadratic function:

$f(x) = w_0 + w_1 \cdot x + w_2 \cdot (x^2)$

How would I know which estimated coefficients are zero by look at a graph?  For example:

I can clearly see that the line starts on y=0 on the above first and thrid graphs. And that both x and y of the line start at 0 for the second and fourth graphs.

But, how would I know which estimated coefficients are zero for each graph?

Disclaimer: I never took linear algebra. Any explanation or even links would be appreciated! Thanks!

$w_0=0$ if and only if the graph starts from the origin $(0,0)$ (true for every polynomials of any degree);
$w_2=0$ if and only if the graph is a straight line (true of for polynomials of degree 2);
$w_1=0$ if and only if the axis parabola is exactly the y-axis or the graph is an horizontal line (degree-2 polynomials).

Remember that you cannot always see this at first blush!

1st Image : $w_2=0$,

2nd Image : $w_0=0$, $w_2=0$

3rd Image : None of $w_0,w_1,w_2$ are zero

4th Image : $w_0$=0


How to use the inverse function theorem to find a local inverse? I have a function $F(x,y)=(x^2 - y^2, xy)$ and I need to show that it has an inverse. How do I find the inverse of this function using the inverse function theorem? I did this a while ago and now I can't find it in my notes. I recall having to make $u = x^2 - y^2$ and $v = xy$ but then I'm lost.

Thank you for your assistance.

The IFT proves the local existence of inverse function, but is not useful for finding the explicit form of the inverse function.
In this case,
$$
\frac{\partial(u,v)}{\partial(x,y)} =
\left|\matrix{2x&y\cr-2y&x}\right|=2(x^2+y^2)\ne 0\iff(x,y)\ne(0,0)
$$
and the inverse exists locally in any point except the origin.

For a concrete formula write, as you say, $u=x^2-y^2$ and $v=xy$. Then $y=v/x$, so that
$$u=x^2-\frac{v^2}{x^2}$$
which implies that
$$x^4-ux^2-v^2=0.$$
Solve the above equation for $x$ using the quadratic formula and substitute back into $y=v/x$ to find the equation for $y.$

Of course care must be taken when choosing signs for particular regions of $\mathbb{R}^2$.


Prove $t^{1\over t-1}$ is rational only when $t=2$ in the context of solving $x^y=y^x$ where $x\neq y$ Prove $t^{1\over t-1}$ is rational only when $t=2$ where $t$ is any real number.
This problem arises from me watching the blackpenredpen video solving $x^y=y^x$ where $x\neq y$. https://www.youtube.com/watch?v=PI1NeGtJo7s, in which he gives the solution $x=t^{1\over t-1}$.
First I considered the integer solutions where $x\neq y$ other than $2^4=4^2$ but realized it is trivial because $1<t^{1\over t-1}<2$ for any $t\geq 2$ and $-1>t^{1\over t-1}>-2$ for any $t\leq-2$.
The next question is then, are there any rational solutions other than the $2,4$ pair. It does not seem trivial to me as inequalities do not help.
My guess is that $t^{1\over t-1}$ cannot be rational other than when $t=2$ but I wan't able to prove.

The range of $f(t)=t^{1/(t-1)}$ is $(1,\infty)$ and the function is continuous and monotonic, hence invertible. So there is an infinite (but countable) number of solutions of
$$t^{1/(t-1)}=q$$ where $q$ is rational.

$$t=-\frac1{\log q}W\left(-\frac{\log q}q\right).$$

We have,
$$t-1=a^{t-1}-1, ~ \text{where} ~ a\in\mathbb {Q^{+}}$$
$$\implies a^x=x+1, x=t-1$$
Substitute $a^x=t \implies x\ln a=\ln t $
$$\implies  t=\dfrac {\ln t}{\ln a}+1$$
Substitute $\ln t=u$
$$\implies e^u=\dfrac{u+\ln a}{\ln a}$$
$$\implies e^{u+\ln a}=\dfrac {a}{\ln a}(u+\ln a)$$
$$\implies (u+\ln a)^{-1} e^{u+\ln a}=\dfrac {a}{\ln a}$$
$$\implies  \left((u+\ln a)^{-1} e^{u+\ln a}\right)^{-1}=\dfrac {\ln a}{a}$$
Substitue $z=u+\ln a$
$$\implies -ze^{-z}=-\dfrac {\ln a}{a}$$
$$\implies W(-ze^{-z})=W\left(-\dfrac {\ln a}{ a}\right)$$
$$\implies z=-W\left(-\dfrac {\ln a}{a}\right)$$
If we take the substitutes backwards
$$t=\dfrac{e^z}{a}$$
We have:
$$\implies t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}$$
So, $a \in\mathbb {Q^{+}}$, where $$\color {gold}{\boxed {\color{blue}{t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}}}}$$
Thus, your conjecture is not true.

Here $W(x)$ is a Lambert function


Prove $e^{\ln{x}} = x$ Is it possible to prove $e^{\ln{x}} = x$ for a student or can you only say that exponentiation is defined to be the inverse of natural logarithm and stop there?

For a student. Let $x>0$. Then

\begin{align*}
e^{ln x} &= x \\
\Leftrightarrow \ln\left(e^{ln x}\right) &= \ln x \\
\Leftrightarrow \ln x \cdot \ln e &= \ln x \\
\Leftrightarrow \ln x \cdot 1 &= \ln x \\
\Leftrightarrow \ln x &= \ln x \\
\Leftrightarrow x &= x.
\end{align*}

$ \therefore e^{ln x} = x \, \forall \, x>0.$

Appealing to the fact that $\log a^b=b \cdot \log a$ is perfectly acceptable, as it is a prerequisite for being in the club of people who are allowed to express logarithms.

As my commenters rightfully state, this identity proof is best shown in both directions,  with the caveat that $x>0$.


Find remainder of $\frac{17^{235}}{ 23}$ I need to find remainder of $\frac{17^{235}}{ 23}$.

This is supposed to be solved using the following method:

$\varphi(23) = 22$
${17}^{235} = (({17}^{22})^{10})\cdot {17}^{15}$
${17}^{22}\equiv 1 \pmod{23}$
Then, according to Euler theorem, $17^{235} \equiv 17^{15} \pmod {23}$

and then i get stuck.

$17^{15}=  17^1\times 17^2\times17^4\times 17^8$.
In the rhs every term is the square of the previous term.

$17^2=289=230+69-10\equiv \mathbf{-10}\pmod{23}$.

$17^4\equiv(-10)^2=100\equiv\mathbf{8}\pmod{23} $

$17^8\equiv 8^2=64\equiv\mathbf{-5}\pmod{23} $

Now multiply $17.(-10).8.(-5)\equiv\mathbf{-8}\pmod{23}$

The answer is 15.

The answer is 15.

You can, by fast exponentiation, check that $17^{11}\equiv-1\mod 23$. Thus, as $23$ is prime, Little Fermat entails that:
$$17^{235}\equiv17^{235\bmod22}= 17^{15}\equiv -17^4=-(17^2)^2\equiv-13^2\equiv -8\equiv 15.$$


Countable or uncountable set 8 signs Let S be a set of pairwise disjoint 8-like symbols on the plane. (The 8s may be inside each other as well) Prove that S is at most countable.

Now I know you can "map" a set of disjoint intervals in R to a countable set (e.g. Q :rational numbers) and solve similar problems like this, but the fact that the 8s can go inside each other is hindering my progress with my conventional approach...

Let $\mathcal{E}$ denote the set of all your figure eights. Then, define a map $f:\mathcal{E}\to\mathbb{Q}^2\times\mathbb{Q}^2$ by taking $E\in\mathcal{E}$ to a chosen pair of rational ordered pairs, one sitting inside each loop. Show that if two such figure eights were to have the same chosen ordered pair, they must interesect, which is impossible. Thus, $f$ is an injection and so $\mathcal{E}$ is countable.

Let us associate each 8 with its center. If two 8's have exactly the same size and are too close together, then they will intersect. The same holds even if they only have almost the same size. That is, there are some $\epsilon,\delta$ such that two nonintersecting 8's of size between $r$ and $(1+\epsilon)r$ cannot be $\delta r$-close (that is, their centers need to be at distance at least $\delta r$.

Let us denote by $S_r$ the set of 8's of size between $r$ and $(1+\epsilon)r$. The circles at radius $\delta r/3$ around the centers are disjoint, and so if we choose for each 8 in $S_r$ a rational point inside the circle at radius $\delta r/3$ around its center, then these rational points will be distinct. This shows that $S_r$ is countable.

The set of all 8's is the union of countably many sets of the form $S_r$ (for example, we can take $r = (1+\epsilon)^z$ for all integers $z$). Since the countable union of countable sets is countable, we conclude that the number of 8's is countable.


What is the highest power of a prime that divides nPr? I know that the highest power of a prime which divides $n!$ is given by
$$\left[\frac np\right]+\left[\frac n{p^2}\right]+\left[\frac n{p^3}\right]...$$
Where $[x]$ is the greatest integer function.

For $^{n}P_{r}$, which equals $\frac{n!}{(n-r)!}$, I can find the highest power of $p$ dividing $n!$ and $(n-r)!$, and then subtract the two to get the highest power of $p$ dividing $^{n}P_{r}$.

Now is there a better method to do this, where we don't need to find the highest power of $p$ dividing $(n-r)!$? Can we directly find the power of $p$ dividing the product of $r$ consecutive integers?

No, because it depends on which $r$ consecutive integers you take (for example, if one of them is $p^{1000}$ you might get a fairly high power). The way to think about computing the $p$-adic valuation of $\binom{n}{r}$ (the first is a fancy name for what you are computing, the second a FAR better notation for your $P$) is this:

Write $r$ and $n-r$ in base $p,$ and then count how many carries you have when adding them.

The largest power of $p$ which divides $n!$ is the number of times $p$ appears in the factorization of $n!$. This is the same as finding how many times $p$ appears in the factorization of $n,n-1,n-2,\cdots,2,1$.

Now using the division algorithm, given an $a \in \mathbb{Z}$, we can find a $q$ and $r$ such that $a=qp+r$, where $0\leq r <p$. Then we have $a/p=q+r/p$. Then taking the greatest integer function of both sides yields
$$
\lfloor a/p\rfloor = \lfloor q\rfloor +\lfloor r/p\rfloor=q+0=q
$$
Of course, $p^2$ may divide one or more of $n,n-1,n-2,\cdots,2,1$, say $i$. Then we can use the same idea as before. So let $n(x,p)$ be the largest power of $p$ that divides $x!$. Then
$$
n(x,p)=\sum_{k \in \mathbb{Z}_+}\sum_{j=1}^x \left\lfloor \frac{x}{p^k} \right\rfloor
$$
Of course, this sum converges as the floor function is $0$ for sufficiently large $k$. In general, this is about as good as a method as one gets without sufficiently advanced methods (that often only lengthen the issue).


Finding a sequence of sets whose intersection is a null set Find a sequence of sets $I_n=\{r:r \in \mathbb{Q}, a_n\le r \le b_n\} $ in $\mathbb{Q}$, where $a_n, b_n \in\mathbb{Q}$ such that $$I_{n+1} \subset I_n\forall n\in\mathbb{N}$$ $\lim_{n \to \infty}(b_n - a_n)= 0 $ but $$\cap_{n=1}^{\infty} I_n= \emptyset$$

Can anyone explain the question and how to go about its solution?

consider a irrational number $i \in \mathbb{R}$...we know that $\mathbb{Q}$ is dense in $\mathbb{R}$....consider a incersing sequence of rationals converging to $i$ as {$a_n$} and consider a decreasing sequence {$b_n$} converging to $i$ and consider $J_n$= {$r \in \mathbb{R}| a_n \leq r \leq b_n$} then $J_n \cap \mathbb{Q} = I_n$ ...then by cantor intersection property $\bigcap_{n} J_n$ is singleton and contain $i$ only...so   $\bigcap_{n} I_n = \phi$

For each $n \in \mathbb{N}$, take $a_n$ a rational number in the interval $ (\pi - \frac{1}{n}, \pi)$, and $b_n$ a rational number in $(\pi, \pi + \frac{1}{n})$. Now,
$$
\bigcap_{n=1}^{\infty} (a_n,b_n) = \lbrace \pi \rbrace.
$$
However, with your definition,
$$
I_n = (a_n,b_n) \cap\mathbb{Q},
$$
then
$$
\bigcap_{n=1}^{\infty} I_n = \bigcap_{n=1}^\infty (a_n,b_n) \cap \mathbb{Q} = \lbrace \pi \rbrace \cap \mathbb{Q} = \emptyset.
$$
Remind that the set of rational numbers is not complete, but the set of real numbers is complete.


What u substitution should I use for this integral, if any, and what is the final result? Evaluate $$\int_1^5 \frac{x}{\sqrt{2x-1}}\mathrm dx$$ by the substitution method.

What u substitution should I use for this integral, if any, and what is the final result?

There are two substitutions that should occur to you as immediate possibilities when you look at this integral: $u = 2x-1$, and $u = \sqrt{2x-1}$. Pick one, and try it; if works, great, and if not, you can still try the other one. You don’t have to get it right on the first try, and you shouldn’t start to worry until you’ve exhausted the obvious, straightforward possibilities. In this case that won’t happen (unless you make a mistake): both substitutions work. In fact, you should try them both to see how they work out, because you’re likely to need both of these types of substitution before you’re done with your course.

try the substitution x=t+3, this will simplify your limits. And you will be able to solve it using properties of definite integrals.


If $f: X \to Y$ is continuous and $E \subseteq X$ is closed and bounded, $f(E)$ is closed and bounded? I'm trying to think of a counterexample to the claim if $f: X \to Y$ is continuous and $E \subseteq X$ is closed and bounded $f(E)$ is closed and bounded.  Here, $X$ and $Y$ are metric spaces.  Obviously, in Euclidean spaces this would be true, since continuity preserves compactness and compactness is equivalent to closed and bounded by Heine-Borel.

Help is appreciated.

Let $X=Y=\Bbb{R}$ and $f$ be the identity. Equip $Y$ with the standard metric and $X$ with the bounded metric $d(x,y)=\min(|x-y|,1)$. Then $f$ is continuous, and $X$ is closed and bounded (as a subset of itself) but its image $Y$ is not bounded.

More examples: $X=(0,1)$, $Y=\Bbb{R}$, $f(x)=1/x$, $E=X$. Here the image is neither closed nor bounded.

$X=(0,1)$, $Y=\Bbb{R}$, $f(x)=x$, $E=X$. Here the image is bounded but not closed.

Actually you need a stronger condition on the metric space, namely total boundedness. This property says that for every $\epsilon>0$, there exists finitely many balls of radius $\epsilon$ such that their union covers the whole space. The proof is this:

Let $(M,d_1)$ total bounded metric space, and $f:M\rightarrow N$ a uniformly continuous function, and $E\subseteq M$ closed and bounded, choose a finite set of pairs $(x_i,\epsilon_i)$ with $x_i\in f(E)$ and $\epsilon_i>0$, in a way that the set of open sets $f^{-1}(B(x_i,\epsilon_i))$ cover $E$, it can be done for the condition of total boundedness and uniformly continuity. This says that $f(E)$ is bounded.

But as Chris Eagle showed, the set $f(E)$ need not be closed.


Fish population growth question: Can someone check my work and answers? I'm reviewing for a math test this Tuesday and just want to make sure I'm doing things right. If someone could check my work that would be great.

Here's the question (work below):

Here's my work:

Thank you in advance!

You only know that the growth rate is $1.2\%$, this is not equal to the exponential constant $r$. To find $r$ do the following:

After one year you know that there must be $n_0 \cdot 1.012$  fish, this has to be equal to $n_0 e^{r}$, thus $r = log(1.012)$

and you get: $n(t) = n_0 e^{log(1.012) t} = 13 \cdot e^{log(1.012) t}$

Your answers are correct but your sketch is not. Because the population is growing exponentially then the derivative of the function will be positive (ie the line will curve upwards.) If you think about it the larger the population is the larger it will grow. Also I might add that the growth rate is 1.021 not 1.21. Apart from that its all good!


Calculating how much light gets through steel mesh (commonly used to make cages) I have expanded steel mesh that I use to make garden cages:

I would like to know how much sunlight the mesh lets through.

I think I need to calculate the area of the mesh's negative space. And then from there I could calculate a percentage of sunlight that is let through.

How can I calculate how much light would get through?

Crop the ruler and binarize:

Then count the white pixels and divide by the sum of white and black to find:

$0.592548$

In Mathematica:

pixels = Flatten[
   ImageData[
     Binarize[
      ImageTake[mesh, {140, 664}, {1, 892}]]]];
N@Count[pixels, 1]/(Count[pixels, 0] + Count[pixels, 1])

$0.592548$

People wanted a different threshold:

I can use Geographic Information Systems (GIS) software to plot and calculate the area of the negative space:

Crop the image down to a more manageable size (in Microsoft Paint).
In GIS software, draw a rectangle around the image to get it's area.
Draw polygons in the negative space.

  Area of image  = 147542 units

  Area of negative space = 92907 units

Calculate the percentage of negative space:

  92907 / 147542 = 62.97%

Therefore, approximately 63% of light would get through this mesh.


Do two paths that merge into one have distinct derivatives at the "merge point"? Let $p_1, p_2: [a, b] \rightarrow \mathbb{R}^n$ be two distinct continuous and differential paths that merge at point $t$.  That is $p_1(t_x) = p_2(t_x)$ for all $t_x \ge t$ and $p_1(t_0) \ne p_2(t_0)$ for all $t_0 < t$.
Does it necessarily follow that
$$
\frac{dp_1(t)}{dt} \ne \frac{dp_2(t)}{dt}
$$
or instead do we have equality
$$
\frac{dp_1(t)}{dt} = \frac{dp_2(t)}{dt}
$$
or is it indeterminate either way?

Assuming $t<b$, then the derivative $p_1'(t)$ can be computed as the limit of $\frac{p_1(s)-p_1(t)}{s-t}$ as $s$ approaches $t$ from the right, and similarly for $p_2'(t)$.  Since $p_1(s)=p_2(s)$ for all $s\geq t$, these difference quotients are the same for $p_1$ and $p_2$, and so $p_1'(t)=p_2'(t)$.

Indeterminate.
Here is one where they are equal:
Curve 1:  $x=t, y=t^2$, when $t<0,$ and $x=t,y=0, t\ge0.$
Curve 2:  $x=t, y=0, \forall t$
Here is a case where they are not equal:
Curve 1:  $x=t, y=t^2$, when $t<1,$ and $x=t,y=1, t\ge1.$
Curve 2:  $x=t, y=2-t^2$, when $t<1,$ and $x=t,y=1, t\ge1.$


Prove that $\sum_{j=1}^n a_{ij} = \lambda$ is an eigenvalue of $A$. Let $\lambda \in \mathbb{R} $ and $A = (a_{ij})_{1 \leq i \leq j \leq n}$ is a $n
 \times n$ matrix.

  Suppose $a_{i1} + a_{i2} + ... + a_{in} = \lambda $.

  For all $i \in \{1,2,...,n\}$, prove that $\lambda$ is an eigenvalue of $A$.

I started by making the sum of the first row so I can use:   $a_{i1} + a_{i2} + ... + a_{in} = \lambda $.

$\begin{align} \det(A - \lambda I_n)= \begin{vmatrix}
a_{11} - \lambda & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} - \lambda & \dots & a_{2n} \\
\vdots &  & \ddots & &  \\
a_{n1} & \dots & & a_{nn} - \lambda
\end{vmatrix} =
\begin{vmatrix}
\sum_{j=1}^{n} a_{1j} - \lambda & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} - \lambda & \dots & a_{2n} \\
\vdots &  & \ddots & &  \\
a_{n1} & \dots & & a_{nn} - \lambda
\end{vmatrix} =
 \begin{vmatrix}
0 & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} - \lambda & \dots & a_{2n} \\
\vdots &  & \ddots & &  \\
a_{n1} & \dots & & a_{nn} - \lambda
\end{vmatrix}
\end{align}$

I don't see how to proceed to prove the sum of $a_{1j}$ is an eigenvalue.

Hint: see what happens when $A$ is multiplied on the right with a column vector of $1$.

you can use induction.
for $n=1$ is abvious, for $n=2 $ you have :

$ \pmatrix{b & a\\ c & d} $ with $ a+b=c+d=K$

so for finding eigenvalue you have to make the determinant of $\pmatrix{a-x & b\\ c & d-x}$ equal zero. i.e

$$(a-x)(d-x)-cb=0$$ $$ad-ax-xd+x^{2}-cb=0$$, we have b=K-a and c=K-d so

$$ad-ax-xd+x^{2}-(K-a)(K-d)$$ so
$$ad-ax-xd+Kd+aK-ad-K^{2}+x^{2}=0$$, so

$$ K=x $$ is eigenvalue .

now you should assume its true for a $n \times n$ matrix , and prove for $ n+1\times n+1$ . it is very easy


How many edges, faces, cells in a $2\times 2 \times 2 \times 2$ hyper cubic lattice? If I have a $2\times 2\times 2\times 2$ hyper cubic lattice, how many corners, edges, faces, and cells will it be composed of? E.g. the 4D analogue of figure below. Assume the faces within the figure are filled in too. This figure has $36$ 2D faces (1x1 each), $27$ vertices, and $54$ 1D edges of length 1. (Some of the faces of the lattice are shared by more than one cell... and I don't want to double-count here).

$\hskip 2in$

We can perform the same counting argument used to count the $k$-cells of an $n$-cube in order to count the $k$-cells of an $\ell_1\times\cdots\times\ell_n$ hyper-rectangle of $n$-cubes. We will do this by counting the number of marked $k$-cells in the figure (a marked $k$-cell is a $k$-cell with one of its corners "marked"). To create a marked $k$-cell, we first pick a vertex, and then pick $k$ of the edges connected to it to determine the $k$-cell the vertex is a corner of, whenever this is possible. The number of marked $k$-cells with given corner vertex $v$ is $\binom{r}{k}$, where $r=\deg(v)$.

We must determine the number of vertices $v$ with vertex degree $\deg(v)=r$ in the figure, for all possible values of $r$. Consider the $\ell_1\times\ell_2\times\ell_3$ case for inspiration. For the purpose of abbreviation, denote $t_i=\ell_i-1$ for $i=1,2,3$. There are

$8$ vertices $v$ with $\deg(v)=3$
$4(t_1+t_2+t_3)$ vertices $v$ with $\deg(v)=4$
$2(t_1t_2+t_2t_3+t_3t_1)$ vertices $v$ with $\deg(v)=5$
$t_1t_2t_3$ vertices $v$ with $\deg(v)=6$.

Consider "faces" of the hyper-rectangle of various dimensions, defined so that for instance a $3$-rectangle has one $3$-face, six $2$-faces, twelve $1$-faces and eight $0$-faces. Each face is a collection of vertices, and the set of all faces is a partition of the vertex set. For $r=n,n+1,\cdots,2n$ a vertex will have degree $r$ if and only if it is in an $(r-n)$-face.

Let $\Lambda_m$ be the set of $m$-faces. For each $m$-subset $I\subset\{1,\cdots,n\}$ let $\Lambda_m(I)$ be the subset of $m$-faces with dimensions $\prod_{i\in I}t_i$. Note that each $\Lambda_m(I)$ has the same number of $m$-faces, by symmetry. There are a total of $\binom{n}{m}$ $m$-subsets of $\{1,\cdots,n\}$, and there are $2^{n-m}\binom{n}{m}$ $m$-faces in the collection $\Lambda_m$, therefore $|\Lambda_m(I)|=2^{n-m}$ for each $I$. And so $2^{2n-r}e_{r-n}(t_1,\cdots,t_n)$ counts the number of vertices $v$ with $\deg(v)=r$. Multiply this by $\binom{r}{k}$ and sum over $r=n,\cdots,2n$ in order to get the number of marked $k$-cells. Since every $k$-cell has $2^k$ possible corners to mark, this overcounts the number of unmarked $k$-cells by a factor of $2^k$. Dividing, we get

$$\# k\textrm{-cells of an }\ell_1\times\cdots\ell_n\textrm{ hyper-rectangle} =\sum_{r=n}^{2n}2^{2n-r-k}e_{r-n}(\ell_1-1,\cdots,\ell_n-1)\binom{r}{k}.$$

Your hypercube is determined by the vertices $(0,0,0,0)$, $(0,0,0,1)$, $(0,0,1,0)$, $(0,0,1,1)$, etc. Thus there are $2^4$ vertices (one for each quadruple).

How many edges? An edge is two points that differ in only one place, such as $(0,\color{red}0,1,0)$ and $(0,\color{red}1,1,0)$. An edge is determined by the three coordinates that don't change. Thus, there are:
$$\dbinom432^3=32\text{ edges};$$
$\binom43$ ways of selecting three edges and $2^3$ ways of giving them each a zero or one.

How many faces? A face is four points that differ in two places, such at $(0,\color{red}0,1,\color{red}0)$, $(0,\color{red}0,1,\color{red}1)$, $(0,\color{red}1,1,\color{red}0)$, and $(0,\color{red}1,1,\color{red}1)$. A face is determined by the two coordinates that don't change. Thus, there are:
$$\binom422^2=24\text{ faces};$$
$\binom42$ ways of selecting two edges and $2^2$ ways of giving them each a zero or one.

Similarly, there are:
$$\dbinom412^1=8\text{ cells}.$$


Integral of $f(x)$ converges but of $f(x)^2$ diverges? Does anybody know a function  $f$ that is continuous on $[0, \infty)$, so that $\int_{0}^{\infty} f(x)dx$ converges but $\int_{0}^{\infty} f(x)^2dx$ diverges? Thanks

EDIT: Sorry! I meant that $f$ is also continuous there.

A simple function with this property is $f(x)=\sin(x)/\sqrt{x}$, whose graph looks like so:

A simple appeal to the alternating series test shows that the integral converges, as
$$\int_0^{\infty} f(x) dx = \sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi} f(x) dx.$$

To see that the integral of $f(x)^2$ diverges, let's take a look at a graph of $f(x)^2$:

Each dashed rectangle that you see lives over an interval of the form
$[k\,\pi+\pi/6,k\,\pi+5\pi/6]$, which has length $4\pi/6$ or $2\pi/3$. The height of each rectangle is
$$\frac{\sin^2(k\,\pi+5\pi/6)}{k\,\pi+5\pi/6} = \frac{1/4}{k\,\pi+5\pi/6}.$$

Thus, the sum total of the areas of those rectangles is
$$\sum_{k=0}^{\infty} \frac{\pi/6}{k\,\pi+5\pi/6} = \infty.$$

As those rectangles lie wholly under the graph of the squared function, the improper integral must diverge as well.

Can someone tell me what's wrong with the following? From @MarkMcClure's comment, there must be something wrong. It could must be a subtle point of improper Riemann integrals.

If $f$ is continuous on $[0,\infty)$ and $\int_0^\infty f < \infty$, then $f(x) \to 0$ as $x \to \infty$, so there's some sufficiently large $R$ such that $f(x) \leq 1$ for $x > R$. Then $\int_R^\infty f^2 \leq \int_R^\infty f$; in particular, $\int_R^\infty f^2 < \infty$ (it exists). Since $f^2$ is the composition of continuous functions and $[0,R]$ is compact, $\int_0^R f^2$ also exists. Hence $\int_0^\infty f^2$ doesn't diverge.


$arg({a \over b})=arg(a)-arg(b)$

So your problem becomes...

$arg(z_3-z_2)-arg(z_3-z_1)=\frac{1}{2}(arg(z_2)-arg(z_1))$

Now let $z_k=re^{i\theta_k}=r(\cos{\theta_k+i\sin{\theta_k}})$

$arg(r(\cos{\theta_3}-\cos{\theta_2}+i(\sin{\theta_3}-\sin{\theta_2})))-arg(r(\cos{\theta_3}-\cos{\theta_1}+i(\sin{\theta_3}-\sin{\theta_1})))=\frac{1}{2}(arg(r(\cos{\theta_2+i\sin{\theta_2}}))-arg(r(\cos{\theta_1+i\sin{\theta_1}})))$

$arctan({\sin{\theta_3}-\sin{\theta_2} \over \cos{\theta_3}-\cos{\theta_2}})-arctan({\sin{\theta_3}-\sin{\theta_1} \over \cos{\theta_3}-\cos{\theta_1}})={1 \over 2}(arctan({\sin{\theta_2} \over \cos{\theta_2}})-arctan({{\sin{\theta_1} \over \cos{\theta_1}}}))$

It can be proven with trig identities that ${{\sin{x}-\sin{y}} \over {\cos{x}-\cos{y}} }={\cot({\frac{x+y}{-2}})}$

$arctan({\cot({\frac{\theta_3+\theta_2}{-2}})})-arctan({\cot({\frac{\theta_3+\theta_1}{-2}})})={1 \over 2}(arctan(\tan{\theta_2})-arctan(\tan{\theta_1})$

It can be shown that $arctan(\cot{x})={\pi \over 2}-x$

$({\pi \over 2} - \frac{\theta_3+\theta_2}{-2})-({\pi \over 2} - \frac{\theta_3+\theta_1}{-2})={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_3+\theta_2}{2}-\frac{\theta_3+\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_3+\theta_2-\theta_3-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_2-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

The statement appears to be proven, but be aware that this makes many assumptions, largest of which is that the angles all exist in the first quadrant. If the angles were all in different quadrants, the proof would be much more involved, and to make such a proof would need a lot of meticulous calculations and studying many different cases - mostly because of the arctan function and its limited range.

Good luck.

I don't think the statement is true. The equation doesn't work for $z_1=z_2=z_3=1$


Proving Continuous Fourier Transform Formulas Given a continuous non-periodic function, its Fourier transform is defined as:

$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk, \ \ \ \ \ \ \ \ \ \ \ \ \ c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$

The problem is proving this is true by evaluating the integral when $c(k)$ is plugged into the equation for $f(x)$.

This ends up with a long integral:

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx' \right) e^{ikx} dk$$

I'm not sure really how to proceed from here. I moved the $e^{ikx}$ into the inner integral, which I figured was fine since it's constant relative to $x'$.

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{ik(x-x')} dx' \right) dk$$

I tried to kill at least one of the integrals by seeing if something evaluated to a Dirac Delta but I can't seem to get that result. I also tried integrating by parts, but that led me nowhere.

Let $f\in L^1(\mathbb{R}^d)$ such that $\widehat{f}\in L^1(\mathbb{R}^d)$ and:

$$
\forall x\in\mathbb{R}^d, \forall \epsilon\in\mathbb{R}, \phi(x)=\pi^{-\frac{d}{2}}\exp(-|x|^2), \phi_\epsilon(x)=\frac{1}{\epsilon^d}\phi\left(\frac{1}{\epsilon}x\right)\\
$$
Lemma:$$
\mathcal{F}(\exp(-a|\cdot|^2))(\xi)=\left(\frac{\pi}{a}\right)^{\frac{d}{2}}\exp\left(-\frac{|\xi|^2}{4a}\right)
$$
Then:
$$
\begin{equation}
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi=\lim\limits_{\epsilon\to0}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi,x\rangle)\widehat{\phi_\epsilon}(x)\widehat{f}(\xi)\mathrm{d}\xi
\end{equation}
$$

By Fubini theorem:
$$
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\int\limits_{\mathbb{R}^d\times\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)f(y)\mathrm{d}\xi\mathrm{d}y
$$
So:
$$
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, y-x)\widehat{\phi_\epsilon}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\left(\frac{4\pi}{\epsilon^2}\right)^{\frac{d}{2}}\exp\left(-\frac{|x-y|^2}{\epsilon^2}\right)=\phi_\epsilon(x-y)
$$
And:
$$
\forall \epsilon>0, (\phi_\epsilon\star f)(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle x, \xi\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi
$$
Let $(\epsilon_n)_{n\in\mathbb{N}}$ be a sequence of reals, $\epsilon>0$, that converge to 0. Then $\lim\limits_{n\infty}\|\phi_{\epsilon_n}\star f-f\|_{L^1(\mathbb{R}^d)}=0$, there is a subsequence such that for allmost all $x\in\mathbb{R}^d$,
$$
f(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi
$$

From your work:
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}x' \mathrm{d}k$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{-ik(x'-x)} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')2\pi \delta(x'-x)\mathrm{d}x'$$
$$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x'-x)\mathrm{d}x'$$


An example showing that a Skolem normal form of $A$ can be not logically equivalent to $A.$ I am trying to learn a little about Mathematical Logic.
Precisely now I am reading about Prenex Normal Forms from E. Mendelson, Introduction to Mathematical Logic, 2nd Edition. I would like to know whether I have correctly worked out exercise 2.80 (which is Exercise 2.87 in the 4th Edition):

  Find a Skolem normal form $B$ for $\forall x\exists yA^2_1(x,y)$ and show that $\not\vdash B\leftrightarrow \forall x\exists yA^2_1(x,y).$

What is the context?

Mendelson is working with a pure predicate calculus, i.e. a predicative calculus without individual constant nor function letters, such that for any positive integer $n$ there are infinitely $n$-ary predicate letters.

What I have done?

I have applied the described algorithm to find a Skolem normal form, and I have found $B:=\exists x \exists y \forall z[(A_1^2(x,y)\to P(x))\to P(z)],$ where $P$ is a $1$-ary predicative variable.
By Goedel's completeness theorem, I have to show the $B\leftrightarrow \forall x\exists yA^2_1(x,y)$ is not universally valid, i.e. I have to find an interpretation $\mathfrak{A}$ s.t. $\mathfrak A\not\models B\leftrightarrow \forall x\exists yA^2_1(x,y).$
I have considered the interpretation, with domain $\mathbb N,$ which assigns to $A_1^2(x,y)$ the relation $x>y,$ and to $P(x)$ the relation "x=1".
If I am not wrong then, for any $s\in\mathbb{N}^\omega,$ I have $\mathfrak A\not\models\forall x\exists y A_1^2(x,y)[s]$ while $\mathfrak A\models B[s].$

As obvious, any feedbak is highly appreciated.

It looks alright to me.

Note that a simpler counterexample interpretation would be to make $P({\cdot})$ always true and $A_1^2({\cdot},{\cdot})$ always false.

The example of S.N. is not correct in my opinion since it states the rule to be:
There exist a $y$ so that $y$ is larger than any $x$, that is not correct in $T_n$ nor what he wanted to represent.

The correct representation is $\forall x\exists y \cdot x\ge y$, that is: for every x there is a larger or equal $y$.
If you get to the "Skolemized" normal form you get not a constant anymore for $y$ but a function of the $x$ you are considering; like that: $x\ge g(x)$.
The above expression holds for any function $g(x)$ mapping $x$ to an equal or greater value than $x$; like $g(x)=x+1$.


What is the Laplace transform of this random variable? Define a random variable that takes only one value for example $$X=c$$ where c is a positive constant.

What does the Laplace of it evaluate to i.e the following $$\mathcal{L}_X(s)= \mathbb{E}[e^{-sX}]= \int e^{-sx} f_X(x)\ dx$$

where $\mathbb{E}$ is the expected value operator.

I have done the following, $$\int ( e^{-sx} ) 1_{x=c} \ dx = e^{-sc}$$

Is my derivation correct?

Yes your analysis is correct. Notice that $X$ is a discrete random variable where $E(e^{X})=\sum_x e^{x}P(X=x)$ where we have $P(X=c)=1$

You can always verify your results by taking the inverse laplace transform of your function.

Here:
$$\mathcal{L}^{-1}(e^{-sc})=\delta(t-c)\text{ where }\delta\text{ is Dirac's Delta Function}$$


Polar equation of an ellipse given the origin coordinates and major and minor axis lengths? I've been trying to create a polar equation that will give me all points on an ellipse with the independent variable being theta and the dependent variable being the radius, but I'm having a great deal of trouble wrapping my mind around how to accomplish such a feat.

The ellipse is going to be defined by it's origin's x and y positions, it's major axis length, and it's minor axis length. It can be assumed that in every case the major axis is perfectly vertical and the minor axis is perfectly horizontal.

Not really relevant to the answer, but just a little more background information. I am trying to program hit detection for a game that uses ellipses for the hitbox boundaries. The reason I need an equation like this is that I am going to be determining whether an object is within the hitbox by comparing the distance from the ellipse's center to a point on the object with the distance from the ellipse's center to the point along the ellipse in the direction of the point on the object.

Any feedback is appreciated.

Divide your major and your minor axes by $2$ to get the "major radius" and "minor radius". Call these $a$ and $b$. (A major and minor axis length don't uniquely specify an ellipse, you'll still have to decide if it's wider than long or longer than wide, so you'll have to make that choice here).

Substitute them into the usual equation of an ellipse.

$$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1$$

Substitute in $x = r\cos(\theta), y = r\sin(\theta)$.

$$
\begin{aligned}
\left(\frac{r\cos(\theta)}{a}\right)^2 + \left(\frac{r\sin(\theta)}{b}\right)^2 &= 1\\
\frac{r^2\cos^2(\theta)}{a^2} + \frac{r^2\sin^2(\theta)}{b^2} &= 1\\
\frac{r^2\cos^2(\theta)b^2}{a^2b^2} + \frac{r^2\sin^2(\theta)a^2}{b^2a^2} &= 1\\
r^2(b^2\cos^2(\theta) + a^2\sin^2(\theta)) &= a^2b^2\\
r^2 &= \frac{a^2b^2}{b^2\cos^2(\theta) + a^2\sin^2(\theta)}\\
r &= \frac{ab}{\sqrt{b^2\cos^2(\theta) + a^2\sin^2(\theta)}}\\
\end{aligned}
$$

HINTS:

For a start we can use standard  parametric equations $ (r,\theta),(x,y) $ and then convert to Cartesian coordinates, translate by $ (h,k) $ re-convert to required polar equation  $ \rho = f( \gamma ).  $

$$
r= p/ (1- \epsilon \cos \theta) ; x= r \cos\theta + h ; y= r \sin \theta +k ; \rho^2 = x^2 + y^2 ; \gamma = \tan^{-1} {\frac{y}{x}};
$$

There are two ellipse tangent limits for $\gamma$ limits each of constant value, two variable radius $\rho$ limits for each intermediate $\gamma$ value as shown.

Mathematica program if helpful:

ecc = 0.6; p = 1.; h = 3.; k = 2.; r[t_] = p/(1 - ecc Sin[t]);
ell = ParametricPlot[ r[t] {Cos[t], Sin[t]}, {t, 0, 2 Pi},
   PlotStyle -> {Thick, Magenta}];
rho[t_] = Sqrt[r[t]^2 + h^2 + k^2 + 2 r[t] ( h Cos[t] + k Sin[t])];
gam[t_] = ArcTan[   r [t] Cos[t] +h, r [t] Sin[t] +k ] ;
displ = ParametricPlot[
   rho[t] {Cos[gam[t]], Sin[gam[t]]}, {t, 0, 2 Pi},
   PlotStyle -> {Thick, Red}];
Show[{ell, displ}, PlotRange -> All, GridLines -> Automatic]

To  find polar angle $ \gamma $ of  tangential contact to radius vector, next use the condition

$$ \frac{d \gamma}{d \rho} =0  $$

and then proceed radial transversal intersection limits...

(If you want to locate target area of displaced ellipses, calculation with pole/polar of ellipse is perhaps best, but it is not mentioned here).


Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$

HINT: you can write your result in the form $$\frac{\ln\left(\frac{5}{3}\right)}{\ln\left(\frac{5}{2}\right)}$$


How to decompose edges of $K_n$ into Hamiltonian paths? Let $K_n$ be a complete graph with n vertices, where n is even

Show that $K_n$ can be decomposed into  $\frac{(n-1)}{2}$ disjointed Hamiltonian paths on edges

My idea was to use Menger's theorem that says:

Let $G$ be a connected graph and let $u$ and $v$ be vertices of $G$. Then the number of $uv$-disjoint paths per edge is equal to the minimum number of edges on a $(u, v)$ -separator

Since we are working with a complete graph the minimum number of edges that there will be in a $(u, v)$ - separator will be $n-1$, so we will have  $n-1$ internally disjoint paths in edges.
However, I don't know how to continue

There is a standard construction for decomposing $K_{2k}$ into $k$ Hamiltonian paths. It is illustrated below:

Formally, if the vertices are numbered $0, 1, \dots, 2k-1$ modulo $2k$ (which we visualize by arranging them in a circle, as I did above) then the $i^{\text{th}}$ Hamiltonian path is
$$
   (i, i+1, i-1, i+2, i-2, i+3, i-3, \dots, i + (k-1), i - (k-1), i+k).
$$
Some checking needs to be done to verify that every edge is used exactly once.
This construction can be extended to decomposing $K_{2k+1}$ into $k$ Hamiltonian cycles. If the extra vertex of $K_{2k+1}$ not appearing in $K_{2k}$ is called $x$, then each of the Hamiltonian paths in $K_{2k}$ is extended to a Hamiltonian cycle in $K_{2k+1}$ by joining both endpoints to $x$. (Note that each vertex in $K_{2k}$ is the endpoint of exactly one Hamiltonian path in the decomposition.)

We can decompose the set of edges of $K_n$ into ($n/2$ edge-disjoint) paths provided $n+1$ is prime. For this it suffices to decompose the set of edges of $K_{n+1}$ into Hamiltonian cycles, then fix a vertex $v$ of $K_{n+1}$ and remove it from each of cycles of the decomposition. This results a decomposition of the set of edges of $K_{n+1}$ without $v$ into Hamiltonian paths.
We can construct a required decomposition of $K_{n+1}$ identifying its set of vertices with a field $\Bbb Z_{n+1}=\Bbb Z/(n+1)\Bbb Z$. To each integer $i$ from $1$ to $n/2$ corresponds a Hamiltonian cycle in $K_{n+1}$, whose edges are $\{j,k\}\subset\Bbb Z_{n+1}$ such that $j-k=\pm i\pmod {n+1}$.


Graphing the sum of two absolute values I have an inequality $$||x|-1|+||x|+2|=3$$. Id like to graph it. I've done with just one absolute value within another absolute value, but I can't seem to grasp how to do this one. Also, I'm having some trouble setting cases for the algebraic route: I've done 6 cases in total - $$x<-2, -2<x<-1, -1<x<0, 0<x<1, 1<x<2, 2<x$$ i think i did it right. Here is how I got to these cases:

|x|={x, x>0 & -x, x<0}
|x-1|={x-1, x>1 & -x-1, x<1}
|-x-1|={-x-1, x<-1 & x+1, x>-1}
|x+2|={x+2, x>-2 & -x-2, x<-2}
|-x+2|={-x+2, x<2 & x-2, x>2}

Did I set my cases correctly? Also, I know how the graph is supposed to look like, I just don't know how to get to it.

Note that:  $||x| + 2| = |x| + 2$ so you can get rid of the second absolute value immediately.

Case 1:

$|x| \lt 1$

Case 2:

$|x| \ge 1$

... and you just continue analyzing in each of these cases ...

Just plot the function or by hand!

The minimum value is $3$
At large $x >0$ the slope is $+2$, since the individual terms each have a slope of $+1$
A large negative $x$, the slope is $-2$, for similar reasons
The transition points must be when $x=\pm 1$


Help with my proof of maximum number of intersection points of n lines being n(n-1)/2 I know that this question is a hoary old chestnut, but I have never seen a proof before working one out myself, so I'd like you to help me see if mine is rigorous enough.

Obviously with $n$ linear two-variable equations, a pair has at most one unique solution, so the maximum cannot exceed the number of pairs that can be chosen is $\frac{N(N-1)}{2}$

Then one proceeds to construct a system in which each pair has a unique solution distinct from the others.

Let $a_k(x)+b_k=y$ for each $k\in [1, N]$. Let $a_k$ be $1/p_k$, the reciprocal of the $k$-th prime. Let $b_k$ be $1/p_k^2$, i.e. the square of the reciprocal of the $k$-th prime.

For the pair $k=$ distinct $i$ and $j$, the solution is $x=-(1/p_i+1/p_j)$. (A bit of algebra will figure that out.) The proof consists of showing that the sum of the reciprocal of two distinct primes cannot be represented by such a sum with two other primes.

$1/p_i+1/p_j=(p_i+p_j)/(p_i*p_j)$, the RHS in its simplest form already since if $GCD(a,b)=1$, $GCD(a+b,ab)=1$ too. (Otherwise some prime divides $ab\Rightarrow$ some prime divides one of $a$ or $b\Rightarrow$ that prime divides the other of $a$ or $b$ $\#$).

Thus for any $p_m, p_n$ that satisfies $1/p_m+1/p_n=(p_i+p_j)/(p_i*p_j)$, it follows that $p_m+p_n=p_i+p_j$, and that $(p_m)(p_n)=(p_i)(p_j)$. If $p_i$ and $p_j$ are known, then the two above equations imply $p_m$ and $p_n$ equal $p_i$ and $p_j$, except in respect of the order. (To be really rigorous on this point, construct a quadratic equation the roots of which are $p_m$ and $p_n$).

Thus for two distinct pairs of equations, the solution cannot overlap.

Thanks for any help on this matter!

So if you pick $n$ lines at random they will intersect in $\binom{n}{2}$ points with probability 1, so there are all sorts of ways you could construct such an example.  I didn't read your exact construction, but here's a simple way:

The result is trivial for $n=1$.  Suppose we have a set of n lines, each of which intersects each of the others in a unique point.  Now:

Pick a slope $m$ which does not match that of any of the other n lines.
Pick a line $\ell$ of slope $m$ which does not pass through any of the $\binom{n}{2}$ points of intersection between the other lines.
Add $\ell$ to our set of lines.

Then our new set consists of $n+1$ lines, with all pairwise intersections unique.


How do you find nth maclaurin/taylor series or polynomial? Like for example, one of my homework problems is
g(x)= x2ex3

and it's asking me to find g(40)(0)
am I supposed to find the derivative 40 times and plug 0 in for x?

I turned the function into a maclaurin series in sigma notation
and not sure what to do

$\sum_{n=0}^{\infty}$ $\frac{x^{3n+2}}{n!}$

We can form a series that is somewhat easier to work with by noting that $ \ \frac{d}{dx} \ [e^{x^3}] \ = \ 3x^2 · e^{x^3} \ \ , $ so that we produce the anti-derivative series
$$ G(x) \ \ = \ \  \frac13 \  e^{x^3} \ \ =  \ \frac13 \ \sum_{n \ = \ 0}^{\infty} \ \frac{1}{n!} \ (x^3)^n \ \ = \ \ \frac13 \ \sum_{n \ = \ 0}^{\infty} \ \frac{x^{3n}}{n!} \ \ ,    $$
with $ \ G'(x) \ = \ g(x) \ = \ x^2 e^{x^3} \ . $  The series for $ \ G(x) \ $ is non-zero only for every third term, that is to say,
$$ G^{(n)}(0) \ \ = \ \ \left\{  \begin{array}{rcl} a_{n} \ \ ,  & \ \text{for} \ n = 3k \ \ , \ \ k \ \ \text{a non-negative integer}   \\ 0 \ \ , & \text{otherwise} \end{array} \right. \ \ . $$
Since $ \ g^{(40)}(0) \ = \ G^{(41)}(0) \ \ , $ and $ \ 41 \ $ is not a multiple of $ \ 3 \ \ , $ we can simply conclude that $ \ g^{(40)}(0) \ = \ 0 \ \ . $  (Had the problem asked for $ \ g^{(41)}(0) \ \ \text{or} \ \ g^{(50)}(0) \ \   $ (as appears in one of the answers -- the original version of the post?), we would have more work to do, as discussed below.
$$ \ \ $$
Another way we can approach the problem is to make use of the "higher-derivatives" version of the Product Rule,
$$ \frac{d^n}{dx^n} \ [f_1(x) · f_2(x)] \ \ = \ \ \sum_{k \ = \ 0}^{n} \ \binom{n}{k} \ f_1^{(k)}(x) · f_2^{(n-k)}(x) \ \ , $$
which behaves like a binomial expansion.  This will be of help to us if we define, as we did above, $ \ G(x) \ = \ \frac{1}{3} · e^{x^3} \ = \ \frac{1}{3} · e^u \ = \ y \ \ , $ so that $ \ g(x) \ = \ G'(x) \ = \ u' · y \ \ . $  We will then set  $ \ f_1 = u' \ $ and $ \ f_2 = y \ \   $  in making our calculations.
This may look daunting for the derivative we wish to determine, but there are features of the function that lead to considerable simplification.  We have $ \ G(0) = \frac13 \ $ and $ \ u = x^3 \ \Rightarrow \ u' = 3x^2 \ , \ u'' = 6x \ , \ u''' = 6 \ \ , $ and $ \ u^{(k)} = 0 \ $ for $ \ k \ge 4 \ \ . $  What this tells us is that the number of terms in the expansion for successive derivatives is quite limited.  Moreover, since we are evaluating the derivatives of $ \ g(x) \ $ at $ \ x = 0 \ $ , the only non-zero value for the $ \ u^{(n)}  $ 's is $ \ u''' = 6 \ \ . $
With this in mind, we see that
$  g(0) \ = \ G'(0) \ = \ y'(0) \ = \ 0 · \frac13 \ = \ 0 \ \ , $
$  g'(x) \ = \ G''(x) \ = \ y'' \ =  \ u''· y \ + \ u'· y' \ \ \Rightarrow \ \ g'(0) \ = \ 0 · \frac13 \ + \ 0 · 0 \ = \ 0 \ \ , $
$  g''(x) \  = \ y''' \ =  \ u'''· y \ + \ 2 · u''· y' \ + \ u'·y'' \ \ \Rightarrow \ \ g''(0) \ = \ 6 · \frac13 \ + \ 2 · 0 · 0 \ + \ 0·0 \ = \ 2 \ \ . $
Here, we attain our first non-zero derivative with $ \ G'''(0) \ = \ g''(0) \ \ . $  With the next derivative, our expressions begin to include derivatives of $ \ u \ $ which are simply zero.
$  g'''(x) \  = \ y^{(4)} \ =  \ u^{(4)}· y \ + \ 3 · u'''· y' \ + \ 3·u''·y'' \ + \ u'·y''' $
$ \Rightarrow \ \ g'''(0) \ = \ 0 · \frac13 \ + \ 3 · \frac13 · 0 \ + \ 3·0·0 \ + \ 0·6 \ = \ 0 \ \ , $
$  g^{(4)}(x) \  = \ y^{(5)} \ =  \ u^{(5)}· y \ + \ 4 · u^{(4)}· y' \ + \ 6·u'''·y'' \ + \ 4·u''·y''' \ + \ u'·y^{(4)} $
$ \Rightarrow \ \ g^{(4)}(0) \ = \ 0 \ + \ 0 \ + \ 6  ·  \frac13  ·  0 \ + \ 4·0·2 \ + \ 0 \ = \ 0 \ \ , $
and
$  g^{(5)}(x) \  = \ y^{(6)} \ =  \ u^{(6)}· y \ + \ 5 · u^{(5)}· y' \ + \ 10·u^{(4)}·y'' \ + \ 10·u'''·y''' \ + \ 5·u''·y^{(4)} \ + \ u'·y^{(5)} $
$ \Rightarrow \ \ g^{(5)}(0) \ = \ 0 \ + \ 0 \ + \ 0 \ + \ 10·6·2 \ + \ 0 \ + \ 0  \ = \ \binom{5}{3}·6·2 \ = \   120 \ \ . $
What becomes clear from this examples is that the non-zero derivatives for non-negative integer $ \ k \ $ are given by the recursion relation
$$ g^{(3k + 2)}(0) \  = \ y^{(3k + 3)}(0) \ =  \ \binom{3k+2}{3k}·u'''·y^{(3k)}(0)  \ =  \ \binom{3k+2}{3k}·6·y^{(3k)}(0) \ \ , \ \ y(0) \ = \ \frac13 \ \ . $$
Hence,
$$ g^{(8)}(0) \  = \ y^{(9)}(0) \ =  \  \binom{8}{6}·6·y^{(6)}(0) \ = \ \binom{8}{6}·6· \left[\binom{5}{3}·6·2 \right] \ = \ 20,160 \ \ ,  $$
$$ g^{(11)}(0) \  = \ y^{(12)}(0) \ =  \  \binom{11}{9}·6·y^{(9)}(0) \ = \ 55·6·20,160 \ = \ 6,652,800 \ \ ,  $$
and so forth.  We observe that while $ \ g^{(41)}(0) \  = \ y^{(42)}(0) \ $ would be enormous, $ \ g^{(40)}(0) \  = \ y^{(41)}(0) \ $ is not in this set of derivatives, and so is equal to zero.
$$ \ \ $$
We might like to have something a bit simpler to work with than the recursion relation for the non-zero derivatives.  If we write out the last example in full, we obtain
$$ g^{(11)}(0) \ \  = \ \ \binom{11}{9}·6·\binom{8}{6}·6·  \binom{5}{3}·6·\binom{2}{0}·6·\frac13 $$ $$ = \ \ \frac{11·10·9·8·7·6·5·4·3·2·1}{2 \ · \ 9 \ · \ 2 \ · \ 6 \ · \ 2 \ · \ 3 \ · \ 2} \ · \ 6^3 \ · \ 2  \ \ = \ \ \frac{11!}{2^3 \  · \ 3^3 \ · \ (3·2·1) \ · \ 2} \ · \ 6^3 \ · \ 2 \ \ = \ \ \frac{11!}{3!} \ \ . $$
The general calculation yields $ g^{(3k + 2)}(0) \  = \ \frac{(3k+2)!}{k!} \ \ , $ for integer $ \ k \ge 0 \ $ ; this completes the derivation characterization  given above
$$ G^{(n+3)}(0) \ \ = \ \ \left\{  \begin{array}{rcl} \frac{(3k+2)!}{k!} \ \ ,  & \ \text{for} \ n = 3k \ \ , \ \ k \ \ \text{a non-negative integer}   \\ 0 \ \ , & \text{otherwise} \end{array} \right. \ \ . $$
Inserting this into the general term for a Maclaurin series, we then produce
$$  g(x) \ \ = \ \ x^2 · e^{x^3} \ \ = \ \ \sum_{m \ = \ 0}^{\infty} \ \frac{g^{(m)}(0) }{m!} \ x^m    $$
[removing the "zero terms"]
$$ = \ \ \sum_{k \ = \ 0}^{\infty} \ \frac{(3k+2)! \ / \ k! }{(3k+2)!} \ x^{3k+2} \ \ = \ \  \sum_{k \ = \ 0}^{\infty} \ \frac{x^{3k+2} }{k!} \ \ = \ \ x^2 \ + \ x^5 \ + \ \frac12 · x^8 \ + \ \frac16 · x^{11} \ + \ \ldots \ \ ,   $$
in agreement with the method of substitution into the series for $ \ e^u \ $ and Gary's comment.


Is $[0,1]$ a countable disjoint union of closed sets? Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?

The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpiński in $1918$ $[1]$. You can find the proof in the book "General Topology" by Ryszard Engelking, but I'll post here since it's not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:

Theorem (Sierpiński). If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.

In order to prove this we'll need the following lemmas:

Lemma $1$. Let $X$ be a continuum. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.

Proof. Let $x_0$ be in $C$. Since $X$ is Hausdorff compact, quasicomponents coincide with components, so $C$ is the intersection of all open-closed sets in $F$ which contain $x_0$. Suppose that $C$ is disjoint from $\text{Bd}(F)$. Then, by compactness of $\text{Bd}(F)$, there is one open-closed set $A$ in $F$ containing $x_0$ and disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Thus the equality $A \cap \text{Bd}(F) = \emptyset$ implies that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. But $A$ is also closed in $X$, and contains $x_0$, so $A=X$. But then $\text{Bd}(F) = \emptyset$, which is not possible since $F$ would be non-trivial open-closed in $X$. $\bullet$

Lemma $2$. If a continuum $X$ is covered by pairwise disjoint closed sets $X_1, X_2, \ldots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and at least two sets in the sequence $C \cap X_1, C \cap X_2, \ldots$ are non-empty.

Proof. If $X_i$ is empty then we can take $C = X$; thus we can assume that $X_i$ is non-empty. Take $j \ne i$ such that $X_j \ne \emptyset$. Since $X$ is Hausdorff compact, there are disjoint open sets $U,V \subseteq X$ satisfying $X_i \subseteq U$ and $X_j \subseteq V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and since $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$

Now we can prove the theorem:

Proof. Assume that at least two of the sets $X_i$ are non-empty. From lemma $2$ it follows that there exists a decreasing sequence $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua contained in $X$ such that $C_i \cap X_i = \emptyset$ and $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first part implies that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and from the second part and compactness of $X$ it follows that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$

The Hausdorff hypothesis is fundamental. For example, consider $X$ a countable infinite set with the cofinite topology. Then $X$ is compact, connected and a $T_1$-space. However, we can write $X$ as a disjoint union of countable singletons, which are closed.

$[1]$ Sierpiński, W: Un théorème sur les continus, Tôhoku Math. J. 13 (1918), 300–305.

A partial solution:

As Sargera began, if such a cover exists, it must be countably infinite. Let $[0,1]=\displaystyle\bigcup_{n=1}^{\infty}[a_n, b_n] = \displaystyle\bigcup_{n=1}^{\infty}I_n$ By the axiom of choice, construct $(x_n)$ by $x_n \in [a_n, b_n].$ Since $[0,1]$ is compact, there exists $(x_{n_k})$ such that $x_{n_k} \to x\in [0,1].$ Then there exists unique $s\in \mathbb{N}$ such that $x\in [a_s, b_s].$ If this interval is non-degenerate then corresponding to $\varepsilon = \frac{1}{2}\min\{x-a_s, b_s-x\}$ there exists $m \in \mathbb{N}$ such that $|x_{n_k}-x|< \varepsilon$ for $k \geq m.$ In particular $x_{n_{m}} \in (a_s, b_s)\subset [a_s, b_s] = I_s$ and since $x_{n_{m}} \in I_{n_{m}}$ by construction, we obtain a contradiction to the fact that the $I_n$ are disjoint.

If $I_s$ is degenerate, I haven't thought of a workaround yet. Maybe someone could suggest one.


A geometry problem (how to find angle x) The solution is $x=50^{\circ}$.

How to prove $x=50^{\circ}$ without trigonometry?

Find $K$ on $\overline{BD}$ such that $\angle KAB = 10^\circ$. Then $\angle AKB = 160^\circ = 2\cdot\angle ACB$; this makes $K$ the center of the circumcircle of $\triangle ABC$. Thus, $\overline{KB}\cong\overline{KC}$, whereupon $\angle KCB = 20^\circ$. Note also that $\triangle ACK$ must be equilateral. (Why?)

We also have that $\angle DKA = 20^\circ = \angle DAK$. This makes $\triangle DAK$ isosceles, so that $\overline{CD}$ bisects $\angle ACK = 60^\circ$. The result follows. $\square$

This is incorrect and just left for reference

The strategy to take appears to be to find all the angles in the centre at $D$  in terms of $x$ and then equate to $360$ but I don't get $x=50$

$\angle BDC=180-20-x=160-x$

$\angle ADB=180-30-10=140$

Use the largest triangle to find $\angle ACD$

$\angle ACD =180-70-30-x=80-x$

This means $\angle ADC =180-40-(80-x)=60+x$

At the centre you have $160-x+140+60+x=360$

The $x$ cancel and I briefly checked that $x=60$ is a solution as well.

See comments below as to why this fails


Convergence/divergence of the series Prove or disprove: The series $$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}$$ converges.

I know its general term approaches zero. But I'm unsure to move further.
Can I have a hint?

A series is convergent if and only if the limit of its partial sums is finite.

Write, for $k < \infty$:
$$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}+\sum_{n=1}^{k}\dfrac{(-1)^n}{n}\text{.}$$
It follows that
$$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}  - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}\text{.}$$

The proof:
Suppose by contradiction that
$$\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}$$
is finite.

We know that $$\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{(-1)^n}{n}$$
is finite, hence
$$
\begin{align}
\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}  - \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{(-1)^n}{n} &= \lim_{k \to \infty}\left[\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} \right] \\
&= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n-(-1)^n}{n} \\
&= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{1}{2n}
\end{align}$$
will also be finite (because the sum of two finite real numbers is also finite), which is a contradiction.

Suppose assume the given series converges. Then

$$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}=\sum_{n=1}^\infty  \frac{1}{2n}+\sum_{n=1}^\infty \frac{(-1)^n}{n}=\frac{1}{2} \sum_{n=1}^\infty  \frac{1}{n}+\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
We already know $\sum_{n=1}^\infty  \frac{1}{n}$ diverges and $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges

Also the difference of convergent series is  convergent. so $$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}-\sum_{n=1}^\infty \frac{(-1)^n}{n}=\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n}$$ converges, which is a contradiction

Hence  $\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}$  diverges


Journals of math history? In a related question to this one, in what journals do math historians publish their article in? Brian M. Scott provided a link to Judy Grabiner's, who is a math historian, home page and it seems that she publishes in general mathematical journals for the most part, such as the American Mathematical Monthly. However there is mention of a journal that seems to be exclusively for math history called "Historia Mathematica". Are there any other journals exclusively for math history?

BSHM Bulletin. Journal of the British Society for the History of Mathematics.
Taylor & Francis, 4 Park Square, Milton Park, Abingdon 0X14 4RN England.
ISSN: 1749-8430.
E-ISSN: 1749-8341.
3 issues/vol./yr.

Revista Brasileira de História da Matemática. An International Journal on the History of Mathematics,
Sociedade Brasileira de História da Matemática Caixa postal 1631 CEP 59.078-970 Campus Universitario Natal RN Brazil.
ISSN: 1519-955X.

Gaṇita Bhāratī. Bulletin of the Indian Society for the History of Mathematics.
Formerly Gaṇita-Bhāratī.
MD Publ. Pvt., Ltd. "MD House'' 11 Darya Ganj, New Delhi 110 002 India.
ISSN: 0970-0307.
2 issues/vol./yr.

Closely related:
International Journal for History of Mathematics Education
http://www.comap.com/historyjournal/


Discover where Bob is sleeping using hidden Markov chains Bob lives in four different houses $A, B, C$ and $D$ that are connected like the following graph shows:

Bob likes to sleep in any of his houses, but they are far apart so he only sleeps in a house adjacent to the one in which he slept the previous night. To clarify, this means that if Bob slept in house $A$ on night $1$, he may sleep in house $A$, $B$ or $C$ on night $2$ (not house $D$). The probability of each case is the same (one third); on each day, Bob takes a random walk from where stayed the previous night (and he might stay put).

Now Bob is a wanted criminal so on a given night the FBI would like to estimate where Bob is sleeping. Data from a satellite gives us the following probabilities of where Bob is sleeping on night $1$ and night $2$ (and any subsequent nights):

         Night 1   Night 2   ...
House A  0.8       0.05      ...
House B  0.1       0.4       ...
House C  0.05      0.05      ...
House D  0.05      0.5       ...

How can we use this data to calculate the probability of where Bob was sleeping on night $2$, for example calculating the probability Bob slept in house A? Could we use that method iteratively to calculate where Bob was sleeping on night $n$ if we continue to receive satellite data for each night?

Note: I made up this problem to understand better how hidden Markov chains work because I am interested in seeing the calculations on a concrete example. Many thanks for any input.

The goal is to update probability of Bob given satellite measurements.

The state of Bob at time $k$ is given by $x_{k}$. The measurement of a satellite is $y_{k}$. Since no prior information is given, we initialize with a uniform distribution
$$ \text{for } i={A,B,C,D},\,\, p(x_{0}=i|y_{0})=1/4.$$

Now that we have initialized, let's use Baye's rule to get $p(x_{1}=i|y_{0},y_{1})$. I'm only going to do this for i=A. Baye's rule yields

$$ p(x_{1}=A|y_{0},y_{1}) = \frac{p(x_{1}=A \cap y_{1}|y_{0})}{p(y_{1}/y_{0})}. $$
Now we just fill in the blanks. First, I like to calculate my "prediction", which is

$$ p(x_{1}=A | y_{0}) = \sum_{x_{0}\in{A,B,C,D}} P(x_{1}=A|x_{0})P(x_{0}|y_{0}).$$

Assuming bob is equally likely to stay in the house, or go to an adjacent house, this value is  3*(1/3)*1/4 = 1/4, and for this iteration at least, the prediction for other states is the same value. Now let's take a look at the Baye's Rule equation we outlined earlier

$$ p(x_{1}=A \cap y_{1}|y_{0}) = p(y_{1}|x_{1}=A)p(x_{1}=A|y_{0})=1/5$$

Finally,

$$ p(y_{1}/y_{0}) =  \sum_{x_{1}\in{A,B,C,D}} p(y_{1}/x_{1})p(x_{1}/y_{0}). $$

This comes out to be 1/4. Finally,

$$ p(x_{1}=A|y_{0},y_{1}) = \frac{1/5}{1/4} = 4/5. $$

You may want to look at this.
http://en.wikipedia.org/wiki/Viterbi_algorithm


Closed subset of a variety is also a variety Suppose that $(X,\mathcal{O}_{X})$ is a variety and $Y\subset X$ is closed and equipped with the induced subspace topology. I want to prove that $(Y,\mathcal{O}_{Y})$ is also a variety, where for every open $V\subset Y$ we have $\mathcal{O}_{Y}(V):= \{f:V\rightarrow k\rvert \forall P\in V, \ \exists U\subset X, \ \exists g\in\mathcal{O}_{X}(V) \ \text{s.t} \ P\in U,  \forall Q\in V\cap U \ \text{we have} \ f(Q) = g(Q)\}$.

My Attempt First one has to show that $(Y,\mathcal{O}_{Y})$ is a $k$-space. But this follows easily from the definition above of $\mathcal{O}_{Y}$. So assuming we know that $(Y,\mathcal{O}_{Y})$ is a $k$-space, then we only have to show that for every $y\in Y$ there is an open $V\subset Y$ with $y\in V$ such that $(V,\mathcal{O}_{Y}\rvert_{V})$ is isomorphic (as $k$-spaces) to $(Z,\mathcal{O}_{Z})$ for some closed $Z\subset \mathbb{A}^{n}$ for some $n$. Since we know that $(X,\mathcal{O}_{X})$ is a variety, we know that there exists a cover of $X$, say $X = \cup_{i\in I} U_{i}$ where $U_{i}$ is an open subset of $X$ such that $(U_{i},\mathcal{O}_{X}\rvert_{U_{i}})$ is isomorpic to $(S,\mathcal{O}_{S})$ for some closed $S\subset \mathbb{A}^{m}$ for some $m$. We notice that $Y = \cup_{i\in I} (U_{i}\cap Y)$, where we notice that the $U_{i}\cap Y$ are open in $Y$. My guess would be that for every $i\in I$ we have $(U_{i}\cap Y,\mathcal{O}_{Y}\rvert_{U_{i}\cap Y})$ is isomorphic to some $(Z_{i},\mathcal{O}_{Z_{i}})$ for some closed $Z_{i}\subset \mathbb{A^{n}}$ for some $n$. If so, then we are done. But I have problems with proving this.

Definition for morphism I use:
Let $(X,\mathcal{O}_{X})$ and $(Y,\mathcal{O}_{Y})$ be $k$-spaces and $\varphi: X\rightarrow Y$. Then $\varphi$ is a morphism of $k$-spaces if $\varphi$ is continuous, and $\forall U\subset Y$ open, and for all $f\in\mathcal{O}_{Y}(U)$, the function $f\circ\varphi : \varphi^{-1}(U)\rightarrow k$ is in $\mathcal{O}_{X}(\varphi^{-1}(U))$.

Hint: a closed subset of a closed subset is again closed.

Bigger hint:

   Apply this to $U_i\cap Y\subset U_i$ and $U_i\cong S \subset \Bbb A^n$.

Giving the whole game away:

   Since $U_i\cap Y$ is closed in $U_i$ and $U_i$ is isomorphic to a closed subset in $\Bbb A^n$, the image of $U_i\cap Y$ in $S$ under the isomorphism $U_i\cong S\subset \Bbb A^n$ will again be a closed subset of $\Bbb A^n$. The sheaf condition should follow quite quickly.


The problem of instant velocity The concept of velocity is by definition the movement divided by the interval of time between initial position and final position.

If $f(t)$ is the position of a particle at time $t$; the velocity in the interval $[t_0;t_1]$ is $\dfrac{f(t_1)-f(t_0)}{t_1-t_0}$

The problem is that in a single instant there is no movement and the time is not changed; so no velocity.

I can consider $\lim_{t_1 \to t_0} \dfrac{f(t_1)-f(t_0)}{t_1-t_0}$, but mathematically it is only the limit of average velocity function and doesn't represent velocity at instant $t_0$

What are your views about this problem ?

Your excellent question is as old as the invention of calculus. As you correctly point out, velocity makes no sense if all you know is what is happening at just that instant of time. Physicists and mathematicians take the limit of the average velocity as the very definition of instantaneous velocity.

That turns out to be a very good definition, since it leads to physics that accurately describes the behavior of the world and mathematics that's consistent and interesting and useful. So people no longer worry about the question in the form in which you've asked it.

Edits to respond to comments. Edited again (as @Polygnome suggests) to incorporate the sense of the comments as well

@pjs36 Yes indeed thanks. The question really does go back to Zeno's paradox of the arrow. On that wikipedia page you can read

  Zeno states that for motion to occur, an object must change the
  position which it occupies. He gives an example of an arrow in flight.
  He states that in any one (duration-less) instant of time, the arrow
  is neither moving to where it is, nor to where it is not.[13] It
  cannot move to where it is not, because no time elapses for it to move
  there; it cannot move to where it is, because it is already there. In
  other words, at every instant of time there is no motion occurring. If
  everything is motionless at every instant, and time is entirely
  composed of instants, then motion is impossible.

@Max says

  In the Newtonian model of the universe, momentum/velocity is something
  that objects have at every instant of time

I didn't know that. Perhaps it's why he could develop calculus reasoning with infinitesimals without addressing the philosophical problem and without formal notion of limits. His assumptions were not universally accepted at the time. The philosopher George Berkeley argued that

  ... forces and gravity, as defined by Newton, constituted "occult
  qualities" that "expressed nothing distinctly". He held that those who
  posited "something unknown in a body of which they have no idea and
  which they call the principle of motion, are in fact simply stating
  that the principle of motion is unknown."

( https://en.wikipedia.org/wiki/George_Berkeley#Philosophy_of_physics )

@leftaroundabout I agree that momentum is a better fundamental notion than velocity (certainly for quantum mechanics, possibly for Newtonian too). I don't think it's better to start calculus there, though.

@Hurkyl notes correctly that there are new mathematical structures - germs - that capture the idea of what happens near but not at a point. But I think the idea of the germ of a function is more technical and abstract than called for by the question.

"The concept of velocity is by definition the movement divided by the interval of time between initial position and final position."

Says who?

As with any other conceptual notion, the definition is whatever we deem it to a) answer a desired question while b) not contradicting itself or other connected definitions.  For example, nobody says non-integer rational exponents are to be considered comparable to roots; that's simply a convenient discovery that syncs up nicely with the already-established (by general principles of the definition of multiplication) properties of integer exponents.  It's the transitive property dropped right in our laps.

In this case, one might simply say that "velocity" has no inherent meaning other than that which we assign to it, and we've collectively decided to treat it as if it were "naturally" the definition we mean when we actually say "average" velocity (probably because we lack the biological ability to extract ourselves from the fourth dimension and thus escape the compulsion to measure that elapsed time as having an unavoidably non-zero amount).  For all we know, we might well already have the instruments to measure instantaneous velocity; we just lack the ability to pull ourselves outside of time and actually do so.

EDIT:  OK, since my answer has rubbed a few the wrong way, let us consider it this way: it is not true to say that in an instantaneous, non-elapsed-time sense, there is "no velocity."

"No" implies "zero."

The velocity is not zero in this case, it is the indeterminate form 0/0.  (Despite several of my students' beliefs to the contrary, 0 ÷ 0 $\not =$ 0 automatically.) It is, at best, unclear; it may be zero, or infinitely large, or some finite non-zero amount. What limits (and, through them, L'Hospital's Rule) explores is how we can explore such singularity-esque quantities only by looking at them indirectly.  My previous comment was phrased merely to suggest that we are as likely to understand instantaneous velocity as a limited case of average velocity as we are to understand average velocity as an extension of instantaneous velocity.


Ratio of product from one point and minimum distance Given points $A_0,A_1,\ldots,A_n$ in the plane, let $m$ denote the minimum distance among any two points. What is the minimum value of $$\dfrac{|A_0A_1|\cdot|A_0A_2|\cdot\ldots\cdot|A_0A_n|}{m^n}?$$

Here $|AB|$ denotes the Euclidean distance between points $A$ and $B$.

[Source: Based on Chinese competition problem]

Lets consider

$\dfrac{|A_0A_1|\cdot|A_0A_2|\cdot\ldots\cdot|A_0A_n|}{m^n}$

as

$\dfrac{|A_0A_1|}{m}\cdot\dfrac{|A_0A_2|}{m}$...$\dfrac{|A_0A_n|}{m}$

We know that for any k:

$\dfrac{|A_0A_k|}{m}\geq1$

Thus the minimum value for

$\dfrac{|A_0A_1|\cdot|A_0A_2|\cdot\ldots\cdot|A_0A_n|}{m^n}$

would be $1$, but we must show that such a configuration can indeed exist on the plane. For this we will just pick three points $A_0,A_1,A_2$ such that they form an equilateral triangle and we get that our expression evaluates to $1$, so we are finished.

In the case where $m=0$ and $2$ points coincied our expression is undefined so it is not at a minimum there either.


According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$

simplify to

$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\sqrt{1 + \dfrac{1}{x^2}} + 1}$$

I understand how he simplifies $\frac{x}{x}$ to 1 but how does he simplify $\frac{\sqrt{x^2 + 1}}{x}$?

Hint: Observe that, for $x>0$,
$$\large\sqrt{x^2+1\strut}=\sqrt{x^2(1+\tfrac{1}{x^2})}=\sqrt{x^2\strut} \;\cdot\;\sqrt{1+\tfrac{1}{x^2}}=x\;\cdot\;\sqrt{1+\tfrac{1}{x^2}}$$

$$
\frac{\sqrt{x^2 + 1}}{x} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \sqrt{\frac{x^2}{x^2} + \frac 1 {x^2}} = \cdots
$$

$\frac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{\sqrt{x^2 + 1} + x}{x}\right)}$ is exactly the same thing as $\dfrac 1 {\sqrt{x^2+1}+x}$.  To get that just multiply the numerator and the denominator by $x$.  Then:
$$
0 < \frac 1 {\sqrt{x^2+1}+x}<\frac 1 {2x} \to 0 \text{ as } x\to\infty.
$$


Analogy between the fundamental theorems of arithmetic and algebra For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:

$$ n = p_1\cdot p_2 \cdots p_k$$

$$ P(z) = z_0\cdot(z_1 -z)\cdot (z_2 -z) \cdots (z_k -z)$$

which makes obvious that the irreducible polynoms of first degree play the same role in $\mathbb{C}[X]$ as do the prime numbers in $\mathbb{Z}$ (which both are unitary rings). It also gives — in this special case — the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.

Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?

In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.

However, there are significant differences. In the case of $\Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.

On the other hand, in the case of $\Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $\Bbb R$, $\Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.

This analogy can be formalise in ring theory. The sets $\mathbb C[X]$ and $\mathbb Z$ are both rings, and more precisely, noetherian rings.

In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.

Which is exactly the meaning of those two theorems.

In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.


For how many integers from 1 through 99,999 is the sum of their digits equal to 10? I am doing some practice in counting, and came to this question:

For how many integers from $1$ through $99999$ is the sum of
their digits equal to $10?$

I am unsure, how to handle this question - anyone that got a idea of, how to do this kind of questions?

Count the number of ways of assigning values (any values) to non-negative integers $x,y,z,u,v$ that add up to $10$, using the algorithm in this answer of mine (with $14$ instead of $20$, and four instead of three). Then subtract $5$ to discard the extraneous solutions $(x,y,z,u,v) = (10,0,0,0,0),(0,10,0,0,0)$ etc.

Hint: There are $5$ digits - So, $a+b+c+d+e = 10$ That is .. there are $5$ boxes and $10$ balls to be put in them. The only condition is that all the balls cannot go into the same box.