Trigonometry word problem involving wheel? A wheel 5 feet in diameter rolls up with an incline of 18 degrees 20 minutes. What is the height of the center of the wheel above the base of the incline when the wheel has rolled 5 ft up the incline?

Does this diagram help? It is not drawn to scale but it can help in visualizing the problem.

So note that $|BD|=5ft$, $|CD|=2.5ft$, $\angle BDC=90^\circ$. Having settled that $$|CM|=\dfrac{2.5}{\sin 72}=2.628655561\ldots$$ we now need to find $$|MD|=\dfrac{2.5}{\tan 72}=.812229924\ldots$$ Since $|BD|=|BM|+|MD|$, then $$|BM|=5-|MD|=4.187700759\ldots$$ Calculating now $$|ME|=|BM|\sin 18=1.294070702\ldots$$ Finally, $$|CE|=|CM|+|ME|.$$

You'll just have to imagine that the radius (Half of the diameter which is 2.5 feet) is aligned with the height of the triangle. So You just have to get the height of the triangle. To get it's height. Simply...

Sin 18.33 = height / 5 feet
Sin 18.33 x 5 feet = height
(0.3145)(5 feet) = height
1.5725 = Height

So, to get the height of the center of the circle to the ground. Simply...

Height of the center of the circle  w/respect to the base = 1.5725 + 2.5
The answer would be...

4.0725 feet


Choosing e and d in RSA I just finished watching a video explaining RSA where the following example was used:

Choose two random prime numbers:

$$P=2, Q=7$$

compute $n = pq$

$$2\times7 = 14$$

Compute the totient $\phi(n) = (p-1)(q-1)$

$$6\times 1 = 6$$

Choose $e$ that $e > 1$ and coprime to $6$

$$e=5$$

choose $d$ to satisfy $de \equiv 1 \pmod {\phi (n)}$

$$5d \equiv 1 \pmod 6$$

The instructor then goes on to say d may have multiple solutions $5,11,17, \ldots$

but the example I read prior to the video indicated that d had a unique solution.  Please help.

It's a unique solution $\bmod 6$, but any value $5+6k$ is from the same equivalence class.

In general, take a prime number such as a fixed one like a Mersenne number or a prime using a primality test.


Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$. Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$.
Are the following containments always valid?.
$(1)    f^{-1} (f(S))⊆ S$
$(2)     S⊆f^{-1} (f(S))$
Attempt
$S ⊆ A$
$f^{-1}(S) = \{a ∈ B : f(a) ∈ S\}$
$ a ∈ f^{-1}(S) ⟺ f(a) ∈ S$
$ f^{-1}(S) ⊆ B$
I just need a little help finishing the proof. Am i wrong in the attempt? Thanks.

There is a problem with your proof. $f^{-1}(S)$ doesn't exist. This is because $S$ is a subset of $A$. It only makes sense to take a preimage under $f$ of a subset of $B$. On the other hand, $f(S)$ makes perfect sense. Indeed, $f(S)=\{f(a)\in B|a\in A\}$. Note that $f(S)$ is a subset of $B$. Then, it now makes sense to write $f^{-1}(f(S))=\{a\in A|f(a)\in f(S)\}$. From this it should now be clear that only one of the two options you presented in your question is always true. Can you tell which one it is?
On the other hand, the other option will be true if we add a further assumption on $f$. Can you tell which assumption we need to add? (hint: recall the notion of an injective/surjective function).

Recall that for a function $ f: X \rightarrow Y $, and two subsets $A\subseteq X, B\subseteq Y$ that
$$ (f \circ f^{-1})(B) = I_Y(B)=B \\ (f^{-1} \circ f)(A) = I_X(A)=A $$
Where $I_X, I_Y$ are the identity functions. Do you think you can continue from here?


Jane saw a police officer and Roger saw one too Statement:

Jane saw a police officer and Roger saw one too

Let

$J(x)$: Jane saw police officer $x$

$R(x)$: Roger saw police officer $x$

Hence,
$$\exists x \exists y \left( J(x) \land R(y) \right)$$
Is this correct?

Your translation is not incorrect, but quite imprecise. What you summarized as one predicate, "$P(x)$ = Jane saw police officer ", still contains an individual (Jane), a relation (saw), and a property (is a police officer). But it is not good practice to import specific individuals (Jane) and additional restrictions (is a police officer) into a predication (z saw y). To exaggerate, you could of course just shove everything into one symbol and define a single zero-place predicate "$P()$ = Jane saw a police officer and Roger saw one too" and be done with it -- technically not wrong, but kinda defeats the point of it all. Atomic formulas should express atomic propositions; what can be decomposed in more detail should be decomposed in more detail.
So analyze the sentene further into the parts mentioned above:

individuals:

$j$ = Jane
$r$ = Roger

properties and relations:

$po(x)$ = $x$ is a police officer
$saw(x,y)$ = $x$ saw $y$

As a rule of thumb,
$$\text{an A (is) B/there is an A that B}$$
translates as
$$\exists x (A(x) \land B(x))$$
and
$$\text{all A (are) B/everything that A B}$$
translates as
$$\forall x (A(x) \to B(x))$$
What you want here is "$\text{ a } \underbrace{\text{police officer}}_{A} \text{ is } \underbrace{\text{seen by Jane}}_{B}$", and the same with Roger. So you end up with
$$\exists x (po(x) \land saw(j,x)) \land \exists y (po(y) \land saw(r,y))$$
If you prefer, you can write the formula in a logically equivalent variant with the quantifiers moved to the front and the conjuncts commuted:
$$\exists x \exists y (po(x) \land po(y) \land saw(j,x) \land saw(r,y))$$
What is important is, as you did, to introduce two different individuals for the police officers and apply the predications to both of them, because the original sentence doesn't say that Jane and Roger necessarily saw the same officer (though the above formula correctly doesn't exclude this possibility, either).

I wouldn't call your solution incorrect per se but it's not an ideal answer. The functions should be predicates, which are essentially clear yes/no questions. The desired predicate (named $P$ in the comments) asks "did $x$ see a police officer", to which "yes" implies $x$ did, while "no" implies $x$ didn't. But your predicate, $J$, asks "did Jane see that police officer" - it's not as informative what yes or no mean here. "No" could mean Jane didn't see the police officer (but someone else did) or it could mean Jane didn't see that police officer (but she did see a different one). This is why $P$ is better.


What does the logical condition "implies" actually mean? $P_n$ and $Q_n$ are propositions which has a truth value $n$ where $n \in (T,F)$

So i want to access the truth value of $P\Rightarrow Q$

So

$P_T$ and $Q_T$, $(P \Rightarrow Q)_T$

$P_T$ and $Q_F$, $(P \Rightarrow Q)_F$

$P_F$ and $Q_T$, $(P \Rightarrow Q)_T$

$P_F$ and $Q_F$, $(P \Rightarrow Q)_T$

I don't really understand the last 2 statements. Can anyone explain them?

Like for  (2), If P is true then Q is false, then  $P \Rightarrow Q$ is false. that is self evidential. But how about (3) and (4) i can't seem to understand

In propositional calculus the "implies" connector means something a little more technical than implication in the cause and effect sense. It is usually the case that $A\implies B$ is shorthand for $\neg A \lor B$.

It's a definition.

$(P \Rightarrow Q)$ means that either (1), (3) or (4) is true.

In other words, $(P \Rightarrow Q) \leftrightarrow [(P \wedge Q) \vee(\neg P \wedge Q)\vee (\neg P \wedge \neg Q)]$

As (2), or $(P \wedge \neg Q)$, make (1), (3) and (4) all false, (2) also makes $(P \Rightarrow Q)$ false.


Extension of maps on $\partial D^n $ to maps on $D^n$ I'm trying to understand the extension of maps on $\partial D^n $ to maps on $D^n$

For a map $f:\partial D^n\mapsto X$, when can we say this map extends to a map $f': D^n \mapsto X$?

For two maps $f_1,f_2: D^n\mapsto X$, if they induce the same map on $\partial D^n$, when can we say that they are thus homotopic maps?

Also, does the above things actually comes from some larger setting rather than just for pair $(\partial D^n, D^n)$? Any help is appreciated.

By definition, a map $f:\partial D^n\to X$ extends to a map $f':D^n\to X$ if and only if $f$ is null-homotopic, i.e. homotopic to a constant map. See, for example, this question.

This is false in general (see this question, for example). However, in some special cases, we have results such as Alexander's Trick, which may be of interest.


Convergence of random variables What is the difference between convergence in distribution, convergence in probability and almost sure convergence?

References

Convergence of random variables

Convergence in distribution means pointwise convergence of the CDFs.  It's not a form of convergence of the random variables per se.

Convergence in probability says that for every $\epsilon > 0$, the probability of the sequence of random variables being more than $\epsilon$ from the limiting random variable goes to $0$. Convergence in probability is also called convergence in measure.

Convergence almost everywhere = almost sure convergence = pointwise convergence of the random variables except possibly on a set of measure zero.

Here is a diagram showing which modes of convergence imply which other modes.

A sequence X1  is. 1,1,2,1,1,2 ... 1,1,2...
A sequence X2 is. 2,1,1,2,1,1,... 2,1,1.

Assume a classical setup for the measure.

X1 and X2 converge in distribution.  Since both random variables output the value "1" with 2/3 probability, and "2" has probability 1/3.

But X1 will never converge in probability to X2 because as the number of samples increases two thirds  of the samples from the random variables will still be different from each other.  Obviously they are not getting close to each other as they should in convergence in probability.


How I can evaluate $\lim_{(x,y) \rightarrow (0,0)} xy(\frac{1+xy}{x^3+y^3})^{1/3}$ I don't have idea how I can evaluate this double limit
$$\lim_{(x,y) \rightarrow (0,0)} xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$$ could you help me please!

I try prove that $f$ is continuous: $f(x,y)=xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$ if $x\not=-y$ and $f(x,y)=0$ if $x=-y$

Hint: replace $1+xy$ by $1$ in the numerator (why?) and reduce to
$$
\lim_{(x,y) \to (0,0)} \left( \frac{x^3 y^3}{x^3+y^3} \right)^{1/3}.
$$
Put $\alpha =x^3$, $\beta=y^3$ and consider the constraint $\alpha + \beta = \alpha^3$ with $\alpha \to 0$. Then
$$
\frac{\alpha \beta}{\alpha+\beta} = \frac{\alpha^4-\alpha^2}{\alpha^3}
$$
which becomes unbounded as $\alpha \to 0$.

Try Polar coordinates maybe $x=r \cos \phi$ and $y=r \sin \phi$ and then check the limit $r \to 0$.


Is '$10$' a magical number or I am missing something? It's a hilarious witty joke that points out how every base is '$10$' in its base. Like,

2 = 10 (base 2)
8 = 10 (base 8)

My question is if whoever invented the decimal system had chosen $9$ numbers or $11$, or whatever, would this still be applicable? I am confused - Is $10$ a special number which we had chosen several centuries ago or am I missing a point?

Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero".

You must distinguish between numbers and representations. A pile of rocks has some number of rocks; this number does not depend on what base you use. A representation is a string of symbols, like "10", and depends on the base. There are "four" rocks in the cartoon, whatever the base may be. (Well, the word "four" may vary with language, but the number is the same.) But the representation of this number "four" may be "4" or "10" or "11" or "100" depending on what base is used.

The number "ten" — the number of dots in ".........." — is not mathematically special. In different bases it has different representations: in base ten it is "10", in base six it is "14", etc.

The representation "10" (one-zero) is special: whatever your base is, this representation denotes that number. For base $b$, the representation "10" means $1\times b + 0 = b$.

When we consider the base ten that we normally use, then "ten" is by definition the base for this particular representation, so it is in that sense "special" for this representation. But this is only an artefact of the base ten representation. If we were using the base six representation, then the representation "10" would correspond to the number six, so six would be special in that sense, for that representation.

This is something extremely aesthetic about the number 10 which I have not heard from anyone.

...........................

...........................

prime-prime intersections when compared with its reversed digits. 10 is the first number which have no such intersections.


All real continuous functions on the closed unit interval are analytic Suppose $f:[0,1] \to \mathbb{R}$ is a continuous function. By Weierstrass' theorem there exist a sequence of polynomials $\{p_i\}$ so that $p_i \to f$ uniformly. Now define the following sequence $\{q_i\}$ by $q_1 = p_1$ and $q_i = p_i - p_{i-1}$. Then we have that $\sum_{i=0} ^n q_i \to f$ uniformly. And thus, $\sum_{i=0} ^{\infty}q_i$ is a sum of polynomials and hence a power series that converges to $f$ so $f$ is analytic.

What is the mistake?

The mistake is that your sum up to $N$ of polynomials may mess with the "first" coefficients when you group them up in a power-series-like way whenever you change $N$. A power series doesn't allow that: whenever you truncate, the coefficients of a smaller truncation are "preserved".

As an analogy, consider a line of people forming outside your house, one day at a time. A power series is just new people coming. Your sum can change the people that were there yesterday.

To see this happen mathematically, take a sequence of polynomials approximating $f(x)=|x|$ in $[-1,1]$.

More concretely, suppose for instance that your $q_0=x$, $q_1=x^2-3x+1$ and $q_2=-2x^2+8x-4$. Then $\sum_{i=0}^1q_i=x^2-2x+1,$ $\sum_{i=0}^2 q_i=-x^2+6x-3$. How do you propose to form a power series if this keeps happening (which is certainly allowed)?

First of all, a power series is the limit of a sequence of the form $f_k(x) = \sum_{n=0}^k c_n x^n$ but here it is $f_k(x) = \sum_{n=0}^{d_k} c_{n,k} x^n$ which is different, so the theorems about power series don't apply.

More generally, the mistake is that if a sequence of complex analytic functions $f_n : \mathbb{C} \to \mathbb{C}$ converges uniformly to $f$ on $U \subset \mathbb{C}$ then :

$f$ is continuous on $U$
if $U$ is open in the complex topology, then $f$ is complex analytic on $U$. But $U = (0,1)$ is open in the real-line topology, not in the complex topology

($U$ is open in the complex topology iff for any $z_0 \in U$, there exists $\epsilon > 0$ such that $\{z \in \mathbb{C}, |z-z_0| < \epsilon\} \subset U$)

An useful example is with the Fourier series : if $f$ is $2\pi$ periodic and continuous then $f_n(x) = \sum_{k=-n}^k c_k e^{ikx}, c_k = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ikx}dx$ is a sequence of analytic functions converging uniformly to $f$. But of course this doesn't mean $f$ is analytic.


Derivative of functions at a point If the function $f:[0,1]\to \mathbb R$ defined by $f(x)=\dfrac{\sin x}{x}$ when $x\neq 0$ and $f(0)=0$, then is $f$ differentiable at $x=0?$

I think by using L'Hospital's rule $f$ is differentiable more than one time. Am I right?

As $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\dfrac{\sin x}{x}=1\ne0=f(0),$$
$f$ is not even continuous in 0, so can't be differentiable.

$f(x)=\frac{sinx}{x}$

To check differentiability at $x=0$ you can use the fundamental theorem of differentiability

R.H.D

$f'(x)=limh \to 0\frac {f(x+h)-f(x)}{h}$

$f'(x)=limh \to 0\frac {\frac{sin(x+h}{x+h}-\frac{sinx}{x}}{h}$

on expanding sin series u get the value of R.H.D=1

again similarly L.H.D=1

since both the values are equal but unequal to the function value @ x=0 hence$f(x)$ is not differentiable at $x=0$.I hope now u find the general way.

THANKS.


Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$ The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?

Let $x= \tan \theta$

$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$

I had to delete the rest of my attempt because it was totally wrong. What are the methods to solve this problem?

As J.G. noted in comments, there are no solutions for $|x|\gt1$. This is because the range of the arcsine function is $[-\pi/2,\pi/2]$, so the left hand side, $2\arcsin\left(2x\over1+x^2\right)$, is never greater than $\pi$ in absolute value, while $\pi|x^3|\gt\pi$ if $|x|\gt1$. We'll come back to this in a moment.

Now after substituting $x=\tan\theta$, the left hand side becomes

$$2\arcsin\left(2\tan\theta\over1+\tan^2\theta\right)=2\arcsin\left(2\sin\theta\cos\theta \right)=2\arcsin(\sin2\theta)$$

It's tempting to "cancel" the the inverse trig function, leaving just $4\theta$.  But that's only correct for $|\theta|\le{\pi\over4}$.  However, that is exactly the domain we need in order to have $|x|\le1$, since $-1\le\tan\theta\le1$ for $-\pi/4\le\theta\le\pi/4$. So we can go ahead and write the equation as

$$4\theta=\pi\tan^3\theta$$

with the understanding that we only care about solutions with $-\pi/4\le\theta\le\pi/4$.

Three solutions jump off the page: $\theta=0$, $\theta=\pi/4$, and $\theta=-\pi/4$. The question is. are there any others?  For this it helps to look at the function $f(\theta)=4\theta-\pi\tan^3\theta$, for which

$$f'(\theta)=4-3\pi\tan^2\theta\sec^2\theta$$

and
$$f''(\theta)=-3\pi(2\tan\theta\sec^4\theta+2\tan^3\theta\sec^2\theta)=-6\pi\tan\theta\sec^2\theta(2\sec^2\theta-1)$$

It's clear that $f''(\theta)\lt0$ for $0\lt\theta\lt\pi/4$, and easy to see that $f'(0)=4\gt0$ while $f'(\pi/4)=4-6\pi\lt0$. Consequently there are no other solutions in $(0,\pi/4)$, hence, by symmetry, in $(-\pi/4,0)$ either.

To sum things up, $x=0$, $x=1$ and $x=-1$ are the only three solutions to

$$2\arcsin\left(2x\over1+x^2\right)-\pi x^3=0$$

Shouldn't the third line be
$ 2\theta = \dfrac{\pi}{2}\tan^3\theta
$?

The arcsin cancels out the sin.

Wolfy says that
the roots in
$[-\pi/2, \pi/2]$
are
$0$
and
$± 0.785398163397448...$.


Find all prime numbers p such that 2p+1 is a perfect cube. $$2p+1=n^3$$
$$ 2p = n^3 - 1$$
$$ 2p = (n-1)(n^2+n+1)$$

The only number that fits the criteria is 13, how can I prove this?

HINT: The factorization of $2p$ shows that $n-1$ must be either $2$ or $p$. (The other two cases, $n-1=1$ and $n^2+n+1=1$, are trivially eliminated.) You already know what happens when $n-1=2$. Otherwise, $n-1$ must be an odd prime $p$. In that case $n$ is even. What does that tell you about $n^3-1$?

Since $p$ is a prime, the only factors possible for $2p$ is $2$ and $p.$
This must correspond to $(n-1$) and $(n^2 + n +1)$. We know using common sense that $n^2 + n +1$ is greater than $n-1,$ also that since $p$ is a prime, $p>2. $
Hence,$n-1=2$ which means $n=3.$
$p=n^2 + n +1$. Here, replace the value of $n$ with $3$(As obtained from the previous step) and you'll get $p=13$


Poisson process with dead time A well studied model for radioactive particle counters (Geiger counters) is the so called Type 1 a.k.a. non-paralyzable model which assumes that the particle detector goes "blind" for a certain time after each detection event. This is called the dead time of the detector.

Let's assume that particles arrive at the detector according to a Poisson process with inter-arrival times $\{X_m\}$ that are i.i.d. exponential with rate $\lambda$. Let $\{V_m\}$ denote the sequence of particle arrival times. Let $\{T_n\}$ denote the sequence of detection times. Let $\{U_n\}$ be the sequence of i.i.d. positive random variables denoting the random dead times.
Also assume that we start off with a particle counted at $V_0 = T_0 = 0$.

I can think of two methods to model the random process that is the events detected by the counter and I'd like to know if they are equivalent.

Method 1

(Grimmett and Stirzaker, Probability and Random Processes, Sec. 10.4)

As soon as the detector exits its last dead time, the time until it sees the next particle is the residual time of an exponential random variable, which, by the memorylessness property is exponential with rate $\lambda$. This allows us to write:
$$
T_n = T_{n-1} + X_n + U_{n-1}.
$$

Method 2

There are two random processes to look at. First, we first explicitly write the Poisson process event times:
$$
V_m = V_{m-1}+X_m.
$$
Next, we model the random process that is the output of the Geiger counter. The counter's $n$th detection time can be written in terms of the $n-1$st detection time as:
$$
T_n = \min_i {\{V_i: V_i \geq T_{n-1}+U_{n-1} \}}.
$$

Question

Is $T_n-T_{n-1}$ i.i.d. between the the two methods and over $n$?

(as requested in comments)

I would have thought $T_n-T_{n-1}$ would be i.i.d. between the the two methods and over time.  My simulation in R seems to confirm this

Let's assume the detector has seen $n-1$ events so far and we are
waiting for the $n$th event.

At $T_{n-1}$ we generate a random dead time $U_{n-1}$. The detector
wakes up at time $T_{n-1}+U_{n-1}=:t(n).$ Since $V_{j}$'s are epochs
of a Poisson process, there exists a unique $k^{*}=k^{*}(n)$ such
that $V_{k^{*}}\leq t(n)\leq V_{k^{*}+1}.$ (For example, in the figure, $k^{*}=j+1$). This implies that
$$
\min\{i:V_{i}\geq T_{n-1}+U_{n-1}\}=k^{*}+1.
$$
so that $T_{n}=V_{k^{*}+1}.$

Defining the $n$th residual time $Y_{n}:=T_{n}-(T_{n-1}+U_{n-1})=V_{k^{*}+1}-t(n)$, we can write
$$
T_{n}-(T_{n-1}+U_{n-1})=Y_{n}
$$
which is identical to Method 1 because $Y_{n}\overset{iid}{\sim}\mbox{Exp}(\lambda)$ by the memoryless property of a Poisson process.


How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root. Let $W(x)$ be a polynomial of degree> 2 having at least three different real roots. How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root?

Whether the assumption of the theorem can be weakened?

Consider the function $\phi\colon\mathbb{R}\to\mathbb{R}$, defined by $\phi(x)=e^{W(x)}$, and calculate
$$
\phi'=W'e^W, \qquad \phi''=(W''+(W')^2)e^W=Pe^W.
$$
Thus we have to prove that $\phi''(d)=0$ for some $d\in\mathbb{R}$.

We know that $\exists a<b<c\in\mathbb{R}$ such that $\phi(a)=\phi(b)=\phi(c)=1$.  Rolle's theorem gives that there exist $a_1\in[a,b]$ and $b_1\in[b,c]$ such that $\phi'(a_1)=\phi'(b_1)=0$. Applying the theorem again to $\phi'$ we get what we need.

Hint: Suppose that $W(x)$ has at least 3 different real roots called $a,$ $b$ and $c$ so $W(x)= (x-a)(x-b)(x-c) g(x)$ where $\deg(g)=\deg(W)-3$ and $g(a),$ $g(b)$ and $g(c)$ is non zeros. Then, you can compute find the expression of $P(x)$ in terms of $(x-a),$ $(x-b)$ and $(x-c).$


Puzzle about six travellers going through bridge above canyon with an oil lamp There is a dark night and there is a very old bridge above a canyon. The bridge is very weak and only 2 men can stand on it at the same time. Also they need an oil lamp to see holes in the bridge to avoid falling into the canyon.

Six man try to go through that bridge. They need 1,3,4,6,8,9(first man, second man etc.) minutes to pass the bridge.

What is the fastest way for those six men to pass this bridge?

A minimal solution is:

1 and 6 cross the bridge; 1 comes back (7)
1 and 3 cross the bridge, 1 comes back (4)
8 and 9 cross the bridge; 3 comes back (12)
1 and 3 cross the bridge, 1 comes back (4)
1 and 4 cross the bridge (4)

Total is $7+4+12+4+4 = 31$. There are minor variations that are still minimal. For example, 3 can return in step 2 instead of in step 3.

I have nothing useful to say about how to find the solution; I programmed the computer to do an exhaustive search.  I only posted this because the other guy was getting upvotes for his wrong answer.

[ By request, my search program is available ]

A minimal solution is:

1 and 3 cross the bridge; 1 comes back (4)
12 and 8 cross the bridge, 3 comes back (15)
1 and 6 cross the bridge; 1 comes back (7)
1 and 3 cross the bridge (3)
Total is 4+15+7+3=29.


$\lim_{n \to \infty}(\lim_{k \to \infty}(\cos(n!\pi x)^{2k}))$ Identify the function $f(x) = \displaystyle \lim_{n \to \infty}(\lim_{k \to \infty}(\cos(n!\pi x)^{2k}))$.

I was confused how to find this limit because cosine oscillates, so I didn't think it was possible to find the limit.

Set $$D_n = \{ x \in \mathbb{Q} \, : \, x=p/q \text{ with } q \text{ integer divisor of } n! \}. $$It is enough to notice that $$\lim_{m \to \infty} [\cos^2 (n! \pi x)]^m=\begin{cases} 1 & \text{if } x \in D_n \\  0 & \text{if } x \in \mathbb{R} \setminus D_n \end{cases}=f_n(x).$$Since $$\bigcup_{n=1}^\infty D_n = \mathbb{Q}$$we have that $$\lim_n f_n(x) = \chi_{\mathbb{Q}} (x)= \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$$

if x \in Q then x = p/q
  then for some very large n  n!/q is even number therefore making
      cos(n! \pi x) = 1
hence making the limit 1

else if x is irrational then (n!\pi x) is never \pi hence
     Z = (cos(n! \pi x) )^2 <1
 implies Z^k=1
for all k


Probability Problems Problem:

The probability that a man who is 85 years. old will die before attaining the age of 90 is $\frac13$. A,B,C,D are four person who are 85 years old. what is the probability that A will die before attaining the age of 90 and will be the first to die ?

The probability that A dies first is 1/4 (since it's the same probability for everyone).

The probability that they all die after the age of 90 is $(2/3)^4 = 16/81$. The two events are independent, so the probability that A dies first and they all die after the age of 90, or equivalently that A dies first and that that happens after he is 90, is $\frac{16}{81}\cdot\frac{1}{4}=\frac{4}{81}$. But if A dies first, then either he dies first and is above 90 or he dies first and is younger than 90, and the two are mutually exclusive. Therefore (or equivalently using the rule that $P(X) = P(X\cap Y) + P(X\cap Y^C)$), we deduce that the probability that A dies first and before the age of 90 is 1/4 - 4/81 = 65/324.

The probability of dying by 90 is the same for all four.  The probability that A dies first is simply 1/4 since we are give no more information.  Since the two events are independent, the probability of their conjunction, i.e., that A dies and is the first to die, is simply the product of the two probabilities, or 1/12.


$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}=C$$

Where $C \approx 1.7579$ (OEIS A072449) is the nested radical constant.  No closed form of this constant is known.

$$\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}=\sigma$$

where $\sigma \approx 1.6616$ (OEIS A112302) is Somos's Quadratic Recurrence Constant.

$\sigma$ has an alternate form (I hesitate to call it a "closed form") of
$$\sigma =
\exp\left[-2^n \frac{\partial \operatorname{Li_n\left(\frac{1}{2}\right)}}{\partial n}\bigg|_{n=0}+\frac{1}{2} \frac{\partial \Phi\left(\frac{1}{2},s,n+1\right)}{\partial s}\bigg|_{s=0}\right]$$

where $\operatorname{Li}_n(z) = \sum^\infty_{n=1} \frac{z^k}{k^n}$ is the polylogarithm and $\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$ is the Lerch transcendent, and

$$\sigma = \exp\left[\int_0^1 \frac{1-x}{(x-2)\log x}\, \mathrm{d}x\right]$$


Show that $f(z)=\sum_{n=-\infty}^\infty \frac{1}{(n-z)^2}$ converges uniformly on compact sets in $\Bbb C - \Bbb Z$. Show that

$$f(z)=\sum_{n=-\infty}^\infty \frac{1}{(n-z)^2}$$

converges uniformly on compact sets in $\Bbb C - \Bbb Z$, and is a continuous function there.

Here is the original dutch version:

My first idea is to write this like:

$$1/z^2+\sum_{n=1}^\infty 1/(n-z)^2+1/(n+z)^2$$

And then use the Weisterstraas M test. But this didn't work.
Any other ideas ?

Any compact set $K$ in $\mathbb C\setminus\mathbb Z$ is contained in some disk $B(0,N)$, $N\in\mathbb N$. At the same time $K$ has a positive minimal distance from $\mathbb Z$, $0<\delta=d(K, \mathbb Z)$, since $d(x,\mathbb Z)=\min_{n\in\mathbb Z}|x-n|$ is a continuous function.

Then for any $z\in K$
\begin{align}
\left|\frac{1}{(z-n)^2}\right|&<\frac{1}{δ^2}&&\text{ for }|n|\le N\\
\left|\frac{1}{(z-n)^2}\right|&<\frac{1}{(|n|-|N|)^2}&&\text{ for }|n|\ge N+1
\end{align}
so that
$$
\sum_{n\in \mathbb Z}\left|\frac{1}{(z-n)^2}\right|\le \frac{2N+1}{δ^2}+2\cdot\frac{\pi^2}6
$$
By the Weierstraß M test convergence is uniform on $K$.

Which means that the idea in the question was right, one just needs to take out a larger middle part.

i think (by taking euler product) that the function is about

$$ \frac{d^{2}}{dx^{2}}log(sin(z)/z) $$

THE infinite Euler product is equal to $ sin(z) $ up to some constant


If $2x^2-3xy-5y^2$ =0 then show that $x^2+y^2\propto xy$ If $ 2x^2 -3xy-5y^2=0$ then show that $$x^2+y^2\propto xy$$
This is the question.I tried it to so using elementary algebra but I can't proceed. Somebody please help me.
Thank you

As such, the claim is false.

Counterexample:

$$(-1,1)\to x^2+y^2=2, xy=-1\\
(5,2)\to x^2+y^2=29, xy=10.$$

Anyway, the constraint can be factored as

$$(x+y)(2x-5y)=0$$ so that on both branches of this degenerate hyperbola,

$$(x,mx)\to x^2+y^2=(1+m^2)x^2,xy=mx^2.$$

Then $$x^2+y^2\propto xy.$$


What would be the cardinality of $S$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying $$f(x)\equiv 1\mod (x-1)$$ and $$f(x)\equiv 0\mod (x-3)$$

What would be the cardinality of $S$?

Second equation means $x=3$ is a root of $f(x)$. That means $f(3)\equiv 1\mod2 $
 and that means $0\equiv 1\mod 2$ which is a contradiction. So $S$ is empty.

Please have a look at the work. Am I right?

Thanks.

Ok, after the very last comment below the question by the OP I think the following can be what was intended: it is given $\;f(x)\in\Bbb Z[x]\;$ s.t. both the following are true:

$$\begin{cases}I\;\;\;f(x)\cong1\mod{(x-1)}\iff f(x)=1+(x-1)p(x)\;,\;\;p(x)\in\Bbb Z[x]\\{}\\
II\;\;f(x)\cong0\mod{(x-3)}\iff f(x)=(x-3)q(x)\;,\;\;q(x)\in\Bbb Z[x]\end{cases}$$

but then we get:

$$\begin{cases}I\;\implies\; f(1)=1=1\pmod 2\\{}\\
II\implies f(1)=-2q(1)=0\pmod2\end{cases}\;\;\;\implies\text{ contradiction}$$

and what the OP did is thus correct: we get that in fact $\;S=\emptyset\;$ .

$f(x)\equiv  1\bmod (x-1)$ is a congruent condition as polynomials, i.e. $f(x)=1+p(x)(x-1)$ for some polynomial $p(x)\in\mathbb{R}[x]$. In particular, $f(3)=1+p(3)(2)$. Note that $p(3)$ (a priori) does not need to be integer, so saying $f(3)\equiv 1\bmod 2$ is not justified unless you can prove that $p(3)\in\mathbb{Z}$ (which would be true if you can prove $p(x)\in\mathbb{Z}[x]$)


Proof of $x^p$ sub/super additive? What is the easiest proof that $x^p$ is sub (or super) additive for  $p>1$ (or $p<1$) and $p\in \mathbb R$.
This is the same as showing for  $a,b\geq 0$
$$a^p+b^p\leq (a+b)^p,$$
when $p\geq 1$. It is possible to take a logarithm and compare the Taylor series to prove the claim, but is there anything simpler?

Let $p>1$. You can define for $x\geq 0$
$$f(x) = (a+x)^p\,;\qquad g(x) = a^p + x^p\,.$$
Clearly $f(0) = g(0)$. Now take the derivatives.
$$f'(x) = p(a+x)^{p-1}\,;\qquad g'(x)=p x^{p-1}\,.$$
Since $a>0$ and $p>1$ it's clear that for all positive $x$ you have $f'(x)\geq g'(x)$. This ends the proof for $p>1$ since for all $x\geq 0$ you'll have $f(x)\geq g(x)$.
For $p<1$ you can proceed in the same way and get the opposite inequality.

We have that the $p$-norm is a norm on $\mathbb{R}^2$ when $1\leqslant p < +\infty$. This means that $\lVert.\rVert_p$ is subadditive. So let's define $a=(x,0)$ and $b=(0,y)$. The subadditivity means that
$$\lVert a + b \rVert = (|x|^p + |y|^p)^{1/p}\leqslant |x|+|y|=\lVert a \rVert + \lVert b \rVert$$
Which is exactly what we wanted:
$$|x|^p+|y|^p\leqslant (|x|+|y|)^p$$


How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$ How to solve this equation?

$$x=\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}.$$

Answer: $x=2$


Why does multiplying a number on a clock face by 10 and then halving, give the minutes? ${}{}$ My daughter in grade 3 is learning about telling time at her school. She eagerly showed me this method she has discovered on her own to tell the minutes part of the time on an analogue clock. I wasn't sure at first because I have never heard about it before but it works really well.

Here is her method:

Look at the minute hand and see what number it is pointing to, let's say it's $3$.
Add a zero at the end to make it $30$.
Halve it to get the minutes, so $3$ becomes $30$ and halving it gives $15$, $6$ becomes $60$ and halving it gives $30$ and so on.

It works well but I am not sure why. What mathematically justifies the method used?

Each number on the clock face is worth five minutes. One good way to multiply by $5$ is to first multiply by $10$, and then divide by $2$. This works because $5=10\div 2$.

From the perspective of the minute hand, the numbers mark the 1/12ths of an hour. But we are usually more interested in the 1/60ths of an hour: the minutes.

Since 12 and 60 differ by a factor of 5 (i.e. $12*5=60$), converting between the two is quite easy. We just multiply by 5.

Or as your daughter prefers, you could multiply by 10 (by adding a 0) and then divide by 2. Since $10/2=5$, it amounts to the same thing.


application of L'Hopital's rule? I am trying to evaluate the following  limit:
$$
\lim_{x \to 0} \frac{e^x}{\sum_{n = 1}^\infty n^k e^{-nx}},
$$
where $k$ is a large (but fixed) positive integer. I am unsure how to proceed. Can this be done using L'Hopital's rule? Just started learning calculus, thanks guys!!

For each $x\lt\frac12$, there is an $n\in\mathbb{N}$ so that $1\lt\frac1x-1\le n\lt\frac1x$. Picking this one term out gives
$$
\sum_{n=0}^\infty n^ke^{-nx}\ge \left(\frac1x-1\right)^ke^{-1}
$$
As $x\to0^+$, $\left(\frac1x-1\right)^ke^{-1}\to\infty$

You don't need L'Hopital's rule; try substituting $0$ for $x$ and see what you get.


disjoint subset of random variables Can someone explain me (possibly with examples) the meaning of the following statement: X and Y are two disjoint subsets of random variables? what this disjoint means here? I understand what disjoint means in terms of events, i.e if And B are two disjoint events then $A \cap B=\emptyset $, but what that means in the context of random variables? Thanks

For what it is worth, I see no reason why this could refer to the hypothesis that every random variable in $X$ should be independent of every random variable in $Y$, nor to the hypothesis that $X\cap Y$ should be empty. If I had to bet, I would bet that this simply refers to two collections of random variables.

Or, one is given previously some set $Z$ of random variables and the hypothesis is that $X\subseteq Z$, $Y\subseteq Z$, and $X\cap Y=\varnothing$.

As per the definition random variable, the values the random variable can take should follow a determinable distribution. Hence this disjoint random variable may mean that the distributions of the two variables from which the two subsets are drawn from. Further the subsets themselves may be disjoint.


Concentrated schemes are closed under finite gluings Let $X$ be a scheme. Assume that $X = \cup_i X_i$ is finite open covering, such that the $X_i$ as well as their intersections $X_i \cap X_j$ are concentrated, i.e. quasi-compact and quasi-separated (see Daniel Murfet's notes on concentrated schemes as well as EGA IV.1). Does it follow that $X$ is concentrated? I'm almost embarrassed to ask this question.

Clearly $X$ is quasi-compact, as a finite union of quasi-compact subspaces. In order to show that $X$ is quasi-separated, it suffices to show that for open affines $U \subseteq X_i$ and $V \subseteq X_j$ their intersection $U \cap V$ is quasi-compact. Well this is clear for $i=j$ since $X_i$ is quasi-separated. But what happens for $i \neq j$? Somehow we have to use that $X_i \cap X_j$ is quasi-compact.

Here is a more abstract idea: We have to show that $\Delta : X \to X \times X$ is a quasi-compact morphism. This can be checked locally on the base, hence it suffices to prove that the morphisms $X_i \cap X_j \to X_i \times X_j$ are quasi-compact. Well, $X_i \cap X_j$ is quasi-compact, hence by (EGA IV, 1.1.2 (v)) it would suffice that $X_i \times X_j$ is separated. But we don't assume that the $X_i$ are separated, they are just quasi-separated.

Note that actually EGA IV.1.2.7.c proves the claim, but I don't understand the proof. Here is my translation: "We know that in order to prove that $\Delta : X \to X \times X$ is quasi-compact, it suffices to prove that the inverse images $X_i \cap X_j$ of the $X_i \times X_j$ are quasi-compact (1.1.1)". Here, (1.1.1) is just the definition of a quasi-compact morphisms, after which it is remarked that it suffices to prove this with respect to an affine(!) open covering. But here, we just have an open covering consisting of quasi-compact schemes. This seems to be a gap?

Or does the following statement hold (which we could then apply to $\Delta$)? If $f : X \to Y$ is a morphism and $Y$ is covered by quasi-compact open subsets $Y_i$ such that each $f^{-1}(Y_i)$ is quasi-compact, then $f$ is quasi-compact. I doubt that this is true without any assumption on $Y$.

Edit. Sorry I think now I can answer my question. The treatment of quasi-compact morphisms is better in EGA I (1970), §6. Specifically, in order to show that $X_i \cap X_j \to X_i \times X_j$ is quasi-compact, it suffices that $X_i \cap X_j$ is quasi-compact and $X_i \times X_j$ is quasi-separated by Prop. 6.1.5. (v), and we have both. Note that EGA VI.1.2.7 is reproduced as EGA I.6.1.12, but still the end just refers to the definition of quasi-compact morphisms, which doesn't seem to suffice?

So what about a more direct proof, showing directly that (with the notation above) $U \cap V$ is quasi-compact? One can spell out the proof above as follows (still quite tricky):

$X_i \cap X_j \to (X_i \cap X_j) \times (X_i \times X_j)$ is a base change of the diagonal of $X_i \times X_j$, which is quasi-compact, hence it is also quasi-compact. Since $U \cap V$ is the preimage of the quasi-compact open subscheme $(X_i \cap X_j) \times (U \times V)$, it is quasi-compact.

(Usually I delete a question when I solve it quickly before an answer arrives, but this time I won't because it has already some upvotes and therefore seems to be interesting for others, too. Besides, I would like to know if you also agree that the EGA proof has a gap.)

The problem is roughly the result in the picture below from EGA I.6.1.12 p.296 (Springer 1971).

I claim that the condition d) is not sufficient to imply a) (so EGA I.6.1.12 is not correct as stated). As a counterexample, let take $X$ to be the infinite dimensional affine space over a field with double origin, as Exercise 10.11 in Algebraic geometry I by Görtz-Wedhorn:and we take all $U_{\alpha}$ to be the quasi-compact scheme $X$ itself, then d) holds but a) does not. See also a comment in [Görtz-Wedhorn, Remark 10.2 (4)] for a related issue on quasi-compactness.

So you don't want to choose the $X_i$ to be affine from the beginning? Maybe that's a work around for your last question if you assume $Y$ to be quasi-compact.

Let $f \colon X \to Y$ be a morphism and $Y = \bigcup_{i \in I} Y_i$ with $Y_i$ quasi-compact and $I$ finite. Let $Y_i = \bigcup_{j \in J} Z_j^{(i)}$ where  the $Z_j^{(i)}$ are affine and $J = J(i)$ can be chosen to be finite. Since
$$ f^{-1}(Y_i) = f^{-1} \Bigl( \bigcup_j Z_j^{(i)} \Bigr) = \bigcup_j f^{-1}(Z_j^{(i)}) $$
you get that
$$
X = f^{-1} \Bigl( \bigcup_i Y_i \Bigr) = \bigcup_i f^{-1}(Y_i) = \bigcup_i \bigcup_j f^{-1} (Z_j^{i}) $$
Now stack the project (Tag 01K4) tells us that it is enough to find some cover by open affines to check that $f$ is quasi-compact.


Solve $\partial_t u= -\partial_x^3 u$ I want to solve the PDE
$$\partial_t  u= -\partial_x^3 u\quad \text{s.t.}\quad u(0,x) = g(x).$$
where $u:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$. I used the Fourier-Transformation $\mathcal{F}_x$ w.r.t $x$ to solve this:

$$\partial_t(\mathcal{F}_xu)(t,\xi) = -i\xi^3(\mathcal{F}_xu)(t,\xi)\quad\text{s.t.}\quad (\mathcal{F}_xu)(0,\xi) = (\mathcal{F}_xg)(\xi)$$
$$\Leftrightarrow\quad(\mathcal{F}_xu)(t,\xi) = (\mathcal{F}_xg)(\xi)\cdot\exp(-i\xi^3t)$$
$$\Leftrightarrow\quad u(x,t) = \frac{1}{\sqrt{2\pi}}(g*\mathcal{F}^{-1}_xh)(x),$$

where $h(t,x) = \exp(-ix^3t)$. Unfortunately I'm not able to calculate $\mathcal{F}^{-1}_xh$ and even WolframAlpha doesn't give me a nice closed form solution. Can this problem be further simplified? Does this equation have a specific name?

HINT : Solution in term of Fourier series.
$$\partial_t  u= -\partial_x^3 u $$
With the method of separation of variables, it is easy to find an infinity of particular solutions on the form :
$$u_{\lambda}(t,x)=A_\lambda\sin(\lambda x+\lambda^3 t)+B_\lambda\cos(\lambda x+\lambda^3 t)$$
where $\lambda$ , $A_\lambda$ , $B_\lambda$ are any constants.

More general solution :
$$u(t,x)=\sum_{\forall \:\lambda}\left(A_\lambda\sin(\lambda x+\lambda^3 t)+B_\lambda\cos(\lambda x+\lambda^3 t) \right)$$
Boundary condition : $ \quad u(0,x) = g(x)$

Expressing $g(x)$ on the form of Fourier series, for example :
$$g(x)=\sum_{n=0}^\infty\left(A_n\sin(n x)+B_n\cos(n x) \right)$$
gives $\quad \lambda=n\quad;\quad A_\lambda=A_n \quad;\quad B_\lambda=B_n\quad$ and the solution :
$$u(t,x)=\sum_{n=0}^{\infty}\left(A_n\sin(n x+n^3 t)+B_n\cos(n x+n^3 t) \right)$$
Of course, depending on the range of $x$ considered, the Fourier series can be computed with other coefficients and different variable, for example $n\pi x$ , or $2n\pi x$ , ..., instead of $nx$.


shadow simulation from buildings is it possible to calculate shadow areas of buildings or simulate shadows of buildings in a city, using the heights of these buildings and the sun angle and azimuth? the basic light tracing concept using trigonometry is known. But the Problem is how to find away to come up with simulated shadow of group of buildings in one place.

I'll assume that the buildings are made up of polygons.

Project each polygon onto the ground to find its shadow.
Merge these shadow polygons (e.g. using Weiler-Atherton).
Find the area of the merged polygon.

if you have the data on the buildings, model the whole scene as tall boxes in OpenGL, render it with a suitable light source, see the shadow.


Surface integral using Stokes' theorem $$
\int_\Gamma y\,dx+z\,dy+x\,dz
$$
when

$\Gamma$ $= \{ (x,y,z): x^2+y^2+z^2=9\}$ $\cap$ $x+y+z=0$

There's a theorem that states:
$\int_S(\nabla \times \vec F)\cdot d \vec S$= $\int_S(\nabla \times \vec F)\cdot \vec n\cdot d \vec S$

So, according to Stokes' theorem, I calculate $ \partial F $= $dx \wedge dy + dy\wedge dz + dz \wedge dx$.
I want the surface $S$ to be the plane $x+y+z=0$ inside the sphere, so I calculate its normal vector: $$\vec n = (\frac {1} {\sqrt3},\frac {1} {\sqrt3},\frac {1} {\sqrt3})$$
If I place $\partial F$ and $\vec n$ in the equation above, I will have $\frac {1} {\sqrt3}\int(x+y+z)\,dS$, and this is supposed to equal $0$ because $x+y+z=0$.

But the answer is not $0$, so  what am I doing wrong here?

Your mistake seems to have been in the computation of the curl of the vector field whose line integral you are evaluating.

Supposing $\vec{F}=\langle y,z,x\rangle$, then

$$\nabla\times\vec{F}=\langle -1,-1,-1\rangle.$$

Let $S= \{ (x,y,z): x^2+y^2+z^2=9\}$. This is the sphere of radius $3$ centered at the origin. The plane $x+y+z=0$ cuts this sphere into two hemispheres. Let $H$ be the hemisphere on the same side of the plane as the normal vector $\vec{n}=\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle$.

$$\int_\Gamma ydx+zdy+xdz=\iint_{H}(\nabla\times\vec{F})\cdot\vec{n}\,\mathrm{d}S=-\sqrt{3}\iint_{H}\mathrm{d}S=-\sqrt{3}(2\pi9)\\=-18\sqrt{3}\pi,$$

where in the second to last step we've simply used the fact that the surface area of a hemisphere of radius $r$ is $2\pi r^2$, or half that of a sphere of radius $r$.


How many labeled trees exist on n vertices with exactly 3 vertices of degree 1? My combinatorics class is covering spanning trees right now and one of the questions being asked is "What is the number of labeled trees on n vertices with exactly $3$ vertices of degree $1$?"

I've tried a couple different ways of doing this and none of them seem to work. I tried examining the degree sequence, looking at the Prüfer code as words from an $[n-3]$ length alphabet, but it seems fruitless.

Any help would be greatly appreciated. Thanks!

Edit: I mis-read the problem, this was first posted with at least 3 vertices of degree 1, but the actual problem states exactly 3 vertices of degree 1.

Here's a way to find the number.  It looks like it might not be the best (Sloane's A055315 has other formulas), but it works.

Let $n$ be the number of vertices and $e=n-1$ be the number of edges.

Claim: There is exactly one vertex of degree $3$, all others have degree $2$ or $1$.  (Assume otherwise, and we violate the Handshaking Lemma, which stipulates the sum of degrees is $2e=2(n-1)$.)

So, the trees look like this:

If we delete the degree-$3$ vertex, we obtain three components.  These three components only have vertices of degree $1$ and $2$, and so must be paths.  Suppose the paths have $a$, $b$ and $c$ vertices.  Then $a+b+c=n-1$ and $a,b,c$ are all positive integers.  Conversely, given three paths of lengths $a \geq 1$, $b \geq 1$ and $c \geq 1$ and $a+b+c=n-1$, we can join their endpoints to a new vertex to create an $n$-vertex tree with exactly three $3$ leaves.

Let $a(n)$ be the number of partitions of $n-1$ into $3$ non-zero parts; this is equivalent to Sloane's A001399 which includes the formula $$a(n)=1+\left\lfloor\frac{(n-1)^2+6(n-1)}{12}\right\rfloor$$ among others.

Given an unlabeled tree, we can label the vertices in $n!$ (not necessarily distinct) ways.  In this way, we generate $a(n)\ n!$ labeled trees.

In some cases (not many), there are symmetries which need to be accounted for.  Because of symmetry, in this case

we count every labeling twice.  This happens whenever $\{a,b,c\}$ have two parts of the same size.

And in this case

we count every labeling six times.  This happens when $\{a,b,c\}$ has all three parts of the same size.

In these cases, we'll need to add a correction.

In the first case: we subtract $$\sum_{a \geq 1:n-1-a \text{ is even and } \geq 2} \tfrac{1}{2}n!=\frac{\lfloor (n-2)/2 \rfloor\ n!}{2}$$ since we have double-counted the $\frac{1}{2} n!$ graphs with $b=c=\tfrac{n-1-a}{2}$.
In the second case: if $n-1 \equiv 0 \pmod 3$, we first "uncorrect" the first correction, so we add back in $\tfrac{1}{2}n!$ and then subtract $\frac{5}{6}n!$ since we originally counted six times the $\frac{1}{6} n!$ graphs with $a=b=c$.

This gives $$\begin{cases} a(n)\ n!-\tfrac{\lfloor (n-2)/2 \rfloor\ n!}{2} & \text{if } n-1 \not\equiv 0 \pmod 3 \\ a(n)\ n!-\tfrac{\lfloor (n-2)/2 \rfloor\ n!}{2}+\tfrac{1}{2}n!-\tfrac{1}{6} n! & \text{if } n-1 \equiv 0 \pmod 3. \end{cases}$$ in agreement with Sloane's A055315.

In fact, the problem can be solved by Prufer code only. Consider first we choose $3$ nodes, they are not in Prufer code and there are $C(n,3)$ ways to choose. Then for last $n-3$ nodes, they must be in the code. It's easily to find that only one of them show 2 times and the other shows exactly one times. Then we can choose the position for the 2 times' number , there are $(n-3)C(n-2,2)$ ways, for the last $n-4$ numbers, we have $(n-4)!$ ways to arrange them.
As all above, we have $C(n,3)(n-3)C(n-2,2)(n-4)!$ such trees.
You can simplify the result, it is same to the other answers.


Poisson Distribution practice problem I have just started to learn Poisson Distribution and I have no idea how to deal with the following practice from my textbook: Suppose the average amount of cars passing on a street per minute is Poisson (8.6). It takes a lady 5 seconds to cross a street and she waits for two cars to pass. What is the probability that she is safe to cross the street ?

The lady waits for 2 cars means:

the first 5 seconds, the lady waits because there is car passing.
the second 5 seconds, the lady waits because there is car passing.
the third 5 seconds, the lady waits because there is no car passing.

there is no car passing is P(0)=48.8% (you can do the math)
there is car passing is 1-P(0)=41.2%

so the probability is 41.2% x 41.2% x 48.8%

THe first step is to calculate the law of the number of car that pass on the street in 5s.

It can be shown that it follow a Poisson law with $\lambda = 5\times\frac{8.6}{60} \simeq 0.7166$

Then the number of car that pass on the street in 5s is

$$P(X = k) = \frac{\lambda^k}{k!}e^{-\lambda}$$

So the probability that no car pass in the street in the next 5s is

$$P(X=0) \simeq e^{-0.7166} \simeq 48.8 \%$$


How to understand reproducibility in statistics? I'm having problems to find info about reproducibility in statistics. The books I have with an starting level in statistics doesn't even mention it.

At college, one of my previous professors explained it like something you just have to remember. She gave me a list of distributions with this property and the parameter that is involved with reproducibility. I don't like that because I have a really bad memory, in fact, I don't remember that list anymore. I think if I can understand properly why this happens and maybe reading a demonstration I will remember it much better and I will be able to use it faster.

I wonder if I'm using the wrong name for this property. I'm talking about:
$$
X_i\sim N(\mu,\sigma^2) \quad\forall i=1,\ldots,n \implies \sum X_i \sim N(n\mu,n\sigma^2)
$$
Some distributions only fulfill this property for one of its two parameters and some others doesn't fulfill it. I would like to know where is this property coming from.

Thanks

I haven't heard it called reproduceability, but the property you are referring to appears to be that of stable distributions:

independent sums of random variables from a stable distribution family will remain in that family, within a location-scale transformation.

I would call that "linearity" rather than "reproducibility".  A function f(x) of a single variable is linear if and only if f(ax+ by)= af(x)+ bf(y).


Prove that $\sqrt{3}$ is not a rational number There is a similar question however that question asks why $3 |p^2$. Here the question is about $ 3 | p^2 \rightarrow 3 | p$.

It is a simple exercise (1.2.1) from Abbot's "Understanding Analysis".

$\nexists p,q \in \mathbb{N} : \left(\dfrac{p}{q}\right)^2 = 3$

$p$ and $q$ have no common factors, otherwise they would cancel each other out.

Contradiction:

$p^2 = 3q^2$

The troublesome part for me is

  From this, we can see that $p^2$ is a multiple of 3 and hence $p$ must
  also be a multiple of 3.

Why is that?

The proof goes on with

$p = 3 r$

$q^2 = 3r^2 \rightarrow q = 3 \lambda$.

It concluses now that $q$ and $p$ have common factors, which invalidates the original statement.

How can I prove $p^2 = 3q^2 \rightarrow p = 3 r$? If this is not true, then I don't see how the contradiction proves the statement, since it only invalidates it for such $p,q$ that have common factors.

What if $p^2 = 3q^2$ and $p \ne 3 r$?

If $p^2=3m$ for some $m\in \mathbb Z$, then $p=3k$ for some $k\in \mathbb Z$. Suppose otherwise. By Division Algorithm, $p=3k+1$ or $p=3k+2$. Squaring $p$, we get either $9k^2+6k+1$ or $9k^2+12k+4$ which are both not multiples of $3$, contradicting the assumption. Thus, $p$ has to be a multiple of $3$.

Alright, let me explain troublesome part.

Notice, we have $$p^2=3q^2\tag 1$$ shows that the $p^2$ is divisible by $3$ which shows that $p$ is also divisible by $3$  (by fundamental theorem)

Hence, we have $$p=3m\tag 2$$ where, $m$ is some integer

Now, substituting $p=3m$ in (1), we get $$(3m)^2=3q^2$$$$\iff 3m^2=q^2$$

now, from above equation,  we conclude that $q^2$  is divisible by $3$ which shows that $q$ is also divisible by $3$ (by fundamental theorem)

then we have $$q=3n\tag 3$$

Now, from (2) & (3), we find that $p$ & $q$ have a common factor $3$ which is a contradiction.


Is it always possible to find a $t>0$, such that $\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|dx<C~~~?$ Is it always possible to find a $t>0$, such that

$$\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$

where $C$ is independent of $n$.

Here is my idea:

We know that

\begin{align}
\sum_{k=1}^{n}\cos kx&=\frac{\sin(n+\frac{1}{2})x}{2\sin\frac{1}{2}x}-\frac{1}{2}\\
&=\frac{\sin nx \cos\frac{1}{2}x+\cos nx\sin\frac{1}{2}x}{2\sin\frac{1}{2}x}-\frac{1}{2}\\
&=\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}+\frac{\cos nx}{2}-\frac{1}{2}
\end{align}
So we have
\begin{align}
\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx\leq\int_{0}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|+1\,dt
\end{align}
The key is that, could we find a constant $C$(just independent of $n$),such that
\begin{align}
\int_{0}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|\,dt<C~?
\end{align}
It is an improper integral. If
$$\lim_{\varepsilon \to 0^+}\int_{\varepsilon}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|\,dt \to \infty$$
Then could we find an interval $[a,b](a>0)$, such that
$$\int_{a}^{b}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$

$$\vert \cos kx \vert \leq 1 \implies \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq n \implies \int_0^t \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq nt \implies nt < C \implies t < \frac{C}{n}$$


Could PI have a different value in a different universe? The value of pi is determined by the circumference of a circle.
Why is it any particular constant number? Would a circle as defined as a perfect circle in any universe lead to a different value of pi?
Would all universes where a circle could be constructed by "people" there also lead to the value of pi?
If it is true then it leads to the conclusion that pi is some sort of constant value constant to all universe. What is the meaning of that?

Science fiction references.
In science fiction pi sometimes has a different value in different universes, for example Greg Bear's "The Way", it says "Gates are capped with cupolas formed from Space-time itself. As distortions in space-time geometry, their nature can be calculated by 21st century instruments laid on their 'surfaces'. The constant pi, in particular, is most strongly affected.".
A message is found encoded within pi, in the novel by Carl Sagan, "Contact" "Ellie, acting upon a suggestion by the senders of the message, works on a program which computes the digits of pi to record lengths in different bases. Very far from the decimal point (1020) and in base 11, it finds that a special pattern does exist when the numbers stop varying randomly and start producing 1s and 0s in a very long string.".

Physically, the ratio of a circle's circumference to its diameter $C/d$ is not really $\pi$. General relativity describes gravity in terms of the curvature of spacetime, and roughly speaking, if you take $(C/d-\pi)/A$, where $A$ is the circle's area, what you get is a measure of curvature called the Ricci scalar.

But even if you're doing general relativity, you don't just go around redefining $\pi$. The thing is, $\pi$ occurs in all kinds of contexts, not just as $C/d$. For instance, you could define $\pi$ as $4-4/3+4/5-4/7+\ldots$, which has nothing to do with the curvature of space.

So if you define $\pi$ as $C/d$, you don't even get a consistent value within our own universe, whereas if you define it as $4-4/3+4/5-4/7+\ldots$, you get an answer that is guaranteed to be the same in any other universe.

Another way of looking at it is that $\pi$ is not the $C/d$ ratio of a physical circle, it's the $C/d$ ratio of a mathematically idealized circle that exists in certain axiomatic systems, such as Euclidean geometry. Viewed this way, it doesn't matter that our universe isn't actually Euclidean.

What if every alternate universe had its own unique frequency f and/or wave number k? Then π = kc/2f. The values generated might be very close to our π, but be different in say, the hundredth or thousandth decimal digit (a 7 instead of a 6). This would have no effect on the observable universe, as most sources say that only the first 40 or so decimal digits are critical in defining the universe. This would allow for variations between alternate universes while keeping all other factors the same.


How do you prove that the following sequence is well defined and convergent How do you prove that the following sequence is well defined and convergent?

$a_{n+1}=\sqrt{2+\sqrt{a_n}}$ with $a_1=0$.

You can also solve a simple quadratic equation.  It should be clear that the sequence converges to solution the positive solution to:
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}},$$
giving
$$x=\sqrt{2+x},$$
or
$$x^2=2+x,$$
whose positive solution is $x=2.$


$8$ billion people in the world, increase each year by $2\%$, how many people will be in $9$ months? I have the following ODE problem :

$8$ billion people in the world, increase each year by $2\%$, how many people will be in $9$ months?

What I did :

The change each year is : $$x'(t)=0.02(x(t))$$

Therefore $$x(t)=Ce^{0.02t}$$

In $t_0$ we know that $x(t_0)=8$, therefore $C=8*e^{-0.02t_0}$

So I got that $$x(t)=8*e^{(0.02(t-t_0))}$$

The number of people in $9$ months are $t-t_0=3/4$

Therefore I got : $$x(t)=8*e^{0.02(0.75)}=8.1209$$

In the text book the right answer is $8.1197$

I don't understand what's wrong with my solution.

Any help will be appreciated, thanks.

You don't have to assume the exponential form to begin with. In fact, if this is an exercise on ordinary differential equations, then you shouldn't in fact assume the form.

Instead, recognise the basic behaviour - this is a continuous growth model where the rate of increase of population $\frac {dP}{dt}$ is proportional to the present population $P$. Let the constant of proportionality be an unknown $k$ (which is to be determined).

So you start with $\frac{dP}{dt} = kP$, solve it to get $\ln P = kt + C$, where $C$ is another constant (of indefinite integration), giving $P = Ce^{kt}$.

Now, you know that when $t=0$ (at the start), the population is $P_0$, the initial population. So $P_0 = C$, and the solution can be written $P = P_0e^{kt}$.

You're also told that when $t=1$, $P = 1.02P_0$ (that's how you use the $2 \%$ increase given in the question).

So $1.02P_0 = P_0e^{k}$, giving $k = \ln{1.02}$.

Your final equation is therefore $P = P_0e^{t\ln{1.02}} = P_01.02^t = (8 \times 10^9)1.02^t$, into which you can substitute $t=0.75$ to get the correct required value.

So, pro tip: don't start out assuming the constants in your ODE, get a general form first then work out the constants using what's given.

There are two types of exponential functions:
$$\text{ordinary: } S=P(1+r)^t \ \ \text{and} \ \ \text{natural: } S=Pe^{rt}.$$
You used natural:
$$S=8\cdot e^{0.02\cdot \frac{9}{12}}=8.1209045,$$
whereas the textbook used ordinary:
$$S=8\cdot (1+0.02)^{\frac{9}{12}}=8.119702.$$
The population grows continuously, so the natural exponential function is more realistic.


Inverse and composite functions. I do not understand the following.

  If $f$ is a bijection, the $f \circ f^{-1}$ is the identity function on $Y$and $f^{-1} \circ f $ is the identity function on $X$.

I just thought that the identity function is essentially $y=x$?

What does it mean for the ' identity function to be on $Y$ or $X$?'

They mean that it becomes a function from $X$ to $X$ (for the second identity function), or from $Y$ to $Y$ (for the first identity function).

For example, for the second one: $(f^{-1} \circ f) (x) = x$ for any $x \in X$, hence it maps any $x \in X$ to $x$ itself. Hence it is the identity function from $X$ to $X$, which they say is a function 'on' $X$ itself: any function from $X$ to $X$ itself is a function 'on' $X$, whether it is an identity function or not.

Identity function maps an element to itself, so it is not possible to exist an identity function from X to Y. The sentence "identity function to be on Y or X" means that it depends on your domain. If your domain is X then you should consider the function f and that is why the identity map is from X to X.


Inertial Frames of Refereence I am told that in Newtonian mechanics, no coordinate system is "superior" to any other. Also, all inertial frames are in a state of constant, rectilinear motion with respect to one another.

So am I right to understand that "inertness of coordinate systems" is an equivalence relation on all the coordinate systems in a space. Furthermore, one should not talk of an inertial coordinate system on its own. In order to talk about inertness one has to choose two coordinate systems and compare them. Finally no equivalence class is superior to another, whatever superior means in its usage in the first paragraph, to which meaning I am not knowledgeable.

If any of this is not true, please include an example as well.

The statement that no coordinate system is superior to any other would appear to be the source of confusion.  I'm not altogether sure what is meant by that statement; it does not reflect how Newtonian mechanics has historically been understood and practiced.  Most likely it is intended to reflect a more recent understanding that reflects the development of special and general relativity, but even in the more modern context, I'm not sure what it is trying to say.

The word "inertial" in the term "inertial frame of reference" indicates that the frame is one in which the law of inertia, also known as Newton's first law, holds.  That is, it is a frame in which a body not acted on by any forces will remain in the same state of motion, experiencing no acceleration.

One therefore does not speak of one frame being "inertial" with respect to another.  Rather, there is a set of preferred frames, the inertial frames, which are all moving with constant rectilinear motion with respect to one another.  In all other frames of reference, a body not acted on by any forces will undergo accelerations.  These can be understood in terms of centrifugal and Coriolis forces, but such forces are considered "fictitious" since they are not caused by interactions with other physical bodies, but instead are due to the choice of coordinate system.

It might seem that Newton's first law is a special case of Newton's second law ($F=ma$) in which the force on a body is zero, but I don't believe that this is how Newton viewed things.  Rather, the first law is making a nontrivial statement about the physical world, namely that inertial frames exist.  Assuming that forces are due to other bodies and that force falls off with distance, a body sufficiently far from all other bodies could be used as a test body to determine whether a frame is inertial.  Once such an inertial frame is found, all frames moving with constant velocity with respect to the first will also be inertial, and all frames that are accelerating with respect to the first will not be inertial.  A big question is whether one can find any inertial frames at all: will all test bodies agree on whether a frame is inertial?  Newton's first law is asserting that this is so.

This is true for both Newtonian mechanics and special relativity.  The transformation between inertial frames is different in the two.  One inertial frame can be much more convenient than others for calculation, but that does not make it superior in theory.


Find the equation of normal line to the graph $y=2(x-1)^3$ Find the equation of normal line to the graph $y=2(x-1)^3$  at the point where $x=\frac12$.

So far, I found the derivative:

$$\frac{dy}{dx}= 6(x-1)^2 $$

What to do next?

The normal line is perpendicular to the tangent line.  Therefore, its slope is the negative reciprocal of the tangent line.

To find the slope of the tangent line at the point where $x = 1/2$, you evaluate the derivative of the equation $y = 2(x - 1)^3$ when $x = 1/2$.

As you found, the derivative is

$$y' = 6(x - 1)^2$$

Hence, when $x = 1/2$, the tangent line has slope

$$y' = 6\left(\frac{1}{2} - 1\right)^2 = 6\left(-\frac{1}{2}\right)^2 = 6\left(\frac{1}{4}\right) = \frac{3}{2}$$

Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the normal line has slope $-2/3$.

When $x = 1/2$,

$$y = 2(x - 1)^3 = 2\left(\frac{1}{2} - 1\right)^3 = 2\left(-\frac{1}{2}\right)^3 = 2\left(-\frac{1}{8}\right) = -\frac{1}{4}$$

so the tangent line and normal line both pass through the point $(1/2, -1/4)$.

If you use the point-slope form of the equation of a line

$$y - y_0 = m(x - x_0)$$

you obtain

$$y + \frac{1}{4} = -\frac{2}{3}\left(x - \frac{1}{2}\right)$$

for the equation of the normal line.


What is the maximum value of $k$ so that $x^2+4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2-x\sec{x}\csc{x})=0$ has real roots. What is the largest value of $k$ such that the equation $$x^2+4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2-x\sec{x}\csc{x})=0$$ has real roots?

I tried to find out the roots in terms of $k$.
But I couldn't.

How do we approach this problem without finding the roots actually?

I would suggest plugging your equation into the quadratic formula to get

$$ x = \frac{1}{2} \sqrt{(-16)(\sin^2 x \tan^2 x + \cos^2 x \cot^2 x +k^2 - x \sec x \csc x)}$$

From here we see we need

$$(-16)(\sin^2 x \tan^2 x + \cos^2 x \cot^2 x +k^2 - x \sec x \csc x) \gt 0$$

to have real roots.  Rearranging $\to$

$$ k^2 \lt  - \sin^2 x \tan^2 x - \cos^2 x \cot^2 x + x \sec x \csc x$$


Has the polynomial distinct roots? How can I prove it? I want to prove that the polynomial

$$
f_p(x) = x^{2p+2} - cx^{2p} - dx^p - 1
$$

,where $c>0$ and $d>0$ are real numbers, has distinct roots. Also $p>0$ is an even integer. How can I prove that the polynomial $f_p(x)$ has distinct roots for any $c$,$d$ and $p$.

PS: There is a similar topic that How to prove that my polynomial has distinct roots?

Here is a partial solution that shows there can be no multiple REAL roots. The proof doesn't work for complex numbers though (and I am not sure the result is even true in complex numbers).

$f' = (2p+2) x^{2p+1} - 2pcx^{2p-1} - pdx^{p-1}$. If $x$ is a multiple root of $f$, then both $f$ and $f'$ vanish at $x$. But $(2p+2) f - xf' = -2cx^{2p}-(p+2)dx^p-(2p+2) < 0$ for all $x \in \mathbb R$ because $p$ is even and $c$, $d$ and $p$ are all positive.

‘All’ you have to do is computing the g.c.d. of $f_p(x)$ and $f'_p(x)$ via the Euclidean algorithm. A polynomial has only simple roots (in $\mathbf C$) if and only if this polynomial and its derivative are coprime.


Best practice book for calculus I tried with every inch in me to not ask a question such as this but I just couldn't resist asking this.

  What is the best Calculus practice book?

I tried looking around but couldn't find a decent consensus on one book or even a few. It would be great if someone could point me towards a book that has both challenging and fun problems and furthermore, allows the student to build a strong foundation while doing those problems.

Thanks a bunch!

Collection of problems in analysis by B.P. Demidovich
Problems in Mathematical Analysis  B.P. Demidovich

These books are very famous here in Brazil. It's very organized and easy to get addicted.

Hard and imaginative problems the whoascum county problem book
the green/red book of math problems.


Find multiple ODE solutions (counter example) I want to find some counter examples for the uniqueness of this ode
$$ \frac{dy}{dt} = t \sqrt{1-y^2}$$

and initial condition: $y(0)=1$
I don't have idea of how I should start to think, may someone give me a (some) clue(s)?

$y=1$ and $y=\sin\left(\frac{1}{2}t^2+\frac{\pi}{2}\right)$ both satisfy the initial condition $y(0)=1$

This is a separable DE:

$$ \frac{dy}{dt} = t \sqrt{1-y^2}$$

$=\frac{1}{\sqrt{1-y^2}} dy=t dt$

$sin^{-1}(y)+d=\frac{t^2}{2}+c$     (by integration)

$y= sin (\frac{t^2}{2} +a)$

Satisfy the equation for any $a$


Calculating the square root of an integer as an integer with the smallest error I was recently asked to write an algorithm to find the square root of a given integer, with the solution expressed as an integer with the smallest error.

It was unclear to me whether this meant I need to return the integer closest to the actual square root, or return the integer such that, when squared, returned the closest value to the target number. When asking for clarification, I was told these are the same.

This got me thinking however, are these really equivalent?

Clearly, when deciding between two real approximations, minimizing the error of the square roots isn't the same as minimizing the error of the squares. For example, if I had to pick between 1.9 and 2.096 as an approximation of the square root of 4, then 2.096 would be the closest to the actual square root (error of 0.096 compared to 0.1), while 1.9 would provide the smallest error when squared (error of 0.39 compared to 0.393216).

For the integer approximation case, it doesn't appear obvious to me why the solutions to the two cases will always be the same. I've experimentally verified it to be so for the first 200 billion integers. Can anyone provide an intuitive or formal proof of this?

Suppose we want to approximate $\sqrt x$ with $n^2<x<(n+1)^2$.

If we minimize the error of the squares, we're going to choose $n$ or $n+1$ according to whether $x>\frac{n^2+(n+1)^2}2 = n^2 + n + \frac12$.

If we minimize the error of the square root, we should choose $n$ or $n+1$ according to whether $\sqrt{x}>n+\frac12$, which is the same as $x > (n+\frac12)^2 = n^2+n+\frac14$.

Because $x$ is an integer, being larger than $n^2+n+\frac12$ is the same as being larger than $n^2+n+\frac14$.

If $k-{1\over2}\lt\sqrt n\lt k+{1\over2}$, then

$$(k-1)^2=k^2-2k+1\lt k^2-k+{1\over4}\lt n\lt k^2+k+{1\over4}\lt k^2+2k+1=(k+1)^2$$

To flesh this out, the fact that $k$ and $n$ are integers implies

$$k^2-k+1\le n\le k^2+k$$

If $n\gt k^2$, then $n-k^2\le k$ while $(k+1)^2-n\ge k+1$.  Likewise, if $n\lt k^2$, then $k^2-n\le k-1$ while $n-(k-1)^2\ge k$.  In either case, $n$ is closer to $k^2$ than it is to either $(k-1)^2$ or $(k+1)^2$.


In the following figure, $AD$ is the bisector of $\angle A$.Prove that: $\angle CBA = \angle DAB$ In the following figure, triangle $ABC$ is inscribed in circle $C$ and $AD$ is the bisector of  $\angle A$.Also it's known that: $AD=BC$.Prove that: $\angle CBA = \angle DAB$

I tried as follows:
It's obvious that $\angle DBC=\angle CAD$ , so $\angle DBC=\angle DAB$.Now it remains to show that: $\angle DBC=\angle CBA$. Maybe triangle $ABC$ and $ABD$ are equall(they have two equal sides and $\angle ADB=\angle ACB$) but I don't see another equal pair of angles between them!

But this is easy. Since $AD=BC$ we have $\angle BAC = \angle ABD$. Since $\angle CAD = \angle CBD$ we have:

$$ \angle CBA = \angle DBA -\angle DBC = \angle BAC -\angle CAD = \angle DAB $$

And we don't need angle bisector.

$\angle ACB= \angle ADB$ since they are angles in the same segment. $BC=AD$ by your ratio condition. Finally since $C,D$ are on the same side of $AB$, $\angle CAB$ and $\angle DBA$ must both be acute or both obtuse. So $\triangle ABC \cong \triangle BAD$, in particular, $\angle CBA=\angle DAB$.


Infinitely many primes of the form $3k-1, \ k\in \mathbb{N}$ How to prove that there exist infinitely many primes of the form $3k-1, \ k\in \mathbb{N}$?

I know the proof that there exist infinitely many primes of the form $3k+1$, but I don't know how to do it in this case.

I tried to use:

Let $p_1,...,p_n$ be primes of the form $3k-1$. Then $N:=(3p_1\cdot \cdot \cdot p_n)-1$ is of the form $3k-1$ and $N$ must have at least one prime divisor of the form $3k-1$.

But $N$ is not divisible by any primes $p_1,...,p_n$.

So $N$ is a prime of the form $3k-1$ outside $p_1,...,p_n$.

This procedure can be continued indefinitely, so there exist infinitely many primes of the form $3k-1$.

Is this correct? Or is there something missing or wrong?

As 3 and 2 are coprime, by Dirichlet's theorem on arithmetic progressions there infinitely many prime numbers of the form 3*n+2, n>=0. By changing n by k-1, you obtain the desired result.


If $F$ is an algebraically closed field, then every maximal ideal of $F[x_1,...,x_n]$ is of the form $(x_1-c_1,...,x_n-c_n)$ I have problem to understand all the proof of the theorem :

  If $F$ is an algebraically closed field, then every maximal ideal of $F[x_1,...,x_n]$ is of the form $(x_1-c_1,...,x_n-c_n)$.

Proof : Let $\mathfrak m\subset F[x_1,...,x_n]=:R$ be a maximal ideal of $R$ and let $$\varphi: R\longrightarrow R/\mathfrak m=:k$$
the projection. Since $\mathfrak m$ is maximal, $k$ is a field. Since $\mathfrak m$ is a proper ideal, $1\notin \mathfrak m$, and thus $F\cap \mathfrak m=(0)$.

Q1) $\mathfrak m$ is a proper ideal of $R$, not of $F$, so how can we have $F\cap \mathfrak m=(0)$ since $\mathfrak m$ is even not in $F$ ?

In particular, $\varphi|_F$ is an injection. Let $\bar x_i$ the residue classes of $x_i$ in $k$. Then, using $\varphi|_F$, we have that $k$ is the extension of $k$ by elements $\bar x_1,...,\bar x_n$.

Q2) I don't understand this fact. I agree that $\varphi|_F$ is a field homomorphism, but why (and what does it mean) that $k$ is the extension of $F$ by the elements $\bar x_1,...,\bar x_n$ ?

Since $F$ is algebraically closed, if $\bar x_i$ is algebraic over $F$, then it's contained in $F$. By reordering the variable, we may assume the the first $j$ are not algebraic, that is $k=F[\bar x_1,...,\bar x_j]$ where $\bar x_i$ is tanscendental of $f$ for $i=1,...,j$.

If $j=0$, then $k=F$. Then, let $c_i$ the residue classes of $x_i$ in $k\cong F$. Then, $\mathfrak m=\ker \varphi\supset (x_1-c_1,...,x_n-c_n)$, and since $\mathfrak m$ is maximal, we get $m= (x_1-c_1,...,x_n-c_n)$.

Q3) I don't understand why $m=\ker \varphi\supset (x_1-c_1,...,x_n-c_n)$.

For the rest, it's fine for the moment.

we know that any ideal of the form $(x_1-c_1,...,x_n-c_n)$ is maximal,
and $F \cong k$ .Let $b_i=x_i(mod m) \in k$ and $a_i=\varphi^{-1}(b_i)$ then $x_i-a_i \in ker\varphi=m $ so $(x_1-c_1,...,x_n-c_n)\subseteq ker\varphi$. and from kronecker theorem: we have that $k$ is extension of $F$. if $F\cap m \ne (0)$ then $m=F$ contradiction.


Divisibility by $2$ and powers of $3$ While dabbling in a similar but different problem, I sort of noticed this.

Is it true that $3^n-1$, for all odd $n\in\mathrm{N}$, is no more than once divisible by $2$? If so, is there a conclusive proof for this statement? In other words, if:
$$3^n-1=2a$$
where $a\in\mathrm{N}$, is it true that $a$ must be odd too?

I attempted to disprove this in the following way-

Consider that $n=2k+1$, and $j$ and $a$ where $k,j,a\in\mathrm{N}$ and:
$$\begin{split}
3^n-1&=3\cdot3^{2k}-1\\
&=2^ja
\end{split}$$
Using induction on $k$ here led to the following:
$$ 3\cdot3^2\cdot3^{2k}-1=2^{j}\cdot3^2\cdot a+8$$
which doesn't really put a bound on $j$. I haven't been able to find a counterexample yet which makes me think that the above isn't correct.

So is the statement correct? Please provide an answer with a proof, if possible.

Well, $3^n-1=2(1+3+3^2+...+3^{n-1})$

When $n$ is odd there are odd number of odd numbers plus together in the bracket.

$3^n-1\equiv 0 \pmod 4$ if $n$ is a power of $2,n\gt1$.


Is $|1-i|$ larger than $|1|$? I am confused about complex numbers. Does $1-i$ lie outside the unit circle? How do I show that the absolute value of $1-i$ is larger than that of $1$?

Let me do something very different for me. Let me give a geometric proof:

Note that $1-i$ is the bottom-right corner of the square whose center is $0$ and whose edges have length $2$. The other corners are $1+i; -1+i; -1-i$.

The diagonal running from $-1+i$ to $1-i$ is a straight line passing through $0$. Its length, by the Pythagoras theorem, is $2\sqrt2=\sqrt8$. Therefore the distance between $0$ and each of the corners is exactly half, i.e. $\sqrt2$.

And it is trivial to see that $\sqrt2>1$.

Here is a drawing:

$\hspace{5cm}$

Here is another different approach:

The unique angle (in $[0,2\pi)$) between the vector $(1,i)$ in the complex plane and the x-axis is $\phi=\frac{7\pi}{4}$

Euler's identity yields now:

$$1-i=L(e^{\frac{7\pi}{4}i})=L\left(\cos\left(\frac{7\pi}{4}\right)+i\sin\left(\frac{7\pi}{4}\right)\right)=L\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)=L\left(\cos\left(\frac{\pi}{4}\right)-i\sin\left(\frac{\pi}{4}\right)\right)=L\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right) $$

Hence $L=\sqrt{2}$


1 = 0 finding the mistake I was shown the following calculation. There is clearly something wrong but I can not see the mistake. Could someone point out the wrong step or something?

$$ \int \frac{1}{x} dx = \int 1 \cdot \frac{1}{x} dx$$

Let's say $ u = \frac{1}{x}$ and $ dv = 1\ dx$.
So $ du = -\frac{1}{x^2} \ dx$ and $ v = x$.
Putting that into per partes formula:
$$ \int \frac{1}{x} dx = \int 1 \cdot \frac{1}{x} dx= x \cdot \frac{1}{x} - \int x \cdot (-\frac{1}{x^2})dx = 1 + \int \frac{1}{x} dx$$
So we can substract $\int \frac{1}{x}dx$ from both sides which then gives us
$$0 = 1$$

You are forgetting the constant of integration. When you write $\int\frac1xdx$, that could represent any of a family of functions. So in this case, the left-hand $\int\frac1x dx$ represents an antiderivative of $\frac1x$ which is $1$ larger than the $\int \frac1xdx$ on the right-hand side. This also means that you cannot "subtract $\int\frac1xdx$" as easily as you might think.

The moment you put bounds on your integrals, everything changes. In that case (assuming the bounds are valid), we get
$$
\int_a^b\frac1xdx = 1\Big|_{x = a}^b + \int_a^b\frac1xdx
$$
and we see that the $1|_a^b$ term disappears, and all is well.


Finding solutions of modulus functions I decided to do some practice with some functions, and was posed with the following question:

So, I sketched the two graphs. For convenience I'll display a photo of them from Desmos.

The blue line is |$3x - 2$| and the red is |$x-5$|.
Now, to find when the two intersect, it the case where the two equations have the same output with the same input is achieved.
However, the modulus sign seemed to trip me up when I was writing |$3x - 2$| = |$x-5$| and begin doing algebra. I couldn't merely solve it like the function had no modulus sign, since that is a different function. So, I figured by inspection that only the left side of the red function makes the intersections, so the following conditions are the only valid ones.
$$-x-5 = 3x-2$$
$$-x-5 = -3x+2$$
And with that I can find the inputs necessary.
However, is there a general way to solve this type of problem? Say, if I didn't have the graph to make the inference that I did? How could I solve this problem if I couldn't graph the two functions?

You need to consider the plus-minus signs of both the equations. In the above you have considered only one. You should solve:

`$\begin{alignat*}{1}
-x-5 &= 3x-2\\
-x-5 &= -(3x-2)\\
-(-x-5) &= 3x-2\\
-(-x-5) &= -(3x-2)
\end{alignat*}$

Just square both side, and you will get a quadratic equation. Factorise it, and you'll get the answer.


Solve $x = \frac{1}{2}\tan(x)$ I did this using trial and error, but I was just wandering if there is an algebraic way of solving this?

I thought about double angle formula but that doesn't work properly does it?

I then tried writing it in the form $x \cdot 2 \cot(x) = 1$ but even then, I can't solve it properly by re-writing $\cot(x)$ because of the $x$ outside the brackets.

Anybody know how I would do this algebraically?

Of course, the roots are transcendantal numbers, which cannot be expressed in terms of a finite number of elementary functions. But they can be expressed on the forme of series :

It cannot be solved algebraically as it is a transcendental equation.


How to find a affine function$\ \leq \ $convex function? let $f:\mathbb R^n \rightarrow \mathbb R$ be a convex function.

for arbitrary $p\in \mathbb R^n$, how can I find an affine function $\ell_p(x)$ such that

$$\ell_p(x)\leq f(x)\ \text{ and }\ \ell_p(p)=f(p)?$$

I could find it for $n=1$, but I don't know what to do when $n\geq2$.

Taking the axial sections we arrive at convex functions of one variable. A convex function of one variable admits one-sided derivatives at every interior point of a domain. That is why all one-sided parial derivatives do exist at $p$. It is enough to take a plane described by @Jack D'Aurizio with, say, left-hand-side partial derivatives at $p$.

EDIT

As @JonnMa said, it is not enough. See my comment below.


Proving left-invariance (and proof-verification for right-invariance) for metric constructed from left-invariant Haar measure $\newcommand{\diff}{\mathrm{d}}$
TL;DR

Having read this I know something about Haar measures, in particular that a left-invariant one exists and is unique on any Lie group $G$.
I know that defining:

$$\langle x,y\rangle_g=\langle(\mathrm{d}_gL_g)^{-1}(x),(\mathrm{d}_gL_g)^{-1}(y)\rangle,$$

where $\langle\cdot,\cdot\rangle$ is any inner product on the Lie algebra $T_eG$ and $L_g$ is the map $L_g(h)=gh$, one obtains a left-invariant metric for $G$.
I know that if $G$ is compact we can define the metric:

$$(u,v)_g=\int\limits_G\langle\diff_gR_h(u),\diff_gR_h(v)\rangle_{gh}\diff\mu(h),$$

where $\mu$ is the unique left-invariant Haar measure on $G$;
I believe the following proves right-invariance for that metric:

  \begin{align*}
(\diff_gR_ku,\diff_gR_kv)_{gk}={}&\int\limits_G\langle\diff_{gk}R_h\diff_gR_ku,\diff_{gk}R_h\diff_gR_kv\rangle_{gkh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(R_h\circ R_k)u,\diff_g(R_h\circ R_k)v\rangle_{gkh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_gR_{kh}u,\diff_gR_{kh}v\rangle_{gkh}\diff\mu(h)=\int\limits_G\langle\diff_gR_\ell u,\diff_gR_\ell v\rangle_{g\ell}\diff\mu(k^{-1}\ell)={} \\
{}={}&\int\limits_G\langle\diff_gR_\ell u,\diff_gR_\ell v\rangle_{g\ell}\diff\mu(\ell)=(u,v)_g.
\end{align*}

  Steps thus justified:

  Definition of the metric.
  $\diff_g(R_h\circ R_k)=\diff_{R_kg}R_h\circ\diff_gR_k=\diff_{gk}R_h\circ\diff_gR_k$ by the chain rule.
  $R_h\circ R_k=R_{kh}$.
  Set $\ell=kg$ and change variables. $h=k^{-1}\ell$.
  This is where I use the invariance. I used it in the form $\diff\mu(k^{-1}\ell)=\diff\mu(L_{k^{-1}}\ell)=\diff\mu(\ell)$. Is that right?
  Definition of the metric again.

So my question is: is the proof in 4. correct? And how do I prove left-invariance?

(For those wishing to delve deeper into the history of this post and its earlier exact duplicate, see edit history)

$\newcommand{\diff}{\mathrm{d}}$
Converting @Daniel's comment to an answer to get this question answered.

The proof of 4 is, he says, correct.

As for left-invariance, I merely have to exploit the fact that $L_g\circ R_k=R_k\circ L_g$, and do the following:

\begin{align*}
(\diff_gL_ku,\diff_gL_kv)_{kg}={}&\int\limits_G\langle\diff_{gk}R_h\diff_gL_ku,\diff_{gk}R_h\diff_gL_kv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(R_h\circ L_k)u,\diff_g(R_h\circ L_k)v\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(L_k\circ R_h)u,\diff_g(L_k\circ R_h)v\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_{gh}L_k\diff_gR_hu,\diff_{gh}L_k\diff_gR_hv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_gR_hu,\diff_gR_hv\rangle_{gh}\diff\mu(h)=(u,v)_g.
\end{align*}

The elimination of $\diff_{gh}L_k$ was done thanks to left-invariance of $\langle\cdot,\cdot,\rangle$. That this be "positive definite" is obvious, since the integrand will be strictly positive on the whole of $G$. As proved here, or with similar arguments, if I plug in smooth fields I get smooth functions in the integrand, and the integral of a smooth function is smooth. So this is indeed a biinvariant metric. Of course, this requires the integral to converge, which is guaranteed by the compactness, but it is not strictly required that the group be compact.

Finiteness of Haar volume of a compact group : let $U$ be an open neighborhood of identity. The open sets $g\,U$ for $g \in G$ form an open cover of $G$. Since $G$ is compact, this open cover has a fintie subcover. However, since the Haar volume is invariant, all the sets $g\,U$ have the same volume. Thus, we have bounded the volume of $G$ by an integer multiple of the volume of $U$, (the volume of $U$ is finite, of course!) -- Salem


Find all real to real function satisfy this functional equation.! $f((x+y)/(x-y))=[f(x)+f(y)]/[f(x)-f(y)]$ Find all real to real function satisfy this functional equation.!

$$f\left(\frac {x+y}{x-y}\right)=\frac {f(x)+f(y)}{f(x)-f(y)}$$

I couldn't get to the final answer but I get $f(0) = 0$ and $f(1) = 1$
and this functional is an odd function

It is easy to see that $f(0) = 0, f(1) = 1, f(-x) = -f(x),$ so we assume this.

We note that $$f\left(\frac{x+a x}{x-a x}\right) = \frac{f(x) + f(x a)}{f(x) - f(x a)}.$$
Factoring out by $x$ on the RHS, we see that $f(x a) = C f(x)$ (just solve on the RHS). In fact, if we set $x = 1,$ we see that $$f(a x) = f(a) f(x).$$ In particular, this means that $f(a^2) = f(a)^2,$ so $f$ is positive on $\mathbb{R}_+.$  What is more, by using the original condition, together with positivity of $f$ on positive numbers, we see that $f$ is order-preserving. Restricting to the positive real axis, we see that $f$ is an order-preserving homomorphism of $\mathbb{R}_+,$ and as is easy to see, this means that $f$ is of the form $x\rightarrow x^c,$ for some $c$ (see this discussion). Now, by plugging into the original constraint, it is easy to see that $c=1,$ so $f(x) = x.$

Have you tried finding very simple functions that satisfy this relation? For example, can you find a quadratic function $f(x) = ax^2 + bx +c$ that works? If so, broaden your horizons! If not, what went wrong, and can you fix it?

I have a hunch that such a function must belong to a very simple class of polynomials, but have yet to verify that these are the only functions that can work. I'm not certain that oddness (in full generality) plays much of a role.

Lastly, I don't see any reason why $f(1) = 1$.


Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ Can somebody find an elementary proof of the following identity:
$$
\arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2
$$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x} + \int \frac{\sqrt{1 - x}}{\sqrt{x}}
$$
where the first equality follows after applying an integration by parts with $f = \sqrt{x}$ and $\mathrm{d} g = 1/\sqrt{1 - x}$. For the sake of simplicity we omit $\mathrm{d}x$ in each integral. Adding the integral to itself:
\begin{align} 2I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} &= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&= -2 \sqrt{1-x}\sqrt{x} + \int\frac{1}{\sqrt{x}\sqrt{1-x}}\end{align}
The last integral on the RHS evaluates to $\arcsin(1 - 2x)$, so $$I = - \sqrt{1-x}\sqrt{x} + \frac{1}{2}\arcsin(1 - 2x) + C.$$
On the other hand, the integral can also be evaluated by applying a $u$-sub with $u = \sqrt{x}$. We find that:
$$
I = 2 \int\frac{u^2}{\sqrt{1 - u^2}} =2 \left(-u \sqrt{1 - u^2} + \int\sqrt{1 - u^2}\right)  = - u\sqrt{1 - u^2} + \arcsin u + C_2
$$
So then it follows that $I$ is also equal to $-\sqrt{x}\sqrt{1-x} + \arcsin{\sqrt{x}} + C_2$. Equate the results of the two methods and plug in a random point to find $C - C_2$ and the subsequent identity.

Here’s an elementary proof. Let $x=\sin^2\phi$. Then, since $\cos(2\phi)=1-2\sin^2\phi$, we have
$$\arcsin(1-2x)=\arcsin(1-2\sin^2\phi)=\arcsin(\cos(2\phi))=\frac{\pi}{2}-2\phi$$
and
$$2\arcsin(\sqrt{x})=2\arcsin(\sin\phi)=2\phi$$
From which the result follows elementarily (without calc).

Let $x=u^2$ with $0\le u\le1$ and rewrite the equation $\arcsin(1-2x)+2\arcsin(\sqrt x)=\pi/2$ as
$$\arcsin(1-2u^2)={\pi\over2}-2\arcsin u$$
Think of each side as an angle, $\theta_L$ and $\theta_R$. Clearly $-\pi/2\le\theta_L\lt\pi/2$, that being the range of the arcsine function.  Because $u$ is nonnegative, we have $0\le2\arcsin u\le\pi$, and hence $-\pi/2\le\theta_R\le\pi/2$ as well.  Thus to prove that $\theta_L=\theta_R$, it suffices to show that $\sin\theta_L=\sin\theta_R$. Obviously
$$\sin\theta_L=\sin(\arcsin(1-2u^2))=1-2u^2$$
while standard trig identities take care of $\theta_R$:
$$\sin\theta_R=\sin\left({\pi\over2}-2\arcsin u \right)=\cos(2\arcsin u)=1-2\sin^2(\arcsin u)=1-2u^2$$


Reflection of a circle Someone submitted this problem to me:

A physicist would argue that light is a wave and that it passes anyway.
However here we are dealing with a problem of pure geometry.
My first instinct was to try and find a way that it does follow some kind of easy rule, but it did not work.
My second attempt was to try and find a pattern when solving analytically the problem for low values of N reflections. Here is the Desmos simulation of my attempt: https://www.desmos.com/calculator/k9cuxxn4x5
It is very basic geometry with tangent calculation (calculation of lines intercepting a circle, calculation of tangent and calculation of reflexions), and the analytic geometry becomes complicated for large N, so I assume there is an easy way that I completely missed.
In this Desmos sheet I called a the distance between the top of the circle and the mirror.

Ray component on x-axis is always to the right.

When gap is zero, angles of incidence are progressively diminishing to converge towards zero  monotonously. Reflected rays are confined to the left of arbelos area and undergo infinitely many bounces but never leave the zero gap tangency point.

A positive gap allows a positive rightward horizontal component by laws of reflection and the rays converge before diverging. Reflected angles at fist reduce and then increase. That is, rays eventually escape. A rough typical ray-trace is sketched having forward components.


How many copies of P3 are there in K10 How many copies of P3 are there in K10?
I can draw both of the graphs, but I don't know how you calculate this and assume there is a method that can be used to make this easier.
Thanks

I assume $P_3$ is the graph

       *---*---*

An embedding of this into $K_{10}$ is simply a labeling of it with three different numbers from $\{1,2,\ldots,10\}$.

To choose such a labeling, first pick the label of the middle node; this can be done in $10$ ways. Then pick the numbers that will label the two end nodes; this can now be done in $\binom92$ ways (since there's an automorphism of $P_3$ that swaps the ends).

Just realised both mine and hennings answers are wrong. Purely because $P_3$ has 4 nodes not 3. It has 3 edges not 3 vertices.

Labelling the nodes of $P_3$  as {$1,2,3$} then there are $3$ 'triangles minus one edge' that you can make.

Using this you can say that for every $K_3$ in $K_{10}$ there are $3P_3$.

So the number of $K_3$ in $K_{10}$ is ($^{10}C_3$) which gives the number of $P_3$ as $3(^{10}C_3)=360=10(^9C_2)$ as Henning said.

$^{10}C_3$ is 10 choose 3


Showing the equation $Ax=b$ is well posed if $x$ is least squares solution of minimal norm The operator equation $Tf=g$ is said to be totally well-posed where $T:H_1\to H_2$ if for each $g\in H_2$.(H_1 and H_2 are Hilbert spaces.)
The equation has a unique solution and this solution depends continuously on both the "data" $g$ and the "model" operator $T$.

It's clear to me that when $A$ is invertible, the equation $Ax=b$ is totally well-posed. However, I have no idea why if we replace the notion of "solution" by "least squares solution of minimal norm", the matrix equation is only well-posed.(Not totally well-posed). I appreciate any hint.

Given $\mathcal{A} \in\mathbb{C}^{m\times n}$, the linear system
$$
 \mathbf{A} x = b
$$
we have existence of a least squares solution when $b\notin\mathcal{N}(\mathbf{A}^{*})$. The least squares minimizers are
$$
  x_{LS} = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A}\right) z, \quad z \in \mathbb{C}^{n}.
$$
The minimizers are an affine space, represented by the dashed red line in the figure below.

When $\mathcal{N}\left( \mathbf{A}\right)$ is trivial, the least squares solution is
$$
 x_{LS} = \mathbf{A}^{\dagger} b = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} b.
$$
When  the nullspace is trivial, $\mathcal{N}\left( \mathbf{A}\right) = \mathbf{0}$, and $\mathbf{A}^{\dagger} \mathbf{A} = \mathbf{I}_{n}$, and the least squares solution is unique. (Fredholm alternative.)


Number of bounces by an object in a unit square with velocity $\vec{(p,q)}$, where p and q are co-prime. Please bear with me.

Imagine the unit square in the plane to be a carrom board. Assume the striker is just
a point, moving with no friction (so it goes forever), and that when it hits an edge, the
angle of reflection is equal to the angle of incidence, as in real life. If the striker ever
hits a corner it falls into the pocket and disappears. The trajectory of the striker is
completely determined by its starting point (x, y) and its initial velocity $$\vec{(p, q)}$$.
If the striker eventually returns to its initial state (i.e. initial position and initial velocity),
we define its bounce number to be the number of edges it hits before returning to its initial
state for the first time.
For example, the traectory with initial state $$[(.5, .5);\vec{(1, 0)}]$$ has bounce number 2 and
it returns to its initial state for the first time in 2 time units. And the trajectory with
initial state $$[(.25, .75); \vec{(1, 1)}] $$has bounce number 4.

(a) Suppose the striker has initial state $[(.5, .5);\vec{(p, q)}]$. If p > q ≥ 0 then what is the
velocity after it hits an edge for the first time? What if q > p ≥ 0?

(b) Draw a trajectory with bounce number 5 or justify why it is impossible.

(c) Consider the trajectory with initial state $[(x, y); \vec{(p, 0)}]$ where p is a positive integer.
In how much time will the striker first return to its initial state?

(d) What is the bounce number for the initial state  $[(x, y);\vec{
(p, q)}]$   where p, q are relatively
prime positive integers, assuming the striker never hits a corner?

I have done (a), (b) and (c) and am stuck on (d).

Answer to (a) is $\vec{(-p,q)}$ and $\vec{(p,-q)}$.

Answer to (c) is $$\frac{2}{p}$$

Can you please tell me how to do it?

Thanks.

The usual way to do this sort of problem is using images.  For example, here we have a straight line from the centre of one lattice cell to another crossing $5$ edges.

This corresponds to a trajectory that goes from the centre of your square to itself after $5$ reflections.


What is the reciprocal of $(-1/2)^k$? What is the reciprocal of $(-1/2)^k$?

The answer is meant to be $2^{-k}$ as if you flip something upside down the power becomes negative. However, I am not sure what happens to the negative in front of the fraction.

The reciprocal of $x$ is simply $\frac1x$. So the reciprocal of $(-\tfrac12)^k$ is $\frac1{(-\tfrac12)^k}$.

You may also recall from the exponentiation laws that $(\frac1x)^y=x^{-y}$, and that the repciprocal of $x^y$ is $\frac1{x^y}=x^{-y}$.
This allows us to rewrite $(-\tfrac 12)^k$ as $(-2)^{-k}$, and then the reciprocal of this is $(-2)^k$. If $k$ is even, this simplifies to $2^k$; if $k$ is odd, it becomes $-2^k$.

The answer would look like $(-1)^k.2^{-k}$. The negative would depend if k is even or odd.


From 1D gaussian to 2D gaussian I read this:

  The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:

  g2D(x,y,$\sigma_1^2 + \sigma_2^2$) = g1D(x,$\sigma_1^2$)g2D(y,$\sigma_2^2$)

saying that the product of two 1 dimensional gaussian functions with variances $\sigma_1^2$ and $\sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.

I tried to deduce this by using:

g1D(x,$\sigma1^2$)g2D(y,$\sigma2^2$) = $\frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-x^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi}\sigma_2}e^{\frac{-y^2}{2\sigma_2^2}}$ = $\frac{1}{2\pi\sigma_1\sigma_2}e^{-(\frac{x^2}{2\sigma_1^2}+\frac{y^2}{2\sigma_2^2})}$

but I fail to obtain

$\frac{1}{2\pi(\sigma_1^2 + \sigma_2^2)}e^{\frac{-(x^2+y^2)}{2\sigma_1^2 + 2\sigma_2^2}}$

which is equal to g2D(x,y,$\sigma_1^2 + \sigma_2^2$).

Someone know how to get there?

Your first expression,

$G(x,\sigma_1^2) G(y,\sigma_2^2) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-x^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi}\sigma_2}e^{\frac{-y^2}{2\sigma_2^2}}$ = $\frac{1}{2\pi\sigma_1\sigma_2}e^{-(\frac{x^2}{2\sigma_1^2}+\frac{y^2}{2\sigma_2^2})}$

is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,

$\frac{1}{2\pi(\sigma_1^2 + \sigma_2^2)}e^{\frac{-(x^2+y^2)}{2\sigma_1^2 + 2\sigma_2^2}}$

is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.

It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $\sigma_1^2$ and $\sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $\sigma_1^2 + \sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.

Hint: You see in the exponential that we have
$$ \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right )=\exp \left (  \frac{-x^2 }{2( \sigma_1^2 + \sigma_2^2 ) } \right )\exp \left (  \frac{- y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right )  $$
The normalization hanging around should be
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) } $$
as you've seen...but under the change of variables
$$ z_1 = \frac{ \sigma_1x}{\sqrt{\sigma_1^2 + \sigma_2^2}}  \quad \& \quad z_2 = \frac{ \sigma_2y}{\sqrt{\sigma_1^2 + \sigma_2^2}}  $$
What do you obtain? Also note that
$$ 2 \pi = \sqrt{2\pi} \sqrt{2 \pi} $$
Edit: We see that
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) }  \int_{\mathbb{R}^2} \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right ) dxdy = $$
$$= \left[ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} }  \int_{\mathbb{R}} \exp \left ( - \frac{x^2 }{2( \sigma_1^2 + \sigma_2^2 ) }  \right ) dx \right] \left [ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} }   \int_{\mathbb{R}} \exp \left ( - \frac{y^2 }{2( \sigma_1^2 + \sigma_2^2 ) }  \right )dy  \right]$$
$$ =\left[ \frac{1}{\sqrt{2 \pi }\sigma_1 }  \int_{\mathbb{R}} \exp \left ( - \frac{z_1^2 }{2 \sigma_1^2   }  \right ) dz_1 \right] \left [ \frac{1}{\sqrt{2 \pi }\sigma_2 }   \int_{\mathbb{R}} \exp \left ( - \frac{z_2^2 }{2\sigma_2^2  }  \right )dz_2  \right] $$