Consecutive binary variables, without using auxiliary variables I have $i = 1,.....,n $ binary variables $x_{i}=\{0,1\}$ that I am using the model a resource allocation problem.

I am struggling with the constraint that if $\exists i\in[1,n]$ such that $x_{i}=1$ then $$\sum_{i=1}^n x_{i} \ge k,$$ $k\in \mathbb{R}$.

I have read through some similar problems like: How to model a consecutive binary constraint? but the suggestion is to add auxiliary variables. The value of $n$ is currently large and I would not like to increase it with more variables.

The way I read (or misread) your problem is that $$z = \sum_i x_i$$ where $z=0$ or $k \le z \le n$. This is the same as saying $z$ is a semi-continuous variable with bounds $k$ and $n$. Most MIP solvers support semi-continuous variables directly. Otherwise you can do:

$$\begin{align} & k\cdot \delta \le z \le n\cdot \delta\\ &\delta \in \{0,1\}\\ & 0 \le z \le n \end{align}
$$

Let $m=\lceil k \rceil$

If $m\leq 2$ and $n \geq m$

$\forall S \subset [1,n], |S| = n - (m - 1)$

$\displaystyle\sum_{i\in S}x_i - x_j \geq 0\;\;\;\forall j\notin S$
describe your problem (with the binary variables constraints)

If $x_j = 1$, $\displaystyle\sum_{i\in S}x_i - x_j \geq 0$
because there is only $m-1$ variables not in $S$

The 0 - vector also verify this constraint

If a vector does not verify your constraint, there is less than $m-1$ variables equal to 1, and there exist $x_j = 1$.

Then, we have $\displaystyle\sum_{i\in S}x_i - x_j < 0 $ where S has no variable equal to 1.

I think those constraints are all facets, except the ones coming from the binary variables.


For any $X\subset \mathbb{R}$, is $|X|<|\mathbb{R}|, |X|\leq|\mathbb{R}|,$ or $|X|=|\mathbb{R}|$? So take for example $(20,35) \subset \mathbb{R}$
$(20,35)$ has an infinite number of elements and is uncountable
$\mathbb{R}$ has an infinite number of elements and is uncountable
Can we find a function $f:(20,35)\rightarrow \mathbb{R}$ that maps every element of the domain to the codomain?
Or is the infinity posed by $|\mathbb{R}|$ bigger than the infinity posed by $|(20,35)|$?
Like, for example, we know that $|\mathbb{N}|<|\mathbb{R}|$ since for any $n \in \mathbb{N}$, we can find an $x \in \mathbb{R}:n<x<n+1$, thus $|\mathbb{N}|<|\mathbb{R}|$
I would like to think that since $(20,35)\subset\mathbb{R}$, that by design, $|(20,35)|< |\mathbb{R}|$ but the uncountability of the reals makes this dangerous to assume.

"Can we find a function f:(20,35)→R that maps every element of the domain to the codomain?" Yes.
If $g(x) = \frac {x-20}{15}$ then $g:(20,35)\to (0,1)$ is a bijjection.
And if $h(x)=\pi x - \frac \pi 2$ then $h:(0,1)\to (-\frac \pi 2, \frac \pi 2)$ is bijection.
And $\tan:(-\frac \pi 2,\frac \pi 2) \to (-\infty, \infty) = \mathbb R$ is a bijection.
So if $f= tan(\frac{x-20}{15}\cdot \pi - \frac \pi 2)=\tan(h(g(x))) $ the $f:(20,35)\to \mathbb R$ is a bijection.
SO $|(20,35)| = |\mathbb R|$.

By definition, a set is infinite if and only if it contains a proper subset with the same cardinality as itself. Put another way:
A set $K$ is infinite $\iff \exists L \subset M, L \neq M : |L|=|M|$
We can see examples of this with $2\mathbb{Z} \subset \mathbb{Z}$
We can find a bijective function $f: 2\mathbb{Z} \rightarrow \mathbb{Z}$ so that every element of $2\mathbb{Z}$ gets mapped to an element in $\mathbb{Z}$
In the same manner, with the uncountable reals,
For any nonempty proper subset $X \subset \mathbb{R},$ where $X$ is not countable,
Then $|X|=|\mathbb{R}|$
Here, also we can find a bijective function $f: X \rightarrow \mathbb{R}$ so that every element of $X$ gets mapped to an element in $\mathbb{R}$


A system of linear equations with solution $x=20$ and $y=20$ I am trying to come up with a system with solution $x=20$ and $y=20$. This is what I have now:

$$\begin{array}{|l} 3(x-3y)-(2x-3y)=-100 \\ 4(x-3y)+2(2x-3y)=-200\end{array}$$
The expected solution is to define $a=x-3y$ and $b=2x-3y$ and solve in this way. Do you think it's okay right now or the solution's obvious? How can I make it more complicated?

Your system is fine.

If you want to make it more complicated make it nonlinear.

For example $$ (2x-y)^2+(x+y)^2=2000, 3(2x-y)+(x+y)^2=1660$$

Simplifying your system we get
$$x-6y=-100$$
$$4x-9y=-100$$
Multiplying the first equation by $-4$ and adding to the second we get
$$150y=300$$ so $y=20$ as you stated.


Is this logic behind the derivative of $x^x$ a coincidence? As many of you probably know, when you take the derivative of $x^x$, you cannot treat it as either a exponential function ($a^x$) or like a function of the form $x^a$.

If you treat it like $x^a$ then you get an aswer of $xx^{x-1}$, which is just $x^x$.

Treat it like $a^x$, and you get $x^xln(x)$.

The correct derivative is $x^x(ln(x)+1)$, which you get by doing implicit differentiation after taking the natural log of both side.

This is the same as $x^xln(x)+x^x$, which happens to be the combination of the "wrong" derivatives you get when treating $x^x$ as the 2 different types of functions I specified earlier.

Is there any real logic behind this or is it coincidence. Also, could this also be applied to other function hybrids like $x^x$?

I think this is the answer you're looking for. Consider the function of two variables
$$f(x,y) = x^y = e^{y\ln x}.$$
Then
$$\frac{\partial f}{\partial x} = \left(\frac yx\right) x^y \quad\text{and}\quad \frac{\partial f}{\partial y}= (\ln x)x^y.$$
By the multivariable chain rule (setting $y=x$ in the second coordinate)
$$\frac d{dx} f(x,x) = \frac{\partial f}{\partial x}\Big|_{(x,x)} + \frac{\partial f}{\partial y}\Big|_{(x,x)}\cdot 1 = \left(\frac xx + \ln x\right)x^y = (1+\ln x)x^y.$$

Regarding "treating like $a^x$ versus $x^a$", the logic is that the $a^x$ rule applies when the base $a$ is constant, and the $x^a$ rule applies only when the exponent $a$ is constant, and so you simply do not apply those rules to $x^x$.

So what do you do? Implicit differentiation is not necessary. Instead, you should start from the very definition of exponentials:
$$y = x^x = e^{x \, \ln(x)}
$$
At this point, it becomes very clear and logical what you should do, namely apply first the chain rule then the product rule:
$$\frac{dy}{dx} = e^{x \, \ln(x)} \frac{d}{dx}(x \ \ln(x)) = e^{x \, \ln(x)} (1 \cdot \ln(x) + x \frac{1}{x}) = x^x (\ln(x) + 1)
$$


Prove that $Z(S_n) = \{(1)\}$ for every $n \geq 3$. Induction I wonder if this questions can be done by induction.

$S_3 = \{(1),(12),(13),(23),(123),(132)\}$

$Z(S_3)$ contains all the elements in $S_3$ that commutes with all the element in $S_3$

We can easily proof the base case for $n=3$

However, I have a hard time proving the induction step.

Assume $Z(S_k) = \{1\}$  then now we need to show it works for $Z(S_{k+1})$

The thing is $S_{k+1}$ contains $(k+1)|S_{k}|$ elements which is $(k+1)k!$

But can I go anywhere from here ?

Here's an induction "proof". I'll explain the scare quotes at the end.

Base Case: verify $S_3$.

Induction: Assume for some $n$ than $Z(S_n) = \{e\}$. We want to show that $Z(S_{n+1}) = \{ e \} $. Let $\sigma \in S_{n+1}$ be some non-identity permutation. If $\sigma \in S_n$ then $\sigma$ is not in the center by the induction hypothesis. Therefore assume $\sigma \not\in S_n$, so $\sigma$ does not leave $n+1$ fixed (here I am identifying $S_{n+1}$ with the symmetry group of the set $\{ 1, 2, \ldots , n+1\}$). Now let $a = \sigma(n+1)$ and let $b$ be some element of $\{ 1, 2, \ldots , n+1\}$ not equal to $a$, which is possible (why can we choose $b$ like this?). Let $\tau = (ab)$. Then:

$$ \tau\sigma(n+1) = \tau(a) = b $$
$$ \sigma\tau(n+1) = \sigma(n+1) = a $$
$$ a \neq b $$

This implies $\sigma\tau \neq \tau\sigma$ (why?). Hence for any $\sigma$ we can choose some $\tau$ which does not commute with it.

This is essentially a proof which does not rely on induction, just replace $n+1$ by some element not fixed by $\sigma$, which is possible if $\sigma \neq e$.

Just because you want to prove a family of statements parametrized by $n$ doesn't mean that the statement for $n+1$ depends directly on the statement for $n$. I'm not sure how natural induction is here, although I could be wrong. Below is the proof approach that appeals to me.

Here are some thoughts to help you start. Two elements of a group $\pi,\sigma\in G$ commute if $\sigma\pi=\pi\sigma$, but this is easily seen to be equivalent to $\sigma\pi\sigma^{-1}=\pi$. So what does conjugation do to a permutation, how does it change how it looks? One of the default ways to represent permutations is with disjoint cycle notation, the building blocks of which are cycles. Every permutation is a product of disjoint cycles. For $m$-cycles, conjugating it changes it like so:

$$\sigma(a_1~a_2~\cdots~a_m)\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_m)).$$

Given an arbitrary $\pi\in S_n$ (which is not the identity) (with $n\ge3$) you want to be able to find or make a permutation $\sigma$ so that $\sigma\pi\sigma^{-1}\ne\pi$. Use $\pi$'s disjoint cycle representation to do this.

Note that reasoning with the disjoint cycle representation also plays a part in the elementary proof that alternating groups are the only nontrivial proper normal subgroups of symmetric groups.


AST is conservative over ZF So this statement that AST is conservative over ZF is an exercise in Ralf Schindler's book, which I need to prove, but I am having trouble with it.

So the formulation of AST(Ackerman's Set Theory) which I will be using, is the following: Our language is $\mathcal{L}_\epsilon$ together with a constant symbol $\dot{v}$. And the axioms are Extensionality, Foundation, Comprehension(the same old ZF one), an axiom saying $\dot{v}$ is both transitive and supertransitive and one reflection axiom for each formula $\varphi(v_1, \dots, v_n)$ of $\mathcal{L}_\epsilon$ stating that:
$$\forall v_1\in\dot{v}\dots \forall v_n\in\dot{v} (\varphi^{\dot{v}} \leftrightarrow \varphi)$$

So in a previous exercise I had proven that AST $\vdash$ ZF. So by reflection, we have that $\dot{v} \models $ ZF. Also since AST $\vdash$ ZF, by using the coding in Devlin's book for example, we can see that the satisfaction relation for sets can be constructed in AST, thus AST proves Con(ZF), since $\dot{v}$ is just another set in AST's universe.

This is my problem.  How can AST be conservative over ZF and also in the mean time prove Con(ZF)?

The accepted answer is incorrect. $\omega$-consistency does not say "If you prove $P(n)$ for each standard $n$, then you prove $\forall n(P(n))$." That would be $\omega$-completeness.

Rather, $\omega$-consistency says "If you prove $\exists x\neg P(x)$, then you must not prove $P(n)$ for each standard $n$." And in particular, both $AST$ and the theory $T$ introduced in that answer are $\omega$-consistent, at least assuming $ZFC$ is to begin with.

Getting back to the main question, it may help to drop $AST$ and consider the simpler fact about $ZFC$ alone, which is itself provable in ZFC:

  $(*)\quad$ For every $M\models ZFC$ there is some structure $A\in M$ such that $A\models ZFC$ ... even if $M\models\neg Con(ZFC)$.

The key point is that "$A\models ZFC$" is interpreted in reality; we may not have $M\models(A\models ZFC)$.

This also explains why the OP's reflection argument breaks down - it's exactly the same reason.

Here's how to prove $(*)$ in ZFC:

If $ZFC$ is inconsistent then $(*)$ is vacuously true.
Suppose $ZFC$ is consistent. Let $M\models ZFC$. If $M\models Con(ZFC)$ then since $ZFC$ proves the completeness theorem we're done.
So suppose $M\models \neg Con(ZFC)$. Let $n\in\omega^M$ be what $M$ thinks is the largest number such that there is no proof of a contradiction from the first $n$ axioms of $ZFC$. If we can conclude that $n$ is nonstandard, we'll be done: by completeness in $M$, any model $A$ of the first $n-1$ axioms of $ZFC$ in the sense of $M$ will in reality be a model of $ZFC$, even if $M$ doesn't think so.

Now here's the cute bit: we internalize the reflection principle. Looking at the usual argument we see in fact that ZFC proves "ZFC proves every finite subtheory of ZFC."  (Note the crucial nested "proves" here.) This means we can next say...

Since $M\models ZFC$, for each standard $k$ we have $M\models$ "The first $k$ axioms of $ZFC$ are consistent." So $n$ is nonstandard and we're done.

(I had deleted this answer as it does not answer the question. I undeleted it for the context of the comments in Noah Schweber's answer, with some modifications.)

I checked Reinhardt's original article, and I found that what he proved is the following statement:

  Theorem (Reinhardt, Results 0.2.) If $\phi$ is a theorem of $\mathsf{ZF}$, then $\mathsf{AST}$ proves $\phi^{\dot{v}}$. Moreover, the converse also holds.

Hence I think Schindler states the exercise problem somewhat misleadingly.

I was not correct at that point. OP's comment on the reflection argument seems to work.

I reached an unexpected conclusion: $\mathsf{AST}$ is not $\omega$-consistent$\omega$-complete by the following argument: we may identify formulas to natural numbers by coding. We know that $\mathsf{AST}\vdash \phi\in\ulcorner\mathsf{ZF}\urcorner\to c\models \phi$ for each 'standard' $\phi$.
If $\mathsf{AST}$ is $\omega$-complete, then we have $\mathsf{AST}\vdash \forall\phi:\phi\in\ulcorner\mathsf{ZF}\urcorner\to c\models \phi$ for each 'standard' $\phi$. This means $\mathsf{AST}\vdash \dot{v}\models\mathsf{ZF}$.
However, we know that $\mathsf{AST}$ and $\mathsf{ZF}$ are equiconsistent.

Here is another example of an extension of $\mathsf{ZF}$ that is not $\omega$-complete (maybe due to Feferman.) Consider the language $\langle \in,c\rangle$, where $c$ is a constant symbol. Let $T$ be the theory given by the following axioms: the usual axioms of $\mathsf{ZF}$, the statement that $c$ is a transitive set, and $\phi^c$ for all axioms $\phi$ of $\mathsf{ZF}$.
By the similar argument for $\mathsf{AST}$, $T$ is $\omega$-incomplete.


Does the principle of explosion depend on the rules of weakening? The principle of explosion is this:

$$ X \land \lnot X \rightarrow Y $$

And similarly, the rule of weakening/monotonicity of entailment is this:

$$ (X \rightarrow Y) \rightarrow (X \land A \rightarrow Y) $$

If in a given logic, the rule of weakening holds, you could substitute $X = P$, $Y = Q$, and $A = \lnot P$ (or $P = \lnot P$ and $A = P$) in the rule of weakening to get something that gets close assuming the law of the excluded middle holds (see line 4) - it matches the principle of explosion at least when $Q = X = \top$:

$$ (P \rightarrow Q) \rightarrow (P \land \lnot P \rightarrow Q) \tag{1} $$
$$ (\lnot P \rightarrow Q) \rightarrow (P \land \lnot P \rightarrow Q) \tag{2} $$
$$ ((P \lor \lnot P) \rightarrow Q) \rightarrow ((P \land \lnot P) \rightarrow Q) \tag{3} $$
$$ Q \rightarrow (P \land \lnot P \rightarrow Q) \tag{4} $$

Similarly, most logics I've found that reject the law of the excluded middle also reject the principle of explosion, either implicitly like minimal logic or explicitly like relevant logic. This pervasive pattern led me to question why that pattern exists in the first place - if nobody is making their logic work with it when they drop the rule of weakening, is it even possible at all (despite the broad opinionatedness)?

My question is two-fold:

Does the principle of explosion depend on the rule of weakening in any case? Or in more formal terms, does proving the rule of weakening to hold also prove the principle of explosion to hold or vice versa? If not generally, then under what constraints? (The above proof sets two constraints, but I'm not fully convinced either one of them are absolute.)
Are there any known non-monotonic logics where the principle of explosion does hold? (This would serve as a counterexample to my hypothesis.)

I find the following form of the principle of explosion more useful in proofs:

  $$A\implies [\neg A \implies B]$$

I don't how much flexibility you have in choosing your axioms, but here is a proof using natural deduction (more straightforward than substitutions):

1) $A$ (assume)

2) $\neg A$ (assume)

3) $\neg B$ (assume)

4) $A \land \neg A$ (intro $\land$, 1, 2)

5) $\neg \neg B$ (discharge, 3)

6) $B$ (elim $\neg \neg$, 5)

7) $\neg A \implies B$ (discharge, 2)

8) $A\implies [\neg A \implies B]$ (discharge, 1)

See my formal proof in HTML format here.


Proof of $\lim_{x \to \infty} \frac{\sin x}{x} =0$ Prove that $$\lim_{x \to \infty} \frac{\sin x}{x} =0$$

For a given $\epsilon \gt 0$ we have

\begin{align}\left |\frac{\sin x}{x} -0\right|&\lt \epsilon\\
\implies \frac{|\sin x|}{|x|} &\lt \epsilon\end{align}

From here how can we get an $M \gt 0$ such that $x \gt M$ $\implies$ $|f(x)-L|\lt \epsilon$

Well, the proof goes like this:

Let $\epsilon>0$. We have to find a $M=M(\epsilon)>0$, such that for every $x>M$, we have that $\left|\frac{\sin x}{x}\right|<\epsilon$.

Note that $|\sin x|\leq1$ for every $x\in\mathbb{R}$. Note also that, from Archimedes-Eudoxus Principle, we can find a $n_0=n_0(\epsilon)\in\mathbb{N}$ such that:
$$\frac{1}{n_0}<\epsilon$$

Let $M=n_0$. Now, since $\frac{1}{x}$ is strictly decreasing and, hence $x>M\Rightarrow\frac{1}{x}<\frac{1}{M}=\frac{1}{n_0}<\epsilon$, we have, for every $x>M>0$:
$$\left|\frac{\sin x}{x}\right|\leq\left|\frac{1}{x}\right|\overset{x>0}{=}\frac{1}{x}<\frac{1}{M}<\epsilon$$
So, the proof is complete.

The function $\sin x$ always lies between $0$ and $1$.
So it will have any of the value between $0-1$, when $x$ is tending to infinity.
So as $x$ tends to infinity $\frac{\sin x}{x}$ will be zero.


One more way: to make yr life easier use the property of yhe logsrithm:

$$
\log_a b = \frac{\ln b}{\ln a}
$$

wher3 ln is a natural log. After a bit of algebra exponentiate both sides.


Infimum and supremum of $\{x\in[-1,4]\mid\sin x>0\}$ Let $S=\{X\in[-1,4]\mid\sin(x)>0\}$. Which of the following is true?
  $\inf(S)<0$,
  $\sup(S)$ does not exist,
  $\sup(S)=π$,
  $\inf(S)=\dfracπ2$.

I was trying to get the infimum and supremum of the set, but I am not able solve it. Please help.

Since sin(x)>0, x can take values only between 0 and π(3.14) as you can see in the graph of sin function. So the set S = {x€[-1,4] | sin(x)>0} will be reduce to S={x€(0,π(3.14)) | sin(x)>0} . Sup(S) is not 4. Therefore, The correct answer is Sup(S) = π.


Ball withdrawn from a box. White ball by kth draw. A box contains m white and n black balls. Balls are drawn at random one at a time without replacement. Find the probability of encountering a white ball by the kth draw.

This is our tutorial sheet's question and I don't know the answer. Thanks in advance.

Edit : I corrected the title and I misunderstood the question. I read 'at' instead of 'by'. With 'by', the question is very easy, still thank you for your answers. If you have solution for 'at', please answer. Again thank you for your efforts.

The probability of encountering a white ball by the $k^{th}$ is equal to $1 - P$ where $P$ is the probability that the first $k$ balls were all black. The latter probability is easier to compute. For example, the probability of the first ball being black is $\frac{n}{m+n}$. The probability that the second ball is also black is $\frac{n-1}{m+n-1}$. So, the probability that the first two balls are both black is $\frac{n(n-1)}{(m+n)(m+n-1)}$. The probability that the first $k$ balls are all black is similarly given by $\frac{n(n-1)...(n-k+1)}{(m+n)(m+n-1)...(m+n-k+1)}$.

Note that if $k > n$, then the probability becomes 1 because clearly you will have to draw a white ball by the $k^{th}$ draw (a version of Pigeon Hole Principle).

The problem is to find the probability of getting a white ball in $k$-th draw. May be for the first time or may be not... so we may have already gotten a white ball in first $k-1$ times. Therefore we have to consider them. If we consider '$i$' as no. of times a black ball can come in those $k-1$ times(as $k$-th is assumed to be a white ball) $i=0,1,2,.....k-1$.

Total probability is : $(m/m+n) + (n/m+n)(m/m+n-1) + (n/m+n)(n-1/m+n-1)(m/m+n-2) +
                      (n/m+n)(n-1/m+n-1)(n-2/m+n-2)(m/m+n-3) + ...+
                      (n/m+n)(n-1/m+n-1)(n-2/m+n-2)...(n-k+2/m+n-k+2)(m/m+n-k+1)$


Is there a difference between the calculated value of Pi and the measured value? The mathematical value of Pi has been calculated to a ridiculous degree of precision using mathematical methods, but to what degree of precision has anyone actually measured the value of Pi (or at least the ratio of diameter to circumference), by actually drawing a circle and then measuring the diameter and circumference?

If these two values differ, is the resulting difference (discounting inaccuracy in measurement) the result of the curvature of the surface on which the circle is drawn, or in the case of a circle in space in zero gravity (as much as that can exist), the curvature of space-time?

There's an underlying error in the question, namely the assumption that being in a curved space would result in a "different measured value of $\pi$".

What happens in a curved space is that the ratio between a circle's circumference and diameter is no longer the same for all circles. More precisely, the ratio will depend on the size of the circle. For small circles (with diameter tending towards 0) the ratio will converge towards the one unchanging mathematical constant $\pi$ -- as circles get larger the ratio will either become larger and smaller according to whether the curvature of space is negative or positive.

However, $\pi$ as the limit of $\frac{\text{circumference}}{\text{diameter}}$ for small circles is the same mathematical constant for all possible curvatures of space.

According to the General Theory of Relativity we live in a slightly curved space. This has been measured directly in the vicinity of Earth by the Gravity Probe B experiment. The experiment didn't actually measure the circumference of a large circle, but the results imply that the geometric circumference of a circle approximating the satellite's orbit around the earth is about one inch shorter than $\pi$ times its diameter, corresponding to $\frac CD\approx 0.9999999984\, \pi$. (The curvature is caused by Earth's mass being inside the orbital circle. A circle of the same size located in empty space would have a $\frac CD$ much closer to $\pi$).

Science fiction authors sometimes get this wrong. For example in Greg Bear's Eon there's a mathematician character who concludes she's in a curved area of space by measuring the value of $\pi$ and getting a nonstandard value. I headdesked -- it doesn't work that way.

By definition, pi is the ratio of the circumference of a circle to its diameter.

So how can others answer that this value is not best found from the most precise measurement we can make?

In the real world, a perfect circle has a definite value for radius. Pi is thought to be a never ending, non repeating decimal. This kind of value cannot truly represent a real perfect circle. This kind of value can only be used for approximations. The formulas found in the link of user4894's post do just that, approximations.

  Atomic-force microscopy (AFM) or scanning-force Microscopy (SFM) is a very-high-resolution type of scanning probe microscopy (SPM), with demonstrated resolution on the order of fractions of a nanometer, more than 1000 times better than the optical diffraction limit.
  AFM - wikipedia

We should really use this to measure a circle! But how would we make a perfect circle to measure?

Circles are abstract, it might not even be possible to make a perfect circle to measure. I cannot find instances of anyone trying to measure a circle with AFM. But C = Pi*D still stands as the definition for pi that all those fancy calculations attempt to approximate. Good question OP!

One good way to approximate the right dimensions for C and D is to fit a polygon inside a circle, and keep subdividing it until you reach an absurd amount of vertices. This polygons perimiter is a good approximation of the circumference of the circle, take the distance between two points that are straight across from one another for diameter. We can reach the highest levels of precision with this method, althrough there are more efficient ways to generate good pi values, it seems the most trustable to me. Very simple.


Given values of X , find fractions Solve the equation expressing $x$ in terms of $y$
$$x^2 - 6xy + 5y^2 = 0$$

Given that area of part $A$ of a circle is
$$3x^2 - 2xy - y^2 \text{ cm}^2$$

Area of circle is $4x^2$ cm^2

Calculate fraction $$\frac{\text{Area of part $A$ of the circle}}{\text{Area of circle}}$$

My workings

$ X= 5y OR y $

Sub X=5y
Area of part A of the circle = $ 3(5y)^2 - 2(5y)(y) - (5Y)^2 = 40 y^2 $

Area of circle = $ 4 (5y)^2 = 100y^2 $

Fraction = $40y^2 / 100y^2 = 2/5 $

I found that my ans is wrong and the ans is 16/25 can anyone help me .. Thanks a lot !

Hint divide first equation by $y^2$ you get a quadratic in $y/x$ thus getting relation between $y,x$. Hope its clear.


Fixed points of: $\dot{x}=\sin(y) \qquad \dot{y}=\cos(x)$ How can you find the fixed points of this system:

$\dot{x}=\sin(y)\\
\dot{y}=\cos(x)$

Normally I would suggest that you find the points when both functions are equal to 0.

As you suggested, to solve the fixed points simply set $\dot{x} = \dot{y}= 0$.

So $$\dot{x} = 0 \implies y = n\pi\quad \forall n \in \mathbb{N},$$
and $$\dot{y} = 0 \implies  x = k\pi + \frac{\pi}{2} \quad \forall k \in \mathbb{N}.$$
Hence your fixed points are all the pairs $(x,y) = \left(n\pi+\frac{\pi}{2}\pi,k\pi\right)$ for $n,k\in\mathbb{N}.$

No, you're thinking of stationary points. I'm guessing this question is to do with the Banach contraction mapping principle. You would rewrite the coupled DEs in the form:

$$X(t)\stackrel{def}{=}\left(\begin{array}{c}x(t)\\y(t)\end{array}\right) = \int_0^t \left(\begin{array}{c}\sin(y(u))\\\cos(x(u))\end{array}\right)\mathrm{d}u$$

i.e. you write it in the form $X = \mathscr{L} X$, where $\mathscr{L}:C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]\to C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]$ is the integral operator defined above. Then you prove that $\mathscr{L}$ is contractive: i.e. $\exists N\geq0,\,0<K<1$ such that $\|\mathscr{L}^N X_1-\mathscr{L}^N X_2\| \leq K \|X_1-X_2\|$ for all $X_1,\,X_2\in C^1[-t_{max},t_{max}]$. Under these conditions, repeated iteration of $\mathscr{L}$ on any "point" (i.e vector of functions) $X_0\in C^1[-t_{max},t_{max}]$ will converge to the unique fixed point $X_\infty\in C^1[-t_{max},t_{max}]$ such that $\mathscr{L} X_\infty = X_\infty$. In effect, I think you're being asked to work through the proof of the local version of the Picard-Lindelöf Theorem for this differential equation. The inequalities $|\sin(x_1-x_2)| < |x_1-x_2|$ and $|\cos(x_1-x_2)| < |x_1-x_2|$ will help you. Also, use the maximum modulus ($\mathcal{L}^\infty$) norm to define the metric space of functions.

You're probably not being asked to explicitly give a closed form for the fixed point (i.e. the solution of the DE), but simply to prove it exists and is unique by the arguiments shown to you on the Wiki page.


Theory of Congruences There is a problem in Elementary Number Theory by David Burton - We need to show that $0, 1, 2, 2^2, 2^3, \dots , 2^9$ form a complete set of residues modulo $11$. I tried this way - we need to show that $11 \nmid 2^r(2^{r-s} - 1)$ where $r > s$ and $r, s = 0, 1, 2, \dots , 9$. It is obvious that $11 \nmid 2^r$ but how to show that $11 \nmid 2^{r-s} - 1$ without actual verifying each value of $2^{r-s} -1$.

Instead of this, use the fact that $2^{10} \equiv 1 \mod 11$, which is Fermat's little theorem. Now, if $2^s \equiv 1 \mod 11$ for some $1 \leq s \leq 9$, then $s$ must divide $10$.(*) (That is, if you don't know this go below the answer).

However, it is easy to see that $s=1,2,5$ don't work out, since these are easy powers of $2$ to compute. Hence, none of $1,2,...,2^9$ leave remainder $1$ upon division by $10$, and hence $2^r(2^{s-r} - 1)$ cannot be a multiple of $11$ if $0 \leq r < s < 10$.

EDIT : If you did not want to use Fermat's little theorem, then you must make do with the observation that $2^5 \equiv -1 \mod 11$ by inspection, since $2^5 = 32$ is very small,  so squaring both sides $2^{10} \equiv (-1)^2 \equiv 1 \mod 11$. Then the above proof works out, since we only needed knowledge about $2^{10}$ to proceed further.

(*) Suppose $s$ is the smallest positive number satisfying $2^s \equiv 1 \mod 11$. Then $s \leq 10$. Let $r$ be the remainder when $10$ is divided by $s$. We see that $$qs + r = 10 \implies 1\equiv 2^{10} \equiv (2^{s})^q 2^r \equiv 1^q2^r \equiv 2^r \mod 11$$, so $r < s$ satisfies the proposition. To prevent a contradiction, we must have $r = 0$ i.e. $s$ divides $10$.

Hint: Note that $(2,11)=1$, so $1,2,\ldots,2^{9}$ in $\operatorname{mod} 11$ occupy $\{1,2,\ldots,10\}$.


Finding matrix with given special eigenvalues Given an arbitrary matrix $A=\left[a_{i,j}\right]\in\mathbb{C}^{2\times2} $

Now define the set
$$ \mathcal K = \left\lbrace z\in\mathbb C : \left|z-a_{1,1}\right|\cdot\left|z-a_{2,2}\right|=\left|a_{1,2}\right|\cdot\left|a_{2,1}\right| \right\rbrace$$

And the set of so-called equiradial matrices

$$\omega(A) = \left\lbrace B\in\mathbb C^{n\times n} : b_{1,1}=a_{1,1}, b_{2,2}=a_{2,2}, \left|b_{2,1}\right|=\left|a_{2,1}\right|, \left|b_{1,2}\right|=\left|a_{1,2}\right| \right\rbrace$$

Now I want to show, that the set of all spektra of those matrices $B$ is equal to $\mathcal K$

Necessarily $B$ is of the form
$$B = \begin{pmatrix} a_{1,1} & \left|a_{1,2}\right|\exp{i\psi_1} \\
\left|a_{2,1}\right|\exp{i\psi_2} & a_{2,2}
\end{pmatrix}$$
where $\psi_1$, $\psi_2$ are some real numbers.

Now fix one of those matrices $B$ and calculate the eigenvalues gives, putting them into the defining equality in $\mathcal K$ demonstrates, that they really belong to $\mathcal K$

  But I am stuck on the other direction. Now I want to fix any $z \in \mathcal K$ and show that there is a corresponding matrix $B$ in $\omega(A)$ that has this $z$ as an eigenvalue. It's quite obvious that there must exist such an $B$ but how to chose $\psi_1$ and $\psi_2$ depending on $z$?


Prove by induction for $n\geq1, n \in \mathbb{N}$ $, 2^{2n+1}\equiv 9n^2-3n + 2\pmod{54}$. Question taken from the book by Andre Weil, titled Introductory number theory, chapter 5, question #V.3.
Prove by induction that, if $n$ is a positive integer, then
$2^{2n+1}\equiv 9n^2-3n + 2\pmod{54}$.
Take base case, $n=1, 2^3\equiv 9-3+ 2\pmod{54}$ $\implies  8\equiv 8\pmod{54}$.
So, the base case should be a value that yields LHS $\geq 54$.
$n=3, 2^7\equiv 9.9-3.3+ 2\pmod{54}$ $ \implies  128\equiv 81-9+2\pmod{54}$ $\implies 128\equiv 74\pmod{54}$ $\implies 20\equiv 20\pmod{54}$.
Assume, it is true for $n=k$, get:
$2^{2k+1}\equiv 9k^2-3k + 2\pmod{54}$.
Need show that it is true also for $n=k+1$.
$2^{2k+3}\equiv 9(k+1)^2-3(k+1) + 2\pmod{54}$
$\implies 2^{2k+1}.2^2\equiv 9(k)^2 +18k +9-3(k)-3 + 2\pmod{54}$
As LHS is proved, so need take the RHS, and show:
$\implies 9(k)^2 +18k +9-3(k)-3 + 2\equiv 9(k)^2 -3(k)+ 2 \pmod{54}$
$\implies 9(k)^2 +18k -3k +8 \equiv 9(k)^2 -3(k)+ 2 \pmod{54}$
$\implies 18k +6 \equiv 0 \pmod{54}$
$\implies 3k +1 \equiv 0 \pmod{9}$
Why got such a wrong result?

Alternatively if induction is not mandatory, $$2^{2n+1}-2=2(4^n-1)$$
$$(1+3)^n-1\equiv\binom n13^1+\binom n23^2\pmod{27}\equiv\dfrac{9n^2-3n}2 $$
$$\implies2(4^n-1)\equiv9n^2-3n\pmod{54}$$


A Set of Linear Equations Equal to Zero If A+B+C+D = 1;

Let A,B,C,D be elements of vector V.

Let there be a 4*4 matrix M as such :

  -5  5  0  0
   4 -8  4  0
   0  5 -7  2
   0  0 10 -10

A property exist that VM = 0.

What is V=[?, ? ,? ,?] ?

I have tried to solve this on matlab and failed. I always reach to conclusion that the vector V are all 0,0,0,0.

Also that solves the equation as well, that is not the answer i am looking for since the SUM of V=1.

The answer should be a non zero value.

These are the set of equations:

4b-5a =0
5a-8b+5c=0
4b-7c+10d=0
2c-10d=0

Are you sure that $vM=0$ has infinitely many solutions for $v$?

The question asked is more of a matlab question than a math question: you need to find the function that returns the space of all solutions (i.e. the left nullspace of $M$), not just one.

Or, you could ask matlab to solve the equation you're really trying to solve:

  /  |1\
v |M |1| = [0 1]
  |  |1|
  \  |1/

Add the first and second equation. This gives $-4b = 0$, $b = 0$, and subsequently using the first equation $a = 0$.
Adding the third and fourth equation gives $c = 0$, $d = 0$.
Therefore $Mv$ does note have non-trivial solutions.


Lang problem about automorphism of polynomial ring STATEMENT: Let $A$ be a commutative entire(integral domain) ring and $X$ a variable over $A$. Let $a,b\in A$ and assume that $a$ is a unit in $A$. Show that the map $X\mapsto aX+b$ induces a unique automorphism of $A[X]$ inducing the identity on $A$. What is the inverse map?

QUESTION: Not really sure how to show the surjective part of the mapping to finish proof that it is an isomorphic map. Some help would be appreciated.

It is not too difficult to guess the inverse map. Since it is a twosided inverse, we get injectivity and surjectivity for free from its existence.


Physical intuition behind inverse of derivative Suppose you do a transformation of variables from (x, a)-> (y, b) like from cartesian to polar.I know that $$\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}$$ This holds true for total derivatives but not for partial derivative. Does anybody has any logical explanation/intuition behind this?

Where did you get that it does not hold true for partial derivatives, which are just the usual single derivatives?

For instance, if you now let $y=f(z,w),$ and moreover if as a function of $z$ alone we have that $y$ is continuously differentiable and $y_z\ne 0,$ then the derivative of the inverse $y\mapsto z$ exists, and is still given by $$\frac{1}{\frac{\partial y}{\partial z}}.$$


How many functions satisfy the property $f(i)<f(j)$ for some $1 \leq i \leq j \leq n$? Let $F$ be the set of one-to-one functions from the set $\{1, 2, ….., n\}$ to the set $\{1, 2, ….., m\}$ where $m\geq n\geq 1$.

How many functions are members of $F?$
How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i,
        1 \leq i \leq n?$
How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for some
        $1 \leq i \leq j \leq n$?

Somewhere it explained as :

A function from $A$ to $B$ must map every element in $A$. Being one-one, each element must map to a unique element in $B$. So, for $n$ elements in $A$, we have $m$ choices in $B$ and so we can have $^m\mathbb{P}_n$ functions.
Continuing from $(1)$ part. Here, we are forced to fix $f(i) = 1$. So, one element from $A$ and $B$ gone with $n$ possibilities for the element in $A$ and $1$ possibility for that in $B$, and we get $n \times ^{m-1}\mathbb{P}_{n-1}$ such functions.
$f(i) < f(j)$ is satisfied by all such functions as we are considering only one-one functions. So, this answer will be same as in $(1)$.

I've understood well part $(1)$ and part $(2)$ . I'm not getting explanation of part $(3)$.

How is it satisfied by one to one function , and how many such functions are there ?

The "somewhere" explanation is not correct. For a counterexample, take $n=m=2$. Then the function
$$ f = \{ 1\mapsto 2, 2\mapsto 1 \} $$
is in $F$ but does not satisfy the property in (3).

There are in general $\binom mn$ strictly decreasing functions that are in $F$ but don't satisfythe condition.

answer of c) is actually C(m,n) because you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing , so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n)


Double Integral Question on unit square Hints on solving following double integral will be appreciated.

$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$

The integral does not converge. By various tricks, you can cause some kind of iterated calculation to give any value you like.

You can cut up the domain into parts. For $x^2 + y^2 \geq 1$ there is no trouble. Furthermore, there is a $\pm$ symmetry across the $45^\circ$ line $y=x.$ So, we consider the part $$ x^2 + y^2 \leq 1, \; \; y \leq x. $$ On this part, the integral is, indeed, $$ \int_0^{\pi/4}  \int_0^{1} \frac{\cos 2 \theta}{r} dr d\theta = \frac{1}{2 }  \int_0^{1} \frac{1}{r} dr = \left.\frac{1}{2 } \log r \; \right\rvert_{r=0}^{r=1} $$ which does not work.

There is surely some good way to write the traditional vertical bar symbol denoting the evaluation step for a one-variable integral. I think it will work if I make the $r$ values into an array with a blank line or two in the middle. That's better, array, with matching \left|  and \right. which produces nothing as the right delimiter; if I had put \right| there would be a second vertical bar.

http://en.wikipedia.org/wiki/Multiple_integral#Multiple_improper_integral

0 by symmetry.Integrand changes sign across the diagonal


how to prove, $f$ is onto if $f$ is continuous and satisfying $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$ let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that
$|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$.
then $f$ is onto.

Hint:

The given inequality implies that $f$ must be one-to-one. Without loss of generality, suppose $f$ is increasing.

Suppose now that $f$ does not take the value $M>f(0)$ with $M$ positive. Then by continuity we must have $|f(x)-f(0)|\le M-f(0)$ for all $x\ge 0$. This, however, contradicts your given inequality (consider the inequality with $y=0$ and $x>M-f(0)$).
Thus, $f$ is not bounded above.

Now argue in a manner similar to the above that $f$ is not bounded below.

So, you have a   continuous function which is neither bounded above nor bounded below...

Assume $f$ is not onto, implying that there exist $y\in Y$, here i denote Y as the codomain of $f$, where no $x\in X$, X is the domain of $f$, satisfies $f(x)=y$. Note that $f$ is continuous in the whole real number line. In another words, for any $y\in Y$ as well as in real number line, due to continunity, there should exist and $x$ s.t.$\lim_{t\rightarrow x}f(t)=y$ implies $x$ satisfies $f(x)=y$ for any $y$ in real which is a contradiction


An alternative approach; compute

$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+3}}{(2n+3)(2n+2)(2n+1)}$$

This has the property that $f'''(x)= \sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$. Integrate once  (partial fractions) to get logarithms. Integrate twice more (by parts) and evaluate at 1 to get the answer $f(1)$.

\begin{align}
&\sum_{n = 0}^{\infty}
{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}
=
{1 \over 4}\sum_{n = 0}^{\infty}
\left({1 \over n + 1/2} - {2 \over n + 1} + {1 \over n + 3/2}\right)
\\[3mm]&=
{1 \over 8}\left\lbrack
\sum_{n = 0}^{\infty}{1 \over \left(n + 1/2\right)\left(n + 1\right)}
-
\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)}
\right\rbrack
\\[3mm]&=
{1 \over 8}\left\lbrack
{\Psi\left(1/2\right) - \Psi\left(1\right) \over 1/2 - 1}
-
{\Psi\left(1\right) - \Psi\left(3/2\right) \over 1 - 3/2}
\right\rbrack
=
{1 \over 4}\left\lbrack - \Psi\left(1 \over 2\right) + 2\Psi\left(1\right)
-
\Psi\left(3 \over 2\right)
\right\rbrack
\\[3mm]&=
{1 \over 4}\left\lbrace
-\Psi\left(1 \over 2\right) + 2\Psi\left(1\right)
-
\left\lbrack 2 + \Psi\left(1 \over 2\right)\right\rbrack
\right\rbrace
=
{1 \over 2}\left\lbrack
-\Psi\left(1 \over 2\right) + \Psi\left(1\right) - 1
\right\rbrack
\\[3mm]&=
{1 \over 2}\left\lbrace
-\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack
+
\left(-\gamma\right) - 1
\right\rbrace
=
\ln\left(2\right) - {1 \over 2}
\\[1cm]&
\end{align}

\begin{align}
\sum_{n = 0}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}
&=
\ln\left(2\right) - {1 \over 2}
\\[3mm]
\color{#0000FF}{\large\sum_{n = 1}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}}
&=
\left\lbrack\ln\left(2\right) - {1 \over 2}\right\rbrack - {1 \over 6}
=
\color{#0000ff}{\large \ln\left(2\right) - {2 \over 3}}
\\[1cm]&
\end{align}

$\Psi\left(z\right)$ is the Digamma function and
$\gamma = 0.57721566490153286060651209008240243104215933593992\ldots$ is the Euler-Mascheroni constant.


Clairaut's Theorem And continuity of a multivariate function Is showing that $f_{xy}$ is not equal to $f_{yx}$ at a given point sufficient/good to show that a function is discontinuous there?

Can you also suggest a clear method to find without graphing a method to tell whether a function is continuous or not for a multivariable function?

Further, but unrelated, does $f_{xy}=f_{yx}$ imply in all cases?
I mean is $f_{xxyz}=f_{xyxz}=f_{zyxx}$...?

First question: It's very important when studying a delicate theorem of calculus to understand exactly what it says. Let's look at the exact statement of the equality is mixed partial derivatives theorem (which is also attributed to Herman Schwartz,so really should be called the Clairaut-Schwartz Theorem).
    The theorem states that if $f \colon \mathbb{R}^n \to \mathbb{R}$

has [[continuous function|continuous]] second [[partial derivatives]] at any given point in $\mathbb{R}^n $, say,  $(a_1, \dots, a_n),$ then for $1 \leq i,j \leq n,$

$$\frac{\partial^2 f}{\partial x_i\, \partial x_j}(a_1, \dots, a_n) = \frac{\partial^2 f}{\partial x_j\, \partial x_i}(a_1, \dots, a_n).\,\!$$

In words, the partial derivations of this function are [[commutative operation|commutative]] at that point.

The statement clearly assumes the second order partial derivatives of each variable of f are continuous at the point chosen, it says absolutely nothing about the continuity of the function itself at that point-nothing about the lower order partials or the linear derivative function either, for that matter!
The classic counterexample is the following (although there are more sophisticated ones):

$$f(x,y) = \begin{cases}
                     \frac{xy(x^2 - y^2)}{x^2+y^2} & \mbox{ for } (x, y) \ne (0, 0)\\
                      0                            & \mbox{ for } (x, y) = (0, 0).
                \end{cases}
$$

Prove to yourself this function does not have equal mixed second order partial derivatives-but it is nevertheless continuous at $(0,0)$ in the plane!

Second question: Well, clearly, if a multivariable function is differentiable in the strong sense, i.e. there exists a linear map that satisfies the limit that defines it for vector valued functions at a point in n-dimensional space, then the function is continuous there. If f is not differentiable there, then it becomes a very tricky game of calculation to find out if the function is continuous there since the limits for each variable holding the others constant must be calculated and they must be equal to the value of he map at that point regardless of which variables are held constant.This is equivalent to finding the value of the function on paths on the n-dimensional Euclidean space that specifies the domain where (n-1) variables are constant and the limit of f along the path "converges" to to the limit point. For example, in the counterexample above, we can randomly choose any line in the plane that intersect the surface defined by f. It can be proven using an epsilon-delta argument that for any such line, f is continuous at (0.0) regardless of "line of approach". Otherwise, we need to use a trial and error method to find if the limit does not exist.  If it doesn't, we should be able to pick 2 lines of approach where we get 2 different one-sided limits of f at (0,0). And be careful here because remember: A function being defined at a point does not guarantee it's continuous there!

Here's a simple example of what I mean:

$$f(x,y) = (x^2 - y^2)$$

It's clear f(0,0) exists and is equal to 0. But this isn't sufficient for continuity at (0,0). Consider taking the limit of f at (0,0) along the line x=0. Then lim  f(x,y) ----> (- y^2) as we approach (0,0) along x = 0.But if we approach along y =0, Then lim  f(x,y) ----> (x^2) and clearly these limits for f don't match along these 2 paths in the plane. So the limit doesn't exist for his f as we approach (0,0) despite the function being defined at (0,0) and being equal to 0!

The third question is obviously true by the statement of the general theorem.

Hope that answered your question!

just a quick comment. On your last example on limits, like you say, one approached -y squared and the other x squared. But both x and y are approaching 0 no matter which path they are coming. So wouldn't the limit be the same, namely 0?


Cover of (0,1) with no finite subcover & Open sets of compact function spaces I just got back from my exam and these questions' solutions eluded me, it would be great to use the rest of my evening figuring these out...

Q1: Find an open covering of the set $(0,1) \subset \mathbb{R}$, say $G =\{U_\alpha\}_{\alpha \in A}$, (where $A$ is some indexing set) such that $G$ has no finite subcover.

Q2: Let $f: [0,1] \to [0,\infty) $ be a continuous function. Let there be some $c\in [0,1]$ such that $f(c)$ is non-zero. Prove that there exists an $\epsilon \gt 0$ such that the set:

$X_1=\{\ x\in[0,1]\ | \ f(x)\gt\epsilon\  \}$

is non-empty, and open.

Q1: think about the sets $(1/n,1)$ for $n\ge1$.

Q2: does it help if you take $\epsilon = f(c)/2$? Do you know something about preimages of open sets under continuous functions?

The questions were asked almost 4 years ago and don't know if there is any interest - had to say something about Q2 since I don't see anything redeemable about it. If we take the statement of the question at face value, the conclusion is wrong if the continuous function is a constant, $f(x)=k, k\in\mathbb{R}$. Choose $\varepsilon < k$, then $X_1=[0,1]$; choose $\varepsilon\geqslant k$, then $X_1=\varnothing$. Suppose $[0,\infty)$ is a co-domain spec; one would hope so since a continuous mapping with compact support can not generate an unbounded range; its range is closed and bounded. So say, $f:[0,1]\to[a,b]$ is continuous and onto with $b>a\geqslant0$. If we now suppose $f$ is one-to-one, the conclusion fails once again. The possible sets are $X_1=\varnothing$, $X_1=[0,1]$, $X_1=[0,p)$ or $X_1=(p,1]$ where $p\in(0,1)$ and $f(p)=\varepsilon$. The conclusion is true for some continuous functions that are not one-to-one. For example, $f(x)=1-x^2$ on $[0,1]$ fits the bill; choose $\varepsilon\in(0,1)$ and $X_1=(-\sqrt{1-\varepsilon},\sqrt{1-\varepsilon})$. The statement, in general, is true for those continuous functions that are not one-to-one, which attain exactly one maximum on $(0,1)$ or if they attain more than one maxima on $(0,1)$, they attain a specific value in $(a,b)$ an even number of times.
Please point out any errors.


Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$
I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:
$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$
So that:
$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$
Then i tried to separate this last integral by partial fractions, my result on this was:
$$\frac {1} {(x+a)(x^2 + 1)} = \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1}\right)$$
And the integral reduces to:
$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$
Then this last expression evaluates to:
$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$
Then integrating from 0 to 1 with respect to $a$ we will get:
$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$
(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)
But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.
Do i need some other approach? Or did i made any mistake?
Any help is appreciated.


Replacement of sentence symbols by $0$-place connectives Suppose we add $0$-place connectives $\top, \bot$ to our language. For each well-formed formula wff $, \phi$ and sentence symbol, $A,$ let $\phi^{A}_{\top}$ be the wff obtained from $\phi$ by replacing $A$ with $\top.$ Similarly for $\phi^{A}_{\bot}.$ Let $\phi^{A}_{*}= (\phi^{A}_{\top} \lor \phi^{A}_{\bot}).$

(i) Suppose $\phi \models\psi$ and $A$ does not appear in $\psi,$ could anyone advise me on how to show $\phi^{A}_{*} \models \psi$ and (ii )how to show $\phi$ is satisfiable iff $\phi^{A}_{*}$ is satisfiable?

Hints will suffice. Thank you.

You must start with the definition of tautological implication (we may use this because we are considering only sentence letters) :

  $\phi \models\psi$ iff, for every valuation $v$ such that $|\phi|_v = t$, also $|\psi|_v = t$.

Now assume that $\phi$ is built from $n$ sentence letters : $A, A_1, ... A_{n-1}$.

Consider now the truth-table (with $2^n$ rows) for $\phi$ : in these table, half of the rows have $A$ evaluated at t and half evaluated at f.

When we build the formula $\phi^{A}_{\top}$, this formula is equivalent to all the rows of the t-t for $\phi$ where $A$ is evaluated to t; the same with $\phi^{A}_{\bot}$, that is equivalent to the rows with $A$ evaluated to f.

So the formula $\phi^{A}_{*} = (\phi^{A}_{\top} \lor \phi^{A}_{\bot})$ may replace all the truth-table for $\phi$.

Now, go back to the tautological impliction :

  $\phi \models\psi$ iff for every valuation $v$ such that $|\phi|_v = t$, also $|\psi|_v = t$,

i.e. for every row in the t-t for $\phi$ that evaluates to t, also  $\psi$ is evaluated to f.

But the rows of the t-t for $\phi$ which evaluate to t may have $A=t$ or $A=f$; both cases are covered by the disjunction $\phi^{A}_{\top} \lor \phi^{A}_{\bot}$ (if the row has $A=t$, this row is "covered" by $\phi^{A}_{\top}$ and if the row has $A=f$, it is covered by$\phi^{A}_{\bot}$.

So the taut implication still hold : of course, $A$ must not be in $\psi$, because substituting $A$ with $\top$ or $\bot$, we have lost "the link" between the two original formuals when we use the valuation $v$.

Now I think that you can easily use this argument to show satisfiability...

I'm assuming that we're talking first-order predicate logic here and that $\models$ refers to logical consequence. You could first show that $\phi_* ^{A} \models \phi$. Because $\models$ is transitive, the result you're looking for follows from that.


Limiting behavior of this sequence of integers For $n\in \Bbb N \setminus \{1\}$ we have $n={p_1}^{k_1} \cdots {p_m}^{k_m}$ uniqely, by the fundamental theorem of arithmetic.

We can now define sequence $\alpha(n)$ to be $\alpha(n)=\sum_{r=1}^m k_r$ and take as a convention that $\alpha(1)=0$.

Then we can define sequence $\beta(m)=\sum_{w=1}^m \alpha(w)$.

  Does there exist real numbers $q_1$ and $q_2$ such that we have $$\lim_{d \to \infty} \frac {\beta(d)}{{q_1} \cdot{d}^{q_2}}=1$$

Let $$\Omega(n) = \sum_{p^k \| n} k$$ Then $$ \beta(N)= \sum_{n \le N} \Omega(n) = \Omega(N!)= \sum_{p^k \le N}  \lfloor N/p^k \rfloor= N\sum_{p^k \le N}  \frac{1}{p^k}+\mathcal{O}(N) = N \log \log N+\mathcal{O}(N) $$
by Mertens 2nd theorem

$\beta(n)$ is OEIS A022559. An asymptotic expression is given there by Charles R. Greathouse IV:
$$\beta(n)=n\log\log n+1.0346n+o(n)$$
So the $q_1$ and $q_2$ as constants in the proposed limit do not exist, since $\beta(n)$ exhibits quasilinear rather than polynomial growth.


Different proofs of uniqueness of the Laplace transform How many different types of proof do you know for the so-called Lerch's theorem, i.e., uniqueness of the Laplace transform?

I have found the following references for proofs. New books, in general, do not include the proof for Lerch's theorem.

In [1, Sections 5 and 6], there is a general proof given for Stieltjes
type Laplace transforms.
In [2, Appendix II] and [3, Section 5], the
proof uses substitution first and then approximation to a continuous
functions by polynomials.
In [4, Section 3], there is a totally
different proof, where the author makes use of the function
$E_{n}(\tau-t):=1-\mathrm{e}^{-\mathrm{e}^{nk(\tau-t)}}$ which
converges to $H(\tau-t)$ as $n\to\infty$, where $H$ is Heaviside's
unit function.

Do you know other proofs for Lerch's theorem.
Can you redirect me to some other proofs?

References
[1] D. V. Widder, The Laplace Transform,
Princeton Mathematical Series, v. 6. Princeton University Press, Princeton, N. J., 1941.
[2] D. L. Kreider, R. G. Kuller, D. R. Ostberg, F. W. Perkins, An Introduction to Linear Analysis, Addison-Wesley Publishing Co., Inc., Reading, Mass.-Don Mills, Ont. 1966.
[3] G. Doetsch, Introduction to the Theory and Application of the Laplace Transformation, Springer-Verlag, New York-Heidelberg, 1974.
[4] E. J. Watson, Laplace Transforms and Applications, VNR New
    Mathematics Library, 10. Van Nostrand Reinhold Co., New York-London,
    1981.

I don't think this question has a well defined answer.  As far as I know, all proofs of the uniqueness of the Laplace transform are essentially corollaries of this one statement:

$\displaystyle \int_{0}^{\infty} f(t) \text{e}^{-st} dt = 0 \implies f(t) = 0$

The number of statements equivalent to this is infinite (e.g. you could add $1$ to both sides of the equation, to get a stupid yet true statement).  Perhaps you mean to restrict the proofs to those which are reasonably simple and useful, but then we start to get into highly subjective territory.


When evaluating $\int x \sin x \, dx$ explain how letting $ u =\sin x $ and $dv = x dx$ makes the solution more difficult to find. Integration by parts. The solution doesn't seem any more difficult for me. I'm just curious on how someone would answer this.

As per Ross Millikan's request, I'll provide an answer:

From integration by parts, we have $\displaystyle\int u \, \mathrm dv = uv-\displaystyle\int v \, \mathrm du$. The entire point of this is to go from a more complicated integral to an easier one. If $\displaystyle\int v \, \mathrm du$ is simpler than $\displaystyle\int u \, \mathrm dv$, you've made the "right" choice for $u$ and $\mathrm dv$.

In this example, you have $\displaystyle\int x\sin(x) \, \mathrm dx$. If you choose $u = x$ and $\mathrm dv = \sin(x) \, \mathrm dx$, you'll get $\mathrm du = \mathrm dx$ and $v = -\cos(x)$. This gives:

$$\int x\sin(x) \, \mathrm dx = -x\cos(x)-\color{blue}{\int-\cos(x) \, \mathrm dx}$$

Notice how $\color{blue}{\displaystyle\int v \, \mathrm du}$ is simple to integrate: it's just $-\sin(x)+C$. Notice that this is because choosing $x$ as $u$ (which is differentiated) will result in the exponent going down, which makes things simpler.

However, what if you pick $u = \sin(x)$ and $\mathrm dv = x \, \mathrm dx$? Then, you'll have $\mathrm du = \cos(x) \, \mathrm dx$ and $v = \dfrac{1}{2}x^2$. This gives:

$$\int x\sin(x) \, \mathrm dx = \frac{1}{2}x^2\sin(x)-\color{blue}{\int\frac{1}{2}x^2\cos(x) \, \mathrm dx}$$

How are you going to integrate $\color{blue}{\displaystyle\int v \, \mathrm du}$ now? By choosing $x \, \mathrm dx$ as $\mathrm dv$ (which is integrated), the exponent goes up. This means you've only made the problem harder. Going back in reverse would just waste two steps, so you might as well have just make the "right" choice to start with.

The same idea applies to other integration by parts problems as well. As it's been pointed out in the comments, the LIATE rule is usually used as the rule of thumb, but once you get used to these problems, you don't really have to rely on mnemonics anymore.

$$ \int x \sin x \, dx = -\int x \,  d\cos x =- x\cos x + \int \cos x \,  dx = \sin x - x\cos x + \textrm{const.}$$

The reason why the other way is more "difficult," is that it does not lead to the solution, but rather, a more compicated expression, as KM101 pointed out in the comments.


Given $f$ is $o(1/x)$ as $x\to \infty$, is it true that $f$ is $o(1/x^{1+\epsilon})$ as well? (for some small enough $\epsilon>0$) I'm basically trying to show that $f$ is summable $1$ to $\infty$ and that $f$ is integrable $1$ to $\infty$. Also, it necessary that $f$ be monotonic or continuous?

No.  Let $f(x)=\frac 1{x \log x}$.  This is $o(\frac 1x)$ but not $o(\frac 1{x^{1+\epsilon}})$

No it's not.

Say $f(x) = \frac{1}{x^{1+\frac{\epsilon}{2}}}$

Think about another approach to show that f is integrable


Find the derivative of this function using chain rule or product rule I am trying to find the derivative of $f(x)$$=$$10^{x^2+2}$
Can anybody help me try to solve this through the chain rule?
I have tried using the chain rule and exponent rule I figured out that
$f(x)=10^x$ the derivative is $f'(x)=ln(10)10^x$
$g(x) = x^2+2$ the derivative is $g'(x) = 2x$
Hence my answer for this derivative of function f(x) was
$f'(x) = (2x)(ln(10))(10^{x^2+2})$
however on the calculator/wolfram alpha it says that the answer is

I am unsure what the next steps to take to confirm which one is the right derivative.

You did it correctly and got the same answer as Wolfram did.
When Wolfram writes "$\log$" it means the natural log.  (That's common, but confusing as hell, practice).
Wolfram wrote $2^{x^2+3}\times 5^{x^2 + 2}x \log 10$ and meant $2^{x^2+3}\times 5^{x^2 + 2}x \ln 10$ and you got $(2x)(ln(10))(10^{x^2+2})$.
what's the difference?
Actually nothing.
$2^{x^2+3}\times 5^{x^2 + 2}x \ln 10 =$
$2\cdot 2^{x^2 + 2}\times 5^{x^2 + 2} x \ln 10=$
$2(2\times 5)^{x^2 + 2}  x\ln 10=$
$2\cdot 10^{x^2+2} x \ln 10$
which, order and parenthesis aside, is exactly what you got.
So you did jes' fine.


Do we really need Choice to pick socks? It is said that you need the Axiom of Choice to pick one sock from each of infinitely many pairs, but that you don't need it for shoes, since you can just pick all the left shoes.

But Choice is equivalent to the statement that every set can be well-ordered, right? And can't be well-order a two-element (or any finite) set without invoking anything related to Choice? If so, doesn't that mean we don't need Choice to pick socks after all?

The axiom of choice is not needed to choose from one pair of socks. You do that every morning when you put on your socks. In fact, even if you have infinitely many socks, choosing one doesn't require the axiom of choice. Neither does choosing five, or ten socks. That's just existential instantiation and induction.

The axiom of choice is needed in order to choose from infinitely many pairs at once. The reason is that between two socks, there is no "left" or "right", and there is no distinguishing property that we can always say that "given a pair of socks, one of them is such and such". And so the axiom of choice is really needed for that.

More formally, Fraenkel, and later Cohen, showed that this argument is mathematically solid. It is consistent that the axiom of choice fails, and there is a set which is the countable union of pairs, but there is no function which chooses a single element from each pair.

Yes you do need the axiom of choice to show that for any set of pairs of socks, there is a choice function that picks one sock from each pair. Each sock from one of those pairs can either function as a left sock or as a right sock so that doesn't give us a way to define in terms of each pair of socks in that set of pairs one sock from that pair. The fact that each pair of socks in that set of pairs can be well ordered is not enough to show that there is a function that assigns to each pair of socks one sock in that pair. To show that such a function exists, you have to assume that the union of all pairs of socks in that set of pairs can be well ordered.

Suppose you have a set of infinitely many pairs of socks. If all those socks were instead shoes and any two socks in a pair were instead a left shoe and a right shoe, then a choice function would exist so how could such a function not exist just because they're socks instead of shoes? That's because to show that the set of pairs of socks has a choice function, you have to assume that that set is a set that can be gotten by turning each shoe in the union of a set of pairs of shoes into a sock and that assumption can't be proven without using the axiom of choice.


Identity relation vs Reflexive Relation So we're starting relations in my discrete structures class this week, and I've probably read this over 10 times by now...I believe I have a good understanding of Identity Relations, but Reflexive Relations seem to have me slightly confused.

From my understanding, an example of Identity relation using set $A = \{1,2,3,4\}$

$R_1 = \{(1,1), (2,2), (3,3), (4,4)\}$ because each element is equal to itself.
$R_2 =\{ (1,1), (2,2), (3,3), (4,4), (1, 4)\}$ would not be an identity relation, as $1 \neq 4$.

What I don't understand is why

  The relation $R_2$ defined by $R_2 = \{(1, 1), (3, 3), (2, 1), (3, 2)\}$ is not a reflexive relation on $A$, since $(2, 2) \notin R_2$.

is not a Reflexive Relation

Could someone give me an example of what a simple reflexive relation is, and isn't?

Thanks all for the input, see below for a good example of a Reflexive Relation

Here's what the book describes both as:

Identity relation.

  Let $A$ be any set. Then the relation $R = \{(x, x) : x \in A\}$ on $A$ is called the identity relation on $A$. Thus, in an identity relation, every element is related to itself only.

  For example, consider $A = \{a, b, c\}$ and define relations $R_1$ and $R_2$ as follows.
  $R_1 = \{(a, a) ,(b, b), (c, c)\}$
  $R_2 = \{(a, a), (b, b), (c, c), (a, c)\}$

Then $R_1$ is an identity relation on $A$, but $R_2$ is not an identity relation on $A$ as the element $a$ is related to $a$ and $c$.

Reflexive relation.

  A relation $R$ on a set $A$ is said to be a reflexive relation if every element of $A$ is related to itself. Thus, $R$ is reflexive iff $(x, x) \in R$ for all $x \in A$.
  A relation $R$ on a set $A$ is not reflexive if there is an element $x \in A$ such that $(x, x) \notin R$.
  For example, consider $A = (1, 2, 3)$. Then the relation $R_1$ defined by $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 3), (2, 1)\}$ is a reflexive relation on $A$.
  The relation $R_2$ defined by $R_2 = \{(1, 1), (3, 3), (2, 1), (3, 2)\}$ is not a reflexive relation on $A$, since ($2, 2) \notin R_2$.
  Remark

  Every identity relation on a non-empty set $A$ is a reflexive relation, but not conversely. Consider $A = \{a, b, c\}$ and define a relation $R$ by $R = \{(a, a), (b, b), (c, c), (a, b)\}$. Then $R$ is a reflexive relation on $A$ but not an identity relation on $A$ due to the element $(a, b)$ in $R$.

Let,
 a,b € Z
(a, a) € R.
Then, a-a=0
=(a-a) is divisible by 2
Z={1,2,3}.
So, Z×Z
=R={(1,1),(2,2),(3,3),(1,3).    ["R" is reflexive relation]
I={(1,1),(2,2),(3,3)}.     [where, "I" is Identity Relation]
So,from the above example  we can notice that :-
Reflexive relation- is a kind of relation which contains the elements related to itself  as well as can contain other pairs too.
Identity Relation- is a kind of relation which contains the elements related to itself only.
**Thus, "Every IDENTITY Relation on a Non-Empty set is a REFLEXIVE Relation but not vice-versa....


Can I Rearrange This Equation? I'm not the best at math, and I'm trying to build an equation for an Excel/Sheets file.
Here's my equation:
$E=(A*B)+((D*B)*C)$
Is it possible to move the $B$ variable to be by itself on the left side of the equation?
I'm at my wits' end trying to figure this out, any help is greatly appreciated!

Elaborating on Gerry Myerson's suggestion, you have successively, using commutativity, associativity and distributivity,
$$
E=(A*B)+((D*B)*C) = (B*A)+(B*C*D) = B*(A+(C*D))
$$
Now, if $A + (C*D)$ is not null, you can write $B = E/(A + (C*D))$.

$$E=(A\cdot B) + ((D\cdot B)\cdot C$$
can be transformed to
$$ E= B\cdot(A+D\cdot C)$$
Thus,
$$ B = \frac{E}{A+D\cdot C}$$


How to find the derivative of $F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt$? For a real number $t>0$, let $\sqrt t$ denote the positive square root of t.  For a real number $x>0$, let $F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt$.  If $F'$ is the derivative of $F$, then

    (a). $F'(\frac \pi 2)=0$

    (b). $F'(\frac \pi 2)=\pi$

    (c). $F'(\frac \pi 2)=-\pi$

    (d). $F'(\frac \pi 2)=2\pi$

I am guessing that I could use Fundamental theorem of integral calculus. But I couldn't construct a function whose derivative is $\sin \sqrt t$. If I through I could find the answer.

Hint:

$$
F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt = \int_{0}^{4x^2} \sin \sqrt t\;\;dt - \int_{0}^{x^2} \sin \sqrt t\;\;dt
$$

We could define $$G(x) = \int_0^ x \sin \sqrt t\;\;dt$$ and rewrite as
$$
F(x) = G(4x^2) - G(x^2).
$$

By Chain Rule,
$$
F^\prime(x) = G^\prime(4x^2)(8x) - G^\prime(x^2)(2x).
$$
By FTC,
$$G^\prime(x) = \sin \sqrt x.$$

Is it clear how to find $F^\prime(\frac{\pi}{2})$ from here?

You do not need to solve the integral because you can use the Second Fundamental Theorem of Calculus.

With that the integral is cancelled by the derivative but you need to use the composition of functions.

If we call the function that is inside the integral f(t) and the upper limit v(x) and the lower limit u(x). The solution is

((derivative v(x)) f(v(x)) ) -  ((derivative u(x)) f(u(x)) )=

(8x)(sin sqrt(4x*2)) - (2x)(sin sqrt(x*2))=

8x sin(2x) - 2x sin(x)

The solution works for any x value.

If you need to evaluate the result at pi/2, you just plug it instead of the x.

4pi sin(pi) - pi sin(pi/2)=

negative pi


How are metalogic proofs valid? I'm reading some materials on mathematical logic. I wonder how we can "prove" metalogical properties (soundness, completeness, etc.)? As at this point, the proof system has not been verified yet. Isn't this a chicken-and-egg question (we may then have meta-metalogic)? What types of proofs are considered valid at this stage?

Short answer: yes, it is essentially a chicken-and-egg problem, or perhaps a hermeneutical circle or a spiral.  The common interpretation of the incompleteness theorems is that this circularity cannot be avoided.

You can use normal mathematical reasoning for the metatheory - which is not formalized in the first place - or you can choose some formal theory. In the latter case, you can choose a stronger metatheory, like ZFC, or a weaker one, like PRA.

In principle, you would then be able to choose a meta-metatheory, a meta-meta-metatheory, etc., but in practice essentially all the interesting issues arrive at the theory/metatheory level, and the higher levels just repeat these issues rather than giving new issues.

Yes, it's all circular. You can have weird cases (maybe not in math), were you are proving metatheoretical stuff in a different logic (many sorted, polivalued, second order, etc.), but your theory is in first order. In math this never happens, as far as i know, but in logic people do these weird combinations. But again in these cases are kind'a cirular because you have to know the theory of your metatheory, so one assumes it.


Dimension of a $G$-variety $X$ that is a finite union of $G$-orbits Suppose that $G$ is an algebraic group acting on a variety $X$, and $X$ is a finite disjoint union of $G$-orbits $\mathcal{O}_i$, $i=1,\ldots,n$, under this action.  Is it true that the dimension of $X$ is the maximum of the dimensions of the $\mathcal{O}_i$?  It seems like this should be true, and is probably a result of something more general.  Thanks in advance for any help with this.

Disclaimer: I only know the following for a fact if we are over an algebraically closed field of characteristic zero.

Let $X=X_1\cup\cdots\cup X_r$ is the decomposition of $X$ into irreducible components and assume without loss of generality that $X_1$ is of maximal dimension.

It is a well-known and basic fact that any $G$-orbit is open in its closure of $G.x$ is the union of $G.x$ and orbits of strictly smaller dimension. See these very nice notes by Michel Brion. Now by assumption, there are finitely many $x_1,\ldots,x_n\in X$ such that $X=\bigsqcup_{i=1}^n \overline{G.x_i}$ (disjoint union). Thus, $X_1 = \bigsqcup_{i=1}^n (\overline{G.x_i}\cap X_1)$. Since $X_1$ is irreducible, we have (again wlog) $X_1 =  \overline{G.x_1}\cap X_1$. Thus,
$$\dim(X)=\dim(X_1)=\dim(\overline{G.x_1}\cap X_1)\le \dim(\overline{G.x_1})=\dim(G.x_1)\le \dim(X).$$

Yes since the relation between the orbits and the variety is finite to one maximum dimension of the orbits is the same with the dimension of X.


Why are these triangles similar ? (Derivation of uniform circular motion equation ) I am studying the equations of uniform circular motion and I am  having difficulties  with the derivation of them.

I quote my book :

  We can find a simple expressio for the magnitude of the acceleration
  in uniform circular motion.We begin with the below diagram,which shows
  a particle moving with constant speed in a circular path of radius $R$
  with center at $O$ . The particle moves a distance $ \Delta s$ from
  $P_1$ to $P_2$ in a time interval $\Delta t$. The figure below also
  shows the vector change in velocity $\Delta \vec{v}$ during this
  interval.  The angles labeled $\Delta \phi $ in the figure are the
  same because $\vec {v_1}$ is perpendicular to the line $OP_1$ and
  $\vec{v_2}$ is perpendicular to the line $OP_2$. Hence the triangle in
  the figure are similar.

I sadly don't see how the angles are equal and how the triangles are similar as a consequence of the fact that $\vec {v_1}$ and $\vec {v_2}$ are perpendicular to $R$.

Can you guys help with a math proof ?

Note that $v_1$ and $v_2$ are equal in magnitude and perpendicular to the respective radii. So they form an isosceles triangle and the angle between them is the same as the angle between the radii. The base angles of the two isosceles triangles are also equal, hence they are similar.

To show the angles are equal translate the configuration at $P_2$ to $P_1$ and work round the angles in that figure. The angle between $v_2$ and $OP_1$ adds to the angle between $OP_2$ and $OP_1$ to give a right angle and similarly with the angle between $v_2$ and $v_1$ so the angles you want are equal.

What helped me is this: If the line $O-P_1$ was rotated by 90 degrees, you would be able to see that the difference between $v_1$ and $v_2$ would also be 90 degrees.

This holds true when the vectors flip 180 degrees as well, so logically (to me at least) it should hold true for any angle between $O-P_1$ and $O-P_2$.


How do you find the modular inverse of $5\pmod{\!11}$ I need to find out the modular inverse of 5(mod 11), I know the answer is 9 and got the following so far and don't understand how to than get the answer. I know how to get the answer for a larger one such as 27(mod 392) but am stuck because they are both low numbers.

11=5 (2)+1

5=1 (5)

In finding a modular inverse, you are trying to solve the modular equation
$$
ax\equiv 1\pmod n.
$$
Ordinarily, you use the Extended Euclidean Algorithm for this to solve the equation $ax+ny=1$. If the numbers $a$, and $n$ are small, then simple trial and error is probably just as fast or faster.

For your example, we have $a=5$ and $n=11$, which means would could just use trial and error.
\begin{align}
1\cdot 5 &\equiv 5\\
2\cdot 5 &\equiv 10\\
3\cdot 5 &\equiv 15\equiv 4 \\
4\cdot 5 &\equiv 20\equiv -2\\
5\cdot 5 &\equiv 25\equiv 3 \\
6\cdot 5 &\equiv 30\equiv -3 \\
7\cdot 5 &\equiv 35\equiv 2 \\
8\cdot 5 &\equiv 40\equiv -4 \\
9\cdot 5 &\equiv 45\equiv 1 \\
10\cdot 5 &\equiv 50\equiv -5 \\
\end{align}

Since $9\cdot 5 \equiv 1$ then we have found the modular inverse to be 9.

When looking at those numbers on the far right side, keep in mind that any multiple of 11 made be added or subtracted to the modulus and it is still equivalent. That is, $-3\equiv 30\equiv 8$ since $-3+3(11)=30$ and $8+2(11)=30$.

Here is a piece of C code that you might find useful:

int Inverse(int n,int a)
{
    int x1 = 1;
    int x2 = 0;
    int y1 = 0;
    int y2 = 1;
    int r1 = n;
    int r2 = a;

    while (r2 != 0)
    {
        int r3 = r1%r2;
        int q3 = r1/r2;
        int x3 = x1-q3*x2;
        int y3 = y1-q3*y2;

        x1 = x2;
        x2 = x3;
        y1 = y2;
        y2 = y3;
        r1 = r2;
        r2 = r3;
    }

    return y1>0? y1:y1+n;
}

Calling Inverse(11,5) returns 9.

Calling Inverse(392,27) returns 363.


Can a coin with an unknown bias be treated as fair? This morning, I wanted to flip a coin to make a decision but only had an SD card:

Given that I don't know the bias of this SD card, would flipping it be considered a "fair toss"?

I thought if I'm just as likely to assign an outcome to one side as to the other, then it must be a fair. But this also seems like a recasting of the original question; instead of asking whether the unknowing of the SD card's construction defines fairness, I'm asking if the unknowing of my own psychology (e.g. which side I'd choose for which outcome) defines fairness. Either way, I think I'm asking: What's the exact relationship between not knowing and "fairness"?

Additional thought: An SD card might be "fair" to me, but not at all fair to, say, a design engineer looking at the SD card's blueprint, who immediately sees that the chip is off-center from the flat plane. So it seems fairness even depends on the subjects to whom fairness matters. In a football game then, does an SD card remain "fair" as long as no design engineer is there to discern the object being tossed?

Here's a pragmatic answer from an engineer. You can always get a fair 50/50 outcome with any "coin" (or SD card, or what have you), without having to know whether it is biased, or how biased it is:

Flip the coin twice.
If you get $HH$ or $TT$, discard the trial and
repeat.
If you get $HT$, decide $H$.
If you get $TH$, decide $T$.

The  only conditions are that (i) the coin is not completely biased (i.e., $\Pr(H)\neq 0, \Pr(T)\neq 0$), and (ii) the bias does not change from trial to trial.

The procedure works because whatever the bias is (say $\Pr(H)=p$, $\Pr(T)=1-p$), the probabilties of getting $HT$ and $TH$ are the same: $p(1-p)$. Since the other outcomes are discarded, $HT$ and $TH$ each occur with probability $\frac{1}{2}$.

By definition, an unbiased coin is fair. So, I suspect the answer is "don't know".

Let $F=${"the coin is fair"} and $B=${"the coin is biased"} and
$P\left ( B\cup \bar{B} \right )=P\left ( B \right ) + P\left (  \bar{B} \right ) = 1$ (obvious). Then $$P\left ( F | B \cup \bar{B} \right ) = \frac{P\left ( F\cap \left (  B \cup \bar{B} \right ) \right )}{P\left ( B\cup \bar{B} \right )}=P\left ( F\cap B \right ) + P\left ( F \cap \bar{B} \right )$$
But, total probability says: $$P\left ( F \right )=P\left ( F| B \right )\cdot P(B) + P(F|\bar{B})\cdot P(\bar{B}) = P(F\cap B) + P(F\cap \bar{B})$$
So $$P\left ( F | B \cup \bar{B} \right ) = P(F)$$ and $B \cup \bar{B}=${"the coin is either biased or not"} (sort of "don't know"). Or, as other people suggested, the fact that we don't know if the coin is biased or not, has no value for determining if the coin is fair or not.


Surface area and volumes, please solve The length of the diagonal of a cuboid is $5\sqrt{5}$ cm and the sum of its length, width and height is $19$ cm. Find its surface area.

As Adam Hrankowski says we are given $$l+w+h=19\\\sqrt {l^2+w^2+h^2}=5\sqrt 5$$
We can square the second to get $$l^2+w^2+h^2=125$$ and we are asked to find the surface area, which is $2lh+2lw+2hw$.  As we have only two equations in three unknowns we do not expect to be able to find $l,h,w$.

One approach is to note that $(5,6,8)$ will solve the system.  I found that because I know that $5^2=25$ and $6^2+8^2=100$.  We can then compute the area as $166$.  If the problem setter has been fair, this should not change for the different sets of $l,w,h$ that solve the problem and we can quit here.

Alternately we can square the first and subtract the squared second.  That gives
$$(l+w+h)^2=361\\l^2+w^2+h^2+2lw+2lh+2hw=361\\2lw+2lh+2hw=166$$

Try drawing a picture. Once you determine the other dimensions, you are well on your way. Pythagoras is your friend here. Here are the two equations you need to solve:

$$l + w + h = 19$$

$$ \sqrt {l^2 + w^2 + h^2} = 5\sqrt{5} $$


Help: rules of a game whose details I don't remember! In a probability course, a game was introduced which a logical approach won't yield a strategy for winning, but a probabilistic one will. My problem is that I don't remember the details (the rules of the game)! I would be thankful if anyone can complete the description of the game. I give the outline of the game, below.

Some person (A) hides a 100 or 200 dollar bill, and asks another one (B) to guess which one is hidden. If B's guess is correct, something happens and if not, something else (this is what I don't remember). The strange point is, B can think of a strategy so that always ends to a positive amount, but now A can deduce that B will use this strategy, and finds a strategy to overcome B. Now B knows A's strategy, and will uses another strategy, and so on. So, before even playing the game for once, there is an infinite chain of strategies which A and B choose successively!

Can you complete the story? I mean, what happens when B's guess correct and incorrect?

Thanks.

In the Envelope Paradox player 1 writes any two different numbers $a< b$ on two slips of paper.  Then player 2 draws one of the two slips each with probability $\frac 12$, looks at its number $x$, and predicts whether $x$ is the larger or the smaller of the two numbers.

It appears at first that no strategy by player 2 can achieve a success rate grater than $\frac 12$.  But there is in fact a strategy that will do this.

The strategy is as follows: Player 2 should first select some probability distribution $D$ which is positive everywhere on the real line. (A normal distribution will suffice.)   She should then select a number $y$ at random according to distribution $D$. That is, her selection $y$ should lie in the interval $I$ with probability exactly $$\int_I D(x)\; dx.$$  General methods for doing this are straightforward; methods for doing this when $D$ is a normal distribution are well-studied.

Player 2 now draws a slip at random; let the number on it be $x$.
If $x>y$, player 2 should predict that $x$ is the larger number $b$; if $x<y$ she should predict that $x$ is the smaller number $a$.  ($y=x$ occurs with probability 0 and can be disregarded, but if you insist, then player 2 can flip a coin in this case without affecting the expectation of the strategy.)

There are six possible situations, depending on whether the selected slip $x$ is actually the smaller number $a$ or the larger number $b$, and whether the random number $y$ selected by player 2 is less than both $a$ and $b$, greater than both $a$ and $b$, or in between $a$ and $b$.

The table below shows the prediction made by player 2 in each of the six cases; this prediction does not depend on whether $x=a$ or $x=b$, only on the result of her comparison of $x$ and $y$:

$$\begin{array}{r|cc}
& x=a & x=b \\ \hline
y < a & x=b &  \color{blue}{x=b} \\
a<y<b &  \color{blue}{x=a} & \color{blue}{x=b} \\
b<y & \color{blue}{x=a} & x=a
\end{array}
$$

For example, the upper-left entry says that when player 2 draws the smaller of the two numbers, so that $x=a$, and selects a random number $y<a$, she compares $y$ with $x$, sees that $y$ is smaller than $x$, and so predicts that $x$ is the larger of the two numbers, that $x=b$.  In this case she is mistaken.  Items in blue text are correct predictions.

In the first and third rows, player 2 achieves a success with probability $\frac 12$. In the middle row, player 2's prediction is always correct.  Player 2's total probability of a successful prediction is therefore
$$
\frac12 \Pr(y < a)  + \Pr(a < y < b)  + \frac12\Pr(b<y)  = \\
\frac12(\color{maroon}{\Pr(y<a) + \Pr(a < y < b) + \Pr(b<y)}) + \frac12\Pr(a<y<b) = \\
\frac12\cdot \color{maroon}{1}+ \frac12\Pr(a<y<b)
$$

Since $D$ was chosen to be everywhere positive, player 2's probability $$\Pr(a < y< b) = \int_a^b D(x)\;dx$$ of selecting $y$ between  $a$ and $b$ is strictly greater than $0$ and her probability of making a correct prediction is strictly greater than $\frac12$ by half this strictly positive amount.

This analysis points toward player 1's strategy, if he wants to minimize player 2's chance of success.  If player 2 uses a distribution $D$ which is identically zero on some interval $I$, and player 1 knows this, then player 1 can reduce player 2's success rate to exactly $\frac12$ by always choosing $a$ and $b$ in this interval. If player 2's distribution is everywhere positive, player 1 cannot do this, even if he knows $D$. But player 2's distribution $D(x)$ must necessarily approach zero as $x$ becomes very large.  Since Player 2's edge over $\frac12$ is  $\frac12\Pr(a<y<b)$ for $y$ chosen from distribution $D$, player 1 can bound player 2's chance of success to less than $\frac12 + \epsilon$ for any given positive $\epsilon$, by choosing $a$ and $b$ sufficiently large and close together. And even if player 1 doesn't know $D$, he should still choose $a$ and $b$ very large and close together.

I have heard this paradox attributed to Feller, but I'm afraid I don't have a reference.

[ Addendum 2014-06-04: I asked here for a reference, and was answered: the source is Thomas M. Cover “Pick the largest number”Open Problems in Communication and Computation Springer-Verlag, 1987, p152. ]

Probability theory is formalized in first order logic so anything you solve using probability theory is solved using logic. Thus such problem can not exist.


Proving $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent I need to prove that  $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent only by Cauchy's test.
Cauchy's test for sequences convergence :
The $\sum \limits _{n=1}^{\infty}a_n$ convergent $\iff$ $\forall \varepsilon >0$ $ \exists$ $ n_0\in \mathbb{N}$ such that $\forall $ $n_0 \le n$ , $\forall$ $p\in \mathbb{N}$ the following holds $\mid S_{n+p}-S_n\mid = \mid a_{n+1}+\dots+a_{n+p}\mid<\varepsilon$

My Attempt:
Let $\varepsilon>0$,$\ p\in\mathbb{N}$. We define $n_0=\frac{p}{\varepsilon}$, and since $-1\le \sin(x)\le 1$ therefore, for each $n\le n_0$: $$\mid \frac{\sin (n+1)}{n+1}+\frac{\sin (n+2)}{n+2}+...+\frac{\sin (n+p)}{n+p} \mid \ \le \ \mid \frac{\sin (n+1)}{n+1}\mid + \mid \frac{\sin (n+2)}{n+2}\mid+...+\mid \frac{\sin (n+p)}{n+p} \mid \le \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+p}\le \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}=\frac{p}{n}<\varepsilon$$
Is my proof correct? if not so how can I prove it only by Cauchy's test as I said earlier, without any further test - such as Dirichlet, Abel...

There have been a couple of requests in the comments to show a way forward after I suggested applying summation by parts.  To that end, we proceed.

Since we asked to prove that the series $\sum_{n=1}^\infty \frac{\sin(n)}{n}$ converges, we begin with the partial sum $S_N$ defined by
$$S_N=\sum_{n=1}^N \frac{\sin(n)}{n}$$
Letting $T_n=\sum_{m=1}^n \sin(m)$ and summing by parts, we have for $N>M+1$
$$\begin{align}
\left|S_N-S_M\right|&=\left|\frac{T_N}{N+1}-\frac{T_M}{M+1}+\sum_{n=M+1}^N \left(\frac1n-\frac1{n+1}\right)T_n\right|\\\\
&\le \max_n (|T_n|)\left(\frac1{N+1}+\frac1{M+1}+\frac1{M+1}-\frac1{N+1}\right)
\end{align}$$
I showed in THIS ANSWER that $\max_n |T_n|\le \frac12(\csc(1/2)+\cot(1/2))\le 2$.
Using this bound for $|T_n|$, we can assert that for any given $\varepsilon >0$, $|S_N-S_M|<\varepsilon$ whenever $N>M+1>\frac{4}{\varepsilon}$.
From Cauchy's criterion, we conclude that $S_N$ converges.  And we are done!

Your proof is wrong because you have
$\sin(n)$ in the numerators
where you should have
$\sin(n+p)$.


Lowest possible vertex density in a graph What is the least amount of vertexes possible in a graph where the distance from any point on or around the graph to any vertex is at most $1$ unit of length?
For example, a map covering $10$ by $10 \text{ km}$ and you want to have waystations distributed on the map so that each waystation is at most $5\text{ km}$ away from any point.

This is the complete graph of arbitrary $V$:


Prove that Petersen's graph is non-planar using Euler's formula Prove that Petersen's graph is non-planar using Euler's formula.
I know that $n - m + f = 2$. But should I count $f$ and prove that the summation does not equal to two or solve to get $f =7$ and argue that it is impossible???

We know the Petersen graph has 15 edges and 10 vertices.  In a planar graph, V+F-E=2.  In Petersen, that would be 10+F-15 = 2, so it would have 7 faces in it's planar embedding. The minimal cycle in Petersen is 5, so it would need to be made from pentagons, hexagons, or larger.

7 pentagons = 35 edges or more.  Half that, rounded down = 17 edges. Each edge can only be used twice, and we've gone over.

Using your notation, we have that $n=10$, $m=15$.  Inspecting the graph, we see that (1) each edge is included in exactly two faces, and (2) each cycle has a length of 5 or greater.  Combining these two facts yields $5f \le 2m.$  Substituting this into Euler's formula, we obtain a contradiction.


Probability of getting at least two cards in the same suit Three cards are drawn from one standard deck of 52 cards. What is the probability of at least two of the cards being in the same suit?

There are a total of $52C3$ sets of 3 cards which could be chosen. You are looking for two positive outcomes: all 3 from the same suit; or two from the same suit and one from another. There are $4 \times (13C3)$ sets of three cards which are all from the same suit. And there are $4 \times (13C2) \times (3\times13)$ sets of cards where two are from the same suit and one is from another. Note here: the $4 \times (13C2)$ comes from choosing a suit (4 options) and then choosing 2 cards from that suit $(13C2)$ options, and then choosing any other card from the other 3 suits ($13\times3$ choices). Together, this gives
$$
\frac{4 \times(13C3) \quad + \quad 4 \times(13C2) \times 39}{(52C3)} = 0.60235...
$$
Alternatively, look at the negation. I pick 3 cards. In how many ways can they all be from different suits? The first card can be anything. The second card can be from any of the other 3 suits (so 39 choices). The third can be from any of the other two suits (so 26 choices). Together, this gives
$$
1 - \frac{52 \times 39 \times 26}{52\times51\times50} = 0.60235...
$$
also giving the same result.
Note, here $(nCr)$ denotes the choice function.

There are in total $C_{52}^{3}$ options if you are going to take random 3 cards from a standard deck.
To have at least 2 cards being in the same suit, you either need to withdraw 3 cards of a same suit, or 2 cards of one suit and one of the other. So there are $$4 \cdot C_{13}^{3} + 4 \cdot C_{13}^{2} \cdot 3 \cdot C_{13}^{1}$$ available options.
Therefore, the probability is $(4 \cdot C_{13}^{3} + 4 \cdot C_{13}^{2} \cdot 3 \cdot C_{13}^{1} )/ C_{52}^{3}$.
Check the combination calculator out if you need the precise result.


Let $M$, $N$, $L$ be midpoints of $AB$, $BC$, $AD$. If $CL\parallel AB$ and $NQ\parallel ML$, find the area of $ABCD$. Let $M$, $N$, and $L$ be the midpoints of $AB$, $BC$ and $AD$, respectively.  Suppose that $CL\parallel AB$.  A point $Q$ on $CL$ is such that $NQ\parallel ML$.    Calculate the area of the quadrangular region $ABCD$ as a function of areas $\Bbb A$, $\Bbb B$, $\Bbb C$.

I tried in so many ways but I couldn' isolate this area in terms of $\Bbb A$, $\Bbb B$, and $\Bbb C$, because I coudn't list the area [CLD] with the areas $\Bbb A$, $\Bbb B$, and $\Bbb  C$.
The answer is
$$[ABCD]=8(\Bbb A+\Bbb B-\Bbb C).$$
I tried to draw AN and divide the areas A and B in $A_1,A_2,B_1,B_2$, and I got the equations:
$$A_2+B_1+H=D,$$
$$A_1+B_2+F=C+G+E.$$
How can I do this?

Let $P=ML\cap AN$, $E=CL\cap DB$, $F=AN\cap BD$, and $R=AE\cap PL$.  Because $AB\parallel CL$ and $L$ biescts $AD$, $E$ is the midpoint of $BC$.  Since $M$ is a midpoint of $AB$, $ME\parallel AD$.  Therefore, $\square ALEM$ is a parallelogram.  Therefore, the diagonals $AE$ and $ML$ bisect each others.  Hence, $R$ is a midpoint of $AE$.  As $N$ is a midpoint of $BC$ and $AB\parallel CL$, we see that $NR\parallel AB$.  Therefore,
$$[MNA]=[MRA].$$
This shows that
$$\Bbb C=[MPN]=[MNA]-[MPA]=[MRA]-[MPA]=[APR].$$

We then have
$$[ARL]=[APL]-[APR]=\Bbb A-\Bbb C.$$
Note that $\triangle ARL\sim\triangle AED$ because $BD\parallel ML$ ($M$ and $L$ are midpoints of $AB$ and $AD$).  Therefore,
$$\frac{[ARL]}{[AED]}=\left(\frac{AL}{AD}\right)^2=\frac14.$$
Since $ED=EB$, we get
$$[AEB]=[AED].$$
So
$$[ABD]=[AEB]+[AED]=2[AED]=2\cdot 4[ARL]=8(\Bbb A-\Bbb C).$$

Note that $\triangle CNQ\sim \triangle CBE$ because $BD\parallel ML$ and $ML\parallel NQ$, so
$$\frac{[CNQ]}{[CBE]}=\left(\frac{CN}{CB}\right)^2=\frac14.$$
Since $L$ bisects $AD$ and $CL\parallel AB$, $E$ also bisects $BD$.  Therefore
$$[CBE]=[CED].$$
Hence
$$[BCD]=[CBE]+[CED]=2[CBE]=2\cdot 4[CNQ]=8\Bbb B.$$
That is,
$$[ABCD]=[ABD]+[BCD]=8(\Bbb A-\Bbb C)+8\Bbb B=8(\Bbb A+\Bbb B-\Bbb C).$$

If all else fails, you can use assign coordinates to the problem.  WLOG, let $A=(0,0)$, $L=(2,0)$, $D=(4,0)$, $M=(0,1)$, and $B(0,2)$.  We can do this because area ratios are preserved under affine transformations.

If $C=(2,2c)$, then $N=(1,c+1)$.  Since $NQ\parallel ML$, $NQ$ is given by the linear equation
$$y-c-1=-\frac{1}{2}(x-1).$$
So $Q=\left(2,c+\frac12\right)$.  The line $ML$ is
$$y=-\frac{1}{2}(x-2)$$
and the line $AN$ is
$$y=(c+1)x.$$
If $P=AN\cap ML$, then
$$P=\left(\frac{2}{2c+3},\frac{2c+2}{2c+3}\right).$$

We have
$$[APL]=\frac{1}{2}\cdot\frac{2c+2}{2c+3}\cdot 2=\frac{2c+2}{2c+3},$$
$$[CNQ]=\frac{1}{2}\cdot 1\cdot\left(2c-c-\frac12\right)=\frac{2c-1}{4},$$
and
$$[MPN]=\frac12\left|\begin{array}{ccc}
1&0&1\\
1&\frac{2}{2c+3}&\frac{2c+2}{2c+3}\\
1&1&c+1
\end{array}\right|=\frac{1}{2}\left(\frac{2c+1}{2c+3}\right).$$
This means $2[APL]-2[MPN]=1$ and $c=2[CNQ]+\frac12$.
However,
$$[ABCD]=\frac{1}{2}\cdot 2\cdot (2+2c)+\frac12\cdot 2\cdot 2c=2+2c+2c=4c+2.$$
That is
$$[ABCD]=4\left(2[CNQ]+\frac12\right)+2=8[CNQ]+4=8[CNQ]+4\big(2[APL]-2[MPN]\big).$$
Hence
$$[ABCD]=8[APL]+8[CNQ]-8[MPN]=8\Bbb A+8\Bbb B-8\Bbb C.$$

Angles $BAD$ and $ABC$ are arbitraries so if the required relation is non constant then the problem is not determined. Consequently in order to verify the relation $[ABCD]=8(\Bbb A+\Bbb B-\Bbb C)$ we choose a convenient quadrilateral for easy calculation.

Let the trapezoid $ABCD$ where $A=(0,0),\space B=(0,2a)\space C=(a,2a)\space D=((2a,0)$ We have $M=(0,a),\space L=(a,0)\space N=(\dfrac a2,2a)$ and $CL\parallel AB$.

$AN$ and $ML$ determine a point of intersection $R=(\dfrac a5,\dfrac{4a}{5})$ and the point $Q=(a,\dfrac {3a}{2})$ is given by intersection of $CL$ and $NQ$.

With this triangles $NQC$ and $ARL$ clearly have an area of $\dfrac{a^2}{8}$ and $\dfrac{2a^2}{5}$ respectively and usually calculation gives the other shadow triangle $MNR$ has an area equal to $\dfrac{3a^2}{20}$.

Now $$8( \dfrac{2a^2}{5}+\dfrac{a^2}{8}-\dfrac{3a^2}{20})=\dfrac{8\cdot15a^2}{40}=3a^2$$

This $3a^2$ is precisely the area of the trapezoid.


Interesting geometry problem (square and two circles) What's the area of the main square? (I think the attached picture defines the problem clearly.)

Let the square be of side $2a$.  Clearly $a > 6$. Then with the origin at the centre of the square, the inside circle can be expressed as $(x-b)^2+y^2=(a-b)^2$

So we know that the intercept of this circle with the positive $Y$ axis must be $a-4$ and the intercept with the $-X$ axis is $6-a$.  Hence
$$b^2+(a-4)^2=(a-b)^2, \qquad (6-a-b)^2=(a-b)^2$$

$$\implies a \in \{3, 8\} \implies a = 8$$

So the area required is $256$.

There are infinite solutions.

Please see the GeoGebra.org figure that I constructed to demonstrate my answer here. (the following text is repeated there, so no need to read it twice)

Here are the steps I took to construct this figure, under the assumption that the given figure in the original post is not to scale:

I drew arbitrary points A1 and B1, connected them by a line, and labeled them "Move me".
I used a circle and perpendicular lines to construct the large square. (Because the construction of the figure is dependent upon the first two points and the hidden circle connecting them, the whole figure scales and moves along with those two points if they are moved)
I found the midpoints of the square's segments and connected them, they cross at the center, point O.
I inscribed the large circle into the square.
I made circles of radius 4 and 6, per the problem, coloring the segments of set length. (This figure does have axes, origin, and absolute distances, but they are only shown in the radii of the two dashed circles)
I used the "circle through 3 points" tool in Geogebra to construct the inner circle using points A, B, and C.

I labeled the given points A,B,C, and O.
I added segments AB, and BC as dashed lines, and used Geogebra to measure the angle ABC. It looks like this might be a right angle, but it turns out that ABC can be almost any angle, depending on the side length of the large square.

Any size square you choose to make using my GeoGebra.org figure fits all of the criteria in the originally given figure, so there are infinitely many possible areas for the large square. Or in other words, constraining the distance between two circles inside of a square (as in the given problem) does not constrain the size of the square.

In case you cannot get the dynamic figure to work, here is the same figure with two arbitrary square side lengths. The given segment lengths remain constant, but the large square and inscribed circle can be almost any arbitrary size.

(In case you cannot get the dynamic figure to work,
here is the same figure with two arbitrary square side lengths. The given segment lengths remain constant, but the large square and inscribed circle can be almost any arbitrary size.)


Find explicit expression of $P_n$ for a given 'graph' Given this image:

For each positive integer $n$ we will define $P_n$ the number of series of length $n$ from the group $\{a,b,c,d,e\}$ so that the first element is $a$ and $2$ consecutive elements are connected by a line (In the image above).

For example: $a-e-d-a-c-a-b-e$ is a legal series of length $8$

Question: Find an explicit form of $P_n$

I tried to think about it, if the first element is $a$ so for the second element we have $4$ options, then we have to choose whether to go back to $a$ or go to the other $3$ options we have.

But I don't think I get it, it has many forms... maybe I alternate between $a$ and $d$ - $n$ times? or maybe go from $a-d-e-a-d-e-a$ etc.. ?

Thank you very much for helping!

The solution below is not related to graph theory, but given the ugly-ish results, it might be the easier one.
Let $A_n$ be the number of path starting and ending at $a$, and $B_n$ be the number of path starting at $a$ but ending at another vertex. So that
$$ P_n=A_n+B_n$$
We have the following recursive equations :
$$\left\{\begin{align}
A_{n+1}&=B_n\\
B_{n+1} &= 4 A_n+2 B_{n}
\end{align}\right.$$
Leading to
$$ B_{n+1} = 2 B_n + 4 B_{n-1}$$
Therefore $B_n$ is of the form
$$ B_n = \lambda r_1^{n-1} + \mu r_2^{n-1}$$
Where $r_1$ and $r_2$ are the solution to $X^2-2X-4=0$, and $\lambda$ and $\mu$ might be found using $B_1=0$ and $B_2=4$.
Finally (I skiped the full calculation, you should be able to do them by yourself), using $P_n=A_n+B_n$ we end up with (note that the expression might be reducible),

$P_1=1$ and for any $n\geq 2$
$$ P_n=\frac{2}{\sqrt{5}}(1+\sqrt 5)^{n-2}(2+\sqrt 5)-\frac{2}{\sqrt{5}}(1-\sqrt 5)^{n-2}(2-\sqrt 5)$$

Yielding the values $(4,12,40,128,416,...)$
Edit The reduced version would be

$P_1=1$ and for any $n\geq 2$
$$ P_n=\frac{-(1-\sqrt 5)^{n+1} + (1 + \sqrt 5)^{n+1}}{4 \sqrt 5}$$

As noted by @bof, we can refer to the Fibonacci sequence, with $\phi = \frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$,
$$P_n=2^{n+1} \frac{\phi^{n+1} - \psi^{n+1}}{4 \sqrt 5} = 2^{n-1}F_{n+1}$$
Edit : Complement on order-$2$ homogeneous linear recurrence with constant coefficients
Let a and b be two constant, and the linear recurrence
$$u_{n + 2} = au_{n+1} + bu_n\quad(R)$$
The polynomial $X^2-aX-b$ is called the characteristic polynomial of the recurrence. It can be shown that $u_n$ verifies
$$u_n = \lambda r_1^{n} + \mu r_2^{n} \quad (*)$$
Where $r_1$ and $r_2$ are the roots of the characteristic polynomial (this works also for higher orders). Indeed if you plug $(*)$ into $(R)$ you get
\begin{align}
u_{n + 2} - au_{n+1} - bu_n &=(\lambda r_1^{n+2} + \mu r_2^{n+2}) - a(\lambda r_1^{n+1} + \mu r_2^{n+1})-b(\lambda r_1^{n} + \mu r_2^{n}) \\
&= \lambda r_1^n (\underbrace{r_1^2-ar_1-b}_{=0}) + \mu r_2^n (\underbrace{r_2^2-ar_2-b}_{=0})\\
&=0
\end{align}
Where the $=0$ statements are directly because $r_1$ and $r_2$ are the roots of the characteristic polynomial.


How to define derivative in Minkowski space My understanding of derivative is like this: it is the unique linear mapping that sends the difference in $x$ to the difference in $f(x)$ when the difference in $x$ is small. To put it more rigorously, $g$ is the derivative of $f$ if and only if for every positive real number $\epsilon$, there exists a positive real number $\delta$ such that
$$\lVert f(x+\Delta x)-f(x)-g(x)(\Delta x)\rVert \le \epsilon \lVert\Delta x\rVert$$
for every $\lVert\Delta x\rVert < \delta$. Is it the same in Minkowski space where $\lVert \Delta x \rVert$ may be equal to $0$?

I've only seen this definition of derivative (The Fréchet Derivative) used for normed spaces - in the case of a semi-norm such as your example, things get very nasty. For example in order for $f$ to be differentiable by this definition we must have $f(x + \Delta x) = f(x) + g(x) \Delta x$ whenever $\Vert \Delta x \Vert \le 0$, i.e. $f$ must be affine outside the light-cone of any particular point. I'm pretty sure that by pasting together this fact at multiple base points you can conclude that $f$ must be affine.

The better abstract definition of derivative to use in this setting is the Gâteaux derivative, which is defined independent of the norm. It thus treats Minkowski space identically to Euclidean space, producing the familiar differential/gradient whenever it exists. If you require the Gâteaux derivative to be linear to call the function differentiable then you recover exactly the classically differentiable functions.

In a Minkowski space such a derivative does not exists in general. To see this focus on direction of $\Delta x$ ,if its direction changes we have a different value for $g$ so we can not define $g$ as function of $x$.