$\sqrt{2}$ cannot represent a rational number I am not asking for a proof that shows me that $\sqrt{2}$ cannot represent a rational number, because I have already seen one by contradiction, which was quite simple, but I have problems in understanding the following proof:

  By Rational Zeros Theorem, the only rational numbers that could possibly be solutions of $x^2 - 2 = 0$ are $\pm1, \pm2$. [Here n = 2, $a_2 = 1$, $a_1 = 1$, $a_0 = -2$. So rational solutions must have the form of $\frac{p}{q}$, where $p$ divides $a_0 = -2$, and $q$ divides $a_2 = 1$.] One can substitute each of the four numbers $\pm1, \pm2$ into the equation $x^2 - 2 = 0$ to quickly eliminate them as possible solutions of the equation. Since $\sqrt{2}$ represents a solutions for $x^2 - 2 = 0$, it cannot represent a rational number.

I have a few questions:

Where does this equation $x^2 - 2 = 0$ come from?
I have read the Rational Zeros Theorem, but I don't understand completely why $\pm1, \pm2$ are the only solutions.
We want to show that $\sqrt{2}$ cannot represent a rational number, that the proof finishes by saying "Since $\sqrt{2}$ represents a solutions for $x^2 - 2 = 0$, it cannot represent a rational number.", which sincerely I am not seeing well the point. Ok, $\sqrt{2}$ is a solution, but that it cannot represent a rational number from this proof, this seems like not connected at all.

I am know I am missing something that you guys understand on the fly, but of course I am not you :D

HINT: $\alpha=\pm\sqrt{2}$ if and only if $\alpha$ is a solution of $x^2-2=0$. The Rational Zeros Theorem states the following:

  If $p/q$ (where $p$ and $q$ are relatively prime integers) is a root of the polynomial $a_0+a_1x+\cdots +a_nx^n$ (where each $a_i$ is an integer), then $p$ must be a divisor of $a_0$ and $q$ must be a divisor of $a_n$

If you apply this result to the polynomial $x^2-2=0$, you get that any rational root of it must be of the form $p/q$ where $p$ is a divisor of $-2$ and $q$ is a divisor of $1$, what are the possibilities for the rationals roots then? can you find a rational root of this polynomial?

This is a consequence of Gauss' Lemma about content of a polynomial with integer coefficients. Here it means, if a monic polynomial with integer coefficient factorizes over the rationals then it factorizes over integers.
So any factorization of $(x^2-2)$ will be a factorization over integers: now see what are the possibilities for the constant term


Is there a geometry in which $\pi$ is a natural number? Is there a geometry in which $\pi$, the ratio of a circle's circumference to its diameter, is an integral number such as $3$ or $4$?

If you use the distance metric of $d=|x_1-x_2|+|y_1-y_2|$ rather than the Pythagorean one then a circle (all points equal distance from a centre) looks like a diamond and the ratio of circumference to diameter is 4.

Interestingly additional fact if you consider all related distance metrics: $d_n=\left(|x_1-x_2|^n+|y_1-y_2|^n\right)^{\frac1n}$ and draw the corresponding circle and then measure the circumference (non-trivial due to different distance metric) and hence calculate $\pi_n$ that the value we use for $\pi$ (in this case $\pi_2$) is a minimum value for $\pi_n$.

In any geometry, in which circles can be defined, either the ratio of circumference of a circle to its diameter is $\pi$ or that ratio is not a constant.  So, no, there is no geometry in which that ratio is constant other than $\pi$.


The complement of a simple non closed curve is path connected? Let $f: [0,1]\rightarrow \mathbb{R}^2$ be a continuous and injective function and $C=f([0,1])$. Is it true that $\mathbb{R}^2\setminus C$ is path connected?

It's pretty obvious that the answer should be positive, but it would be nice if there existed a not quite involved proof of it.

My attempt so far: For every $x\in C$ there exists an $\varepsilon_x>0$ such that $B(x,\varepsilon_x)\cap C$ is connected. (Here by $B(x, \varepsilon)$ we denote the open ball centered at $x$ with radius $\varepsilon$). This is pretty straightforward because $[0,1]$ and $f([0,1])$ are homeomorphic.

The collection $  \{ B(x, \varepsilon_x): x\in C\}$   is an open cover of the compact set $C$, so there exist $x_1, \ldots, x_n$ such that $C\subseteq \bigcup_{i=1}^n B(x_i, \varepsilon_{x_i})=A$. Lets also assume that the $x_i$'s are "consecutive" with $x_1$ and $x_n$ be the starting and ending points of the curve respectively.
We notice in particular that both of the sets $B(x_1, \varepsilon_{x_1})\setminus C$ and $B(x_n, \varepsilon_{x_n})\setminus C$ are path connected. This is a nice property that we will need.

An easy (?) inductive argument shows that $A\setminus C$ is path connected. This is all due to $B(x_1, \varepsilon_1)\setminus C$ being path connected combined with the fact that the set $ B(x_2, \varepsilon_2) \cap C^c $ consists of two connected components, both of which intersect with $B(x_1, \varepsilon_1)$ and so on. From this it follows that $\mathbb{R}^2 \setminus C$ is path connected as well.

I think the proof is ok, but it uses a lot of stuff. Am I missing something obvious here? I also tried to go by contradiction. I supposed that $C$ cuts the plane into some path connected components and then I tried to close the curve avoiding self intersections. The plan was to create more than 2 path connected components and reach a contradiction by  the Jordan Curve Theorem. This seemed quicker, but it didn't  go through.

Consider the following theorem:

  Theorem: Let $n$ be fixed. Suppose that $Y$ is a compact space with the property that $\widetilde{H}_*(S^n-f(Y))=0$ for every embedding $f:Y \to S^n$. Then $I \times Y$ also has this property. (*)

The proof of the above theorem is not difficult to follow (given Mayer-Vietoris and knowledge of the basic properties of homology). From it, it follows by induction that

  Corollary: If $f:D^r \to S^n$ is an embedding then $\widetilde{H}_*(S^n-f(Y))=0.$ In particular, $S^n-f(D^r)$ is connected.

If you have $f:I \to \mathbb{R}^2$, consider $\widetilde{f}=f \circ i$, where $i: \mathbb{R}^2 \hookrightarrow S^2$ is the "inclusion by steoreographic projection". By the corollary, $S^2-\widetilde{f}(I)$ is connected. Since it is open, when you take out a point it will remain connected (**), and it follows that $\mathbb{R}^2-f(I)$ is connected.

(*) - Reference: Bredon

(**) - One can see this in several ways. For instance, it is an open connected subset of a manifold, hence path-connected. Taking out a point of a path-connected manifold of dimension greater than $1$ can easily seen to result on a path-connected connected manifold. In particular, connected.

Another way is computing $H_0(X-\{p\})=H_0(X-\{p\}, D-\{p\})=H_0(X,D)=H_0(X),$ where the first equality is by the long exact sequence of homology, the second is excision and the last also by the long exact sequence of homology, and $D$ is a small disk around $p$ ($X$ is a $2$-dimensional manifold).

Hint for an easier proof: Choose $M > 0$ such that $-M <f(x) <M$ for $x\in [0,1].$ If $z\in U = \mathbb R^2\setminus Gr(f),$ then there is a vertical line segment in $U$ connecting $z$ to one of the lines $y=-M, y=M.$


Transform square region to triangular region How do you express x and y in terms of u and v so that the region $\{(u,v): 0\le u, v\le 1\}$ is mapped to the triangular region in the $xy$-plane with vertices $(0,0)$, $(1,0)$, and $(0,1)$?

Now, this is similar to a question I've asked before, but I still don't understand the answer I got on a similar question.

Also, can anyone suggest tutorials that address how to perform these type of questions?

Thanks.

$$f(u,v)=(u,u*v) $$
$$x:=u,y:=u*v$$
is such a function witch is continous.

you can think of it as shrinking every vertical line in the squere   to a line(or a point) from zero to the line $u=v$,witch means,until the u coordinate of that line.

an other function can be a reflection through the line $v=-u+1$:

$$ f(u,v)=(u-|u-v|,v-|u-v|)$$
$$x:=u-|u-v|,y:=v-|u-v|$$(i'm not sure this is the function.can someone check?)

that is,switching the coordinates if we are above the line,and no change otherwise.

of course,the shapes are homeomorphic, so you can also find a function that is a homeomorphism,but it is less geometricly intuitive.

According to this transformation from triangle $0<s<1,0<t<1-s$ to square $-1<\xi,\mu<1$ is done by:

$$
s = \frac{1 + \xi}{2} \\
t = \frac{1}{4}(1-\xi)(1+\mu)
$$


$N_1$ and $N_2$ are cyclic submodules Please help me to prove the following result:

Let $R$ be a ring with $1$ and $M$ a noetherian  left $R$-module. Suppose that if $N_1$ and $N_2$ are cyclic submodules of $M$ than $N_1+N_2$ is also cyclic.

  Show that $M$ is cyclic

Thanks

An often useful equivalent property to the ascending chain condition is the maximal condition:

  the module $M$ is noetherian if and only if every nonempty set of submodules of $M$ has a maximal element

(maximality is with respect to set inclusion.

Since $M$ is noetherian, the set of its cyclic submodules has a maximal element $N$; since $\{0\}$ is cyclic, this set is not empty. If $N\ne M$, then take $x\in M$, $x\notin N$. By assumption, $N+xR$ is a cyclic submodule, where $xR$ denotes the cyclic submodule generated by $x$: a contradiction to maximality, because $N+xR\supsetneq N$ as $x\in N+xR$. Hence $N=M$.

@ Tobias Kildetoft , Is it right if i answer like the following:

Let $m_0\in M$ and assume that $m_0$ does not generate $M$. We pick an $x_1$ not in the submodule generated by $m_0$ and let $m_1$ be a generator for the submodule generated by $m_0$ and $x_1$. We repeat the argument to get larger and larger submodules. We are picking elements not in the submodule generated by various elements in order to get the next one and  we are using the fact the the sum is cyclic to pick a generator at each step. Thus we obtain an ascending chain
$$(m_0)⊂(m_1)⊂(m_2)⊂....$$
of submodules of $M$. Since $M$ is noetherian there exists some integer $n_0$ such that $(m_n)=(m_{n+1})$ for all $n\ge n_0$.  This means we have $M=(m_{n_0})$ and thus $M$ is cyclic.


How to measure the radius of a sphere? It is easy enough to determine the radius of a circle, using a minimal number of "tools". For instance, one could take any two points on the circle, build a right angle triangle and, by Thales's theorem, the hypotenuse of said triangle would also be the diameter of the circle.

However, I would like to know if there are any simple methods of accurately measuring the diameter/radius of a sphere, without cutting it or altering it in any other way, and using a minimum number of tools and intermediate measurements (to reduce the final error).

One method that comes immediately to mind is to pin down the sphere between two flat surfaces (blocks) and measure the separation between these blocks; but that method introduces many possible sources of measurement errors, and requires the use of additional equipment (the blocks themselves).

tl;dr : Is there a simple way to determine the radius of a sphere without using anything else than a (bendy) ruler?

Are you aware of Eratosthenes' experiment?

A variation of this would involve taking a wedge formed by two planes and putting the sphere between the planes and measuring the distance from the corner of the wedge. Could you work the math for this example?

It is easy enough to determine the radius of a circle. There is another experiment rather

Using a compass, draw four identical circles, each having a flat radius $r$ (known value), on the surface of the sphere with unknown radius say $R$ such that these circles exactly touches one another at six different points on the spherical surface. (see in the diagram)

In this case of tangency of four circles, the radius of sphere $R$ is simply given as
$$R=r\sqrt{\frac{3}{2}}$$


Is the function $f(x)=1^x=1$ considered an exponential function? I am confused about the following:

The exponential function (by definition) is a function of the form $f(x)=a^x$ where $a>0$. However, when $a=1$, we get the constant function $f(x)=1^x=1$. Is the constant function $f(x)=1^x=1$ still considered an exponential function even though it does not have behave like an exponential function? Is the definition of the exponential function that I gave above (that I read in many textbooks) not entirely correct? Should we define the exponential function by:"a function of the form $f(x)=a^x$ where $a>0$ and $a\neq 1$"? I welcome any answer. Thanks!

According to the definition you give, $f(x) = 1$ is indeed an exponential function. As you point out, it is indeed of the form $a^x$ for some $a>0$. Wikipedia's definition explicitly rules out a base-1 exponential as being an exponential function, and this is understandable, as it is somewhat of a degenerate case if you consider it with other exponential functions. But it does satisfy the properties $f'(x) = k f(x)$ for some $k\in\mathbb{R}$ and $f(x+y) = f(x)f(y)$, so from those perspectives, it is also natural to include $f(x) = 1$ as an exponential function.
At the end of the day, it's a matter of convention what you decide to call an exponential function.

Well, an exponential function is one whose rate of change is proportional to its current $y$-value. i.e. one satisfying the differential equation $\frac{dy}{dx}=ky$ for some constant $k$.

$\frac{d}{dx}[a^x]=a^x\ln(a)=ka^x$ for $a \neq 0$.

Now, if $a=0$, i.e. if we want to evaluate $\frac{d}{dx}[1^x]$, we see that $\frac{d}{dx}[1^x]=1^x\ln(1)=1^x(0)=0 \neq k(1^x)$ (unless $k=0$, in which case $ \frac{dy}{dx}$ is not proportional to $y$).

So, essentially, your way is the way to go! As far as definition is concerned, an exponential function is any $f(x)=a^x$, where $a\in \Bbb{R^+} \setminus \{1\}$.


How many ways may 9 different chocolate bars be given to 3 children’s such that each child gets 3 chocolate bars? Assume order is important How many ways may 9 different chocolate bars be given to 3 children’s such that each child gets 3 chocolate bars? Assume order is important.

I’m not sure how to solve this question or where to even start ive tried a few ways but don’t get the correct answer

There are $\binom{ 9}{3, 3, 3} = \frac{ 9!} { 3! 3! 3!} = 1680 $ ways to distribute 9 distinct chocolate bars to the 3 children, each receiving 3 of them. This is the Multinomial theorem.

If you care about the order, then you can multiply by the order that you care about (IE order just to an individual child, or order of all 9).

It seems to me that the division among the three children is just dust in the eyes if the order is important. Let me unpack.

Let's call the chocolate bars: 1, 2, ..., 9.

Suppose given a list of these bars you give the first three to the first kid, second three to the second kid, last three to the third kid. Let's depict it this way:

c1 c2 c3 | c4 c5 c6 | c7 c8 c9,

where ci is the ith chocolate bar. If the order is important, i.e.

1 2 3 | 4 5 6 | 7 8 9

and

2 1 3 | 4 5 6 | 7 8 9

count as two different ways of giving out bars, then the question really is: "How many permutations of the 9 chocolate bars exist?". And the answer is 9!.

If this is not the answer, then the question should be better posed.


Proving $E^{x}[|B_s-B_t|^4]=n(n+2)|t-s|^2$ Prove that:

$$E^{x}\left[ \left| B_{t}-B_{s}\right| ^{4}\right] =n\left( n+2\right) \left| t-s\right| ^{2},$$
where $B$ is a brownian motion.

Now there are two methods I have tried, but I am clearly missing something vital as my attempts are incorrect. Would someone please let me know where I am going wrong?

Attempt 1:

Useful equations:

$$E[B_t^2]=t,\ E[B_t^3]=0,\ E[B_t^4]=3t^2.$$

First idea, as the assumption that B being a brownian motion with mean(0) and variance(t or s)

$$(B_t-B_s)^4=-4 B_s^3 B_t+6 B_s^2 B_t^2-4 B_s B_t^3+B_s^4+B_t^4.$$

Now substituting the useful equations using I don't get the right answer, hence either my calculations are wrong or my assumptions,(understanding) is wrong.

Attempt 2:

Useful equations 2

$$E^x\left[(B_t-x)^2\right]=nt,\\
E^x\left[((B_t-x)(B_s-x))^2\right]=n·\min(s,t),\\
E^x\left[(B_t-B_s)^2\right]=n(t-s).$$

Mini question when stating min(s,t) I understand that it takes the min of (s,t) but if t and s represent time and t>s does that if both start at t=s=0 that $B_t$ being a vector has more elements in it then $B_s$?

$$\left((B_t-x)-(B_s-x)\right)^4.$$

Now expanding this out and using Useful equations 2 I still not get the right answer, hence I am doing something wrong with the calculations or my attempt in itself is flaud.

Would someone please take me through the steps?

It is an exercise of a book "Stochastic Differential Equations" by Øksendal, isn't it? I have also considered about it a couple of days. You seem to confuse 1-dimensional solution and n-dimensional solution.

In n-dimensional case, $ |B_{t}-B_{S}| $ is
$$ |B_{t}-B_{S}|=\sqrt{\left(B_{t}^{(1)}-B_{s}^{(1)}\right)^{2}+\left(B_{t}^{(2)}-B_{s}^{(2)}\right)^{2}+\ldots+\left(B_{t}^{(n)}-B_{s}^{(n)}\right)^{2}} $$

where $ B_{t}^{(i)} $ and $ B_{s}^{(i)} $ are i-th elements of $ B_{t} $ and $ B_{s} $, respectively. Accordingly,
$$ |B_{t}-B_{s}|^{4}=\left\{\left(B_{t}^{(1)}-B_{s}^{(1)}\right)^{2}+\left(B_{t}^{(2)}-B_{s}^{(2)}\right)^{2}+\ldots+\left(B_{t}^{(n)}-B_{s}^{(n)}\right)^{2}\right\}^{2} $$

Since elements of i and j are independent, the expectation with an initial location, x, is
$$ E^{x}\left[|B_{t}-B_{s}|^{4}\right]=\sum_{i}^{n}E^{x}\left[\left(B_{t}^{(i)}-B_{s}^{(i)}\right)^{4}\right]+\sum_{i}^{n}E^{x}\left[\left(B_{t}^{(i)}-B_{s}^{(i)}\right)^{2}\right]\sum_{j\neq i}^{n}E^{x}\left[\left(B_{t}^{(j)}-B_{s}^{(j)}\right)^{2}\right] $$
$$ =n3(t-s)^{2}+n(t-s)(n-1)(t-s) $$
$$ =n(n+2)(t-s)^{2} $$

I am not a mathematician originally, therefore, this may include mistakes. It could be also helpful that some hints for exercises of Øksendal book are provided by http://www.quantsummaries.com/oksendal.pdf.

First you don't write gaussian brownian motion as brownian motion is a gaussian process. (See Oksendal's "Stochastic Differential Equations", chapter 2 section 2.) (Thx for having edited your question.)

Second, you seem to use a one-dimensional brownian motion so that your $n$ would be equal to $1$, which I assume from now.

Last, by definition of $B$ you know that $B_t - B_s$ (with $t \geq s$) has the law of a centered normal distribution with variance $t-s$. Up to a change of variables, calculating the expectation of $(B_t - B_s)^4$ amounts to calculate a sum of the form $\int_{\mathbf{R}} x^4 e^{-x^2 / 2} dx$. Integrating by parts four times and knowing the (well-known) value of $\int_{\mathbf{R}} e^{-x^2 / 2} dx$ will allow you to conclude.


How many significant figures in 0.0 So I know all the significant figures rules, but an edge case that doesn't quite make sense to me is 0. Say I measured the mass of an object to 0.000g.  How many significant figures is that measurement to?  As close as I can tell, it would be 0 significant figures by the rules I know, but it doesn't make sense that a measurement that precise would have no significant figures.

Although there are explicit rules for counting significant figures, they are really a rough idea for the accuracy of a number.  You have shown that the mass is less than $0.000\ 5$ gram, but it could be $0.000\ 000\ 005$ gram for all you know.  In the sense that one significant figure indicates that you know the value to about $\pm 10\%$ (it ranges from $5\%$ if the figure is $9$ to $50\%$ if the figure is $1$), it makes sense to say you have no significant figures at all because you have no bound on the fractional error that might be made.

If you write a number that is between 0 and 1, calculate how many significant figures it has. Then delete the last digit and there is one less significant figure. If you keep deleting the last digit until only zeros remain, there will be zero significant figures.

Example:
$0.020356$ has 5 significant figures,
$0.02035$ has 4 significant figures,
$0.0203$ has 3 significant figures,
$0.020$ has 2 significant figures,
$0.02$ has 1 significant figure,
$0.0$ has 0 significant figures.

It will work with anything that is between 0 and 1. So $0.003725$ and $0.020080$ will work similarly.


$\sum\limits_{n=1}^\infty \log(1+a_n)$ converges absolutely $\iff\sum\limits_{n=1}^\infty a_n$ converges absolutely. $$\sum\limits_{n=1}^\infty \log(1+a_n) \text{ converges absolutely}
 \Leftrightarrow \sum_{n=1}^\infty a_n \text{ converges absolutely}.$$

How to prove this,

Suppose $$\sum_{n=1}^\infty a_n \text{ converges absolutely}.$$ Let $u_{n}=a_{n}$ and $v_{n}=\log(1+a_n)$, then $$\lim_{n\to\infty} \frac{u_{n}}{v_{n}}=1>0 \implies\sum_{n=1}^\infty \log(1+ a_n) \text{ converges absolutely}.$$ How to prove the converse part?

Hint: From the definition of $\ln'(1),$ we have

$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.$$

Thus there is $a>0$ such that

$$\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}$$

for $u\in (-a,a),u\ne0.$

My way:
$$\sum_{n=1}^\infty \log(a_n+1) = \log\prod_{n=1}^\infty (a_n+1)$$
since the series converges absolutely so does the product. Hence $a_n +1 \rightarrow 1$ and so $a_n \rightarrow 0$.
Now the nontrivial step: $a_n \rightarrow 0$, and in proximity of $0$ you have $\log(1+x) = x + o(x^2)$ so dividing by $x$ for any $\epsilon > 0$ exists $n$ such that
$$\frac{|\log(a_n+1)|}{|a_n|} \leq 1+ \epsilon$$
or in other words
$$(1-\epsilon)|a_n| \leq |\log(a_n+1)| \leq (1+\epsilon)|a_n|$$
Apply the sum to the last inequality and you'll prove the proposition both ways.


Uniform convergence of $\sum\limits_{n=1}^{\infty}(1-x)x^n$ I want to study the uniform convergence of $\sum\limits_{n=1}^{\infty}(1-x)x^n  \rm{~~for}~~ x$ on $[0, 1]$. This is my attempt:

First I study convergence on $[0, 1)$: by Ratio test sum is convergent when $|x|< 1$. Hence, $\sum\limits_{n=1}^{\infty}(1-x)x^n$ on $[0, 1)$. When $x = 1$, $f_n(x) = 0$ and sum of series is convergent. Thus, sum of functional series is convergent on $[0, 1]$.

Is my reasoning correct, and if so can it be considered as a rigorous proof? Thanks.

Hint: $(1-x)x^n=x^n-x^{n+1}$. Just use the partial sums, then everything gets simple as these are partial sums of the geometric series.

Though it is a overkill but i will mention it as alternative approach :

Dini's Theorem : If $f_n$ is a sequence of real valued function converging pointwise to a continuous limit function f on a compact set $S$ and if $f_n(x) \ge f_{n+1}(x)$ for each $x\in S$ and for each $n=1,2,3..$  then $f_n \to f $ uniformly .

In your problem set $f_n=\sum_{1}^{n} (1-x)x^n=x-x^{n+1}$ . $ \lim f_n(x)=0$ for all $x\in [0,1]$ ,a compact set . Hence by dini's theorem $f_n\to f$ uniformly .


Find the density function of a random variable that depends on two other random variables with a given joint distribution. For example,

  The joint density of $X$ and $Y$ is given by
  $$f(x, y) = \begin{cases}e^{-x-y}&\text{ if } 0<x<\infty, 0<y<\infty\\ 0 &\text{ otherwise}\end{cases}.$$
  Find the density function of the random variable $Z = X/Y$.

How would I do or even approach a problem like this?

Source http://dept.stat.lsa.umich.edu/~ionides/425/notes/joint_rvs.pdf

There are a couple ways. First, it's usually easiest to attack these things through the CDF. We have $$ F_Z(z) = P(Z\le z) = P(X/Y\le z) = P(X\le zY).$$ Then we can express the final probability as an integral. We want to integrate the joint PDF of $X$ and $Y$ over the region $x\le zy.$ This can be set up as $$ \int_0^\infty \int_0^{zy}e^{-(x+y)}dx\; dy = \int_0^\infty e^{-y}\int_0^{zy}e^{-x}dx\; dy \\= \int_0^\infty e^{-y}(1-e^{-zy})dy \\= \int_0^\infty (e^{-y}-e^{-(z+1)y})dy \\= 1-\frac{1}{z+1}.$$

Now, the PDF is the derivative of the CDF, $$ f_Z(z) = \frac{d}{dz}\left(1-\frac{1}{z+1}\right) = \frac{1}{(z+1)^2}.$$

Another way is, if you're familiar with variable transformations. We can set $Z=X/Y$ and $W = XY$ and then do a transformation of the PDF to get $f_{Z,W}(z,w)$ and then integrate over $w$ to get the marginal PDF of $Z.$

Finally, there is a general formula for the quotient of two continuous random variables (that can be derived from either of these two methods: $$ f_Z(z) = \int uf_{X,Y}(u,zu) du$$

I believe the procedure is to first rewrite the inequality, sketch the area that the inequality represents in $\mathbb R^2$ and then integrate the joint pdf over this area.

Rewrite the inequality:

Your first instinct might be to cross-multiply:

$$P(Z := \frac X Y \le z) = P(X \le Yz)$$

But wait, can we cross-multiply inequalities? Yes if $Y > 0$ surely. What if $Y > 0$ only almost surely, i.e. $P(Y > 0) = 1$ but technically we don't have surely that $Y > 0$ (Precisely there are $\omega$'s s.t. $Y(\omega)<0$, but they are too small to count)?

Well then, we'll use total probability:

$$P(\frac X Y \le z) = P(\frac X Y \le z \cap Y > 0) + P(\frac X Y \le z \cap Y < 0)$$

$$ = P(X \le Yz \cap Y > 0) + P(X \ge Yz \cap Y < 0)$$

Now consider the latter term, $P(X \ge Yz \cap Y < 0)$. Observe that $\{X \ge Yz \cap Y < 0 \} \subseteq \{Y<0 \}$. By monotonicity of probability, since $P(Y<0)=0$, (*) we have that $P(X \ge Yz \cap Y < 0) = 0$.

Hence,

$$P(\frac X Y \le z) = P(X\le Yz)$$

Sketch the area.

You may have noticed in the beginning that the expression $'\frac X Y \le z'$ looks something like an area below or above a line. That's because it is. But which area? Le't take cases.

Case 1: $z<0$

Since $P(X<0)=0$ for the same symmetric reasoning as $P(Y<0)=0$ (*), we have that $P(X \le zY)=P(0<X \le zY < 0) = P(\emptyset) = 0$ because $\{0<X \le zY < 0\}=\{0<0\}=\emptyset$. Nothing to sketch here.

Case 2: $z=0$

Again, since $P(X<0)=0$, we have that $P(X \le zY)=P(X \le 0)=P(X<0)=0$. Nothing to sketch here.

Case 3: $z>0$

Finally, we have something to sketch, namely the set $\{X \le zY\}$. Observe that $X = zY$ is a line, if we plot $Y$ vertically and $X$ horizontally. Imagine $z=5$. Consider any point in $\mathbb R^2$, say $(0,1)$. Is that in the set? Yes. Let's check: $0 = X \le z(1) = z = 5$ is true, and this extends to any $z > 0$. Thus, $\{X \le zY\}$ is all the points above the line $X \le zY$ as seen in $(0,1)$, which is above $X \le zY$. We have shown that $Y>0$, $X>0$ almost surely, hence, by total probability,

$$P(X \le zY) = P(0 < X \le zY)$$

Integrate

Case 1: 0

Case 2: 0

Case 3:

Let $y_0 < 0, y_1 > 0, x_0 < 0, x_1 > 0$. Try to sketch a rectangle with these 4 lines as boundaries and then draw $x=zy$ as a diagonal of the rectangle. Then,

$$P(X \le zY) = \lim_{(y_0,y_1,x_0) \to (-\infty,\infty,-\infty)} \int_{y_0}^{y_1}\int_{x_0}^{zy} f(x,y) dx dy$$

$$\implies P(X \le zY) = P(0 < X \le zY) = \lim_{(y_0,y_1) \to (-\infty,\infty)} \int_{y_0}^{y_1}\int_{0}^{zy} f(x,y) dx dy$$

Because $Y>0$ almost surely, we have $$P(X \le zY) = \lim_{y_1 \to \infty} \int_{0}^{y_1}\int_{0}^{zy} f(x,y) dx dy$$

For reversed order of integration, we have:

$$P(X \le zY) = \lim_{(x_1,y_1,x_0) \to (\infty,\infty,-\infty)} \int_{x_0}^{x_1}\int_{\frac x z}^{y_1} f(x,y) dy dx$$

$$\implies P(X \le zY) = P(0 < X \le zY) = \lim_{(x_1,y_1) \to (\infty,\infty)} \int_{0}^{x_1}\int_{\frac x z}^{y_1} f(x,y) dy dx$$

I'll do the one without the fraction in the bound for integration:

$$P(X \le zY) = \lim_{y_1 \to \infty} \int_{0}^{y_1}\int_{0}^{zy} f(x,y) dx dy$$

$$ = \lim_{y_1 \to \infty} \int_{0}^{y_1}\int_{0}^{zy} e^{-x}e^{-y}1_{x>0}1_{y>0} dx dy$$

$$ = \lim_{y_1 \to \infty} \int_{0}^{y_1}\int_{0}^{zy} e^{-x}e^{-y} dx dy$$

$$ = \lim_{y_1 \to \infty} \int_{0}^{y_1}e^{-y}\int_{0}^{zy} e^{-x} dx dy$$

$$ = \lim_{y_1 \to \infty} \int_{0}^{y_1}e^{-y} [- e^{-zy} + e^{0}] dx dy$$

$$ = \lim_{y_1 \to \infty} \int_{0}^{y_1}e^{-y} [- e^{-zy} + 1] dx dy$$

(*) Pf that $P(Y<0)=0$, i.e. $P(Y > 0)=1$, i.e. $Y>0$ almost surely.

$$P(Y<0) = \int_{y<0} f_Y(y) dy = \int_{y<0} \int_{\mathbb R} f_{X,Y}(x,y) dx dy = \int_{y<0} \int_{\mathbb R} e^{-x}e^{-y}1_{x>0,y>0} dx dy$$

$$ = \int_{y<0} \int_{\mathbb R} e^{-x}e^{-y}1_{x>0}1_{y>0} dx dy = \int_{y<0} e^{-y} 1_{y>0} \int_{\mathbb R} e^{-x}1_{x>0} dx dy = \int_{y<0} e^{-y} 1_{y>0} \int_{x>0} e^{-x} dx dy$$

$$ = \int_{y<0} e^{-y} 1_{y>0} (1) dy = \int_{y<0} e^{-y} 1_{y>0} dy = \int_{\mathbb R} e^{-y} 1_{y>0} 1_{y<0} dy = \int_{\mathbb R} e^{-y} (0) dy = \int_{\mathbb R} 0 dy = 0$$

QED


Area of a region bound by three circular arcs, why doesn't this approach work? Three circular arcs of radius $5$ units bound the region shown. Arcs AB and AD are quarter-circles, and arc BCD is a semicircle.
I tried to find the answer by calculating the total area of the circle if it was a full one,and then I subtracted the area of the two quarter circles:
$$
\pi 5^{2} - 1/4\pi 5^{2} - 1/4 \pi r 5^{2} = 39.25
$$
but the answer should be $50$, why doesn't this approach work
${\large ?}$.

To answer the question as to why it doesn't work: draw the entire circle. Your cuts are not actually removing the entirety of two quarter circles, but only a section thereof.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

In order to evaluate the area $\ds{\mathcal{A}}$ of the $\ds{\color{blue}{\mbox{blue region}}}$, the below evaluation subtracts $\ds{\color{red}{twice}}$ the $\ds{\color{maroon}{maroon}}$ area from the total area $\ds{25\pi}$ of the "big circle". Namely,
\begin{align}
\mathcal{A} & \equiv
\bbox[5px,#ffd]{25\pi - \color{red}{2}\,\,\,
\overbrace{\iint_{\large\pars{0,5}^{2}}
\bracks{x^{2} + y^{2} < 25}\bracks{\pars{x - 5}^{2} + \pars{y - 5}^{2} < 25}
\dd x\,\dd y}^{\ds{\color{maroon}{Maroon}\ \mbox{Area Value}}}}
\\[5mm] & =
25\pi
\\[2mm] & - 2\iint_{\large\pars{0,5}^{2}}
\bracks{0 < r < 5}
\bracks{r^2 - 10r\cos\pars{\phi} -10r\sin\pars{\phi} + 25 < 0}
r\,\dd r\,\dd\phi
\\[5mm] & =
25\pi
\\[2mm] & - 2\iint_{\large\pars{0,5}^{2}}\bracks{0 < r < 5}
\bracks{r^{2} - 10\root{2}r\cos\pars{\phi - {\pi \over 4}} + 25 < 0}
r\,\dd r\,\dd\phi
\\[5mm] & =
25\pi
\\[2mm] & - 2\iint_{\large\pars{0,5}^{2}}\bracks{0 < r < 5}
\bracks{r > 5\root{2}\cos\pars{\phi - {\pi \over 4}} -
5\root{\sin\pars{2\phi}}}\ \times
\\[2mm] & \phantom{-2\iint_{\large\pars{0,5}^{2}}\,\,\,}
\bracks{r < 5\root{2}\cos\pars{\phi - {\pi \over 4}} +
5\root{\sin\pars{2\phi}}}r\,\dd r\,\dd\phi
\end{align}

$$
\mbox{However,}\quad 5\root{2}\cos\pars{\phi - {\pi \over 4}}
\color{red}{\large +}
5\root{\sin\pars{2\phi}} \color{red}{> 5}\quad
\mbox{when}\quad \phi \in \pars{0,{\pi \over 2}}
$$

Then,
\begin{align}
\mathcal{A} & \equiv
\bbox[5px,#ffd]{25\pi -\color{red}{2}\iint_{\large\pars{0,5}^{2}}
\bracks{x^{2} + y^{2} < 25}\bracks{\pars{x - 5}^{2} + y^{2} < 25}
\dd x\,\dd y}
\\[5mm] & =
25\pi - 2\iint_{\large\pars{0,5}^{2}}
\bracks{ 5\root{2}\cos\pars{\phi - {\pi \over 4}} -
5\root{\sin\pars{2\phi}} < r < 5}r\,\dd r\,\dd\phi
\\[5mm] &  =
25\pi - 4\int_{0}^{\pi/4}
\int_{5\root{2}\cos\pars{\phi} - 5\root{\cos\pars{2\phi}}}^{5}
r\,\dd r\,\dd\phi
\\[5mm] &  =
25\pi - 2\int_{0}^{\pi/4}
\braces{25 - \bracks{5\root{2}\cos\pars{\phi} - 5\root{\cos\pars{2\phi}}}^{2}}\,\dd\phi
\\[5mm] &  =
{25\pi \over 2} + 50\ \underbrace{\int_{0}^{\pi/4}
\bracks{\root{2}\cos\pars{\phi} - \root{\cos\pars{2\phi}}}^{2}\,\dd\phi}
_{\ds{1 - {\pi \over 4}}}\ =\
\bbx{\large 50} \\ &
\end{align}


How can I expand $f(x)$ in powers of x? $f(x)=\frac{1-x}{1+x}$. The closest thing I know to this would be $\sum_{k=0}^\infty x^k=\frac{1}{1-x}$ but I don't know how to use it to write $f(x)$

$f(x)=\frac{1-x}{1+x} = \frac{2}{1+x} - \frac{1+x}{1+x}
= 2g(-x) - 1$ where $g(x) =\frac{1}{1-x}$.

You should be able to take it from there.


Primitive and derivative of little $o$ For $x\in {\rm I\!R}$ in a neighborhood of $0$ and $n>1$, we define the function $f(x) = o(x^n)$ and one of its primitives $F(x) = \int_0^x f(t)dt$.

Is it true that:

$F(x) = o(x^{n+1})$
$f'(x) = o(x^{n-1})$

How to prove it? Here are my attempts:

I need to prove that:

$$
\lim_{x\rightarrow 0} \frac{F(x)}{x^{n+1}} = 0
$$

I tried to develop the fraction :

$$
\frac{F(x)}{x^{n+1}} = \int_0^x \frac{f(t)}{x^{n+1}}dt
$$

But I couldn't go further...
A related question seems to add a condition to get this true but I coundn't really understand the comments.

I would use L'Hôpital's rule :

$$
\lim_{x\rightarrow 0} \frac{f'(x)}{x^{n-1}} = \lim_{x\rightarrow 0} \frac{f(x)}{\frac{x^n}{n}} = 0
$$

Is it correct ?

In general the integration of little $o$ is okay, but differentiation is not.

Differentiation
Consider the function
$$f(x) = \begin{cases}
x^3\sin{(1/x^4)}\,, & x\neq0\,;\\
0\,, & x = 0\,;
\end{cases}
$$
which is $o(x^2)$ as $x\to0$ (so it satisfies your condition that $n>1$), but $f'(x) = 3x^2\sin{(1/x^4)} - 4x^{-1}\cos{(1/x^4)}\,,$ which is clearly not $o(x)$ as $x\to0\,$.

The moral is that the asymptotic bounds cannot control the wild oscillations.

Integration
Suppose $f(x)=o(g(x))$ as $x\to0\,,$ and define $F(x)=\int_0^x f(t) \,\mathrm{d}t\,$, and $G(x)=\int_0^x g(t) \,\mathrm{d}t\,$. By definition of $f(x)=o(g(x))\,$, given any $\epsilon >0\,$, we have that there exists some $\delta>0$ such that
$$
\big|f(x)\big| \leq \epsilon \,g(x)
$$
whenever $0<x<\delta\,$. We can then estimate
$$
\big|F(x)\big| \leq \int_0^x \big|f(t)\big| \, \mathrm{d}t \leq \epsilon\int_0^x g(t)\,\mathrm{d}t = \epsilon \,G(x)\,.
$$
As such we have proved that $F(x) = o(G(x))$ as $x\to0\,$.

In differntiation you misused L'Hopital's rule because:
the limit of the fraction of the derivatives first has to exist, if you can prove it's existance only then will it be equal to the limit of the fraction of the main functions.


Sum of normal bundle and tangent bundle. Would somebody be able to prove that the whitney sum of the normal and tangent bundles of a submanifold of $\mathbb{R}^n$ is trivial? Would apreciate a detailed proof...I'm struggling a little.
Tina

Let $f:M^m\rightarrow \mathbb{R}^n$ be an embedding, let $\tau(M)$ be the tangent bundle of $M$, $\nu (M)$ the normal bundle of the embedding $f$.  Then we have the formula

$\tau(M)\oplus\nu(M)\cong f^*\tau(\mathbb{R}^n)$

where $f^*\tau(\mathbb{R}^n)$ denotes the pullback of the tangent bundle of $\mathbb{R}^n$ along $f$ (which you could think of as the tangent bundle of $\mathbb{R}^n$ restricted to $f(M)$).

$\tau(\mathbb{R}^n)$ is canonically isomorphic to $\mathbb{R}^n \times\mathbb{R}^n$, and so its pullback is also trivial.

You may use the Serre-Swan theorem to prove the result: It says there is an exact equivalence between the category of finite rank projective $C^{\infty}(X)$-modules and maps of modules and the category of finite rank real smooth vector bundles on $X$ and maps of vector bundles.
If $di: T(X) \rightarrow i^*T(Y)$ is the tangent mapping and $N_X(Y):=coker(di)$, the Serre-Swan theorem proves the following: Let
$$di^*: T(X)^* \rightarrow (i^*T(Y))^*$$
be the corresponding map of projective $C^{\infty}(X)$-modules. Since $T(X)^*$ is projective and $di^*$ is injective it follows $di^*$ splits and you get an isomorphism
$$S1.\text{  }T(X)^* \oplus N_X(Y)^* \cong (i^*T(Y))^*$$
and $(i^*T(Y))^*$ is a trivial $C^{\infty}(X)$-module of rank $n$ since $T(Y)$ is the trivial vector bundle of rank $n$. It follows there is an isomorphism of vector bundles
$$T(X) \oplus N_X(Y) \cong (i^*T(Y)).$$
Note: If $R$ is a commutative unital ring and $P$ is a finite rank projective $R$-module it follows for any surjection $\phi: M \rightarrow N$ of left $R$-modules and any map $g:P \rightarrow N$ there is a lift $g^*: P \rightarrow M$ with $g^* \circ \phi=g$. A similar result holds dually for injections. Apply this result to the map $di^*$ to get the splitting in $S1$.
Note: The Serre-Swan theorem is a classical result in differential geometry relating finite rank vector bundles on manifolds to finite rank projective modules on commutative rings. Similar result hold for complex holomorphic vector bundles on complex manifolds and algebraic vector bundles on algebraic varieties.
https://en.wikipedia.org/wiki/Serre%E2%80%93Swan_theorem


Using the Well Ordering Principle to Show that n^3 - n is divisible by 6. A Guide On Well-Ordering Principle

MIT OCW Guide on WOP

I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the  following problem:

Use the well-ordering principle to show that n^3 - n is divisible by 6.

Any help would be greatly appreciated!

Note: After factoring out (n-1)^3 - (n-1) it took the form of n^3 - 3n^2 + 2n and the only difference between that and n^3 - n is "3n^2 - 3n" which I proved was divisible by 6 via induction. Is there a better way to solve this problem?

Edit: Show that $n^3-n$ is divisible by $6$ using induction, post is useful however I not supposed to use induction for the main part of the proof.

The basic proof strategy using the WOP is that you need to show that there are no exceptions to the claim that:

$P(n): n^3-n$ is divisible by $6$

and you do this by a proof by contradiction:

Assume there are exceptions. Then let $k$ be the smallest number for which the claim does not hold, i.e. assume that $\neg P(k)$.  Now, if we can now show that $\neg P(k-1)$, then we have our contradiction, because $k$ was supposed to be the smallest number for which the claim did not hold.  That is, you need to show that:

$\neg P(k) \Rightarrow \neg P(k-1)$

(Note: this assumes $k>0$, or, if you start with $1$, that $k>1$. ... if $k=0$ (or $k=1$), you show there is a contradiction by showing that in fact we do have $P(0)$.  So note how WOP is really equivalent to induction, where you show $P(0)$, as well as $P(k-1)\Rightarrow P(k)$, which of course is just the contrapositive of $\neg P(k) \Rightarrow \neg P(k-1)$)

Well, you can show that $(k-1)^3-(k-1)=(k^3-k)-(3k^2-3k)$

So, given the assumption that $\neg P(k)$, i.e. that $(k^3-k)$ is not divisible by $6$, then (start of inside proof by contraposition) if $(k-1)^3-(k-1)$ were to be divisible by $6$, then $(3k^2-3k)$ cannot be divisible by $6$. OK, but obviously $(3k^2-3k)=3(k^2-k)$ is divisible by $3$. And for the remaining term $(k^2-k)$ we have that $(k^2-k)=k(k-1)$, and obviously either $k$ or $k-1$ is even, and so that term is divisible by $2$, and so $(3k^2-3k)$ is divisible by $6$ after all: Contradiction! So, $(k-1)^3-(k-1)$ is not divisible by $6$, under the assumption that $(k^3-k)$ is not divisible by $6$

$n^3-n=n(n-1)(n+1)$ which is product of three consecutive numbers.

Well if you have three consecutive numbers one of them is a multiple of two and one is a multiple of three so the product is a multiple of six.


Causality of a discrete recursive system I am new in this site and this my first question.
How can I mathematically prove that the system with a transfer function like the below one is causal?
$$H(z)=\frac{B(z)}{A(z)}=\frac{b_{0}+b_{1} z^{-1}+b_{2} z^{-2}+\cdots+b_{N} z^{-N}}{1+a_{1} z^{-1}+a_{2} z^{-2}+\cdots+a_{M} z^{-M}}$$
P.S:
I know that for a system to be causal there has to be past inputs in the equation of difference and that the inpulse resoponse has to be zero for negative $n$ ($h[n]=0 , n<0$) but the thing is that when i try to prove the second one by putting values into $n$ and writing down the result of $H[n]$, I always get a reslut that depents on a past value of $H[n]$

How can I mathematically prove that the system with a transfer function like the below one is causal?

You can't.
The transfer function $H(z)$ in itself doesn't tell you if the filter is causal or not.
More precisely. An LTI is causal if $h[n] =0$ for $n<0$. But $H(z)$ is not enough to determine $h[n]$. You need also the ROC (region of convergence of the Z transform).
For example, $H(z)=\frac{1}{1+z^{-1}}$ (a particular case of your transfer function) is the Z transform both of a causal response function  and an anti-causal one (details here).
Specifically: for a LTI to be causal, the ROC of the $Z$ transform must extend "to the outside", ie., must be of the form $|z| > R_0$ or $|z| \ge R_0$.
For a given rational funtion, $R_0$ would be the largest absolute value of all poles.

Just write down the corresponding difference equation. Then it is obvious that the system is causal. There is no need to compute the pulse response. The difference equation reads

$y(n) + a_1 y(n-1) + \ldots = b_0 x(n) + b_1 x(n-1) + \ldots$

Clearly, if you solve for $y(n)$, the right hand side only depends on previous $y(n)$ and previous $x(n)$ so it is causal.

It would not be causal, if any exponent in the numerator or denominator of the transfer function is positive (e.g. there is some term with $z^2$, instead of $z^{-2}$ for instance). The difference equation would then contain $x(n+2)$ in the right hand side for $y(n)$. That is clearly not causal.

In an abstract way, all systems are causal that have transfer functions which is rational polynomial in $z^{-1}$ with a nonzero lowest order coefficient in the denominator.


How many real roots does $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$ have? A multiple choice questions asks me to find the number of real roots of $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$.
A brute force approach involving the factor theorem and algebraic long division gives $3$ real roots.
Is there a quicker way?

I was thinking Descartes rule of signs tells us that since there are $4$ sign changes, there are at most 4 positive roots. We can also use the rule on $f(-x)$ to say there is at most one negative root. However, I can't see how to continue along this path.
I tried the derivative and deduced that there at least two turning points on the curve $y=f'(x) = 10x^4 - 12x^3 + 3x^2 - 8x - 6$, since $10 > 0$ and the $y$ intercept is negative. But that doesn't rule out anything since the turning points could be above the axis.
Of course I could work this out in more detail, but I am looking for efficiency.

We only need to use Descartes' rule of signs.
When $x\le0$ you have worked out that $f(x)$ has one negative root.
As $f(x+1)=2x^5+7x^4+9x^3+x^2-13x-6$, there is one sign change so $f(x)$ has one positive root when $x\ge1$.
Consider $g(x)=x^5f(1/x)$. As $g(x+1)=4x^5+14x^4+12x^3-7x^2-17x-6$, there is one sign change so $g(x)$ has one positive root when $x\ge1$. Hence $f(x)$ has one positive root when $0<x<1$.
Therefore $f(x)$ has three real roots.


Doubts about Goedel Completeness Theorem My book (Mendelson) states this theorem the following way:

  (1) A logically valid formula of a first order theory is a theorem.

On Wikipedia the statement is a little more general:

  (2) For any first-order theory T with a well-orderable language, and any sentence s in the language of the theory, there is a formal proof of s in T if and only if s is satisfied by every model of T.

Now “well-orderable language” is implicit in (1), and the “only if” part is fairly obvious. My doubt is about the hypothesis of “validity” used in (1) which is stronger than “true in any model” used in (2).

Because you can have an interpretation of a 1st order language which satisfies all the logic axioms, but doesn’t satisfy proper axioms, so isn’t a model of the theory. Basically valid means true under any interpretation, and there are more interpretations than models.

Am I understanding this right? Is the stronger hypothesis of (1) strictly required? Or it is also possible to prove the theorem:

  (1’) A formula true in any model of a first order theory is a theorem.

Rephrasing your question a bit more clearly, you're asking why the seemingly-weaker completeness theorem

  Every validity is provable from the empty theory

implies the seemingly-stronger completeness theorem

  For every theory $T$ and every sentence $\varphi$ true in every model of $T$, we have $T\vdash\varphi$,

or in symbols why "$\emptyset\models\varphi\implies\emptyset\vdash\varphi$" implies "For all $T$, $T\models\varphi\implies T\vdash\varphi$."

You are right to be suspicious: on the face of it you cannot immediately deduce the stronger version from the weaker version. However, they are nonetheless equivalent:

First, we handle the case where $T$ is finite. If $\varphi$ is a formula of $T$ then $\varphi\wedge\psi$ is a validity (where $\psi$ is the conjunction of the axioms in $T$), hence by the weak version we have $\emptyset\vdash\varphi\wedge\psi$; this in turn gives $\{\psi\}\vdash\varphi$, and that gives $T\vdash \varphi$ since we have $T\vdash\psi$.
Next, we handle the case where $T$ is infinite. Here we cannot "put all of $T$ into one sentence." However, the compactness theorem fixes things: if $\varphi$ is true in every model of $T$, then there is some finite subtheory $T_0\subseteq T$ such that $\varphi$ is true in every model of $T_0$. (HINT: note that $T\cup\{\neg\varphi\}$ is unsatisfiable ...) Hence $T_0\vdash\varphi$, and a fortiori $T\vdash\varphi$.

However, this isn't really necessary: the proof of the seemingly-stronger version of completeness is basically the same as the proof of the seemingly-weaker version.

You are confused.  A first-order language includes both the logic axioms and the proper axioms; it is not defined by the logic axioms alone.  So an 'interpretation' which satisfies the logic axioms but not the proper axioms simply isn't an interpretation of that language, period.  An interpretation is a mapping between the 'concepts' (i.e predicates, functions, constant terms, etc) of the language and corresponding objects in the model - ALL of them.  In that sense, 'interpretation' and 'model' are synonymous terms.


Riesz's Lemma on $l_\infty$ Riesz's Lemma says that for a Banach space $X$, a proper closed subspace $Y\subset X$, and $\epsilon\geq 0$ there exist $x\in X$ with $\|x\|=1$ and $d(x,Y)\ge 1-\epsilon$.

I have to find a closed subset in $l_\infty$ that demonstrates that the above result is not true for $\epsilon=0$. That is, construct $Y$ such that for all $x\in l^\infty$ with $\|x\|_{l^\infty}=1$ it holds $d(x,Y)<1$.

The problem is that I'm not suposed to use weak topology, so I can't use Jame's Theorem (I've read a solution using it).

We proceed in two steps: first show that we can construct such a subspace using a non-norm attaining functional, second construct such a functional for $l^\infty$.

Assume that $g\in X^*$ is a functional such that for all $x$ with $\|x\|_{X}=1$ it holds $|g(x)|<1$. I.e., the  supremum in the definition of the functional is not attained.

Define $Y:=\ker g$. Let $x$ with $\|x\|_X=1$ be given. Assume wlog $g(x)>0$. Then there is $z$ with $g(x)<g(z)<1$.
Define $y:=x - g(x)g(z)^{-1}z$. Then $g(y)=0$, thus $y\in Y$, and
$$
\|y-x\| = g(x)g(z)^{-1} \|z\|<1.
$$
This shows $d(x,Y)<1$ for all $x$ with $\|x\|_X=1$.

It remains to construct such a functional $g\in (l^\infty)^*$.
Let $\phi\in (l^\infty)^*$ be an extension of the limit functional. Let $(a_k)$ be a sequence of positive numbers such that $\sum_{k=1}^\infty a_k=1$.
Define
$$
g(x)=\sum_{k=1}^\infty a_kx_k - \phi(x_k).
$$
First let us show $\|g\|=2$. For given $n$ set
$$
x_k = \begin{cases} 1 & k\le n\\
-1 & k >n
\end{cases}
$$
Then
$$
g(x)=\left(\sum_{k=1}^n a_k - \sum_{k=n+1}^\infty a_k\right) +1 \to\left( \sum_{k=1}^\infty a_k\right)+1 = 2\quad \text{ for } n\to\infty.
$$
If $x$ is constant one, then $g(x)=0$. If $x$ is of unit norm, but $x_k<1$ for some $k$, then $g(x) < g(x + (1-x_k)e_k)$. This shows that $|g(x)|<\|g\|$ for all $x$ with $\|x\|_{l^\infty}=1$.

So we do not use James' theorem, but rather construct a not norm-attaining functional explicitly. The construction in James original paper uses Banach limits as well, there I got the inspiration from.

How about $Y$ as the set of all sequences in $\ell_\infty$ that converge to $0$, and $x = (1,1,1,1,1,\ldots)$.


Must a basis for an $n$-dimensional vector space have $n$ vectors? Does a basis for an $n$-dimensional vector space have to have $n$ vectors? For example, if I form a basis for $\mathbb{R}^n$, do I need at least $n$ vectors in my basis set?

In other words, can I form a basis for $\mathbb{R}^n$ by using only $n-1$, or less, vectors?

Note that, in this question, we only consider the whole vector space not creating a basis for a subspace.

Yes, that is precisely the definition of the dimension. The number of vectors you need in a basis

The definition of basis of $\Bbb R^m$ is a set of vectors that are both linearly independent and spans $\Bbb R^m$. Assume that there are n vectors. Those n vectors in $\Bbb R^m$ form a matrix $[n_1 n_2 ...... n_n]$. If those vectors are linearly independent, that means that there must be n pivots in each column(because if there is a column without a pivot, there is a free variable, which leads to linear dependence).

That shows n <= m.(if n>m, there will be columns without pivots and again linear dependence) If you are not convinced, write a $3 \times 2$ or $4 \times 2$ matrix and see if they are linearly dependent.
Analyzing the second definition--those vectors span $\Bbb R^m$.

If that's the case, then the matrix will have a pivot in each row(if there is a row without a pivot, which has the form $[0,0,\cdots , 0]$, then the matrix can't span $\Bbb R^m$.) That gives us that $n \geq m$ .(again, try out some matrices that $n < $ m to see whether they span IR^m).
combine those two conclusions, we have $n\leq m $ and $n \geq m$. So n must equal to m.
Therefore a basis has to have the form of $n \times n$.


Complex analysis problem, using limsup Im trying to solve the following problem.

Let $\Omega \in \mathbb{C}$ an open bounded set, let $f\colon \Omega \to \mathbb{C}$ be holomorphic, and supose there exists a constant $M \geq 0$ wich satisfies that if $(z_n)_{n\geq 1}$ is a convergent sequence of points in $\Omega$ such that if $\lim\limits_{n \to \infty} z_n \in \partial\Omega$ then $\limsup\limits_{n \to \infty} |f(z_n)| \leq M $.

Prove that for all $z \in \Omega$ we have $|f(z)| \leq M$.

My attempt: First supose $f$ is not constant because if it is constant i have nothing to prove, because the lim sup is equal to $f(z)$ in any point and it would be true. Then, as we said, supose $f$ is not constant, and supose the condition doesnt satisfies, that is there exists some $z$ such that $|f(z)| > M$, then the all the $z$ that do this cannot be a discrete set, because I can pick the maximum of the set and would get that the function is constant by the Maximum Modulus Principle. So the set $A =$ {$z$ such that $|f(z)| > M$} is infinite, now if $\overline A = \Omega$ i would reach a contradiction, because I can pick a sequence $(z_n) \subset A$, convergent to the border which would yield that $M < limsup |f(z_n)| \leq M$. The thing is if A is not dense in $\Omega$ then Im a bit lost.

You don't need that $A$ is dense, all you need is that there is at least one sequence converging to (a point of) $\partial\Omega$ with $\limsup \lvert f(z_n)\rvert > M$.

On the path you started on, you can modify $A$, choose an $\varepsilon > 0$ such that $\lvert f(z)\rvert > M+\varepsilon$ for some $z\in\Omega$, and use $\tilde{A} = \{z \in \Omega : \lvert f(z)\rvert > M + \varepsilon\}$. When only demanding $\lvert f(z)\rvert > M$, you cannot deduce that $\limsup \lvert f(z_n)\rvert > M$ when $z_n$ is a sequence in $A$ converging to a point of $\partial \Omega$, therefore we only consider points where the value of $\lvert f(z)\rvert$ is larger than something fixed that is larger than $M$.

Then the problem is to deduce that there is at least one sequence in $\tilde{A}$ converging to a point of $\partial \Omega$. Now, if there weren't such a sequence, then $\tilde{A}$ would have a positive distance from $\partial \Omega$, that means there is a $\delta > 0$ such that

$$\tilde{A}\subset \Omega_\delta = \{ z \in \Omega : D_\delta(z) \subset \Omega\}.$$

But then $f$ would be holomorphic on $\Omega_\delta$ and continuous on $\overline{\Omega}_\delta$, and

$$\sup_{z\in\partial \Omega_\delta} \lvert f(z)\rvert \leqslant M+\varepsilon < \sup_{z\in\tilde{A}} \lvert f(z)\rvert$$

contradicts the maximum modulus principle.

Take the closure of the curve union lim sup.
Take max of closure.
By Max mod principle , the max is on the boundary.
Only lim sup could be on the boundary.
QED


Sum of the areas of an arbitrary subdivision of a rectangle equals the area of the original rectangle Let $R$ be a $3\times 5$ rectangle divided as shown in the figure below, where the division points are arbitrary. The the sum of the areas of the subrectangles is always $15$. Prove that this is similarly true for a general rectangle and an arbitrary subdivision.
I found this https://math.stackexchange.com/a/771043/533661, but I haven't learnt the Green's Theorem yet. Are there other ways to prove this statement? Thanks.

For any such subdivision, extend all the horizontal and vertical segments to the boundary of the rectangle. Then you have a sum of areas of rectangles arranged in a grid.
The dimensions of the small rectangles will be differences of the $x$ and $y$ coordinates of the given points. If you write out the sum carefully you can make it telescope to give you the area of the containing rectangle.
The algebra and notation will be tedious but straightforward.

Here, as the OP requests, are arbitrary division points (not some given set):
$$A = \color{red}{x_2 y_1}\color{black} + \color{purple}{x_1 (h - y_1)}\color{black} + \color{blue}{(w - x_2) y_2}\color{black} + \color{green}{(w - x_1)(h - y_2)}\color{black} + \color{yellow}{(x_2 - x_1)(y_2 - y_1)}\color{black} = w h .$$

The same principle holds within any component rectangle as well, of course, yielding the same total area.


Is there any way to solve equations like these? Is there any formula or method to solve equations like

$ ax + b^x = c $

Where $ a, b, c $ are real numbers

I mean exact solutions not approximations

Had any mathematician already worked on these?

The Maple command simplify(solve(a*x+b^x=c,x)); gives

$$\frac{c\ln(b)-a\,\mathrm{LambertW}\left(\dfrac{\ln(b)\,b^{c/a}}{a}\right)}{a\ln(b)}$$

So, there is a solution in terms of the Lambert W function, a special function.
I don't know enough about it to explain it though but thought you might be interested.

Solving $f(x)=c$ involves intersection of curve $y=f(x)$ and a line $y=c$. This intersection can be computed with a graphs $(x,f(x))$ and $(x,c)$ in 2d space in such way that points $(x,y)$ are in the intersection if $y=f(x)=c$.

Your problem also has $a$ and $b$ so it requires some currying:
 $(a,b) \rightarrow (x) \rightarrow f(x,a,b)$. I.e. once you know values of $a$ and $b$, the equation solving problem can be attempted.


Equivalent condition of $f$ to be $L^2$ Let $f$ be Lebesgue measurable function on $[0,1]$. Prove that $f\in L^2[0,1]$ iff $f\in L^1[0,1]$ and there is a monotone increasing function $g$ such that for all closed intervals $[a,b]\subset[0,1]$, $$\left| \int_{a}^b f(x)dx \right|^2 \le (g(b)-g(a))(b-a)$$

My attempt: Since $[0, 1]$ is a finite measure space, it is clear that $L^2\subset L^1$. But how should I prove that there is a monotone increasing function $g$ satisfying the above inequality? Also I have to prove the converse too, but I have no idea for proving the converse, either.

Any hints or advices will help a lot!

For the converse: You can approximate $f \in L^1$ in $L^1$-norm by functions of the form $\sum_{k=1}^n c_k 1_{[a_k,b_k)}$ with disjunct intervals $[a_k,b_k)$. This convergence is also uniformly in the total-variation norm, i.e. $|\int_a^b f_n \, \rm{d}x - \int_a^b f \, \rm{d}x| \rightarrow 0$ uniformly for all $0 \leq a < b \leq 1$.

Now, use the given property for the intervals $[a_k,b_k)$, that is
$$c_k^2 = \left|\int_{a_k}^{b_k} f_n \rm{d} x \right|^2 \leq 2\varepsilon_k^2+2(b_k-a_k)(g(b_k)-g(a_k)),$$
where
$$\varepsilon_k := \int_{a_k}^{b_k} |f_n-f| \, \rm{d} x$$
and sum it
$$\int_0^1 f_n^2 \, \rm{d} x = \sum_{k=1}^n c_k^2 \leq 2\sum_{k=1}^n  \varepsilon_k^2+ 2\sum_{k=1}^n(b_k-a_k) (g(b_k)-g(a_k)) $$
Using a Telescoping Sum argument and the monotonicity of $g$, the last term can be bounded by $2(b-a) (g(b)-g(a))$. Since $\varepsilon_k \leq \int_0^1 |f-f_n| \, \rm{d} x$, we can choose $n \geq N$ such that $\varepsilon_k \leq 1$. Thus $\varepsilon_k^2 \leq \varepsilon_k$ and
$$\sum_{k=1}^n \varepsilon_k^2 \leq \sum_{k=1}^n \varepsilon_k \leq \int_0^1 |f-f_n| \, \rm{d} x.$$
Chosing a $\lambda$-a.s. convergent subsequence, one can apply the Fatou-Lemma in order to get
$$\int_0^1 f^2 \rm{d} x \leq 2(b-a)(g(b)-g(a))$$

$\Rightarrow$: To construct $g$ you may just use the Hölder-inequality:
$$\left| \int_a^b f(x) \, \rm{d} x \right|^2 \leq (b-a) \int_a^b f(x)^2 \, \rm{d} x,$$
i.e. one can define
$$g(y) := \int_0^x f(x)^2 \, \rm{d} x.$$

$f \in L^2[0,1] \Rightarrow f^2 < \infty \text{ a.s.}$
then $f < \infty \text{ a.s.}$ and
$$ \lvert f \rvert \leq 1_{\{f^2 \leq 1 \}} + f^2 $$

Hence
$$\int \lvert f \rvert \mathrm{d}\lambda \leq 1 + \int f^2 \mathrm{d}\lambda < \infty,$$ i.e. $\lvert f \rvert \in L^1[0,1]$, then $f$ is also.

The converse is not true.
Take for example $$f = \sum\limits_{n=1}^{\infty}n^{1-\varepsilon} \cdot 1_{\left[\frac{1}{n+1}, \frac{1}{n}\right)} $$.

$f \in L^1[0,1]$, since the series $\sum\limits_{n=1}^{\infty}\frac{n^{-\varepsilon}}{n+1}$converges for all $\varepsilon$, but $f^2 \notin L^1 \Rightarrow f \notin L^2$.


Issue with modelling word problems as cubic equations I am having problems with a specific set of word problems, which are meant to be modeled as cubic equations in order to be solved.

I will give some examples to specify where I can't solve it.

The first one is:
"In a rectangular piece of cardboard with perimeter 20ft, three parallel and equally spaced creases are made. The cardboard is then folded t make a rectangular box with open square ends.

Show that the volume of the box is     $V(x) = x^2(10 - 4x)"$

So, I know that the volume of a box is height times width times length. I sub in $x$ for width and another $x$ for height and have my $x^2$. Here, though, is my problem. The given answer for length, as far as I understand, is $(10-4x)$. I don't see where this comes from. I'll give another example, continued example here:

"In a rectangular piece of cardboard with perimeter 30in, two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equilateral triangles.

Show that the volume of the prism is $V(x) = (\frac{\sqrt3}{2}x^2)(15-3x)$"

I can see the pattern here of the answer taking the format of (area of end shape)(half the perimeter - number of folds) but I don't understand where the half the perimeter comes from. I've always had quite a lot of difficulty translating word problems to mathematical forms and this has thrown me. The questions after this scale up in difficulty reasonably quickly, so any advice or resources where to get good explanations of how to to about translating these kinds of problems would be greatly appreciated.

As to my attempts thus far, I think I'm making a mistake now that the shape isn't open. For instance a problem like "In a sheet of length y and width z, a square of side x is cut out. Find the maximum volume of the open-top box which results" seems reasonably straightforward to me, whereas a problem such as:

"A 10cm by 20cm piece of sheet metal is cut and folded to make a box with a top.

Show that the volume of the box is $V(x) = x(10-x)(10-2x)$"

This is similar to the ones I find difficult above. The way I solved it, possibly incorrectly, was to draw the shape and observe that, where the box folded upwards, it could only have gone a maximum of 10-x, or else the box would not have a closed top to match the bottom. I'm not sure this is correct reasoning as I find it difficult to apply the same structure to the two problems given above, which seem structurally to be similar.

Any help on where to focus my thinking on these types of problems would be greatly appreciated.

There seems to be an error in the 2nd question:
The area of the base of the prism = 1/2 x base x height
Take the base to be x, and since all 3 sides of the base are equal, using Pythagoras' theorem, make the height of the triangle, h
Then    h^2 +〖(1/2 x)〗^2 = x^2
so the height h = sqrt(x^2 - [(1/2)x]^2  which = sqrt(3{x^2}/4)
the length of the prism would have been calculated the same way as given by Andre Nicolas for the 1st question: (30-3x-3x)/2 = 15-3x
So volume = area of base x length of prism
= 1/2 (x)[sqrt({3x^2}/4)(15-3x)
= (sqrt 3)(x^2)(15-3x)/4 rather than divided by 2 as given in the question


almost surely convergence of a random variable I came across the following problem: Let $\xi\in L^{2}(\mathbb{P})$. Can we obtain  that $\xi/T^{\epsilon}$ converges almost-sure to zero, for any $\epsilon$ as $T$ tends to infinite?
Thus, is $\frac{\xi}{T^{\epsilon}}\overset{a.s.}{\rightarrow }0$ right?
Thanks for your consideration.

$X=\lim_{T\to\infty}\xi/T^{\epsilon}$ exists and by the dominated convergence theorem $$0=\lim_{T\to\infty}E(\lvert\xi/T^{\epsilon}\rvert)=E(\lvert X\rvert)$$ so $X=0$ almost surely.


How to prove volume of parallelepiped? Let's say that three consecutive edges of a parallelepiped be a , b , c . Then how to show that volume is = [a b c]
Or = a.(b × c) ?

Also how to prove this  this too?

Let $\vec a$ and $\vec b$ form the base. Then the area of the base is,

$$|\vec a \times \vec b|$$

The height of the parallelogram is orthogonal to the base, so it is the component of $\vec c$ onto $\vec a \times \vec b$ which is perpendicular to the base,

$$\text{comp}_{\vec a \times \vec b}\vec c=\frac{|c. (\vec a \times \vec b)|}{|\vec a \times \vec b|}$$

Multiplying the two together gives the desired result.

area of base of parallelepiped (parallelogram) = $\mathbf b \times \mathbf c$

the vector $\mathbf b \times \mathbf c$ will be perpendicular to base

therefore:
$$
\begin{align}
\text{volume of parallelopiped} &= \text{area of base} \times \text{height}\\
                              &= (\mathbf b \times \mathbf c) \times A \cos \theta\\
                              &= \mathbf a\cdot(\mathbf b \times \mathbf c)
\end{align}
$$


How to find the normal to a curve at a specific point? Can someone walk me through how to find the normal to a curve at a specific point? Taking one example from my Calculus Textbook, it states Find the equation of the normal to the curve $sqrt(y)+xy^2=5$ at the point (4,1)?

Now, I know a Normal is perpendicular to a tangent line, and has a reverse-signed reciprocal slope to our tangent. But beyond the general knowledge there, how do I go about solving this as asked?

There are two basic approaches here:

Rewrite the defining equation for your curve to make one coordinate a function of the other. In this case it is easy to reach $$ x = \frac{5-\sqrt y}{y^2}$$ which you can differentiate symbolically to find a tangent vector and then produce the equation for the normal (goes through the specified point, and perpendicular to the tangent).
Know that the normal to a level curve $f(x,y)=c$ is always in the direction of the gradient of $f$, so compute $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ at $(4,1)$ construct a line through $(4,1)$ in that direction.

We call $ \mathbf {r} $ the curve. To find the normal vector in a point $ p $ of the curve you derive the vector$ \mathbf {r'}/||\mathbf {r'}||$  and calculate in $ p $ the derivate. In your case first find a parametrization of the curve.


Find gradient of this implicit function How to find a gradient of this implicit function?
$$
xz+yz^2-3xy-3=0
$$

EDIT: Abhinav pointed out a mistake, which has been corrected for posterity.

To find $\frac{\partial z}{\partial x}$ by implicit differentiation means to differentiate both sides of the equation with respect to $x$, while remembering that $z$ is implicitly a function of $x$ (we didn't write it with function notation such as $z(x)$).  Don't forget to use the product rule when differentiating a product!
$$
\begin{align}
xz + yz^2 - 3xy - 3 &= 0 \\
\tfrac{\partial}{\partial x} \left[ xz + yz^2 - 3xy - 3 \right] &= \tfrac{\partial}{\partial x} \left[ 0 \right] \\
\tfrac{\partial}{\partial x} \left[ xz \right] + \tfrac{\partial}{\partial x} \left[ yz^2 \right] - \tfrac{\partial}{\partial x} \left[ 3xy \right] - \tfrac{\partial}{\partial x} \left[ 3 \right] &= 0 \\
\left[ 1 \cdot z + x \cdot \tfrac{\partial z}{\partial x} \right] + y \cdot 2z \tfrac{\partial z}{\partial x} - 3y - 0 &= 0 \\
z + x \tfrac{\partial z}{\partial x} + 2yz \tfrac{\partial z}{\partial x} - 3y &= 0
\end{align}
$$
This yields
$$
\frac{\partial z}{\partial x} = \frac{3y - z}{x + 2yz}.
$$

Try to find $\tfrac{\partial z}{\partial y}$ in an analogous way, and post your result in the comments.

By using implicit differentation, determine the gradient of curve $3xy+y^2= -2$ at the point $(1,-3)$


If I want to avoid quantifiers? In mathematics quantifiers always is used with restrictions $\forall x\in A$ etc and mathematicians often write:
$$x\in A\implies p(x)\;\;\;\text{instead of}\;\;\;\forall x\in A:p(x).$$
Is there a counterpart expression for $\exists x\in A: p(x)$ without quantifier?

Is writing like this logical unobjectionable?

I'm interested in trying to replace 'predicate logic' in mathematics with a 'mathematical propositional algebra', that is, only use propositions of the type $x\in A\implies p(x)$. For such propositions $P_i$, there would be a "mathematical ring" with elements
$$\displaystyle
\bigoplus_i\bigwedge_j p_j^{n_i}.
$$
I guess it won't work but in the process I might learn something about logic and semantics.

The fact that the universal quantifier on $x$ is omitted doesn't mean it's not supposed to be there. The correct parsing of $\forall x\in A: p(x)$ is $\forall x(x\in A\rightarrow p(x))$.

For existential quantifier this would be $\exists x(x\in A\land p(x))$. You may feel it is okay to omit the outer quantifier, but you have to remember it's still there.

Considering the other answers, I conclude that it really is possible to exclude quantifiers with limitations of the form $x\in X$ ($\forall x\in X$ and $\exists x\in X$) without loss of information, when properly distinguishes between bounded and free variables. Obviously

$\forall x\in X:P(x)\quad\Leftrightarrow\quad a\in X\Rightarrow P(a)$

and here there are no problems with the interpretation of the alternatives. Less obvious  is

$\exists x\in X:P(x)\quad\Leftrightarrow\quad a\in X\wedge P(a)$

but successive reversible transformations shows:

$\exists x\in X:P(x)\;\Leftrightarrow\;\neg\forall x\in X:\neg P(x)\;\Leftrightarrow\;
\neg(a\in X\Rightarrow\neg P(a))\;\Leftrightarrow\;\neg(a\notin X\vee\neg P(a))\;\Leftrightarrow\;$$(a\in X\wedge P(a))$.

Not all the steps are equally easy to interpret, but never the less they are logical equivalents and it is more a psychological issue that it may be difficult to interpret a specific step in the chain. But the logical machinery works consistent. If one is interested of the interpretation of the statement one have to transform the statement to a more readable form ($\mathbb N_m=\{x\in\mathbb N|x>m\}$ and $\oplus$ is exclusive or below):

$1\oplus(\varepsilon\in\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)(n\in\mathbb N_m)\oplus$
$(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
(1+(n\in\mathbb N_m))\oplus(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
(n\notin\mathbb N_m)\oplus(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\vee (m\in\mathbb N\wedge n\notin\mathbb N_m)\vee
(m\in\mathbb N\wedge|x-x_n|<\varepsilon)\Leftrightarrow$
$\varepsilon\in\mathbb R_+\Rightarrow m\in\mathbb N\wedge(n\in\mathbb N_m\Rightarrow |x-x_n|<\varepsilon)\Leftrightarrow$
$\forall\epsilon\in\mathbb R_+\exists k\in\mathbb N:n\in\mathbb N_{k}\Rightarrow |x-x_n|<\epsilon\Leftrightarrow \displaystyle \lim_{i\rightarrow\infty}x_i=x$.

Thus, the logic in (non meta) mathematics is substantially propositional logic.

In general predicate logic it is not always possible to eliminate quantifiers.


Katie wants to make a list of songs from her phone , in how many ways can she make one? Katie has $m$ songs in her phone, she wants to make a playlist consisting of $s$ songs such that every song is played at least once, and  a song can be played again if and only if at least $k$ other songs have been played in between.

In how many ways can she create such a playlist?

attempt: Let Katie have $m$ songs in her phone such that there are $s$ songs to be played . Suppose $k \leq t$  so that we select $t$ songs from $m$ songs, so that we have $m \choose t$ , then at least $k$ songs implies we have $k$ or more, so we have $m \choose k $ + $m \choose k+1$ + ...+ $m \choose m$ = $\Sigma_{t = k}^m $ $m \choose t$ total ways.
Can someone please verify this is correct? and if it's not, please provide some feedback. Thank you.

Let $m,s$ be positive integers, and let $k$ be a nonnegative integer.

Let $f(m,k,s)$ be the number of playlists of length $s$ from a collection of $m$ songs, with the restriction that if a given song is replayed, there must be at least $k$ other songs between the play of the given song and the replay.

Then $f$ is given by the following piecewise formula:

$$
{\small{\bullet}}\;\;\text{If $s \ge k+1$ then}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\qquad\qquad\qquad\qquad\qquad
$$

$$
f(m,k,s) =
\begin{cases}
\left({\large{\frac{m!}{(m-k-1)!}}}\right)(m-k)^{\large{{s-k-1}}} & \text{if $m >k$}\qquad\qquad\qquad\;\;\;\,\\[6pt]
\;\;0 & \text{otherwise}\\[6pt]
\end{cases}
$$

$$
{\small{\bullet}}\;\;\text{else if $s < k+1$ then}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\qquad\qquad\qquad\qquad\;\;\,
$$

$$
f(m,k,s) =
\begin{cases}
m!/(m-s)! & \text{if $m \ge s$}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\[6pt]
\;\;0 & \text{otherwise}\\[6pt]
\end{cases}
$$

For m = 3, s = 5 and k = 1. The above answer gives 48. But the correct answer is 42. If the three songs are A, B and C. Then the playlists like ABABA and ACACA should be subtracted from the count since not every song is played at least once.

P.S I do not have credit to add a comment for the above answer. That's why added as a separate answer.


Direct sum of indecomposable modules Prove or disprove: There exist indecomposable modules $V,U,W$ such that $\displaystyle{ V \oplus U \cong V \oplus W }$ and $\displaystyle { U \not \cong W}$.

I think that this is true, but I have no idea how to construct such indecomposable modules.

Any ideas?

Modules over what?  If we are looking at $R$-modules for $R$ a field, for example, then only one-dimensional spaces are indecomposable and there are no such examples.

Over $\mathbb{Z}$ there should be some interesting examples (hint: Hilbert's Hotel cannot be modeled directly, but there are some infinite groups that might behave similarly...).  In general, this is a hard and interesting problem without assumptions on $R$.  This paper, for example, says some interesting things in the case where $R$ has dimension 1.

HINT: Take the quotient of both sides with respect to V.


Evaluate conditional expectation of $X^2$ given $\sin(X)$ and uniform distribution of $X$ over $[0,\pi]$ I want to find the following conditional expectation: $\mathbb{E}(X^2\mid\sin(X))$ where $X$ is distributed uniformly on $[0, \pi]$.

I guess if $\sin(X)=y$, then $X^2$ can be either $\arcsin(y)$ or $\pi-\arcsin(y)$. And maybe the formal proof should be based on this?

Let $X$ have the pdf $\rho_X$. Then
$$\rho_{g(X)}(y) =
\int_{\mathbb R} \delta(g(\tau) - y) \rho_X(\tau) d\tau =
\sum_{\tau: \,g(\tau) = y} \mu(\tau),
\quad \mu(\tau) = \frac {\rho_X(\tau)} {|g'(\tau)|},\\
\rho_{X, g(X)}(x, y) =
\int_{\mathbb R} \delta(\tau - x) \delta(g(\tau) - y) \rho_X(\tau) d\tau =
\sum_{\tau: \,g(\tau) = y} \mu(\tau) \delta(\tau - x), \\
\mathbb E(f(X) \mid g(X) = y) =
\frac {\int_{\mathbb R} f(x) \rho_{X, g(X)}(x, y) dx} {\rho_{g(X)}(y)} =
\frac
 {\sum_{\tau: \,g(\tau) = y} \mu(\tau) f(\tau)}
 {\sum_{\tau: \,g(\tau) = y} \mu(\tau)}, \\
\mathbb E(f(X) \mid g(X)) =
\frac
 {\sum_{\tau: \, g(\tau) = g(X)} \mu(\tau) f(\tau)}
 {\sum_{\tau: \, g(\tau) = g(X)} \mu(\tau)}.$$
If $\rho_X(x) = [0 < x < \pi]/\pi$, $g(x) = \sin(x)$ and $X$ takes a value in $(0, \pi)$, the sums have two non-zero terms, corresponding to $\tau = X$ and $\tau = \pi - X$.

Continue with your approach and note $y=\sin x$ has two solutions, i.e. over either [0, $\pi/2$] or [$\pi/2, \pi]$. The density function for $y$ is then,

$$ f(y) = \frac2\pi \frac{dx}{dy}
= \frac2\pi \frac1{\sqrt{1-y^2}}$$

Thus,

$$E[X^2|\sin(X)]= \int_0^1 \arcsin^2(y)f(y)dy
= \frac2\pi \int_0^1 \frac{\arcsin^2(y)}{\sqrt{1-y^2}}dt=\frac{\pi^2}{12}$$


Is there any known relationship between Goldbach's comet G(n) and the prime counting function (${\pi(n)}$)? The "extended" Goldbach conjecture defines R(n) as the number of representations of an even number n as the sum of two primes, but the approach is not related directly with ${\pi(n)}$, is there any kind of Goldbach-${\pi(n)}$ (I will call it G${\pi(n)}$ for short) function?

http://mathworld.wolfram.com/GoldbachConjecture.html

I have tried an approach to that idea as follows.

Calculate ${\pi(n)}$
Calculate ${\pi(n/2)}$
I defined then G${\pi(n/2)}$ as the subset of primes p from ${\pi(n/2)}$ that are symmetrical on n/2, so they have a counterpart prime pc = n-p in [n/2,n-2] so n=p+pc.
I did a test for the first 2000 even numbers (my computer slows down very much after that point) and prepared a graph showing ${\pi(n)}$, ${\pi(n/2)}$, G${\pi(n/2)}$ and the linear interpolation of the value of G${\pi(n/2)}$ (a kind of average value of the growing G${\pi}$ set of primes.
Then I played with lower values of the ${\pi}$ function looking for a lower bound of the linear interpolation. Finally ${\pi(n/10)}$ seemed a good value to define a lower bound for G${\pi(n/2)}$ because the linear interpolation is always over that value (at least in the test it seems so).

Here is the graph.

When I checked the results, I wondered if there is a way to work on the conjecture through a relationship between Goldbach's comet value for n, and ${\pi(n)}$, specially if the density of the subset of primes in [2,n/2] that are symmetrical in n/2 is always greater than the density of primes in a lower subset of ${\pi}$ function (e.g. ${\pi(n/10)}$).

So the question is: is there any relationship already known or being researched with ${\pi(n)}$ or only the extended Goldbach conjecture R(n) function is the correct approach to a solution of the Goldbach conjecture?

I don't believe there's any deep relationship between $\pi(n)$ and your function $R(n).$ For one thing, their expected 'bulk' rates of growth are different: $R(n)$ should grow something like $n/\log^2(n),$ while $\pi(n)\sim n/\log(n).$ So dividing $\pi(n)$ by 10 isn't going to be enough -- you'll need to divide by more and more as $n$ grows. For example, $\pi(10^{14})=3204941750802$ but $R(10^{14})=90350630388$ and so their quotient is already 35.

Second, $\pi(n)$ grows smoothly, in the sense that $\pi(n)\le\pi(n+1)\le\pi(n)+1.$ But $R(n)$ grows wildly, more even than your graph so far suggests. Essentially, $R$ is sensitive to the small prime divisors in $n$, while $\pi$ doesn't care. So $R(30n)$ grows differently from $R(30n+1)$, while the same is not true for $\pi(30n)$ and $\pi(30n+1).$

There is a relationship between Goldbach's comet G(n) and the prime counting function π(n). This sum gives the exact result for G(n) (A002372)
$$G(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$ where p(n) is the nth prime.


How do I find how tall a building is? How tall is a tower if it casts a shadow 65 ft. long at the same time that a building 160 ft. tall casts a shadow 130 ft. long?
I tried using the hypotenuse formula but didn't get the right answer.
Help.

As Arturo said, you need to use similar triangles. Here’s a picture to get you started:

The building and its shadow, together with the hypotenuse joining their ends, form a triangle similar to the one formed by the tower and its shadow. You know the lengths of both shadows and the height of the tower; now just use the fact that the large triangle is just a scaled up version of the smaller one: every side has been multiplied by the same amount.

Toss a firecracker on to the street from the roof. It will fall at 30 feet per second, the speed of gravity. When it hits the ground the startled pedestrians below will likely swear in surprise. Their swear words will float back up to you at 1100 feet per second, the speed of sound. Therefore the length of time for you to hear swearing after you chuck the firecracker will be:

height/30 seconds + height/1100

=

1100 x height / 33000 + 30 x height / 33000

=

1130 x height / 33000

therefore

the building's height = 33000/1130 x the amount of seconds between when you dropped the cracker and when you heard people swearing at you.

Sometimes maths can be fun. But usually only when it involves things that explode.


$G$ conex planar graph, the total number of faces Let $G$ be a conex planar graph with n more or equal than $3$ vertices, so that each face will contain in its frontier a circuit of length at least $5$.So, the number of faces will be at most $(2n-4)$$/$$3$.
Can someone pleae help me with this problem?

Lemma: If $G$ is a planar graph with $V$ vertices, $E$ edges, and finite girth $g$ then, $$E \leq \frac{g(V−2)}{V−2}$$
Look here: girth in planar graph

So $g=5$ and thus $3E\leq 4V-10$. Using Euler theorem: $$V-E+F =2$$ we have $$3V+3F -6 \leq 4V-10$$ so $$F\leq {V-4\over 3}$$
which is much better bound.

From the given, $5f\le 2e$. Use $n+f=e+2$ to eliminate $k$.


Binary arithmetic - overflow and carryout at same time? In binary arithmetic, When you subtract 2 signed numbers you must discard the carry out. My question is, is it possible for overflow to occur and a carry out? So, on paper there would be two extra bits and you would have to discard one?

Also, if your subtracting 2 signed numbers, how do you differentiate between overflow and a carry out?

Overflow and carry out are philosophically the same thing. Both indicate that the answer does not fit in the space available.  The difference is that carry out applies when you have somewhere else to put it, while overflow is when you do not.  As an example, imagine a four bit computer using unsigned binary for addition.  If you try to add $1010_2+111_2$ without the word length restriction, you get $10001_2$  The high bit does not fit in our word.  If this word is all the space allocated to this variable, it represents overflow as the result is too large to represent. If we have a two word space allocated to this variable, it becomes a carry out and is stored in the higher word.  We would then represent the addition as $0000\ 1010_2+0000\ 0111_2=0001\ 0001_2$ and the carry is explicit.

In processors, overflow flag indicates that sign bit has been changed during adding or subtracting operations But carry flag means adding or subtracting two registers has carry or borrow bit.


a function with differentiable partial derivatives but unequal mixed derivatives I am looking for an example of a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are both differentiable at some point, say the origin, but $\frac{\partial^2 f}{\partial x\partial y}\neq\frac{\partial^2 f}{\partial y\partial x}$ at that point.

If $f_{xy}$ and $f_{yx}$ are continuous, then they are necessarily equal.
The only examples you will find where the mixed partial derivatives are $not$ equal will be just like the standard example, $f(x,y)=x \cdot y\cdot \displaystyle \frac{x^2-y^2}{x^2+y^2}$ (and $f(0,0)=0$), and this example works precisely because $f_{xy}$ and $f_{yx}$ are not continuous at the origin. In this case $f_{xy}(0,0)=1$ and $f_{yx}(0,0)=-1$. At all other points, the mixed partials are equal.

Let $z = \arctan(y/x)$, then $\frac{d}{dy}\left(\frac{dz}{dx}\right) \not= \frac{d}{dx}\left( \frac{dz}{dy}\right)$


'compactness' of a category with respect to a family of functors I was wondering if the following idea is a notion that has been studied. I thought of it while thinking of functorial invariants (such as (co)homology functors) and wondering what kind of category has a 'compactness condition' for a collection of functors that are able to distinguish non-isomorphic elements in a category. I tried searching for some likely terminology with no luck.

Definition: Let $\mathcal{C}$ be a category, and $\mathfrak{F} = \{ F_i \}_{i \in I}$ be a family of functors $F_i: \mathcal{C} \to \mathcal{D}_i$ for all $i \in I$ for some index set $I$ (where $I$ may be infinite). Then we say that $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$ if for all $a, b \in \mathcal{C}$, with $a$ not isomorphic to $b$, there exists $F_j \in \mathcal{F}$ such that $F_j(a) \neq F_j(b)$.

Definition: $\mathcal{C}$ is functorially compact if for all families of functors $\mathfrak{F}$ that distinguish objects of $\mathcal{C}$ there exists a finite subset $\mathfrak{G} \subseteq \mathfrak{F}$ that distinguishes objects of $\mathcal{C}$.

Question: What sort of category can be 'functorially compact'?

Thoughts: The identity functor always distinguishes objects of $\mathcal{C}$, so we know that every category has at least one family of functors that distinguishes objects in it. I believe categories with a finite number of objects would always be functorially compact as well. I imagine that this is a rather strict property that wouldn't allow sophisticated categories, but I'd like to be proven otherwise. But what about categories with an infinite number of objects (small or large)?

Note: These definitions may not be perfect - maybe some restriction must be put on what kind of functor is allowed to make things interesting, but I don't know enough about category theory to know where to begin with that.

Edit: I made corrections as per Qiaochu's comments below.

I think that a category is "functorially compact" iff it is finite.

It is easy to see that finite $\Rightarrow$ functorially compact: for any family of functors $\mathfrak{F}$, we construct a finite sub-family by picking one distinguishing functor from $\mathfrak{F}$ for each pair of objects in $\mathcal{C}$.

To show that functorially compact $\Rightarrow$ finite, consider the family $\mathfrak{F} = \{F_A\}_{A \in Obj(\mathcal{C})}$ indexed by the objects of $\mathcal{C}$, where each $F_A : \mathcal{C} \rightarrow \mathbf{2}$.  We take $\mathbf{2}$ to be the category consisting of two objects $X$ and $Y$, and two (non-identity) morphisms $f : X \rightarrow Y$ and $g : Y \rightarrow X$.  We define $F_A$ to be the functor that sends $A$ to $X$ and every other object to $Y$ (this map uniquely determines where the functor maps the morphisms of $\mathcal{C}$).  $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$, but there can be no finite subfamily that does the same, unless $\mathfrak{F}$ was finite to begin with (if there exist objects $A$ and $B$ such that neither $F_A$ nor $F_B$ belong to this subfamily, then $A$ and $B$ are not distinguished).


Maximum number of isomorphic graphs Given a simple graph on $n$ vertices, how many graphs are atmost there that are isomorphic to the given graph? Is it $\Theta(n^2!)$ or $\Theta(n!)$ which is number of permutations of rows or columns?

Let $V=\{1,\ldots,n\}$ and let $G$ be a graph on vertex set $V$.  I assume you are asking for the number of graphs on vertex set $V$ that are isomorphic to $G$.  The symmetric group $S_n$ acts on all graphs on vertex set $V$.  We shall apply the orbit-stabilizer lemma.  The number of graphs on $V$ isomorphic to $G$ is the size of the orbit of $G$ in this action.  The stabilizer of $G$ is the automorphism group of $G$, which we shall denote by $Aut(G)$.  By the orbit stabilizer lemma, the order of the group $S_n$ is the product of the size of the orbit of $G$ and the size of the stabilizer.  Hence, the size of the orbit is $n!/|Aut(G)|$.

The answer is exactly $n!$, no $\Theta$ required. Every other isomorphic graph is uniquely determined by an isomorphism, which is a bijective function from the set of vertices to itself. There are $n!$ such functions.


How do I solve $f(x) - f(x - 1) = x$? Is it possible to solve the functional equation $f(x) - f(x - 1) = x$ by algebraic means?

First note that we can add a constant to $f$ and it will still be a solution, so we might as well define $f(0)=0$  Then we can find $f(1)=1, f(2)=3, f(3)=6$ and so on.  We can take finite differences and find that a quadratic polynomial fits all of these, then find that $\frac 12x(x+1)$ satisfies our equation.  The full solution is then $f(x)=\frac 12x(x+1)+f(0)$

Yes, just take any $f(0)$, and $f(n) = f(0)$ plus some summation.


If $n$ is odd, then $n/2 + 1/2$ is always even? If $n$ is odd, prove that $n/2 + 1/2$ is even.

Context: I'm a Statistician and the term $n/2 + 1/2$ showed up in the index of a summation when deriving the pdf of some Order Statistic:
$$
\sum_{j = (n+1)/2}^{n}...
$$
I realized that $n/2 + 1/2$ is always even if $n$ is odd, but I couldn't prove the result to myself (well, I have no training in Number Theory).
What I've tried:
Suppose $n$ is odd. Then $n + 1$ is even (by the successor function?). Then $n + 1 = 2k$ for $k \in \mathbb{N} \Rightarrow (n+1)/2 = k $. But this doesn't show that $k$ is even.

No.
Consider $n=1$.
I add this sentence because my answer is too short.

Odd numbers are $2$ units apart, so that the half odd numbers are $1$ unit apart and induce both parities.


Existence of a numerable family given a element of a $\sigma-$algebra I'm trying to prove the next stament:

Let $E\subset\mathcal{P}(X)$ fixed. Then for all $A\in\sigma(E),$ there exists a numerable subfamily $E_{0}\subset E$ such that $A\in \sigma(E_{0}).$

My attempt is based on the follow: Let $S=\bigcup\{\sigma(E^{'}):E^{'}\subset E \quad\text{is numerable}\}.$ I've proved that $S$ is a $\sigma-$algebra. Then, I'd like to show that $E \subset S$ to prove that $S=\sigma(E)$ and finishing the proof, but I'm stuck in this.

Any kind of help is thanked in advanced.

The punchline is that ${S}$ contains ${E}$ because for any $A \in {E}$, $A \in \sigma(A)$ hence $A \in S$ by definition of $S$.


Can there exist $3$, $4$ and $5$-faceted shapes with congruent flat sides in $\mathbb{R}^3$? I rose this question in my discrete math class (the unit on probability) not too long ago. For instance, a two-sided shape (like a coin) can be one with any geometrical shape as its "side," such as a circle (as with a coin) or a square, or any other connected shape in $\mathbb{R}^2$, and in tossing this shape there is a $1/2$ "probability" it will land on one side as opposed to another. A cube (like a dice) is a shape with $6$ sides, all congruent flat squares, and upon tossing this object into the air, there is a $1/6$ chance of landing on any given side.

Now suppose I wished to toss a $3$-sided object such that all facets of this shape are congruent, as in the penny or dice, resulting in a $1/3$ chance of landing on any given side.

Can there exist solids in $\mathbb{R}^3$ (or elsewhere) with $3,4$ and $5$ flat facets, such that all facets are congruent? Naturally these shapes may exist with curved sides, but I'm more concerned with solids of flat facets. I'm certain one of $3$ sides cannot exist, but perhaps some of you can enlighten me on how I might prove these cases?

What about a solid with $n$-faces? Perhaps this is related to packing problems (and hence I will add that tag)?

One of the dies you want you can make. The others you can fake.

The tetrahedron is the first platonic solid. It has four triangular faces.

The platonic solids give you fakes for probabilities of 1/3 (label the faces of a cube with 1, 2, and 3 each twice) and 1/5 (label the faces of an icosahedron with 1 through 5, each four times).

You're right that a real 3 is impossible. I'm pretty sure 5 is too. Maybe I'll come back to this with proofs, if noone else provides them.

For some larger values of n, check out the wikipedia page for deltahedra:
http://en.wikipedia.org/wiki/Deltahedron

What you are describing are the platonic solids (link to Wikipedia).


To prove a sequence is Cauchy I have a sequence:
$ a_{n}=\sqrt{3+ \sqrt{3 + ... \sqrt { 3} } } $ , it repeats $n$-times.

and i have to prove that it is a Cauchy's sequence.
So i did this:
As one theorem says that every convergent sequence is also Cauchy, so i proved that it's bounded between $ \sqrt{3}$  and $ 3 $ (with this one i am not sure, please check if i am right with this one.)And also i proved tat this sequence is monotonic. (with induction i proved this: $ a_{n} \leq a_{n+1} $
so if it's bounded and monotonic, therefore it is convergent and Cauchy.
I am just wondering if this already proved it or not? And also if the upper boundary - supremum if you wish - is chosen correctly.
I appreciate all the help i get.

Yes, correct ideas.

For boundedness, you can use induction:

$\sqrt 3<3$, good.
Suppose $a_n<3$ then $a_{n+1}=\sqrt{3+a_n}<\sqrt{3+3}=\sqrt6<3$.

${ a }_{ n+1 }=\sqrt { a_{ n }+3 }  $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ as $n\rightarrow \infty $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ $\quad x^{ 2 }=x+3\quad \Rightarrow $ $x^{ 2 }-x-3=0 $ $and\quad it\quad$ convergents to the $x=\frac { 1+\sqrt { 13 }  }{ 2 } $


Prove that simple graph $G$ is orientable I am a student in an introductory class for graph theory and this question is on our first homework assignment and I don't know where to start to prove the following:

  Suppose that $n \geq 3$ and a simple graph G has $n$ vertices and at
  least $(n^2 - 3n + 6)/2$ edges. Prove that G is orientable.

Thank you for your time. Any hints or help will be greatly appreciated.

EDIT: A graph g is orientable if it is the underlying graph of a strongly connected digraph. A digraph is strongly connected if for all vertices $v$ and $w$ there exists a walk $v-w$.

By Robin's Theorem, a graph is orientable if and only if it is connected and has no bridge.

So try to prove:

Step 1: Any graphs with $(n^2 - 3n + 6)/2$ edges is connected.

Step 2: Any graphs with $(n^2 - 3n + 6)/2$ edges cannot have bridges.

Step 1 is pretty standard and easy. For step 2, assume by contradiction that $G$ has a bridge $b$ then, by erasing this bridge you get a graph with two disjoint components, one with $k$ vertices and the other one with $n-k$ vertices.

What is the maximum number of vertices in each component?

First thing which come to my mind is to suppose that this is non-orientable graph.

So if we suppose that this is non-orientable, then it follows to 2 cases (maybe more):

1) It is a path graph.
2) It has a bridges in it (so cycles, as well).

What is the number of edges it should have, if it is a path graph?
What if it has bridges+cycles?

Keep in mind it is simple graph, so no loops and multiple edges.


Logic and implication negation I am new to predicate logic and there is something I can't seem to understand.
Let's have an implication:
If I have a sister, I have a sibling.
Now, it's negation should be:
I have a sister and I do not have a sibling.
But I can't understand why the following is not a valid negation:
If I have a sister, I do not have a sibling.
It should be a negation because the first part of the implication happened and the second did not.

First note that
$$P\rightarrow Q\equiv\lnot P\lor Q$$
Here's how to properly negate the above statement
$$\lnot(P\rightarrow Q)\equiv\lnot(\lnot P\lor Q)\equiv P\land\lnot Q$$
Now let me show you why the following is not a negation of the original statement
$$P\rightarrow\lnot Q\equiv\lnot P\lor\lnot Q$$
Notice that this only negates $Q$, however we are seeking to negate the whole statement.
Also note that if $P\equiv\mbox{false}$, then we have
$$P\rightarrow Q\equiv P\rightarrow\lnot Q\equiv\mbox{true}$$
Which is clearly not a negation of the original statement. I hope this helps you understand.

The structure of your statement is $A\Rightarrow B$ where $A=\text{"I have a sister"}$ and $B=\text{"I have a sibling"}$. Then the negation is $\lnot B\Rightarrow \lnot A$ which is "If I don't have a sibling I don't have a sister".


Treat Differentials As Fractions I am readying in this website for a long time today similar questions and the answers provided about the issue I am asking now but I have to say I am more baffled than I was originally.

My original question would be:

I have read in many textbooks when they teach integration that they emphasize not to treat dx as a quantity but only as a notation that shows which is the variable of integrations. The same textbooks after a while they start talking about differentials and based on the following expressions

$$
df = f'(x)dx\quad df = f_{x}dx+f_{y}dy
$$

they say that now that differentials have been introduced we can treat them as independent variables and use them freely as fractions.

Now I was always working with that idea, but after some digging I start seeing a lot of different answers, especially in here, and most of them were conflicting.

There were people mentioning Non-Standard Analysis and Differential Forms and were trying to justify that we can treat differentials as fractions. Also there were other people saying that every time we treat them as fractions we actually apply other theorems and so we don't actually treat them that way.

So the questions I have to ask now is:

Can we actually treat differentials as fractions in EVERY framework, or is there today a debate about it?

I don't want personal opinions or what we can do in a specific framework. I saw a lot of people mentioning that in single variable calculus you may treat them like that because you won't do mistakes. Can you actually treat them like that strictly mathematically in EVERY framework or you can just do it for ease but you actually apply different theorems behind the scenes?
If differentials can't be treated in EVERY framework as a fraction then is it defined and proven that they can be always treated as a fraction in some frameworks?
Given that we have defined differentials as

$$
df = f'(x)dx\quad df = f_{x}dx+f_{y}dy
$$

then didn't we automatically defined than they can be used as independent quantities, so they can be seen as fractions also? Shouldn't we accept them as fractions from the moment of that definition existed?

The reason the definition of a differential doesn't allow you to treat derivatives as fractions is because you're not automatically allowed to divide by everything you see in an equation. You can't divide by a three dimensional vector, for example, and this example is just like that. Differentials are vectors.

In nonstandard analysis in a single variable derivatives are the standard parts of quotients of infinitesimals. There's no disagreement about that. But the infinitesimals are not differentials.

While it's useful sometimes to treat derivatives as fractions, it can lead to error. There's a famous example, I forget the exact circumstances, where you have functions $u, v, w$ with
$$\frac{\partial u} {\partial v} \frac{\partial v} {\partial w} \frac{\partial w} {\partial u} =-1$$
which proves that you can't treat multivariable derivatives as fractions and expect to always get the right answer.

This question comes a lot, and the best answer, to when you are new to learning calculus is to simply regard $\frac{dy}{dx}$ as non-sense which gives you a correct answer. Think of it as notation, with its own rules, that when you learn how to use it, will lead to the correct answer. Do not try to understand it.

I would suggest to learn calculus both the "non-sense" way and the precise way. The precise way you will avoid using this notation and instead rely on the properties of the derivative. This way you would be able to convince yourself that the trick-in-differential-notation leads to the answer that you would get if you did it precisely.


Combinations: How many handshakes? I need help with the following question which I cannot seem to solve:

17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?

My approach: no. of ways $ = 1 + 2 + 3 + ... + 14 = 7(15) = 105 $.

Apparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?

Each of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are

$$\frac{17\times 14}{2} = 119$$

handshakes (dividing by 2 to account for symmetry, as you would otherwise count "$A$ shaking hands with $B$" and "$B$ shaking hands with $A$" as distinct events).

I will explain for 6, same logic applies for all the numbers.

first,
if 6 persons shakes will all in the party, then number of shakes will be:

A  B  C  D  E  F

A can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5

B can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4

C can shake with only his right sides(D,E,F) ppl(not with A&B bcos,A&B already shook with C just now. so no repeat) = 3

D can shake with only his right sides(E,F) ppl(not with left side ppl  bcos, they already shook with D just now. so no repeat) = 2

E can shake with only his right sides(F) ppl(not with left side ppl  bcos, they already shook with E just now. so no repeat) = 1

as all of the persons shook hand with F, He dont shake with any one now.

So total possible shakes for 6 persons are = 5+4+3+2+1 => 15

now in questions it is given that, No one shakes hand with neighbors.

Imagine ppls sitting around table

        A   B
      F       C
        E   D

Please treat above image as 6 Ppl around circle :
So now,

Invalid Shakes:

A cannot shake with B,F . I am representing it as below :
  A  -no- B & F    => 2 invalid shakes.
  B  -no- C only. Bcos  B -no- A is covered as part of first case.
                   =>  1 invalid shake
  C  -no- D only.  =>  1 invalid shake
  D  -no- E only.  =>  1 invalid shake
  E  -no- F only.  =>  1 invalid shake
  F dont shake with A is already covered as part of first invalid shakes.

so total invalid shakes are 2+1+1+1+1 = 6

so for 6 ppl:  out of  total 15 , 6 are invalid. So 15-6 = 9

on the same line :

for 17 ppl:  out of  total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119

shortcut :   For N ppl, Total shakes are  (N-1)(first term + last term)/2  Arithmetic progression sum. out of which Invalids are N.
So valid shakes =>  (Total - N)