Is it possible to have a field without an additive identity? If I drop the axiom that Zero is the identity of an addition what consequences does this entail? What do I need to change to my axiomatization?

By definition it is not possible, but are there mathematical structures like a field but without an additive identity?

I don't have a concrete example/issue but I am interested in this question.

The usual definition of a field is:

Definition. A set $F$ together with two binary operations $+$ and $\times$ is a field if and only if:

$+$ is associative;
$+$ is commutative;
$+$ has a neutral element $0$;
For every $a\in F$ there exists $b\in F$ such that $a+b=0$.
$\times$ is associative;
$\times$ is commutative;
$\times$ distributes over $+$;
There is an element $1\neq 0$ such that $a\times 1=a$ for all $a\in F$;
For each $a\in F$, if $a\neq 0$ then there exists $x\in F$ such that $a\times x = 1$.

Now, you can certainly ask whether the axioms are independent. For example, it is easy to see that if you drop the assumption of commutativity for $+$, then you can deduce it from the other eight axioms; on the other hand, the real quaternions show that you cannot drop the assumption that $\times$ is commutative and deduce it from the other axioms.
So you could be asking whether Axiom 3 is independent of the other axioms. The main difficulty with dropping Axiom 3 is that without it Axiom 4 becomes unintelligible, and Axiom 9 is also problematic. So before we drop Axiom 3, we need to replace Axioms 4 and 9 with something else that, together with Axiom 3 give a field, but which make sense in the absence of Axiom 3.
There are plenty of ways of defining "abelian group" without specifying the existence of a neutral element (e.g., "for every $a,b\in F$ there exists $x\in F$ such that $a+x=b$") replacing 4 with something like this will automatically imply the existence of a $0$. Do you have something specific in mind?
(For a similar question, see for example this sci.math post by Dave Rusin where he discusses the independence of the axioms of a vector space, where he faces a similar problem with dropping "existence of $0$")
Added. While I was writing this, the title of the post was changed to "without an identity" instead of "without a $0$". You run into similar problems: if you drop Axiom 8, you need to replace Axiom 9 with something that still makes sense; depending on what you replace it with, it may or may not imply the existence of a multiplicative identity in the presence of the other axioms. Again, the question is whether you have something in mind or not.

If the question is just a poorly phrased way of asking if the one element ring is a field, the answer is that it is not considered to be a field. There are good reasons for this, even though the one element ring satisfies all the axioms except for the $1\neq 0$ clause of Axiom 8.

If I understand the definition of a field via abelian groups and distributivity as in Asaf's answer correctly, the perhaps surprising answer is no, you don't need to have an identity element in your axiom set for a field.  The trick lies in that you don't need an identity element in the axioms for a group.  Group theory, in fact, can get founded upon a single axiom, or several axioms which make no reference to a neutral element, if you really want that.  I've only glanced at things here, so see 1 2 and 3 for particular axiom sets.  Once you pick an appropriate axiom set for groups, you use the first part of Asaf's procedure for defining a field, which implies no mention of an identity element in the axiom set for fields.  The existence of inverses also doesn't mean one needs to talk about the identity element in the axiom set.  One could rewrite such existence statements via unary functions as one might do in the context of universal algebra, and this does capture the concept here, since inverses can get proved unique.


Calculus II Professor will not accept my correct integral evaluation that uses a different method, should I bring this up further? I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:

Evaluate the following integral:

$\int_0^4\sqrt{16-x^2}dx$

My answer: 4$\pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $\frac{1}{4}$$\pi$$r^2$ and wrote my answer of 4$\pi$.

The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:

  To receive full credit, you would have had to evaluate an integral, as the instructions indicated.

Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?

Thank you.

I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.

You could argue your case.  Your professor could argue back that solving the integral by trig substitution does not require the formula $A = \pi r^2$, and he did not permit the use of that formula.  He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $\int_{-r}^r \sqrt{r^2-x^2}dx$).

It could go either way for you.  But I think it would be a waste of your and your professor's time.

I love your solution! It reminds me of the following question. What is ${{\int}_0}^{2\pi}{\rm{cos}^2}x{\rm{d}}x$? Answer: $\pi$. The reasoning is that as ${\rm{cos}^2}x + {\rm{sin}^2}x = 1$, as the areas under ${\rm{sin}^2}x$ and ${\rm{cos}^2}x$ in the interval $[0,2{\pi}]$ are equal, and as ${{\int}_0}^{2\pi}{{\rm d}}x = 2{\pi}$, we conclude that $${{\int}_0}^{2\pi}{cos^2}x{\rm{d}}x ={{\int}_0}^{2\pi}{sin^2}x{\rm{d}}x = {\pi}$$ The whole point of mathematics to me is beautiful pieces of inspiration like the one you had on your exam. You should get full credit.


Basic Question on Definition of Brownian Motion I am quite new to discrete and continuous stochastic processes. It seems there is something I don`t understand about definition of Brownian motion.

Let $\Omega, \mathcal{F}, \mathbb{P}$ be a probability space and $B_t$ be a standard continuous Brownian motion $B_t: \Omega \rightarrow C(\mathbb{R}^+, \mathbb{R})$.

Then the definition requires that $B_t \sim N(0,t)$ and $B_t - B_s \sim B_{t-s} \sim N(0,t-s)$. I don`t see how the first condition is compatible with the last. We know that $B_t \sim N(0,t)$ and $B_s \sim N(0,s)$ so $B_t - B_s \sim N(0,t+s)$ (seen as a sum of two normally distributed random variables) which seems to be in contradiction with independent increments. I guess I am missing something very basic but since I am new to this topic I cant see it. Thanks

update: based on answers below, yes it was indeed very basic, $B_t$ and $B_s$ are not independent as random variables so that is why the usual rule for sum of independent normally distributed random variables dont apply

Answer: The values of a Brownian motion path at different times are not independent. The variance of $B_t-B_s$ may at first seem like it should be $t+s$, but we can think of its variance being reduced due to $B_t$'s dependence on $B_s$.

Elaboration: We are only looking at a single Brownian motion process here, $B$. The value of $B$ at time $t$ is exactly its value at time $s$ plus some random fluctuation: $B_t(\omega)=B_s(\omega) + $ "the change in value from $s$ to $t$ for path realization $B(\omega)$". The part in quotes is declared to be independent from $B_s$ but also to be normally distributed with mean zero and variance equal to the length of time it covers: $t-s.$

Of course $B_t=B_s+(B_t-B_s)$ with $B_s$ and $B_t-B_s$ being independent. Given that $\text{Var}(B_t)=t$ and $\text{Var}(B_s)=s$, we can deduce that $\text{Var}(B_t-B_s)=t-s$.

If we were allowed to plug in distinct $\omega$'s to evaluate $B_t$ and $B_s$ independently, then the variance would be $t+s$.


Find deviance and canonical link function $g(\mu)=\eta$ where $Z$ is the response variable of Pareto distribution $f(y,\theta)=\theta y^{-\theta-1}$ Find the deviance and the canonical link function $g(\mu)=\eta$ where $Z$ is the response variable of Pareto distribution $f(y,\theta)=\theta y^{-\theta-1}$, where $z=\log y$

I brought the distribution into the exponential family form:

$\mid dy/dz \mid f(y,\theta)=f(z,\theta)$

Hence $f(z,\theta)=\exp \{-\theta z + \log \theta +d(z) \}$

$E(Z)=\mu=\frac{1}{\theta}$

Link function $g(\mu)$:

My notes suggest to use $g(\mu)=-\theta=\eta$ hence $g(\mu)=-1/\mu$

As for Deviance,
$D=-2\log \frac {L(\omega)}{L(\Omega)}$ where $w$ is the reduced model.

For the full model I have: $\hat \mu_i=z_i$ and the reduced model has $\hat \mu_i=\hat \mu_i$

$$D=-2(l(\hat \mu_i)-l(z_i))=-2(\{-\theta z + \log \hat \mu_i +d(z) \}-\{-\theta z + \log z_i +d(z) \})=-2\log \frac{\hat \mu_i}{z_i}$$
Is this correct?

The answer is incorrect because $\theta$ depends on $\hat \mu_i$ in each likelihood.

Thus instead of $\theta z$ you need $z/\mu$ (which in the case of the full model is -1).

Similarly you have jumped straight from $\log\theta$ in the likelihood to $\log\mu$ in the deviance. This should instead be $\log(1/\mu)$ or $-\log\mu$.

Also, as mentioned in the comments, the sum is forgotten.


A question on metric space defined on $\mathbb{Q}$. Consider $\mathbb{Q}$ be the set of all rational numbers. Defined $d(p,q)=|p-q|
$. Then which of the following statements are true?

$\{q \in \mathbb{Q} : 2<q^2<3\}$ is closed.

$\{q \in \mathbb{Q} : 2 \leq q^2 \leq 4\}$ is closed.

$\{q \in \mathbb{Q} : 2 \leq q^2 \leq 4\}$ is compact.

$\{q \in \mathbb{Q} : q^2 \geq 1\}$ is compact.

So I was thinking about it, where option 4. is not true because this is not bounded. So, not compact follows from the unboundedness. So If we can show that here the set in 4. And I think no 1. is closed, since it's complement is $\mathbb{Q}$ union some open set in $\mathbb{R}$.
For the other statement we may use the general criteria that "A metric space is compact iff it is complete and totally bounded". But I need some help to do this.

We can write 1. as $\left((\sqrt{2}, \sqrt{3}) \cup (-\sqrt{3}, -\sqrt{2})\right) \cap \mathbb{Q}$ which is a real-open set (open intervals are open) intersected with $\Bbb Q$, so that set is open in $\Bbb Q$. It is also closed in $\Bbb Q$ because we can also write it as $\left([\sqrt{2}, \sqrt{3}] \cup [-\sqrt{3}, -\sqrt{2}]\right) \cap \mathbb{Q}$, which is closed for similar reasons.
2 is closed as we can write it as $\left([\sqrt{2}, 2] \cup [-2, -\sqrt{2}]\right) \cap \mathbb{Q}$ and as its element $2$ is not an interior point of it, it's not open.
The set under 3 is just the same as under 2 so is indeed closed, as we saw, so it could be compact, as it's also bounded. But in fact it is not, as we can pick any irrational $p$ "within" the set (say $\sqrt{3}$ will do) and find a sequence of rationals $q_n$ in the set that converges to $p$ in the reals (this can always be done). But then the sequence $(q_n)_n$ is Cauchy (it's convergent in the reals after all) but not convergent in $\Bbb Q$ (as the only point it could converge to doesn't lie in the set). So the set is not compact. A deeper reason why it is not compact( which you've probably not covered yet) is that a compact countable set in a metric space must have an isolated point, and this set has none. But the non-completeness (or the related fact that we have a sequence without a convergent subsequence) can be used to refute compactness at a more elementary level.
For 4, in all metric spaces we know that "$A$ compact $\implies$ $A$ closed and bounded;  Heine-Borel is the inverse implication which holds in subsets of $\Bbb R^n$ in the Euclidean metric. The "force" of it is to quickly prove compactness. But the always valid implication can be used to easily refute compactness, and 4 is an example: not bounded so non-compact is a valid deduction in any metric space.

True. Split the set into $\\{q \in \mathbb{Q} : 2 < q^2 < 4$ and $q < 0\\}$ and $\\{q \in \mathbb{Q} : 2 < q^2 < 4$ and $q > 0 \\}$. I claim each of these is closed; thus, their union is closed. For $\\{q \in \mathbb{Q} : 2 < q^2 < 4$ and $q > 0 \\}$ is the intersection of all sets of the form $\\{q \in \mathbb{Q} : a \leq q \leq b\\}$ where $a, b$ range over all rational values s.t. $a < \sqrt{2}$ and $b > \sqrt{3}$. The intersection of any family of closed sets is closed. And similarly with $\\{q \in \mathbb{Q} : 2 < q^2 < 4$ and $q < 0\\}$.

True. It's the inverse image of the real interval $[2, 4]$ under the continuous map $q \mapsto q^2$; the inverse image of a closed set under a continuous map is closed.

Edit: misread this the first time. 3 is in fact false. For consider the collection of open sets of the form $\\{q \in \mathbb{Q} : q^2 > a\\}$ where $a$ is some rational s.t. $a^2 > 2$. This collection provides an open subcover of the set but clearly does not have a finite subcover.

False. A compact metric space is always bounded; this one is not bounded. That is, we may consider the open cover $\{(-a, a) : a \in \mathbb{Q}\}$; this does not have a finite subcover since if it did, the set $\{q \in \mathbb{Q} : q^2 \geq 1\}$ would be bounded.


Using differentiation to find the power series for $x/(1+2x)^2$ So I am not sure how to go about this problem: finding the series representation of
$$f(x) = \frac{x}{(1+2x)^2}$$
I used $1/(1+2x)$ and created the power series
$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$
and differentiated both sides. My derivative was
$$\sum_{n=0}^\infty (n+1)(-1)^{n+1}(2x)^n = \frac{-2}{(1+2x)^2}$$
I found that I needed to multiply my power series by $-x/2$ in order for the power series to be equivalent to $x/(1+2x)^2$. What I don't understand is why this is wrong. Can someone help point me in the right direction?

First, how I would approach this:
You'll want to start the geometric series, $|x| < 1$,
$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
Differentiate both sides with respect to $x$, and then multiply both sides by $x$. You will obtain
$$\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$
Replace $x$ with $-2x$:
$$\sum_{n=0}^\infty n(-2x)^n = \frac{-2x}{(1+2x)^2}$$
Multiply both sides by $-1/2$:
$$- \frac 1 2 \sum_{n=0}^\infty n(-2x)^n = \frac{x}{(1+2x)^2}$$
You can bring the $-1/2$ into the summation since $(-2x)^n = (-1)^n 2^n x^n$, causing some simplification:
$$\sum_{n=0}^\infty n(-1)^{n+1} 2^{n-1} x^n= \frac{x}{(1+2x)^2}$$
Of course, $(-1)^{n+1} = (-1)^{n-1}$, so for a slightly cleaner-looking final answer, you can use that fact to recombine the $-1$ and $2$:
$$\sum_{n=0}^\infty n(-2)^{n-1} x^n= \frac{x}{(1+2x)^2}$$

Your work, and where your error seems to arise:
You begin by claiming, via the geometric series,
$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$
Differentiating with respect to $x$ on both sides, you should get
$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \frac{-2}{(1+2x)^2}$$
The summation simplifies a bit to
$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \sum_{n=0}^\infty n(-2)^n x^{n-1}$$
which doesn't at all seem like the result you claimed of
$$\sum_{n=0}^\infty (n+1)(-1)^{n+1} (2x)^n$$
So I assume a calculation error arises here.

$\frac{1}{1+2x} = \sum _0^{inf} \left(-2x \right )^n$
$-\frac{2}{\left ( 1+2x \right )^2} = \sum _1^{inf}{n \times \left( -2\right )^n \times x^{n-1}}=\sum _0^{inf}{\left( n+1\right ) \times \left( -2\right )^{n+1} \times x^{n}}$


Solve in $\mathbb{R^{3}}$ : $\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$ Solve the system in $\mathbb{R^{3}}$ :

  $$\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$$

My try :

By imaging I see $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ is a solution!

From a first equation :

$$x=\frac{1}{2}\sqrt{\frac{4z^{2}}{y}-1}$$

So by second equation :

$$y=\frac{1}{2}\sqrt{\frac{\frac{4z^2}{y}-1}{z}-1}$$

But after applied I get difficult equation for $z$ ?

We see that all are nonnegative.

If one of them is $0$ then all are $x=y=z=0$. Now suppose all are $>0$.

Now $1+4x^2\geq 4x$ so $$4z^2=y(1+4x^2)\geq 4xy\implies z^2\geq xy$$

In the same manner we get $y^2\geq xz$ and $x^2\geq yz$ so, if say $x^2>yz$ we get $$x^2y^2z^2>xyxzzy$$
a contradiction. So $x^2=yz$ and similary for other, so we have equality sign every where and thus $$1+4x^2 = 4x\implies x={1\over 2} = y=z$$


Is the space generated by $f(x)=\sin(x)$ a vector space? In my textbook it says that set of all the functions on a given interval is a vector space.
Does that mean that space generated by any function is a vector space?
If not then explain what it means with example.
If yes then how can one say that space generated by $\sin(x)$ is  a vector space because it's not closed under multiplication
$\sin(90)=1$
so $1$ belongs to vector space
$10*\sin(90)=10$ should also belong to vector space but it does not.

The vector space generated by $\sin{x}$ over $\mathbb{R}$ in the sense you mention would be
$$ \{ k\sin{x} : k \in \mathbb{R} \}. $$
It is isomorphic as a vector space to $\mathbb{R}$ by the linear map $k \sin{x} \mapsto k $.

What the book should actually mean is that
$$ \{ f : (f: [a,b] \to \mathbb{R} ) \} $$
is a vector space over $\mathbb{R}$. This is easy to check: clearly if $f,g$ are such functions, and $a,b \in \mathbb{R}$, then
$$ a f + b g $$
is another such function.

It means that the set of functions $f(x)$ where $f(x)$ has domain and range $a$ is a vector space with vector addition defined as the composition of functions. $f(x)= \sin (x)$ is not a vector spaces for the reason you gave.


Finding function with discontinuity at $x = 0$ I am trying to find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous everywhere except $x = 0$. The only example that comes to mind is $f(x) = \frac{1}{x}$, but this isn't even a candidate, I think, because it is undefined at $x = 0$, so I can't really talk about "continuity" at $x = 0$. Another option is a function with a jump discontinuity at $x = 0$: something like $f(x) = x$ for $x \neq 0$ and $f(x) = 1$ when $x = 0$. Then I can take a sequence of points, either from above or below, converging to $0$, but not to $f(0)$.
Is this incorrect? Is there a more elegant solution than this?

The "jump discontinuity" approach is very common.
$$ f(x) = \frac{1}{x}$$
is indeed not defined at $0$, so it is not an example of the desired function.
The canonical example is called a $\delta$ function, which there are two (Dirac and Kronecker). In a set-theoretic application, the Kronkecker delta function is also called an indicator function.
Take
$$
\delta_a(x) =  \begin{cases}1 & x = a \\ 0 &  x \neq a\end{cases}
$$
In topology, the "topologist's sine curve" can be extended onto all of $\mathbb R$ by
$$
T(x) = \begin{cases} \sin\left( \frac \pi x\right) & x \neq 0 \\ 0 & x = 0
\end{cases}
$$
$T$ has some interesting topological properties when restricted to $(0,1]$.

What about
$$f(x)=\begin{cases}\frac{\sin x}x,&x\neq0\\{}\\-4,&x=0\end{cases}\;\;\;?$$


$\zeta_n^k$ is primitive if and only if $(k,n) = 1$ Show that for $k \in \mathbb{Z}$:

Is $\zeta_n$ a primitive $n$-th root of unity, then $\zeta_n^k$ is primitive if and only if $(k,n) = 1$.

I only need the backwards direction:

$\zeta_n^k$ is primitive $\Rightarrow$ $(k,n) = 1$

First of all notice that $\zeta_n^k$ is of the form
$$
e^{2\pi i\frac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(\zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,b\in\Bbb Z$ and $a,b\ge1$. Thus
$$
\zeta_n^k=e^{2\pi i\frac {ad}{bd}}=e^{2\pi i\frac d{b}}.
$$
From here we deduce that $(\zeta_n^k)^b=1$ and since $b<n$ (in fact $0\le k\le n-1$) our root cannot be primitive.


How to find GCD of those two complex polynomials? I have a polynomial

$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$

I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $f(x)$ and its derivative $f'(x)$.

$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^4+3x^2+1)-8x^3$$

$$f'(x) = i(6x^5-12x^3+6x)+(6x^5+12x^3+6x)-24x^2$$

I know I should find the GCD as $(x-i)(x-1)$ but this is where I'm stuck. What would the next step be?

It's best not to split the polynomials up using $i$, but collect terms in powers of $x$.

$$ \eqalign{f_0 = f(x) &=  \left( 1+i \right) {x}^{6}+ \left( 3-3\,i \right) {x}^{4}-8\,{x}^{3}+
 \left( 3+3\,i \right) {x}^{2}+1-i \cr
f_1 = f'(x) &=  \left( 6+6\,i \right) {x}^{5}+ \left( 12-12\,i \right) {x}^{3}-24\,{x
}^{2}+ \left( 6+6\,i \right) x
}
$$
The remainder of $f_0$ on division by $f_1$ is
$$  f_2 = f_0 - (x/6) f_1 = \left( 1-i \right) {x}^{4}-4\,{x}^{3}+ \left( 2+2\,i \right) {x}^{2}+
1-i
$$
Then take remainder of $f_1$ on division by $f_2$, etc.

By inspection, $f(1)=0$ so $x-1$ is a factor of $f(x)$.

Since $f'(x)= 3i(x^2-1)^2(2x)+3(x^2+1)^2(2x)-24x^2$, we also have that $f'(1)=0$, so $x-1$ is a factor of $f'(x)$.

This means $(x-1)^2$ is a factor of $f(x)$.

One also has that by inspection $f(i)=0$ so $x-i$ is a factor of $f(x)$.

We also have that $f'(i)=0$, so $x-i$ is a factor of $f'(x)$.

So we also have that $(x-i)^2$ is a factor of $f(x)$.


If $P(A \ \cup \ B) = P(A) + P(B)$, is it the case that $A$ and $B$ are disjoint? I know that if $A$ and $B$ are disjoint events, then $P(A \cup \ B) = P(A) + P(B)$. However, is the converse true as well? Thanks.

No, all you can deduce is that $P(A \cap B) = 0$, because $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ It doesn't mean that $A \cap B$ is empty.

EDIT: This is incorrect, disregard. The comments have valuable discussion and counterexamples.

Yes. Assuming $A$ and $B$ are independent, by definition, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If we have $P(A \cup B) = P(A) + P(B)$, then $P(A \cap B) = 0$, which is the definition of disjoint events.


Determining the limit of $\bigl(1 + \frac{1}{x+1}\bigr)^x$ as $x\to\infty$ $$L = \lim_{x\to \infty} \biggl(1+\frac{1}{x+1}\biggr)^x $$

This one has to solved using $$\lim_{x\to \infty} \biggl(1+\frac{1}{x}\biggr)^x = e.$$

I did this
\begin{align}L &= \lim_{x\to \infty}\biggl( \biggl(1+\frac{1}{x+1}\biggr)^{x+1}\biggr)^{\frac{x}{x+1}}\\
& = e^{\frac{x}{x+1}}
\end{align}

I get the limit to be (I figured it out this is wrong) $$\frac{1}{e}$$
while the answer is $e$.

Your solution is good until the final step.
$$\lim_{x\to \infty}{x\over x+1}=\lim_{x\to \infty}{x\over x(1+{1\over x})}=1$$

Well that new question is entirely different.

remember if $\lim b_n = B$ and $\lim c_n = C \ne 0$ then $\lim b_n/c_n = B/C$

$\lim (1 + 1/x)^x = e$

so $\lim (1 + {1 \over {x + 1}})^{x+1} = e$

and $\lim (1 + {1 \over {x + 1}}) = 1$

so $\lim (1 + {1 \over {x + 1}})^x = $

$\lim \frac { (1 + {1 \over {x + 1}})^{x+1}}{1 + {1 \over {x + 1}}} =$

$\frac {\lim (1 + {1 \over {x + 1}})^{x+1}}{\lim (1 + {1 \over {x + 1}})}  = {e \over 1} = e$

==== original answer for the original (miswritten) question below ====

Well, I thought my way was simple.

For $1 + 1/x = \frac {x + 1}{x}$

So $\lim_{x \rightarrow \infty}\frac {x + 1}{x} = e \ne 0$

So $\lim_{x \rightarrow \infty}\frac {x}{x + 1} = {1 \over e}$


How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve

$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$

I've chosen the substitution variables

$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$

However, I am completly stuck at

$$\int\sqrt{1+xu} dx$$

Which let me believe I've chosen wrong substitution variables.

I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.

Would someone please give me an hint on this ?

Thanks.

I thought about Wolfram and Maple too, needs a little help. But I was wrong.

$$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$$

\begin{gathered}
  y = {x^2} + 2 \\
  dy = 2xdx \\
\end{gathered}

\begin{gathered}
  \int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right) \\
   = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2}  + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right) \\
\end{gathered}

Now Wolfram: This is an identity:

$$\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right)$$

Resubstituting gives me:

$$\frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right) = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2}  + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right)$$

Differentiation:

$$\frac{d}{{dx}}\left( {\frac{2}{{15}}\sqrt 2 {{\left( {\sqrt {{x^2} + 2}  + 2} \right)}^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right)} \right) = \sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } $$`

And:

$$\sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} }  \ne \sqrt {1 + x\sqrt {{x^2} + 2} }$$

Very poor! For me. Made a mistake by myself.

Consider the integral
\begin{align}
I = \int \sqrt{1 + x \sqrt{x^{2} + 2}} \, dx
\end{align}
Make the substitution $x = \sqrt{2} \, csch(t)$ to obtain the integral
\begin{align}
I = - \sqrt{2} \, \int \sqrt{1 + 2 \, csch(t) \, coth(t)} \cdot csch(t) \, coth(t) \, dt.
\end{align}
Now Wolfram can calculate the integral and provides the result
\begin{align}
- \frac{I}{\sqrt{2}} &= \frac{1}{2} \sinh(t) \sqrt{2 coth(t) csch(t)+1} \left[\frac{g(t)}{\sqrt{4 cosh(t)+cosh(2 t)-1}} - csch^2(t) \right]
\end{align}
where
\begin{align}
g(t) = \ln\left(\tanh^2\left(\frac{t}{2}\right)\right) - \ln\left(1 + \tanh^2\left(\frac{t}{2}\right) + \sqrt{-\tanh^4\left(\frac{t}{2}\right) + 2 \tanh^2\left( \frac{t}{2}\right) +1} \right)
\end{align}
By making a careful backward substitution the result becomes
\begin{align}
I &= \frac{1}{\sqrt{2}} \left[ x \sqrt{1+x\sqrt{x^{2}+2}} + \ln\left( 1 + \frac{(x + \sqrt{x^2 + 2})^{2}}{2} + \sqrt{\frac{(x + \sqrt{x^2 + 2})^{4}}{4} + (x + \sqrt{x^2 + 2})^{2} - 1} \,  \right) \right]
\end{align}


Can someone show by vector means that any inscribed angle in a semicircle is a right angle Could someone explain how to prove any angle inscribed in a semicircle is a right angle using vectors. I understand that the dot product of two vectors is 0 is they are perpendicular but I don't know how to show this in a semicircle.

Wlog. we can assume the semicircle to be the upper unit semicircle with centre in the plane origin $(0, 0)$. Let $(x, y)$ be a fixed point on the semicircle with $y>0$.

Consider the vector $v_1$, which is the difference between the positional vectors $(x, y)$ and $(-1, 0)$, and the vector $v_2$, which is the difference between the positional vectors $(x, y)$ and $(1, 0)$.

Now note that the angle inscribed in the semicircle is a right angle if and only if the two vectors are perpendicular. Using the scalar product, this happens precisely when $v_1\cdot v_2=0$.

So just compute the product $v_1\cdot v_2$, using that $x^2+y^2=1$ since $(x, y)$ lies on the unit circle.

A semicircle is one half circle.  Define 3 vectors r(sub 1), r(sub 2) and r(sub 3). From the center point of the semicircle radiate these three vectors such that say r(sub one) radiates from the center to the leftmost endpoint of the semicircle; r(sub 2) radiates from the center to the rightmost endpoint of the semicircle; and r(sub 3) radiates from the center of the circle to a locus at any point on the circle.  We then know the magnitude of r(sub 1), r(sub 2) and r(sub 3) are equal

Further two vectors, one emanating from the rightmost endpoint to the locus, and one emanating from the leftmost endpoint to the locus will form an angle; an angle which we must prove is a right angle.

Call r(sub 1), r(sub2) and r(sub 3) vectors with origins at the center. The vectors that join the endpoints (of the semicircle) to the locus, let's call U and V (these are all vectors, please keep this in mind). Say vector U emanates from the left semicircle endpoint, and terminated at the locus. Vector V emanates from the right semicircle endpoint and terminates at the locus.

We can then redefine vector U in terms of r(sub 1), and r(sub 3) or U + r(sub  1) = - r(sub 3) and vector V can be redefined as V + r(sub 2) = - r(sub 3) {I THINK} However r(sub 1) = -r(sub 2) ...

And here, I unfortunately, lose it.  We have to show either U dot V equals zero, or the magnitude of U cross V equals one.  I suspect the cross product is an easier proof; but I keep banging my head on the dot product ... I am just a stubborn old coot.

Then by these definitions, r(sub 1) equals -r(sub 2) (Negative r(sub 2).  we can


proving that if $f(x)$ odd then $f(0)=0$ Need to prove that if $f(x)$ is an odd function that defined in the point: $x=0$,

So $f(0)=0$.

I know that odd function is: $f(-x)=-f(x)$

And that $f(x)=0$ is an odd function but dont know how to prove.

Thanks.

For $x=0, f(0)=-f(-0)=-f(0)$, but this is saying that one number is equal to its negative. This is only true for $0$.

$a=-a\implies a=0$

Take that definition you have for an odd function and use the fact that -0=0,  then treat f(0) as a variable to solve for like an algebra equation.


Find angle using the sine and cosine formulas Given that $\cos A = \frac{8}{17}$ find

a) $\sin 2A$

b) $\cos 2A$

Using $\cos A$ I can tell that $\sin A = \frac{15}{17}$

a) $\sin 2A = 2\sin A\cos A = \frac{240}{289}$

Is this correct?

b) $\cos 2A$

For this part, there are 3 formulas for expanding $\cos 2A$, is there any one I should favour over the others?

Say I chose $1 - 2\sin^2 A$, how could I find $A$?

For a)

No, it is not correct. If $\cos A=8/17$, then from $\cos^2A+\sin^2A=1$
$$\sin A=\color{red}{\pm}\sqrt{1-\left(\frac{8}{17}\right)^2}=\color{red}{\pm}\frac{15}{17}$$
and so
$$\sin 2A=2\sin A\cos A=\color{red}{\pm}\frac{240}{289}$$

For b)

You can use whichever you like :
$$\cos 2A=2\cos^2A-1=1-2\sin^2A=\cos^2A-\sin^2A$$
By the way, note that $\cos^2A=(\cos A)^2$ and that $\sin^2A=(\sin A)^2$ : you don't need to find $A$.

Draw a right angled triangle. This example uses Pythagorean triplets so integral sides occur.Only positive sign of $cos$ considered here. For negative sign change sign.

$$ c = 8/ 17 ;\ s  = 15/17 $$

$$ s_2 = 2 \cdot s \cdot c = =\frac {240}{289} = \pm 240/289 $$,

$$ c_2 =  c^2 -s^2 =  \frac {8^2-15^2}{289} = \pm \frac {-161}{289} .  $$


Showing that the closure of the closure of a set is its closure I have the feeling I'm missing something obvious, but here it goes...

I'm trying to prove that for a subset $A$ of a topological space $X$, $\overline{\overline{A}}=\overline{A}$.  The inclusion $\overline{\overline{A}} \subseteq \overline{A}$ I can do, but I'm not seeing the other direction.

Say we let $x \in \overline{A}$.  Then every open set $O$ containing $x$ contains a point of $A$.  Now if $x \in \overline{\overline{A}}$, then every open set containing $x$ contains a point of $\overline{A}$ distinct from $x$.  My problem is: couldn't $\{x,a\}$ potentially be an open set containing $x$ and containing a point of $A$, but containing no other point in $\overline{A}$?

(Also, does anyone know a trick to make \bar{\bar{A}} look right?  The second bar is skewed to the left and doesn't look very good.)

How do you define the closure of a set $A$? If it's not already your definition, it might be a useful thing to prove that the closure of a set is precisely the intersection of all closed sets containing it.

A set is closed iff every convergent sequence in the set has its limit in that set. The closure operation adds a set's limit points to the set. A closed set already contains its limit points, so repeating the operation doesn't change the answer (idempotent).

For Example:

The sequence 1/n; 0 < n < inf goes like this..

1, 1/2, 1/3, 1/4, ... , 1/465674564, ...

It's limit is 0, which is not part of the sequence, so closure would add the 0.

NOTE - The above example is just meant to illustrate how closure works.  As far as I know, this sequence is not a subset of any metric space.


Closed form of the sequence $a_{n+1}=a_n^2+1$ If $$a_{n+1}=a_n^2+1,$$ with initial $a_1=\frac{1}{2}$. How to solve this sequence problem, i.e., how to represent $a_n$ in closed form?

EDIT: The answer below assumes $a_1$ is an integer, and as such is not directly applicable to your question, but I am leaving
  it here in case people can make use of the results mentioned in the
  paper below.

Aho and Sloan proved that for sequences like yours, there is a constant $k$ such that

$$a_n = \lfloor k^{2^n} \rfloor$$

for sufficiently large $n$. $k$ can be defined as a limit of a sequence using $a_n$ itself. If you include $k$ as one of your closed form constants, you are done!

See their paper for details: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf

Of course, for your special case, one might still be able to find a different "closed form" which might be more appealing to you.

Perhaps let $a_n=\tan \theta_n$.  Then $\tan \theta_{n+1}=\sec^2 \theta_n$.   So, $\tan \theta_2=\sec^2 \theta_1$,  $\tan \theta_3=1+\sec^4 \theta_1$, etc. I am not sure if this procedure will produce a closed form.


Real numbers and rational number I know I am getting somewhere wrong but my question is how rational numbers are real numbers(Q is subset of R) .Rational numbers don't have least upperbound property and lub property is necessary condition for being real numbers

A bounded subset of the rational numbers has a least upper bound in the real numbers, but not necessarily in the rational numbers.

Many subsets of the real numbers do not have the least upper bound property (within themselves).  For instance, the irrational numbers do not have the least upper bound property.  That is, there are bounded sets of irrational numbers that do not have a least upper bound in the irrationals.  Of course they still have a least upper bound in the reals (but that least upper bound would be rational).

Every rational real is the lub of itself.


A question about using the squeeze theorem? If you have the inequalities not equal to, but the function is just strictly between two other functions, can you use the squeeze theorem?

Remember,  if you have a fact "A", then you also have the fact "A or B",  for any fact B.   So, in this case,  if you have "$x<y$" then you have $x\le y$ because that's just shorthand for $x<y$ OR $x=y$,  and since we have $x<y$, then we have $x\le y$.  So yes,  strict inequality is just fine, as that's even better than less than or equal to.

Sorry for the late answer but just answering in case anyone else has the same doubt.
by looking at the question ,it seems that the op is wondering if $f(x) < g(x) < h(x)$ how can it be possible that the limits are equal as the functions are not allowed to be equal
The fact is $$f(x) < g(x) \implies \lim_{x\to a}f(x) \leq \lim_{x\to a}g(x) $$
this can be easily proved by taking h(x)=f(x)-g(x) and h(x)<0,now apply $\epsilon-\delta$ definition ,u cant rule out the possibility of $lim_{x\to a} h(x)=0$ and now u can use limit law to split f(x) and g(x) and hence proving the original statement
Several examples showing such scenario are provided in comments section.
Now coming to squeeze theorem,now u are possibly understanding that even if u have strict inequalities for the functions,it translates to having non strict inequalities for the limit and hence when the lower and upper bound for $\lim_{x \to a} g(x)$ are equal and it is allowed to take that value,it must take that value
alternatively, if u want to prove squeeze theorem for strict inequality from scratch ie. from $\epsilon -\delta $  definition , then proceed as shown in https://en.wikipedia.org/wiki/Squeeze_theorem#Proof and just remove equality wherever necessary


Question about the problem that $P^n$ admits a field of tangent $1$-planes if and only if $n$ is odd. I want to ask the problem 4-C in the characteristic classes written by John W. Milnor.

  Problem [4-C]. A manifold $M$ is said to admit a field of tangent $k$-planes if its tangent bundle admits a sub-bundle of dimension $k$.
  Show that $P^n$ admits a field of tangent $1$-planes if and only if $n$ is odd.

From right to left, I can prove it by using the fact that $S^n$ has a non vanishing vector field when $n$ is odd. But from left to right, I try to use the Whitney product theorem to prove.

Can anybody give me a hint or a proof?
Thank you.

An odd dimensional projective space admits of a nowhere zero vector field
The trick is to embed the corresponding sphere into complex numerical space: $S^{2n-1}\subset \mathbb C^n      $.
We then have a nowhere vanishing vector field $S^{2n-1}\to TS^{2n-1}: v\mapsto iv$ which, being invariant under the antipodal map $S^{2n-1}\to S^{2n-1}:v\mapsto -v$,  descends to the quotient manifold $\mathbb P^{2n-1} $ as a nowhere zero vector field $s:\mathbb P^{2n-1}\to T\mathbb P^{2n-1}$
The existence of such a nowhere tangent field is equivalent to the existence of a trivial rank one subbundle of the tangent bundle: here the subbundle $L\subset T\mathbb P^{2n-1}$ is given by $L(v)=\mathbb R\cdot s(v)$

An even  dimensional projective space does not admit of a nowhere zero vector field

A nowhere vanishing vector field on $\mathbb P^{2n}$ would lead to the existence of a trivial rank one subbundle $L\subset T\mathbb P^{2n}$ and thus to a (non-canonical) splitting $T\mathbb P^{2n}=L\oplus V$ where $L\cong \underline {\mathbb R}$
But then the top Stiefel-Whitney class of $T\mathbb P^{2n}$ should vanish: $w_{2n}(T\mathbb P^{2n})=0$
However this is false: the total  Stiefel-Whitney class of $T\mathbb P^{2n}$ is $(1+h)^{2n+1}$ so that $$w_{2n}(T\mathbb P^{2n})=\binom {2n+1}{2n}h^{2n}=(2n+1)\cdot h^{2n}=h^{2n}\in H^{2n}(\mathbb P^{2n},\mathbb Z/2)=\mathbb Z/2\cdot h^{2n}$$ This contradiction shows that there is no rank one trivial subbundle of $T\mathbb P^{2n}$

Edit
As commented by PVAL,  in the even dimensional case I have proved that $T\mathbb P^{2n}$ doesn't contain a trivial line bundle but haven't proved that it doesn't contain a non trivial one .  Here is a proof of that last fact:
If $T\mathbb P^{2n}$ contained a non-trivial line bundle, it would be the tautological line bundle $\gamma$ (the only non trivial line bundle on $\mathbb P^{2n}$) and we would have $(1+h)^{2n+1}=(1+h)\cdot w(V)$ since $T\mathbb P^{2n}=\gamma\oplus V$.
Hence we would deduce that $w(V)=(1+h)^{2n}$ because, as PVAL judiciously comments, $1+h\in H^\star (\mathbb P^{2n},\mathbb Z/2)$ is an invertible element .
But then we deduce $w_{2n}(V)=h^{2n}\neq0\in H^{2n}(\mathbb P^{2n},\mathbb Z/2)$: this is absurd because $V$ has rank $2n-1$ and cannot have a non-zero Stiefel-Whitney class of degree $2n$.

$\mathbb{P}^n(\mathbb{R}) \simeq S^n/{\pm 1}$, that is $S^n$ is a two-fold cover of $\mathbb{P}^n$. Fields of tangent $1$-planes are $1$-dimensional sub-bundles of the tangent bundle. Consider such a field on $\mathbb{P}^n$. Pull back to $S^n$ and get a field on $S^n$ ( invariant under $x\mapsto -x $). Since $S^n$ is simply connected ($n\ge 2$) this $1$- bundle is orientable  and  from it we  get a nonzero vector field on $S^n$ (once the direction is chosen, just take the vector of unit length).

So this is possible only for odd $n$.


Interception with $x$-axis - not so trivial? I want to find the interception with the x-axis of the following function:
$f(x) = \frac{1}{4}x^4-x^3+2x$.

So putting $0 = \frac{1}{4}x^4-x^3+2x$ I would get $0 = x(\frac{1}{4}x^3-x^2+2)$ but how to check for the other solutions properly?

$f(x)=\frac14 x^4-x^3+2x=\frac14 x(x^3-4x^2+8)$.

Let $g(x)=x^3-4x^2+8$. Since $g(2)=0$, we know that $x-2$ divides $g(x)$, so $g(x)=(x-2)(x^2+kx-4)$ for some constant $k$. Equating coefficients of $x$ gives $-2k-4=0$ so $k=-2$. Now $g(x)=(x-2)(x^2-2x-4)$.

Let $h(x)=x^2-2x-4=(x-1)^2-5$. Then the zeroes of $h(x)$ are $1+\sqrt 5$ and $1-\sqrt 5$. Hence the zeroes of $f(x)$ are $0$, $2$, $1+\sqrt 5$ and $1-\sqrt 5$.

Also $x=2$ is a root, so $f(x)=\frac{1}{4}(x-2)x(x^2-2x-4)$.  wolframalpha is your friend.


Interesting examples where $\frac 0 0$ is defined. After giving up some nice algebraic properties we could give examples where $\frac a 0$ is defined for nonzero $a$, for example consider the real line with one point compactification.
Just for curiosity, are there any interesting examples in which we give up the least amount of good algebraic properties and make the expression $\frac 0 0$ defined?(that is, there is a unique element $k$ such that $0\cdot k$ is $0$) By interesting I mean nothing like the trivial ring.
I am thinking I should consider multiplication where $0\cdot x$ is only zero for $x=k$ but I’ve run out of examples.

Some of these are not your exact $\frac 00$ case, but they are still wondrous:
1.Here is wheel theory. This establishes some rules for division by zero.
2.The function $x^x$ has the limit for x=0 as $0^0\to 1$. Graph
3.The function $\frac 1x$ has the limits to 0 from the positive side as $\infty$ and $-\infty$ from the other side. Here $\frac 10$ can be defined by complex infinity
4.Here are some other possibilities and their discussion of this case. It has rules for indeterminate, shown by ¿, complex infinity, shown by an infinity with a tilda over it, and more!
5.Here is a Wikipedia page discussing division by zero here.
6.This case in your title appears in many limits such as these.
7.Best for last, the function $x^{{{\frac 1x}^ {\frac 1x} }^{…}}$ evaluated at x=0 gives:$$0^{{{\frac 10}^ {\frac 10} }^{…}}\to 0$$
Here is the graph of this function

Here is a nice limit. You should see it in the linked answers about a limit. If you apply direct substitution you get $$\frac{1^\frac 10-1^\frac 10}0\ \to \frac {3e^2}4$$

I could go on forever. I will add more. As always, please correct me and give me feedback!

Perhaps dy/dx is an interesting example of what happens when we reach 0/0.
As you will know it is well defined for most differentiable algebraic formula, but with a wide variation in the particular answer you get.
Also try considering if it is identical to (1/0)*(0/1) and variants on that theme.
E.g. If infinity (1/0) is important to study, why aren't infinitesimals just as interesting, etc.


How do I make this simple proof better (and more correct?) Let $x$ and $y$ be real numbers. If $x\cdot{y}>\frac{1}{2}$, then $x^2+y^2>1$.

Proof: We will prove with the direct method. Let $x$ and $y$ be real numbers. Since
$$ x\cdot{y}>\frac{1}{2} $$
it follows that
$$ 2xy>1,$$
which means
$$x^2+y^2 \geq 2xy.$$
Therefore,
$$ x^2+y^2>1. $$

There is a much simpler proof.

Since $xy > \frac{1}{2}$, $x$ and $y$ must be non zero numbers, which are both positive or negative.

Suppose $x>0$ and $y>0$.

If $x>1$, then also $x^2>1$, and $x^2 +y^2>x^2>1$ is obvious.

Take $0<x\leq 1$ and $y>0$.

Since $0<x\leq 1$, the function $f$ with $f(x)=\sqrt{1-x^2}$   is well defined.  Let  $g$ be the function with $g(x)=\frac{1}{2x}, \ 0<x\leq 1$.

It can be easily proved that the function h with $h(x)=-x+\sqrt{2}, \ 0<x\leq 1$
      has a graph which is a line tangent to the graphs of $f$ and $g$ at the point
      $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

One can also verify easily that $f$ is a concave function and that $g$ is a convex funtion. Therefore the tangent line $h$ is above the graph of $f$ and below the graph of $g$.
      So if  $0<x\leq 1$,the previous remark implies that  $f(x)\leq h(x)\leq g(x)$ and  $\sqrt{1-x^2}\leq \frac{1}{2x}$.

Now, the proposition we want to prove is obvious: If $xy > \frac{1}{2}$, then $y>\frac{1}{2x}$,
      and $\frac{1}{2x} \geq \sqrt{1-x^2}$ from the previous inequality, hence
      $y> \frac{1}{2x} \geq \sqrt{1-x^2}$. Consequently, $y>\sqrt{1-x^2}$, which implies
      $y^2>1-x^2$ and therefore $x^2+y^2>1.$

The case $x<0$ and $y<0$ is similar.


If $G/Z(G)$ is Abelian and $\{e\}\ne H\triangleleft G$, then $H\cap Z(G)\ne\{e\}$ I am somewhat unfamiliar with commutator subgroups, so I am not sure about the last couple of lines of this proof.

  Let $G$ be a group such that $G/Z(G)$ is Abelian, and let $\{e\}\ne H\triangleleft G.$  Show that $H\cap Z(G)\ne\{e\}.$

Consider the commutator subgroup $G'<G.$ For any $N\triangleleft G$ with $G/N$ Abelian, we have that $G'\subset N.$ Thus since $Z(G)$ is a normal subgroup of $G$ and the quotient group is Abelian by assumption, $G'\subset Z(G).$ Now let $g\in G$ and $h\in H$. Then
$$ghg^{-1}h^{-1} = (ghg^{-1})h^{-1}\in H\cap G'.$$ Now if $ghg^{-1}h^{-1}\ne e$ for some $g,h,$ then we're done. But if $ghg^{-1}h^{-1} = e$ for all $g\in G,\; h\in H$, then this implies $H\subset Z(G)$.

We start with a
Lemma Let $H \unlhd G$ with $H \cap G’=1$. Then $H \subseteq Z(G)$.
Proof Let $h \in H$ and $g \in G$ be arbitrary. Then the commutator $[g,h]=g^{-1}h^{-1}gh=(g^{-1}h^{-1}g)h \in H$, since $H$ is normal. But also $[g,h] \in G’$, hence since $H$ and $G’$ intersect trivially we get $[g,h]=1$ and $H$ must be central.
Now assume that $G/Z(G)$ is abelian. This is equivalent to $G'\subseteq Z(G)$. If $1 \neq H \unlhd G$, then we have two cases: either $H \cap G' = 1$ and then by the Lemma we have $H \subseteq Z(G)$, so certainly $H \cap Z(G)=H \neq 1$. If $H \cap G' \neq 1$, then since $G’\subseteq Z(G)$, we get $1 \neq H \cap G’\subseteq H \cap Z(G)$ and we are also done.
Note that $G/Z(G)$ being abelian implies that $G$ is nilpotent. In general, if $G$ is nilpotent and $1 \neq H \unlhd G$, then $H \cap Z(G) \neq 1$.

The commutator subgroup $G'$ of $G$ is defined as $$G' = \{ghg^{-1}h^{-1}: g, h \in G\}.$$ The commutator of two elements measures how non-commutative that they are. For instance, if $g$ and $h$ commute then their commutator $ghg^{-1}h^{-1}$ is trivial.

So, to discuss the proof. (We will verify each claim.)

If $N$ is normal and $G/N$ is abelian then $G' < N$. This is true since if $a, b \in G$ then $aNbN = bNaN$ so $a^{-1}bNaN = NbN = NNb$ so letting the elements in the first two $N$'s be the identity gives that there is $x \in N$ such that $a^{-1}ba = xb$, so $N \ni x = a^{-1}bab^{-1}$. However, $a$ and $b$ were arbitrary so by replacing $a$ with $a^{-1}$ we see that $N$ contains all the commutators.

$(ghg^{-1})h^{-1} \in H\cap G' \subset H \cap Z(G)$ since $H$ is normal and the entire product is a commutator.

If $ghg^{-1}h^{-1} \neq e$ for some choice of $h \in H$ and $g\in G$ then we have that $H \cap Z(G) \neq \{e\}$.

If $ghg^{-1}h^{-1} = e$ for all $h \in H$ and $g\in G$, then this means that $H$ commutes with every element of $G$ so $H \subset Z(G)$, so $H \cap Z(G) \neq \{e\}$ since $H$ is non-trivial.


Is there a natural number between $0$ and $1$? Is there a natural number between $0$ and $1$?

A proof, s'il vous plaît, not your personal opinion. (Assume the Peano Postulates.)

Every natural number $m$ is either $0$ or $s(n)$, where $n$ is a natural number.

Proof: It can't be both, because $s(n)$ can't be $0$. Set of all natural numbers which are either $0$ or $s(n)$ for some $n$ satisfies induction principle, so it contains all natural numbers.

Direct consequence: Every natural number is either $0$, or $s(0)$ or $s(s(n))$ for some natural number $n$.

Suppose there is $m$ such that $0 < m < s(0)$. Either $m$ is $0$, $s(0)$ or $s(s(n))$. First two cannot hold, so you have $s(s(n)) < s(0)$, i.e., $s(n) < 0$.

One is defined as $\{\emptyset\}$. That a number n is between 0 and 1 means that $0\in n$ and $n\in 1$. Since $n\in 1$, it follows that $n=0$, a contradiction.


Can someone explain what this question is asking me? (Calculus II) How to define the area of the region bounded by $y = \sqrt{1-x^2}$ and $y = 0$?

I am assuming we will be using Riemann's sum however I am not sure if we are given $[a,b]$ or is that what we are given?

Hint:

The graph of $y=\sqrt{1-x^2}$ is the unit circle $x^2+y^2=1$ with range $0\leq y\leq 1$.

The region is exactly $\frac12$ of the total area of the unit circle.

the resultant area is the area inside the circle and above x-axis

so area is equal to $\pi r^2$/2 where r=1

and hence the area is $\pi$/2


Deducing results about continuous time random walks from the corresponding discrete time result Is there any standard way to prove results about continuous time random walks from the corresponding results for discrete time random walks?

Specifically, my problem is that I was reading Lawler and Limic's Random Walk: A Modern Introduction and there is a result that bounds the transition probability which feels like is should be true still for continuous time random walks.

The result states $\exists c>0$ such that $\forall x,n$

$$p_n(x)\leq\frac{c}{n^{d/2}}.
$$

So my question is, can this be used to deduce the analogous result with a continuous time parameter $t$ in place of $n$?

You can discretize a continuous 1- dimensional random walk $X_t$:
let $p=\mathbb P(X_t \mbox {hits 1 before hitting -1}|X_0=0)$ and $q=1-p$. Define a
simple discrete random walk $Y_n$ with transition probabilities p and q.
This gives you $t_1<t_2<..$ such that $X_{t_n}=Y_n$ and some version of the mentioned result must hold for $X_t$.

I think I read this in Durrett's "Probability theory and applications".


Solve the system of equations? What is this format of an answer? So I solved this system without using matrices, just by (sort of) reverting to high school math instincts.
$$w+x=-5$$
$$x+y=4$$
$$y+z=1$$
$$w+z=8$$
From this, I got $w=-9,x=4,y=0$ and $z=1$. How would I convert this back to what appears to be a matrix form?


Finding number of solutions to a congruence equation Let the congruence equation be:

$$ 51813x \equiv 14593 \pmod{119472}$$

How can you determine the number of roots that exist for this equation? Is there also a way to find the smallest/largest roots? I am under the impression I should use the chinese remainder theorem but I am not sure where. Any help is appreciated.

The equation $ax \equiv b \bmod m$ is equivalent to $ax=b+my$.

Therefore, $ax \equiv b \bmod m$ has solutions iff $\gcd(a,m)$ divides $b$.

Try that for your equation.

Hint: There are no solutions for $x$. Compute the gcd(51813,119472).


Matrix diagonalization and operators Let $V=\Bbb F^{m\times n}$
$T: V\to V$ , by $T(B)=P^{-1}BP$ , for any $B$ in $V$ , where $P$ is an invertible matrix.
prove that if $A$ is an eigenvector of $T$, with eigenvalue $\lambda$
and $A$ is a diagonalizable matrix, then $\lambda=1$.

I know that $A$ and $\lambda A$ are similar, then they have the same eigenvalues.
then if $a$ is an eigenvalue of $A$, $(\lambda*a)$ is an eigenvalue of $(\lambda A)$,
I'm not sure that I can claim then $(\lambda*a)=a$, because we don't know they have same eigenvectors.

Thank you in advance.

Suppose that $A \neq 0$ is diagonalizable, then there exists an invertible (unitary) matrix $U$ such that $A = U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}$.

Then $\lambda A = T(A) = P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$.

We get $\lambda U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1} =  P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$ (*)

On taking the trace of the (*) and using the fact $\mathrm{Tr(AB)} = \mathrm{Tr(BA)}$, we should be able to force the condition we need on $\lambda$. Better yet, when are two diagonal matrices of the same size similar?

I'm sorry, I think I found a contradiction. Let

$$A= \left(\begin{array}{ccc}1 & 0 & 0\\0 &-1& 1\\0& 0 &0 \end{array}\right)$$

A is diagonalizable.

Then if

$$P= \left(\begin{array}{ccc}0 & 1 & -1\\1 &0& 1\\0& 0 &1 \end{array}\right)$$

We get $P=P^{-1}$, and $P^{-1}AP=-A$, therefore $T(A)=-A$, and $\lambda=-1$.


Clustering analysis of a weighted graph My data consists of a large weighted undirected graph of $n$ nodes. I need to group the nodes into $m$ clusters ($m < n$), such that nodes in a cluster are connected with heavy weights. What algorithms are available for this problem?

Some keywords and starting references:

Spectral clustering
Community detection

Walktrap and the Louvain algorithms work very well with weighted undirected graphs.
You may find them in most modern graph libraries.


Equivalence of two definitions of Almost Complex Structure. Let $M$ be a smooth manifold of even dimension $2n$. I would like to understand the equivalence of the following two standard definitions for an almost complex structure on $M$.

Denote by $TM$ the tangent bundle of $M$. Then the first definition I wish to consider is the following.

  Definition 1 An almost complex structure on $M$ is a vector bundle morphism $J:TM\rightarrow TM$ satisfying $J^2=-id_{TM}$.

Now $TM$ is a real vector bundle of rank $2n$ over $M$ and is thus classified up to bundle isomorphism by the homotopy class of a map $\tau_M:M\rightarrow BGl(\mathbb{R}^{2n})$, where $BGl(\mathbb{R}^{2n})$ is the classifying space of the Lie group $Gl(\mathbb{R}^{2n})$. The complex general linear group $Gl(\mathbb{C}^n)$ also has a classifying space $BGl(\mathbb{C}^n)$, and the map $\mathbb{C}^n\rightarrow \mathbb{R}^{2n}$, $\underline z=\underline x+i\underline y\mapsto (\underline x,\underline y)$, corresponds to a homomorphism $r:Gl(\mathbb{C}^n)\rightarrow Gl(\mathbb{R}^{2n})$ given by

$$r(X)=r(A+iB)=\begin{pmatrix}A & -B\\B & A\end{pmatrix}$$

where $A$, $B$ are real matrices. This homomorphism subsequently induces a map of classifying spaces $Br:Gl(\mathbb{C}^n)\rightarrow BGl(\mathbb{R}^{2n})$ with homotopy fibre $Gl(\mathbb{R}^{2n})/Gl(\mathbb{C}^n)$.

  Definition 2 An almost complex structure on $M$ is a lift through $Br$, up to homotopy, of the classifying map $\tau_M$. $$\begin{array}{ccccccccc}
    & & BGl(\mathbb{C}^n) \\
    &  \nearrow & \downarrow{Br}\\
    M & \xrightarrow{\tau_M} & BGl(\mathbb{R}^{2n}).
    \end{array}$$

Remark that $BGl(\mathbb{R}^{2n})$ is modelled by the Grassmannian $Gr_{2n}(\mathbb{R}^{\infty})$ of $2n$-planes in $\mathbb{R}^\infty$, and a corresponding statement holds for $BGl(\mathbb{C}^{n})$. Since $M$ is finite dimensional the map $\tau_M$ will factor through the inclusion $Gr_{2n}(\mathbb{R}^{2N})\hookrightarrow Gr_{2n}(\mathbb{R}^{\infty})$ for some $N\geq n$, and $Gr_{2n}(\mathbb{R}^{2N})$ may be given the structure of a smooth manifold and $\tau_M$ represented by a smooth map.  Similarly $Gr_{N}(\mathbb{C}^{N})$ carries the structure of a complex manifold and $Br|$ may be chosen to be a smooth map between these spaces.

Now my question: How are these two definition equivalent?

The key point is to understand how the classifying map works. Here's one way to:

One can produce a bundle-embedding $TM \hookrightarrow M \times \bf R^\infty$ (which can moreover factor through some trivial bundle with finite-dimensional fibers), in which case there is a Gauss map $\mathscr{G} : M \to \text{Gr}_{2n}(\mathbf{R}^\infty)$ defined by sending each point $p \in M$ to the tangent space $T_p M \subset \mathbf{R}^\infty$. Modelling $B\text{GL}_{2n}(\mathbf{R})$ on $\text{Gr}_{2n}(\mathbf{R}^\infty)$, this map $\mathscr{G}$ is exactly the classifying map of the tangent bundle of $M$. This map does not depend on the bundle-embedding upto homotopy and one can recover $TM$ as $\mathscr{G}^* \gamma_{2n}$ where $\gamma_k \to \text{Gr}_k(\mathbf{R}^\infty)$ is the tautological $k$-plane bundle.

If $M$ admits a complex structure, then the fibers of $TM$ admit a (fiberwise continuously varying) complex structure. This enables one to construct a bundle embedding $TM \hookrightarrow M \times \bf R^\infty$ such that if $\mathbf{R}^\infty$ is realized as a "$\infty/2$"-dimensional complex vector space $\mathbf{C}^\infty$, then the bundle-embedding respects the fiberwise complex structure, i.e., $T_p M \subset \mathbf{C}^\infty$ is a $n$-dimensional complex subspace. Once that is done, let $\mathscr{G}$ be the corresponding Gauss map.

Then $\mathscr{G}$ has a natural lift to a map $M \to \text{Gr}_n(\mathbf{C}^\infty)$, factoring through the map of Grassmannians $\text{Gr}_n(\mathbf{C}^\infty) \to \text{Gr}_{2n}(\mathbf{R}^\infty)$ sending a complex $n$-subspace of $\mathbf{C}^\infty = \mathbf{R}^{2\infty}$ to it's underlying real $2n$-subspace. Modelling $B\text{GL}_n(\mathbf{C})$ as $\text{Gr}_n(\mathbf{C}^\infty)$ once again, this is the same as Definition 2. If the classifying space were modeled otherwise, one would have to do the checking that the map $Br : B\text{GL}_n(\mathbf{C}) \to B\text{GL}_{2n}(\mathbf{R})$ you wrote down is naturally isomorphic to the Grassmannian-level map I wrote down, by a commutative diagram.

This shows Definition 1 implies Definition 2. Can you try the converse?

1 implies 2. Suppose defined an almost complex structure on $M$. Consider an atlas $U_i$ of $M$ where $U_i$ is contractible and diffeomorphic to an open subset of $
\mathbb
{R} ^n$. The restriction $TM_i$ of the tangent bundle $TM$ to $U_i$ is trivial. Let $J_i$ be the restriction of $J$ to $U_i$, there exists an isomorphism of pseudocomplex bundles $g_{ij} $ between the restriction of $(TM_j, J_j)$ and $(TM_x, J_x)$ to $U_i\cap U_j$. The family of morphisms $g_{ij} $ is a trivialisation of $TM$.  Remark that $g_{ij}(x) $ is a complex automorphism of $(TM_x, J_x)$. Thus it is an element of $Gl(n, \mathbb{C})$ and $TM$ has a $Gl(n, \mathbb{C})$ reduction.

2 implies 1. The existence of $\tau_M$ is equivalent to saying that $TM$ has a $Gl(n, \mathbb{C})$ reduction, which is equivalent to saying that we can suppose that $g_{ij} (x) \in Gl(n, \mathbb{C})$. Denote by $J_i$ the standard complex structure on $U_i$ and $f_i$ a partition of unity associated to $U_i$, write $J=\sum f_iJ_i$.


Find the smallest value of the function $F:\alpha\in\mathbb R\rightarrow \int_0^2 f(x)f(a+x)dx$ Let $f -$ fixed continuous on the whole real axis function which is periodic with period $T = 2$, and it is known that the function $f$ decreases monotonically on the segment $[0, 1]$ increases monotonically on the segment $[1, 2]$ and for all $ x \in \mathbb R $ satisfies the equality $f(x)=f(2-x)$. Find the smallest value of the function
  $$F:\alpha\in\mathbb R\rightarrow \int_0^2 f(x)f(\alpha+x)dx$$

I have no clue how to start. Any kind of help will be appreciated.

We intend to prove that the smallest value of the function $F$ defined by
\begin{equation*}
F({\alpha}) = \int_0^2f(x)f(x+{\alpha})\, dx\tag{1}
\end{equation*}
is $F(1)$.
At first we give $f$ the additional property to be differentiable with a continuous derivative. At the end we will fill that gap.

The function $F$ will be continuous and have the period $T = 2$. Thus there exists a smallest value of $F(\alpha).$

If $f$ is differentiable then $F$ will also be differentiable.

We intend to prove that
\begin{equation*}
F'(\alpha) \le 0 \text{ for } 0 \le \alpha \le 1. \tag{2}
\end{equation*}
Since
\begin{equation*}
f(x) = f(2-x) = f(-x)\tag{3}
\end{equation*}

then $F(\alpha) = F(2-\alpha)$. Then (2) implies that
\begin{equation*}
F'(\alpha) \ge 0 \text{ for } 1 \le \alpha \le 2.
\end{equation*}
Then we know that $F(1)$ is the smallest value.

We start by studying (1). If we change $x$ to $x+1$ and then $x$ to $1-x$ we get
\begin{gather*}
\int_1^2f(x)f(x+\alpha)\, dx = \int_0^1f(x+1)f(x+1+\alpha)\, dx = [(3)]\\ = \int_0^1f(1-x)f(1-x -\alpha)\, dx = \int_0^1f(x)f(x-\alpha)\, dx.
\end{gather*}
Now we can rewrite $F(\alpha)$ as
\begin{equation*}
F(\alpha) = \int_0^1f(x)(f(x+\alpha)+f(x-\alpha))\, dx.
\end{equation*}
Via differentiation under the integral sign  (we now use that $f$ is differentiable) followed by integration by parts we get
\begin{gather*}
F'({\alpha}) = \int_0^1f(x)(f'(x+{\alpha})-f'(x-{\alpha}))\, dx = \left[f(x)(f(x+{\alpha})-f(x-{\alpha}))\right]_0^1 \\- \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx = f(1)\underbrace{(f(1+{\alpha})-f(1-{\alpha}))}_{= 0}-f(0) \underbrace{(f({\alpha})-f(-{\alpha}))}_{= 0} \\- \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx = - \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx.
\end{gather*}
Since $f$ is decreasing and differentiable on $[0,1]$ then $f'(x) \le 0$ in the integral above. It remains to prove that
\begin{equation*}
g(x,\alpha) = f(x+{\alpha})-f(x-{\alpha}) \le 0
\end{equation*}
on the square $0 \le x \le 1, \,  0 \le \alpha \le 1.$ But \begin{equation*}
g(\alpha,x) = f(\alpha +x) - f(\alpha-x) = [(3)] = f(x+{\alpha})-f(x-{\alpha}) = g(x,\alpha).
\end{equation*}
Consequently it is sufficient to examine $g$ on the triangle $0 \le \alpha \le x \le 1.$ To do that we study $g$ on that part of the straight line $\alpha = x-k$, that runs inside the triangle. Here $k$ is constant in $[0,1]$.
To be more precise look at
\begin{equation*}
g(x,x-k) = f(2x-k)-f(k), \quad 0 \le k \le 1, \: k \le x \le 1.
\end{equation*}
We split the $x$-interval into two. If $k \le x \le \dfrac{k+1}{2}$ then $k \le 2x - k \le 1$. Thus $2x-k \in[0,1]$ and $k\in[0,1]$. But on $[0,1]$ the function $f$ is decreasing. Since $2x-k \ge k$ then $f(2x-k) \le f(k)$ and $g(x,2x-k) \le 0.$

The second part of the $x$-interval is $\dfrac{k+1}{2} \le x \le 1$. Then $1 \le 2x-k \le 2-k \le 2$. Furthermore $f(k) = f(2-k)$ and $1 \le 2-k \le 2.$ But on the interval $[1,2]$ the function $f$ is increasing. Since $2x-k \le 2-k$ then $f(2x-k ) \le f(2-k)$ and $g(x,2x-k) \le 0.$

We have proved that $F(1)$ is the smallest value under the additional assumption that $f$ is differentiable with a continuous derivative.

Now we will prove that the same is true if $f$ only is continuous. Define a smooth substitute $\tilde{f}$ for $f$ via
\begin{equation*}
\tilde{f}(x) = \dfrac{1}{2h}\int_{x-h}^{x+h}f(t)\, dt = \dfrac{1}{2h}\int_{-h}^{h}f(x+t)\, dt.\tag{4}
\end{equation*}
Then $\tilde{f}$ will be differentiable with the continuous derivative $\dfrac{f(x+h)-f(x-h)}{2h}$. Furthermore it inherits a lot of properties from  $f$. It is obvious that $\tilde{f}$ will be periodic with $T=2$. It will also fulfil the condition $\tilde{f}(x) = \tilde{f}(2-x) = \tilde{f}(-x).$ It will also be decreasing on $[0,1]$. To realize that we need a little argument. Assume that $0 < x_1 < x_2 <1$. If $h$ in (4) is small enough then $|t|$ will be so small that
\begin{equation*}
0 <x_1+t<x_2+t <1.
\end{equation*}
From (4) we then get that $\tilde{f}(x_1) \ge \tilde{f}(x_2)$. We have proved that $\tilde{f}$ is decreasing on the open interval $0 < x < 1$. But since $\tilde{f}$ is differentiable it will also be decreasing on the closed interval $0 \le x \le 1$ (use the intermediate value theorem).
Analogously we prove that $\tilde{f}$ is increasing on $1\le x\le 2.$

Since $f$ is continuous and since we work on a compact interval $f$ will be uniformly continuous. Thus to a given $\varepsilon $ there exists a $\delta $ such that
\begin{equation*}
 |f(x_1)-f(x_2)| < \varepsilon \text{ if } |x_1-x_2|< \delta .
 \end{equation*}
 With these $\varepsilon$ and $\delta$ and with $0<h<\delta$ we get
 \begin{equation*}
 |f(x)-\tilde{f}(x)| \le \dfrac{1}{2h}\int_{-h}^{h}|f(x)-f(x+t)|\, dt \le  \dfrac{1}{2h}\int_{-h}^{h}\varepsilon\, dt = \varepsilon .\tag{5}
 \end{equation*}

We are now prepared to study the function

\begin{equation*}
\tilde{F}(\alpha) = \int_0^2\tilde{f}(x)\tilde{f}(x+\alpha)\, dx.
\end{equation*}
According to what we have done above we know that
\begin{equation*}
\tilde{F}(\alpha) \ge \tilde{F}(1)
\end{equation*}
and that $\tilde{F}(1)$ is the smallest value.

From (5) we get
\begin{gather*}
|F(\alpha) - \tilde{F}(\alpha)| \le \int_0^2|f(x)f(x+\alpha)-\tilde{f}(x)\tilde{f}(x+\alpha)|\, dx \\= \int_0^1|(f(x)-\tilde{f}(x)) f(x+{\alpha})+\tilde{f}(x)(f(x+{\alpha})-\tilde{f}(x+{\alpha}))|\, dx \\
\le \int_0^2(\epsilon C + C\varepsilon)\, dx = 2C\varepsilon
\end{gather*}
where  $\displaystyle C = \max_{0 \le x\le 2}|f(x)|$.
Consequently
\begin{equation*}
F({\alpha}) = F({\alpha}) -\tilde{F}({\alpha}) +\tilde{F}({\alpha}) \ge -2C\varepsilon +\tilde{F}(1).
\end{equation*}
But $\varepsilon$ can be arbitrarily small.
 In the limit we have that
 \begin{equation*}
 F({\alpha})\ge F(1)
 \end{equation*}
 i.e. $F(1)$ is the smallest value of $F(\alpha).$

I welcome shorter solutions.

Let $F(\alpha)=\int_0^2{f(x)f(\alpha+x)dx}$, changing in variable by letting $\alpha+x=u$, we get $F(\alpha)=\int_{\alpha}^{\alpha+2}{f(u-\alpha)f(u)du}$. Since $f$ is continuous then $F$ is differentiable and
\begin{align*}
F'(\alpha)=f(2)f(\alpha+2)-f(0)f(\alpha)
\end{align*}
but since $f(0)=f(2)$ and $f(x+2)=f(x)$, for all $x\in \mathbb{R}$. Then
\begin{align*}
F'(\alpha)=0,\,\,\,\, \forall \alpha
\end{align*}
which means that $F(\alpha)$ is constant over all $\mathbb{R}$.


Open problems in General Relativity I would like to know if there are some open mathematical problems in General Relativity, that are important from the point of view of Physics.

Is there something that still needs to be justified mathematically in order to have solid foundations?

Let me sneak in an answer before this gets closed, because the answer to this question is likely to be very very different depending if you ask a mathematician or a physicist.  :) So here is a long list of questions that a mathematician considers to be important open problems in general relativity, and why they are physically relevant.

Cosmic censorship

Let me first get the biggie out of the way. There are actually two different cosmic censorship hypotheses, which are minimally related to each other. The Weak Cosmic Censorship (WCC) hypothesis states that (a fairly modern formulation)

  For generic initial data to the evolution problem in general relativity, there cannot be naked singularities.

Unfortunately, if you ask me to define the terms in that hypothesis, I can just shrug and tell you, "no can do". This question is so open as an open problem that we don't even know the correct formulation of the statement! If you are interested, you may want to consult Christodoulou, D. "On the global initial value problem and the issue of singularities", Classical Quant. Grav., 1999, 16, A23. Let me give just a short digression on what the difficulty of formulating the problem is.

The first is the problem of what do you mean by a naked singularity? In certain situations, the definition of a naked singularity is very clear cut (for example, if we assume mathematically that the space-time is spherically symmetric). But in general, the problem of what a naked singularity is is very difficult to define accurately in mathematics. The main problem being this: the equations of motion given by the Einstein's equations are hyperbolic. As such, we cannot necessarily meaningfully extend the notion of a solution to points which can "see" the singularity. (Sort of a catch 22: morally speaking, if you can see the singularity, you are at a (possibly different) singular event, so it doesn't make much sense to use the physics definition of a naked singularity being one that can be observed by a far away observer.) (This is not to say this is completely impossible: there are singular solutions which the solution can extend continuously to the singular boundary, and so technically you can exist all the way up to the moment you see the singularity, and then poof, you are gone.) So we have to probe at the notion of a naked singularity using alternative definitions that are more stable and can be mathematically formulated. And that is hard. (Once physical intuition fails, things get hard.)

The second problem is generic. It is already known that there are examples of solutions in general relativity which contains a naked singularity for all reasonable definitions of "naked singularity". So it is in fact impossible to prove a statement that says "naked singularity cannot exists. Period.". Any statement you prove must be a statement that is true away from some exceptional set. The hope is that the bad set is positive codimension in the space of all initial data. But we don't know.

The other cosmic censorship hypothesis is the Strong Cosmic Censorship (SCC). Despite the name it is independent of WCC. The SCC states that

  A generic solution to the Einstein's equation cannot be continued beyond the Cauchy horizon.

Again, this is from the view of an evolution equation. A funny thing about general relativity as an evolution equation is that you build your manifold and your solution on the manifold at the same time, which causes an interesting problem. Roughly speaking, usually when you solve an nonlinear wave equation, and you extended your solution as much as you can, then you either exhausted the entire space-time, or you ran into a singularity. But in general relativity, "running out of space-time" is not a hard obstruction to extending the solution further. (This goes back to the hyperbolicity of the equation.) Basically, you can "run out of space-time that can be completely predicted by your initial data", while still have a possible extension of your solution "to infinity and beyond". The classic examples of this are the Reissner-Nordstrom and the Kerr-Newman black hole solutions. If you solve it as an initial value problem in partial differential equations, you get a black hole solution. But if you start from this black hole solution, you can actually analytically continue the solution (similar to how analytic continuation is done in complex analysis) to beyond what can be defineable from the inital data. The continuation is in some sense "outside" our universe, so the existence of this type of solutions causes some, at the very least, epistemological difficulties. The SCC posits that these type of solutions are rare. Just like the case of the WCC, the correct mathematical statement for this problem is also not known, and therefore it is hard to say what techniques of mathematics will be required to resolve it.

Final state conjecture

The final state conjecture in classical relativity states that (assuming the universe is not expanding) if you let the universe run long enough, because gravity is an attractive force, and is the strongest force over cosmological distances, we expect the entire universe to either collapse into a black hole, or become "mostly" empty (things drifting away from each other). Or slightly more jargony: the end state of the universe is either a single black hole or vacuum.

Mathematically this requires deriving the asymptotic behaviour of solutions to the Einstein's equations in general relativity. The Einstein equations are a highly nonlinear system of partial differential equations, and it is very difficult to study these things asymptoticly. There are however, some special cases that have either been worked out or are receiving a lot of specialist attention.

Stability of Minkowski space

One of the first things to consider, in terms of the final state conjecture, is that whether, if we start with a universe that is already very sparse, we can guarantee that this will end up asymptotic to Minkowski space. The answer to this, as it turns out, is "yes". The first work was due to D. Christodoulou and S. Klainerman, with extensions of the result and improvements by H. Lindblad and I. Rodnianski, L. Bieri, N. Zipser, J. Speck, etc.

Uniqueness of black holes

Now, if we posit that the final state of the universe is either Minkowski space or a black hole, we better show that a black hole is the only possible stationary solution! (Any other stationary solution would be another steady state!) The problem of black hole uniqueness is almost, but not completely, resolved.

In the case where we assume some additional symmetry (that the universe is either axially symmetric or rotationally symmetric), the uniqueness of black holes is known. In the case where we assume that all the objects we are dealing with are real analytic, we also know that black holes are unique.

The remaining case, which is expected to be true, is when we relax the regularity assumption to just infinitely differentiable, but not real analytic. In this case we only have some partial results, recently due to A. Ionescu, S. Alexakis, S. Klainerman, plus some additional contributions by P. Yu and yours truly. So far we have two major class of results: one is that if we allow only small perturbations of a stationary black hole, then there are no other stationary solutions that are approximately a known black hole solution without being one. The other is that if you assume certain special structures on the event horizon (roughly speaking, assume that the event horizon looks like the one for a known black hole solution), you cannot have other stationary exteriors.

Stability of Kerr-Newman black holes

Now, assuming we accept that the known Kerr-Newman family of black holes form the unique stationary state of general relativity, the next problem is to prove that they are actually stable under perturbations. That is, if we start out with initial data very similar to a Kerr-Newman black hole, is the evolution going to "track" a Kerr-Newman black hole? So far, there are quite some progress made in the linearised problem, due to contributions from (in no particular order) I. Rodnianski, M. Dafermos, P. Blue, A. Soffer, W. Schlag, R. Donniger, D. Tataru, J. Luk, J. Sterbenz, G. Holzegel, and many others. (It is quite an active field at the moment.) But the way forward for the full nonlinear problem is still somewhat elusive.

Penrose inequality and friends

Recently between H. Bray and the team of Huisken and Illmanen, what is known as the Riemannian Penrose inequality was proved. The Riemannian Penrose inequality sits in a large class of mass inequalities for space-time manifolds. These types of inequalities all take the form

  Mass of the space-time is bounded below by Blah.

The first, and most famous, of these type of results is the positive mass theorem of Schoen and Yau, which states that the mass is bounded below by 0. The Riemannian Penrose Inequality states that the mass is bounded below by number related to the size of the black hole (when there is one). In other words, a black hole must contribute a certain amount of mass to the universe, the amount depending on its surface area.

There are further refinements of the known statements conjectured. One is the version called the Gibbons-Penrose inequality, which gives some improved lower bounds when there are multiple black holes (the attracting force between the multiple black holes will creating some contribution to the mass at infinity). And there are also other versions which also factors in electromagnetic forces and so forth. Most of them are open questions.

A somewhat related problem is the Hoop Conjecture of Kip Thorne. The conjecture roughly states that

  If you try to squeeze lots of mass into a small space-time volume, what determines whether the mass will collapse and form a black hole is the "circumference" of the space-time region.

This problem is being actively pursued by M. Khuri and others, and is still open to my knowledge.

Some other open problems:

Low regularity evolution Can we solve the equations of motions assuming very rough (in the sense of having very few derivatives) initial data? These types of question are important because (a) when comparing to physical measurements, the rougher the data the theory can accomodate, the less sensitive the solution is to small perturbations, and so the better we can figure out the "error bar" due to theory (b) the problem of shockwaves. Physical systems can exhibit shock waves, and it is important to have a theory that can deal with that.

Higher dimensional cases In some cosmological/string theories, one is lead to consider space-time manifolds of higher dimensions. On the one hand, radiative decay of solutions is stronger on higher dimensions, and it would enhance stability. On the other, there are more degrees of freedom, which will increase possibilities of instability. There are numerical evidence to suggest that certain types of higher dimensional black holes are in fact unstable, and could be poked and perturbed into other types of black holes.

Evolution of apparent horizon Under the banner of dynamical horizon is a collection of topics related to how the apparent horizon (the "boundary of a black hole") will evolve in general relativity. It is important to note that a lot of the physics literature on the evolution of apparent horizon (such as black hole evaporation) is based on a linear analysis, which to first order assumes that the horizons do not move. (Yes, by assuming that the horizons do not move (much), they show that quantum effects will make the horizons shrink.) The true nonlinear versions of the evolution is not yet well understood.

Geodesic hypothesis One of the postulates in studying general relativity is that point particles with negligible mass will travel along geodesics of the space-time. This question has a surprisingly long history (it was first considered by Einstein in the 1920s), and is still not completely resolved. The main problem is how to make the process of "taking the negligible mass limit" rigorous. For a physical object in general relativity, when it moves, its motion will cause "ripples" in the space-time caused by gravitational backreaction of its own presence. While the three body problem is difficult in classical mechanics, even the two body problem in full generality is still unresolved in general relativity.

I hope the above provides some keywords or phrases for you to look up.

[Addendum: 18.09.2013]

Here are some more open problems in mathematical relativity that concern Cosmology. To start with, let me first quickly say what I mean by cosmology here. To me, cosmology is the study of the large scale behaviour of the universe. In other words, it is the "blurry" picture of what the universe would look like if we ignore the small-scale structures such as stars and galaxies. As such, in the study of cosmology we often assume the Cosmological Principle holds. The cosmological principle is a strengthening of the Copernican principle, and says that

  On the large scale, the universe can be modeled by a solution to Einstein's equations which admits a foliation by Cauchy hypersurfaces each of which is homogeneous and isotropic.

In layman's words, it says that we can pick a (possibly preferred) notion of cosmological time, such that at every instant of cosmological time, space looks identical in all directions, and you cannot distinguish between any two points. (If we factor in the small-scale structure, this is quite obviously false: we live on a planet we call Earth orbiting a star we call Sol, that quite clearly distinguish us from many other points in the universe. But remember that the cosmological principle is only supposed to apply "in the large".) One can get into an argument about whether this principle holds in practice (cf. the cosmic microwave background anisotropy), but it is under this principle most of the computations in cosmology are made.

This leads to some immediate open problems:

Stability of Cosmological Solutions

Part of the cosmological principle assumes that the universe is well-modeled by the spatially homogeneous and isotropic cosmological solutions. In particular, this means that we are assuming the cosmological solutions are stable (that if we perturb the cosmological solutions a tiny bit, say, by now factoring in the small-scale structure, the evolution as governed by Einstein's equations will be, in the large, not very different from the evolution of the cosmological solutions).

Since in general this requires studying a large and complicated set of PDEs around a special solution, this is generally tackled on a case-by-case basis and there are still many interesting cosmological models to be studied.

Cosmological principle as a derived consequence

It would be much more satisfying, however, if the cosmological principle can be derived as a consequence of general relativity (under suitable assumptions on the cosmological constant and what not), rather than something assumed a priori. Namely, we ask the question:

  Can we find some conditions such that the evolution of arbitrary data in the cosmological setting will eventually converge to one for which the cosmological principle holds?

Mathematically, we are asking whether homogenisation and isotropisation can be derived as a consequence of Einstein's equations.

We remark here that it is generally accepted that the sufficient conditions will involve the space-time being expanding. Heuristically speaking if the space-time is expanding then the relative strength of the small-scale structure should get weaker as time goes along (since they get relatively smaller and smaller) and so one should converge toward a cosmological space-time. (On the other hand, homogenisation and isotropisation are not expected to hold in contracting space-times [either toward the big-bang singularity in the past or possibly toward a big-crunch singularity in the future], because if space-time contracts, the relative size of the small-scale features get bigger...)  Luckily for us, the recent astrophysical observations seem to suggest that our universe is in a regime of accelerated expansion, so it is okay to not consider the other alternatives. :-)

Geometrisation

One can conjecture that something even stronger than homogenisation and isotropisation should hold. This conjecture is backed up by all the currently known results in this direction.

From a powerful result in three dimensional topology, we expect that spatial slices of the universe (assuming that the universe as a compact topology) can be written as the sum of some graph manifolds with some hyperbolic three-manifolds.

Then the conjecture is the following:

  In the expanding direction, as time goes to infinity, the limits of the spatial slices is such that we can rescale the metric in such a way that (a) the rescaled metric on the hyperbolic pieces converge to the standard complete hyperbolic metric and (b) the rescaled metric on the graph manifold parts collapse.

In special situations or under special assumptions, this conjecture has been confirmed by

Lars Andersson and Vincent Moncrief by showing the stability of the Milne model
Vincent Moncrief and Yvonne Choquet-Bruhat in their work studying the $U(1)$ symmetric space-times (under their assumptions the spatial slices must be Seifert fibered spaces, and they do in fact show the desired [rescaled] collapse).
The work of Michael Anderson, Martin Reiris, and possibly others, which show the desired conclusion if one were to assume some strong a priori bounds on the geometry.

[Added 23.01.2014]

There are, of course, many, many interesting problems concerning the initial data for Einstein's equations. Just a quick review: the strong ties to geometry of Einstein's equation in general relativity means that the system of equations is somehow simultaneously over- and underdetermined. The latter problem can be cured by "gauge fixing"; but the former problem manifests at the level of initial data.

One is used to be able to freely specify "initial conditions" when attacking an evolution equation. In the case of Einstein's equations (and many other geometric partial differential equations), the initial data must satisfy certain initial constraints to be compatible with the global geometry: here we take the naive view that the initial data should contain the metric tensor on the initial slice and also its first time derivative.  (This review by Bartnik and Isenberg gives a pretty high-powered review if you are already familiar with Lorentzian geometry; if you are not then I suggest looking at standard textbooks such as Wald's General Relativity.)

Several natural questions arise concerning these constraint equations.

How rigid are the constraints?

By writing down the constraint equations (which can be expressed as 4 partial differential equations), we immediately see that relative to the number of degrees of freedom we have, the constraint equations are under-determined. Therefore we expect "lots" of solutions. But natural questions like "how many solutions are there given a set of prescribed asymptotics" still need to be answered. This is especially so since as part of the proof of the Positive Mass Theorem is included the following rigidity statement:

  ... the ADM mass of a complete asymptotically flat initial data set under assumption of dominant energy vanishes if and only if the initial data is a space-like initial data slice of Minkowski space.

Therefore we know that certain asymptotics for the initial data are very rigid.

An interesting development is the gluing technique due to Justin Corvino and Rick Schoen from a few years back. A corollary to the gluing technique is that once the ADM mass is allowed to be positive, the asymptotics become a lot less rigid for restricting the initial data. In particular, they were able to construct many initial data sets with asymptotics virtually identical to that of the Schwarzschild and Kerr black holes, but with different topology. The full force of this gluing technique and its implication is still under investigation today.

On the other hand, the success of the positive mass theorem (which is essentially a result on scalar curvature comparison) has led to many current research topics concerning various aspects of the rigidity of asymptotically flat manifolds with non-negative scalar curvature. These questions are mostly Riemannian geometric in nature and practice, and in addition to the Riemannian version of the Penrose inequality mentioned above, just to give a taste the kind of results obtained and questions asked (since I am not really too much of an expert in this direction), one can check out this recent result of Michael Eichmair and Simon Brendle which does not admit an easy physical interpretation.

How to parametrise the initial data?

Giving that the initial data sets need to satisfy the constraint equations, one is led to ask whether there exists an alternative way of parametrising the initial data (as opposed to the naive initial metric + first time derivative splitting) such that

The data can be freely prescribed. For this we need a method to start with some tensors and a method of transformations which guarantees us to recover geometric data that satisfy the constraint equations.
Is exhaustive; that is, every solution of the constraint equations can be generated from free data using this method.

This dream dates back to at least Andre Lichnerowicz and is still not entirely settled. A part of the story is the immense success of his conformal method in constructing and parametrising initial data sets of general relativity; it turns out that however, this conformal method is very well suited for a certain situation (data that has constant mean curvature or almost constant mean curvature), but may (the understanding on this issue is a bit muddy at the moment; but this is a post on open questions after all) be very much less well suited for other situations (data that has strongly varying mean curvature). If you are interested, I would highly recommend the two videos of David Maxwell's lectures from September of last year.

There are some problems related to initial data, such as if gravity can be "insulated", or if there exist generic set of spacetimes which do not admit CMC spacelike hypersurface.


Necessary condition for the integral equation $u(x) = f(x)+\lambda\int K(x,t)u(t)dt$ to have a continuous solution? Is the following condition necessary for the integral equation $$u(x) = f(x)+\lambda\int K(x,t)u(t)dt$$  to have a continuous solution: $f(x) \neq 0$, is real and continuous in the interval $[a,b]$?

When $f(x) = 0$, the integral equation will become homogeneous. So I think this should be a necessary condition.  Am I right? Please suggest me

Why would $f$ have to be nonzero or real? If $f = 0$ there is certainly a continuous solution, namely $0$.  There may be nonzero solutions as well.  You'll probably want to assume $K(x, \cdot)$ to be in $L^1(a,b)$ so the
integral exists for all $u \in C[a,b]$.  Then under some conditions on $K$ (e.g. if $K$ is continuous) $\int_a^b K(x,t) u(t)\ dt$ will be continuous for all continuous $u$.
If so, $f(x) = u(x) - \lambda \int_a^b K(x,t) u(t)\ dt$ will have to be continuous in order for there to be a continuous solution, and will have to be real if $K$ is real and there is a real solution.

It has a unique continuous solution.It's solution is given by Lioville-Neumann Series.
          http://en.wikipedia.org/wiki/Liouville-Neumann_series


From prime to prime by squaring the digits I took prime $131$, squared digits of it and wrote them in natural order as they appear, from left to right, and obtained $191$, then I obtained $1811$ by the same procedure, and then $16411$ and then $1361611$, and $131,191,1811,16411$ are primes and $1361611$ is not.

To illustrate how to arrive at the next number in sequence from previous one, take, for example, $16411$.

We have: $1^2=1$ and $6^2=36$ and $4^2=16$ and $1^2=1$ and $1^2=1$ so we obtain $1361611$ from $16411$.

  Can we generate in this way as large a number of different (to avoid loops like one that starts with $11$) primes as we want? Or there is/are some law/laws that do not allow that?


Show that matrix $A_k$ is also symmetric and positive definite. Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix that is positive definite.

Exercise: show that the matrix $A_k$ that consists of the first $k$ rows and columns of $A$ is also symmetric and positive definite:

My solution: We know that $A_k$ is symmetric because $A$ is symmetric. Furthermore, $A$ is SPD (symmetric positive definite) so every leading principal minor of $A$ is positive. This means that every leading principal minor of $A_k$ is positive, since $\{x:x \text{ is a leading principle minor of } A_k\} \subset \{y:y \text{ is leading principal minor of A}\}$. Because every leading principal minor of $A_k$ is positive, we have that $A_k$ is SPD.

EDIT: Matrix $A$ is SPD $\Rightarrow$ all its leading principal minors are positive: Let matrix $A$ be SPD and assume that there exists at least one leading principal minor that is nonpositive. This means that at least one of the eigenvalues of the corresponding leading principal submatrix is nonpositive and hence this leading principal submatrix is not SPD. This contradicts the fact that $A$ is SPD, since all the principal submatrices of a SPD matrix are SPD. So if $A$ is SPD all the leading principal minors are positive.

My question: is my solution correct or is it too intuitive? Should I be more elaborate when concluding that all leading principal minors of $A_k$ are positive?

I think you are correct, but our professor prefers to use leading principal submatrix rather than leading principal minor, which sounds kind of broad.


Approximating $\log x$ with roots The following is a surprisingly good (and simple!) approximation for $\log x+1$ in the region $(-1,1)$:
$$\log (x+1) \approx \frac{x}{\sqrt{x+1}}$$

Three questions:

Is there a good reason why this would be the case?
How does one go about constructing the "next term"?
Are the any papers on "generalized Pade approximations" that involve radicals?

Let's rewrite both sides in terms of $y = x + 1$: we get

$$\log y \approx \sqrt{y} - \frac{1}{\sqrt{y}}$$

on, let's say, the interval $\left( \frac{1}{2}, 2 \right)$ (I hesitate to discuss the entire interval $(0, 2)$; it seems to me that the approximation is not all that good near $0$). The RHS should look sort of familiar: let's perform a second substitution $y = e^{2z}$ to get

$$2z \approx e^z - e^{-z} = 2 \sinh z$$

on the interval $\left( - \varepsilon, \varepsilon \right)$ where $\varepsilon = \frac{\log 2}{2} \approx 0.346 \dots$. Of course now we see that the LHS is just the first term in the Taylor series of the RHS, and on a smaller interval than originally. Furthermore, the Taylor coefficients of $2 \sinh z$, unlike the Taylor coefficients of our original functions, decrease quite rapidly. The next term is $\frac{z^3}{3}$, which on this interval is at most

$$\frac{\varepsilon^3}{3} \approx 0.0138 \dots$$

and this is more or less the size of the error in the approximation between $\log 2$ and $\frac{1}{\sqrt{2}}$ obtained by setting $y = 2$, or equivalently $x = 1$.

With the further substitution $t = \sinh z$, the RHS is just the first term in the Taylor series of the LHS. To get the "next term" we could look at the rest of the Taylor series of $\sinh^{-1} t$. The next term is $- \frac{t^3}{6}$, which gives

$$z \approx \frac{e^z - e^{-z}}{2} - \frac{(e^z - e^{-z})^3}{48}$$

or

$$\log y \approx \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right) - \frac{1}{24} \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)^3.$$

I don't know if this is useful for anything. The series to all orders just expresses the identity

$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$

$$\log(x+1)=\lim_{n\to\infty}n(\sqrt[n]{x+1}-1).$$

In the case $n=2$,
$$2(\sqrt{x+1}-1)=2\frac{x}{\sqrt{x+1}+1}\approx\frac x{\sqrt{x+1}}$$
for small $x$.

The approximation works better as it has a vertical asymptote at $x=-1$.


Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$ I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much

Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$.

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.

Easy proof by set theory: (I'm a beginner and thought of this, please correct if me if I'm wrong)
Let $a=xy, b=xz$.
So
$a = \{x,y\}$ and $b = \{x, y\}$,
where $x$ is the product of the common prime factors of $a, b$ which is $\gcd(a,b)$ and the product $xyz$ produces $\text{lcm}(a,b)$
By Set Theory,
$P(a \cup b) \cdot P(a \cap b) = P(a) \cdot P(b)$.
$xyz \cdot x = xy \cdot xz$
lcm(a,b) * gcd(a,b) = ab


not an Ideal of F[x] Let $F$ be a field, and $R$ be the polynomial ring $F[x]$, and $f,g,d\in R$ be monic. Let $\Gamma=$ {monic $d\in R : d|f$ and $d|g$}. Show $\Gamma$ is not ideal, and it has a unique element of maximal degree: $gcd(f,g)$. I don't even know where to begin :(.

First let $d \in R$ be an element of $R$ which is a divisor of $f$ and $g$. Now find an element $r$ in $R$ for which $d*r$ or $r*d$ is not in $\Gamma$.


Derivative of t per unit step - meaning of $t\delta (t)$ I derived the function ($u(t)$ is the unit step):

$$x(t)=t \space u(t)$$

$$dx/dt=1 \space u(t)+t \space \delta(t)$$

What is the graphical meaning of the second term?

Thank you very much.

The function $x(t)$ is a ramp and is not differentiable at $t=0$ (the left and right derivatives differ). The term $t\,\delta(t)$ accounts for that non-differentiability (the Heaviside step takes a finite value at $t=0$).

Graphically speaking, you can picture it as two infinitely close Dirac deltas with opposite signs (which implements a differentiation operator by convolution).

$\delta (t)$ is the Dirac delta function (well... distribution). This "function" assumes the value $\infty$ at zero and is zero everywhere else. The funny thing is

$$\int_{\mathbb{R}} \delta (t) \ dt = 1 $$

I suggest consulting the Wiki page. The Dirac delta is like a point pulse.


Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $ I've got in my assignment to show if the following series converges or diverges.
  $$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$

Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) &=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)
\end{align*}

At this point I'm lost. I tried using D'Alambert as follows:

$$\lim \frac{a_{n+1}}{a_n}=\lim\frac{\sqrt[n+1]{e}-1-\frac{1}{n+1}}{\sqrt[n]{e}-1-\frac{1}{n}}$$

which I tried to simplify with basic limit laws (hopefully correctly):

$$\lim\frac{\sqrt[n+1]{e}-1}{\sqrt[n]{e}-1} = 1$$

I don't know where to go from here. Thank you for all your help in advance.

Just use comparison test. Using this Proving Schwarz derivative $\frac{f''(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule? you have $$\lim_{n\to\infty}{n^2}\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) =\lim_{x\to 0}\frac1{x^2}\left(e^x  -1 -x\right) =\lim\limits_{x\to 0} \frac{\frac{e^x -1}{x}-1}{x}= \frac12$$

Hence, for n large enough we have  $$ \left|n^2\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) -\frac1{2}\right|\le \frac14$$

That is $$ \frac1{4n^2}\le \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) \le \frac3{4n^2}$$
But $$\sum_{n=1}^\infty\frac1{n^2}<\infty$$
 That is
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right)$$
converges too

Thank you for the comment. By using the comparison test I easily proved the sum converges.

$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$