Mathematical precise definition of a PDE being elliptic, parabolic or hyperbolic what is the general definition for some partial differential equation being called elliptic, parabolic or hyperbolic - in particular, if the PDE is nonlinear and above second-order.

So far, I have not found any precise definition in literature.

The short answer is: there is none.
The slightly less short answer is: there are some nice descriptions of ellipticity for higher order equations and systems. There are some nice descriptions of hyperbolicity for higher order equations and systems. There is generally no description for parabolicity. And elliptic, parabolic, and hyperbolic do not come close to exhausting the possible PDEs that can be written down beyond second order scalar equations.
The long answer is, well, long.

To scratch at the surface of this problem, we need to journey back to 1926, when the definitions of elliptic, parabolic, and hyperbolic PDEs are given by Jacques Hadamard. In his study of scalar linear partial differential equations of second order (the work has since been compiled and published as Lectures on Cauchy's problem in linear partial differential equations by Dover publications in 1953), Hadamard made the following definitions. (As an aside, it is also in those lectures that Hadamard made the first modern definition of well-posedness of the Cauchy problem of a PDE.)
Given a linear partial differential operator of second order with real coefficients, its principal part can be represented by a (symmetric) matrix of coefficients $a^{ij}\partial^2_{ij}$. The operator is then said to be

Elliptic if $a^{ij}$ is positive definite or negative definite
Parabolic if $a^{ij}$ is positive or negative semi-definite, and admits precisely one 0 eigenvalue.
Hyperbolic if $a^{ij}$ is indefinite, but is non-degenerate; that is, for every vector $v_j$, there exists some vector $w_i$ such that $w_ia^{ij}v_j \neq 0$. In other words, it has no zero eigenvalue.

Already you see that the classification is incomplete: any operator with principle part having nullity $>1$, or having nullity $= 1$ but with indefinite sign is not classified. Furthermore, the definition of hyperbolic is different from the modern one. Indeed, Hadamard made the additional definition that the operator is

Normal hyperbolic if $a^{ij}$ is hyperbolic, and furthermore all but one of its eigenvalues have the same sign.

The wave operator is a normal hyperbolic operator. Whereas what now-a-days we call the ultrahyperbolic operators (in order to distinguish them from hyperbolic ones) were nominally considered to be hyperbolic by Hadamard's standards.
Hadamard was able to show that linear, normal, hyperbolic PDEs admit well-posed initial value (Cauchy) problems for data prescribed on hypersurfaces over which tangent space the restriction of $a^{ij}$ is elliptic.
Hyperbolicity redefined
Let us first talk about hyperbolic equations (because I'm more familiar with its history).
The next advance came around the mid 20th century, with contributions by people like Jean Leray and Lars Gårding. Rather than using Hadamard's geometric definition of hyperbolicity, in order to deal with higher order equations and systems of equations, hyperbolicity becomes more-or-less synonymous with "having a well-posed initial value problem". This recognition does not mean the definition is separate from the geometry of the coefficients. Here I give a theorem (which you can find in Lars Hörmander's Analysis of Linear Partial Differential Operators):
Theorem A linear partial differential operator of order $m$ with smooth coefficients admit a well-posed Cauchy problem for arbitrary smooth data if and only if its symbol is a hyperbolic polynomial.
Basically, a linear partial differential operator $P = \sum_{|\alpha| \leq m} (i)^{|\alpha|}a^{\alpha}\partial_\alpha$ (where $\alpha$ is a multi index) has corresponding symbol $p(\xi) = \sum_{|\alpha| \leq m} a^\alpha \xi_\alpha$. The symbol $p$ is said to by a hyperbolic polynomial in $\xi\in\mathbb{R}^n$ if there exists a vector $\tau\in\mathbb{R}^n$ such that the $\sum_{|\alpha| = m} a^\alpha \tau_\alpha \neq 0$, and that there exists some number $s_0 > 0$ such that for all $s < s_0$, the polynomial
$$ p(\xi + i s \tau) \neq 0 \qquad \forall \xi\in\mathbb{R}^n $$
To apply to a system of equations, We consider the system given by
$$ P_{ij}u^j = 0 $$
and $P_{ij} = \sum (i)^{|\alpha|} a^{\alpha}_{ij} \partial_\alpha$ is a matrix of scalar partial differential operators, the condition is analogous, with the symbol defined instead by
$$ p(\xi) = \det( \sum_{|\alpha| \leq m} a^{\alpha}_{ij} \xi_\alpha ) $$
via the determinant.
The problem with this definition is that it is not-so-easy to check, and does not generalize easily to non-linear equations.
As it turns out, to consider quasilinear equations, one realizes that what is really important for the well-posedness of the initial value problem is only the principal part of the operator (that is, the part with the highest number derivatives). There have been many attempts to provide sufficient conditions on the principal part of the operator that guarantees the well-posedness of the Cauchy problem, and all of these conditions are named as "(blank) hyperbolicity". Notice that none of them are necessary conditions: they are all strictly stronger assumptions than the conditions given above in the theorem.
I will not attempt to give all the definitions here. I'll refer you to some keywords for searching in textbooks or literature. First there is the notion of strict hyperbolicity. This is the direct analogue of the above characterisation using hyperbolic polynomials in the context of principal symbols. In fact, we have that
Theorem If the principal part of an operator is strictly hyperbolic, than regardless of what the lower order terms are, the operator is hyperbolic.
The definition of strict hyperbolicity is as follows: construct the principal part of the symbol as above, but just using the highest order terms $p_m(\xi) = \sum_{|\alpha| = m} a^\alpha\xi_\alpha$, so it is a homogeneous polynomial. It is strictly hyperbolic if there against exists some vector $\tau$ such that for any $\xi\neq 0$, the $m$-th degree polynomial in $s$ given by $p(\xi + s\tau)$ has $m$ distinct real roots.
A rather different definition of hyperbolicity is called symmetric hyperbolicity. This relies on re-writing your higher-order equation/system as a first order system of conservation laws. It requires being able to represent/convert the system into a special algebraic form, with certain symmetry and positivity conditions on its principal part. There are many, many good books written about this subject.
One interesting facet of symmetric hyperbolic systems is that sometimes the usual/simple way of reducing a system does not suffice to exhibit the symmetric hyperbolic structure, and one may be forced to "augment" the system with dummy dependent variables. Much of this is discussed in C. Dafermos's book Hyperbolic conservation laws in continuum physics, and has been used by, e.g. Denis Serre to show that complicated systems such as the Maxwell-Born-Infeld model of non-linear electrodynamics is hyperbolic. (Though the same can be shown, perhaps slightly more easily, using the next method; a proof can be found in Jared Speck's recent paper.)
A third method of defining hyperbolicity is the notion of regular hyperbolicity due to Demetrios Christodoulou in his book Action Principle and Partial Differential Equations (though the germ of the idea was already considerd by Hughes, Kato, and Marsden in the 1970s). This is a method specially adapted for studying second order quasilinear systems of differential equations. I gave a quick discussion in a recent paper of mine. In particular, it is noted in that paper that there is a large gap between regular hyperbolicity, which is perhaps the least restrictive in the context of second order systems of equations of the three above, and ellipticity.
There are also other more special cases where the "time direction" is distinguished in the system of equations. In particular there is a large class of equations of the form
$$ \partial_t^{2m}u = F(u,\partial u, \ldots, \partial^{2m-1} u) + A u $$
which can easily to be seen as hyperbolic as long as $A$ is a quasilinear elliptic operator of order $2m$.
In general a principal feature of hyperbolic equations is that their solution tends to rely on the existence of $L^2$ energy estimates.
Ellipticity
Compared to hyperbolicity, ellipticity is better understood (but unfortuantely, not by me. So I'll point you to references).
For scalar nonlinear equations, there are very well developed theory available in various textbooks. See, for example, Gilbarg and Trudinger, Elliptic partial differential equations of second order (pay attention to the last few chapters), or perhaps Caffarelli and Cabre, Fully nonlinear elliptic equations for development in the second order case. For higher order operators, say $P$ is of order $2m$, a genearally used criterion is that its principal symbol satisfies the elliptic inequality $p_{2m}(\xi) \geq \lambda |\xi|^{2m}$.
For systems, on the other hand, ellipticity is also a tricky business. In so far as second order systems are concerned, a very useful condition is the so-called Legendre-Hadamard condition, otherwise known as  rank-1 convexity condition. This condition goes back to (if I am not mistaken) Jesse Douglas and C.B. Morrey in their works on the Plateau problem in Riemannian geometry. The effectiveness of this condition is still a much studied subject related to nonlinear elasticity and calculus of variations. You can learn quite a bit about it by looking at some of John Ball's survey articles.

If you have a system with $dim = 2$
then it is based on the discriminant in the quadratic formula
(ie
$Au_{xx} + Bu_{xy} + Cu_{yy} = 0$
or
$Ax^2 + Bxy + Cy^2 = 0$) then discriminant is $B^2 - 4AC$

If
$$B^2 - 4AC > 0 $$
2 Real Roots
= Hyperbolic

$$B^2 - 4AC < 0 $$
2 Complex Roots
= Elliptic

$$B^2 - 4AC = 0 $$
2 Identical Real Roots (or one Real Root, one zero root)
= Parabolic


Consider $(X, \tau)$. Then $X=A \sqcup B$ disconnected and $A,B$ connected implies $A,B \in \tau$ My thought for solving this was to consider $C,D \in \tau$ such that $X = C \sqcup D$ as $X$ is disconnected. Then I thought about trying to prove that either $C=A, D=B$ or $C=B, D=A$. But that would imply that there's only one way to separate $X$, which is not true in general, correct?

Do you have any advice on how to proceed?

As $X$ is not connected. It's equal to $C\sqcup D$ for some nonempty $C,D\in\tau$. As $A$ and $B$ are connected, $A\subseteq C$ or $A\subseteq D$. Similarly, $B\subseteq C$ or $B\subseteq D$. If both $A$ and $B$, say, lie in $C$, then $D=\emptyset$, contrary to the choice of $D$. Hence $A\subseteq C, B\subseteq D$, or $B\subseteq C, A\subseteq D$. As $A$,$B$ disjoint, $C$,  $D$  disjoint, $A=C,B=D$ or $A=D,B=C$. Thus $A,B\in \tau$.


cofree coalgebra: explicit description from its universal property Let $k$ be a commutative ring. There is a forgetful functor
$$U : \mathsf{Coalg}_k \to \mathsf{Mod}_k$$
from $k$-coalgebras to $k$-modules. This has a right adjoint $R : \mathsf{Mod}_k \to \mathsf{Coalg}_k$, called the cofree coalgebra on a $k$-module. There are several ways to prove its existence, but let us assume that we have chosen an abstract approach (for example, by using some adjoint functor theorem). If necessary, let us assume that $k$ is a field. Can we derive from the universal property how the cofree coalgebra looks like explicitly?

So, given some $k$-module $M$ and a coalgebra $R(M)$ equipped with a natural bijection $\hom(C,R(M)) \cong \hom(U(C),M)$ for coalgebras $C$ , how can we make explicit what $R(M)$ is, i.e. what is the underlying $k$-module and its comultiplication? For example, by taking the coalgebra $C=(k \cdot x,\Delta(x)=x \otimes x,\varepsilon(x)=1)$, we see that
$$\{a \in R(M) : \Delta(a)=a \otimes a, \, \varepsilon(a)=1\} \cong \hom(C,R(M)) \cong \hom(k \cdot x,M) \cong |M|.$$

It does not follow exactly from the universal property, but on page 6 of this paper it is reported that in Richard E. Block, Pierre Leroux, "Generalized dual coalgebras of algebras, with applications to cofree coalgebras" it is shown that the cofree coalgebra of a vector space $V$ can be realized as the space of $k$-linear functions $f : k[t] \to T (V)$ (where $k[t]$ denotes the polynomial algebra and $T (V) = \bigoplus_{n=0}^{\infty} V^{\otimes n}$ the tensor algebra) that are of degree zero (i.e. $f(t^n) \in V^{\otimes n}$) and representative (i.e. for which there is a finite family $g_i, h_i: k[t] \to T (V)$ such that, for every $n, m \in \mathbb{N}$ we have $f(t^{nm}) = \sum_{i=1}^k g_i(t^n)h_i(t^m))$.


Expressing limit of sum definite integral Evaluate limit by expressing it as a definite integral.

$$\lim_{n\to\infty}\frac{\pi}{2n}\left[\cos\left(\frac{\pi}{2n}\right)+\cos\left(\frac{\pi}{n}\right)+\cos\left(\frac{3\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)\right]$$

I do not know how to write this formula out first as a sum formula, any help would be appreciated!

Note that

  $$\cos\left(\frac{n\pi}{2n}\right)=\cos\left(\frac{\pi}{2}\right)=0$$

So

$$\cos\left(\frac{\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)=\cos\left(\frac{\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)+\cos\left(\frac{n\pi}{2n}\right)$$

And hence

$$\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n-1}\cos\left(\frac{k\pi}{2n}\right)=\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)$$

and then

  $$\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)=\frac{\pi}{2}\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)=\frac{\pi}{2}\int_0^1\cos\left(\frac{x\pi}{2}\right)dx$$

The sum of limits can be solved for a given integral equation, i.e.:
$$\int_a^b f(x)dx =\lim_{n \rightarrow \infty} \frac{1}{n} [f(a)+f(a+h)+...+f(a+n-1)h],$$ where $h=\frac{b-a}{2}.$


Any infinite set of a compact set $K$ has a limit point in $K$? I'm reading principles of mathematical analysis and have a question about a theorem 2.37.

Theorem 2.37

  If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

The proof is

  If no point of $K$ were a limit point of $E$, each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$.
  It is clear that no finite subcollection of $\{V_q\}$ can cover $E$.
  The same is true of $K$, since $E \subset K$.
  This contradicts the compactness of $K$.

I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.

Thanks in advance.

A set with no limit points is necessarily a closed set.  Being a closed subset of a compact space, it is compact.  On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.

Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$.  That set is open, since as noted above, $E$ is closed.  Now you've got an open cover of $K$.  It must therefore have a finite subcover.  But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction.

Suppose $\overline{E}$ doesn't have a limit point. Then $\forall e\in E,\exists N_{r_e}(e)\ s.t\ N_{r_e}(e)\cap E=\{e\}$.

Consider the open cover of the compact set ${\overline E}$:
$$\bigcup_eN_{r_e}(e)\rightarrow\bigcup_{i=1}^nN_{r_{e_i}}(e_i)$$
but
$$\bigcup_{i=1}^nN_{r_{e_i}}(e_i)\cap E=\bigcup_{i=1}^n(N_{r_{e_i}}(e_i)\cap E)=\bigcup_{i=1}^ne_i\ne E$$
which is absurd.


Relation of two definite integrals For a differentiable continuous function $g(u)$ over a finite region $R$, I know that
$$\int_R e^u g\; u\; du = 1$$
Is there any way to determine from this the value of
$$\int_R 10^u g\; u\; du$$

The second integral is

$$J=\int 10^u g(u)du =$$

$$=\int e^{uln(10)}g(u)du=.$$

put $v=u ln(10)$

then

$$J=\frac{1}{ln(10)}\int e^v g(\frac{v}{ln(10)})dv.$$

So, in general, we cannot get tha value of $J$

from $I=\int e^u g(u)du=1$.

if $g$ is linear  then $J=\frac{1}{(ln(10))^2}$.

No, it depends on $g$ look at the two cases $g(x)=x$ and $g(x)=x^2$.


Exponential distribution problem, solution clarification Problem:

  Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with parameter $λ = 1/20$. Smith has a used car that he claims has been driven only $10,000$ miles. If Jones purchases the car, what is the probability that she would get at least $20,000$ additional miles out of it?

Solution
$$\begin{align}P(X \geq 10,000+ 20,000 \mid X >10,000) &= P(X \geq 20,000)\\[1ex] &= \int_{20}^{\infty} e^{-x/20} \end{align}$$

Why does the integral go from $20$ to infinity? My thinking is that is should go from $20,000$ to infinity.

That is because exponential distributions have the memoryless property, which means that a car with 10,000 miles has the same probabillity of lasting $x$ more miles as a new car (that is what the first equality states in the solution). In this case the probability of the car lasting another 20,000 miles is therefore
$$
\begin{align}
P(X \geq 30,000 \mid X > 10,000) &= P(X \geq 20,000)\\
&= \int_{20}^{\infty} {1\over20}\, e^{-x/20}\, dx \\
&= e^{-20/20}\\
&= 36~\%
\end{align}
$$
--- there was a slight error in the stated solution; the density is ${1\over20}\, e^{-x/20}$.

(this is true of any exponential distribution : the probability of exceeding the mean is $1/e=0.36\approx1/3$).

Consider the integrand!
But it's still wrong…

It's $$P(X \ge 20000) = \int_{20000}^\infty \lambda e^{-\lambda x} dx = 120 \int_{20000}^\infty 120 e^{-120 x} dx$$

now you could substitute $z=\frac{x}{1000}$ and get:

$$P(X \ge 20000) = 120000 \int_{20}^\infty e^{-0.12x} dx$$

But the given "solution" is wrong.


Characteristic of Field For which prime $p$, the polynomial $x^4 + x + 6$ have a root of multiplicity $> 1$ over a field of characteristic $p$?

Options:

2
3
5
7

Hint: You can check by hand all the answers

Hint:

If characteristic is 2 or 3 then the equation will be $ x(x^3 + 1 ) $
So consider $ x^3+1 $ in case of $ char \space 2 $ one is a root of $ x^3+1 $, so factor $ x^3+1 $ and check if 1 is root of factor, if not check for zero.

If above thing fails, so the same with $ Char \space3 $, here 1 is not a root of $ x^3+1 $, so only options are 2 and 0. In fact 2 is a root of $ x^3+1 = 8+1 = 0 $ so factor  $ x^3+1 $ with $ \pmod 3$ and check.

For 5 and 7 you can minimize calculation overhead by considering $ { -2, -1, 0, 1 ,2} $ and  $ { -3,-2, -1, 0, 1 ,2, 3} $


Alternative proof : If E and F have positive measures then E+F contains an interval The statement is following :

  $E, F$ are measurable subsets in $\mathbb{R}$ which have positive
  measure. Then prove that $E+F=\left\{ x+y\ :\ x\in E,\ y\in F\right\}$
   contains an interval.

This is a problem 30 in Stein's Real analysis. I already read the proof of this by using convolusion which is here.(This problem is so-called Steinhaus theorem) Moreover, I proved that the previous problems number 28 and 29 in the book.
My goal is applying the same technique to above one, using the technique in problem number 29 :

First, by problem number 28, $ \exists I^F$ and $I^E$ such that $$m(E\cap I^{E})\geq{\displaystyle \frac{9}{10}m(I^{E})} \text{ and }  m(F\cap I^{F})\ge{\displaystyle \frac{9}{10}m(I^{F})}.$$ Let $E_{0}= E\cap I^{E}$ and $F_{0}= F\cap I^{F}$ and let $e_{0}, f_{0}$ : center of $I^{E}, I^{F}$, respectively.

Suppose $E_{0}+F_{0}$ does not contain an interval centered at $e_{0}+f_{0}$. Then $\exists a$ : small such that $E_{0}\cap(e_{0}+f_{0}+a-F_{0})=\phi$. But, $E_{0}\cup(e_{0}+f_{0}+a-F_{0})\subset I^{E}\cup(e_{0}+f_{0}+a-I^{F})$. Measure of left hand side is more than $\frac{9}{10}(m(I^{E})+m(I^{F}))$ but the measure of Right had side is slightly less than $\max(m(I^{E}),m(I^{F}))$.

And that's it. I stopped here. I can't make any contradiction.
I thought that if I can make measure of $I^E and I^F$ are equal, then I prove the problem but this one is also a big problem.

Anyone can comment on this?

This proof uses the  Lebesgue density theorem.  If $e$ and $f$ are points of density of $E$ and $F$, and $\epsilon > 0$, there is $\delta > 0$ such that $m([e-\delta,e+\delta] \backslash E)/(2\delta) < \epsilon $ and
$m([f-\delta,f+\delta] \backslash F)/(2 \delta) < \epsilon$.
I claim that $e + f + [-\delta/2,\delta/2] \subseteq E + F$.

Consider any $z \in e + f + [-\delta/2, \delta/2]$.  Let $s$ be chosen from the interval $e + [-\delta/2, \delta/2]$.  Thus $z-s \in f + [-\delta, \delta]$.  The set of $s \in e + [-\delta/2,\delta/2]$ for which $s \notin E$ has measure $< 2 \delta \epsilon$.  The set of $s \in e + [-\delta/2, \delta/2]$ for which $z-s \notin F$ has measure $< 2 \delta \epsilon$.  If $4 \delta \epsilon < \delta$, removing these leaves a set of positive measure (and in particular at least one) in which $s \in E$ and $z-s \in F$, so that $z \in E + F$.

actually I think your idea could be extended by substituting $e+f$ to any real number, then you could move Ie until Ie is contained in another


$a \equiv b \pmod n$ solveable - which necessary and sufficient conditions ? For given $n\in \Bbb N, n \ge2, a,b \in \Bbb Z$ give necessary and sufficient conditions such that $$a\equiv b\pmod n$$
  is solveable.

First I thought that it would be sufficient if $b$ would be a multiple of $a$, so for $c\in \Bbb Z$ we'd have $a= b \cdot c$.

For the necessary part I have no clue, I mean if the $gcd(a,n)=1$ so it's solveable for each $b \in \Bbb Z$, but this had been already be talked about in the lecture, so it's not that what it's asked for.

Any hint for which part I should pay attention to ?

The congruence holds if and only if the gcd of $a$ and $n$ is a divisor of $b$.


Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.

I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$  Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?

Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):

$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$

$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$

$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$

$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$

Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 +
  a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$

And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
 }}.$$

So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$

In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$.  In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.

Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$

First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?

If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)

(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)

Thank you!

I don't have much time for elaborating now, but you definitely can do this yourself (feel free to add your own answer).

Just observe that substitutions $x = r \cos \theta$ and $y = r \sin \theta$ to $z = a \left( x/y + y/x \right)$ turn this equation to $z = \tfrac{a}{\cos \theta \cdot \sin \theta}$ which clearly does not depend on $r$. This means that for each $\theta$ the ray $\theta = const$, $r>0$ lies on the surface. Now you can use Corollary 18.6. from Chapter 18 of A.Gray, E.Abbena and S.Salamon "Modern Differential Geometry of Curves and Surfaces with Mathematica" stating that "A straight line that is contained in a regular surface is necessarily an asymptotic curve".

Alternatively you may try to use the differential equation for asymptotic curves from here.

I believe this surface can be described by two networks of lines. ( Shown in Red and Cyan). Make the substitution $nt$ for $u$ , and separately $nt$ for $t$ ,
so that the new parametrization might look like : $<t,nt,a(1/n + n/1)> $ and/or $<nt,t,a(1/n+ n)>$ these are the asymptotic lines. Then switch the sign of n to pick up the surface below the z axis. I've used $n$ as an integer , but every line in the surface chosen in this way is an asymptotic curve, and a geodesic.

(Added a guess for the solution of the equations based on a study of the surface.)


I do not see how $ \frac { 1 }{ 1-x } = \sum _{ n=0 }^{ \infty }{ { x }^{ n } } $ I do not see how $$ \frac { 1 }{ 1-x } =\quad \sum _{ n=0 }^{ \infty  }{ { x }^{ n } }  $$

for example, When x = 10
and n = 1

Then $$ \frac { 1 }{ 1-10 } \neq \quad { 10 }^{ 1 } $$

What am overlooking? What is clear way to show? $$ \frac { 1 }{ 1-x } =\quad \sum _{ n=0 }^{ \infty  }{ { x }^{ n } }$$

First of all, you can't choose just one $n$ to plug into the formula. $n$ isn't a variable, it's an index: you sum over all $n$, meaning
$$
\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots,
$$
when the right hand side converges (meaning you can only plug in certain $x$). If $\left|x\right| < 1$, you can show that this is true without much trouble: let $S_N = \sum_{n = 0}^N x^n$. Then $x S_N = \sum_{n = 0}^N x^{n+1}$. Then $S_N - x S_N = \sum_{n = 0}^N x^n - \sum_{n = 0}^N x^{n+1} = 1 - x^{N+1}$, and hence
$$
S_N = \frac{1 - x^{N+1}}{1 - x}.
$$
Taking the limit, we have
$$
\lim_{N\to\infty} S_N = \lim_{N\to\infty}\frac{1 - x^{N+1}}{1 - x} = \frac{1}{1 - x},
$$
because $x^N\to 0$ as $N\to\infty$ if $\left|x\right| < 1$. By definition, $\sum_{n = 0}^{\infty} x^n = \lim_{N\to\infty} S_N$, so you have your formula.

$\sum _{n=0}^\infty x^n=1+x+x^2+\ldots $  which is a G.P. series with common ratio $x$

Hence $1+x+x^2+\ldots =\dfrac{1}{1-x} \text{if |x|<1}$


Pointwise functions on vector space (reference request) I'm just looking for a reference or anything related to the question I have outlined below. I have personally never come across anything like this -

Suppose for $\mathbb{R}^n$ we define a continuous pointwise function as a function $F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ that can be written in the form $$F(v) = (f_1(v_1), f_2(v_2), f_3(v_3),\cdots,f_n(v_n))$$

w.r.t some basis, and where the $f_i$'s are continuous functions from $\mathbb{R} \rightarrow \mathbb{R}$.

Now my question is if we consider the set of functions from $\mathbb{R}^n \rightarrow \mathbb{R}^n$ that are generated by composing these continuous pointwise functions, is that set of functions equal to, or dense in set of all continuous functions from $\mathbb{R^n} \rightarrow \mathbb{R^n}$?

Note that when composing two continuous pointwise functions, they do not have to be pointwise w.r.t the same basis.

Composing 2 of your continuous pointwise functions just gives another continuous  pointwise function . If $ F=(f_1,...,f_n) ,and G=(g_1,...g_n) then FoG=((f_1)o(g_1)...(f_n)o(g_n))$


The 2-genus surface is not contractible. I'm trying to prove that the 2-genus surface is not contractible.
My first idea was to prove that the circle and the 2-genus are homotopy equivalent using an homotopy retract, but since I don't know how to triangulate the 2-genus surface I'm not able to do this.
Then, I have tried to compute the fundamental group of the 2-genus surface, but I haven't studied yet The Seifert-Van Kampen Theorem.

Can someone give me a hint please? I have started studying algebraic topology this year, so my knowledge is small. Thank you.

Follow the hint of @iwriteonbananas.

First, define a retraction (not a deformation retraction, which would be impossible. Just a retraction) of the surface $\mathbb{S}$ to $S^1$. You can do this by considering the surface embedded in $\mathbb{R}^3$, taking a loop $L$ on the plane $xy$ around the two holes (which will be $S^1$), a point $P$ inside a hole and defining the composition $p\circ \pi$, where $\pi$ is the projection $\mathbb{R}^3 \to \mathbb{R}^2$ and $p$ is the projection to the loop with respect to the point $P$.

Having constructed a retraction, use the fact that $\pi_1( \cdot)$ is functorial to conclude that $i_*$, the map induced by the inclusion $L \hookrightarrow \mathbb{S}$, is injective. This will show that $\pi_1(\mathbb{S},*)$ cannot be trivial.

Hint: are there any non-contractible loops in your surface? By definition, a space is contractible if it is homotopy equivalent to a point. What do homotopy equivalences do to the fundamental group?


Get solutions of an equation given its output Lets assume I have a function $f(x_1, x_2, x_3)$ that takes $x_1, x_2, x_3$ as inputs. For the sake of simplicity, lets assume that this $f(x)$ is a linear equation.

My question is- if I decide that the output of my function has to be an arbitrary value- $v$, can I generate all or a fixed number of the possible solutions or combinations of values that when passed to $f$ will generate $v$?

If yes, how do I achieve this?

Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $\mathbf{a}=(a_1,a_2,a_3)$, you have to find all $\mathbf{x}\in\mathbb{R}^3$ such that $\mathbf{a}.\mathbf{x}=v'$.

To find $\mathbf{x}$, we can do a coordinate transformation so that $\mathbf{a}$ is aligned with the $x$-axis and then $\mathbf{x}$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $\mathbf{x}$ in the original coordinate system.

I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.

Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.

In general, for $f:\mathbb{R^n}\to\mathbb{R^m}$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.

For the particular case of $f:\mathbb{R^3}\to\mathbb{R}$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve

$$a_1x_1+a_2x_2+a_3x_3=v$$

which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$

$$x_1=\frac{v-a_2s-a_3 t}{a_1}$$


How can I calculate the time two people will meet if they are paddling towards each other on a lake? Here's the question I'm having trouble with. Ken and Kara are 30 miles apart on a calm lake paddling towards each other. Ken paddles at 4 mph, while Kara paddles at 7 mph. HOw long will it take them to meet?

I suppose that you know how to find variable $t$ (time) from system below :

$$\begin{cases}
 x + y = 30 \\
 x=4\cdot t \\
 y=7 \cdot t
\end{cases}$$

You can use relative velocities, the difference between their velocities is 3mph, and the distance is 30 miles, so it would take 10 hours for them to cover the same distance.


Intersecting angles on trigonometric circle after solving

$|\sin(2 x - \pi/12)|\le\sqrt(2)/2$

I get two interval solutions which intersect eachother,

$5 \pi/12 + k \pi \le x \le \pi/6 + k \pi$
and
$11\pi/12 + k \pi \le x\le 2\pi/3 + k \pi$

Since the solutions I get arent the basic, I don't have a picture in my mind how they look and intersect.

Note that $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$. Also note that when $\sin$ is negative, the angle is also. So that means $$-\frac{\pi}{4}\le2x-\frac{\pi}{12}\le\frac{\pi}{4}$$

Can you solve it from here?

$$\left|\sin\left(2x-\dfrac\pi{12}\right)\right|\le\dfrac{\sqrt2}2$$

$$\iff2\sin^2\left(2x-\dfrac\pi{12}\right)\le1$$

Using $2\cos2A=1-2\sin^2A$

$$\iff\cos\left(4x-\dfrac\pi6\right)\ge0$$

$$\implies2m\pi-\dfrac\pi2\le4x-\dfrac\pi6\le2m\pi+\dfrac\pi2$$ where $m$ is any integer


The map $f:S^1 \to S^1, f(z) = z^k$ is well-defined only if $k \in \mathbb{Z}$ Is the map $f:S^1 \to S^1, f(z) = z^k$ well-defined only if $k$ is an integer?

I believe that $k$ needs to be at least a rational number, because $(-1)^k$ is not really defined when $k$ is irrational. However, I don't know how to continue from here.

You need $k$ to be an integer for $f(e^{0i})=f(e^{2\pi i})$. And this equality needs to hold for well-defined-ness.


Decide whether or not there exists $f(f(x))=5x-\lfloor 5x\rfloor$ Decide whether or not there exists  $f;[0,1)\to [0,1)$ such
  $$f(f(x))=5x-\lfloor 5x\rfloor=\{5x\}$$

Today I suddenly thought of an interesting question, but I have not found any function of symbolic conditions so far.

Yes, there is such a function. (But I'm going to use the Axiom of Choice to show that, so I don't have any easy formula to offer - but I'm also going to use a general method that solves other problems too)

An easy way to see this is to throw out all the structure that is irrelevant to whether or not a functional square root exists. In particular, let us make the following definitions (which are not a standard definition or names):

  A system is a pair $(S,f)$, where $S$ is a set and $f:S\rightarrow S$ is a function.

  A system is square if it is of the form $(S,f\circ f)$, where $f:S\rightarrow S$ is a function.

  Two systems $(S,f)$ and $(T,g)$ are isomorphic if there exists a bijection $\pi:S\rightarrow T$ such that $\pi\circ f = g\circ \pi$.

  If we have two systems $(S,f)$ and $(T,g)$ with disjoint domains $S$ and $T$, we can define their union $(S,f)\cup (T,g)$ as a pair $(S\cup T,f\cup g)$, where $f\cup g$ is the function taking the values of $f$ in the component of $S$ and the values of $g$ in the component of $T$. Arbitrary unions can be defined similarly.

These definitions tell us to think about functions $S\rightarrow S$ as directed graphs with vertices $S$ and an edge $(x,f(x))$ for every $x\in S$ and to consider isomorphisms of those graphs. Here four examples, showing this graph for some functions with $S=\{0,1,2\}$, namely $f(x)=\max(x-1,0)$ and $f(x)=\min(x+1,2)$ and $f(x)=2$ and $f(x)=x+1\text{ mod }3$:

Note that the first two are basically mirror images of one another. The notion of isomorphism captures how these functions have the same structure, whereas the other two functions are different.

Now, we can show a lemma:

  Let $(S,f)$ and $(T,g)$ be isomorphic by some map $\pi:S\rightarrow T$. Assume $S$ and $T$ are disjoint. Then, $(S,f)\cup (T,g)$ is square.

This is because we can define $h:S\cup T\rightarrow S\cup T$: $$h(x)=\begin{cases}\pi(f(x)) & \text{if }x\in S\\
\pi^{-1}(x) & \text{if }x\in T
\end{cases}.$$
Then, for $x\in S$, we have that $$h(h(x))=h(\pi(f(x))=\pi^{-1}(\pi(f(x))=f(x).$$ For $x\in T$ we have $$h(h(x))=h(h(x))=h(\pi^{-1}(x))=\pi(f(\pi^{-1}(x)))=g(\pi(\pi^{-1}(x)))=g(x).$$ Thus, $h$ is a functional square root of $f\cup g$.

Pictorally, we can do an example. We could take a pair of isomorphic functions like the following, where the isomorphism is shown:

Then, we essentially weave back and forth between the isomorphic parts, in such a way that we take a step "forwards" each time. The result is a functional square root that looks like the following:

  Let $\{(S_i,f_i\circ f_i)\}_{i\in I}$ be a collection of square systems with the $S_i$ pairwise disjoint. Then, $\bigcup_{i\in I}(S_i,f_i\circ f_i)$ is square.

This follows by noting that the functional square root of this system is $\bigcup( S_i,f_i)$.

Now, we can apply this to our given function $f(x)=5x-\lfloor 5x\rfloor$ by writing $f$ as the union of other functions. In particular, we define the following:

  A component of a system $(S,f)$ is a set of the following form, for some $x\in S$: $$\langle x\rangle=\{y\in S:f^n(y)=f^m(x)\text{ for some }n,m\}.$$

These are exactly the components of the underlying directed graph. A system is always the union of its restrictions to each component. Moreover, each component is a directed graph with either zero or one directed cycles.

Now, we can finish by considering what the components of the given $f$ are. Note that every point in $[0,1)$ has exactly $5$ preimages under $f$. First, there are uncountably many components of the form $\langle x\rangle$ for irrational $x$, and none of these may contain a cycle, since $x,\,f(x),\,f^2(x),\ldots$ is not eventually periodic for irrational $x$. These components all consists of isomorphic systems corresponding to infinite bidirectional trees where every vertex has $5$ predecessors. We may use the axiom of choice to pair them up, and take the square root of every pair.

Then, for rational $x$, the component $\langle x\rangle$ must have a cycle, since $f$ never increases the denominator of a value. The component must therefore consist of an $n$-cycle, where each element of the cycle has $4$ preimages outside of the cycle, then everything outside of the cycle has $5$ preimages. The isomorphism class of this component only depends on $n$. Example components of this form are shown below:

Now, to finish, we just show that there are an even number of components with an $n$-cycle, since then we can pair these up and take their square root individually, then piece all the pairs back together to get a functional square root of the whole thing.

To see this, note that there are $5^n-1$ elements $x$ satisfying $f^n(x)=x$ - these are just those numbers of the form $\frac{c}{5^n-1}$ for $0\leq c < 5^{n}-1$. Then, we can use the Mobius inversion formula to get that there are
$$\sum_{d|n}\mu(d/n)(5^d-1)$$
elements that are members of an $n$-cycle (and no smaller cycle), where $\mu(x)$ is the Mobius function, equalling $0$ if $x$ is not square-free, otherwise equalling $1$ if $x$ is a product of an even number of primes and $-1$ if it is a product of an odd number of primes.

The number of $n$ cycles is just this quantity over $n$, thus we can do our pairing argument exactly if $2n$ divides $\sum_{d|n}\mu(d/n)(5^d-1)$ for all $n$. Note that, except for $n=1$, this sum can also be written as $\sum_{d|n}\mu(d/n)\cdot 5^d$, since the constant terms will cancel.

Showing this final step is basically an exercise in number theory. If you look mod each prime power and group the given sum by pairs of the form $\{n/d,n/(pd)\}$ where $p$ does not divide $d$, the result falls out fairly easily by noting the exponent of the multiplicative group mod $p^n$. Then, we can repeat this trick on mod powers of $2$, noting that the multiplicative group mod $2^n$ only has exponent $2^{n-2}$ for $n\geq 2$, which let us get the result we want.

Yes, it it exists as the following example shows.The function $\{5x\}$ in the interval $[0,1)$ is explicitly defined by
$$\begin{equation}\{5x\}=\begin{cases}5x\space \text{ for }0\le x\lt 0.2\\5x-1\space \text{ for }0.2\le x\lt 0.4\\5x-2\space \text{ for }0.4\le x\lt 0.6\\{5x-3}\space \text{ for }0.6\le x\lt 0.8\\5x-4\space \text{ for }0.8\le x\lt 1\end{cases}\end{equation}$$ so we can calculate an example of $f(x)=\sqrt5\space x+b$ such that $f(f(x))$ coincides with the funcion above. Calculation gives

$$\begin{equation}f(x)=\begin{cases} \sqrt 5\space x\quad\text{ for }0\le x\lt 0.2\\\\\sqrt 5\space x-\dfrac{\sqrt5-1}{4} \text{ for }0.2\le x\lt 0.4\\\\\sqrt 5\space x-\dfrac{(\sqrt5-1)}{2}\space \text{ for }0.4\le x\lt 0.6\\\\\sqrt 5\space x-\dfrac{3(\sqrt5-1)}{4}\space \text{ for }0.6\le x\lt 0.8\\\\\sqrt 5\space x-(\sqrt5-1)\space \text{ for }0.8\le x\lt 1\end{cases}\end{equation}$$ It follows
 $$\color{red}{f(f(x))=\{5x\}}$$ as it is easy to verify.


Counting graphs with even degrees How many non-isomorphic distinct labeled 5-vertex graphs with even degrees are there? The answer is $2^6$, but I don't seem to be able to solve the problem.

P.S. It's not a homework. I'm just studying for an exam.

We are looking, essentially, for the cycle space of $K_5$, the set of all graphs formed from disjoint unions of cycles.  This fact follows by strong induction on the number of edges: since all degrees are even, the components of the graph cannot be trees, so it must contain a cycle if it is nonempty.

The cycle space is the orthogonal complement of the cut space in the edge space (which is a vector space over $\mathbb F_2$).  The order of the edge space is easy enough to calculate, and it corresponds to all possible graphs with $5$ vertices (not up to isomorphism).  It has order $2^{10}$ since for any graph $G$, each of $10$ edges of $K_5$ can be in $E(G)$ or not in $E(G)$.

The cut space contains the empty graph, $5$ copies of $K_{1,4}$ and $10$ copies of $K_{2,3}$, so it has $16$ graphs in total.  Since it is a subspace of $\mathcal E(G)$, it has a basis of order $4$.  So the cycle space has $2^{10} / 2^4 = 2^6$ members.

What do you mean by a non-isomorphic labelled graph?

In terms of just 5-vertex graphs, there is one up to isomorphism with no edges, 1 with 3 edges, 1 with 4 edges,  1 with 5 edges, 1 with 6 edges, 1 with 7 edges, 1 with 10 edges.  A total of 7

Edit:  Sorry forgot the 5-cycle :)


Finding the value of $(-1+i)^{10}$ by using the trigonometric form's expression I am trying to find out what is the value of $(-1+i)^{10}$ by using the trigonometric expression, which should simplify it by avoiding to multiply in binomial form. So far I got this (sin==sen, and excuse my shoddy writing):

But every other calculator I used to test the result says that it's -$32i$, and not $32i$, which is driving me mad! I can only guess that I messed u on the argument($\alpha$)

Any help is welcome!

Edit: Solved, I hadn't added the fractions properly, see comment chain below post

Hint:

Use https://en.m.wikipedia.org/wiki/Atan2#Definition_and_computation

arg$(-1+i)=\arctan\dfrac1{-1}+\pi=\dfrac{3\pi}4$

arg$(-1+i)^{10}=10\cdot\dfrac{3\pi}4=\equiv-\dfrac\pi2\pmod{2\pi}$


Probability sitting at a round table What is the probability that 6 individuals will be seated correctly in order of their height when at a round table?

Is it 1/6!?

1 Because it can only be ascending or descending, and 6! on the bottom for all 6 possibilities?

We fix the first person and arrange all others. we have $5!$ possibilites for this arrangement- only two of these yield the wanted arrangement, thus the solution is $\frac{2}{5!}$

Start with the tallest person. This person can sit in one of $6$ positions. Now, the next tallest can go either left or right of that person, so that's $2$ options there. The rest is now fixed.

So, it is $6 \cdot 2= 12$ out of $6!$, i.e. $\frac{12}{6!}$


Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$

So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$

Then:

$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$

Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).

On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$  ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.

Why is there a problem and how can I get the correct answer?

Edit: Here is Wolfram's answer:

Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$

The substitution is incorrect : the tangent is not bijective on the interval $[0,2\pi]$. First, you need to restrict yourself to an interval on which the tangent behaves better. Using the $\pi$-periodicity of the function you want to integrate, you can show that:

$$\int_0^{2 \pi} \frac{1}{a \sin^2 (\theta)+b \cos^2 (\theta)} d \theta = 2 \int_{-\pi/2}^{\pi/2} \frac{1}{a \sin^2 (\theta)+b \cos^2 (\theta)} d \theta,$$

and go from there.

Note that this is a good warning about using Wolfram (or any formal computation system) : the formula for the indefinite integral is good, but it holds only on each interval $(k\pi -\pi/2, k\pi+\pi/2)$, which the program does not tell you.

You have everything right up to
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))
$$
 Now $\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))$ is $2\pi$ because the $\arctan$ and the $\tan$ are inverse functions.

So we get
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))=\frac{1}{ab}2\pi-0
$$
or
$$
\frac{2\pi}{ab}
$$


Independent Event Complements I have the following homework assignment that I've already finished, but am confused on whether I've gotten right/wrong, and was hoping someone could help explain so I understand the problem better.

  An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian projet is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)=0.4 and P(B)=0.7

  a.) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.

  b.) What is the probability that at least one of the two projects will be successful?

  c.) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?"

Here is what I've gotten for each part:

$$P(A \cap B) = P(A)P(B) = (.4)(.7) = .28 $$

a.) $P(B^c) = 1 - 0.7 = 0.3 $

b.) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = .4 + .7 - .28 = .82$

c.) $P(A) - P(A \cap B) = .4 - .28 = .12 $
    $.12/.82 = .146$

I am confused in that the two events are independent of each other and the book states that for part a the answer should be .126 instead of what I got. Am I doing these problems correctly or am I committing some error?

Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.

Note that even though it is quite intuitive that "$A$ and $B$ are independent $\rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.
(You can find some proofs here, though I bet you would succeed in proving it by yourself.)

(a) This question belongs to "conditional model"
But since A and B are independent
You may take directly p(B not)=0.3

Reason: independent means

P(A .B)=P(A).P(B)


Mixed Nash equilibria in $n$-player games I'm reading up on Game Theory.

So far, I feel like I have a pretty good understanding on two-player games and their properties.

Consider a two person game where the payoff matrices are $A_{m\times n}$ and $B_{m\times n}$. A pair of strategies $(x, y)$ is a Nash equilibrium if and only if

$\forall 1\leq i \leq m, x_i > 0 \implies (Ay)_i = max_k(Ay)_k$
$\forall 1\leq j \leq m + n, y_j > 0 \implies (x^TB)_j = max_k(x^TB)_k$

Where $(x, y)$ are probability distributions.

As said above, I have a pretty good understanding why this is true.

Unfortunately, the definition above describes an equilibrium for the two-player case.

Is there a general definition for the n-player case?
I can't seem to find one.

A mixed action profile $\alpha ^{\ast }$ is a mixed
equilibrium (or mixed Nash equilibrium) of game $G$ if, for all $i\in I$,
\begin{equation*}
\alpha _{i}^{\ast }\in \arg \max_{\alpha _{i}\in \Delta \left( A_{i}\right)
}u_{i}\left( \alpha _{i},\alpha _{-i}^{\ast }\right)
\end{equation*}

Where $\alpha_i$ is a mixed action for player $i$, $\alpha = \prod_i \alpha_i$ is a profile of mixed actions for each player, $\alpha_{-i} = \prod_{j\ne i} \alpha_j$ is the profile of mixed actions for all players other than $i$ and $\Delta(A_i)$ is the set of all probability distributions over $A_i$, which is the set of player $i$'s pure strategies.

I think that's everything -- is that clear? Basically, it's very much like the 2-player definition you give, except every player is best responding to all other players. You may think this is stated slightly differently than the version you have. This just reflects stylistic differences between pure game theory and algorithmic game theory. Consider it an exercise to translate the above pure game theory definition into an equivalent one that looks more like the algorithmic presentation.


How to prove that $\limsup_{n\to\infty}n M_n\geq \frac{1}{\ln 2}$ for the following sequence? Let $(x_i)_i$ be a sequence of distinct numbers in $[0,1]$.
Note that $[0, 1] \setminus \{x_1, \cdots, x_{n-1} \}$ can be written as a disjoint union of non-empty and non-singleton intervals $C_{n, k}$.
Let $M_n \equiv \max_k |C_{n,k}|$.

How do I prove that $\limsup\limits_{n \to \infty} n M_n\geq 1/\ln 2$?

If the sequence is strictly increasing then the limit is always $\infty$.

So the sequence is convergent and Cauchy.

Let $|C_{n,0}|=s_1-0=s_1>0$.

Then there exist an integer $N$ such that for each $l,l-1>N$, then

$|C_{n,l}|=|s_l-s_{l-1}|<|C_{n,0}|$

So

$|M_n|=max\{|C_{n,k}|: k<N\}\geq |C_{n,0}|=s_1$ for each $n\in \mathbb{N}$

This means

$n|M_n|\geq ns_1\to \infty$

In general I think it is not true and
 a counter-example can be the following:

You consider the succession:

$\{1, \frac{1}{2},\frac{1}{4},\frac{3}{4}, \frac{1}{8},\frac{3}{8},\frac{5}{8}\dots\}$

You can observe that, fixed $k\in \mathbb{N}$ , for $n=k+2^k$ the set $[0,1]$ is divided into $2^k$ parts of diameter $\frac{1}{2^k}$, for example:

k=0:

Then $n=1$, that means

$[0,1]/ \{1,\frac{1}{2}\}=[0,\frac{1}{2})\cup(\frac{1}{2},1)$

k=1:

Then $n=3$, that means

$[0,1]/\{1,\frac{1}{2},\frac{1}{4},\frac{3}{3}\}=[0,\frac{1}{4})\cup (\frac{1}{4},\frac{1}{2})\cup (\frac{1}{2},\frac{3}{4})\cup (\frac{3}{4},1)$

Thus if you consider the subsequence

$\{s_{k+2^k}\}_{k\in\mathbb{N}\cup\{0\}}$

you have

$|C_{k+2^k, i}|=\frac{1}{2^k}$ for each $i$ so

$|M_{k+2^k}|=\frac{1}{2^k}$

By contradiction, if

$\lim sup_{n\to \infty}n|M_n|\geq \frac{1}{ln(2)}$

Then each subsequence verifies that inequality, but

$\lim sup_{k\to \infty}(k+2^k)|M_{k+2^k}|=$

$= \lim sup_{k\to \infty}\frac{(k+2^k)}{2^k}=0+1=1<\frac{1}{ln(2)}$


Solving a homogeneous quadratic equation in three variables over the integers. Is there any procedure for determining if an infinite amount of solutions exist for an equation of the type $x^2 = ay^2 + byz + cz^2$ for arbitrary integer constants $a, b, c$ and variables $x, y, z \in \mathbb{Z_+}$? If not, does knowing at least one non-trivial solution of the equation help determine if an infinite amount of solutions exist?

Example (if it helps): let $x^2 = 202 y^2+14yz+9z^2$. Here one solution is $y=z=1$ and $x=15$, do there exist infinitely many other solutions?

TLDR; If there exists one nonzero solution, then there exist infinitely many solutions, paramatrized by $\Bbb{Z}^2$. Popular candidates to test are triplets $(0,y,z)$ where $y\mid c$ and $z\mid a$, though such solutions need not exist.

Every integral solution to your equation with $x\neq0$ yields a rational solution to
\begin{equation}aY^2+bYZ+cZ^2=1,\end{equation}
by setting $Y:=\tfrac  yx$ and $Z:=\tfrac zx$. Conversely, every rational solution to this equation yields an integral solution to your equation by multiplying out the denominators. This also shows that multiplying an integral solution through by an integer yields another integral solution.

Given a nonzero rational solution $(Y_0,Z_0)$ to your equation, also
$$(aY_0+bZ_0,-aZ_0),$$
is a rational solution, and moreover for every $k\in\Bbb{Z}$ also
$$\left((aY_0+bZ_0)k^2+2cZ_0k-cY_0,-aZ_0k^2+2cY_0k+bY_0+cZ_0\right),$$
is a rational solution. These are in fact all rational solutions. This then in turn yields all integral solutions, except those with $x=0$. For these we have
$$Y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Z,$$
so such solutions exist if and only if $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ are rational, i.e. if and only if $ax^2+bx+c$ has a rational root, which is easy to test.


Assuming that the sum of the angles of any triangle is 180, how can I deduce Euclid's 5th postulate? I already did the reverse, namely, if we assume Euclid's 5th postulate, then the sum of the angles of any triangle is 180 degrees.
Now I need to show the converse, but I don't really know how to approach this way. Any tips on how to get started would be appreciated.

  Euclid's Fifth Postulate:  If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

There is some confusion about these questions. So I state here some facts:

The axioms of Euclid (minus 5) prove that parallels exist.
Playfair's postulate is that through a point not on a line there is ONLY ONE parallel. This is equivalent to the statement of Euclid's fifth postulate.
$\Delta=\pi$ (that is the sum of the angles of a triange is $\pi$ radians) DOES NOT imply the parallel postulate without the Archimedean axiom.
If one assumes that Archimedean axiom then $\Delta=\pi \Rightarrow $Playfair.

The proof of the last statement is as follows:
Let $A$ be a point not on line $l$, drop perpendicular from $A$ to $l$ at $B$.
Take the line $m$ through $A$ perpendicular to $AB$, this is parallel to $l$ as Euclid shows. If there were a second line through $A$ parallel to $l$, say making angle $\beta$ with $m$, then take a point $C$ on $l$ such that $\angle ACB <\beta$ (this is where the Archimedean axiom is used). Now the right angle triangle $ABC$ as angle sum $<\pi$ as one sees form the picture.

Assume that the sum of the angles of any triangle is $180$ degrees.

Let $L_1$, $L_2$ be two straight lines that intersect in the angle $\gamma > 0$. Now, let $\bar{L_3}$ be a line segment that intersects both $L_1$ and $L_2$. Assume that $L_1$ and $L_2$ meet on the side of $\bar{L_3}$ where the two interior angles:$\alpha, \beta$ on that same side formed by $\bar{L_3}$ intersecting both $L_1$ and $L_2$, add to form a sum that is greater than or equal to $180$ degrees. Then the three angles $\alpha$, $\beta$, $\gamma$ form a triangle whose angle sum is strictly greater than $180$ degrees.

Thus, $L_1$ and $L_2$ meet on the side where the two side-sharing interior angles formed by the intersection of $\bar{L_3}$ with both $L_1$ and $L_2$ sum to less than $180$ degrees.


discrete metric, both open and closed. I've checked several answers though, still don't understand last bit.

Taking radius r = 1/2 then every subset is singleton and it is open.

But then how do you deduce it is also closed?

Well, a subset is closed if its complement is open... but then 'every' subset is open hence its complement is empty then closed as empty is both open and closed....??

Thanks.

ps. I haven't learnt about topology space, only metric space.

Note that if every subset is open, then every subset is closed: Given $A \subset X$, then the complement $A^c = X \setminus A$ is a subset, therefore open, and $A^c$ open is equivalent to $A$ is closed.

If you want to be concrete, you can view the complement of a single point as the union of the balls of radius $1/2$ centered on $y$, as $y$ ranges across every point other than $x$.  This union is evidently the complement of $\{x\}$.

let A is subset of discrete metric space.then to show A is closed then A contains all of its limit point.if r=1 clousre of A =A let if c is a limit point which is outside of A let r1 choose such that r1=(min (c,ai_)then open sphere centered with radius r1 not intersect A then A is closed


Fallacy - where is the mistake? Could anyone help me to find the mistake in this fallacy? Because the actual result for $I$ is $\pi/2$

\begin{equation}
I = \int_{0}^{\pi} \cos^{2} x \; \textrm{d}x
\end{equation}

\begin{equation}
I = \int_{0}^{\pi} \cos x \cos x \; \textrm{d}x
\end{equation}
substitution:
\begin{equation}
\sin x = u
\end{equation}

\begin{equation}
\cos x \; \textrm{d}x = u
\end{equation}

\begin{equation}
\cos x = \cos (\arcsin x)
\end{equation}
and the limits:
\begin{align}
\begin{split}
x &= 0 \quad \Rightarrow \quad u=0 \\
x &= \pi \quad \Rightarrow \quad u=0
\end{split}
\end{align}
so
\begin{equation}
I = \int_{0}^{0} \cos (\arcsin u) \textrm{d}u
\end{equation}

\begin{equation}
I=0
\end{equation}

The problem is in using $\cos x = \cos \sin^{-1} x$. Arcsine is not a single valued function on $x=[0,\pi]$. In the conventional sense, it's not defined outside of $[-\pi/2,\pi/2]$. You can only do a $u$ substitution using a one-to-one function.

if $\sin(x)=u$ we get $x=\arcsin(u)$ and $dx=\frac{1}{\sqrt{1-u^2}}du$ thus we get
$\int \cos(x)^2dx=\int 1-\sin(x)^2dx=\int \sqrt{1-u^2}du$


Average of square roots's sum vs. square root of an average I was watching a video on youtube about how colors work in computers, and found this statement:

  "The average of two square roots is less than the square root of an
  average"

The link to the video where I found this: here

There's even an image in the video with the corresponding algebraic expression, even though in the image it's using a <= (less or equal) sign, instead of a < (less than) sign.

Could someone proof how this makes sense?

P.S. Sorry I don't include the expression in this question, I don't know how to.

This is from the comment of André Nicolas, which I will just attempt to clarify.
Assume $x,y\ge0$.
We want $${[\sqrt x + \sqrt y]\over 2 } \le \sqrt{(x+y)\over 2}$$

We proceed with inequalities equivalent to the first and to each other:

$${[\sqrt x + \sqrt y]\over 2 } \le {\sqrt{(x+y)}\over \sqrt 2}$$
$${\sqrt x + \sqrt y } \le {2\sqrt{(x+y)}\over \sqrt 2}$$
$${\sqrt x + \sqrt y } \le {\sqrt2\sqrt{(x+y)}}$$
squaring
$$x+2\sqrt{xy}+y\le 2x+2y$$
$$x+y-2\sqrt{xy}\ge0$$
$${{[\sqrt x - \sqrt y]}}^2\ge0$$ which we know to be true. Thus, its equivalent inequality, the one we are trying to prove is also true.

As Bill says, the square root is concave so just apply Jensen’s (usually stated for convex, so flip the inequality in the usual statement) for the expectation in the discrete case.


How do I convince my students that the choice of variable of integration is irrelevant? I will be TA this semester for the second course on Calculus, which contains the definite integral.

I have thought this since the time I took this course, so how do I convince my students that for a definite integral

$$\int_a^b f(x)\ dx=\int_a^b f(z)\ dz=\int_a^b f(☺)\ d☺$$

i.e. The choice of variable of integration is irrelevant?

I still do not have an answer to this question, so I would really hope someone would guide me along, or share your thoughts. (through comments of course)

NEW EDIT: I've found a relevant example from before, that will probably confuse most new students. And also give new insights to this question.

Example: If $f$ is continuous, prove that

$$\int_0^{\pi/2}f(\cos x)\ dx = \int_0^{\pi/2}f(\sin x)\ dx$$

And so I start proving...

Note that $\cos x=\sin(\frac{\pi}{2} -x)$ and that $f$ is continuous, the integral is well-defined and

$$\int_0^{\pi/2}f(\cos x)\ dx=\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx $$

Applying the substitution $u=\frac{\pi}{2} -x$, we obtain $dx =-du$ and hence

$$\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx=-\int_{\pi/2}^{0}f(\sin u)\ du=\int_0^{\pi/2}f(\sin u)\ du\color{red}{=\int_0^{\pi/2}f(\sin x)\ dx}$$

Where the red part is the replacement of the dummy variable. So now, students, or even some of my peers will ask: $u$ is now dependent on $x$, what now? Why is the replacement still valid?

For me, I guess I will still answer according to the best answer here (by Harald), but I would love to hear more comments about this.

Draw a graph of the function on the blackboard, showing $a$ and $b$ and a crosshatched area representing the integral. Put an $x$ on the horizontal axis. Erase the $x$ and put a $z$ there. Does that change the area? Erase the $z$ and put a smiley face there. Does the area change? Why/why not?

I think something to think of here is the fact that in teaching calculus (especially the definite integral; which from my experience is a second semester calculus class) the students should already be aware of the fact that x is a variable and the choice of letter (or symbol) is completely arbitrary. I would be more worried about the students that can't see this and would see it as them being in the wrong class. If they don't understand the fundamental basis of a variable they probably shouldn't be in a calculus class.

I would venture to say that this shouldn't be/isn't a problem for a university level calculus class.

If you truly are worried about this, I would suggest not focusing on the topic but instead showing it through examples; continually substitute in whimsical pictographs in place of the constant x/y/z. It would be entertaining to the students and would drive the point home without you actually having to focus on it.


$(\lambda-a)^{-1}$ as limits of 'polynomials' For a unital $C^*$-algebra $\mathcal{A}$ the spectral permanence gives \begin{equation}
\sigma_{\mathcal{B}}(a)=\sigma_{\mathcal{A}}(a)
\end{equation} for any unital $C^*$-subalgebra $\mathcal{B}$.

It is natural to look at the smallest such subalgebras, namely, the $C^*$-subalgebra generated by $1,a$ and $a^*$. Then the permanence says if $\lambda-a$ is invertible, then $\lambda-a$ is in the closed linear span of products of $1,a$ and $a^*$ (although order of multiplications matters here and it is not actually a polynomial).

I am wondering whether there is some canonical way to construct these 'polynomials'. That is, given $a\in\mathcal{A}$ invertible, how can one find explicitly the linear span of products of $1,a$ and $a^*$ that converges to $a^{-1}$?

Thanks!

My feeling (not formally justified) is that that there is no canonical choice. Take a look at the simplest example: let
$$
a=\begin{bmatrix}2&0\\0&3\end{bmatrix},
$$
$\lambda=1$.
So $a$ is selfadjoint, and of course
$$
(a-\lambda)^{-1}=\begin{bmatrix}1&0\\0&1/2\end{bmatrix}=p(a-\lambda)
$$
for an appropriate polynomial.
Now, what is the canonical polynomial that takes $1,2$ to $1,1/2$? Let us assume we want the minimum degree possible (i.e. $2$); this is already arbitrary. You could ask that the polynomial be monic, and in this case
$$
p(t)=t^2-\frac72\,t+\frac72;
$$
or you could want $p(0)=0$, in which case
$$
p(t)=-\frac34\,t^2+\frac74\,t.
$$
I don't really see a reason that makes one more canonical than the other.

To answer the question in your title:

If $\lambda -a$ is invertible you can write its inverse as a power series in $a$:

$$\frac{1}{\lambda - a} = \frac{1}{\lambda(1 - \frac{a}{\lambda})} = \sum_{n = 0}^\infty \frac{a^n}{\lambda^{n+1}}.$$


Balls of 3 colours in a bag - Probability There are $12$ balls in a bag. $3$ of them are red, $4$ of them are green, and $5$ of them are blue. We randomly take out $3$ balls from the bag at the same time. What is the probability that all three balls are of the same colour?

My answer: $(3/12)^3 + (4/12)^3 + (5/12)^3$. Is this correct?

EDIT: My explanation is that since there are the odds of getting a red ball are 3/12, I simply multiplied it 3 times since I consider multiplication to be the "and" operator. Similarly, I consider addition to be the "or" operator. So, the way I think of it is:

(1 red and 1 red and 1 red) or (1 green and 1 green and 1 green) or (1 blue and 1 blue and 1 blue).

There are  ${5\choose{3}} + {4 \choose 3} + {3 \choose 3}$ total ways of having a success.  There are a total of $12 \choose 3$ different ways of choosing 3 balls.

So the answer is $$\dfrac{{5\choose{3}} + {4 \choose 3} + {3 \choose 3}}{12 \choose 3} = \dfrac{3}{44} $$

Combination ($\binom n r$) can only be used when we can differentiate the balls even when they have the same colour, but if they are all of a particular colour they cannot be distinguished and hence the number of ways to pick up $r$ balls from a set of $n$ balls of a particular colour is only $1$ and not $binom n r$, and the probability  is $\dfrac r n$.


Find all the functions $f: \mathbb{N} \to \mathbb{N} $ such that $(m+f(n))(n+f(m))$ is a perfect square for all $m,n$ Let $N$ be the set of natural number. Find all functions $f: \mathbb N \to \mathbb N$ , such that the number $(m+f(n))(n+f(m))$ is perfect square for all natural numbers $m$ and $n$.

I was unable to solve this problem so I need a solution to it, I know that the condition that $(m+f(n))(n+f(m))$ is a perfect square would imply that if $n = m$ then $f(n) = f(m)$ but how would that help me in solving the problem?

When we have two numbers $a, b$ whose product is a perfect square, we can trace two observations on them:
Obs. 1: If $a$ is a square then $b$ is a square too.
Obs. 2: If $p$ is a prime and $p^k||a$ (short for "$p^k|a$ but $p^{k+1}\not|a$") and $p^l||b$, then $k+l$ is even; in particular, if $k$ is odd, then $l$ is odd (and it implies an obvious but useful thing: odd numbers can't be zero).

Obviously, constant functions aren't solutions for our problem. And by simple verification $f(x)=x+C$ with C natural works. We want to show there are no others.
Lemma 1: for every prime $p$ and integers $a,b$  such that $p|a-b$, there exists a constant $k$ such that there are odd numbers $\alpha, \beta$ such that $p^\alpha||a+k$ and $p^\beta||b+k$.
Proof:
Let $z$ the smallest natural number such that $p|a+z$. Obviously, $p|b+z$ too. Now some case-by-case analysis:

$p^2|a+z$ and $p^2|b+z$. That way, $a+z+p \equiv p \not\equiv 0 \pmod{p^2}$ but $a+z+p \equiv 0 \pmod{p}$, and we can choose $k=z+p$, and $\alpha=1,\beta=1$.

$p^2 \not |a+z$. More cases!

$p^2 \not| b+z$. Well, just add $p^3$ to both! It will give us $\alpha=1,\beta=1$.
$p^3 \not| b+z$. Now we can choose $k=C \cdot p^3-b-z$ with $C$ a very large non-multiple of $p$ (in order to $k>0$). That way, we have $a+z+k=Cp^3+a+z$ is multiple of $p$ but not $p^2$, and $b+z+k=Cp^3$ multiple of $p^3$ but not $p^4$. That way we can take $\alpha=1, \beta=3$.
$p^4 \not| b+z$. Just add $p^5$, it will give us $\alpha=1, \beta=3$.
$p^4 | b+z$. Just add $p^3$, it will give us $\alpha=1, \beta=3$.

Now we need to show that $f(n+1)-f(n)=1$ for any $n$.
If there was a prime $p$ such that $p|f(n+1)-f(n)$, then by the Lemma abiove we can choose $k$ such that $p^\alpha||k+f(n+1)$ and $p^\beta||k+f(n)$ with $\alpha, \beta$ odd.
But $(k+f(n+1)) \cdot (f(k)+n+1)$ is a perfect square. By the Obs. 2, $p|(f(k)+n+1)$. The same way, $p|(f(k)+n)$ too, and then $p|1$, contradiction.
Then $p$ does not exist, and so $f(n+1)=f(n)+1$ or $f(n+1)=f(n)-1$
If there was an $N$ such that $f(N+1)=f(N)-1$, then $(N+f(N+1)) \cdot (N+1+f(N)) = (N+f(N)-1) \cdot (N+f(N)+1) = (N+f(N))^2-1$ would be a perfect square, and it can occur if and only if $N+f(N)=0$, or $f(N)=0, N=0$.
Then we can split it in two cases:

$f(0) = 0$. Then $f(N+1) = f(N)-1$ only if $N=0$. That way, $f(1)=0-1<0$, contradiction.

$f(0) \not = 0$. Then $f(N+1) \not= f(N)-1$ for every $N$, and then $f(N+1) = f(N)+1$ for every $N$.

Problem solved! By naive induction, $f(n)=n+C$ with $C=f(1)-1$.

There are only two ways that $(n+f(m))(m+f(n))$ can be a perfect square.

$(n+f(m))$, $(m+f(n))$ are perfect squares for every natural number $n,m$.
$(n+f(m))=A=(m+f(n))$ in which case $(n+f(n))(m+f(n))=A^2$

The first way cannot happen, because if $(n+f(m))$ is a perfect square $$((n+1)+f(m))=(n+f(m))+1$$ isn't. You cannot add one to a perfect square and get another perfect square here is a list to convince you:

$$1,4,9,16,25,36,49,...$$

So the only way $(n+f(m))(m+f(n))$ is a perfect number is if$$n+f(m)=m+f(n)$$  Set $m=n-1$ then we get

$$n+f(n-1)=(n-1)+f(n)$$

solving for $f(n)$ gives us
$$f(n)=f(n-1)+1$$
Notice that

$$f(n)=f(n-1)+1=f(n-2)+2=...=f(1)+(n-1)$$

What this means is given any natural number $c$. $$f(n)=c+(n-1)$$ is a solution.
Here is a check:

$$(m+f(n))(n+f(m))=(m+[c+(n-1)])(n+[c+(m-1)])=(m+n+c-1)^2$$


Why do we reverse inequality sign when dividing by negative number? We all learned in our early years that when dividing both sides by a negative number, we reverse the inequality sign.
Take
$-3x < 9$
To solve for $x$, we divide both sides by $-3$ and get
$$x > -3.$$
Why is the reversal of inequality? What is going in terms of number line that will help me understand the concept better?

Dividing by a negative number is the same as dividing by a positive number and then multiplying by $-1$. Dividing an inequality by a positive number retains the same inequality. But, multiplying by $-1$ is the same as switching the signs of the numbers on both sides of the inequality, which reverses the inequality:
$$
\tag{1} a\lt b\quad\iff -a\gt -b.
$$
You should be able to convince yourself why the above is true by looking at the number line and considering the various cases involved.

Seeing why (1) is true is not too hard.

Here is the hand waving approach I suggested above:

Consider, for example, in (1), the case when $a$ is negative and $b$ is positive.
We have $a<b$. Then $-a$ is positive and $-b$ is negative. Thus,  we have $-b<-a$.

As another case, suppose $a$ and $b$ are both negative with $a<b$.  Switching the signs here
makes the resulting numbers both positive with $-a>-b$ (you can see this by drawing the points on the number line and noting that with the given conditions, $b$ is closer to the origin than $a$):

).

The other cases can be handled similarly.

But, perhaps a bit of rigor is needed here.

Recall that
$$a<b\quad\text{ if and only if }\quad  b-a\quad \text{ is positive.}$$
Now, $b-a$ is positive if and only if $(-a)-(-b) =-a+b=b-a$ is positive. So
 $a<b$ if and only if  $-a> -b$.

It's because negative numbers appear bigger but are actually smaller. Take this example: Which is bigger, $-7$ or $-9999$? First instinct, and absolute value tells us that the latter is the larger, but as per human definition of negatives, $-7$ is to the right of $-9999$ on the number line, therefore making it larger.

Consider the following:
$$7<9999$$
$$7(-1)<9999(-1)$$
$$-7 < -9999$$
Wait, the last statement is not true! We need to flip the sign to make it true. $-7$ appears smaller than $-9999$ but it really isn't. For negative numbers, things are flip-flopped.

NOTE: Multiplying by a negative is multiplying by a positive (signs still same) then multiplying by -1 (Signs now flipped). Dividing is multiplying by a fraction, and dividing by a negative number is multiplying by a negative fraction. Also, if you think about it, this also holds when you reciprocate both sides.


Proof of a limit for a recursively-defined sequence I have a sequence defined by
$$
a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2.
$$
Now, if $\lim_{n\to \infty}a_n=g$ then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}$ so $g=\sqrt{2}$ or $g=-\sqrt{2}$ but $a_n>0$ so $g=\sqrt{2}$.

Now, how do I prove that it has an actual  limit? Also, it can be proven that $1\le a_n\le2$, and it's not monotonic because $a_4 \gt a_5 \lt a_6$.
Also, it's not monotonic after some $N\in\mathbb N$

As you mention, it is straightforward to prove by induction that $1 \le a_n \le 2$, for all $n \ge 1$.

Let $L = \limsup_{n\to\infty} a_n$, $l = \liminf_{n\to\infty} a_n$, we have $1 \le l, L\le 2$. For any $\varepsilon>0$, we know that $l - \varepsilon \le a_n \le L + \varepsilon$, if $n$ is sufficiently large. Thus, for $n$ large enough
$$ \frac{2}{L+\varepsilon} \le a_n \le \frac{2}{l-\varepsilon}. $$
Since $\varepsilon$ can be chosen to be arbitrarily small, we get
$$ L = \limsup_{n\to\infty} a_n \le \frac{2}{l}, \,\, l = \liminf_{n\to\infty} a_n \ge \frac{2}{L},$$
hence $L \cdot l =2$.

Now, we may choose a subsequence $\{n_k\}$ such that, as $k\to\infty$,
$$ a_{n_k + 1} \to L,\, a_{n_k} \to l_1,\, a_{n_k - 1} \to l_2,\, a_{n_k - 2} \to l_3, $$
for some $l_1, l_2, l_3 \in [l,L]$. By the definition of the sequence $a_n$, we have
$$ \frac{2}{l} = L = \frac{1}{l_1} + \frac{1}{l_2},\quad l_1 = \frac{1}{l_2} + \frac{1}{l_3}. $$
The left equality implies $l_1 = l_2 = l$, and then, using $\frac{2}{L} = l = l_1$, the right equality gives $l_2 = l_3 = L.$ Therefore, $l = L$, and since $L \cdot l = 2$, we find that the limit is $\sqrt{2}$.

To be able to do $$g=\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}=\frac{2}{g}$$ you should prove first that the sequence is bounded and some how it gets arbitrarily close to the candidate for limit, and therefore convergent. For example, let $c_n$ be a sequence of positive real numbers, if for all $n\in\mathbb{N}$ $c_n\leq M$ for some $M\geq 0$ and for all $N\in\mathbb{N}$ exists $n_0\in\mathbb{N}$ such that $c_n\geq c_{N}$ for all $n\geq n_0$, then $c_n$ is convergent, not necessarily to $M$. Only then you're able to manipulate limits as you wish to find the actual limit.

Otherwise you may end up with stuff like this: Take $b_1=1$ and $b_n=1+b_{n-1}^{2}$, if $\lim_{n\to\infty}b_n=g$ then $$g=\lim_{n\to\infty}b_n=\lim_{n\to\infty}1+b_{n-1}^{2}=1+g^{2}$$ so $g=1\pm i\sqrt{3}$ which clearly doesn't make sense.


The two-daughter-problem When hearing about the two-daughter problem, I first thought it to be quite clear (after, of course, at first falling into the trap like many of us), but on the second glance, I encountered some serious problems with my understanding.

The original problem seems to be quite easy: Assume that the only thing you know about a man with two kids is that at least one of the kids is a daughter. What is the probability that the other kid is a daughter as well? (Boys and girls are assumed to be born equally often.)

After the first impulse ("1/2 of course!"), it becomes clear that it is only 1/3. The problem can be mapped to a situation where from the multitude of families with two children, only those with M/M are ruled out, while the equally often cases F/F, F/M and M/F remain, making F/F only one third of all remaining cases.

But now, meet Mr. Smith. I don't know much about him (except that he has two children), but when he approached me, he told me: "I am so happy! Victoria just got the scholarship she wanted!"

Now what is the probability that Victoria has a sister?

Since I only know that Mr. Smith has two children, and one is obviously a girl, I am tempted to map this onto the two-daughter-problem, leading to the answer "1/3".

But wait! What if I ask Mr. Smith first, if Victoria is his elder daughter? Assume his answer is yes (and ignore any problems with twins - even then one is typically a few seconds "older" than the other). So now I know that from the cases (F/F, F/M, M/F), M/F also drops out. And now, the probability for F/F just rose to 1/2.

Okay, but what if his answer is no? Then Victoria is the younger one, and F/M drops out. Again, the probability rises to 1/2.

So I'm going to just ask him: "Well, Mr. Smith, is Victoria your elder daughter? Wait - don't answer, because whatever you may answer, it doesn't matter. The probability just rose from 1/3 to 1/2."

Or, even better, I do not even have to ask him, just thinking about the question will shift probabilities to 1/2, which means that the original probability for Victoria having a sister must already have been 1/2. But then the mapping to the two-daughter-problem is obviously false.

Where is my error?

Making things worse, I could also create a setup where Mr. Smith just tells me: "I have two kids, and at least one of them is a girl." I then ask him: "Oh, can you give me a name of a daughter of yours?" and he answers: "Sure. Victoria."

(Side note: I have a gut feeling that this has something to do with how to assume probability distributions behind situations, similar to the Two envelopes problem, but I can't figure this out completely yet.)

-------- UPDATE --------

It seems that my error is that the question "Is Victoria the older child?" does not change the probabilities. If I know for sure that Mr. Smith was picked from an equally distributed (M/F, F/M, F/F) sample, then the knowledge that Victoria is the older child does not change anything, as was pointed out  here, and the probability for her having a sister is 1/3.

But it is very interesting that solely from the sentence "Victoria just got the scholarship she wanted!" I can NOT infer that Mr. Smith is indeed chosen from this uniform distribution.

Imagine that all kids have the same chance to get a scholarship, and the happy father will tell us if it is the case. Then it is actually twice as probable that Mr. Smith will tell us about his daughter's success if he has two girls, so the weighting of the four possibilities (M/M, F/M, M/F, F/F) is (0, 1, 1, 2). And in this case, the probability of Victoria having a sister is 1/2.

So another problem in my reasoning is the mapping of Mr. Smith's statement to the two-daughter-problem. Simply put, without knowing more about the circumstances that led to Mr. Smith telling me about Victoria, I simply can't say if the probability is 1/3 or 1/2.

Now I've got a headache...

I think the confusion arises because the classical boy-girl problem is ambiguous:

'You know that Mr.Smith has two kids, one of which is a girl. What is the chance she has a sister?'

The ambiguity here is that from this description, it is not clear how we came to know that 'Mr.Smith has two kids, one of which is a daughter.'

Consider the following two scenarios:

Scenario 1:

You have never met Mr. Smith before, but one day you run into him in the store. He has a little girl with him, which he tells you is one of his two children.

Scenario 2:

You are a TV producer, and you decide to do a show on 'what is it like to raise a daughter?' and you put out a call for such parents to come on the show. Mr.Smith agrees to come on the show, and as you get talking he tells you that he has two children.

Now notice: the original description applies to both cases. That is, in both cases it is true that you know that 'Mr.Smith has two children, one of which is a daughter'.

However, in scenario 1, the chance of Mr. Smith having two daughters is $\frac{1}{2}$, but in scenario 2 it is $\frac{1}{3}$. The difference is that in the first scenario one  specific child has been identified as female (and thus the chance of having two daughters amounts to her sibling being female, which is $\frac{1}{2}$), while  in the second  scenario no specific child is identified, so we can't talk about 'her sibling' anymore, and instead have to consider a conditional probability which turns out to be $\frac{1}{3}$.

Now, your original scenario, where you don't know anything about Mr. Smith other than that he has two children, and then Mr.Smith says 'I am so happy Victoria got a scholarship!' is like scenario 1, not scenario 2. That is, unless Mr. smith has two daughters called Victoria (which is possible, but extremely unlikely, and if he did one would have expected him to say something like 'my older Victoria'), with his statement Mr.Smith has singled out 1 of his two children, making it equivalent to scenario 1.

Indeed, I would bet that most real life cases where at some point it is true that 'you know of some parent to have two children, one of which is a girl' are logically isomorph to scenario 1, not scenario 2. That is, the classic two-girl problem is fun and all, but most of the time the description of the problem is ambiguous from the start, and if you are careful to phrase it in a way so that the answer is $\frac{1}{3}$, you will realize how uncommon it is for that kind of scenario to occur in real life. (Indeed, notice how I had to work pretty hard to come up with a real life scenario that is at least somewhat plausible).

Finally, all the  variations of whether Victoria is the oldest, youngest, or whether you don't even know her name ('Mr. Smith tells you one his children got a scholarship to the All Girls Academy') do not change any of the probabilities (as you argued correctly): in most real life scenarios, the way you come to know that 'Mr.Smith has two children, one of which is a girl' (and I would say that includes your original scenario) means that the chance of the other child being a girl is $\frac{1}{2}$, not $\frac{1}{3}$.

So, when at the end of you original post you ask "where is my error?" I would reply: your 'error' is that you assumed that the correct answer should be $\frac{1}{3}$, and that since your argument implied that is would be $\frac{1}{2}$, you concluded that there must have been an error in your reasoning. But, as it turns out, there wasn't! For your scenario, the answer is indeed $\frac{1}{2}$, and not $\frac{1}{3}$. So your 'error' was to think that you had made an error!

Put a different way: you were temporarily blinded by the pure math ( and I say 'temporarily', because you ended up asking all the right citical questions, and later realized that the classic two-girl problem is ambiguous: good job!). But what I mean is: we have seen this two-girl problem so often, and we have been told that the solution is $\frac{1}{3}$ so many times, that you immediately assume that also in your descibed scenario that is the correct answer... When in fact that is not case because the initial assumptions are different: the classic problem assumes a Type 2 scenario, but the original scenario described in your post is a Type 1 scenario.

It's just like the Monty Hall problem ... We have seen it so often, that as soon as it 'smells' like the Monty Hall problem, we say 'switch!' ... when in fact there are all kinds of subtle variants in which switching is not any better, and sometimes even worse!

Also take a look at the Monkey Business Illusion: we have see that video of the gorilla appearing in the middle of people passing a basketball so many times that we can now surprise people on the basis of that!

For me, probability theory is not applicable here in the "pure form": you cannot ask about probabilities of actual facts of life.

What is the probability that Napoleon had blue eyes according to what you know? There is no "probability" here: he either had blue eyes, or he hadn't, regardless of what you know.  What is the probability I am typing this from my office on Saturday evening?  What is the probability that Higgs boson exists?  What is the probability that your wife/husband is cheating on you?

The problem would make more sense in the following form (for example): suppose you know that someone has had a daughter, what is the probability that their next child, if there will be one, will be a daughter as well?

You might argue that we can view the problem practically, as if someone tells you that someone else has two children, of which one is a daughter, and asks you to place a bet on the sex of the other child. But in this settings surely you should take into account the motives of the person who asks you this, because probably they wouldn't be asking if there were not good chance that you would lose your bet.

More formally: the probability space in such questions about past events or actual facts of life is not well defined, and most definitions would be practically irrelevant.  For past events and actual facts of life one uses statistics, not probability theory.

Update.

After much criticism my point of view and interpretation of the problem have received, let me mention that I am also aware of the following simple probabilistic interpretation.

Consider the totality $X$ of all people with 2 children of which at least one is a daughter.  This is about 3/4 of the totality of all people with 2 children.  Now choose a person in $X$ at random.  Then the probability that both their children are girls is approximately 1/3.


Limit of a function with square roots I've got the following limit to solve:

$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$

I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?

Try putting $t=\sqrt s$ to get $$\frac {t-t^4}{1-t}=\frac {t(1-t^3)}{1-t}=t(1+t+t^2)$$You can notice this without the substitution, of course, but sometimes a substitution like this helps to clarify what is going on.


Proof about linear functionals without reference to a basis. Recall the following: Let $V$ be a vector space, and let $v$ be in $V$.  If $\lambda(v) = 0$ for all $\lambda \in V^*$, then $v = 0$.

Does anyone know of a proof that makes no reference to a basis?  And if not, is there reasonable intuition behind why a choice of basis is necessary in proving the result?

Thanks in advance.

It is known that the axiom of choice (AC) is equivalent to the statement that all vector spaces have a basis, and that these bases cannot be "written down" in general. If $V$ has a basis, then your statement, which says that the canonical map $e : V \to V^{**}$ is injective, is clearly true; but actually something weaker will suffice, namely that all $1$-dimensional subspaces are direct summands. This statement follows from AC, but is probably weaker than AC, and cannot be proven in ZF. That is, there are models in ZF where there a vector spaces $V$ such that $e : V \to V^{**}$ is not injective.

My intuition is that infinite-dimensional vector spaces are no 'reasonable' spaces to study because their properties often depend on set theory. Notice that $V \to V^{**}$ is an isomorphism for finitely generated vector spaces $V$, and that you can prove this in ZF. Also, Banach spaces and Hilbert spaces are often far more well-behaved than infinite-dimensional vector spaces, basically because we can make sense of infinite sums.

By the Riesz Representation Theorem: corresponding to each $\lambda \in V^*$ is an element $u \in V^{**} \supseteq V$ such that for any $x \in V$ one has $\langle x,u\rangle = \lambda(x)$.

So, we choose $\lambda_v \in V^*$ such that for any $x \in V$ one has $\lambda_v(x) = \langle x,v\rangle$. By the statement in your question, it follows that

$$0 = \lambda_v(v) = \langle v,v \rangle = \|v\|^2$$
implying that $v \equiv 0$ by norm positivity.

(For a proof of the Riesz Representation Theorem, please see Folland's Real Analysis pg.212 or 223 for a version which does not rely on a basis for the space)


To find the number of Elements of order 4 in Galois Field $GF(p^n)^*$, If we consider $p$ be a prime number such that $p \pmod4 =3.$ How many elements of order $4$ are in $GF(p^n)^*?$
If modulo were $p \pmod 4 = 1.$ Then I could easily solve the question. Since $GF(p^n)^*$ is isomorphic to $Z_{p^{n}-1}$ which is cyclic. And $4$ divides $p-1$ and can easily show that $4$ divides $p^n-1.$
And using the Theorem: If $d$ is the divisor of $n$,  then number of elements of order $d$ in a cyclic group of order $n$ is $\phi(d)$.
So I can say that there are $2$ elements of order $4.$
But above question please provide your hint.

The multiplicative group of the field is cyclic of order $p^n-1$. So since $p\equiv 3 \mod 4$, if $n$ is odd then it contains no subgroups of order $4$ by Lagrange and if $n$ is even it contains one subgroup of order $4$ hence $2$ elements of order $4$.

It comes down to whether $4|p^n-1$.
Say $p=7$.  Then since $4|p^2-1$, and the multiplicative group of a finite field is cyclic, we get $\varphi(4)=2$ for $\Bbb F_{49}$.
But for $\Bbb F_7$ there are none, since $4\not|7-1=6$

Now, when $n=2k$ is even, we have $p^n-1=p^{2k}-1=(p^k+1)(p^k-1)$, which is divisible by $4$ as the product of two even numbers.
On the other hand, when $n=2k+1$ is odd, we have $p^n=(4l+3)^n\cong (4l+3)^{2k}(4l+3)\cong 4l+3\cong3\bmod4$, by Euler's theorem.  So $p^n-1\cong2\bmod4$. Thus $4\not|p^n-1$.


How many functions (Counting problem) Let $A$ be a set of size $25$ and let $B$ be a set of size $17$. How many functions $f\colon A\to B$ are there?

$17!$
$\frac{25!}{17!}$
$17^{25}$
$25^{17}$

I think the answer is 3. Is it?

A set is a group of unique elements (in opposed to a multi-set).

A function $f : A \rightarrow B$ maps every element in the domain $A$ to a single element in the range $B$.

You have $25$ elements in the domain $A$ and $17$ elements in the range $B$.

Each element in the domain $A$ must be mapped to an element in the range $B$:

The 1st element in the domain $A$ has $17$ mapping options in the range $B$
The 2nd element in the domain $A$ has $17$ mapping options in the range $B$
The 3rd element in the domain $A$ has $17$ mapping options in the range $B$
...
The 25th element in the domain $A$ has $17$ mapping options in the range $B$

So the total number of different mappings (i.e., functions) from $A$ to $B$ is: $$17 \times 17 \times 17 \times \dots \times 17 = 17^{25}$$

Yes: (cardinality of codomain) to the power of (cardinality of domain) is right.


How can I integrate this rational trigonometric function? $$\int^\pi_0 \frac{(\sin x+\cos x)^2}{(1+\sin2x)^{1/2}}\operatorname d x$$
I have managed to simplify this expression to:
$$\int_0^\pi\frac{(\sin x+\cos x)^2}{|\sin x+\cos x|}\operatorname d x\text.$$
Please help me move forward from here onwards using the limits.

You can notice that $(\sin x+\cos x)^2=1+\sin2x$. Now substitute $2x=\pi/2-2t$, so your integral becomes
$$
\int_0^\pi\sqrt{1+\sin2x}\,dx=-\int_{\pi/4}^{-3\pi/4}\sqrt{2}\lvert\cos t\rvert\,dt
$$

$HINT$:
We Know That,
$\int^\pi_0 \frac{(\sin x+\cos x)^2}{(1+\sin2x)^{1/2}}$ = $\int_0^\pi\frac{(\sin x+\cos x)^2}{|\sin x+\cos x|}.$
From o to $3pi/4$ denominator is positive and from $3pi/4$ to $pi$ it is negative,
Then you can proceed through simple calculations,


Why solutions of $y''+(w^2+b(t))y=0$ behave like solutions of $y''+w^2y=0$ at infinity Assume $w>0$ and $b(t)$ be continuous on $[0,+\infty)$ and $\int_0^\infty |b(t)| dt <\infty$
show that $y''+(w^2+b(t))y=0$ has solution $\phi(t)$ such that $$\lim_{t\to\infty} [(\phi(t)-\sin(wt))^2+(\phi'(t)-w\cos(wt))^2]=0$$

thanks

Here is an outline of an approach that almost gets there (with some pieces left to be filled in). I will show that we can find solutions $\phi(t)$ that are arbitrarily close to $\sin wt$ for all sufficiently large $t$ under the assumption that $\lim_{t\to\infty}b(t) = 0$.

Write the solution to the ODE as $\phi(t) = \sin w t + \delta y$ and impose the initial conditions $\delta y(t_*) = \delta y'(t_*) = 0$ for some $t_*$ to be fixed later on. Then

$$\delta y'' + (w^2+b)\delta y = -b\sin w t$$

This system is equivalent to a ball rolling without friction attached to a spring with varying spring-constant and we have some driving force $b(t)\sin wt$.

This physics-analogy motivates us to try to find an 'energy equation' for this system. Multiply by $\delta y'$ to find

$$\frac{d}{dt}(\delta y'^2 + w^2\delta y^2) = -b(t)\left(2\sin w t\frac{d\delta y}{dt} + \frac{d\delta y^2}{dt}\right)$$

Integrating over $[t_*,t]$ yields

$$(\delta y'^2 + w^2\delta y^2) \leq \max_{t'\in [t_*,t]}|b(t')|\left(2|\delta y| + \delta y^2\right)$$

Now since $y$ is bounded (I will just assume this here, see the other answer for the proof. Physically this is because the energy is bounded.) and $b\to 0$ (Edit: $b\to 0$ does not follow from continuity, but I will need to assume this here) we get that for any $\epsilon > 0$ there exist a $t_*$ s.t. for $t>t_*$ we have

$$\delta y'^2 + w^2\delta y^2 < \epsilon$$

and since $w$ is a constant it also follows that we can find a $t_*$ s.t.

$$\delta y'^2 + \delta y^2 < \epsilon$$

for all $t>t_*$. Writing out this equation we get

$$(\phi(t)-\sin wt)^2 + (\phi'(t)-w\cos wt)^2 < \epsilon$$

for $t>t_*$. Thus there exists solutions that are arbitrarely close to $y=\sin w t$ for all sufficiently large $t$, however I have not yet found a way to show that we can find one particular solution where the difference goes to zero at infinity.

i think you need integral representation of the solution $y$ and an application of the gronwall's inequality.

rewrite the $y^{''} + \omega^2 y = -by$ a particular solution is $$y(t) = {1 \over \omega} \cos (\omega t)\int_0^t \sin( \omega s) b(s) y(s) ds- \sin(\omega t) \int_0^t \cos(\omega s) b(s) y(s) ds$$

taking the absolute value $|y(t)| \le K\int_0^t |b(s)||y(s)| ds.$  here i am not too sure but the gronwall inequality now should give $|y| < |y(0)|exp(\int_0^t |b(s)|ds$
which is bounded above by $$|y(t)| \le |y(0)|exp\left(\int_0^\infty |b(t)|dt\right).$$


If $(x - \frac {1}{x})^2 = 2$ and $x > 0$ what is $x^4 - \frac{1}{x^4}$? If $(x - \frac {1}{x})^2 = 2$ and $x > 0$ what is $x^4 - \frac{1}{x^4}$?

I tried factorising the $x^4 - \frac{1}{x^4}$ expression and I got $(x^2 - \frac {1}{x^2})(x - \frac {1}{x})(x + \frac {1}{x})$ but I'm not sure what to do from here.

Thanks!

Note that by expanding we have $2 = \left(x - \frac{1}{x} \right)^2 = x^2 - 2\frac{x}{x} + \frac{1}{x^2} = x^2 + \frac{1}{x^2} - 2$. So we have $x^2 + \frac{1}{x^2} = 4$, and squaring this again gives $x^4 + 2 + \frac{1}{x^4}=16$. So our answer is $14$.


Can an undirected graph cycle have only two vertices? I've always seen cycles in graphs described as containing three or more vertices.  I had a question posed to me today that I now pose to you: Is it valid to consider a pair of vertices connected by a single edge in an undirected graph to be a cycle?

More concretely, given the following undirected graph:

a---b
 \ /
  c
  |
  d

Obviously, (c, b, a, c) is a cycle.  Is (c, d, c)?  Why or why not?

once this doubt occurred to me. It seams trivial, but when I am on doubt I always look for strong definitions, so this is what I found.

I will cite the definition of cycle in the book of Diestel, Graph Theory [http://www.flooved.com/reader/3447?no-redirect#16]

Page 8: "If $P=x_0 ... x_{k-1}$ is a path and $k \geq 3$, then the graph $C:=P+x_{k-1}x_0$
is called a cycle."

Thus $cdc$ is not a cycle because $cd$ is not a path with more than two vertices.

But what about $cdcdc$? It would be a path if $cdcd$ would. However, according Diestel definition of path (page 6) all vertices of a path are different.


Find all $3 \times 3$ magic square matrices $M$ such that $M^2$ is also magic A magic square matrix $M$ is a square matrix with real entries such that the sum of the entries in each column, each row, and each main diagonal is the same.
The problem is to characterize all $3 \times 3$ magic square matrices $M$ such that $M^2$ is also a magic square matrix.
Supposedly, there is an elegant way to do this besides writing out variables and bashing out the multiplication, but I haven't yet found such a solution.

First we look at the properties of the matrix and its eigenvalues. Now since the matrix $M$ has rows that add up to the same number, lets say $s$, then we know that $$\begin{pmatrix} 1\\1\\1 \end{pmatrix}$$
is an eigenvector with eigenvalue $s$.

If we take the transpose of the matrix $M^T$ we can then apply the same principle and find that $s$ is a repeated eigenvalue.

Now for the third eigenvalue we notice that the sum of the trace is equal to the sum of the eigenvalues. However we know the sum of the trace to be $s$ so doing this would mean that we get the thrid eigenvalue to be $-s$.

Now by eigenvalue decomposition we know that $M^2 = S\Lambda^2S^{-1}$ and if it were magic its sums, $s_2$ its eigenvalues would be $s_2$ repeated twice and $-s_2$. However, notice how $\Lambda^2$ will have only positive entries. Therefore there does not exist a magic square $M$ such that $M^2$ is also magic.