arithmetic progression, finding the nth term. The sum of the 1st n terms, of an AP is $S_n=n^2-3n$. Write down the 4th term and find an expression for the $n$th term.

Will the 4th term be $t_4= a+3d$?

Let the arithmetic sequence be

$ a, a + d, a + 2d, \cdots $

The sum of the first $ n $ term is

$ a + (a + d) + \cdots + a + (n - 1)d $

Remember the sum is simply the (first term + the last term) * number of terms divide by 2, so we have

$ n^2 - 3n = S_n = \frac{(a + a + (n - 1)d)n}{2} = \frac{(2a - d)n + n^2d}{2} $

So match coefficients we can get $ \frac{d}{2} = 1, \frac{2a - d}{2} = -3 $

Now it is obvious that $ d = 2 $, $ a = -2 $.

The 4th term is $ a + 3d = 4 $, the nth term is $ a + (n - 1)d = 2n - 4 $.


Abstractly constructing splitting fields I have a series of exercises where I have to determine the degree of various splitting fields. I am freely using the following observation, which I feel is intuitively true, but I am asking here to make sure it's true.

  Let $f(x) \in \mathbb Q[x]$. Let $a_1,\dots,a_n$ be roots of $f$ in $\mathbb C$. Then the splitting field is $\mathbb Q(a_1,\dots,a_n)$.

Clearly $f$ splits over this extension, but I'm not sure how to "rigorously" check that this is indeed the smallest extension.

It's by definition. Recheck your definition of what $\mathbb{Q}(a)$ means! It's already the smallest such field.

The proposed minimal spliff (splitting field) is indeed minimal because it by definition is the minimal extension of the rationals which contains the roots.


Proof explanation of Every nonnegative function can be written as the increasing limit of a sequence of simple functions Consider the following Theorem (ii)

  In the part that says by passing from $f_k$
   to $f_{k+1}$ each subinterval is divided in half. It follows that $f_k\le f_{k+1}$.

I can see that indeed each subinterval is divided in half but I cannot see how that implies that $f_{k}$ is monotone increasing.

Could you explain please?

  In the part that states  wherever $f$ is finite, we have $$0\le f-f_k\le 2^{-k}$$

Why ?

If $f$ is finite then $f_k=\frac{j-1}{2^k}$

and where does $0\le f-f_k\le 2^{-k}$ come from?

Note this is not a duplicate with If $f$ is measurable function then there exist a sequence of $φ_n$ measurable functions such that $φ_n\to f$ pointwise because the presented proofs are different.

The proof is, in effect, doing a base-$2$ version of looking at the decimal expansions of the functions values. For example, suppose $f(x)=\pi=3.14159265\ldots$. Then we can approximate $f$ with a sequence of simple functions $g_k$ whose values never exceed $k$ and never have more than $k$ decimal digits:

$$\begin{align}
g_1(x)&=1.0\\
g_2(x)&=2.00\\
g_3(x)&=3.000\\
g_4(x)&=3.1415\\
g_5(x)&=3.14159\\
g_6(x)&=3.141592\\
&\,\,\vdots
\end{align}$$

Written in binary, we have $\pi=11.0010010000111111\ldots$, and so the proof's sequence increasing to the value $f(x)=\pi$ is

$$\begin{align}
f_1(x)&=1.0\\
f_2(x)&=10.00\\
f_3(x)&=11.000\\
f_4(x)&=11.0010\\
f_5(x)&=11.00100\\
f_6(x)&=11.001001\\
&\,\,\vdots
\end{align}$$

In general we have $f_k(x)\le f_{k+1}(x)$ because $f_{k+1}$ looks at more binary digits of the value $f(x)$, and, once $k\gt f(x)$, we have $0\le f(x)-f_k(x)\lt2^{-k}$ because $f_k(x)$ is looking at the first $k$ binary digits of $f(x)$.

Put differently, in step $k$, we partition $[0,\infty)$ as
$$\tag1 [0,\infty)=[0,2^{-k})\cup [2^{-k},2\cdot 2^{-k})\cup \ldots \cup [(j-1)2^{-k},j2^{-k})\cup \ldots \cup [k-2^{-k},k)\cup [k,\infty)$$
and let $f_k(x)$ be the lower end point of the unique interval on the right of $(1)$ containing the value $f(x)$. Now note that the partition for $k+1$ is a refinement of the partition for $k$. Hence, if $f(x)\in[a,b)\subseteq [c,d)$, where $[a,b)$ is an interval of step $k+1$ and $[c,d)$ an interval of step $k$, we have $f_{k+1}(x)=a\ge c=f_k(x)$.

Regarding the second part, the claim $f(x)-f_k(x)<2^{-k}$ does not hold as generally as written, but rather only eventually, namely, when $k\ge f(x)$. For then $f(x)$ is in one of the $k2^k$ first intervals in $(1)$, and as these are of length $2^{-k}$, the claim $0\le f(x)f_k(x)<2^{-k}$ follows. Of course, the merely eventual nature of the inequality does not disturb the ultimately desired claim about convergence.


If $(u_n)$ is bounded and $\lim u_n^2+u_n-u_{n+1}=0$ then $u_n \to 0$ Let $(u_n)$ be a real bounded sequence. Suppose that $\lim\limits_{n \to \infty} (u_n^2+u_n-u_{n+1})=0$. Prove that $u_n \to 0$.

I was able to develop a prove, looking at the map $x \mapsto x^2+x$ and proving that $0$ is the only possible limit point of $(u_n)$. But the proof has several cases... Not something very straightforward.

Do your have any idea of something simple?

Suppose $u_n$ did not go to zero. Then since the sequence is bounded, $L = \limsup_{n \rightarrow \infty} |u_n| > 0$. Let $\{n_k\}_{k = 1}^{\infty}$ be such that $\lim_{k \rightarrow \infty} u_{n_k} = L$ or $-L$. First we consider the $L$ case. Then one has
$$\lim_{k \rightarrow \infty} u_{n_k + 1}= \lim_{k \rightarrow \infty} (u_{n_k} +  u_{n_k}^2) + \lim_{k \rightarrow \infty}
(u_{n_k +1} - (u_{n_k} +  u_{n_k}^2))$$
$$= L + L^2 - 0$$
$$= L + L^2$$
Since $L + L^2 > L$, this contradicts that $L$ is the limsup of the absolute values.

Now suppose the subsequence has limit $-L$. This time we observe that
 $$-L = \lim_{k \rightarrow \infty} u_{n_k} = \lim_{k \rightarrow \infty} (u_{n_k - 1} +  u_{n_{k}-1}^2) + \lim_{k \rightarrow \infty}
(u_{n_{k}} - (u_{n_{k}-1} +  u_{n_{k}-1}^2))$$
$$= \lim_{k \rightarrow \infty} (u_{n_k-1} +  u_{n_k-1}^2)$$
Passing to a convergent subsequence $u_{n_k-1}$, if the limit of the subsequence is denoted by $M$ then by the above we have
$$M + M^2 = -L$$
Thus $M < -L$, contradicting that the negative number $L$ is the limsup of the absolute values of the sequence.

Thus regardless of which case we are in, we get a contradiction, so the limit must be zero.

suppose that $u_{n}$ doesn't converge to zero. So there is a positive number $r$ for which we have $\lvert u_{n} \rvert$ > r, for each n. Define $f(n)= u_{n}^2 + u_{n}- u_{n+1}$ and consider $f(n)+f(n+1)+...f(n+k)= u_{n}^2+u_{n+1}^2+...+u_{n+k}^2+u_{n}-u_{n+k+1}$. Since $u_{n}$ is bounded, there is a positive number M so that we have $\lvert u_{n} \rvert$ < M. Now we can see:
$\lvert f(n)+f(n+1)+...f(n+k) \rvert$ > $\lvert u_{n}^2+u_{n+1}^2+...+u_{n+k}^2 \rvert$ - $\lvert u_{n} \rvert$ - $\lvert u_{n+k+1} \rvert$ > $kr^2-2M$. Choose $k$ big enough so that $kr^2-2M=A>0$. On the other hand
lim $\lvert f(n)+f(n+1)+...f(n+k) \rvert=0$ according to the assumption when n goes to infinite which contradicts to $\lvert f(n)+f(n+1)+...f(n+k) \rvert$ >$A$ >$0$.
( I showed both converging and converging to zero at the same time )


Finding the area of a fraction of a circle defined by points on a grid. A grid is shown below, where the shortest distance between any two points is 1. Find the area of the circle that passes through points $X,$ $Y,$ and $Z.$

Can it be assumed that the arc created over each of the chord lengths shown will create a semicircle when put together? How can I determine what fraction of the circle passes through points X,Y, and Z to calculate the area?

Note that the distance $XY$ is $3$ units, where a unit is the length of the diagonal formed by the smallest square in the lattice, and $YZ$ is $4$ units. Does that strike anything?

As we know that a unique circle passes through three given points. Hence, if we assume the points $X, Y, Z$ as $(3, 3), (0, 0), (4, -4) $ the unique circle will be like this

Note: $X, Z$ are diametrically opposite points. Now, you can get answer.


How do we find the general term from an infinite nested radical I am struggling on how we definite infinite nested radical.

Consider $ {\sqrt  {2+{\sqrt  {2+{\sqrt  {2+{\sqrt  {2+\cdots }}}}}}}}$

From what I have researched, I know we can consider the above expression as the limit of a certain squence. I know how to find this value. However, I would like to find the general term of the corresponding sequence but I do not know where to start. I tried and error but failed.

Is there any formula/theorem that are specialised for finding the general terms of such thing?

Many thanks.

If you google for the value of trigonometric functions of angles $\theta_n=\frac \pi {2^n}$ (for $n=2,3,4,5$, you will find them in http://mathworld.wolfram.com, you will notice
$$\cos\left(\frac \pi {4}\right)=\frac 12 \sqrt 2$$
$$\cos\left(\frac \pi {8}\right)=\frac 12 \sqrt{ 2+\sqrt 2}$$
$$\cos\left(\frac \pi {16}\right)=\frac 12 \sqrt{ 2+{\sqrt {2+\sqrt 2}}}$$
$$\cos\left(\frac \pi {32}\right)=\frac 12 \sqrt{ 2+{\sqrt {2+{\sqrt {2+\sqrt 2}}}}}$$

Do you see the pattern ?

Hint: Set $x=\sqrt{2+\sqrt{2\ldots}}$. Then $x^2 = 2 + x$.


How to show that $\frac{f}{g}$ is measurable Here is my attempt to show that $\frac{f}{g}~,g\neq 0$ is a measurable function, if $f$ and $g$ are measurable function. I'd be happy if someone could look if it's okay.

Since $fg$ is measurable, it is enough to show that $\frac{1}{g}$ is measurable.
$$
\left\{x\;\left|\;\frac{1}{g}\lt \alpha \right\}\right.=\left\{x\;\left|\;g \gt \frac{1}{\alpha} \right\}\right., \qquad g\gt 0,\quad \alpha \in \mathbb{R},$$
which is measurable, since the right hand side is measurable.

Also, $$
\left\{x\;\left|\;\frac{1}{g}\lt \alpha \right\}\right.= \left\{x\;\left|\;g\lt\frac{1}{\alpha} \right\}\right.,\qquad g\lt 0,\quad \alpha \in \mathbb{R},$$ which is measurable since the right hand side is measurable.
 Therefore, $\frac{1}{g}$ is measurable, and so $\frac{f}{g}$ is measurable.

Using the fact that $fg$ is a measurable function and in view of the identity $$\frac{f}{g}=f\frac{1}{g}$$ it suffices to show that $1/g$ (with $g\not=0$) is measurable on $E$. Indeed, $$E\left(\frac{1}{g}<\alpha\right)=\left\{\begin{array}{lll} E\left(g<0\right) & \quad\text{if }\alpha=0\\ E\left(g>1/\alpha\right)\cup E\left(g<0\right) & \quad\text{if }\alpha>0\\ E\left(g>1/\alpha\right)\cap E\left(g<0\right) & \quad\text{if }\alpha<0 \end{array}\right.$$ Hence, $f/g$ ($g$ vanishing nowhere on $E$) is a measurable function on $E$.

For a better understanding of what is going on, I suggest to plot the function.

As you said, it suffices to show that $\dfrac{1}{g}$ is measurable.

Set $h(x)=\frac{1}{g(x)}$ if $g(x) \neq 0$ and $h(x)=0$ if $g(x)=0$

Let $a=0$ and $B=\{g=0\}$

Then $\{h>0\}=  \{x:\frac{1}{g(x)}>0\} \cap B^c$

If $a>0$ then $\{h>a\}=B^c \cap \{x:\frac{1}{g(x)}>a\}$

If $a<0$ then $\{h>a\}=(B^c \cap \{x:\frac{1}{g(x)}>a\} )\cup B$

In either case the sets are measurable because $g$ is measurable.


Help find hard integrals that evaluate to $59$? My father and I, on birthday cards, give mathematical equations for each others new age.  This year, my father will be turning $59$.

I want to try and make a definite integral that equals $59$.  So far I can only think of ones that are easy to evaluate. I was wondering if anyone had a definite integral (preferably with no elementary antiderivative) that is difficult to evaluate and equals $59$?  Make it as hard as possible, feel free to add whatever you want to it!

compact : $$\int_0^\infty \frac{(x^4-2)x^2}{\cosh(x\frac{\pi}2)}\,dx$$


How to construct a line with a given equal distance from 3 Points in 3 Dimensions? Important: I'm now convinced that 4 points are needes in order to reduce the solutions to a finite number. (Which is necessary because I need ALL solutions)

In a computer science context I need to solve a geometrical problem which states:

  Given three points in three-dimensional space, find a line $L$ (in any form, e.g. specify two points which lie on the line) so that the distance between each of the points and $L$ are equal to a given distance $d$, if possible.

By distance between a point $P$ and a line $L$, the euclidean distance between $P$ and the foot of the perpendicular on $L$ that passes through $P$ is meant.

In two dimensions (given two points) this is rather simple as it involves only a few trigonometric functions, altough I really struggle with it in three dimensions. The reason surely is that I'm not from a math background and don't know too much about linear algebra (which I believe is involved here).

It would be ideal if there is a solution for $N$ dimensions (I think that $N$ Points are needed then), altough I would be very happy if someone could give at least some hints about the three dimensional problem (maybe some sort of heuristic that i didn't thought of could also work). :)

EDIT: Clarification

The distance between the points and the line is a given constant. Example:
Given three Points and a distance $d=3$, I want to find the line which has a distance of $d$ (in this case $3$) to every given point, if possible (of course there are many cases where such a line does not exist). And I am aware of the fact that this line is not unique (several, or in case of colinearity of the three points, infinitely many lines exist)

EDIT: Clarification II

It seems that my wording causes much confusion about exactly WHAT properties the line should have. A picture showing the two-dimensional case follows:

In this case the Points $P_1$ and $P_2$ are given and the task was to find the line $g$ such that every given Point has the same shortest distance $r_B$ (which is preset) to $g$. (In this context the line was specified by a point $C$ and the angle $\alpha$, altough I'm happy with any kind of parameterization).

Now I have given 3 three-dimensional points and want to find a line with the described properties and I do not have any idea how to do this.

EDIT III:

I should also have mentioned that I should be able to find all solutions (I am 99% convinced that there is only a finite number of solutions for ordinary cases)

Maybe the following will help. It's not an answer, but it's too long for a comment.

Suppose the given distance is $d$. Then $L$ must be simultaneously tangent to three spheres of radius $d$, one centered at each of the given points.

Consider the case where the distances between some pair of the points is $D>2d$. The radius $d$ spheres can be nestled inside a ruled surface of this shape (so long as it's sized appropriately):

Any line of the ruled surface is tangent to both spheres.

You can also fit two spheres with centers distance $D$ apart into a ruled surface that's more (or completely) cone-like, or into one that's less cone-like and more (or exactly) cylindrical. Any of these ruled surfaces will yield an infinite number of lines $L$ tangent to these first two spheres.

If the sphere of radius $d$ about the third point intersects any of these ruled surfaces (hyperboloids of one sheet with axis containing the first two spheres' centers - the cone and the cylinder being degenerate ones), it seems like there will be either one or two lines $L$ from that ruled surface where $L$ equidistant from the three points.

I think that the third sphere will intersect one of these ruled surfaces if and only if it intersects either the cone or the cylinder, so maybe only the degenerate hyperboloids need to be considered.

If none of the points are more than $2d$ units apart, the first two spheres will intersect in a circle. The families of lines tangent to both will comprise a cylinder and some hyperboloids, but you can't twist them completely into a cone. Instead, there will be a different degenerate ruled family of tangent lines: those in the plane through the spheres' circle of intersection that are also tangent to that circle. Again, if the third sphere intersects any of these families of lines (and I think in this case it must), you'll get solutions.

There is a line for which each point has the same distance from the three points, but that distance will depend on which point of the line it was.
There is either zero, one or two points on the line that are a distance $d$ from the three points.
Given the points $(a,b,c),(e,f,g),(h,k,m)$, and a point equidistant $(x,y,z)$, then
$$(x-a)^2+(y-b)^2+(z-c)^2=d^2\\(x-e)^2+(y-f)^2+(z-g)^2=d^2\\(x-h)^2+(y-k)^2+(z-m)^2=d^2$$
Subtract the first equation from the other two
$$2x(a-e)+2y(b-f)+2z(c-g)=a^2-e^2+b^2-f^2+c^2-g^2\\
2x(a-h)+2y(b-k)+(2z(c-m)=a^2-h^2+b^2-k^2+c^2-m^2$$
This is two linear equations in three unknowns.  I hope you know how to solve it; it usually gives a line.
To get a common distance $d$, feed your answer back into the first equation, and you get a quadratic in $d$, which has zero, one or two solutions.


How big is infinity? This might be more philosophy than math, but it’s been bothering me for a while.

  Question: If there’s an infinite amount of real numbers between $ 0 $ and $ 1 $, shouldn’t there be twice the amount of real numbers between $ 0 $ and $ 2 $? Wouldn’t that be $ 2 \times \infty $? What about the amount of real numbers between $ 0 $ and $ \infty $? Wouldn’t that be $ \infty \times \infty $?

I guess the whole concept of infinity just kind of evades me. Any help would be appreciated!

I answered a similar question on May 13th:

  There are many different things that could be called "the infinite" in mathematics.  None of them is a real number or a complex number, but some are used in discussing functions or real or complex numbers.

  There are things called $+\infty$ and $-\infty$.  Those appear in such expressions as
  $$
  \lim_{x\to-\infty}\frac{1}{1+2^x} = 1 \text{ and }\lim_{x\to+\infty}\frac{1}{1+2^x}=0.
  $$
  There is also an $\infty$ that is approached as $x$ goes in either direction along the line.  That occurs in
  $$
  \lim_{x\to\infty} \frac{x}{1-x} = -1\text{ and }\lim_{x\to1} \frac{x}{1-x}=\infty.
  $$
  The second limit above could be said to approach $+\infty$ as $x$ approaches $1$ from one direction and $-\infty$ if from the other direction, but if one has just one $\infty$ at both ends of the line, then one makes the rational function a continuous function at the point where it has a vertical asymptote.  This may be regarded as the same $\infty$ that appears in the theory of complex variables.
  There are "points at infinity" in projective geometry.  This is similar to the "infinity" in the bullet point immediately preceding this one.  Two parallel lines meet at infinity, and it's the same point at infinity regardless of which of the two directions you take along the lines.  But two non-parallel lines pass through different points at infinity rather than the same point at infinity.  Thus any two lines in the projective plane intersect exactly once.
  There are cardinalities of infinite sets such as $\{1,2,3,\ldots\}$ (which is countably infinite) or $\mathbb R$ (which is uncountably infinite).  When it is said that Euclid proved there are infinitely many prime numbers, this sort of "infinity" is referred to.
  One regards an integral $\int_a^b f(x)\,dx$ as a sum of infinitely many infinitely small quantities, and a derivative $dy/dx$ as a quotient of two infinitely small quantities.  This is a different idea from all of the above.
  Consider the step function $x\mapsto\begin{cases} 0 & \text{if }x<0, \\  1 & \text{if } x\ge 0. \end{cases}$  One can say that its rate of change is infinite at $x=0$.  This "infinity" admits multiplication by real numbers, so that for example, the rate of change at $0$ of the function that is $3.2$ times this function, is just $3.2$ times the "infinity" that is the rate of change of the original step function at $0$.  This is made precise is the very useful theory of Dirac's delta function.
  There is the "infinite" of Robinson's nonstandard analysis.  In that theory, we learn that if $n$ is an infinite positive integer, then evern "internal" one-to-one function that maps $\{1,2,3,\ldots,n-3\}$ into $\{1,2,3,\ldots,n\}$ omits exactly three elements of the latter set from its image.  Nothing like that holds for cardinalities of infinite sets discussed above.
  I'm sure there are other examples that I'm missing here.

Infinity is so large, it can contain multiple copies of itself, within itself.  It's not at all like any of the other numbers, so many of the "common sense" approaches to infinity don't make a lot of sense.

That answer that pointed you to Cantor's Diagonalization Argument...  yes, that's the way to go...


Integration of $\frac{f'(x)}{f(x)}$? $\frac{f'(x)}{f(x)}$ integrated must be
$$\int\frac{f'(x)}{f(x)}dx=\ln\rvert f(x)\rvert+c.$$
But when trying to show this with partial integration I get another result:
$$
\begin{align*}
&\int\frac{\frac{d}{dx}f(x)}{f(x)}dx \\
=\quad&\int\frac{d}{dx}(f(x))\cdot (f(x))^{-1}dx \\
=\quad&f(x)\cdot (f(x))^{-1}-\int -f(x)\cdot (f(x))^{-2}dx \\
=\quad&1+\int (f(x))^{-1}dx \\
=\quad&1+\ln\rvert f(x)\rvert+c
\end{align*}
$$
Is there a mistake or can I summarize $1+c$, because $1$ is a constant?

$$
1+\int f^{-1}(x)dx = 1+\ln\rvert f(x)\rvert+c
$$
Wait!
$$
\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c$$
but
$$
\int \frac1{f(x)} dx \ne\ln|f(x)| + c
$$

Any function of arbitrary constants is also another arbitrary constant.


Limit problem: $\lim\limits_{n\to ∞} \sqrt[n]{7^n+3^n}$ $$\lim_{n\to ∞} \sqrt[n]{7^n+3^n}$$

The answer should be $7$, but I have no idea how to get there.

Factor out the biggest term:
$$L = \lim_{n \to +\infty} \sqrt[n]{7^n\left(1 + \left(\frac37\right)^n\right)} = \lim_{n \to +\infty} 7\sqrt[n]{1 + \left(\frac37\right)^n} = 7,$$
since $3/7 < 1$ and $(3/7)^n \to 0$.

You can easily generalize the above to prove:
$$\lim_{n \to +\infty} \left(a_0^n + \cdots + a_k^n\right)^{1/n} = \max\{a_0, \ldots, a_k\},$$
with $a_0,\ldots,a_k \in \mathbb R^+$.


Standard LP vs Mixed-Integer LP Optimal solution. I have a question about standard LP optimal solution vs Mixed Integer LP optimal solution.

Let $z*$ denote the optimal value for $c \cdot x$ and $z^{int}$ to be the optimal value $c \cdot x$ for the mixed interger LP. Why is it that $z* \ge z^{int}$?
$$
 \max  c \cdot x  \\
\mathrm{s.t.} \quad Ax \leq b\\
x \geq 0 \\$$

$$
 \max  c \cdot x  \\
\mathrm{s.t.} \quad Ax \leq b\\
x \geq 0 \\
x_{i} \in Z, i \in I $$

$z*\ge z^{int}$ because $z*$ is the optimal solution of the relaxation of the associated MILP. In other words it has less constraints, so the feasible space is larger: there are more potential good solutions.

Note that for a minimization problem, you would obviously have $z*\le z^{int}$.

It can be proved the other way that $z^{int}$ can never be greater than $z*$ by contradiction since the space of $z^{int}$ is a subset of the space of $z*$. So, $x^{int}$ corresponding to $z^{int}$ would also be a solution for $z*$ too.


How to solve $z^3+z+1=0$? I have troubles with solving an equation
$$z^3+z+1=0$$
in complex numbers. I know that in real numbers we can use Cardano's formula and it's clear to me. But what about the situation when $z \in \mathbb{C}$?

Given the (real) cube roots $A,B$ in Cardano's mthod, so that the real root of your cubic is $A+B,$ the other two roots are complex conjugates, given by
$$  A \omega + B \omega^2 $$
$$  A \omega^2 + B \omega $$
where
$$ \omega = \frac{-1 + i\sqrt 3}{2}$$
and
$$ \omega^2 = \bar{\omega}  $$
and $$\omega^3 = 1 $$

$$ z = re^{i \theta}$$
Or,
$$ r^3 \left[  e^{i3 \theta} \right] + re^{i \theta} + 1 =0$$
$$ r^3 \left[ \sum_{k=even}^{2}\binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta+ \sum_{k=odd}^{3} \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta \right] + r e^{i \theta} + 1 =0$$
Equating real and imaginary part to zero,
$$ r^3 \left[ \sum_{k=odd}^3 \binom{3}{k} (i)^{3-k} \sin^{3-k} \theta \cos^k \theta  \right] + r \cos \theta + 1 = 0$$
Or,
$$ r^3 \left[ \sum_{k=0}^1 \binom{3}{2k+1}  (-1)^{1+k} \sin^{2-2k} \theta \cos^{2k+1} \theta  \right] + r \cos \theta + 1 = 0 \tag{1}$$
And,
$$  r^3 \left[\sum_{k=even}^2 \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta  \right] + r i \sin \theta = 0$$
Or,
$$  r^2 \left[\sum_{k=0}^1 \binom{3}{2k} \cos^{2k} \theta (-1)^{k+1}  \sin^{3-2k} \theta  \right] +  \sin \theta = 0 \tag{2}$$
I'll leave you with solving the trignometric equation of (1) an d (2)


How can vectors with different units (position, speed, ...) "share" the same space? My question is too stupid to be googled, so I'll ask it here (because I didn't get any answers from google).

Context:
I have a three dimensional space and the units for $x,y,z$ are given in meters. On the plane $xy$ there is an area $A$ with fluid coming at out of it at velocity $V$ ($3\,\mathrm m/\mathrm s$). My textbook just draws this velocity vector with a magnitude of $3$ on top of area $A$.

My question is:
Why is it valid to draw this vector in a system with no reference to time (because velocity is m/s)? Why is it just assumed that its magnitude is $3$ on this coordinate system?

Thanks.

You might say that you have several vector spaces overlapping, one space $A$ for spacial locations (with an arbitrarily picked origin and units in meters, say), one space $B$ for velocities with units in m/s, say (and add spaces for acceleration, impulse, force, ... if needed). Principally, these are totally distinct spaces and it makes no sense e.g. to add a vector $v\in B$  to a vector $r\in A$. However, for any given timespan $t$, we have a map $v\mapsto tv$ from $B$ to $A$. It is convenient to "draw" both spaces in the same 3d space - actually this is not much different from the convention to "draw" a 3d sapce on the falt page of a textbook in the first place. Some means of distinction should be used, maybe different colours; or one agrees upon a standard timespan to map the velocities space into the dislpacement space.

You may have noticed that I mentioned the arbitrariness of choice of origin for $A$, but said nothing about the origin of $B$. Well, actually the most correct way of handling the situation would be to attach a copy of $B$ to every single point in $A$; one speaks of tangential space and hs to concider topological peoperties to relate velocities attached to differente (nearby) points. But I'm afraid that would be even more confusing than the illustration you already complain about ...

Because size of the velocity vector is the value of the velocity without units, and direction is the angle it creates with the determined beginning of axis


Count the number of isosceles triangles How can I count the amount of isosceles triangles fitting by the following criteria: the base of the triangles equals 1, all the angles are natural numbers and are less than $75^\circ$. (The equilateral triangle is counted only once)

If each of the base angles is $n^\circ$ and the third angle is $(180-2n)^\circ$ then you want

$n$ an integer
$n \lt 75$
$180 -2n \lt 75$, i.e. $n \gt 52.5$

If the angle opposite the base is $x$ degrees, the angles at the base are $\frac{180-x}{2}$. Do a little arithmetic to determine when that is less than $75$, that will give you bounds on $x$ then it's simple to figure out how many integer values $x$ can take.


Using Newton-Raphson to Approximate $\pi$ Is it possible to use the Newton-Raphson method to approximate $\pi$? If you use this to find the root of $\sin{x}$, could you use:

\begin{align}
x_0 &= 3\\
x_{n+1} &= x_n - \tan{x_n}
\end{align}

When I plug this into my calculator, it converges on $\pi$ quickly. Is this a real approximation of $\pi$ though? Does it require prior knowledge of the value of $\pi$?

Yes, it is a real approximation of $\pi$.  Whether it "requires prior knowledge of the value of $\pi$" depends on what you're using to calculate $\tan x_n$.  A typical arbitrary-precision method of calculating the $\tan$ function might use the Maclaurin series for $\sin$ and $\cos$ in a neighbourhood of $0$, and trig identities to reduce other values to those in that neighbourhood.  Those reductions do require the value of $\pi$.  However, they are not the only way.
You could use the Maclaurin series for $\sin$ and $\cos$ even for values of $x$ near $\pi$: it would just take a very large number of terms.

The approximations that you get using tangent or sine functions are real in the sense that they do give a valid sequence that converges to $\pi$. But the problem with using trig functions with the N-R method is that the approximations are just as irrational (transcendental actually) as $\pi$ itself. It is rather like the chicken and egg dilemma. The usual way of approximating $\pi$ is with infinite series like Taylor series or infinite products like Wallis's formula.


Calculation double Integral over Ball (optical size) I hope that someone can help me with the following problem.
I have to show that
$$\int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy=4\pi^2~,$$
with $B_1(0)\subset\mathbb{R}^3$.
I have no idea how to calculate those integrals, the common(for me) tricks won't help.

The integral is connected with the mean inverse squared distance between two points within a unit ball.
Following the method used by Christian Blatter in this post we have that,
$$\begin{align*}
\int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy
&=|B_1(0)|^2
\int_0^1\int_0^1\int_0^\pi \frac{f_R(r) f_S(s)f_\Theta(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\
&=\left(\frac{4\pi}{3}\right)^2
\int_0^1\int_0^1\int_0^\pi \frac{3r^2\cdot 3s^2\cdot1/2\sin(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\
&=8\pi^2
\int_0^1\int_0^1r^2 s^2\left(\int_0^\pi \frac{\sin(\theta) \ d\theta}{r^2+s^2-2rs\cos\theta}\right)\ ds\ dr\\
&=4\pi^2
\int_0^1r\left(\int_0^1 s\ln\left(\frac{(r+s)^2}{(r-s)^2}\right)\ ds\right)\ dr\\
&=4\pi^2
\int_0^1r\left(\ln\left(\frac{1+r}{1-r}\right)(1-r^2)+2r\right)\ dr\\
&=4\pi^2.
\end{align*}$$
and we are done.

I assume that $B_1(0)$ is the ball of radius $1$ about the origin.
Since this space has nice spherical symmetry, it will probably be
easier to convert to spherical coordinates. Let our change of coordinate
map be
$$ H(\rho,\theta,\phi)
   = (\rho\cos\theta\cos\phi,
      \rho\sin\theta\cos\phi,
      \rho\sin\phi), \qquad
      \rho \in [0,1], \theta \in [0,2\pi], \phi \in [-\pi/2,\pi/2] $$
so we wish to integrate
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over |r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2}
   \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$
Now absolute value is usually defined as $|x|=\sqrt{x^2}$, so we have
$$ \begin{align}
|r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2
   &= r^2\cos^2\theta\cos^2\phi
      + r^2\sin^2\theta\cos^2\phi-2r^2\cos^2\phi\sin\theta\cos\theta \\\
   &= r^2\cos^2\phi(1- 2\sin\theta\cos\theta)
\end{align}$$
Then our integral is
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over r^2\cos^2\phi(1- 2\sin\theta\cos\theta)}
   \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$
easily simplifying to
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over \cos\phi}{1 \over (1- 2\sin\theta\cos\theta)}
   \thinspace drd\theta d\phi $$
Clearly $r$ integrates out. By the standard tricks,
$$ \int {1 \over \cos x}\thinspace dx = \log(\tan x + \sec x) $$
and
$$ \int {1 \over 1-2\sin x \cos x} = {\sin x \over \cos x - \sin x} $$
So the antidifferentiation results in
$$ \begin{align}
   &\log(\tan\phi-\sec\phi)\bigg|_{-\pi \over 2}^{\pi \over 2}
   \times {\sin \theta \over \cos \theta - \sin\theta}\bigg|_0^{2\pi} \\
\end{align} $$
which is not defined. Hence the integral does not exist.


Complex solutions of $(x+1)^{10}+(x-1)^{10}=0$ Solve $$(x+1)^{10}+(x-1)^{10}=0$$

What method is available to solve this equation?

As pointed out by Gerry Myerson's comment , put $\frac{x+1}{x-1}=z$ for some complex $z,x\neq1$ Then, $x=\frac{z+1}{z-1}$. Now your equation reduces to $z^{10}=-1$, which is easily solvable by using Euler's/De-Moivre's Theorem as $z=e^{i\phi},\phi=\frac{(2k+1)\pi}{10},k\in\mathbb{Z^+\cup 0}:k<10$ from which $x$ can be calculated

For $x\in \mathbb{R}$ $(x+1)^{10}+(x-1)^{10}=0$ don't have any solution. This is true by the argument Dr.Sonnhard Graubner.

So let's rewrite question, $$(z+1)^{10}+(z-1)^{10}=0\text{ for }z\in \mathbb{C}.$$

Then suppose if $(z-1)^{10}=0,$then $\implies z=1$ , but $(1+1)^{10}\not=0$ hence $(z-1)^{10}\not=0$.
\begin{gather}
\Rightarrow\frac {(z+1)^{10}}{(z-1)^{10}}+1=0 \\ \frac{z+1}{z-1}=\sqrt[10]{-1}
\\
\end{gather}
Hence $z=\frac{1+\sqrt[10]{(-1)}}{\sqrt[10]{(-1)} -1}$.


Exsitence of extrema and boundedness of a function Let $f:[0, \infty) \longrightarrow \mathbb R$ be a continuous function which converges for $t \rightarrow +\infty$.

What can you say about existence of extrema and boundedness of the function f?

If $\Vert f \Vert = \sup_{x \in [0,\infty)} \frac{|f(x)|}{|x|} < \infty$ Holds it is bounded. If the image of $f$ is a compact set it attains its maximum and minimum


Series Solutions Near an Ordinary Point I am attempting to solve this problem for practice: $y"-(x-3)y' - y = 0$ at $x_{0} = 3$.

But it appears as though I don't have an idea of the best approach to employ to go about solving it.

Can someone help me understand the steps in solving a DE such as this?  I know that a power series is required, but the garden variety method of solution doesn't appear to be working for me.

Any suggestions?

It looks much simpler than that. Try differentiating $\exp(-3x+\frac{1}{2}x^2)$.

This seems to be attracting down votes. So maybe I should add a little more. If you do that you find that it is a solution to $y''-(x-3)y'-y=0$. Note also that the equation is linear, so you can take any constant multiple and the solution still works. Of course, there is another solution which is harder to find. But you have not specified the initial conditions clearly, so you may not need it.


There are apparently $3072$ ways to draw this flower. But why? This picture was in my friend's math book:

Below the picture it says:

  There are $3072$ ways to draw this flower, starting from the center of
  the petals, without lifting the pen.

I know it's based on combinatorics, but I don't know how to show that there are actually $3072$ ways to do this. I'd be glad if someone showed how to show that there are exactly $3072$ ways to draw this flower, starting from the center of the petals, without lifting the pen (assuming that $3072$ is the correct amount).

First you have to draw the petals.  There are $4!=24$ ways to choose the order of the petals and $2^4=16$ ways to choose the direction you go around each petal.  Then you go down the stem to the leaves.  There are $2! \cdot 2^2=8$ ways to draw the leaves.  Finally you draw the lower stem.  $24 \cdot 16 \cdot 8=3072$

The presumption is wrong!

Most answers here start like this:
"First you have to draw the petals."

This is a mistake. You could start drawing the flower from two points: The centerpoint of the blossom OR the endpoint at the lower stem.

Result is:
There are 2 x 3072 = 6144 ways to draw the Flower!

I think it might have to do with the way you asked the question. ;)


Integral of $ \int \frac {\tanh(x) dx}{\tanh(x)+\operatorname{sech}(x) }$ My attempt at solution:

Also I got a question, is there a way this can be solved without using hyperbolic tangential half angle substitution? Because I don't get how you can treat hyperbolic functions as if they were trigonometric and deducing sinx and cosx from a right triangle

The integral can also be seen as the following:
\begin{align}
\int \frac{\tanh(x) \, dx}{\tanh(x) + sech(x)} &= \int \frac{\sinh(x) \, dx}{\sinh(x) + 1} = \int \left( 1 - \frac{1}{1 + \sinh(x)} \right) \, dx \\
&= x + \sqrt{2} \, \tanh^{-1}\left( \frac{1}{\sqrt{2}} \, \left(1 - \tanh\left(\frac{x}{2}\right) \right) \right)
\end{align}


The only differentiable function $f \colon \mathbb R \to \mathbb R$ such that $f^\prime(x)=f(x)$ is $f(x)=ce^{x}$ Prove that the only differentiable function $f \colon \mathbb R \to \mathbb R$ such that $f^\prime(x)=f(x) \mspace{1ex} \forall x\in \mathbb R$ is $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

Let $g:\mathbb R \to \mathbb R$, $g(x)=f(x)/e^{x}$. Then $$g^\prime(x)=\frac{e^{x}f^\prime(x)-f(x)e^{x}}{e^{2x}}=\frac{f^\prime(x)-f(x)}{e^{x}};$$

by hypothesis $f^\prime(x)=f(x)$ hence $g^\prime(x)=0$, and therefore $g(x)=c$ for some $c\in \mathbb R$.

Then $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

I would like you to tell me if this proof is correct thank you :)

$f(x) = 0$

Is also a function with this property.


Seeking help to understand a simple Kripke model I'm reading A Brief Introduction to the Intuitionistic
Propositional Calculus, at page 7, there is a simple Kripke model represented by a graph, I interpret it as:

$W = \{w_1, w_2\}$
$w_1 \ge w_2$
$w_2 \models \alpha$

In this simple Kripke model, the author states that
$\lnot \lnot \alpha \Rightarrow \alpha$ (double negation elimination) fails at $w_1$, which I don't follow.

I attempted to prove by contradiction, but failed:

To show $w_1 \models \lnot \lnot \alpha \Rightarrow \alpha$, we want:
$\forall v \in W, v \ge w_1, \text{if } v \models \lnot \lnot \alpha, \text{ then } v \models \alpha$

The only instance that satisfies $\forall v \in W, v \ge w_1$ is $w_1$ itself,
but since $w_2 \le w_1$ and $w_2 \models \alpha$, we know by definition of Kripke model that $w_1 \models \alpha$. So the conclusion $
v \models \alpha$ always holds.

I don't think this conclusion is right, so I look into the precondition $v \models \lnot \lnot \alpha$, of which the only instance is $w_1 \models \lnot \lnot \alpha$. This holds if and only if $\forall v \ge u, v \not\models \lnot \alpha$.

I got stuck here because I can't find a rule to apply on $v \not\models \alpha$ from page 6.

So far I can only show the law of the excluded middle fails at $w_1$ (not sure if it is correct though) but have no idea about this double negation elimination nor Peirce's law. Since in page 4 it says that triple negation reduction is an intuitionistic tautology, I want to test if $\lnot \lnot \lnot \alpha \Rightarrow \lnot \alpha$ fails, but failed to do so due to a similar reason.

Is there something I missed or am I doing something totally wrong?

I will use the verb "forces" to refer to $\models$, and I will say $v $ "extends" $w$ if $v \geq w$.

It is not the case that $w_1 \models \alpha$. The only way for a world to satisfy a propositional variable is for the variable to be true in that world, and they have set it up so that neither $\alpha$ nor $\lnot \alpha$ is true in $w_1$.

At the same time, we do have that $w_1 \models \lnot \lnot \alpha$. This is because there exists an extension of $w_1$ (namely, $w_2$) that does not force $\lnot \alpha$. We have that $w_2 \not \models \lnot \alpha$ because there is an extension of $w_2$ (namely, $w_2$ itself) that does force $\alpha$.  Note that, in general, a node $x$ forces $\lnot \phi$ if and only if no extension of $x$ (including $x$ itself) forces $\phi$.

Therefore, combining the previous paragraphs, $w_1 \not \models \lnot\lnot \alpha \to \alpha$.

One difficulty for many people the first time they run into the definition of the forcing relation $\models$ is in how it deals with negation:

$w$ forces $\lnot \phi$ if and only if no extension of $w$ forces $\phi$.
$w$ does not force $\lnot \psi$ if and only if some extension forces $\psi$
$w$ forces $\lnot \lnot \theta$ if and only if no extension $v$ forces $\lnot \theta$, if and only if, for every extension $v$ of $w$ there is an extension $v'$ of $v$ that forces $\theta$.  We usually abbreviate this as: $w$ forces $\lnot \lnot \theta$ if and only if the set of worlds that force $\theta$ is dense above $w$.

It is also worth knowing that the forcing relation is often written $\Vdash$ instead of $\models$.

I use Polish notation.  The formation rules run as follows:

All lower case letters of the Latin alphabet are formulas.
If x is a formula, then Nx is a formula.
If x and y are formulas, then so are Cx y, Ax y, and Kx y.

One can axiomitize intuitionistic propositional calculus as given in your reference as follows under the rules of uniform substitution, and implication elimination/detachment: {Cx y, x} $\vdash$ x, as follows.

C p Cqp.
C CpCqr C Cpq Cpr.  (1 and 2 together allow us to prove the Deduction Meta-Theorem)
C p C q Kpq (conjunction introduction)
C Kpq p (conjunction elimination left).
C Kpq q (conjunction elimination right).
C p Apq (disjunction introduction left).
C p Aqp (disjunction introduction right).
C Apq C Cpr C Cqr r (disjunction elimination).
C Cpq C CpNq Np (negation introduction).
C Np Cpq (negation elimination).

Using Mace4 I found the following 3-valued model of those axioms and the rule of inference which allows to check the claim here.  "2" is the designated element.  Also "1" corresponds to "falsum" of two-valued logic:

A  0  1  2  C  0  1  2  K  0  1  2  N
0  0  0  2  0  2  1  2  0  0  1  0  1
1  0  1  2  1  2  2  2  1  1  1  1  2
2* 2  2  2  2  0  1  2  2  0  1  2  1

Notice that C NN0 0=C N1 0=C20=0.  Thus, CNNpp does not hold in intuitionistic logic.  Now, we only need to check all 10 claims above for each value of one variable.

So, for C p Cqp, we have C 0 Cq0.  C 0 C00=C02=2, C 0 C10=C02=2, C 0 C20=C02=2.  Thus, C 0 Cq0.  Notice that $\forall$x, C1x=2.  Thus substituting x with Cq1, we have that C 1 Cq1=2.  Notice that $\forall$y, C2y=y.  Thus, substituting y with Cp2 we have C2Cq2=Cq2.  We do have that $\forall$q, Cq2=2.  So, we have that C 2 Cq2=2.  Thus, since we have C 0 Cq0=2, C 1 Cq1=2, and C 2 Cq2=2, it follows that C p Cqp=2.
C C2Cqr C C2q C2r= C Cqr Cqr, since $\forall$x, C2x=x.  Note that $\forall$x Cxx=2.  Substituting x with Cqr we obtain C Cqr Cqr.  Thus, C C2Cqr C C2q C2r=2.  C C1Cqr C C1q C1r=C 2 C22=C22=2.  C C0Cqr C C0q C0r... C C0C2r C C0q C02=C C0C2r C C0q 2=C C0C2r 2=2.  C C0C1r C C01 C0r=C C0C1r C 1 C0r= C C0C1r 2=2. C C0C00 C C00 C00= C C0C00 C C00 2= C C0C00 2=2.  C C0C01 C C00 C01=C C0C01 C C00 1=C C0C01 C21=C C0C01 1=C C01 1=C11=2.  So, C C0C0r C C00 C0r, C C0C1r C C01 C0r, C C0C2r C C02 C0r all hold leading to C C0Cqr C C0q C0r.  And thus we obtain C CpCqr C Cpq Cpr=2.
C 0 C q K0q... C 0 C 0 K00=C 0 C 0 0, which is an instance of C p Cqp.  So, C 0 C 0 K00=2.  C 0 C 1 K01=C02=2.  C 0 C 2 K02=C 0 K02=C00=2.  Thus, we can infer that C 0 C q K0q=2.  C 1 C q K1q=2 by the above.  Note that $\forall$y K2y=y.  Thus, C 2 C q K2q=C 2 Cqq.  Cqq=2, and thus C 2 Cqq=C22=2.  Since we've covered all three cases, C p C q Kpq=2.
C K0q 0... C K00 0=C00=2.  C K01 0=C10=2.  C K02 2=C00=2.  So, C K0q 0=2.  Note that $\forall$y, K1y=1.  So, C K1q 1=C11=2.  C K2q 2=2 by the above.  Therefore, C Kpq p=2.
Notice that $\forall$x$\forall$y Kxy=Kyx.  Thus, since C Kpq p=2, C Kqp p=2.  Therefore, C Kpq q=2.
C 0 A0q... C 0 A00=C00=2, C0 A01=C00=2, C0 A02=C02=2.  Thus, C 0 A0q.  C 1 A1q=2 by the above.  Note that $\forall$y, A2y=2.  Thus, C 2 A2q=C22=2.  Therefore, C p Apq=2.
Notice that $\forall$x, $\forall$y, Axy=Ayx.  Thus, since C p Apq=2, it follows that C p Aqp=2.
C Apq C Cp0 C Cq0 0 ... C A0q C C00 C Cq0 0=C A0q C 2 C Cq0 0=C A0q C Cq0 0... C A00 C C00 0=C 0 C C00 0, which is an instance of CpCqp.  So, C A00 C C00 0=2.  C A01 C C10 0=C 0 C C10 0... which is an instance of CpCqp.  So, C A01 C C10 0=2.  C A02 C C20 0=C 2 C20 0.  This is an instance of C p C Cpq q.  This is provable from the axioms CpCqp, and CCpCqrCCpqCpr under detachment.  Since if Cpq=2, as well as p=2, then q=2 holds by the above table, it thus follows that C p C Cpq q=2.  Putting those three cases together we have C A0q C Cq0 0=2, and thus C A0q C C00 Cq0 0=2.  C A1q C C10 C Cq0 0....  Note that $\forall$x A1x=x.  Thus, C A1q C C10 C Cq0 0=C 1 C C10 C Cq0 0.  C 1 C C10 C Cq0 0 is an instance of C p C q C Cpr r, which is a theorem of the implicational calculus here.  Thus, C A1q C C10 C Cq0 0=2.  C A2q C C20 C Cq0 0=C 2 C C20 C Cq0 0=C C20 C Cq0 0=C 0 C Cq0 0... another instance of CpCqp.  Thus, C A2q C C20 C Cq0 0.  Putting C A2q C C20 C Cq0 0, C A1q C C10 C Cq0 0=2, and C A0q C C00 C Cq0 0 together we obtain C Apq C Cp0 C Cq0 0.  C Apq C Cp1 C Cq1 1=C Apq C Np C Nq 1=C Apq C Np NNq... C A0q C N0 NNq=C A0q C 1 NNq=C A0q 2=2.  C A1q C N1 NNq=C A1q C 2 NNq=C A1q NNq... C A10 NN0=C 0 NN0=C0 N1=C02=2.  C A11 NN1=C 1 NN1=C 1 N2=C11=2.  C A12 NN2=C 2 NN2=NN2=N1=2.  Thus, C Apq C Np NNq=2, and so C Apq C Cp1 C Cq1 1=2 also.  C Apq C Cp2 C Cq2 2=C Apq C Cp2 2=C Apq 2=2.  Now since we have C Apq C Cp0 C Cq0 0=2, C Apq C Cp1 C Cq1 1=2, and C Apq C Cp2 C Cq2 2, we can infer that C Apq C Cpr C Cqr r=2.
C C0q C C0Nq N0=C C0q C C0Nq 1... C C00 C C0N0 1=C C00 C C01 1=C C00 C11=C22=2.  C C01 C C0N1 1=C 1 C C0N1 1=2.  C C02 C C0N2 1=C 2 C C01 1=C 2 C11=C22=2.  Thus, C C0q C C0Nq N0.  C C1q C C1Nq N1=C C1q C C1Nq 2= C C1q 2=2.  C C2q C C2Nq N2=C C2q C C2Nq 1... C C20 C C2N0 1=C C20 C C21 1= C C20 C11=C C20 2=2.  C C21 C C2N1 1=C 1 C C2N1 1=2.  C C22 C C2N2 1=C 2 C C21 1=2.  Thus, C C2q C C2Nq N2.  Therefore, C Cpq C CpNq Np=2.
C 0 C N0 q=C 0 C1q=C02=2.  C 1 C N1 q=2.  C 2 C N2 q=C 2 C1q=C22=2.  Therefore, CpCNpq=2.

And that confirms that the above model does indicate that all the axioms, and the rule of inference {Cx y, x} $\vdash$ y do hold for this model, but CNNpp does not hold.  Therefore, CNNpp is not a theorem of intuitionistic logic.


Why, intuitively, is the order reversed when taking the transpose of the product? It is well known that for invertible matrices $A,B$ of the same size we have $$(AB)^{-1}=B^{-1}A^{-1} $$
and a nice way for me to remember this is the following sentence:

  The opposite of putting on socks and shoes is taking the shoes off, followed by taking the socks off.

Now, a similar law holds for the transpose, namely:

$$(AB)^T=B^TA^T $$

for matrices $A,B$ such that the product $AB$ is defined. My question is: is there any intuitive reason as to why the order of the factors is reversed in this case?

[Note that I'm aware of several proofs of this equality, and a proof is not what I'm after]

Thank you!

One of my best college math professor always said:

  Make a drawing first.

Although, he couldn't have made this one on the blackboard.

My professor always called them the shoe-sock theorems or the likes, the idea basically being that you put on your socks first then the shoes, but to do the reverse you must get off the shoes first and then the socks.


If I want to distribute 50 identical candies to 100 children, what is the expected number of candies a child with at least one candy has? If I want to distribute 50 identical candies to 100 children, what is the expected number of candies a child with at least one candy has?
For example, if I give 24 candies to child A and 26 candies to child B, and don't give any candy to other 98 children, since only two children has nonzero candies, if this way the only way how I can distribute 50 identical candies to 100 children, the quantity I'm looking for would be 25.
I tried for an hour, and in the end came up with the following "solution":
Let $Q(k)$ be the average number of candies a child with at least one candy has provided that we distribute the candies only to $k$ children. Then I assumed that the average number of candies a child from this set has $50/k$, and there are $\binom{100}{k}$ different ways of selecting these set of children, so doing a weighted average, I got
$$
\frac{
\sum_{k=1}^{50} 50*(100!) / (k * (k!) * (100-k)!)} { \sum_{k=1}^{50} 100! / ((k!) * (100-k)!)} \approx 1.08481.
$$
Is my solution correct? If not, could you provide me with an detailed answer about how you solved it?

The probability that any given child would not get any candy is (99/100)^50, about .6050, so 100*(1-.6050), 39.499, would get 1 or more: on average 50/34.999 or 1.26584.


Pizza Hut Math (equal-sides sets of numbers with same sum and product) Pizza Hut, in honor of $\pi$ day, posted this problem:

  Our school’s puzzle-club meets in one of the schoolrooms every Friday after school.

  Last Friday, one of the members said, “I’ve hidden a list of numbers in this envelope that add up to the number of this room.” A girl said, “That’s obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?”

  He (after scribbling for some time): “No.” She (after scribbling for some more time): “well, at least I’ve worked out their product.”

  What is the number of the school room we meet in?”

Is there a non-brute-force way of doing this problem? Additionally, I've found multiple same-sized sets of numbers with the same product and sum, so I imagine I am utterly misunderstanding the problem anyway (e.g. 2 triples that sum to 23 and multiply to 360, 2 triples that sum to 22 and multiply to 360, 2 triples that sum to 21 and multiply to 240, etc.).

Well, here is my solution (tl;dr my answer is 12)

First, we are looking for a number. We can write all possible unique "sum representations" of that number with different length. Every representation is a list of sorted numbers. For example 3 can be represented as [[3], [1 2], [1 1 1]]. Next, for every representation we can calculate product of the elements in it, for 3 it will be [3, 2, 1]. Also for every representation we can find number of elements in it: [1, 2, 3]. We in the "sum representations" of a number we want to find such representations that have same product of elements, but only one. I am sorry, I don't know how to explain it better (English is not my native). But let me explain why 23 is not an answer.

Indeed, [4 9 10] and [5 6 12] are two sum-representations of 23 with product equals to 360. But, there is another representations of 23: [1 2 8 12] and [2 2 3 16] that give product 192. Thus, girl would not be able to guess the product in case of room number 23, because there is a lot of groups of representation (226) that will give the same product.

Long story short, I was not able to find a number lower then 12 that would give me two representations with the same product. 12=1+3+4+4=2+2+2+6. 13 already have two different repeated products: 36 ([1 6 6] and [2 2 9]) and 48 ([1 1 3 4 4] and [1 2 2 2 6]). As you may notice, any other number bigger than 13 will have at least two representations with the product 36 and at least two with the product 48. So, 12 is the only solution.

I am sorry for quite complicated explanation. I am not doing math on professional level, so I do not know all terminology of the number theory. I will be glad to answer any questions.

Its important to capture two things here 1] number representation in sets, 2] product of those numbers.

If you know the product and the corresponding number of numbers in the set then you can actually determine the Room number. For example lets say the product is 12 the set becomes [1, 12], [4,3] [2,2,3]. Now here we have two, 2 number sets both having different sum so can't predict the room number but given that there are three numbers we can predict the room number as 7

So, as you increase the product (from 12 to 13, 14, 15, 16 and so on) you may get different number of sets with different distinct sum and thats endless sequence.

The only possible sequence looks here is for product value 2 and the set as [1,2] (since we have more than one number as said in question) with the sum as 3 which seems to me as the room number as well.


Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$

I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.

Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks


Find the slope of a line given a point and an angle I'm trying to figure out this problem and feel like it's something that must be so simple that I could've done in high school no problem, but for some reason my brain is frozen this morning.  I would really appreciate any help, and want to say thanks in advance. I tried to draw a picture below; I want to find the slope of a line given a point $(x,y)$ and $\theta$.

$\tan \left( \tan^{-1}\left(\frac{y}{x}\right) - \theta\right)$ is the slope $m$.

Then use "point slope formula" (if you want an equation of the line, that is...)

$y-y_1 = m(x - x_1)$

For variety, I'll explain.

Labeling the origin "$O$" and the point $(x, y)$ "$P$", the segment $\overline{OP}$ makes an angle of $\tan^{-1} \left(\frac{y}{x}\right)$ with the positive x-axis. But this is the sum of $\theta$ and the angle $\phi$ that your line makes with the positive x-axis (since we have opposite interior angles).

So $\tan^{-1}\left(\frac{y}{x}\right) - \theta = \phi$.

Finally, $\tan \phi = m$.

find slope of the line you know as m1. m2 will be the slope of the unknown line and Ө the angle. The eqn tan(Ө) = (m2 - m1)/ (1 + (m1*m2))


A number that is even and prime. So I have the following sentence : There exists a unique prime number that is even.

I have the following predicates :

Prim(x) : x is a prime number.
Even(x) : x is an even number.

My answer : Prim(Even(x))

I don't know, but I think that this would be better : Ǝ Prim(Even(x))

Would this make sense ? Even though we know that there is only one such number that is prime and even.

Thank yoU!

"There is exactly one even prime number" can be expressed formally as
$$ \exists x: [(Even(x) \land Prim(x)) \land \forall y : [Even(y) \land Prim(y) \implies x=y]]. $$
The first part says there is at least one even prime number. The second part says there is at most one even prime number.

For the statement "There exists a unique Prime number that is Even."
it's rather more simple.
$\exists  x: [\operatorname{Even}(x)\land\operatorname{Prim}(x)]$
My explanation for this is that the existential quantifier $\exists x$  means:-

There exists an 'x'
For some 'x'
For at least one 'x'

Here it is used to reference that "there exists a unique 'x' satisfying both $\operatorname{Even}(x)$ and $\operatorname{Prim}(x)$".


Why does cross product give a vector which is perpendicular to a plane I was wondering if anyone could give me the intuition behind the cross product of two vectors $\textbf{a}$ and $\textbf{b}$. Why does their cross product $\textbf{n} = \textbf{a} \times \textbf{b}$  give me a vector which is perpendicular to a plane?

I know I can just check this by using dot product but I'm not totally satisfied with "it just works" answer =)

Thank you for any help! =)

The determinant formula isn't so mysterious. Consider the cross product $\mathbf{v} = \langle a,b,c \rangle \times \langle d,e,f \rangle$ as the formal determinant

$$ \det \left(\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a & b & c \\ d & e & f \end{array} \right) $$

where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the standard basis vectors. If instead one considers $\mathbf{i}, \mathbf{j}, \mathbf{k}$ as indeterminates and substitutes $x, y, z$ for them, this determinant computes the dot product $\mathbf{v} \cdot \langle x, y, z \rangle$. But letting $\langle x, y, z \rangle$ be $\langle a, b, c \rangle$ or $\langle d, e, f \rangle$ gives a zero determinant, so $\mathbf{v}$ is perpendicular to the latter two vectors, hence to the plane they span, as Omnomnomnom says.

I would start by something simpler.
What is i x j ?
answer: k
but why?  Well, it's by definition.  The cross product of a unit vector in the x-direction (i) and a unit vector in the y-direction (j) is a perpendicular vector in the z-direction  (k).
Given the above, one can easily see that:
2i x j = 2k
In 2 dimensions, if you want to multiply a x b

Simply breakdown vectors a and b into their i and j components.
Multiply the i from the first vector (a) with the j in the second vector (b).
Multiply the j from the first vector (a) with the i in the second vector (b).
Add them together, remembering that j x i is NEGATIVE k.

so step 2 and 3 work opposite of each other - if you will.
That's it.  Now think about this in 3 dimensions remembering that:
j x k = i
k x i = j
and that the reverse cross product gives a negative sign.


Exists $C$ where $\int_{\mathbb{R}^n} {{u^2}\over{|x|^2}}\,dx \le C \int_{\mathbb{R}^n} |Du|^2\,dx$, $u \in H^1(\mathbb{R}^n)$? For each $n \ge 3$, does necessarily exist a constant $C$ so that$$\int_{\mathbb{R}^n} {{u^2}\over{|x|^2}}\,dx \le C \int_{\mathbb{R}^n} |Du|^2\,dx$$for all $u \in H^1(\mathbb{R}^n)$?

Ideas. I probably want to use the fact that$$\left|Du + \lambda {x\over{|x|^2}}u\right|^2 \ge 0$$for each $\lambda \in \mathbb{R}$ somehow, but I do not what do from here on out. Could anybody help?

Let $u\in C_0^\infty(\mathbb{R}^n)$. Note that
$$0\leq \left|Du + \lambda {x\over{|x|^2}}u\right|^2=|Du|^2+\lambda^2\frac{|u|^2}{|x|^2}+2\lambda\sum_{i=1}^nu_{x_i}\frac{x_i}{|x|^2}u$$
Integrating over $\mathbb{R}^n$ (here we use the compact support of $u$), we get
$$0\leq \int_{\mathbb{R}^n} |Du|^2\;dx+\lambda^2\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx-\lambda n\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx+2\lambda\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx$$

Taking $\lambda$ such that $\lambda(n-2-\lambda)>0$ (here we use $n\geq 3$) we get
$$\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx\leq C \int_{\mathbb{R}^n}|Du|^2\;dx$$
with $C=\frac{1}{\lambda(n-2-\lambda)}$.

As we can see here, a density argument extends the result for $u\in H^1(\mathbb{R}^n)$.

You can take $C = 4/(n-2)^2$. This is Hardy inequality.


What is the Fisher information for a Uniform distribution? If X is U[$0$,$\theta$], then the likelihood is given by $f(X,\theta) = \dfrac{1}{\theta}\mathbb{1}\{0\leq x \leq \theta\}$. The definition of Fisher information is $I(\theta) = \mathbb{E} \left[ \left(\dfrac{d \log(f(X,\theta))}{d\theta}  \right)^2 \right]$. How can this be calculated when $\log f(X,\theta) $ is not defined for $\theta < X$? I understand that we also have $f(X,\theta) = 0$ for $\theta < X$ but can we ignore this when taking the expectation? If so, why?

it is $n^2/\theta$.

We get this from calculating the log-likelihood first which is $-n \log(\theta)$, then taking its derivative, we will get $\frac{-n}{\theta}$. squaring it and take its expectations we will have $\frac{n^2}{\theta^2}$


Prove that $(2m+1)^2 - 4(2n+1)$ can never be a perfect square where m, n are integers I could prove it hit and trial method. But I was thinking to come up with a general and a more 'mathematically' correct method, but I did not reach anywhere. Thanks a lot for any help.

Suppose to the contrary that the difference is a perfect square. Note that the difference is odd, so we would have
$$(2m+1)^2-4(2n+1)=(2q+1)^2.$$
This can be rewritten as
$$4(2n+1)=4(m^2+m)-4(q^2+q).\tag{1}$$
Note that both $m^2+m$ and $q^2+q$ are always even, so the right-hand side of (1) is divisible by $8$. But the left-hand side is not, and we have reached a contradiction.

an odd^2 is always of the form 8k+1.now if it is a ssquare then it wiil be of the form 8k+1.again (2m+1)^2 will be of the form some 8t+1 i.e. 4(2n+1) will be divisible by 8 which leads to contradiction


Simple triangulation over flat torus This is somewhat of a computational question: let me know if it is inappropriate.

I have a flat torus with sone random points marked.

I would like to compute a triangulation of said torus such that my points are the vertexes of the triangulation.

A bit of googling has not given me any results. My naive idea would be to start from the fully connected graph, and every time two edges cross, simply remove one of them. Unfortunately this looks like it's going to be quite expensive, as I have $n^2$ edges each of which can potentially intersect all of the others. I also looked into the Delauney triangulation, but I am not sure it would work on an generic torus, and moreover I have no idea on how to implement it successfully. It also seems rather overkill, since I do not need my triangulation to have any special property.

Is there a simple, greedy way to get a triangulation in non-prohibitive time?

Thank you.

It is possible to construct a triangulation of $n$ points on the flat torus in $O(n)$ time by "brute force" as follows, at least if the points are in general position.  I'll assume we're working on the square flat torus, obtained by identifying opposite sides of an axis-aligned unit square.

First sort the points by their $x$-coordinates and connect them into a polygonal cycle $P$.  Then we can triangulate above $P$ by repeatedly cutting off a triangle $pqr$ whenever $p$ lies below the geodesic segment $pr$.  Similarly, we can triangulate below $P$ by repeatedly cutting off a triangle $pqr$ whenever $p$ lies above the geodesic segment $pr$.  This triangle cutting can be implemented in linear time using a stack, similar to the Graham scan convex hull algorithm.

After both these phases, we have triangulated the points on the square flat cylinder obtained by identifying the left and right sides of the unit square.  The top and bottom edges of this triangulation are closed horizontal geodesics, passing through the "highest" point $a$ and the lowest point $z$.  To complete the toroidal triangulation, we can extend two diagonals upward from $a$ to $z$.

In the following figure, the heavy black segments show the initial polygonal cycle $P$, the edges of the upper triangulation of $P$ are red, the edges of the lower triangulation of $P$ are green, and the final two diagonals between the highest and lowest points of $P$ are blue.

The flat torus can be constucted by identifying the opposite sides  of a square in the plane.

Take an element in center of such a square and an element at the middle of each side. Draw a segment between the element in the center and the corners of the square, and between the element in the center and the elements in the middle of each side. This induces a triangulation of the torus.

You can generalize this for the $n$-torus which can be constructed by identifying opposite face of a parallepiped, triangulate each face  such that no point is identifyied in the quotient and draw a segment between a point in the center and the vertices of the triangulation on the faces.


What values of $p$ make $d$ a metric? I'm trying to determine what values of p make the following a metric:
$$d(x,y)=|x-y|^p$$ for x,y∈R.
Obviously, it's not difficult to show that this satisfies the first two conditions for most values of $p$. I'm just not seeing how to prove the triangle inequality one way or the other for some values of $p$, specifically values in $(-1,1)$ and for positive non-integers. Any guidance would be appreciated.

HINT: The triangle inequality applied to the points $0,1$, and $2$ already imposes a significant condition on $p$. Notice also that

$$|x-z|^p\le|x-y|^p+|y-z|^p\quad\text{for all }x,y,z\in\Bbb R$$

if and only if

$$|u+v|^p\le|u|^p+|v|^p\quad\text{for all }u,v\in\Bbb R\;,$$

and you may find the latter version easier to think about and work with.

Try using minkowski's inequality. You then can use it to prove that $(a+b)^p<a^p+b^p$ with $0<p<1$. Finally work with this to get the triangle inequality, dunno if this helps, Im a first timer in analysis.


I lost my love of math; I'm getting it back. How can I determine if math is actually right for me? This question has been on my mind for a very long time, and I thought I'd finally ask it here.

When I was 6, my dad pulled me out of school.  The classes were too easy; the professors, too dull.  My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background.  He wanted me to have a different life and taught me math accordingly.  When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university.  I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all).  However, around this time, I came to the conclusion that math was not for me.  I decided to try a different path.

Why, you might ask, did I do this?  The answer was simple: I didn't believe I could be a great mathematician.  While I thrived taking the courses, I never turned math into a lifestyle.  I didn't come home and do complex questions on a white board.  I didn't read about Euler in my spare time.  I also never felt I had a great intuition into problems.  Once you showed me how to solve a problem, I was golden.  But start from scratch on my own?  It seemed like a different story entirely.  To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby.  Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else.

Over time, I came to agree with this statement.  Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there.  This crushed me at the time; my dream of studying math at one of the great institutions had ended.  Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned).  Being at an engineering school, I thought I'd give aerospace a shot.  It had lots of math, right?  Not really, or at least not enough for my taste.  I went into CS.  This was much better, but still didn't feel quite right.  At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS.

My question is, how do I know I'm not making a mistake?  There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career?  I apologize for the long and possibly quite subjective post.  I'd just really like to hear from math people who know their stuff.  Thanks a bunch in advance.

There is a lot of hype surrounding "math competitions", since "winning" and "competition" can easily be understood in broad cultural terms, whether or not we collectively think of these as either highest goals or as legitimate formative principles. Most professional mathematics and its practice does not resemble competition-math at all, specifically, serious projects have substantial background requirements, may take months or years to complete or partially complete, and have meaning beyond "getting the result before anyone else" (although some people've been so indoctrinated in math-as-game that they never recover, and find no other meaning in it).

Also, there is much hype about "the other people" being incredible geniuses, while few sane people will look at themselves and see "a genius". :)  But this is mostly gossip or mythology, in part generated to create a superficial excitement where otherwise there'd be mostly hard work without pop-style-glamour. :)  And, of course, there is a common style of "bluffing" and/or never admitting weakness or ignorance, but this is mostly a facade, whose maintenance is most enthusiastic among those most worried about the "game" aspect of mathematics or anything else. Such stuff should not be unquestioningly believed.

To make a living as a mathematician, and to make reasonable contributions, it is not necessary to be a larger-than-life romantic-heroic figure. :)  Possibly many of us, occasionally, wish that we were such a figure, but that more comic-book or video-game reality than any human-lifetime reality.

More important than hype is the concrete reality of how one spends one's days: if you like thinking about mathematical things, perhaps teaching mathematical things, then being a mathematician is a happy job, with or without rich-and-famousness. The purported glamor of "being a great [whatever]" is not a reliable thing to aim toward, since the "process" of practicing the [whatever] is how one will spend one's days. It is a reasonable analogy, I think, to say that "great musicians" occur by accident among the class of people who really, really enjoy practicing and "jamming" (in whatever genres), rather than people who wish for celebrity but hate practicing. Of course there're the people who pose as never practicing, but reality belies that cute P.R. pose...

This would be my personal answer. I have medals in international physics olympiad. What do I do now?

I am a businessman.

Yes. I love Math. I love money and women more.

No I will not be a great mathematician or physicist. Why would I?

I still love Math. I read calculus, accounting. But then again, besides programming skills, most truly advance Math is useless in real life. Eventually it's money that turns me on.

People say do what you love rather than make money. That's true. But keep an eye on money.

As of now, I really don't feel like learning more Math. I want to learn more about biz, having the right mindset, law of attraction, evolutionary psychology, and so many things that are actually useful. Yea I can derive most college level and high school level formula. Why invent? Just learn makes me happy. Why discover new things for others?

I am still able to derive all Math formulas all the way to colleges I think on top of my head. I taught Math when I was poor and I love it. But then, I have no time for it. In a sense I still love Math. Being a businessman means I no longer have time to learn it. I wish I can pass on my skills in learning to others but alas, I'll die with it too I guess.

I once use my math skills to taught high school children Math. They are impressed that I can derive all those formulas. But that's it. A $100 a month job. Needless to say I made far more money in biz. I honestly do not think people like me are meant to get a job anyway. So yea, that's how useful Math is. Interpersonal skills? Applied Math? Programming? Internet marketing? Now that gets me very far.

If anyone wants to verify my IPHO claim privately and know how to do so, let me know. Really. It's my life skills learned through experiences, rather than Math is how I got where I am. It's what gets me cash. It's what gets me laid. It's what gets me a wife (that I am now divorcing because my math shows that marriage is inefficient).


Order of the Elements in $D_4$ I am having trouble finding the order of each element in $D_4$.

I know $D_4 = \{R_0, R_{90}, R_{180}, R_{270}, V, H, D, D'\}$ and $|D_4| = 8$.

How would I go about finding the order of the elements?

Start with the definition of order, which loosely means in this context: "how many times can you iterate one of the operations before everything is back to how it started?" For instance, $R_{90}$ refers to a rotation by 90 degrees. If we do that 4 times, our object is back to how it started, so the order of that element is 4.

The Dihedral Group D4 has representation $ \{I,A,B,C,D,E,F,G\} $ which
$ I=
\begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix} $,
$ A=
\begin{bmatrix}
    0 & -1 \\
    1 & 0
  \end{bmatrix} $,
$ B=
\begin{bmatrix}
    -1 & 0 \\
    0 & -1
  \end{bmatrix} $,
$ C=
\begin{bmatrix}
    0 & 1 \\
    -1 & 0
  \end{bmatrix} $,
$ D=
\begin{bmatrix}
    -1 & 0 \\
     0 & 1
  \end{bmatrix} $,
$ E=
\begin{bmatrix}
    0 & 1 \\
    1 & 0
  \end{bmatrix} $,
$ F=
\begin{bmatrix}
     1 & 0 \\
     0 & -1
  \end{bmatrix} $,
$ G=
\begin{bmatrix}
    0 & -1 \\
   -1 & 0
  \end{bmatrix} $
By multiplication of Matrices, the order of each element is
$ |I|=1 $, $ |A|=4 $, $ |B|=2 $, $ |C|=4 $, $ |D|=2 $, $ |E|=2 $, $ |F|=4 $, $ |G|=2 $.


Probability that last child is a boy Johnny has 4 children. It is known that he has more daughters
than sons. Find the probability that the last child is a boy.

I let A be the event that the last child is a boy, P(A) = $\frac{1}{2}$.
and B be the event that he as more daughters than sons. But im not sure how to calculate P(B) and what are the subsequent steps to take after.

Appreciate any help.

Thanks

If he has more daughters than sons, Below are the 5 possible cases:

D D D D --> All Daughters
S D D D --> 3 Daughters
D S D D
D D S D
D D D S

So probability of having last child as son is = 1/5.

They are mutually exclusive events. How many son/daughters the couple already has is irrelevant. Hence, the probability of the last child (or rather any child) being boy is 1/2 (some would say 1/3 too).


How to solve this quasilinear parabolic evolution equation (result of curve shortening flow)? In the classic paper by Hamilton and Gage (see http://intlpress.com/JDG/archive/1986/23-1-69.pdf), they give the PDE problem:

Find $k:S^1 \times [0,T) \to \mathbb{R} \text{ s.t }$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\text{(i) } k \in C^{2+\alpha, 1+\alpha}(S^1 \times [0,T-\epsilon]) \text{ for all } \epsilon > 0$$

$$\text{(ii) }\frac{\partial k}{\partial t} = k^2 \frac{\partial^2 k}{\partial \theta^2} + k^3$$

$$\;\;\;\;\;\;\;\;\;\text{(iii) }k(\theta, 0) = \psi(\theta) \text{ satisfies }$$

$$\qquad \;\;\;\;\;\;\text{(a) } \psi \in C^{1+\alpha(S^1)}$$

$$\qquad \text{(b) } \psi(\theta) > 0$$

$$\qquad \;\;\;\;\;\;\;\;\;\;\;\;\text{(c) } \int_0^{2\pi} \frac{\cos(\theta)}{\psi(\theta))} = 0$$

And they say that is a shown from standard results. Every paper that I've seen that references this paper says that H-G prove this (but they don't, they just state it), and the ones that talk about it give references to papers that are difficult (maximal regularity and Volterra conditions and I don't know which spaces to use as it's quite an abstract result (by Herbert Amann)) that may not be necessary to use anyway. Does anyone know what the standard results are? I have a book on quasilinear equations but it looks like this equation doesn't fit the conditions that are required for the existence result it gives. It's all very confusing because there are lots of different requisites in different sources...

I appreciate any help.

Gage and Hamilton prove this fact in the paper "The Heat Equation Shrinking Convex Plane Curves" see page 80 of this paper.. I confess that I didn't understand the argument too..
I can't access the link you furnished..


If $T$ and $T^2$ have same image, then $N(T) \cap Im(T)={0}$ Let V be a finite dimensional vectorial space and $T:V \rightarrow V$ a linear operator. Show that if  $T$ and $T^2$ have same image, then $N(T) \cap Im(T)={0}$.

My first attempt to solve this problem was think in the matrix associated to the linear transformation. I tried to use the fact that the span of the column vectors of the matrix T was equal to the span of vectors of the matrix T^2, but as I don't know the dimension of the matrix T, this problem got very complex.

Consider the endomorphism $T$ restricted to its image. This gives an endomorphism $T'$ of $T(V)$. The assumption that $T^2$ and $T$ have the same image implies that $T'$ is surjective. Hence, since $T(V)$ is finite-dimensional, $T'$ is also injective. This implies that $\mathrm{Ker} \ T \cap T(V)=0$.

The proof is indeed rather sketchy: If $w$ belongs to $N(T) \cap \mbox{Im} T$, then $w=T(v)=0$, for some $v$ in $V$. Using the hypothesis $\mbox{Im} T=\mbox{Im} T^2$, we get
the sequence of identities
$$w=T(w)=T(0)=0$$
so that $w\in \{0\},$ as desired.


Integral $I(a,b)= P\int_{0}^{\pi}\frac{d\theta}{a-b\cos\theta}$ Hi I am trying to calculate this integral
$$
I(a,b)= P\int_{0}^{\pi}\frac{d\theta}{a-b\cos\theta},\quad 0
<a<b,\quad a,b\in \mathbb{R}.
$$
We can first write
$$
I(a,b)=\frac{1}{2} P\int_{-\pi}^{\pi}\frac{d\theta}{a-b\cos\theta}
$$
since integrand is even.  Now we will use $z=e^{i \theta}$, $dz=ie^{i\theta} d\theta$, $d\theta=dz/(iz)$.  We can write the $\cos$ function as
$$
\cos \theta=\frac{1}{2}(z+z^{-1}).
$$
We can now write the integral as
$$
I(a,b)=\frac{1}{2i}  P \oint_{|z|=1} \frac{dz}{z(a-\frac{b}{2}(z+z^{-1}))}=\frac{1}{i} P\oint_{|z|=1} \frac{dz}{2az-bz^2-b}=-\frac{1}{ib} \oint_{|z|=1} \frac{dz}{z^2-2z\frac{a}{b}+1}.
$$
But I am starting to get stuck here.... We can define $\gamma= a/b < 1$ and now calculate the poles of the integrand.  The poles are given by
$$
z^2-2\gamma z+1=0 , \ z_{\pm}=\gamma +\frac{1}{2} (4\gamma^2-4)^{1/2}=\gamma \pm i(1-\gamma^2)^{1/2}.
$$
If we take the squared magnitude of roots we obtain
$$
|z_{\pm}|^2=\gamma^2 + (1-\gamma^2)=1
$$
thus we can see that $z_{\pm} \in |z|=1$.
How can we proceed , it seems we just have simple poles?  Thanks

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$\ds{{\rm I}\pars{a,b}
     \equiv\pp\int_{0}^{\pi}{\dd\theta \over a - b\cos\pars{\theta}}\,,
     \qquad 0 < a < b\,,\quad a, b\in {\mathbb R}}$.

  \begin{align}
{\rm I}\pars{a,b}&
=\Re\ \overbrace{\int_{0}^{\pi}{\dd\theta \over a - b\cos\pars{\theta} + \ic 0^{+}}}
^{\ds{t \equiv \tan\pars{\theta \over 2}}}\
=\ \Re\int_{0}^{\pi}{2\,\dd t/\pars{1 + t^{2}}
\over a - b\pars{1 - t^{2}}/\pars{1 + t^{2}} + \ic 0^{+}}
\\[3mm]&=2\Re\int_{0}^{\infty}
{\dd t \over a\pars{1 + t^{2}} - b\pars{1 - t^{2}} + \ic 0^{+}}
={2 \over a + b}\Re\int_{0}^{\infty}
{\dd t \over t^{2} - \mu^{2} + \ic 0^{+}}
\\[3mm]&\mbox{where}\quad \mu \equiv \root{b - a \over b + a} > 0.
\end{align}

\begin{align}
{\rm I}\pars{a,b}&
={2 \over a + b}\pp\int_{0}^{\infty}{\dd t \over t^{2} - \mu^{2}}
={2 \over a + b}\lim_{\epsilon \to 0^{+}}\pars{%
\int_{0}^{\mu - \epsilon}{\dd t \over t^{2} - \mu^{2}}
+\int_{\mu + \epsilon}^{\infty}{\dd t \over t^{2} - \mu^{2}}}
\end{align}

  $$
\int{\dd t \over t^{2} - \mu^{2}}
={1 \over 2\mu}\int\pars{{1  \over t - \mu} - {1 \over t + \mu}}\,\dd t
={1 \over 2\mu}\,\ln\pars{\verts{t - \mu \over t + \mu}}
$$

\begin{align}
{\rm I}\pars{a,b}&
={2 \over a + b}\,{1 \over 2\mu}\lim_{\epsilon \to 0^{+}}\bracks{%
\ln\pars{\epsilon \over \verts{2\mu - \epsilon}}
-\ln\pars{\epsilon \over 2\mu + \epsilon}}
\end{align}

  $$\color{#00f}{\large%
{\rm I}\pars{a,b}
     \equiv\pp\int_{0}^{\pi}{\dd\theta \over a - b\cos\pars{\theta}} = 0}
$$

Note that you made an incorrect assumption. If $a/b\le 1$, then the integral is not defined and so one must assume that $a/b>1$. Then you have two simple poles, one inside the unit disk and the other outside. Now just compute the residue for the one inside the unit disk.


Tree of groups $(\mathcal{G},T)$ Let $G$ be act on $\Gamma$ with a fundamental domain $T$ where $T$ is tree. We construct tree of groups $(\mathcal{G},T)$ with the following structure:
$$\text{for every } v\in V(T),\,\,G_v=\operatorname{Stab}_G(v) $$

$$\text{for every } e\in E(T),\,\,G_e=\operatorname{Stab}_G(e) $$

Assume that $G_T$ is the direct limit of the system $(\mathcal{G},T)$. With using the universal property of the definition of the direct limit, we get the map $\phi\colon G_T\mapsto G$.

Now my question is:

If $\Gamma$ is connected, then why can we conclude that $\phi$ is a surjective map?

This is precisely the statement of Lemma 4 in Chapter 4.1 of Serre's Trees book. In terms of your specific problem in the lemma is a little too general and you can set $Y=T$. However, the previous Lemma 2 which is being generalized is too specific. This book is a beautiful read and strongly recommended!

Note that this is an answer to your question in terms of Bass-Serre-Theory. That is also why Lee Mosher's answer is incorrect, since in his example there exists no fundamental domain.

We cannot conclude that $\phi$ is surjective, because it is not true in general. Here is a counterexample.

$\Gamma = \mathbb{R}$ with vertex set $\mathbb{Z}$.
$G = \mathbb{Z}$ acting on $\mathbb{R}$ by translation.
$T = [0,1]$ is a fundamental domain.

The stabilizer of each vertex $0$ and $1$ of $T$ is trivial, and the stablizer of the unique edge $[0,1]$ in $T$ is trivial. So $G_T$ is the trivial group. There is no map from the trivial group onto $\mathbb{Z}$.


Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).

I've tried completing the square,

$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
 \bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1  \\
z= \frac{a^2 +b^2}{2ab}  \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ ,  \\
$$
which doesn't nearly suggest the simple end result.

I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.

Context.

Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$

Indeed, one can see that

$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$$

Going from the RHS to the LHS is called fraction decomposition.

Your result should be clearer now:

$$z^2 - z\left(\frac{a^2 + b^2}{ab}\right) + 1 = z^2 - z\frac{a}{b} - z\frac{b}{a} + 1 = \left(z - \frac{a}{b}\right)\left(z - \frac{b}{a}\right)$$

it must be $$z_{1,2}=\frac{a^2+b^2}{2ab}\pm\sqrt{\left(\frac{a^2+b^2}{2ab}\right)^2-1}$$
and we can factorize to $$-\frac{(a-bz)(az-b)}{ab}$$ and $$ab\ne 0$$


How to solve congruence system What is the step how can I solve following system of congruences (that is one system):

  $7x-8y≡5 \pmod {11}$
  $2x+5y≡9 \pmod {11}$

Adding the first equation with twice the second one we obtain

$$7x-8y+2(2x+5y)≡5+2(9)  \implies 11x+2y\equiv23 \implies2y\equiv 1 \mod 11$$

$$\implies 6\cdot 2y\equiv 6\cdot 1 \implies 12y\equiv6\implies y\equiv 6 \mod 11$$

Then from the second equation

$$2x+5y≡9  \implies 6\cdot 2x+6\cdot 5y \equiv 6\cdot 9 \implies12x+4\equiv 10 \implies x\equiv 6 \mod 11$$


Definition of a linear equation in two variables General form of linear equation using two variables I always read can be written:
$$
 Ax + By + C = 0
$$
where $A$ and $B$ cannot both be $0$.

So does it means like
$$
 1x + 0y + c = 0
$$
is a linear equation of two variables i.e basically what I am asking is that both $A$ and $B$ cannot be zero but can either $A$ or $B$ be $0$ like:
$$
  0x +4y  -5 =0
\quad\text{or}\quad
  3x +0y -15 =0
$$
Are the above equation a Linear equation of 2 variables?

Please clarify my doubt .

Let's discuss the equation $0x+4y-5=0$.  Geometrically this equation specifies a horizontal line whose $y$ intercept is $(0,\frac 54)$.

Technically speaking this equation only has $1$ variable -- and interestingly it's not $y$.  $y$ is completely determined so it is a constant.  The variable is actually $x$.  Even though $x$ doesn't appear anywhere in the simplified equation $4y-5=0$ it is still the only variable in this equation.  But the number of "variables" in this sense isn't really that important.  What's important in determining what this equation really means is the "space" that this equation exists in.  In this case that "space" is the plane.

Perhaps it would be instructive to parametrize this line. Parametrization means finding a formula for the points on the line (rather than an equation relating the $x$ and $y$ values) in terms of other variables called parameters.

Because $y$ is completely determined in this case we don't need to introduce a parameter for it.  But $x$ can take any value. So set $x=t$ for real number $t$ (this is called the parameter). Then we see that every point on the line has the form $(x,y)=(t,\frac 54)$. If you plug in any value of $t$ you like you'll get a point that's on your line.  So even though no $x$'s appear in the simplified equation $4y−5=0$, we can still say that this is an equation relating both $x$ and $y$.

$x=o$  is the equation of the $y$-axis; and every pt. on it is (0,t), where t is a real number...

This is indeed a linear equation in two variables.


Help in proving that $\bigl (n^\alpha x^n (1-x) \bigr)_{n=0}^{\infty}$ is convergent. I need some help in the following question:

Let $\ 0<\alpha \in \mathbb R$. If $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ a  sequence of functions as $\forall n\in\mathbb N \ \ \ f_n:[0,1]\to \mathbb R$:$$\forall n\in\mathbb N, \ \ \forall x\in\mathbb [0,1]: \ \ \ f_n(x)=n^\alpha x^n (1-x)$$

Q:  1. Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is pointwise convergent in $[0,1]$.

Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is uniformly convergent iff $\alpha<1$.

My thought till now: For the first part I tried to prove $\lim_{n\to\infty}n^\alpha x^n (1-x)=0$ for a constant $x$ but the $n^\alpha$ is confusing me and I don't know how to continue.
Thanks in advance

Hint:
Just show the logarithm of
$$\ln(n^\alpha x^n)=\alpha\ln n+n\ln x \to -\infty.$$

We can write estimation:
$$|f_n(x)|=|n^\alpha x^n (1-x)| \leqslant n^\alpha$$
from which for $\alpha < 0$ you have as 1 so 2.
For case $0 < \alpha < 1$ It's easy find, that we have maximum in  $x=\frac{n}{n+1}$ and then  use criterion
$$\lim_{n \to \infty} \sup_{}|n^\alpha x^n (1-x)|=\lim_{n \to \infty} n^{\alpha} \left( \frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}=0$$


Why is second homology/fundamental group of $RP^3$ trivial? I'm looking at a space that I believe is the real projective space $RP^3$. It seems that cutting the space in half with a surface should furnish a non-trivial second homology element and a non-trivial second fundamental group element. By "cutting in half" I am thinking of $RP^3$ as a three ball with opposite points identified, and then slicing this guy down the middle. So I don't see how such a surface can be the boundary of a volume, nor do I see how to contract it. All the references seem to agree that both these groups are trivial so I'm obviously missing something. Perhaps I am being derailed by the intuition from $RP^2$ where it seems clear that $H_1(RP^2)=\pi_1(RP^2)=Z_2$.

Edit: just to clarify, I'm looking for a geometric intuition here, not a calculation, unless a calculation can show me for example an explicit homotopy to the trivial element or an explicit volume for which this surface is the boundary of.

Another Edit: I've just realized that this "cutting the space in half" produces a surface that is not a 2-sphere (which is what lives in the second fundamental group) but actually it is a $RP^2.$ So even if this "cut" is not contractible (as i believe it is not) it says nothing about $\pi_2(RP^3).$

Write the Serre exact sequence attached to $Z/2\rightarrow S^3\rightarrow RP^3$, you have $\pi_2(S^3)\rightarrow \pi_2(RP^3)\rightarrow \pi_1(Z/2)$, since $\pi_2(S^3)=\pi_1(Z/2)=1$, you have $\pi_2(RP^3)=1$.

For the homology see the link, the standard way to compute the homology of a projective space is to use a cellular decomposition.

http://topospaces.subwiki.org/wiki/Homology_of_real_projective_space


Picturing different functions in the complex plane Just wanted to ask a quick question. Which is, if I am taking a complex function for example: $z = 2x + iy$ then say I wanted to plot it for $x \in (-\infty,\infty)$ where x is a real variable, and $y=0$. What I am wondering is what would that look in in the complex plane ? Would it just be a line along the real axis ? I know it wouldn't be a typical thing someone may do. But I am asking this because I had the idea if, using the complex plane as the vertical axis as the imaginary axis, and the horizontal axis as the real axis.  If my intuition is true then could you also extend this to all real functions such that all real functions be just a flat line that lays along the real axis when plotted in the complex plane ? I do apologize if this is completely trivial but I found it interesting in if this is the case how even complicated real functions would be the same/similar as simple real functions when plotted in the complex plane. If anyone has any suggestions for this please let me know as I find this interesting.

edit: added in the fact the x-variable is a real variable.

This function would simplify to $z=2x$ if you set $y=0$, so it would look like a straight line with slope $2$ in the $z$–$x$ plane (or the $Re(z)$–$x$ plane, if you're thinking of $z$ as complex).

If you want more general visual intuitions for what functions from the complex plane to the complex plane look like, I put together some Python code that helped me a lot—let me know what you think!


If $\limsup\limits_{n\to\infty}a_n=+\infty$ and $\limsup\limits_{n\to\infty}b_n\in\mathbb{R}$, then $\limsup\limits_{n\to\infty}(a_n+b_n)=+\infty$? There is the following exercise in a book:

$\limsup \limits_{n \to \infty} (a_n + b_n) \leq \limsup \limits_{n \to \infty} a_n + \limsup \limits_{n \to \infty} b_n $

And the author(Kazuo Matsuzaka) writes as follows in his solution without proofs:

If $\limsup \limits_{n \to \infty} a_n = +\infty$ and $\limsup \limits_{n \to \infty} b_n = +\infty$, then $\limsup \limits_{n \to \infty} (a_n + b_n) = +\infty$.
If $\limsup \limits_{n \to \infty} a_n = +\infty$ and $\limsup \limits_{n \to \infty} b_n \in \mathbb{R}$, then $\limsup \limits_{n \to \infty} (a_n + b_n) = +\infty$.

Please tell me how to prove the above two statements if they are true.
Please tell me counter examples if the above two statements are false.

Hint:

You just have to prove that, for all $n\in\mathbf N$, one has:
$$\sup_{k\ge n}(a_k+b_k)\le\sup_{k\ge n}a_k+\sup_{k\ge n}b_k $$
then use the basic results on (ordinary) limits.

The first is easy to prove:

Since,for all $m \in \mathbb{N}$:

$$
\limsup_{n\to\infty}a_n \leq \sup_{k\geq m} a_k
$$
If:
$$
\limsup_{n\to\infty}a_n = +\infty
$$
Then:
$$
\sup_{k\geq m} a_k = +\infty
$$
Therefore if:
$$
\limsup_{n\to\infty}a_n = +\infty\\
\limsup_{n\to\infty}b_n = +\infty
$$
Then, for all $k \in \mathbb{N}$:
$$
\sup_{k\geq m}a_n = +\infty\\
\sup_{k\geq my}b_n = +\infty
$$
Therefore:
$$
\sup_{k\geq m}a_n > 0\\
\sup_{k\geq m}b_n > 0
$$
Therefore:
$$
\sup_{k\geq m}a_n + \sup_{k\geq m}b_n > \sup_{k\geq m}a_n
$$
By the linearity of $\sup$:
$$
\sup_{k\geq m}(a_n + b_n) > \sup_{k\geq m}a_n
$$
Taking limits on $k$(allowable since the inequality is valid for all $k \in \mathbb{N}$):
$$
\lim_{k\to\infty}\sup_{k\geq m}(a_n + b_n) > \lim_{k\to\infty}\sup_{k\geq m}a_n
$$
Thus, by the definition of $\limsup$:
$$
\limsup_{n\to\infty}(a_n + b_n)  > \limsup_{n\to\infty}a_n
$$
Since:
$$
\limsup_{n\to\infty}a_n = +\infty
$$
Therefore:
$$
\limsup_{n\to\infty}(a_n + b_n) = +\infty
$$

I'll have to think about the second statement.


Problem working out a second derivative I am studying maths as a hobby and feel I am having problems working out second derivatives.
The problem is as follows:
Find the maximum or minimum values of the function $y = (2x - 5)^4$
Here is my working:
$$dy/dx = 4(2x - 5)^3\cdot 2 = 8(2x -5)^3 ,$$
Which is zero when $x = 2\frac {1}{2}$.
To find whether this is a maximum or a minimum I find the second derivative:
$$d^2y/dx^2 = 24(2x - 5)^2.2 = 48(2x - 5)^2.$$
But this is where I feel I must have gone wrong because this is zero when $x = 2\frac {1}{2}$.

The second derivative test, that is, computing the second derivative of $f$ does not always tell you whether or not a point is a local maximum or local minimum. It is only always true that if $f''(a) > 0$, then $f$ has a local minimum at $a$.
Hence, when you encounter (as in this case) $f''(a) = 0$, a simple alternative would be to fall back to inspecting the first derivative in an interval around the critical point. For instance, if $x < a \Rightarrow f'(x) < 0$ and $x > a \Rightarrow f'(x) > 0$, then we can safely conclude that the critical point at $x=a$ is a local minimum.
The computations should be trivial given that you have worked out the value of $x = 2\frac{1}{2}$ to be a stationary point.

If the second derivative test fails, I usually check the first derivative at points enclosing the point of interest, e.g. 2 and 3
since the first derivative at x=2 is negative and that at x=3 is positive, I will conclude that it is a minima at x=2.5


Subgroups of a group If we have a group $G$ of order $10$, and $H$ is a subgroup of $G$ then Lagrange's theorem states each $H$ may be of order $1, 2, 5$ or $10$. Now we have the trivial subgroup $\lbrace e\rbrace$ of order $1$ and the subgroup of $G$ itself. Clearly both of these contain the identity element ($e$). Do all subgroups of a group contain the identity element?

As mentioned, part of the definition of a subgroup is that $e \in H$. Actually there is a deeper reason for this requirement:

We want $H$ to be a group. Then $H$ must have an identity, lets call it $e'$. Now we have in $H$

$$e'e'=e' $$

But this identity must hold in $G$, and multiplying by $e'^{-1} \in G$ we get that
$$e'=e $$

It follows that as long as we ask for $H$ to be a group, with the same operation as in $G$, it's identity must be the identity of $G$.

This is the reason why the definition always asks for $e \in H$.

P.S. If $G$ is finite, as long as $H$ is a non-empty subset of $G$ such that for all $x,y \in H$ we also have $xy \in H$, it follows that $H$ must contain $e$.

Do all subgroups of a group contain the identity element?
YES!  A subgroup is a subset of a group which is a group in its own right.... do all groups require an identity?  YES!!!


Showing limit does not exist at c ≠0. Define $f(x)$ by $$f(x) = \left\{\begin{array}{cc}x,&x\in\mathbb{Q}\\0,&x \notin \mathbb{Q}.\end{array}\right.$$

Show that $\lim_{x \to 0} f(x)=0$ but $\lim_{x \to c} f(x)$ does not exist for any $c \neq 0$.

My attempt:

At $x = c$, the right hand limit (RHL) is
$\begin{align}
\lim_{x \to c^{+}}f(x) & = \lim_{h \to 0} f(c + h) \\
& = \lim_{h \to 0} c + h \\
& = \lim_{h \to 0} c + 0 \\
& = c. \tag{3}\label{3}
\end{align}$

The left hand limit (LHL) is
$\begin{align}
\lim_{x \to c^{-}}f(x) & = \lim_{h \to 0}f(c - h) \\
& = \lim_{h \to 0} 0 \\
& = 0. \tag{4}\label{4}
\end{align}$

From $\ref{3}$ and $\ref{4}$,

$$\lim_{x \to c^{+}}f(x) \neq \lim_{x \to c^{-}}f(x),$$

so the limit does not exist at $x = c$.

Regarding the first part, for any $x \in \mathbb{R}$ it should be clear that $|f(x)|\leq |x|$ (note this is trivially true for irrationals), hence

$$\lim_{h\to 0} |f(0+h)| \leq \lim_{h \to 0} |0+h| = 0$$
so $\lim_{h\to 0} f(0+h) = 0 = f(0)$, hence $f$ is continuous at $x=0$.

On the other hand, take any $c \neq 0$. If $f$ were continuous at $c$ then we would need to show that for any $\epsilon > 0$ there exists some $\delta>0$ so that if $|x-c|<\delta$ then $|f(x)-f(c)|<\epsilon$.

To show that this doesn't work take any $\delta>0$, if $c$ is rational then choose $x$ irrational in $(c-\delta, c+\delta)$, if $c$ is irrational then choose $x$ rational in $(c-\delta,c+\delta)$ so that $|c|<|x|$, in either case, given that $c\neq 0$ we have
$$|f(x)-f(c )| = \begin{cases}|c| & c\in\mathbb{Q}\setminus \{0\}\\|x| & c\not\in\mathbb{Q}\\\end{cases}
\geq |c| > 0$$
so $\epsilon$ cannot be made smaller than $|c|$, so there is no continuity at any $c\neq 0$.

Short answer:

In any neighborhood of $c>0$, $\min_\limits x f(x)=0$ and $\max_\limits x f(x)\ge c$, hence $|f(x)-L|<\delta$ is not possible.


Homeomorphic manifolds have the same dimension So I want to prove: If two manifolds $M$ and $N$ are homeomorphic then $dim(M) = m = n = dim(N)$.

My idea was to use the property of the manifolds that they are locally homeomorphic to the $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively. So I want to take an open subset $U \subset M$ which is homeomorphic to the $\mathbb{R}^m$, but my problem is how do I know that the image $f(U) = V \subset N$ is homeomorphic to the $\mathbb{R}^n$. I probably don't, but do we even have the existence of such $U$ and $V$? If so, why? Is this even the right approach to the proof?

Let $M,N$ be manifolds with dimension m, n respectively. Suppose $\Phi:M \to N$ is homeomorphism. Let $\psi_1: M \supset D \to U \subset \mathbb R^m$ be a map (that is homeomorphism of some open subset of $M$ and open subset of Euclidean space). $U$ is open so we can choose open some ball in it and restrict $\psi_1$ to subset that is mapped onto this ball. So without loss of generality you can assume some subset of $M$ is homeomorphic to open ball in $\mathbb R^m$. Analogously we construct homeomorphism $\psi _2$ from open subset of $N$ to open ball in $\mathbb R^n$. Therefore $\psi_1 \circ \Phi \circ \psi_2 ^{-1}$ maps ball in $\mathbb R^n$ to ball in $\mathbb R^m$. That is what I meant before editting. Open ball in $\mathbb R ^n$ is homeomorphic to $\mathbb R^n$ so final result is that $\mathbb R^n $ and $\mathbb R^m$ are homeomorphic. I assume you know that this implies $n=m$.

Consider a neighbourhood $O_M$ in $M$. Since $M$ is a manifold, it's locally homeomorphic to $\mathbb{R}^m$ for some integer $m$. Let $f_1$ be the homeomorphic mapping from $\mathbb{R}^m$ to $O_M$. Now consider the homeomorphism $f_2$ from $M$ to $N$. Since $f_2$ is a homeomorphism, $f_2(O_M)$ is a neighbourhood in $N$, call it $O_N$. And finally, $O_N$ is homeomorphic to $\mathbb{R}^n$ for some integer $n$ and call that homeomorphism $f_3$.

Now composing the three homeomorphisms, we get $f_3 \circ f_2 \circ f_1$, which is a homeomorphism from $\mathbb{R}^m$ to $\mathbb{R}^n$. Now if $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^m$, then $n=m$, by some theorem whose name I've forgotten.