nature of diophantine solutions in general I had a question in my mind from many years.  we generally present the trivial solutions  to  Diophantine equations. Diophantine equations usually always have some sort of trivial solution (if you allow zero even Fermat’s equation has trivial solutions: $0^n+x^n=x^n$). I suppose the hard question is whether there are non-trivial solutions? The question of non-trivial solutions is already tough for polynomial Diophantine equations and requires detailed knowledge in algebraic and arithmetic geometry, modular forms, etc (see Wiles’ proof of Fermat’s last theorem).

What I am looking for, can we find non-trivial solutions to any polynomial Diophantine equations? If yes, what are the method so far existing?

Thanks in advance to all members of mse.

This question is Hilbert's 10th problem, from Hilbert's famous list of 23 problems promulgated 112 years ago in 1900.

The work of Julia Robinson, Martin Davis, Hillary Putnam, and Yuri Matijasevich, culminating in 1970, showed that it lacks an algorithmic solution.

A polynomial Diophantine equation is an indeterminate polynomial equation whose solutions are restricted to be polynomials in the indeterminate. A Diophantine equation, in general, is one where the solutions are restricted to some algebraic system, typically integers

So you can refer to this wiki article and also this one for a background. Apart from that there is another article that serves a good start. For those who can access IEEE journals, this one is a very nice one.

Thank you and all the best.


Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:

Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)

I know one of solutions is $5776=(76)^2$.

We need $D^2\equiv D\pmod{10}$ hence $D(D-1)$ must be a multiple of $10$. This implies that $D\in\{0,1,5,6\}$.

Next, the tens digit of $(CD)^2=(10\cdot C+D)^2=100\cdot C^2+20\cdot C\cdot D+D^2$ is determined by the ones digit of $2\cdot C\cdot D$ and the tens digit of $D^2$.

For $D=0$ we need $2\cdot 0\cdot C\equiv C\pmod {10}$, so $C=0$.
For $D=1$ we need $2\cdot 1\cdot C\equiv C\pmod {10}$, so $C=0$.
For $D=5$ we need $2\cdot 5\cdot C+2\equiv C\pmod {10}$, so $C=2$
For $D=6$ we need $2\cdot 6\cdot C+3\equiv C\pmod {10}$, so $C=7$

Thus the full list of answers is
$$00^2=0000\quad 01^2=0001\quad 25^2=0625\quad 76^2=5776 $$
and possibly you won't count the first three as valid.

If a programmatic solution (Python 2.7) is acceptable, then:

for n in range(0,100):
    if (n*n)%100 == n:
        print "%04d"%(n*n)

Gives:

0000
0001
0625
5776


Complex Numbers: How do I know which plane to use? I'm new to complex numbers and I want to know precisely when I need to use which plane for graphing and a general idea of what the plot would look like.

$$z=a+bi$$

When it's in the form above, I know to interpret that as the point $(a,b)$. I would plot this on the complex plane using the Imaginary and Real axes as a point.

$$z=re^{i\theta}$$

When it's in this form, I assume that I'd have to also use the complex plane. I know that the radius is $r$ and that you go counter-clockwise around the origin according to the value of $\theta$. However, I'm not sure how to actually draw it. Would it be a ray or a line? And for either one, does it include the origin?

Point $z=re^{i\theta}$ is given in polar coordinates. It means that you have a radius $r$ and an angle $\theta$. You can always go back to cartesian coordinates using the formula:

$$re^{i\theta}=r\cdot \cos(\theta)+r\cdot i\sin(\theta)$$

There's a nice representations of both coordinate systems at Wikimedia:

$re^{i\theta}=r\cos\theta+ir\sin\theta$

Hence
$$a=r\cos\theta$$
$$b=r\sin\theta$$
and you can plot it in exactly the same way as the $a+bi$ example, by first calculating the values $a$ and $b$ using the above formulae.


4 color theorem equivalent to cubic planar bridgeless are 3 edge colorable To follow on
[How to prove Tait's theorem about planar cubic bridgeless graph being 3-edge-colorable?

The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.

I disagree with the solution given (As stated in my comment). The provided links does not proove the equivalence. It shows
1) from 4 color-theorem, how to build a 3-edge coloring for bridgeless cubic graph
2) from a 3-edge-coloring, how to build a 4 face coloring for the same graph

The theorem by Tait is much more powerful. If I can 3-edge color any cubic bridgeless planar graph, then I can 4-color ANY planar graph (not just cubic bridgeless planar).

Any idea how to prove the equivalence. I cannot find the original paper from Tait. Lots of reference but never the actual proof.
The implication 4CT $\Rightarrow$ 3-edge-coloring for bridgeless planar cubic graph is easy.
The other implication is the one missing :
$$ \{ \forall G, \text{ cubic, planar, bridgeless}, \exists \text{ a 3-edge coloring}\}$$
$$\Rightarrow$$
$$\{\forall G, \text{ planar,} \ \exists \text{a 4-vertex-coloring}\}$$

Suppose we want to 4-color the vertices of a planar graph.

We may assume it's simple, because loops are forbidden and multiple edges don't affect coloring.

We may assume it's a maximal planar graph (that is, a planar triangulation), because adding more edges to triangulate the graph only makes the problem harder.

All planar triangulations on at least 4 vertices are 3-connected.

Instead of coloring the vertices of this graph, we can color the faces of its dual. The dual is another planar graph. It is cubic (because we started with a triangulation) and it is 3-connected (because it's the dual of a simple planar 3-connected graph) so in particular it is bridgeless.

So we have reduced the problem to coloring the faces of a planar cubic bridgeless graph, which is the kind of graph that Tait's theorem applies to.

Why do you say that the explanation in the link does not prove the equivalence? You can decide to color the faces of the map OR you can color the edges of the same map. Once you have finished coloring one or the other (faces or the edges) you can switch to the other the way described in the link. The difficulty of three coloring the edges or four coloring the faces is the same. Since one has already been proved the other is proved too.

About the theorem of Tait, there is no theorem of Tait, at least not about four coloring.

The equivalency only states that for the same map (not any map) if you find one coloring (faces of edges) you can switch to the other coloring, it does not say that it is easier to find one or the other coloring. To go from one coloring to the other (related to the same map) the algorithm is described in the link. The algorithm is the proof of the equivalency.

UPDATE (25/Jan/2019): The proof should be here:

P.G. Tait, ”On the colouring of maps”, Proceedings of the Royal Society
of Edinburgh Section A, 10: 501-503, 1878-1880.

https://www.cambridge.org/core/journals/proceedings-of-the-royal-society-of-edinburgh/article/4-on-the-colouring-of-maps/6385395D337371EE9ED16C6AA6694AC0


Density in the unit sphere Prove that $\left\{ ( \sin a \cos b , \cos a\cos b , \sin b)\mid a , b \in \mathbb{N} \right\}$ is dense in the unit sphere.

Any help will be appreciated.

Because $\cos(n)$ and $\sin(n)$ (for n an integer) are dense on the interval $[0,1]$ (because $\pi$ is irrational) and those map continu to your coordinates.


In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$? Given dual numbers, what would be the value of $0^\varepsilon$ and $\varepsilon^\varepsilon$?

I suspect that neither one of your expressions has a canonical value.  Exponentials can be tricky because the operation of exponentiation conflates two (and a half) loosely-related concepts: the family of power functions $f_n(x):x\to x^n$ that raise $x$ to the $n$th power (and more abstractly, the notion of 'exponential object' in a category that considers $A^B$ as the collection of all the functions from $A$ to $B$) and the (canonical) exponential map $\operatorname{exp}(x):x\to e^x$ that (in a very abstract sense) maps 'local to global'.

For any ring $\mathcal{R}$, the family of functions $f_*$ has a straightforward, canonical definition as a set of functions from $\mathcal{R}$ to $\mathcal{R}$: $f_n$ is just the $n$-times multiplication of $x$ with itself.  (On the other hand, these functions don't necessarily have inverses: $f_2$ - that is, the squaring function - is a map from $\mathbb{Z}\to\mathbb{Z}$, but there's no $n\in\mathbb{Z}$ with $f_2(n)=2$.)  More broadly, this idea leads to the aforementioned notion of an exponential object in a category as a means of organizing functions from $A$ to $B$ - in this sense, $n^2$ just counts the way you can pick two (ordered, not necessarily distinct) numbers between $1$ and $n$, and likewise $2^n$ counts the number of ways you can pick $n$ 'bits'.

On the other hand, the exponential map is a fundamentally distinct concept; whereas the power maps have range a subset of their domains (since multiplication is a map $\mathcal{R}\times\mathcal{R}\to\mathcal{R}$), the range of an exponential map can be entirely distinct from its domain.  For instance, the exponentiation map on a Lie algebra takes elements of that algebra to their 'exponentiations' in a Lie group; see http://en.wikipedia.org/wiki/Exponential_map#Lie_theory for some details on this case.  While the two structures are related, they're fundamentally distinct; in a sense the Lie algebra studies local structure, while the Lie group looks at global structure, and exponentiation serves as a sort of 'integration' to show how the local structure extends out into a global structure.

A classic example of this is rotation (for simplicity, about the origin): on a 'local' (or 'instantaneous') level a rotation in three dimensions picks out an axis (the axis being rotated around) and a velocity; in two dimensions it just picks out a velocity (the speed of rotation).  The exponential map then takes this to a transformation of the underlying two-dimensional space; it serves as a map (e.g.) $\operatorname{rot}:\mathbb{R}\to SL_2(\mathbb{R})$ from real numbers to $2\times 2$ matrices of real numbers with determinant $1$ that integrates out the 'instantaneous rotation' to the resulting transformation after one (abstract) unit of time.  Note that even though this can be written as $e^X$ (and that it happily can be computed using the power series for exponential in this case), it's not a function that takes the two arguments $e$ and $X$; and raises one to the power of the other; it's fundamentally a function that takes a single argument $X$ and gives you its exponential.

The fact that these two distinct concepts of exponentiation (essentially) happen to agree over the real numbers (and that there's a real number $e$ such that $\operatorname{exp}(x) = e^x$) is, in some sense, a deep coincidence; it's not something that you can expect to happen anywhere else.  In particular, this means that you shouldn't expect expressions like $\varepsilon^\varepsilon$ to be defined on the dual numbers, because the definition of exponentiation that shows up in the duals is essentially the exponential map rather than an exponential object; its exponentiation is a one-argument function, not a two-argument one, so the expression itself doesn't make sense.

I suspect that $0^\epsilon=0$.

If we define the exponent $\exp(x)$ in the usual way $\exp(x):= 1 + x +\frac{x^2}{2!} + \ldots$ then in the duals for $ k \cdot \epsilon$ is $\exp(k\, \epsilon) = 1 + k\,\epsilon $.
http://en.wikipedia.org/wiki/Dual_number

Then if we use $a^b= \exp (\log(a) b) $ we get $\epsilon^\epsilon= 1+ \log (\epsilon) \epsilon$.
However, I am not sure that $\log( \epsilon)$ is well defined.


How to compare periodically and continuously compounding interest, when time is in-between periods? I am working on a textbook problem, and I think I disagree with the solution. The problem is

  (10.6) Bruce deposits 100 into a bank account. His account is credited interest
  at a nominal rate of interest of 4% convertible semiannually. At the same
  time, Peter deposits 100 into a separate account. Peter’s account is credited
  interest at a force of interest of δ. After 7.25 years, the value of each account
  is the same. Calculate δ.

The solution provided is $\delta = .0396$, which I can get by setting these two accumulation functions equal:
$$
(1+\dfrac{.04}{2})^{2 \times 7.25}=e^{7.25 \delta}
$$
The reason I feel this is incorrect is that the semiannual accumulation function should not permit a time value of $t=7.25$, which is between periods. Instead I feel that a value of $t=7$ should be used, being the end of the most recent period:
$$
(1+\dfrac{.04}{2})^{2 \times 7}=e^{7.25 \delta}
$$
This yields $\delta \approx 0.0382$.

The same principal invested for the same period of time yields the same accumulated value. So over $1$ year we must have the equivalence
$$
\left(1+\frac{i^{(2)}}{2}\right)^2=\mathrm e^\delta
$$
and then
$$
\delta=2\ln\left(1+\frac{i^{(2)}}{2}\right)\approx 0.0396
$$

Usually you use the $\texttt{simple interest}$ for the remaining three months.

$$\left(1+\frac{0.04}{2}\right)^{2 \cdot  7}\cdot \left(1+\frac{0.04}4\right)=e^{7.25 \cdot \delta}$$

$$7.25 \cdot\delta=\ln\left[1.02^{14}\cdot 1.01 \right]$$

$$7.25 \cdot\delta=0.287187$$

$$\delta=\frac{0.287187}{7.25}=0.039612\approx 3.96\%$$


Identifying the series $\sum\limits_{k=-\infty}^{\infty} 2^k x^{2^k}$ I came across following  bi-infinite sum:

$\sum_{k=-\infty}^{\infty} 2^k x^{2^k}$

Is this a known series? After some plotting I have the feeling that it could be equal or very similar to

$-\frac{1}{\ln(2) \ln(x)}$

for all $x\in(0,1)$.
Is my assumption correct?

Note that $f(x^2)=\frac12f(x)$, $x\in(0,1)$. Thus with $g(x)=1/f(e^{-x})$ we have $g(2x)=2g(x)$, $x\in(0,\infty)$. And with $h(x)=\ln g(e^x)-x$, $x\in\Bbb R$, we have $h(x+\ln 2)=h(x)$, which must therefore be periodic.

Numerically (summing $f$ only from $k=-50$ to $50$, but that should be pretty good when computing $h(x)$ for $0\le x\le\ln 2$), $h$ seems to make only a (sine-like?) oscillation between $\approx-0.3665228$ and $\approx-0.366503$ (both $\approx \ln\ln 2$). So if $h(x)=\ln\ln 2+\epsilon(x)$ with $\epsilon(x)\approx 0$, then $g(x)=e^{\ln\ln 2+\epsilon(\ln x)+\ln x}=v(x) x \ln 2$ with $\upsilon(x)=e^{\epsilon(\ln x)}\approx 1$, and then
$$f(x)=\frac1{g(-\ln x)}=\frac{-1}{v(-\ln x)\ln 2\ln x}.$$
The numerical results seem to indicate that $|\epsilon(x)|<9.885\cdot 10^{-6}$ so that $|\upsilon(x)-1|<9.885\cdot 10^{-6}$, i.e.,
$$f(x)\approx -\frac1{\ln 2\ln x} $$
with a relative error $<10^{-5}$. On the other hand, the oscillation does not seem to disappear when computing more summands of $f$, which means that this bound for the relative error
is pretty sharp.

That sum can be written as $\sum_{k=-\infty}^\infty \left(2x^2\right)^k$ which I would then break into two parts: $\sum_{k=-\infty}^0 \left(2x^2\right)^k+ \sum_{k= 0}^\infty \left(2x^2\right)^k- 1$ (The -1 is there because k= 0 occurs twice and one has to be subtracted off.)

$\sum_{k= 0}^\infty \left(2x^2\right)^k$ is a geometric series with "common ratio" $2x^2$. That has sum $\frac{1}{1- 2x^2}$.  $\sum_{k= -\infty}{0} \left(2x^2\right)^k= \sum_{k= 0}^\infty \left(\frac{1}{2x^2}\right)^k$ is also a geometric series with "common ratio" $\frac{1}{2x^2}$.  That has sum $\frac{1}{1-\frac{1}{2x^2}}= \frac{2x^2}{2x^2- 1}$


Area Between Three Circles of Differing Radii From the link in wikipedia

http://web.gnowledge.org/wiki/index.php/Area_Between_Three_Circles_of_Differing_Radii

OPEN QUESTION:

What is the equation, in three variables, relating the radii of three circles to the area between them, when each is tangent to the other two?

But wiki says that it is an open question  and I am interested what is main reason of it? or is the author simply trying to understand if anybody can solve it (while it's answer is known?)

As a follow-up to Gerry's answer and Day Late Don's comment there, let's consider the triangle formed by the centers of the circles and subtract the sectors.  Call the circles 1, 2, and 3, with radii $r_1$, $r_2$, and $r_3$, respectively, and let $\theta_1$, $\theta_2$, and $\theta_3$, be the interior angles of the triangle formed by the centers of the circles, at the centers of circles 1, 2, and 3, respectively.

Applying the Law of Cosines (and simplifying some) to find $\theta_1$, $\theta_2$, and $\theta_3$ gives:
$$\theta_1=\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$$
$$\theta_2=\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$$
$$\theta_3=\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$$
so the sector areas are
$$\frac{1}{2}\theta_1r_1^2=\frac{1}{2}r_1^2\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$$
$$\frac{1}{2}\theta_2r_2^2=\frac{1}{2}r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$$
$$\frac{1}{2}\theta_3r_3^2=\frac{1}{2}r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$$

Applying Hero's formula to find the area of the triangle (and simplifying) gives $$K_\triangle=\sqrt{r_1r_2r_3(r_1+r_2+r_3)}.$$

So the area of the target region is
$$\sqrt{r_1r_2r_3(r_1+r_2+r_3)}-\frac{1}{2}\left(\begin{align}
r_1^2&\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)
\\&+r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)
\\&+r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)
\end{align}\right).$$

In the special case of all 3 circles of the same radius $r$, the area is given by the equation $k r^2$  where $k=0.161254481$.


Show the next inequality Show that:
$$
\frac{3}{8}\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+x}}dx\leq \frac{\sqrt{3}}{4}.
$$
I have
$$m(b-a)\leq\int_{a}^{b}f(x)dx\leq M(b-a),$$
but the function does not present critical points in its derivative.

And how can I show the following?
$$\frac{|a+b|-|a-b|}{ab}=\frac{2}{\max\{|a|,|b|\}}$$

For the first part, there seems to be some typing mistake.

For the second, presume that $\left|\frac{b}{a}\right| \le 1$. Then
$$\frac{|a+b|-|a-b|}{ab}=\frac{|a||1+b/a|-|a||1-b/a|}{ab} = |a|\frac{1+b/a-1+b/a}{ab}=\frac{2|a|}{a^2}=\frac{2}{|a|}.$$
Can you figure out what happens when $\left|\frac{a}{b}\right| \le 1$?

If you have
written it correctly,
$\frac{3}{8}
\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\leq \frac{\sqrt{3}}{4}
$
is the same as
$\frac{3}{8}\sqrt{3}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{\sqrt{3}}{4}\sqrt{3}
$
or
$\frac{3\sqrt{3}}{8}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{3}{4}
$.

$\sqrt{1-x}$
is monotonic decreasing,
so,
on $[0, \frac12]$,
$1 \ge \sqrt{1-x} \ge \frac{\sqrt{2}}{2}$.
Therefore
$\frac1{2\sqrt{3}}
\approx 0.2886
\geq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\geq \frac{\sqrt{2}}{4\sqrt{3}}
\approx 0.20412
$.

If the denominator
is actually $x+2$,
then
$\sqrt{\frac{1-x}{x+2}}
$
is again monotonic decreasing
so,
on $[0, \frac12]$,
$\sqrt{\frac12}
\ge\sqrt{\frac{1-x}{x+2}}
\ge \sqrt{\frac{\frac12}{\frac52}}
= \sqrt{\frac15}
$
so the integral
is within half of these bounds.

For the second question,
use
$\min(a, b)
=\frac12(|a+b|-|a-b|)
$
and
$\max(a, b)\min(a, b)
=ab
$.


Galois theory : Intermediate fields of $L = \mathbb{Q} ( \sqrt 2 , \sqrt 3 , \sqrt 5)$ I want to get all the intermediate fields of $L=\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$.

Please help me and give me advice.

Let $L=\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$. Then $L/\mathbb Q$ has degree 8. By the way, element of Galois group has to through permute roots of $x^2-2$, permute roots of $x^2-3$ and roots of $x^2-5$. Then all element has order $2$, and thus $$\text{Gal}(L/\mathbb Q)=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z.$$

Find the subgroup of $\text{Gal}(L/\mathbb Q)$ and conclude using the Galois correspondance theorem.

The Galois group of the extension is $G=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, now:

$1)$ Show that besides trivial subgroups, $G$ has $6$ subgroups of order $2$, and $6$ subgroups of order $4$.

$2)$ Conclude that these are all the intermediate fields of $L/\mathbb{Q}$.

$$\mathbb{Q},$$$$\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{5}),\mathbb{Q}(\sqrt{6}),\mathbb{Q}(\sqrt{10}),\mathbb{Q}(\sqrt{15}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}(\sqrt{2},\sqrt{5}),\mathbb{Q}(\sqrt{3},\sqrt{5}),\mathbb{Q}(\sqrt{5},\sqrt{6}),\mathbb{Q}(\sqrt{10},\sqrt{3}),\mathbb{Q}(\sqrt{15},\sqrt{2}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$$


Find the arc length of the cardioid: r = 3-3cos θ This is what I have so far:

Using the formula $\mathrm ds = \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm dθ}\right)^2}$

$$\frac{\mathrm dr}{\mathrm d\theta} = 3\sin\;\theta $$

$$r^2 = 9 - 18\cos\;\theta + 9\cos^2\theta$$

$$\mathrm ds = \sqrt{9 - 18\cos\;\theta + 9\cos^2 \theta + 9\sin^2 \theta}$$

using $\cos^2 \theta + \sin^2 \theta = 1$,

$$\mathrm ds = \sqrt{18(1-\cos\;\theta)}$$

Then I have

$$\int_0^{2\pi} \sqrt{18(1-\cos\;\theta)}\mathrm d\theta$$

But I'm not sure how to integrate this.

So that this does not remain unanswered:

Exploiting the trigonometric identity (which can be obtained from the cosine's double-angle formula):

$$1-\cos\;\theta=2\sin^2\frac{\theta}{2}$$

your integral turns into

$$6\int_0^{2\pi} \sin\frac{\theta}{2}\mathrm d\theta$$

which you should be able to handle.

Alternatively, since the cardioid is symmetric about the horizontal axis, you can instead start with the integral

$$2\int_0^{\pi} \sqrt{18(1-\cos\;\theta)}\mathrm d\theta$$

This answer is correct and equal to 24.

Hint: Also note,

$$1−\cos(\theta)=2\sin^2\left(\frac{\theta}{2}\right)$$


Trigonometry: Law of Cosines How to solve using rule of cosines? I can solve using law of sines but trying to check using rule of cosines is tripping me up, can anyone help clear things up?

By writing down the law of cosines for this situation, we have
$$ 5^2=a^2+4^2-2\cdot a\cdot 4\cdot \cos 65^\circ$$
i.e.
$$ a^2-3.3809 a-9\approx 0.$$
That is a quadratic in $a$, hence you'll find two solutions, but one will be negative.

I'm new to using latex so please excuse the displayed equations. The underlying mathematics should be accurate. Feel free to correct it if you know how.

The interior sum of angles of a triangle equal $180^\circ$. Therefore we can use [Law of Sines][1] to find $\angle A$ opposite a by deduction.

Let B be the angle opposite 4 and C=$65^\circ$. Then we have c=5.

$$\frac a{\sin A^ \circ}=\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac 4{\sin B^\circ}=\frac 5{\sin 65^\circ}$$
$$\frac 1{\sin B^\circ}=\frac 5{4\sin 65^\circ}$$
$${\sin B^\circ}=\frac {4\sin 65^\circ}5$$
$$B^\circ=\sin^{-1} ({\frac {4\sin 65^\circ}5})$$
$$B=46.47269373^\circ$$

Now we can find the third angle by [Sum of Angles of triangle][1] = $180^\circ$

$$180^\circ=A+B+C$$
$$180^\circ-(A+C)=B$$
$$B=68.53^\circ$$

Now you can use either [law of cosines] or [law of sines] to find $a$

Law of Cosines states:
$$a^2=b^2+c^2-2bc\cos A^\circ$$
$$a=\sqrt{b^2+c^2-2bc\cos A^\circ}$$
$$a=\sqrt{5^2+4^2-2(5x4)\cos 68.53^\circ}$$
$$a=4.91$$


A way to write derivatives There are many notations when it comes to derivatives such as $f'(x)$ and $\frac{dy}{dx}$. As of the moment, I can't find a reference that uses $dy/dx$ instead of $\frac{dy}{dx}$.
Is it acceptable to write $\frac{dy}{dx}$ as $dy/dx$ or not?

Yes, absolutely. Writing $\frac{dy}{dx}$ as $dy/dx$ is much the same as writing $\frac{3}{5}$ as $3/5$. There is no mathematical difference between the two notations—only a typographical one.

Writing $\frac{dy}{dx}$ or $dy/dx$ is a matter of choice, mathematical speaking they are both ratios, so I can even write as $[ dy : dx ]$
But when compiling a good math text, in textbooks or some type-settings, it's better to write as $\frac{dy}{dx}$ because it looks more like function than variable
Also consider when we want to write higher derivatives
$$\frac{d^7y}{dx^7}$$
$$d^7y/dx^7$$
vote to choose


Flipping coins- percentages of heads vs tails If I flip a coin multiple times and count the number of time it fell on heads and the number of times it fell on tails and keep a track of them. In how many flips on average will the delta between percentage of heads and percentage of tails will be less than 0.1%?

1) Interpreting the revised question as asking how many tosses $n$ of a fair coin are required to have the probability that $|H-T| \lt 0.001 n $ greater than or equal to $0.99$, where $H$ is the number of heads and $T$ the number of tails with $H+T=n$, this is equivalent to asking for $n$ such that $\mathbb{P}\left(-0.0005n \lt H-\frac12{n} \lt 0.0005n\right) \ge 0.99$ since if $H=\frac12{n}+0.0005n$ then $T=\frac12{n}-0.0005n$ and $H-T= 0.001n$.

It is typical to use a normal approximation to the binomial distribution for questions like this.  $H$ has a mean of $\frac12{n}$, a variance of $\frac14n$ and a standard deviation of $\frac12\sqrt{n}$.  So using $\Phi^{-1}$ as the inverse cumulative distribution function of a standard normal random variable, we have $\mathbb{P}\left(-\Phi^{-1}(0.995) \lt \dfrac{H-\frac12{n}}{\frac12\sqrt{n}} \lt \Phi^{-1}(0.995)\right) \gtrsim  0.99$ or  $\mathbb{P}\left(-\Phi^{-1}(0.995)\frac12\sqrt{n} \lt H-\frac12{n} \lt \Phi^{-1}(0.995)\frac12\sqrt{n}\right) \gtrsim 0.99$.

So we need to solve  $0.0005n \gtrsim \Phi^{-1}(0.995)\frac12\sqrt{n}$ which gives  $n \gtrsim (1000\Phi^{-1}(0.995))^2$ and since $\Phi^{-1}(0.995)\approx 2.575829$ this gives  $n \gtrsim 6634897$.

But this is only an approximation. We could use computer calculations of the binomial distribution to get exact answers, but there is an issue that due to rounding every thousand, the probability is not a monotonic function of $n$.  I believe that the smallest $\mathbb{P}\left(\left|\frac{H-T}{n}\right| \lt \frac1{1000}\right) \approx 0.9900006$ for $n=6633001$ but about $0.989995$ for $n=6636000$, with any greater $n$ giving over $0.99$.  So the normal approximation gives a reasonable answer.

2) The original question is actually slightly different, asking "In how many flips on average will the delta between percentage of heads and percentage of tails will be less than $0.1\%$?" Interpreting this as asking for the expected cumulative number of times in the $n$ tosses it is true that $\left|\frac{H-T}{H+T}\right| \lt \frac1{1000}$, it is possible to use linearity of expectation and to add up the probabilities of being in this interval.  For small values of $H+T$, it makes sense to use the binomial distribution as rounding matters: for example with $n \le 1000$ the probability is zero for odd $n$.

As I said in a comment while the question was closed, my calculations suggest that for $1$ flip, the expectation is $0$; for $10$ flips the expectation is ${2 \choose 1}\frac1{2^{2}}+ {4 \choose 2}\frac1{2^{4}}+ {6 \choose 3}\frac1{2^{6}}+ {8 \choose 4}\frac1{2^{8}}+ {10 \choose 5}\frac1{2^{10}}\approx   1.71$; for $100$ flips about $7.04$; for $1000$ flips about $24.25$; for $10^4$ flips about $540.19$; for $10^5$ flips about $16663.55$; for $10^6$ flips about $483950.29$.  For small $n$ the expected number is much smaller than $n$, but for very large $n$ most cases will fall in the relative small interval around the mean.

Great question! So let's look at the binomial distribution. We can represent the standard deviation with the equation $\mu_x = \sqrt{npq}$ and $p$ in our case is the probability of heads. This is $.5$. Same with tails. $n$ is the number of tosses. So $\mu_x = \sqrt{.5 * .5 * n} = .5*\sqrt{n}$

Our mean here is zero and our standard deviation is $.5\sqrt{n}$. Because we can apply the same rules as the normal distribution, if we had, let's say $10,000$ tosses, 68% will fall within 1 standard deviation, so $.5\sqrt{10,000}$ which is $50$. For 99% confidence, we would do 3 standard deviations, so for this would be $150$. So we can be 99% confident that the difference is equal to or less than 150 tosses. You can apply this logic to your problem to get the answer.

Note: I'm not an expert in this field so please do feel free to edit anything that isn't proper terminology.


On a zero-sum game betting market I'm trying to come up with the rules for a betting game. My problem is to figure out under what constraints that game has zero-sum game properties: in other words, I want to make sure that no money is created nor destroyed, but is rather exchanged between participants of the game.

I think I'm starting off with an ill-posed problem, so don't hesitate to point out the modeling errors you may find, or if you have any idea to better translate the problem into maths.

The Rules

People can bet on the outcome of a particular future event for a certain amount of money. We suppose the event has two different outcomes, $k$ and $\neg k$ and from now on, let's call $P_k$ and $P_{\neg k}$ the respective probabilities of this event to end with those outcomes.

Let's call $S_{k}$ and $S_{\neg k}$ the amounts and $N_{k}$ and $N_{\neg k}$ the count of bets on each of our two outcomes. $S$ is the sum of the amounts on either of the outcomes, and $N$ is the sum of the bets in either of the outcomes. In other words, $S=S_{k}+S_{\neg k}$ and $N=N_{k}+N_{\neg k}$.

Now, let's introduce the notion of time: it's a turn-based game and at every step, a player can buy a $x$ new bets on one of the two issues for a certain price. And he can also sell back existing ones. We note the current step $t$, with $t \in \{0,\dots, N-1, N\}$, N being the last step during which the outcome is revealed. So all the previous notations can be enriched with the $t$ parameter:

$N_{k}(t)$ is the number of bets on outcome $k$ at the time $t$.

$S(t)$ is the total amount of money in play at the time $t$.

If the players could only buy new bets and the money could only be redistributed when the outcome is known, it would be pretty simple to ensure the zero-game property. When the outcome is known, we would have to give $S(t)$ divided by the number of bets on each bet on the winning outcome, and nothing to all the bets on the losing outcome.

But instead, at each step, the players can buy a new bet whose price depends on the ratio of the previous bets: the higher the bets on an issue are, the more expensive they get. And on top of that, players can also sell their current bets and make a profit without even waiting for the issue to be revealed. For now, there's no other assumption on the way the price for buying a new bet is calculated, except that it grows with the $N_{k}(t)/N(t)$ ratio.

The Question

A trivial way to ensure the zero-sum condition would be to make sure that every purchase of a winning bet matches the sale of an opposite one, like in stock exchange markets.

I'm trying to find a way to ensure that without the need to pair opposite bets.

If I assume that my bet price can vary between $0$ and $1000$, and that I give $1000$ to every winning bet, then my gain expectancy should be something like:

$$E=\mbox{S}\left( t \right)-P_{k}\left( t \right)\cdot 1000\cdot N_{k}\left( t \right)-P_{\neg k}\left( t \right)\cdot 1000\cdot N_{\neg k}\left( t \right)$$

If $E=0$ then I have a zero-sum game, right?

Now, if I use the number of bets on each outcomes to approximate the probability of the outcome like this:

$$P_{k\left( t \right)}=\frac{N_{k}\left( t \right)}{N\left( t \right)}$$

I can deduct this expression:

$$\mbox{S}\left( t \right)=\frac{1000}{N\left( t \right)}\left[ N_{k}\left( t \right)^{2}+N_{\neg k}\left( t \right)^{2} \right]$$

And to model the act of buying or selling bets, I can write:

$$S(t+1) = S(t)+M$$

$M$ being the price to sell or buy $x$ bets on a particular outcome. Then if I calculate the price with $M=S(t+1)-S(t)$ at every step, then I should have a zero-sum game, right?

But according to my simulations, I don't. I give more money that what has been put in the game. Any idea why? What am I missing?

I think it would need to operate like The Tote in horse racing. Where the players buy tickets for an outcome (each ticket worth the same value), and the winners are paid out dividends from the pool. The winners would get a percentage of the pool based on the number of tickets they held.


Expected Value Problem (Q-function...inside a function) I'm working through my textbook for a communications course I'm taking, and this problem is confusing me big time. Like always, the math questions give me the most problems. Maybe I should take the hint. Anyways, I'll just get to it I guess.

Given a carrier phase synchronization error $x$, it can be shown that the probability of bit error for coherent BPSK is:
$\displaystyle P(e) = Q \left( \sqrt{\frac{2E_b}{N_0}cos^2(x)} \right)$
If $x$ is a random variable uniformly distributed over $-a \leq x \leq a$, where $0 < a \ll \pi$, determine the average probability of bit error as a function of $a$.

The question itself seems pretty straightforward. Bit error is a function of a random variable. I need to find the average bit error. That means I need to figure out an expression for the expected value of bit error. Here's my brief attempt of a solution:
$\displaystyle E[P(e)]=\int_{-\infty}^{\infty}P(e)f_x(x)dx = \frac{1}{2a}\int_{-a}^{a}Q \left( \sqrt{\frac{2E_b}{N_0}cos^2(x)} \right)dx$
Beyond this, I really don't know how to approach the problem. I don't think I can really do anything with the Q-function as-is, so I was thinking to express it as an integral and hoping that I could do something with that.
$\displaystyle E[P(e)]=\frac{1}{(\sqrt{2\pi})(2a)}\int_{-a}^{a} \int_{\sqrt{\frac{2E_b}{N_0}cos^2(x)}}^{\infty} exp \left( -\frac{u^2}{2}\right)du \thinspace dx$
This is obviously pretty messy and I'm not sure if this is even the right way to be going with it. I haven't used the information that $0<a \ll \pi$ yet, so if I had to guess, there might be some sort of trickery I could apply to that double integral and work from there, but I'm not seeing it at this point. If $0<a \ll \pi$ is implying $a < \frac{\pi}{10}$, I could make some (crazy) assumption that $a = 0$, but that would completely break the question and make it immediately solvable from the start with a solution not in terms of $a$, not to mention that it's a pretty broad assumption to make. If it's just suggesting that $a$ is closer to $0$ than to $\pi$, I have no idea how I can use that information to make anything easier. The only possible thing I can think of is that $cos^2(x)$ is decreasing on that interval, but again, no idea how that might help. Maybe I can go back to my previous expression (with the Q-function still intact) and say something about the integral being over a small region of the Q-function and make some assumption there? Again, my knowledge isn't really up to par here and I don't know if any of this is possible.
So that's about it. I'm honestly completely stumped at this point. Any help would be much appreciated.

All you really need to do here is use the fact that $a$ is very small.  You have everything else written down OK:

$$E[P(X)] = \frac{1}{2 a} \int_{-a}^a dx \: Q\left(\sqrt{\frac{2 E_b}{N_0}} |\cos{x}|\right)$$

When $a$ is small, you can approximate the integral by $2 a$ times the integrand evaluated at $a$.  Thus

$$E[P(X)] \approx Q\left(\sqrt{\frac{2 E_b}{N_0}} \cos{a}\right)$$

You can make further approximations on $\cos{a}$, but I am not sure how much that will further simplify things.


Calculation to find the draft value of a fantasy football player I'm a programmer working on an online fantasy football draft board. In this I would like to find the best and worst draft pick per team based on the projected number of fantasy points for the drafted player and the round that he was taken.

It seemed like a relatively easy problem to solve (and it probably is for y'all), but I am unable to :(.

So here are my thoughts:

| Round | Fantasy Points | Draft Value |
----------------------------------------
|   1   |       300      |      85     |
|   2   |       250      |      86     |
|   3   |       200      |      86     |
|   4   |       210      |      93     |
|   5   |       150      |      85     |
.....
|   12  |       250      |      99     |

So basically, round 1 you will obviously get a player with a high projection of points, but that doesn't necessarily make it the best draft value. If you were to, say, in the 12th round draft someone with a projected 250 fantasy points, then that would be a much higher draft value than a number 1 pick of 300.

Make sense?

Is there a formula for doing this?

You could compare the projection with the average value of the positional draft slot. So, say the 5th QB taken has over the past 5 seasons scored 290 points, if the projection is 300, then the value would be +10. You could then compare that surplus value to that of the next RB#, WR#, and TE# on the board. Just keep in mind that not all projections are equal (for example, QB scoring is more predictable than RB scoring).


Confusion in proof of theorem ($2.7$) in Rudin's Real and complex analysis I am not able to fill the gap in proof of following theorem which is stated as...

Let $U$ be an open set in a locally compact hausdorff space $X$, $K\subset U$ and K is compact. Then there exists an open set $V$ with compact closure such that :
$$K\subset V\subset \overline{V}\subset U$$

Proof : Let $p\in K$ as $X$ is locally compact there exists an open set $V_p$ such that $\overline{V_p}$ is closure. See that this collection $\{\overline{V_p}\}_{p\in K}$ is an open cover for $K$. as $K$ is compact this cover has a finite collection which covers $K$.
Suppose $V_1,V_2,\cdots,V_n$ covers $K$. I now set $V=\bigcup_{i=1}^n V_i$ see that $V$ is open being finite union of open sets and $K\subset V$

Now $\overline{A\cup B}=\overline{A}\cup \overline{B}$ so, for similar reasons we have  $\overline{V}=\overline{\bigcup_{i=1}^n V_i}=\bigcup_{i=1}^n \overline{V_i}$ and see that $\overline{V}$ is compact being finite union of compact sets.

So, I have an open set $V$ of $X$ with compact closure such that $K\subset V\subset \overline{V}$

Suppose my $U$ is $X$ then i am done as i need not check if $\overline{V}$ is in $X$

Suppose not then i am not sure if this closure is in $X$

Just to not get confused i now denote the union $\bigcup_{i=1}^n V_i$ as $G$

As $U^c\neq \emptyset$ I denote this $U^c$ as $C$.

I understood that given $p\in C=U^c$ there exists an open set $W_p$ such that $K\subset W_p$ and $p\notin \overline{W_p}$

I do not understand why $\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is empty..

Help me to fill this gap..

I guess i got the solution and i thought it is not a better idea to edit the question..
So, I am writing this..
Suppose  $\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is non empty...
Then i have $x\in \bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$
In particular, $x\in C\cap \overline{G}\cap \overline{W_x}$
But then We have the condition that if $x\in C$ then $x\notin \overline{W_p}$
Thus we have a contradiction... so,$\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is empty.

By local compactness, for each $p \in K$, there exists an open neighborhood $V_p$ of $p$ with $\bar{V_p}$ compact and $\bar{V_p} \subset U$.

By compactness of $K$, there is a finite subset $F$ of $K$ for which $K\subset \bigcup_{p\in F} V_p$. Let $V = \bigcup_{p\in F} V_p$.

Certainly $V$ is open.

Finally, since $F$ is finite, $\bar{V} = \bigcup_{p \in F} \bar{V_p} \subset U$, and $\bar{V}$ is compact.


Intuition behind independence and conditional probability I have a good intuition that $A$ is independent of $B$ if $P(A \vert B) = P(A)$, and I see how you can easily derive from this that it must hold that $P(A,B) = P(A)P(B)$.

But the first statement is not normally taken as a definition; instead the second is.

What is the intuition, or even derivation behind defining $A$ and $B$ as independent iff $P(A, B) = P(A)(B)$?

The kind of explanation I am looking for would be one similar to that given by Jaynes for the definition of conditional probability in the first chapter of Probability: The Logic of Science, or even a Kolmogorov axiomatic explanation would help.

Arguing from the intuitive idea of probability (be it frequentist, Bayesian, or a la Jaynes), what can we say about $P(AB)$? Let us assume that $P(A)\le P(B)$. Since $AB\subseteq B$ we can safely deduce that $P(AB)\le P(B)$. By looking at well-known and elementary examples it is easy to be convinced that $P(AB)$ can attain any value between $0$ and $P(B)$. But examining these cases shows that extreme values, close to $0$ or close to $P(B)$ are obtained when information about $B$ having occurred either severely conflicts with $A$ occurring (to get close to $0$), or strongly correlates with $A$ occurring (to get close to $P(B)$).

Now, more mathematically, one value in the range of $P(AB)$ that appears naturally is, of course, $P(A)P(B)$, so it is natural to investigate when that would occur. Notice that this value is symmetric in $A$ and $B$. Since the exact location of $P(AB)$ in its possible range seems to be highly sensitive to whether, and how, $A$ and $B$ influence each other we must conclude that the special value $P(A)P(B)$, being symmetric in the arguments, means that the mutual influences are neutral. That neutrality is another way of thinking about independence. Thus, we turn the intuition into a definition and say that $P(AB)=P(A)P(B)$ holds if, and only if, $A$ and $B$ are independent.

As a definition of independence, $P(A,B) = P(A)P(B)$ uses intuitively simple concepts involving the probability both events happen and the probabilities each of then happens.  It may not be intuitive for everyone why this is the definition, but it is intuitive what it is.

$P(A \vert B) = P(A)$ uses the less intuitively simple concept of conditional probability, which needs both definition and understanding.


Is there analytical solution to the following integral? I have an integral
$$I=\int_0^{a}\frac{a-x}{c^m+(R^2+x^2)^{m/2}}\,\mathrm{d}x~,$$
where $a>0, R>0$ and $m>0$ are reals. Is there any analytical solution possible for this integral. Solution in terms of Hypergeometric function would do.

Well, you can split the integral into two pieces, as a first thing.

$$a\int_0^a \frac{\text{d}x}{b + (R^2 + x^2)^{\beta}} - \int_0^a \frac{x\ \text{d}x}{b + (R^2 + x^2)^{\beta}}$$

Where I called $b = c^m$ and $\beta = m/2$.

The second integral is analytically solvable in terms of Hypergeometric function as you sighed for. By setting

$$R^2 + x^2 = z ~~~~~~~ \text{d}z = 2x\ \text{d}x$$

We can easily rewrite the second integral as

$$-\frac{1}{2}\int_{R^2}^{R^2+a^2} \frac{\text{d}z}{b + z^{\beta}}$$

whose solution is

$$-\frac{1}{2(\beta-1)} \left\{\left| R\right| ^{2-2 \beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left| R\right| ^{-2 \beta}\right)-\left(\frac{1}{b}\right)^{-1/\beta} b^{-1/\beta} \left(a^2+R^2\right)^{1-\beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left(a^2+R^2\right)^{-\beta}\right)\right\}$$

For what concerns the first part, well it's tougher than what it could seem.
I am still thinking about it but, for example, we can play a bit.

For example, if we assumed that $b << (R^2 + x^2)^{\beta}$, then the first integral would approximate to

$$a\int_0^a \frac{\text{d}x}{(R^2 + x^2)^{\beta}}$$

whose solution is well known and again in terms of Hypergeometric function:

$$a R^{-2 \beta} \, _2F_1\left(\frac{1}{2},\beta;\frac{3}{2};-\frac{a^2}{R^2}\right)$$

So you have a part with an exact solution, and another one with an approximate solution.
Of course we could play the same by considering

$$b >> (R^2+x^2)^{\beta}$$

But that integral would be trivial then.


Is probability determined by perspective? My question: is probability determined by perspective?

The scenario that raised the question for me:

Initial condition: The Monty Hall problem.  We know the contestant’s original choice of door #1 (of 3 total) is only correct 33% of the time.  After Monty Hall reveals door #3 is incorrect the contestant is asked if he would like to switch his answer to door #2.  We know he should choose to change his answer to the other unopened door (#2) which has a 66% chance of being correct.  He does so.

However, let’s say once he decided to switch to door #2, and before either door is revealed, another contestant enters the room.  She does not have any knowledge of what has just transpired on stage.  She is offered a choice to pick which door the car behind of the 2 remains closed doors and also randomly chooses door #2.

Are the probabilities of being correct different for each contestant?  Seemingly they are.  Contestant 1 had 3 doors to choose from, giving him a probability of 33% that door #1 is the answer.  Contestant 2 only had 2 doors to choose from, giving her a probability of 50% after choosing the same door contestant #1 choose.

If we repeat the experiment 1000 times, what will the numbers turn out to be for door #1?  333 or 500?

The other answers have adequately explained this variation of the Monty Hall problem, so I'm going to focus on the question:

  My question: is probability determined by perspective?

The answer is: It depends.

Broadly speaking, there are two schools of thought about what a "probability" is and how it should be used.  The frequentist believes that probability describes something like a roulette wheel, which can be spun many times but always behaves the same (i.e. it always produces the same distribution of results).  When we say $P(X) = \frac{1}{6}$, we mean that, if a given event is tried many times, in the long run the outcome $X$ will happen one sixth of the time.  This convergence is guaranteed by the law of large numbers.  Probability is thus an objective fact about the universe, and not something subject to a person's perspective.  In this worldview, the probabilities in the Monty Hall problem are always 1/3 and 2/3, regardless of whether we know which is which.

The Bayesian, on the other hand, sees probability as a degree of belief.  You might think of this like in a court of law: In order to convict the defendant of some crime, we need to be 99% certain that the defendant committed the crime.  Seeing incriminating evidence may raise our subjective belief in the defendant's guilt, while exculpatory evidence would lower it, both according to Bayes' theorem.  When we're making a verdict, we ask ourselves whether the defendant has at least a 99% probability of having committed the crime.  In the frequentist worldview, this is a nonsensical question; either the defendant is truly guilty or the defendant is truly innocent, and the probability is accordingly either 100% or 0% (we just don't know which).  Similarly, in the Bayesian worldview, the Monty Hall problem is nonsensical unless you specify the person whose worldview we are following and their subjective prior probabilities for each door.  Bayesian reasoning, then, could give you a 50-50 split for your hypothetical second contestant, but only if she started with 33-33-33 priors, and then only if there is no other evidence allowing her to distinguish between the two remaining doors.

It is also important to recognize that the frequentist and Bayesian approaches are not mathematically distinct as both probabilities are subject to the same mathematics (i.e. each system admits both the law of large numbers and Bayes' theorem).  What differs is how the math is applied to the real world.  Because the frequentist deals in objective probability, they cannot tell you "the probability that candidate X wins the election."[1]  Because the Bayesian deals in subjective probability, they cannot tell you much of anything without a set of prior probabilities,[2] which are necessarily tied to a particular observer at a particular point in time and space.  Ultimately, both systems inevitably require making certain assumptions about how your data relates to the real world.  So you should examine those assumptions with care before blindly accepting the result which they have produced.

[1]: The election happens only once; it doesn't make sense to ask how often candidate X wins.  Imagining the election being re-held many times doesn't work either because elections are deterministic, so the same people will vote or not-vote in the same ways every time, and you will get the same result.  Instead, you have to engage in a far more roundabout investigation of the likely level of errors in the polls, which gives a less obviously meaningful value as its final output.

[2]: In cases like Monty Hall, some set of priors is typically "obviously correct" (e.g. "All three doors are equally likely to conceal the car").  However, this still has to be explicitly stated as an assumption of the Bayesian method.  Many circumstances, including elections, have no obviously correct set of priors (though betting markets may be a good first step).  In cases like the court of law, it may be desirable to begin with a set of priors which is "obviously wrong" (we must assume the defendant is probably innocent, even though most criminal defendants are probably guilty).

Is probability determined by perspective?

The answer is undoubtedly yes, but I don't think the Monty Hall problem illustrates the possibility of different perspectives well. Rather I like Bertrand's Paradox.

  Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?

Bertrand then gave three answers depending on different probability distributions. The whole point was that for the problem to have a particular solution then the probability distribution for the chord length distribution must be specified in the problem statement since different distributions lead to different answers.


Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet? Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet."

I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true?  Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.  Therefore, it isn't necessarily given that the monkeys would write Hamlet.

Could someone who's better at this kind of math than me tell me if this is correct?  Or is there more to it than I'm thinking?

I found online the claim (which we may as well accept for this purpose) that there are $32241$ words in Hamlet.  Figuring $5$ characters and one space per word, this is $193446$ characters.  If the character set is $60$ including capitals and punctuation, a random string of $193446$ characters has a chance of $1$ in $60^{193446}$ (roughly $1$ in $10^{344000}$) of being Hamlet.  While very small, this is greater than zero.  So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet.  But don't hold your breath.  It doesn't even take an infinite number of monkeys or an infinite number of tries.  Only a product of $10^{344001}$ makes it very likely.  True, this is a very large number, but most numbers are larger.

Here's a rather interesting  article  discussing the probability of a monkey producing Shakespeare's works and uses a random letter generator to demonstrate some results.


Is $0$ divisible by $0$? I know that there's resources out there but my professor told me that nothing is divisible by $0$ because you can't divide by $0$ which makes it a NaN

The class is related to programming we are defining a "divisible" function that takes two arguments and I have to return true or false, so that's the context. I tried to explain to him what the "Let $a,b∈\mathbb{Z}$. Then $a$ is divisible by $b$ if and only if there exists an integer $k$ such that $a=kb$" as one of the answerers said but the professor said that just because $a$ implies $b$ doesn't mean $b$ implies $a$.

Also, there is a strong hint that we are supposed to use a modulus standard built in function in the programming language and this mod function returns undefined if you ask it to do $0 \mod 0$ so there's also that.

Yes, it is. Formal definition of divisibility is the following:

  Let $a,b \in \mathbb{Z}$. Then $a$ is divisible by $b$ if and only if there exists an integer $k$ such that $a = kb$.

Now, in your case, we have $a = b = 0$, that is, $0 = k\cdot0$ and there exists infinitely many such $k$ so yes, $0$ is divisible by $0$.

If you use algebra, the answer is zero. If A / B = C then C times B = A. For this example, A, B, and C are all zero.

To sum it up in a word problem: A birthday cake will be divided evenly among children at a party. The party got cancelled. There was no birthday cake and no children showed up. How big was the slice of cake each child got?
Answer: Zero grams.

Another example: An electrical wire contains zero watts and zero volts. How many amps of electricity is in the wire? Amps = Watts / Volts.
Answer: Zero amps.


Leibniz rule implies that an operator only depends on the values of a function at a point Let $U$ be an open subset of $\mathbb R^n$ and let $\phi : C^\infty(U) \rightarrow C^\infty(U)$ be a linear map satisfying the Leibniz rule, $\phi(fg) = f\phi(g) + g\phi(f)$. I am trying to prove that if $f =g$ on a neighborhood of $x\in U$, then $\phi(f)(x) = \phi(g)(x)$. Since $f$ is linear, it will suffice to show that if $f \equiv 0$ in a neighborhood of $x$, then $\phi(f)(x) = 0$, but I cannot seem to find a way to do this. I tried using bump functions, but it did not seem to give me the result I need. Any suggestions?

Let $f\equiv 0$ in an open neighborhood $V$ of $x$. Take a smaller neighborhood $W$ of $x$ and take a bump function $\zeta$ such that
$$ W \prec \zeta \prec V.$$
(This means that $\zeta$ takes values in $[0, 1]$, is identically $1$ in $W$ and is supported in $V$). Now consider the function $\overline{\zeta}=1-\zeta$. Since $\overline{\zeta}$ is identically $1$ outside $V$, we have
$$f=\overline{\zeta}f.$$
Applying $\phi$ to both sides of this identity and using Leibniz Rule we get
$$\phi(f)=\overline{\zeta}\phi(f)+\phi(\overline{\zeta})f, $$
which evaluated at $x$ yields
$$\phi(f)(x)=\overline{\zeta}(x)\phi(f)(x)+\phi(\overline{\zeta})(x)f(x)=0+0.$$

Well if $\phi$ is linear, to show that $\phi(0)=0$ is trivial:

$\phi(0) = \phi(0 + 0) = \phi(0)+\phi(0) \Rightarrow \phi(0)=0.$


Decidability and Subformula Property How does one prove the decidability of intuitionistic propositional logic, given the subformula property if a formula $A \rightarrow B$ is a subformula of itself?

Two of the texts I'm reading say that this proof is immediate, but what prevents one from repeatedly applying a rule like:

$\Gamma \vdash A \rightarrow B \\
----- \\
\Gamma \vdash A \rightarrow B$

Is it that decidability is not actually immediate from the subformula property alone and these authors are implicitly appealing to the specific rules of the logic in question, or am I missing something?

In other words, does decidability only follow from a slightly stronger property like the proper subformula property, or some other strictly decreasing order on sequents?

I am looking specifically at Girard's Proofs and Types, in which he says decidability is an "immediate corollary" of the weak normalization theorem on page 24.

Proofs in sequent calculi can be represented as directed acyclic graphs instead of lists. In this case, there are no repetitions, and so there is a finite number of proofs that contain only subformulas of the derived sequent. Decidability follows from this finite search space.


Are there real-life relations which are symmetric and reflexive but not transitive? Inspired by Halmos (Naive Set Theory) . . .

For each of these three possible properties [reflexivity, symmetry, and transitivity], find a relation that does not have that property but does have the other two.

One can construct each of these relations and, in particular, a relation that is
symmetric and reflexive but not transitive:
$$R=\{(a,a),(a,b),(b,a),(b,b),(c,c),(b,c),(c,b)\}.$$
It is clearly not transitive since $(a,b)\in R$ and $(b,c)\in R$ whilst $(a,c)\notin R$. On the other hand, it is reflexive since $(x,x)\in R$ for all cases of $x$: $x=a$, $x=b$, and $x=c$. Likewise, it is symmetric since $(a,b)\in R$ and $(b,a)\in R$ and $(b,c)\in R$ and $(c,b)\in R$. However, this doesn't satisfy me.
Are there real-life examples of $R$?
In this question, I am asking if there are tangible and not directly mathematical examples of $R$: a relation that is reflexive and symmetric, but not transitive. For example, when dealing with relations which are symmetric, we could say that $R$ is equivalent to being married. Another common example is ancestry. If $xRy$ means $x$ is an ancestor of $y$, $R$ is transitive but neither symmetric nor reflexive.
I would like to see an example along these lines within the answer. Thank you.

$\quad\quad x\;$ has slept with $\;y$ ${}{}{}{}{}$

$x$~$y$:$x$ is the classmate of $y$.

I hope this example work.


What Jordan curve of length $1$ maximizes the expected length of a chord between two uniformly picked points on it? For a (sufficiently nice) Jordan curve $\sigma :[0,1] \to \mathbb{R}^2$ with unit length and natural (w.r.t. arc length) parametrization, we denote the expected value of the distance between two points that are picked from the path independently and uniformly by $c(\sigma)$. So

$$c(\sigma) = \mathbb{E}[ \space|\sigma(T_1)- \sigma(T_2)| \space],$$

where $T_1, T_2 \sim U([0,1])$ are independent.

  What path maximizes $c$?

For example, for $\sigma =$ circle (of radius $\frac{1}{2\pi}$ to have unit perimeter),
$$c(\sigma) = \frac{2}{\pi^2}$$

(It is known that for unit circle we get $\frac{4}{\pi}$ so scaling leads to the value $\frac{2}{\pi^2}$.)
Is circle the maximizing path?

PS. Here is a small application for testing different kinds of paths. The circle seemed to be the best among all paths I tried out.

I'm pretty certain that a straight line segment is the optimizing path.

For then the expected length appears to be $\frac{1}{3}$ (if my back-of-the-envelope computation is correct), which is much larger than $\frac{2}{\pi^2}$.


Dominated convergence theorem arctan(nx) Calculate $$\lim\limits_{n\to\infty}\int_a^{\infty}\frac{n}{1+n^2x^2}\,d\mu$$
  where $\mu$ is the Lebesgue measure and $a\geq0$.

First is easy to see that $\arctan(nx) ' = \frac{n}{1+n^2x^2}$ so I concluded that the limit is $\begin{cases}0 &\text{if $a>0$}\\\pi/2&\text{if $a=0$}\end{cases}$.

But the idea is to use the dominated convergence theorem, so if $a>0$ is easy to see that $\frac{n}{1+n^2x^2} < \frac{1}{x^2}$, so we conclude that the limit is $0$. My problem is when $a=0$. I integrate by parts getting

$$\int_0^{\infty}\frac{n}{1+n^2x^2} = \int_0^{\infty}\frac{2n^3x^2}{(1+n^2x^2)^2}$$

On the right side separate the domain on $[0,1]\cup(1,\infty)$ y I demonstrate that
$\frac{2n^3x^2}{(1+n^2x^2)^2} < \frac{1}{x}\in L^1([0,1],\mu)$ and is easy to notice that $\frac{2n^3x^2}{(1+n^2x^2)^2} < \frac{2}{x^2}\in L^1([1,\infty),\mu)$, but the dominated convergence theorem implies that the limit is $0$, not $\pi/2$ as I concluded before.

Where is my mistake?

Do a change of variables.
$$\int_a^\infty{n\,dx\over 1 + n^2x^2} = \int_{na}^\infty {n\cdot (dx/n)
\over 1 + x^2} = \int_{na}^\infty {dx\over 1 + x^2} = {\pi\over 2} - \arctan(na). $$
The last term goes to $0$ at $\infty$.  If you wish to use dominated convergence, it applies in the integral just before the final evaluation.

Here is some $f_n(x)$
I think $f_n(x)\to f$
uniformly in (0,infinity)


Distance between a point and a plane I was just working on some review textbook problems in James Stewart's Multivariable Calculus when I encountered a problem that looked like the following:

  Find the distance between the point $(1,2,3)$ and the $xz$-plane.

I've seen questions that ask for the shortest distance between a point and a plane, but I haven't seen this sort of problem yet.

The way I tackled this problem is by using the distance formula like this:

I called the point $A(1,2,3)$ and called the $xz$-plane as $T(x,0,z)$ and then did $||AT||=\sqrt{(x-1)^2+(-2)^2+(z-3)^2}=\sqrt{(x-1)^2+4+(z-3)^2}$. I'm not sure how to proceed after this. (Maybe this is as far as it goes, because there are infinitely many distances from a point to a plane, depending on what $x$ and $z$ are?)

Any hints would be appreciated. Thanks.

In $\mathbf{R}^3$, the distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is simply $$\frac{|ax_1 + by_1 + cz_1 +d|}{\sqrt{a^2+b^2+c^2}}$$ This is derived in Stewart and is quite easy using the projection of any vector from $(x_1,y_1, z_1)$ to any point in the plane onto a normal vector to the plane. As noted in the comment above, distance usually means shortest distance.

$$
{\cal F}\left(\vec{r}\right)
=
{1 \over 2}\,\left(\vec{r} - \vec{P}\right)^{2} - \mu\,\vec{r}\cdot\hat{y}\,,
\quad
0={\partial{\cal F}\left(\vec{r}\right) \over \partial\vec{r}}
=
\left(\vec{r} - \vec{P}\right) - \mu\,\hat{y}
\quad\Longrightarrow\quad
\vec{r} = \vec{P} + \mu\,\hat{y}
$$

$$
0 = \vec{r}\cdot\hat{y} = \vec{P}.\cdot\hat{y} + \mu
\quad\Longrightarrow\quad
\mu = -\vec{P}\cdot\hat{y}
\quad\Longrightarrow\quad
\vec{r} = \vec{P} - \vec{P}\cdot\hat{y}\ \hat{y}
$$

$$
?
=
\left\vert\,\vec{r} - \vec{P}\,\right\vert
=
\left\vert\,\vec{P}\cdot\hat{y}\ \hat{y}\,\right\vert
=
\left\vert\,P_{y}\,\right\vert
=
\color{#ff0000}{\Large 2}
$$


Calculus question with circle, and string tracing an area The figure shows a piece of string tied to a circle with a radius of one unit. The string is just long enough to reach the opposite side of the circle. Find the area of the region, not including the circle itself that is traced out when the string is unwound counterclockwise and continues counterclockwise until it reaches the opposite side again.

I'm mostly condensing what Eric Weisstein wrote in this MathWorld article, citing a 1998 paper by Hoffman, The Bull and the Silo: An Application of Curvature. You can solve this by finding the area of the semicircle the string traces (before the string begins to wrap) and then do some fancy calculus to find the area of the regions formed when the string begins to wrap around the circle.

By the calculations in the article, given a circle of radius $a$ and a string of length $L$, the area $A$ will be
$$A = \frac{\pi L^2}{2} + \frac{L^3}{3a}\;.$$

So in your case where $a=1$ and $L=\pi$ we have $A = \frac{5}{6}\pi^3 \;\approx\; 25.838564$.

Well, I cheated, and solved it numerically: it comes up at about 19.55.

Unfortunately that does not match up with Mike Pierce's answer of $\frac56 \pi$...

Here is a graph of the interesting part, and here is the Matlab code.


Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$

I start by multiplying with the conjugate:

$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\right)\right]$$

$$\lim_{x\to-\infty} {x^2 - (x^2 + 2x)\over x - \sqrt{x^2+2x}}$$

$$\lim_{x\to-\infty} {-2x\over x - \sqrt{x^2+2x}}$$

divide by highest power of denominator

$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x - ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$

$$\lim_{x\to-\infty} {-2\over 1 - \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$
but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.

Notice that since $x$ is tending to $-\infty$ then $x=-\sqrt{x^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation. That is:

$$\frac{-2}{\frac{x-\sqrt{x^{2}+2x}}{x}}=\frac{-2}{1-\frac{1}{x}\sqrt{x^{2}+2x}}=\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$$

which tends to $-1$ as $x$ tends to $-\infty$.

Me I think, negative infinity minus negative infinity is not indeterminate, it has a meaning, therefore the limit does not exist. You only multiply with the conjugate if the limit are indeterminate


Let $A$ be a square matrix of order $n$ over $\mathbb{R}$ such that $A^2-3A+I_n=0_n$. Then $A^{-1}=3I_n-A$ Let $A$ be a square matrix of order $n$ over $\mathbb{R}$ such that $A^2-3A+I_n=0_n$. Then $A^{-1}=3I_n-A$.

The answer given is true.

Suppose $A^{-1}$ exists.
Then $A^{-1}(A^2-3A+I_n=A^{-1}(0_n)$ which implies $A-3I_n+A^{-1}=0_n$. So clearly, $A^{-1}=3I_n-A$.

But I can't show that the inverse of $A$ exists.

For square matrices, a one-sided inverse is automatically the (unique) inverse. Let's exhibit a one-sided inverse... $3\,{\rm Id}_n-A$, say? We check: $$A(3\,{\rm Id}_n - A) = 3A - A^2 \stackrel{(\ast)}{=} {\rm Id}_n,$$where in $(\ast)$ we use the hypothesis.

i guess this method's easier. Just expand the 'I'


Calculus (Derivatives) prove the given question Let $f(x)$ and $g(x)$ be two continuous and differentiable functions defined on the interval $[0,2]$. Suppose $f(0)>g(0)$ and $f(1)<g(1)$, show that there exists a number $u$ in the interval $(0,1)$ such that $f(u)=g(u)$. Suppose further that $f(2)>g(2)$. Show that there is a number $w$ in the interval $(0,2)$ such that $f'(w)=g'(w)$. (Hint: consider $h(x)=f(x)-g(x)$.)

I haven't really made any progress with this question. I did try applying the Intermediate value theorem, Rolle's theorem and the Mean value theorem but I'm not very sure how to go about it also I don't get why $h(x)$ would be $f(x) - g(x)$. Any hints would be appreciated!

Since the function f(x) is first increasing and then decreasing we get that it must intersect g(x) somewhere between 0 and 1. Also it becomes a monotonous function in between 0 and 2 hence there must be at least one point where the derivates(slopes) are equal hence the result


Let $w, x, y, z$ be natural numbers. Find the correct alternative. Let $w$, $x$, $y$, $z$ be four natural numbers such that their sum is $8\cdot m+10$, where $m$ is a natural number. Given $m$ which of the following is possible:

The max. possible value of $w^2+x^2+y^2+z^2$ is $6\cdot m^2+40\cdot m+26$.
The max. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+28$.
The min. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+28$.
The min. possible value of $w^2+x^2+y^2+z^2$ is $16\cdot m^2+40\cdot m+26$.

The answer given is option 4.

I tried to expand $(w+x+y+z)^2$. This will be $0\ (\operatorname{mod}\ 4)$. Now the only way sum of four integers give even number is if 1. All even, 2. All odd, 3. 2 odd and 2 even. For the first 2 cases if we expand the expression everything except $w^2+x^2+y^2+z^2$ will be a multiple of four. So, $w^2+x^2+y^2+z^2$ should be divisible by $4$. Only case 3 requires $w^2+x^2+y^2+z^2$ to be divisible by $2$. Now if we put $m=1$ and take $w, x, y, z$ as $1, 3, 5, 9$ we can rule out first 2 options of the question. From here, can we argue that option 4 is the possible choice? Since option 4 gives $2\ (\operatorname{mod}\ 4)$ case.

Is there any other general way to prove this? This is a question from a chapter of number theory. But solution in any form is heartily welcome.

You can use the AM-QM inequality;

$$ 2m + \frac 52= \frac {w + x + y + z}4 \le \sqrt{\frac{w^2 + x^2 + y^2 + z^2}4}$$

Hence you easily get $w^2 + x^2 + y^2 + z^2 \ge 16m^2 + 40m + 25$

Your book reports $26$ instead of $25$; this can be fixed by noting that to have equality you should have $w = x = y = z$, but his would imply $8m + 10 = 4w$. The left hand side is not divisible by $4$ though, so we cannot have equality and the minimun attainable is at lest one above the previous bound, that is to say $16m^2 + 40m + 26$.

Of course one should now prove that you can actually reach this; you can do so by following the general line guide "the more the numbers are close to each other, the less their quadratic mean will be" (if the arithmetic mean is fixed, of course)

So following this idea you can try to set $x = w = 2m + 2$, $y = z = 2m + 3$. You can easily see that with this choice $$\begin{cases} w + x + y + z = 8m + 10 \\ w^2 + x^2 + y^2 + z^2 = 16m^2 + 40m + 26\end{cases}$$ and it is indeed the minimum for what we have said before

The simplest aproach would be using Arithmetic Mean - Geometric Mean Inequality.

Let, $A=w+x+y+z = 8m+10$ . Therefore, $A^2 = 64m^2 + 160m +100 \; $

Also, $$A^2=w^2+2 w x+2 w y+2 w z+x^2+2 x y+2 x z+y^2+2 y z+z^2 $$
$$A^2 \le 4w^2 + 4x^2 + 4y^2 + 4z^2 \; \; \;   *$$

Therefore, $$w^2+x^2 + y^2 + z^2 \ge \frac{1}{4} A^2 = 16m^2 + 40m + 25$$
The equality is possible, if and only if $w=x=y=z=2m+2.5$ .

Therefore,

  $$min(w^2+x^2 + y^2 + z^2) = 16m^2 + 40m + 26$$

$* \; \; \; \;  \; 2wx \le w^2 + x^2, \; 2wy \le w^2 + y^2, \; and \; etc.$


Confused about part of Milne's notes on the Galois theory of étale algebras over a field Let $F$ be a perfect field, let $F^c$ be an algebraic closure and let $G=\text{Gal}(F^c/F).$

To each finite set $S$ with continuous $G$-action, Milne associates the $F$-algebra

$$A(S):=\{f:S \to F^c \; | \; f(\sigma\cdot s)=\sigma(f(s)) \;\; \forall s \in S \; \; \forall \sigma \in G\}$$

of $G$-equivariant maps $S \to F^c.$

He then claims that this is an étale $F$-algebra.

It is clear to me that this algebra is reduced but not clear that it is finite-dimensional over $F.$

Confusion: For example, if $G$ acts transitively on $S,$ then surely $A(S)$ is in bijection with $F^c$ and this isn't finite-dimensional over $F$...

To begin with, the functor $A$ from the category of finite sets with a continuous $G$-action into the category of $F$-algebras takes coproducts to products, i.e. $A(S\sqcup T)=A(S)\times A(T)$ holds for any finite $G$-sets $S$ and $T$. This is not hard to show. It is therefore sufficient to consider a transitive $G$-set $S$, noting that every finite $G$-set decomposes into a finite coproduct of such transitive $G$-sets.

Now fix any point $s\in S$ and let $H\subset G$ be the stabilizer of $s$, which means that we find $S\cong G/H$ as $G$-sets by the orbit-stabilizer theorem. Observe that $H$ is a closed subgroup of finite index in $G$. Hence $A(S)\cong A(G/H)$ and I claim that the latter is isomorphic as an $F$-algebra to the fixed field of $F^c$ by $H$ by virtue of the $F$-algebra homomorphism
$$A(G/H)\rightarrow (F^c)^H,~f\mapsto f(H).$$
This is not hard to show, I'll let you work out the details by yourself. Finally, infinite Galois theory tells us that the right-hand side is a finite subextension of $F^c/F$ and therefore étale by separability. Hence $A(S)$ is étale over $F$, which proves the claim.

I assume $F$ is finite. Since $S$ is finite, $G/H$ is finite where $H$ is the kernel of the homomorphism $G\rightarrow Aut(S)$. This implies that the image of every element of $A(S)$ is contained in $L$ where $L$ is the subfield of $F^c$ fixed by $H$. Since $L$ is finite, the result follows.


Are there any explanations for these patterns in the Collatz sequences? I've been messing around with the Collatz sequences a bit, and have come across a few patterns - I was wondering if there are any known explanations for these patterns.

The first is the plot of stopping times (sequence lengths) for all Collatz sequences beginning with the first $100,000$ integers. As confirmed by both the wikipedia page and by this quick Python script/Excel plot, this strange line-pattern exists in the stopping times:

Another pattern was in the number of "hits" for each number. By that, I mean the number of unique sequences a specific number exists in. Generated by this similar Python script, its showing very strange patterns (if anyone wants the excel file generated, its here- just make sure you don't select the ads that look like download buttons). Specifically, one of the patterns I immediately noticed was this counting patterns - every third integer has a countdown number of hits. For example, here are the first 20 integers:

$$
        \begin{array}{c|c}
1 &0 \\
2 &99998 \\
3 &16 \\
4 &99997 \\
5 &93742 \\
6 &15 \\
7 &19146 \\
8 &99996 \\
9 &14 \\
10 &93741 \\
11 &44375 \\
12 &14 \\
13 &47646 \\
14 &19145 \\
15 &13 \\
16 &99995 \\
17 &46031 \\
18 &13 \\
19 &17685 \\
20 &93724 \\
        \end{array}
$$

Notice how everything else looks random, but every third integer counts down. This pattern remains throughout the whole of the $100,000$ integers (though the spaces between each different number in the countdown increase dramatically).

If anyone knows an explanation for these patterns, or can direct me towards further reading about these patterns, it'd be grealy appreciated.

Hmm, I'm not sure that I got you right. But it seems to me, that this is simply a consequence of any tree-structure, that there are numbers near the root through which many pathes go and that there are other numbers, far away from the root, through which less pathes go. And now, the Collatz-transformation-rule can be written in an inverted tree, say
$$ \begin{array}{}
     & 1 \to 2 & \to 4 & \to 8 &\to 16 &\to 32 &\to \cdots\\
     &         &       &       & \to 5 &\to 10 & \to 20 & \to 40 & \to \cdots\\
     &         &       &       &       &       &        & \to 13 & \to 26 & \to \cdots\\
     &              &  &       &       & \to 3 & \to 6 & \to 12 &\to \cdots\\
   \end{array} $$
or much better visible in this (very old) link then it is obvious that the number $2$ is "on the path" of many transformations, and $3$ or $13$ much less - where "many" and "less" is a bit sloppy for comparisions of infinite numbers, but it may be taken as more or less when the numbers are only taken by some finite subset.
The graphs in my link might be a bit difficult to read - I just left out the even numbers because... well, they might be seen as "trivial" and ... they didn't complain ;-)

Number 3 you will never find, after you go just a little away from the start (of counting hits) and there is a way to prove it (I can not prove, because I am not good with proofs, by a professor proved it once for me). That is the reason 6 only show a few times, because it is 3*2.

The same logic applies every 3 numbers, because they are divided by 3, and this won't show it anymore, once it gets on even function, they will go to the odd function and it will never have an ouput with 3 as factor ever again from that on.

I guess it just goes with the fact, that when you get a number on odd function, it goes 3n+1 and then is never divisible from 3. You only need to prove that. And that is your question.

Sorry for not having a definite answer, as why is this.

If you want a pattern to work on: take a look at one I found it about collatz.
Collatz Pattern


Field fixed by the conjugation automorphism in a Galois extension I am unable to see whether this statement is true or false, any help would be appreciated.

Let $E$ be the splitting field of a polynomial over $\Bbb{Q}$ which has both real and complex roots. Then the conjugation automorphism is an element of $Gal(E/\Bbb{Q})$ of order 2. Let $\Omega$={$ \alpha_1, ..., \alpha_n, \beta_1, ..., \beta_{2m} $} be the set of roots of the polynomial, where the $\alpha_i$ are the real roots are the $\beta_i$ the complex ones, with $\beta_{2i-1}$ and $\beta_{2i}$ conjugates.

Is it true that $\Bbb{Q}(\alpha_1, ..., \alpha_n, \beta_1+\beta_2, ..., \beta_{2m-1}+\beta_{2m})$ is always the field fixed by the conjugation automorphism? The $\subseteq$ inclusion is obvious, but I can't figure out $\supseteq$.

The fixed field of the complex conjugation is $M=E\cap\Bbb{R}$, and by Galois theory we always have $[E:M]=2$. The field $K$ generated by the real zeros and the real parts of the complex zeros (which is what you get when you use sums of complex conjugate pairs) is obviously contained in $M$, but may be smaller.

Daniel Fischer's example $f(X)=(X-1)(X^2+3)(X^2+5)$ is a simple example of this phenomenon. As a product of two roots $\sqrt{15}=\sqrt{-3}\sqrt{-5}$ is an element of $M$. But all the irrational zeros of $f(X)$ are purely imaginary so we have $K=\Bbb{Q}$.

I want to add examples of irreducible polynomials such that $K$ is a proper subfield of $M$. If we didn't have the extra requirement that $f(X)$ must also have real zeros, then $f(X)=X^4+4X^2+2$ would do well. It is irreducible by Eisenstein, $p=2$. The zeros of the quadratic $q(X)=X^2+4X+2$ are $-2\pm\sqrt2$, both negative, so as $f(X)=q(X^2)$, all the zeros of $f(X)$ are purely imaginary, again implying $K=\Bbb{Q}$.

This was really reusing Daniel's idea. A problem is that the above example polynomial does not meet the requirement that it should also have real zeros. Let's modify the situation marginally, and use the cubic $q(X)=X^3+4X^2-2$ instead of a quadratic. It is still irreducible by Eisenstein as is $f(X)=q(X^2)=X^6+4X^4-2$. A difference is that $q(X)$ has a single positive zero and two negative ones: $q(-1)>0>q(0)$ and $\lim_{x\to\pm\infty}q(x)=\pm\infty$. Therefore $f(X)$ has two real zeros and four purely imaginary zeros. If $\alpha_1$ is one of the real zeros, then $-\alpha_1$ is the other. Consequently $K=\Bbb{Q}(\alpha_1)$ is a degree six extension of $\Bbb{Q}$.

We observe that the discriminant of $q(X)$ is $d=404$, a non-square. Therefore the Galois group of $q(X)$ is $S_3$, and its splitting field, call it $L$, is a degree six extension. Let's denote the zeros of $q(X)$ by $\beta_1=\alpha_1^2>0$, $\beta_2$ and $\beta_3$. The Vieta relations tell us that $\beta_1\beta_2\beta_3=2$. As the zeros of $f(X)$ are $\pm\sqrt{\beta_i},i=1,2,3$, we can deduce that $\sqrt2\in M$.
The discriminant gives the quadratic subfield  $\Bbb{Q}(\sqrt{d})=\Bbb{Q}(\sqrt{101})$ of $L$. As $L\subset M$ we see that $\Bbb{Q}(\sqrt{101},\sqrt2)\subseteq M$, implying that $4\mid [M:\Bbb{Q}]$.

Consequently $K$ is a proper subfield of $M$ also in this case.

I didn't double check everything, but a study of the cycles structure of Frobenius elements in $G=Gal(E/\Bbb{Q})$, when viewed as a group of permutations of the six roots of $f(X)$ seems to imply that $G\simeq C_2\wr S_3$. In other words $[E:\Bbb{Q}]=48$, and consequently $[M:K]=4$.

Let $\sigma$ be the complex conjugation. Write $P=\Pi_{i=1}^{i=n} (X-a_i)=P^{\sigma}=\Pi_{i=1}^{i=n} (X-\sigma(a_i))$ implies that $\sigma$ preserve the roots.


Completeness of $\mathbb{R}$ In the proof below I don't understand how did Terence Tao construct $L$ integer $\frac{L}{n} < x_0$?

Given a nonempty set $E$ of reals, an element $x_0$ of $E$, and a positive integer $n$, OP wants to know how to construct an integer $L$ such that $L/n<x_0$. In the comments, OP insists on $L$ being nonnegative.

But this is clearly impossible, if $x_0\le0$ (and there is nothing in the problem statement forbidding such an $x_0$). For if $x_0\le0$, then, since $n\ge1$, we have $nx_0\le0$, so if $L/n<x_0$ then $L<nx_0\le0$, contradicting the condition that $L$ be nonnegative.

For any $\epsilon > 0$ and $x \in \mathbb{R},$ there is $r$ such that $r\epsilon > x$ is what you state your version of the Archimedean property is. Let's see how this implies the variant Tao uses, which is that there is an integer $K$ such that $K/n \geq M.$

By your Archimedean property, there is a real number $L \in \mathbb{R}$ such that $L \cdot (1/n) \geq M.$ The integers are unbounded in $\mathbb{R}$ (proving this will require you to be more precise on how you define $\mathbb{R}$), and so there is some $K \in \mathbb{Z}$ such that $K > L.$ Hence, there is an integer $K$ with the desired property.


Finding the inner product given the norm Possible Duplicate:
  Norms Induced by Inner Products and the Parallelogram Law

So suppose we are given a norm on a vector space.

If the Parallelogram law holds does that automatically mean we have the inner product which we can find using the Polarisation identity? Or is showing the Parallelogram law holds not sufficient to show that there exists an associated inner product?

Also, given that the Parallelgram law fails, e.g. $\Vert(x_1,x_2)\Vert_1 = |x_1| + |x_2|$ in $\ell^1(2)$, is there any significance in considering the Polarisation identity?

Well, you know the definition of angle in an IPS? You can use this to find the i.p. if you know the norm and angle between two vectors: $$x\cdot y=||x||\cdot||y||\cdot\cos\theta$$ with $\,\theta\,=$ the angle between vectors $\,x\,,\,y\,$


What is the order of this pole? $$f(z)=\frac 1{\cos(z^4)-1}$$

$z=0$ is a pole of $f$, and I believe that the Laurent series centred at $0$ is $-\frac 2{z^8}-\frac 16+...$, which looks like the pole is of order $8$, but why does Wolfram Alpha claim that the pole is of order $2$?

You are right. And Wolfram seems to agree.

By Taylor expansion $\cos z=1-z^2/2+O(z^4)$ so
$$
\cos(z^4)-1=-\frac{z^8}{2}+O(z^{16})=-\frac{z^8}{2}(1+o(1)).
$$
Hence
$$
f(z)=\frac{1}{\cos(z^4)-1}=\frac{1}{-\frac{z^8}{2}(1+o(1))}=-\frac{2}{z^8(1+o(1))}\sim-\frac{2}{z^8}
$$
So $\lim_0 z^8f(z)=-2\neq 0$. This means that $g(z)=z^8f(z)$ is holomorphic at $0$ (in a neighborhood of $0$) with $g(0)=-2$. So a power series expansion of $g$ will yield a Laurent expansion of $f$ starting by $-\frac{2}{z^8}$.

So indeed, $0$ is a pole of $f$ of order $8$.

The order of the pole is the number of elements in the finite (!) principal part of the Laurent series of your function (or the highest negative power).
You have 4 poles (solve the equation $(\cos \left(z^4\right)-1)$: $\left\{\left\{x\to -\sqrt[4]{2 \pi } \left(\sqrt[4]{n}\right)\right\},\left\{x\to -i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to \sqrt[4]{2 \pi } \sqrt[4]{n}\right\}\right\}$.
If you expand the initial function into Laurent series you will get (for example for the last root):
$\frac{5 \left(z-\sqrt[4]{2 \pi }\right)}{32 \sqrt[4]{2} \pi ^{9/4}}+\frac{3}{16\ 2^{3/4} \pi ^{7/4} \left(z-\sqrt[4]{2 \pi }\right)}+\frac{1}{16 \left(\sqrt{2} \pi ^{3/2}\right) \left(z-\sqrt[4]{2 \pi }\right)^2}+\left(-\frac{1}{6}-\frac{19}{128 \pi ^2}\right)+O\left(\left(\text{z}-\sqrt[4]{2 \pi }\right)^2\right)$
The order of this pole is 2. One can conclude that you have 4 poles, each having order 2. So I guess Wolfram Alpha is correct :).


Doing Some Asymptotic Integration: Problem with a Taylor Series I evaluated this integral (with some help from you guys):

  $$\int_0^{\arcsin\left(\frac{a}{b}\right)} \exp\left(-\beta\left(b\cos(x)-\sqrt{a^2 - b^2 \sin^2(x)}\right)^2 \right) \, dx$$

and got an answer in terms of an F1 Appell hypergeometric series (with some sums), so I tried using a Taylor expansion to simplify things...

I only need the integral to be valid for small $b\cos(x)-\sqrt{a^2-b^2 \sin^2(x)}$ so I Taylor expand:

  $$
\int_{0}^{\arcsin\left(\frac{a}{b}\right)}\left(1-\beta\left(b\cos(x)-\sqrt{a^{2}-b^{2}\sin(x)}\right)^2\cdots\right)dx$$

  $$
=\arcsin\left(\frac{a}{b}\right)-\beta\left[\left(a^{2}x+b\cos(x)\sin(x)-b\sin(x)\left(\sqrt{a^{2}-b^{2}\sin^{2}(x)}\right)-a^{2}\arctan\left(\frac{b\sin(x)}{\sqrt{a^{2}-b^{2}\sin^{2}(x)}}\right)\right)\right]_{0}^{\arcsin\left(\frac{a}{b}\right)}$$

according to Mathematica. Notice the $a/0$ in the $\arctan$ term at the upper limit, which makes the integral undefined at this point so the approach doesn't work. I know this integral has a definite value (from the hypergeometric version), so what's gone wrong with this approach?

Following the approach like Difficult Gaussian Integral Involving Two Trig Functions in the Exponent: Any Help?, you will get the two key types of integral: $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m}x\left(a^2-b^2+b^2\cos^2x\right)^m~dx$ and $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ , where $m$ and $n$ are any non-negative integers

For $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m}x\left(a^2-b^2+b^2\cos^2x\right)^m~dx$ , where $m$ and $n$ are any non-negative integers, it can expand to the polynomial of $\cos x$ , so the indefinte integral will have $x$ term and polynomial of $\sin x$ and $\cos x$ , substitution of $0$ and $\sin^{-1}\dfrac{a}{b}$ should be no problem.

For $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ ,

$\int\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$

$=\int\cos^{2n-2m}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~d(\sin x)$

$=\int(1-\sin^2x)^{n-m}\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~d(\sin x)$

According to http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions, the indefinte integral will have polynomial of $\sin x$ and $\sqrt{a^2-b^2\sin^2x}$ and $\sin^{-1}\dfrac{b\sin x}{a}$ term, substitution of $0$ and $\sin^{-1}\dfrac{a}{b}$ should be no problem.

How you get the $\tan^{-1}\dfrac{b\sin(x)}{\sqrt{a^2-b^2\sin^2x}}$ term? It should be not contained!

When you substitute the very last arctan with arcsin, you should remember:

$\sin^2(\arcsin(a/b))=(\sin(\arcsin(a/b)))^2=(a/b)^2$

So, $a/0$ appears as the following:

$\frac{b \sin(\arcsin(\frac{a}{b}))}{\sqrt{a^2-b^2 \sin^2(\arcsin(\frac{a}{b}))}}$=$\frac{b(\frac{a}{b})}{\sqrt{a^2-b^2(\frac{a}{b})^2}}$=$\frac{a}{0}$


Permutations , Probablity and Combinations A group of 25 friends were discussing a large positive integer. "It can be divided by 1," said the first friend. "It can be divided by 2," said the second friend. "And by 3," said the third friend. "And by 4," added the fourth friend. This continued until everyone had made such a comment. If exactly two friends were incorrect, and those two friends said consecutive numbers, what was the least possible integer they were discussing?

Note that if the friend who said 3 is wrong, then any friends who say any multiple of 3 (such as 6 or 9) will also be wrong. This limits the consecutive integers to be greater than 12 (so their multiples are not in 25).

If the friend who said 6 is wrong, then there the friend who said 2 is wrong or the friend who said 3 is wrong, and 2 or 3 is not consecutive to 6. This limits the integers to be powers of a single integer or primes.

We can easily find the two consecutive integers following the above conditions by checking. They are 16 and 17.

Thus, the integer is LCM(1,2,...,15,18,19,...,25)=787386600.

Any number is divisible by $1$ therefore, divisible by $1$ is a true statement. So that number must be divisible by $2$ and so the number is $2$.


A handy integration trick a couple of days ago I stumbled upon a handy integration trick which goes like this

$$ \int_{}^{} f(x)^n dx = \int_{}^{f(x)} x^n(f^{-1}(x))'dx $$

where $f^{-1}(x)$ is the inverse of $f(x)$.

This can easily be proven by taking the derivative of both sides. You then get $\frac{d}{dx}\int_{}^{} f(x)^n dx = f(x)^n$

and $\frac{d}{dx}\int_{}^{f(x)} x^n(f^{-1}(x))' dx = \frac{f(x)^nf'(x)}{f'(f^{-1}(f(x)))} = f(x)^n$. Since $(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$.

This can be used to find hard integrals like $\int \tan(x)^3 dx$.

$n=3$, $f(x) = \tan(x)$, $f^{-1}(x) = \arctan(x)$ and $(f^{-1}(x))' = \frac{1}{1+x^2}$

$$ \int_{}^{} \tan(x)^3 dx = \int_{}^{\tan(x)} \frac{x^3}{1+x^2} dx= \int_{}^{\tan(x)} x-\frac{x}{1+x^2} dx = [\frac{1}{2}\left(x^2-\ln(1+x^2)\right)]^{\tan(x)} = $$
$$ \frac{1}{2}\left(\tan(x)^2 - \ln(1+\tan(x)^2)\right) + C $$

I have never seen this trick before, so I was quite amazed when I saw how powerful it could be. Solving higher exponents of $\tan(x)$ was not a big challenge.

Let's take a new example. $n=1$, $f(x) = \arccos(x)$, $f^{-1}(x) = \cos(x)$ and $(f^{-1}(x))' = -\sin(x)$.

$$ \int \arccos(x) dx = \int_{}^{\arccos(x)} -x\sin(x) dx = [x\cos(x)-\sin(x)]^{\arccos(x)} =$$

$$ x\arccos(x) - \sin(\arccos(x))+C = $$

$$ x\arccos(x) - \sqrt{1-x^2} + C$$

So I don't have a question, just wanted to show it to you. Please let me know what you think!

$$\int(\tan x)^3dx=\int_{}^{\tan x}y^3d(\arctan y)=\int_{}^{\tan x}\frac{y^3}{1+y^2}dy=\int_{}^{\tan x}\left(y-\frac{y}{1+y^2}\right)dx$$
$$=\left(\frac{y^2}{2}-\frac{1}{2}\ln |{1+y^2}|\right)^{\tan x}+C=\frac{1}{2}\tan ^2x+\ln |\cos x|+C$$
$$\int \arccos xdx=\int^{\arccos x} yd(\cos y)=y\cos y-\int^{\arccos x}\cos ydy=y\cos y-\sin y+C$$
$$=x\arccos x-\sqrt{1-x^2}+C $$


What is the probability that a rational prime remains prime in $\mathbb Z[i,\sqrt{-3}]$? Using Chebotarev's density theorem, asymptotically, what is the probability that a rational prime remains prime in $\mathbb Z[i, \sqrt{-3}]$?

If you're referring to $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$, the ring of algebraic integers of $\mathbb{Q}(\zeta_{12})$ (see its page in the LMFDB), then the answer is that no prime from $\mathbb{Z}$ remains prime in $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$.

The intermediate fields are $\mathbb{Q}(i)$, $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$. For a prime from $\mathbb{Z}$ to be prime in both $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$, it must be $p \equiv 5 \bmod 12$. But then this means that $p \equiv 1 \bmod 4$ and you know what that means for $\mathbb{Z}[i]$...

P.S. A Naiade discovered the density theorem twelve centuries before Chebotarev.

Based on Euler results and Chebotarev's density theorem, we have that, asymptotically:

i) One half of all rational primes split in $Z[i]$, and these are the primes $p = 1$ mod $4$. Let's call this set $Split(Z[i])$ and,

ii) One half of all rational primes split in $Z[\sqrt{-3}]$, and these are the primes $p = 1$ mod $3$. Let's call this set $Split(Z[\sqrt{-3}])$

Therefore, the rational primes that split in $Z[i,\sqrt{-3}]$ are the union of $Split(Z[i])$ and $Split(Z[\sqrt{-3}])$.
We have then, by set rules, that:

$Split(Z[i,\sqrt{-3}]) = Union$ ($Split(Z[i])$,$Split(Z[\sqrt{-3}])$) := {primes| $p = 1$ mod $4$ OR $p = 1$ mod $3$}.

And the rational primes that remain inert in this ring $Inert(Z[i,\sqrt{-3}])]$ are in the complement set $C[Split(Z[i,\sqrt{-3}])]$, and by set rules:

$Inert(Z[i,\sqrt{-3}])]$ ⊊ $C[Split(Z[i,\sqrt{-3}])]  = C[Union (Split(Z[i]),Split(Z[\sqrt{-3}]))] :=$ {primes | NOT ( $p = 1$ mod $4$ OR $p = 1$ mod $3$) } = {primes| NOT ($p = 1$ mod $4)$ AND NOT ($p = 1$ mod $3$) } = {2,3,primes|  $p = 3$ mod $4$ AND $p = 2$ mod $3$ }

Finally, by Chebotarev's density theorem and probability rules, asymptotically, the probability $P$ that a prime $p = 3$ mod $4$ AND $p = 2$ mod $3$ is equal to the product of the probabilities for each condition:

$P$ (prime $p = 3$ mod $4$ AND $p = 2$ mod $3$) = $P$ (prime $p = 3$ mod $4$) * $P$ (prime $p = 2$ mod $3$) = $1/2 * 1/2 = 1/4$

Since the primes that ramify in $(Z[i,\sqrt{-3}])$ form a finite set, we have that $P(p$ is inert in $Z[i,\sqrt{-3}]$)=$P(p$ doesn't split in $Z[i,\sqrt{-3}]$), therefore $1/4$ of all rational primes remain prime in $Z[i,\sqrt{-3}]$


Showing sequence is Cauchy by Definition Question: I have to show that sequence $(x_n)$ defined by

$x_n=\frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$.

Is Cauchy sequence using definition only.

My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $\mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I   considered

$|x_n-x_m|= |\frac{n+(-1)^n}{2n-1}-\frac{m+(-1)^m}{2m-1}|$

$=|\frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$

$=|\frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$

but I am unable to proceed further :-( please Help me...

Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|\frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|\frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.


Trying to define $\mathbb{R}^{0.5}$ topologically A few days ago, I was trying to generalize the defintion of Euclidean spaces by trying to define $\mathbb{R}^{0.5}$.

Question: Is there a metric space $A$ such that $A\times A$ is homeomorphic to $\mathbb{R}$?

I am interested also in seeing examples of $A$ which are only topological spaces

Edit: If there exists a topological space $A$ such that $A\times A\cong \Bbb R$, then $A\times \{a\}$ is a subspace of $A\times A$ ($a\in A$). Hence $A\times\{a\}$ can be embedded in $\mathbb{R}$, since $A\cong A\times \{a\}$. Thus $A$ can be embedded in $\mathbb{R}$. Therefore $A$ is metrizable.

Thank you

No, no such space exists. Suppose $A$ is a topological space such that $A\times A\cong \mathbb R$, with the Euclidean induced topology on $\mathbb R$. If $A$ is disconnected then so is $A\times A$, but that would contradict the connectivity of $\mathbb R$, so it follows that $A$ is connected. But, since $A$ is connected it follows that $A\times A$ with a single point removed is still connected. However, $\mathbb R$ with a single point removed is not connected. Contradiction.

This was intended as a comment on the answers above, but I'm a new user, so I don't have that as an option, so I'll need to post it as a new answer instead.

The other answers here show that $\mathbb{R}$ cannot be written as a "square" of sorts.  The situation is not unlike that of the real number -1, which cannot be written as a square of real numbers.

But one can extend the real numbers to create a square root of $-1$.  Similarly, one could extend the notion of topological space to create a "square root" of $\mathbb{R}$.  One way to do so, in analogy with the complex numbers, would be as follows.  Define a "complex topological space" to be an ordered pair $(A,B)$ of topological spaces.  Define the direct product of two such ordered pairs by

$(A,B) \times (C,D)$ equals (disjoint union of $A\times C$ and $\mathbb{R}\times B \times D$, disjoint union of $A\times D$ and $B\times C$)

Identify a topological space $Y$ with the ordered pair $(Y, \varnothing)$.

Then $(\varnothing$, one-point set$)\times (\varnothing$, one-point set) = ($\mathbb{R}$, empty set).


Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.

How? What's the proof? Or maybe it is self apparent just looking at the above?

PS: This problem is known as "The sum of the first $n$ positive integers".

Let $$S = 1 + 2 + \ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \ldots + 2 + 1.$$
Add the two equations, term by term; each term is $n+1,$ so
$$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1).$$
Divide by 2: $$S = \frac{n(n+1)}{2}.$$

Well if we graph the numbers $1,2,3,4,.. $ we get a line. Graphing $(1,1)$, $(2,2)$, $(3,3)$ ...I mean. And these numbers occur in a linear fashion so the average of these numbers from $1$ to $n$ is $\frac{n+1}{2}$ and we multiply that by $n$ numbers. Pretty cool huh?

Maybe this one will interest you too:

$$1+3+5+7+9....+(2n-1)=n^2$$


Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)? Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is
$$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$
I am wondering why this is non-negative?

Using taylor series:
$$\frac {\ln(1-x)}{x}=-\frac {x+\frac {x^2}2+...}x=-
(1+\frac x2+...)$$
Which is decreasing, similarly:
$$\frac {\ln(x)}{1-x}=-\frac{\ln(1-(1-x))}{1-x}=-(1+\frac{1-x}2+...)$$
Which is increasing, then its reciprocal $\frac {1-x}{\ln(x)}$ is decreasing.
Since these two are negative, then they multiplied together is increasing.

(Not a complete answer but hopefully close enough)
You want to show that
$$ x \log(x)+ (1-x)\log(1-x) \le - \log(x) \log(1-x)$$
Divide on both sides by $\log(x) \log(1-x)$ and this is
$$ \frac{x}{\log(1-x)} + \frac{1-x}{\log(x)} \le -1 $$
This is of the form
$$ f(x) + f(1-x) \le c$$
where $f(x) = \frac{x}{\log(1-x)}$. I don't know if there is a nice way to show this.
But the not-so-nice way is to Taylor expand
$$ \frac{x}{\log(1-x)} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dotsm $$
$$ \frac{1-x}{\log(x)} = -1 + \frac{1-x}{2} + \frac{(1-x)^2}{12} + \frac{(1-x)^3}{24} + \dotsm $$
Now you add them up, you see that the RHS becomes
$$ -2 + \frac{1}{2} + \frac{x^2 + (1-x)^2}{12} + \frac{x^3 + (1-x)^3}{24} + \dotsm $$
$$ \le -2 + 1 + \frac{1}{12} + \frac{1}{24} + \dotsm $$
But then showing that the last sum is $\le -1$, seems clear to me but maybe you might want more concrete justification...


Complicated rational exponents How is this $\frac{m}{n}$ equals to  $m \times \frac{1}{n}$? Any logical proof for this? Which draws this conclusion:

  A fractional exponent like $\frac{m}{n}$ means do the $m^{\text{th}}$ power, then take the $n^{\text{th}}$ root or take the $n^{\text{th}}$ root and then do the $m^{\text{th}}$ power.

I assume we're taking $m, n \in \mathbb{Z}$, $n \neq 0$.

First, note that $$m\times \frac{1}{n} = \frac{m}{1}\times\frac{1}{n} = \frac{m\times 1}{1\times n} = \frac{m}{n}.$$

Now $x^{\frac{m}{n}} = x^{m\times\frac{1}{n}}$ as $\frac{m}{n} = m\times\frac{1}{n}$. Now, using the rule $x^{a\times b} = (x^a)^b$ we see that

$$x^{\frac{m}{n}} = x^{m\times\frac{1}{n}} = (x^m)^{\frac{1}{n}}.$$

Now we use the rule that $x^{\frac{1}{n}} = \sqrt[n]{x}$ and see that $x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}} = \sqrt[n]{x^m}$.

To see that we also have $x^{\frac{m}{n}} = (\sqrt[n]{x})^m$, note that multiplication is commutative (the order doesn't matter), so $m\times\frac{1}{n} = \frac{1}{n}\times m$. Therefore $$x^{\frac{m}{n}} = x^{m\times\frac{1}{n}} = x^{\frac{1}{n}\times m} = (x^{\frac{1}{n}})^m = (\sqrt[n]{x})^m.$$

If you are making a number bigger by multiplying by $m$, then making it smaller by dividing it by $n$, it will be the same as making it smaller by $n$ then making it bigger by $m$. I guess it's just how numbers work.