simply connected covering of a path connected space (II) Let $p:\overline{X}\rightarrow X$ be a simply connected covering of a path connected space $X$ and $A\subset X$ be a path connected set.

Show that the inclusion induced homomorphism $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is injective iff each path component of  $p^{-1}(A)$ is simply connected.

First of all i do not think i understand what does it mean to say $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is injective..

Is it like if i have a loop in $A$ which is not nullhomotopic in $A$ (existence of $H:I\times I\rightarrow A$) then it is not nullhomotopic (existence of $H:I\times I\rightarrow X$) in $X$.

Let me know if this is what it actually mean...

Assume $i_{\sharp}$ is injective and let $\omega$ be a loop in $p^{-1}(A)$... See this as a loop in $\overline{X}$... as $\overline{X}$ is simply connected $\omega$ is nullhomotopic in $\overline{X}$ i.e., we have
$H:I\times I\rightarrow \overline{X}$ such that

$H(t,o)=\omega(t)$ and $H(t,1)=w(0)$ for all $t\in I$..

Compose this with $p$ to get $I\times I\xrightarrow{p\circ H}X$ with $(p\circ H)(t,0)=(p\circ \omega)(t)$ and $(p\circ H)(t,1)=(p\circ \omega)(0)$..

So, this $p\circ \omega$ is null homotopic in $X$ so it has to be null homotopic in $A$ as well.. As $p\circ \omega$ is nullhomotopic in $A$ i belive this would imply $\omega$ is nullhomotopic in $p^{-1}(A)$..

I could not think of any ideas about converse part and how to prove if $p^{-1}(A)$ is not actually path connected...

Please give only hints..

You are right about the first part. If $p\omega$ is nullhomotopic in $A$, then the homotopy to the constant loop lifts to a homotopy between $\omega$ and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map $p:\tilde X\to X$ restricts to a covering map $p^{-1}(A)\to A$ for every subset $A$ of $X$.

Regarding the second part, let every path component of $p^{-1}(A)$ be simply-connected.  If $\omega$ is a loop in $A$, nullhomotopic as a loop in $X$, then what can you say about its lift $\tilde\omega$. Recall that every homotopy of paths lifts to a homotopy between their unique lifts at a given point $\tilde x$ lifting the starting point of the paths.

The Statement is not true. Für example $X=\overline{X}={\mathbb R}^2$, which is simply connected, p the Identity, and $A=S^1$. The Fundamental Group of A is nontrivial, so $i_\sharp$ is not injective, but $p^{-1}A=A$ is connected.

By the way, the Statement as above has no meaning anyway. The Path component is path connected anyway, you must mean something else.


Discrete Probability Problem: determining probability mass function and cumulative distribution function Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let X = the number of people who arrive late for the seminar.

a) Determine the probability mass function of X. [Hint: label the three couples #1, #2, and #3, and the two individuals #4 and #5.]

b) Obtain the cumulative distribution function of X, and use it to calculate P(2 <= X <= 6).

Let $L_i$ be an indicator function, equal to 1 if #i is late. Then
$$
    X = 2 L_1 + 2 L_2 + 2 L_3 + L_4 + L_5
$$
where each $L_i$ is independent and follows Bernoulli distribution with probability $p=0.4$. The easiest way to find probability mass function is through the probability generating function:
$$
  \mathcal{P}_X(s) = \mathbb{E}( s^X ) = \left(\mathcal{P}_L(s^2)\right)^3 \cdot \left(\mathcal{P}_L(s)\right)^2 = (1-p+p s^2)^3 (1-p + p s)^2
$$
Now expand this polynomial. Then
$$
    \mathbb{P}(X= k) = [s]^k \mathcal{P}_X(s)
$$

If you can use excel simply use the binomial distribution function: BINOM.DIST(x,n,p,cumulative).

Where x is the number of late people, n is the total number of people, p is the probability of someone being late, and cumulative is a bool that says whether or not to calculate the cumulative probability. Consider couples to be one person, since they arrive together.

So for the pmf use BINOM.DIST with cumulative set to false on each of the x values,
then to find the cdf repeat with cumulative set to true.

This will work up to 5 then you have to use some logic to find the rest from the values you already have.


Quicksort with Trivalued Logic Does anyone know a way to do a quick sort with trivalued logic?
The problem I’m trying to solve is this: I’m trying to display a view of a complex 3d object from a given viewing angle. I’ve broken the object into many 2d surfaces that I can draw separately, but to display the image properly, I need to determine the z-order of the surfaces – a classic computer drawing problem.  It’s guaranteed that none of the surfaces intersect, so the problem is solvable. It would be simple if on comparing any two surfaces, I could always determine which one is in front – then a simple mergesort would suffice. But very often, if I compare two surfaces, it’ll turn out that, with the angle I’m viewing from, there’s no overlap at all. One surface is over here, and the other surface is over there, so it’s impossible to say which one is in front.
In mathematical terms, what I’m trying to do is sort a set of entities - call them $a$, $b$, $c$, etc. Transitivity is guaranteed:  If $a < b$ is true and $b < c$ is true then $a < c$ is always true. But the complicating factor is the trivalued logic: $a < b$ could be unknown. A consequence is the final sorted list may contain small sets of elements within which the order doesn’t matter, e.g., The result may be   $a < (b, c) < d$, etc.
Note that even if $a < b$ is unknown, other comparisons may indirectly force a certain ordering for $a$, $b$. E.g., If $a < b$ is unknown, but it turns out that $a < c = \mbox{true}$ and $b < c$ is false, then the sorted order must be $a < c < b$.
I can solve the problem with a bubble sort, but that’s bad because $O(N^2)$ comparisons, and each comparison is very expensive (since it involves figuring out whether two surfaces can block each other when viewed from a certain angle).  Is there a way to solve this with a faster sort? (eg. Some adaptation of a mergesort)?

What you have is not really a trivalued logic, but a partially ordered set (aka poset).  There is a fairly large body of research on sorting partially ordered sets (a quick googling for "poset sorting" gives some good hits).  In particular, you may want to look for something called a "chain merge data structure".  Also, a paper called "Sorting and Selection in Posets".

Though your surfaces are sometimes incomparable (this is the third value in you hoped for trivalent logic), I think it is not necessary to worry about it here. If for two surfaces, neither is in front of the other for the given viewpoint, then you can draw either one before the other, it doesn't matter. So when you sort your surfaces you can just arbitrarily pick one (or the other) to be drawn before the other and it won't mess up the drawing.


What operator/matrices can be written in Jordan normal form? The wikipedia article is too bad in explaining. Is there any theorem that summarises which operators or matrices can be written in Jordan normal form?

Every square matrix can be written in Jordan Canonical form. This asserts that every linear operator has a basis of generalized eigenvectors.


Partial average of normal distribution suppose we have a normal distribution $ N(\mu,\sigma^2)$, then

$$
CDF(x) = \int_{-\infty}^{x} f_X(t)d_t
$$

which can be shown as

$$
CDF(x) = \frac{1}{2}[1+erf(\frac{x-\mu}{\sigma\sqrt{2}})]
$$

I am wondering if there is any function for partial averaging of normal distribution such as:
$$
\int_{-\infty}^{x} tf_X(t)d_t
$$

Sure there is.  When you are taking the expected value in this way, the integral is easy in closed form for $\mu=0$, and contains an erf term for non-zero $\mu$:

$$
f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$

$$E(X | X<x_0) = \int_{-\infty}^{x_0} xf(x)\,dx = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0} x e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx =  \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0-\mu} (t+\mu) e^{-\frac{t^2}{2\sigma^2}}dt = \mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}+
\left[
-\frac{\sigma}{\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}
\right]_{-\infty}^{x_0}=
\mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}-\frac{\sigma}{\sqrt{2\pi}} e^{-\frac{x_0^2}{2\sigma^2}}
$$


Verifying Carmichael numbers I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:

A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$.

Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.

Solution given:

Apply Fermat's Little Theorem to each prime divisor of $m$:
\begin{align*}
a^2 &\equiv 1 \pmod{3}\\
a^{10} &\equiv 1 \pmod{11}\\
a^{16} &\equiv 1 \pmod{17}
\end{align*}
This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.

I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).

Can somebody clarify this for me?

Thanks!

Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So,
\begin{align*}
a^{80}= (a^2)^{40} &\equiv 1^{40} = 1\pmod{3},\\
a^{80}= (a^{10})^{8} &\equiv 1^8 = 1 \pmod{11},\\
a^{80}= (a^{16})^5 &\equiv 1^5 = 1\pmod{17}.
\end{align*}

By the Chinese Remainder Theorem, the system of congruences
\begin{align*}
x&\equiv 1\pmod{3}\\
x&\equiv 1\pmod{11}\\
x&\equiv 1\pmod{17}
\end{align*}
has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then
the two solutions we found must be congruent. That is,
$$a^{80}\equiv 1\pmod{561}.$$

(Added. Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$).

Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular,
$$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$
as desired.

You will have to consider two cases though. Case 1 will be that $a$ is not relatively prime to $561$ and case two will be that $5$61 and $a$ are relatively prime.


How to prove that $4x^3+6x^2+4x+1$ is not a fourth power of an integer, for any $x\in\mathbb N$? How do I prove that for all positive integers $x$ it's true that, $4x^3+6x^2+4x+1$ is not a fourth power of an integer?

I've tried doing modulo 3 and 5 checks, and it didn't really go far from there

If $4x^3+6x^2+4x+1=w^4$, then:
$$ x^4 + w^4 = (x+1)^4 $$
but such an identity contradicts the case $n=4$ of the Fermat's last theorem (there is an elementary proof of that case in terms of Gaussian integers, so the full power of the general FLT is not needed).

Note we can write:
y⁴ = 4x³ + 6x² + 4x + 1
as:
y⁴ = (x + 1)⁴ - x⁴
and as mentioned is a form of Fermat's Last Theorem.
→ x⁴ + y⁴ = (x + 1)⁴
This instance of FLT was put to rest long before Wiles finally put the problem to rest.  Check out this simple proof:
http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-4.html?m=1


Is this a Valid proof for $(2n+1,3n+1)=1$? Is this a valid proof for $(2n+1,3n+1)=1$?

$\exists \ d \in \mathbb{Z}$

$d | 2n+1$ means $2n+1 \equiv 0$ (mod $d$)
$d | 3n+1$ means $3n+1 \equiv 0$ (mod $d$)

Multiply 1. by $3$ and 2. by $2$:

$6n+3 \equiv 0$ (mod $d$)

$6n+2 \equiv 0$ (mod $d$)

Subtract the second from the first to get:
$ 1 \equiv 0$ (mod $d$), which means $d|1$, which means $d=1$.

Thanks!

Yes, nice job! That's a very nice application of the method that I described in your prior question. Notice again how simple the problem becomes once it is reduced to equational (congruence) form.

@Bill has already commented on the truth of your proof, here is another one.

Suppose that $(2n+1,3n+1)=d$.

$$d|3n+1, d|2n=1$$

$$d|3n+1-(2n+1)\implies d|n\implies d|2n$$

$$d|2n+1-2n\implies d|1\implies d=1$$


Help with a cumulative distribution function question. This is a question I want to solve:

The random variable $X$ has cdf:
$$
F_X(x) = \begin{cases}
\begin{align}
&0 &&x <0\\
&0.5 + c\sin^2\left(\frac{\pi x}{2}\right) &0 \leq\; &x \leq 1\\
&1 &&x>1
\end{align}
\end{cases}
$$

(a) What values can $c$ assume?

(b) Plot the cdf.

(c) Find $P[X > 0]$.

I assume that $x$ is between $0$ and $1$ so
$$
0.5 + c \sin^2(\pi\times x/2) = 0
$$

then $x=1$ and
$$
c \sin^2(\pi\times 1/2) = -0.5
$$

since $c = -0.5$

Is that correct and I need help with the nother parts please.

You have no reason to assume that $0.5 + c \sin^2(\pi\times x/2) = 0$.
In fact the definition of $F$ tells you that
$$F_X(0) = 0.5 + c \sin^2(\pi\cdot 0/2) = 0.5 \neq 0,$$
so your assumption is false for at least one $x$ in the interval $0 \leq x \leq 1$.

There is also no reason to assume that $c = -0.5$.
In fact it cannot be $-0.5.$

I would suggest looking carefully at what happens around $x=1.$
It is possible for a cdf to be discontinuous, but only some kinds of discontinuity
are possible.

I think the CDF given in the question is wrong as in your problem it is neither a continuous CDF nor a discrete one as LHL (left hand limit) approaching 0 and RHL (right hand limit) approaching 1 do not converge to the same value. Which in this case is not true for F(x) at $0\leq x \leq 1$.


Proving a claim using a counter example $p,q\in\mathbb{Q}$ such that $(q\neq 0)$, then $\frac { p }{ q } $ is rational.

Steps I took and my thoughts on this:

This seems awfully obvious, but yet I can't seem to organize my thoughts in any way to construct a formal mathematical proof. The most I can think of is as follows:

let $p=\frac { a }{ b } $ and $q=\frac { m }{ n }$; such that $a,b,m,n\in \mathbb{Z}$ and $(m\neq 0$, $b\neq 0$, $n\neq 0)$

$$\frac { p }{ q } =\frac { \frac { a }{ b }  }{ \frac { m }{ n }  } =\frac { a }{ b } \cdot \frac { n }{ m } =\frac { an }{ bm } $$

$an$ and $bm$ are both integers because $a,b,m,n$ are all integers.

I imagine that this is way off. Please put me on the right path, and forgive my amateur attempt at a proof. I have a lot to learn about writing mathematical proofs, and I am just starting.

That's not way off.  It's close.  In fact you are done and you were correct.

First figure out what rational means, we did that right: p = a/b, q = m/n a,b,m,n in Z, b, n ne 0.  Then you identified p/q = qm/bn and as qm, and bn are in Z and bn ne 0, p/q is rational.  You did it perfectly.

The only thing I might add would be using closure of integers for multiplication to show that the new numerator and denominator are themselves integers.


I know this DE is solvable... I need help with a seemingly simple looking diff equ

$$
x\frac {d^{2}y} {dx^{2}}+2y=0
$$

$$
\rightarrow \frac {d^{2}y} {dx^{2}}+2\frac {y} {x}=0
$$

$v= (\frac {y} {x})$ substitution isn't working as it eventually shows:
$$
xv''+v'+2v=0
$$

which isn't any easier.  That variable is making my blood pressure go up.

I know I'm looking past something stupid.

Don't feel bad about not being able to solve this.  Many (maybe, most) easily expressed differential equations cannot be solved in terms of elementary functions.  In spite of this, standard existence and uniqueness theorems often guarantee the existence of solutions.  When the solution of an important class of differential equations cannot be expressed in terms of elementary functions, we give the solution a name, in this case, the Bessel functions.  Functions generated in this fashion are generally called special functions.

Many mathematical software packages know about special functions.  For example, if you type this differential equation into WolframAlpha, you'll get all kinds of information about the solution.

Text entered: x*y'' + 2y = 0

Output:

Note that the solution is expressed in terms of functions denoted $J_1$ and $Y_1$, called Bessel functions.  Also, in spite of the fact that there's no simple formula, several plots are generated.

I thought the wolfram alpha page on Bessel function lacked definition, therefore I will provide the full derivation of it, so lurkers like me can get review from time to time, and since some applications require the full notation.

Starting with a function similar to:
$$
y''+f(x)y'+g(x)y=0
$$
You're looking for two functions$f(x)$ and $g(x)$ that satisfies the differential equation and $y$ is the expression.

Starting with $y$:
$$
y=x^{a}J_{\pm}(bx^{c}) = x^{a}J(z);  z=bx^{c}
$$

Taking derivatives of $x$, in which $J$ is a function of $z$, therefore implementing the chain rule:
$$
\frac {dy} {dx}=ax^{a-1}J+x^{a}(\frac {dJ} {dz})(\frac {dz} {dx})
$$
$$
\Rightarrow \frac {dy} {dx}=ax^{a-1}J+x^{a}(\frac {dJ} {dz})bcx^{c-1}
$$
$$
\Rightarrow \frac {dy} {dx}=ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})
$$
Now taking the second derivative
$$
\Rightarrow \frac {d^{2}y} {dx^{2}}=a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}\frac {d^{2}J} {dz^{2}}
$$
Now replace $\frac {d^{2}} {dz^{2}}$ in the last equation into the differential equation
$$
J''=-(\frac {1} {z}J'+(1-\frac {m^{2}} {z^{2}})J)
$$

Now determine$f(x)$ and $g(x)$ such that
$$
y''+f(x)y'+g(x)y=0
$$
$$
\Rightarrow [a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}\frac {d^{2}J} {dz^{2}}]+f(x)[ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})]+g(x)[x^{a}J(z)]=0
$$

$$
\Rightarrow [a(a-1)x^{a-2}J+bcx^{a+c-2}(2a+c-1)\frac {dJ} {dz} + b^{2}c^{2}x^{a+2c-2}[-\frac{1}{bx^{c}}(\frac {dJ} {dz})-(1-\frac {m^{2}}{bx^{c}})J]+f(x)[ax^{a-1}J+bcx^{a+c-1}(\frac {dJ} {dz})]+g(x)[x^{a}J(z)]=0
$$

And group all of the terms into two term, ie:
$$
()\frac {dJ} {dz}+()J=0
$$

With each set $()=0$

This gives:
$$
f(x)=\frac {(1-2a)} {x}
$$
and
$$
g(x)=-(\frac {c^{2}m^{2}-a^{2}} {x^{2}})+b^{2}c^{2}x^{2c-2}
$$

And thus, once simplified, substitution reveals the famous Bessel Function:
$$
\therefore \frac {d^{2}y} {dx^{2}}+\frac {1-2a} {x}\frac {dy} {dx}-[(\frac {c^{2}m^{2}-a^{2}} {x^{2}})+b^{2}c^{2}x^{2c-2}]y=0
$$
with solutions:
$$
y=k_1x^{a}J_m(bx^{c}+k_2x^{a}(bx^{c})
$$

Therefore, given the DE:
$$
xy''-2y=0
$$

comparing the Bessel Function, we can see that
$$
1-2a=0;
$$
$$
b^{2}c^{2}=2;
$$
$$
c^{2}m^{2}-a^{2}=0;
$$
$$
2c-2=-1
$$

Set up a system of equations using these, solve for the constants, and plug into the Bessel solution: $y=k_1x^{a}J_m(bx^{c}+k_2x^{a}(bx^{c})$, for the answer


Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.

I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.

Hint: We have $$\sin 3x = 3\sin x - 4\sin^3 x\neq 3\sin x\cos x$$

So you get (let $\sin x = \alpha$) $$3\alpha - 4\alpha^3 = 4\alpha^2 \iff 4\alpha^3 + 4\alpha^2 - 3\alpha = 0$$

Which is a simple cubic in $\alpha$. Find the roots of the cubic, back-substitute and find the corresponding values of $x$.


On simple groups $G$, where $2\mid |G|$, $4\not\mid |G|$ The (old) exam I'm looking at has the following problem:

  Suppose the order of $G$ is even, but is not divisible by $4$. Prove that $G$ is not simple.

A group with $2$ elements is clearly a counter-example to that.

Are those the only counterexamples?

(If yes, how would one prove that?)

It is indeed the only counterexample. Here's how you go about showing that.

Let $|G|=2n$ where $n$ is odd, and let $x\in G$ have order $2$. The permutation $G\to G$ given by multiplication (say on the left) by $x$ is an involution, and yet it fixes nothing. Thus its cycle decomposition is comprised of $n$ $2$-cycles, whence $x$ induces an odd permutation.

The set of elements of $G$ which induce even permutations forms a subgroup $H$. For any $g\in G\setminus H$, we note that $xg$ is an element of $H$, so $g\in x^{-1}H$. This means that the index of $H$ in $G$ is $2$, and and as result $H\lhd G$.

If $|G|>2$, then $H$ is a nontrivial normal subgroup of $G$.

I learned this argument in the text Finite Group Theory by Martin Isaacs, pg. 35, and I have hardly changed it here.

Suppose G is a nonabelian group of order 2m where m is odd, and it has a subgroup A = {1,a} of order 2 which is not normal. Note that $a^2$ = 1. Then $\exists$ y $\in$ G such that

(1) $y^{-1}ay$  =$b \neq$ a. Then

$b^2$ = $y^{-1}ayy^{-1}ay$ = 1
$a^2$ = $b^2$ so that

aa = bb $\Rightarrow$
$ab^{-1} =ba^{-1}$ $\Rightarrow$ ab= ba

Then the set {1,a,b,ab} is a group.  You can see that by inspection (multiplying together any two elements). For example bab = bba = a.

But 4 does not divide 2m, so there cannot be a subgroup of order 4 in G.  So there is no y as in (1).


Calculate limit with floor function How can I proceed to find the following limit:

$$\lim_{n \rightarrow \infty }(n\sqrt 2-\lfloor n\sqrt2 \rfloor) $$

where $n$ is a natural number.
 Please if there is no limit, would you provide a proof.

Let $a_n=n\sqrt 2-\lfloor n\sqrt2 \rfloor$. Note that $a_n>0$ for all $n\in\mathbb{Z^+}$.

Suppose the sequence converges to a limit, $L$, and choose $\epsilon=0.01$, then we have for all $n\ge N$ (for some choice of $N\in\mathbb{N}$),

$$0.01<a_n-L<0.01$$

Then

$$\begin{align}
a_{n+1}-a_n&=((n+1)\sqrt 2-\lfloor (n+1)\sqrt2 \rfloor)-(n\sqrt 2-\lfloor n\sqrt2 \rfloor) \\[2ex]
           &=\sqrt 2+\lfloor n\sqrt2 \rfloor-\lfloor (n+1)\sqrt2 \rfloor
\end{align}$$
The quantity on the right-hand side is equal to either $\sqrt2-1$ or $\sqrt2-2$.

In the first case

$$\begin{align}
a_{n+1}=a_n+(\sqrt2-1) &\implies \\[2ex]
-0.01+(\sqrt2-1)<a_{n+1}-L<0.01+(\sqrt2-1) &\implies \\[2ex]
a_{n+1}-L>0.4
\end{align}$$

and in the second case

$$\begin{align}
a_{n+1}=a_n+(\sqrt2-2) &\implies \\[2ex]
-0.01+(\sqrt2-2)<a_{n+1}-L<0.01+(\sqrt2-2) &\implies \\[2ex]
a_{n+1}-L<-0.5
\end{align}$$

So for any positive $\epsilon<0.01$, if for any supposed limit $L$ we have $|a_n-L|<\epsilon$, then we have $|a_{n+1}-L|>\epsilon$, so the limit does not exist.

I don't know if one can do that more formally, but this is how I would show it:

Denote the sequence by

$\{x_n\}_{n\in\mathbb{N}},\,x_n = n\sqrt 2-\lfloor n \sqrt{2}\rfloor$.

You're interested in the limit of this sequence.

Note: For $a\in\mathbb{R}$ you have that $a-\lfloor a\rfloor$ is the decimal-part of $a$. This decimal-part is always in $[0,1)$. Especially does that mean, that $\{x_n\}\subset [0,1)$.

Take some point $x\in [0,1]$ (if your series converges, it converges to some point in [0,1] ).

We need to show, that $\forall\varepsilon>0\exists n_0\in\mathbb{N}$ s.t. $|x_n-x|<\varepsilon,\,\forall n\geq n_0$ (or that this doesn't hold)

W.l.o.g we consider $\varepsilon < \frac{1}{2}$. If it fails to find $n_0$ for this type of epsilon, that's enough to prove that the series does not converge.

So, by restricting the choice of $\varepsilon$, either $x-\varepsilon$ or $x+\varepsilon$ is also in $[0,1)$.

W.l.o.g we have $\tilde{x} = x+\varepsilon\in[0,1)$.

$\sqrt{2}$ is irrational. This means that for our $x$, we can find an $n_0^x\in\mathbb{N}$ such that

$|n\sqrt{2}-\lfloor n\sqrt{2}\rfloor - x|<\frac{\varepsilon}{2}$.

However, we can also find an $n_0^\tilde{x}\in\mathbb{N}$ such that

$|n\sqrt{2}-\lfloor n\sqrt{2}\rfloor -\tilde{x}|<\frac{\varepsilon}{2}$.

HERE COMES THE VERY IMPORTANT PART: Again, as $\sqrt{2}$ is irrational, the decimal part of $n\sqrt{2}$ "bounces" in the interval (0,1). Especially, you will NOT find $m\in\mathbb{N}$ s.t. $m\sqrt{2}-\lfloor m\sqrt{2}\rfloor \in \{ 0,1\}$. Using this, we can always chose $n_0^\tilde{x} > n_0$. But the same argument also holds in the other direction, so we can always chose $n_0^x > n_0^\tilde{x}$.

You see:

$\forall \varepsilon<\frac{1}{2},\forall x\in[0,1] \not\exists n_0\in\mathbb{N}:\,|x_n-x|<\varepsilon,\,\forall n\geq n_0$.

This suffices to see, that $x_n$ does not converge.


Defining i,j,k,l,.. for complex number in a 3D space, in 4D space, etc Real number is often used to represent a point in a 1-dimensional number line. Real numbers are written as $a$ where $a \in \mathbb{R}$.

Complex number is often used to represent a point in a 2-dimensional plane. Complex numbers are written as $a + bi$ where $a,b \in \mathbb{R}$ and $i^2 = -1$.

What about higher dimensional space?

We can extend the concept of complex number, and write a point in a 3-dimensional space as $a + bi + cj$ where $a,b,c \in \mathbb{R}$, but how do we define i and j?

Certainly, this also applies to 4-dimensional space. We can write a point in 4D as $a + bi + cj + dk$, how do we define $i,j,k$?

How about even higher n-dimensional space? Is there a system to define i,j,k,... that will extend to arbitrary dimensions? Is there a name for it?

It depends upon what structure you want.  In $\mathbb{R}^n$ we do not have the operation of multiplying two elements, as we do on $\mathbb{C}$. If you want that, you probably want a normed division algebra, which are available in four and eight dimensions, but no more.

If you just want a vector space they are basis vectors but you can't multiply two elements.

If we take complex number in 3d plan than we found 2 condition,
let we take 3 axis x y and z:

x is real and yand z are imaginary, that is: $x+i(y+z)$
x and z is real but y is imaginary, that is: $x+z+iy$


Probability that a Mersenne number is prime Let $p$ be a prime and let $M_p = 2^p-1$ be a (Mersenne) number. Is there any known result on the probability that $M_p$ is prime? In particular is it known whether the probability tends to $1$ as $p \to \infty$? If so, is there any known lower bound for the asymptotic rate with which this probability tends $1$? Thank you.

I'm pretty sure there is no such result: it is not even known that infinitely many $M_p$ are primes, or that infinitely many are composite.

For the already mentioned reason, this probability question can't be definitively solved right now.  But for fun you could assume one of the conjectures about the distribution of Mersenne primes (I think Wagstaff conjectured an asymptote of C*loglog(x) for some funky constant C?) and see what that brings.


How many configurations of a 10x10x10 Rubik's cube? I was wondering how many possible configurations there are of a Rubik's cube of size greater than $3\times3\times3$ (e.g. for $4\times4\times4$, $5\times5\times5$, $\dots 10\times10\times10)$? We know that for a $3\times3\times3$-cube there are about $4.3 \times 10^{19}$ configurations, what about  the bigger cubes?

I'll give an upper bound.

We are considering a 10 by 10 cube here. A cube has 6 faces. Assuming that the cube has six colors, there should six options for each square on a face. There are a 100 squares per face. That means there are $6^{100}$ permutations per face. Multiplied by the six faces of the cube gives an upper bound of $6^{600}$ possibilities.

In an attempt to lower the bound, assume that after the first face, there are only 5 options left for each remaining square. After forming the permutations on the second face there would only be 4 options, etc. multiplying that out we get, $(720)^{100}$ as an estimate


Prove that $\log(n) = O(\sqrt{n})$ How to prove $\log(n) = O(\sqrt{n})$?

How do I find the $c$ and the $n_0$?

I understand to start, I need to find something that $\log(n)$ is smaller to, but I m having a hard time coming up with the example.

Less sophisticated: $\log(x) < x$ for all $x > 0$ (as is easily seen from its integral representation).  Therefore $\log(x) = 2 \log \sqrt{x} < 2 \sqrt{x}$.

consider n=2^100

then log(n)=Log(2^100)

           =100*log2
           =100

and now apply this to sqrt(n)

then Sqrt(n)=sqrt(2^100)

           =2^50

we can clearly say that sqrt(n) is larger then logn.


Confusion about the definition of "expected number" in probability When you roll a dice $6$ times, the "expected number" of $6$'s that should come up is $1$. I used to think that this is because by the time you have rolled the dice $6$ times, the chance of having rolled a $6$ is over $50\%$. However, further inquiry has led me to believe that this is not in fact what the "expected number" means.

The chance of rolling a $6$ after $n$ rolls = $1 - (5/6)^n$

After $4$ rolls, the chance of rolling a $6$ is $1 - (5/6)^4 = 51.8\%$ (1 d.p.)

So by then, the probability has already risen to over $50\%$. So what does the "expected number" of $6$'s actually mean?

The expected value is just a weighted average of the values of the outcomes. For instance, the expected value when rolling a fair dice is $3.5$, as:
$$
\frac{1}{6}\cdot 1 +\frac{1}{6}\cdot 2 +\frac{1}{6}\cdot 3 +\frac{1}{6}\cdot 4 +\frac{1}{6}\cdot 5 +\frac{1}{6}\cdot 6 = 3.5
$$

Each outcome is a realization of a random variable (eg, ''six'' when rolling a dice), and the weights are the corresponding probabilities (note that the sum of the probabilities for all possible outcomes should be exactly one, that's why the coefficients of the outcomes are actually weights.

The expected number of sixes when you roll six fair dice at once (or six times a fair dice) is exactly 1. To see why, consider an equivalent experiment with smaller numbers: rolling four 4-side dice (tetrahedral shape). The expected number of fours is exactly 1 as well. Why? Well, when you roll four dice, there are 256 possible outcomes: $(1,1,1,1), (1,1,1,2), \dots (4,4,4,3), (4,4,4,4)$. Here, the number of fours is the outcome (not four itself, but the number of them), and the probabilities are as follows:

Probability that the number of fours is zero is 81/256 $\approx$ 0.316
Probability that the number of fours is one is 108/256 $\approx$ 0.422
Probability that the number of fours is two is 54/256 $\approx$ 0.211
Probability that the number of fours is three is 12/256 $\approx$ 0.047
Probability that the number of fours is four is 1/256 $\approx$ 0.004

Therefore, the expected value of the number of fours is the weighted average:
$$
0.316 \cdot 0 + 0.422\cdot 1+ 0.211 \cdot 2 + 0.047 \cdot 3 + 0.004 \cdot 4 = 1.000
$$

Note that, because the random variable is the number of fours, the result wouldn't change if you compute the expected value of the number of twos or the number of threes. You will always get exactly 1.

The expected number of something is pretty subtle. If you have a function $X$ which outputs numbers $\{x_1, \dots, x_n\}$, then its mathematical expectation is defined to be:

$$E(X) = x_1 Pr(X = x_1) + x_2 Pr(X = x_2) + \dots + x_n Pr(X = x_n)$$

In your case, $X$ would be the "number of sixes in a roll of six dice". You'd get:

$$E(X) = 1 Pr(X = 1) + 2 Pr(X = 2) + \dots + 6 Pr(X = 6)$$

The probability of rolling ``$k$ sixes'' would be: $Pr(X = k) = \binom{6}{k} (5/6)^{6-k} (1/6)^{k}$. If you calculated out that sum, you get one. (Here it is on WolframAlpha.)

It might help to imagine what happens if you roll seven dice. In that case, you get $E(X) = 1.1666.$ You can never quite get a fractional number as an output, but you would "expect" that many outputs of six.

Does that help?


Jointly Gaussian uncorrelated random variables are independent Let $X,Y$ be jointly normally distributed and uncorrelated. Why are they independent?

In short, they are independent because the bivariate normal density, in case they are uncorrelated, i.e. $\rho =0$, reduces to a product of two normal densities the support of each one ranges from $(-\infty, \infty)$. If the joint distribution can be written as a product of nonnegative functions, we know that the RVs are independent. Moreover, we know, and can show, that each marginal density is normal on its own.

That is easy to see in the bivariate density below:

$$f(x,y)= \frac{1}{2 \pi \sigma_1  \sigma_2  \left( 1-\rho^2 \right)^{1/2}} \exp\{-q/2 \}, \quad -\infty<x<\infty,\quad -\infty<y<\infty $$

where $$q= \frac{1}{1-\rho^2} \left[ \left( \frac{x-\mu_1}{\sigma_1} \right)^2-2\rho \left(\frac{x-\mu_1}{\sigma_1} \right) \left(\frac{y-\mu_2}{\sigma_2} \right)+\left(\frac{y-\mu_2}{\sigma_2} \right)^2 \right]$$

Put $\rho=0$. There is also a nice proof involving mfgs. Is that what you were looking for?

Not true
http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent
http://probability.ca/jeff/teaching/uncornor.html

Consider a simple example. Let X have a standard normal distribution. Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=+1] = Pr[Z=-1] = 1/2). Let Y = X Z (the product of X and Z). Then it is easy to see that Y also has a standard normal distribution, and that Cov(X,Y) = 0. On the other hand, clearly X and Y are not independent: indeed, it always holds that |X| = |Y|. The point is that, just because each of X and Y has a normal distribution, that does not mean that the pair (X,Y) has a bivariate normal distribution, nor even that (X,Y) is jointly absolutely continuous, nor does it mean that zero covariance implies independence.


Simple Probability Question about Combinations If someone could please point me in the right direction on these. I get lost on how to think about them.

In a game there are four holes with values 0, 1, 2, and 4. You are given 6 balls to shoot into the holes.

What is the sample outcome:
This is $4^6$. Any ball can go into any hole.

Probability that at least one of the holes has three or more balls?
First off, what is the complement here. I know it might not be used, but I would like to know. I thought it might be none of the holes have three or more balls. So all the holes have less than 3 balls, so at most 2. This is where I get lost on how think about where to put the balls into the holes.

Prob at least one of 4 holes has no ball?
Here I said prob of 1 hole having no balls is $1/6$. Prob of 2 holes is $1/36$. Prob of 3 holes having no balls is $1/6^3$. Lastly prob of 4 holes having no balls is 0. So I add up these correct?

If score 20, win another game. What is Prob of a new game?
Here I said, 6 balls into 4, get 24. 5 balls in 4 get 20, so have one ball left can go into any 3 holes, so have 3 ways. Lastly 4 balls in 4, need 2 balls in 2. So there are 5 chances of a new game. So $5/6^4$ I said, which seems very small. Order does not matter in this case correct?

There are $4^6$ equally probable microstates in the sample space.   Each distinct ball can have $4$ distinct states.

Probability that at least one of the holes has three or more balls?

The complement of this event is that all holes contain less than three balls.  Which will be:

(Case $a$) 3 holes each containing 2 balls, and 1 hole empty, or
(Case $b$) two holes with two balls and two holes with one ball.

$$1-\dfrac{{4\choose 3}{6\choose 2,2,2}+{4\choose 2}{6\choose 2,2,1,1}}{4^6}$$

Prob at least one of 4 holes has no ball?

Use the principle of inclusion and exclusion, to count the ways at least one hole can be empty.

Count the ways to select one hole, and put the balls in the other three, minus the ways to select 2 holes and put the balls into the other two, plus the ways to select 3 holes and put the balls into the remaining hole.

$$\frac{{4\choose 1}3^6-{4\choose 2}2^6+{4\choose 3}1^6}{4^6}$$

We use the Principle of Inclusion and Exclusion to avoid over counting intersections.

The set of arrangements where hole 1 is empty intersects with set of arrangements where hole 2 is empty.  The intersection is the set of arrangements where holes 1 and 2 are empty. Et cetera.

So let $E_x$ represent the arrangements where hole $x$ is empty, $E_{x,y}$ those where hole $x$ and $y$ are both empty and likewise $E_{x,y,z}$.

$$\begin{align}
|\bigcup_x E_x| & = \sum_x |E_x| - \sum_{x<y} |E_{x,y}| + \sum_{x<y<z} |E_{x,y,z}|
& \text{P.I.E.}
\\ & = 4 |E_\ast| - {4\choose 2}|E_{\ast,\ast\ast}| + {4\choose 3} |E_{\ast,\ast\ast,\ast\ast\ast}|
& \text{each set of the same arity has the same cardinality}
\\ & = 4\cdot 3^6 - 6\cdot 2^6 + 4\cdot 1^6
\end{align}$$

If score 20, win another game. What is Prob of a new game?

Yes.  Considering the cases needed to score 20 or more gives:

Six balls score $4$.
Five balls score $4$, one balls scores $0$, $1$, or $2$.
Four balls score $4$, two balls each score $2$.

Can you complete?

Note Do not use stars and bars, or counting visible outcomes, as we must assume each ball is distinct for the purpose of evaluating probabilities.

Although the balls appear to be indistinguishable they are in fact distinct entities; they have hidden-distinctness.  Several arrangements of such entities can result in the same outcome.  To measure the probabilities of an event it is not sufficient to merely count the number of distinct outcomes, we must actually measure the probabilities of each distinct outcome by considering the arrangements of hidden-distinct entities that contribute to them.

As demonstrated below:

If I take 2 marked balls and toss them into 3 distinct holes, the $3^2$ equally possible outcomes are:

$$\begin{array}{|c|c|c|}\hline
AB & &
\\\hline  & AB &
\\\hline & & AB
\\\hline A & B &
\\\hline B & A &
\\\hline A & & B
\\\hline B & & A
\\\hline & A & B
\\\hline & A & B
\\\hline
\end{array}$$

All these outcomes are equally likely if each ball has an independent $1/3$ probability of landing in any given hole.

So for example, the probability that the balls land in two different boxes is: $2/3$.

Now, should a coloured sheet of plastic be placed between you and the hole so that the marks become indistinct, then the visible outcomes become:
$$\begin{array}{|c|c|c|}\hline
XX & &
\\\hline  & XX &
\\\hline & & XX
\\\hline X & X &
\\\hline X & & X
\\\hline & X & X
\\\hline
\end{array}$$

However, the probabilities that the balls land in two different boxes does not magically become $1/2$.  The individual balls still have the same probabilities of landing where they may even if you can no longer tell them apart.

The following configurations and respective permutations are possible:

(6,0,0,0): can occur $\frac{4!}{3!} = 4$ ways.
(5,1,0,0): can occur $\frac{4!}{2!} = 12$ ways.
(4,2,0,0): can occur $\frac{4!}{2!} = 12$ ways.
(4,1,1,0): can occur $\frac{4!}{2!} = 12$ ways.
(3,3,0,0): can occur $\frac{4!}{2!2!} = 6$ ways.
(3,2,1,0): can occur $4! = 24$ ways.
(3,1,1,1): can occur $\frac{4!}{3!} = 4$ ways.
(2,2,2,0): can occur $\frac{4!}{3!} = 4$ ways.
(2,2,1,1): can occur $\frac{4!}{2!2!} = 6$ ways.

There are 84 possible arrangements.

The complement is "none of the holes has 3 or more balls" which occurs with the (2,2,2,0) and (2,2,1,1) configurations. So, the probability of the complement is 10/84. As a result the probability that at least one of the holes has three or more balls is 74/84.
Probability that at least one of 4 holes has no ball is given by the complement of the event "all four holes have balls". This event is represented by the (3,1,1,1) and (2,2,1,1) configurations and its probability is 10/84. As a result the probability of your event of interest is 74/84.
The 5 arrangements that produce a score of 20 or above seems to be correct. The probability of a new game is 5/84.


Finding a curve from its evolute Consider the evolute given by
$$ \gamma: I \subset \mathbb{R} \to \mathbb{E}^2: t \mapsto (\cos(t),\sin(t))$$
Now, how do I find all the curves $\alpha$ that have $\gamma$ as their involute?
I tried using the fact that an involute is given by
$$\gamma(t) = \alpha(t) + \frac{1}{\kappa(t)}N(t)=(\cos(t),\sin(t)) $$
so that also
$$\gamma'(t) = \left(\frac{1}{\kappa(t)}\right)'N(t)= (-\sin(t),\cos(t))$$
and then using the Frenet-Serret formula's, but I haven't gotten very far...

Hint: The involute of an evolute is the original curve. and the evolute of an involute is the original curve.


Integral of $e^{\frac{y}{x}}$ How can we evaluate the following?

$$\int e^{\frac{y}{x}}\ \mathrm dy$$

An explanation of the answer would be helpful.

The answer I got is $ x  e^{y/x}$.

But not sure about the steps used for obtaining the answer...

Since this a task on finding a primitive, you can always check you candidate answer by taking (partial in your case) derivative.

First note that
$$\int a^t\ \mathrm dt=\frac{a^t}{\ln a}+C$$
Therefore
$$\int e^{\frac{y}{x}}\ \mathrm dy=\int \left(e^{\frac{1}{x}}\right)^y\ \mathrm dy$$
$$=\frac{\left(e^{\frac{1}{x}}\right)^y}{\ln e^{\frac{1}{x}}}+C=xe^{\frac{y}{x}}+C$$


Calculate center of mass with respect to radius I want to calculate the center of mass of a half circle with respect to the radius to avoid using x,y coordinates directly. Such that I can write $S_x=S_r\cdot \cos(S_{\phi})$

See here on Wolfram Alpha:

https://www.wolframalpha.com/input/?i=(2%2Fpi)+integrate+r%5E2+dphi+dr,+phi%3D0..pi,+r%3D0..1

The result is obviously wrong. Is there a mistake in my formula or is it just not possible to calculate the center of mass with respect to radius/angle, whether in polar or spheric coordinates?

This can be readily solved in the complex plane without reference to the Cartesian or even polar coordinates. We begin with the definition of the centroid

$$
A=\frac{1}{2}\int\Im\{z^*\dot z\}d\theta\\
Z_c=\frac{1}{3A}\int z\ \Im\{z^*\dot z\}d\theta
$$

We the consider a semicircle of unit radius

$$
z=e^{i\theta},\quad \theta\in[0,\pi]\\
\dot z=ie^{i\theta}\\
z^*\dot z=i\\
\Im\{z^*\dot z\}=1\\
\begin{align}
Z_c
&=\frac{1}{3A}\int e^{i\theta} ~d\theta,\quad A=\pi/2\\
&=\frac{2}{3\pi}\frac{e^{i\theta}}{i}\biggr|_0^{\pi} \\
&=\frac{4 i}{3\pi},\quad (\text{i.e., on the } y \text{-axis})
\end{align}
$$

as expected.

You need a $sin(\phi)$ in the integration.

See: https://www.wolframalpha.com/input/?i=(2%2Fpi)+integrate+sin(phi)+r%5E2+dphi+dr,+phi%3D0..pi,+r%3D0..1

That gives $4\over3\pi$.


show this inequality $\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\frac{1}{x_{i}}+n\right)^n$ Let $x_{i}\ge 1$,show that
$$\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n\right)^n$$

or
$$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$
and it seem use AM-GM inequality?
$$\sum_{i=1}^{n}x_{i}\ge n\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$$
$$\sum_{i=1}^{n}\dfrac{1}{x_{i}}\ge \dfrac{n}{\sqrt[n]{x_{1}x_{2}\cdots x_{n}}}$$ let $\sqrt[n]{x_{1}x_{2}\cdots x_{n}}=t$,since
$$\Longleftrightarrow \left(\dfrac{t+1}{\frac{1}{t}+1}\right)^n\ge t^n$$But I can't it

We need to prove that
$$\frac{\sum\limits_{k=1}^nx_k+n}{\sum\limits_{k=1}^n\frac{1}{x_k}+n}\geq\sqrt[n]{\prod\limits_{k=1}^nx_k}$$ and by  the Vasc's EV Method (see here:

http://emis.ams.org/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf ,

Corollary 1.7(a))

it's enough to prove our inequality for $$x_1=b^n\leq x_2=...=x_n=a^n$$

where $a\geq1$ and $b\geq1$ or
$$\frac{(n-1)a^n+b^n+n}{\frac{n-1}{a^n}+\frac{1}{b^n}+n}\geq a^{n-1}b$$ or
$$ab^{n-1}\left((n-1)a^n-na^{n-1}b+b^n\right)\geq a^n-nab^{n-1}+(n-1)b^n.$$
Let $a=xb$.

Hence, $x\geq1$ and we need to prove that
$$b^nx\left((n-1)x^n-nx^{n-1}+1\right)\geq x^n-nx+n-1$$ and since $b\geq1$ and by AM-GM
$$(n-1)x^n-nx^{n-1}+1\geq0,$$ it's enough to prove that
$$(n-1)x^{n+1}-nx^n+x\geq x^n-nx+n-1$$ or $f(x)\geq0$, where
$$f(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-(n-1).$$
Now,
$$f'(x)=(n+1)(n-1)x^n-n(n+1)x^{n-1}+n+1$$ and
$$f''(x)=n(n+1)(n-1)x^{n-1}-(n-1)n(n+1)x^{n-2}\geq0.$$
Thus, $$f'(x)\geq f'(1)=0$$ and from here
$$f(x)\geq f(1)=0$$ and we are done!

we have to prove  :
$$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$

Let $x_i\geq 1$ be real numbers so we have :
$$\frac{1}{\sum_{i=1}^{n}S_i}\left(\sum_{i=1}^{n}S_i[(-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n]\right)\ge \prod_{i=1}^{n}x_{i}$$

Where :

$$S_i=\frac{\prod_{i=1}^{n}x_{i}}{|((-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n)-1|}$$

And :

$$S_{min}=\frac{n\prod_{i=1}^{n}x_{i}}{|((-x_{min}+\frac{(\sum_{i=1}^{n}x_{i})+n}{(\sum_{i=1}^{n}\frac{1}{x_{i}})+n})(n)x_{min}^{n-1}+x_{min}^n)-1|}$$

After that we use the theorem 5 of this link :

You just have to replace the function by $\phi(x)=x^n$ with domain $[x_{min};x_{max}]$ and $d=\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n}$ and remark that we have :
$x_{min}\leq d \leq x_{max}$ and put $S_i=W_i$ to get the LHS.


$P[X_1<X_2]$ for independent exponentially distributed $X_1$ and $X_2$ Given $X_1$ and $X_2$ are exponentially distributed with parameter $\lambda_1$ and $\lambda_2$ respectively, we can see
$P[X_1 < X_2] = \int_0^{\infty}P[X_2 > x | X_1 = x]P[X_1=x]=\frac{\lambda_1}{\lambda_1+\lambda_2}$.

Given the simplicity of this result is there a way to prove this more intuitively like just from the first principles that
$P[X_i \in (x,x+dx)] = \lambda_idx, i=1,2$

My acknowledgements to André for his discussions that helped me figure out this answer. Let $A$ denote the event $min(X_1, X_2) \in (x,x+dx)$.

\begin{eqnarray*}
P(X_1 < X_2) &=& \int_0^{\infty}\left[P(X_1 < X_2 | A)P(A)\right] \\
&=& \int_0^{\infty}\left[\frac{\lambda_1dx}{\lambda_1dx+\lambda_2dx}P(A)\right] \\
&=& \frac{\lambda_1}{\lambda_1+\lambda_2}\int_0^{\infty}P(A) \\
&=& \frac{\lambda_1}{\lambda_1+\lambda_2}
\end{eqnarray*}

Similar arguments show that, given two coins with probabilities $p_1$ and $p_2$ for head, the probability that the first tail happens simultaneously in a series of infinite tosses is $\frac{(1-p_1)(1-p_2)}{1-p_1p_2}$ without needing to calculate the infinite sums.

the best way to solve this is like this
$$\begin{align}P(X_1<X_2)
&=\int^\infty_0P(X_1<X_2|X_1=x)f(x)dx\\
&=\int^\infty_0P(x<X_2|X_1=x)f(x)dx \\
&=\int^\infty_0P(X_2>x|X_1=x)f(x)dx\end{align}$$
because $X_2>x$ and $X_1=x$ are independent, so that

$$\begin{align}P(X_1<X_2)
&=\int^\infty_0P(X_2>x)f(x)dx\\
&=\int^\infty_0(1-1+e^{-\lambda_2x})\lambda_1e^{-\lambda_1x}dx\\
&=\int^\infty_0\lambda_1e^{-\lambda_1x-\lambda_2x}dx\\
&=\frac{\lambda_1}{\lambda_1+\lambda_2}\\
\end{align}$$


Demonstration of $\int_{a}^b f(x) \,dx= 0 \Rightarrow f(x)\equiv0 $ Good morning,

Can you give me a help to demonstrate this proposition:

$f$ is a continuous and not negative function on the interval $[a,b] \ a,b \in \Re $,  Demonstrate:

$$\int_{a}^b f(x) \,dx= 0 \Rightarrow f(x)\equiv0  $$

Let

$$\Phi(t)=\int_a^tf(x)dx$$
then $\Phi'(t)=f(t)\ge0$ hence $\Phi$ is non-decreasing but since $\Phi(a)=\Phi(b)=0$, this means $\Phi$ is constant and $\Phi'(t)=f(t)=0$.

If $f$ is Riemann integrable on $[a,b]$, then $f$ is Lebesgue integrable on $[a,b]$, and the Riemann and Lebesgue integrals are equal. It is fairly straightforward to show that a nonnegative measurable function has Lebesgue integral zero if and only if it is zero almost everywhere. Since we have the additional assumption that $f$ is continuous, if $f$ is zero almost everywhere, then necessarily $f$ must be identically zero.

I don't understand why I have 3 down votes for this answer and no comments...


Show that every monotonic increasing and bounded sequence is Cauchy. The title is kind of misleading because the task actually to show

Every monotonic increasing and bounded sequence $(x_n)_{n\in\mathbb{N}}$ is Cauchy

without knowing that:

Every bounded non-empty set of real numbers has a least upper
bound. (Supremum/Completeness Axiom)
A sequence converges if and only if it is Cauchy. (Cauchy
Criterion)
Every monotonic increasing/decreasing, bounded and real
sequence converges to the supremum/infimum of the codomain (not sure
if this is the right word).

However, what is allowed to use listed as well:

A sequence is called covergent, if for $\forall\varepsilon>0\,\,\exists N\in\mathbb{N}$ so that $|\,a_n - a\,| < \varepsilon$ for $\forall n>N$. (Definition of Convergence)
A sequence $(a'_k)_{k≥1}$ is called a subsequence of a sequence $(a_n)_{n≥1}$, if there is a monotonic increasing sequence $(n_k)_{k≥1}\in\mathbb{N}$ so that $a'_{k} = a_{n_{k}}$ for $\forall k≥1$. (Definition of a Subsequence)
A sequence $(a_n)_{n≥1}$ is Cauchy, if for $\forall\varepsilon>0\,\,\exists N=N(\varepsilon)\in\mathbb{N}$ so that $|\,a_m - a_n\,| < \varepsilon$ for $\forall m,n>N$. (Definition of a Cauchy Sequence)
(Hint) The sequence $(\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}$ is unbounded for $\varepsilon>0$. (Archimedes Principle)

Would appreciate any help.

If $x_n$ is not Cauchy then an $\varepsilon>0$ can be chosen (fixed in the rest) for which, given any arbitrarily large $N$ there are $p,q \ge n$ for which $p<q$ and $x_q-x_p>\varepsilon.$

Now start with $N=1$ and choose $x_{n_1},\ x_{n_2}$ for which the difference of these is at least $\varepsilon$. Next use some $N'$ beyond either index $n_1,\ n_2$ and pick $N'<n_3<n_4$ for which $x_{n_4}-x_{n_3}>\varepsilon.$ Continue in this way to construct a subsequence.

That this subsequence diverges to $+\infty$ can be shown using the Archimedes principle, which you say can be used, since all the differences are nonnegative and there are infinitely many differences each greater than $\varepsilon$, a fixed positive number.

1)Every monotonous increasing/decreasing, bounded and real
sequence converges to the supremum/infimum of the codomain (not sure if this is the right word).

2)A sequence converges if and only if it is Cauchy. (Cauchy
Criterion).

You wrote it yourself.


How to define mathematically the integer part of an average? Scenario: Suppose I have three numbers: 1, 3 and 4. The average of the three numbers are 2.67 (to 2 decimal places). It many real-world applications, there is the need to consider only the "whole number", that is, the integer part, hence the value 2 only.
Question: How do I define using mathematical expression this particular issue, that is, consider only the whole number result of the average expression?
Thank you.

You can define the integral part using Greatest integer function(represented as |_ _|).

So, greatest integer(2.67) =  |_ 2.67 _| = 2.

And, in almost all of the programming languages there exists methods/functions defined in mathematical libraries to achieve the same.

In Java, you can achieve the same using double floor(double d) method.

Ex :- Math.floor(2.67) = 2.0 and then you can convert it into int using down-casting OR down promotion.

int res = (int)(Math.floor(2.67));

As mentioned by Evert, if you're performing operations with negative average, then  you need to use smallest-integer function(ceil function).

int res = (int)(Math.ceil(2.67));

So, your code would be :-

if(avg>=0)
Math.floor(avg);
else
Math.ceil(avg);

Mathematically,
     Avg(n1,n2,...nm) = p/q for lowest values (no more divisible) for all p belongs to natural number and q =1.


Does infinite time = time with no end = never? Say an object is to travel from point A to point B, a finite distance of 2 meters. Say the object travels at 1m/s.

After 1 meter the speed of the object is halved. After another half of the previous distance the speed is halved again, and so on.

So for each term in the infinite series 1 + 1/2 + 1/4 + 1/8 + ..., the term represents the distance the object travels in meters before the speed is halved.

Now apparently the object does eventually reach point B, but it takes an infinite amount of time to do so.

Yet infinity has no end.

So if the the object does ever reach point B, wouldn't that mark the end of the infinite length of time that it took to reach it and wouldn't that contradict with the definition of infinity?

Yet if we say it never reaches point B, which is to say that there is no point in time that it reaches point B because that point in time would mark an end to infinity, then isn't it consistent with what we mean by infinity?

The key to cracking Zeno's paradox is the realization that an infinite series can have a finite sum. You mention an example of such a series:

$$\sum\limits_{n=0}^\infty \frac{1}{2^n} = 1 +\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 2$$

That yields the 2 meters from your example. You can divide that number infinitely, as Zeno did, but if you add those divisions back up, you still get 2: a finite number. Because it's a finite distance, it can be crossed in finite time.

You still have to cross each partial distance, but each of these distances is also finite (in fact, you can subdivide each of the the same way you divided up the original distance), so they can all be crossed in finite time too. If you move at the same speed through each one, they form the very same sort of infinite series as the distances do. In fact, you don't even have to move at the same speed through each one; it just makes the math easier if you do.

Proof by assumption:
1. Assume there is an object C which is twice as far away as object B in the same direction.
2. Since the velocity is constant, and object B is in the path between object A and object C, then by the same process you can see after the first term in the new sequence (for A and C) that object A does in fact reach object B.


Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ?

$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$

$$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace  $$

So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 -  3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence
$\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3}) $.

Another solution:
Clearly it has to be a subfield of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$. Thus its degree has to be one of these numbers: $1,2,4$. We just have to rule out it is of degree $2$. If it is, then there is a minimal polynomial of degree two s.t $\sqrt{2}+\sqrt{3}$ is its root. However, if there was such polynomial, say $x^2+bx+c$ then $5+c+2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$ which is impossible as $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$ and the equation implies it is linearly dependent!


Balance chemical equations without trial and error? In my AP chemistry class, I often have to balance chemical equations like the following:

$$ \mathrm{Al} + \text O_2 \to \mathrm{Al}_2 \mathrm O_3 $$

The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side.

A solution:

$$ 4 \mathrm{Al} + 3 \mathrm{ O_2} \to 2 \mathrm{Al}_2 \mathrm{ O_3} $$

When the subscripts become really large, or there are a lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced?  (Do such equations exist?) I tried one for a long time only to realize the problem was wrong.

My teacher said trial and error is the only way. Are there other methods?

Yes; it's possible to write a system of equations that can be solved to find the correct coefficients. Here's an example for the given formula.

We're trying to find coefficients A, B, and C such that

$A (\mathrm{Al}) + B (\mathrm{O_2}) \rightarrow C (\mathrm{Al_2 O_3})$

In order to do this, we can write an equation for each element based on how many atoms are on each side of the equation.

for Al: $A = 2C$
for O: $2B = 3C$

This is an uninteresting example, but these will always be linear equations in terms of the coefficients. Note that we have fewer equations than variables. This means that there's more than one way to correctly balance the equation (and there is, because any set of coefficients can be scaled by any factor). We just need to find one integral solution to these equations.

To solve, we can arbitrarily set one of the variables to 1 and we'll get a solution with (probably fractional) coefficients. If we add $A=1$, the solution is $(A,B,C) = (1,\frac{3}{4},\frac{1}{2})$. To get the smallest solution with integer coefficients, just multiply by the least common multiple of the denominators ($4$ in this case), giving us $(4,3,2)$.

If the set of equations has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced.

Given that:

$$Al+O_2\to Al_2O_3,$$

introduce variable coeffecient i.e:

$$sAl+cO_2\to eAl_2O_3.$$

Comparing coeffecient on both sides

Comparing A: $1s=2e$;
Comparing O: $2c=3e$.

When $e=2$, $S=2\cdot2=4$. Again, $2c=3\cdot2\implies c=3$. Hence $e=2$,  $s=4$ and $c=3$:

$$4Al+3O2\longrightarrow 2Al_2O_3.$$


Solution Verification: Differentiate $\left(x-1/2\right)^2+\left(y-1/4\right)^2=5/16$ with respect to $x$ Differentiate $$\:\left(x-1/2\right)^2+\left(y-1/4\right)^2=5/16$$

With respect to $x.$

The above simplifies to

$$x^2-x+y^2-\frac{y}{2} = 0$$

I know that you have to take the implicit derivative, such that

$$\frac{d}{dx}(x^2-x+y^2-\frac{y}{2}) = \frac{d}{dx}(0)$$

Thus

$$2x-1+2y-\frac{1}{2}=0$$

and

$$y=-x+\frac{3}{4}$$

So $$y'=-x+\frac{3}{4}$$

Have I made any errors, and where can I improve on this solution? In particular, I am worried because the last two equations are the same except the difference between $y$ and $y'.$ I mean, both can't be assumed to be true, right?

And I am asked to find the equation of the tangent of the circle that intersects at the point $(1, 0).$ In this case, do I write the equation in the form $y=mx+b$ (but with numbers replacing $m$ and $b$) as well, in which case I would have two $y$ variables in my solution?

Your error is first forgetting that $y$ depends on $x$ (I assume because implicit derivatives are mentioned and $x(y)$ would be pretty mean) and then getting confused and solving for $y$ and not $y'$.

You are right that you need to take
$$
\frac{d}{dx}(x^2-x+y^2-y/2)=0
$$
But $y$ is a function of $x$ so this should be by the chain rule
$$
2x-1+2yy'-1/2y'=0
$$
Can you solve for $y'$ now?

Now as for the tangent line to the circle described by this curve, you now have the equation for the slope of the tangent line to the curve at any point and a given point. So your $m=y'(1)$ and you can use precalculus methods to find the equation of the line in point slope form.

It should be $2x-1+\left(2y-\frac{1}{2}\right)y'=0$


Is the perimeter of a square isometric to the circle? Homeomorphic? An isometry is defined as a function that preserves distances between points two metric spaces.

If make a circle with the same perimeter/circumference as the square, is it possible to find an explicit isometry between the two sets?

As for homeomorphic, I can visualize a homeomorphism, but again, I cannot find an explicit function between the square and circle that is bicontinuous and bijective.

If the metric is minimal distance along the curve, then you find a isometric mapping by simply walking each curve with the same speed.

Isometric? No. Homeomorphic? Yes. Remember isometry is a distance preserving bijection.


Non-degenerate symmetric bilinear form; dimension formulae. Let $E$ be a vector space endowed with a non-degenerate symmetric bilinear form. Show $\dim F+\dim F^{\perp}=\dim E=\dim\left(F+F^{\perp}\right)+\dim\left(F\cap F^{\perp}\right)$

Lang uses this formula in proposition 1.2, page 573 of his book "Algebra" (Graduate).

I did not find a straight forward argument to show it, the only argument I found was really long, and not very enlightening.
But I assume if he stated it without proof

Here is what I want to show:

Let $E$ be a finite-dimensional vector space over a field K.
Let $g$ be a symmetric/alternating/hermitian form (it suffices to me if you prove it for symmetric and hopefuly I can go from there).
Assume g is non-degenerate.
Let F be a subspace of E.
I want to show the following:
$
\dim F+\dim F^{\perp}=\dim E=\dim\left(F+F^{\perp}\right)+\dim\left(F\cap F^{\perp}\right)
$
It may be that not all hypothesis are necessary, he just uses this formula in a proof but doesn't derive it.

Thanx to anyone who helps!
cheers

For arbitrary subspaces $U,V$ of an arbitrary vector space $W$, we have the dimension formula

$$\dim U + \dim V = \dim (U+V) + \dim (U\cap V).$$

(Take a basis $B_1$ of $U\cap V$, and extend it by systems $B_2$ to a basis of $U$, and $B_3$ to a basis of $V$. Then $B_1 \cup B_2 \cup B_3$ is a basis of $U+V$.)

That yields the

$$\dim F + \dim F^\perp = \dim (F + F^\perp) + \dim (F\cap F^\perp)$$

part. It remains to see that $\dim F + \dim F^\perp = \dim E$. Let's denote the non-degenerate bilinear form by $\beta$. $\beta$ induces a linear map $\Phi \colon E \to E^\ast$ ($E^\ast$ is the dual space of $E$, the space of all linear maps $E\to K$) via

$$\Phi(x)(y) = \beta(x,y).$$

The non-degeneracy of $\beta$ is equivalent to the injectivity of $\Phi$. Since the spaces are finite-dimensional, we have $\dim E^\ast = \dim E$, and thus $\Phi$ is an isomorphism. For a subspace $F \subset E$, the image of $F^\perp$ under $\Phi$ is the annihilator of $F$,

$$\Phi(F^\perp) = F^0 = \{ \lambda \in E^\ast : F\subset \ker\lambda\}.$$

If you already know that $\dim F + \dim F^0 = \dim E$ for all subspaces of finite-dimensional spaces, that's it. Otherwise, to see that, choose a basis $B_0 = \{v_1,\dotsc, v_f\}$ of $F$, extend it to a basis $B = B_0 \cup B_1 = \{v_1,\dotsc,v_f,v_{f+1},\dotsc,v_e\}$ of $E$, and consider the dual basis $B^\ast = \{\lambda_1,\dotsc,\lambda_e\}$ of $E^\ast$, defined by

$$\lambda_i(v_j) = \delta_{ij} = \begin{cases}1 &, i = j\\ 0 &, i \neq j. \end{cases}$$

It is then easy to see that $F^0 = \operatorname{span} \{\lambda_{f+1},\dotsc,\lambda_e\}$, whence $\dim F^0 = e - f = \dim E - \dim F$.

Have you read about complements of a vector space? Finite-dimensional vector spaces have the good property that every subspace is of finite dimension and has a complement. If you have a subspace $F \subseteq E$, then you get the complement $F^\perp$ why considering all elements of $E$ that are ortogonal with respect to your bilinear form, ie. all $x \in E$ with
$$\langle x,y \rangle = 0 \text{ for all } y \in F$$
Then you get a decomposition of $E$ with respect to your subspace $F$ and your bilinear form $\langle,\rangle$ into
$$ E = F \oplus F^\perp$$
So the dimension forumlar should be clear. If the bilinear form degenerates, the sum coudl be not direct so you have to take the dimension of "the rest" $F \cap F^\perp$ into account.

You can read the Wikipedia article on orthogonal complements to get informed about this subject as well.

Hope, that helps you!


Convergence of the sequence $\left(\frac{1}{n}\right)^{\sin\left(\frac{1}{n}\right)}$ I tried using the logarithmic method. But I couldn't work with it as I am not sure we can apply L'Hospital to the indeterminate form -infinity/infinity.
Please help me prove the convergence.

For any $x\in(0,1)$ we have $x\sin(1) \leq \sin(x) \leq x$ by concavity, hence
$$ \lim_{n\to +\infty}\frac{1}{n^{\sin(1/n)}} = \lim_{n\to +\infty}\sqrt[n]{1/n}=1 $$
holds by squeezing.


Is x*(1/x)=1 propositional logic? My book says variables cannot be used in propositional logic. Is this an exception? It can't be false so I know it is a tautology, but my book doesn't categorize tautology into a type of logic.

No, this is not a formula of propositional logic.

In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.

(Note that in the jargon of mathematical logic, something like $x\cdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).

This is not a propositional logic. It would be a propositional logic if it was:
$$x\neq0 \Rightarrow   x\frac{1}{x} = 1$$


How do $\lim_{n\to\infty} \frac{\tan(n)}{2^n} =0 $? How do one show $\lim_{n\to \infty} \frac{\tan(n)}{2^n}=0 $
But, I didn't compute $\lim_{n\to \infty} \frac{\tan(n)}{2^n} $, as $\lim_{n\to\infty} \tan(n) $ doesn't exist even in extended real line. So, I am unable to use L'Hospital's rule.
What should I do?
My other question is what is the limit of $\lim_{n\to\infty} |\tan(n)| $, is it $\infty$ or doesn't exist infinitely?
Sorry to everyone for edit, i am now curious about this limit. That's why I edit it.
I knew, I asked a different question before in this post, but please don't do vote for closing it, I didn't know much about this $\lim_{n\to \infty} \frac{\tan(n)}{2^n}$ because of the tan(n) term, so , please help me how should I deal with it?

For b), if the sequence diverged, then $\forall L>0 \ \ \exists n_0$ s.t.$ \tan(n)>L \ \forall n>n_0$. Yet $\tan (\pi k)=0 \ \forall k=2n, n \in \mathbb{N}$. Hence the sequence doesn't diverge. Use the same logic to show sequence doesn't converge either. Hence the limit doesn't exist.


Show that $g$ is not topologically conjugate to the tent function Question: Define the function $g : [0, 1]\mapsto [0, 1]$ by

$g(x)=\begin{cases}3x&\text{if}\;\;0\le x\le \frac{1}{3}\;\;\\{}\\
2-3x&\text{if}\;\;\frac{1}{3}\le x\le \frac{2}{3}\;\;\\{}\\
3x-2&\text{if}\;\;\frac{2}{3}\le x\le 1\;\;\\{}\\\end{cases}$

Show that $g$ is not topologically conjugate to the tent function.

Def.: The functions $f : X\mapsto X$ and $g : Y\mapsto Y$ (and the
dynamical systems defined by them) are said to be topologically conjugate if
there exists a homeomorphism $h : X\mapsto Y$ such that $g\circ h = h\circ f$. The
function h is called a topological conjugacy between $f$ and $g$.

If I must show that there is no $h(x)$ such that $g\circ h = h\circ f$, I don't have any clue how to guess some $h(x)$ at all; even though, it's not a proof it's 'many examples'.

Also, I don't know which of tent functions writer is mentioning; i.e., for what $n$ in $T^n$. The pictures for $T^1$ and $T^2$ follows:

Thank you.

Consider the solutions of $g\circ h=\frac{1}{2}$ and $h\circ T=\frac{1}{2}$ (or $h\circ T^n=\frac{1}{2}$ if you want or need to prove it for any number of tents).

Solving $g\circ h=\frac{1}{2}$:

$g=\frac{1}{2}$ at $3$ points in the interior of $[0,1]$. These points are mapped by $h^{-1}$ to three interior points of $[0,1]$.

Solving $h\circ T=\frac{1}{2}$:

$h^{-1}$ maps $\frac{1}{2}$ to an interior point $a\in[0,1]$. Then $T=a$ (or $T^n$) at an even number of points.


Using sequences to show that limits of certain functions as certain values are approached do not exist I vaguely recall from undergraduate analysis that there is a relationship that allows
sequences to be used to show that certain limits don't exist. For example, I recently
came across a limit problem, specifically the limit of $\sin(1/x)$ as $x$ approaches zero.
The person who posted the solution used sequences to show that this limit does not exist.
I was wondering if someone would be kind enough to refresh my memory and generalize this
process along with any theorems that may be necessary or perhaps refer me to another internet resource that would do so, as I was
foolish enough to sell my analysis text, and I don't believe a basic undergraduate calculus text would cover this topic.

Thanks

For any fixed $a\in[-1,1]$, the equation $f(x)=\sin(1/x)=a$ has infinitely many solutions on every interval $(0,\delta)$, from which you can construct a sequence $x_n\to 0$ such that $\lim_{n\to\infty} f(x_n)=a$. Since you can reach any $a\in[-1,1]$, the limit as a function does not exist. This applies to any oscillating singularity.


Practical applications of Graeffe's root finding method What are the practical applications of Graeffe's root finding method? I searched a lot but couldn't find any. I found that it is used in aerodynamics and electric circuit analysis. But I don't know much about that.

The Graeffe iteration itself is used in other root finding schemes as a means to compute correct inner and outer root radii. See for a quite graphical example Dedieu/Yakoubshohn on the Bisection-Exclusion algorithm in the complex plane.

Schönhage's circle splitting method uses it to find areas with many roots and to find their factor in the polynomial. (But then you may ask, where is the circle splitting method actually implemented for wide use...)

It would also be interesting to know if the "Tangent Graeffe iteration" by Malajovich/Zubelli (2001) using a special data type of dual numbers, with a renormalization to drastically extend the range of the floating point exponent, has seen any practical application.

I used this method to find practical approximation in the problem of the comparison of 2 binomial samples.


Improper integral $\int_0^\infty x\sin(x^3) dx$ I need to check if the following integral converges / diverges conditionally.  $$\int_0^\infty x\sin(x^3) dx$$
I have tried integrating by parts and it didn't work.
Will appreciate any help :).

The convergence can be based on the alternating series test.
Let $0 < \alpha < \beta$. Then
$$
\int_{\alpha}^{\beta} x \sin(x^3) \, dx
= \int_{\alpha^3}^{\beta^3} t^{1/3} \sin(t) \cdot \frac13 t^{-2/3} \, dt
= \frac13 \int_{\alpha^3}^{\beta^3} t^{-1/3} \sin(t)\, dt
$$ by change of variables, so your integral converges if and only if
$$
\int_{2\pi}^{\infty} t^{-1/3} \sin(t)\, dt
$$
converges; it does since the subsequent integrals
$$
\int_{2\pi}^{3\pi} t^{-1/3} \sin(t)\, dt, \int_{3\pi}^{4\pi} t^{-1/3} \sin(t)\, dt, \cdots
$$
form an alternating sequence which decreases monotonically in absolute value and converges to zero.

Ok so applying Yves' suggestion I get:
$\int_0^\infty x\sin(x^3) dx = \int_0^\infty\sin(t)/t^{-3}dt$ which converges. Thank you!


Why do the equations $Ax + By + Cz= D$ represent planes in $\Bbb R^3$ and not lines? I was wondering why $$Ax+By+Cz=D$$ represents a plane in $\mathbb{R}^3$, because from my perspective it seems that it could simply also represent a line in  $\mathbb{R}^3$. And, yes, I do acknowledge that lines aren't three-dimensional, but surely we can have lines in three-dimensional space.

If you know about dot product, the above equation can be written as (assume $C\neq 0$)

$$ \langle A, B, C\rangle \cdot \langle x, y, z - \frac{D}{C}\rangle = 0$$

Thus $(x, y, z)$ satisifies $Ax+By+Cz = D$ if and only if the vector from $(0,0,\frac{D}{C})$ to $(x, y, z)$ is perpencidular to $\langle A, B, C\rangle$. The set of all such $(x, y, z)$ will be a plane passing through $(0, 0, \frac DC)$ with normal $\langle A, B, C\rangle$.

If we fix a value of x and y, we get only one value of z. i.e x and y can take any real value but then z will be fixed.This is the algebraic way of stating a plane.


Finding the closed-form expression of x I am struggling to find a closed-form expression for $x$ in the following equation. I feel that the Lambert W function may be useful for finding an answer. I have been working on this for hours; however, I could not find any solution. I would really appreciate it if you could help me solve this. Thanks.
$$\frac{ax+b}{(cx+d)(kx+d)}-\ln\left(\frac{kx+d}{cx+d}\right)=y$$

Assuming that all constants are non $0$, $c\neq k$, considering $$y=\frac{ax+b}{(cx+d)(kx+d)}-\ln\left(\frac{kx+d}{cx+d}\right)$$ let $$t=\frac{kx+d}{cx+d}\implies x=\frac{d(1- t)}{c t-k}$$ making
$$y=\frac{\left(b k^2-a d k\right)+ (a c d+a d k-2 b c k)\,t+ \left(b c^2-a c
   d\right)\,t^2 }{d^2  (c-k)^2 \,t}-\log(t)$$ The roots of the numerator are
$$t_1=\frac kc\qquad \text{and} \qquad  t_2=\frac{b k-a d}{b c-a d}$$ which means that $y$ write in the form of

$$y=\frac{\alpha(t-t_1)(t-t_2)}t-\log(t)$$ which means that $t$ "can" be expressed in terms of the generalized Lambert function.

While I don't want to bash out an exact solution, I think I have an idea to help you solve the problem.

We can rewrite your equation as $\dots $

$$ \frac{ax+b}{(cx+d)(kx+d)}  - \left(ln(kx+d) - ln(cx+d)\right)=  y $$

Let $f(x) = ln(kx+d) - ln(cx+d)$. Since $\frac{d}{dx} (ln(x)) = \frac{1}{x}$, we can write

$$f'(x) = \frac{d}{dx} f(x) = \frac{k}{kx+d} - \frac{c}{cx+d} = \frac{d(k -c)}{(kx+d)(cx+d)}.$$

Therefore if we have $a = 0$ and $b = d(k-c)$, we have an equation of the form $\dots$

$$f'(x) - f(x) =y $$

which has solutions of the form (https://www.wolframalpha.com/input/?i=f%27(x)+-+f(x)+%3D+y)

$$y = c_1 e^x - y$$

Hope this helps find some solutions!


How to prove $\lim_{n \to \infty}\prod_{k=1}^n(1 - ma_k) = 0$ if $\sum a_k = \infty$, $1/a_1 \geq m > 0$ As mentioned above, we have a sequence $\{a_k\}$, $k=\{1, 2, \ldots \}$, $a_k \in (0, +\infty)$ is decreasing and $\sum a_k = \infty$, $\lim_{k\to \infty} a_k = 0$.

$$\frac 1 {a_1} \geq m > 0.$$

Then, how to prove that

$$\lim_{n \to \infty}\prod_{k=1}^n(1 - ma_k) = 0$$

I think we can use log. Also, the condition $\sum a_k = \infty$ is necessary.

Taking the log on both sides, we wish to prove
$$
\sum_{k=1}^\infty\log(1-ma_k) = -\infty
$$
Note that the choice of $m$ ensures that $ma_k\le ma_1\le 1$, so the logarithm is well-defined (except if $ma_1=1$, but that case is trivial).

We use the inequality $\log x\le x-1$:
$$
\sum_{k=1}^\infty\log(1-ma_k)
\le \sum_{k=1}^\infty -ma_k
= -m\sum_{k=1}^\infty a_k
= -\infty
$$

The assumption that $a_k\to 0$ is not actually necessary. In fact, if $a_k\to c>0$, then the claim is trivial by an argument similar to David Sillman's answer. (I.e. $\prod (1-ma_k)$ $\le \prod (1-mc)$ $= (1-mc)^n$ $\to 0$).

As stated, $m\in(0,1/(a_1)]$. Keep in mind, as well, that $(a_k)\searrow$. This means that the greatest value of $ma_k$ occurs at $k=1$. Due to the limit condition on $a_k$, we have that $\inf (ma_k)=0$ AND that $\inf (ma_k) = 0\notin (ma_k)$ because $a_k$ cannot take the value 0. Therefore, the product decays which can be shown using that $\sup (ma_k) = 1$ and $\sup (ma_k) = 1 \notin (ma_k)$ due to the limits on $m$. (Specifically, try comparing it to $\prod (1-ma_1)$, which is necessarily greater than or equal to our product, $P$, due to the decreasing property of $(a_k)$. HINT: $(1-ma_1)\in(0,1)$ via what we've shown above).


Example of infinite game without any Nash equiblibria I have to find an example of a game that does not admit (mixed strategy) Nash equilibria.

Consider a game in normal form. Let $N=\{1,2\}$ be set of players and $S_i=\mathbb{R}$ a set of possible strategies for each player (uncountably many), $i=1,2$.

The rules of this game are as follows: each player chooses one real number (his strategy). Player who chose a bigger number than the other player wins $1$, player who chose smaller number gets $0$. If both players choose the same number, they both get $0$.

Formally, payoff functions of each player one looks like this:
$$v_1(s_1,s_2)=\left\{\begin{matrix} 1 \ \mathrm{if}\ s_1>s_2 \\ 0 \ \mathrm{if}\ s_1\leq s_2 \end{matrix}\right. $$
$$v_2(s_1,s_2)=\left\{\begin{matrix} 1 \ \mathrm{if}\ s_2>s_1 \\ 0 \ \mathrm{if}\ s_2\leq s_1 \end{matrix}\right. $$

Do you think it's a good example?

There is none, unless the problem is wrongly planned. By definition, every game has at least one equilibrium in mixed strategies.


What is the logic underlying this proof? Proposition: A metric space $X$ is connected if, and only if, every continuous function $f:X\to (\{0,1\},d_D)$ is a constant function, where $d_D$ is the discrete metric on the set $\{0,1\}$.

  Proof: Suppose $X$ is disconnected. Then $X$ has a nonempty proper subset $A$ which is both open and closed. Define the function $f:X\to (\{0,1\},d_D)$ by $$f(x)=\left\{\begin{array}
&0 & \text{if } x\in A\\
1 & \text{if } x\notin A \end{array} \right.$$ There are only four open sets in $\{0,1\}$, and it is easy to check that the inverse image of each one of them under the function $f$ is an open set in $X$. Hence $f$ is continuous, and clearly not constant.

  Conversely, suppose there is a continuous function $f:X\to (\{0,1\},d_D)$ which is not constant....

Can you please explain the logic behind the first part of the proof given above? This is my attempt to understand the logic:

Let the statement "every continuous function $f:X\to (\{0,1\},d_D)$ is a constant function" be denoted $A$; and the statement "$X$ is connected" be denoted $B$. The first part of the proof has proved that $A\wedge\neg B$ is false when $B$ is false. Hence $A$ is false (I'm not sure if this is relevant, but it does mean that $A\implies B$ is true, because $A$ is false).

If the author is trying to prove the contrapositive, I think that he failed because he did not prove $\neg B \implies \neg A$ for all cases of non-constant continuous functions, just one. What I usually expect is a proof by contradiction. If this were a proof by contradiction, what is being implied at the start are the words, "Suppose $X$ is disconnected and suppose every continuous function $f:X\to (\{0,1\},d_D)$ is constant." Why would a contradiction then lead to the conclusion that $X$ must have been connected? Why do we not simply conclude that there exists a non-constant function $f:X\to (\{0,1\},d_D)$ when $X$ is disconnected? What is so special about the statement $A$ that it must be possible to be true under one of the conditions '$X$ is connected' or '$X$ is disconnected'?

The author proves $\neg B \implies \neg A$, so $A\implies B$.

Indeed $\neg A$ reads "there is a continuous non-constant function $X\to \{0;1\}$", which is what the author proves.

The proposition is $B\iff A$. The first part of th eprove starts with "Suppose $\neg A$" and deduces $\neg B$ from that, thus shows $\neg A\implies \neg B$ (or equivalently $B\implies A$)


A straight line through origin meets parallel lines $4x+2y=9$ and $2x+y+6=0$ at P and Q. Find PQ segment ratio by origin Let the equation of line be $y=mx$

Then
$$2x+y=\frac 92$$
$$2x+mx=\frac 92$$
$$x=\frac{9}{2(2+m)}$$

$$y=\frac{9m}{2(2+m)}$$for the second line
$$x=\frac{-6}{m+2}$$
$$y=\frac{-6m}{m+2}$$
Let (0,0) divide it in the ratio k:1
$$0=k\frac{9}{2(m+2)}+\frac{-6}{m+2}$$
$$k=\frac 43$$
But the answer given is $\frac 34$.

Now this isn’t simply a perspective problem, because both options 4/3 and 3/4 were present. How do we tell which one is right?

Note that the question refers to P and $4x+2y=9$ first, and Q and $2x+y +6=0$ second. So, it is logical to assume that the ratio refers to $\frac{OP}{OQ}$.

Also, the ratio can be conveniently found as the ratio of the distances $d_P$ and $d_Q$ between the origin and the two parallel lines, because the distance line, PQ and the two parallel lines form a pair of similar right triangles. Thus,

$$\frac{OP}{OQ}=\frac{d_P}{d_Q}=\frac{\frac{|0+0-9|}{\sqrt{4^2+2^2}}}{\frac{|0+0+6|}{\sqrt{2^2+1^2}}}=\frac{\frac92}6=\frac34$$

First, do you see that the origin lies between P and Q?  It can't divide PQ in ratio "4/3" because that is larger than 1 and implies that that the origin is NOT in the interval.


How to Form a Row-Complete Latin Square This is one of the descriptions I've seen online:

For any even $n$, say $n = 2m$, a row complete Latin square of order $n$ can be formed by writing down

$$0, 1, 2m - 1, 2, 2m - 2, 3,\ldots, m + 1, m$$

as the first row and then developing subsequent rows by adding $1$ modulo $n$.

I'm not quite clear on how that goes, such as what the difference is between $2m$ and $m$.  $2m$ stands for $n$, does $m$ stand for modulo?  I'd like to see and example with an even number or two, such as $4$, and $6$.

I tried, $2m = 4$

$$0, 1, 3, 2, 2\ldots$$
wait, that can't be right.

It should turn out with non repeating sequences, and non recurring pairs between rows.
I can outline the other methods I've found (but don't understand) if this one is not the best.

As I have essentially asked the same question somewhere else, I wanted to share the answer I was given there by Gage:

  See here for more information [on row-complete Latin squares] http://personal.maths.surrey.ac.uk/st/H.Bruin/MMath/LatinSquares.html

The reference outlines a fairly straight forward algorithm and I wrote a a quick MATLAB script to build me an order-n square: https://gist.github.com/hcc23/274b76edc3dd82d5de1a


Proof limit by definition Prove: $\lim_{x \to 2}x^2=4$

We need to find $\delta>0$ such that for all $\epsilon>0$ such that if $|x-2|<\delta$:

$$|x^2-4|=|(x-2)(x+2)|=|x+2|\cdot|x-2|\implies|x-2|<\frac{\epsilon}{|x+2|}$$

We will take $\delta=\frac{\epsilon}{|x+2|}$

is it valid?

If $x \in \Bbb{R}$, then
$$
|x^{2}-4| = |x-2||x+2|;
$$
if in addition $|x-2| < 1$, then $|x| - 2 \leq |x-2| < 1$, implying $|x| < 3$, implying $|x+2| \leq |x|+2 < 5$, implying
$|x-2||x+2| < 5|x-2|$;
given any $\varepsilon > 0$, we have $5|x-2| < \varepsilon$ if in addition $|x-2| < \varepsilon/5$. In conclusion, for every $\varepsilon > 0$, if $|x-2| < \min \{ 1, \varepsilon/5 \}$ then $|x^{2}-4| < \varepsilon$.

Note that the choice of $\delta$ you made is not gonna work, for it depends on $x$; recall the definition of limit.

You can cheat a little bit, as long as it doesn't appear in your final solution: At $x = 2$, the derivative of $x^2$ is $4$. That means that $\delta$ must be, for small $\epsilon$, less than $\frac\epsilon4$.

For small $\delta$, $x$ is close to $2$, which makes $|x+2|$ close to $4$. That means you're on the right track, but you have to eliminate the $x$ from that expression in some way, probably by saying something like $\delta = \min(\frac14, \frac\epsilon5)$. (I am quite certain that this is good enough, by the way, but you would have to check it; I only went with what felt right here.)


L1 minimization problem with nested sums as LP problem I've been trying to solve this problem but I have an issue with the fact that there is a sum under each absolute value. I'm trying to convert this minimization problem (with respect to $x, y_1, \dots,y_n$)

$$
\sum_{i=1}^{m} \left| x - \sum_{j=1}^{n}\left| y_j - a_{ij} \right|  \right|
$$

to a linear programming problem. It's similar to this question, however that problem hasn't got the nested sums.

Does it suffice to just do the same substitution as in the linked question or should I make other substitutions and constraints as well?

You cannot convert this to a linear program. The objective function is not convex (consider the special case $m = n = x = a_{11} = 1$, draw the function and you'll see). You cannot use epigraph models when the expression you outer bound enters in a nonconvex fashion (which the inner absolute value does).

EDIT: Giving a more complete example which shows the failure of the LP proposed in another answer. Assume $a = (2,0,-1)$. An optimal solution to this is $x = y = 1$ (it is not unique) with objective value 1. An optimal solution to a flawed LP model would be $x=2, y=0, z = (0,0,0), \omega = (2,2,2)$. The LP objective is thus $0$, but if you plug the solution $x = 2, y = 0$ into the original nonlinear objective, it has objective value 3. In other words, the LP is incorrect. Here is a MATLAB/YALMIP model to reproduce example

a = [2;0;-1];
sdpvar x y
Objective = sum(abs(x - abs(y-a)));
optimize([],Objective)
[value(x) value(y) value(Objective)]

z = sdpvar(3,1);
w = sdpvar(3,1);
Model = [-z <= x - w <= z, -w <= a-y <= w];
optimize(Model,sum(z))
[value(x) value(y) value(sum(z)) value(Objective)]

First of all, as pointed out by Johan Löfberg, this is not a convex function, so it is vain to search for a pure linear program. Therefore, the following does not work in the general case :

Step 1:
$$
\sum_{i=1}^m z_i
$$
subject to
$$
x-\sum_{j=1}^n|y_j-a_{ij}| \le z_i\\
-x+\sum_{j=1}^n|y_j-a_{ij}| \le z_i
$$

Step 2:
$$
\sum_{i=1}^m z_i
$$
subject to
$$
x-\sum_{j=1}^n \omega_{ij} \le z_i\\
-x+\sum_{j=1}^n \omega_{ij} \le z_i\\
y_j-a_{ij} \le \omega_{ij}\\
-y_j+a_{ij} \le \omega_{ij}\\
$$

I am convinced however that a MILP very close to the above program should do the trick. Still working on it, trying to use binaries to ensure tightness.


Show that there is no invertible continuous function $f:[0,1) \to (0,1)$ This is a problem I came across recently and have been trying to figure out, but I'm not having a lot of luck!

By considering the restriction of $f$ to $(0,1)$ or otherwise, show that there is no invertible continuous function $f:[0,1) \to (0,1)$.

Hint: there are bijections which map $[0,1)$ onto $(0,1)$. Can you construct one?

The main idea is that such a function has to be always increasing or always decreasing. Assume that there exist three points $a<b<c$ such that $f(a) \leq f(b)$ and $f(b) \geq f(c)$.

If either $f(a)=f(b)$ or $f(b)=f(c)$ we can conclude that the function is not injective hence cannot be invertible. Now we can assume $f(a) < f(b)$ and $f(b) > f(c)$ Then $\exists p \in [f(a),f(b)]\cap[f(c),f(b)]$.

Then by IVP $\exists m_{1},m_{2}$ such that $a<m_{1}<b$ and $b<m_{2}<c$ with $f(m_{1})=f(m_{2})=p$. Thus $f$ is not injective hence not invertible. This proves that f has to be always increasing or always decreasing.

Assume $f$ is always increasing and $f(0)=t\in (0,1)$ what maps to the rest of the values in $(0,t)$? This shows that f cannot be bijective (because it is not surjective) hence not invertible.

a theorem states: if f:U->V is a continuous function, and A is an open set in V, then f^-1(A) is also open. the same for closed set. (c.f. Baby Rudin Chap.4 theorem 4.8.)now (0,1) is obviously open and [0,1) is obviously not open.

Edit:Note:This answer is incorrect(see the comments).


Is my proof correct? 'let a,b∈ Z. We write A | B if A divides B. Is the relation |, symmetric, transitive and/or reflexive?' The relationship is not symmetrical. When a relationship is symmetrical: if xRy implies yRx for all x, y ∈ A (where A is a non-empty set, and R is a relation in A)

If a, b ∈ Z, and as a | b means that there is an integer k such that b = a k ∙. When we take a = 1 and b = 2 applies: b = a ∙ k → k = 2. So for a = 1 and b = 2 holds a | b, for k ∈ Z.
If a, b ∈ Z, and if b | a is that there is an integer k such that a = b ∙ k. When we take again a = 1 and b = 2, a = b ∙ k → k = 1/2. So for a = 1 and b = 2 is not that b | a ┤ because k ∉ Z.
This implies that f is not symmetrical.

The relationship is transitive. When a, b, c ∈ Z, and as a | b and b | a, is that there is an integer k such that b = a ∙ k and l is an integer that there exists such that c = b ∙ l. This means that
c = a b ∙ ∙ l. Because a, b, c ∈ Z means that the b ∙ l ∈ Z (an integer multiplied by an integer indicates an integer). This means that there exists an integer m (where m = a ∙ b) such that c = m ∙ A. It follows that a | c, so the relationship is the transition.

The relationship is reflexive when (∀a ∈ Z)a|a. because every a ∈ Z can be divided by itself, we say that it is reflexive.

After Looking it through again I now have my proof like this:

(a) Since 0 does not divide 0,  "|" is not reﬂexive.
(b) 2 divides 4 so 2 | 4. But 4 does not divide 2, so 4 does not divide 2. Thus, "|" is not symmetric.
(c) To see that  is transitive, let a, b, c be integers. Suppose that a| b and b |c. Thus,
a divides b and b divides c so there exist integers k and l such that b = ak and c = bl. This
gives c = bl = (ak)l = a(kl). Therefore, a divides c so a | c.

Is this prove correct?

The relation "|" is reflexive, since $n|n$ holds for all $n$, even $n = 0$.

The relation "|" is not symmetric. Notice that $2$ divides $6$ (as $6 = 2 \cdot 3$), but $6$ does not divide $2$ (there is no integer $k$ such that $2 = 6 \cdot k$).

Transitive ? Suppose $a|b$ and $b|c$. By definition, we have $k_1, k_2$ such that $b = a \cdot k_1$ and $c = b \cdot k_2$.

We want to show that $a|c$, or that $c = a \cdot k$ for some $k$. Can you find a $k$ that works?

The relationship is transitive. A relation is transitive if xRy and yRz implies that xRz for all a,b,c∈A ( with A as a non empty set and R as a relation to A).

When a,b,c∈ ℤ, and if a|b and b|b, is that there exists an integer k so b= a ∙ k and that there exists an integer l so c= b ∙ l. This means that

c = a ∙ b ∙ l. Because a,b,c∈ ℤ this means that b ∙ l ∈ ℤ (an integer multiplied by an integer produces an integer). This means that there exists an integer m (where m = a ∙ b) so that c = m ∙ a. It follows that a|c, so the relationship is transitive.


Sum of a power series with a parameter Does anybody know the sum

$$ \sum_{n=0}^{\infty}\frac{x^{n}}{(n+a)n!}=f(x,a)$$

here $a$ is a number $ a>0 $. A hint please ? :D

If $ a=1$ I believe $ f(x,1)=(e^{x}-1)/x $

For $a\gt0$,
$$
\begin{align}
af(x,a)
&=a\sum_{n=0}^\infty\frac{x^n}{(n+a)n!}\\
&=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+a}\right)\frac{x^n}{(n-1)!}\\
&=e^x-x\sum_{n=0}^\infty\frac{x^n}{(n+a+1)n!}\\
&=e^x-xf(x,a+1)\tag{1}
\end{align}
$$
Reversing $(1)$ yields
$$
f(x,a)=\frac{e^x-(a-1)f(x,a-1)}{x}\tag{2}
$$
Evaluating $(1)$ directly gives
$$
\begin{align}
f(x,1)
&=\sum_{n=0}^\infty\frac{x^n}{(n+1)n!}\\
&=\frac1x\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\\
&=\frac{e^x-1}{x}\tag{3}
\end{align}
$$
$f(x,a)$ can be computed for higher $a\in\mathbb{Z}$ using recursion and $(2)$.

I think you should rewrite your function in a better way...
$$
f(x,a) = \sum \frac{x^n}{(n+a)n!} = \sum \frac{x^n}{n \cdot n!} + \sum \frac{x^n}{a \cdot n!} = S + \frac{1}{a}\sum \frac{x^n}{n!} = S + \frac{e^x}{a}.
$$
If you differentate $S$, you'll find that $S' = \sum \frac{n \cdot x^{n-1}}{n \cdot n!} = \sum \frac{x^{n-1}}{n!}$, so $xS' = \sum \frac{x^n}{n!} = e^x$.
Now ... since $xS' = e^x, S = \int e^x / x = Ei(x)$.
So $f(x,a) = \sum \frac{x^n}{(n+a)n!} = e^x/a + Ei(x)$.


How to factorize $(x-2)^5+x-1$? This is a difficult problem.
How to factorize this?

$$(x-2)^5+x-1$$

we can't do any thing now and we should expand it first:

$$x^5-10x^4+40x^3-80x^2+81x-33$$

but I can't factorize it.

Using the substitution $t=x-2$, it suffices to factor $t^5 + t+1$.

But $t^2+t+1$ divides $t^5+t+1$. Why?

$t^2+t+1 = (t-\zeta)(t-\zeta')$, where $\zeta,\zeta'$ are the primitive $3rd$ roots of unity, and thus $\zeta^5 + \zeta + 1 = \zeta^2 + \zeta + 1 = 0$, similarly for $\zeta'$.

Now, the quotient $(t^5+t+1)/(t^2+t+1)$ is easily determined via polynomial division.

Alternative solution. Using $t^3-1=(t^2+t+1)(t-1)$, we get:
$$t^5+t+1=(t^5-t^2)+(t^2+t+1)=(t^3-1)t^2 + (t^2+t+1)$$
$$=(t^2+t+1)(t-1)t^2 + (t^2+t+1) = (t^2+t+1)((t-1)t^2+1)$$


If a fiber bundle with a connected total space admits a section, is the fiber connected? Question. If a fiber bundle with a connected total space admits a section, is the fiber connected?

(Since there's a section the bundle is surjective, whence the base is connected as a continuous surjective image of a connected set. Thus the isomorphism type of the fiber is constant over the base.)

For covering maps this reduces to the following.

Fact. For a covering map with connected total space, it admits a section iff it's a homeomorphism.

To prove this one can show the image of any section is clopen upstairs. This approach does not generalize to the case of non-discrete fibers, since sections are generally far from having open image (consider sections of product projections).

As long as we're dealing with a sane base (paracompact Hausdorff), a fiber bundle is a Serre (even a Hurewicz) fibration and therefore has an associated long exact sequence of homotopy groups. This sequence shows that when the total space is connected and the base is simply connected, the fibers must also be connected. Thus an example must involve a non-simply connected base.

I think this follows from the long exact sequence of homotopy groups for a fibration (assuming our bundle is locally trivial).

Let $F\stackrel{\iota}{\to} E\stackrel{\rho}{\to} X$ be a fibre bundle such that $E$ is path-connected, and let $s\colon X\to E$ be a section, i.e. $\rho\circ s = id_X$. Then for each $n$ we have $\pi_n(\rho)\circ \pi_n(s) = id_{\pi_nX}$, so in particular $\pi_n(\rho)$ must be surjective for all $n$. Therefore the connecting maps $\pi_n X \to \pi_{n-1} F$ in the long exact sequence are all $0$ and so $\pi_n(\iota)$ is injective for all $n$, so in particular $\pi_0F \hookrightarrow \pi_0 E$ so $F$ must be connected.

Not always, a principal bundle has a section if and only if it is trivial.

https://en.wikipedia.org/wiki/Principal_bundle#Trivializations_and_cross_sections


Confusing probability question I have got a task, which seems a quite confusing for me. It is simple: In a market, they sell eggs in egg holders, they store $10$ of them in each. There is $60$% chance, that all of the eggs are ok, $30$% chance, that exactly $1$ of them is broken, and $10$% chance, that exactly $2$ of them are broken(it is random, which one is broken).

We buy an egg holder, and after we grab our first egg, we are sad, because it is broken. What is the probability, that there is one more broken egg in our holder?

The "logical" way would be: $30$% of them have $1$ broken egg, $10$% of them have $2$, so, to have $2$ broken, the chance must be $\frac14$. But I am not really sure if that is the correct approach, since the broken egg can be anywhere, getting a broken one for first may be not that easy, or is that independent?(Maybe, I could use Bayes Theorem somehow)?

Any help appreciated.

I see this as analogous to the pancake problem ... specifically, I would argue that having randomly observed a broken egg, the chances that you are in case C have disproportionately increased.

Imagine that you had $100$ of your cartons, $60$ of type A, $30$ of type B, and $10$ of type C.  You select one egg randomly from each.  You'll get $3$ broken ones from the Bs and $2$ from the Cs.  Hence your probability is $\frac 25$.

Per the OP.

  What is the probability, that there is one more broken egg in our holder?

thanks to @drhab for the diagram that shows this intuitively :)

Clarifying Edit

1) What the OP is asking is somewhat equivocal and perhaps this question is better suited to Puzzling Stack Exchange. See points 2) and 3) following.

2) Because it is given that we have a holder with (at least) one broken egg, the proportion of holders without broken eggs is irrelevant to the answer.

3) Because it is given that there can be only a maximum of 2 broken eggs/holder, finding the broken egg in the first slot (or for that matter the 2nd slot,... 9th slot) is irrelevant to the answer.

4) Given the description in the OP of the holders with broken eggs, two questions may be asked:

i) What proportion of the egg holders with broken egg(s) have 2 broken eggs? Answer = 25%
ii) What proportion of broken eggs come from holders with 2 broken eggs? Answer = 40%

You'll excuse me if that I took the OP to be asking question i) when others took it to be asking question ii).  ESL and all that... and thanks to @kviiri for pushing me until I understood.


Needle problem in Probability A grid of lines distance $a$ apart and another grid of lines distance $b$ apart are perpendicular, dividing plane into rectangles. A needle of length $l$=min$\{a,b\}$ is thrown at random onto the grid. What is the probability it intersects lines in both grids simultaneously?

I have no idea how to approach such a problem. Any help would be appreciated.

Assume that the vertical lines are $a$ apart and the horizontal lines $b$ apart, $\ell=a\leq b$. The needle $N$ has a random inclination $\phi$ with repect to the horizontal, whereby we may assume that $\phi$ is uniformly distributed in $\bigl[0,{\pi\over2}]$, making use of symmetry. Given $\phi$, the probability that $N$ intersects a vertical line is ${a\cos\phi\over a}$,
 the probability that $N$ intersects a horizontal line is ${a\sin\phi\over b}$, and these two events are independent. It follows that the overall probability $P$ that $N$ intersects lines of both kinds is given by
$$P={2\over\pi}\int_0^{\pi\over2}\>{a\cos\phi\over a}{a\sin\phi\over b}\>d\phi={a\over\pi b}\ .$$

Hint. For one set of parallel lines this is Buffon's Needle Problem. There are many solutions on the internet.

That should get you started dealing with two sets of parallel lines.

https://en.wikipedia.org/wiki/Buffon%27s_needle

http://www.cs.umb.edu/~eb/piday/
http://www.cs.umb.edu/~eb/piday/whypi.pdf


Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.

I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$

Then, I started with the following statement and manipulated it to match the integrand.
$$
\begin{align*}
x^4 +1 &\gt x^4 \\ \sqrt{x^4+1} &\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &\lt\frac{1}{x^2}
\end{align*}
$$

It is a well-known result that
$$
\int_1^{\infty}\frac{dx}{x^2}
$$
converges. So would it be correct to say that, by comparison, that

$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges as well?

In the regime $|x| \ge 1$, you can compare $(x^4+1)^{-1/2}$ against $x^{-2}$, but in the neighborhood of $0$, the latter is unbounded so you are best off noting that $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx < 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx  + \int_{x=1}^\infty x^{-2} \, dx \right).$$  The first term is obviously finite since $0 < (x^4 + 1)^{-1/2} \le 1$ on $[0,1]$; and the second is also finite as you stated.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x  \over \root{1 + x^{4}}}} =
     2\int_{0}^{\infty}{\dd x  \over \root{1 + x^{4}}}}$. Let
$\ds{t \equiv {1 \over 1 + x^{4}}}$ such that
$\ds{x = \pars{{1 \over t} - 1}^{1/4}}$.

Then,

\begin{align}
\ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x  \over \root{1 + x^{4}}}}} & =
2\int_{1}^{0}t^{1/2}\,{1 \over 4}\,\pars{{1 \over t} - 1}^{-3/4}
\pars{-\,{1 \over t^{2}}}\,\dd t
\\[3mm] & = \half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-3/4}\,\dd t =
\half\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/4 + 1/4}}
\\[3mm] & =
\color{#f00}{{1 \over 2\root{\pi}}\,\Gamma^{2}\pars{{1 \over 4}}} \approx
3.7081
\end{align}

since $\Gamma\pars{1/2} = \root{\pi}$.