Finding surface area - integral of $\sqrt{1+z^2}$ Sorry about this, this is more of a "am I going the right way" question, there's a surface it goes:

$$x^2+y^2-z^2=1$$

Now this is nice because
$x^2+y^2=r^2=1+z^2$ thus $r=\sqrt{1+z^2}$ (I want the positive side)

(BORING PART - I am sure this is right)

So the shape of this thing has rotational symmetry about z (the vertical axis) some sort of hyperbola comes to mind because it has (for r) two parts (the + and -) and it "tends to linear" as z gets large.

That's not analysis reasoning, just for large $z$ $z^2$ will be pretty big compared to one, so $\sqrt{1+z^2}$ will be slightly bigger than $z$.

Anyway, my task is to find the surface area between the planes z=1 and z=0.

So.... the circumfrence of a circle is $2\pi r$and $r$ is as above. The height of each 'lil cylinder is $dz$ (This was once fine, Analysis makes me feel bad for saying this!) so $dA = 2\pi\sqrt{1+z^2}dz$ It makes sense to sum from $z=0$ to 1, hence:

ACTUAL QUESTION
$$A = 2\pi\int^1_0\sqrt{1+z^2}dz$$

I am pretty sure I am being dippy by not seeing it, I can see that:

using $\cosh(\theta)^2-\sinh(\theta)^2=1$ one can easily get to:
$\cosh(\sinh^{-1}(z)) = \sqrt{1+z^2}$

The question is supposed to be easy, just wondering if there's a quicker way, someone to go "yes that's right!" would be great.

Your analysis is incorrect, although your approach is certainly in the right spirit.  The "height" of a cylinder is not $dz$, but $ds$, or the arc length element along the surface.  In this case

$$ds = \sqrt{1+\left ( \frac{\partial r}{\partial z}\right)^2} dz = \sqrt{1+\frac{z^2}{1+z^2}}dz = \sqrt{\frac{1+2 z^2}{1+z^2}}dz$$

Then the integral for the surface area is instead

$$2 \pi \int_0^1 dz \, \sqrt{1+2 z^2}$$

You can then make a substitution $u=z \sqrt{2}$ to get

$$\sqrt{2}\pi \int_0^{\sqrt{2}} du \, \sqrt{1+u^2}$$

Substituting $u=\sinh{v}$ produces a simpler integral to evaluate.  The result I get for the surface area is

$$S =  \frac{\pi}{\sqrt{2}} \left [\sqrt{6} + \log{(\sqrt{2}+\sqrt{3})}\right]$$

For future people looking for an answer, the integral in the question is totally wrong. I'd gone the wrong direction.

To see an answer that uses the rotational symmetry about the z axis of the question (ie elementary .... tapered disks (NOT cylinders)) please see Ron Gordon's answer.

To have your memory jogged (subtly getting my excuse in there) about surface area, read on.

It ought to be known that a curve in $\mathbb{R}^3$ can be parameterised by a map $\boldsymbol{r}:\mathbb{R}\rightarrow\mathbb{R}^3$
and by analogy a surface $S$ in $\mathbb{R}^3$ by:

$\boldsymbol{r}:\mathbb{R}^2\rightarrow\mathbb{R}^3$

The surface $S =\lbrace\boldsymbol{r}((u,v))|(u,v)\in\Omega\rbrace$
From now on I'll write r(u,v) as r of the vector (u,v), save some brackets.

and: $\Omega=\lbrace(u,v)|u\in U, v\in V\rbrace$

tl-dr parametric surface.

Now neglecting questions of existence (a smooth way of pretending I'm not sidestepping answering those questions) the following all happens:
with the substitution involving root 2, it should be 1 over root 2du not root2du
consider the parallelogram (with curvy edges maybe, but for small enough delta u and v will exist such that the four 'edges' do not cross, which is what is important):

$\boldsymbol{r}(u,v)$

$\boldsymbol{r}(u+\delta u,v)$

$\boldsymbol{r}(u,v+\delta v)$
and

$\boldsymbol{r}(u+\delta u,v+\delta v)$

the area of a parallelogram can be found quite easily by showing that (looking at it so one of the sets of parallel lines is 'an x axis' and reflecting is (isometry) as required so the slanty edge is to the right, ie the top now horizontal line ends to the right of the base) you can 'cut off' a triangle that starts at the right end of the base, defined by going vertically upwards from that, then right to the end of the top edge,

and that it fits with the bottom right vertex being the top vertex of a similarly constructed triangle on the other side.

This sounds so much harder than my simple picture, but just use an angle "theta" from the base to the left slanty edge, if you continue the base, there's another angle theta from that base to the right slanty edge. cos defines the 'overhang' it's very easy to show they're similar triangles.

Thus you can turn the parallelogram into a rectangle, of width |b| (b for the base vector) and height |a|sin(theta) where a is the vector that is the slant, and |a| the length of it

So the area is |a||b|sin(theta)

Now you should know that a x b = |a||b|sin(theta)n where n is a unit normal vector to a, b (perpendicular to them both)

so |a x b| = | |a||b|sin(theta)n | = |a||b|sin(theta) |n| = |a||b|sin(theta)

this means the magnitude of the cross product of the (direction) vectors that represent the edges of the parallelogram is the area of that parallelogram.

So going back to our parallelogram, we can see that:

$\boldsymbol{r}(u+\delta u,v)-\boldsymbol{r}(u,v)\approx\frac{\partial\boldsymbol{r}}{\partial{u}}\delta u$

and

$\boldsymbol{r}(u,v+\delta v)-\boldsymbol{r}(u,v)\approx\frac{\partial\boldsymbol{r}}{\partial{v}}\delta v$

Which are the edges of the 'parallelogram'

the area of which is ($\approx$):

$\lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\delta u\times\frac{\partial\boldsymbol{r}}{\partial{v}}\delta v\rVert=\lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\times\frac{\partial\boldsymbol{r}}{\partial{v}}\rVert\delta u\delta v$

Thus: $\int\int_S ds=\int\int_\Omega \lVert\frac{\partial\boldsymbol{r}}{\partial{u}}\times\frac{\partial\boldsymbol{r}}{\partial{v}}\rVert dudv$

I really really .... want to stop typing now, so skipping working out all that you get:

$A=\int^{u=1}_{u=0}{\sqrt{1+2u^2}\int^{v=2\pi}_{v=0}{dv}du}$

Clearly the integral involving v is just 2pi, the second one, I just said, let sinh(t)=root(2)u and got a new integral with limits arcsinh(root(2)) and 0, blah blah blah, answer is:

$A=\pi[\sqrt{3}+\frac{1}{\sqrt{2}}\ln(\sqrt{2}+\sqrt{3})]$


understand the solution to the unbiased estimator of area of circle when given n independent radius $R$ measurement with error $\sim N(0,\sigma^2)$ $S =  \pi R^2$
$E(\bar{X}^2) = Var(\bar{X})+(E(\bar{X}))^2=\sigma^2/n + R^2$
then it states thats an unbiased estimator of $\sigma^2/n = \frac{1}{n(n-1)} \sum_{i=1}^n(X_i-\bar{X})^2$.
However, when i approached it by finding the unbiased estimator of $\sigma^2$ in $N(\bar{X},\sigma^2)$ from its MLE, the result is $\frac{n}{n-1}(X_i-\bar{X})^2$. Thus I don't understand where the extra division by $n$ is coming from.

Suppose $R_1,\ldots,R_n\sim N(R,\sigma^2)$ are $n$ independent measurements of your radius $R$. Set $A_i=\pi R_i^2$ and $S^2=\frac{1}{n-1}\sum_{i=1}^n(R_i-\bar{R})^2$. Notice that $S^2$ is an unbiased estimator for $\sigma^2$ and $$E(A_i)=\pi E(R_i^2)=\pi \Big[\sigma^2+\big(E(R_i)\big)^2\Big]=\pi E(S^2) + \pi R^2$$ If we take $\bar{A}=\frac{1}{n}\sum_{i=1}^nA_i$ it's easy to see that $E(\bar{A})$ also equals $\pi E(S^2)+\pi R^2$ and so $\hat{A}=\bar{A}-\pi S^2$ is an unbiased estimator for $A=\pi R^2$.
Remark: If you want to know why $S^2$ is an unbiased estimator for $\sigma^2$, note $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$. Hence $$E\Big(\frac{(n-1)S^2}{\sigma^2}\Big)=n-1 \iff E(S^2)=\sigma^2$$


The fundamental group of the Möbius strip What is the fundamental group of the Möbius strip?
Is it given by $\{-1,1\}$ as the lemma of Synge supposes, or am I wrong and it does not apply there?

The moebius strip is homotopy-equivalent to the circle, so has the same fundamental group which is $\mathbb Z$.

The fundamental group of the moebius strip is $\{a,b|a^2=b^2\}$.
Cf. http://www2.math.ou.edu/~forester/5863S14/fsol.pdf


A question about automorphism and inner automorphism of $\mathbb{Z}$ $\mathbb{Z}$  is a group under addition. I wish to find $\text{Aut}(\mathbb{Z})$ and $\text{Inn}(\mathbb{Z})$

I know the only automorphism groups of $\mathbb{Z}$ under addition are$\{f(x)=x, f(x)=-x\}$ but what about the inner atomorphisms $\text{Inn} (\mathbb{Z})$?

Refers this question, you can see what $Aut(\mathbb{Z})$ is.

And note that for the ring $R$, the inner automorphisms of R form a normal subgroup of the automorphism group of $R$.


Show limit for recursive sequence by induction Let $(a_n)_{n \in \mathbb N}$ be a recursive sequence. It is defined as $a_1=1, \quad a_{n + 1} = \frac{4a_n}{3a_n+3}$.

I have to show by induction, that it is limited for $a_n \ge \frac{1}{3}$.

So as an induction step I have $n \to n + 1$. And the idea is to get $a_{n+2}$ and show somehow $a_{n+2} \ge \frac{1}{3}$

$$a_{n+1} = \frac{4a_n}{3a_n + 3} $$
$$a_{n+2} = \frac{4 \frac{4a_n}{3a_n + 3}}{3 \frac{4a_n}{3a_n + 3} + 3} = \frac{ \frac{16a_n}{3a_n+3} }{ \frac{12 a_n}{3a_n + 3} + 3}$$

Question: How should I go on or am I on the wrong path?

Edit: You can find other answers for the exact same question here.

You wonder if $a_{n+1}\geq {\frac{1}{3}}$, so if

$$\frac{4a_n}{3a_n+3}\geq {\frac{1}{3}}$$ but this is equivalent to $$ 4a_n \geq a_n+1$$
or $$3a_n\geq 1$$
which is true by induction assumption.

To show that $a_{n+2}\ge1/3$, given that $a_{n+1}\ge1/3$, I would just plot the formula
$$
a_{n+2}=\frac{4a_{n+1}}{3a_{n+1}+3}
$$
as a function of $a_{n+1}$. This is clearly a strictly increasing function for $a_{n+1}\ge1/3$, which has limit $4/3$, and it takes on its minimum value ($=1/3$) at $a_{n+1}=1/3$. So $a_{n+2}\in[1/3,4/3)$. So your claim holds and the induction step from $n+1$ to $n+2$ is proved.


If $A$ is compact, $\lambda\ne0$ and $y$ is fixed, is there a unique $x$ with $(\lambda-A)x=y$? If $B$ is a compact linear operator on a Banach space $X$, we know that $\dim\mathcal N(1-B)\in\mathbb N_0$ and $\mathcal R(1-B)$ is closed.
Thus, if $\lambda\ne0$ and $$B:=\frac A\lambda$$ such that $\dim\mathcal N(1-B)=0$, then for all $y\in X$, there is at most one $x\in X$ with $(\lambda-A)x=y$. But can we guarantee that there always is such a $x$?

A bounded operator $T$ is said to be Fredholm if $\text{dim}(\mathcal N(T))<\infty $,  and
$\text{codim}(\mathcal R(T))<\infty $.  In this case the index of $T$ is defined by
$$
  \text{ind}(T)=
  \text{dim}(\mathcal N(T))-\text{codim}(\mathcal R(T)).
  $$
It is a well known result,  sometimes called the Fredholm alternative,  that $I-B$ is a Fredholm
operator with
index zero  for every compact operator $B$.
Thus,  if $\mathcal N(I-B) = \{0\}$,  it follows that $\mathcal R(I-B) = X$ so, yes, the desired $x$ always exists.

Let $ \lambda \ne 0$. If $ \lambda $ is an eigenvalue of $A$, then
$$0< \dim \mathcal{N}(\lambda -A) < \infty,$$
since $A$ is compact. If $y \in \mathcal{R}(\lambda -A)$, then there is $x_0 \in X$ with $(\lambda -A)x_0=y$.
Then we have $(\lambda -A)(x+x_0)  =y$ for all  $x \in  \mathcal{N}(\lambda -A).$


is "$a^0 = 1$" a definition or there exists a proof? I'm interested in knowing whether $a^0 = 1$ ('$a$' not zero) is a definition.
If not, can anyone please help me with proving this?

It's a definition. A convenient definition.

We know from our early encounter with mathematics that $a^m\times a^n=a^{m+n}$ if $m,n\in \mathbb{N}$ (not including zero) because it's very natural: "Multiplying $m$ times and multiplying $n$ times, and then multiplying those values should be the same as multiplying $m+n$ times".

What if $m$ or $n$ is $0$? We'd like that formula to still be true, i.e. we'd like

$$a^0\times a^n = a^{0+n}=a^n$$

However, the only number that satisfies this is $1$ so it's necessary to define $a^0:=1$ in order to keep this property.

This is exactly the same reason why we define

$$a^n:=\frac{1}{a^{-n}}$$

if $n$ is a negative integer, because we want laws as

$$\frac{a^m}{a^n}=a^{m-n}$$

to hold even if $m$ is not greater than $n$.

$${a^b\over a^c} = a^{b-c} $$and vice versa.     {Index rule}

Therefore,  $$a^0 = a^{x-x}$$ for any x

$$= {a^x\over a^x}   ,$$ using the aforementioned index rule.

$$= 1 $${since any thing divided by itself is 1 except 0)


Can a continuous function have a non-continuous derivative? If $f(x)$ is continuous throughout its domain, can its derivative, g(x) be non-continuous for any point?

If so, what effect does a non-continuous derivative have on the function?

For $x\in [0,1]$, $f(0)=0$ and $f(x)=x^2\sin\frac{1}{x}$ for $x\neq 0$.
Then $f(x)$ is differentiable at any point in $[0,1]$, but $f'$ is not continuous at $x=0$.

Yes, consider the function $$f(x) = \begin{cases} 0 \qquad x\leq0 \\ x \qquad x>0\end{cases}.$$

It's gradient is defined everywhere except $x=0$

$$f^\prime(x) = \begin{cases}0 \qquad x<0\\1\qquad x>0\end{cases}.$$

If you want a function which is continuous but no-where differentiable Brownian Motion/Wiener Process satisfies this.


If $p | a$, prove that $p \nmid (a+1)$ Suppose that $p | a$.  Prove that $p \nmid a + 1$.

$a = p l$

$a + 1 = (pl) + 1$

$1 \in (0, p)$

As such, $p \nmid a + 1$ by definition of divisibility.

Does this look correct?

If the number $p$ is supposed to be a prime, then your proof is correct. I'd write $0\le 1<p$, rather than $1\in(0,p)$.

Your proof appeals to uniqueness of the remainder, rather than definition of divisibility, and you should mention it.

A different proof is also easy: if $a=pl$ and $a+1=pk$, then
$$
1=(a+1)-a=pk-pl=p(k-l)
$$
which implies $p\mid 1$.

I shall try it by contradiction....

Let p|a and p|a+1
Let q be a prime divisor of p....then q also divides a and a+1....
Hence it divides ua + v(a+1); for u and v integers.....
In particular q|(-a+a+1)....
i.e. q|1
But q was a prime..... Hence q >=2.....
Hence q cannot divide 1(in integers) ....
A contradiction...... Hence p can not divide any two consecutive integers until and unless p =1.....

U can also try it using congruency....


Trying to model an equation for this problem: A drug is administered once every four hours. Let Dt be the amount the drug in blood system at t-th interval. The body eliminates a certain fraction p of the drug during each time interval. If the amount administered is Do find Dt and limit t->infinity Dt.

Basically I am looking for a equation for Dt. which can tell me at given time interval t how much of drug is still left in body.

Assume that the amount of the drug administered is a constant $A$ and that the amount of drug in the body at the start of the interval from $t$ to $t+1$ before the drug is administered is $D_t$. Then at the start of the interval from $t$ to $t+1$ after the drug is administered we have $D_t+A$ in the body and at the end of the interval we have:
$$
D_{t+1}=(D_t+A)-p\times(D_t+A)
$$

Now as $t\to \infty$ we should get $D_{t+1}\to D_t$ (that is the amount in the body at the end of an interval approaches that in the body at the end of the previous interval). Hence after a large number of intervals (that is for large $t$):
$$
D_t\approx (D_t+A)-p\times(D_t+A)
$$
Solving this gives:
$$
D_t\approx A(1-p)/p
$$
This approximation becomes better and better as the number of intervals increases, so that in the limit:
$$
D_{\infty}= A(1-p)/p
$$

The model I think is

$$D_t = t*D_0-p*D_0*(t-1)$$

The (t-1) depends on when the withdrawl happens before or after the drug has been administered into the body on any given time interval.  If it is after, then the model is
$$D_t = t*D_0-p*D_0*t$$


Combinatorics problems that can be solved via infinite descent I'm looking for high school problems that can be solved with the method of infinite descent. Usually, those problems are from number theory, but I would be very happy if someone could provide a problem(s) from combinatorics and/or any other field of mathematics. Here are some problems from number theory:

Prove that a following equations have no nontrivial solutions in $\mathbb{Z}$:

$a^3+2b^3 = 4c^3$

$2a^2+3b^2 = c^2+6d^2$

$x^2 + y^2 + z^2 = 2xyz$

$x^4+y^4 = z^2$

There is no regular pentagon whose vertices are on the lattice points.

Suppose that $ABCDE$ is a regular pentagon where $A,B,C,D,E$ are on the lattice points. Then, let $A',B',C',D',E'$ be the intersection point of $BD$ with $CE$, $AD$ with $CE$, $AD$ with $BE$, $AC$ with $BE$, $BD$ with $AC$ respectively.

Then, since $ABA'E$ is a parallelogram, we see that $A'$ is also on the lattice point. Similarly, we see that $B',C',D',E'$ are on the lattice points.

So, we see that $A'B'C'D'E'$ is a smalller regular pentagon whose vertices are on the lattice points.

The Fundamental Theorem of Arithmetic

Existence

The following can be proven using the Principle of Infinite Descent - see this link.

Proposition 1: Every integer $n \ge 2$ can be factored into primes:
There exists a nonempty family $(p_i)_{\, 1 i \le i \le u }$ of primes such that

$$\tag 1 n = p_1 p_2 \dots p_u$$

Uniqueness

Proposition 2: If an integer $n \ge 2$ is written as

$$\tag 2 n = p_1 p_2 \dots p_u$$

and

$$\tag 3 n = q_1 q_2 \dots q_v$$

with two nonempty families of primes, $(p_i)_{\, 1 i \le i \le u }$ and $(q_i)_{\, 1 i \le i \le v }$,
then $u = v$ and the families can be brought into a $1:1$ correspondence.

Proof
To get a contradiction, assume the proposition is false.

An easy factoring argument presents us with an integer $n \ge 2$ satisfying $\text{(2)}$ and $\text{(3)}$ with $p_i \ne q_j$ for any subscripts. Also, both $u$ and $v$ must be greater than $1$ and we can insist that $p_1 \lt q_1$.

The number

$$\tag 4 n' = (q_1 - p_1) q_2 \dots q_v \ge 2$$

is divisible by $p_1$. If $q_1 - p_1 \ge 2$, by proposition 1 we can factor it into primes and it is impossible for $p_1$ to appear in this factorization. Since we also have $p_1 \ne q_i$ for $i \ge 2$, the integer $n' \lt n$ can now also be presented as an integer with two different prime factorizations.

So by the Principle of Infinite Descent, the proposition is true. $\quad \blacksquare$

Combined Statement

Theorem 3: Every integer greater or equal to $2$ has one and only one
(up to a rearrangement) prime factorization.


Multipicative group and Additive groups Maybe it's a silly question but how exactly do you find a multiplicative group and additive group? Is it always a modulo function?

How do you find the multiplicative group and additive group of real numbers?

An example or maybe a few, would be very helpful! Thank you!

We tend to call the group operation multiplication regardless of whether it is multiplication or not.

Addition modulo $n$ will form a group.

Multiplication modulo $n$ is a little trickier.  If we take the integers that are co-prime with $n$ we can form a group with the multiplication as the group operation.

But, our group operation may not be addition or multiplication.  It may be something else entirely, it may be function composition, for example.

i.e. the set of bijective functions from $A\to A$ forms a group, with group operation as composition.

The group may represent geometry.  The set of rigid motions in the plane forms a group.

Any set with a binary operation that satisfies the group axioms is a group.


How do you find the time of the collision of circles moving under uniform circular motion? I am working on a simulation involving circular pucks moving in circular orbits at constant angular speed. I found the equation determining when the pucks collide to be intractable so have been calculating the time of collisions in a given time delta via a binary search which has worked adequately; however, I am by no means an expert in numerical techniques.
I do believe that a closed form symbolic solution to this problem either does not exist or is so large and unwieldy it is worthless, but I wonder if there is a technique for finding a numerical solution that is more direct than the binary search approach I am using. If I am wrong about a symbolic solution being intractable I'd also be interested in that, obviously.
The problem can be formalized as below. One puck is revolving around the origin with angular speed $\alpha_1$, initial angular position $\theta_1$, and radius of revolution $r_1$. The other puck is revolving around $(c_x,c_y)$ with analogous properties. $d$ is the sum of the of the pucks' radii; $t$ is time; we are solving for time:
$$ [r_1 \cos(\theta_1 + t \alpha_1) - (c_x + r_2 \cos(\theta_2 + t \alpha_2))]^2 + [r_1 \sin(\theta_1 + t \alpha_1) - (c_y + r_2 \sin(\theta_2 + t \alpha_2))]^2 = d^2$$

I am afraid that a numerical solution is also difficult. None of the methods mentioned by you and others will work correctly if you don't have correct initial approximations of the solution, and you could easily miss the first solution. As the function is a combination of trigonometric functions with incommensurable pulsations, the behavior of the distance function is cahotic. And you can have double or near-double roots, which are hard to find.

So you have to sample the function in a systematic way (in constant steps), staying below the Nyquist frequency of both signals, and look for crossings of the minimum distance. At the same time, look for changes of sign of the derivative, and consider the first change event.
If you find a change of sign of the derivative, refine that zero and check if it causes a crossing of the function itself, compared to the other sampled points. I would recommend regula falsi, as it is a safe method to refine a root inside an interval.
When you have found a crossing, refine by regula falsi.

There are plenty of equation solving algorithms that a google search will turn up. You are solving for the zeros of a function mapping one variable to one variable. Look at bisection, interpolation methods, Newton-Raphson, etc.


Flatness of finitely generated (/finitely presented) module carries to power series module? Let $M$ be a finitely generated flat module over a commutative ring with unity $R$ . Then is $M[[x]]$ also flat over $R[[x]]$ ? If this is not always true , then what if we also assume $M$ is finitely presented and flat over $R$ ; is $M[[x]]$ flat over $R[[x]]$ then ?

I think this is true. Let $0\to K\to N$ be an inclusion of $R[[x]]$ modules. Then, $0\to K\otimes_R M\to N\otimes_R M$ is exact, since $M$ is flat over $R$. Now, $K\otimes_R M=K\otimes_{R[[x]]} R[[x]]\otimes_R M=K\otimes_{R[[x]]} M[[x]]$ and similarly for $N$, proving flatness of $M[[x]]$ over $R[[x]]$.


Solve a special non-linear Backward SDE It is straigtforward to solve a linear Backward SDE. i.e.
$dY_t=Z_tdW_t+ aY_tdt+bZ_tdt$ with $Y_T=\xi$ (where a  and b are constants, $\xi$ is bounded Randon Variable.)

How can I solve $dY_t=Z_tdW_t+ a(Y_t)^+dt +bZ_tdt $ with $Y_T=\xi$ ?

($(Y_t)^+$ means the positive part of $Y_t$)

This question is very important for me. I have got stucked in it for more than one week, after trying several methods. I will be very grateful if anyone can give me some hint about possible ways to solve it! Please, any suggestion is very important for me.

I am also interested in solving $dY_t=Z_tdW_t+ a(Y_t)^+dt +b(Z_t)^+dt $ with $Y_T=\xi$ ?

I am a starter in Backward SDE and wish to explore deeper in how to solve different BSDEs. Can anyone suggest me some related books?

Thank you so much!!

The easiest way is to assume Y is positive and solve it. If the solution is positive, then it is the actual solution. Another way is that if the BSDE is Markovian, you can solve the corresponding PDE(Feynman-Kac formula). In general, the solution should depend heavily in choosing $\xi$. Therefore, one may not get general solution for any $\xi$.


How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that

$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?

I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being between 1 and 2).

Any suggestion is welcomed.

Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$.

For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \frac{k}{n}}{1+0} = 1 + \dfrac{k}{n}$.

Therefore, $\displaystyle\dfrac{n}{n+1} \cdot \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right) \le \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$.

Now, use the fact that $\displaystyle\lim_{n \to \infty}\dfrac{n}{n+1} = 1$ and that $\displaystyle\dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$ is a Riemann sum for $\displaystyle\int_{0}^{1}(1+x)\,dx$ to get the result.

I think the integral becomes $\int_0^1 1+x dx$.
The effect of $k$ in the denominator is to change the sum by a factor between 1 and $1+1/n$, so in the end it does not change the limit.


What's this number called and what are its properties? What is the following number called and what are its known properties?
$$.12233344445555566666677777778888888899999999910101010101010101010...$$
(I think you get the pattern.)

P.S.Also, see what I did with the integer part there? :)

It shares thematic similarity w/ the "look and say sequence" described by John Horton Conway in the mid 1980s. That's a pyramid where every row beneath the first one 'describes' the previous row (e.g. The second row '1 1' describes the first row '1' when spoken as "one one").


How to solve an exponential and logarithmic system of equations? $$
\left\{\begin{array}{c}
e^{2x} + e^y = 800
\\ 3\ln(x) + \ln(y) = 5
\end{array}\right.$$

I understand how to solve system of equations, logarithmic rules, and the fact that $\ln(e^x) = e^{\ln(x)} = x$. However, any direction I seem to go with this problem causes me difficulties. For example, when solving for y, I can get $y = \ln\left(800 - e^{2x}\right)$ or $y = \frac{e^5}{x^3}$. Both directions seem to lead me into a dead end. Although system of equations seems to be a frequent question on here, I haven't found anything helpful towards this problem.

Having  set your equation as Harish Chandra Rajpoot answered $$f(x)=e^{2x}+e^{(e^5/x^3)}-800$$ each piece varies very fast. So, hoping that only one solution exist, try inspection $$f(1)\approx 2.85112\times 10^{64}$$ $$f(2)\approx 1.13992\times 10^8$$ $$f(3)\approx-152.665$$ So, the solution is very close to $3$. Since we have a "reasonable" guess, let us use Newton method which, starting at the guess $x_0=3$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For your case $$f'(x)=2 e^{2 x}-\frac{3 e^{\frac{e^5}{x^3}+5}}{x^4}$$

Applying the method, the successive iterates are $$x_1=2.71403$$ $$x_2=2.79701$$ $$x_3=2.85671$$ $$x_4=2.88250$$ $$x_5=2.88649$$ $$x_6=2.88657$$  which is the solution for six significant figures.

Edit

As Mark Dickinson commented, the first derivative of the function cancels only once for $x_*\approx 3.059$ and for this value $e^{2x_*}+e^{(e^5/x_*^3)}\approx 632.498$ which means that equation $e^{2x}+e^{(e^5/x^3)}=a$ would not show any solution if $a < 632.498$. Building a Taylor expansion around $x=3$, we have $$f(x)=\left(-800+e^6+e^{\frac{e^5}{27}}\right)-\frac{1}{27} \left(e^5
   \left(e^{\frac{e^5}{27}}-54 e\right)\right) (x-3)+\frac{e^5 \left(2916 e+36
   e^{\frac{e^5}{27}}+e^{5+\frac{e^5}{27}}\right)
   }{1458}(x-3)^2+O\left((x-3)^3\right)$$ the roots of which being $\approx 2.62606$ and $\approx 3.47307$ which are very good guesses of the solution given above and of the extra solution identified by @user190080.

It would have been a better idea to look at the roots of function $$g(x)=\log\Big(e^{2x}+e^{(e^5/x^3)}\Big)-\log(800)$$ which is much better conditionned (at least from a numerical point of view).

You have a set of 2 logarithmic equations, but using the rules of logarithms, you can turn them into a set of simplier equations.

$ln(e^{2x} + e^{y}) = ln(800)$ for the first equation and $e^{3ln(x) + ln(y)} = e^{5}$ for the second.

Solving using your logarithmic rules the gives you a set of two equations which you can solve any way you choose.

New system, $2x + y = ln(800)$ and $x^3 + y = e^5$, gives you a real solution at $x \approx  5.34162$ and$ y \approx -3.99862$ along with two complex solutions.

edit: my mistake, not a set of linear equations


HINT: Prove there does not exist a rational number that satisfies $x^3=p/q$ I would like to request a hint for a problem I am working on form Hardy's a Course of Pure Mathematics.

Question
Prove generally that a rational fraction $p/q$ in its lowest terms cannot be the cube of a rational function unless $p$ and $q$ are perfect cubes.

My Work

Claim: $\nexists x\in \mathbb Q$ such that $x^3=\dfrac{p}{q}$ unless $\sqrt[3]{p} , \sqrt[3]{q} \in \mathbb Z$

Attempt: Since $x$ is a rational number, it can be written as the ratio of two integers, $w$ and $v$. $x=w/v$ Therefore, we have:

$\dfrac{w^3}{v^3}=\dfrac{p}{q}$

Since $p$ and $q$ are in lowest possible terms, we can infer that both cannot be even.

Rearranging the above, we get:

$w^3 \times q=v^3 \times p$

Based on this, we can identify two possible cases:

1) $w^3$ and $p$ are even and $v^3$ and $q$ are odd.

2) Vice-verse

My problem is that I cannot seem to carry the proof from here.

Another potential root I was thinking of going was basing a proof on the fact that $p$ and $q$ are integers and then seeing if I can find a contradiction, but I am at a loss on how to begin it this way.

We may assume that $p$ and $q$ are relatively prime. Suppose there are relatively prime integers $a$ and $b$ such that $\left(\dfrac{a}{b}\right)^3=\dfrac{p}{q}$.
Then
$$a^3q=b^3p.\tag{$1$}$$

It is easy to see that $a^3$ and $b^3$ are relatively prime. For if they are not, then there is a prime $r$ that divides both. But then $r$ divides $a$ and $b$.

Now argue that  $a^3$ divides $p$ and $p$ divides $a^3$. Since you asked for a hint, we leave this part out.

Conclude from the above result that $p=\pm a^3$.

A similar argument shows that $q$ is the cube of an ineger.

What about $x=1, p=1$, and $q=1$? I'm pretty sure $1$ is a rational number, and $1/1$ is in its lowest terms.


geometric meaning of complex cubic polynomial coefficients A complex cubic polynomial arranged as $f(z)=z^3-3az^2+3b^2z-c^3$ with coefficients $a,b,c\in\mathbb{C}$ can be represented as the product of (unknown) factors $f(z)=(z-p)(z-q)(z-r)=z^3-3\left(\frac{p+q+r}{3}\right)z^2+3\left(\sqrt \frac{pq+qr+rp}{3}\right)^2z-\left(\sqrt[3]{pqr}\right)^3$ with roots $p,q,r\in\mathbb{C}$.

Thus we can read the centroid $a=\frac{p+q+r}{3}$ of the Steiner ellipse of the triangle $p,q,r\in\mathbb{C}$ right from the coefficients (e.g. reasoning with Marden’s theorem).

Do the coefficients $b=\sqrt \frac{pq+qr+rp}{3}$ and $c=\sqrt[3]{pqr}$ have similar nice intuitive geometric interpretation, e.g. describe characteristic points or features of the triangle or its Steiner ellipse?

Vieta's Formulae

\begin{align}
  f(z) &= z^3-3az^2+3b^2z-c^3 \\
  &= (z-p)(z-q)(z-r) \\
  a &= \frac{p+q+r}{3} \\
  b^2 &= \frac{pq+qr+rp}{3} \\
  c^3 &= pqr
\end{align}

Foci of Steiner ellipse

\begin{align}
  f'(\lambda) &=0 \\
  \lambda &= a\pm \sqrt{a^2-b^2} \\
  &= \frac{p+q+r \pm \sqrt{(p+q+r)^2-3(pq+qr+rp)}}{3}
\end{align}

Resolvents
\begin{align}
  u &=
  \sqrt[3]{\frac{c^3+3ab^2-2a^3}{2}+
           \sqrt{(b^2-a^2)^3+
           \left( \frac{c^3+3ab^2-2a^3}{2} \right)^2}} \\
  v &=
  \sqrt[3]{\frac{c^3+3ab^2-2a^3}{2}-
           \sqrt{(b^2-a^2)^3+
           \left( \frac{c^3+3ab^2-2a^3}{2} \right)^2}} \\
  0 &= f(a+u\, \omega^n+v\, \omega^{2n})  \tag{$\omega=e^{2\pi i/3}$}
\end{align}

  Centroid
  $$a=\frac{p+q+r}{3}$$
  Linear eccentricity of Steiner ellipse
  $$\sqrt{|a^2-b^2|}=\sqrt{|uv|}$$
  Semi-major axis of Steiner ellipse
  $$\frac{|u|+|v|}{2}$$
  Semi-minor axis of Steiner ellipse
  $$\frac{||u|-|v||}{2}$$

The geometric meaning of roots is that they are points on coordinate axis where the derivative or slope of a given graph becomes $0$ or the place where graph touches axes .  Hope i have interpreted it correctly.


Conjecture function $g(x)$ is even function? Let $f,g:R\to R\setminus\{0\}$ and $\forall x,y\in R$,such
$$\color{crimson}{f(x-y)=f(x)g(y)-f(y)g(x)}$$ I have  prove  the function $\color{crimson}f$ odd function.

because let $y=0$ we have
$$f(x)=f(x)g(0)-f(0)g(x)\tag{1}$$
Let $x=0,y=x$ we have
$$f(-x)=f(0)g(x)-f(x)g(0)\tag{2}$$
by $(1),(2)$ ,then $$f(x)=-f(-x)$$

Conjecture： $\color{crimson}{g(x)}$ is even function

For this function $g(x)$ is even problem ,I don't have any idea to prove it.But I think is right,because
$$\color{blue}{\sin{(x-y)}=\sin{x}\cos{y}-\sin{y}\cos{x}}$$
$$\color{crimson}{\sinh{(x-y)}=\sinh{x}\cosh{y}-\sinh{y}\cosh{x}}$$

The conjecture is false.

Consider $f(x) = x$ and $g(x) = 1+x$.
Now, for all $x,y \in \mathbb{R}$,
$$f(x-y) = x-y$$
and
$$f(x)g(y) - f(y)g(x) = x(1+y) - y(1+x) = x + xy - y - xy = x-y,$$
so $f(x-y) = f(x)g(y) - f(y)g(x)$. However, $g$ is not even.

You're in the right direction when you got $Eq-2$ by putting $x=0$ and $y=x$. From that you get $g(x) = \frac{f(-x) + f(x)g(0)}{f(0)}$. Similarly, putting $x=0$ and $y=-x$ can give you $g(-x)$ which is same as $g(x)$. Thus proving $g(x)$ is even. Note that dividing by $f(0)$ is safe because the range of $f$ does not include 0.

BTW, the function range for odd functions usually include 0, unless you don't want to define your function to be defined at 0. You could argue that $\lim_{x \rightarrow 0} f(x) = 0$ but then I am talking things that I am not sure about :D


Limit theorem of Markov chains applied to higher order Markov chains I have a second order Markov chain with 4 states {A,T,C,G} (the 4 DNA nucleotides).

the transition matrix looks like this:

    A    T    C    G
AA[0.1, 0.6, 0.2, 0.1]
AT[0.3, 0.1, 0.5, 0.1]
AC[0.5, 0.3,  0,  0.2]
AG[..., ..., ..., ...]
TA[..., ..., ..., ...]
TT[..., ..., ..., ...]
TC[..., ..., ..., ...]
TG[..., ..., ..., ...]
CA[..., ..., ..., ...]
CT[..., ..., ..., ...]
CG[..., ..., ..., ...]
GA[..., ..., ..., ...]
GT[..., ..., ..., ...]
GC[..., ..., ..., ...]
GG[..., ..., ..., ...]

I wanted to calculate the stationary probability vector for the 4 states to which this matrix converges. The Markov chain is regular.

In case of first order Markov chains this is easily done by calculating the limit of $P^n$ with $n\rightarrow \infty$.

I do not know how to approach the problem in case of second order Markov chains.

Also, having a limited dataset from which to determine the transition matrix, can I consider the stationary distribution of the 4 nucleotides as being the theoretical distribution I would have if I had a much larger pool from which to draw (with the same transition matrix)?

In other words, can I consider the stationary distribution like an estimation of the theoretical nucleotide frequency given the transition matrix obtained from limited data?

A second order Markov chain is a random process $(X_n)_n$ on an alphabet $A$, whose distribution is specified by its transition probabilities $Q(x\mid y,z)=P[X_n=x\mid X_{n-1}=y,X_{n-2}=z]$, for every $(x,y,z)$ in $A\times A\times A$ (and by an initial distribution on $A\times A$).
A stationary distribution of $(X_n)$ is a probability measure $\pi$ on $A\times A$ such that, if $\pi(x,y)=P[X_n=x,X_{n-1}=y]$ for every $(x,y)$ in $A\times A$ and some $n$, then $\pi(x,y)=P[X_{n+1}=x,X_{n}=y]$ for every $(x,y)$ in $A\times A$.

Thus, one asks that, for every $(x,y)$ in $A\times A$,
$$
\pi(x,y)=\sum_{z\in A}Q(x\mid y,z)\pi(y,z).
$$
As in the first order case, this linear system, together with the normalizing condition
$$
\sum_{(x,y)\in A\times A}\pi(x,y)=1,
$$
fully determines $\pi$ as soon as $(X_n)_n$ is irreducible.
A new feature, absent of the first order case, is that every stationary distribution $\pi$ has identical marginals, that is, for every $x$ in $A$,
$$
\varrho(x)=\sum_{y\in A}\pi(x,y)=\sum_{y\in A}\pi(y,x).
$$
Finally, the MLE of $\pi$ based on $(X_k)_{0\leqslant k\leqslant n}$ is $\hat\pi_n$ defined by
$$
\hat\pi_n(x,y)=\frac1n\sum_{k=1}^n\mathbf 1_{X_k=x,X_{k-1}=y}.
$$
The MLE is consistent, that is, $\hat\pi_n(x,y)\to\pi(x,y)$ almost surely, for every $(x,y)$ in $A\times A$, when $n\to\infty$. In particular, the frequency of $x$ in $A$ stabilizes, since
$$
\frac1n\sum_{k=1}^n\mathbf 1_{X_k=x}=\sum_{y\in A}\hat\pi_n(x,y)\to\varrho(x).
$$

Since your second-order Markov chain is regular, you can still compute $\lim_{n\to \infty} Q^n$, where

$$
Q^n(x|y,z) = \mathbb{P}(X_{n+1}=x|X_1=y,X_0=z).
$$

In the limit, you will have $\lim_{n\to \infty}Q^n(x|y,z)=\varrho(x)$ for every pair $(y,z)$. More information can be found in Chapter 7 of this book (especially equation 7.1.3).

Note however, that regularity or irreducibility of $Q$ do not imply that a unique stationary distribution $\pi(x,y)$ exists:

Consider a second-order Markov chain on $\{1,2,3,4\}$. Consider further, that there are two possible classes of cycles this Markov chain may go through: 1-2-3-4-1 and 1-2-3-1 (to break periodicity), or 1-4-3-2-1 and 1-3-2-1. From all pairs of states, the Markov chain moves to any of these two cycles and remains in them. One can show that this second-order Markov chain is regular and that $Q^n$ converges to a matrix with identical rows with entries $\varrho(x)$. But there is no unique invariant distribution $\pi(x,y)$, since there are two recurrent classes of tuples $(x,y)$ of states.

More generally, as mentioned in this book on page 173, if $x$ is a recurrent state for $(X_n)$, it need not be the case that a state $(x,y)$ for $(X_n,X_{n+1})$ is recurrent.


Find the locus of a point P? I don't understand how to go on with this question:

"As the point $R$ moves on the line $x+y=1$ from $(1,0)$ to $(0,1)$, the point $P$ moves such that it has the same $x$-coordinate as $R$, and its $y$-coordinate is equal to the square root of that of $R$. Describe the locus of $P$ and draw the loci of both $R$ and $P$ on the same set of axes."

Any help would be appreciated,
Thanks

First, just graph the line first on the Cartesian plane. And then,
determine the relationship between $R$ and $P$. Any point $R$ would
be of the form $\left(t,1-t\right)$, where $0\leq t\leq1$.

This means that $P$ will have coordinates $\left(t,\sqrt{1-t}\right)$.
So actually what you have is
\begin{eqnarray*}
f:R & \rightarrow & P\\
\left(t,1-t\right) & \mapsto & \left(t,\sqrt{1-t}\right).
\end{eqnarray*}
After some changes in the coordinates, what you have is $\left(1-s,s\right)\mapsto\left(1-s,\sqrt{s}\right)$,
where $0\leq s\leq1$. Just run through all the points and you'll
eventually see a graph, which is actually $y=\sqrt{1-x}$ on $\left[0,1\right]$.

First draw the line hence it will be between $y = 1$ and $x = 1$ and then use the $1/x$ method and it will increase hence the curve will go above the line and create a parabola shape.


Can someone explain how $\frac{\tan x}{\sec x}=\sin x$? What identities are used to get $\sin x$ from $\tan x \operatorname{/} \sec x$? I was looking at an example in my textbook and the problem went from $\tan x \operatorname{/} \sec x$ to $\sin x$. I don't understand how.

Hints:

  $\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$

$\tan(\pi/2) = \infty$ and $\sec(\pi/2)$ is undefined whereas $\sin(\pi/2) = 1$ so the given equation is incorrect.


How to determine $\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } }?$ How to compute

$\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } }?$

Wolframalpha gave me the solution $0$ but I dont get how to determine that.

Hope somebody can help.

$$\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } } = \lim \limits_ {n \rightarrow \infty} \left(\frac{31^ { \frac{1}{5} }}{17^{ \frac{1}{2} } }\right)^n = \lim \limits_ {n \rightarrow \infty} \left(\frac{{\left(31^2\right)}^ { \frac{1}{10} }}{{\left(17^5\right)}^{ \frac{1}{10} } }\right)^n  = \lim \limits_ {n \rightarrow \infty} \left(\frac{{31^2}}{17^5}\right)^{n\over10} = 0 $$

since $\left|\frac{{31^2}}{17^5} \right|< 1$


Line segment connecting points in a convex set A subset $A$ of $\mathbf{R}^d$ is called $convex$ if for all $\mathbf{a,b} \in A$, the line segment in $\mathbf{R}^d$ connecting $\mathbf{a}$ and $\mathbf{b}$ lies entirely in $A$. Prove:

For any $\mathbf{a}$, $\mathbf{b}\in \mathbf{R}^d$, the line segment connecting $\mathbf{a}$ and $\mathbf{b}$ is the set of all points

$\mathbf{r}(t)=t\mathbf{a}+(1-t)\mathbf{b},0\le t\le 1$.}

I'm just looking for a hint to get started on this.

Let $r(t_1)=t_1a + (1-t_1)b$ and $r(t_2)=t_2a+(1-t_2)b$ be two points on the line connecting a and b. I claim $tr(t_1)+(1-t)r(t_2)$ also takes the above form of a point on the line connecting and b. So

$tr(t_1)+(1-t)r(t_2)=t(t_1a+(1-t_1)b)+(1-t)(t_2a+(1-t_2)b)=(tt_1+(1-t)t_2)a+(t(1-t_1)+(1-t)(1-t_2))b$.

Since you asked for a hint i did not complete the computation but you start this way and you must continue until you get the equation into the same form as r(t) given in the question. You must check that the two coefficients also add up to 1 and that each coefficient of $a$ and $b$ is in $[0,1]$.

Assume that vector a and b represents the endpoints of the line segment. Now the line made from this line segment will go through direction b-a and to get all the point multiply it by t which gives t(b-a). This line goes through origin. We can shift it by adding b which gives b + t(b-a). Now change t from 0 to 1 and draw out the vectors on paper and you can see why it forms a line segment with a, b as endpoints.

To learn about line in vectors.
https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/vectors/v/linear-algebra-parametric-representations-of-lines


Show that $2f'(0)=\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-f(-\zeta)}{\zeta^{2}} d\zeta,$ whenever $0<r<1.$ Suppose $f: \mathbb{D} \rightarrow \mathbb{C}$ is holomorphic. Show that
$$2f'(0)=\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-f(-\zeta)}{\zeta^{2}} d\zeta,$$
whenever $0<r<1.$

So far my understanding of this problem is that I need to prove that we have $f(z)=\frac{f(z)-f(-z)}{2}$ for $z$ satisfying $|z|<1$, since the rest is an application of Cauchy's Integral Formula. I noted that the result is true if an only if we have $f(z)=-f(-z)$, for all $z$ satisfying the said condition. Here is where I am stuck so far. Could you please provide a small hint to point me in the right direction, please provide no complete solution, only small hints.

By extension Cauchy's Integral Formula ,we have

$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{{(\zeta-z_0)}^{n+1}} d\zeta,$$

provided $f(\zeta) is \ analytic\ in\ |\zeta|\ = \ r$

So,

$$2f'(0)=\frac{2*1!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{\zeta^{2}} d\zeta$$
$$\quad =\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-(-\zeta)}{\zeta^{2}} d\zeta \quad (given)$$

And ,Since $f(\zeta)$  is  analytic in $|\zeta| = r $  ,
thus it is continuous and differentiable in that region,

So ,$2f(\zeta)$ and  $f(\zeta) - f(-\zeta)$ are two continuous and differentiable functions with same antiderivatives

So , Two functions must be equal.

Hence,$$2f(\zeta) = f(\zeta) - f(-\zeta)$$
$$f(\zeta) =  - f(-\zeta)$$


How can I calculate the variance of a minimum of two random variables with different pdf and cdf? I am trying to solve a problem where I need to find the variance of min (a,b). a is actually a function of a uniformly distributed r.v. while b is another r.v. with pdf and cdf as f and F. The support for both is [0,inf]. Any suggestions please?
Thanks in advance.

If $a$ is a function of a uniformly distributed r.v. then it is just another r.v.

Are your r.v.s indepentent?

I think it should work to write everything in terms of indicator functions:

$c = \min(a,b) = a\cdot 1_{\{ a<b \}} + b \cdot 1_{\{ b\leq a \}}
= a\int_a^\infty f(x) dx + b\int_0^a f(x) dx$.

Now you have to find the variance of the sum. Since your r.v.s live $[0,\infty]$, I would suggest computing the Laplace transform $\mathbb{E}[\exp(-sc)]$ (note I named $c = \min(a,b)$). Of course, the outcome depends on $a$ and $f$. If you have this, you can compute the first and second moment (which you need for the variance) by differentiation w.r.t. $s$.

Hope this helps.


Counting number of license plates How many license plates can be made using 4 digits followed by 3 letters or 4 letters followed by 3 digits?

My attempt is the following. First, there are $9^4 \times 26^3$ license plates using 4 digits followed by 3 letters. Second, there are $26^4 \times 9^3$ license plates using 4 letters followed by 3 digits. Thus, in total there are:

$$9^4 \times 26^3 + 26^4 \times 9^3$$

possible combinations. Is this correct?

Since there are 10 digits (0-9) and 26 letters (a-z), the total number of valid combinations equals:

$$10^4 \cdot 26^3 + 26^4 \cdot 10^3 = 458,803,904$$

I suppose there is a slight error
The first digit is not zero hence 9 ways but for the rest there should be 10 ways
Hence correct answer will be
9×10^3×26^3+26^4+9×10^2


Approximating $\int_1^\infty x^a (x-1)^b e^{cx}dx$ After lengthy calculations, I arrived at
\begin{align*}
\int_1^\infty x^a (x-1)^b e^{cx}dx,
\end{align*}
which cannot be solved in closed-form. I thus seek to approximate the integral, potentially including special functions. The constants $a,b,c$ may be complex but are  chosen such that the integral exists.
Any ideas how to approximate the integral? I thought of rounding the real and imaginary part of $a$ and $b$ to integers. Then, simplifying the first two terms and plenty of partial integrations should give a solution. But that seems to be a very naive idea.

Different elements of solution.
I will consider only real values of $a,b$, with $c$ real and negative for evident convergence concerns.
If $a,b$ are positive integers, there is an easy answer because a function of the form $e^{cx}P_n(x)$ has a primitive function of the same form $e^{cx}Q_n(x)$ with degree$(P_n)  \ = \ $ degree$(Q_n)=n$ which is:
$$\int e^{ax}P_n(x)dx  = e^{ax}\sum\limits_{k = 0}^n (-1) ^k \frac{1} {a^{k+1}}\frac{d^k}{dx^k}\left(P_n(x)\right)  + C\tag{*}$$
Proof: First part of this answer (https://math.stackexchange.com/q/155808) by Robjohn.
See the remark at the bottom for a proof using a less "symbolic" approach.

In the general case, the different tracks I have obtained using the research tool https://approach0.xyz/ (especially efficient on formulas) were dealing with bounds either different from $1$ or $\infty$. Nevertheless, one of the main indications is the closed form expression of the indefinite integral (= primitive function), as found in the answer by Harry Peter here
$$\int e^{ax}x^b(1-x)^c~dx=\dfrac{x^{b+1}\Phi_1(b+1,-c,b+2;x,ax)}{b+1}+C$$
where $\Phi_1$ can be defined in the following way:
$$\text{For} \ \Re \,c>\Re \,a>0, \ \ \ \Phi _{1}(a,b,c;x,y):=$$
$${\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}\,\mathrm {d} t.$$
(see here).
With definite bounds, one finds, from the same Harry Peter here, a formula for
$$\int_0^\infty x^{a-1}(1-x)^{b-1}e^{t-cx}~dx=$$
$$=\dfrac{e^t\Gamma(a)\Gamma(b)M(a,a+b,-c)}{\Gamma(a+b)}-(-1)^be^{t-c}\Gamma(b)U(b,a+b,c)\tag{1}$$
(with an interesting splitting of the integral $\int_0^1+\int_1^{+\infty}$)
For $t=0$, formula (1) is nothing else than the Laplace Transform of $x^a (x-1)^b$, easy to find directly.
It remains to subtract to (1) the value of $\int_1^\infty x^a (x-1)^b e^{cx}dx$ as indicated here.
Remark: A more classical approach to formula (*):
If $I_n:= \int x^n e^{ax}dx$, we have: $$(-1)^na^n\frac{I_n}{n!}=e^{ax} \sum_{k=0}^n(-1)^ka^{k-1}\frac{x^k}{k!}$$
(proofs by Qiaochu Yuan and Olivier Oloa here). Please note that the summation above is a kind of truncated exponential.

$$I(a,b,c)=\int_1^\infty x^a(x-1)^b e^{cx}dx$$
$$I_c=I(a+1,b,c)$$
$$I_{ca}=(a+1)I\tag{1}$$
this is a pde so you could try and solve this, initial conditions with $c=0$ or $b=0$ should make this problem simple. However notice that this integral appear to diverge for $c>0$


Basic Logarithm Equation $\log_2(x) = \log_x(2) $

Using the change of base theorem: $\dfrac{\log(x)}{\log(2)} = \dfrac{\log(2)}{\log(x)}$

Multiplied the denominators on both sides: $\log(x)\log(x) = \log(2)\log(2)$

I kind of get stuck here. I know that you can't take the square root of both sides of the equation, but still, $x = 2$ seems to be an obvious solution to the equation. I've missed $2^{-1}$ or $\frac12$ as another answer to the equation, which I am struggling to get to.

Any help will be greatly appreciated, thanks in advance.

You have it, actually.
\begin{align}
(\log(x))^2 &= (\log(2))^2 \\
\log(x) &= \pm \log(2)
\end{align}

For the "$+$" case, you've already solved it.

In the "$-$" case, you have $\log(x) = -\log(2) = \log(2^{-1}) = \log(\frac12)$, from which you can get $x=\frac12$.


Probability of intersection in conditional probability Suppose there are seven keys:

5 red keys.
2 green keys.

Only one of those seven keys opens a lock.

If for the experiment "Choose a random key and see if it opens the lock", I define the following events:

A: "The chosen key opens the lock".
G: "The right key is green".

And knowing that:

The probability of A is: 1/7.
The probability of G is: 2/7.
The probability of A knowing that the key is green is: 1/2

According to the relationship in conditional probability, the probability of A intersection G is: 1/7

Because p(A|G) = A intersection G / p(G)

But according to the probability of an intersection of independent events, the probability of A intersection G should be: 1/7 * 2/7 = 2/49

I do not understand why I get different values.

This is not quite a random experiment as it is determined in advance which key opens the lock. So, your calculations involving G are illegal.

The problem is you are assuming in your last calculation that $A$ and $G$ are independent.  In general,
$$\mathbb{P}(A \cap G) = \mathbb{P}(A \mid G) \cdot \mathbb{P}(G) \neq \mathbb{P}(A) \cdot \mathbb{P}(G).$$
The fact that $\mathbb{P}(A \mid G) = \frac{1}{2}$ and not $\frac{1}{7}$ is telling you the two events are not independent, and that's the reason you're getting different values.


Tell about derivatives $f(x,y)$. I have $f(x,y)=e^{-(x^2+y^2)^{-1}}$. Also $f(0,0)=0$. What I can tell about differentiable?

Of course I know $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

E.g. $$\frac{\partial f}{\partial x}=\frac{2x\cdot e^{-\frac{1}{x^2+y^2}}}{(-x^2-y^2)^2}$$

What I can tell about $\frac{\partial f}{\partial x}$ near $(0,0)$?

Thanks.

We have that

$$f_v(0,0)=\lim_{(h,k)\to(0,0)}\frac{e^{-(h^2+k^2)^{-1}}}{\sqrt{h^2+k^2}}=\lim_{u\to0^+}\frac{e^{-\frac1u}}{\sqrt{u}}=\lim_{v\to\infty}\frac  {\sqrt{v}}{e^{v}}= 0$$

then $f_x=f_y=0$.

Moreover since as $(x,y)\to (0,0)$

$$\frac{\partial f}{\partial x}=\frac{2x\cdot e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2}\to 0 \quad \frac{\partial f}{\partial y}=\frac{2y\cdot e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2}\to 0$$

by "Differentiability theorem" since all the partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.

The function $\phi (t)=e^{-1/|t|}$ for $t \neq 0$, $\phi (0)=0$ is an infinitely differentiable function from $\mathbb R$ into itself. Since  $f(x,y)=\phi (x^{2}+y^{2})$ it follows that $f$ is differentiable.


let $A$ be a finite open subset of a metric space $M$ prove that every point in $A$ is an isolated point of $M$ Is the following approach correct?

Since $A$ is a open subset (finite) of the metric space $M$, we can use the following theorem(s) which states:

Theorem I. A subset of a metric space is open i.f.f it contains, along with any point $x$, some neighborhood of $x$
Theorem II. A subset of a metric space is open i.f.f it is expressible as a union of open balls.

Therefore, since we already have the information that $A$ is a open subset, it can be
I. Expressible as a union of open balls [not assumed to be disjoint]
II. Every point $x$ in $A$ has a neighborhood.

''Definition'' picked from WIKIPEDIA In mathematics, a point $x$ is called an isolated point of a subset $S$ (in a topological space $X$) if $x$ is an element of $S$ but there exists a neighborhood of $x$ which does not contain any other points of $S$.

"Therefore, every $x$ in A is an isolated point of $M$ since every $x$ according to above has a neighborhood (balls)..." [not conclusive enough]
From here, I do struggle a bit concluding.
Does anyone have any suggestions?

You can write in a simpler way :

Let $p \in A$. Because $A$ is finite, $A \setminus \lbrace p \rbrace$ is finite so it's closed in $M$. So $M \setminus \left( A \setminus \lbrace p \rbrace \right)$ is open. Now note that
$$\lbrace p \rbrace = A \cap \left( M \setminus \left( A \setminus \lbrace p \rbrace \right) \right)$$

is the intersection of two open sets, so it is open. So $p$ is isolated in $M$.

Let X=$\{x_1, ..., x_n\}$ be the points of your open subset. Since your space is metric, it is separated and so $\forall i != j , \exists B_{i,j} , B_{j,i}$ two open balls such that $i \in B_{i,j}, j \in B_{j,i}, B_{j,i} \cap B_{j,i} = \emptyset$, then define $\forall i, B_i = \bigcap_{j=1}^n B_{i,j}$, then $\forall i != j , x_j \not \in B_i$, and as a result $B_i \cap X - \{x_i\} = \emptyset$ and every point of X is isolated.


Extending a continuous function on the unit sphere to the unit ball I'm trying to solve a problem in Lee's topology book. In this book, a closed n-cell $D$ is any space homeomorphic to the closed unit ball. Then $\mathrm{Int} D$ , resp. $\partial D$ are the images of the open unit ball, resp. the boundary of the closed unit ball under such a homeomorphism. Given a continuous function $f: \partial D \rightarrow \left [ 0,1 \right ]$, he wants one to show that $f$ extends to a continuous function $F: D \rightarrow \left [ 0,1 \right ]$ that is strictly positive in $\mathrm{Int} D$. Obviously, it suffices to show this for a closed unit ball and I was thinking something like $$\left \|x  \right \|f\left(\frac{x}{\left \| x \right \|}\right) + \dfrac{1 -\left \| x \right \|}{2},$$ but that does not quite do it. Tietze's extension theorem is not in that book (and in fact is a weaker result). Any ideas? Thanks in advance.

F is not defined at the origin. By defining F(0) = 1/2, F becomes a solution to the problem.


Differentiation of subtraction I've got an exercise to do and I don't really know what to do.

Exercise : We've got function $f$, where $f(a) = 0$ and $f'(a)$ exists. Also we got function $g$ which is continuous. Does exist $(f-g)'(a)$? Explain it.

My opinion is that exists, but I've got no idea how should I explain it. Some help? Thank you.

$f(a)=0$ and $\exists f'(a)$.

The expression : $(f-g)'(a)$ is : $f'(a) - g'(a)$.

You cannot say that this derivative exists, if $g$ is not differentiable at $a$. In your question's body, you have only stated that $f$ is differentiable and that $g$ is only continuous. Thus, no, you cannot say generally that $\exists (f-g)'(a)$.

Is $g$ differentiable?  That information is not given.  It is possible for $g$ to be continous but not differentiable.

Anyway differentiation is "linear", in that $(f-g)'(a)$ = $f'(a) - g'(a).$
So, if $g'(a)$ exists (and $f'(a)$ exists which is given) then $(f-g)'(a)$ exists.

But if $g'(a)$ does not exist, then $(f-g)'(a)$ does not exist.


Relation between Variety of $(I\cap J)$ and Variety of $(I)$ $\cap$ Variety of $(J)$ I was wondering whether a relationship exists between
$V(I\cap J)$ and $V(I)\cap V(J)$. Where $I$ and $J$ are ideals of the ring $R=K[x_1,...,x_n]$.

The correct correspondence is

$$V(I) \cap V(J) = V(I+J)$$

The intuition here is that the algebra and geometry are dual to one another. If I add more polynomials to my ideal there are fewer points that vanish on the whole ideal. If I take away polynomials, there are more points which vanish on the whole ideal. With that in mind, you're not going to get an identity where both the algebraic and geometric objects are getting smaller.

As an example, take $I = (x)$ and $J = (y)$ in $K[x,y]$. The intersection of $V(x)$ and $V(y)$ is the origin. But the intersection of $I$ and $J$ is $(xy)$, where $V(xy)$ is the union of the coordinate axes.

If we take $I = (x-3y)$ and $J = (y)$ then $V(I) \cap V(J)$ is still the origin. But $V(I \cap J)$ will be neither the origin nor the union of the coordinate axes. This shows that we cannot deduce the identity of $V(I \cap J)$ from our knowledge of $V(I) \cap V(J)$.

We know $V(I) \cap V(J) \subseteq V(I\cap J)$.


Confusion with notation of square bracket and round bracket of indices of a tensor Refer to the following picture:

I am confused with the last notation. Say if I got a tensor ${T^{abc}}_{de}$ and I would like to denote a new tensor which is defined by permuting the indices $a$ and $c$. But you just can't add an open bracket before $a$ and a closed bracket after $c$, because in the notation, it means permuting $a,b,c$ instead of just $a,c$ only.

So how should I do? Thank you.

Suppose, $t^{abc}_{de}=U^{ac}⊗V^{b}_{de}$ with $U∈⊗^{2}T_{x}$, $V∈⊗T_{x}⊗^{2}T^{*}_{x}$ with $x∈M$, $M$ some underlying manifold. Then, $\widetilde{t}≑ U^{(ac)}⊗V^{b}_{de}$ and,
$\overline{t}≑ U^{[ac]}⊗V^{b}_{de}$.


Two identical dice are thrown simultaneously. Find probability of getting a $3$ and a $2$. Two identical dice are thrown simultaneously. Find the probability of getting a $3$ on
  one of the dice while a $2$ on the other.

I'm pretty sure the answer is $\frac{2}{36}$ but my friend says the answer should be $\frac{1}{21}$.

He argues that as the dice are identical, so sample space comprises only $21$ combinations.

I'm unable to explain it to him but I don't agree with his answer of $\frac{1}{21}$.

Can someone tell me the correct answer and provide a short explanation?

There might be 21 possible outcomes ( (1,1), (1,2),....(6,6) ) , but they don't have equal probability, 1+1 can only occur in one way, but 2+3 can occur two different ways.  There are 6 ways to throw (1,1), (2,2), (3,3), etc, and 15 different pairs (1,2)/(2,1), (1,3)/(3,1), etc, making 36 possible combinations. We can get a 2 and 3 in two different ways, so the probability is 2/36 as you say.

As the outcomes of dices should be $2$,$3$. So $2$ on one dices and $3$ on another as dices are identical we can't distinguish them so only one way that is $(2,3)$. Now, the sample space for this will be $21$ not $36$ because the dices are identical, i.e. $(1,2)$ is not different from $(2,1)$ etc.
So sample space = $21$ and probability = $1/21$.


Explicit sum formula for repeated integers I know that $\sum_{i=0}^{n-1} i = \frac{n(n-1)}{2}$, but I am having troubles with finding an explicit sum formula for:

$ 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3\dotsc$

The sum sequence of which would be:

$ 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 27, \dotsc$

I thought about converting my sequence to $0,1,2,3,\dotsc$ by doing something with $\lfloor\frac{i}{8}\rfloor$, summing over that and then converting back, but I can't manage to work it out. Any thoughts?

I love this question, because i've studied those periodic shifts myself and led me toward tons beautiful methodes of summation.

Anyway the general answer is if you start at 0:
$$\sum_{n=0}^{dp} \sum_{z=0}^{d-1}f(n-z)\sum_{k=}^{d-1} (e^{\frac{2i\pi k}{d}})^{n-z}=\sum^p_{n=0} d^2 f(nd)$$

However i'd prefer to start at 1:

$$\sum_{n=1}^{dp} \sum_{z=0}^{d-1}f(n+z)\sum_{k=0}^{d-1} (e^{\frac{2i\pi k}{d}})^{n+z}=\sum^p_{n=1} d^2 f(nd)$$

Your periode of d=8 and f(n)=n will give the new function.
$$1/64\sum_{z=0}^{7}f(n+z)\sum_{k=0}^{7} (e^{\frac{2i\pi k}{8}})^{n+z}
$$
You will get 1,1,1,1,1,1,1,1,2...
Take the summation over that and you have your answer. If you wish to get an exact soltution there are serveral ways to solve these, but since your periode is 8, i'm not going to write it down. What i found the easiest myself was to take the delta of an answer you'd expect. ( like $n*(e^{\frac{2i\pi k}{d}})$) And work back. Since this is a simple example, i assume you should be able to solve it now.

I'll provide you an example in case d=2.

$$\sum_{n=1}^{m} (2n+1-(-1)^n)/4=$$

$$m^2/4+m/2-(-1)^m/8+1/8=$$

Nevermind, I found it:

$S(n) = \lfloor\frac{n}{8}\rfloor \cdot (n - 4\lfloor\frac{n}{8}\rfloor-4)$

(Note that the indices are really shifted by 1 compared to the original question, but that is actually what I wanted.)


What does 'dissipative PDE' means? Can you give me an idea what is meant with dissipative partial differential equations?

I am no phycist (and do not know the difference between initial energy to final energy), but wikipedia told me that

  A dissipative process is a process in which energy (internal, bulk flow kinetic, or system potential) is transformed from some initial form to some final form; the capacity of the final form to do mechanical work is less than that of the initial form. For example, heat transfer is dissipative because it is a transfer of internal energy from a hotter body to a colder one.

Now I wonder what a dissipative PDE means (mathematically).

I think dissipative typically refers to a PDE with a highest spatial derivative of even number eg heat equation while dispersive refers to one with an odd derivative eg KdV see for example Whitams book


Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $

I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.

However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}

I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?

If I take the derivative of your second answer (call it $g(x)$), I get:
\begin{eqnarray*}
\frac{dg}{dx}
& = & -\frac{-\sin x}{\cos x} + \frac{\sin x}{2(1-\cos x)} + \frac{-\sin x}{2(1+\cos x)}\\
& = & \frac{\sin x\left(1-\cos^2 x + \frac{1}{2}\cos x(1+\cos x) - \frac{1}{2}\cos x(1-\cos x)\right)}{\cos x(1-\cos x)(1+\cos x)}\\
& = & \frac{\sin x\left( 1- \cos^2 x + \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x\right)}{\cos x(1-\cos^2 x)}\\
& = & \frac{\sin x}{\cos x\>\sin^2 x} = \frac{1}{\cos x\sin x}.
\end{eqnarray*}
So I'm not sure why Mathematica says the second method is not "the right answer".

$$\int \frac{1}{\sin x\cos x} dx  = \int( \tan x+\cot x)\, dx= \int\frac{\sin x}{\cos x}\, dx+\int\frac{\cos x}{\sin x} \,dx$$ from where it should be fairly simple....(choose correctly the variables though)


Don't see the point of the Fundamental Theorem of Calculus. $$\frac{d}{dx}\int_a^xf(t)\,dt$$

I would love to to understand what exactly is the point of FTC.  I'm not interested in mechanically churning out solutions to problems.    It doesn't state anything that isn't already known.  Prior to reading about FTC, the integral is defined as the anti-derivative.  So, it's basically an operator.  "Take the anti-derivative by figuring out whose derivative this is!"  Simple.  So, what is so "fundamental" about redundantly restating the very definition of the integral? (The derivative of the anti-derivative is the function).  This to me is like saying $-(-1) = +1$.  Not exactly earth shattering.

Am I missing something with regard to the indefinite vs. definite integral?

If we look at a simple example,
$$\frac{d}{dx}\int_1^xt^2 \, dt = \cdots =x^2$$

Can we discuss what exactly this is representing?

Why would you even write this? Why would you take the rate of change
of an area under the curve? Why would you want to take the
derivative of an integral? Or, is this just done to prove something
else?  When would you even come across this situation in Math?  Taking the rate of change of the area under a curve and/or total displacement?   (derivative of the definite integral)
Also, what is the significance of using $t$ as a variable?
Why would you integrate from a constant to a function in the first
place?   (take area under the curve or compute total displacement)

I don't understand what exactly things FTC even allows anyone to do.  Without FTC, I can already evaluate definite integrals.  Without FTC, I can already take derivatives.  So, with FTC, I can take an integral then take a derivative?  So, what's even the point of FTC?  I really don't see anything "fundamental" whatsoever about this redundant self-evident "theorem". This is like taking the inverse of an inverse.  Right back to f(x), but that's simply a "neat trick" vs. a "Fundamental Theorem of Algebra".

I am guessing that you have been taught that an integral is an antiderivative, and in these terms your complaint is completely justified: this makes the FTC a triviality.

However the "proper" definition of an integral is quite different from this and is based upon Riemann sums.  Too long to explain here but there will be many references online.

Something else you might like to think about however.  The way you have been taught makes it obvious that an integral is the opposite of a derivative.  But then, if the integral is the opposite of a derivative, this makes it extremely non-obvious that the integral can be used to calculate areas!

Comment: to keep the real experts happy, replace "the proper definition" by "one of the proper definitions" in my second sentence.

The only reason you know the definition of an integral as an antiderivative is BECAUSE of the FTC.

And there are all sorts of situations where it is useful in the real world. For example, if you integrate force along a path, you get energy. Let's say that that force is gravity. FTC says that know matter whether you push a boulder straight up the side of a mountain or take a bunch of switch-backs, the rock still has the same amount of gravitational potential energy at the top of the mountain.


Sum of two co-prime integers I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $\gcd(a,b)=1$.

I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).

I would appriciate it a lot if someone could give a hand here.

Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.

In this particular problem, we can break down cases into the residue classes $\bmod 4$ in order to hunt for patterns:

1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $k\geq 3$.

2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?

I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.

We know that $n>6$ and we need to prove that any integer $n>6$ can be written as the form of two co-primes. So it has a very simple proof.
We know that any integer $n$ and $n-1$ are always co-prime, this means that their gcd is $1$.
So,$$n = (n-1) + 1\\
         = a + b.$$  ( Where a is n-1 and b is 1)
So gcd (a,b) = 1 , so  I say that any integer n>1 can be written as the sum of two co primes.


Greedy algorithm minimizes the sum There are $n$ positive numbers (not necessarily distinct). In every step, we
can select two numbers and replace each of it with the sum of the two
numbers. Prove that in any amount of steps, selecting the two least
number in every step minimizes the sum of the $n$ numbers.

When the $n$ numbers are different. We assume that they are $x_1,x_2,..,x_n$. We will also assume that at every step two of of $x_i$ has changed its value so there's no need to define other variables. We can characterize every step by only using the indices, so let step $\{a,b\}$ denote changing each of $x_a$ and $x_b$ to $x_a+a_b$. We also used an arrow ($\rightarrow$) to indicate the order of the steps. Let any arbitrary $k$ steps be $\{i_1,j_1\} \rightarrow \{i_2,j_2\} \rightarrow ... \rightarrow \{i_k,j_k\}$. We can finish the problem by showing that by only changing the first step to select the two least number, will result to smaller than or equal sum. I think there will be induction involved here. Any idea?
Edit: I think I've solved it using the idea above.

Let the points be $x_1, x_2, .. x_n$, and let $S = \sum_{i = 1}^n x_i$. Now let any two distinct points $u$ and $v$ ($u \neq v$) be replaced by their sums. Denote the new sum by $S'$.
Then $S' = S + u + v$.
This implies $\min{S'} = \min{u + v} + S$. Hence to get the minimum sum in this step, we choose $u$, $v$  s.t. $u+v$ is minimum. It is trivial to see this minimum occurs for the least $2$ elements of the set.
Above was for a single step. Consider any sequence $n$ of steps. For each such sequence, the greedy algorithm produces a sum which is less than the original sequence in each step, and hence less overall (formally prove by induction on $n$)


How fast can one move around an ellipse with bounded acceleration? Given a smooth closed planar curve $\Gamma$, I'm looking for its periodic parametrization $\phi : \mathbb{R}\to\Gamma$ such that

the second derivative $\phi''$ is bounded by $1$ in the norm: $|\phi''|\le 1$
the period $T$ of the parameterization is as small as possible.

The "real world" motivation is auto racing: set the best lap time $T$ given that the car's acceleration is limited.

Question: what is the smallest $T$ for an ellipse with semiaxes $a,b$? (Auto racing tracks often look like an ellipse, although they are not really.)

Special cases

The special case $a=b$, a circle, is easy. Assume it centered at the origin. Then $|\phi|^2\equiv a^2$. Differentiate twice: $|\phi'|^2+\phi\cdot \phi'' =0$. This yields $|\phi'|\le \sqrt{|\phi||\phi''|} \le \sqrt{a}$, hence $$T\ge 2\pi a/\sqrt{a} = 2\pi\sqrt{a}$$
This lower bound is attained by $\phi(t) = a(\cos (t/\sqrt{a}), \sin (t/\sqrt{a}))$.

The degenerate ellipse, $b=0$, is also easy to handle: the car has to stop at the endpoints of the interval $[-a,a]$, and then have full acceleration up to the midpoint $0$. This yields $T=4\sqrt{2a}$, because one loop consists of $4$ segments of constant acceleration or deceleration.

In general, the smallest period $T(a,b)$ scales like $T(\lambda a,\lambda b)=\sqrt{\lambda}T(a,b)$ because the function $\lambda \phi(t/\sqrt{\lambda})$ has the same top acceleration as $\phi$.

The major and minor axes cut an ellipse into four congruent segments. It will suffice to find the optimal parametrization one of these segments, for the parametrization can be extended by symmetry. I will assume the ellipse has equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
$$
with $a\geq b$, and describe a parametrization from $(a,0)$ to $(0,b)$.

Because the curvature of the ellipse decreases as we move from $(a,0)$ to $(0,b)$, the optimal parametrization has a concrete description: start out as fast as possible, and accelerate as much as possible. More precisely, start out at a velocity such that the entire unit of acceleration must be directed inward to stay on the track, and at each point, choose a tangential component of acceleration to be as large as possible (given the bound on total acceleration).  I will give an argument this parametrization is optimal at the end of this post.

I will use a parameter $0\leq w\leq \pi/2$, with a point on the ellipse written as $(a\cos w,b\sin w)$. Write $t$ for time (starting at $t=0$), $v$ for speed, and $K$ for the curvature of the ellipse. I will also write $s$ for arc length along the ellipse. Each of these quantities is a function of $w$.

First some auxiliary calculations. It is convenient to introduce the function $C(w):=\sqrt{a^2\sin^2 w+b^2\cos^2 w}$. A formula for $K$ is given by
$
K(w) = {ab}/{C(w)^3}
$. We have $ds/dw=C(w)$. And of course $ds/dt=v$. The inward acceleration needed to stay on the track is $K v^2$, so our tangential acceleration will be
$$
\frac{dv}{dt}=\sqrt{1-K^2v^4}.
$$
We can combine these equation to get a differential equation for $v$:
\begin{align*}
\frac{dv}{dw}&=\frac{dv}{dt}\left(\frac{ds}{dt}\right)^{-1}\frac{ds}{dw}\\
&=\frac{\sqrt{1-\frac{a^2b^2}{C(w)^6}v^4}}{v}C(w).
\end{align*}
This is a non-linear first order ordinary differential equation. The initial condition will have $K(0)v(0)^2=1$, so $v(0)=b/\sqrt{a}$. I have no reason to believe there is a closed form solution. But if a solution $v(w)$ is found numerically, then the period $T$ can be computed as
$$
T=4\int_0^{\pi/2}\frac{ds}{v}=4\int_0^{\pi/2}\frac{C(w)}{v(w)}dw.
$$

Proof of optimality: Write $s_1$ for our speed function, as suppose $s_2$ is the speed function for some other parametrization. By the way we chose acceleration, we have $s_1'>s_2'$ at every point where $s_2>s_1$. So in particular, at the point where $s_2-s_1$ is maximized, we have $s_1'=s_2'$, and so $s_2\leq s_1$ at that point. Thus there is no parametrization that is faster at any point on the ellipse than the one we describe, so ours gives the minimal period.

I see the problem as follows. For an ellipse with semi-major/minor axes $(A,B)$, the trajectory is parameterized in the complex plane as

$$z=A\cos(t/\sqrt{a})+iB\sin(t/\sqrt{a})$$

Here, $a$ must have the dimensions [$\text{t}^2$], and, of course $(A,B)$ are [L].

The velocity and acceleration are given by

$$
v=\dot z=\frac{1}{\sqrt{a}}\left[-A\sin(t/\sqrt{a})+B\cos(t/\sqrt{a})\right],\quad \text{[L/t]}\\
\dot v=\ddot z=\frac{1}{a}\left[-A\cos(t/\sqrt{a})-B\sin(t/\sqrt{a})\right]=-\frac{z}{a},\quad \text{[L/t$^2$]}\\
$$

And finally, the instantaneous arc length, if needed, is given by

$$s=\int_0^t |\dot z|~dt=\frac{B}{\sqrt{a}}\int_0^t \sqrt{C\sin^2(t/\sqrt{a})+1}~dt,\quad C=\left(\frac{A^2-B^2}{B^2}\right)\\$$

The integral is an incomplete elliptic function of the second kind.

This should be sufficient to complete your task.


Repeating Decimals and Rounding I have stumbled across the proof that $0.\overline{9}=1$. The proof is as follows.

Let $x=0.\overline{9}$
$10\cdot x = 9.\overline{9}$
$10\cdot x = 9 + 0.\overline{9}$
Now that $0.\overline{9}=x$, $10\cdot x = 9 + x$.
We get $9x=9$ and $\therefore x=1$.

With this proof, we know that $0.\overline{9}=1$. Is there any scenario in math where using $0.\overline{9}$ instead of $1$ offers an easier solution to a problem?

I don't know of any case where using this representation offers an easier solution to a given problem. But in one case it is useful is Cantor's diagonal argument for the proof of uncountablity of the reals using decimal number system. There you want to make sure that while listing out the reals, that you always consider one representation of such recurring $9$s vs $0$s consistently. Otherwise someone might claim that oh well! the number you constructed is not in our list, but an alternate representation might have it and other more subtle issues.So yeah! without this basic fact, proving uncountability of real becomes harder by Cantor's beautiful agrument.


Where to start with Algebraic Geometry? I'm looking into studying Algebraic Geometry alongside some Algebraic Topology and was wondering what book would be useful. I was looking along the lines from Igor's Basic Algebraic Geometry, Cox, Little and O'Shea's Ideals, Varieties, and Algorithms, and Justin Smith's Introduction to Algebraic Geometry. I've heard that Igor's book Basic Algebraic Geometry is fantastic but the problems are known to be impossible.
What is the best text to use for study on algebraic geometry?
I have an understanding in Real Analysis (only single variable), Topology, Abstract Algebra (up to Ring theory).

Thomas Garrity's Algebraic Geometry: A Problem Solving Approach.

The book starts with establishing the equivalence of conics in the complex projective plane and then moves on smoothly to discussing tangents and singularities, elliptic curves, Bezout's theorem, Riemann-Roch, affine and projective varieties, and -- finally -- a brief intro to sheaves and cohomology.

You don't have enough background in commutative algebra. Try to finish Atiyah & McDonald or Eisenbud first. Then uyou can start Liu Qing or Hartshorne.


Prove that for each $y$ the y-section set $A_y=\{x\in \mathbb{R}: (x,y)\in A\}$ is a Borel measurable subset of $\mathbb{R}$. Suppose $A$ is a Borel measurable subset of $\mathbb{R}^2$. Prove that for each $y$ the y-section set
  $A_y=\{x\in \mathbb{R}: (x,y)\in A\}$
  is a Borel measurable subset of $\mathbb{R}$.

I feel like that is not easy to show $A_y$ is closed in $\mathbb{R}$. Any other methods?

The claim is wrong.

Pick any set $B\subset \mathbb{R}$ which is not measurable.

Let $A = \{0\} \times B \subset \{0\}\times \mathbb{R}$, where the latter is Lebesgue measurable in $\mathbb{R^{2}}$, with zero measure, therefore the first is also Lebesgue measurable in $\mathbb{R^{2}}$ (Lebesgue measure is complete).

By the choice of $B$, we get $A_0$ is not measurable in $\mathbb{R}$.

Hint: If a property holds on the generating set, then the property holds on all Borel sets. The open boxes $(a,c)\times(b,d)$ generate $\mathcal{B}(\mathbb{R}^2)$.


Why are the solutions of polynomial equations so unconstrained over the quaternions? An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers.  But it may have an uncountable set of zeroes in the quaternions.  For example, $x^2+1$ has two zeroes in $\mathbb C$, but in $\mathbb H$, ${\bf i}\cos x + {\bf j}\sin x$ is a distinct zero of this polynomial for every $x$ in $[0, 2\pi)$, and obviously there are many other zeroes.

What is it about $\mathbb H$ that makes its behavior in this regard to be so different from the behavior of $\mathbb R$ and $\mathbb C$?  Is it simply because $\mathbb H$ is four-dimensional rather than two-dimensional? Are there any theorems that say when a ring will behave like $\mathbb H$ and when it will behave like $\mathbb C$?

Do all polynomials behave like this in $\mathbb H$? Or is this one unusual?

When I was first learning abstract algebra, the professor gave the usual sequence of results for polynomials over a field: the Division Algorithm, the Remainder Theorem, and the Factor Theorem, followed by the Corollary that if $D$ is an integral domain, and $E$ is any integral domain that contains $D$, then a polynomial of degree $n$ with coefficients in $D$ has at most $n$ distinct roots in $E$.

He then challenged us, as a homework, to go over the proof of the Factor Theorem and to point out exactly which, where, and how the axioms of a field used in the proof.

Every single one of us missed the fact that commutativity is used.

Here's the issue: the division algorithm (on either side), does hold in $\mathbb{H}[x]$ (in fact, over any ring, commutative or not, in which the leading coefficient of the divisor is a unit). So given a polynomial $p(x)$ with coefficients in $\mathbb{H}$, and a nonzero $a(x)\in\mathbb{H}[x]$, there exist unique $q(x)$ and $r(x)$ in $\mathbb{H}[x]$ such that $p(x) = q(x)a(x) + r(x)$, and $r(x)=0$ or $\deg(r)\lt\deg(a)$. (There also exist unique $q'(x)$ and $s(x)$ such that $p(x) = a(x)q'(x) + s(x)$ and $s(x)=0$ or $\deg(s)\lt\deg(a)$.

The usual argument runs as follows: given $a\in\mathbb{H}$ and $p(x)$, divide $p(x)$ by $x-a$ to get $p(x) = q(x)(x-a) + r$, with $r$ constant. Evaluating at $a$ we get $p(a) = q(a)(a-a)+r = r$, so $r=p(a)$. Hence $a$ is a root if and only if $(x-a)$ divides $p(x)$.

If $b$ is a root of $p(x)$, $b\neq a$, then evaluating at $b$ we have $0=p(b) = q(b)(b-a)$; since $b-a\neq 0$, then $q(b)=0$, so $b$ must be a root of $q$; since $\deg(q)=\deg(p)-1$, an inductive hypothesis tells us that $q(x)$ has at most $\deg(p)-1$ distinct roots, so $p$ has at most $\deg(p)$ roots.

And that is where we are using commutativity: to go from $p(x) = q(x)(x-a)$ to $p(b) = q(b)(b-a)$.

Let $R$ be a ring, and let $a\in R$. Then $a$ induces a set-theoretic map from $R[x]$ to $R$, "evaluation at $a$", $\varepsilon_a\colon R[x]\to R$ by evaluation:
$$\varepsilon_a(b_0+b_1x+\cdots + b_nx^n) = b_0 + b_1a + \cdots + b_na^n.$$
This map is a group homomorphism, and if $a$ is central, also a ring homomorphism; if $a$ is not central, then it is not a ring homomorphism: given $b\in R$ such that $ab\neq ba$, then we have $bx = xb$ in $R[x]$, but $\varepsilon_a(x)\varepsilon_a(b) = ab\neq ba = \varepsilon_a(xb)$.

The "evaluation" map also induces a set theoretic map from $R[x]$ to $R^R$, the ring of all $R$-valued functions in $R$, with the pointwise addition and multiplication ($(f+g)(a) = f(a)+g(a)$, $(fg)(a) = f(a)g(a)$); the map sends $p(x)$ to the function $\mathfrak{p}\colon R\to R$ given by $\mathfrak{p}(a) = \varepsilon_a(p(x))$. This map is a group homomorphism, but it is not a ring homomorphism unless $R$ is commutative.

This means that from $p(x) = q(x)(x-a) + r(x)$ we cannot in general conclude that $p(c) = q(c)(c-a) +r(c)$ unless $c$ commutes in $R$ with $a$. So the Remainder Theorem may fail to hold (if the coefficients involved do not commute with $a$ in $R$), which in turn means that the Factor Theorem may fail to hold So one has to be careful in the statements (see Marc van Leeuwen's answer).  And even when both of them hold for the particular $a$ in question, the inductive argument will fail if $b$ does not commute with $a$, because we cannot go from $p(x) = q(x)(x-a)$ to $p(b)=q(b)(b-a)$.

This is exactly what happens with, say, $p(x) = x^2+1$ in $\mathbb{H}[x]$. We are fine as far as showing that, say, $x-i$ is a factor of $p(x)$, because it so happens that when we divide by $x-i$, all coefficients involved centralize $i$ (we just get $(x+i)(x-i)$). But when we try to argue that any root different from $i$ must be a root of $x+i$, we run into the problem that we cannot guarantee that $b^2+1$ equals $(b+i)(b-i)$ unless we know that $b$ centralizes $i$. As it happens, the centralizer of $i$ in $\mathbb{H}$ is $\mathbb{R}[i]$, so we only conclude that the only other complex root is $-i$. But this leaves the possibility open that there may be some roots of $x^2+1$ that do not centralize $i$, and that is exactly what occurs: $j$, and $k$, and all numbers of the form $ai+bj+ck$ with $a^2+b^2+c^2=1$ are roots, and if either $b$ or $c$ are nonzero, then they don't centralize $i$, so we cannot go from $x^2+1 = (x+i)(x-i)$ to "$(ai+bj+ck)^2+1 = (ai+bj+ck+i)(ai+bj+ck-i)$".

And that is what goes wrong, and there is where commutativity is hiding.

If we have the equation $x^2+1=0$, then it is enough to consider the subspace $x=(x_1,x_2,0,0)$ since we know that there is a solution in $\mathbb{C}\subset\mathbb{H}$. The additional structure, i.e. when $x_3,x_4\neq 0$, creates infinitely many solutions.

And speaking about additional structure: the conjugate of a real number is the number itself. Thus, the equation $x\overline{x}-1=0$ has only one solution in $\mathbb{R}\subset \mathbb{C}$, but there are infinitely many in $\mathbb{C}$.


Field with four elements If $F=\{0,1,a,b\}$ is a field (where the four elements are distinct), then:

1.What is the characteristic of $F$?

2.Write $b$ in terms of the other elements.

3.What are the multiplication and addition tableau of these operations?

The addition: $\;a+b=0\implies a=-b=b\;$ , since $\;\text{char}\, F=2\;$ . It also can't be $\;a+b=a\;,\;\;a+b=b\;$ , else $\;b=0\;$ or $\;a=0\;$ . Thus it must be $$\;a+b=1\implies b=1-a=1+a\;$$

Generalize the above and get the addition table.

As for multiplication $\;ab\neq 0,a,b\;$ , else $\;a=0\;$ or $\;b=0\;$ , or $\;a,b=1\;$ and none of this is true, thus it must be $\;ab=1\iff b=a^{-1}\;$ . Generalize now for the table.

Hint : What can u get of this $$F=\frac {Z_2[x]}{<x^2+x+1>}$$


How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ Question:
Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}>0$, and for all $n \in \mathbb{N}$ we have:

$$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$

Prove that :

$\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.

My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x>0$;
then we have
$$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$
But I can't see how to prove the limit of $\{a_{n}\}$ exists.
Thank you for you help!

Obviously, it is enough to show that

$$
\limsup a_n \leq 4 \leq \liminf a_n.
$$

To prove this, first remark that the sequence $M_n = \max\{4,a_n, a_{n-1}\}$ is non-increasing.
Indeed, the trivial lower bound $M_n \geq 4$ yields $a_{n+1} \leq 2\sqrt{M_n} \leq M_n$ ; we conclude with
$$
M_{n+1} = \max\{4, a_{n+1}, a_n\} \leq \max\{4, M_n, M_n\} = M_n.
$$
As a consequence, an upper bound for $a_n$ is $\max\{4, a_1, a_2\}$. In the same way, we can prove the lower bound $a_n \geq \min\{4, a_1, a_2\}$.

These bounds show that both $\liminf a_n$ and $\limsup a_n$ are finite and positive. Furthermore we deduce from $a_{n+1} = \sqrt{a_n} + \sqrt{a_{n-1}}$ that
$$
\liminf a_n \geq   2\sqrt{\liminf a_n},\qquad \limsup a_n \leq 2\sqrt{\limsup a_n}
$$
The conclusion follows because $x \geq 2\sqrt{x} \implies x \geq 4$ as well as $x \leq 2\sqrt{x}\implies x \leq 4$ for every positive real number $x$.


What does $y_i=f(\sum_j w_{ij}y_j)$ mean (in an artificial neuron model)? While trying to understand artificial neural networks, I came upon an equation for finding the net input of an artificial neuron. Can someone explain this to me and what it means? Here is the original article

$$y_i=f\left(\sum_j w_{ij}y_j\right)$$

Thanks in advance, ell.

So hopefully you understand what the $\sum$ symbol means, it's just a summation across all of those indices. In this case we have our weight vectors

$w_1,w_2,\dots,w_n$

and our vector $y$. The neuron then is going to compute the new value of $y_i$, based on a dot product of $w_i$ with $y$. Hence

$\sum_{j} w_{ij}y_j$

is just dot product of $w_i$ and $y$, where both are $n$-dimensional vectors.

Then $f$ is an activation function, it could be a preceptron or it could be a linear node. So it may output the value of the net, or it may output a 0 or a 1 based on the value of the summation. Or it could have some non-linear neuron it all depends on the network.

What you are looking at is a simplistic model of the neuron. In particular, the inputs are dendrites and the outputs are the synapse.


Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove $\exists b \in G $ that $a = b^m$. Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove that $\exists b \in G $ that $a = b^m$.

Well, I'm not sure how to approach this, any hints are welcomed.

Due to gcd($m,n$)=1, $\exists$ $c,d \in \mathbb{Z}$ $|$ $1=cm+dn$ $\Rightarrow$ $a^1=a^{cm+dn} \Rightarrow a^{cm+dn}=a^{cm}a^{dn}=(a^{c})^m (a^n)^{d}=(a^c)^m e^d=(a^c)^m$.

Now, define as $b:=a^c$, and that's all!!

Since $a^n=e$, then $G$ is a cyclic group generated by a. If $(m,n) = 1$,then $(-m,n)=1$ and $a^{-m}$ is another generator of the group $G$. Denote $k := -m$.

For the proof, suppose by contradiction. Suppose that $a^k$ is not a generator of $G$. So, the order of the element $a^k$ is not $n$, i. e. there exists a natural number $n_1 < n$ such that $({a^k})^{n_1}=a^{kn_1}=e$. Since the order of the element $a$ is $n$, then $kn_1=nq$ and since $(n,k)=1$, then $n\mid n_1$, i. e. $n_1\ge n$. This is a contradiction and as a result $a^k$ is a generator of the group $G$: $a^k=a^{-m}=b$, i. e. $b^m=a$. Choose $b = a^{-m}$.


Find a formula for a sequence of number A sequence starts at $n=1$: $\{1, 4, 13, 40, 121, 364... \}$.

Find an explicit formula that generates these numbers.

Thanks a lot!

A good technique to employ on such problems is to calculate the differences between consecutive terms. Here you get:

$$\begin{array}{rcc}
n:&1&2&3&4&5&6\\
a_n:&1&4&13&40&121&364\\
a_n-a_{n-1}:&&3&9&27&81&243
\end{array}$$

The numbers in the last line should look very familiar: they’re consecutive powers of $3$. It appears that $a_n=1+3+3^2+\ldots+3^{n-1}$; to get your answer, just find the sum of the geometric series.

Since you included the (recursion) label, you might also note that since the bottom line of the table seems to show that $a_n-a_{n-1}=3^{n-1}$, the sequence is apparently generated by the recurrence and initial datum

$$\left\{\begin{align*}
&a_1=1\\
&a_n=a_{n-1}+3^{n-1}\text{ if }n>1\;.
\end{align*}\right.$$

(The recurrence could just as well start with $a_0=0$.)


Hausdorff dimension of graph of function This question came up on an exam

Decide the Hausdorff dimension of the graph of the following function for $x>0$
$$y = \log(1+x)\sin\frac{1}{x}$$

In the course, we only touched upon the subject and the example given in the course was of the Cantor set. How does one go about solving this problem or more specifically, where do you start when you want to decide the Hausdorff dimension of some arbitrary graph $y = f(x)$?

Seems like the dimension is $1$. Note that the word "hence" below often means you have some details to work out:

If $Q$ is a square in the plane let $\ell(Q)$ be the side length of $Q$. If $I$ is an interval let $|I|$ be the length of $I$. Say $G$ is the graph.

If $G\subset\bigcup Q_j$ then $(0,\infty)\subset \bigcup I_j$, where $|I_j|=\ell(Q_j)$. Hence $\sum\ell(Q_j)=\infty$. So $h_1(G)=\infty$.

For $n\in\Bbb Z$ let $I_n=(2^{n-1},2^n]$. Let $G_n=\{(x,f(x))\,:\,x\in I_n\}$.

Suppose $n<0$. Then $\log(1+2^n)\sim 2^n$. So $G_n$ is contained in the union of boundedly many squares of side length $2^n$. Hence $$h_1\left(\bigcup_{n=-\infty}^0G_n\right)<\infty,$$so $$h_\alpha\left(\bigcup_{n=-\infty}^0G_n\right)=0\quad(\alpha>1).$$

On the other hand, $f'(x)$ is bounded for $x>1$. Hence $$h_\alpha\left(\bigcup_{n=0}^\infty G_n\right)\le ch_\alpha((1,\infty))=0\quad(\alpha>1).$$

The graph of $y=f(x)$ where $x\in \mathbb{R}^m$ and $f(x)\in \mathbb{R}^n$ is a set of $\mathbb{R}^{m+n}$. If $f$ is smooth then Hausdorff dimension coincide with the euclidian dimension. Otherwise, I think you need do for each case, considerating coverings and using the definitions and auxiliar results.

A good book is Falconer, Fractal Geometry.


Why is the sequence $x(n) = \log n$ **not** Cauchy? I read in the book Applied Analysis by Hunter and Nachtergale that

  the sequence $x(n)=\log(n)$ is not Cauchy since $\log(n)\to\infty$

But that seems to be irrelevant to the definition of a Cauchy sequence which I understand is as follows:

  A sequence $x(n)$ is said to be Cauchy if for every $\epsilon > 0$  there exists an $N$ such that $\lvert x(m)-x(n)\rvert < \epsilon$ for all $m,n>N$.

This sequence $(\log n)$ seems to meet the definition. So how come it is not considered Cauchy?

Hint. Look at the difference
$$
\log(2n) - \log(n) .
$$

Because it can not converge. "$x_n$ is a Cauchy sequence if and only if it is convergennt".
($x_n$ is a real sequence).


Homogenous equation to higher order ODE Hello I have a quick question in regard to general form of the solution to

$$y^{(4)}-2y^{(3)}+y''=0$$

I had thought to find this solution we would consider

$r^{4}-2r^{3}+r^{2}=0$

which factors as $r^2(r-1)^{2}$

that is we have $r_{1,2}=0$ and $r_{3,4}=1$

So what from what I had thought I knew, I thought this implied a solution of form

$$y=c_1e^{0x}+c_2xe^{0x}+c_3e^{x}+c_4xe^{x}=c_1+c_2x+c_{3}e^{x}+c_{4}xe^{x}$$ with $c \in \mathbb{R}$

But wolfram says it is $$y(x)=e^x(c_{2}x+c_{1}-2c_{2})+c_{4}x+c_{3}$$

So where am I going wrong? Thanks all

PS: I hear it is correct, and I would by wondering about solving

$y^{(4)}-2y^{(3)}+y''=x$

Would I be able to use method of undetermined coefficients for this?

And if so, because r=0 is a double root of the equations, would by assumed form for a particular solution need to be $x^2({A_{o}x+A_{1}})$? If I did it with this, could I get a correct answer? Im  not sure exactly

you have $$y^{(4)} - 2y^{(3)} + y'' = 0 $$ if we set $u = y'',$  we get $$ u''-2u' + u = 0\to u = e^x, xe^x$$  solving $$y'' = e^x \to y' = e^x, xe^x \to y = e^x, xe^x, x^2e^x $$ and solving $$y'' = xe^x$$ you will not add anything new. but  $$y'' = 0  $$ adds $1, x$ to the basis of solutions found already. a fundamental set of solutions is $$\{1, x, e^x, xe^x \}.$$

it should be $$y \left( x \right) ={\it \_C1}\,{{\rm e}^{x}}+{\it \_C2}\,{{\rm e}^{x}
}x+{\it \_C3}+{\it \_C4}\,x
$$
for the particular solution of the second equation make the ansatz $$y_p=Ax^3+Bx^2$$


Set cardinality, function onto, open unit square maps into real number set I have a question in my homework that I have trouble solving it. I'm not sure if I understand the question actually. I'll attach the question below and hope someone could give me any hints.

Consider the open interval $(0,1)$, and let $S$ be the set of points in the open unit square; that is, $S = \{ (x,y): 0 < x, y < 1 \}$. Use the fact that every real number has a decimal expansion to produce a $1-1$ function that maps $S$ into $(0,1)$. Discuss whether the formulated function is onto.

edited: there's a hint in the question "keep in mind that any terminating decimal expansion such as .235 represents the same real number as .2349999... ."

I think the "formulated function" is the $f: S \to (0,1)$. I don't know how to use the decimal expansion concept.

Thanks

Do you know the extended problem of the Hilbert hotel? Suppose you have a hotel with infinitely many rooms numbered $R_1\cdots R_\omega$ which are all full. Suppose infinitely (but countably) many new guests arrive and you have to accommodate them all. The problem you have is very similar if you think of a point in $S=(a,b)$ as being the two sets of guests (the ones in the rooms already and the newly arrived ones) and the points in $(0,1)$ as each being a Hilbert hotel.You will ofcourse have to also deal with repeating decimals but you can manage that by just adding another set of guests to your hotel.

Consider the two sequences of decimal expansions of $x$ and $y$. How do you combine them into one sequence injectively?


Why should a physicist care about measurability of random variables? I'm working with some physicists at the moment, and I made the remark that random variables are, technically, defined as measurable functions on some "background" probability space, hence requiring the choice of two sets and two sigma algebras.   They immediately asked where the term "measurable" came from (I gave the brief history about measuring lengths and areas), and then asked if it had anything to do, in general, with whether or not something was "measurable in a laboratory" (I said I wasn't sure).  They then proceeded to say that it was nonsensical to consider any "non-measurable" random variables because such things are, almost by definition, not "interesting" to the physicist.

So, how can I explain to my physicist colleagues that we actually do care about measurability?  Is there an example of a physical quantity which would otherwise be interesting but which turns out to be non-measurable (in the mathematical sense), and which might justify my mathematical nitpickiness to them?  I'm OK with examples from ordinary probability or stochastic process theory.

This is easily explained

Suppose there are three states of the world a, b c.
And suppose we receive a signal $s$ from $\{a,b,c\}$ to $\mathbb{R}$ such that $s(a)= s(b)> s(c)$.

A random variable is measurable on the algebra generated by $s$ is precisely those that can't distinguish between states $A$ and $b$. So measurability is what you can measure in a lab if what you are measuring depends exclusively on the data given by the signal generating the $\sigma$-algebra.

When there is a finite number of states in a measure space $\Omega$ every algebra of events is defined by a signal $s\colon \Omega\to \mathbb{R}$. A a random variable $f\colon \Omega \to \mathbb{R}$ is measurable only if $f= g\circ s$ for some function $g\colon \mathbb{R} \to \mathbb{R}$. So measirability is precisely measurability based on available data.

With an infinite state space, say $[0,1]$ an event $E\subset [0,1]$ is non-measurable if it is inconceivable for you to describe how you would distinguish in any conceivable laboratory if a state lies in $E$ or its compliment when the signal you receive is $s(a) = a$ for all $a\in [0,1]$. You cannot conceivably imagine an experiment that can answer this question.

Of course, if the physicist wants to integrate a function that is not measurable; then there is no way of doing this for the reasons above.

Let me tell you a story.

Certain dynamic light scattering techniques characterize their samples by finding the underlying variance of the motion distribution. That is to say they collect realizations of a random process. From this they can get things like diffusion coefficients, etc. Problem here is that sometimes the underlying physics of the model might dictate a Cauchy distribution which, as we know doesn't have a variance. A the end of the day they're trying to make sense of what the meaning of their experimentally measured variance means when the underlying model doesn't have a variance term...