Complex Number Question - $|z^{z}|$ Find all possible values of
$$\mid z^{z} \mid$$
using the polar for of $z$.

I have tried putting it into polar form but nothing comes out that seems easy to work with/looks like a reasonable simple answer.

Assuming $\;\arg z=\theta\;,\;\;x:=z+iy\notin\Bbb R_-\cup\{0\}\,,\;$ and we choose the principal branch of the logarithmic function:

$$z^z=e^{z\,\text{log}\,z}=e^{z\left(\log|z|+i\theta\right)}=e^{\frac x2\log(x^2+y^2)-y\theta+i\left(x\theta+\frac12y\log(x^2+y^2)\right)}\implies$$

$$\implies |z^z|=e^{\frac x2\log(x^2+y^2)-y\theta}$$

Using the polar form:

$$z=|z|e^{i\theta}\implies z^z=|z|^ze^{iz\theta}=e^{z\log|z|+iz\theta}$$

which is the same as before (look after the second equality sign above)

You can write $a^b = e^{b\text{Log}(a)}$.  We need to have some understanding of notation before we go any further:

\begin{align}
\text{Log}(a):& \text{Most general natural logarithm--gives all complex values}\\
\log(a):& \text{Principle logarithm--gives value using the arg from } [0, 2\pi) \text{ (may be same as ln)}\\
\ln(a):& \text{real valued natural logarithm.  Returns a single, real value and it must be that } a \text{ is real and } a > 0
\end{align}

So now we can write $z^z = e^{z\text{Log}(z)}$.  So we first need to find $\text{Log}(z)$:

$$
\text{Log}(z) = a + bi \\
z = e^{a + bi} = r_ze^{\theta_z i + 2\pi ni} \\
e^ae^{bi} = r_ze^{\theta_z i}\rightarrow r_z = e^a \rightarrow a = \ln(r_z) = \ln\left(|z|\right) \\
b = \theta_z + 2\pi n \\
\text{Log}(z) = \ln\left(r_z\right) + (\theta_z + 2\pi n)i
$$

Now we just need to find $z\cdot \text{Log}(z)$.  I'm going to stick with $r_z$ (which remember is really $|z|$) and $\theta_z$, so that we have:

\begin{align}
z =& r_z\left(\cos(\theta_z) + i\sin(\theta_z)\right) \\
z\text{Log}(z) =& r_z\left(\cos(\theta_z) + i\sin(\theta_z)\right)\left(\ln\left(r_z\right) + (\theta_z + 2\pi n)i\right) \\
z\text{Log}(z) =&r_z\left(\cos(\theta_z)\ln(r_z) - \sin(\theta_z)(\theta_z + 2\pi n) + \text{imaginary terms}\right)
\end{align}

Remember $\left|e^{a + bi}\right| = e^a$ because it really is $e^ae^{bi}$ and $\left|e^{bi}\right| = 1$.  Thus the magnitude of $z^z$ would be:

\begin{align}
\left|z^z\right| =& e^{r_z\left(\cos(\theta_z)\ln(r_z) - \sin(\theta_z)(\theta_z + 2\pi n)\right)} = \frac{e^{r_z\cos(\theta_z)\ln(r_z)}}{e^{r_z\sin(\theta_z)(\theta_z + 2\pi n)}} \\
\left|z^z\right| =& \frac{r_z^{r_z\cos(\theta_z)}}{e^{r_z\sin(\theta_z)(\theta_z + 2\pi n)}}
\end{align}

I don't really want to write it any other way, I mean $r_z = |z|$, $r_z\cos(\theta_z) = \text{Re}(z)$, and $r_z\sin(\theta_z) = \text{Im}(z)$, but you still have a $\theta_z$ by itself and it's misleading to write that as $\theta_z = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)$ because the inverse tangent won't give the correct $\theta_z$ in general.  I could write $\theta_z = \arg(z)$, but then why not just write $\theta_z$?


Number of maximum components of a graph once any one vertex is removed Let $G$ be an arbitrary graph with $n$ nodes and $k$ components. If a vertex is removed from $G$, the number of components in the resultant graph must necessarily lie between$\ldots$?

I figured that in worst case number of components would be $k - 1$ if the vertex removed was a component in itself.

For the best case, I reasoned that removal of a vertex from a component might divide the component into two (if the vertex is a kind of a cut-vertex of the component), making the total number of components equal to $k + 1$.

However the answer given says that the number of components lie between $k - 1$ and $n - 1$. I don't understand the $n - 1$ part. Please point me in the right direction.

Removing a single vertex $v$ can cut a component into $\deg(v)$ components, not just two. What if the original graph has $k-1$ isolated vertices and one star? (By a star I mean a complete bipartite graph $K_{1,m}$ for some $m$.)

take a case where all vertices are connected to one vertex and thus removing this vertex will leave n-1 component


Does every convergent infinite series have a closed-form value? Does every convergent infinite series have a closed-form value?

I apologize if this question seems totally crazy to some of you. There's a ton of series that converge, but only a fraction of them have a closed-form value, so how can I even ask such a question? Isn't it obvious that the answer is no?

Well, not so fast. My intention with this question is more like "Does the possibility exist that every convergent infinite series have a closed-form value?"  or have we proven that it's impossible? A "simple" way to prove this would be to find an infinite convergent series and then show that it would be impossible for it to have a closed-form value. But how would you do that?

Sure, there is an enormous amount of series that converge, for which there is not a known closed-form value, e.g. one of the series I asked about earlier, and one may use this as an argument for the answer no, but that's not enough for me.

I hope this provokes some discussion and that you don't think this questions is completely dumb.

I think the definition of "closed-form value" should include that it is expressible as a finite combination of some known constants via some algebraic (or transcendental, but still elementary) formula. Therefore, a series that converges to a number that is not computable does not have a closed form value. Since most real numbers are not computable, most series do not converge to a closed form value.

A number that is not computable is one where there is absolutely no computer program that will output all of the digits of the number, even given infinite time. In fact, since there are only countably many computer programs, only countably many real numbers are computable, even though it is, by definition, difficult to produce even one that is not. Any number with a closed form expression would be computable, and we could compute it by applying the formula.

Depends on what you understand by "closed form". If you mean "expressible in a finite number of (give a list of functions and constants)", absolutely not. For a few examples, the values of Riemann's zeta function:

$\begin{align*}
\zeta(s)
   = \sum_{n \ge 1} n^{-s}
\end{align*}$

are known only for even values of $s$, their values for odd values of $s$ are a complete mystery.

Simpler, even: It is easy to see that there the set of convergent series is non-denumerable, even if you restrict yourself to rational terms; but the set of finite formulas made up of algegraic numbers, a handful of "well-known" trascendentals, and a finite set of functions is denumerable. There must be series with no "closed form".


Rate of convergence of fixed-point iteration in higher dimensions Consider the fixed-point iteration process in $\mathbb{R}^n$.

Given a sufficiently smooth function $f:\mathbb{R}^n\to\mathbb{R}^n$ and an initial value $x_0\in\mathbb{R}^n$, define the iteration sequence $x_{k+1}=f(x_k)$. Suppose that
$$\lim_{k\to\infty}x_k=x^*,$$
then apparently $x^*$ is a fixed point of $f(x)$. I'm familiar with the case in $\mathbb{R}^1$ where the sequence is (generally) linearly convergent with rate
$$\lim_{k\to\infty}\frac{|x_{k+1}-x^*|}{|x_k-x^*|}=|f'(x^*)|<1.$$
And I thought the analogy of this constant $|f'(x^*)|$ in $\mathbb{R}^n$ case would be $\|J_f(x^*)\|$, where $J_f(x^*)$ denotes the Jacobian matrix $(\partial f_i(x^*)/\partial x_j)_{n\times n}$ and $\|\cdot\|$ the operator norm induced by vector norm.

But the results of my numerical experiments proved me wrong, and I found the following claim on some website:
$$\lim_{k\to\infty}\frac{\|x_{k+1}-x^*\|}{\|x_k-x^*\|}=\rho(J_f(x^*))<1,$$
where $\rho$ the spectral radius of a matrix. Indeed the claim fits well my experiment results.

It does surprise me that the rate of convergence is independent of the vector norm, but I could not find a proper proof either by myself or by online materials.

Any help or link on its proof would be appreciated.

As in many similar situations in higher dimensional spaces, it helps to look at the simplest case where the function can be decoupled. That is, the function requires no interaction between the variables. For two dimensions this is $f(x,y) =(f_1(x),f_2(y))$. For even more simplicity, assume $f(x,y) = (ax,by)$. Now the rate of convergence is obviously dominated by $\max(|a|,|b|)$ which corresponds to the spectral radius.

It's well known that  \ru(J) <=||J|| for every norm. so the the fixed point iteration will be convergence if the second relation be valid. The proof can be found in basic linear algebra books.


Correlation problem involving diagram Here's a difficulty I am facing to understand the solution of this problem.The first pic is the question and the second one is the solution given in the book.
I am not getting the process of getting the marginal density of $Y$.That is how will I find $h(y)$.
Please help

More or less by definition, the marginal pdfs $f_X$ and $f_Y$ satisfy\begin{align*}
f_Y(y) = \int_{-\infty}^{\infty} \;f_Y(y|x)f_X(x)\, dx.
\end{align*}
(I'm calling this essentially a definition, but compare it to the discrete case or consider Fubini's theorem and related results.) Thus
\begin{align*}
h(y)
  &= \int_{-\infty}^{\infty} \;f_Y(y|x)g(x)\, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \;f_Y(y|x)\, dx,
\end{align*}
and churning through the integral as indicated gives $h$.

The joint density $f(x,y)$ is uniform on a region that's symmetric about the $y$-axis:

$$\begin{align}f(x,y)&=g(x)\,f_Y(y|x)\\
&=[-\frac{1}{2}<x<\frac{1}{2}]\,\left([-\frac{1}{2}<x<0][x<y<x+1]+[0<x<\frac{1}{2}][-x<y<-x+1]\right)\\
&=[-\frac{1}{2}<x<0][x<y<x+1]+[0<x<\frac{1}{2}][-x<y<-x+1]
\end{align}$$

using Iverson bracket notation. Thus $f(x,y)$ is everywhere either $0$ or $1$. Let $S$ denote the region in the $(x,y)$-plane where $f(x,y)=1$; i.e., $S$ is the shaded region in the posted diagram.

Then
$$h(y)=\int_\mathbb{R}f(x,y)\,dx= \int_{S_y} 1\,dx$$
where $S_y$ denotes the portion of $S$ intersected by a line parallel to the $x$-axis at a given $y$-coordinate.

To evaluate the integral explicitly, consider whether & how a line parallel to the $x$-axis at a given $y$-value intersects $S$. Inspection shows that there are four regions to consider -- three regions where such a line intersects some part of $S$ and therefore contributes to the integral, and the remaining region where such a line misses $S$ and therefore does not contribute to the integral:

If $\frac{1}{2}<y<1$, then such a line intersects $S$ giving a segment of $x$-values in an interval $y-1<x<1-y$, whose length is $(1-y)-(y-1)=2(1-y)$:
$$\int_{S_y}1\,dx = \int_{y-1}^{1-y}1\,dx= 2(1-y).$$
If $0<y<\frac{1}{2}$, then such a line intersects $S$ giving a segment of $x$-values in an interval $-\frac{1}{2}<x<\frac{1}{2}$, whose length is $\frac{1}{2}-(-\frac{1}{2})=1$.
$$\int_{S_y}1\,dx = \int_{-\frac{1}{2}}^{\frac{1}{2}}1\,dx= 1.$$
If $-\frac{1}{2}<y<0$, then such a line intersects $S$ giving two segments of $x$-values, namely, the intervals $-\frac{1}{2}<x<y$ and $-y<x<\frac{1}{2}$, whose lengths are, respectively, $y-(-\frac{1}{2})=y+\frac{1}{2}$ and $\frac{1}{2}-(-y)=y+\frac{1}{2}$, for a total length of $1+2y$:
$$\int_{S_y}1\,dx = \int_{-\frac{1}{2}}^{y}1\,dx + \int_{-y}^{\frac{1}{2}}1\,dx= 1+2y.$$
If $y<-\frac{1}{2}$ or $y>1$, then such a line does not intersect $S$:
$$\int_{S_y}1\,dx =0.$$

NB: Because $f$ is symmetric in $x$ (i.e., $f(-x,y)=f(x,y)$ for all $x,y$), the marginal density $h(y)$ is not needed to answer the questions in the exercise. If $\phi(x,y)$ is any function that's antisymmetric in $x$ (i.e., $\phi(-x,y)=-\phi(x,y)$ for all $x,y$, then
$$\begin{align}E[\phi(X,Y)]&=\int_\mathbb{R}\phi(x,y)f(x,y)\,dx\,dy\\
&=\int_A \phi(x,y)f(x,y)\,dx\,dy +  \int_B \phi(x,y)f(x,y)\,dx\,dy\\
&=\int_A \phi(x,y)f(x,y)\,dx\,dy +  \int_A \phi(-x,y)f(-x,y)\,dx\,dy\\
&=\int_A \phi(x,y)f(x,y)\,dx\,dy -  \int_A \phi(x,y)f(x,y)\,dx\,dy\\
&=0
\end{align}$$
where $A$ and $B$ are the portions of $S$ to the right and left, respectively,  of the $y$-axis.

Thus, $E[XY]=0$ and $E[X] = 0$, so $X$ and $Y$ are uncorrelated (i.e., $E[XY]-E[X]\cdot E[Y]=0-0\cdot E[Y]=0$).

NB: There's a typo in the book's answer, which should say that $X$ and $Y$ are not independent. Without having to find the marginal density $h$, this follows immediately from the fact that the support of $f$ (i.e., the set $S$) is not a Cartesian product. In order to have $f(x,y)=g(x)h(y)$ for all $x,y$, it's necessary that the support of $f$ be the Cartesian product of the supports of $g$ and $h$. If the latter are intervals (as they are here), then the former must be a rectangle, unlike the non-rectangular shape of $S$.


Understanding the proof of $\sup(AB)=\sup A\cdot \sup B$. I need help understanding this:
Let $A, B\subseteq [0,\infty)$ be bounded from above and non-empty sets. Let $AB=\{ab, a\in A, b\in B\}$ Prove that $\sup(AB)=\sup A\cdot \sup B$.

If $\sup A=0 \Rightarrow A=\{0\}$ so $AB=\{0\}$ so the statement is true.
Suppose $\sup A>0$ and $\sup B>0$.
Then for $a\in A$ and $b\in B$: $ab\leq \sup A\cdot \sup B$. $\Rightarrow \sup A\cdot \sup B$ is an upper bound of $AB$.
Let $0<\varepsilon<\sup A\cdot \sup B$.
Then for  $\varepsilon_1=\frac{\varepsilon}{2\sup A}>0$ and $\varepsilon_2=\frac{\varepsilon}{2\sup B}>0$ there exists $a\in A$ and $b\in B$ such that $a>\sup A-\varepsilon_1$ and $b>\sup B-\varepsilon_2$.
$\Rightarrow ab>(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\sup B -\varepsilon +\frac{\varepsilon^2}{4\sup A\sup B}>\sup A \sup B-\varepsilon$

I don't understand this part:
$$(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\sup B -\varepsilon +\frac{\varepsilon^2}{4\sup A\sup B}$$
As far as I know: $$(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\cdot \sup B -\frac{\varepsilon\cdot\sup A}{2\sup B}-\frac{\varepsilon\cdot \sup B}{2\sup A}+\frac{\varepsilon^2}{4\sup A\sup B}$$
Why is $$\frac{\varepsilon\cdot\sup A}{2\sup B}+\frac{\varepsilon\cdot \sup B}{2\sup A}=\varepsilon$$?

I think that it should be $(sup A-\epsilon_2)(sup B-\epsilon_1)$, then the answer falls out.

As $AB=\{xy:x\in A, y\in B\}$, $A$ and $B$ are subsets of positive real numbers such that $\sup A$ and $\sup B$ exists.
Now Assume that $\sup A=a$ and $\sup B=b$.
Then it is easy to see that
$xy\le ab$, for all $xy\in AB$.
So $ab$ is an upper bound for $AB$.
Now let $e>0$, take $ab-e<ab$,
so $b-\frac e a<b$,
since $b$ is supreme of $B$ so there exist some $u\in B$ such that $b-\frac e a<u$.
That is $ab-e<au$ so that is
$ab-\frac e u<a$ and
since $a$ is $\sup A$ so there exist some $v\in A$ such that
$ab-\frac e u<v$ that is,
$ab-e<uv$,
since $vu\in AB$, so there exist an element $uv$ in $AB$ such that
$ab-e<uv$.
Hence $\sup(AB) =ab=\sup A\sup B$.


Prove almost sure convergence related to i.i.d random variables I have following question and stuck at the 1st one.

  $\left\{X_n\right\}$ are i.i.d. random variables with non-zero finite mean. Let $S_n=X_1+X_2+\dots+X_n$. Prove
  1. $|X_n|/n \xrightarrow{a.s.}0$;
  2. $(\max_{1\le k \le n}|X_k|)/n \xrightarrow{a.s.} 0$
  3. $(\max_{1\le k \le n}|X_k|)/(1+S_n) \xrightarrow{a.s.} 0$

I tried to use Borel-Cantelli lemma and Markov inequality, but failed.
1st Try:
It's equivalent to show for any $\epsilon > 0$, we have $P(|X_n|/n > \epsilon \text{ i.o.})=0$

$$\sum_{n=1}^{\infty} P(|X_n|/n > \epsilon) \le \sum_{n=1}^{\infty}E[|X_n|]/n\epsilon=\infty$$
I figured it didn't work probably because i.i.d. condition was not used.
2nd Try: $$\begin{align}P(|X_n|/n > \epsilon \text{ i.o.}) &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n > \epsilon) \\ &= \lim_{m \rightarrow \infty}1-P((\bigcup_{n=m}^\infty |X_n|/n > \epsilon)^c) \\ &=\lim_{m \rightarrow \infty}1-\prod_{n=m}^\infty P(|X_1|/n \le \epsilon)\end{align}$$ Stuck again.

The Markov inequaltiy is too weak in this case. Instead, use the fact that the $X_n$ are all identically distributed and the integral comparison test.

$$\sum \limits_{n = 1}^\infty P(|X_n|/n > \varepsilon) = \sum \limits_{n = 1}^\infty P(|X_1|/\varepsilon > n) \le 1 + \int_0^\infty P(|X_1|/\varepsilon > x) \, dx$$

Now note that for any nonnegative random variable $Y$ the equation
$$\begin{align*}
\int_0^\infty P(Y > x) \, dx &= \int_0^\infty \int_0^\infty I\{y > x\} \, dP^Y(y) \, dx = \int_0^\infty \int_0^\infty I\{y > x\} \, dx \, dP^Y(y)\\
&= \int_0^\infty y \, dP^Y(y) = E[Y]
\end{align*}$$
holds.

I guess following is the answer.
$$\begin{align}P(|X_n|/n > \epsilon \text{ i.o.}) &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n > \epsilon)\\ &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_1|/n > \epsilon)\\ &= \lim_{m \rightarrow \infty}P(|X_1|/m > \epsilon) \\ & \le \lim_{m \rightarrow \infty} E(|X_1|)/n \epsilon \rightarrow 0\end{align}$$


What is the difference between the geometric and trigonometric definition of an angle? I vaguely remember reading that there is a difference between the geometric definition of an angle and the trigonometric definition of an angle. I've tried to search everywhere I can think of but I can't find what I read.

So what I would like to know is:

Is there a difference between the two?
and if there is, what are the two definitions?

I have never heard of such a dichotomy, but I can imagine that one might call "two rays with a common vertex" the "geometric definition of an angle" and the "numeric angle measure" the "trigonometric definition of an angle".
The usual terminology is just "angle" for the figure and "angle measure" for the number.

Geometric right angle = $90^\circ$, trigonometric right angle = $\pi/2$?


A ring with ACC on prime ideals whose spectrum is non-noetherian. I am currently working on the converse of the exercise #12 on chapter 6 of Atiyah-Macdonald's book on commutative algebra.

  The problem is asking whether there is a ring $A$ which satisfies the ascending chain condition on prime ideals (that is, every chain of prime ideals of a ring $A$ stabilizes) such that $\mathrm{Spec}(A)$ is a non-noetherian space.

What I've tried so far was first, struggling with the proof that spec($A$) is noetherian when $A$ satisfies ACC on prime ideals, but I got stuck because I failed to extract ascending chains of prime ideals.

So I tried to find a counterexample. So, I looked up some examples of non-noetherian rings so that I can have non-noetherian spectrum. (It was tough for most cases since non-noetherian rings can have a noetherian spectrum. For example, the integral closure of $\mathbb{Z}$ in $\mathbb{C}$ is not noetherian. However, I couldn't understand whether it's spectrum is noetherian or not.) But, in every cases, I also couldn't find an appropriate counterexample.

Is there any hint or comment about what I should consider? Any help will be appreciated.

Hint. Take $A$ a countable direct product of copies of $\mathbb F_2$.

Perhaps $\mathbb{Z}$ is the easiest counterexample.


Definition of Sobolev space $H^s$ and domain of $-\Delta^s$ The spaces below are on $\partial\Omega$, the boundary of a bounded smooth domain $\Omega$. I read this in the book on page 141.

  Define $H^2 := \{ u \in L^2 \mid (-\Delta u) \in L^2\}$.

And then:

  We can define $H^s$ ($s \in \mathbb{R}$) as the domain of $(-\Delta)^s$. If $\lambda_j, e_j$ are the eigenpairs of $-\Delta$, then
  $$H^s = \{ u \in (C_0^\infty)' \mid  \sum_j\lambda_j^{2s}|(u,e_j)_{L^2}|^2 < \infty \}$$

Now I'm confused, shouldn't the power of the eigenvalues be $s$, not $2s$?

For example take $u \in H^1$. Then $u=\sum (u,e_j)e_j$, so $-\Delta u = \sum \lambda_j(u,e_j)e_j$ in the weak sense, so $\langle -\Delta u, u \rangle = \sum \lambda_j |(u,e_j)|^2 < \infty$. But with the definition of $H^1$ above, we would need $\sum \lambda_j^2|(u,e_j)|^2 < \infty$. Why this discrepancy?

You're looking at the square of the norm.  In the last line of example 4.17, note the square root of the sum of squares.

It's natural to see the squares of the eigenvalues in the square of the norm.  Both the norm and the square are going to be finite, and can be used to define $H^s$, but the square isn't going to be homogeneous (like you're expecting).

EDIT: again, what you're looking at is the square of a norm.  In the example you cite, $\langle -\Delta u, u \rangle$ is the square of a norm on $u$ in $H^{1/2}$ on the boundary.

Consider an analogy in $R^n$ with a symmetric,   positive-definite matrix $A$ with the $A$-norm $\left\|x\right\|_A^2 = x^TAx = \sum \lambda_i (x\cdot u_i)^2$ where the $\lambda_i$ are the (positive) eigenvalues of $A$ and $u_i$ the corresponding eigenvectors.  Note that $\lambda_i=\left(\lambda_i^{1/2}\right)^2$.  Another way to rewrite this is $\left\|x\right\|_A = \left\|A^{1/2}x\right\|$.

EDIT2: Or is the confusion from the interplay between the interior and the boundary?  $\langle -\Delta u, u \rangle$ isn't the square of a norm on $H^1$ on the boundary (but is on the interior).

EDIT3: There is a bounded, linear operator -- the trace operator -- that maps from $H^1(\Omega)$ to $H^{1/2}(\partial\Omega)$.  It is the operator that finds an extension of a function defined on the interior of a domain to its boundary.  The integration by parts formula you cite is part and parcel of the proof that such a map exists.

Note that a function $u\in H^1(\Omega)$ is not guaranteed to have an extension $u|_{\partial\Omega}$ in $H^1(\partial\Omega)$.  That is, the values of the extension on the boundary are slightly less regular than the values of the function on the interior.  In the other direction, the interpolation construction in Oden's book is necessary to show that the extension is in $H^{1/2}(\partial\Omega)$ and not just $L^2(\partial\Omega)$.


How do you prove that the invariant subspaces of a diagonal matrix are direct sums of eigenspaces? Given $A$ an $n \times n$ diagonal matrix with distinct non-zero values on the diagonal ($A_{ii} = A_{jj} \rightarrow i = j$)
Let $S$ be the set of eigenspaces of $A$.
How do I show that for any subspace $U$ invariant under $A$ there exist some $T \subset S$ such that
$$
U =  \sum_{V \in T} V
$$
(Where by sum I mean the direct sum, not sure which notation makes the most sense here really.)
I think I understand a few ways I could conceptualize the problem, but when I try to write it down I have a hard time keeping it from getting out of control, and it seems like maybe I am overcomplicating it.
One idea I have tried is that for a subspace $U$ with dimension $d$ I can take a basis $f$ for that space. I am not sure how to "state it" but it seems to me that the subspace $U$ is invariant if and only if the basis $f$ can be represented with no more than $d$ of the eigenvectors. I thought I would take a vector $x \in U$ rewrite it with $f$ and again with the $\textbf 1$ unit eigenvectors. Then show that $Ax \in U$ if it can be represented by $f$... I am not quite sure how to "close the loop" on this idea though.
My second idea was to show that for an invariant subspace $U$ with dimension $d$ and any point $x \in U$ then for any positive integer $k$ we have $A^kx \in U$. Then I thought I could show that if there are more than $d$ of the unit eigenvectors in the representation of $x$ then the points $Ax$, $A^2x$ ... $A^{d + 1}x$ would form $d + 1$ linearly independent points in $U$ contradicting the original definition of $d$ as the dimension of $U$. Yet I ended up getting totally lost trying to show that the points are linearly independent.
Is there a simpler approach that I am missing? Is there any easy way to complete one of these that I missed?

It is unnecessary to require all of the diagonal entries to be nonzero. Also, with a slight modification to the definition of the subspace $W$, this proof will work for any matrix/operator with $n$ distinct eigenvalues, not just diagonal matrices.
You have a decomposition into $n$ eigenspaces (since all eigenvalues are distinct):
$$V = S_{1} \oplus \cdots \oplus S_{n}$$
where the $S_{i}$ are the eigenspaces (each 1-dimensional). We will write $\lambda_{j}$ for the eigenvalue associated with $S_{j}$, and note that these eigenspaces are orthogonal (when the matrix/operator is diagonal; the only place orthogonality is used is the definition of the subspace $W$ below).
Assume that $U$ is an invariant $d$-dimensional subspace. We can prove our result by induction on $d$: If $d=1$, then $U$ is clearly an eigenspace.
If the result holds for all subspaces of dimension less than $d$ then we can find a $1 \leq j \leq n$ so that $U$ is not contained in the hyperplane $H= \sum_{i \neq j} S_{i}$.
Then $U \cap H$ has dimension $d-1$ and is invariant (since $H$ and $H$ are both invariant), thus $U \cap H$ is a direct sum of eigenspaces by our inductive hypothesis.
Finally, you can argue that $U = S_{j} \oplus (U \cap H)$ using a dimension argument. We only need to show that $U \cap S_{j} \neq \{0\}$.
Put $W = U^{\perp} \cap H$; since $H$ is a sum of eigenspaces, and $U \cap H$ is a sum of eigenspaces, $W$ is also, so it is clear that $W$ is invariant under $A$. Then
$H = (U \cap H) \oplus W$ and
$V = S_{j} \oplus (U \cap H) \oplus W$.
If we have $u \in U \setminus H$ then $u$ can be written uniquely
$u = s + h + w$, where $s \in S_{j}\setminus\{0\}$, $h \in H \cap U$, $w \in W$. If $w = 0$ then $s \in U\cap S_{j} \setminus\{0\}$ and we are done, so assume $w \neq 0$; also, since $h \in U$ we can replace $u$ with $u-h$ if necessary to WLOG put $u = s+w$ with $s,w$ both nonzero.
Now $Au = As + Aw = \lambda_{j}s + w^{\prime} \in U$,
where $w^{\prime} \in W$ and (since eigenvalues are distinct) $w^{\prime} \neq \lambda_{j} w$.
Then
$$\lambda_{j} u - Au = (\lambda_{j}w - w^{\prime}) \in U,$$
but this is impossible since
$(\lambda_{j}w - w^{\prime}) \in W \setminus \{0\}$ which has trivial intersection with $U$.
Therefore $s \in (S_{j} \cap U)\setminus\{0\}$, so $S_{j} \subseteq U$, and $U = S_{j} \oplus (U \cap H)$ is a sum of eigenspaces.

A nearly correct completion of my second approach.
Suppose $U$ is an invariant subspace under $A$ then ...
Let $d$ be the dimension of $U$.
Suppose also that $U$ is not spanned by a set of $d$ eigenvectors then ...
Let $v \in U$ such that $v$ has the maximum number of non-zero coefficients in it's representation relative to the basis $e$.
If the number of non-zero elements of $v$ is not greater than $d$ we get a contradiction, because either:

All vectors in $U$ have a subset of the same non-zero coefficients, and thus $U$ is actually spanned by $d$ eigenvectors.

There exists another vector in $U$ with at least one non-zero coefficient that corresponds to a zero coefficient in $v$, but then there is a linear combination of these vectors with more non-zero elements than $v$.

Therefore $v$ has at least $d + 1$ non-zero elements.
$X = \{ v, Av, A^2v \; ... \; A^dv \}$ is a set of $d + 1$ vectors in $U$ because $U$ is an invariant space.
The vectors of $X$ are linearly dependent because the dimension of $U$ is $d$.
Let $\alpha$ be the coeifficients of $v$ relative to $e$ the simple unit vectors. So that $v = \sum \alpha_i e_i$.
Then $A^nv = \sum \alpha_i (\lambda_i)^n e_i$.
Form a new $d + 1 \times d + 1$ matrix $M$ with the non-zero coefficients of $A^nv$ forming each column. That is $M_{ij} = \alpha_{k_i} (\lambda_{k_i})^j$ where $k$ indexes the non-zero elements of $v$.
$$
M = \begin{bmatrix}
\alpha_{k_1} & \alpha_{k_1} \lambda_{k_1} & ... & \alpha_{k_1} (\lambda_{k_1})^{d + 1} \\
\alpha_{k_2} & \alpha_{k_2} \lambda_{k_2} & ... & \alpha_{k_2} (\lambda_{k_2})^{d + 1} \\
. & . & & . \\
. & . & & . \\
. & . & & . \\
\alpha_{k_{d + 1}} & \alpha_{k_{d + 1}} \lambda_{k_{d + 1}} & ... & \alpha_{k_{d + 1}} (\lambda_{k_{d + 1}})^{d + 1} \\
\end{bmatrix}
$$
We can see that M is the product of a Vandermonde matrix and a diagonal matrix, and therefore it has rank $d + 1$. But then the columns are linearly independent and the vectors of $X$ are linearly independent, contradicting the earlier contrary conclusion.
Therefore the supposition that $U$ is not spanned by a set of $d$ eigenvectors is incorrect.
All together we have that any subspace $U$ invariant under $A$ is spanned by a set of $d$ eigenvectors.


Why are the number of pivot variables equal in A and the transpose of A? This was a fact used to prove the column rank of A was equal to the row rank of A in my textbook.

But why is this true?

The number of pivots of $A$ is its column rank, say $k$. The
column rank of $A$ is equal to the row rank of A, which is the column rank of $A^T$.

Hence, $A^T$ must have the same number of pivots as $A$.


Proving convergence of an improper integral If $f$ is a non negative-valued continuous function with domain $[1,\infty)$ and $\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$ exists, then prove the improper integral $\int_{1}^\infty f(x)\,dx$ converges.

I know that for the first part to have a limit that exists, it must mean that it itself is convergent since it is monotone and bounded by the limit. Not sure where to go from there.

Let $\lim_{n\to \infty} \int_1^n f = L.$ For $x > 0,$ we have by the nonnegativity of $f$ that

$$\tag 1 \int_1^{F(x)} f \le \int_1^{x} f \le \int_1^{F(x) + 1} f.$$

Here $F$ denotes the floor function. As $x\to \infty,$ both $F(x), F(x) + 1\to \infty$ through integer values. Hence both the left and right side of $(1)$ tend to $L$ as $x \to \infty.$ By the squeeze theorem, the middle term must also $\to L.$ That is the very definition of $\int_1^\infty f = L.$

Note that $\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$ exists, hence, let $N=\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$, for some real $N$. Also, since $f$ is continuous on the domain $[1,\infty)$,

$$\int_{1}^\infty f(x)\,dx=\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx=N$$

In this case, $\int_{1}^\infty f(x)\,dx$ converges to $N$.

Note The following definitions of improper integrals of Type 1:

$(a)$ If $\int_{a}^n f(x)\,dx$ exists for every number $n\ge$$a$, then $\int_{a}^\infty f(x)\,dx=\lim\limits_{n \to \infty} \int_{a}^n f(x)\,dx$ provided that this limit exists (as a finite number).

$(b)$ If $\int_{n}^b f(x)\,dx$ exists for every number $n\le$$b$, then $\int_{-\infty}^b f(x)\,dx=\lim\limits_{n \to -\infty} \int_{n}^b f(x)\,dx$ provided that this limit exists (as a finite number).

The improper integrals $\int_{a}^\infty f(x)\,dx$ and $\int_{-\infty}^b f(x)\,dx$ are called convergent if the corresponding limits exist and divergent if the limits do not exist.


Calculate all parameters of a triangle from some known partial lengths and angles I have the problem described in this image (not to scale): http://i.imgur.com/owWVUmj.png

A (partial cathetus), B (small, vertical cathetus) and C (which forms a 90º angle with hypotenuse) are known lengths (A=3, B=8, C=2 meters, inches, or whatever). I need to obtain the exact values of all sides and angles. The most important length I need is X, but I could calculate it myself once I know the at least one of the angles or lengths.

Could someone please tell me how to calculate at least one of the sides or angles?

Thank you very much.

Let $Y$ be the rest of the leg on which $A$ lies. Then
$$\begin{align}
\frac{C^2}{Y^2} = \frac{B^2}{B^2 + (A+Y)^2}
&\iff B^2Y^2 = C^2(B^2 + A^2 + 2AY + Y^2)\\
&\iff (B^2 - C^2)Y^2 - 2AC^2Y - C^2(A^2 + B^2) = 0.\end{align}$$

Using the quadratic formula,

$$\begin{align}
Y & = \dfrac{2AC^2 \pm \sqrt{4A^2C^4 + 4(B^2-C^2)C^2(A^2+B^2)}}{2(B^2-C^2)}
\\ & = \dfrac{AC^2 \pm \sqrt{A^2C^4 + A^2B^2C^2 + B^4C^2 - A^2C^4 - B^2C^4}}{B^2-C^2}
\\ &= \dfrac{AC^2 \pm C \sqrt{A^2B^2 + B^4  - B^2C^2}}{B^2-C^2}
\\ &= \dfrac{AC^2 \pm BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}.
\end{align}$$
We know $B>C$ since $C$'s triangle is similar to the overall triangle and embedded in it, so the radical is always real.  Also $BC \sqrt{A^2 + B^2 - C^2} > ABC > AC^2$, and $Y$ is positive, so the plus case gives us the only solution:

$$Y = \dfrac{AC^2 + BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}.$$

Then $\dfrac{Y}{A+Y} = \dfrac{X}{B}$, so $X = \dfrac{BY}{A+Y}$.

May be this can help you in simplifying your problem


Proof that $2^{ab} - 1$ is divisible by $2^a - 1$ Could you please comment on the correctness and quality of the following proof?
I'm a bit suspicious of the line $n = ab, a = 1, r = 2^a$. Should it be $r = 2$? And are there any "off by one errors? E.g. $n = ab$ vs $n = ab -1$?
To prove:
For $a, b > 1$
$2^{ab} - 1$ is divisible by $2^a - 1$
Explanation:
$2^{ab} - 1$ can be expressed as $\sum_{i=0}^{ab-1}2^i$
Considering the formula for the sum of a geometric series
$S_n = a \frac{r^n-1}{r-1}$
with
$n = ab, a = 1, r = 2^{a}$
we have
$S_n = \frac{(2^a)^b-1}{2^a-1}$
From which it can be seen that $2^a - 1$ is indeed a factor of $2^{ab} - 1$.
$\square$

I'm going to be picky:
You state that $2^{ab}-1=\sum_{k=0}^{ab-1} 2^k$ but never actually use it. So leave it out; it just confuses things.
You should state explicitly that $S_n = \sum_{k=0}^{b-1} (2^a)^k$; as it is written I use confusing it with $\sum_{k=0}^{ab-1}=2^{ab}-1$ which is and entirely different completely irrelevant sum-- I was confused by the first line.
Then I'd like the, admittedly trivial but key, statement that: $S_n$ is an integer (as a finite sum of integers).

I.E. the proof should be:  $S_n= \sum_{k=0}^{b-1} (2^a)^k$ is integer. But $S_n = \frac {(2^a)^b -1}{2^a - 1}$.  So $2^a - 1| (2^a)^b - 1= 2^{ab}-1$.

Otherwise the proof is good.
======
Although......
For the purpose of a stronger proof.
If we can prove that $x-1$ always divides $x^n - 1$ we will be done by letting $x = 2^a$ and $n = b$.
And the formula for the geometric series states precisely that:  That $\sum_{k=0}^{n-1} x^k$ which is clearly an integer is equal to $ \frac{x^n-1}{x-1}$.  So $x-1|x^n -1$.
Of course, it could be that this exercise is expecting you to prove the formula for a geometric series.
Which you should be prepared to do.
Pf:  $(x- 1)(\sum_{k=0}^{n-1} x^k) = x\sum_{k=0}^{n-1}x^k - \sum_{k=0}^{n-1} x^k=$
$\sum_{k=0}^{n-1} x^{k+1} - \sum_{k=0}^{n-1} x^k=$
$\sum_{k=1}^n x^k - \sum_{k=0}^{n-1} x^k=$
$ ((\sum_{k=1}^{n-1} x^k) + x^n)- (x^0 + \sum_{k=1}^{n-1} x^k)=$
$x^n - 1$.
So $\sum_{k=0}^{n-1} x^k = \frac {x^n - 1}{x-1}$.
(Assuming, of course, $x \ne 1$... Obviously if $x = 1$ then $(1-1)\sum_{k=0}^{n-1} 1^k= 0*n = 0 =1^n-1$ which is fairly pointless)
====
More interesting is proving that if $2^m -1$ is not divisible by $2^a-1$ unless $m$ is divisible by $a$.

If we pick the common ratio to be $2^a$,
\begin{align}
\sum_{i=0}^{ab-1}2^i = \sum_{t=0}^{b-1}\sum_{i=0}^{a-1}2^{at+i} =\sum_{t=0}^{b-1}(2^a)^t \left(\sum_{i=0}^{a-1}2^i\right)
\end{align}
the first term is $\left(\sum_{i=0}^{a-1}2^i\right)$ and we have $b$ terms in total.
That is $$2^{ab}-1=\sum_{t=0}^{b-1}(2^a)^t \left(2^a-1\right)$$
$$\frac{2^{ab}-1}{2^a-1}=\sum_{t=0}^{b-1}(2^a)^t$$
Hence $2^a-1 | 2^{ab}-1$
I usually prove it in the following way:

$x^b-1$ is clearly divisible by $x-1$ since $1^b-1=0$.

Now, we just have to let $x=2^a$


Let $A,B \subset \mathbb R$ with $m^*(A)=0$. Prove $m^*(A \cup B)=m^*(B).$ Let $A,B \subset \mathbb R$ with $m^*(A)=0$. Prove $m^*(A \cup B)=m^*(B).$

$m^*(A \cup B)=m^*(A) + m^*(B)$ if A and B are disjoint.

In general, $$m^*(A \cup B) \leq m^*(A) + m^*(B)=0+m^*(B)=m^*(B)$$ but $m^*(A \cup B) < m^*(B)$ is impossible so $m^*(A \cup B)=m^*(B).$

Is that valid or am I missing something?

First, since $m^*(A) =0$. Then, $$m^*(A \cup B) \leq m^*(A) + m^*(B) \leq m^*(B).$$
It suffices to show that $m^*(B) \leq m^*(A \cup B)$. We need to consider three cases.

If $A = \emptyset$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.

If $\emptyset \neq A \subseteq B$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.

If $\emptyset \neq A \nsubseteq  B$, then $B \subset A \cup B$ and $m^*(B) \leq m^*(A \cup B)$.

As you can see all the three cases gives $m^*(B) \leq m^*(A \cup B)$ as required.


A question about the Theta functions ( I have seen this question: How to start with elliptic functions but it has no answer. )
In Mathematica ( this is not a Mathematica question ) I calculated the following sum:
Sum[E^(-\[Pi] n^2 x), {n, 1, Infinity}]

The answer is:
1/2 (-1 + EllipticTheta[3, 0, E^(-\[Pi] x)])

In my Mathematics studies I never came across this EllipticTheta function.
Which standard university mathematics course introduces this function?
What is your suggested book on the subject and in which standard ( graduate ? ) course is it taught?

EllipticTheta[1] to EllipticTheta[4] are also called Jacobi theta functions:
$$
\vartheta_3(z,q) = \sum _{n=-\infty}^{\infty }{q}^{{n}^{2}}{{\rm e}^{2\,inz}}
$$
So Bellman may be OK.  The book on Ramanujan theta functions probably assumes you already know about Jacobi theta functions.
Probably Mathematica has documentation on its functions ... Here is the reference that Maple's documentation recommends:

Chapter 16, "Jacobian Elliptic Functions and Theta Functions" of Handbook of Mathematical Functions edited by Abramowitz and Stegun


Finding explicit and/or recursive formula for the simple sequence $6, 15, 18, 19, ...$ Given the sequence $6,15,18,19,...,$ what is the recursive/explicit formula?

This is clearly not an arithmetic or geometric series but it does seem to satisfy:

\begin{align*}
u(1) &= 6 \\
u(n) &= u(n-1) + 9^{1/n}
\end{align*}

A general question:  Given only a few terms I would guess that my solution is not unique.  Is there any way to prove this other than by counter-example?

Are there any good strategies for determining these formulae other than intuition and trial&error?

Thank you,
Chris

A sequence is just an ordered list of things; there doesn't necessarily have to be a single "nice" formula that goes with it. To show that your solution is not unique, just start listing other numbers. If we trust your recursive formula for the first few terms, the sequence could be
$$
  6,15,18,19,0,0,0,0,0,\dotsc
$$
given by the rule
$$\begin{cases}
u(1) = 6 \\
u(n) = u(n-1) + 9^{1/n} \quad\forall x \in \{2,3,4\}\\
u(n) = 0 \quad\text{otherwise}
\end{cases}$$

Note that is $+9,+3,+1$. You can complete as you want... For example $+9,+3,+1$ recursively, or $a_n=a_{n-1}+3^{4-n}$, etc


Let $f : A\subset \mathbb{R}^{n+1} \to \mathbb{R}$, what does mean that $f$ is a submersion? I am trying to answer the following question:

Let $M_a := \{ (x^1,\ldots,x^n,x^{n+1}) \in \mathbb{R}^{n+1} : (x^1)^2 + \cdots +(x^n)^2 - (x^{n+1})^2 = a\}$. For which values of $a$, $M_a$ is a submanifold of $\mathbb{R}^{n+1}?$

I have to use the regular value theorem, but I think I don't know in fact how to do it... I thought the following:

Let $f: \mathbb{R}^{n+1} \to \mathbb{R}$ defined as:
$f(x^1,\ldots,x^{n+1}) = (x^1)^2 + \cdots+ (x^n)^2 - (x^{n+1})^2.$ Then if $f$ has constant rank, then every set level is a manifold... How to relate this with $a$? I am not seeing the connection...

If $f$ has constant rank, you have a submersion and $a$ does not matter. But it is not the case (as noticed by Tsemo Aristide):
$$\nabla f(x^1,...,x^{n+1})=(2x^1,...,2x^n,-2x^{n+1}).$$
Therefore, $(0,...,0)$ is not a regular point.

In the theorem, the dependence of the point $a$ appears when you need to check regularity (in this case): $M_a$ is a submanifold if $\nabla f(x)$ is non-zero for all $x\in M_a$. So, if you change $a$, you change the points where you have to check the condition on the gradient/differential.

Note that this argument guarantee that $M_a$ is a submanifold for $a\neq 0$.

You have to compute $df =2x_1dx_1+...2x_ndx_n -2x_{n+1}dx_{n+1}$. Let $a$ be a non zero real. for every $x=(x_1,...,x_n)\in f^{-1}(a)$, there exists $i$ such that $x_i\neq 0$. Let $e_i$ be the vector of $R^{n+1}$ whose $i$-coordinate is $1$ and its other coordinates are $0$.  $df_x(0,..,0,1,0..)=x_i\neq 0$. Thus $df_x$ is surjective, and $f^{-1}(a)$ is a manifold.

If $a=0$, $0_{n+1}$ the vector of $R^{n+1}$ whose coordinates are $0$ is in $f^{-1}(0)$. $df_{o_{n+1}}=0$. Thus the tangent space of $f^{-1}(0)$ is not defined at $0_{n+1}$. Thus $f^{-1}(0)$ is not a submanifold.


How to show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ So as title says I wanna show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$

So I know that the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ is $x^{4}-6x^{2}+7$ the zeroes of which are $\pm \sqrt{3-\sqrt2}, \pm \sqrt {3+\sqrt 2}$.

So far I have got as far as:

$\sqrt{3-\sqrt2}\sqrt {3+\sqrt 2} = \sqrt7$.

Any tips on how to find $\sqrt7$ in terms of the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ ?

Since $2,7$ and $2\times 7$ are all nonsquares
in $\mathbb Q$, we see that $K={\mathbb Q}[\sqrt{2},\sqrt{7}]$ has
degree $4$ over $\mathbb Q$. If we put
$\theta=\sqrt{7}+\sqrt{3+\sqrt{2}}$, we have

$$
\theta^2+4=14+\sqrt{2}+2\sqrt{7\times(3+\sqrt{2})}=
\sqrt{2}+2\sqrt{7}\theta \tag{1}
$$

So $\theta$ is a root of the polynomial
$P=X^2-(2\sqrt{7})X+4-\sqrt{2}$. The discriminant of $P$ is
$d=3+\sqrt{2}$. If $d$ were a square in $K$, we would have
$x,y\in{\mathbb Q}[\sqrt{7}]$ such that $(x+y\sqrt{2})^2=3+\sqrt{2}$,
hence $x^2+2y^2=3,2xy=1$. This yields $y=\frac{1}{2x}$,
$x^2+\frac{2}{4x^2}=3$, so $2x^4-6x^2+1=0$ and hence
$x^2=\frac{3\pm\sqrt{7}}{2}$ which can be seen to be impossible in
${\mathbb Q}[\sqrt{7}]$.

We have therefore shown that $d$ is a nonsquare in $K$, so
$[K(\theta):K]=2$. Let $L={\mathbb Q}(\sqrt{7},\theta)$.
By (1) above, $L$ contains $\sqrt{2}=\theta^2-2\sqrt{7}\theta+4$,
so $L$ contains $K(\theta)$. We deduce $L=K(\theta)$ as the other
inclusion is trivial. Thus $[L:{\mathbb Q}]=8$.

Finally, if we had $\sqrt{3+\sqrt{2}}\in M$ where
$M={\mathbb Q}[\sqrt{3-\sqrt{2}}]$, we would have
$\sqrt{7}\in M$ as explained in the OP, so $\theta\in M$
and eventually $M=L$, contradicting $[L:{\mathbb Q}]=8$.

Let us try this. Suppose $\alpha=\sqrt{3+\sqrt{2}}$. Then $\alpha^{2}=3+\sqrt{2}$, and $\alpha^{4}=11+6\sqrt{2}=11+6(\alpha^{2}-3)$ and hence $\alpha$ is a root of the polynomial $p(X)=X^{4}-6X^{2}+7$. This polynomial is irreducible, simply by attempting quadratic formula. Thus the degree $[\mathbb{Q}(\sqrt{3+\sqrt{2}}):\mathbb{Q}]=4$. Now suppose $\beta=\sqrt{3-\sqrt{2}}\in\mathbb{Q}(\sqrt{3+\sqrt{2}})$. Then

\begin{equation}
[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]
\end{equation}

Using the procedure similar for $\alpha$, you can show that $[\mathbb{Q}(\beta):\mathbb{Q}]=4$. Therefore $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]$, i.e. $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$. But any field isomorphism $\phi$ that maps $\alpha$ to $\beta$ cannot work: suppose $\phi(\alpha)=\beta$ up to $\pm$ sign.
\begin{equation}
\phi(\alpha)^{2}=3+\sqrt{2}=\beta^{2}=3-\sqrt{2}\Rightarrow \sqrt{2}=-\sqrt{2}
\end{equation}
a contradiction. Therefore our supposition is false.


Proof involving Fermat's Theorem. I believe I know the answers to most of $a$.

i. the quantifier is signifies for all and in this problem it is the symbol $ \forall $.

ii. the variables are $n, a, b$ and $c$ with $n>2, c> 0, b>0$ and $c>0$.

iii. the predicate is that $a^n + b^n\ne c^n$. (not sure about this)

However, I am lost on part b, because I do not understand how this has to do with Fermat's theorem because Fermat's theorem is focused around numbers.

An implication "$P \implies Q$" is always valid if $P$ is false. This is because an implication is somewhat like a promise: "given that the left side is true, I guarantee the right side is true". Thus if the left side is not true, all bets are off, because no one can give you a proof of the left side, and so you can promise whatever you want. "Given that pigs can fly, I can prove Fermat's Last Theorem": no one will ever be able to disprove this claim, because first they would have to demonstrate pigs flying.

the statement is correct but it doesn't prove the Fermat Last Theorem. For to prove it, it should come from well established axioms,lemmas, and theorems.. truth of the statement does not necessarily imply proof of the implication.


Markov Chain $3$-step state $X_3 \mid X_0=1$ conditional probabilities, mean and variance. Consider the Markov chain with one-step transition probability matrix:
$$P=
\begin{bmatrix}
1/3 &1/3  &1/3 \\
3/4 &1/8  &1/8\\
1/4 &1/2  &1/4
\end{bmatrix}$$ and initial probabilities $p_{0} = p_{1} = p_{2} = 1/3.$

Compute the three-step transition probability matrix.
Find $ P (X_{0} = 1, X_{3} = 0).$
Find $P (X_{3} = 0).$
Find $P (X_{0}= 1\mid X_{3}= 0)$.
Find $E[X_3 \mid X_0 = 1]$ and $Var(X_3 \mid X_0 = 1).$

I need help to solve to solve part 5.

In part 1. you have calculated the matrix $$P^{(3)}=P^3=(p^{(3)}_{ij})_{i,j \in \Bbb X}$$ where $\Bbb X=\{0,1,2\}$. So, for $k\in \Bbb X$ you have that $$P(X_3=k\mid X_0=1)=p^{(3)}_{1k}$$ (these will be the entries of the first line of the matrix $P^{3}$). These values completely determine the pmf of the random variable $X_3 \mid X_0=1$ and the computation of $E[X_3 \mid X_0=1]$ and $Var(X_3 \mid X_0=1)$ are then standard.

Just to compare your result: $$P^3=P=
\begin{bmatrix}
193/432 & 529/1728 & 427/1728\\
\dots & \dots &\dots\\
\dots &\dots  & \dots
\end{bmatrix}$$ Hence $$E[X_3\mid X_0=1]=0\cdot\frac{193}{432}+1\cdot\frac{529}{1728}+2\frac{427}{1728}$$ and $$Var(X_3\mid X_0=1)=E[X_3^2\mid X_0=1]+\left(E[X_3\mid X_0=1]\right)^2$$ with $$E[X_3^2\mid X_0=1]=0^2\cdot\frac{193}{432}+1^2\cdot\frac{529}{1728}+2^2\frac{427}{1728}$$Can you finish the calculations?


Find the area between $r = 1 - \sin θ$ and the circle $r = \sin θ$? I have found the points of intersection but am not sure how I would split this into two integrals or why?

As you can see from Desmos, if you want the area between both we can do a little bit of symmetry to just get the area of one of the lobes, giving us
$$I = 2\cdot\frac{1}{2}\int\limits_{\text{first lobe}} r^2\:d\theta = \int_0^{\frac{\pi}{6}}\sin^2\theta\:d\theta + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(1-\sin\theta)^2\:d\theta$$
from looking at the bounds in the picture. Can you take it from here?

Use polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$ to change your problem to Cartesian coordinates. Then you can use a geometrical approach, even integration, to find the number.
Edit: You could try making graphs on dependent variable $r$ for independent variable $\theta$ without going to Cartesian (since the equations are so simple).


L'hospital rule for two variable. How to use L'hospital rule to compute the limit of the given function

  $$\lim_{(x,y)\to (0,0)} \frac{x^{2}+y^{2}}{x+y}?$$

There is no L'Hopital's Rule for multiple variable limits. For calculating limits in multiple variables, you need to consider every possible path of approach of limits.

What you can do here:

Put $x=r\cos\theta$ and $y=r\sin\theta$, (polar coordinate system) and $(x,y)\to (0,0)$ gives you the limits $r\to 0$ and no limits on $\theta$.

Now we need to substitute these in your problem which then becomes $$\lim_{r\to 0}\dfrac{r^2}{r(\cos\theta+\sin\theta)}=\lim_{r\to 0}\dfrac{r}{(\cos\theta+\sin\theta)}$$ Now for paths  where $\cos\theta\to-\sin\theta$ or $\cos\theta=-\sin\theta$, the denominator $\to0$ or $=0$ while the numerator is just tending to $0$ but not exactly $0$.Since the second case exists $\implies$ the limit doesn't exist.

If extra condition that $(x+y)\neq 0$ is applied, even then the limit does not exist as for path $\theta=0$, limit is $0$ while for path $\sin\theta=-\cos\theta+r\cos^2\theta$, limit is something which can be non-zero.

Since there are two variables, the equation is a surface (2 dimensional) therefore, the limit is also a 2 dimensional patch. You should apply hopital rule in both directions, but since the equation is symmetric, that shouldn't be a problem.


Regular polygon diagonal lengths Suppose that a regular $n$-gon has integer side length $m$. Is the lengths of its diagonals always algebraic numbers? If yes and if $n,m$ are given, is there an easy way to compute the diagonal lengths as a root of some polynomial?

All diagonals are either diameters, or sides of a triangle whose other two legs are segments uniting the center of the polygon to the diagonal's two extremities. As such, their lengths can be computed using the generalized Pythagorean theorem, also known as the law of cosines. The angle between these two sides is a rational multiple of π, hence its sine and cosine are both algebraic, being the real and imaginary part of the n complex roots of unity, per de Moivre's and Euler's formula. Now all that's left to show is that these radii are themselves algebraic: Indeed, $r^2+r^2$$-2rr\cos\dfrac{2\pi}n=m^2$ implying $r=\dfrac m{\sqrt{2\Big(1-\cos\frac{2\pi}n\Big)}}\in\mathbb{A}$. Then $d_j=r\sqrt{2\bigg(1-\cos\dfrac{2k\pi}n\bigg)}\in\mathbb{A}$. QED.

To find the shortest diagonal of a regular $n$-gon, it's
$$
s\sqrt{\frac{n}{2}}
$$
For a triangle you'll just get the side length.


Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.

Using induction I have the following:

Base case:

$n=0$:   $13^0-6^0 = 1-1 = 0,   7|0$

so, generally you could say:

$7|13^k-6^k , n = k \ge 1$

so, prove the $(k+1)$ situation:

$13^{(k+1)}-6^{(k+1)}$

$13 \cdot 13^k-6 \cdot 6^k$

And then I'm stuck....where do I go from here?

We write
$n=1$ and get
$13 - 6 =7$( a multiple of $7$)
Moving to the induction hypothesis,
We assume that   $n=k$, therefore
$13^k - 6^k = 7b$($b$ shows that $7$ is a multiple of $13^k -6^k$ by $b$ times)
Taking  $k=1$ from above ($n=k=1$) then we get :

$13^1 - 6^1=13 - 6=7$.

Take one side of the equation and use the equivalent to its multiple in each in  the form of $13-7$ or  $7+6$ : $\pm$ the divisor($7$).

Either  $13^k(7+6) - 6^k(6)$ or $13^k(13) - 6^k(13-7)$.

Taking $  13^k(13) - 6^k(13 -7)$ gives
  $13^k(13) - 6^k(13) + 6^k(7)$.

Change the power $k$ on $6^k(13)$ to $13$ ;$13^k(6)$
         $13^k(13) - 13^k(6) + 6^k(7)$

Factor out $13^k$ on the LHS
 $13^k(13-6) + 6^k(7)$ giving

$13^k(7) + 6^k(7)$. This can be compressed to:

$7(13^k + 6^k)$
Therefore now $b= 13^k + 6^k$  making it $7|(13^n - 6^n)$ valid for all positive integers values of $n$.


Is $\mathbb{Q}$ isomorphic to $\mathbb{Z^2}$? Most of us are aware of the fact that $\mathbb{C}$ is isomorphic to $\mathbb{R^2}$, as we can define $\mathbb{C}$ as follows :

$$\mathbb{C} := \left\{z : z=x+iy \ \ \ \text{where} \ \ \langle x,y \rangle \in \mathbb{R^2}\right\}$$

Now this implies

$$\mathbb{C} \cong \mathbb{R^2} $$

But, by this same argument, can't we make the same argument for $\mathbb{Q}$, the set of rational numbers and  $\mathbb{Z^2}$? We can define $\mathbb{Q}$ as follows:

$$\mathbb{Q} := \left\{\frac{m}{n} :  \langle m,n \rangle \in \mathbb{Z^2} \ \ \land \ n \neq 0 \right\}$$

Which would imply

$$\mathbb{Q} \cong \mathbb{Z^2} $$

And we could go further and explain that $\{\mathbb{Q}\} = \mathbb{Z}r\mathbb{Z} - 0$, where $r$ is a relation on $\mathbb{Z}$.

So is $\mathbb{Q}$ isomorphic to $\mathbb{Z^2}$?

You have to be more specific about what you mean by "isomorphic". It depends on what category you are working in. For example, as sets $\mathbb{Z}^n \cong \mathbb{Q}$ for any natural number $n \geq 1$, since both sets are countable. On the other hand, it is certainly false that $\mathbb{Z}^2 \cong \mathbb{Q}$ as groups.

Perhaps more shockingly, if you've taken linear algebra, all finite dimensional vector spaces are isomorphic as sets! On the other hand, they are obviously not isomorphic as vector spaces. So obviously bijection is the wrong notion of isomorphism if you're studying linear algebra,

Here's the general idea. Depending on what you are interested in studying, you define a category of objects and morphisms in that category. Usually (but not always) the objects will be presented by sets with some structure and the morphisms will be functions that preserve the structure. So if you are studying groups, your objects will be groups and the morphisms will be group homomorphisms. If you are studying topological spaces the objects will be spaces and the morphisms will be continuous maps. Anyway, now that you've decided on a category to study, the isomorphisms will be the invertible morphisms in that category.

If you want to learn more about this structural perspective on mathematics, check out Lawvere and Schanuel's book Conceptual Mathematics: A First Introduction to Categories. It's very accessible for beginners and teaches you some very important tools and ideas for doing mathematics.

No. An equivalence relation is missing to do the "isomorphism" (or representation, visualization) correctly.

$Q$ is defined by axioms, and this way of expressing $Q$ as $Z ^ 2$ and the relation $R$ is one of the multiple ways of "visualizing" the axioms of $Q$.

The equivalence relation $R$ that is missing is that two elements $(a, b)$ and $(c, d)$ are realigned, if and only if, $a * d = b * c$.

Then $Z ^ 2$, under the equivalence ralation mentioned above, is isomorphic to $Q$. (Defining correctly the two operations of Q, addition and multiplication, on all the pairs $(a, b)$).


Probability that total number of sixes in $n$ dice tosses Let X be the total count of fives in $n$ dice ups and downs s.t. it is in between  $\frac 1 k n +/- \sqrt n$  with k=6. Calculate Pr(X).

I personally think that I  should integrate the probability density function of n such that it is in between the two values? I dont know how to find that pdf but I know for n throws, you expect to get n/6 sixes. any hin would be helpful thanks!

Hint:

Let $Y \sim Bin(n, \frac16)$
Use Chebyshev's inequality on $Pr \left(|Y-\mathbb{E}(Y)|\ge \sqrt{n} \right)$

Hint: $P(a< X\leq b) = P(X\leq b) - P(x\leq a) = F_X(b) - F_X(a)$


Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$) Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$).

I know by Fermatt's Little Theorem, $x^{58} \equiv1$ (mod 59), and I know that $x^{58} = (x^{11})^5(x^3)$, but I dont know where to go from here. What do I do next?

Continuing rtybase's answer,
A LDE can be formed : 27t + 59y =1, where t = x^9
You can now deuce that t=35 is the only solution when 0<= t < 59,
     hence t is equivalent to 35 (mod 59)


Partitions of a set with n elements (proof) I was reading a textbook about combinatorial mathematic which claimed that we can calculate the exact possible partitions of a set with n elements .
I searched it on wikipedia and I read about bell number and summation formulas;

The first one which The Bell numbers satisfy a recurrence relation involving binomial coefficients :
$$B_{n+1}=\sum\limits_{k=0}^{n} \binom{n}{k} B_k.$$

The second one is A different summation formula represents each Bell number as a sum of Stirling numbers of the second kind :

$$B_n=\sum\limits_{k=0}^{n} \left\{n \atop k\right\}.$$

Can you help me where the first one come from ?

And help me prove it through?

For the first equation, it might be easier to see what's going on if we rewrite the equation as
$$B_{n+1}=\sum\limits_{k=0}^{n} \binom{n}{n-k} B_k$$
(Remember that $\binom{n}{k}=\binom{n}{n-k}$ by matching each subset with its complement) and then rewrite is again as
$$B_{n+1}=\sum\limits_{k=0}^{n} \binom{n}{k} B_{n-k}.$$
(by reversing the order of the sum).

This allows us to contextualise the right hand side of the equation as the follows. In order to construct a partition of the set of $n+1$ elements, we will:

First, decide how many other elements, $k$, are in the same partition as the element $n+1$. $k$ can run from $0$ ($n+1$ is in a part by itself) to $n$ (all of the other $n$ elements are in the same part as $n+1$: that part is the entire set).
Second, decide which subset of $k$ elements we want to include in the part with $n+1$. There are $\binom{n}{k}$ ways to pick this subset.
Finally, choose the partition of the remaining $n-k$ elements.

A partition of $n+1$ elements is uniquely determined by the part containing $n+1$ and how the rest of the elements are partitioned so this method gives us all possible partitions of $n+1$ elements.

The Stirling number of the second kind $S(n,k)$ is the number of partitions of a set with $n$ elements into $k$ nonempty subsets. Summation over all $k$ gives the Bell number.


Comparing logarithms with different bases $\log_3 4$ and $\log_7 10$: which of these two logarithms is greater?

I figured out that both are between $1$ and $2$, then between $1$ and $1.5$. And then $\log_34$ is greater than $1.25$, and $\log_710$ is smaller than $1.25$. However, that method doesn't work for every example, and I wonder if there's a easier way to solve this?

Here is a proof using no numbers larger than $128$. We have $81 < 125 < 128$, i.e. $3^4 < 5^3 < 2^7$, therefore $\log_35 > \tfrac{4}{3}$ and $\log_25 < \tfrac{7}{3}$, therefore $\log_25 < 1 + \log_35$, therefore:
$$
\log_210 < 2 + \log_35 = \log_345 < \log_349 = 2\log_37,
$$
therefore $\log_410 < \log_37$, therefore $\log_34 > \log_710$. $\square$

The last step uses the general proposition that $\log_ab > \log_cd$ if and only if
$\log_bd < \log_ac$, which can be proved by rewriting all the logarithms in terms of logarithms to a single base (e.g. $\log_ab = \ln b/\ln a$, etc.).


Conditional expectation of a functional of an Itô's semimartingale under its equivalent martingale measure Consider a probability filtered space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfying the habitual conditions and is generated by $1 d $- Brownian Motion (with $\mathcal F_T = \mathcal F$).

Also, consider a process $X = (X_t)_{t \in [0,T]}$ given by
 $$X_t= X_0 + \int_0^t  \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t  \geq 0$$

where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.

Suppose that there is a measure $\mathbb Q \sim \mathbb P$ and a $\mathbb Q$-brownian motion  $W^{\mathbb Q}$ such that
$$X_t= X_0+\int_0^t \sigma_s ~dW_s^{\mathbb Q}\quad , t  \leq T$$

I want to evaluate the function $p$ defined as by

$$ p(t,x) := \mathbb E^{\mathbb Q} \left [ (X_T-\kappa X_1)^+ | X_t=x \right] \quad \text{for} \ (t,x) \in [0,1]\times(0,\infty)$$

where $\kappa >0$ and $T>1$.

For this, I was claimed to show the  following relation (that I failed to demonstrate)

$$ p(t,x) = x F(1,\kappa, \int_1^T \sigma_s ^2 ds) \quad \text{if} \ t\in [0,1]$$

where, for $y, K, \gamma^2 >0$
$$F(y,K,\gamma^2) = \mathbb E \left [ (ye^Y -K)^+ \right] \quad \text{with} \ Y \sim \mathcal N(-\gamma^2/2, \gamma^2) \  \text{under} \ \mathbb P$$

I have tried to explore the fact that $ ( X_T-\kappa X_1)^+ =X_1(X_T/ X_1-\kappa )^+$. It can possibly help those that have some financial mathematical background, to know that the motivation behind this is problem is the pricing of a Forward start option.

I would appreciate any advice. Thanks in advance.

Assume that $\kappa=x=0$, $\sigma_t=1$ and $\mu_t=0$ for every $t\geqslant0$, then $\mathbb Q=\mathbb P$, $(X_t)_{t\geqslant0}$ is a Brownian motion, and $p(t,0)=\mathbb E[(X_T)^+\mid X_t=0]=\mathbb E[(X_{T-t})^+\mid X_0=0]=\sqrt{T-t}\cdot\mathbb E[Z^+]$ where $Z$ is standard normal.

In the same setting, $p(t,0)=\sqrt{T-1+(1-\kappa)^2(1-t)}\cdot\mathbb E[Z^+]$ for every $\kappa$.

Note that
\begin{align} p(t,x) &:= \mathbb E^{\mathbb Q} \left [ X_1 (X_T/ X_1-\kappa)^+ | X_t=x \right]  \\&= \mathbb E^{\mathbb Q} \left [ X_1 \mathbb E^{\mathbb Q} \left [(X_T/ X_1-\kappa)^+ | \mathcal F_1\right] | X_t=x \right]  \\&= x \mathbb E^{\mathbb Q} \left [ (X_T/X_1-\kappa )^+ | X_t=x \right]    \end{align}

for all $t<1$ and  $x \in (0, +\infty)$, since $X$  is a $\mathbb Q$ -martingale. Now, if we include the aditional condition on $\sigma$ that $\sigma_t := \tilde\sigma(X_t) ~X_t$, we can conclude that
\begin{align} p(t,x) &= x F(1,\kappa, \int_1^T \tilde\sigma(X_s)^2 ds),  \end{align}

where $F$ is defined as in the question BUT under $\mathbb Q$ and not under $\mathbb P$ as  it says.


the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a]$ I have to show that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a], a>0$.
$$$$
That's what I have tried:

$w_n=\frac{(-1)^n}{\sqrt{n}} \sin1 \cos(\frac{x}{n})+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})=\frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})+1-1)+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})= \frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})-1)+\frac{(-1)^n}{\sqrt{n}} \sin1 +\frac{(-1)^n}{\sqrt{n}}  \cos1 \sin(\frac{x}{n})=\sin1 \cdot g_n(x)+ \cos1 \cdot  f_n(x)+\frac{(-1)^n}{\sqrt{n}} \sin1 $ .

We set $d_n(x)=w_n-\frac{(-1)^n}{\sqrt{n}} \sin1=\sin1 \cdot g_n(x)+ \cos1 \cdot  f_n(x)$

$|g_n(x)|=|\frac{(-1)^n}{\sqrt{n}}(\cos(\frac{x}{n})-1)|=\frac{1}{\sqrt{n}} (\cos(\frac{x}{n})-1) (\cos(\frac{x}{n})+1)=\frac{1}{\sqrt{n}}(\sin^2(\frac{x}{n})) \leq \frac{1}{\sqrt{n}} \cdot \frac{x^2}{n^2} \leq \frac{1}{\sqrt{n}} \cdot \frac{a^2}{n^2}=\frac{a^2}{n^{\frac{5}{2}}}$

$\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}< +\infty$.

$|f_n(x)|=|\frac{(-1)^n}{\sqrt{n}} \sin(\frac{x}{n})| \leq \frac{1}{\sqrt{n}} \frac{x}{n} \leq \frac{a}{n^{\frac{3}{2}}}$

$\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}< +\infty$.

Also,from Dirichlet's test,we know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ converges,let $b:=\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

Therefore,we have $d_n(x) \leq \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1  \frac{a}{n^{\frac{3}{2}}}$.

$\sum_{n=1}^{\infty} \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1  \frac{a}{n^{\frac{3}{2}}}$ converges,as $\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}$ converges  and $\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}$ converges too.

So, $\sum_{n=1}^{\infty} d_n(x)$ converges uniformly to a $s(x)$.
$\sum_{n=1}^{\infty} d_n(x)=\sum_{n=1}^{\infty}  (w_n-\frac{(-1)^n}{\sqrt{n}} \sin1)= \sum_{n=1}^{\infty} w_n-b \overset{uniformly}{=}s(x) \Rightarrow \sum_{n=1}^{\infty} w_n=b+s(x):=q(x):[-a,a] $

So,the series converges uniformly to $q(x)$.

Could you tell me if it is right?

Hint You may use Dirichlet's criterion, since $\sin (xn^{-1})$ is eventually decreasing in $[0,M]$ and $1-\cos (xn^{-1})$ is eventually decreasing in $[0,M]$ too, and both go uniformly to zero, and $$\sum_{n\geqslant 1} (-1)^{n}n^{-1/2}$$ converges.


Formal notation related to a sequence or a set My question is quite naive...

I just want to represent a finite sequence of Natural number, is it the best way to write it like this?:

$\langle a_0, \ldots, a_n \rangle $, where $\forall i \in [ 0, n ], a_i \in \mathbb{N}$

I have also seen somewhere $\{ a_i \}$, does it mean a set (or a sequence) of variables? is it finite or infinite? where could i add the constraint like $\forall i \in [ 0, n ], a_i \in \mathbb{N}$?

Also, is there any difference between "tuple" and "sequence"?

Hope my question is clear, I just want to make sure what I write matches the convention...

Thank you very much

There are several usual notations:

Let $\bar a = \langle a_i\mid i<n\rangle \in \mathbb N^{<\mathbb N}$... (where $\mathbb N^{<\mathbb N}$ means the collection of all finite sequences of natural numbers, and $\mathbb N^\mathbb N$ is the collection of all infinite sequences)
Let $\bar a = \langle a_i\mid i<n\rangle$ be a finite sequence of natural numbers...

And you can always replace $\langle a_i\mid i<n\rangle$ by $\langle a_0,\ldots,a_{n-1}\rangle$.

Note that $[0,n]$ can be read as the real numbers between $0$ and $n$.

The important thing is to be clear in your intention, and consistent in your notation.

And lastly, the difference between a "tuple" and a "sequence" is the context, usually tuple is used for finite length, while sequence is for infinite ones. Also, tuples are usually for a constant length, while sequences are a variable length.

The notation $\{a_i\}$ means that this is a set whose elements are distinct. Naturally it says $i\in I$ for some index set $I$, which in turn might be that this is just a sequence (or a directed set). The meaning is usually clear from context, read more - a lot more - and things will slowly clarify.

Sometimes I see $a_1, a_2, ...$ without the brackets.


Forced wave equation If a system satisfies the equation $$\nu^2 {\partial^2 \psi\over \partial x^2}={\partial^2 \psi\over \partial t^2}+a{\partial \psi\over \partial t}-b\sin\left({\pi x \over L}\right)\cos\left({\pi \nu t\over L}\right)$$
subjected to conditions: $\psi(0,t)=\psi(L,t)={\partial \psi(x,0)\over \partial t}=0$ and $\psi(x,0)=c\sin\left({\pi x\over L}\right)$,

how might I solve this? I can solve the equation $$\nu^2 {\partial^2 \psi\over \partial x^2}={\partial^2 \psi\over \partial t^2}+a{\partial \psi\over \partial t}$$ by separation of variables. But I don't know how to deal with the $$b\sin\left({\pi x \over L}\right)\cos\left({\pi \nu t\over L}\right)$$ term. Also, what is the "forced component" of $\psi(x,t)$?

Thanks.

There are many methods that work. I prefer to do eigenfunction expansion in $x$ first, but I think you can do Laplace transform in $t$ first too.

To use eigenfunction expansion, make sure you are expanding the variable that is finite. Also, make sure your non-homogeneous boundary conditions have been taken care of before applying the eigenfunction expansion. Fortunately, this problem has homogeneous boundary conditions, so we don't need to deal with that.

First, find all eigenfunctions of the differential operators in the direction with finite domain, which is $x$ in this case. The differential operator in $x$ is

$$M = \frac{\partial^2}{\partial x^2}.$$

To find eigenfunctions, solve the ODE

$$M f(x) = \lambda f(x), f(0) = 0, f(L) = 0.$$

(Note that we NEED homogeneous boundary conditions here.) The conditions $f(0) = f(L) = 0$ will force $\lambda$ to be negative. It is customary to rewrite this as

$$M f(x) = -\lambda^2 f(x), f(0) = 0, f(L) = 0.$$

It turns out $\left(\frac{n\pi}L, \sin\left(\frac{n\pi x}L\right)\right)$ will be your $(\lambda, f)$ pairs for $n = 1, 2, 3, \ldots$. I'll name them $\lambda_n = \frac{n\pi}L$ and $f_n(x) = \sin\left(\frac{n\pi x}L\right)$. Now we know

$$M f_n = -\lambda_n^2 f_n, \qquad n = 1, 2, 3, \ldots.$$

By Sturm-Liouville theory (or just the theory of Fourier series in this case), we can approximate any (nice enough) function $g(x)$ that satisfies $g(0) = g(L) = 0$ by a series of $f_n$. Hence, we assume that all functions in $x$ are expanded as series in $f_n$: Let the solution be expressed as

$$\psi(x, t) = \sum_{n = 1}^\infty \psi_n(t)f_n(x).$$

Also, the non-homogeneous forcing term should be expanded in terms of $f_n$:

$$g(x, t) = -b\sin\left(\frac{\pi x}L\right)\cos\left(\frac{\pi\nu t}L\right) = \sum_{n=1}^\infty g_n(t)f_n(x)$$
$$\text{where } g_n(t) = \frac{2}{L}\int_0^L g(x, t) f_n(x) dx.$$

Fortunately, you are given $g(x, t)$ that matches the frequency of $f_1$, so you can get $g_1(t) = -b\cos\left(\frac{\pi\nu t}L\right)$ and $g_n(t) = 0$ for $n \ne 1$ by just comparing coefficients.

Anyway, now you just plug the expansion back into the PDE

$$\nu^2 M \psi = \frac{\partial^2 \psi}{\partial t^2} + a\frac{\partial \psi}{\partial t} + g(x, t)$$

to get

$$-\nu^2\sum_{n=1}^\infty\lambda_n^2\psi_n(t)f_n(x) = \sum_{n=1}^\infty\psi_n''(t)f_n(x) + a\sum_{n=1}^\infty\psi_n'(t)f_n(x) + \sum_{n=1}^\infty g_n(t)f_n(x).$$

By orthogonality of $f_n$, we can just compare coefficients:

$$-\nu^2\lambda_n^2\psi_n(t) = \psi_n''(t) + a\psi_n'(t) + g_n(t), \qquad n = 1, 2, 3, \ldots$$

This is just an ODE for each $n$. Rearranging gives

$$\psi_n''(t) + a\psi_n'(t) + \nu^2\lambda_n^2\psi_n(t) = -g_n(t).$$

The initial conditions of the PDE must be expanded in terms of $f_n$ to give initial conditions of these ODEs:

$$\psi(x, 0) = \sum_{n=1}^\infty\psi_n(0)f_n(x) = c \sin\left(\frac{\pi x}L\right)$$
$$\therefore \psi_n(0) = \begin{cases}
c & ; n = 1 \\
0 & ; n \ne 1
\end{cases}$$

$$\frac{\partial\psi}{\partial t}(x, 0) = \sum_{n=1}^\infty\psi_n'(0)f_n(x) = 0$$
$$\therefore \psi_n'(0) = 0.$$

Since $\psi_n(0) = \psi_n'(0) = 0$ and $g_n(t) = 0$ when $n > 1$, the uniqueness theorem tells us that $\psi_n(t) = 0$. The only non-trivial case is $n = 1$:

$$\psi_1''(t) + a\psi_1'(t) + \left(\frac{\nu\pi}L\right)^2\psi_1(t) = b\cos\left(\frac{\nu\pi t}L\right), \psi_1(0) = c, \psi_1'(0) = 0.$$

I'll let you solve for $\psi_1$. The final solution will be $\psi(x, t) = \psi_1(t)f_1(x) = \psi_1(t)\sin\left(\frac{\pi x}L\right)$.

Note that the final form of $\psi_1(t)$ depends heavily on $a$. In particular, you need to split into cases: $a > \frac{2\pi\nu}L$, $a = \frac{2\pi\nu}L$, and $a < \frac{2\pi\nu}L$. I believe in most physical problems, you'd have $a < \frac{2\pi\nu}L$. (I assume $a \ge 0$ always.)

It is convenient to assume that $\nu = 1 ,L=\pi$.And $\psi (x,t)= c \phi (t) sin(x)$. So, the equation becomes $\ddot{\phi} + a \dot{\phi} + \phi = b cos(t)$, with initial value $\dot{\phi}(t=0)=0$; $\phi (t=0) = 1$, and this equation maybe easy to solve now.


Basis on vector space $V$ If $S_i$ is a set of linearly independent vectors of vector space $V$ and $S_g$ a set of generators of $V$. Prove that it exist $S'_g\subset S_g$ that $S_i\cup S'_g$ is a basis of $V$.

Notice that $V$ is not necessarily finite dimensional.

If you are asking about finitely-dimensional spaces, this is exactly Steinitz exchange lemma.

If you are interested in infinitely-dimensional space, I think that you only need to modify a little the standard proof of existence of Hamel basis based on Zorn Lemma. This proof can be found in many places. (Or maybe I should say that it is more similar to the proof of the fact, that every linearly independent subset of $V$ is contained in a Hamel basis of $V$.)

We will work with the system
$$\mathcal P=\{S\subseteq S_g; S\cup S_i\text{ is linearly independent}\}.$$
We check that for any chain $\mathcal C\subseteq\mathcal P$ we also have $\bigcup\mathcal C\in\mathcal P$. So the poset $(\mathcal P,\subseteq)$ fulfills the assumptions of Zorn Lemma.

By Zorn Lemma there exists a maximal $S'_g\in\mathcal P$. It only remains to show that $S'_g\cup S_i$ generates $V$. (We will use maximality of $S'_g$ in the proof.)

Say $S_i = \{\vec v_1, \vec v_2, ... \vec v_n \}$.

Assume that there does not exist any vector $\vec v \in S_g$ such that $\{ \vec v \cup S_i \}$ is linearly independent.
  Now for every vector $\vec v \in S_g$ there exists $a, a_1, a_2, ...$ such that $ a\vec v + a_1 \vec v_1 + a_2 \vec v_2 + ... + a_n \vec v_n = \vec 0 $.
  However, by the linear independence of $S_i$,  $a \neq 0$.
  Now we have a expression of $\vec v$ in terms of $ \vec v_1, \vec v_2, ... \vec v_n $, and so we have shown $S_i$ itself generates $V$.

Else, there exists a $\vec v \in S_g$ such that $\{ \vec v \cup S_i \}$ is linearly independent. Then just add this $\vec v$ to $S_i$.

By repeating this procedure, we will end up with a linearly independent extension of $S_i$ that generates $V$, because $S_g$ generates $V$ so even in the worst case, we can add all of $S_g$ to $S_i$.

The vectors of $S_g$ we have added to $S_i$ forms the desired subset $S_g' \subseteq S_g$.


Abelian quotient group I'm stuck on the following practice problem.   Any hints would be appreciated.

Suppose $N$ is a normal subgroup of  $G$ such that every subgroup of $N$ is normal in $G$ and
$C_{G}(N) \subset N$. Prove that $G/N$ is abelian.

I'm not sure how to use the fact that $C_{G}(N) \subset N$.

Thanks

Let $n\in N$, and consider the action of $G$ on $\langle n\rangle$. This embeds $G/C_G(\langle n\rangle)$ into $Aut(\langle n\rangle)$, an abelian group. Doing this for all cyclic subgroups of $N$ gives an embedding of $G/C_G(N)$ into a direct product of abelian groups. We are done then, because that means $G/C_G(N)$ is abelian, and $G/N$ is a quotient of that group.

First of all, don't get stuck on what is given. This is the wrong place to look when you start on a proof. Rather, you should look at what you need to prove. In this case, we want to show that $G/N$ is abelian. What does it mean for a group to be abelian?

Well, the definition states that a group $G$ is abelian if for all $g, h \in G$ we have $gh = hg$. So this means we need to pick any two elements from $G/N$ and show that they commute under the group's operation.

I'll let you think about it from there. Let me just emphasize that whenever you write a proof, you need to start with the definition of what you are trying to prove. This almost always gives you a guide as to how to start your proof.


Are limits commutative? Generally speaking, is the following true:
$$\lim_{x\to a}f'(x)=\lim_{x\to a}\left(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right)=\lim_{h\to 0}\left(\lim_{x\to a}\frac{f(x+h)-f(x)}{h}\right)$$

Your statement boils down to the continuity of $f'.$

However, the derivative, even if it exists for any point $x\in\mathbb{R},$ does not have to be continuous. A classic counterexample is $f(x)=x^2\sin(1/x)$ for $x\neq 0,$ and $f(0)=0.$

If the limit of derivative exists at a point $a$, then function can be approximated by a line at points really close to left of $a$, now if the limit of derivative exists, then another line with same slope as previous line approximates the points really close to right of $a$. That was about LHS of the equation, now coming to RHS, your first limit means you approach $a$ from any side, and fix an $x$ really close to $a$, with second limit you are talking about derivative at that $x$, which may not exist, but as you get closer to $a$,    the derivative is defined at those points close to $a$, because your LHS(limit of derivative exists), meaning derivative exists around a neighborhood of $a$.

As for examples for equality,you can have $y=x$, at non zero $x$, and make this function discontinuous at zero.


Area of the intersection of two triangles. Let $\triangle{ABC}$ be a triangle with $AB=5$, $BC=7$, and $CA=4$. Define $D$, $E$, and $F$, to be the midpoints of $AB$, $BC$, and $CA$ respectively. Let $G$ the intersection of the medians of $\triangle{ABC}$, and let $H$, $I$, and $J$ be the midpoints of $AG$, $BG$, and $CG$ respectively. Find the area of the hexagonal region common to both $\triangle{DEF}$ and $\triangle{HIJ}$.

This is too tough for me to even start. I suppose you would start with similar triangles? I know for a fact that the areas of both $\triangle{DEF}$ and $\triangle{HIJ}$ are both a quarter the area of $\triangle{ABC}$, but I have no idea how to proceed. Can somebody help?

We can start by finding the length of the median, using Apollonius' Theorem, which yields $AB^2 + AC^2 = 2(AE^2 + BD^2) \rightarrow 25+16 = 2(AE^2 + 12.25) \rightarrow AE = \frac{\sqrt{33}}{2}$. It is easy to see that the ratio of $AG : GE = 2:1$ , and since $H$ is the midpoint of $AG$, we know that $AE$ is trisected by $H, G$. Repeat this for the other medians, and then it should be easy to go from there.

The hint.

Let $AE\cap IJ=\{T\}$, $IJ\cap ED=\{M\}$ and $EF\cap IJ=\{P\}$.

Thus, $T$ a midpoint of $EG$, which says $$ET=\frac{1}{2}\cdot\frac{1}{3}AE=\frac{1}{6}AE.$$
Also, since $ADEF$ is a parallelogram and $IJ||DF,$ we obtain $ET$ is a median of $\Delta MEP.$

But $\Delta MEP\sim\Delta CAB,$ which gives
$$S_{\Delta MEP}=\frac{1}{36}S_{\Delta ABC}.$$
Id est, the needed area is equal to
$$S_{\Delta EFD}-3S_{\Delta MEP}=\frac{1}{4}S_{\Delta ABC}-3\cdot\frac{1}{36}S_{\Delta ABC}=\frac{1}{6}\sqrt{8\cdot1\cdot4\cdot3}=\sqrt{\frac{8}{3}}.$$


Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$

and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n$ becomes sufficiently large. I am wondering whether showing the ratio between the two terms approaching $1$ is sufficient to show the difference is $0$; if so, how can this be formalized? If not, how can it more directly be shown that the difference approaches $0$?

Given $$\displaystyle \lim_{n\rightarrow \infty}\left[(n+1)^{\frac{1}{7}}-n^{\frac{1}{7}}\right]\;,$$ Now let $$\displaystyle n= \frac{1}{y}\;,$$ Then $y\rightarrow 0$

So $$\displaystyle \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{y^{\frac{1}{7}}} = \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{(1+y)-1}\cdot \frac{y}{y^{\frac{1}{7}}} = \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{(1+y)-1}\cdot \lim_{y\rightarrow 0}y^{\frac{6}{7}} = \frac{1}{7}\times 0 = 0$$

Above we used the formula $$\displaystyle \lim_{x\rightarrow a}\frac{x^n-a^n}{x-a} = na^{n-1}$$


If A ∪ (B ∩ C) is equivalent to (A ∪ B) ∩ (A ∪ C), why are the possible element locations stated more explicitly in the second set than the first? If x ∈ (A ∩ C) then x ∈ A, which is an option given by A ∪ (B ∩ C), so saying that x ∈ (A ∩ C) logically holds for both sides of the equation. However, it feels significant that I can say x ∈ (A ∩ C) when looking at (A ∪ B) ∩ (A ∪ C), but not when looking at A ∪ (B ∩ C), yet these are supposed to be equivalent forms of the same logic. Shouldn't the exact same information be available from both versions if they are just different ways of writing the same thing? Or am I missing something?

I think this just comes down to the fact that some notations are simpler (for certain purposes) than others. For example, compare prime factorization and decimal representation of natural numbers. It's easier to tell which of $13124$ or $32581$ is bigger, whereas that could be difficult to determine using prime factorization; conversely, prime factorization makes it trivial to compute greatest common factors.

Incidentally, I don't find either notation much more explicit than the other; this sort of thing is pretty subjective. I'd say that the LHS makes it easier to tell when an element is in the set, while the RHS makes it easier to tell when an element is not in the set.

If x ∈ (A ∩ C) then x ∈ A

Right here is where you are missing something. Neither of these expressions guarantees $x \in A \cap C$. Not sure where you got that from.

As an example, let $A = \{1,2,3\}$, $B = \{3,4,5,6\}$, and $C = \{5,6,7\}$

Then $A \cap C$ is empty, whereas both sides of the equation evaluate to $\{1,2,3,5,6\}$.


Evaluating $\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(3x)\ldots \sin(nx)\sin(n^{2}x) \over x^{n + 1}}\,dx $ How can we calculate
$$
\int_{0}^{\infty}{\sin\left(x\right)\sin\left(2x\right)\sin\left(3x\right)\ldots
\sin\left(nx\right)\sin\left(n^{2}x\right) \over x^{n + 1}}\,\mathrm{d}x ?
$$

I believe that we can use the Dirichlet integral

$$
\int_{0}^{\infty}{\sin\left(x\right) \over x}\,\mathrm{d}x =
{\pi \over 2}
$$

But how do we split the integrand?

We have (theorem $2$, part $(ii)$, page 6) that:

  If $a_{0},\dots,a_{n}
 $ are real and $a_{0}\geq\sum_{k=1}^{n}\left|a_{k}\right|$, then $$\int_{0}^{\infty}\prod_{k=0}^{n}\frac{\sin\left(a_{k}x\right)}{x}dx=\frac{\pi}{2}\prod_{k=1}^{n}a_{k}.$$

So it is sufficient to note that if we take $a_{0}=n^{2},\, a_{k}=k,\, k=1,\dots,n
 $ we have $$a_{0}=n^{2}\geq\frac{n\left(n+1\right)}{2}=\sum_{k=1}^{n}a_{k}
 $$ hence

  $$\int_{0}^{\infty}\frac{\sin\left(n^{2}x\right)}{x}\prod_{k=1}^{n}\frac{\sin\left(kx\right)}{x}dx=\frac{\pi n!}{2}.$$

If you know the answer can you verify with my answer

$$\frac12[\pi(n^2 n!)-\frac{\pi}{2^n(n+1)!}((1+2+..+n+n^2)-2)^2]$$

If you think it is not correct, then verify it by reducing the terms


Prove that $(p \to q) \lor (q \to p)$ is a tautology using natural deduction I have to prove that '  (p ⊃ q) ∨ ( q ⊃ p) ' is a tautology.I have to start by giving assumptions like     a1 ⇒ p ⊃ q and then proceed by eliminating my assumptions and at the end i should have something like ⇒(p ⊃ q) ∨ ( q ⊃ p) but could not figure out how to start.

As a quick overview: You should be able to show $p\vdash q\supset p$ as well as $\neg p\vdash p\supset q$, and then by case distinction $p\lor\neg p\vdash (p\supset q)\lor (q\supset p)$.

$(p\rightarrow q)\lor (q\rightarrow p)\equiv (\lnot p\lor q)\lor (\lnot q\lor p)\equiv \lnot p\lor (q\lor\lnot q)\lor p\equiv \lnot p\lor T\lor p\equiv \lnot p\lor p\equiv T$


Is $n \choose k$ defined when $k < 0$? What about $n < k$? I know that ${n \choose 0} = 1$, and this makes sense to me based on my understanding of combinatorics.

But what about ${n \choose -1}$? My instinct is that this is undefined, since it is equivalent to $\frac{n!}{k!(n - k)!}$, which when $k = -1$ equals $\frac{n!}{(-1)!(n + 1)!}$, and factorials are only defined for non-negative integers.

Similarly, ${5 \choose 6} = \frac{5!}{6!(5 - 6)!} = \frac{5!}{6!(-1)!}$ should also be undefined. However, intuitively it seems to me that "5 choose 6" could also equal $0$, since there are no combinations of 6 objects to be made from a set of only 5 objects.

So, is $n \choose k$ defined when $k < 0$? What about $n < k$? And in either case, what's the reasoning?

It is customary to define $\binom{n}k=0$ for $0\le n<k$. If you restrict $n$ and $k$ to the non-negative integers and think of $\binom{n}k$ as the number of $k$-element subsets of an $n$-element set, then $\binom{n}k$ should certainly be $0$ when $n<k$. If instead you use the more general algebraic definition of the binomial coefficient,

$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}$$

for real $x$ and non-negative integer $k$, you automatically get $\binom{n}k=0$ when $n$ is a non-negative integer less than $k$. (Here $x^{\underline k}$ is a falling factorial.)

Not everyone defines $\binom{n}k$ for negative integers $k$, but when it is defined for them, it’s defined to be $0$. This ensures that the Pascal triangle identity,

$$\binom{n}k=\binom{n-1}{k-1}+\binom{n-1}k\;,$$

holds even for $k=0$ (and for integers $k<0$ as well, of course); this is convenient in some induction proofs, for instance.

As one of the other users suggested, you can indeed extend $\binom{n}{k}$ to cases where $k<0$ using the Gamma function. However, there are more ways to do it. For example, if we want to have $n$ be negative. Let $r=|n|$ and $f=r+k-1$, then

\begin{align}\binom{-r}{k} &= \frac{-r\cdot-(r+1)\dots-(r+k-2)\cdot-(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&=(-1)^k\;\frac{r\cdot(r+1)\cdot(r+2)\cdots(f-2)\cdot(f-1)\cdot f}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&=(-1)^k\binom{f}{k}\\
&=(-1)^k\left(\!\!\binom{f-k+1}{k}\!\!\right)\\
&=(-1)^k\left(\!\!\binom{r}{k}\!\!\right)\;,\end{align}
where $\left(\!\!\binom{r}{k}\!\!\right)=\frac{r^{(k)}}{k!}=\frac{r\cdot(r+1)\cdot(r+2)\cdots(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}$. These types of things arise naturally in the study of multisets. In fact, one uses these often in Mathematical Biology while studying genetics. However, this is beyond "normal" probability/combinatorics. The reason it is often the case that $0\leq k$ and $0 \leq n$ is $k$ and $n$ often represent a number of objects, which must be nonnegative. The reason that $k\leq n$ is just as you suggest, one would obtain negative factorials (which require the Gamma function). Moreover, in the case where $2=n<k=3$, how does one take 3 total objects from a possible 2? It can't be done nor does it make sense. So it is natural to only consider $k\leq n$.


How to calculate all the prime factors of $lcm$ of $n$ numbers? I have $n$ numbers stored in an array $a$, say $a[0],a[1],.....,a[n-1]$ where each $a[i] \leq10^{12}$ and $n < 100$ . Now,I need to find all the prime factors of the $lcm$ of these $n$ numbers i.e., $lcm$ of $\{a[0],a[1],.....,a[n-1]\}$

I have a method but I need more efficient one.

My method :

 First calculate all the prime numbers up to $10^6$ using sieve of Eratosthenes.

 For each a[i]
      For all p prime <= sqrt(a[i])
             bool check_if_prime=1;
             if a[i] % p == 0 {
                store p
                check_if_prime=0
             }
             if check_if_prime
                store a[i]     // a[i] is prime since it has no prime factor <= sqrt(n)
  Print all the stored primes

Is there any better approach to this problem ?

This is the link to the problem:

http://www.spoj.pl/problems/MAIN12B/

Apparently you are only interested in the set of prime factors, not their multiplicity in the LCM. In other words you need those prime numbers that divide at least one of the $a[i]$. Then you could just traverse the prime numbers $p$ just once, in increasing order, and divide all the $a[i]$ by $p$; whenver the division is exact you can replace $a[i]$ by $a[i]/p$ (and repeat, so the all fectors $p$ get eliminated), and if this happens at least once, $p$ can be retained as prime factor of the LCM. If you are interested in the multiplicity of $p$ after all, this is easy to record during this loop as well: it is the maximum number of factors $p$ found in one same $a[i]$. Once one of the $a[i]$ becomes less than $p^2$ (either by division or because $p$ grows) it can itself be added to the the list of prime numbers, and removed from the collection of $a[i]$ to test. Keep a linked list of candidates that are left over, and stop when it becomes empty.

The advantage of this approach is that for $a[i]$ that are large but contain a lot of small prime factors, they are reduced in size rapidly and hence can be removed from the search relatively fast. You can get the same advantage in your proposed approach if you divide $a[i]$ by $p$ once you find that te remainder is $0$. However, some work is then still required to prune multiply stored prime factors.

Perhaps this Maple program is good solution for your problem :

L:=[4*10^7,3*10^9,2*10^6,10^8]:
LCM:=1:
for n in L do
LCM := ilcm(LCM,n):
end do;
ifactors(LCM);


Helmholtz equation Inequality (potential theory) I was reading the chapter about surface potential. I came across an inequality
$$|e^{ik|x_1-y|}-e^{ik|x_2-y|}|\leq k|x_1-x_2|,$$
$k$ is a complex number here. Since $k$ is a complex number here, I am a bit confused. I am wondering what is the correct way to interpret this ineuqality?

The map $z \mapsto e^z$ maps complex numbers to complex numbers. Hence $e^{ik|x_1-y|}$ is a complex number and
$$|e^{ik|x_1-y|}-e^{ik|x_2-y|}|\leq k|x_1-x_2|$$ is an inequality between complex numbers modulus.


The boundedness of an integral Is there a constant $C$ which is independent of real numbers $a,b,N$, such that

  $$\left| {\int_{-N}^N \dfrac{e^{i(ax^2+bx)}-1}{x}dx} \right| \le C?$$

These integrals are indeed uniformly bounded. As the only effect of changing the signs of $a,b,N$ on the integral is a sign change and/or complex conjugation, we can restrict to $a,b,N\gt0$. Setting $M=Nb$ and $\alpha=a/b^2$, we have
$$
\int_{-N}^N\left(e^{i(ax^2+bx)}-1\right)\frac{dx}x=\int_0^M\left(e^{i(\alpha x^2+x)}-e^{i(\alpha x^2-x)}\right)\frac{dx}x.
$$
However, the integrand can be written as $2ie^{i\alpha x^2}\frac{\sin x}x$, so is bounded by $2$ in absolute value. Therefore, fixing any constant $K\gt0$, the integral is bounded by
$$
\begin{align}
2K+\int_K^{K\vee M}e^{i(\alpha x^2+x)}\frac{dx}x-\int_K^{K\vee M}e^{i(\alpha x^2-x)}\frac{dx}x.&&{\rm(1)}
\end{align}
$$
I'll prove that this is bounded with the help of a lemma.

  Lemma: (van der Corput lemma) If $f\colon[A,B]\to\mathbb{R}$ is convex with $\lvert f^\prime(x)\rvert\ge\lambda\gt0$ and $g\colon[A,B]\to\mathbb{C}$ is differentiable then,
  $$\left\lvert\int_A^Be^{if(x)}g(x)dx\right\rvert\le\frac2\lambda\left(\lvert g(B)\rvert+\int_A^B\lvert g^\prime(x)\rvert dx\right)$$

As I will only be interested in intervals $[A,B]\subset[K,\infty)$ with $g(x)=1/x$, the bound in the lemma can be written as
$$
\begin{align}
\left\lvert\int_A^Be^{if(x)}\frac{dx}x\right\rvert\le\frac2{\lambda K}&&{\rm(2)}
\end{align}
$$
Taking $f(x)=\alpha x^2+x$, this shows that the first integral in (1) is bounded by $2/K$. For the second integral, take $f(x)=\alpha x^2-x$, fix any $0\lt\epsilon\lt1/2$, and first look at value of the integral with integration range restricted to $[0,\epsilon/\alpha]$. In this range, we have $f^\prime(x)\le f^\prime(\epsilon/\alpha)=-(1-2\epsilon)$. So, by inequality (2), this part of the integral is bounded by $2K^{-1}(1-2\epsilon)^{-1}$.

Next, for any fixed $\gamma\gt1/2$, the value of the last integral in (1), restricted to the range $[\epsilon/\alpha,\gamma/\alpha]$ is bounded by
$$
\int_{\epsilon/\alpha}^{\gamma/\alpha}\frac{dx}{x}=\log(\gamma/\epsilon).
$$
Finally, look at the last integral in (1) restricted to the range $[\gamma/\alpha,\infty)$. As $f^\prime(x)\ge f^\prime(\gamma/\alpha)=(2\gamma-1)$ in this range, inequality (2) shows that this part of the integral is bounded by $2K^{-1}(2\gamma-1)^{-1}$.

Putting these together shows that the set of integrals in the question is bounded above by a constant. The upper bound obtained here is
$$
2K+\frac2K+\frac2{K(1-2\epsilon)}+\frac2{K(2\gamma-1)}+\log(\gamma/\epsilon)
$$
for arbitrary positive constants $K,\epsilon,\gamma$ with $\epsilon\lt1/2\lt\gamma$.

we can use theorem
for $a \le b$

$ |\int _a^b f(x)dx|\le \int_a^b|f(x)|dx$

$\int _{-N}^N|(e^{i(ax^2+bx)}-1)\frac1x|dx\le \int _{-N}^N|(e^{i(ax^2)}-1)\frac1x|dx\ $
you can play around with the integral to make it more intergreable


GCD of two numbers divided by their greatest common divisor is 1 Im trying to prove that, given $a,b$ with at least one of $a,b \neq 0$,
$$
\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1
$$
I have tried to prove the identity
$$
\gcd(c\cdot a, c\cdot b) = c\cdot \gcd(a,b)
$$
with $c =  \dfrac{1}{\gcd(a,b)}$

However I'm having trouble understanding the proof.

Thank you for your time.

Let $c = \gcd\left( \dfrac a {\gcd(a,b)}, \dfrac b {\gcd(a,b)} \right).$

Then $c$ divides $\dfrac a {\gcd(a,b)}$ and $c$ divides $\dfrac b {\gcd(a,b)}.$

Thus $c\cdot\gcd(a,b)$ divides $a$ and $c\cdot\gcd(a,b)$ divides $b.$

So $c\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.

If $c>1$ then $c\cdot\gcd(a,b)$ is a greater common divisor than the greatest common divisor.

Let g be a common divisor of both a and b, and let g' be a common divisor of both a/g and b/g. Then (a / g) / g' = a / (gg') is an integer, and so is (b / g) / g' = b / (gg'). Therefore g*g' is a common divisor of a and b.

If g is the greatest common divisor of a and b, then g*g', which is also a common divisor, is not greater than g, and therefore g' = 1.


Multiple-choice: sum of primes below $1000$ I sat an exam 2 months ago and the question paper contains the problem:

  Given that there are $168$ primes below $1000$. Then the sum of all primes
  below 1000 is

  (a) $11555$ (b) $76127$ (c) $57298$ (d) $81722$

My attempt to solve it: We know that below $1000$ there are $167$ odd primes and 1 even prime (2), so the sum has to be odd, leaving only the first two numbers. Then I tried to use the formula "Every prime can be written in of the form $6n-1$,$6n+1$ except $2$ and $3$.", but I got stuck at that.

The sum of the first 168 positive integers is $\frac{168^2+168}{2}=14196$, which is greater than answer (a). The sum of the first 168 primes must be even greater than that.

The methods provided previously are very correct. However, I would like to share my approach as I believe this is even simpler. It's less calculative, and more logical.
Sum is odd ⇒ answer is either (a) or (b)
Taking the number in (a) 11555 we divide it by 168 [to get the average]
11555/168≈69
69<168/2
An average of 168 distinct numbers can be less than 168/2(=84) iff at-least some of them are negative.
Which is clearly not possible since we are dealing with whole numbers.
⇒ only possible answer remain (b).


If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ has coordinates $f^1 \ldots f^m$, and each $f^i$ is differentiable at $0$... ...then does it follow that $f$ is differentiable at 0?

My motivation for asking this is as follows: in Spivak's Calculus on Manifolds, in theorem 2.9, he uses this with the additional condition that each $f^i$ is continuously differentiable in a nbh of 0, to conclude that $f$ is differentiable and I don't think is necessary.

Namely, if each $f^i$ has derivative $Df^i$, I claim the matrix with $i^{th}$ row $Df^i$ will serve as $Df$. Indeed, let $v_j$ be a sequence tending to 0 in $\mathbb{R}^n$,  we have (by the triangle inequality, if you wish)$$\frac{| f(v_j) - f(0) - \sum_i Df^i(v) |}{|v_j|} \leqslant \frac{\sum_i |f^i(v_j) - f^i(0) - Df^i(v_j)|}{|v_j|}$$Taking the limit as $j \rightarrow \infty$, each summand goes to 0 by the differentiability of $Df^i$ (and there's only $m$ of them), hence the limit is 0.

Is this wrong? Thanks in advance!

EDIT: btw, conditional on the above proof being right and/or the claim being right, does anyone know maybe what Spivak was going for?

In case this is useful to anyone else, let me record the comments of Georges Elencwajg and Dylan Moreland-the answer is yes it's true, and the condition in Spivak is superfluous.

The proof is the one liner I wrote above, albeit with better notation: If $\lambda^i$ are the derivatives of the $f^i$ at 0, I claim the matrix with $i^{th}$ row $\lambda^i$ will serve as $Df(0)$. Indeed we have $$\frac{|f(v) - f(0) - \lambda^i(v)e_i|}{|v|} \leqslant \sum_i \frac{|f^i(v) - f^i(0) - \lambda^i(v)|}{|v|}$$Taking any $v_i \rightarrow 0$, applying the above inequality, and using the differentiability of each $f^i$ at 0 gives the result.

Thanks all for the assistance!

The "continuously differentiable" condition for the partial derivatives is essential to showing that $f$ is differentiable $0$. This is the canonical counterexample $$f(x,y) = \left\{
     \begin{array}{lr}
       \frac{x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\
       0 & \text{if } x = (0,0)
     \end{array}
   \right.$$


Kolmogorov's maximal inequality and convergence of random series. Let $(X_n)_{n\ge 1}$ be a sequence of mutually independent random variables, on the same probability space, with expectation 0 and finite variance. Let $S_n = \sum_{l=1}^n X_l$. Prove that for any $\lambda > 0$,

$$\lambda^2 \displaystyle P \left(\max_{1\le k\le n} \big| S_k \big| \ge \lambda \right) \le Var(S_n)$$

Prove that if $\sum_l Var(X_l) < \infty$, then $(S_n)_{n\ge 1}$ converges almost surely.

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There's a more elementary proof which avoids the use of martingales$^1$. Observe that the event $\{\max_{1\le k\le n}|S_k|\ge\lambda\}$ is the disjoint union of the events $A_k=\{|S_k|\ge\lambda,|S_j|<\lambda\text{ for }j<k\}$, $k=1,\ldots,n$. Moreover, we have

\begin{align*}
\mathbb P(A_k) &\le\lambda^{-2}\mathbb E[S_k^2\mathbf1_{A_k}]\\
&\le\lambda^{-2}\mathbb E[\big(S_k^2+(S_n-S_k)^2\big)\mathbf1_{A_k}]\\
&=\lambda^{-2}\mathbb E[\big(S_k^2+2S_k(S_n-S_k)+(S_n-S_k)^2\big)\mathbf1_{A_k}],
\end{align*}
where the last line follows since $S_k\mathbf1_{A_k}$ and $S_n-S_k$ are independent and $\mathbb E[S_n-S_k]=0$. Thus we have

$$\mathbb P(A_k)\le\lambda^{-2}\mathbb E[\big(S_k+(S_n-S_k)\big)^2\mathbf1_{A_k}]=\lambda^{-2}\mathbb E[S_n^2\mathbf1_{A_k}]$$

and so summing over $k$ we find

$$\mathbb P\left(\max_{1\le k\le n}|S_k|\ge\lambda\right)\le\lambda^{-2}\mathbb E[S_n^2\mathbf1_{\max_{1\le k\le n}|S_k|\ge\lambda}]\le\lambda^{-2}\mathbb  E[S_n^2].$$

$1$. I have always found the use of martingales here unsatisfactory. Kolmogorov's inequality is often used to prove the Strong Law of Large Numbers which I imagine most would encounter before martingales in a first course in rigorous probability theory. Moreover, one of the existing answers here lifts directly from Wikipedia - aside from the blatant plagiarism, that proof has some issues. (E.g. it assumes both $\mathbb E[S_i]=0$ and $S_i\ge0$, it does not justify why $Z_n$ is a martingale, etc.)

For the Inequality:

This argument employs discrete martingales. The sequence $S_1,S_2,...,S_n$ is a martingale. Without loss of generality, we can assume that $S_{0}=0$ and $S_{i}\geq 0\forall i$.

Define: $(Z_i)_{i=0}^{n}$ as follows: Let $Z_{0}=0$, and

$$
Z_{i+1} =
\begin{cases}
S_{i+1} & \text{if } \max_{1\leq j\leq i}S_{j}<\lambda \\
Z_{i} & \text{otherwise}
\end{cases}
$$

$\forall i$. Then $(Z_{i})_{i=0}^{n}$ is also a martingale.

Since $S_{i}-S_{i-1}$ is indepenent and mean $0$,

$$\sum_{i=1}^{n}\mathbb E\left[\left(S_{i}-S_{i-1}\right)^2\right]=$$

$$\sum_{i=1}^{n}\mathbb E\left[S_{i}^{2}-2S_{i}S_{i-1}+S_{i-1}^{2}\right]=$$

$$\sum_{i=1}^{n}\mathbb E\left[S_{i}^{2}-2(S_{i-1}+S_{i}-S_{i-1})S_{i-1}+S_{i-1}^2\right]=$$

$$\sum_{i=1}^{n}\mathbb E\left[S_{i}^2-S_{i-1}^{2}\right]-2\mathbb E\left[S_{i-1}\left(S_{i}-S_{i-1}\right)\right]=$$

$$\mathbb E\left[S_{n}^2\right]-\mathbb E\left[S_{0}^{2}\right]=\mathbb E\left[S_{n}^{2}\right]$$

The same is true for $\displaystyle \left(Z_{i}\right)_{i=0}^{n}$. Thus,

$$\mathbb P\left(\max_{1\leq i\leq n}S_{i}\geq \lambda\right)=\mathbb P\left[Z_{n}\geq\lambda\right]$$

$$\leq \frac{1}{\lambda^2}\mathbb E\left[Z_{n}^{2}\right]=\frac{1}{\lambda^2}\sum_{i=1}^{n}\mathbb E\left[\left(Z_{i}-Z_{i-1}\right)^2\right]$$

$$\leq\frac{1}{\lambda^2}\sum_{i=1}^{n}\mathbb E\left[\left(S_{i}-S_{i-1}\right)^2\right]=\frac{1}{\lambda^2}\mathbb E\left[S_{n}^2\right]=\frac{1}{\lambda^2}\mathrm{Var}\left[S_{n}\right]$$
by Chebyshev's inequality.

Multiplying through by $\lambda$, we get
$$\lambda^2\mathbb P\left(\max_{1\leq k\leq n}S_{i}\geq \lambda\right)\leq \mathrm{Var}\left(S_{n}\right)$$

To show convergence:

A series converges if and only if it satisfies the Cauchy criterion. To check the latter, consider $N$ and consider

$$\mathbb P\left(|S_n-S_N|>\delta \hspace{1mm} \text{for some}\hspace{1mm} n\geq N\right)=\lim_{m\rightarrow \infty}\mathbb P\left(|S_n-S_N|>\delta\hspace{1mm}\text{for some}\hspace{1mm} N\leq n\leq N+m\right)$$

Thus, for fixed $N$, $m$, we must estimate the probability of the event

$$\displaystyle \delta<\max_{1\leq k\leq m}|S_{N+k}-S_{N}|$$

For a fixed $k$ we can use Chebyshev's inequality to get

$$\mathbb P\left(\delta<\max_{1\leq k\leq m}|S_{N+k}-S_{N}|\right)\leq \delta^{-2}\mathrm{Var}\left(X_{N}+X_{N+1}+...+X_{N+m}\right)$$

However we don't have a technique for controlling the maximum of $\displaystyle |S_{N+k}-S_{N}|$ over $k=1,2,...,m$

Hence we need Kolmogorov's maximal inequality:

Using the inequality, we obtain:

$$\mathbb P\left(|S_{n}-S_{N}|>\delta\hspace{1mm}\text{for some}\hspace{1mm} N\leq n\leq N+m\right)\leq \delta^{-2}\sum_{k=N}^{N+m}\mathrm{Var}(X_{k})\leq \delta^{-2}\sum_{k=N}^{\infty}\mathrm{Var}(X_{k})$$

The right hand side goes to zero as $N\rightarrow \infty$, thus we conclude that for any $\delta>0$,

$$\lim_{N\rightarrow\infty}\mathbb P\left(|S_n-S_N|>\delta\hspace{1mm} \text{for some}\hspace{1mm}n\geq N\right)=0$$

This implies that:

$$\limsup S_{n}-\liminf S_{n}\leq \delta$$

almost surely. Take intersection over $\delta+\frac{1}{k}$, $k=1,2,...$ to get that $S_{n}$ converges almost surely.