What is the meaning of this question? Exercise: Consider the question of representing integers with the base $a$ . In order to name the integers in this system we need words for the digits $0$, $1, \ldots, a –1$ and for the various powers of $a$: $a$, $a^2$, $a^3, \ldots$. How many different number words are needed to name all numbers from zero to one thousand, for $a = 2$, $3$, $4$, $5,\ldots, 15$? Which base requires the fewest? (Examples: If $a = 10$, we need ten words for the digits, plus words for $10$, $100$, and $1000$, making a total of $13$. For $a = 20$, we need twenty words for the digits, plus words for $20$ and $400$, making a total of $22$. If $a = 100$, we need $100 + 1$.)

PS : I don't need solution , I just need a brief explanation on what this question is asking about as my English is not that good enough to understand it

It's the number of words we need to define the number system with a specific base. Since an integer with base $a$ is represented by $x_1\times1+x_2\times a+x_3\times a^2$ and so on. We need words to represent $x_1$,$x_2$,$x_3$... Those are the symbols from $0$,$1$,$2 \ldots,a-1$ that is a number of words and also the words for the powers of $a$ since the basic set of symbols with value below a is being multiplied by the powers of $a$. Also there is a constraint here, numbers up to only, so the limit and the question of which uses least number of words. But in case of absence of constraints, it would be infinite number of words in each case.


Why does $x^{\frac{1}{a}} = \sqrt[a]{x}$? I was attempting to answer this question, and came across the fact that $x$ to a non integer power, $x^\frac{1}{b}$ is equal to $\sqrt[b]{x}$. Why does this work?

Similarly, does $x^{\frac{a}{b}} = a\sqrt[b]{x}$  ?

It's because however we define fractional exponents, we want to make sure they behave according to the same algebraic rules as integer exponents.

For example, for integer exponents we have the property
$$(x^a)^b = x^{ab}$$
and we want to extend that property to make sure it also holds when one of the exponents is a fraction.  That means we want
$$(x^{1/a})^b = x^{a/b}$$
for all $a,b$ (with $a \ne 0$).  In particular, we need
$$(x^{1/b})^b = x^{b/b} = x^1 = x$$
which means (by definition) that $x^{1/b}$ is a $b$th root of $x$, i.e. that $x^{1/b} = \sqrt[b]{x}$.

It basically is a definition;

Consider $\sqrt{2}$, this can be written as $2^{\frac{1}{2}}$ which is same as $\sqrt[2]{2}$


If f: $[0,\infty)\to \mathbb{R}$ is continuous and uniformly continuous in $[k,\infty]$, then it's uniformly continuous in $[0,\infty]$. Let f: $[0,\infty]\to \mathbb{R}$ continuous. Prove that, if f is uniformly continuous in $[k,\infty]$ for some k, then f is uniformly continuous in $[0,\infty]$ I don´t know how to do this, I would appreciate your help

(S.A. Understanding Analysis. pp 119 question 4.4.8)

Hint: It's very relevant that $[0, k]$ is compact, and every continuous function on a compact set is uniformly continuous. Can you combine uniform continuity on $[0, k]$ with uniform continuity on $[k, \infty)$?

(http://answers.yahoo.com/question/index?qid=20090401115020AACCrH3)

S.A. Theorem 4.4.8 pp 118 :
A continuous function on a closed and bounded interval is uniformly continuous.

Let $e > 0$ be given. Since f is uniformly continuous on $[b, \infty)$, this means
there exists $d > 0 : $ $\forall x, y \in [b, \infty)$, $|x-y| < d \implies |f(x) - f(y)| < e$.
We can assume that $\color{red}{d < 1}$. If you aren't solid on this, think about why.

$\color{red}{ \text{If anyone is solid on this, expatiate please. } }$

The question posits f is continuous on $[0,\infty).$ Ergo it's continuous on any subset of it.
Ergo consider $[0, b+1]$, a closed and bounded interval.   $\color{red}{ \text{I think this step is the nub. I don't know how to presage it feyly. Demystify please... } }$
By the agency of Theorem 4.4.8, ergo, $f$ is uniformly continuous there.
We conclude that there is $D > 0$ with the property that
whenever $x,y \in [0, b+1]$ satisfy $|x-y| < D$, then $|f(x) - f(y)| < e.$

Let $m = \min(d, D)$ (or their common value if d = D).
Suppose $x, y \in [0, \infty)$ satisfy $|x-y| < m \le \color{red}{d < 1}.$ Then x and y are less than one unit apart.
It follows that either they both lie in $[0, b+1]$ or they both lie in $[b,\infty).$

Why? See picture beneath. If one of them, posit $s$, were $\in [0,b+1]$ but $\not \in [b,\infty)$, then $s< b$.
If the other, posit $t$, were $\not \in [0,b+1]$ but $\in [b,\infty)$, then  $t > b+1$.
And then $|x-y| = |s-t| > 1$.

If x and y are both in $[0, b+1]$; then we know, because $|x - y| < m \le D$, that $|f(x) - f(y)| < e$.

If x and y are both in $[b,\infty)$, then we know because $|x - y| < m \le d$ that $|f(x) - f(y)| < e.$

In both cases, we have deduced that whenever $|x-y| < m$ we have $|f(x) - f(y)| < e$.
So f is uniformly continuous.


Solving equations of the form $c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x+p_1\right)=0$ How do I solve this equation?
$$c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x\right)=0$$
where $c_0$, $p_0$, and $c_1$ are fixed constants? What about the more general form:

$$c_0 \sin\left(x+p_0\right)+c_1\sin\left(2x+p_1\right)=0$$

where $c_0$, $p_0$, $c_1$, and $p_1$ are constants?

WLOG, $c_1=1$.

The equation can be rewritten under the form

$$s \sin x+c\cos x+\sin x\cos x=0.$$

Then we use the substitution

$$\sin x=\frac{2t}{t^2+1},\cos x=\frac{t^2-1}{t^2+1}$$
and obtain the quartic equation

$$t^4+2\frac{s+1}ct^3+2\frac{s-1}ct-1=0.$$

I don't think that there will be a simplification in the expressions of the roots and you must solve this using the general formulas. There will be two or four real roots in $t$, i.e. an even number of roots per period, as was expected.

By the theory of quartic equations, the number of roots is two or four depending on the sign of the discriminant
$$-27(b^4+d^4)-4(bd)^3+6(bd)^2-192bd-256$$ where $b$ and $d$ are the cubic and linear coefficients above.

Letting $u:=\cos x$ and $v:=\sin x$, the given equation is that of an equilateral hyperbola, which passes through the origin,

$$(u+s)(v+c)=cs,$$ and we are intersecting it with the unit circle

$$u^2+v^2=1.$$

Original post

If either $c_0=0$ or $c_1=0$ (or both) the problem is trivial.

Hence we assume the opposite and let $a=\frac{c_1}{c_0}$. Furthermore we let $x=y-p_{1}/2$ and $q=p_{0}-p_{1}/2$. Then your generalized equation becomes

$$ \sin(y+q) + a \sin(2y) = 0\tag{1}$$

Since $\sin(y+q) = \sin(y) \cos(q) + \cos(y) \sin(q)$ and $\sin(2y) = 2 \sin(y) \cos(y)$  we get

$$ \sin(y) \cos(q) +  \cos(y) \sin(q) +  2 a \sin(y) \cos(y) = 0\tag{2}$$

Now letting $z = \sin(y)$ we have $\cos(y) = \sqrt{1-z^2}$ and $(2)$ becomes

$$ z \cos(q) +  \sqrt{1-z^2} \sin(q) +  2 a z \sqrt{1-z^2}  = 0\tag{3}$$

This results is an algebraic equation of 8th degree in $z$. But it depends only on $z^2$ so that it is 4th degree in $s=z^2$. Let the four solutions be $s_{k}$, $k=1..4$.

Then the solutions $y$ are given by $y_{k}= \arcsin(\sqrt{s_{k}})$, $k=1..4$

EDIT

Here are some plots of the LHS of $(1)$ for $x \in [0,2\pi)$ for randomly chosen parameters $q\in (0,2\pi)$ and $a\in(-2,2)$

There are 2 roots (figs. 6,8,9), 3 roots (two have merged into a double root, figs. 5 and 7), in all other cases there are 4 roots.


Why does two terms immediately adjacent "mean" multiply? I am currently teaching a GED math class. While learning about the order of operations, the students asked why does a number next to a parentheses mean multiplication?

I understand the rule that two terms next to each mean multiply those two terms together, regardless as to what those terms look like. But did this notation that we now all understand come from shorthand? Or was it easier to read? Or is there some other reason entirely?

Any insight that you can provide will be appreciated. I would get very angry with teachers in my life who explained things by saying, because that is the way it is. If this is the case, then I at least want to be able to tell them that honestly.

I would argue a case for how we describe multiples using natural language. When I say, "I have three boxes", I don't need any other words between 'three' and 'boxes'. Similarly, "I have three $x$", in shorthand, becomes simply $3x$.

Contrast this with addition: "I have an apple and two oranges". The 'and' separates objects that are together but are not necessarily of the same form. "I have an $x$ and two $y$s" becomes $x+2y$.

Teophile's answer is great.

I would like to add that multiplication is the "natural" mathematical operation, not addition, as some people might think. A group has multiplication, but no addition. Addition is only introduced in rings, once we already have a multiplication.

Also, function concatenation is a multiplication, and mathematicians like to write this without an operator (look e.g. into the Lambda calculus), unless we talk about function application. But even then, mathematicians do not need an additional symbol - note how there is no extra symbol on the right hand side.

$$
  (fg)(x) = f(g(x))
$$

(you might more be used to write $(fg)(x)$ as $f\circ g(x)$ ($\circ$ here being function concatenation))


Change of variables involving two uniformly distributed random variables We know the familiar theorem for the change of variables involving random variables (see for example here, Theorem 4.1):

  Suppose $T:(x,y) \mapsto (u,v)$ is a 1-1 mapping from some domain $D\subseteq \mathbb{R}^2$ to some range $R \subseteq \mathbb{R}^2$. Define the Jacobian $J$ as a function of $u$ and $v$ by $J(u,v) = \left| x_u y_v - x_vy_u\right|$. Assume the partials exist and are continuous. If $X,Y$ have joint probability density function $f_{X,Y}$, then the random variables $U,V$defined by $(U,V)  =T(X,Y)$ are jointly continuous with joint probability density function $f_{U,V}$ given by $f_{U,V}(u,v) = f_{X,Y}(x(u,v), y(u,v))J(u,v)$ if $(u,v)\in R$ and $f_{U,V}(u,v) = 0$ otherwise.

Okay, now suppose $X,Y \sim \mathrm{Unif}[0,1]$ independent random variables, let $T:\mathbb{R}^2 \longrightarrow \mathbb{R}^2$, be given by $(x,y) \mapsto (x+y, x-y) =: (u,v)$. We have $D = [0,1]^2$, $R = [0,2]\times[-1,1]$ and $x(u,v) = \frac{u+v}{2}$ as well as $y(u,v) = \frac{u-v}{2}$. We obtain our Jacobian as $J(u,v) =\left| \frac{1}{2} \frac{-1}{2} - \frac{1}{2}\frac{1}{2}\right| = \left| \frac{-1}{4} - \frac{1}{4} \right| = \frac{1}{2}$. As clearly $f_{X,Y}(x,y) = f_X(x)f_Y(y) = (1)(1) = 1$ by independence, we get by our formula that $f_{U,V}(u,v) = \frac{1}{2}$ on $R$ and $0$ everywhere else. A simple integration gives us $\int_R f_{U,V}(u,v)\mathrm{d}u\mathrm{d}v = 2$ which is clearly a contradiction as p.d.f's integrate to $1$.

Can someone help me and tell me where I made a mistake? I feel like this is such an easy question, but I can't really see what I'm  doing wrong... Any help will be much appreciated!

The joint pdf of $U$ and $V$ does not have support $R = [0,2]\times [-1,1]$ as you claim it does; the support is a square region where the square has vertices $(0,0)$, $(1,1)$, $(2,0)$ and $(1,-1)$. The area is $2$ and the pdf has value $\frac 12$ on the region.

$$f_{U,V}(u,v) = f_{X,Y}(\frac{u+v}{2},\frac{u-v}{2}).\frac{1}{2}$$
$$f_{X,Y}(x,y) = f_X{(x)}.f_Y{(y)}$$
$$ f_{U,V}(u,v) = \frac{1}{2}, 0\le u \le2, -1\le v\le1$$
$$ \left[\int_{-1}^{0}\int_{-u}^{2+u} +  \int_{0}^{1}\int_{u}^{2-u}\right]\frac{1}{2}dvdu = 1; $$

It is indeed 1.

The mistake is in your limits, you took u to be from 0 to 2 and v from -1 to 1 and you got 2 two's and a pdf of one half resulting in 2.


Evaluate $\int\limits_{|z|=1}\frac{\sin{1/z}}{(z-2)^2}dz.$ This is a question from an old exam. The two integrals are seemingly similar, but the first one seems quite tedious compared to the other one. It seems, according to the prof solution, that the hard part in the first integral is computing the residue at $z=0.$

  Evaluate the intergals

  a) $$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$

  b) $$\int\limits_{|z|=3}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$

In the first, we have $2$ poles, first one is $z_1=0$ and the second one is a pole of order $2$, which is $z_2=2.$ However, only $z_1$ is inside our unitcircle so we only need to compute $\text{Res}_{z_1}(f(z))$ and apply the residue theorem.

Since our function is of the form $f(z)=\frac{g(z)}{(z-z_1)^k}$ the residue is given by

$$\text{Res}_{z_1}(f(z))=\frac{g^{k-1}(z_1)}{(k-1)!},$$

and here $k=2$ and $(\sin(1/z))'=-\cos(1/z)/z^2$, so

$$\text{Res}_{z_1}(f(z))=-\frac{\cos\frac{1}{z_1}}{z_1^2}=...\text{this is the moment I realised I'm screwed.}$$

Howver, computing the residue at $z_2=2$ for the other integral I can use conventional methods, but not here.

Can someone break down the main difference between these integrals and show how to find the residue at $z_1=0?$

EDIT:

I need to solve $a)$ using Laurent series.

We substitute $w = 1/z$ and taking the explicit parametrization $z=e^{it}$ we see that the integration path changes the orientation, this brings a minus in front of the integral, so
we have:
$$
\begin{aligned}
J_1
&=
\int_{|z|=1}\frac{\sin(1/z)}{(z-2)^2} \; dz
\\
&=
-\int_{|w|=1}\frac{\sin w}{\left(\frac 1w-2\right)^2} \; \left(-\frac 1{w^2}\right)\;dw
\\
&=
\int_{|w|=1}\frac{\sin w}{(1-2w)^2} \; \;dw
\\
&=2\pi\; i
\operatorname{Residue}_{w=1/2}
\frac{\sin w}{(1-2w)^2}
\\
&=2\pi\; i
\cdot\frac 14\operatorname{Residue}_{w=1/2}
\frac
{\sin\left( \left(w-\frac 12\right)+\frac 12\right)}
{\left(w-\frac 12\right)^2}
\\
&=2\pi\; i\cdot\frac 14\cdot\cos\frac 12
\ ,
\end{aligned}
$$
because in the expansion $\sin\left( \left(w-\frac 12\right)+\frac 12\right)
=
\sin\left(w-\frac 12\right)\cos\frac 12+
\cos\left(w-\frac 12\right)\sin\frac 12
$
only the $\sin\left(w-\frac 12\right)$ brings the needed odd power that contributes to the residue.

Let us check the above numerically, pari/gp:

? intnum( t=0, 2*Pi, sin( exp(-I*t) )  / ( exp(I*t)-2)^2 * I * exp(I*t) )

%1 = 2.2522481052807061131814259376142973690 E-58
   + 1.3785034646766524887891155026097526287*I
? Pi/2*cos(1/2.)
%2 = 1.3785034646766524887891155021962680546

(Result was manually rearranged.)

The other integral is converted by the same trick into one over $|w|=1/3$, and the contour avoids the pole $1/2$.

The integrand has a third-order zero at infinity, therefore the residue at infinity is $0$. Assuming that the contour goes around the origin counterclockwise, the first integral is $-2 \pi i$ times the residue at $z = 2$ and the second integral is $0$.


Graphing polynomials Sketch a graph of the polynomial $P(x)=(x-2)^2(x+1)^3$. You must plot and label the x and y intercepts and these should be the only points you plot.

How do I sketch the graph of a polynomial?

Hint.  Basically you need to know two things.

What is the general shape of a quintic graph?
What does the graph of a polynomial look like near a double or triple root?

Edit.  For this kind of question I think it is more helpful to post hints than a complete solution.  However I am adding the following because IMO many of the answers posted here have completely missed the point of the question.

So, what is the point of graph sketching?  IMO:

  The point of graph sketching is to obtain a reasonably accurate picture of the curve with minimum work.

The "minimum work" requirement is one reason (not the only reason) why plotting lots of points is (usually) a really bad idea.  I am sure that "minimum work" is also what this question means you to do, in view of the specification that you must not plot any points except the intercepts on the axes.  What I suggest for the present example is this.

The graph is a quintic and the coefficient of $x^5$ is positive.  Since we know the general shape of a quintic we can immediately say that the graph increases from $-\infty$ to $\infty$, maybe with some turning points in between.
The factor of $(x+1)^3$ shows that the graph goes through the $x$ axis at $x=-1$, making a "cubic wriggle" as it does so.  (If you don't know what a "cubic wriggle" looks like, then you should be looking at the graph of $y=x^3$ before attempting this problem.)
The factor of $(x-2)^2$ shows that the graph touches the $x$ axis at $x=2$ and then stays on the same side of the axis.  (If this is not clear, think of the graph of $y=x^2$.)
The graph cuts the $y$ axis at $y=4$.  (I wouldn't actually have bothered with this, but the question asked for it.)

So we get the following graph.  I post a hand-drawn sketch to try to make the point of how totally irrelevant graphing software should be for this kind of problem.  The graph literally took me $15$ seconds to draw, including thinking time.

Final point: the graph clearly has a turning point somewhere between $-1$ and $2$.  We could work it out if we really needed to know it, but it would add little or nothing to the general accuracy of our graph.  Remember: minimum work!!!

Take some time to compute a number of $P(x)$ values, say every unit in range $[-10,10]$. Plot them by hand and adjust the scaling. Where it seems that the function changes behavior, try intermediate values.

You'll learn more this way.


Sketching on the complex plane $|z + 1| - |z - 1| = 4$ Sketch the set of points that satisfy $|z + 1| - |z - 1| = 4$ on the complex plane.

Wolfram alpha gives me an empty graph, I end up with the equation of an ellipse but with the condition $x>4$. The ellipse is the same as $|z + 1| + |z - 1| = 4$ but I had to be careful with the square roots.

So is it true that no complex number satisfies this?

Triangle inequality:

$$\lvert z+1\rvert = \lvert (z-1)+2\rvert \leqslant \lvert z-1\rvert + 2 \implies \lvert z+1\rvert - \lvert z-1\rvert \leqslant 2.$$

  So is it true that no complex number satisfies this?

Yes, that is true.

That is not true. Take $z=2$, then $$|z+1|+|z-1|=3+1=4$$

I believe this is the equation of a hyperbola, not an ellipse. The subtraction means it will be a hypomolobro

I can't comment or upvote, but whoever just said it's a hyperbola is my hero. #upvotes


Irrationals forming rationals Can we obtain every rational number from the multiplication of two irrational numbers? If not, which ones can we not obtain?

Let $r$ be a non-zero rational number.  Then $\pi$ and $r/\pi$ are irrational, and $\pi \cdot \frac{r}{\pi}=r$ is rational.  To see why $r/\pi$ is irrational, suppose for the sake of a contradiction that it were rational, so $r/\pi=r'$, where $r'\in \mathbb{Q}$; in fact, $r'$ is non-zero since the quotient of two non-zero numbers is non-zero.  But then $r=r'\pi$ and $r/r'=\pi$.  The quotient of two rational numbers is rational, so this would show that $\pi$ is rational, which is not the case.  Thus, $r/\pi$ must be irrational.

However, the same cannot be said about $0$, for if $ab=0$, then at least one of $a$ and $b$ must also be $0$.  The above proof that $r/\pi$ is irrational fails because in this case $r'=0$, and we could not divide by it.

Take the square root of a rational.  It's likely irrational.  Multiply that square root by itself, you have a rational.  Thus you have a big set of irrational * irrational = rational numbers.


$\lim_{x \to \infty} f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$ and the function is not uniformly continuous " Does there exist a continuous function $f: \mathbb{R} \to \mathbb{R}$  so that  $\lim_{x \to \infty} f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$ and the function is not uniformly continuous on $\mathbb{R}$?"
I wanted to prove that this is impossible because it implies that the function has a horizontal asymptote.
I tried to prove:
Take $\epsilon$>0 randomly.
Because $\lim_{x \to \infty} f(x)=1$ there exist an $N\in \mathbb{R}$ so that $|f(x)-1|<\epsilon/2$ for $x>N$.
Because $\lim_{x \to -\infty} f(x)=-1$ there exist an $M\in \mathbb{R}$ so that $|f(x)+1|<\epsilon/2$ for $x<M$.
Because $f$ is continous in $\mathbb{R}$ we take $a \in \mathbb{R}$. There exist a $\delta>0$ so that for $x \in \mathbb{R}$ , $|f(x)-f(a)|<\epsilon$ for $|x-a|<\delta$.
And this is were it gets hard. I thougt you can define a $\delta_{2}=\max\{N,M\}$.
Take $x,y$ random in $\mathbb{R}$ so that $|x-y|<\delta_{2}$. Then $|f(x)-f(y)|=|f(x)-1+1-f(y)|=|f(x)-1|+|1-f(y)|<\epsilon$
I don't know if I defined my $\delta_{2}$ correctly.
Can someone help me how to prove this correctly?

The function $\displaystyle\frac{x^{1/3}}{1+|x^{1/3}|}$ which has a vertical tangent line at the origin.

Ok, this one does not deserve credit, it is wrong, but will edit this to make it right... if possible
So let me make it right. There is no such function.
Proof. Take any $\varepsilon>0.$ Take $M>0$ such that
$|f(x)+1|<\frac\varepsilon3$ if $x\le-M$, and such that
$|f(x)-1|<\frac\varepsilon3$ if $x\ge M$.
Since $[-M,M]$ is compact (i.e., closed and bounded interval in this context), $f$ is uniformly continuous on this interval, and there is $\delta>0$ such that for
$x,x'\in[-M,M]$ if $|x-x'|<\delta$ then we have $|f(x)-f(x')|<\frac\varepsilon3$.
Then for any $x,x'\in(-\infty,\infty)$ if $|x-x'|<\delta$ we have
$|f(x)-f(x')|<\varepsilon$, that is, $f$ must be uniformly continuous on $(-\infty,\infty)$.
The inequality $|f(x)-f(x')|<\varepsilon$ is easy to verify considering cases... will not list them all, but one of the interesting ones (though they all are easy) is say $x\le M \le x'$ (and we may also assume that $\delta<M$ for convenience and without loss of generality).
So assume $x\le M \le x'$ and $|x-x'|<\delta$ (and hence $0<x$).
Then $|f(x)-f(x')|\le$
$\le|f(x)-f(M)|+|f(M)-f(x')|\le$
$\le|f(x)-f(M)|+|f(M)-1|+|f(x')-1|<$
$<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon$.
Thus $|f(x)-f(x')|<\varepsilon$.
Now that I posted my answer and read more of the comments, I see that what @questmath put in a comment, starting with
"$f$ is uniformly continuous on $[M,N]$..."
already provides (a clear enough hint for) the answer. I realize I only provided the details of a similar answer (with my notation being a bit different than what OP was asking, but I will keep it this way).
For that matter I also missed the very first comment, by @Surb (just read it, and understood what it says), already voted up three times because it is right and has what is needed...


Limit of the sum of a factorial series Is there any function $f$ over the positive integers such that

$$\lim_{n\rightarrow\infty}\frac{\sum_{i=0}^{n} {\frac{n!}{(n-i)!}}}{f(n)} = 1$$

and

$$f(n)\not\equiv\sum_{i=0}^{n} {\frac{n!}{(n-i)!}}$$
What is the definition of the function if it exists, or if it doesn't exist, why not?

(updated)

As it stands the question has an obvious answer: take $f(n) = \sum\limits_{i=0}^{n} \frac {n!} {(n-i)!}+1$.

PS If $a_n \to \infty$ there are lots of sequences $f(n)$ such that $\frac {a_n} {f(n)} \to 1$; for example $f(n)=a_n+c\sqrt {a_n}$ (where $c$ is a constant) is one such.


Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy

$$
  \textrm{det } X=
  \begin{vmatrix}
  x &2 &2 &2\\
  2 &x &2 &2\\
  2 &2 &x &2\\
  2 &2 &2 &x
  \end{vmatrix}\\
$$

My teacher does this by adding the three bottom rows to the top row

$$
\textrm{det } X=
  \begin{vmatrix}
  x+6 &x+6 &x+6 &x+6\\
  2 &x &2 &2\\
  2 &2 &x &2\\
  2 &2 &2 &x
  \end{vmatrix}\\
$$

and then subtracting a row of $2$'s from the bottom three rows

$$
\textrm{det } X=
  (x+6)
  \begin{vmatrix}
  1 &1 &1 &1\\
  0 &x-2 &0 &0\\
  0 &0 &x-2 &0\\
  0 &0 &0 &x-2
  \end{vmatrix}.
$$

The answer is

$$
x\in \{-6,2\}.
$$

I think I understand the operations (although subtracting an arbitrary row of numbers from a matrix/determinant row is something I've never seen before, but I don't see why that wouldn't be allowed. Just like you can subtract arbitrary coefficients on both sides of an equation, right?), my main issue is why they are performed.

Why can't I just in the same way subtract a row of $2$'s from the three bottom rows in the first determinant? If I do that I get a different answer.
I know I want a column of all zeroes except one column-element, but why do I need to perform the first operation beforehand? Is it somehow necessary that all the top row elements to be the same, $(x+6)$?

I think you should give a look at gaussian elimination.

You can only do certain manipulation the the matrix without changing it's determinant (or at least only changing it's sing or by some scalar). This manipulations are as follows:

Swap two rows multiplies the determinant by $-1$
Multiplying a row by a non zero scalar $\lambda$, multiplies the determinant as well by the same scalar
Adding or subtracting multiples of one row to another leaves the determinant untouched

The algorithm for gaussian elimination goes something like this

If the first row has first element zero than change it with some other row that has fist element not zero, if there aren't go to step 3
Fore every row $A_i$, not counting the first, $(i
\gt 1)$ multiplay the first row by some coefficient so that the sum of the first row and the row $A_i$ has fist element zero. Substitute the $i$-th row with the sum just calculated
Now every element of the fist column, except many for the first one, are zero. Now go to the sub-matrix obtained by eliminating the first row and column and repeat this steps on every submatrix.
At the end of this you should have a triangular or diagonal matrix

"Subtracting a row of 2s" is NOT a valid matrix operation.  But subtracting one row from another is.  Perhaps what your teacher did (or meant to do) is, first subtract the third row from the second to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 2 & 2 & x & 2 \\ 2 & 2 & 2 & x \end{array}\right|$.

Now, subtract the fourth row from the third to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2 & 2 & 2 & x \end{array}\right|$.

Finally, you can subtract the first row from the fourth to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2- x & 0 & 0 & x- 2 \end{array}\right|$.

That's almost an "upper triangular" matrix.  We can calculate the determinant reasonably easily by "expanding on the first column:
$x\left|\begin{array}{cc} x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \\ 0 & 0 & x- 2 \end{array}\right|$$- 2\left|\begin{array}{cc} 2 & 2 & 2 \\ x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \end{array}\right|$.

That first determinant is $x(x- 2)^3$.  For the second we can further expand that three by three determinant on the first column to get $-2\left(2\left|\begin{array}{cc}2- x & 0 \\ x- 2 & 2- x\end{array}\right|- (x- 2)\left|\begin{array}{cc}2 & 2 \\ x- 2 & 2- x\end{array}\right|\right)= -4(2- x)^2- (x- 2)(4- 2x- 2x+ 4)= -4(2- x)^2- (x- 2)(8- 4x)= -4(2- x)^2+ 4(2- x)^2= 0$.

So the determinant of the given matrix is $x(x- 2)^3$.


Is Hausdorffness necessary condition for the Stone-Weierstrass theorem? Following Rudin's argument, I cannot find where it used Hausdorffness condition.

=======

Rudin PMA p.162

Let $X$ be a compact space.
Let $\mathscr{A}$ be a subalgebra of $C(X,\mathbb{R})$.
If $\mathscr{A}$ separates points and vanishes nowhere, then $\mathscr{A}$ is dense in $C(X,\mathbb{R})$ with respect to the uniform topology.

======

The proof is divided into four steps.

$\forall f\in \overline{\mathscr{A}}, |f|\in \overline{\mathscr{A}}$
$\forall f,g\in \overline{\mathscr{A}}, \max(f,g),\min(f,g)\in \overline{\mathscr{A}}$.
$\forall ( f\in C(X,\mathbb{R}),x\in X, \epsilon >0), \exists g\in \overline{\mathscr{A}}$ such that $g(x)=f(x)\bigwedge g(x)>f(x)-\epsilon$.
$\forall ( f\in C(X,\mathbb{R}), \epsilon>0), \exists h\in \overline{\mathscr{A}}$ such that $||h-f||_\infty <\epsilon$.

I believe Steps 1&2 use only uniform continuity and classical Weierstrass theorem and Steps 3&4 only use compactness.

Where did I go wrong?
Where in the argument use the Hausdorff condition?

Wikipedia's statement of the theorem is

  Suppose $X$ is a compact Hausdorff space and $A$ is a subalgebra of $C(X,\mathbb R)$ which contains a non-zero constant function. Then $A$ is dense in $C(X,\mathbb R)$ if and only if it separates points.

(Emphasis added.)

What you've quoted from Rudin asserts only the "if" direction (separating points implies dense); for that statement, we can omit the assumption that $X$ is Hausdorff because (as noted by others here already) the assumption that $A$ separates points already implies that $X$ is Hausdorff.  (If, as suggested in comments, it is thought misleading to omit this condition — as Rudin does — I would strongly prefer merely to add a remark to the effect that the hypothesis implies Hausdorffness, not to add redundant hypotheses, which only make theorems harder to use.)

However, for the "only if" direction (dense implies separating points) the assumption that $X$ is Hausdorff cannot be dropped.  (Example: if $X$ has at least two points and the trivial topology, i.e., only $\emptyset$ and $X$ are open, then $C(X,\mathbb R)$ is a dense subalgebra of itself but doesn't separate points.)  I can't think of a use for this direction off the top of my head; I suspect it's much less important than the other direction.

In fact, if a topology is not Hausdorff, it makes no means. We can always assume that any topology is Hausdorff except in pure theory of topology.


How to remove minimum number of points of two sets such that their respective convex hulls are disjoint?? Suppose we are given two set of points A and B in a 2D plane, and their convex hulls are not disjoint, how to remove minimum number of points so that they become disjoint?

Let $c(A)$ denote the convex hull of the set of points $A$, then remove the points $(c(A)\cap c(B))\cap A$ or $(c(A)\cap c(B))\cap B$, whichever has lesser points.


Is $ 2.7182818281828...$ a semiconvergent of e? Euler's number $e=2.71828 18284 59045... $ can be approximated by the rational number:
$$
x=\frac{271,828-27}{100,000-10}= \frac{271,801}{99,990} =2.7182818281...
$$

Also, $e$ has the well-known continued fraction expansion
$$
e = [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...].
$$

Is the fraction $x$ a semiconvergent of $e$? That is, does there exist $ a $ such that
$$
x=\frac{h_{n-1} +a h_n}{k_{n-1}+a k_n},
$$
where $h_{n-1}, h_n$ and $k_{n-1}, k_n$ are successive numerators and denominators of convergents of $e$.

Background

It is known that $y=355/113$ is a close rational approximation to $\pi$ which is also easy to remember (113 355). I want to know if there is a similar way to relate $2.718281828...$ to $e$.

Yes!

In Mathematica, define

Semiconvergent[a_, b_, k_: 1] := (Numerator[a] + k Numerator[b])/(Denominator[a] + k Denominator[b])

Then

es = Convergents[E, 15]
Semiconvergent[es[[13]], es[[14]], 5]

Gives

271801/99990


Inverse matrix's eigenvalue? It's from the book "linear algebra and its application" by gilbert strang, page 260.

$(I-A)^{-1}$=$I+A+A^{2}+A^{3}$+...

Nonnegative matrix A has the largest eigenvalue $\lambda_{1}$<1.

Then, the book says, $(I-A)^{-1}$ has the same eigenvector, with eigenvalue $\frac{1}{1-\lambda_{1}}$.

Why? Is there any other formulas between inverse matrix and eigenvalue that I don't know?

A matrix $A$ has an eigenvalue $\lambda$ if and only if $A^{-1}$ has eigenvalue $\lambda^{-1}$. To see this, note that
$$A\mathbf{v} = \lambda\mathbf{v} \implies A^{-1}A\mathbf{v} = \lambda A^{-1}\mathbf{v}\implies A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$$

If your matrix $A$ has eigenvalue $\lambda$, then $I-A$ has eigenvalue $1 - \lambda$ and therefore $(I-A)^{-1}$ has eigenvalue $\frac{1}{1-\lambda}$.

If you are looking at a single eigenvector $v$ only, with eigenvalue $\lambda$, then $A$ just acts as the scalar $\lambda$, and any reasonable expression in $A$ acts on $v$ as the same expression in $\lambda$. This works for expressions $I-A$ (really $1-A$, so it acts as $1-\lambda$), its inverse $(I-A)^{-1}$, in fact for any rational function of $A$ (if well defined; this is where you need $\lambda_1<1$) and even for $\exp A$.


linear independent rows of a matrix If I have a matrix
$$
\begin{bmatrix}
1 & 2 & 1 \\
1 & 2 & 1 \\
1 & 2 & 1 \\
\end{bmatrix}
$$

I can reduce it to

$$
\begin{bmatrix}
1 & 2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
$$

Do I have one linearly independent row here or zero independent rows?

$$
\begin{bmatrix}
1 & 2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
$$

First think about what it means to be linearly independent. Linearly independent means that every row/column cannot be represented by the other rows/columns. Hence it is independent in the matrix.

When you convert to RREF form, we look for "pivots"

Notice that in this case, you only have one pivot.

A pivot is the first non-zero entity in a row.

Here is your pivot:

$$
\begin{bmatrix}
\color{red}{1} & 2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
$$

Therefore, since you only have one pivot, then two of your columns are not linearly independent, and one of them is. Therefore, we say that the $\dim(A)=1$, where $A$ is the matrix.

We say that the rows are not independant.

The space spanned by the rows has dimension 1.


Partially ordering finite graphs What are some interesting partial orders on the set of all finite graphs (identified up to isomorphism), apart from the usual (induced) subgraph relation and the (topological) minor relation, and why are they interesting?

"interesting" is subjective, but you could come up with several examples of partial orders (or total orders for that matter).

For example $\preceq_{\omega}$ which I define as $G\preceq_{\omega} H$ iff $\omega(G)\leq \omega(H)$ (where $\omega(G)$ denotes the clique number of the graph $G$, i.e. the maximal number of vertices in a complete subgraph of $G$)

Note that $\subseteq$ (the subgraph relation) is a subset of $\preceq_{\omega}$.  That is to say, if $G\subseteq H$ then $G\preceq_{\omega} H$.  The proof of which should be clear since if $K$ is a maximal clique of $G$ with $|K|=\omega(G)$, then $K\subseteq G\subseteq H$ and so $H$ has $K$ as a complete subgraph implying $\omega(H)\geq |K|=\omega(G)$.  It is also worth noting that this is a total order, unlike $\subseteq$, since all graphs can be compared.

You might try also playing with the similar idea of using instead $\alpha(G)$, the independence number of a graph and see how that relates to the subgraph relation.


Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem :

In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$

My approach :

Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get:
\begin{equation}
a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i})
\end{equation}

By using projection formula which states: $b=a\cos C +c \cos A $

(i) will become :

$$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here?

Law of sines states that

$$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$

which means that $c \sin{A} - a \sin{C} = 0$.  The desired result follows.

Which means that $c \sin(B+C)=a \sin(A+B)$

$$c\sin(B+C)-a\sin(A+B)=O$$


Compact image is compact? Let $f: X \to Y$ be a function between metric spaces $X, \phi$ and $Y, d$ such that $f^{-1}(U)$ is open in $X$ for every subset $U$ open in $Y$. Prove that if $C$ is a compact subset of $X$, then $f(C)$ is a compact subset of $Y$.

My attempt: Let $C$ be a compact subset in $X, \phi$. Let $\{V_{\alpha}\}$ be a collection of open sets such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha}$. Then there exist $\alpha = 1, 2, ..., n$ such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha_n} : n \in \mathbb{N}$.

I want to show this subcover is open and thus it has an open image, but I'm not sure as if that's correct because only the converse is true. Is it true?

I need way better notation. Also, how to continue - any suggestions?

You wish to show the image of a compact set is compact under a continuous mapping.  This can be done by taking finite sub-covers.  Let $\{\mathcal{O}_\alpha\}_{\alpha\in A}$ be an open cover of $f(C)$. Then $\{\mathcal T_\alpha\}_{\alpha \in A}$ is an open covering of $C$, where $T_\alpha \equiv f^{-1}(\mathcal{O}_\alpha)$ for each $\alpha \in A$.

Now choose $\alpha_1 \ldots \alpha_n$ for some $n \in \mathbb{N}$ such that $C \subset \cup_{i=1}^n T_{\alpha_i}$.  This can be done since $C$ is compact.

I will leave it to you to show that $\cup_1^n f(T_{\alpha_i}) = \cup_1^n \mathcal{O}_{\alpha_i} \supset f(C)$.

Since we have taken an arbitrary open cover of $f(C)$ and found a finite subcover, then $f(C)$ must be compact.

Since we have chosen an arbitrary compact set $C$ and shown $f(C)$ is compact, then the image of any compact set is compact under $f$.

And finally, since $f$ is an arbitrary continuous function, we have shown that continuous functions map compact sets to compact sets.

We can see that map $f$ is a continuous , for a continuous i have some equivalent conditions in metric space:

$a)$ $f$ is a continuous map

$b)$ The image of an open set is an open set

$c)$ The image of an close set is an close set .

So $f$ is continuous , i will show that if $C$ is a compact set then $f(C)$ is a compact set , I choose an any infinity sequence $f(x_1),f(x_2),..f(x_n),,,$ in $f(C)$ , because $C$ is compact so i can choose $(x_{k_n})$ of $(x_n)$ that converging and $f$ is continuous then $f(x_{k_n})$ is converging . Finally i have $f(C)$ is compact set


Let $x=e^a$, $y=e^b$, $z=e^c$ $\;$

and  $a\ge b \ge c$ (WLOG)

Our inequality is equivalent to
$$3e^{a+b+c} + e^{3a}+ e^{3b} + e^{3c} \ge 2 \left( e^{3(a+c)/2} + e^{3(a+c)/2} + e^{3(a+b)/2} \right)$$

If $a+c\le 2b$ ,

This inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec  \left( 3a,3b,(a+b+c),(a+b+c),(a+b+c),3c \right)$$

The majorization holds because,

$$3a\ge \frac{3(a+b)}{2}$$

$$3a+ 3b \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$

$$3a+ 3b + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$$3a+ 3b + (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$$3a+ 3b + (a+b+c) + (a+b+c)+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}+ \frac{3(b+c)}{2}$$

$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$

$3^{rd}$ and $4^{th}$ inequalities by $a+ c \le 2b$. $\;$ Other inequalities are obvious.

If $a+c\ge 2b$ ,

The desired inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec  \left( 3a,(a+b+c),(a+b+c),(a+b+c),3b,3c \right)$$

The majorization holds because,

$$3a\ge \frac{3(a+b)}{2}$$

$$3a+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$

$$3a+ (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$\dots$

$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$

$3^{rd}$ inequality comes from $a+ c \ge 2b$. $\;$ Other inequalities are obvious.

First, Schur's inequality provides that
$$
x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x).
$$
Then,
$$
xy(x+y)\ge 2xy\sqrt{xy}
$$
and hence
$$
x^3+y^3+z^3+3xyz\ge 2\big(xy\sqrt{xy} +yz\sqrt{yz} +zx\sqrt{zx} \big)
$$


Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$

I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$

I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.

I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$

Cant find a way to proceed.

Thanks for helping.

So we have to deal with

$$ S=\sum_{k\geq 1}(-1)^{k+1} \sum_{n=\binom{k}{2}+1}^{\binom{k+1}{2}}\frac{1}{n} = \sum_{k\geq 2} (-1)^k A_k $$
and to prove convergence it is enough to show that $\{A_k\}_{k\geq 1}$ is decreasing (from some point on) and convergent to zero. The last claim is straightfoward to prove, since $A_k\geq 0$ but
$$A_k\leq \frac{k}{\binom{k}{2}+1}\sim\frac{2}{k}. $$
The first claim can be easily proved by induction or a convexity argument:
$$ A_k = H_{\binom{k+1}{2}}-H_{\binom{k}{2}}, $$
hence $\{A_k\}$ is decreasing iff $k\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function. Since:

$$\begin{eqnarray*} H_n=\sum_{k=1}^{n}\frac{1}{k} &=& \sum_{k=1}^{n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)+\sum_{k=1}^{n}\log\frac{k+1}{k}\\&=&\gamma+\log(n+1)-\sum_{k>n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)\end{eqnarray*}$$
and $\left(\frac{1}{k}-\log\frac{k+1}{k}\right)$ is bounded between $\frac{1}{2k(k+1)}$ and $\frac{1}{2k^2}$, we have:
$$ H_{\binom{k+1}{2}}-H_{\binom{k}{2}} = \log\frac{k^2+k+2}{k^2-k+2}+O\left(\frac{1}{k^4}\right)$$
so $x\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function from some point on.

We may also notice that the value of our series depends on the integral:
$$ I =\int_{0}^{1}\frac{1+2\sum_{k\geq 1}(-1)^k x^{\binom{k}{2}}}{x-1}\,dx.
 $$

1.in case of non absolute convergence, grouping is not a good idea.
2.Sum $S, \quad \frac {1}{2}<S<1$, using Bolzano-Weisrass theorem convergence follows


if $x_{n+1}=ax_{n}+b$ then How find this $a,b,c$? Question:

  Let there be a sequence $x_{n}$ such that $$x_{n+1}=ax_{n}+b,,x_{1}=c$$
  where $a,b$ and $c$ are positive integers.

Suppose, too, that for $n,m\in \mathbb{Z^{+}}$ we have that
$$
\text{(*)   } ~~~n\mid m\implies x_{n}|x_{m}
$$

  show that:$$b=c$$

My try: if $a=1$, then we have
$$x_{n+1}-x_{n}=b,x_{1}=c\Longrightarrow x_{n}=c+(n-1)b=bn+c-b$$
So  if
$n|m$ then let $m=nk$, so
$$\dfrac{x_{m}}{x_{n}}=\dfrac{bkn+c-b}{bn+c-b}=k+\dfrac{c-b-kc+kb}{bn+c-b}\in N^{+}$$
So how do I find $a,b$ and $c$?

Case $2$: if $a\neq1$, then we have
$$x_{n}+\dfrac{b}{a-1}=a(x_{n}+\dfrac{b}{a-1})$$
So
$$x_{n}=\left(c+\dfrac{b}{a-1}\right)a^{n-1}-\dfrac{b}{a-1}$$

By the way:

Sometime ago, I solved the following hard problem: (IMO shorsits)

If $a,b\in \mathbb{Z^{+}}$, and for any positive integer $n$ we have
$$\dfrac{a^n-1}{b^n-1}\in N^{+}$$

Then we also must have that there exists some positive integer $k$ so that $a=b^k$

Then I can't, thank you

This problem was created by a China IMO team student.

I'm going to restate the problem, and give a partial answer:

Find all positive-integer triples $a, b, c$ such that the arithmetic sequence
$$
x_n = a x_n + b; x_1 = c
$$
has the property that $n | m \implies x_n | x_m$.

Suppose that the sequence has this property. Then consider the case $n = 1$, which divides any other number $m$; we get that $x_1 | x_m$ for every $m$. In particular, the number $c = x_1$ must divide $x_2 = ac + b$; that means that
$$
c | b.
$$
What about $x_3$? That's
$$
x_3 = a(ac + b) + b = a^2 c + ab + b
$$
which is also divisible by $c$ (since $ac$ and $b$ both are). A simple induction argument shows that $c | x_n$ for every $n$ once we know that $c | b$.

The stuff above is a cleaner version of what I wrote earlier. But $c | b$, although necessary, is not a sufficient condition. Let's examine further. We have
$$
x_4 = a^3 c + a^2 b + ab + b = a^2(ac + b) + (a+1)b
$$
which must be divisible by $x_2 = ac + b$. Thus we must have $(ac + b) | (a+1) b$. Writeing $b = sc$, we get $(ac + sc) | (a+1) sc$, so $(a + s) | (a+1)s$, so $(a + s) | (as + s)$.

So just using the first few terms, we find that

(1) $c | b$. Writing $b = cs$, we also have

(2) $(a+s) | (as + s)$

So those are a couple of constraints on $a, b, c$.


Lusin's Theorem for finite measure spaces Lusin's Theorem. Let $f$ be a real-valued measurable function on $E$. Then for each $\varepsilon>0$,  there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\sim F)<\varepsilon$.

can anybody guide me the proof of the theorem or can advice me some book from where i can go through the proof.

Here is a proof of the Lusin's Theorem where $f:[a,b]\to \mathbb{C}$.

Lusin's Theorem: If $f:[a,b]\to \mathbb{C}$ is Lebesgye measurable and $\epsilon > 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) < \epsilon$ and $f|E$ is continuous. (Note I will use Egoroffs theorem, Theorem 2.26, theorem 1.18, and Corollary 2.32 in Folland's Real Analysis, I am sure there are the same theorems in your book)

Proof: Let $f:[a,b]\to \mathbb{C}$ be Lebesgue measurable and $\epsilon >0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\to f \in L^1$$
Then by corollary 2.32 there is a subsequence $\{g_{n_j}\}$ of $\{g_n\}$ such that $g_{n_{j}}\to f$ a.e. Now by Egoroff's theorem, for any $\epsilon > 0$ there exists a set $F\subset [a,b]$ with $\mu(F) < \epsilon/2$ such that $g_{n_j}\to f$ uniformly on $F^{c}$.

Now by theorem 1.18, since $\mu([a,b]) < \infty$, there is $E$ compact subset of $[a,b]$ such that $E\subset F^{c}$ and $$\mu(F^c) - \epsilon/2 < \mu(E) \leq \mu(F^c)$$ So $F\subset E^c$ and we have
\begin{align*}
\mu(E^c) &= \mu(F) + \mu(E^c\setminus F)\\
&= \mu(F) + \mu(E^c\cap F^c)\\
&= \mu(F) + \mu(F^c\setminus E)\\
&= \mu(F) + (\mu(F^c) - \mu(E))\\
&\leq \epsilon/2 + \epsilon/2\\
&= \epsilon
\end{align*}
Note since $E\subset F^c$ and $g_{n_j}\to f$ uniformly on $F^c$, we have that $g_{n_j}\to f$ uniformly on $E$. Since, for all $j$, $g_{n_j}$ is continuous, we have that $f$ is continuous on $E$, that is $f|E$ is continuous.

Hope this helps.


Maple - PDE substitution Lets say I have the heat equation:

$$u_t-u_{xx}=0.$$

I want to apply the following substitution to the heat equation in Maple.

$$\zeta=t/x^2$$
$$u(x,t)=x^cF(t/x^2)=x^cF(\zeta)$$

How can i achive this?

The result should be:

$$cF-c^2F+\frac{d F}{d\zeta}-6\zeta\frac{d F}{d\zeta}+4c\zeta\frac{d F}{d\zeta}-4\zeta^2\frac{d^2 F}{d \zeta^2}=0$$

In Maple, you could do it like so,

eq := diff(u(x,t),t) - diff(u(x,t),x,x) = 0:
s1 := t/x^2 = xi(x,t):
s2 := u(x,t) = x^c*F(t/x^2):

eval(eq, [s2, s1]):
normal( simplify( %, {s1} ) ):
simplify( %/x^(c-2), power ):

And, depending on how you want xi and the derivatives represented,

convert( subs(xi(x,t)=xi,%), diff );

And terms may be grouped together,

#simplify(%,size);
collect(%,diff,factor);

                                     /  2        \
                    / d        \     | d         |   2
(4 c xi - 6 xi + 1) |---- F(xi)| - 4 |----- F(xi)| xi  - F(xi) c (c - 1) = 0
                    \ dxi      /     |    2      |
                                     \ dxi       /

I don't see how Maple is involved in this.

Nevertheless, I agree with the expected result, as shown below :


The sum of the largest $n$ integers is greater than the sum of the remaining $n+1$ Let $a_1\geq a_2\geq\ldots\geq a_{100n}$ be positive integers such that if we choose any $2n+1$ numbers from them, the sum of the largest $n$ is strictly greater than the sum of the remaining $n+1$. Prove that

$$(n+1)(a_1+\ldots+a_n)>a_{n+1}+\ldots+a_{100n}.$$

For the case $n=1$, this becomes $a_1\geq\ldots\geq a_{100}$, and we need to prove $2a_1>a_2+\ldots+a_{100}$. We have $a_1>a_2+a_3, a_2>a_3+a_4, \ldots$, and it's not clear how the desired inequality follows.

$$
2a_1 > 2a_2+2a_3>a_2+2a_3+a_3+a_4 = a_2+a_3+a_4+2a_3
$$
Similar for $a_3$ and you will gain desired inequality. Then just use induction.


Probability dependent on two independent factors If we know that $P(A|B) = R_1$ and $P(A|C) = R_2$ with $R_2 > R_1$ what is the probability of $A$ given both $B$ and $C$ are true (perhaps denoted $P(A|B \wedge C)$)

Intuitively it would seem like it is greater than $R_1$ and less than $R_2$ but I can't seem to figure out what it is exactly.

I apologize if this is a duplicate, I might just not know the terminology to search for this correctly

Both are false $P(A|B\cap C)$ could take any value from 0 to 1:

Consider that we choose an uniformly randomly number $X$ from $\{1,2,3\}$

Let $A = \{ X= 1 \ \mathrm{or} \ 2\}$, $B=\{X=2 \ \mathrm{or} \ 3 \}$, $C=\{X=1 \ \mathrm{or} \ 3 \}$

Well $P(A|B\cap C)=0$ and $P(A|B) = P(A|C)= \frac 12$

Now let's say $A=\{X=3  \}$

Well $P(A|B\cap C)=1$ and $P(A|B) = P(A|C)= \frac 12$

I think it must be larger than both $R_1$ and $R_2$.

For example, suppose we roll a fair die. Let $A$ be the event that we get a $6$, $B$ the event that we get a multiple of $2$, $C$ the event we get a multiple of $3$.

Then $\Bbb P (A|B)=\dfrac 13$ and $\Bbb P (A|B)=\dfrac 12$ where $\dfrac 12 > \dfrac 13$ as required.

However, $\Bbb P (A|B \cap C)=\Bbb P (\text{roll a }6|\text{multiple of 6})=1$

EDIT:

Note that we need the condition that $B \cap C \neq \emptyset$ since  $\Bbb P(A| \emptyset )$ is undefined.


Revisit "How can I visualize the nuclear norm ball" Revisiting How can I visualize the nuclear norm ball?

Two eigenvalues are reproduced as following:
$$ s_{1,2}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2\pm|x+z|\sqrt{(x-z)^2+4y^2}}. $$
According to the following (from a paper)

If a symmetric matrix: $$ A=\left( \begin{array}{cc} x & y\\ y & z\end{array} \right)$$
is rank $1$, then $y=\sqrt{xz}$, which comes from the fact that $vv^T$ is rank $1$ and any rank $1$ matrix can be represented in this form.
$$\left[\begin{array}{cc} v_1\\ v_2\end{array}\right]\left[\begin{array}{cc} v_1 & v_2\end{array}\right]=\left( \begin{array}{cc} v_1^2 & v_1v_2\\ v_1v_2 & v_2^2\end{array} \right)$$

My question: how to explain the red circle in figure (b) is the $2\times 2$ symmetric unit-Euclidean-norm rank $1$ matrix?
This is a circle in $3$-D, how to get the equation of this circle through the rank $1$ matrix provided above?

I believe just replace $y=\pm\sqrt{xz}$ in $s_{1,2}$ and can get the answer.
So I choose the larger one of $s_{1,2}$:
$$ s_{\max}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2 + |x+z|\sqrt{(x-z)^2+4y^2}}= \sqrt{2(x+z)^2}=1 (\text{unit norm})$$

The answer to the linked question shows that the curved side of the "cylinder" satisfies the equation
$$(x-z)^2+4y^2=1$$
and its planar caps satisfy
$$(x+z)^2=1.$$
The red curves are the intersection of the two surfaces, so they satisfy both equations. Therefore, they must also satisfy their sum,
$$x^2+4y^2+z^2=2.$$
Thus the curves lie on the intersection of the ellipsoid $x^2+4y^2+z^2=2$ and the planes $x+z=\pm1$, so they are ellipses (but not necessarily circles).

$A$ must be rank-$1$ and symmetric. Thus
$$A=vv^T$$
This implies that
$$y^2=xz$$
In this problem, we also have $\|v\|=1$, so
$$x+z=1$$
All of these result in
$$y=\pm\sqrt{x(1-x)},\quad z=1-x;$$


Question about an integrable singularity I've been trying to understand a concept of an integrable singularity. So far what I have discovered was that the case of the singularity occurs when there is a point where the integration becomes infinite. Am I right? Also, is there anything else that would add to the definition of the integrable singularity?
Thanks.

You can think of integrable singularity as a singularity of integrand that will disappear under suitable change of variables. Consider for example:

$$
   \int_{0}^1 \frac{1}{\sqrt{1-x} } \mathrm{d} x
$$
Then for $1-x = y^2$ and $-\mathrm{d}x = 2 y \, \mathrm{d} y$:

$$
   \int_{0}^1 \frac{1}{\sqrt{1-x} } \mathrm{d} x =
     \int_{0}^1 \frac{1}{\sqrt{y^2} } 2 y \, \mathrm{d} y = 2

$$

Why you take $1-x=y^{2}$ in a substitution ?? is something wrong in further calculations If i take $1-x=y$


Show that if $ f'(0)=1, f'(x)=e^x $ with the law of exponents I'm really stuck on this one. Show that, if $f'(0)=1$ is known, one can deduce
$
f'(x)=e^x
$ with the aid of the law of exponents.
I started here and assumed that I should go towards L'hospital's rule but I get stuck after.

$\lim \limits_{x \to 0} \frac{f(x) - f(0)}{x}
$

The wording exactly:
Let $f(x)=e^x$. Explain why $ f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$. Show that, if $f'(0)=1$ is known, one can deduce $f'(x)=e^x$ with the aid of the laws of exponents.

The first part was trivially easy but the way the problem is worded makes it seem like two separate sets of facts to deduce from.

EDIT: To clarify for you, the problem is asking you to derive the fact that the derivative of $e^x$ is $e^x$ from the basic laws of exponents, rather than as an inherent rule of derivatives.

Given $f(x) = e^x$ we see that $f'(a) = \lim\limits_{x \to a} \dfrac{e^x - e^a}{x-a}$. Therefore $f'(0) = \lim\limits_{x \to 0} \dfrac{e^x - e^0}{x-0} = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x}$. Given that $f'(0) = 1$, we naturally see $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$.

So we now know $f(x) = e^x$ and $f'(0) = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$. Use these facts to show $f'(x) = e^x$. This can be done as follows:

The general form of a derivative $f'(x) = \lim\limits_{h \to 0}\dfrac{f(x+h) - f(x)}{h}$. In this case, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^{x+h}-e^x}{h}$. By the properties of exponents, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^xe^h-e^x}{h}=\lim\limits_{h \to 0}\dfrac{e^x(e^h - 1)}{h}$.

Because the limit is with respect to $h$ (not $x$), $e^x$ is a constant, and can be pulled out of the limit such that $f'(x) = e^x \lim\limits_{h \to 0} \dfrac{e^h - 1}{h}$. Substituting $h$ for $x$ in the given fact that $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$ we see $f'(x) = e^x (1) = e^x$.

There exist an infinite number of functions whose derivative, at x= 0, is 1. It is certainly NOT true that the derivative must be f'(x)= e^x.


Riemann $\zeta$ and Chebyshev's estimates I would like to shred some light concerning the relations between properties of the Riemann $\zeta$ function and the weak version of the Prime Number Theorem proven by Chebyshev. More precisely, we have the following known facts :

the pole at $s=1$ is equivalent to the infinitude of primes
the non-existence of zeros on $\sigma = 1$ is equivalent to the PNT in the form $\pi(x) \sim x / \log x$
the known (or conjectured) zero-free region implies estimates for the error term

However, the proofs I find concerning the Chebyshev's result
$$\pi(x) \asymp \frac{x}{\log x}$$

do not use any analytic property of the Riemann zeta function but rather computations with orders of arithmetic functions (I do not know whether or not the proof I have in mind is Chebyshev's).

  Is there any proof using analytic properties of $\zeta$?

My attention has been caught by a remark in Montgomery-Vaughan's book (Multiplicative Number Theory), stating that Chebyshev's estimates have been derived from the behavior of $\log \zeta(s)$ when $s \to 1^{+}$.

How about this for a start? It shows $\limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} \ge 1$.

For $\sigma > 1$, $$\log \zeta(s) = \log \prod_p (1-p^{-s})^{-1} = \sum_p \sum_{m \ge 1} \frac{1}{mp^{ms}} = \sum_p \frac{1}{p^s}+O_{s \to 1}(1).$$ Now let $s=1+\epsilon$ for small $\epsilon > 0$. Then $$\log \frac{1}{\epsilon} \sim \log \zeta(s) = \sum_p \frac{1}{p^s} = \lim_{x \to \infty} \sum_{p \le x} \frac{1}{p^s} = \lim_{x \to \infty} \frac{\pi(x)}{x^s}+s\int_1^x \frac{\pi(t)}{t^{s+1}}dt = (1+\epsilon)\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt.$$

Now if there were some $\delta > 0$ so that $\pi(t) \le (1-\delta)\frac{t}{\log t}$ for all large $t$, then $$1 = \lim_{\epsilon \downarrow 0} \frac{\log(1/\epsilon)}{\log(1/\epsilon)} = \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt \le \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}(1-\delta)\int_2^\infty \frac{1}{t^{1+\epsilon}\log(t)}dt = 1-\delta,$$ a contradiction.

A simple Google search for
"proof of prime number theorem"
turned up this,
which appears to be what you want:

/https://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf

Here is the title
and first paragraph:

Newman's Short Proof of the Prime Number Theorem

D. Zagier

Dedicated to the Prime Number Theorem on the occasion of its 100th birthday

The prime number theorem, that the number of primes < x is asymptotic to x/log x, was proved (independently) by Hadamard and de la Vallee Poussin in 1896. Their proof had two elements: showing that Riemann's zeta function $\zeta(s)$ has no zeros with Re(s) = 1, and deducing the prime number theorem from this. An ingenious short proof of the first assertion was found soon afterwards by the same authors and by Mertens and is reproduced here, but the deduction of the prime number theorem continued to involve difficult analysis. A proof that was elementary in a technical sense-it avoided the use of complex analysis-was found in 1949 by Selberg and Erdos, but this proof is very intricate and much less clearly motivated than the analytic one. A few years ago, however, D. J. Newman found a very simple version of the Tauberian argument needed for an analytic proof of the prime number theorem. We describe the resulting proof, which has a beautifully simple structure and uses hardly anything beyond Cauchy's theorem.


convergence and divergence of series with a summand Does $\sum_{n \in \mathbb N}\frac{n!}{(2n)!}$ converge?

I tried the ratio test however confused especially with the summand before it

Let $a_k=\frac{k!}{(2k)!}$. For the Ratio Test, we examine
$$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.\tag{1}$$
In our case, we have
$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!(2n)!}{n!(2n+2)!}.\tag{2}$$
The above expression simplifies to
$$\frac{1}{2(2n+1)},$$
which has limit $0$ as $n\to\infty$. Thus the series converges. Fast.

We do the Ratio Test on the series as follows.
$$\lim_{n \to \infty}\left|\frac{\frac{(n + 1)!}{(2n + 2)!}}{\frac{n!}{(2n)!}}\right|$$
$$= \lim_{n \to \infty}\left|\frac{(n + 1)!}{(2n + 2)!} \times \frac{(2n)!}{n!}\right|$$
$$= \lim_{n \to \infty}\left|\frac{n + 1}{(2n + 2)(2n + 1)}\right|$$
$$= \lim_{n \to \infty}\left|\frac{1}{2(2n + 1)}\right|$$
$$= 0.$$

Since this limit is $0 < 1,$ the series must converge.


Check this measure theory argument Here's a problem from Cohn's Measure Theory.

  [Cohn, 2.1.3] Let $f$ and $g$ be continuous real-valued functions on $\mathbb{R}$. Show that if $f =g$ for $\lambda$-almost every $x$, then $f = g$ everywhere.

Here $\lambda$ denotes Lebesgue measure.

Here's what I do.

Let $h: \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose that $h \neq 0$. Then for some $x \in \mathbb{R}$, there is $n \in \mathbb{N}$ such that $|h(x)| > 2/n$. Then by continuity for some $\delta > 0$, we may ensure that whenever $|y - x| < \delta/2$, we have
$$
|h(y)| \geq |h(x)| - |h(y) - h(x)| > 2/n - 1/n = 1/n.
$$
Hence, $\lambda(\{h \neq 0\}) \geq \lambda(\{|h| > 1/n\}) \geq \delta > 0$. Taking the contrapositive, we just showed if $h$ is a continuous real-valued function on $\mathbb{R}$, then
$\lambda\{h \neq 0\} = 0$ implies $h = 0$ (everywhere). The result follows now by setting $h = f - g$, which is continuous.

Question. Is this basically the simplest way to do it? Other ways?

I would say that $\{x:h(x)\ne0\}$ is an open set in $\Bbb R$. (Since it is
the inverse image of the open set $\Bbb R\setminus\{0\}$ under the continuous
function $h$). Each nonempty open set contains a nonempty open interval,
which has positive Lebesgue measure. So if $h$ is continuous, and not identically
zero, then it cannot be almost everywhere zero (w.r.t. Lebesgue measure).

(For completeness, following Lord Shark the Unknown's advice.)

Let $h: \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose that $h \neq 0$. Thus, the set $\{h \neq 0\}$ is nonempty and open, and hence has positive Lebesgue measure. Taking the contrapositive, if $h$ is a continuous real-valued function on $\mathbb{R}$, then $\lambda\{h \neq 0\} = 0$ implies $h = 0$ (everywhere). Now take $h = f - g$. The result follows.


What is geometry, algebra, or topology? I have trouble grasping the notion of geometry, algebra, and topology. An example is when someone might say, "I study the geometry of jet spaces" or "I study the lie algebras". What does it mean for a problem to be geometric, or algebraic, or topological? What separates an algebraic problem from a geometric or topological one? I feel like if I better understood these words I would understand the big picture more. There are probably more to add besides geometry, algebra, and topology but I couldn't think of any more.

This is a bit simplified, but it's my view on these areas:

Geometric studies consider planes and curves, and how they bend and twist.
Algebraic studies consider structures of arithmetic and of symmetry.
Topologic studies consider continuity of functions.
Analysis studies differentiation and integration, series and limits.

These can also be mixed. For instance:

In algebraic topology you construct a structure of continious functions that you can combine in some specified way, and you study the arithmetical structure that arises (there are loads of ways to do it, and they all have different names like fundamental group, homology groups and cup product).
In differential geometry you use analysis (differentiation) to analyse how curves curve, or to study the possible vector fields in a space (an example of a vector field on a space is the globe of the Earth, with details on how the wind is blowing on each point).

Try these simpler words; Algebra is the use of simple equations and variables to solve a problem. Geometry uses these equations and variables and applies then with various theorems to shapes. Algebra II is more complex equations and can deal with several variables in one problem whilst general topology is based solely on set theory and concerns itself with structures of sets. Hope these basic mental sets help.


Adding covariance to a multi-dimension random vector equivalence For a 1D random variable X, we know that $\mathrm{Var}(cX) = c^2 \mathrm{Var}(X)$ and $\mathrm{Var}(X+c) = \mathrm{Var}(X)$.

I'm reading some notes that talk about a model defining a joint distribution on $(x,z)$ as
$$z \sim \mathcal{N}(0,I) \\
x|z \sim \mathcal{N}(\mu + \Lambda z, \Psi)$$

where the parameters of my model are the vector $\mu \in\mathbb{R}^n$, the matrix $\Lambda \in \mathbb{R^{n\times k}}$ and the diagonal matrix $\Psi \in \mathbb{R}^{n\times n}$.

They say that this is equivalent to the model
$$z \sim \mathcal{N}(0,I)\\ \epsilon \sim \mathcal{N}(0,\Psi)\\ x = \mu + \Lambda z + \epsilon
$$

where $\epsilon$ and $z$ are independent. They reason that this is true because we first simulate $z$, then simulate $x$ given $z$ and then ADD a covariance $\Psi$ noise to $\mu + \Lambda z$.

I don't get how it's equivalent, and why we "add" the noise when adding constants to a variance keeps it unchanged.

Can someone clarify how they're equivalent as well? with an intuition or possibly a mathematical proof?

If $\epsilon$ was not added to $x$, then $x$ could not have been a random variable given $z$, since $z$ would have determined $x$ un-randomly. To say why these forms are equivalent, note that if $$x=\mu+\Lambda z+\epsilon$$then $$x|z{\sim \mathcal{N}(\mu+\Lambda z+\mu_{\epsilon},\sigma^2_{\epsilon})\\\sim \mathcal{N}(\mu+\Lambda z,\Psi)}$$since given $z$ the quantity $\mu+\Lambda z$ serves as a deterministic number. In contrast if $$x=\mu+\Lambda z$$then $$x|z\sim\mathcal{N}(\mu+\Lambda z,0)$$which is a zero variance random variable or technically, a fixed scalar.

Alternative Intuition

Another intuition justifying why there should be two sources of randomness ($z$ and $\epsilon$) is that the random variable $x|z$ passes two steps; first: $z$ is randomly determined (first source of randomness); second: after determining $z$, the PDF of $x$ can be determined from $\mathcal{N}(\mu+\Lambda z,\Psi)$. Now $x$ will be determined obeying this PDF randomly (second source of randomness). Any equivalent equation for $x$ then, must have 2 sources of randomness, intuitively.

To show that both are equivalent - we first show that they both follow a multi-variate normal distribution. Then we show they have the same mean and covariance matrix.

It is trivial to see that both the distributions follow a normal distribution.

For the first representation, lets use the expectation and variance decomposition formulas:

The mean of a random variable $X$ is
$$\Bbb E(X) = \Bbb E_z(\Bbb E(X| Z=z)) = \Bbb{E}_z(\mu+\Lambda z) = \mu$$

The covariance of the random variable $X$ is
\begin{align}
\text{var}(X) & = \Bbb E(\text{var}(X|Z)) + \text{var}(E(X|Z)) \\ & = \Bbb{E}(\Psi) + \text{var}(\mu + \Lambda z) \\ & = \Psi + \Lambda \Lambda^T
\end{align}

For the second representation,
the mean is $\mu$ because $z$ and $\epsilon$ are independent and expectation is a linear operator. The covariance for $X$ is $\Psi + \Lambda \Lambda^T$ because the variance of two independent variables is sum of their individual variances - here the indepedent variables are $\Lambda z$ and $\epsilon$.

Hence they are equivalent.


Subspace of $\mathbb{R}^3$ $\{(x,y,z)\mid 2x^2-8y^2-9z^2\le 1\}$

Why is this not a subspace of $\mathbb{R}^3$?

I know the zero vector is in this set because $ 2(0) - 8(0) - 9(0) \le 1$

But I can't seem to verify if the set is closed under linear combination.

Note: This is NOT for a test/marked assignment, simply practice.

Note that $\displaystyle v=\left(\frac{1}{\sqrt{2}},0,0\right)$ is a vector in this collection. Is $2\cdot v$ in this collection?

I think it's because it's not under R^n by scalar multiplication. If u choose

2(1)-8(1)-9(1) = -15

And if u then multiply by a scalar -1, you get 15 which is not defined in the collection.


"multiplicative inverse in the modulo of the larger number" what does that mean? while I was reading this artical I have read the following paragraph:

  The interesting thing is that if two numbers have a $\gcd$ of $1$, then the smaller of the two numbers has a multiplicative inverse in the modulo of the larger number. It is expressed in the following equation:

and he gives the following example:

Lets work in the set $\mathbb{Z_9}$, then $4\in\mathbb{Z_9}$ and $\gcd(4,9)=1$.

Therefore $4$ has a multiplicative inverse (written $4^{−1}$) in $\bmod9$, which is $7$.

And indeed, $4\cdot7=28\equiv1\pmod9$.

But not all numbers have inverses.

For instance, $3\in\mathbb{Z_9}$ but $3^{−1}$ does not exist!

This is because $\gcd(3,9)=3\neq1$.

but what I do not understand is what does he mean by:

  then the smaller of the two numbers has a multiplicative inverse in
  the modulo of the larger number.

and how he got the $7$

The two numbers in his example are $4$ and $9$. The statement is that $4$ has a multiplicative inverse in the integers modulo $9$, or in other words, there is an integer $n$ such that $4 \cdot n \equiv 1 \mod 9$. The $7$ can be obtained by some trial and error (you only need to check the integers $1$ through $9$). He then gives an example of an integer that does not have a multiplicative inverse modulo $9$, namely $3$.

A multiplicative inverse of a number $a\in R$ for a ring $R$ is a number $a^{-1}$ such that $aa^{-1}=1$.

What he means by saying "modulo the bigger number" which we can denote $b$ is a multiplicative inverse of $a$ (the smaller number) in the ring $\mathbb{Z_b}$ which is the ring of the integers under the identification $\forall c,d \in \mathbb{Z} \ \ \bar{c}=\bar{d}\in\mathbb{Z}_b \  \iff c\equiv d(\mbox{mod} b) $.

Note this is intereseting because in general $\mathbb{Z}_b$ is a ring where we do not know if we have multiplicative inverses.

$\mathbb{Z}_b$ is a field (commutative ring with inverses) $\iff \ b$ is a prime number.


Extension of $|\cdot|_\infty$ on $\mathbb R$ to $\mathbb C$ Let $|\cdot|$ be the usual absolute value on $\mathbb C$. My question is:

  Is the only extension of $|\cdot|$ on $\mathbb R$ to $\mathbb C$ $|\cdot|$ itself?

I'm not sure about the uniqueness. I want to show any two extensions of $|\cdot|$ to $\mathbb C$ induce same topologies, which means that one is a positive power of the other (and so are equal).
Is it true any two absolute values on $\mathbb C$ induce the same topology (as $\mathbb C$ is a finite dimensional vector space over $\mathbb R$)?

Thank you!

Any absolute value on $\mathbf{C}$ restricting to the usual absolute value on $\mathbf{R}$ is in particular a norm on $\mathbf{C}$ as a two-dimensional real vector space. All norms on a finite-dimensional vector space induce the same topology. Thus your absolute value $|\cdot|$ on $\mathbf{C}$ restricts to a continuous homomorphism $S^1\to \mathbf{R}^+$, which must be trivial as $\mathbf{R}^+$ has no nontrivial compact subgroups, so you must have $|re^{i\theta}|=|r|$ for all $r$ and $\theta$.

Another elementary argument;

For roots of unity, it's clear they have norm 1. Now let $c$ be with irrational angle times pi, then take powers of it so that the angle is super close to $0$ mod $2 \cdot pi$, then use the triangle inequality in both directions its norm is close to norm $1$.

EDIT:
Since many people seem to be confused, I will add more details:

If $x$ is a root of unity of order $n$, $|x|^n=|x^n|=1$, proving $|x|=1$. In paricular this is true for $x=i$. Now let $x$ be on the unit circle, with angle $r \cdot 2pi$. Then it suffices to show that for arbitarily large $n$, $|x|^n$ is arbitarily close to $1$ (since if the norm was more than $1$, it'd explode, and if it was less, it would tend to $0$).

It is well known that if $a$ is irrational, then $na$ mod $1$ is gets arbitarily small. Applying this on our $a=r$, we find $x^n$ with $x^n-1=c+d\cdot i$ with $c$ close to $1$, and $d$ close to $0$. Then since we know $|i|=1$, subtracting $d\cdot i$ can only change the norm by $d$, so the norm is $c$ up to $+-d$, finishing the proof.


Finding number of roots using Rolle's Theorem, and depending on parameter I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?

Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one  $c \in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.

You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is

$$ax^3+bx^2+cx+d=0$$

so in your case

$$a=1,\ b=3,\ c=-m,\ d=5$$

In general, the discriminant is

$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$

You can substitute in your values and get a cubic expression in $m$. Then your equation has

three distinct real roots, if $\Delta>0$.
one or two distinct real roots, if $\Delta=0$.
one real roots (and two complex roots), if $\Delta<0$.

To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.

Note that this did not need Rolle's theorem. Do you really need to use it?


Smooth Reformulation of NonSmooth Constraints If I have something like :
\begin{align}
\min_x \max_i f_i(x)
\end{align}
I can reformulate this nonsmooth formulation as:
$$\min_x z$$
$$z\geq f_i(x)$$
and I have a smooth formulation of the problem.

Now, Suppose I have :
$$\min_x \min_i f_i(x)$$
Can I reformulate it as a smooth version?

I tried:
$$\min_x z$$
$$z\leq f_i(x)$$
but then, $z$ can always take the value $-\infty$.

My question is: Is a smooth formulation of min of mins even possible? My gut feeling is no but I wanted to be sure.

In the first case you put a lower bound for a minimization problem. In the second case you put an upper bound for a minimization problem which is practically non-sense.


Simple yet confusing probability problem. A bag contains x green candies and y red candies. A candy is selected at random from the bag and it's color is noted. It is then replaced into the bag with an additional ten candies of the same color. A second candy is then randomly chosen. Find the probability that the second candy is red.
I think that the answer is y/(x+y+10) but am unsure.

Also where would you recommend to get more practice on probality problems

We have x green candies and y red.

  Case 1-

Green candy is picked in first attempt.

$\frac x{x+y}$

Now green candies x + 10.

Red candy is picked in second attempt.

$\frac {y}{x+y+10}$

Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} $

  Case 2-

Red candy is picked in first attempt.

Probability = $\frac y{x+y}$

Now red candies is y + 10.

Red candy is picked in second attempt.

$\frac {y+10}{x+y+10}$

Probability = $\frac {y}{x+y} × \frac {y+10}{x+y+10} $

Total Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} +\frac {y}{x+y} × \frac {y+10}{x+y+10} $

Also where would you recommend to get more practice on probality problems.

Play Dungeons and Dragons.  If that is too much of a commitment.  Play Backgammon.

As for your question.

Would you believe that it is exactly the same as the chance that the first candy is red?

The expected number of red candies added is proportional to the number of red candies.  Adding candies in this way doesn't change the probabilities.

If that is too abstract:

The candies could be Green followed by red.
or red followed by red.

(Green)(red) + (red)(red) = $\frac {x}{x+y}\frac {y}{x+y+10} + \frac {y}{x+y}\frac {y+10}{x+y+10} = \frac {y(x+y+10)}{(x+y)(x+y+10)} =  \frac {y}{x+y} $


How many points in a line segment? My teacher said that in the circumference of circle there are infinite points. When I was learning more about circle, I came to this picture:

My question is: When we unroll the circle, then the length of the circle and line segment are the same. For this reason, I think that the unrolled line segment should also have infinite points! But there are only two end points. Can anyone explain to me where am I going wrong?

You are conflating end points and points.

The circumference of a circle does not have any end points but infinitely many points.
The line segment has two end points and still infinitely many points.

Sidenote: The length of the segment is not as important as you may think. We cannot distinguish the number of points on a line segment of length $π$ from those on a line segment of length $2π$.

Any legitimate question depends on whether the terminology is understood correctly. The problem is, the definition of a "point" is not understood correctly. Your teacher believes a point has no size because most, if not all, of the geometry books, old and new, say so. It is an error of immense proportions, wasting people's time and even warping their thinking. Geometry is supposed to help us understand the physical world around us. If a point has no size, then it has zero size. Basic math tells us that no matter how many times you add zero, it will always equal zero.

The correct definition of a POINT is "the smallest possible size". From there, everything will make sense mathematically, and logically. Both a circle, and a line segment, have an "indefinite" amount of points, but surely finite.


Guessing the other dice number, given two dice were rolled, and one of them rolled a 3 An argument i had with a friend followed this question:
"Given two dice were rolled, and one of them rolled a 3. What would you bet the other dice rolled?"

The phrasing is just to enhance that there might be some number that has an higher probability of showing on the other dice. One of us said that since 7 is the most likely number to be rolled by two dices, then 4 should be the answer. The other said that once the 3 was set on one dice, we should not be looking at anything other than the other cube - so any number would be an equal guess.

This kind of question is very sensitive to how the roller decides what to say.  Maybe the dice are different colors and he will tell you the number on the green die.  You know nothing about the other die, so all numbers are equally probable.  Maybe he will tell you the higher number.  Now the only rolls possible are $13,23,33,31,32$ so you should bet on $1$ or $2$.  Maybe he will pick a random die and tell you the number on it.  Now he could have rolled $33$ and been forced to say $3$ or rolled $3x$ or $x3$ and chosen to say $3$.  You have the same chance of $x$ as $3$ for the other die so bet on anything you like.  If you don't know how the roller chooses what to say it is not a mathematical question. There is not enough information to answer.

Since your question doesn't mention any selection criteria or bias in the dice used, the answer is clearly that there is no bias in the values of the second die.

When you roll two dice there are 36 possible permutations of answers that can arise.  If the first die shows a 3 then all permutations that involve the first die having any other value are eliminated.  You've eliminated 30 possible permutations and there are only 6 possible permutations remaining.

This is because probability is only ever about what you don't already know, or what has not already happened.  Flip a coin 9 times.  What's the probability of the next flip being heads? 50%.  What if all 9 flips were heads?  Still the same: 50%.  Flip 10 coins together, look at the first 9.  The 10th still has a 50% chance of being heads.

And yes, the most common sum rolled on two dice is 7, at 6 out of 36 permutations.  6 and 8 are 5/36 each, 5 and 9 are 4/36 each, and so on.  But that's only important when you have two unknowns.  When only one is unknown then the probability is still even.  You can easily write down the permutations and count them to see what's up.

Ross Millikan's answer adds additional information, but is correct when he says:

  You have the same chance of $x$ as 3 for the other die so bet on anything you like.

ConMan's answer however misses something.

If the selection of which die to reveal is random, and the revealed number is 3, then there are in fact 12 different possible results - 6 for each non-revealed die - with 2 results each for the number on the other die.  So each possible number has exactly 1/6 probability.  His answer that there are only 11 possible results is ignoring the fact that the [3,3] result would have two chances of being the chosen permutation.

The big issue is that our intuition is generally quite poor at picking the right answer here.  We tend to mix up situations where the probabilities change as we progress - card games, lottery numbers, etc - with ones where they don't.  We tend to assume that the probability of an event that happened yesterday is the same as it happening tomorrow, and that things that have happened affect things that will happen regardless of a lack of causal links.

Casinos make bank on us making those mistakes.


What is $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2} $ Find the following limit
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2}$$

This question has been bugging me for some time. Couldn't find it anywhere on the Internet

You can cancel a common factor:
$$\frac{x^3-y^3}{x^2-y^2}=\frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)}=\frac{x^2+xy+y^2}{x+y} $$
(This also continuously extends the function to most of the line $x=y$).

Along any straight line $(x,y)=(at,bt)$, $t\to 0$, this is $\frac{a^2 +ab+b^2}{a+b}\cdot t$ and tends to $0$. But this of course does not work for the case $b=-a$ where the expression is undefined.
However, convergence along straight lines is not convergence!
If we consider $(x,y)=(t+t^2,-t)$, $t\to 0$, we arrive at
$$\frac{(t+t^2)^2+(t+t^2)(-t)+(-t)^2}{t+t^2-t} =\frac{t^2+t^3+t^4}{t^2}=1+t+t^2\to 1$$

i think the limit is 0
switch to polar coordinates and you will see it. you can also simplify the equation: x^3 - y^3 = (x-y)(x^2+xy +y^2)
good luck


The sum of consecutive integers is $50$. How many integers are there? I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.

If your $n$ is odd, then the middle number has to be $50/n$.  The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$

If $n$ is even, then $50/n$ is the half-integer between the middle two numbers.  So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.

If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$

If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$

So there are 4 answers: $n=1, 4, 5, $ and $20$.

Edit:  As Bill points out, I missed the divisors $25$ and $100$, which give two more answers:  $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.

Note that each solution with negative integers is related to an all-positive solution.  From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.

First, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.


Is $0 $ radians an acute angle? I know that an angle less than $\frac {\pi}{2} $ radians is called acute, but under this definition, is an angle that is $0$ radians also considered acute?

According to mathcaptain, which is a website maintained by the publisher Pearson (which publishes math textbooks), an angle of $0$ is called a "zero" angle, an angle of $\pi$ is called a "straight" angle, and an angle of $2\pi$ is called a "complete" angle.

Since you wrote

  I know that an angle less than $\frac {\pi}{2} $ radians is called acute

then obviously an angle of zero radians is acute, simply because $0<\frac {\pi}{2}$.


Let $f$ be a twice differentiable function on $\mathbb{R}$. Given that $f''(x)>0$ for all $x\in \mathbb{R}$.Then which is true? Let $f$ be a twice differentiable function on $\mathbb{R}$. Given that $f''(x)>0$ for all $x\in \mathbb{R}$.Then which  of the following is true?

1.$f(x)=0$ has exactly two solutions on $\mathbb{R}$.
2.$f(x)=0$ has a positive solution  if $f(0)=0$ and $f'(0)>0$.
3.$f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)>0$.
4.$f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)<0$.

My thoughts:-
(1) is not true as $f(x)=x^2+1$ is a counter example.
Now suppose the conditions in (2) holds.then $f'(x)$ is increasing everywhere.so $f'(x)$ is never zero for all positive $x$.so $f(x)$ can not have a positive solution otherwise $f'(x)$ have a zero between $0$ and $x$ by Rolle's Theorem.
so (3) is true.
Are my arguments right?

Your arguments are correct. For (2), a counterexample would have been enough. However, your procedure was more efficient, since it dealt simultaneously with (2) and (3).

You left out analyzing (4), which is not always true. For example, let $f(x)=-2x+x^2$.

But there are functions for which (4) holds. For instance, let $f(x)=\frac{1}{1+x}-1$.

Correct answer for above question is option no.3.Second option is not correct because it has a counter example.$$f(x)=e^x-1$$ f satisfy condition $f(0)=0,f'(0)>0$but $0$ is the only solution of $f(x)=0$.Hence $f(x)=0$ has no positive solution.


An irreducible polynomial of degree coprime to the degree of an extension is irreducible over this extension I'm having a hard time showing this:

  If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.

I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.

Any help would be appreciated.

Let $u$ be a root of $f(x)$. Then $[\mathbb{Q}(u):\mathbb{Q}] = n$. Now consider $K(u)$. We know that $[K(u):K]\leq n$, since the monic irreducible of $u$ over $K$ divides $f$. Since
$$[K(u):\mathbb{Q}] = [K(u):K][K:\mathbb{Q}] = m[K(u):K]\leq mn$$
and
$$[K(u):\mathbb{Q}] = [K(u):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}] = n[K(u):\mathbb{Q}(u)]$$
then $[K(u):\mathbb{Q}]$ is at most $nm$, is a multiple of $n$, and is a multiple of $m$. Since $\gcd(n,m)=1$, it follows that $[K(u):\mathbb{Q}]=nm$, so the degree $[K(u):K]$ is equal to $n$.

What does that imply about the monic irreducible of $u$ over $K$ and about $f$?

Other Argument:

Suppose on contrary $f $ is reducible over $K$ So $\exists u\in K$ such that $f(u)=0$

$F\subset F(u)\subset K$

By Tower theorem $[F(u):F]|[K:F]$ i.e $n|m$ which is contradiction


Equivalence of countable choice for subsets of the reals and "second countable $\implies$ Lindelöf" Looking for a proof that in a second countable space every open cover has a countable subcover -- i.e. every s.c. space is a Lindelöf space -- I bumped into this question. That answered my question. I then left a couple of comments on both the then present answers (which are, to this date, the only ones) and received this reply, stating the axiom of countable choice for subsets of the reals is actually equivalent to every s.c. space being Lindelöf. Now it seems incredible that a special case of the axiom of countable choice, which in its general form states any countable collection of sets has a choice function, so for all countable collection of sets $A_n$ there is a function associating to each set an element of that set, provided of course the sets are non-empty, and in its particular form says so for sets which are subsets of the reals, be equivalent to s.c. implies Lindelöf. At most, I would expect the general form to be equivalent to that, since that s.c. implies Lindelöf is proven with a form of countable choice. So how can it be proved that the two results are equivalent?

This is not that incredible if you understand what's going on.

Given a second-countable space $X$, fix a countable basis $\{U_n\mid n\in\Bbb N\}$. Now you can observe that every open set is uniquely determined by the basis elements it contains, namely the function $U\mapsto\{n\in\Bbb N\mid U_n\subseteq U\}$ is an injective function. And since the real numbers can be thought of as subsets of $\Bbb N$, this is an injective function from the topology of $X$ into $\Bbb R$.

So if we are given an open covering, $\{V_i\mid i\in I\}$ we may assume that $|I|\leq|\Bbb R|$ by the above consideration, and for each $n\in\Bbb N$ we can look at the set $A_n=\{V_i\mid U_n\subseteq V_i\}$.

Not every $A_n$ is non-empty, but at least one of them is, of course. By the translation between open sets and real numbers, this is reduced to a countable family of sets of reals. Using $\sf AC_\omega(\Bbb R)$ we can choose from each non-empty $A_n$, and we can prove that this is a countable subcover by noting that $\{U_n\mid A_n\neq\varnothing\}$ is an open cover as well.

On the other hand, if $\{A_n\mid n\in\Bbb N\}$ is a family of pairwise disjoint sets of reals without a choice function. We may assume that each $A_n$ is a subset of $[n,n+1)$ by using canonical bijections of $[\inf A_n,\sup A_n+1)$ and $[n,n+1)$.

Now consider the open covering of $\Bbb R$ defined by $\{(-r,r)\mid\exists n: r\in A_n\}$. Since the $A_n$'s grow arbitrarily large this is an open covering indeed, so by the assumption it has a countable subcover, which gives for each $A_n$ a countable subset, so by diluting this subcover we may assume that each $A_n$ gives at most a single real number.

And therefore we found a partial choice function for the $A_n$'s. Now it is enough to show that it is indeed enough to consider partial choice functions.

Indeed recall that $\Bbb R^{<\omega}$, the finite sequences of real numbers have the same cardinality as $\Bbb R$ itself. Given a sequence of sets $X_n$ which are non-empty, consider $A_n=\prod_{k\leq n}X_n$, then the $A_n$'s are non-empty sets of finite sequences of real numbers, so by uniform encoding we may assume these are sets of real numbers. If $I\subseteq\Bbb N$ is infinite and $\sigma$ is a function choosing from $\{A_i\mid i\in I\}$, then it defines a choice function from the $X_n$ by looking at $i=\min\{j\in I\mid j>n\}$ and looking at $\sigma(i)(n)$ (recall that $\sigma(i)$ is a choice function which is defined on $X_n$).

Let $\mathcal B$ be the countable base of $X$. Then for any open cover $\mathcal U=\{U_\alpha:\alpha \in \Omega\}$, then for any $x\in X$ and $x \in U_\alpha$, there exists $B_x\in \mathcal B$ such that $x\in B_x \subset U_\alpha$. Then $\{B_x:x\in X\}$ is the subcover and countable.


How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

Proof:
Consider the function $f(t) = e^{-it}(\cos t + i \sin t)$ for $t \in \mathbb{R}$. By the product rule
\begin{eqnarray}
f^{\prime}(t) = e^{-i t}(i \cos t - \sin t) - i e^{-i t}(\cos t + i \sin t) = 0
\end{eqnarray}
identically for all $t \in \mathbb{R}$. Hence, $f$ is constant everywhere. Since $f(0) = 1$, it follows that $f(t) = 1$ identically. Therefore, $e^{it} = \cos t + i \sin t$ for all $t \in \mathbb{R}$, as claimed.

An easy way out to answer the question is when we try to extend the definition of exponential to complex plane in a "nice" way (read "nice" as holomorphic) we then end up with this definition. And as you are probably aware there is only at most one such extension. So if we want to define our exponential in the complex plane such that the exponential function is holomorphic, and it matches with the conventional exponential function on the real line, we end up with $e^{it} = \cos(t) + i \sin(t)$.

All the other answers I think are circular in some sense.

For instance, how do we know that we can expand $e^{i x}$ as a power series. All we know is that we can expand $e^{x}$ as a power series when $x$ is real. We do not know apriori that the expansion can be carried forward if $x$ happens to be complex.


Gamma & Zeta Summation $\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!}=0$ According to Gamma Summation & Zeta Summation:
$$ \sum_{n=0}^{\infty} {(-1)^n \frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=\Gamma(s-1) \quad : \space Re\{s\}<2 $$

  Show that:
  $$ \sum_{n=0}^{\infty} {\frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=0 \quad : \space Re\{s\}<1 $$

In other words, the Even & Odd parts are convergent series, equaling sums, and different signs:
$$ \sum_{n=0}^{\infty} {\frac{\Gamma(2n+s) \zeta(2n+s)}{(2n+1)!}}=-\sum_{n=0}^{\infty} {\frac{\Gamma(2n+1+s) \zeta(2n+1+s)}{(2n+1+1)!}}=\frac{\Gamma(s-1)}{2} \space : \space Re\{s\}<1 $$

note that $\frac{\Gamma(s+n)}{(n+1)!} = \frac{\Gamma(n+s)}{\Gamma(n+2)} = \mathcal{O}(n^{s-2})$, and since $\zeta(s+n) \to 1$, it converges for $Re(s) < 1$
but since $$\Gamma(s) \zeta(s) = \int_0^\infty \frac{x^{s-1} }{e^x-1}dx, \qquad \color{red}{\text{only for }} Re(s) > 1$$

(for proving it : $\int_0^\infty x^{s-1} e^{-nx} dx = n^{-s}\Gamma(s), Re(s) > 0$ and $\frac{1}{e^x-1} = \sum_{n=1}^\infty e^{-nx}, x > 0$)
we have to consider the following regularized version
$$\begin{eqnarray}\sum_{n=0}^\infty z^n \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!}  &=& \int_0^\infty \sum_{n=1}^\infty z^n \frac{x^n}{n!}\frac{x^{s-2} }{e^x-1}dx \\
 &=& \int_0^\infty (e^{zx}-1)\frac{x^{s-2} }{e^x-1}dx \\
& =& -\zeta(s-1)\Gamma(s-1)+
\int_0^\infty x^{s-2}\sum_{n=1}^\infty e^{-(n-z)x}dx \\
 &=& \Gamma(s-1)(-\zeta(s-1)+\sum_{n=1}^\infty (n-z)^{1-s}) \\
&=& \Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1)) \\ & &\qquad\qquad\qquad  \color{red}{(Re(s) > 2, |z| < 1)}\end{eqnarray}$$
where $\zeta(s,a) = \sum_{n=1}^\infty (n+a)^{-s}$ is the Hurwitz zeta function, having a nice analytic continuation too, such that $\zeta(s) = \lim_{a\to 0} \zeta(s,a)$ also for $Re(s) < 1$. (**)

Hence, assuming the analytic continuation (*) of two variables $z,s$ works well  (keeping $|z| < 1$) this stays true for $Re(s) < 1$ where we can extend to $|z| = 1$ and get $$\sum_{n=0}^\infty \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!} = \lim_{z \to 1^-}\sum_{n=0}^\infty z^n \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!} $$ $$\qquad =\lim_{z \to 1^-} \Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1)) = 0\quad (Re(s) < 1, -s \not \in \mathbb{N})$$

the $\lim_{z \to 1^-} $ is justified by the fact the LHS converges absolutely when $|z| \le 1, Re(s) < 1$ (so it is continuous in $z$), and the fact the RHS is continuous in $|z| \le 1$ too

(*) and it does since the RHS $\Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1))$ is obviously holomorphic/analytic/meromorphic in $s$,
while the LHS is also holomorphic in $s$, by showing $\sum_{n=0}^\infty z^n \frac{\frac{d}{ds}\Gamma(s+n)\zeta(s+n)}{(n+1)!}$ converges absolutely when $|z| < 1$, and when $|z|= 1, Re(s) < 1$

(**) integrating by parts $\zeta(s,a) = \sum_{n=1}^\infty (n+a)^{-s} = \int_{a-\epsilon}^\infty \sum_{n=1}^\infty \delta(x-n-a)x^{-s}dx = s \int_a^\infty \lfloor x+a \rfloor x^{-s-1} dx$ $  $ $= \frac{s a^{-s}}{s-1}+a^{-s}+s \int_a^\infty (\lfloor x+a \rfloor-x-a) x^{-s-1} dx \ \ (Re(s) > 0)$, and we can integrate by parts several times the same way

The proposer wants $R(1,s-1)$ of the function defined below:
$$R(x,a)=\sum_{n=1}^\infty \dfrac{x^n}{n!}\Gamma(n+a)\,\zeta(n+a).$$
The $\Gamma\cdot\zeta$ product has a well-known integral relationship, as alluded to by Jack D'Aurizio.  Switch integration and summation $-$ the summation is an exponential $-$ and one gets
$$R(x,a)=\int_0^\infty du \,u^{a-1} \,\dfrac{e^{x\,u}-1}{e^{u}-1 }=\Gamma(a)(\zeta(a,1-x)-\zeta(a)). $$
(Need $a>0$ for the integral to converge, but the right-hand side is an analytic   continuation.)
The Hurwitz $\zeta$ reduces to the Riemann $\zeta$ for $x=1,$ and the formula is proved.  Incidentally, the Hurwitz $\zeta$ also has closed forms in terms of the Riemann $\zeta$ for $x=-1,$ as the proposer has noted, but also for $x=1/2.$


ideals in rings of algebraic integers are finitely generated I am trying to write about rings of algebraic integers $\mathcal{O}_K$ in a number field $K$ without introducing to much field theory. I want to show that these rings are Dedekind. First of all I want to show that these rings are Noetherian. Is there any simple way to show that every ideal is finitely generated?

The argument I found so far is to consider $\mathcal{O}_K$ as an $\mathbb{Z}$-module. After deducing that $\mathcal{O}_K$ has a finite basis as such, one deduces that every subgroup of this free abelian group is finitely generated and hence that this implies that every ideal of $\mathcal{O}_K$ has to be finitely generated since the underlying group is. Or this is how I understand the argument anyway. The conclusion that $\mathcal{O}_K$ has a finite basis as a $\mathbb{Z}$-module involves a lot of work in the book I am using, I would appreciate if someone wants to make that argument clear here. Or correct me if I misunderstood any other part of the argument. Or give another argument leading to the desired conclusion.

$K$ is a number field of a certain degree, say $n$, hence is the $\mathbb Q$-span of a set which has $n$ elements. Then $K$ has an integral basis made up of $n$ algebraic integers of $K$. The $\mathbb Z$-span of that integral basis is $\mathcal O_k$ hence $\mathcal O_k$ is finitely generated as a $\mathbb Z$-module. Any ideal of $\mathcal O_k$ is generated by a subset of that integral basis hence is finitely generated. Therefore $\mathcal O_k$ is noetherian.


Proving that $f(x) = x^{1/3}$ is absolutely continuous. Here is a trial:

We will divide our interval into 2 intervals $[-1, 0)$ and $[0,1]$:\

(1) for the interval $[-1, 0)$, our function is increasing on $(-1,0)$ then it is differentiable a.e. on $(-1,0)$ by Lebesgue theorem and its derivative is $f'(x) = \frac{1}{3}x^{-2/3}$ and

our function $f$ is continous. \

(2)for the interval $[0, 1]$.\

We will do this with the aid of problem 37 . Let $\epsilon > 0.$ taking $\{[c_{i}, d_{i}]: 1 \leq i \leq n\}$ to be a non-overlapping collection of intervals in $[0,1]$ such that $\sum_{i=1}^{n} (d_{i} - c_{i}) <  \epsilon^3.$ Choose $a = \frac{\epsilon^3}{8}.$ Now we break the sum $\sum_{i=1}^{n} |f(d_{i}) - f(c_{i})|$ into two parts, those intervals that are in $[0,a]$ and those that are in $[a,1].$ i.e. we break the interval at $a$  . Let $a = d_{m}$ for some $m.$ \

Now consider the sum over the intervals that are in $[0,a],$ $$\sum_{i=1}^{m} |f(d_{i}) - f(c_{i})| = \sum_{i=1}^{m} |d_{i}^{1/3} - c_{i}^{1/3}| \leq a^{1/3} = \epsilon /2. $$

This follows from the fact that $ x^{1/3}$ is an increasing function. monotonicity ensures that the function does not oscillate widely.\

Now we consider the sum of the intervals that are in $[a,1].$

$$\sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})| = \sum_{i=m+1}^{n} |d_{i}^{1/3} - c_{i}^{1/3}| = \sum_{i=m+1}^{n} |d_{i}^{1/3} - c_{i}^{1/3}| \times \frac{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|}{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|} $$\
$$ = \sum_{i=m+1}^{n} \frac{(d_{i} - c_{i})}{|(d_{i}^{2/3} + (d_{i} \times c_{i})^{1/3}+  c_{i}^{2/3})|}  $$\

So,$\sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})|\leq \sum_{i=m+1}^{n} \frac{(d_{i} - c_{i})}{8 a^{2/3}} = **\frac{1}{8 a^{2/3}}**\sum_{i=m+1}^{n}(d_{i} - c_{i}) < \frac{1}{2 \epsilon^2} . \epsilon^3 = \frac{\epsilon}{2}. $

We used $(x-y) = (x^{1/3} - y^{1/3})  (x^{2/3} + (xy)^{1/3}+  y^{2/3}).$
Combining these two sums we see that

$\sum_{i=1}^{n} |f(d_{i}) - f(c_{i})| \leq (\sum_{i=1}^{m} |f(d_{i}) - f(c_{i})| + \sum_{i=m+1}^{n} |f(d_{i}) - f(c_{i})|) < \epsilon.$\

My question is:

Note the trial above is based on Royden fourth edition "Real analysis."

1-Also, the general idea of the proof for interval $[0,1]$ is not clear for me, could anyone explain it for me, please?

EDIT:
I have answered all my questions.

Being that $x^{\frac{1}{3}}$ is differentiable with derivative $\frac{1}{3\sqrt{x}}$. We know that:

$$x^{\frac{1}{3}} = \int_0^x \frac{{\rm d}t}{3\sqrt{t}}$$

and the derivative is Lebesgue-integrable (since the above integral is improper), which implies that, $x^{\frac{1}{3}}$ is absolutely continuous.