$x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove $x_n\leq\frac{1}{2^{n-2}}$. $x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove

    $x_n\leq\frac{1}{2^{n-2}}$.

I proved $0<x_{n+1}<x_n$ by contradiction, and I also get $x_n\geq\frac{1}{2^{n-1}}$ by induction. I tried $\ln(1+x)\geq x-\frac{1}{2}x^2$, but it did not work. Could anyone help me? Thanks!

I tried a new way

$\forall \epsilon>0, \ln(1+x)>(1-\epsilon)x$, then $x_n>(2-\epsilon)x_{n+1}$, thus $x_{n+1}<\frac{1}{(2-\epsilon)2^{n-2}}\to \frac{1}{2^{n-1}}$.

Is it right?

well, I think I misunderstand this question, sorry.

as you said

$x_{n+1}\ge ln(1+x_{n+1})\geq x-\frac{1}{2}x^2$

we can get $x_n\in(\frac{1}{2^{n-1}}, 2-\sqrt{4-2x_{n-1}})$

Thus we need to show $2-\sqrt{4-2x_{n-1}}\le \frac{1}{2^{n-2}}$

in fact, the above $2-\sqrt{4-2x_{n-1}}$ is determined by $ 2-\sqrt{2\sqrt{2\sqrt{2...}}}$ which you can check with $\frac{1}{2^{n-2}} $


Does this common PDE have a name? What is the name of this common 2D PDE:

$$\nabla^2 u - k^2 u=0$$

where $k\in\mathbb{R}$.

And where can I find a general solution and the way to derive it? THanks!

Edit:

Some people have suggested that this is the Helmoltz equation. My concern is the following: the Helmoltz equation does admit a solution with products of trigonometric functions and corresponding bessel J functions:

$$f(r,\phi)= A_n\sin(n\phi)J_n(\lambda r)$$

The one I wrote, with a minus instead of a plus, does not admit such solutions. Infact admits exponential solutions:

$$f(r,\phi)= \exp\left[\frac{\lambda}{\sqrt{2}} (r\cos\phi+r\sin\phi)\right]$$

because of its structure.

If you define $\nabla^2$ as $\nabla^2 u = -\sum u_{x_ix_i}$, this is the Helmholtz equation.  If you then give it Dirichlet boundary conditions, it is the equation of Dirichlet eigenvalues, and it will have solutions.  If, however, you are defining $\nabla^2 u = \sum u_{x_ix_i}$, the Dirichlet problem for your equation (i.e. zero boundary condition) will have no nontrivial solutions by the spectral theorem, so it isn't very interesting.

It is called Helmotz equation. When derived from the wave equation, it appears with $+k^2\,u$ instead of $-k^2\,u$.


Prove that $x^p\equiv 1$ (mod $p$) has only one solution. I know that said solution is $x\equiv 1$ (mod $p$). However, I'm having difficulty proving this result.

So far, I've tried $x^p\equiv 1$ (mod $p$) $ \Rightarrow $ $p\mid (x^p-1) \Rightarrow p\mid(x-1)(x^{p-1} + x^{p-2} + \cdots + x + 1)$.

From here, it's clear that the objective is to somehow show that $p\mid(x^{p-1} + x^{p-2} + \cdots + x + 1)$ also yields $x\equiv 1$ (mod $p$), but I've been unsuccessful in showing this after $(x^{p-1} + x^{p-2} + \cdots + x)\equiv-1$ (mod $p$).

$(x-1)^p \equiv (x^p - 1) \mod p$  by binomial theorem. So, if $p$ divides $(x^p - 1)$, it certainly divides $(x-1)^p$. Therefore, by fermat's last theorem, $p$ divides $x-1$. Hence, you get the desired result.

Using Frobenius isomorphism you receive that $x^p-1 = (x-1)^p $.
$\mathbb Z_p $ is a field, $\implies \mathbb Z_p [x]$ is a PID, therefore an integral domain, from which you can conclude that $(x-1)^p = 0 \iff (x-1)=0 \iff x = 1$


How to plot $N$ points on the surface of a $D$-dimensional sphere roughly equidistant apart? Let's say I have a $D$-dimensional sphere with a radius $R$. I want to plot $N$ number of points evenly distributed (equidistant apart from each other) on the surface of the sphere. It doesn't matter where those points are exactly, just that they are ROUGHLY equidistant from each other. How would I do this?

If $N \le D + 1$ all points must be located on $N-1$ dimension equilateral triangle.

UPDATE

So, I thought today  at this problem and invented method, contains following steps:

Generating $N$ random points on sphere $R$ (that is generating points with coordinates with Gaussian distribution).
Building Convex Hull (or triangulation on hypersphere even better) with generated on first step points. This step can be solved with MIConvexHull library if you familar with C#, similar library on your favorite language or your own code.
Using Genetic algorithm, Simulated annealing or another method of global optimization. This method then have to be applied to variance value of all edge lengths from convex hull from step 2.


Are there any good examples from other math fields or intuition supporting $\int_0^1\frac1xdx=\int_1^\infty\frac1xdx$? This question is related to the potential possibilities of classification of divergent integrals more precisely than just "divergent to infinity" and the like. Improper divergent integrals can have many distinguishing properties, like germ at infinity, rate of grow and regularized value.
A separate question is whether we can talk about equivalence (in whatever sense) of integrals divergent at infinity and those divergent at a singularity or pole.
For instance, formally applying the Laplace transform we can establish the following formal relations (for $n>0$):
$$\int_0^1 x^{-1+n} dx=\int_1^\infty n!x^{-1-n}dx$$
and
$$\int_1^\infty x^{-1+n} dx=\int_0^1 n!x^{-1-n}dx$$
This works for convergent integrals:
$$\int_0^1 dx=\int_1^\infty \frac1{x^2}dx=1$$
If generalized to divergent integrals, it also seems to work well, the equality
$$\int_1^\infty dx=\int_0^1 \frac1{x^2}dx$$
seems to be supported well by the hyperfunction theory.
But with $n=0$ we have (in the both formulas):
$$\int_0^1 \frac1x dx=\int_1^\infty \frac1x dx$$
This looks counter-intuitive at the first glance, as the figures under the integrals are identical up to a reflection relative the $y=x$ axis, except the first one has an additional $1\times 1$ square.
Of course, since the integrals are divergent, it should not be an obstacle to the identity (as the positions of the figures and the limits of integration are different). The both integrals are regularizable to the value $0$.
So, I wonder, if there is any intuition or example from other fields or mathematics or physics where these integrals demonstrate equivalence in a more strict sense than being both divergent to infinity?
For instance, from geometry or probability theory.
UPDATE
It seems, the equality should not hold, and the more correct one would be $$\int_0^1 \frac1x dx=\gamma+\int_1^\infty \frac1x dx$$

Your paradox, as I understand it, concerns the correspondence between the divergent integral and the geometric area.  Let's be a little more careful and define finite regions which are symmetric about $y=x$ before considering the limit as the regions approach the $x$- and $y$-axes.
There are two such areas we can define. First, there is $A_1$, between $y=1$ and $y=1/x$ on the interval $\epsilon < x< 1$, given by
$$A_1 = \int_\epsilon^1 dx\left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x} +\epsilon-1.$$ Here, $0<\epsilon<1$ is a small number.
If we reflect this region about $y=x$, we get
$$ A_1 = \int_1^{\epsilon^{-1}} dx\left(\frac1x-\epsilon\right) = \int_1^{\epsilon^{-1}} \frac{dx}{x} + \epsilon-1.$$
A second area can be defined as that between $x=1,$ $x=\epsilon^{-1}$, $y=1/x$, and the $x$-axis:
$$
A_2 = \int_1^{\epsilon^{-1}} \frac{dx}{x}.$$
The reflection about $y=x$ is
$$
A_2 = \int_0^\epsilon dx (\epsilon^{-1}-1) + \int_\epsilon^1 dx \left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x}.$$
So, in both cases, the geometric areas agree after reflection about $y=x$ if and only if the divergent integrals agree as claimed.
I believe the confusion arises because there are two relevant geometric areas, and which one you get depends on how you take the limit. If you imagine replacing the integral with a finite sum of rectangles, the height of the rectangles may or may not diverge depending on how they are oriented, and this leads to two different answers.


Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$

The qty in bracket tends to 1 as x→infinte and power tends to infinity u can easily prove that Lt(x→a){f(x)}^(g(x)) if f(a)→1 and g(a)→∞ then its equal to limit of e^(f(x)-1)(g(x)) as x→a so here it is.. e^(2/(x^2-1))(x^2) limit as x→ ∞ giving e^2 .. !


Cardinality of set of real continuous functions I believe that the set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$, the cardinality of the continuum. However, I read in the book "Metric spaces" by Ó Searcóid that set of all $[0, 1]\to\mathbb{R}$ continuous functions is greater than $\mathfrak c$:

"It is demonstrated in many textbooks that $\mathbb{Q}$
is countable, that $\mathbb{R}$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions deﬁned on  $[0,1]$ is of a greater cardinality than $\mathbb{R}$, and that there are sets of greater and greater cardinality."

I understand that (via composition with the continuous function $\tan$ or $\arctan$) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this?

The cardinality is at least that of the continuum because every real number corresponds to a constant function.  The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $\mathbb R^N$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.
The Schroeder-Bernstein theorem now implies the cardinality is precisely that of the continuum.
Note that  then the set of sequences of reals is also of the same cardinality as the reals.  This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.

I have a somewhat simple answer to this question. It is not as elaborate as the other but perhaps it will add some intuition.

let's look at the number of functions from $\mathbb R^n \to \mathbb R$.

For every element in $\mathbb R^n$ we need to choose a corresponding image in $\mathbb R$. There are $c$ elements in $\mathbb R$, and so if there are $\alpha$ elements in $\mathbb R^n$, there are $\alpha c$ functions (not continuous! just functions) from $\mathbb R^n \to \mathbb R$.

Let's find alpha: For the first element in an $n$ vector we have $c$ choices, for the second one $c$ choices, for the third one $c$ choices etc...overall $c^n=c$ choices.

So overall there are $c^{n+1}=c$ functions from $\mathbb R^n \to \mathbb R$.

since the continuous functions are a subset of this set, there are AT MOST $c$ continuous functions.

Now let's look at the function $f(x)=\xi ||x||_2$ where $\xi$ is some real number, $x$ is an $n$ vector, and $||x||_2$ is the euclidean norm of the vector $x$.

This function $f$ is continuous no matter which $\xi$ we pick. We have $c$ options to choose from, and so the set of continuous functions includes all the functions of the form $f(x)=\xi ||x||_2$ and so it is AT LEAST $c$.

Since it is at most $c$, and at least $c$, the conclusion is that it is exactly $c$.


If a covering map has a section, is it a $1$-fold cover? If $q: E\rightarrow X$ is a covering map that has a section (i.e. $f: X\rightarrow E, q\circ  f=Id_X$) does that imply that $E$ is a $1$-fold cover?

It follows from your assumptions that $q$ is a 1-sheeted and is a homeomorphism.  I'm going to call the map $\pi$ instead of $q$ for the rest of this post.

Assume we have a covering $\pi:X\rightarrow Y$ and $f:Y\rightarrow X$ with $\pi\circ  f = Id_X$.

I claim that $f(Y)$ is both open and closed in $X$.

To see it, for any $\hat{p}\in X$, let $p = \pi(\hat{p})$.  Choose a neighborhood $U$ around $p$ for which $\pi$ trivializes: $\pi^{-1}(U) = \coprod V_\alpha$ with $\pi|_{V_\alpha}$ a homeomoprhism. and let $V$ be the particular $V_\alpha$ containing $\hat{p}$.

Now, if $\hat{p}\in f(Y)$, then $V\subseteq f(Y)$.  This follows from considering the inclusion $i:U\rightarrow Y$.  Since both $f|_{U}$ and $\pi^{-1}|_{U}$ are lifts of this inclusion agreeing at $\hat{p}$, they must agree on all of $U$.  It follows that $V=\pi^{-1}(U) = f(U)$ as claimed.  This shows $f(Y)$ is open.

If, on the other hand $\hat{p}\notin f(Y)$, a very similar argument shows that $V\cap f(Y) = \emptyset$, showing that $f(Y)^c$ is open, that is, that $f(Y)$ is closed.

Putting this together, $f(Y)$ is open and closed.  Hence, it is a connected component of $X$.  If $X$ itself is connected, this implies $f(Y) = X$ which implies that $\pi$ is a homeomorphism with inverse $f$ so, is in particular, 1 sheeted.

A connected covering space $f:E\rightarrow X$ admits no section ( global section) unless $f$ is a homeomorphism.

Edit: looking @Andy's post I'm not so sure of what I said now.


Is it true to say that "it's not logically possible to prove something can't be done"? A friend of mine asked me if I could explain this statement: "It's not logically possible to prove that something can't be done". The actual reason is the understanding of this strip:

Since I'm not an expert on logic, but at least know the basics, I told him that the statement is false. To prove it I gave him the counterexample of the impossibility of trisecting an angle by using only straightedge and compass, which was proved by Pierre Wantzel. This shows clearly that it's possible to prove that something can't be done.

When I told him this, he explained to me that this is sort of naive because Asok, the guy who makes the statement, is a graduate from IIT and also I read that he's always a brilliant character in the strip.

Can you guys please tell me what you think about this? Also I'd like to know how you would interpret this dialog.

What the strip most likely was (awkwardly) referring to is not that it's impossible to prove something can't be done, but rather that it's impossible to prove within a (sufficiently powerful) formal system that something can't be proven in that system — in other words, letting $P(\phi)$ be the statement 'there exists a proof of $\phi$', then it's impossible to prove any statement of the form $\neg P(\phi)$ (i.e., the statement $P(\neg P(\phi))$ can't be true).

The reason behind this has to do with Godel's second incompleteness theorem that no formal system can prove its own consistency - in other words, the system can't prove the statement $\neg P(\bot)$ (or equivalently, $P(\neg P(\bot))$ is false).  The logical chain follows from the statement 'false implies anything': $\bot\implies\phi$ for any proposition $\phi$.  By the rules of deduction, this gives that $P(\bot)\implies P(\phi)$, and then taking the contrapositive, $\neg P(\phi)\implies\neg P(\bot)$ for any proposition $\phi$; using deduction once more, this gives $P(\neg P(\phi))\implies P(\neg P(\bot))$ — if the system can prove the unprovability of any statement, then it can prove its own consistency, which would violate Godel's second incompleteness theorem.

Note the keyword 'sufficiently powerful' here - and also that we're referring to a formal system talking about proofs within itself.  This is what allows statements like the impossibility of angle trisection off the hook: the axioms of ruler-and-compass geometry aren't sufficiently strong for Godel's theorem to apply to the system, so that geometry itself can't talk about the notion of proof, and our statements of proofs of impossibility within that system are proofs from outside of that system.

By Godel's theorem maths is incomplete or inconsistent, logically. Since we can't accept proofs that are inconsistent we must accept that maths cannot be proven, logically. We need more axioms. If we can't prove maths we certainly can't prove programming.

The halting problem is algorithmic and so has more implicit axioms related to achievability rather than logic. Compare: "it is impossible for me to count to infinity" and "it is logically impossible for me to count to infinity".


Prove that every integer is either prime or composite In the book I'm reading, the following proof is given for the stated theorem:

Let n be any integer that is greater than 1. Consider all pairs of positive integers
$r$ and $s$ such that $n = rs$. There exist at least two such pairs, namely $r = n$ and $s = 1$ and $r = 1$ and s = n. Moreover, since $n = rs$, all such pairs satisfy the inequalities
$1 ≤ r ≤ n$ and $1 ≤ s ≤ n$. If $n$ is prime, then the two displayed pairs are the only ways
to write $n$ as $rs$. Otherwise, there exists a pair of positive integers $r$ and $s$ such that $n = rs$ and neither $r$ nor $s$ equals either 1 or $n$. Therefore, in this case $1 < r < n$ and $1 < s < n$, and hence $n$ is composite.

When I attempted the proof myself, I came up with the following reasoning:
Consider all integers $1<k<n$. Now, either $n$ is divisible by one of them or not. If not, $n$ is prime; if it is, $n$ is composite. Therefore, every integer is either prime or composite.

I was wondering if my proof is incorrect or less rigorous in some sense. Isn't the textbook complicating things unnecessarily? Or is that a better demonstration of mathematical thinking? Some thoughts are appreciated.

P.S.: I was also not sure whether this question belonged to proof-writing or proof-strategy.

What do you mean exactly by "composite"? Disregarding $0$ and $\pm1$ for the sake of simplicity, if you define composite as divisible by something else then $1$ and itself (i.e. "non-prime") the proof you need is just an obvious appealing to the principle of tertium non datur.

If, on the other hand, you define composite a number which can be written as a product of at least two primes you actually need some argument to show that primes and composites exhaust all numbers.

Certainly, your suggestion that if $n$ is not prime then there's a number $1<k<n$ that divides $n$ is correct, but it is not the end of it since $n=k\cdot(n/k)$ needs not to be a decomposition into prime factors. I'd say that you need a little inductive argument to come to a conclusion.

I think the proof is right, an integer number is either $1$, a prime or a product of some primes, so it is either prime or composite, or $1$.


Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must

be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$

My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula

$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$

$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$

$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$

Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.

Help is required.

Thanks

First Answer:
\begin{align*}
2\alpha +\beta = -\frac{3b}{a} &\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
&\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\
&\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\
&\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\
&\Rightarrow 2a{\alpha}^3+4b{\alpha}^2+2c\alpha=0\tag{1}
\end{align*}
and
\begin{align*}
{\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.\tag{2}
\end{align*}

By subtracting (1) and (2) we have
 $$b\alpha^2+2c\alpha+d=0\Rightarrow ab\alpha^2+2ac\alpha+ad=0.$$
Now from (0) we have
$$ab\alpha^2+2b^2\alpha+bc=0$$
Now subtract two last equations: $(2ac-2b^2)\alpha+ad-bc=0$
Thus
$$\fbox{$\alpha=\frac{bc-ad}{2(ac-b^2)}$}$$

Second Answer for alpha:

You can continue like this
$$
2\alpha +\beta = -\frac{3b}{a} \Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
$$
put it in the last one:
\begin{align*}
\left(-\frac{3b}{a}-2\alpha\right)\alpha^2=-\frac{d}{a}&\Rightarrow-2\alpha^3-\frac{3b}{a}\alpha^2=-\frac{d}{a}\\
&\Rightarrow 2a\alpha^3+3b\alpha^2-d=0\\
\end{align*}
on the other hand $\alpha$ must satisfy the original equation, thus
$$a\alpha^3+3b\alpha^2+3c\alpha+d=0$$
Then add two last equations:
$$3a\alpha^3+6b\alpha^2+3c\alpha=0$$
Since $\alpha\ne0$ we have : $a\alpha^2+2b\alpha+c=0$, thus
$$\alpha=-b+\sqrt{b^2-ac}\quad \text{ or }\quad \alpha=-b-\sqrt{b^2-ac} $$

I made a formula to solve a cubic equation with all equal roots.

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

To find such an equation you will have to check the condition $b=\sqrt{3ac}$ if it is so then we will use this formula $\frac{-b+\sqrt[3]{b^3 - 27a^2d}}{3a}$.


Does there exist an onto map from $[0,1]$ to $\mathbb Q$? Does there exist an onto map from $[0,1]$ to $\mathbb Q$?
I am unable to come up with a proof or a counter example.
If we take $f$ to be an onto mapping $[0,1]$ to $Q$, then $f([0,1])=\mathbb Q$.
I am unsure if it will give any contradiction. If not then what will be the map?
Can someone please help me out?

If by "map" you mean a "function" then yes. First take any surjection $g:[0,1]\to\mathbb{R}$, e.g. variant of $g(x)=\tan(x)$. Now pick any $x_0\in\mathbb{Q}$ and define
$$f:\mathbb{R}\to\mathbb{Q}$$
$$f(x)=\begin{cases}
x &\text{if }x\in\mathbb{Q} \\
x_0 &\text{otherwise}
\end{cases}$$
It is surjective which I leave as an exercise. You can then compose that $f$ with  $g$ to obtain surjection $[0,1]\to\mathbb{Q}$.
If, on the other hand, by "map" you mean "continuous function" then the answer is "no": $\mathbb{Q}$ is (totally) disconnected while $[0,1]$ is connected.

Note that to show that the positive rationals are countable, we can use the mapping $$\frac{p}{q} \mapsto 2^p3^q$$ which is onto. It is easy to see then that
$$f(x)=\begin{cases}
\frac{p}{q} &\text{if }x=\frac{1}{2^p3^q}&\text{with }p,q &\text{integers}\gt0\\-
\frac{p}{q} &\text{if }x=\frac{1}{2^p3^q\sqrt{2}}&\text{with }p,q &\text{integers}\gt0\\
0 &\text{otherwise}
\end{cases}$$
is onto from $[0,1]$ to the rationals.


How to solve $|-2x^2+1+e^x+\sin(x)|=|2x^2-1|+e^x+|\sin(x)|$? How to solve $|-2x^2+1+e^x+\sin(x)|=|2x^2-1|+e^x+|\sin(x)|$ ?

I've solved equations like $|a|+|b|=|a+b|$ where the condition must be that $a$, $b$ must be of same sign. But in case of three terms like the one above what should be the condition?

If $x$ is a real number, then all three quantities must be positive and intersection of all three sets will be final value of $x$ i.e $2x^2-1>0$ & $\sin(x)>0$. So $x\in[\frac{1}{\sqrt{2}},\frac{\pi}{2}] \cup [2n\pi,2n\pi+\frac{\pi}{2}]$ and same for the negative values of $x$.

Hope this is helpful

To avoid calculations and achieve at a direct answer, what you could do is compare the domains of the expressions on the left hand side and right hand side of the equations and then take out the common values of x from both the domains. That would be your answer.


Computing the integral of the sine integral remainder function Define $\mathrm{si}(x) := \displaystyle\int_x^\infty \frac{\sin(t)}{t}\mathrm{d}t $ for all $x>0$. I have showed, by integration by parts, that this function has convergent integral over $\mathbf{R}^*_+$, but I now have trouble actually computing
$$ \int_0^\infty \mathrm{si}(x)\mathrm d x. $$
Anyone got any pointers ?

Update: For detail, I'll show's $\mathrm{si}$ integral converges. We have :

$$\mathrm{si}(x) = \cos(x)/x - \int_x^\infty \frac{\cos(t)}{t^2}\mathrm d t = \frac{\cos(x)}{x} + \frac{\sin(x)}{x^2} - 2\int_x^\infty\frac{\sin(t)}{t^3}\mathrm d t. $$
The second integral is smaller than the integral of $1/t^3$ over the same interval, which is $1/2x^2$. Thus
$$ \mathrm{si}(x) = \frac{\cos(x)}x + \frac{\sin(x)}{x^2} + O\left(\frac{1}{x^2}\right) = \frac{\cos(x)}x + O\left(\frac{1}{x^2}\right). $$

Integrating by parts, $$ \begin{align} \int_{0}^{b} \left(\int_{x}^{\infty} \frac{\sin t}{t} \, dt \right) \, dx &=  x \left( \int_{x}^{\infty} \frac{\sin t}{t} \, dt \right)\Bigg|^{b}_{0}+ \int_{0}^{b} \sin(x) \, dx \\ &= b \int_{b}^{\infty} \frac{\sin t}{t} \, dt  -\cos(b) + 1 . \end{align}$$

But you showed that  $\int_{b}^{\infty} \frac{\sin t}{t} \, dt \sim \frac{\cos b}{b} + \frac{\sin b}{b^{2}} $ as $b \to \infty$.

So as $b \to \infty$, $b \int_{b}^{\infty} \frac{\sin t}{t} \, dt  -\cos(b)$ tends to zero, showing that the integral does indeed evaluate to $1$.

Strangely, Wolfram Alpha (unlike Mathematica) says that the integral doesn't converge.  But if you make the upper limit very large, it returns a value very close to $1$.

$\int_0^\infty \mathrm{Si}(x)\mathrm dx$
$\int_0^\infty \int_x^\infty \mathrm{\frac{sint}{t}}\mathrm dtdx$

Then by changing the order of integration we get

$\int_0^\infty \int_0^t \mathrm{\frac{sint}{t}}\mathrm dxdt$

$\int_0^\infty \mathrm{sinx}\mathrm dx$

Which diverges


Why does the Euclidean algorithm always terminate? Why does the Euclidean algorithm always terminate? Can we make this effective by bounding the number of steps it takes in terms of the original integers?

It always terminates because at each step one of the two arguments to $\gcd({}\mathbin\cdot{},{}\mathbin\cdot{})$ gets smaller, and at the next step the other one gets smaller.  You can't keep getting smaller positive integers forever; that is the "well ordering" of the natural numbers.  As long as neither of the two arguments is $0$ you can take it one more step, but it can't go on forever, so you have to reach a  point where one of them is $0$, and then it stops.

As for bounds, a very crude and easily established upper bound on the number of steps is the sum of the two arguments. One of the arguments is reduced by at least $1$ at each step, and you can't reduce $n$ repeatedly by $1$ more than $n$ times without bringing it to $0$.

The worst case is $\gcd(m,n)$ where the ratio of $m$ to $n$ is the ratio of two consecutive Fibonacci numbers.  For now I'll leave the proof of that as an exercise.

If you take gcd (x,0), the process doesn't terminate. In effect because r=x

This causes quite the show if you're using a physical adding machine

https://www.alltechbuzz.net/mechanical-calculator-dividing-by-zero/

(Link shows adding machine stuck in an endless loop)

So if the process never ends, you would conclude one of the integers in the Euclidean Algorithm would have to be 0.


Are there real numbers that cannot be uniquely expressed with a finite number of symbols? Are there real numbers that cannot be uniquely expressed with a finite number of symbols? Is this the same thing as an uncomputable number?

I can show that if there is one such number then there are infinitely many, and the set of all these numbers is not compact if it is nonempty, but I don't know if any such numbers exist.

Suggestions for tags are welcome.

Edit: I want to add that by "symbols" I mean numbers, mathematical symbols, or English language descriptions (or any language really). So for π, you could write it as "the ratio of a circle's circumference to its diameter" and that would count.

Internally, most of them cannot be so expressed, assuming you have a finite alphabet. The set of all finite strings from a finite alphabet is countable, but the reals are uncountable. Consequently, every function from the set of strings to the set reals is missing uncountably many reals.

Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using. Here is a related question asking whether, not just each individual real number, but every set can be defined.

I'd appreciate it if a set-theorist chimes in to shred this answer but here is a half-serious attempt:

Apply a well-ordering to the real numbers.

Define $$E = \{x \in \mathbb R : \text{$x$ cannot be uniquely expressed with a finite number of symbols}\}.$$

Assume $E \not= \emptyset$. Then $E$ has a least element $e$. Thus $e$ cannot be uniquely expressed with a finite number of symbols, yet we can unambiguously state

  $e$ is the least element of $\mathbb R$ (unique, under the given well-ordering) that cannot be uniquely expressed with a finite number of symbols

which uniquely expresses $e$ with a finite number of symbols. This is a contradiction which leads to $E = \emptyset$.

p.s. My own guess is that the argument fails because the well-ordering itself is not expressible with finitely many symbols.


The elementary question on sign of Rational numbers The image below can be used to show that $+\frac{8}{3}=\frac{+8}{3}$.

Similarly we can show $-\frac{8}{3}=\frac{-8}{3}$. Now how can we show that
$$-\dfrac{8}{3}=\dfrac{8}{-3}?$$


When is the number of lattice paths from $(1, 1) \to (x, y)$ divisible by $3$? Let $S$ be the set of $\{(1,1), (1,−1), (−1,1), (1,0), (0,1)\}$-lattice paths which begin at $(1,1),$ do not use the same vertex twice, and never touch either the $x$-axis or the $y$-axis. Let $S_{x,y}$ be the set of paths in $S$ which end at $(x,y).$ For how many ordered pairs $(x,y)$ subject to $1\leq x,y \leq 31$, is $|S_{x,y}|$ a multiple of $3$?

This is an old contest question from Brilliant.org.

This is a cute problem.

Let $D_n$ be the diagonal which contains $(n,1)$ and $(1,n)$

For any two points $P$ and $Q$ on $D_n$, we must have that $|S_P| = |S_Q| = T_n (\text{say})$. (Can be easily shown, I will omit it).

Then we get the recurrence

$$T_{n+1} = 2nT_n + (n-1)T_{n-1}$$

with $T_1 = 1$, $T_2 = 2$.

You can compute your answer now. There are probably ways to simplify it further to ease the computation (so we can do it manually), but the above should be doable with a quick program.

Isn't $|S_{x,y}| = \infty$?

The set of steps allow you to go arbitrarily far and come back, without repeating vertices, or touching the axes.

Perhaps you are missing something from the problem statement?

For instance, are the paths restricted to stay within the square $1 \le x,y, \le 31$ and not just end there?


Obtain the equation of the sphere which passes through the points $(1,0,0),(0,1,0),(0,0,1)$ and has its radius as small as possible. Obtain the equation of the sphere which passes through the points $(1,0,0),(0,1,0),(0,0,1)$ and has its radius as small as possible.

Let  the sphere passes through $(x_1,y_1,z_1)$
Then i obtained the equation of sphere as $(x^2+y^2+z^2)(x_1+y_1+z_1)-(x_1^2+y_1^2+z_1^2-1)(x+y+z)+x_1^2+y_1^2+z_1^2-x_1-y_1-z_1=0$
I am stuck here.I do not know how to minimize the radius.The answer given in my book is $3(x^2+y^2+z^2)-2(x+y+z)-1=0$

If the center of the sphere is $(a,b,c)$, then the square of the distance from $(a,b,c)$ to each of the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ is the same. Solve these equations to get a simple relation among $a$, $b$, and $c$. Then minimize the distance from the center to any one of the points.

Let the equation of sphere is  $x^2+y^2+z^2+2ux+2vy+2wz+d=0$
It's given that the sphere passes through the points $(1,0,0),(0,1,0),(0,0,1)$
Therefore we get $u=v=w=-1/2(d+1)$
We know that $r^2=u^2+v^2+w^2-d$
                     Put the values of $u, v,w$:
                    $r^2=3/4((d+1/3)^2+8/4)$
We have to make $r$ as small as possible
Therefore $r$  will be smallest if $d=-1/3$ hence put the values of u, v, W, and d in equation of sphere and solve it.
We get the required result


$\lim\limits_{N \to +\infty} \sqrt{N+1} \log\left (1+\frac{x}{N+1}\right)$ I have to compute the limit

$$
\lim\limits_{N \to +\infty} \sqrt{N+1} \log \left(1+\frac{x}{N+1}\right)
$$
where $x \ge 0$ is fixed. I tried to see the previous as

$$
\log \lim\limits_{N \to +\infty} \left(1+\frac{x}{N+1}\right)^{\sqrt{N+1}}
$$
and to change variable, but it doesn't work. Intuitively, this limit is 0, but I have no clue on how to solve it. Can you help me?

The "+1" doesn't matter.

For large $N$ and fixed $x$

$\sqrt{N} \log (1+\frac{x}{N})
=\sqrt{N} (\frac{x}{N}+O((\frac{x}{N})^2))
=\frac{x}{\sqrt{N}}+O(\frac{1}{N^{3/2}})
\to 0
$.

This holds with
$N^{c}$
(instead of $\sqrt{N}$)
for all
$0 < c < 1$.

If
$c=1$,
we have
$N \log (1+\frac{x}{N})
=N (\frac{x}{N}+O((\frac{x}{N}))^2)
=x+O(\frac{1}{N})
\to x
$.

hint: use taylor expansion for the Log term


Showing probability no husband next to wife converges to $e^{-1}$ Inspired by these questions:

Probability of Couples sitting next to each other (Sitting in a Row)
Probability question about married couples
Four married couples, eight seats. Probability that husband sits next to his wife?
In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?
No husband can sit next to his wife in this probability question

the more general question of the probability that seating $n$ couples (i.e. $2n$ individuals) in a row at random means that no couples are sat together can be expressed using inclusion-exclusion as

$$\displaystyle\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$$

which for small $n$ takes the values:

n   Probability of no couple together
1   0           0
2   1/3         0.3333333
3   1/3         0.3333333
4   12/35       0.3481481
5   47/135      0.3428571
6   3655/10395  0.3516114
7   1772/5005   0.3540460
8   20609/57915 0.3558491

This made me wonder whether it converges to $e^{-1} \approx  0.3678794$ as $n$ increases, like other cases such as the secretary/dating problem and $\left(1-\frac1n\right)^n$ do.  So I tried the following R code (using logarithms to avoid overflows)

couples  <- 1000000
together <- 0:couples
sum( (-1)^together * exp( log(2)*together + lchoose(couples,together) +
     lfactorial(2*couples - together) - lfactorial(2*couples) ) )

which indeed gave a figure of $0.3678794$.

How might one try to prove this limit?

I observe that each term with $i$ fixed approaches a nice limit. We have

$$ 2^i \frac{n(n-1)(n-2)\cdots(n-i+1)}{i!} \frac1{(2n-i+1)(2n-i+2)\cdots(2n)} $$

or

$$ \frac1{i!} \frac{2n}{2n} \frac{2(n-1)}{2n-1} \cdots \frac{2(n-i+1)}{(2n-i+1)} \sim \frac 1{i!} $$

This gives you the series, assuming the limits (defining terms with $i>n$ to be zero) may be safely exchanged,

$$\lim_{n\to\infty} \sum_{i=0}^\infty [\cdots] =  \sum_{i=0}^\infty \lim_{n\to\infty} [\cdots] = \sum_{i=0}^\infty (-1)^i \frac1 {i!} \equiv e^{-1}$$

Justifying the limit interchange I haven't thought about but I suspect this can be shown to be fine without too much effort... Edit: You can probably use the Weierstrass M-test.

Without loss of generality, suppose everyone is sitting in a circle instead of a row and that there is one unoccupied seat.  For each wife, the probably that her husband is not sitting next to her on either side is $1 - \frac{2}{2 \cdot n}$ and there are $n$ wives, so the probability that no wife is sitting next to her husband is $(1- \frac{1}{n})^n$, whose limit is the definition of $\frac{1}{e}$.


Is there a sequence $a_n$ for which $\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n}|$ doesn't exist, but $\sum a_n$ is convergent? The question is in the title:

Is there a sequence $a_n$ for which:

$\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n}|$ doesn't exist (so it's not defined, which means we're looking for a case when it's now even $+\infty$ nor $-\infty)$
$\sum_{n=1}^{+\infty} a_n$ is convergent

According to my workbook, such a sequence exists but I cannot find it. Any hints?

Take for example the sequence

$$a_n=\frac{\sin n}{n^2}$$

There are different types of non-existent limits (divergent limits). Some are infinitely divergent, and some simply fail to converge. Still others converge to different values depending on direction of approach.

I would have to say that infinite divergent ratios make a sum divergent. As does any convergent ratio larger than 1. Any ratio limit, convergent or divergent, that is larger than 1, proves the sum is divergent.

As far as infinite series go, there is only one direction from which to approach the sum.

BUT... if the ratio simply fails to converge - such as due to oscillations - then I do not believe this says anything about the series.


There are total of $16$ people. How many ways can you make two committees of $8$ people in each committee? You have $16$ people in a class. How many ways can you make two committees of $8$ people in each committee?

How do I do this problem?

How many ways are there to choose the first person on your committee? 16. How about the second person? 15, because there is one person who is not available. So we have $16*15*14*13*12*11*10*9$ ways. But we've overcounted because we've chosen the same committees of people in several different orders. How many times did we count each committee? This is $8!=8*7*6*5*4*3*2*1$ because there are this many ways to order $8$ people. (8 ways to choose the first, 7 ways to choose the second...) So we divide $16*15*14*13*12*11*10*9$ by $8!=8*7*6*5*4*3*2*1$ to account for the overcounting. Assuming that we don't care to which committee a certain group of $8$ people is assigned, we must also divide by 2 because it's the same to choose a group of people or its complement.

The final answer is $\frac{16*15*14*13*12*11*10*9}{8*7*6*5*4*3*2*1}= \frac{16!}{2*8!8!}=\frac{1}{2}\binom{16}{8}$.

An alternative way of thinking about this problem is that you line 16 people up in a line where each position decides what committee they will be placed on. Then we must divide out the redundant combinations. so $ \frac{16!}{8!*8!} $


How can we know it is better to compress some data with an algorithm based on our previous experience of a similar type of data? In practice, people use the same type of compression algorithm for similar data from a similar source. Why can we compress different pictures with the same compression algorithm, and it works for all? Is it because of a statistical or information-theoretic property of the data, and if it is, what is that property?

There's compresson and then there's compression. Say a compression algorithm is lossless if there exists a decompression routine that will return exactly te original file, for any file. Conversely, an algorithm is lossy if it  is not lossless. Typical flavors of jpeg are lossy, for example:  When you convert an image to jpeg in a graphics program you get to choose a compression ratio;  more compression gives a smaller jpeg but it looks  less like the original.

Never mind lossy compression for now; an explanation of why that works at all would involve talking about how visual perception works. Regarding lossless compression: How can we know it works on all files? We can't know that, in fact we do know just the opposite:

    Obviousness. There does not exist a lossless compression algorithm that decreases the size of every non-empty file.

Proof: say $C(f)$ is the compressed version of the file $f$. Suppose $f$ and $g$ are  one-byte files with $f\ne g$. Then $C(f)$ and $C(g)$ are both zero-byte files, so $C(f)=C(g)$; hence $C$ is not one-to-one.

So why do compression algorithms work in practice? They take advantage of  redundancies common to the files being compressed.

For example, the sort of image files  that are often converted to gif typically involve blocks of solid color. Which is to say there's a lot of repetition in thhe bit pattern, making "run-length encoding" useful:

Say the file contains the string "0000000000111110000000". One can easily concoct a format that says "ten zeroes, then five ones, then seven more zeroes" in  less than 22 bytes.


Generate a random direction within a cone I have a normalized $3D$ vector giving a direction and an angle that forms a cone around it, something like this:

I'd like to generate a random, uniformly distributed normalized vector for a direction within that cone. I would also like to support angles greater than pi (but lower or equal to $2\pi$), at which point the shape becomes more like a sphere from which a cone was removed. How can I proceed?

I thought about the following steps, but my implementation did not seem to work:

Find a vector normal to the cone axis vector (by crossing the cone axis vector with the cardinal axis that corresponds with the cone axis vector component nearest to zero, ex: $[1 0 0]$ for $[-1 5 -10]$)
Find a second normal vector using a cross product
Generate a random angle between $[-\pi, \pi]$
Rotate use the two normal vectors as a $2D$ coordinate system to create a new vector at the angle previously generated
Generate a random displacement value between $[0, \tan(\theta)]$ and square root it (to normalize distribution like for points in a circle)
Normalize the sum of the cone axis vector with the random normal vector times the displacement value to get the final direction vector

[edit] After further thinking, I'm not sure that method would work with theta angles greater or equal to pi. Alternative methods are very much welcome.

I'm surprised how many bad, suboptimal and/or overcomplicated answers this question has inspired, when there's a fairly simple solution; and that the only answer that mentions and uses the most relevant fact, Christian Blatter's, didn't have a single upvote before I just upvoted it.

The $2$-sphere is unique in that slices of equal height have equal surface area. That is, to sample points on the unit sphere uniformly, you can sample $z$ uniformly on $[-1,1]$ and $\phi$ uniformly on $[0,2\pi)$. If your cone were centred around the north pole, the angle $\theta$ would define a minimal $z$ coordinate $\cos\theta$, and you could sample $z$ uniformly on $[\cos\theta,1]$ and $\phi$ uniformly on $[0,2\pi)$ to obtain the vector $(\sqrt{1-z^2}\cos\phi,\sqrt{1-z^2}\sin\phi,z)$ uniformly distributed as required.

So all you have to do is generate such a vector and then rotate the north pole to the centre of your cone. If the cone is already thus centred, you're done; if it's centred on the south pole, just invert the vector; otherwise, take the cross product of the cone's axis with $(0,0,1)$ to get the direction of the rotation axis, and the scalar product to get the cosine of the rotation angle. Or if you prefer you can apply your idea of generating two orthogonal vectors, in the manner Christian described.

I've thought of another approach that seems to produce good results from quick visualization tests. I'd love to get feedback on this.

The 2D case is much simpler, I just have to add an angle within [-theta, theta] to the "cone" axis vector's angle and create a vector from the new angle. The 3D case is like a solid of revolution. I can generate an angle within [0, theta] and then move it in 3D space to a random rotation about the cone axis vector. I end up with the following algorithm:

axis = cone axis vector
angleAroundAxis = random(0, 2 * pi)
angleAwayFromAxis = random(0, theta)

matrix1 = rotation matrix of 'angleAroundAxis' around 'axis'
anyNormal = any vector that is normal to 'axis'

normal = anyNormal * matrix1
binormal = normal x axis

matrix2 = rotation matrix from 'axis', 'normal' and 'binormal' axes

result = [cos(angleAwayFromAxis) sin(angleAwayFromAxis) 0] * matrix2

Sorry if this is using non-standard notation, I'm not very familiar with formal math notation.


Non-surjective but injective real polynomial functions $\mathbb{R}^n\to \mathbb{R}^n$ Over algebraically closed fields $K$, the Ax–Grothendieck theorem (see also this thread) states that injective polynomial functions $K^n \to K^n$ in $n$ variables are surjective. Is there a simple counterexample for this statement for real polynomial functions i.e. $K=\mathbb{R}$?

The statement of the theorem holds even for $k=\mathbb{R}$. See the article

Białynicki-Birula, A., Rosenlicht, M.: Injective Morphisms of Real Algebraic Varieties.

A polynomial function $f:\mathbb R\to\mathbb R$ is either of odd degree and surjective, or of even degree and not injective.


Strong Induction Proof: Every natural number = sum of distinct powers of 2 The question I'm looking at, is to show that every positive integer $n$ can be written as a sum of distinct powers of two.

I can see that you can form any number based on the highest $2^t$ that is less than the number, plus some combination of $2^j<n$'s. And that you can make the number odd, by adding $2^0$ at the end.

I just don't know how to create the formula for the proof. I'm trying to figure out my base case, and then my inductive formula to figure out $k+1$, and I've got nothing.

The statement is obviously true for $n=0$.

Assume that we are given an $n\geq1$ and that it is true for all $m$ with $0\leq m<n$.

When $n=2m$ then $m<n$ and therefore $m=\sum_k 2^{p_k}$ with finitely many $p_k$, all of them different. It follows that $n=\sum_k 2^{p_k+1}$ with all $p_k+1$ different.

When $n=2m+1$ with an $m$ as before then $n=2^0+\sum_k 2^{p_k+1}$ with all $p_k+1$ different and different from $0$.

The subtle thing is the exact wording of the proposition.

If you want to prove:

$P(n) = $ $n$ can be expressed as a distinct sum of powers of $2$.

Initial case: $n = 1$ $P(1)$ is provable.

Induction step: $P(k) \implies P(k+1)$

then this is impossible.  $P(k) \not \implies P(k+1)$ and there is simply no way to show it does.

But if instead you try to prove.

$Q(n) = $ of all $j;  0 < j < 2^n$, $j $ can be written as a distinct product of powers of $2$

then $Q(k) \implies Q(k+1)$ is pretty easy to prove.

(if $0 < j \le 2^{n+1}$ then either $j = m$ or $j = m + 2^n$ where $0 < m < 2^n$ so $m$ can be distinctly written as $\sum 2^i$.  As $m < 2^n$ none of its powers of $2$ are to the $n$ power.  So  $m + 2^n = 2^n + \sum 2^i $ is a sum with distinct powers of $2$.  So $j$, whethe $j = m$ or $j = m + 2^n$ can be written as a sum of distinct powers.)

...

Another way to view this, albeit, maybe not as rigorous.  Is that induction need not be a direct $P(n) \implies P(n+1)$ progression.  And $P(n) \implies P(f(n))$ where $\cup_{i \in \mathbb N} F_i= \mathbb N$ where $F_0 = \{1\text{ or } 0\}$ and $F_{i+1} = \{f(a)|a \in \cup_{k \le i}F_i\}$, qualifies as induction as well.

Here $f(n) = n + 2^k$ where $k$ is the smallest power where $2^k > n$.

(although the bit $\cup_{i \in \mathbb N} F_i= \mathbb N$ most likely will boil down to a $Q(n) \implies Q(n+1)$.  e.g.  How do we know $\{0,1\}, \{0,1, 2, 3\},\{0, 1, 2, 3,4,5,6,7\} ....$ will span $\mathbb N$?  Probably by induction.)


Sum $S(n,c) = \sum_{i=1}^{n-1}\dfrac{i}{ci+(n-i)}$ Consider the sum $$S(n,c) = \sum_{i=1}^{n-1}\dfrac{i}{ci+(n-i)}$$ where $0\le c\le 1$.

When $c=0$, $S(n,c)$ grows asymptotically as $n\log n$.

When $c=1$, $S(n,c)$ grows asymptotically as $n$.

What about when $0<c<1$? Can we calculate $S(n,c)$ exactly? What about asymptotics? Can we find upper/lower bounds?

We have
$$S(n;c) = \sum_{k=1}^{n-1} \dfrac{k}{ck+(n-k)} = \sum_{k=1}^{n-1} \dfrac{k/n}{1+(c-1)k/n}$$
Hence,
$$\dfrac{S(n;c)}n = \sum_{k=1}^{n-1} \dfrac{k/n}{1+(c-1)k/n} \dfrac1n \sim \int_0^1 \dfrac{xdx}{1+(c-1)x} = \dfrac{(c-1)-\log(c)}{(c-1)^2} \text{ for }c \in (0,1)$$
Hence,
$$S(n;c) \sim \dfrac{(c-1)-\log(c)}{(c-1)^2} n \text{ for }c \in (0,1)$$
Better approximations can be obtained using Euler Maclaurin formula.

There is a closed form, thanks to mathematica (using digammas):

$\frac{c n+n \psi ^{(0)}\left(\frac{c}{c-1}+\frac{n}{c-1}-\frac{1}{c-1}\right)-n
   \psi ^{(0)}\left(\frac{n
   c}{c-1}+\frac{c}{c-1}-\frac{1}{c-1}\right)-n}{(c-1)^2}$

There are, therefore, asymptotics (also thanks to mathematica), on which MathJax seems to choke, here is a link to the screen grab:
https://www.evernote.com/shard/s24/sh/e1d1abb1-b8a4-4f4f-bad8-7e476f73f017/7700044225f8b6d0d008346a44a2fc5d

These can be simplified, if one knew the sign of $c,$ but I leave this to you.


Check if $f(z) = 2xy + i(x^2 - y^2) $ is analytic or not Check whether the function defined by $f(z) = 2xy + i(x^2 - y^2) $ is analytic or not.

Sufficient condition for a function being analytic: The sufficient condition for a function $f(z) = u + iv$ to be analytic at all the points in region $R$ are
$\displaystyle (a)\; \frac{\partial u}{\partial x}  = \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  = -\frac{\partial v}{\partial x} $
$\displaystyle (b)\; \frac{\partial u}{\partial x}  , \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  ,\frac{\partial v}{\partial x} $ are continuous functions of $x$ and $y$ in region $R$ .
Here, $u=2xy$ , $v = x^2-y^2$
$\displaystyle \frac{\partial u}{\partial x} = 2y$,
$\displaystyle \frac{\partial v}{\partial y} = -2y$
$\displaystyle \frac{\partial u}{\partial y} = 2x$,
$\displaystyle \frac{\partial v}{\partial x} = 2x$
If this is analytic function then,
$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ & $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
So,$x = y = 0$The partial derivatives are continuous .
So, I think function is analytic only at $(0,0)$ but the answer says that it is not analytic.Am I wrong or the answer in the book is incorrect?

For $z=x+iy$  we have $f(z) = 2xy + i(x^2 - y^2) = i\bar{z}^2$  is not analytic.

Since $\partial_{\bar{z}}f \neq 0$.

Hint: You've concluded that for the C-R equations to hold, you need to have $x = y= 0$. So at $x = 1, y = 0$, is the function analytic? No.

Alternatively, you may have made a mistake in checking the C-R equations. I leave it to you to decide which.


Proving the geometric series $\sum_{i=0}^n r^i = \frac{1-r^{n+1}}{1-r}$ by induction for $n\geq 1$ Let $P(n)$ be the statement
$$
P(n) : \sum\limits^{n}_{i=0}r^i = \dfrac{1-r^{1+n}}{1-r}\text{ for all }n \in \mathbb{N}\text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = \dfrac{1-r^2}{1-r}\text{.}$$

I am stuck as to how I can show $P(1)$ is true.

Note: First note that you are really summing the geometric series ($a,r\in\mathbb{R}, a\neq 0, r\neq 1$)
$$
a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}=a\frac{1-r^{n+1}}{1-r}
$$
with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.

Claim: Let $a$ and $r$ be real numbers with $a\neq 0$ and $r\neq 1$. Prove that for each integer $n\geq 1$ that
$$
a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}.
$$
Proof. Fix $r\in\mathbb{R}, r\neq 1$, and let $S(n)$ denote the following statement for $n\geq 1$:
$$
S(n) : 1+r+r^2+\cdots+r^n=\frac{r^{n+1}-1}{r-1}.
$$
In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $n\geq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $r\neq 0$ without loss of generality.

Base step ($n=1$): The statement $S(1)$ says that $1+r=\frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $r\neq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=\frac{r-1}{r-1}$.]

Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and suppose that $S(k)$ is true; that is, assume that
$$
S(k) : 1+r+r^2+\cdots+r^k=\frac{r^{k+1}-1}{r-1}
$$
is true. We must show that $S(k+1)$ follows where
$$
S(k+1) : 1+r+r^2+\cdots+r^k+r^{k+1}=\frac{r^{k+2}-1}{r-1}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &= \color{red}{1+r+r^2+\cdots+r^k}+r^{k+1}\tag{by definition}\\[1em]
&= \color{red}{\frac{r^{k+1}-1}{r-1}}+r^{k+1}\tag{by $S(k)$}\\[1em]
&= \frac{r^{k+1}-1}{r-1}+\frac{r^{k+1}(r-1)}{r-1}\tag{common denom.}\\[1em]
&= \frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}\tag{combine like terms}\\[1em]
&= \frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}\tag{expand}\\[1em]
&= \frac{r^{k+2}-1}{r-1},\tag{simplify}
\end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)\to S(k+1)$, completing the inductive step.

By mathematical induction, for all $n\geq 1, S(n)$ holds true. $\blacksquare$

Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $\sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.

$$1+r=\frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
I don't need to tell you more.


Problems with Cantor's diagonal argument and uncountable infinity Cantor's diagonal argument:
As a starter I got 2 problems with it (which hopefully can be solved "for dummies")

First: I don't get this: Why doesn't Cantor's diagonal argument also apply to natural numbers?
If natural numbers cant be infinite in length, then there wouldn't be infinite in numbers. By using a randomly ordered list, you wouldn't end with an endless sequence of 0's you have to change.
Also it initially goes for "set of numbers". It is applied to the "right" side (fractional part) to prove "uncountability" but can't be used for the "left" side (integer part) because of "reasons" (I simply do not get it).

Second: the way is is used so many times, would just work in the case that the length and width of the list equals. Just listing all natural numbers where $0<=x<100$ will have a width of 3 but a length of 100. At base x each increase of digits will increase the length by x times. At base 10, all 4 digit numbers will create a list with 1000 entries. The length increases exponentially while the width does linearly. This is wrong in so many ways but: Doing this infinitely makes it a square?!? (Natural numbers are a "part" of integers but as you can map both with each other they are considered being the same size aka 2 different countable infinities always have the same "size" while one can be just a part of the other)

As the last part: lets assume we divide all real numbers in 2 parts. The integer part which defines the "set" we use. (there will be "countable" infinite of them)
Now, all we need to do is mapping the fractional part. Just use the list of natural numbers and flip it over for their position (numeration). Ex 0.629445 will be at position 544926. You could argue that this isn't possible for numbers like $\sqrt{2}$.
Lets pick $\pi$:

3.1415926535897932384626433832795… will be
3rd set at position: …5972383346264832397985356295141

There is no reason you cannot pick the next digit and put it in front for the position. There is no limit in length for natural numbers -> you can write a natural number which is the index for just that fractional part. Simply put: you can map EVERY number with $0<=x<100$ with a natural number. And a countable infinite amount of sets containing countable infinite entries still is countable.

So there are 3 Questions (I probably need to split this question):

Why doesn't Cantor's diagonal argument also apply to natural numbers? (for dummies: why you can't simply use it to the left)
If the count of digits equals the the length of the list, doesn't it just proves that this construction cannot contain all possibilities?
Wouldn't the construction of a set like in "the last part" be a prove that all real numbers are countable infinite? (Which part can't be done / is invalid?)

The most important part would be the third question. (If it only qualifies for one answer) Thanks in advance.

An additional big "thanks" in advance for correcting all the spelling, orthography and typos...

"If natural numbers can't be infinite in length, then there wouldn't be infinite in numbers." You are extrapolating properties of the natural numbers to what is called "potential infinity." But that is not what Cantor means by Aleph0.$\newcommand\N{\mathbb{N}}$

"Potential Infinity" is a hypothetical number that represents the hypothetical end to the sequence 1,2,3,... . Nobody treats it as a real object, but only in hypothetical evaluations of expressions where its impact on the result diminishes to nothing. For example, f(x)=3/(2+x) approaches 3/2 "as x goes to potential infinity." The point is that it is treated like a number with the same properties as "regular" numbers, but you can never reach it. "Actual infinity" is a well-defined mathematical concept, but it is nothing like a "regular number."

The "cardinality" of the set {1} is the cardinal number 1. That is, the set has one member. The cardinality of the set {1,2} is the cardinal number 2. In general, the cardinality of the finite set {1,2,3,...,n} is the largest number in it, n.

The Axiom of Infinity says that there is a set $\N$ with a similar form, but that has no largest member. Its cardinality is Aleph0, which can be called "Actual Infinity." It is not a member of the set it describes. It is not a "regular number" in any way. Even though we can extend the definition of numbers to include it as an "irregular number," it does not have the all the same properties. One that is different, is that you can't represent it with a string of digits. Not even an infinite string whose value you might think you could claim is "Potential Infinity."

Diagonalization does not work on natural numbers because it requires a digit for every member of $\N$, and that does not represent a natural number.

About your second problem: you can't consider an $\N$ by $\N$ array of characters to be square. "Square" means that you can only pair rows with columns one a 1:1 basis. While you can make such a pairing with an $\N$ by $\N$ array, you can also pair them on a 1:2 basis. Or 3:5. Or 14:9. The point of using Aleph0 for a cardinality, is that it represents any countably infinite set the same way.

For the first question-how natural numbers are infinite in number but every natural number has finite digits?
I will be expressing every natural number in base 2 for simplicity.
Consider a binary tree in which every node has 2 child nodes.Top most node is .0 its child nodes are 0.0 and 1.0.Child nodes for 0.0 are 00.0 and 10.0 and for 1.0 are 01.0 and 11.0 similarly for 00.0 - 000.0 and 100.0 and for 10.0 - 010.0 and 110.0 .........
by using this infinite binary tree we can make an infinite list - (by mentioning the elements of 1st row than 2nd row than 3rd row,4th row...... )                                                 .0
0.0
1.0
00.0
10.0
01.0
11.0
000.0
100.0
010.0
110.0
001.0
101.0
011.0
111.0
.
.
.

clearly this infinite list will contain every natural number (more than ones).
 While making of this list the next element on the list would had only come into the list by being in one of the node of the infinite tree but that means that number would have some leftmost digit but since an infinite number like -'........11110' does not have any leftmost digit it would not be present on the tree and therefore would not have transferred to the list.
ex - just like their is no last digit of e or pi or in the number 0.333333........ (=1/3).
The formal argument that every natural number is finite/has finite digits can be made using induction.

If you are looking for some intuitive understanding for why their are much more real numbers than natural numbers or rational number  here's a one that i developed for myself-
If you think about the number of real numbers by using permutation -
Consider numbers between 0 and 1 all of them can be represented like this-
0.xxxxxxxxxxxxxxxxxxxxxxxxxxx.....
each "x" can have 2 possibility (if binary) and their are aleph null- "x" so total number = 2^aleph null.
Now since numbers from 0-1 can be mapped to entire real number line so they too have 2^aleph null possibilities.
But since natural numbers have only finite digits for permutation so aleph null possibilities.
Also since every rational number can be mapped to natural number so they too have only aleph null possibilities.


Is a certain restriction of an open map open? Let $p:X\rightarrow Y$ be an open map and let $A$ be a subspace of $X$. Then, is it true that $p|_A:A\rightarrow p(A)$ is open?

I think so, but am struggling to show it.

My thoughts: Let $O$ be an open set in $A$. Then, since $A$ has the subspace topology, there exists a $U$ open in $X$ such that $U\cap A=O$. But then, $p(O)=p(U\cap A)$. Yet, I only know that $p(O)=p(U\cap A)\subseteq p(U)\cap p(A)$, which doesn't really help.

If I assume $p$ is also a quotient map, does this help? Or can I do it without that assumption?

The result is not true. See here.

I am not flagging this as duplicate because of the second part of your question.

Also not true for quotient map. Take $p: \mathbb{R} \rightarrow S^1$ (usual topology on the reals, quotient by integers). Then let A be [0.1,0.2]. Then $p(A) $ is not open in $S^1$.

Let X be the real line, Y its quotient by the integers and p be the quotient map. Let A=[0, $\infty)$ and consider the open set [0, 1) in A. Its image is not open in p(A) (which is the whole of Y).


affine variety from ideal I have a problem with this excercise:

*Take a field $F$, algebraic closed, $charF\neq 2$.

1)Consider the following ideal $I=(xy+yz+zx-\frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?

2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*\

How can I check, that it is variety?

I think, that my solutions are incorect, or noncomplete. For now I have:\

1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.\

2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$

(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
   We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
  K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
            =K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.

If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.

(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$

Since $$u=-x-y-z
            =K[x,y,z]/(x^2+y^2+z^2)$$

and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.


How to compute the measure induced by a linear functional The Riesz representation theorem states that for any linear functional $\varphi\in (C_0(K))^*$ there exists a regular complex Borel measure $\mu$ such that
$$
\varphi(f)=\int fd\mu,\ \ \forall f\in C_0(K).
$$
I would like to compute, for a Borel set $E\subset K$, the measure $\mu(E)$, not explicitily, but in terms of $\varphi$. I know that
$$
    \mu(E)=\int \chi_Ed\mu,
$$
but it is not necessarily true that $\chi_E\in C_0(K)$. So I could try to approximate $\chi_E$ by continuous functions $f\in C_0(K)$ such that $Imf\subset[0,1]$.

Define $\mathcal I=\{L: L\subset E\mbox{ and $L$ is compact}\}$. Then $\mathcal I$ is a directed set, if we equip it with the order "$\subset$". Also define $\mathcal J=\{U: E\subset U\mbox{ and $L$ is open}\}$. This is a directed set, if we equip it with the order "$\supset$". Then, the set $\mathcal I\times\mathcal J$ equipped with the order
$$
(L,U)\leq(L',U') \Leftrightarrow L\subset L'\mbox{ or }(L\not\subset L'\mbox{ and }U\supset U'),
$$
is a directed set.

For each $(L,U)\in\mathcal I\times\mathcal J$, we have that $L\subset U$ and, using Urysohn's Lemma, I may fix a function $f_{(L,U)}$ such that
$$
   Im f_{(L,U)}\subset[0,1],\ f_{(L,U)}|_{L}\equiv1\ \mbox{ and }\ f_{(L,U)}|_{U^c}\equiv0,
$$
and consider the net $(f_{(L,U)})_{(L,U)\in\mathcal I\times\mathcal J}$.

It is easy to check that $\chi_{L}\leq f_{(L,U)} \leq \chi_U$. Suppose that $\mu$ is positive measure, then it follows that
$$
\mu(L)\leq \int f_{(L,U)}d\mu = \varphi\left(f_{(L,U)}\right) \leq \mu(U),\ \ \forall (L,U)\in\mathcal I\times\mathcal J.
$$
Since $\mu$ is regular,
$$
\mu(E)=\sup\{\mu(L):L\in\mathcal I\} = \inf\{\mu(U):U\in\mathcal J\}.
$$
Then, by doing some boring calculations, we get that
$$
\liminf_{(L,U) \in \mathcal I\times\mathcal J} \varphi\left(f_{(L,U)}\right) \geq \sup\{\mu(L):L\in\mathcal I\} = \mu(E),
$$
and
$$
\limsup_{(L,U) \in \mathcal I\times\mathcal J} \varphi\left(f_{(L,U)}\right) \leq \inf\{\mu(U):U\in\mathcal J\} = \mu(E).
$$
Therefore,
$$
\lim_{(L,U) \in \mathcal I\times\mathcal J} \varphi\left(f_{(L,U)}\right)=\mu(E).
$$
So this is true when $\mu$ is positive. But we may obtain a decomposition $\mu = (\mu_r^+-\mu_r^-)+i(\mu_i^+-\mu_i^-)$, with $\mu_r^\pm$ and $\mu_i^\pm$ positive measures, so this result follows for the general case.

My question is: Is this right?

That's the right idea.   You can approximate it as well as you would like as follows: for a given $\varepsilon>0$ use regularity to obtain an open set $U$ and a compact set $K$ with $K\subset E\subset U$ and
$$\mu(U)-\varepsilon<\mu(E)<\mu(K)+\varepsilon.$$
 (Or apply regularity for positive and negative variations of the real and imaginary parts of $\mu$ if it is complex, and vary the $\varepsilon$ in the above expression accordingly) and use Urysohn to obtain a continuous $f$ with $\chi_{K}\leq f\leq \chi_{U}$, and thus $|\mu(E)-\int f\ d\mu|<\varepsilon$.

I found one small error: in the line
$$Im f_{(L,U)}\subset[0,1],\ f_{(L,U)}|_{L}\equiv1\ \mbox{ and }\ f_{(L,U)}|_{U}\equiv0,$$
it should be $f_{(L,U)}|_{U^c}\equiv0$.


How to prove this tautology using only logical equivalences? I'm having trouble with this problem.

  Using logical equivalencies prove that $(p \land q)\implies (p \lor q)$ is a tautology.

According to the equivalence $(a\rightarrow b) \Longleftrightarrow \neg a\vee b$, the De Morgan's laws, the associativity and commutativity of the logical operator $\vee$, one has

\begin{align*}
(p\wedge q) \longrightarrow (p\vee q) & \Longleftrightarrow \neg(p\wedge q)\vee(p\vee q) \Longleftrightarrow (\neg p\vee \neg q)\vee(p\vee q)\\\\
& \Longleftrightarrow (\neg p\vee p)\vee(\neg q\vee q) \Longleftrightarrow t\vee t \Longleftrightarrow t
\end{align*}

where $t$ represents a tautology.

A tautology is a statement that is always 'True'.

In your case you have two variables, each may take the value of either 'true' or 'false'.

You may simply plug these four cases in your expression and get 'True' in each case you try. Note that your logical expression is a composition of three operations, i.e. 'and', 'or' and 'implies'. Please, check with Wikipedia for their truth tables.

https://en.wikipedia.org/wiki/Modus_ponens


What is $\lim\limits_{k \rightarrow \infty}{(-2)^k}$? What is the $\lim\limits_{k \rightarrow \infty}{(-2)^k}$?
Can this limit even exist? Isn't the function divergent?
How do I compute $\sum_{k=0}^\infty{(-2)^k}=-2+4-8+16\dots$ ?
For the second part I tried rearranging: $\sum_{k=0}^\infty{(-2)^k}=\sum_{k=0}^\infty{2^{2k-1}}$ I still don't know what this is, first I thought of the geometric series but that only works for $|q|<1$ and 2 obviously isn't smaller than 1. Besides this series is divergent, but the geometric series is konvergent.

If we choose $a_n=2n$ and $b_n=2n+1$, then $a_n\to \infty$ and $b_n\to \infty$. Since
$$\lim_{n\to \infty}(-2)^{a_n}=\infty,$$
$$\lim_{n\to \infty}(-2)^{b_n}=-\infty,$$
we can conclude that the limit doesn't exist.

The limit does not exist, because the sequence has one subsequence that converges to $\infty$ and another subsequence that converges to $-\infty$.

We also know:

  If $\sum_{n=1}^\infty a_n$ exists, then $\lim_{n\to\infty} a_n=0$.

Therefore, we also know that the series is divergent (because the limit, in this case, is not $0$ - since it does not exist).


Question 10 page 43 measure theory for Cohn Prove that if $A$ is Lebesgue measurable set then $\{(x,y):x\in A \}$ is Lebesgue measurable subset of $\mathbb{R}^2$. I tried to prove by contradiction but I did not get the result.

Thanks.

if $m(A)=0$ then $m(A \times R)=0$ also

if $m(A)$ is finite then $m(A \times R) := \infty $

if $m(A) : = \infty$ so $m(A \times R)$ because $m(R) :=\infty$.


Find $\lim_{x\to0}(1+x^{2})^{\cot^2{x}}$ without lhopital I need to find $\lim_{x\to 0}(1+x^{2})^{\cot^2{x}}$.

I started like this:

$\lim_{x\to0}(1+x^{2})^{\cot^2{x}} = \lim_{x\to0}e^{\ln(1+x^{2})^{\cot^2{x}}} = \lim_{x\to0}e^{\cot^2{x} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})}$

Now I'm not sure about this one but can I say:

$\lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})}$?

if so than

$ \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{1 \cdot \ln(1)} = e^0 = 1$

Is this valid? Is there a better way

No, you can't say that.  $\cos^2 x$ and $x^2$ are not very much alike close to $0$.  The limit is actually $e$.  But your tag says "limits without L'hospital", and I don't see a way to get $e$ without it.

Take the log we get
$$\cot^2 x \ln(1+x^2)=\cos^2 x \frac{\ln(1+x^2)}{\sin^2 x}$$
write as

$$\cos^2 x \frac{\ln(1+x^2)}{x^2}\frac{x^2}{\sin^2 x}\to 1$$
by standard limits. Easy peasy.


$V$ is a linear space. Need to compute $T^n$ Given: $V$ is a linear space $(\dim V = n)$ and there is a linear transformation $T: V \rightarrow V$ that $T^n = 0$ and $T^{n-1} \ne 0$ , also there's $u \in V$ that $T^{n-1}(u) \ne 0$
Prove that $B = (u,T(u),T^2(u),\dots,T^{n-1}(u))$ is a base for V.

Well, I tried saying that there are scalars $a_0,a_1,\dots,a_{n-1}$ and:
$a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$
, So if I prove that $a_0=a_1=a_{n-1}$ = 0 then I can say because there are n elements and also it's linearly independent. Thus, B is a base.
Well, I tried using the function $T^{n-1} ,T^{n-2},...,T^0$ like that:
[Explanation]:
First, I use the function $T^{n-1}$ on $a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$
which means it will look like this:  $a_0T^{n-1}u + a_1T^n(u) = 0$
(Because, all the other elements will have a dimension higher than $DimV = n$ so I erased them)
But, $T^n = 0$ Thus,  $a_0T^{n-1}u = 0$ , From here I say $T^{n-1} \ne 0$ So $a_0 = 0$.
Now I use $a_0 = 0$ and I use the function $T^{n-2}$ on $ a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ And so on.. Until I get all the scalars equal to zero. Then, I can say it is a base.
What do you guys think? Is my explanation correct?

[The answer]:
First, I use the function $T^{n-1}$ on $a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$

which means it will look like this:  $a_0T^{n-1}u + a_1T^n(u) = 0$

(Because, all the other elements will have a dimension higher than $DimV = n$ so I erased them)

But, $T^n = 0$ Thus,  $a_0T^{n-1}u = 0$ , From here I say $T^{n-1} \ne 0$ So $a_0 = 0$.

Now I use $a_0 = 0$ and I use the function $T^{n-2}$ on $ a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ And so on.. Until I get all the scalars equal to zero. Then, I can say it is a base.


Hard number theory problems that can be easily solved using algebraic geometry techniques I'm very interested in number theory, but it happens that now I'm taking an Algebraic Geometry course.

I was wondering what kind of number-theoretic problems could be easily solved using some skills and machinery from the algebro geometer theory.

I am not completely sure how to read the "hard" and the "easily" in your question, but as a starter you may consider the following (well-known) proposition.

There are infinitely many non-equivalent pythagorean triples.

This can be proven really easily (a brilliant high school student will understand the argument) observing that the set of inequivalent pythagorean triples is basically in one-to-one correspondence with the set of points with rational coordinates in the circle $x^2+y^2=1$ and the latter set is in one-to-one correspondence with the set of lines with rational slopes through the point $(1,0)$.

This is a basic application of the (geometric!) fact that a non-singular conic is isomorphic to the projective line, with a bit of arithmetic thrown in.

The most crazy example: Fermat's crazy Theorem.


Circle-separable colorings of finite set of points in the plane This is problem 12093 of the American Math Monthly, published a few months back:

Let $S$ be a finite set of points in the plane no three of which are collinear and no four of which are concyclic. A coloring of the points of S with colors red and blue is circle- separable if there is a circle whose interior contains all the red points of $S$ and whose exterior contains all the blue points of $S$. Determine the number of circle separable colorings of $S$.

It seems the number of colorings is, remarkably, independent of the configuration of the points. Let $n=|S|$. In addition to the trivial colorings where red circles contain only one point or none at all (in total $(n+1)$ of them), it seems that every pair of points uniquely determine a circle-separable coloring, and every three out of these $n$ also determine uniquely a circle-separable coloring.

Below, whenever we are considering a circle determined by two points or three points, we mean that the radius of the circle is just large enough to cover the two or three points respectively.

For $n=1$, there are clearly two trivial colorings; for $n=2$, there are clearly four colorings. For $n\geq 3$, we show there are ${{n}\choose{0}}+{{n}\choose{1}}+ {{n}\choose{2}} + {{n}\choose {3}}$ colorings, where $n=|S|$ is the cardinality of $S$.

Clearly, for any point P belonging to the set $S$, we can get a circle with a small enough radius, centered at $P$, so that the interior of the circle contains only $P$. In this way we get ${{n}\choose{1}}$ circle-separable colorings. Also, there are ${{n}\choose{0}}$ colorings where no point is labelled red .

Now choose any two points $A,B$ out of the $n$ points, and consider the line joining these two points. Extend this line $ \overleftrightarrow{AB}$ in both directions. Since no four points are concyclic and no three points are collinear, we must have some $k_{AB}$ of the remaining $(n-2)$ points on one side of the line segment, the remaining $(n-2-k_{AB})$ ones on the other side, and exactly none on the line segment $\overleftrightarrow{AB}$. Since none of the points are concyclic, all the points on one side of $\overleftrightarrow{AB}$ must subtend distinct angles on the segment $\overline{AB}$. The points on the other side of the line should also subtend distinct angles on $\overline{AB}$.

It is not very difficult to see that each of the $(n-2)$ points on either side of $\overleftrightarrow{AB}$ determines exactly one circle (where a circle is determined by $A,B$ and this third point all on the circumference of this circle), while there is one more 'extraneous' circle that is always obtained which either contains only $A,B$ on it's boundary, or is one containing $A,B$ on the boundary and all the points on one side of $\overleftrightarrow{AB}$ in it's interior.

As  we choose any two out of our $n$ points, clearly each of the circles determined by the three points as above are counted thrice. Thus, the total number of actual distinct colorings obtained this way is just the sum of the number of all triangles ${{n}\choose{3}}$.

For two circles each determined by only two points, they can encircle exactly the same number of points (and hence essentially they're the same circle for our purposes) if both these pairs of points are each pairwise consecutive on the convex hull of our entire point set. But it is not hard to see from the previous paragraph that if we take two consecutive points on the convex hull of our point set, the 'extraneous' circle defined by these two points contains only these two points, which is a contradiction. Thus, for each of the ${n}\choose{2}$ pairs of points, we get a unique 'extraneous' circle.

Hence the total number of colorings is ${{n}\choose{0}}+{{n}\choose{1}}+ {{n}\choose{2}} + {{n}\choose {3}}$.

Let the number of circle seperable colourings for $n$ points be $f_n$. By rearranging, we see that to show $f_n$ is as described, we just need to show $f_{n}-f_{n-1}={{n}\choose 2}+1$

For this, fix a point $P$, and note that we have a bijection between $n-1$ circle seperable colourings without $P$, and $n$ circle seperable colourings where $P$ has no choice in its colour. Thus, we may identify $f_{n+1}-f_n$ with the number of CSCs where $P$ has a choice in its colour. The space of circles/lines which seperate two sets of points in the plane is connected, so we have that if $P$ has a choice in its colour given a CSC on the rest, it can be taken to lie on the seperating circle (intermediate value theorem), and perturbing the circle yields the two choices of colour.

Given a partition of the points without $P$ admitting a seperating circle passing through $P$, then sometimes we can colour the partition red/blue in both ways, if our circle's interior can be taken to contain either part of our partition. This is determined by the side on which $\infty$ is on, viewing this as taking place on $\mathbb{CP}^1\cong S^2$. So apply a mobius transformation $\mu$ taking $P$ to $\infty$, now we have $n-1$ points in the plane, and a new point $\mu(\infty)$, and CSCs admitting a circle passing through $P$ are just lines seperating this set. This set has no three points collinear by our hypothesis, and thus it suffices to show that the number of seperating lines in ${n\choose 2}+1$. But this is a simple induction, adding an extra point $Q$ going from $n-1$ to $n$ adds $n$ new seperating lines, corresponding to the $n$ potential seperating lines that pass through $Q$.

There is also probably a cleverer non inductive proof for this last point, but I think the main trick is to mobius transform a point to $\infty$.


A vector that is orthogonal to the null space must be in the row space Simple question.

We know from the fundamental theorem of linear algebra that the nullspace of a matrix is the orthogonal complement of its row space.

I can write this as:

Let $M$ be a matrix. The following two conditions are equivalent:
(i)  $u$ is orthogonal to the null space of $M$.
(ii) $u$ is in the row space of $M$.

That (ii) implies (i) is trivial, pretty much by definition.

Question: But why is it also true (or even obvious) that (i) implies (ii)?

First, I'll prove/outline/mention a few preliminary results.

Lemma 1: If $V$ is a finite-dimensional real-vector space and $W$ is a subspace of $V,$ then for all $v\in V,$ there exist unique $w\in W,w'\in W^\perp$ such that $v=w+w'$.

Proof: It is readily seen that existence implies uniqueness, since if $w_1,w_2\in W$ and $w_1',w_2'\in W^\perp$ such that $w_1+w_1'=w_2+w_2',$ then $w_1-w_2=w_2'-w_1',$ but $w_1-w_2\in W$ and $w_2'-w_1'\in W^\perp,$ so since $W\cap W^\perp$ is the zero subspace (the zero vector is the only self-orthogonal vector), then $w_1-w_2=w_2'-w_1'=0,$ so $w_1=w_2$ and $w_1'=w_2'$. To prove existence, we can use the Gram-Schmidt process, starting with a basis for $W,$ to make an orthonormal basis for $W,$ which we then extend to an orthonormal basis for $V$ (possible in finite dimensions), and the added vectors will be an orthonormal basis for $W^\perp.$ $\Box$

Lemma 2: If $V$ is a real-vector space and $W$ is a subspace of $V,$ then $(W^\perp)^\perp\supseteq W$. [Readily seen by definition.] $\Box$

Lemma 3: If $V$ is a finite-dimensional real-vector space and $W$ is a subspace of $V,$ then $(W^\perp)^\perp=W.$

Proof: Take any $v\in(W^\perp)^\perp,$ and write $v=w+w'$ as in Lemma 1. Since $v\in(W^\perp)^\perp$ and $w'\in W^\perp,$ then $$0=v\cdot w'=(w+w')\cdot w'=w\cdot w'+w'\cdot w',$$ and so $0=w'\cdot w'$ since $w\in W$ and $w'\in W^\perp.$ Only the zero vector is self-orthogonal, so $w'=0,$ so $v=w\in W$, giving us the reverse inclusion of Lemma 2. $\Box$

Lemma 4: Given a finite dimensional vector space $V$ and subspaces $W$ and $X$ of $V$, we have that $W^\perp=X^\perp$ if and only if $W=X$.

Proof: One implication is trivial. For the other, suppose $W^\perp=X^\perp.$ Take $x\in X$. By Lemma 1, there exist unique $w,w'$ such that $x=w+w',$ $w\in W$, and $w'\in W^\perp=X^\perp.$ Then $$0=x\cdot w'=(w+w')\cdot w'=w\cdot w'+w'\cdot w'=w'\cdot w',$$ so $w'=0,$ whence $x=w\in W$, and so $X\subseteq W$. By symmetrical arguments, we likewise have $W\subseteq X$, so $W=X$. $\Box$

Now, because of Lemmas 3 and 4, to show that (i) implies (ii) is equivalent to showing that every vector in the null space of $M$ is orthogonal to every element of the row space of $M$. Indeed, Say that $M$ is an $m\times n$ matrix, and writing $M=[r_1^T\:\cdots\:r_m^T]^T$ where the $r_j$ are the rows of $M,$ we note that for any $n$-dimensional vector $x$ we have $$Mx=\left[\begin{array}{c}r_1\\\vdots\\r_m\end{array}\right]x=\left[\begin{array}{c}r_1x\\\vdots\\r_mx\end{array}\right]=\left[\begin{array}{c}r_1^T\cdot x\\\vdots\\r_m^T\cdot x\end{array}\right].$$ In particular, if $x$ is in the null space of $M$, then $r_j^T\cdot x=0$ for $1\le j\le m,$ meaning that $x$ is orthogonal to each row of $M$, so since the row space of $M$ is spanned by these rows then $x$ is in the orthogonal complement to the row space of $M,$ as desired.

First show that: if u $\in$ Row(M), then $u\in Ker(M)^\perp$ which you already know how to do. For the other direction, consider any $v \in Ker(M)^\perp$. Then $vx=0$ for any $x \in Ker(M)$. Now write $v=v_1+v_2$, where $v_1 \in Row(M)$, $v_2 \in Row(M)^\perp$ by using the Lemma(i) pointed out by Cameron. Now $(v_1+v_2)x=0$ which implies $v_1x+v_2x=0$ which inturn implies $v_2x=0$  (notice $v_1x= 0$, row and ker ). So $v_2 \in Ker(M)^\perp$. Since $v_2\in Row(M)^\perp$, it is also in Ker(M). Thus $v_2 \in Ker(M)^\perp \cap Ker(M)$. Thus $v_2$ is the zero vector. Thus $v \in Row(M)$.
(OK sorry about the formatting earlier, now fixed).


Solving Recurrence Relation 5 How to solve the recurrence relation given by the equation below

$$T(n)=T(n-2)+T(n-4)+T(n-6)+...+T(0)$$

It seems to me that $T(n)$ will be exponential but i don't know how to proceed on this problem.


Quick question about formal definition of a representation So I know that a linear representation is defined as $\rho : G \to GL(V)$ over some finite group $G$. So if we define the action of some group ring, say F[G] over some representation V, is this representation V, the same thing as the homomorphism $\rho$?

Taking G to be any group for example, when we say that V is the trivial representation of G over F, do we mean that V is the subspace left invariant by the action of the trivial representation, in this case, just the multiplicative group $F^{*}$? So what exactly does it mean to define the action of G on V in this case then?

So we are given a group $G$ and a field $F.$ Then there is a "natural" correspondence between linear representations of $G$ over $F$ and linear representations of $F[G]$ over $F.$ This correspondence works as follows.

Suppose $V$ is an $F$-vectorspace. Suppose, we are given a linear representation $\rho:G\rightarrow GL(V).$ Then we can extend $\rho$ to a homomorphism $\sigma:F[G]\rightarrow End(V)$ by defining $\sigma(g) = \rho(g)$ for every basis element $g\in G\subseteq F[G]$ and linear extension. Here, $End(V)$ is the $F$-algebra of $F$-linear maps from $V$ to $V$. It's "very" easy to check that $\sigma$ is indeed a homomorphism of $F$-algebras.

Now, suppose we are given a linear representation $\sigma: F[G]\rightarrow End(V).$ Then we get a group homomorphism $\rho:G\rightarrow GL(V)$ by defining $\rho(g) = \sigma(g)$ for every $g\in G\subseteq F[G].$ Since $g\in G$ is invertible in $F[G],$ $\sigma(g)\in End(V)$ is invertible, and thus we have indeed $\rho(g) \in GL(V).$

These correspondences $\rho \leadsto \sigma$ and $\sigma\leadsto\rho$ are inverse to each other in the sense that in the concatenation $\rho \leadsto \sigma \leadsto \rho',$ we have $\rho = \rho'$ and in the concatenation $\sigma \leadsto \rho \leadsto \sigma',$ we have $\sigma = \sigma'.$

About the trivial representation: to me, "the trivial representation" of a group $G$ over a field $F$ is $V = F$ (one-dimensional vector space) and $g\mapsto 1 \in F.$

If we consider an arbitrary linear representation $\rho:G\rightarrow GL(V)$ of $G$ over $F,$ the we have the fixed point space
$$
V^G := \{v \in V\ |\ \forall g\in G:\rho(g)\cdot v = v \}.
$$
One often says that $G$ "acts trivially" on $V^G.$ This makes perfectly sense, since every $g \in G$ acts as the identity on $V^G.$ Of course, $V^G$ is a direct sum of one-dimensional trivial representations as described above.

Ok, I'm gonna try to explain this. Please correct me if wrong...

The mapping from the group to a representation of the group is called a homomorphism because it "preserves the structure". A group action is defined as a function of a group element on elements from some set. The representation has nothing to do with any such action but is only a remapping of group elements to matrices and the group operation to matrix multiplication such that they behave in the same way.

You can view the representation as a kind of "implementation" of a group if you will. Then it may be that any group action you want to define for the group and some set may also be expressible with matrices, but that is something different from the group representation.


Is there a name for $(x, f(x))$? Given a function $f$, and an $x$ from its domain, is there a name for the pair $(x, f(x))$?
Is there a defined terminology for one (any) such pair? I think the set of all $(x, f(x))$ is called the graph of the function, but I am asking is there's a name for one (or any) point of the graph.

I would call this an element of the graph of $f$, or an element of $\Delta(f)$ where $\Delta$ sends symbols to their denotations, or an element of $f$ as a set.
This amounts to the same thing as calling it an element of $f$, as suggested in the comments, except that it makes explicit the act of sending a function to its corresponding set. This operation is a no-op if you think of a function as just a set of pairs. It is not a no-op if you think of it as an ordered triple with its domain, codomain, and graph as elements, or if you think of a function as some kind of non-set entity.

It's called an ordered pair and it is usually written as $(x,y)$ instead of $(x,f(x))$ because $y=f(x)$. It's called an ordered pair because it's a pair of numbers, and the order matter. For example, $(2,3)$ is different from $(3,2)$.
You may read more about it here
https://en.wikipedia.org/wiki/Ordered_pair
Or watch some videos on this topic here
https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-negative-number-topic/cc-6th-coordinate-plane/v/plot-ordered-pairs
https://www.youtube.com/watch?v=YlT726odQcM


Proving number of roots of a function using Rolle's theorem I am given a function $f(x) = x^3 + 3x - 1$, and I am asked to prove that $f(x)$ has exactly one real root using the Intermediate Value Theorem and Rolle's theorem. So far, I managed to prove the existence of at least one real root using IVT.

Note that $f(x)$ is continuous and differentiable for all $x \in \Bbb R$. By inspection, since $f(-1) = -5$ and $f(1) = 3$, by the IVT, there exists a $c \in [-1, 1] $ such that $f(c) = 0$. Therefore, $f(x)$ must have at least one root. I am not sure if this is the best way to prove that it has at least one root - if there is a better way, please let me know. I don't know how I can use Rolle's theorem to prove that there is only one root. I had an idea of integrating the polynomial and then proving that it has just one maximum using Rolle's theorem but that did not work out so well.

Between any two distinct real roots, there is, by Rolle's Theorem, a root of the derivative. But the derivative has no roots.

There is a perhaps somewhat better way to use IVT to show the existence of a root. Don't bother to find explicit $a$ and $b$ such that our function is negative at $a$ and positive at $b$. It is obvious that when $x$ is large enough negative, our function is negative, and that when $x$ is large enough positive, our function is positive. Thus any cubic has a real root. By the same argument, any polynomial of odd degree has a real root.

Rolle's Theorem always fails because there are no roots. Therefore, the maximum number of roots must be less than 2, making the maximum number of roots possible 1.


It is the set $V(y-\sin(x))\subset k^2$ a variety? It is the set $V(y-\sin(x))\subset k^2$ a variety?

I know that $V(y-\sin(x))=\{...,-4\pi,-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi,4\pi,...\}$. Could it be a set of points a variety? My intuition tells me what not. How can I do a formal test, what property should I use? Thank you very much.

Suppose $X = V(y-\sin(x))\subset \Bbb A^2$ were a variety. Then $X$ also admits a description as $V(f_1,\cdots,f_n)$ for some finite list of nonzero polynomials $f_i(x,y)$, each of which vanishes on $V(y-\sin(x))$.

By examining $f_i(x,0)$, each single-variable polynomial $f_i(x,0)$ should vanish on $V(y-\sin(x))\cap V(y)$. On the other hand, as you (intended to) identify in your post, $V(y-\sin(x))\cap V(y)$ is infinite, which is a problem (Why? Try to identify for yourself before mousing over the following spoiler).

   Any nonzero polynomial in one variable may have at most finitely many solutions. As $f_i(x,0)$ has infinitely many solutions, it must be the zero polynomial, which implies that $y|f_i$ for each $f_i$, or that $V(y-\sin(x))$ contains the $x$-axis, which it does not by your identification of $V(y-\sin(x))\cap V(y)$.

$V=V(y-\sin(x))=\{(x,y); y=\sin(x) \}=\{(x,\sin(x)); x\in \Bbb R \}$  is not variety because no polynomial in $\Bbb R[x,y]$ has zeroes the set $V$.


How irrational quantities physically exist in nature? We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural.
But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ??
Isn't it in contrary with the laws of Physics??

I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $x\pm y$. For example, a length might be measured to be $1.3\pm 0.5~\rm mm$.

Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~\rm mm$ away from the value $1.3~\rm mm$.

Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.

But wait, you say, there are circles, and for those we can prove the circumference over diameter is $\pi$. So $\pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?

And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.

Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $\pi$.  Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.


Closed knight's tour (minimum board size) I'm writing a small java program which calculates all possible knight's tours with the knight starting on a random field on a 5x5 board.

It works well, however, the program doesn't calculate any closed knight's tours which makes me wonder. Is there an error in the code, or are there simply no closed knight's tours on a 5x5 board?

If so, what is the minimum required board size for the existence of at least one closed knight's tour?

No closed knight's tour is possible on a board with an odd number of squares, because each move changes the colour of the knight's square. So after an odd number of moves, you can't be back at the starting square, because it's the wrong colour.

The definition of a 'closed' knights tour on a $m \times n$board, is a sequence of steps from a starting square $a_1$ to another square $a_{mn}$ , such that every square is visited exactly once, and the last sqaure is only one knight step away from $a_1$. Having said that, it is obvious, that for $mn $(mod2) $= 1$, there exists no closed tour.

Short proof:

Suppose $a_1$ is black. Clearly then for any $i; i\le mn , i$(mod 2)$=1, a_i$ must be black. Since $mn$ is odd, $a_{mn}$ must be black, which implies that it cannot be a square one knight step away from $a_1$. Thus there exists no closed tour for odd $mn$.


Name for a semiring minus multiplicative identity requirement Is there a name for a theory that has all axioms of a semiring but an axiom that mandates multiplicative identity?

Semiring theory is plagued by a mess of vocabulary the remains unresolved. Still, there are two books I always check when dealing with them.

In Golan's Semirings and Their Applications, a semiring without identity is called a hemiring.

A newer, also fairly comprehensive book on semirings is Graphs, Dioids and Semirings: New Models and Algorithms by Gondran and Minoux. They list semirings without identity as pre-semirings.

Hemiring in particular gets thousands of hits, although due to the mess of vocabulary in the theory, I can't guarantee what percentage of those is what you seek.

While I don't think those terms are going to be universally used, I think the references in both these books would help you track down specific information about such structures.

This would be the fusion of semirings and pseudo-rings (or rngs), so they should be called pseudo-semirings (or semirngs). But with google I couldn't find any usage for this (only with other meanings).


The space of functions that vanish at infinity is a Banach space Prove that $C_0(X) = \{ f \in C(X) \mid \forall \varepsilon >0 \ \exists K \subset X \text{ compact such that } |f(x)| < \varepsilon \text{ for } x \notin K \}$ is a Banach space with the norm $\|f\|_{\infty}$.

I'm trying to prove that $C_0(X)$ is closed subset of $C(X)$, therefore I suppose $f \in \overline{C_0(X)}$ so there exist a sequence $f_n \in C_0(X)$ such that $f_n \to f$. I am stuck here: why $f \in C_0(X)$?

This is a typical $\varepsilon/2$-argument. Fix $\varepsilon>0$. Then there exists  $n$ such that $\|f-f_n\|<\varepsilon/2$. Since $f_n\in C_0 (X) $, there exists $K\subset X $, compact, with $|f_n (x)|<\varepsilon/2$ for all $x\in X\setminus K $. So, outside $K $,
$$
|f (x)|\leq|f(x)-f_n (x)|+|f_n (x)|<\frac\varepsilon2+\frac\varepsilon 2=\varepsilon.
$$

As $C(X)$ is a Banach space, you only need to prove as you noticed that $C_0(X)$ is closed in $C(X)$.

Suppose that $f_n \to f$. $f$ is continuous because $C(X)$ is a Banach space. You only need to prove that $f$ vanishes at infinity.

Take $\epsilon > 0$. By hypothesis you can find $N \in \mathbb{N}$ with $\Vert f-f_N \Vert_\infty < \epsilon/2$. As $f_N \in C_0(X)$, there exists a compact $K \subset X$ such that $\vert f_N(x) \vert < \epsilon/2$ for all $x \in X \setminus K$.

Hence for $x \in X \setminus K$:
$$\vert f(x) \vert \le \vert f_N(x) \vert + \Vert f-f_N \Vert_\infty \le \epsilon$$

Proving that $f$ vanishes at infinity. And you're done.


Proving Riemann Sums via Analysis Exercise $\bf 5.1.7$: Suppose $f:[a,b]\to\Bbb R$ is Riemann integrable. Let $\epsilon\gt0$ be given. Then show that there exists a partition $P=\{x_0,x_1,\ldots,x_n\}$ such that if we pick any set of numbers $\{c_1,c_2,\ldots,c_n\}$ with $c_k\in[x_{k-1},x_k]$ for all $k$, then $$\left|\int_a^b f-\sum_{k=1}^n f(x_k)\Delta x_k\right|\lt\epsilon.$$

I'm trying to do some example problems in analysis, as I'm struggling to make some mental connections with some of our topics. Anybody able to give me some hints or even an example proof of this problem? This seems to be a problem attempting to prove the validity of riemann sums. Listed below is the definition of riemann integrable that I am working with.

A function is Riemann Integrable if it's lower and upper Darboux integral are equal to each other.

The definitions of Riemann integral a la Darboux and as a limit are equivalent. You ask for a proof in one direction.

The proof is typical and uses the simple fact:

if $\quad \alpha\le c \le \beta \quad$ and $\quad \alpha\le d \le \beta \quad$ then $\quad |c-d| \le \beta-\alpha \;\;$.

So, if $P$ is a partition of $[a,b]$, however tagged, from the theory, following the Darboux definition, one has $$\left|\int_a^b f(x)\,dx-R(f,P) \right|\le U(f,P)-L(f,P)$$

Let $\varepsilon>0$. Then the (Darboux) integrability criterion assures the existence of a partition $P$ of $[a,b]$ such that $$U(f,P)-L(f,P)<\varepsilon$$ Combine everything to get the result.

Hint: If $\mathcal{P}$ is a partition such that $U(f,\mathcal{P}) - L(f,\mathcal{P}) < \epsilon$ then you know $$U(f,\mathcal{P}) - \int f + \int f - L(f,\mathcal{P}) < \epsilon$$ as well as $$U(f,\mathcal{P}) - \int f \geq 0 \quad \text{and}\quad \int f - L(f,\mathcal{P}) \geq 0$$
Try to conclude why this partition $\mathcal{P}$ will work for your cause. Here $U(f,\mathcal{P})$ and $L(f,\mathcal{P})$ are the upper and lower sums respectively.


Example where $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective Can anyone come up with an explicit example of two functions $f$ and $g$ such that: $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective?

I tried the following: $$f:\mathbb{R}\rightarrow \mathbb{R^{+}}
$$ $$f(x)=x^{2}$$

and  $$g:\mathbb{R^{+}}\rightarrow \mathbb{R}$$
$$g(x)=\sqrt{x}$$

$f$ is not injective, and $g$ is not surjective, but $f\circ g$ is bijective

Any other examples?

The simplest example:

$X=\{0\},Y=\{0,1\}$, $f(0)=0$, $g(0)=g(1)=0$.

(Here $g\circ f$ is the bijection, since I inadvertently reversed the names of the functions.)

Everything else is an elaboration of one of this idea.

X={0},Y={0,1}, f(0)=0, g(0)=g(1)=0.
more info could be found at here.