Need help with $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$ I need you help with this integral:
$$\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx.$$
Mathematica says it does not converge, which is apparently false.

Huge pain in the rump, but in the end, fairly straightforward.  First observe that

$$\begin{align}\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} &= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_1^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)}\\ &= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_0^{1} dx \, x^3 \frac{\log{(1+x)}-\log{x}}{(1+x^2)(1+x^3)}\\ &= \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} - \int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)}\end{align} $$

The first integral may be evaluated by subbing $x=\tan{t}$:

$$\begin{align} \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} &= \int_0^{\pi/4} dt \, \log{(1+\tan{t})}\\ &= \int_0^{\pi/4} dt \, \log{(\sin{t}+\cos{t})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \int_0^{\pi/4} dt \, \log{(\sqrt{2} \cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \frac{\pi}{8} \log{2} + \int_0^{\pi/4} dt \, \log{(\cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \frac{\pi}{8} \log{2}\end{align}$$

The second integral is messy but still doable.  First, we may use partial fractions.  Then we will get a series of zeta-like sums.  Some of them we will immediately recognize.  To the rest, we may apply the residue theorem.

$$\begin{align}\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} &= \frac12 \int_0^1 dx \left [\frac{1-x}{1+x^2} - \frac{1-x-x^2}{1+x^3} \right ] \log{x}\\ &= \frac12 \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{2 k}-x^{2 k+1}-x^{3 k}+x^{3 k+1}+x^{3 k+2} \right ) \log{x}\\ &= -\frac12 \sum_{k=0}^{\infty} (-1)^k \left [ \frac1{(2 k+1)^2} - \frac1{(2 k+2)^2} - \frac1{(3 k+1)^2}\\ + \frac1{(3 k+2)^2}+\frac1{(3 k+3)^2}\right ] \\ &= \frac{5}{72} \frac{\pi^2}{12} - \frac12 G + \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \end{align} $$

where $G$ is Catalan's constant.  As for the sum:

$$\begin{align} \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] &= \frac14  \sum_{k=-\infty}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \\ &= -\frac{\pi}{4} \frac1{3^2} \left [\operatorname*{Res}_{z=-1/3} \frac{\csc{\pi z}}{(z+1/3)^2} \\- \operatorname*{Res}_{z=-2/3} \frac{\csc{\pi z}}{(z+2/3)^2} \right ]\\ &= \frac{\pi^2}{36} \left [\frac{\cos{\pi/3}}{\sin^2{\pi/3}} - \frac{\cos{2 \pi/3}}{\sin^2{2 \pi/3}} \right ]\\ &= \frac{\pi^2}{27}\end{align} $$

Therefore

$$\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} = \frac{5 \pi^2}{864} - \frac12 G + \frac{\pi^2}{27} = \frac{37 \pi^2}{864} - \frac{G}{2}$$

and finally...

$$\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} = \frac{G}{2} + \frac{\pi}{8} \log{2} - \frac{37 \pi^2}{864} \approx 0.307524\cdots$$

Hint

This would probably work. Let $u = \ln(1+x)$ then $du = \frac{dx}{1+x}$ and factor the bottom $1+x^3 = (1+x)(\ldots)$.


Infinite Abelian subgroup of infinite non Abelian group example My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $\langle g\rangle$, for some $g$ in $\mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
\bigg[
\begin{matrix}
1&0\\0&2
\end{matrix}
\bigg]
$. Then $g^n$ will be in the form $ g^n=
\bigg[
\begin{matrix}
1&0\\0&2^n
\end{matrix}
\bigg]
$. This is obviously infinite since $g^n=e \Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!

The simplest example is $G=\mathbb Z \times S_3$ and $H=\mathbb Z$.

Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $\mathrm{diag}(x,y)^{-1} = \mathrm{diag}(x^{-1}, y^{-1})$.

This subgroup is isomorphic to $F^\times \oplus F^\times$, which is abelian and infinite if $F$ is.


Show that a prime ring is primitive $R$ is a prime ring in which the zero ideal is the intersection of finitely many maximal left-ideals.

Show that $R$ is primitive.

It was suggested to me that the best method to use is proof by contradiction.

My proof has run like this so far:

Suppose $R$ is not primitive; then

$\exists A_{i} \triangleleft R$ non-zero with $A_{i} \subseteq M_{i}$,

where $M_{i}\cap  M_{2} \cap ... M_{n} =  <0_{R}>  $.

Then, $A_{1} A_{2} ...A_{n} \subseteq A_{1}\cap A_{2}\cap ...A_{n}$, and
$A_{1}\cap A_{2}\cap ...A_{n} \subseteq M_{i}\cap  M_{2} \cap ... M_{n} =<0_{R}>$.

But if R is prime, then $A_{1}A_{2}...A_{n} \neq <0_{R}>$.

Is this contradiction valid?

Yes, that is basically exactly what I would have said: good job.


Rational with finite decimals values for sine, cosine, and tangent What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?

I am aware of the below two cases.

$\sin(x) = 0, \cos(x) = 1, \tan(x) = 0$ is a trivial case for angle measure zero
$\sin(x) = 0.6, \cos(x) = 0.8, \tan(x) = 0.75$ is another case, for an angle of $37$ degrees (approximately)

Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?

Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....

Any Pythagorean triple will give you rational $\sin, \cos, \tan$ values, and any triple of rational trig values will give you a Pythagorean triple.

The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
$$
(u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
$$
and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $\sin$ and $\cos$ pairs are all possible values of
$$
\frac{u^2-v^2}{u^2+v^2}\text{ and }\frac{2uv}{u^2+v^2}
$$

Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $\pi$. It may also work better in this case, as the desired result are finite decimals.


Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number.

Actually, I know a way to solve this, but even if it is very large and cumbersome, and would like to know if you have an easier way, or if I am applying wrong.
Setting by $\left[\frac{b}a \right]$ the quotient of $b$ with $a$, we have too $E_p(m)$ the largest exponent power $p$ dividing $m$, and found the demonstration of a theorem that says (that in my text says it was discovered that Lagendre) $$E_p(n!)=\left[\frac{n}p \right]+\left[\frac{n}{p^2} \right]+\left[\frac{n}{p^3} \right]+\;...$$always remembering that there will be a number $s$ such that $p^s\geq n!$ which tells us that $$\left[\frac{n!}{p^s} \right]=0$$ thus making the sum of a finite $E_p(n!)$. So that I can address the first question I asked, really have to get all the cousins $(p_1,p_2,...,p_k)$ and make all $$E_{p_1},\;E_{p_2},\;E_{p_3},\;...,\;E_{p_k}$$ with $p$ and cousin $1<p<100$. And to find the zeros have to see how we both exponents in numbers 5 and 2.

  Example $$10!=2^83^45^27\\p<10\\E_2(10!)=5+2+1=8\\E_3(10!)=3+1=4\\E_5(10!)=2\\E_7(10!)=1\\$$

It's possible to demonstrate that if N is a multiple of 100, N! ends with (N/4)-1 zeroes.


To show $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ I know this question has been asked before, but this is (hopefully) not a duplicate. Dummit and Foote's Algebra asks us to show $$|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$$
using the following definitions, and nothing else:

$(\mathbb{Z}/n\mathbb{Z})^{\times} := \{\bar a \in \mathbb{Z}/n\mathbb{Z} : \exists \bar c \in \mathbb{Z}/n\mathbb{Z}, \bar a \bar c = \bar 1\}$

Here, $\bar a$ denotes the equivalence class of $a$ in $\mathbb{Z}/n\mathbb{Z}$, namely one of $\bar 0,\bar 1,...,\overline{n-1}$. $\bar a\bar c$ refers to the usual multiplication, i.e. we take any elements from the equivalence classes of $\bar a$ and $\bar c$, multiply them, and then reduce the result $\bmod n$. We have shown this works in an earlier proposition, i.e. modular multiplication is well-defined. So, $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is just the collection of residue classes of $\mathbb{Z}/n\mathbb{Z}$ which have a multiplicative inverse in $\mathbb{Z}/n\mathbb{Z}$.
$\varphi(n)$ is Euler's totient function, as is popularly known. It is not defined using $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ in this text. Rather it is defined as $\varphi(n) = |\{a\in\mathbb{N}: a\leq n, (a,n)=1 \}|$, i.e. number of positive integers less than and co-prime with $n$. $(a,n) = 1$ stands for $\operatorname{g.c.d}(a,n) = 1$.
We have not yet shown that

$(\mathbb{Z}/n\mathbb{Z})^{\times} = \{\bar a \in \mathbb{Z}/n\mathbb{Z} : (a,n) = 1\}$

so please do not use this in your proof. Knowing this, it is trivial, but in a later exercise, we are supposed to prove this using $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ - so let's avoid circular arguments.
Could someone please point me in the right direction? Thanks!

HINT. Note that a solution of
$$
{\displaystyle ax\equiv 1{\pmod {m}}.}
$$
exists if and only if $\gcd(a, m) = 1$.
Proof: Let $x$ be the solution. This is the same thing as saying $ax + my = 1$ for some $y \in \mathbb{Z}$. But if $a$ and $m$ have a common factor $d$, then clearly $d$ divides $1$. This indicates that $d = \pm 1$, so the $\gcd$ of $a$ and $m$ must be $1$. More information here.

Unequivocally the key here is that we have Bézout's identity, which gives equivalence between the statements $(a,b)=1$ and the existence of $x,y$ such that $ax+by=1$.
The nontrivial direction is proved by using the extended Euclidean algorithm.


Questions about product measure On Planetmath, product measure
is roughly defined as follows:

Let $(E_i, \mathbb{B}_i, u_i)$ be
measure spaces, where $i\in I$ an
index set, possibly infinite.

When each $u_i$ is totally finite,
there is a unique measure on the
product measurable space $(E,
    \mathbb{B})$ of $(E_i, \mathbb{B}_i,
    u_i)$. such that "taking measure"
and "taking product" can be
"exchanged" for any $B=\prod_{i \in
    I} B_i$ with $B_i \in \mathbb{B_i}$
and $B_i=E_i$ for all $i \in I$
except on a finite subset $J$ of
$I$.

I was wondering if there is also a
unique measure on the product
measurable space, such that "taking
measure" and "taking product" can be
"exchanged" for any $B=\prod_{i \in
    I} B_i$ with $B_i \in \mathbb{B_i}$,
without requiring "$B_i=E_i$ for all
$i \in I$ except on a finite subset
$J$ of $I$"? This is not used in the
definition of product measure, and is
it only because the product might be for
infinite number of terms?

  If $I$ is infinite, one sees that the
  total finiteness of $u_i$ can not be
  dropped. For example, if $I$ is the
  set of positive integers, assume
  $u_1(E_1) < \infty$ and
  $u_2(E_2)=\infty$ . Then $u(B)$ for
  $$B:=B_1 \times \prod_{i>1} E_i=B_1\times E_2 \times \prod_{i>2} E_i,$$
  where
  $B_1 \in \mathbb{B}_1$  would not be
  well-defined (on the one hand, it is
  $u_1(B_1)<\infty$ , but on the other
  it is $u_1(B_1)u_2(E_2)=\infty$ ).

I don't understand the example.
Specifically how does the last
sentence in parenthesis show that
the measure $u$ is not well-defined
on $B$?

Thanks and regards!

Let  $(\beta_i)_{i \in I} \in [0, +\infty]^{I}$ be an arbitrary
family of non-negative real numbers. We set $ln(0)=-\infty$.

Definition 1. A standard  product of the family of numbers
$(\beta_i)_{i \in I}$ denoted by ${\bf (S)}\prod_{i \in
I}\beta_i$ is defined as follows:

~${\bf (S)}\prod_{i \in I}\beta_i=0$ if ~$ \sum_{i \in
I^{-}}\ln(\beta_i)=-\infty$, where $I^{-}=\{i:ln(\beta_i)<0\}$ ,
and ${\bf (S)}\prod_{i \in I}\beta_i=e^{\sum_{i \in
I}\ln(\beta_i)}$ if $\sum_{i \in I^{-}}\ln(\beta_i) \neq -\infty$.

Now we will try to give  answers to  Stefan's questions when a set
of induces $I$ is countable.

Question 1(Stefan). Is product measure only defined for
probability measures?

Let $(E_i,\mathbb{B}_i,u_i)_{i \in I}$ be a family of totally
finite, continuous measures.

Theoretically there are possible the following three cases:

Case 1. ${\bf (S)}\prod_{i \in I}u_i(E_i)=0.$

In that case we define $\prod_{i \in I}u_i$ as  zero measure, i.e.
$$
(\forall X)(X \in \prod_{i \in I}\mathbb{B}(E_i) \rightarrow
(\prod_{i \in I}u_i)(X)=0).
$$

Case 2. $0< {\bf (S)}\prod_{i \in I}u_i(E_i)< +\infty.$

In that case we define $\prod_{i \in I}u_i$ as follows:

$$
(\forall X)(X \in \prod_{i \in I}\mathbb{B}(E_i) \rightarrow
(\prod_{i \in I}u_i)(X)= ({\bf (S)}\prod_{i \in I}u_i(E_i))\times
(\prod_{i \in I}\frac{u_i}{u_i(E_i)})(X)).
$$

Case 3. ${\bf (S)}\prod_{i \in I}u_i(E_i)= +\infty.$

In that case  we define $\prod_{i \in I}u_i$ as a standard product
of measures $(u_i)_{i \in I}$ construction of which is given
below:

Without loss of generality, we can assume that $u_i(E_i)\ge 1$ when $i \in I$.

Let $L$ be a set of rectangles $R:=\prod_{i \in I}R_i$ where $R_i
\in \mathbb{B}(E_i)(i \in I)$ and $0\le{\bf (S)}\prod_{i \in
I}u_i(R_i)<+\infty$

Note that a rectangle $R$ with $0<{\bf (S)}\prod_{i \in
I}u_i(R_i)<+\infty$ exists because $u_i$ is continuous and
$u_i(E_i)\ge 1$.

Let $\mu_R$ be a measure defined on $\prod_{i \in
I}\mathbb{B}(R_i)$ as follows

$$
(\forall X)(X \in \prod_{i \in I}\mathbb{B}(R_i) \rightarrow
\mu_R(X)= ({\bf (S)}\prod_{i \in I}u_i(R_i))\times (\prod_{i \in
I}\frac{u_i}{u_i(R_i)})(X)).
$$
For each $R \in L$ we have a measure space $(R,S_R(:=\prod_{i \in
I}\mathbb{B}(R_i)),\mu_R)$. That family is consistent in the
following sense: if $R=R_1 \cap R_2$ then
$$
(\forall X)(X \in S_R \rightarrow
\mu_R(X)=\mu_{R_1}(X)=\mu_{R_2}(X)).
$$

If a measurable subset $X$ of $\prod_{i \in I}E_i$ is covered by a
family $\{R_k : R_k \in L ~\&~k=1,2, \cdots\}$  then we set
$$
\Lambda(X)=\mu_{R_1}(R_1 \cap X)+\mu_{R_2}((R_2\setminus R_1)\cap
X)+\cdots+\mu_{R_n}([R_n\setminus \cup_{1 \le i \le n-1}R_i]\cap
X)+ \cdots.
$$
If a measurable subset $X$ of $\prod_{i \in I}E_i$ is not covered
by a countable family of elements of $L$, then we set
$\Lambda(X)=+\infty$.

Note that $\Lambda$ is measure on $\prod_{i \in I}\mathbb{B}(E_i)$
and $\Lambda(R)={\bf (S)}\prod_{i \in I}u_i(R_i)$ for each $R \in
L$.

This measure is called standard product of measures $(u_i)_{i \in
I}$ and is denoted by ${\bf (S)}\prod_{i \in I}u_i$.

As we see product can be defined for totally finite continuous
measures. Here we need no a requirement of totally finiteness(they
may be infinite (i.e. $u_i(E_i)=+\infty$) as well we do not
require their sigma-finiteness.

I think that it gives a partially solution of that problem when
$card(I)=\aleph_0$ and the measure ${\bf (S)}\prod_{i \in I}u_i$
is well defined on $\prod_{i \in I}E_i.$

P.S. I agree with Mister Stefan Walter remark that there may be a situation when product measures are not defined uniquelly.

Indeed, let $(n_k)_{k \in N}$ be a family of strictly increasing natural numbers such that $n_0=0$ and $n_{k+1}-n_k \ge 2$. We set $\mu_k=\prod_{i \in [n_k,n_{k+1}]}u_i$.
Let us consider ${\bf (S)}\prod_{k \in N}\mu_k$. Then that measure will be defined
on $\prod_{i \in I}\mathbb{B}(E_i)$ and  $({\bf (S)}\prod_{k \in N}\mu_k)(R)={\bf (S)}\prod_{i \in I}u_i(R_i)$ for all $R \in L^{+}$, where
$$
 L^{+}=\{ R:R \in L~\&~0<{\bf (S)}\prod_{i \in I}u_i(R_i)<+\infty\}
$$

Note that the measure  ${\bf (S)}\prod_{k \in N}\mu_k$ is called $(n_{k+1}-n_k)_{k \in N}$-standard product of measures $(u_i)_{i \in I}$.

It is natural that both measures  $({\bf (S)}\prod_{i \in I}u_i)$ and
$({\bf (S)}\prod_{k \in N}\mu_k)$ can be considered as products of measures $(u_i)_{i \in I}$ but they(in general) are different.

Indeed, let $u_i=l_1$ for $i \in I$, where $l_1$ denotes a linear Lebesgue measure on real axis. Let $n_{k+1}-n_k=2$ for $k \in N$. Consider a set $D$ defined by
$$
D=[0,2]\times [0,\frac{1}{2}]\times [0,3]\times [0,\frac{1}{3}]\times \cdots.
$$
Then
$({\bf (S)}\prod_{i \in I}u_i)(D)=0$ and $({\bf (S)}\prod_{k \in N}\mu_k)(D)=1.$

You can found answers on yours questions in the following articles:

[1] G.Pantsulaia , On ordinary and standard products of infinite family of $\sigma$ -finite measures and some of their applications. Acta Math. Sin. (Engl. Ser.) 27 (2011), no. 3, 477--496

[2] G.Pantsulaia , On Strict Standard and Strict Ordinary Products of Measures and Some of Their Applications, Georg. Inter. J. Sci. Tech., Nova Science Publishers, Volume 2,  Issue 3 (2010).


What's wrong with the method I used to prove this limit doesn't exist? The limit is

$\lim_{x\to0,y\to0}\frac{x^2y}{x^2+y^2}$

Here is the way to prove the limit is $0$:

For arbitrary $\epsilon$, there exists a $\delta $ ($ 0 < \sqrt{x^2+y^2} < \delta$) that can satisfy the equation $| {\frac {x^2y} {x^2+y^2}}|<\epsilon$ as $x^2 \le x^2+y^2$ hence $|\frac{x^2y}{x^2+y^y}| \le |y| = y^2\le \sqrt {x^2+y^2}$ if we choose $\delta = \epsilon$.

Nevertheless, I try to choose (x, y) approaching (0,0) along a particular curve such as $x^2=-y^2+y$ and $x^2=-y^2+y^3$, which give me the limits are $0$ and $-1$ respectively that can prove the limit doesn't exist.

The question I wanna ask is what's wrong the curve I chosen and what is the rule to or not to choose a particular curve.

Any help will be appreciated, thank you!

You can do this also:

Put $$x=r\cos{t}$$ $$y=r\sin{t}$$

Then:

$$\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r \to 0} \frac{r^3 \cos^2{t}\sin{t}}{r^2}=\lim_{r \to 0}r \cos^2{t}\sin{t} = 0$$,
$\forall t \in [0,2\pi)$

Thus the limit is zero.

In general it is a very useful technique to use polar coordinates for computing multivariable limits or to examine the existence of a limit.

For instanse if the result  as $r \to 0$ depends on values of $t$ then the limit does not exist.

As an exercise for you try to prove that $\lim _{(x,y) \to (0,0)} \frac{x-y}{x+y}$ does not exist,using polar coordinates.


Evaluate $\lim_{n \rightarrow \infty} \frac{1}{n} [(1/n)^2+({2}/{n})^2+ \cdots +({3n}/{n})^2]$ $$\lim_{n \rightarrow \infty} \frac{1}{n} \Bigg[\Bigg(\frac{1}{n}\Bigg)^2+\Bigg(\frac{2}{n}\Bigg)^2+ \cdots +\Bigg(\frac{3n}{n}\Bigg)^2\Bigg]$$

Express this sum as an integral. I know $\Delta x = \frac{1}{n}$ and $f(x) = x^2$.

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{n} \Big(\frac{k}{n}\Big)^2$$ The $3n$ is tripping me up. I know $$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{n} \Big(\frac{k}{n}\Big)^2 = \int_{0}^{1}x^2 dx$$
Any help is appreciated.

Hint: Break the sum into three parts.

Here is the second part:

$$\frac1{n} \sum_{k= n+1}^{2n}\left(\frac{k}{n}\right)^2= \frac1{n} \sum_{k= 1}^{n}\left(\frac{n+k}{n}\right)^2=\frac1{n} \sum_{k= 1}^{n}\left(1 + \frac{k}{n}\right)^2\\ \to\int_0^1(1+x)^2\,dx = \int_1^2x^2 \, dx$$

Hint: What is $\frac{3n}{n}$?

Everything else looks good.


Notation (or Name) of the function which is 0 at x=0 and 1 otherwise Is there a traditional notation or name of this function:

$$
\epsilon (x)  = \begin{cases}
0 & \textrm{ if }x = 0 \\
1 & \textrm{ if }x \neq 0
\end{cases}
$$

I know one can use  Indicator function $1_{A}(x)$ for $A = \mathbb{R} - \{0\}$, but I still want to know if there is a traditional, simpler, notation for it.

The Kronecker delta function $\delta_{ij}$ is $1$ when $i=j$ and $0$ otherwise, so $\epsilon(x)=1-\delta_{0,x}$.

Here's one possible definition, though extremely dependent on how one defines the value $0^0$.

If you happen to subscribe to the belief that $0^0=0$ (disclaimer, I personally define $0^0=1$), then
$$\epsilon(x)=x^0$$


Maclaurin Series Terms and Convergence or Divergence This is one of the HW questions I'm trying to solve

Find the first four terms of this Maclaurin Series of $$f(x)=\cos\left(\frac{x}{x+1}\right)$$

I tried using $\sum_{n=0}^\infty (-1)^n  \frac{x^{2n}}{2n!} = \cos x $ and   $ \sum_{n=0}^\infty (-1)^n x^n= \frac{1}{1+x}$ but no luck! I can't properly substitute the derivative of the series in the Maclaurin Series.

Also related to this question is its second part$$ \sum_{} (n)^p (\cos (\frac{1}{n-1})-\cos(\frac{1}{n})$$
for what values of $p$ it converges or diverges.

I would really appreciate it if someone could type in a detailed solution so I can improve my understanding of this topic.

HINTS
The route you want to go notes that if $g(x) = \frac{x}{x+1}$, then $f(x) = \cos(g(x))$ and expansion into series gives
$$
f(x)
 = \cos(g(x))
 = \sum_{n=0}^\infty(-1)^n \frac{g(x)^{2n}}{(2n)!}
 = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \left(\sum_{m=0}^\infty (-1)^mx^m \right)^{2n},
$$
which is very messy to deal with.
You are better off writing the first four terms using the definition, i.e. compute $f'(x), f''(x)$ and $f'''(x)$ and evaluate at $0$. You may find it helpful to write $g(x) = 1 - \frac{1}{x+1}$.

As for the second part, note that what governs the conergence of the sum is what happens when $n$ is large, in which case $\frac{1}{n}$ and $\frac{1}{n-1}$ become very small. Can you use the Maclaurin series to expand the cosine around zero and approximate the cosine using the expansion which includes the first meaningful term which depends on $n$?
Happy to guide you further if you post progress on these in the comments below.

UPDATE
Your first attempt is correct, you can see Wolfram Alpha computes the Maclaurin series to be
$$
\cos \left(\frac{x}{x+1}\right)
 = 1 - \frac{x^2}{2} + x^3 - \frac{35}{24} x^4 + \frac{11}{6} x^5 + \mathcal{O}\left(x^6\right)
$$
As for the second part, note that for small $x$ we have
$$
\cos x \approx 1 - \frac{x^2}{2}
$$
so $\cos(1/n) = 1 - n^2/2$ and you can find $\cos(\frac1{n-1})$ similarly and subtract, then finding the limit of the difference as $n \to \infty$.

There's only one sensible way to find such expansions (and I guess it's the way Mathematica or Wolfram Alpha do it): Let $$c(x)=\cos\left(\frac{x}{x+1}\right)=c_0+c_1\,x+c_2\,x^2+c_3\,x^3+\ldots\tag1$$ and $$s(x)=\sin\left(\frac{x}{x+1}\right)=s_0+s_1\,x+s_2\,x^2+s_3\,x^3+\ldots\tag2$$
Obviously, $c_0=c(0)=1$ and $s_0=s(0)=0$. Differentiating (1) and (2), we find
$$c'(x)=-\frac1{(1+x)^2}\,s(x)\tag3$$ and $$s'(x)=\frac1{(1+x)^2}\,c(x)\tag4.$$ Since $$\frac1{(1+x)^2}=1-2\,x+3\,x^2-4\,x^3+\ldots,$$ (3) and (4) become
$$c_1+2\,c_2\,x+3\,c_3\,x^2+\ldots=-(1-2\,x+3\,x^2-\ldots)(s_0+s_1\,x+s_2\,x^2+\ldots)\tag5$$ and $$s_1+2\,s_2\,x+3\,s_3\,x^2+\ldots=(1-2\,x+3\,x^2-\ldots)(c_0+c_1\,x+c_2\,x^2+\ldots)\tag6.$$ Comparing coefficients in (5) and (6), we obtain
$c_1=-s_0$, $2\,c_2=-s_1+2\,s_0$, $3\,c_3=-s_2+2\,s_1-3\,s_0, \ldots$ and
$s_1=c_0$, $2\,s_2=c_1-2\,c_0$, $3\,s_3=c_2-2\,c_1+3\,c_0, \ldots$, giving successively $c_1=0, s_1=1\to c_2=-\frac12, s_2=-1\to c_3=-1,\ldots$ It's simple to obtain $$c(x)=1-\frac12\,x^2+x^3+\ldots,\tag7$$ (and quite a few more terms, if necessary) this way. But (7) is sufficient for the second part of the question, because $$\cos\left(\frac1{n-1}\right)=\cos\left(-\frac1{n-1}\right)=c\left(-\frac1n\right)=1-\frac1{2n^2}-\frac1{n^3}+\ldots$$


Show that a ring with disconnected spectrum is a product of two subrings. It's an exercise from the book introduction to commutative algebra by Atiyah and Macdonald. If $\operatorname{Spec}(A)$ is disconnected, I'm asked to show that $A$ is a product of two subrings.

I know that $A/R$ is a product of two rings, where $R$ is the nilradical of $A$. What should I do? Also, my friend told me that it can be solved using the fact(I don't know it) that the category of affine schemes is isomorphic to the opposite category of rings, I don't understand how would this fact help. I want an elementary solution to this exercise. Thanks for any help.

Suppose the spectrum $X=\mathrm{Spec}(A)$ is disconnected. This means that it is the disjoint union of two closed sets (which are also open). Closed sets of the spectrum are, by definition, the subsets of the form $V(I)$ for $I$ an ideal of $A$. So the assumption is that $X=V(I)\cup V(J)$ for some ideals $I,J$ with $V(I)\cap V(J)=\emptyset$. Now, in general, from the definition, one has $V(I)\cap V(J)=V(I+J)$, because a prime ideal of $A$ contains both $I$ and $J$ if and only if it contains $I$ and $J$, if and only if it contains $I+J$. So $V(I+J)=\emptyset$. This means that $I+J$ does not lie in any prime ideal of $A$, which means it cannot be a proper ideal: $I+J=A$. On the other hand, again from the definition, one has $V(I)\cup V(J)=V(IJ)$. Because $I+J=A$, one has $IJ=I\cap J$ (this is proved early on in A and M). In fact, for such ideals $I$ and $J$ (sometimes called comaximal ideals), A and M prove that the natural ring map $A\rightarrow(A/I)\times(A/J)$ is an isomorphism (this is an instance of the Chinese remainder theorem). So indeed $A$ decomposes as a product.

EDIT: As is pointed out by the OP in the comments, my argument is not complete. It only shows what the OP already apparently knew, that $A$ modulo its nilradical decomposes as a product. One needs to argue that idempotents lift modulo the nilradical. This is proved in both the answers to If $\mathop{\mathrm{Spec}}A$ is not connected then there is a nontrivial idempotent

Every connected component $V$ of $\operatorname{Spec}(A)$ is a closed set. Therefore $\operatorname{Spec}(A/I(V))$ is just that component. Therefore one can partition $\operatorname{Spec}(A)$ by the $\operatorname{Spec}(A/I(V))$. Every ring $B$ that has morphisms to each $A/I(V)$ has a universal morphism to $\prod A/I(V)$. Is this a quotient, when $B=A$? What's the kernel? In the other category, the morphism from $A$ induces a natural map from the spectrum of the product to $A$, whose topology certainly matches that of $A$.

I'M AWAKE, I'M AWAKE:

So, what I was hoping would make the statement elementary was that if $\operatorname{Spec}(A)$ had subspaces that matched the product's spectrum, the disjointness of the subspaces would imply the only topological difference there could be was whether all the subspaces of $\operatorname{Spec}(A)$ were closed like they are in the product spectrum. Since I wasn't doing the math, I forgot that if $A$ wasn't reduced it could have the same topology.

The surjectivity condition on the universal map reinforces the connection. That the kernel is the nilradical can checked several ways, but see Kidwell's answer for proving the nilradical is $\{0\}$.

W.l.o.g., let $f, g$ be quotient maps from $A$. Then the universal map $<f,g>:A\rightarrow A/\ker f \prod A/\ker g$ is surjective iff

$$\not \exists <[y]_f,[z]_g>: f^{-1}(y) \bigcap g^{-1}(z)=\emptyset$$
$$\iff \forall y, z \in A, \exists i \in \ker f, j \in ker g : y+i = z+j$$
$$\iff \forall y, z, y-z \in \ker f+ \ker g.$$

Thus surjectivity of the universal map holds iff the kernels are comaximal. Furthermore, comaximality implies disjoint varieties, and the radical of the sum of the kernels must be $A$ - which is just the sum of the kernels, if one is initially defining the kernels as $I(V)$ for $V$ a connected component.


Additional solutions to quadratic equations which don't match the formula answer. I'm hoping for link to some resource which can explain why the following is true.

$$ x^2 + 104x - 896 = 0 $$

Using the quadratic formula we pull a = 1, b = 104, c = 896. Putting that into the formula for the discriminant we get $ 104^2 - 4.1.896 $ which is 10816 - 3584 = 7232.

Using the quadratic formula the discriminant is 7232 and using the quadratic formula the answer is

$$ -104 \pm \sqrt{7232} \over 2 $$

This simplfies to $ -72 \pm 4 \sqrt{113} $.

The problem I have is I have not found anything which explains why if you plug x = 8 into the equation it also balances out. What I have found out infers x = 8 being an answer to the equation but not a factor and so not a solution but I don't really get the point.

Any links that explain the distinction would help. I have an infinite number of these equations which I'm looking for integer answers to so if said link also pointed out how you can obtain the integer answers like the 8 in this case instead of the irrational provided by the quadratic formula that would be great.

The quadratic formula is $$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}.$$
Thus in your problem, $$x = \frac{-104\pm\sqrt{104^2 - 4(-896)}}{2}
=\frac{-104 \pm\sqrt{14400}}{2} = \frac{-104 \pm 120}{2}.$$
This gives $x = 8$ or $-112$.

It seems the problem in your computation was you forgot the negative on $896$ when you plugged it into the quadratic formula.

obviously 8 is one of the solutions with no doubts at all.


Calculate, or estimate, $\max_{t\in (a,b]\\ x,y\in\mathbb R} \left|F(x,y,t)+k\sqrt{t} \, e^{ixy}\, \operatorname{sinc}(xt) \right|$. Let $F$ be a complex-valued function. Let also $0<a<b$ and $k>0$.

Given the
$$\max_{t\in (a,b]\\ x,y\in\mathbb R} \left|F(x,y,t)\right|\, ,$$
I would like to calculate (or estimate) the following
$$\max_{t\in (a,b]\\ x,y\in\mathbb R} \left|F(x,y,t)+k\sqrt{t} \, e^{ixy}\, \operatorname{sinc}(xt) \right|\, ,$$
where, for $x\neq 0$
$$\operatorname{sinc} x = \frac{\sin x}{x}$$
and $\operatorname{sinc}(0)=1$.

My attempt is a simple estimation, where I used the well-known inequality
$$\left||z|-|w|\right|\leq |z+w|, \quad z,w\in\mathbb C\, .$$
Any suggestions, please? Are there better ways?

Thanks.

This's my best shot (but i'm not sure about the first 2 row) and in advance sorry for the bad language :
http://www.mediafire.com/view/1jiff26xniifl1t/1.jpg


Is there a matrix that converts the gradient of any function to gradient of other function? The study of hamiltonian mechanics brought me to the following question.

Let $n$ be a natural number ($n>1$).

Let $A(\mathbf{x})$ be a $n\times n$ matrix consisting of functions $a_{ij}(\mathbf{x})$ ($a_{ij}:\mathbb{R}^n\to\mathbb{R}$):
$$
A(\mathbf{x})=
\begin{pmatrix}
a_{11}(\mathbf{x})& \cdots& a_{1n}(\mathbf{x})\\
\vdots&\ddots&\vdots\\
a_{n1}(\mathbf{x})&\cdots& a_{nn}(\mathbf{x})
\end{pmatrix}.
$$
Let $A(\mathbf{x})$ be so, that for any $F(\mathbf{x})$ ($F:\mathbb{R}^n\to\mathbb{R}$):
$$
\begin{pmatrix}
a_{11}(\mathbf{x})& \cdots& a_{1n}(\mathbf{x})\\
\vdots&\ddots&\vdots\\
a_{n1}(\mathbf{x})&\cdots& a_{nn}(\mathbf{x})
\end{pmatrix}
\begin{pmatrix}
\frac{\partial F}{\partial x_1}\\
\vdots\\
\frac{\partial F}{\partial x_n}
\end{pmatrix}
=
\begin{pmatrix}
g_1(\mathbf{x})\\
\vdots\\
g_n(\mathbf{x})
\end{pmatrix}
=
\begin{pmatrix}
\frac{\partial G}{\partial x_1}\\
\vdots\\
\frac{\partial G}{\partial x_n}
\end{pmatrix}
$$
for some $G(\mathbf{x})$ ($G:\mathbb{R}^n\to\mathbb{R})$.

In other words, if we multiply fixed $A(\mathbf{x})$ by the gradient of any $F(\mathbf{x})$ we necessarily get the gradient of some $G(\mathbf{x})$.

Can we say anything about such $A(\mathbf{x})$? I would be glad if the only opportunity is that $A(\mathbf{x})=cE$, where $E$ is the identity matrix and $c$ is some real number. Is it correct? Is it possible to prove it?

All the functions are considered to be "good enough" ("smooth enough").

I have found the solution (with the help of Mathoverflow users).

See here: https://mathoverflow.net/questions/239275/is-there-a-matrix-that-converts-the-gradient-of-every-possible-function-to-gradi


number of n-vertex graphs degree $\leq$m How many cycle-free connected graphs are there on n vertices, none of which has degree >m?

e.g for m=1 there is 1 graph for n=1,2, 0 otherwise. For m=2 there is always 1 graph.

I am particularly interested in the m=4 case, as I was originally trying to count isomers of straight-chain alkanes.

OEIS sequence A000602 seems to be exactly what you want: ‘Number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers’, with extensive references. There’s a table of the values for $n=0$ through $n=60$ here. This sequence is discussed in this paper by Rains & Sloane and in P. Flajolet and R. Sedgewick, Analytic Combinatorics, p. 478; the book is pretty heavy going, but it’s available for free download here. There does not appear to be any nice formula.

The $n=3$ case is OEIS A000672, which also appears to have no nice formula.

I see that
If $B(z)$ is the OGF of alkanes (EIS A000602), which are unrooted trees, then
$B(z) = 1 + z + z^2 + z^3 + 2 z^4 + 3 z^5 + 5 z^6 + 9 z^7 + 18 z^8 + 35 z^9 + \ldots$

But what is the lower or upper bound on the coefficient of $z^n$?
It would be really helpful if you could tell something about it.


Prove that if $M$ is finitely generated then it is Artinian. Let $M $ be a semisimple $R$-module. Prove that if $M$ is finitely generated then it is Artinian.

To show this we have to prove that every non-empty collection of sub-modules of $M$ has a minimal element. Let $C$ be a non-empty collection of sub-modules of $M$.

Also $M$ is finitely generated say by $\{x_1,x_2,\ldots,x_n\}$. How to arrive at the proof?

Without any assumptions about $R$, this is false. For example, consider $\mathbb{Z}$ as a $\mathbb{Z}$-module. It's finitely generated (the set $\{1\}$ is a generating set), but the collection of submodules
$$\mathbb{Z}\supset 2\mathbb{Z}\supset 4\mathbb{Z}\supset 8\mathbb{Z}\supset\cdots$$
has no minimal element.

I am not sure I fully understand the question. As Hoot said, "is $R$ artinian?" to which you replied as "yes". But you also have the finitely generated assumption. In which case the assumption that $M$ is semisimple is completely unnecessary. A finitely generated module over an Artinian ring is Artinian. Because, we have the exact sequence,
$$ 0 \to K \to R^n \to M \to 0 $$
Since $R^n$ is Artinian (your assumption) then so are both $K$ and $M$.


Can the negation of an implication statement be written in terms of implication operators? I know that $A\Rightarrow B$ is equivalent to $\neg A \lor B$. Also that $\neg (A\Rightarrow B)$ is given by $A \land \neg B$. But can $\neg (A\Rightarrow B)$ be written in terms of $A$, $B$, $\Rightarrow$, and $\neg$ alone (i.e. without the use of $\lor$ and $\land$)?

I think you are asking for the principal connective to be an implication rather than a negation. If that's right, a suitable equivalent to $\lnot (A \Rightarrow B)$ is $(A \Rightarrow B) \Rightarrow \lnot(A \Rightarrow A)$.

Note that it is impossible to write $\lnot(A \Rightarrow B)$ in terms of implication alone. If $\phi(A, B)$ is equivalent to $\lnot(A \Rightarrow B)$ , then $\phi(A \Rightarrow A, A)$ is equivalent to $\lnot A$, so as $\lnot$ and $\Rightarrow$ are functionally complete, if $\phi$ is constructed with $\Rightarrow$ alone, then $\Rightarrow$ on its own would be functionally complete, but by Post's characterisation of functionally complete sets of connectives, it is not (see link above).

I use Polish notation.

The formation rules go:

All lower case letters of the Latin alphabet qualify as well-formed formulas.
0 consists of a well-formed formula standing for the constant false proposition.
If $\alpha$ and $\beta$ qualify as well-formed formulas, then so does C$\alpha$$\beta$.  C stands for implication, or equivalently conditional.

Now let us adopt the definition that the negation of a proposition Np, comes as abbreviation for Cp0, since C00 = 1 and N0 = 1, and C10 = 0 and N1 = 0.

Thus, NCab abbreviates CCab0.

More formally, we can take as a definition:

C$\delta$Np$\delta$Cp0 where $\delta$ consists of a functioral variable of one argument (in other words $\delta$ comes as a variable which ranges over the four unary-truth functors of two-valued logic).

Substituting p with Cab and dropping $\delta$ or equivalently substituting $\delta$' with the apostrophe symbol ' where the apostrophe symbol ' consists of the argument of $\delta$ we have

CNCabCCab0

as a thesis obtained from the definition.  Thus, given NCab by definition we obtain CCab0.  From 1. and the rule of uniform substitution for $\delta$ and the rule of uniform substitution for propositional variables we can also obtain NCab from CCab0.


Why did mathematicians take Russell's paradox seriously? Though I've understood the logic behind's Russell's paradox for long enough, I have to admit I've never really understood why mathematicians and mathematical historians thought it so important.  Most of them mark its formulation as an epochal moment in mathematics, it seems to me.  To my uneducated self, Russell's paradox seems almost like a trick question, like a child asking what the largest organism on Earth is, and getting back an answer talking about a giant fungus.

"Well, OK..."

The set of all sets admits all sorts of unmathematical objects right? Chickens, spoons, 2 $\times$ 4s, what have you...I can't imagine a situation where a mathematician solving a concrete problem would invoke a set like that.  The typical sets one sees in everyday mathematics are pretty well defined and limited in scope: $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{R}^n, \mathbb{Q}_p$, subsets of those things.  They are all built off each other in a nice way.

But clearly important people think the result significant.  It prompted Gottlob Frege, a smart and accomplished man, who Wikipedia tells me extended the study of logic in non-trivial ways, to say that the "foundations of [his] edifice were shaken."  So I suppose Russell's paradox is "deep" somehow?  Is there a practical example in mathematics that showcases this, where one might be led horribly astray if one doesn't place certain restrictions on sets, in say, dealing with PDEs, elliptic curves, functional equations?  Why isn't Russell's paradox just a "gotcha"?

Russell's paradox means that you can't just take some formula and talk about all the sets which satisfy the formula.

This is a big deal. It means that some collections cannot be sets, which in the universe of set theory means these collections are not elements of the universe.

Russell's paradox is a form of diagonalization. Namely, we create some diagonal function in order to show some property. The most well-known proofs to feature diagonalization are the fact that the universe of Set Theory is not a set (Cantor's paradox), and Cantor's theorem about the cardinality of the power set.

The point, eventually, is that some collections are not sets. This is a big deal when your universe is made out of sets, but I think that one really has to study some set theory in order to truly understand why this is a problem.

Fortunately, set theory isn't that important to modern math.  But category theory is, and an analogue of Russell's paradox stops you from constructing what would otherwise be one of the most obvious things to construct, the category of categories.

In practice, I've seen a bunch of expositions that go "Let $\mathbf{Cat}$ be the category of all (small) categories..." with a bit of a wink and a nudge to get you on the right track.  So I don't think this is a terribly important issue -- but it does give a natural place for a working mathematician to get accidentally involved in set theory.

By Russell's paradox proper, there's also no set of groups, rings, topological spaces, and so on, but this can be remedied by introducing proper classes, and as far as I can tell, you'd never want to topologize the class of topological spaces or something like that.

Edit: I realized your question was more historical.  Russell came at a time of tremendous vagueness in math, and his work was largely an effort to eliminate such vagueness.  It was surprising that such a project could possibly have nontrivial results.  Frege, specifically, had been working on a massive treatise on logic that was totally destroyed by Russell's paradox, because it relied so heavily on sets whose elements could come from anywhere.  If you're interested, Logicomix is a really good graphic novel about Russell's life and work -- addressed to non-mathematicians, but a good read and helps motivate issues like these.


How many triples of positive integers satisfy $x * y * z = 9 * 2^{20}$? How many triples of positive integers satisfy $x * y * z = 9 * 2^{20}$?

The order in which the numbers are considered does not matter.

I figure the total number of permutations is $9(\frac{22!}{20!}2!)$.

I tried to solve it by first considering all the triples with two equal numbers such as $2^0 * 2^0 * 9$... up to $2^{10} * 2^{10} * 9$, for a total of $11$ classes (and $11 * 3$ if we consider the order).

Then I figured that since $9$ can be written as $3 * 3 * 1$, I should also count in triples like $(2^0 * 3) * (2^0 * 3) * 2^{20}$, and there are other 11 of these.

And lastly I calculated the following : (Tot - $66$) : $6$, and added $22$.
However, it didn't work so well. What am I getting wrong? (probably more than one thing)

Hint: The product must contain 2 factors of 3, and 20 factors of 2.  This means that you have to choose where to put the two 3's (i.e. $x$ contains 0, 1 or 2, $y$ contains 0, 1 or 2, $z$ contains 0, 1 or 2) and where to put the twenty 2's.  The number of ways of doing that is the same as the number of ways to write 20 (or 2 respectively) as the sum of 3 non-negative integer variables.  And that question has been answered a number of times here.

The answer is $\binom{22-1}{3-1} = 210$.
Its a classic combinatorics (stars and bars) problem.


Determining number of integer solutions for expression of perfect square How many positive integer values of n are there such that $2^n + 7^n$ is a perfect square?

I am not sure how to approach this question given that there are two different bases 2 and 7

The only instance is $n=1$.

Let $a_n=2^n+7^n$.

Of course $a_1$ is a square.  Assume that $n>1$.

We see that $a_n\equiv (-1)^n \pmod 4$ so $n$ must be even (this is where we use $n>1$).

Similarly we see that $a_n\equiv 2^{n+1}\pmod 5$ which implies that $n$ is odd, and we are done

This the first attempted solution not only on this post but also my overall first.

N can take either an even value or an odd value.
Im attempting only one part-easier one.

Case I(N is even)

The last digit of powers of 2 and 7 repeat in groups of four i.e (2,4,8,6) and (7,9,3,1).For N=4k+2 and N=4k both the last digit of the overall expression turns out to be (4+9=)3 and (6+1=)7,both of which can never be last digits of a perfect square.

(or)

Simply take modulo 3 when n is even,which tells us that the whole expression is 2(mod 3),never a perfect square.

Case II-N is odd
When n is odd and greater than 1(for which prob. statement holds true), taking modulo 4 gives a remainder of 3,neither 1 nor 0,hence it wont be a perfect square.


Condition for inequality of sum equation and sum For given two sets of $n$ numbers satisfying:
$$x_1^2+x_2^2+\cdots+x_n^2 ≥ y_1^2+y_2^2+\cdots+y_n^2$$

Here ($x_i^2$ presents the square of $x_i$) and
$x_i, y_i$ are not negative numbers.

What is the condition of $x_i, y_i$ in which the following statement is true?
$$x_1+x_2+\cdots+x_n ≥ y_1+y_2+\cdots+y_n$$

One condition that comes right away is that $$\sum_{i\ne j}x_ix_j\ge \sum_{i\ne j}y_iy_j$$


How to generate a general case algorithm operator counter I have this question for a homework assignment.

Consider the following algorithm segment:

    for i=1 to n
       for j=i to n
           X=X+5;

What is the number of additions (+) performed by this code segment? Prove your answer.

It clearly wants you to create and solve a recurrence relation based on this algorithm although I have no idea how to convert the algorithm into a recurrence relation that can then be solved.

I know that the first for loop (i) will run (n-1) times, but I don't understand how to generalize how many times the second for loop will run given that it will run a different number of times each time the first loop runs given that i will have been incremented.

I'm not looking for a solution to this question but rather the method for converting this type of algorithm, into a general case that can then be calculated.

I'm assuming the limits are inclusive of $1$ and $n$(you might want to use $n-1$ instead). [Not relevant to the counting, but note that $X$ is not initialized.]

The iterative nature of the algorithm is naturally representable in tabular form, recursive relation is not necessary unless your homework explicilty asks for one. Each line below lists out all the values $j$ takes for every iteration of the outer loop. Each time $j$ takes a value from $i$ upto $n$(or $n-1$ if you use that as the limit) there is an addition.

When $i=1$: $j$ takes values $1,2,3...n$ for a total of $n$ additions.

When $i=2$: $j$ takes values $2,3...n$ for a total of $n-1$ additions.

.

.

When $i=n$: $j$ takes the value $n$ for a total of 1 addition.

That should be enough to figure out the total.

for i = 1 to n
   for j = i to n
      x = x + 5;

So here, the outer loop runs $(n-1+1) = n$ times.
For each iteration of  outer loop inner loop also iterates from i to n.

Thus, to calculate total number of additions, we've to calculate the number of times inner loop has been iterated. And it is as follows:

i = 1
   j = 1 -> n
i = 2
   j = 2 -> n
   .
   .
   .
i = (n-1)
   j = (n-1) -> n

Summing up we'll get:
$$(n-1) + (n-2) + \dots + 2 + 1 = $$
$$\frac{(n-1)[(n-1)+1]}{2} = \frac{n(n-1)}{2}$$

And rest you can solve.

[EDIT]
Also, if you want to see more on a computer's perspective then it should also be kept in mind that each iteration requires one addition.
For example, this code segment in C++ can be written as:

for (int i = 1; i <= n; i++)
   for (int j = 1; j <= n; j++)
      x += 5;

Here, at each iteration one addition is also there (i++ and j++).
In such a case except that answer, there will also be same number of additions in the j loop and also $n$ additions in i loop. So, final answer would be:
$$\frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n = $$
$$n(n-1)+n = n^2.$$


Methods for proving an equivalence relation I'll be taking introductory abstract algebra in the fall, and so to prepare, I'm working through Pinter's text.  Chapter 12 includes a number of exercises asking the student to prove that something is an equivalence relation and to describe the associated partition.  For example:  In $\mathbb{Q}$, $r \sim s$ iff $r - s \in \mathbb{Z}$.

I think I'm okay with most (?) of this.  I'd show this is an equivalence relation like so:  If $x, y, z \in \mathbb{Q}$, then $x - x = 0 \in \mathbb{Z}$, and so $x \sim x$.  Second, if $x \sim y$, then $y - x = -(x - y) \in \mathbb{Z}$, and so $x \sim y \implies y \sim x$. Finally, if $x \sim y$ and $y \sim z$, then $x - z = (x - y) - (z - y) \in \mathbb{Z}$, and so $x \sim z$.

The two parts I'm not sure about:  First, that last f on the iff.  Essentially, this means I have to prove that if $r \sim s$, then their difference is an integer, right?  But I thought this is merely how this particular equivalence relation is defined.  How do I know that $r \sim s$ until I look at their difference?

Second, I have a basic idea of what the partition is, but I'm not sure how to form the statement.  The equivalence class $[q] = \{k + q : k \in \mathbb{Z}, q \in \mathbb{Q} \}$.  But if anything, that seems as though it would be the definition for a single equivalence class, not the description of the partition.  (Now that I look at it again, it also leaves out the fact that $k$ is arbitrary but $q$ is fixed.)

If anyone can offer any hints, I'd very much appreciate it.  (Though I'm guessing the second of my questions might be more amenable to a No-this-is-how-you-do-it than to a hint, per se.)

Hint

Prove that the associated partition is
$$\{[q]\, |\, q\in[0,1)\}$$

This post is to answer the question above:

How do I know that r∼s until I look at their difference?

Answer:
Use an indexing number m, with m = r - g(r), with g(r) being the next lower integer from r. Then each class can be represented as Am (I don't know how to subscript). m will be some rational number between zero (inclusive) and 1 (non-inclusive).


Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$ I need to determine whether the next limit exists:

$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$

Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of maclaurin series of $\cos$, Where:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...=1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)$$

So we can simplify the expression:

$\begin{align} \lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}&=\lim_{(x,y)\to(0,0)}\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)-1-\frac{x^2}2}{x^4+y^4}\\
&=\lim_{(x,y)\to(0,0)}\frac{-x^2+\frac{x^4}{4!}+O(x^6)}{x^4+y^4} \end{align}$

This limit does not exist.

So, after solving it the 'hard' way, I was thinking:

If I place $y=mx$ and then try to solve the limit I get:

$$\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4(1+m^4)}=\infty$$

Is that enough to prove that the limit does not exists? I mean, If I find at least one limit that does not 'converge' is it enough to say that a multi-variable function limit doesn't exist?

I think you can assume the following limit as well to show that it doesn't exist:

$$\lim_{n\to\infty}f(x,y_n),~~y_n=\pi/n$$

HINT

Find the limit as $x \to 0$ when $y=x$ and show that it is not equal the limit as $x\to 0$ when $y=0$.


How do I solve Problem 1 of the International Olympiad of Metropolis? The question is: Find all $n$ so that there exists $n$ consecutive numbers whose sum is a square!

My method to solve the problem: I would try to look at the values modulo $n$, and there I see there is no possible square for $n=2x>2$, so all even numbers bigger than 2, and that all odd numbers work...

I can also see that the sum of the consecutive numbers is $$\dfrac{((x+1)+(x+n))
\cdot n}{2}$$ So that has to give a square...

But how do I do that? I cannot put my method in application... Am I totally wrong, and I shouldn't do it like that?

As you noted in the question, if $x +1$ is the first number in the sum, then the sum of the $n$ consecutive numbers is
$$ \frac{n}{2} \cdot (x + 1 + x + n). $$

If $n$ is odd, let $n = 2k + 1$, and take $x = k$. Then the sum of the $n$ consecutive numbers $(k + 1), (k + 2), \dots, (k + n)$ is
$$
  \frac{2k+1}{2} \cdot (k + 1 + k + 2k + 1) = (2k + 1)^2
$$
which is a square, so all odd numbers work.

Now suppose that $n$ is even. Let $n = 2k$ for some $k$.

Then the sum is equal to
$$ k(2x + 2k + 1) $$.

The term in brackets is odd, so for this to be a square, the largest power of $2$ which divides $k$ must be even. i.e. We must have that $k = 2^{2m} \cdot d$ where $m$ is some natural number, and $d$ is an odd natural number.

The sum then reduces to
$$ 2^{2m} \cdot d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$

This is a square if and only if
$$ d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
is a square, so it is enough to choose $x$ so that the term in brackets is a square multiple of $d$. That is, we try to choose $x$ so that
$$ 2x + 2^{2m + 1} \cdot d + 1 = a^2 \cdot d $$
for some natural number $a$.

This is equivalent to
$$ 2x + 1 = \left( a^2 - 2^{2m + 1} \right) \cdot d $$
and it is easy to see that this has a solution for x as long as $a$ is an odd square greater than $2^{2m + 1}$. Thus we have a solution in this case since there are arbitrarily large odd square numbers.

We see that if $n$ is even, then $n$ works if and only if $n$ is divisible by $2$ an odd number of times.

For any odd $n$, sum of $n$ consecutive numbers starting from $\frac {n+1}2$ is

$$\frac n2 (n+1+(n-1)*1) = n^2$$

Thus for every odd $n$, there exist $n$ consecutive numbers whose sum is a perfect square.

More details - This becomes simpler if you know Arithematic Progression. The formula for sum of $n$ numbers starting from $a$, which are $d$ apart (difference between successive numbers is $d$) is

$$S_n=\frac n2(2a+(n-1)d)$$

Here $d=1$ and $a$ can be calculated to be $\frac {n+1}2$

Thank you @Litho for pointing it out.


Evaluating $\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{\pi}{2})(1-\sin x)}{(1+\tan\frac{\pi}{2})(\pi-2x)^3}$ Evaluate $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan x)(1-\sin x)}{(1+\tan x)(\pi-2x)^2}$

Question about this..
Shouldn't the answer be infinity..as I have read $x$ as $\frac{\pi}{2} + h$ where..$h$ tends to zero..

If the limit you want to solve is $\lim_{x \rightarrow \frac{\pi}{2}} \frac{(1-\tan(x))(1-\sin(x))}{(1+\tan(x))(\pi-2x)^3}$, then you could evaluate it in the following way;
$$
\lim_{x \rightarrow \frac{\pi}{2}} \frac{(1-\tan(x))(1-\sin(x))}{(1+\tan(x))(\pi-2x)^3}=\lim_{x \rightarrow 0} \frac{(1-\tan(x+\frac{\pi}{2}))(1-\sin(x+\frac{\pi}{2}))}{(1+\tan(x+\frac{\pi}{2}))(-8x^3)}
$$
using the trigonometrical identity $\sin(a+b)=\sin a \cos b+\sin b \cos a$;
$$
\lim_{x \rightarrow 0} \frac{(1-\tan(x+\frac{\pi}{2}))(1-\cos(x))}{(1+\tan(x+\frac{\pi}{2}))(-8x^3)} \sim \lim_{x \rightarrow 0} \frac{(1-\tan(+\frac{\pi}{2}))}{(1+\tan(x+\frac{\pi}{2}))(-16x)}
$$
$$
\lim_{x \rightarrow 0} \frac{\cos(x+\frac{\pi}{2})-\sin(x+\frac{\pi}{2})}{(\cos(x+\frac{\pi}{2})+\sin(x+\frac{\pi}{2}))(-16x)}=\pm \infty
$$
In fact the limit doesn't exist, since you get $+ \infty$ from the right and $- \infty$ from the left.

As we have
$$\frac{1-\tan (x)}{1+\tan (x)}=\frac{\cos (x)-\sin (x)}{\sin (x)+\cos (x)}\to -1,\text{ as } x\to\frac{\pi}{2}$$
we can write the limit as
$$\lim_{x\to \frac{\pi }{2}} \, \frac{\sin (x)-1}{(\pi -2 x)^2}$$
using L'Hopital rule twice we get the limit that is $-\dfrac{1}{8}$


Proving that $\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$ Let $f,g\in C^1([a,b])$ with $a<b$ then prove that

  $$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$$

It smells  like there is some mean value theorem going around. But I tried it as follows:

Indeed, it springs from mean value theorem that There exists  $c_1,c_2\in (a,b)$ such that

$$\frac{f(b)-f(a)}{b-a} = f'(c_1)~~~ and
~~~~\frac{g(b)-g(a)}{b-a} = g'(c_2)$$

  Then I have  $$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2= f'(c_1)+(g'(c_2))^2\le \max_{t\in [a,b]}\{f'(t)\}+\max_{t\in [a,b]}\{(g'(t)^2)\}$$

Which is however not the required inequality.

Can anyone help? how can I improve this ?

Treat $\max$ as $\sup$, then apply then fact that $\sup(A+B)=\sup(A)+\sup(B)$, for $A,B$ be nonempty bounded subset of $\mathbb R$ and $A+B=\{a+b:a\in A,b\in B\}$


What is the answer to $ \log_5 (25\sqrt 5)$? I have started to do this question, but not sure if I am on the right track or not (or even allowed in math)

Can someone say if what I am doing is right?

Yep your solution is correct except for the fact that you have multiplied when opening the logarithm. This is the correct identity [log(ab)]=loga+logb and not loga.logb

Alternatively:

$25= 5^2$ and $\sqrt 5 = 5^{\frac 12}$ and

$25\sqrt{5} = 5^2*5^{\frac 12} = 5^{2+\frac 12}=5^{\frac 52}$.

There are many ways to interpret this: $(5^{\frac 52} = \sqrt{5^5}=\sqrt{3125}=25\sqrt 5$ or $5^{\frac 52}=(\sqrt 5)^5 = (\sqrt 5)^2*(\sqrt 5)^2*\sqrt 5 = 5*5*\sqrt 5 = 25\sqrt 5$.

So all of this should make sense.

So $\log_5 25\sqrt 5 = \log_5 5^{\frac 52} = \frac 52$.


Proof-verification: $(x_n)$ bounded and $y_n \to 0$ implies $(x_ny_n)\to 0$ Proof-verification: $(x_n)$ bounded and $y_n \to 0$ implies
  $(x_ny_n)\to 0$

My attempted proof:

Let $(x_n)$ be bounded, hence there exists $A\in \mathbb{R}$ such that $|x_n|\leq A$. Furthermore let $y_n$ be a null-sequence; thus for an arbitrary $\varepsilon >0$ there exisits $N\in \mathbb{N}$ such that $|y_n|<\varepsilon$ for all $n\geq N$.

$$|x_n\cdot y_n|\leq |A\cdot y_n|=|A|\cdot |y_n|<|A|\cdot \varepsilon$$ Since $\varepsilon$ does not depend on $|A|$, one can conclude that the claim is indeed true. $\Box$

is that okay?

If we correct the obvious typo then this is fine; exactly this sort of argument is given all the time.

But it's somewhat "informal", in that the conclusion does not look like what's required by the definition. A more formally correct version would read "Since $y_n\to0$, given $\epsilon>0$  there exists $N$ such that $|y_n|<\epsilon/A$ for all $n>N$..."

So whether it's actually "right" depends on the context. In a research paper, or some other context where the point is to convince the  reader it's true, what you wrote is fine - people do write this sort of thing all the time. In a calculus class it's different - there the point is to convince the reader that you know exactly how to prove it, and in that context you're much better off getting a literal $|x_ny_n|\dots<\epsilon$ at the end.

Note: Of course @pre-kidney is also correct in saying that the last sentence is missing the point; that issue doesn't come up if we change it as suggested.

Yes, your proof is okay. Some other users obviously don't share my view. It is my personal opinion they are being too pedantic. I find it amusing that of all answers, this one of mine is so vehemently downvoted and opted for deletion. C'est la vie.


Prove the inverse of this differentiable function is differentiable? Suppose we have a differentiable function $ g $ that maps from a real interval $ I $ to the real numbers and suppose $ g'(r)>0$ for all $ r$ in $ I $.
Then I want to show that $ g^{-1}$ is differentiable on $g(I). $

Intuitively this makes sense but I can't come up with a neat proof. I was thinking to use the mean value theorem but I'm not sure if that would get me anywhere.

By definition of derivative we have

$$g'(r)=\lim_{x\to r}\frac{g(x)-g(r)}{x-r}=\lim_{x\to r}\frac{g(x)-g(r)}{g^{-1}(g(x))-g^{-1}(g(r))}$$

Since $g'(r)>0$, this means

$$\lim_{x\to r}\frac{g^{-1}(g(x))-g^{-1}(g(r))}{g(x)-g(r)}=\frac{1}{g'(r)}$$

Let $y\in g(I)$, say $g(x)=y$ for some $x\in I$, and let $y_n\to y$. Then for each $n$ there is a unique $x_n\in I$ with $g(x_n)=y_n$. Further, one can show that if $f$ is continuous and strictly monotone on an interval $I$, then $f^{-1}$ is continuous on $f(I)$, hence we have $x_n\to x$, it follows that

$$\lim_{n\to\infty}\frac{g^{-1}(y_n)-g^{-1}(y)}{y_n-y}=\lim_{n\to\infty}\frac{x_n-x}{g(x_n)-g(x)}=\frac{1}{g'(x)}$$

Hence $g^{-1}$ is differentiable.

If $g'(r)>0$ then $g(r)$ in injective.

Exist $g^{-1}(r)$.

Then use the chain rule for derivate $g^{-1}(g(r))=r$.


Is continuity at a point only defined for points in the domain? I'm using Michael Spivak's Calculus, 3rd edition textbook. Without ado, I'll state the definition given for continuity at a point:

  DEFINITION$\;\;\;\;$The function $f$ is continuous at $a$ if: $$\lim_{x\to a}f(x) = f(a)$$

And I might as well give the definition for the limit at a point:

  DEFINITION$\;\;\;\;$The function $f$ approaches the limit $l$ near $a$ means: for every $\epsilon>0$ there is some $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$.

It is clear, then, that in order to verify whether a function is continuous at a point $a$, we need to compute $f(a)$, and this is only possible if $a$ is in the domain of $f$. So, what happens when a point $a$ is not in the domain of $f$?

On this matter, I've had some people telling me that statements about continuity only make for points in the domain of $f$. Spivak himself also mentions in the Continuity chapter of his textbook that "We also often simply say that a function is continuous if it is continuous at $x$ for all $x$ in its domain."

But that means functions like $f(x) = \frac{1}{x}$, $f(x) = \frac{1}{x^2-1}$ and $f(x) = \sin{\frac{1}{x}}$ can all be said to be continuous. It just doesn't seem right, in fact I always thought these were textbook examples of discontinuous functions. It also means that we can't say that $f(x) = \frac{1}{x}$ is discontinuous at $0$, because we don't even know the value of $f(0)$ in order to decide whether it's continuous or not.

Are all of these conclusions correct, or did I misinterpret the definitions? I'd love if someone could shed some light on this, I'm having a hard time wrapping my head around it.

There are two different definitions of continuity.  One is used at the elementary calculus level, and the other in more advanced texts.  I am assuming that you are familiar with the first definition and therefore will skip the epsilon, delta quantifiers to simplify the presentation:

(1) a function $f$ is continuous at $c$ for all $x$, if $|x-c|<\delta$ then $|f(x)-f(c)<\epsilon$.

(2) a function $f$ with domain $D$ is continuous at a point $c\in D$ if for all $x\in D$, if $|x-c|<\delta$ then $|f(x)-f(c)<\epsilon$.

The second definition is more complicated because it speaks explicitly about the domain of the function, and limits $x$ to this domain.  The advantage of the second approach is that you don't need to deal with special cases such as one-sided continuity when you want to define what it means for a function to be continuous on a closed interval.

The existence of two slightly different definitions is a constant source of confusion and your question is well taken.

When a function is called continuous, it is often the case that it is continuous on all real numbers. It is reasonable to call a function continuous the set of points where it satisfies the definition of continuity is understood.

Personally, for undergraduate calculus, I prefer to call a function $f:D\rightarrow\mathbb{R}$ (where $D$ is a subset of, or is equal to, $\mathbb{R}$) discontinuous if its graph cannot be drawn with a single penstroke. In which case $f(x)=1/x$ would be called discontinuous. However, $f(x)=1/x$ is definitely continuous on its domain ($x\neq0$). It's not continuous on $\mathbb{R}$ $-$ it's not even defined on all of $\mathbb{R}$.

The examples $f(x)=1/x$ and $g(x)=1/x^2$ are not continuous at $x=0$! They do not satisfy the definition of continuity there, so we call it discontinuous at the origin. Anyone who calls it continuous at the origin is wrong, in general.

However, we can define the extended real numbers: $\overline{\mathbb{R}}=\{-\infty\}\cup \mathbb{R}\cup \{\infty\}$ and define:
$$
g(x)=\begin{cases}
\frac{1}{x^2} \text{ if } x\neq 0, \pm\infty \\
\infty \text{ if } x = 0 \\
0 \text{ if } x=\pm\infty,
\end{cases}
$$
and $g:\overline{\mathbb{R}}\rightarrow \overline{\mathbb{R}}$ can be said to be continuous on the extended reals. Of course, you have to carefully consider how to define continuity in this context.


Does $\mathbb{Z}_5(\sqrt[3]{3})$ make sense? Or, can we always extend a field by a root of a reducible polynomial? I'm preparing assignment questions for a course in ring/field theory.  We'll shortly be looking at extension fields, and the students are meant to understand what notation such as $\mathbb{R}(i)$ means.

Two examples of questions that might arise:

If the question asked to describe the elements in $\mathbb{Z}_5(\sqrt{3})$, the student would be expected to recognise that the polynomial $x^2-3$ has no root in $\mathbb{Z}_5$, and thus is irreducible in $\mathbb{Z}_5[x]$, and thus define some element $\omega$ as a root of $x^2-3$, and set $\mathbb{Z}_5(\sqrt{3})=\mathbb{Z}_5(\omega)$, from which they can identify the elements.
If the question asked to describe the elements in $\mathbb{Z}_7(\sqrt{2})$, the student would be expected to recognise that the polynomial $x^2-2=(x-3)(x-4)$, and thus is reducible in $\mathbb{Z}_7[x]$, and thus recognise that $\mathbb{Z}_7(\sqrt{2})=\mathbb{Z}_7$ (via: if we define $\omega$ as a root of $x^2-2$ then $\omega \in \{3,4\}$, and so $\omega \in \mathbb{Z}_7$).

However, it is possible to have a "partially reducible" polynomial (i.e., one that does not factor entirely into linear factors in $F[x]$).  Which leads to the following question:

  Does $\mathbb{Z}_5(\sqrt[3]{3})$ make sense?  (I.e. The extension field containing $\mathbb{Z}_5$ and $\sqrt[3]{3}$.)

If we use the above approach, we identify that the polynomial $x^3-3=(x-2)(x^2+2x+4)$, and thus is reducible in $\mathbb{Z}_5[x]$.  If we define $\omega$ as a root of the polynomial $x^3-3$, then:

$\omega$ might be $2$, in which case $\omega \in \mathbb{Z}_5$ and $\mathbb{Z}_5(\omega)=\mathbb{Z}_5$.
$\omega$ might instead be a root of $x^2+2x+4$, in which case $\mathbb{Z}_5(\omega) \neq \mathbb{Z}_5$.

In order to reconcile this problem, it looks like there should be some restriction placed on the notation $F(\omega)$, such as "this is semantically incorrect, unless $\omega$ is derived from an irreducible polynomial in $F[x]$", but this would ruin the second question above.

Douglas,

this is not the correct usage and could be very misleading. It seems to have confused you, so it must be even more confusing for students who are just learning.

Here is the way it is used most of the time:

Let $K$ be a field contained in a larger field $L$. Let $\alpha\in L$ be an element. Then $K(\alpha)$ denotes the smallest subfield of $L$ that contains both $K$ and $\alpha$. Obviously, if $\alpha\in K$, then $K(\alpha)=K$. Otherwise, if $\alpha$ is not algebraic over $K$, then $K(\alpha)\simeq K(x)$. If $\alpha$ is algebraic over $K$, then there exists an irreducible polynomial $p\in K[x]$ such that $p(\alpha)=0$ and then $K(\alpha)\simeq K[x]/(p)$. This latter construction works even without specifying $L$ if we start by $p$.
(I'm sure you know all of this, but I need to establish the background.)

Now when one says $K(\sqrt 2)$, then they actually mean (or should mean) that they want to consider the extension of $K$ with a root of $x^2-2$. In truth this is a sloppy shortcut and one has to be careful how to use it.
In fact, it is not a very good notation exactly because of the issue you brought up. If the polynomial is not irreducible, then it is not clear what field one would want to define. Imagine a field in which $x^d-\lambda$ is a product of two irreducible polynomials without roots, but which are different. Extending by $\sqrt[d]\lambda$ could result in two completely different field. So, this should be avoided.

In other words there is a restriction on the meaning of $F(\omega)$ (as you suspect). And it "ruins" your problem as stated. Well, actually it's more like this is not well-phrased problem. In my opinion even the one with $\mathbb Z_7(\sqrt 2)$ is not a good problem either, because it suggests and encourages improper use.

What you could do instead is to ask for the splitting field of those polynomials. For that they would have to adjoin roots as long as there is a factor that is not linear.

No. $\mathbb{Z}_5(\sqrt[3]{3})$ does not make sense, since it is not determined uniquely.


Nonexistence of a limit What does it mean for limit not to exist?

By negating the definition of limit we have:
$$\exists \epsilon > 0, \quad \forall \delta >0, \exists x \in I \setminus\{c\}$$ such that $0 < |x - c| < \delta$ but $|f(x) - L| \geq \epsilon$.

I understand what this means. But what does $L$ mean here? $L$ does not even exist and in fact we are using it as a part of the definition.  Can someone clarify?

The definition of limit should start with $\exists L$. Consequently the negation should start with $\forall L$

$L$ is your candidate for the limit of your function. This does not mean that your function $f$ does not converge, but it does mean that $L$ is not the limit it's approaching (whether $f$ does converge or not is irrelevant). But if you take into consideration EVERY real number $L$ and then negate the definition it means that $f$ diverges as there is no limit $L$.


Intuition behind the Definition of Conditional Probability (for 2 Events) What is some intuitive insight regarding the conditional probability definition: $P(A\mid B) = \large \frac{P(A \cap B)}{P(B)}$ ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask.

Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).

Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.

Hence \[\mathbb P(A\mid B) = \frac{\text{# occurrences of A and B}}{\text{# occurrences of B}} = \frac{\mathbb P(A \cap B)}{\mathbb P(B)}\]

because the "total number of possibilities" in the expressions for $\mathbb P(B)$ and $\mathbb P(A \cap B)$ cancel.

Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers $\{1,2,3,4,5,6\}$ are less than $4$ but only a third of the numbers in our subsection $\{2,4,6\}$ are.

As we are given that $B$ occurred, we reduce the sample space to $B$. Now that, we have reduced the sample space to $B$ we look at what is the probability $A$ has also occurred which leads to the intersection of both i.e $A \cap B$.


If two continuous functions are equal almost everywhere on $[a,b]$, then they are equal everywhere on $[a,b]$ Suppose $f$ and $g$ are continuous functions on $[a,b]$. Show that if $f=g$ almost everywhere on $[a,b]$, then, in fact, $f=g$ on $[a,b]$. Is a similar assertion true if $[a,b]$ is replaced by a general measurable set $E$?

I have known that the set $A=\{x \mid f(x) \neq g(x)\}$ has measure zero and we want to show that A is empty. Now let's assume $A$ is not empty. I am stuck in getting the contradiction.

Thanks for your hints and answers.

You can do it even more simply:

$f,g$ continuous on $[a,b]$ implies $(f-g)$ continuous.  Thus:

$$(f-g)^{-1}(0, + \infty)\cup(0, -\infty) = A$$

is open, thus $A=\emptyset$ as that is the only open set of measure zero.  QED. No epsilonics required.

If $f = g$ almost everywhere on $[a,b]$ then $|f - g| = 0$ almost everywhere on $[a,b].$ Let $\mu$ denote the Lebesgue measure on $[a,b].$ Then $$\int_{a}^{b} |f-g|\ dx = \int_{[a,b]} |f-g|\ d\mu = 0.$$ Since $f,g \in C[a,b]$ it follows that $|f - g| \in C[a,b]$ and moreover $|f-g| \geq 0$ on $[a,b].$ So $$\int_{a}^{b} |f-g|\ dx = 0 \iff |f-g| \equiv 0 \iff f\equiv g.$$


Equation of latus rectum of parabola If parabola $\left ( y+1 \right )^{2}=k\left ( x-2 \right )$ passes through a point (1,-2), then equation of its latus rectum and directrix ?
My attempt

Is this correct answer?
because this answer is not mention in given options.

The parabola is $$(y+1)^2=(2-x)~~~~(1)$$
The most general form of parabpla is $$L_1^2=4AL_2$$ if $L_1$ and $L_2$are non-parallel. If these two are parallel and normalised $(L=\frac{ax+by+c}{\sqrt{a^2+b^2}}),$ Then length of its latus rectum is $4A$, the Eq. of directrix is $L_2=-A$, Eq. of its latus rectum is $L_2=A$. Eq. of its axis is $L_1=0$ tangent at vertex is $L_2=0$, focus is given by $(L_2=A, L_1=0)$ vertex is $(L_2=0, L_1=0).$
Eq. of directrix for (1) is $2-x=-1/4 \implies x=9/4$ and Eq. of latus rectum is $2-x=1/4 \implies x=7/4$

Vertices of parabola are (-1,2)》(h,p)
Find the k value by substituting the given point..
Now we get k=-1
General equation of parabola( Y-p)^2=-4a(X -h)
4a=k  .. a=1/4
Negative sign Denotes that parabola opens towards negative X axis
Now focus 》(2-1/4,-1)》(7/4,-1)》(h-a,p)
Which means equation of lateus rectum is x=7/4 since it passes through focus perpendicular to axis of parabola..
Now directrix passes through》(2+1/4,-1)》(9/4,-1)》(h+a,p)
Means direct passes through (9/4,-1) perpendicular to axis of parabola then its equation is x=9/4


Help with calculating the Curl of [f(r)r_vec] I need to show that the $\partial_x [f(r)(\vec{r})]=0$

Is the way to start this by saying that $\vec{r} = (x,y,z)?$

Then $r = \sqrt{x^2+y^2+z^2}?$

I am not sure if this is the way to start because we dont know what the function is?

It is not clear what you are trying to calculate: the title suggests $ \nabla \times ( f(r) \vec{r}) $ but you write $ \partial_x ( f(r) \vec{r} ) $. Since it is not generally true that $ \partial_x ( f(r) \vec{r} ) = 0 $, I'm assuming you are trying to show $\nabla \times ( f(r) \vec{r}) $. Do you know that $ \nabla \times ( \nabla g) = 0 $ for any $ g $? And as a special case, you can try $ g(r) = \int_0^r f(r') r' dr' $. Otherwise you can do it brute force by looking at (http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates) for the formula for $ \nabla \times $ in spherical coordinates.


Why does the ordering function invert itself? I notice in my R console that x=sample(1:n); any(order(order(x)) == x) always evaluates TRUE, for any n. Just to assure you I'm not on the wrong SE site here, I know exactly what the code means, but this still makes my brain hurt. Can anyone draw a picture or give an explanatory proof to show what's going on?
Programming background
In R, sample scrambles a sequence, sampling without replacement, and order gives indices such that x[order(x)] is ascending. == does element-wise comparison, returning a vector of logical values. any returns TRUE for a list that's all TRUE and FALSE if the list contains any FALSE elements.

In general, when $x$ is a list of $n$ elements, $\text{order}(x)$ is a permutation of $\{1,2,\ldots,n\}$ such that $x_{\text{order}(x)_i} \le x_{\text{order}(x)_j}$ whenever $i<j$.  (The elements of $x$, taken in the order given by $\text{order}(x)$, are non-decreasing.)  When $x$ is itself a permutation of $\{1,2,\ldots,n\}$, this constraint uniquely defines the order function, and indeed we can say that $x_{\text{order}(x)_i}=i$.  (The elements of a permutation of $\{1,2,\ldots,n\}$, taken in non-decreasing order, are just $1,2,\ldots,n$.)
So, in the group of permutations under the composition operator, $\text{order}(x)=x^{-1}.$  But this immediately gives the desired result:
$$
\text{order}(\text{order}(x))=\text{order}(x)^{-1}=\left(x^{-1}\right)^{-1}=x.
$$

any is a quantifying operator. If test whether one sequence is identic independent of order to the other. So x=sample(1:n); any(order(order(x)) == x) has always to be true. order() is idempotent. That is, if applied once there is no more change in the next application of order() to the sequence it is applied to.
From order You get the information the order() has the argument decreasing. Per default that is set to FALSE. To invert the order the command has to be
x=sample(1:n); order(order(x),decreasing=TRUE) == x)

Then the any() is obsolete. any implements "Are Some Values True?". So
in x=sample(1:n); any(order(order(x)) == x) the identity check == is applied elementwise on the cross of the two lists x and order(order(x)) and then the complete misses are discarded in favor of the few matches.
So
x=sample(1:n); order(order(x)) == order(x) is TRUE always verifying the idempotence of order().


How to prove that a $3 \times 3$ Magic Square must have $5$ in its middle cell? A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.

How to prove that a normal $3\times 3$ magic square must have $5$ in its middle cell?

I have tried taking $a,b,c,d,e,f,g,h,i$ and solving equations to calculate $e$ but there are so many equations that I could not manage to solve them.

The row, column, diagonal sum must be $15$, e.g. because three disjoint rows must add up to $1+\ldots +9=45$. The sum of all four lines through the middle is therefore $60$ and is also  $1+\ldots +9=45$ plus three times the middle number.

The proof are in the rules (or conditions) about how a magic square is built...

In this case a 3x3 grid, you need a sequence of nine integer and positive numbers to make a magic square. You must order that numbers by size... the number that goes in the center of the grid is always the middle number of that group you ordered.

For example, you have this group of nine numbers to make a $3\times 3$ MS: $15, 33, 12, 45, 18, 48, 27, 30$ and $42$. After you ordered them by size: $12,15,18,27,30,33,42,45,48$.

The one in the middle of this ordered group of nine numbers is $30$.

30 will be the number that goes to the center of the $3\times 3$ grid.

No need of any formulas, just follow the MS (Magic Square) rules.


Another double fractional part integral Today I'm trying to defeat another fractional part integral but it seems quite difficult… That is $$\int_0^1 \int_0^1 \left\{\frac{x+y}{x-y}\right\}\,\mathrm{d}x\,\mathrm{d}y$$ Probably, a way to solve is noticing that for the definition of fractional part function $$\left\{\frac{x+y}{x-y}\right\}=\left\{1+\frac{2y}{x-y}\right\}=\left\{\frac{2y}{x-y}\right\}$$ and it seems somewhat easier. I suppose that the answer is in evaluating the fractional integrand but I have some difficulties with it and the substitutions that I tried weren't very useful. Any suggestion?

The integrand has interesting behavior around $y=x$, so let's split the integral there into two pieces. Let's convert the bottom piece into an integral in polar coordinates:

$$\int_0^{\frac{\pi}{4}} \int_0^{\sec \theta} \left\{\frac{\cos\left(\theta - \frac{\pi}{4}\right)}{\cos\left(\theta + \frac{\pi}{4}\right)}\right\}r\:dr\:d\theta = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\{\tan\theta\}\csc^2\theta\:d\theta$$

Now let $u = \cot\theta$ to get

$$\frac{1}{2}\int_0^1 \left\{\frac{1}{u}\right\}\:du = \frac{1-\gamma}{2}$$

which is a somewhat fairly known result, depending on who you ask. The top integral should follow similarly, I'll let you try it on your own (aka left to the reader as an exercise!)

$\textbf{EDIT}$: Given that the inside of the curly braces is negative on the upper piece as @EricTowers noted, we can use

$$\{-x\} = 1 - \{x\}$$

for noninteger $x$ to get that the integral of the upper piece equals

$$\iint_{\text{upper}} \left\{\frac{x+y}{x-y}\right\}\:dx\:dy = \iint_{\text{lower}}1 - \left\{\frac{x+y}{x-y}\right\}\:dx\:dy$$

which implies the sum of the two pieces is simply

$$\iint_{\text{triangle}} dA = \frac{1}{2}$$

Hint:  Exchanging $x$ and $y$ negates the integrand, $$ \left\{ \frac{x+y}{x-y} \right\} = -\left\{ \frac{y+x}{y-x} \right\} \text{,} $$
so split the region of integration along the line $x = y$.


Asymptotics of coefficients This is a question that asks the reader for a $strategy$ to solve a particular problem. I cannot solve this problem myself so I am looking around for general methods one might use to confront it with. Suppose
$$f(x)=a_0+a_1x+...,  g(x)=b_0+b_1x_...$$
and given
$$\lim\limits_{x\to 1^-}\frac{f^{(n)}(x)}{g^{(n)}(x)}=1$$
With the additional
$$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$$ and $$\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}=1 $$
for all natural $n$. Prove
$$\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$$

I would like to know your general thoughts about approaching this type of problem. Any insights however small will be much appreciated. Although this is seemingly a problem in real analysis, I have tagged complex analysis because solutions may well involve it.

Under the given assumptions,

$$\lim_{n\to\infty} \frac{a_n}{b_n} = 1$$

need not hold. Counterexample:

$$\begin{align}
f(x) &= \frac{1}{1-x^2} = \sum_{k=0}^\infty x^{2k},\\
g(x) &= f(x) + e^x.
\end{align}$$

We have $\dfrac{a_n}{b_n} = 0$ for all odd $n$, but nevertheless

$$\lim_{x\to 1} \frac{f^{(n)}(x)}{g^{(n)}(x)} = \lim_{x\to 1} \frac{f^{(n)}(x)}{f^{(n)}(x)+e^x} = 1$$

for all $n$, since $\lim\limits_{x\to 1} \lvert f^{(n)}(x)\rvert = +\infty$, while $e^x$ remains bounded.

You need some further conditions to reach the conclusion.

Firstly suppose that $f(x)$ and $g(x)$ are two polynomials, and let their degree be $c$ and $d$, respectively. Suppose $ d \gt c$, you get
$\lim\limits_{x\to 1^-}\frac{f^d(x)}{g^d(x)} = \frac{0}{d!\cdot b_d} = 0 \ne 1$, so it's impossible that $ d \gt c$; obviously you can use the same proof to prove that it's impossible that $ c\gt d$.

Now, since $c=d$, $\lim\limits_{x\to 1^-}\frac{f^c(x)}{g^c(x)} = \frac{c!\cdot a_c}{c!\cdot b_c} = \frac{a_c}{b_c} = 1$ , so $a_c$ = $b_c$.

Similarly, $\lim\limits_{x\to 1^-}\frac{f^{c-1}(x)}{g^{c-1}(x)} = \frac{(c-1)!\cdot a_{c-1}+c!\cdot x\cdot a_c}{(c-1)!\cdot b_{c-1}+c!\cdot x\cdot b_c} = \frac{a_{c-1}+a_c}{b_{c-1}+b_c} = 1$, which means that

$a_{c-1}+a_c = b_{c-1}+b_c$

$a_{c-1} = b_{c-1}$.

Clearly you go on with the same "strategy", and you'll obtain that $a_i$ = $b_i$ for all $i$ such that  $0\le i \le c$, so $\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$.


Criterion for fundamental solutions to Pell's equation of the form $X^2-DY^2=C$ We are trying to codify in terms of modern algorithm the works of the ancient Indian mathematician Udayadivakara (CE 1073). In his work Sundari, he quotes one Acarya Jayadeva who has given methods to solve Pell's equations. In these methods, one can find the the cyclic Chakravala method to deal with $X^2-DY^2=1$ wrongly attributed to Bhaskara. He also gives the method to solve $X^2-DY^2=C$ for any integer $C$.

His algorithm starts off by finding the nearest square integer $>D$ named $P^2$. Then $a=P^2-D$.
Now some $b$ is chosen in such a way that $Db^2+Ca$ is some perfect square $Q^2$.
Then the $X$ and $Y$ solutions can be found by using $Y=\frac{Q\pm P b}{a}$ and $X=PY \mp b$.
This procedure can continue indefinitely to find all the solutions.
Coming to the question of the fundamental solution i.e. the solution with which Bhavana has to be performed repeatedly to get other solutions (related to the modern automorphism group of the quadratic form), Prof. K.S. Shukla who first translated the work from Sanskrit to English, in his example says that it should be chosen "appropriately".
Our primary question is then what is the criterion to derive this fundamental solution? Is there a way to derive such a criterion?
The whole procedure seems to resemble Conway's topograph method which has been posted here several times Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?
It is quite fascinating to think that some wonderful mind came up with this algorithm about 1000 years ago and the optimality of it is equally amazing.

P.S: If anyone so wishes, we would be happy to provide a version of the original paper written by Prof. Shukla in 1950 dealing with this!

In the case of $X^2-DY^2=C$, this applies.

$C$ is the product of short and special primes, and some $Z^2$ relative to base $D$.
$Z=\gcd(X,Y)$ can be any number, but $Z^2\mid C$.
Where $D = 1 \pmod{4}$, then $X$ and $Y$ can be integer-halfs (ie both Z+½).

Let's see what this means.

In base $D$, the period of 1/p will divide $(p-1)/2$ an even (short) or odd (long) number of times.  For example, in decimal, short primes are of the form $40n\pm 3^x$, so 3, 13, 31, 37 are 'short primes'.  This means that the equation has a solution in terms of $C=111$ or $C=1209=3 * 13 * 31 $, for example, since these are the product of short primes relative to 10.

The long primes never occur as solutions unless they divide both $X$ and $Y$, in which case $p^2 \mid C$.

The special primes, are generally taken as the set of 2 and the divisors of $D$.  These will necessarily accompany certain primes.  This give for example, that $3$ will not occur alone, but requires a $2$ or $5$ to be with it.  So there is no solution to $X^2-10Y^2=3$, but there is for $X^2-10Y^2=6$.

General Theory

Algebraically, these equate to values of the form $a \pm b\sqrt{D}$.  This is an integer-system, in that every number of this form divides some integer $C=a^2-D.b^2$.  Many of these integer systems are not completely factorable in themselves, but rely heavily on 'multiplication-half' systems.

For example, the decimal case $a\pm b\sqrt{10}$ has a half-multiples of the form $a\sqrt 2 \pm b \sqrt 5$.  $a$ and $b$ can be integer-halfs when the difference of the numbers under the square roots is divisible by four, ie $4\mid (5-2)$ here.

Note also that there can be a unit in the integer-half system.  $D=10$ does not have one, but there is certainly one in $D=42$ as $\sqrt 7 + \sqrt 6$.  This system also has multiplication-halfs at $a\sqrt 2 + b\sqrt{21}$, and $a\sqrt 3+b \sqrt{14}$.

Note that any given prime occurs in a particular multiplication-half, and requires its special prime(s) to find a value of $C$.  So in decimal, the prime $41$ can occur as a direct solution, whereas $37$ requires an additional special half-factor (2 or 5) to be a solution.  An additional multiplication-half will do, so $13*37=481=10.7^2-3^2$ works.

The case of $D=130$ is interesting.  It has four half-systems (1,130), (2,65), (5,26) and (10,13), which work on the XOR principle (ie a prime in (2,65) times one in (10,13), requires a prime from (5,12) to make the square (2×10×5 is square).  But the unit for this system has no half-form.

I believe that if you take the primitive roots of the primes, and make a table of even and odd indexes (ie $g^i = n\pmod p$), then these rules apply.

If the sum of indexes for $D \pmod p$ is even, then $p$ can occur as a single-power divisor of $C$.
The actual parity of the individual divisors of $D$ determine which 'half' it falls in.  For example, relative to $D=30$, we have the prime 13, as long in binary, short in -3, and long in 5.  This gives 101, so the period of 13 in base 30 has a length dividing 6.  But the unit for $x+y \sqrt{30}$ has a half falling in $\sqrt 6 + \sqrt 5$.  So to make these all even, it's not in the set 000,110,001,111, (which require no helper prime), but in 010, 100, 011, 101, which needs either a $2$ or $3$.  So we find for example, that 26 and 39 are possible candidates for $C$, but not 13 itself.


Find bisector equation using a point the bisector goes through and an angle I am given a triangle with vertices $A=(0,0)$, $B=(-7,0)$ and $C=(0,4)$. This is a right triangle. One might think from graphing that the angle bisector cuts from point B and point $(0,2)$ but this is not the case. I am having trouble coming up with the equation of the bisector given that the bisector goes through point B and has angle of $29.7$ degrees. How does one go about find the slope for this? I know once I find the slope I can plug it into the point slope formula to get my equation. I have searched this problem on here but all I found was something that said I had to get the inverse of $tan$ and subtract this from my $29.7$ degree that I previously found but when I do this, I get something that is not even close to the correct answer.

EDIT I found this but I am still getting the wrong answer.

$BC=\sqrt{AB^2+AC^2}$

$BC=\sqrt{(-7)^2+4^2}=\sqrt{65}$

$BCC'$ is an isosceles triangle (by construction) $\implies BC=BC'$.

$C'=(\sqrt{65}-7,0)$

Let $M$ be the midpoint of $CC'$.

$M=\left(\dfrac{\sqrt{65}-7}{2},2\right)$

Bisector $\overleftrightarrow{BM}$

$\left| \begin{array}{} x & y & 1 \\ -7 & 0 & 1 \\ \dfrac{\sqrt{65}-7}{2} & 2 & 1 \\ \end{array} \right| =0$

$4x-(\sqrt{65}+7)y+28=0$

$$ a=\sqrt{49+16}= \sqrt{65},\, b=4,\, c=7\,$$

Using a known formula bisector length =  Product/AM

$$d= (2 \sqrt{65}\, 7)/ (7+\sqrt{65})\approx 7.49367$$

$$ B/2=\cos^{-1} \frac{7}{d}\approx 20.9133 ^{\circ}$$

EDIT1:

Equation of bisector

$$ \frac{x}{-7}+\frac{y}{d}=1$$


Does Riemann integrable imply Lebesgue integrable? Suppose a definite integral exists in the Riemann sense. Does that mean the integral exists as a Lebesgue integral, and do we get the same result either way? ------- BTW: I have a MS in Electrical Engineering and a strong interest in math. I had one semester of real analysis 25 years ago, I tried to learn Lebesgue integration on my own by reading a book on real analysis, and that was a few years ago. Hence, I don't have a solid grasp of the subject.

This is true for "properly" Riemann integrable functions $f: [a,b] \rightarrow \mathbb{R}$, a fact which is established in all standard treatments of the Lebesgue integral.

However, there are improperly Riemann integrable functions $f: [0,\infty) \rightarrow \mathbb{R}$ which are not Lebesgue integrable.  The most standard counterexample has already been discussed on this site: see here.

Answered in the comments: Yes.

If it is improperly Riemann integrable, that means it is rather a limit of Riemann integrals, and it is a limit of Lebesgue integrals in the same manner.  The difference is that for Riemann integrals, that is the only way to define an integral in the "improper" case, whereas for Lebesgue integration there is a definition that in general might converge without having to restrict the domain and then take the limit.   Pete's answer points out why the distinction needs to be made.


If $ab$ is an element of $G$ and $a\ast b=b\ast a$, we say that $a$ and $b$ commute. Suppose 3 and 2 commute. Does $3^{-1}$ and $2^{-1}$ commute I think it will also commute because it is under multiplication.

Yes they are, since (I took $a=3$ and $b=2$ for clearer presentation)

$a^{-1}b^{-1} = (ba)^{-1} = (ab)^{-1} = b^{-1}a^{-1}$

The point here is the formula $(ab)^{-1}=b^{-1}a^{-1}$ (you can check it by hand just applying the definition of inverse and using its unicity) that allows you to apply the hypothesis of commutation on $a$ and $b$.

If $a*b = b*a$ (i.e. $a$ and $b$ commute), then $a*b$ and $b*a$ have the same inverse.  In other words, $(a*b)^{-1} = (b*a)^{-1}$.  But then, by properties of the inverse, this is

$b^{-1}*a^{-1} = (a*b)^{-1} = (b*a)^{-1} = a^{-1}*b^{-1}$

By a similar argument, it can be shown that

$a$ and $b$ commute if and only if $a^{-1}$ and $b^{-1}$ commute.

That being said, without knowing what $G$ is, it is unclear if this is the statement you need or not.


Inequality regarding diameter, maximum order and number of vertices. Suppose I have a connected graph on $n$ vertices with maximum degree $x$. What is the minimum value of the Diameter $D$?

I'm coming late to this discussion, and perhaps you already know the answer, but anywyay, here it is: Let $n$ be the number of vertices of the graph, $D \geq 2$ its diameter, and $\Delta \geq 2$ the maximum degree. If $\Delta=2$, then $D \geq (n-1)/2$. For $\Delta > 2$ we have that $$D \geq \log_{\Delta-1}[(n-1)(\Delta-2)+\Delta] - \log_{\Delta-1}\Delta.$$ This comes by solving for $D$ in the Moore bound (see "Moore graphs and beyond: A survey in the Degree/Diameter Problem", by Mirka Miller and Josef Siran, Elec. J. of Combinatorics, Dynamic Survey 14).

If $G$ is k-regular, then the diameter is $log_{k}(n)$. So with a maximum degree, we get a lower bound as to the diameter using the same approximation: $diam(G) \geq log_{x}(n)$.


Factorial Calculation for Non-Integers? I was playing with numbers on calculator and to my amaze i could see that calculator calculated $(4.5)!$ or any real numbers but factorial is defined for integers how is this done any advanced function. (Note I am grade $11$ student)

In general, $~n!~=~\displaystyle\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx,~$ which for $~n=\dfrac12~$ yields $~\Big(\tfrac12\Big)!~=~\displaystyle\int_0^\infty e^{-x^2}~dx.~$

But the value of the Gaussian integral is known to be $\sqrt\pi~,~$ implying that $~\Big(\tfrac12\Big)!~=~\dfrac{\sqrt\pi}2,~$

since the integrand is even. Now all that's left to do is to repeatedly employ the well-known

factorial property $(n+1)!=(n+1)~n!~$ for $~n+1=4+\dfrac12,~$ and the result follows.

Along with the gamma function, it is much easier to approximate to good accuracy using Stirling's approximation.

It is defined as:

$$n!\approx \sqrt{2\pi n}\left(\frac ne\right)^n$$


If a triangle has 2 sides of equal length, is it isosceles? I know that if the 2 angles of a triangle are the same, it is isosceles. But what if the two sides are the same? Can we conclude that the corresponding angles are the same and it is isosceles?

Yes. Assume $AC=BC$. Then triangles $ABC$ and $BAC$ are congruent (SAS), hence $\angle A=\angle B$.

Yes, you can use the fact that given a triangle A,B,C if $\alpha$ is the opposite angle of BC, $\beta$ the opposite angle of AC and $\gamma$ opposite to AB it holds $$\frac{BC}{\sin {\alpha}}=\frac{AC}{\sin {\beta}}=\frac{AB}{\sin {\gamma}}$$ but i bet there is a simpler proof that doesn't use trigonometry. By the way i thought that having two equal sides was the definition of isosceles; while it's equivalent i've never heard your definition.


Show almost every element of R can be written as an element from a set of positive measure plus a rational. Let $S$ be a subset of $\mathbb{R}$ with strictly positive Lebesgue measure. Prove that almost every (with respect to Lebesgue measure) real number can be written as the sum of an element of $S$ and an element of $\mathbb{Q}$.

So, I remember proving a long time ago that $S-S$ contains a neighborhood of 0, and I thought that it would be possible to use this to prove the problem, but I was unable to do so.
I know that $S + \mathbb{Q}$ is dense in the reals and has infinite measure, but I can't figure out why it must be $\mathbb{R}$ minus a measure zero set.
Thanks in advance :).

Let $S$ be a Lebesgue measurable set on $[a,b]$ with $m(S)>0$. Then
$$
\bigcup_{q \in \mathbb{Q}}(q+S)=\Bbb{R}-N
$$
where $m(N)=0$.

Proof: for any small $\epsilon>0$, there  an interval $I$ such that
$$
m(S\cap I)>(1-\epsilon)m(I)\tag1
$$
For if not, there is a $\epsilon>0$ that for any interval $I$ that
$$
m(S\cap I)\leqslant(1-\epsilon)m(I)
$$
Then for $x\in S$
$$
1_S(x)=\lim_{\delta\to0}\frac{m(S\cap(x-\delta,x+\delta))}{m((x-\delta,x+\delta))}\leqslant 1-\epsilon<1
$$
This means $m(S)=0$ by Lebesgue density theorem, contradicting $m(S)>0$.

Let $B=\bigcup_{q \in \mathbb{Q}}(q+S)$. Clearly $m(B)>m(S)>0$. Suppose $m(B^c)>0$ too. By $(1)$, there are intervals $I,J$ and $m(I)=m(J)$ such that
$$
m(B\cap I)>r_1m(I)\quad\text{and }\quad m(B^c\cap J)>r_2m(J)
$$
where $r_1>2/3,\: r_2>2/3$. Then there exists a $p\in \mathbb{Q}$ that $I+p=J$. So
$$
m(B+p\cap J)=m(B+p\cap I+p)=m(B\cap I)>r_1m(I)=r_1m(J)
$$
But since $B+p\subset B$, there is
$$
m(B\cap J)\geqslant m(B+p\cap J)>r_1m(J)
$$
And since $(B\cap J)\cap (B^c\cap J)=\varnothing$
$$
m(J)=m((B\cap J)\cup (B^c\cap J))=m(B\cap J)+m(B^c\cap J)>(r_1+r_2)m(J)>m(J)
$$
which is a contradiction. So $m(\Bbb{R}-B)=m(B^c)=0$.

Let $R=\mathbb{R}-(S+\mathbb{Q})$. Assume its measure is strictly positive. Then there are $a,b\in\mathbb{R}$ with $a<b$ such that $(a,b)\subset{}R$. Since the measure of $S$ is strictly positive as well, the same is true for $S$, so $(c,d)\subset{}S$. That clearly contradicts $R=\mathbb{R}-(S+\mathbb{Q})$, since it's easy to find a rational $q$ for which $\gamma+q=\alpha$ with $\alpha\in(a,b)$ and $\gamma\in(c,d)$.


Is there another simpler method to solve this elementary school math problem? I am teaching an elementary student. He has a homework as follows.

  There are $16$ students who use either bicycles or tricycles. The total
  number of wheels is $38$. Find the number of students using bicycles.

I have $3$ solutions as follows.

Using a single variable.

Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.

$$
2\times x + 3 \times (16-x) = 38
$$

The solution is $x=10$.

Using $2$ variables.

Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively.
It implies that

\begin{align}
x+y&=16\\
2x+3y&=38
\end{align}

The solution is $x=10$ and $y=6$.

Using multiples

The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$

The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$

The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:

$\hspace{6cm} (32,6)$ but there are $18$ students

$\hspace{6cm} (26,12)$  but there are $17$ students

$\hspace{6cm} (20,18)$ there are $16$ students

$\hspace{6cm} (14, 24)$ there are $15$ students

$\hspace{6cm} (8, 30)$ there are $14$ students

$\hspace{6cm} (2,36)$ there are $13$ students

Thus the correct combination is $10$ bicycles and $6$ tricycles.

My question

Is there any other simpler method?

If everybody had bicycles, there would be $16 \times 2 = 32$ wheels.  There are actually $38$, so that's $6$ additional wheels.  Each one must be on a different tricycle.
So there are $6$ tricycles, and the other $10$ students have bicycles.

Take variable $x$ for bicycles and $(16-x)$ for tricycles.

Now, consider the equation:

$$ 2x+3(16-x)=38$$

On solving this, we get $x=10.$ So, $10$ bicycles and 6 tricycles.


How do you compute the number of reflexive relation? Given a set with $n$ elements

I know that there is $2^{n^2}$ relations, because there are $n$ rows and $n$ columns and it is either $1$ or $0$ in each case, but I don't know how to compute the number of reflexive relation. I am very dumb. Can someone help me go through the thought process?

A relation is reflexive if and only if every entry on the main diagonal of its matrix is $1$; that’s the only restriction. Fill in $1$’s on the diagonal, and you can put either $0$ or $1$ freely into every other entry in the matrix and have the matrix of a reflexive relation. Thus, the number of reflexive relations on a set of $n$ elements is $2^m$, where $m$ is the number of entries that are not on the diagonal. There are $n^2$ entries altogether, and $n$ of them are on the diagonal, so how many are not on the diagonal? And then how many reflexive relations are there?

A relation $R$ on $A$ is a subset of $AXA$.
If $A$ is reflexive, each of the $n$ ordered pairs $(a,a)$ belonging to $A$ must be in $R$.
So the remaining $n^2-n$ ordered pairs of the type $(a,b)$ where $a!=b$ may or may not be in R.

So each ordered pair has now 2 choices, to be in $R$ or to not be in $R$.
Hence number of pairs = 2($n^2-n$)