Limit as $x \to \infty$ of $\frac{x^5+x^3+4}{x^4-x^3+1}$ Suppose we have to find the following limit
$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}$

Now, if we work with the De L'Hopital rule with successive differentiations we get $L=+\infty$

But if we work like this instead: $$L=\lim_{x\to\infty}\frac{x^5(1+\frac{1}{x^2}+\frac{4}{x^5})}{x^5(\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^5})}$$ then $L$ does not exist.

What is correct and what is false here? I'm a little confused.

First you should avoid division by 0 when calculating limit, as it will tell nothing about the limit. $\frac{1}{0}$ is not infinity, it is undefined. $$\lim_{x\to 0}{\frac{1}{x}}$$ is different scenario. Here you get limit is infinity. In your example you have to apply L'Hopital's rule $4$ times to get $$\lim_{x\to \infty}{\frac{120x}{24}}=\infty$$

First of all, $\frac 1 0 = \infty$, meaning if $f(x)\to 1$ and $g(x)\to 0$ (and $g(x)$ is positive) then $\frac {f(x)} {g(x)}\to +\infty$. It's not an indeterminate form.

However, even if it were an indeterminate form, that's not a problem. Obtaining an indeterminate form doesn't mean the limit doesn't exist, it just means "the method you used doesn't work, try something else". For example, the constant function $1$ definitely has a limit as $x\to\infty$, but if I do something crazy like rewrite it as $\frac x x$, then I get the indeterminate form $\frac \infty \infty$. Every limit can be made to look like an indeterminate form if you try to calculate it in the right way.


Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$ We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$

Starting from the second expression I get :
$$ f(x) = 3x^2 - (x - 1)^2 \ln(1 - x) - (x + 1)^2 \ln(1 + x) $$
Using power series of $\ln(1 + x)$ and $ \ln(1 - x)$ does not lead to the result wanted.
I do not know how to proceed to get the first expression.

Partial fraction decomposition of the coefficient yields to

$$\frac1{n(n+1)(2n+1)}=\frac1n+\frac1{n+1}-\frac4{2n+1}$$

Therefore we may write the original series as

$$\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=\sum_{n=1}^\infty\left[\frac1n+\frac1{n+1}-\frac4{2n+1}\right]x^{2n+2}$$

The easiest way from hereon is to recognize well-known series representations. To be precise we can further conclude that

\begin{align*}
&(1)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n}=x^2\sum_{n=1}^\infty\frac{x^{2n}}{n}=x^2(-\log(1-x^2))\\
&(2)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)+x^2\\
&(3)&&-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)-x\right]
\end{align*}

Now combining these three series we can easily see that

  $$\therefore~f(x)=\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)$$

Of course, as HAMIDINE SOUMARE pointed out within his answer, we have to check beforehand whether these series converge or not and therefore if we are even allowed to split up the sum like this. However, I will leave this to you.

EDIT

Important to notice is that the given logarithm series starts at $n=0$ and not at $n=1$. Thus, we have to consider the first summand extra. Moreover I realised that I made two contrary mistakes which overall did not affect the solution. To be precise my given equations $(2)$ and $(3)$ both contain an error I which I will correct now. For $(2)$ it should rather be

$$(2)~~~\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)\color{red}{-x^2}$$

Hence the series expansion of $\log(1-x^2)$ is given by

$$\log(1-x^2)=-\sum_{\color{blue}{n=1}}^\infty \frac{x^{2n}}n=-x^2-\sum_{n=2}^\infty \frac{x^{2n}}n\implies \sum_{n=2}^\infty \frac{x^{2n}}n=-\log(1-x^2)-x^2$$

From hereon I could clear my doubts concerning the last sum. The sum should correctly be given by

$$(3)~~~-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)\color{red}{-2x}\right]$$

Rather easy this can justified by exploiting the series expansion of $\log\left(\frac{1+x}{1-x}\right)$, which the same as $2\arctan(x)$, given by

$$\log\left(\frac{1+x}{1-x}\right)=2\sum_{\color{blue}{n=0}}^\infty\frac{x^{2n+1}}{2n+1}=2x+2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\implies 2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}=\log\left(\frac{1+x}{1-x}\right)-2x$$

Now note that by adding up these three sums we obtain a closed-form expression of the original sum, to  be precise

\begin{align*}
f(x)&=\underbrace{[x^2(-\log(1-x^2))]}_{(1)}+\underbrace{[-\log(1-x^2)-x^2]}_{(2)}+\underbrace{\left[-2x\left[\log\left(\frac{1+x}{1-x}\right)-2x\right]\right]}_{(3)}\\
&=[-x^2+4x^2]+[-x^2\log(1-x^2)-\log(1-x^2)]+\left[-2x\log\left(\frac{1+x}{1-x}\right)\right]\\
&=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)
\end{align*}

Which is, in fact, the desired solution this time correctly derived.

Hint: $\frac{1}{n(n+1)(2n+1}=\frac{a}{n}+\frac{b}{n+1}+\frac{c}{2n+1}$. Find $a$, $b$ and $c$. Rewrite your series as sum of such series, justify convergence of those series and compute to get the needed function.

Recall: $\sum_{n=1}^{\infty}{\frac{t^n}{n}}=\log(1-t), \forall t, |t|<1$


Prove the following set of vectors is a subspace Let $\,S\,$ be the set of vectors in $\,\mathbb R^3$ that are of the form $\;\begin{bmatrix} a \\ b  \\ a-2b  \\ \end{bmatrix} \;$ where $a$ and $b$ are real numbers.

Prove $\,S\,$ is a subspace of $\,\mathbb R^3$.

So I know in order to show that a set of vectors is a subspace the $0$ vector must be an element of the set, vector addition must be closed, and scalar multiplication must be closed.

I started off showing the $0$ vector is an element of the set by having:
$\;0\cdot a+0\cdot b+0\cdot \left(a-2b\right)=0\,$ which it does so that part is proven.

Now how do I show the closure?
For addition I wrote:

\begin{align}
a &= c_1a+c_2b+c_3\left(a-2b\right) \\
b &= d_1a+d_2b+d_3\left(a-2b\right)
\end{align}

and then combined $\,a+b\,$ to be $\,\left(c_1+d_1\right)a + \left(c_2+d_2\right)b + \left(c_3+d_3\right)\left(a-2b\right)\,$ and since these are linear combinations of one another it is closed under addition? Is this correct? And how do I show closed under scalar multiplication?

Let $\vec{v}_1, \vec{v}_2$ be two arbitrary vectors of this form:

$$
\vec{v}_1=(a_1,b_1,a_1-2b_1)^T,\\
\vec{v}_2=(a_2,b_2,a_2-2b_2)^T.
$$

What we need to show is that their arbitrary linear combination $x\vec{v}_1+y\vec{v}_1$ is again of such form:
$$
x\vec{v}_1+y\vec{v}_1=(xa_1+ya_2,xb_1+yb_2,x(a_1-2b_1)+y(a_2-2b_2))^T=(xa_1+ya_2,xb_1+yb_2,xa_1+ya_2-2(xb_1+yb_2))^T
$$

You seem to be a little confused. You need to show that $$S=\{\begin{pmatrix} a\\ b\\a-2b \end{pmatrix}:a,b\in \mathbb R\}$$ is a subspace.

The zero vector $\begin{pmatrix} 0\\ 0\\0 \end{pmatrix}$ is in $S$ since you may let $a=b=0$ to obtain it.

Closure law holds as $$\begin{pmatrix} a_1\\ b_1\\a_1-2b_1 \end{pmatrix}+\begin{pmatrix} a_2\\ b_2\\a_2-2b_2 \end{pmatrix}=\begin{pmatrix} a_1+a_2\\ b_1+b_2\\a_1+a_2-2(b_1+b_2) \end{pmatrix}\in S$$ by letting $a=a_1+a_2$ and $b=b_1+b_2$.

Similarly for scalar multiplication you may select $a$ to be $\alpha a$ and $b$ to be $\alpha b$.


Deduction of $\forall y \exists x (x=y)$ Give a deduction of $\forall y \exists x (x=y)$

My thinking is I can prove $\exists x(x=y)$ then use generalization theorem，
which is equivalent to prove $\neg \forall x \neg (x=y)$ but I got stuck proving this...
thanks in advance

Sorry that I did not specify what deduction system that I want to use here, it is the deduction sequence  where ai is either Axiom or original hypothesis or obtained from MP

Hint:

$$\matrix{ y=y \\ \hline \exists x.x=y \\
\hline
\forall y.\exists x.x=y } $$

For any arbitrary value $v$, because we always witnesses $v=v$ by the Law of Identity, therefore ...

Since this is so for any arbitrary value , then we deduce ...

$\blacksquare$.


Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square. Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.

What I have done:

This has to either be done with contradiction or contraposition, I was thinking contradiction more likely.

If $b^2-4ac$ was a perfect square then the polynomial $ax^2+bx+c$  would have some rationals $\frac {p_1}{q_1}, \frac {p_2}{q_2}$ as roots($\frac{p_i}{q_i}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$). Therefore $(q_1x-p_1)(q_2x-p_2)=ax^2+bx+c$.
So $q_1,q_2,p_1,p_2$ are odd integers (since $q_1q_2=a,p_1p_2=c$) and $q_1p_2+q_2p_1=-b\Rightarrow\Leftarrow.$

If we let $a$, $b$, and $c$ be $2x+1$, $2y+1$, and $2z+1$ respectively, We will get a general from of a multiple of $8$ minus $3$.
Since one portion of our from can have an $8$ factored out and the other portion be $4$ multiplied by the product of consecutive integers which is $4$ multiplied by an even number, giving us a multiple of $8$.
So our expression is $-3 \pmod 8$ or $5 \pmod 8$, but note that any number squared is $0, 1, \text { or } 4 \pmod 8$.
This is because any number is $0, 1, 2, 3, \dots \text { or } 7 \pmod 8$. Squaring all those values $\pmod 8$ will satisfy the statement.
And so we are done.


Why can ALL quadratic equations be solved by the quadratic formula? In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.

Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

I would like to prove the Quadratic Formula in a cleaner way.  Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.

Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)

We want to solve the equation
$$ax^2+bx+c=0,$$
where $a \ne 0$.  The usual argument starts by dividing by $a$.  That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.

Instead, multiply both sides by $4a$.  We obtain the equivalent equation
$$4a^2x^2 +4abx+4ac=0.\tag{1}$$
Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
$$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
So our equation can be rewritten as
$$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$
or equivalently
$$(2ax+b)^2=b^2-4ac. \tag{3}$$
Now it's all over. We find that
$$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$
and therefore
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$
No fractions until the very end!

Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest  strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.

Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots.  Ideally, each type of proof should be presented, since each connects to an important family of ideas.  And a twice proved theorem is twice as true.

If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant  is called  completing the square).  The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.

So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B \pm \sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.


Not complete but minimal sufficient statistic Let $X =(X_1,\ldots, X_n), ~ X_i \mathrm{iid} \sim \mathcal N ( \theta, \theta^2), ~ \theta \in \Theta = \mathbb R \setminus \{0\}, ~ T(X)=(\sum_{i=1}^n X_i, \sum_{i=1}^n X_i^2)$.

I figured out that $T(X)$ is sufficient. To show that it's not complete I checked the function $g$ with $g(u,v) = 2u^2 - (n+1)v$ and noticed that $E_{\theta}[g(T(X))] = 0 ~ \forall \theta \in \Theta$ - but why is $P_{\theta}(g(T(X)) = 0) < 1$?

And how to show that $T(X)$ is minimal sufficient? I know that one can choose a subfamiliy $\mathcal P_0 \subset \mathcal P = \{f_{\theta} \mid \theta \in \Theta \}$ where $P$ is a family of densities with the same support and that's enough to show that $T$ is minimal sufficient for the subfamiliy. And if we have such a subfamily, $T^{*}(X) = \left(\frac{f_{\theta_1}(x)}{f_{\theta_0}(x)}, \ldots, \frac{f_{\theta_k}(x)}{f_{\theta_0}(x)}\right)$ is minimal sufficient (for $\Theta = \{\theta_0, \ldots, \theta_k\}$). But how to apply on the present case?

G(x) = (n+1)/2 ∑xi^2  – (∑xi )^2

E(G(x)) = 0 doesn't mean G(x)= 0

hence not complete


Find the limit : $\lim _{x \to 0}x^{\sqrt{x}}=?$ Find the limit :
  $$\lim _{x \to 0}x^{\sqrt{x}}=?$$

My Try :

$$f(x)=x^{\sqrt{x}}$$

$$\ln f(x)= \sqrt{x}\ln x$$

Now what ?

You have added the tag limit-without-lhospital, therefore my answer is according to that.

Let $\sqrt{x}=u$. Thus your limit $$\lim_{x \to 0} \sqrt x \ln x= \lim_{u \to 0}   u \ln u^2= 2 \lim_{u \to 0} u \ln u $$

Thus, all you need to find is $\displaystyle \lim_{u \to 0^+}u\ln(u)$.

Let $u=e^{-t}$ and note that as $u \to 0^+$, we have $t \to \infty$. Hence,
$$L = \lim_{u \to 0} u \ln(u) = \lim_{t \to \infty} -te^{-t} = -\lim_{t \to \infty} \dfrac{t}{e^t}$$
Now recall that $e^t \geq \dfrac{t^2}2$. Hence, we have
$$\lim_{t \to \infty} \dfrac{t}{e^t} \leq \lim_{t \to \infty} \dfrac 2t = 0$$

Thus you get $$\color{red}{\lim_{x \to 0} \sqrt x \ln x= 0}$$

Use L'Hopital:

$$\lim_{x\to0} \sqrt{x}\ln x = \lim_{x\to0} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x\to0} \frac{\frac1x}{-\frac1{2x^{3/2}}} = -2\lim_{x\to 0} \sqrt{x} = 0$$

Now you have

$$\lim_{x\to 0} x^{\sqrt{x}} = \lim_{x\to 0} e^{\sqrt{x}\ln x} = e^0 = 1$$

using the continuity of exponentiation.


Solving an equaiton which includes $log$ as both base and exponent Q:
If
$$9x = x^{\log_3x}$$
then what is $x$ ?

I can't solve it. I have tried to use identities in my book but i think they are useless for this question. I need a hint

Hint: Try taking logarithm to the base $3$ on both sides. You'll get a quadratic in $\log_3x$.

Take $\log_3$ of both sides. Then
$$ \log_3{9} + \log_3{x} = (\log_3{x}) (\log_3{x}), $$
using $\log{ab}=\log{a}+\log{b}$ and $\log{a^b} = b\log{a}$. This is a quadratic equation for $\log_3{x}$. Solve that and exponentiate $3$ with the answer to find $x$.


Prove that if $P(A) = P(B) = \frac23$ , then $P(A|B) ≥ \frac12$ Prove that if $P(A) = P(B) = \dfrac23$ , then $P(A|B) ≥ \dfrac12$.

Well I thought that because $P(A) + P(B)> 0$, then they are independent. So I used $P(A|B)= P(A)$ which I can use due to independence. However I have doubts because to be independent they don't necessarily have to be bigger than $0$ but I see no other reason or way to prove that that given probability is bigger than $1/2$.

Hint: Here are the relations which you should use

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$P(A\cap B)=P(A|B)\cdot P(B)$
$P(A\cup B)\leq 1$

All it is left is to put them together.

We have P(A) = P(B) = 2/3.

Take P(A|B) and P(A|not B).

We know that P(A) = P(B) * P (A|B) + P(not B) * P(A|not B).

2/3 = 2/3 * P(A|B) + 1/3 * P(A|not B)

We have 0 ≤ P (A|not B) ≤ 1. Therefore

2/3 * P(A|B) + 1/3 * 0 ≤ 2/3 ≤ 2/3 * P(A|B) + 1/3 * 1

P(A|B) ≤ 1 ≤ P(A|B) + 0.5

0.5 ≤ P(A|B) ≤ 1.

(Intuitively: In 2/3rds of all cases we have B, in 1/3rd we have not B. If we don't have A in at least half the cases where B, the numbers of cases with A don't add up to 2/3rds, even if we have A in all cases where not B. )


Is there a simple way to prove the Four Colour Theorem? The four colour theorem says that:

Given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colours are required to colour the map so that no two adjacent regions have the same colour.
From http://en.wikipedia.org/wiki/Four_colour_theorem

The theorem has been proved using computers. But I am wondering, is there a more simple proof than using computers to do it? I feel that using brute force to prove something is like the last resort. So, is there a simple proof of the Four Colour Theorem, or is it yet to be proved?

A completely computer-free proof has yet to exist. However there are easy proofs for the five(and up) color theorems.

The four colour theorem was supposedly proved by 'Appel and Haken' by using a computer as a tool. They used the tool to supposedly go through every example possible. However there is no 'proof' that the proof actually happened and exists.  I guess it can be proved by hand but it would take you years.. So the answer is no there is no simple proof of the four colour theorem.


How do I write down a curve with exactly one rational point Let $g\geq 1$. I would like to write down (for all $g$) a smooth projective geometrically connected curve $X$ over $\mathbf{Q}$ of genus $g$ with precisely one rational point.

Is this possible?

For which $g$ is this possible?

I think for $g=1$ this is possible. I just don't know an explicit equation, but I should be able to find it. (We just write down an elliptic curve without torsion of rank zero over $\mathbf{Q}$.)

For $g\geq 2$ things get more complicated for me.

I would really like the curve to be of gonality at least $4$, but I'll think about that later.

The smooth plane projective  curve $E$ defined  over $\mathbb Q$ by the equation $y^2z=x^3+2z^3$ has its point at infinity $[0:1:0]$ as its only rational point: $E(\mathbb Q)=\lbrace [0:1:0]\rbrace $. Indeed:

a)  The torsion group of the curve $y^2z=x^3+az^3 $ is  zero  as soon as $a$ is a sixth-power free integer which is neither a square nor a cube nor equal to $-432$.
(Despite appearences I'm not making this crazy theorem up,  but I am quoting theorem (3.3) of Chapter 1 in Husemöller's Elliptic Curves !).

b) On the other hand the curve $E$ has rank $0$, which means that its group of rational points is  torsion (this is stated in the table following the theorem I just quoted).

The  two results  a) and b) prove the assertion in my introductory sentence.

Harry
  When x is rational, the y coordinates are not rational for the majority of the x values.
   If you look at (x^2 + 5) mod (6 - x) the value is only 0, at x = 5; similarly for the other example (x^2 + 203) mod (114 -x) only as 0 at x = 47 and x = 113.  It is the same for larger numbers(100 of digits).


To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3 Let $G$ be a finite group whose order is not divisible by $3$.
  suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian.

Let$ $G be a finite group of order $n$. As $n$ is not divisible by $3$ ,$3$ does not divide $n$ thus $n$ should be relatively prime to $n$. that is gcd of an $n$ should be $1$.
  $n =  1 ,2 ,4 ,5 ,7 ,8 ,10 ,11, 13 ,14 ,17,...$
 further I know that all groups upto order $5$ are abelian and every group of prime order is cyclic. when it remains to prove the numbers which are greater than $5$ and not prime are abelian.

Am I going the right way?
 please suggest me the proper way to prove this.

First note that given condition says that $ f: G \to G$ defined as $x \to x^3$ is an injective homomorphism of $G$.

Further Note that
$$
\forall a,b \in G: \quad ababab = (ab)^{3} = a^{3} b^{3} = aaabbb.
$$
Hence,
$$
\forall a,b \in G: \quad baba = aabb, \quad \text{or equivalently}, \quad (ba)^{2} = a^{2} b^{2}.
$$
Using this fact, we obtain
\begin{align}
\forall a,b \in G: \quad (ab)^{4} &= [(ab)^{2}]^{2} \\
                                  &= [b^{2} a^{2}]^{2} \\
                                  &= (a^{2})^{2} (b^{2})^{2} \\
                                  &= a^{4} b^{4} \\
                                  &= aaaabbbb.
\end{align}

On the other hand,
\begin{align}
\forall a,b \in G: \quad (ab)^{4} &= abababab \\
                                  &= a (ba)^{3} b \\
                                  &= a b^{3} a^{3} b \\
                                  &= abbbaaab.
\end{align}
Hence, for all $ a,b \in G $, we have $ aaaabbbb = abbbaaab $, which yields
$$
f(ab) = a^{3} b^{3} = b^{3} a^{3} = f(ba).
$$
As $ f $ is injective, we conclude that $ ab = ba $ for all $ a,b \in G $.Hence $G$ is an abelian group

Added: I think it's worth mentioning that  there exist nonabelian group $G$ for which $x \to x^3$ is a group homomorphism.Smallest such example is Heisenberg group of order $27$ which can be thought of as all $3 \times 3$ upper diagonal matrices with $1's$ on the diagonal and other entries in the field of order $3$.As $G$ is  of exponent 3 ( i.e. $x^3=1$ for all $x \in G$) hence $f$ is a homomorphism and $G$ is clearly  non abelian because for example following two matrices don't commute:
$$\left(\begin{array}{ccc}
1 & 1 & 0\\
0 & 1 & 2\\
0 & 0 & 1
\end{array}\right),$$ $$\left(\begin{array}{ccc}
1 & 1 & 0\\
0 & 1 & 1\\
0 & 0 & 1
\end{array}\right)$$

In particular this also shows that the condition that $3$ does not divide order($G$) is necessary.

I have followed the similar method as previous answer with a few changes(without defining the f, thought it would be easier:) )
As $(ab)^3=a^3b^3$ for all $a,b\in G$
We have $$ababab=aaabbb$$
$$\Rightarrow baba=aabb$$
$$\Rightarrow (ba)^2=a^2b^2$$
Consider,
$$(ab)^4=((ab)^2)^2$$
$$=(b^2a^2)^2$$
$$=a^4b^4$$
$$=aaaabbbb$$
Also $$(ab)^4=abababab$$
$$=a(ba)^3b$$
Therefore, we get $aaaabbbb=a(ba)^3b$
$$\Rightarrow (ba)^3=(ab)^3$$
$$\Rightarrow (ab)^{-3}(ba)^3=e$$
Where e is identity in G
$$\Rightarrow [(ab)^{-1}(ba)]^3=e$$
Now for $x=(ab)^{-1}(ba)$ ,$|x|$ divides 3 by which $|x|$ can be 3 or 1.
Now $|x|$ can not be 3 (as by Lagrange's theorem if $|x|$=3 then 3 divides |G| which is not true.)
Thus, $|x|$=1 $$\Rightarrow x=e$$
$$\Rightarrow (ab)^{-1}(ba)=e$$
Multiplying by ab from left,
$$\Rightarrow ba=ab$$ for all $a,b\in G$
Thus, G is abelian


Prove that if $a(a^{pq-p-q+2} -b) $ is not divisible by $pq,$ then $a-b$ is not divisible by $pq$. Let $a$ and $b$ be positive integers and p and q be two distinct primes. Prove that if $a(a^{pq-p-q+2} -b) $ is not divisible by $pq,$ then $a-b$ is not divisible by $pq$.

I was trying to do a proof by contradiction:

Suppose that $pq\mid a-b$. We have that $p\mid a-b$ and $q\mid a-b$ and so:

$a \equiv b \pmod{pq} \\
a \equiv b \pmod{p}
\\a \equiv b \pmod{q}$

I notice that I can refactor it as $a((a^{q-1})^{p}(a^{-q})(a^2) -b))$ or $a((a^{p-1})^{q}(a^{-p})(a^2) - b))$, so possibly Fermat's little theorem could be helpful.

Then I am trying to show that

$a(a^{pq-p-q+2} -b) \equiv a_1 \pmod{p} \\
a(a^{pq-p-q+2} -b) \equiv a_2 \pmod{q}.$

Then I use the Chinese Remainder Theorem to show that $a(a^{pq-p-q+2} -b) \equiv 0 \pmod{pq}$

but I have not had any success with doing this, any help is appreciated.

Contrapositive is: $\bmod m\!=\!pq\!:\,\ a\equiv\color{#c09} b\,\Rightarrow\, 0\equiv a(a^{\large \phi(m)+1}\!-\color{#c00}b)\equiv \color{#0a0}{a^{\large 2}}(a^{\large \phi(m)}\!-1),\, $ which is true more generally  for all integers $m$ whose prime factors occur to power at most $\rm\color{#0a0}{two},$ by

Theorem $ $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{\large e_{1}}\cdots\:p_k^{\large e_k}\ $ and suppose also that for all $\,i,\,$  $\ \color{#0a0}{e_i\le e}\ $ and $\ \phi(p_i^{\large e_{i}})\mid f.\ $ Then $\ m\mid \color{#0a0}{a^{\large e}}(a^{\large f}-1)\ $ for all $\: a\in \mathbb Z.$

Proof $\ $ Notice that if $\ p_i\mid a\ $ then $\:p_i^{\large e_{i}}\mid \color{#0a0}{a^{\large e}}\ $ by $\ \color{#0a0}{e_i \le e}.\: $ Else $\:a\:$ is coprime to $\: p_i\:$ so by Euler's phi theorem, $\!\bmod q = p_i^{\large e_{i}}:\, \ a^{\large \phi(q)}\equiv 1 \Rightarrow\ a^{\large f}\equiv 1\, $ by $\: \phi(q)\mid f\, $ and  modular order reduction. Thus since all prime powers $\ p_i^{\large e_{i}}$ divide $\,  a^{\large e} (a^{\large f} - 1)\ $ so too does their lcm = product = $m$.

Examples $\ $ You can find many illuminating examples in prior questions, e.g. below

$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$

$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$

$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$

$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$

$pq-q-p+2=p(q-1)-q+2=(q-1)(p-1)+1$

If $a=b$ mod $p$ and $a=b$ mod $q$

Little fermat applied to $p$ implies that $a^{pq-q-p+2}=a^{(p-1)(q-1)+1}=a^{q-1+1}=a$ $mod$ $p$, this implies that $a^{pq-p-q+2}-b=a^{pq-p-q+2}-a=0$ $mod$ $p$,

a similar argument shows that $a^{pq-p-q+2}-b=0$ $mod$ $q$.


Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$ I'm trying to prove that :

$$\frac{100!}{50!\cdot2^{50}}$$

is an integer .

For the moment I did the following :

$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$

But it still doesn't quite work out .

Hints anyone ?

Thanks

for the number of multiples of $2$:

$$100 = 52 + 2(n_1 - 1)$$
$$n_1 = 25$$

for the number of multiples of $4$:

$$100 = 52 + 4(n_2 - 1)$$
$$n_2 = 13$$

for the number of multiples of $8$:

$$96 = 56 + 8(n_3 - 1)$$
$$n_3 = 6$$

for the number of multiples of $16$:

$$96 = 64 + 16(n_4 - 1)$$
$$n_4 = 3$$

for the number of multiples of $32$:

$$96 = 64 + 32(n_5 - 1)$$
$$n_5 = 2$$

for the number of multiples of $64$:

$$64 = 64 + 64(n_6 - 1)$$
$$n_6 = 1$$

To get the number of $2$'s as factors, add all the $n$'s up, which yields $50$.

This should cancel the $2^{50}$ at the denominator, and thus prove that the expression is an integer.


Find the remainder when $10^{400}$ is divided by 199? I am trying to solve a problem

Find the remainder when the $10^{400}$ is divided  by 199?

I tried it by breaking $10^{400}$ to $1000^{133}*10$ .

And when 1000 is divided by 199 remainder is 5.

So finally we have to find a remainder of :

$5^{133}*10$

But from here I could not find anything so that it can be reduced to smaller numbers.

How can I achieve this?

Is there is any special defined way to solve this type of problem where denominator is a big prime number?

Thanks in advance.

You can use Fermat's little theorem. It states that if $n$ is prime then $a^n$ has the same remainder as $a$ when divided by $n$.

So, $10^{400} = 10^2 (10^{199})^2$. Since $10^{199}$ has remainder $10$ when divided by $199$, the remainder is therefore the same as the remainder of $10^4$ when divided by $199$. $10^4 = 10000 = 50*199 + 50$, so the remainder is $50$.

Since 199 is prime and gcd(10,199) = 1

So, $10^{198} \equiv 1 (mod 199)$

Squaring the both side: $10^{396} \equiv 1 (mod 199)$

Now: $10^3 \equiv 5(mod199)$

$10^{4} \equiv 50 (mod 199)$

$10^{400} \equiv 50 (mod 199)$

So, the remainder is 50. This method is known as Euler's Totient


If $a < b$ and $b = \infty$, then $a < \infty$? A very simple question, but I am not sure for this moment.

I have a strict inequality $a < b$. And I prove that $b = \infty$, say $b$ is an integral. Does this prove that $a < \infty$, that is, $a$ is finite? Question seems simple but things can easily be messed up with infinity.

To be more clear, say $a = \int f dx$ and $b = \int g dx$ and I have $a < b$. If $\int g dx = \infty$, can we say that $\int f dx < \infty$?

If one can construct pathological counterexamples involve limits etc, I will be very happy.

Talking about $\int fdx$ makes no sense, as that is a concrete function, and not a number.

You might want to talk about the definite integral, $\int_0^\infty fdx$ and $\int_0^\infty gdx$. You may also want to discern between showing $\int_0^\infty fdx<\int_0^\infty gdx$ (in which case your claim would be correct), and the case where $f(x)<g(x)$ for all $x$ in the domain over which you are integrating.

In the latter case take $f(x)=1$ and $g(x)=2$.

This assumption is only correct if $ (a,b) \in \mathbb{R}\times\mathbb{R} $.

With $\infty$ in the game, this is no more correct. The concept of $\infty$ has some interesting properties among which is a lot of fluidness.

Actually: $\infty \lt \infty$, $\infty /\infty \ne 1$ and bunch of other crazy things are possible with $\infty$.


$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :

  $$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$

My Try :

$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$

Now what do I do ?

Let $x=2y$

$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$

$$=\dfrac12\lim_{y\to0}\left(\dfrac1{\sin^2y}-\dfrac1{y^2}\right)$$

$$=\dfrac12\lim_{y\to0}\left(1+\dfrac{\sin y}y\right)\lim_{y\to0}\left(\dfrac{y-\sin y}{y^3}\right)\left(\lim_{y\to0}\dfrac y{\sin y}\right)^2$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion for the middle limit

$$\lim_{x \to 0}\frac{1}{1-(1-\frac{x^2}{2!}+O(x^4))} - \frac{2}{x^2} = \frac{2}{x^2}-\frac{2}{x^2} = 0$$(higher order terms neglected as the become very small as $x  \to0$)


How to prove non-regularity of a language from the non-regularity of another language? How can I prove that $L_1=\{a^nb^m\mid n\ne m\}$ is not regular based on the fact that the language $L_2=\{a^nb^n\mid n\in\Bbb N\}$ is not regular?

Thank you

HINT: If $L$ is a regular language over the alphabet $\Sigma$, then $\Sigma^*\setminus L$, the complementary language consisting of words in $\Sigma^*$ that are not in $L$, is also regular. (This is quite easily proved using DFAs.)

Here is some facts:

1)A language L is regular if and only if L-complement is regular.

2)Intersection of two regular languages is regular.

In your question, L1-complement is L2 union all string containing the sub-string 'ba'. L2= L1-complement intersect with 'a* b*'. We know that 'a* b*' is regular(Because we gave an explicit regular expression for it) . Therefore, L1 is regular implies L2 is regular by the above two facts contradiction. Thus, L1 must not have been regular.


How to prove that the limit of a function is $0$ at every point I have the following function defined on $\mathbb{R}$:

$$f(x) = \begin{cases}
    0 & \text{if $x$ irrational} \\
    1/n & \text{if $x = m/n$ where $m, n$ coprime}
  \end{cases}$$

I want to show that $f$ is continuous at every irrational point, and has a simple discontinuity at every rational point. I was able to show the first and partially the second (I showed that $f$ has a discontinuity at every rational point, but got stuck on showing that the discontinuity is simple).

However, I've realized that I can simply show that $\lim_{t \rightarrow x}f(t) = 0$ for every $x$, and both of the things I want to show follow from this. How can I show this?

Take any $x$. Let $\epsilon > 0$, and find $N$ such that $\frac1 N < \epsilon$.

Now, I claim that there is a $\delta>0$ such that if $y \in (x-\delta,x+\delta) \backslash \{ x\}$, then $y$ is not of the form $\frac mN$ for any integer $m$. Suppose not. Then, considering $\delta$ going to zero and repeatedly contradicting the statement, we get a sequence $y_n = \frac {m_n}N$ such that $y_n \to x$. Now, $m_n \to Nx$. However, remember that $m_n$ are integers, and convergence can only happen in the integers if the sequence is eventually constant! (because different integers are at least $1$ distance from each other). So, $y_n$ is eventually constant. You can work it out from here.

With this lemma, surely the proof is not too far away, is it?

HINTS: observe that there are a finite number of rationals $n/m<1$ for any fixed $m$ (observe that Im not taking into account if $n$ and $m$ are coprime or not).

Now remember that exists infinite rationals of the kind $n/p<1$ where $p$ is prime.

What happen with $m$ when you approximate a number $x$ with a rational $n/m$?


Undoing anonymous donations All the students in a class are planning to do a trip. Not all of the students can afford it, and it is considered shameful to reveal their poverty. So it is suggested that anyone can donate anonymously to a fund. If the fund becomes big enough to cover the trip, the trip happens. If not, the donators gets their money back, preserving anonymity of who donated and who didn't.

Is this possible?

Sure. Create a computer system where students can enter a password of their choice and then donate as much money as they choose. If there is not enough total money at the end, students simply re-enter their passwords and the machine gives back that much money.

Yes it is possible if the paiement is done via Paypal / Online paiement.

Bank transactions are validated and done only if the fund limit is reached. So, while the fund is not reached, nobody loose money and everybody stay anonymous.


Proving Riemann integrability using sequences of Riemann sums I am trying to prove the following:

Suppose $ f:[a,b]\rightarrow\mathbb{R} $ is bounded.  Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions $\{P_n\}$ with $\{\mu(P_n)\}\rightarrow0$, the sequence $\{S(P_n,f)\}$ is convergent

,where $\mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.

My attempt at a solution:

Suppose  for each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$  $ \{S(P_n,f)\}$ converges.

Let $\epsilon>0$ be given.  Then there is an $A\in\mathbb{R}$ and $N\in\mathbb{N}$ such that when $n>N$, there exists $\delta$ such that $\mu(P_n)<\delta\implies |S(P_n,f)-A|<\epsilon$

Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.

Now suppose $f$ is integrable.  Then given $\epsilon>0$, there exists $A\in\mathbb{R}$ such that there exists $\delta$ for which $\mu(P)<\delta\implies |S(P,f)-A|<\epsilon, \forall P$.

Then for each sequence each sequence of marked partitions $\{P\}$ with $\{\mu(P_n)\}\rightarrow0,$ $\mu(P_n)<\delta$.

Then, $|S(P_n,f)-A|<\epsilon$ which means that $\{S(P_n,f)\}$ converges to A.  Also by the theorem below, $A=\int f dt$

Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.

And by the way, you need to define what convergence as $\{\mu(P_n)\}\to 0$ and what marked partition mean.


Composition series and chief series of $p$-group composition series and chief series of $p$-group. How to solve the following Problem? Thanks.

Let $G$ be a group of order $p^n$, $p$ prime. Prove every chief factor and every composition factor is of order $p$.

Is every composition series a chief series of $G$?

Hints:

Let $\,G\;,\;\;|G|=p^n\;,\; p\;$ a prime,  be a group, then use that $\,|Z(G)|>1\;$ and a little induction to show that for any $\,0\le k\le n\;,\;\;G\;$ has a normal subgroup of order $\;p^k\;$ . This already answers (almost...) questions (1)-(2) .

As for the last question: take the dihedral group $\;G:=\{s,t\;;\;s^2=t^4=1\;,\;sts=t^3\}\;$ of order 8, and check the series

$$1\le\langle t^2\rangle\le\langle t\rangle\le G\;\ldots$$

Hint for the first question: If $G$ is a finite $p$-group and $G\neq1$, then $G$ contains a normal subgroup of order $p$.

Hint for the second question: find a counterexample.


Evaluate $\sqrt{2x} \cdot \sqrt x$ So I have this problem: $\sqrt{2x} \cdot \sqrt x =\ldots$

I already have the answer which is $x\sqrt2$, but I just can't understand it. Someone that could help?

$$ \sqrt{2x}\sqrt{x}=\sqrt{2}\sqrt{x}\sqrt{x}=\sqrt{2}(\sqrt{x})^2=\sqrt{2}x=x\sqrt{2} $$

since the square and square-root cancel each other out.

you need to convince yourself that (for non-negative reals, say)
$$ \sqrt{a}\sqrt{b} = \sqrt {ab}
$$
note that this is a special case of the fact that for any nonzero real $\rho$ the map $x \to x^{\rho}$ is an automorphism of the abelian multiplicative group structure on $\mathbb{R^+}$


Calculating the matrix corresponding to linear map How does one go about converting a linear map in functional form to a matrix; for instance:

  For a fixed unit vector $\hat{n} \in \mathbb{R}^{3}$, define the map $f:\mathbb{R}^{3}\to\mathbb{R}^{3}$ by: $$f(\vec{v})=\vec{v}-2(\hat{n}\cdot\vec{v})\hat{n}$$ Work out the matrix $\mathbf{A}$ describing $f$ relative to the basis $\{\hat{\mathrm{i}},\hat{\mathrm{j}},\hat{\mathrm{k}}\}$ and show that $\mathbf{A}^{2}=\vec{1}$

The question then goes on to ask how to work out a matrix describing $f$ relative to a different basis set $\{\hat{u}_{1},\hat{u}_{2},\hat{n}\}$ and I'm not sure how to approach these problems.

Recall what it means to find the matrix $A$ of a linear transformation $T$ (with respect to a given basis $B$): it means to find $A$ such that $T([v]_B)=A[v]_B$. (Here, $[v]_B$ means the coordinates of $v$ with respect to the basis $B$.)

In other words, $A$ is the matrix representation of $T$ if the "action" of $T$ on an input vector in $B$-coordinates is just multiplying that input vector on the left by the matrix $A$. Looking at this example might help.

You have the linear transformation $f:\mathbb{R}^3\to\mathbb{R}^3$, $f(\mathbf{v})=\mathbf{v}-2(\mathbf{n}\cdot \mathbf{v})\mathbf{n}$ and you are working with the standard basis on $\mathbb{R}^3$. Let $\mathbf{v}=(a,b,c)$ and $\mathbf{n}=(n_1,n_2,n_3)$. Then
\begin{align}
f(\mathbf{v})&=\mathbf{v}-2(\mathbf{n}\cdot \mathbf{v})\mathbf{n}\\
f((a,b,c))&=(a,b,c)-2\left((n_1,n_2,n_3)\cdot(a,b,c)\right)(n_1,n_2,n_3)\\
&=\begin{bmatrix} a-2an_1^2-2bn_1n_2-2cn_1n_3\\
b-2an_1n_2-2bn_2^2-2cn_2n_3\\
c-2an_1n_3-2bn_2n_3-2cn_3^2
\end{bmatrix}\\
&=\underbrace{\begin{bmatrix} 1-2n_1^2 & -2n_1n_2 & -2n_1n_3\\
-2an_1n_2 & 1-2n_2^2 & -2n_2n_3\\
-2n_1n_3 & -2n_2n_3 & 1-2n_3^2
\end{bmatrix}}_A
\begin{bmatrix}
a\\ b\\ c\end{bmatrix}.
\end{align}

In your case $n=3$, $v_1:=\hat{i}$, $v_2:=\hat{j}$, $v_3:=\hat{k}$, $T:=f$.

Step 1: Evaluate each element of the basis $\{v_1,\ldots,v_n\}$ in the transformation.

Step 2: expand each of these values in the basis $$T(v_i)=\alpha_{1i}v_1+\ldots\alpha_{ni}v_n.$$

Step 3: Put these coefficients as columns of a matrix, in the same order as the basis $\{v_1,v_2,\ldots,v_n\}$. To get $$\begin{bmatrix}\alpha_{11}&\alpha_{12}&\ldots&\alpha_{1n}\\\alpha_{21}&\alpha_{22}&\ldots&\alpha_{2n}\\\vdots&\vdots&\vdots&\vdots\\\alpha_{n1}&\alpha_{n2}&\ldots&\alpha_{nn}\end{bmatrix}$$

Doing the first column:

$f(\hat{i})=\hat{i}-2(\hat{n}\cdot \hat{i})\hat{n}=\hat{i}-2\hat{n}(1)\hat{n}=(1-2\hat{n}(1)^2)\hat{i}-2\hat{n}(1)\hat{n}(2)\hat{j}-2\hat{n}(1)\hat{n}(2)\hat{k}$.

So, the first column of the matrix is $\begin{bmatrix}1-2\hat{n}(1)^2\\-2\hat{n}(1)\hat{n}(2)\\-2\hat{n}(1)\hat{n}(2)\end{bmatrix}$,

where $\hat{n}(1)$, $\hat{n}(2)$, $\hat{n}(3)$, are the components of $\hat{n}$ in the basis $\hat{i}$, $\hat{j}$, $\hat{k}$, which I am assuming to be the standard basis of $\mathbb{R}^3$.


Why are fractions the same as divisions? Ever since I learned about fractions in Elementary School, I've known how to work with them. The problem is, although I remember the mathematical rules, I don't remember how I assimilated the equality between fractions and divisions when I was younger. This is really troubling to my mind because it feels like I partly understand this concept. I know that the denominator represents the amount of parts a whole was divided into and that the numerator represents the amount of parts I'm working with. Let's say, I want to divide 3 by 4. The result of this division is 3/4. But, in this case, I'm dividing 3 wholes by 4. Doesn't this come into conflict with the definition of the denominator, which is the amount of parts in which 1 whole is divided into?
  This confusion is really bugging me and I'd really appreciate if someone could clear it up. Thanks for the attention.

You are right that there are two things going on here, and it doesn't seem obvious that they are the same.

On the one hand you have $3$ units and you take a fourth of that (that's dividing $3$ by $4$).
On the other hand you have a unit, cut it in four parts and keep only three of these parts (that's the fraction "three fourths").

The reason why they are the same is that you can achieve the first operation by cutting each of your three units into four parts, and take one part from each. By doing that, you took a fourth of your three units, and at the same time what you have in your hands is three "fourths of a unit".

Three over four is literally another way of writing "thre divided by four", which is 0.75.

When you see three over four "on its own" that means "0.75 'of one' ", which is of course 0.75.

Three over four, times four is of course three.  Three over four times a hundred is of course 75.  Three over four times a New York pizza is of course six slices of pizza. Three over four of your income is what the tax man takes. And so on.

If you see three over four "on it's own", it's just a convention that it means "times one".

  " I'm dividing 3 wholes by 4. Doesn't this come into conflict with the definition of the denominator, which is the amount of parts in which 1 whole is divided into?..."

Where you wrote "1 whole" just above - near the end - you meant "the whole thing"...

That is to say "the whole thing in question"...

This is completely non-mysterious: in English it's normal to say "one whole thing" where the "one whole thing" might be "a swimming pool", "the USA", "1", "88", "the color blue," or whatever.

(The English phrase "one whole thing" happens to have a the word "one" in it.  Much as one may say, "One must carefully brush one's hair before meeting one's grandmother." It has no connection to "1.0".)

The "whole thing" here is indeed ..... "3".  That's all there is to it.

Just as you say, the denominator ("4.0" here) is what you divide in to the "whole thing".  Here the "whole thing" is 3.

That's all you're seeing there!


Convergence of $\sum_{n=1}^\infty \frac{1}{n}-\ln(1+ \frac{1}{n})$ I'm having trouble to prove/disprove the Series $$\sum_{n=1}^\infty\left( \frac{1}{n}-\ln(1+ \frac{1}{n})\right)$$ being convergent or divergent.

I tried the ratio test but it seems to be inconclusive. I think I should use here some reference for the comparison test but I couldn't find any matching Series.

Also tried to make a common denominator and received $\sum_{n=1}^\infty \frac{1-n\ln(1+ \frac{1}{n})}{n}$ so I have to find something smaller then $n\ln(1+ \frac{1}{n})$, and I couldn't figure out the matching alternative.

Would appreciate your advice

Note that $\ln (1+x)=x-\frac12x^2+O(x^3)$. Hence $\frac1n-\ln(1+\frac 1n)=\frac1{2n^2}+O(n^{-3})$ and so for $n$ large enough we have $0<\frac1n-\ln(1+\frac 1n)<\frac1{n^2}$.

In case you are "afraid" of the big-O or did not learn about Taylor expansion yet, it is sufficient to show the following:
$$ x-x^2\le \ln(1+x)\le x\qquad\text{for }x\ge0.$$
The right hand side is an immediate consequence of $e^x\ge 1+x$ (the mother of all inequalities for the exponential). The left hand side follows from
$e^{x^2-x}\ge 1+x^2-x>0$ so that
$$e^{x-x^2}=\frac1{e^{x^2-x}}\le \frac1{1-x+x^2} =\frac{1+x}{1+x^3}\le 1+x$$

Just use the limit comparison test .
Use the series $\sum_{n=1}^\infty(1/n)$.
The limit comes out to be 1 and hence both the series converge or diverge together.


Average number of $5$-card draws before all $52$ cards in a deck are drawn. So an interesting question was brought to me today and I'm not sure how to formulate the equation to answer it. A person draws $5$ cards from a deck, writes the cards down, puts the cards back in the deck and reshuffles. What is the average number of draws they have to perform before all $52$ cards are drawn?

What about before before $45$ out of the $52$ cards are drawn?

Finally, how many draws if red cards were twice as likely to be drawn as black cards?

If you could show the equation for each it would be much appreciated.

This sounds like the coupon collector problem.
https://en.wikipedia.org/wiki/Coupon_collector%27s_problem

While the 5 card draw does make it easier, for a single card hand it will take 236 draws on average to collect all 52 cards.

So as an upper bound for the 5 card version I would say 48 hands.


Very simple predicate logic deduction question I am very new to logic and currently taking a course about it but unfortunately it's a weekend now so I can't get the answers I need!

Basically I am wondering a very basic thing. I want to prove something with natural deduction and let's say I have this premise:

Ok so let's start the question, here's an example but not a full example, just a short bit:

$$\forall x\forall y\big(A(x)\to B(y)\big)\qquad\text{(premise)}$$

So we got these two variables $x$ and $y$ and then just get rid of the quantifiers and replace the variables with two arbitrary constants just like the rule says:

$$A(c)\to B(d)$$

Okay now for the question... Let's also say that I have another premise, or perhaps just an assumption even, that says:

$$A(d)$$

Would it be usable with $A(c)$? Like, could I use the modus ponens rule like this:

$$A(d)\quad A(c) \to B(d)$$

The two constants $d$ and $c$ are different. I'mma guess the answer to this question is in fact "no" but it's something that keeps bothering me (because I always go like "Hmm but I can do this to appl--Oh... Guess not...)" and I just want to 100% make sure it's not possible since I am absolutely horrible at this subject!

In natural deduction, you may deduce $P(t)$ from $\forall x\: P(x)$ for arbitrary terms $t$.

Thus, in your case, you can deduce $A(d) \rightarrow B(d)$ from $\forall x\forall y\: (A(x)\rightarrow B(y))$, and then use the premise $A(d)$ to deduce $B(d)$.

You can generalize that by instead deducing $A(d) \rightarrow B(t)$ from $\forall x\forall y\: (A(x)\rightarrow B(y))$ ($t$ is again some arbitrary term), and then use $A(d)$ to deduce $B(t)$.

Couldn't you just get rid of the quantifiers with $A(d) \to B(d)$ as well?  I mean, it's true for ALL $x$ and $y$... but presumably $A(c)$ and $A(d)$ are different, so you couldn't just use what you wrote.  Luckily your initial premise is very, very flexible!


Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?

Since every term but the last is divisible by $a$, so we only need to deal with $(2x)^n \bmod \left(x^2-x+1\right)$. Let's start from small ones, and try to find patterns:
$$
\begin{aligned}
((2 x)^1 &\bmod \left(x^2-x+1\right))=\color{gray}{2x} \\
((2 x)^2&\bmod \left(x^2-x+1\right)) =\color{red}{-4+4 x} \\
((2 x)^3 &\bmod \left(x^2-x+1\right)) = \color{blue}{-8}\\
((2 x)^4 &\bmod \left(x^2-x+1\right)) = \color{green}{-16x}\\
((2 x)^5 &\bmod \left(x^2-x+1\right)) = 32-32x\\
((2 x)^6 &\bmod \left(x^2-x+1\right)) = 64\\
((2 x)^7 &\bmod \left(x^2-x+1\right)) =\color{gray}{128x}\\
((2 x)^8 &\bmod \left(x^2-x+1\right)) = \color{red}{-256+256x}\\
((2 x)^9 &\bmod \left(x^2-x+1\right)) = \color{blue}{-512} \\
((2 x)^{10} &\bmod \left(x^2-x+1\right)) =\color{green}{-1024 x}
\end{aligned}
$$
What do you find? Hint: Divide into groups by $n \bmod 6$. Prove by induction separately.

Great job so far! Notice that all terms of the form $\binom{n}{i}a^{n-i}$ are divisible by $a$ if and only if $i<n$. This means that the remainder is simply $(2x)^n$.


A person went to market to buy 1.5 kg of dried peas having 20% water content. He went home and soaked them for some time and the water ... A person went to market to buy 1.5 kg of dried peas having 20% water content. He went home and soaked them for some time and the water content in the peas become 60%. Find the final weight of the soaked peas .

My approach :

Since 1.5 kg contain 20% water , it means 1.5 x 20/100 = 300gm of water content.

Now I am not getting idea how to use this or may be i m wrong here.. please guide will be of great help. Thanks.

1.5 kg leads to 300 grams of water being 20%. So, we have 1.2 kg of dried material.
By soaking we get this 1.2 kg of dried material to equate to 40% of our total mass ( 100% - 60% water weight). $$1.2 \text{kg} /40\% =3 \text{kg}$$ So, that means we have 3 kg of mass total.

This of course assumes the percent water content is by mass and not by volume.

Direct Proportionality implies that increasing mass $m$ is linked to increasing water content $w$, $w \sim m$ or $w = c ~ \cdot ~m$, where $c$ is a constant (direct proportionality factor).
We are given $$(w_1;m_1)=(0.2;1.5)$$
From what's given, we can find $$c = w_1/m_1 = 2/15$$
If we like to find $m_2$ in $(0.6;m_2)$, then we use $c$ and the equation $ w = c ~ \cdot ~m$ to get
$$m_2 = w_2/c=0.6/(2/15)=4.5 ~\text{kg}.$$
EDIT:
Apparaently it's not that easy, because $4.5 ~\text{kg}$ is too much weight? So, we forget about direct proportionality, it's of no help in this case...
Let's check the chemistry and mechanics of the problem. The water content has many different definitions and it can be defined as the gravimetric water content (, which also has many types of sub-definitions depending on the application and on the point of time of investigation of the water content) $$u'=m_{\text{water}}/m_{\text{wet peas}}.$$
If $m_{\text{wet peas}}=1.5 ~\text{kg}$ of dried peas have $u'=20~\%$ water content, then we use above formula: $$m_{\text{water}}=u' \cdot m_{\text{wet peas}} = 20~\% \cdot 1.5~\text{kg}= 0.3~\text{kg}.$$
Now the water mass is just the difference of the masses of the wet peas and the dried peas:
$$m_{\text{water}}=m_{\text{wet peas}}-m_{\text{dried peas}}.$$
Solve above equation for $m_{\text{dried peas}}$ and plug in the numbers for $m_{\text{wet peas}}$ and $m_{\text{water}}$: $$m_{\text{dried peas}} = m_{\text{wet peas}} - m_{\text{water}}= 1.5~\text{kg}-0.3~\text{kg}=1.2~\text{kg}.$$
If now the water content becomes $u'=60\%$, then we still have $m_{\text{dried peas}} =1.2~\text{kg}$, but:
$$ m_{\text{wet peas}} = m_{\text{dried peas}}/(1-u') =\color{red}{3~\text{kg}} .$$
This result is achieved using the definition of the gravimetric water content before drying, which is: $$ u'=m_{\text{water}}/m_{\text{wet peas}}=(m_{\text{wet peas}}-m_{\text{dried peas}})/m_{\text{wet peas}}.$$
EDIT
The easiest and quickest method to solve this problem is by embracing the condition, that the dry mass of the peas before soaking ($=80\% \cdot 1.5~ \text{kg}$) and the dry mass of the peas after soaking ($=40\% \cdot m_{\text{wet peas}}~ \text{kg}$) still didn't change. This condition leads immediately to:
$$80\% \cdot 1.5=40\% \cdot m_{\text{wet peas}} \Rightarrow m_{\text{wet peas}} = (80\% \cdot 1.5)/40\%  = 3~\text{kg}.$$
$$$$


how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$

I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$

By Fermat I got

1)  $x^{5} \equiv 2 \pmod{13}$

2)  $x^{3} \equiv 2 \pmod{11}$

Now I'm stuck..

By the Chinese remainder theorem, you have to solve first
$$\begin{cases}x^{113}\equiv2\mod13\\x^{113}\equiv2\mod11\end{cases}$$
Now Little Fermat says for any $x\not\equiv 0\mod 13\enspace(\text{resp. }11)$, one has $x^{12}\equiv 1\mod 13$, resp. $x^{10}\equiv 1\mod 11$. Hence the system of congruences is equivalent to
$$\begin{cases}x^{113\bmod 12}\equiv x^5\equiv2\mod13,\\x^{113\bmod 10}\equiv x^3\equiv2\mod11.
\end{cases}$$

Let's examine the different possibilities for $x$.

We can eliminate the values $0$ and $1$. Hence,

modulo $13$, let's draw a table, using the fast exponentiation algorithm:
$$\begin{matrix}
x& \pm 2&\pm 3&\pm4 &\pm 5 & \pm 6\\
\hline
x^2&4&-4&3&-1&-3\\
x^4&3&3&-4&1&-4\\
\hline
x^5&\pm6&\pm6&\mp3&\pm5&\pm 2
\end{matrix}$$
Thus the solution is $\;\color{red}{x\equiv6\mod 13}$.
modulo $11$, we have
$$\begin{matrix}
x& \pm 2&\pm 3&\pm4 &\pm 5\\
\hline
x^2&4&-2&5&-1\\
\hline
x^3&\mp3&\pm5&\mp 2&\mp5
\end{matrix}$$
Here the solution is $\;\color{red}{x\equiv-4\mod 11}$.

To solve this system of simultaneous congruences, we need a Bézout's relation between $13$ and $11$:
$$6\cdot 11-5\cdot13=1.$$
The solutions of the system of congruences is
$$x\equiv\color{red}{6}\cdot 6\cdot 11-(\color{red}{-4})\cdot5\cdot 13\equiv656\equiv \color{red}{84 \mod 143}.$$

The first congruence tells you $x \equiv 6 \pmod {13}$ and the second tells you $x \equiv 7 \pmod {11}$, now apply the Chinese remainder theorem to get your answer.


Is the empty set a subset of itself? Sorry but I don't think I can know, since it's a definition. Please tell me. I don't think that $0=\emptyset\,$ since I distinguish between empty set and the value $0$. Do all sets, even the empty set, have infitinite emptiness e.g. do all sets including the empty set contain infinitely many empty sets?

There is only one empty set.  It is a subset of every set, including itself.  Each set only includes it once as a subset, not an infinite number of times.

An empty subset can be selected from any set, including the empty set itself, using selection criteria that cannot be satisfied, e.g. selecting those elements $x$ such that $x\neq x$.
See my formal proof (in DC Proof format) at: http://dcproof.com/ExistenseOfNullSet.htm
There is only one empty set. If $\phi_1$ and $\phi_2$ are empty sets, then $\ \phi_1 = \phi_2$.
See my formal proof (in DC Proof format) at: http://dcproof.com/UniquenessOfNullSet.htm

Edit: For any set $X$ we can select an empty subset $S$ such that:
$\forall a:[a\in S\iff a\in X\land a\neq a]$
or
$\forall a:[a\in S\iff a\in X\land a\notin X]$


Prove that the general linear group, GL(V) is a group If V is a vector space and GL(V) is the set of all linear transformations from V to V that are bijections, prove that GL(V) is a group with operation composition.

I am out of practice with algebra, and perhaps this is too abstract for me, but isn't the fact that the linear transformations are bijections, isn't associativity and inverse proved? But to prove the existence of an identity element?

Let $T \in GL(V)$. We wish to show that $S := T^{-1}$ is a linear map.

Proof: Let $w, w' \in V$. Then, since $T$ is a bijection, there exists $v, v' \in V$ such that $w = T(v)$ and $w' = T(v')$. Then, $S(w) * S(w') = S(T(v)) * S(T(v')) = v*v' = S(T(v*v')) = S(T(v)*T(v')) = S(w*w')$.


Why don’t numbers with a difference of 2 have any common factors beside 1 and possibly 2? For example, if x is odd, that means x-1 and x+1 are both even.  And from what I’ve seen, because x+1 and x-1 have a difference of 2, they have no other common factors beside 1 and because they are both even, 2.  If x is even, then x+1 and x-1 are both odd and therefore have only 1 as a common factor.

My questions are:

Is this true in all cases?
If so, why is this true?

Yes, it is true in all cases. Hint: if $m$ is a common factor of $a$ and $b$, then $m$ divides $a-b$.

If $x,y$ are numbers with $d$ as common factor, then $|x-y|$ is also a multiple of $d$. Hence either $x=y$ or $|x-y|\ge d$. Therefore, numbers with $|x-y|=2$ cannot have a common factor $d$ that is $>2$. Two even such numbers certainly have at least and hence exactly $2$ as common factor. Two odd numbers cannot have an even common factor, hence $1$.


$\lim _{x\to 0}\left(\frac{e^x-1}{\:x^3}\right)$ I apply L'Hopital thrice and I get 1/6 but here they've stopped at infinity.

What is the correct answer?

Note that L’Hopital rule is used only for limits of $\frac00$ or $\frac{\infty}{\infty}$ form.

Right now, on substituting $x=0$, we get a $\frac00$ form. But, if we differentiate it once, we get: $$\lim_{x \to 0} \frac{e^x}{3x^2}$$ which is not of $\frac00$ form. So, we cannot use L’Hopital rule here.

Evaluating the above limit, the answer is obviously $\infty$.

I don't know how you used L’Hopital rule but look(assume I'm taking the limit all the time):

$$\frac{e^x-1}{x^3}=\frac{e^x}{3x^2}$$

This is a limit in the form of $\frac1{0}$, which in this case is determinate form, it goes to $+\infty$ because $\lim_{x\to0}x^2$  goes to $0$ from the positive direction

The answer is infinity


Can we treat logic mathematically without using logic? I'm reading Kleene's introduction to logic and in the beginning he mentions something that I have thought about for a while. The question is how can we treat logic mathematically without using logic in the treatment? He mentions that in order to deal with this what we do is that we separate the logic we are studying from the logic we are using to study it (which is the object language and metalanguage, respectively). How does this answer the question? Aren't we still using logic to build logic?

And I have a feeling that the answer is to some extent that we use simpler logics to build more complex ones, but then don't we run into a paradox of what 's the simplest logic, for won't any system of logic no matter how simple be a metalanguage for a simpler language?

We use logic to study logic, not to create logic.  Our study is usually not intended to justify some logic but rather to understand how it works. For example, we might try to prove that, whenever a conclusion $c$ follows from an infinite set $H$ of hypotheses then $c$ already follows from a finite subset of $H$. Many logical systems have this finiteness property; many others do not. And that's quite independent of the logic that we use in studying this property and trying to prove or disprove it for one or another logical system.

Here's an analogy: Suppose a biologist is writing a paper about the origin of trees.  He could use a wooden pencil to write the paper.  That pencil was made using wood from trees, so its existence presupposes that the origin of trees actually happened.  Nevertheless, there is nothing circular here.  The pencil that is being used probably consists of wood quite different from that in prehistoric trees.  And even if it wasn't different, there's no problem with using the pencil to describe those ancient trees.

Similarly, there's no problem using ordinary reasoning, also called logic, to describe and analyze the process of reasoning.

You can use arithmetics to do logic.

true is 1

false is 0

$a \cdot b \Leftrightarrow$ logical and

$1-a \Leftrightarrow$ logical not

$a+b>0 \Leftrightarrow$ logical inclusive or

$a+b=1 \Leftrightarrow$ logical exclusive or


Proving the expected value of the square root of X is less than the square root of the expected value of X How do I show that $E(\sqrt{X}) \leq \sqrt{E(X)}$ for a positive random variable $X$?

I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 \leq E(X^2)E(Y^2)$, but I'm not sure how.

The Cauchy-Schwarz inequality tells us that if $A=\sqrt{X}$ and $B=1$, then
$$E[\sqrt{X}]^2=(E[AB])^2\leq E[A^2]E[B^2]=E[X]$$
so
$$E[\sqrt{X}]\leq \sqrt{E[X]}$$

Let us assume a variable x and its probability p. Then we calculate         (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.


How many squares are in this image? Is there a method to check? In this image I have counted 14 but others say 18.

Is there a method to check exactly?

[I will give two methods.  The SECOND one is the better one.]

Method?  Well, sort of, you can look at each "atom" piece and count how many squares that is the upper right hand corner of.

A)Top right square -> top right square; engulf the the rectangles for a 3 by 3, the whole 4x4 = 3.

B) top middle rectangle -> 2x2 square; 3x3 square = 2.

C) top left square -> top left square = 1.

D) middle side rectangle -> 2x2 square; 3x3 = 2.

E) next square in the (2,2) spot -> 1x1;2x2;3x3 = 3

F) (3,2) spot ->1x1; 2x2 = 2

G) (4,2) spot -> 1x1 = 1

H) (3,2) spot -> 1x1; 2x2 = 2

I) (3,3) spot -> 1x1 = 1

J) (4,1) spot -> 1x1 = 1

K) low middle rectangle = 0

L) (4,4) spot -> 1x1 = 1

So total: 18

=======

This i a simplification of simply figuring out the squares in a complete grid and subtracting the one that rectangles make impossible.

A 4x4 grid will have:  16 1x1 squares; 9 2x2 squares (as there are 3 squares in each of the top 3 rows that can be an upper right hand corner of a 3x3 square), 4 3x3 squares, and 1 4x4 square.

So an n x n grid will have $\sum k^2$ total squares.

In this case 16 + 9 + 4 + 1 = 30.

The first top rectangle eliminates 2 1x1 squares and 1 2x2 square.  So only 27 possible.

The left side rectangle eliminates 2 1x1 squares  and 1 2x2 square. So only 24 possible.

The right side triangle eliminates 2 1x1 squares (1 2x2 that was already eliminated) and a 3x3. So only 21 possible.

The bottom rectangle eliminates 2 1x1 squares, 2 2x2 squares.  So only 18 possible squares.

19  are yall skipping the big one on the outside? Can't be an even number as only one outside square. I counted 19


Inverse Mellin transform of $\Gamma(s)$ $\zeta(s/2)$ How would you find an approximation of the Inverse Mellin transform of $\Gamma(s)$ $\zeta(s/2)$ near $x=0$?

$\mathcal{M}^{-1}\left\{\Gamma(s)\zeta\left(\dfrac{s}{2}\right)\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{\Gamma(s)}{\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{\frac{s}{2}-1}}{e^x-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s-1}\Gamma\left(\dfrac{s}{2}\right)\Gamma\left(\dfrac{s+1}{2}\right)}{\sqrt\pi~\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{s-2}}{e^{x^2}-1}~d(x^2)\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^\frac{s-1}{2}e^{-x}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~d(x^2)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty\dfrac{x^{s-1}}{2^{s-1}}e^{-\frac{x^2}{4}}~d\left(\dfrac{x}{2}\right)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2}{\sqrt\pi}\int_0^\infty x^se^{-\frac{x^2}{4}}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\dfrac{2}{\sqrt\pi}\int_0^\infty\dfrac{xe^{-\frac{x^2}{4t^2}}}{t^2(e^{t^2}-1)}~dt$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/mellin.pdf)

$=-\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2}{4t^2}}}{e^{t^2}-1}~d\left(\dfrac{1}{t}\right)$

$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2t^2}{4}}}{e^\frac{1}{t^2}-1}~dt$

$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2t^2}{4}-\frac{1}{t^2}}}{1-e^{-\frac{1}{t^2}}}~dt$

$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\sum\limits_{n=0}^\infty e^{-\frac{x^2t^2}{4}-\frac{n+1}{t^2}}~dt$

$=2\sum\limits_{n=0}^\infty e^{-x\sqrt{n+1}}$ (according to How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$)

$=2\sum\limits_{n=1}^\infty e^{-x\sqrt n}$


How to find the number of perfect matchings in complete graphs? In wikipedia FKT algorithm is given for planar graphs. Not anything for complete graphs. I need to find the number of perfect matchings in complete graph of six vertices.

It's just the number of ways of partitioning the six vertices into three sets of two vertices each, right? So that's 15; vertex 1 can go with any of the 5 others, then choose one of the 4 remaining, it can go with any of three others, then there are no more choices to make. $5\times3=15$.

Gerry was correct (sort of) in his first statement, saying that it is the number of ways to partition the six vertices into three sets of two. However, the answer of number of perfect matching is not 15, it is 5. In fact, for any even complete graph G, G can be decomposed into n-1 perfect matchings. Try it for n=2,4,6 and you will see the pattern.

Also, you can think of it this way: the number of edges in a complete graph is [(n)(n-1)]/2, and the number of edges per matching is n/2. What do you have left for the number of matchings? n-1.


GCD to LCM of multiple numbers If I know the GCD of 20 numbers, and know the 20 numbers, is there a formula such that I input the 20 numbers and input their GCD, and it outputs their LCM? I know that
$$\frac{\left| a\cdot b\right|}{\gcd(a,b)} = \text{lcm}(a,b).$$
So is it$$\frac{\left| a\cdot b\cdot c\cdot d\cdot e\cdot f\right|}{\gcd(a,b,c,d,e,f)}?$$If not, what is it?

There can be no formula that computes $\text{lcm}(a,b,c)$ using only the values of $abc$ and $\gcd(a,b,c)$ as input: that's because $(a,b,c) = (1,2,2)$ and $(a,b,c) = (1,1,4)$ both have $abc=4$, $\gcd(a,b,c)=1$, but they don't have the same lcm.

However, there is a straightforward generalization of the $2$-variable formula. For instance,
$$\text{lcm}(a,b,c,d) = \frac{abcd}{\gcd(abc,abd,acd,bcd)}.$$

The correct gcd to take is not of the individual terms $a,b,c,d$ but the products of all the complementary terms (which looks the same in the two-variable case).

LCM can be calculated using:
$\text{GCD}(a,b)\cdot \text{LCM}(a,b)=ab$ and
$\text{LCM}(a,b,c,d,...n)=\text{LCM}(\text{LCM}(a,b),c,d,...,n)$

Example
$\text{LCM}(2,4,5,7)=\text{LCM}(4,5,7)=\text{LCM}(20,7)=140$

or more concisely
$$
\begin{align}
2&,4,5,7 \\
4&,5,7 \\
20&,7 \\
140&
\end{align}
$$

http://mathhelpforum.com/peer-math-review/229498-lcm-b-c-abc.html


Area between $y=x^4$ and $y=x$ The problem I'm having some trouble solving is this: calculate the area between $y=x^4$ and $y=x$. The points are $a = 0$ and $b = 1$, but the definite integral is negative. What am I doing wrong here?

$\int_{0}^{1}\int_{x^4}^{x}dy$ would give you your answer.

Or in simpler form

$\int_{0}^{1}(x-x^4)dy$

I would have even attached the graph of the two functions but I don't know how to.

NOTE: In $[0,1]$, $x\geq x^4$

Always draw the graphs of the corresponding functions before solving such questions. Helps a lot. :)


Stronger than Nesbitt inequality For $x,y,z >0$, prove that
  $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \sqrt{\frac94+\frac32 \cdot \frac{(y-z)^2}{xy+yz+zx}}$$

Observation:

This inequality is stronger than the famous Nesbitt's Inequality
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32 $$ for positive $x,y,z$
We have three variables but the symmetry holds only for two variables $y,z$, resulting in a very difficult inequality. Brute force and Largrange Multiplier are too complicated.
The constant $\frac32$ is closed to the best constant. Thus, this inequality is very sharp, simple AM-GM estimation did not work.

Update: As point out by Michael Rozenberg, this inequality is still unsolved

Here's a proof.

Using Lagrange's identity leads to
$$\sum_{cyc}\frac{x}{y+z}=\frac32+\frac12\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \qquad (*)$$

We get this by letting $a = \sqrt{x+y}$, $b = \sqrt{y+z}$, $c = \sqrt{z+x}$,
$d = \frac{1}{\sqrt{x+y}}$, $e = \frac{1}{\sqrt{y+z}}$, $f = \frac{1}{\sqrt{z+x}}$, and by Lagrange's identity,
$$(a^2+b^2+c^2)(d^2+e^2+f^2) = (ad+be+cf)^2 + (ae-bd)^2+(af-cd)^2+(bf-ce)^2$$
which gives
$$2 (x+y+z)(\sum_{cyc}\frac{1}{y+z}) = 9 + \sum_{cyc} \left(\frac{\sqrt{x+y}}{\sqrt{y+z}} -  \frac{\sqrt{y+z}}{\sqrt{x+y}}\right)^2$$
or
$$3 +\sum_{cyc}\frac{x}{y+z} = \frac92 + \frac12\sum_{cyc} \frac{\left({x+y} - (y+z)\right)^2}{({y+z})(x+y)} $$
which is the desired equation $(*)$.

Squaring both sides of the inequality then gives

$$
\tag{1} \frac16 \left[ \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)}  \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$

We will follow two paths for separate cases.

Path 1:
Omitting the square term   it suffices to prove
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant  \frac{(y-z)^2}{xy + yz + xz}$$

Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq  \frac{(y-z)^2}{xy + yz + xz} (x+y) (y+z) (z+x)
$$

Using $(x+y)(y+z)(x+z) = (x+y+z)(xy + yz+ xz) - x y z$
it suffices to show

$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq  {(y-z)^2} (x+y+z)
$$
or
$$
(x-y)^2 (x+y) +  (z-x)^2 (z+x) \geq  (y-z)^2 x
$$

Since
$$
 (y-z)^2 = (y-x + x- z)^2 = (y-x)^2  + (x- z)^2 + 2 (y-x)(x-z)
$$
this translates into

$$
(y-x)^2 y +  (x-z)^2 z \geq  2 x(y-x)(x-z)
$$
For the two cases $y\geq x ; z\geq x $ and $y\leq x  ; z\leq x $ the RHS $\leq 0$ so we are done. For the other two cases, by symmetry, it remains to show the case  $y>  x  ; z <  x $.

Rearranging terms, we can also write
$$
(y-x)^3 - (x-z)^3 +x ((y -  x) + (z-x))^2 \geq 0
$$

This holds true at least for $(y-x)^3 \geq  (x-z)^3$ or $y+z\geq 2 x$.

So the proof is complete other than for the case $y+z < 2x$ and [ $y>  x  ; z <  x $ or  $z>  x  ; y <  x $ ].

Path 2.

For the remaining case  $y+z < 2 x $ and  [$y>  x ; z <  x$ or  $z>  x  ; y <  x $]  we will follow a different path. Again, by symmetry, we must inspect only  $y+z < 2x$ and  $y>  x > z$.

A remark up front: In the following, some high order polynomials of one variable have to be inspected. MATLAB is used, also for plotting behaviours of these polynomials. There is no case in spending effort for further analytical work on polynomials where their  behaviour is obvious. Still, what follows contains some "ugly" parts.

In the squared version (1) of the inequality, we can use a further inequality which has been proved here:
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}$$

So it suffices to prove

$$
\frac16 \left[ \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2} \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$

Some numerical inspection shows immediately that the first term cannot be ommitted.

Clearing some denominators gives

$$
\tag{2} \frac16 \left[ \frac{27}{8}  \right ]^2 \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4} + \\
 (xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} - (y-z)^2 (x+y)(y+z) (z+x) \geqslant 0 \quad
$$

By homogeneity, we set  $y=1+z$.

The condition $y+z < 2x$ then translates into $1+2z < 2x$, hence we further set $x = z + (1 +q)/2$ where $0\leq q \leq 1$ since also $x = z + (1 +q)/2 < y = 1 +z$.

Inserting $y=1+z$ and $x = z + (1 +q)/2$ into (2) is straightforward, the result is lengthy (not displayed here).

Let us start with focussing  on the first term in (2), calling that fraction $F$:
$$
F= \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4}
$$
With the setting $y=1+z$ this  can be simplified to
$$
F  = \frac{(xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4}
$$

Since by the settings $y=1+z$ and $x = z + (1 +q)/2$ both x and y are linear in z,  the numerator of $F$ is of fifth order in $z$, whereas the denominator is of fourth order in $z$. In leading order, the whole term is therefore of first order in $z$ and will therefore rise with $z$ for large enough $z$. This motivates to show that  indeed
 $$F(q,z) \geq F(q,z=0) = 2 \frac{(q+1)^2}{(q+3)^2}$$
for all $z$ and $q$.
Showing  that directly requires a condition

$$
G = (xy + yz + xz) (x+y)(y+z) (z+x) (q+3)^2 - 2 (q+1)^2 (x+y+z)^4  \geq 0
$$

Inserting $y=1+z$ and $x = z + (1 +q)/2$ into $G$, and expanding  the brackets,  gives a very lengthy expression which however contains only positive terms, so the condition is proved immediately:

$$
G = (z(2q^6z + 2q^6 + 22q^5z^2 + 51q^5z + 22q^5 + 80q^4z^3 + 358q^4z^2 + 357q^4z + 100q^4 + 96q^3z^4 + 960q^3z^3 + 1828q^3z^2 + 1206q^3z + 260q^3 + 864q^2z^4 + 3672q^2z^3 + 4788q^2z^2 + 2484q^2z + 450q^2 + 2592qz^4 + 7344qz^3 + 7398qz^2 + 3159qz + 486q + 2592z^4 + 5832z^3 + 4806z^2 + 1701z + 216))/4 \geq 0
$$

Hence it suffices, instead of (2), to prove the following:

$$
\tag{3} \frac16 \left[ \frac{27}{8}  \right ]^2 2 \frac{(q+1)^2}{(q+3)^2} +
 (xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} \\ -  (x+y)(y+z) (z+x) \geqslant 0 \quad
$$

After insertion of $y=1+z$ and $x = z + (1 +q)/2$, the factors $(x-y)^2$ in the cyclic sum will not be functions of $z$.  Hence,
the LHS  is a third order expression in $z$ with leading (in $z^3$ ) term $( 1 + 3 q^2) z^3$, so for large enough $z$ it is rising with $z$. A remarkable   feature of this expression is that for the considered range $0\leq q \leq 1$ it is actually monotonously rising for all $z$. To see this, consider whether there are points with zero slope. The first derivative of the expression with respect to $z$ is

$$
q^4/4 + (7q^3z)/2 + (7q^3)/4 + 9q^2z^2 + 9q^2z + (5q^2)/4 - (7qz)/2 - (7q)/4 + 3z^2 + 3z + 1/2
$$
Equating this to zero gives
$$
z_{1} =
-(2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)\\
z_{2} =
-(-2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)
$$
One now shows that in the range $0\leq q \leq 1$, there are only negative solutions $z_{1,2}$. So we will have no zero slopes for $z\geq 0$. Since the polynomials  are (roots of)  sixth order in q, we investigate the following figures:

So monotonicity (rising with $z$) is established for all $q$.

Hence, to show the inequality it suffices to inspect  (3) at the smallest $z=0$. This gives

$$
((q + 1)(8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243))/(64(q + 3)^3) \geqslant 0 \quad
$$
or

$$
\tag{4} 8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243 \geqslant 0 \quad
$$

An even stronger requirement is
$$
h(q) =
432q^3 - 216q^2 - 405q + 243 \geq 0
$$
In the considered range $0\leq q \leq 1$, $h(q)$ has a minimum which is obtained by taking the first derivative,
$$
1296 q^2 - 432 q - 405
$$
and equating to zero, which gives $q = 3/4$, and the above $h(q)$ then gives
$$
h(q = 3/4) = 0
$$

This establishes the inequality. $ \qquad \Box$

Very long partial answer :
As in my other answer we have using abc method or uvw method the inequality :
Let $a,b,c>0$
$$\sum_{cyc}\frac{a}{b+c}\geq \sqrt{\frac{9}{4}+\frac{7}{4}\frac{\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)}{ab+bc+ca}}$$
See WA factorization for the case wich is sufficient to show I mean ,$b=c=1$ (because it's homogenous).
As partial answer  we have the constraint :
$$\frac{7*2}{4*3}(a^{2}+b^{2}+c^{2}-ab-bc-ca)\geq (a-b)^2$$
Hope it inspires you !
Update 22/04/2021
Let $a,b,c\in[0.5,1]$ then we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
Proof :
The RHS is equivalent to :
$$0.5\frac{(a-b)^2}{ab+bc+ca}\geq 0$$
For the LHS we can use BW easily .
We can go further as we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
With the constraint :
$$(0.5(a+b-2c)(a^2-ab+b^2-c^2))\geq 0$$
Again I go further :
We have
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+\frac{44}{100}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \sqrt{\frac{9}{4}+\frac{132}{100}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}} \geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
With the first constraint :
$$ (28 a^3 - 3 a^2 b - 47 a^2 c - 3 a b^2 + 44 a b c - 25 a c^2 + 28 b^3 - 47 b^2 c - 25 b c^2 + 50 c^3)\geq 0$$
And the second :
$$\frac{132}{100}\frac{2}{3}(a+b+c)(ab+bc+ca)\geq(a+b)(b+c)(c+a)$$
Update 23/04/2021 :
A general way would be :
We have
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+y\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \sqrt{\frac{9}{4}+3y\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}} \geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}\quad(I)$$
With the constraint :
$$-(0.5 (2 a^3 y - 2 a^3 - 2 a^2 b y + a^2 b + 2 a^2 c y + a^2 c - 2 a b^2 y + a b^2 - 4 a b c y + a c^2 + 2 b^3 y - 2 b^3 + 2 b^2 c y + b^2 c + b c^2 - 2 c^3))\geq 0$$
And :$$3y\frac{2}{3}(a+b+c)(ab+bc+ca)\geq(a+b)(b+c)(c+a)$$
So to be true we have :
$$\frac{\left(a\left(a+b\right)\left(a+c\right)+b\left(b+c\right)\left(b+a\right)+c\left(a+c\right)\left(c+b\right)-1.5\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)}{\left(a-b\right)^{2}\left(a+b+c\right)}-\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(a+b+c\right)\left(ab+bc+ca\right)}\geq 0$$
We have to establish an assumption because like this the inequality is false ($a,b,c>0$ is not sufficient)
A temporary and sufficient assumption is $a\in[0.25,0.5)$ and $b,c\in[0.5,1]$
Using the first inequality and using the variable $y$ as in the inequality $I$
We need to show for $a\ge c\ge b$ and $2c-2b\le a$ :
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq 0\quad (C)$$
Update 24/04/2021
We have the inequality for  $1\ge a\geq c\geq b>0$ and $\frac{\left(a-b\right)^{2}}{ab+bc+ca}\le 1$ or $\frac{\left(a-b\right)^{2}}{ab+bc+ca}\ge 1.2$ and $2c-2b\ge a$:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge 1.5+\frac{\left(\sqrt{\frac{15}{4}}-1.5\right)\left(\left|a-b\right|\right)^{1.75}}{\left(ab+bc+ca\right)^{\frac{1.75}{2}}}\geq  \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}} $$
Update 25/04/2021
To prove it we remark :
$$f(c)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5$$
Is increasing with $a\leq 2c-2b$
$$g(c)=\left(ab+bc+ca\right)^{\frac{1.75}{2}}$$
Is increasing .
So the product of two positives increasing functions is also an increasing function always with constraints above .
It remains to show the case $a=2c-2b$ . As it's homogenous we get an inequality with one variable or a long polynomial .The RHS is trivial .
Update 26/04/2021 :
To prove partialy the inequality $(C)$ we have for $a\ge 2c\ge c\ge b$ and $2c-2b\le a$  :
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq \frac{2(a^2+b^2+c^2-ab-bc-ca)}{3}\geq\frac{\left(a-b\right)^{2}}{2}\quad $$
Wich is trivial using the $uvw's$ method and the assumptions on $a,b,c>0$
For the case $2c\geq a \geq c\geq b$ we add the assumptions $c+b\leq a $ and $a\geq 2c-2b$ remarking that :
$$f(c)=\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)$$
Is a decreasing function always with the assumptions above .Remains to show the cases $a=c+b$ or $a=2c-2b$ wich is a one variable inequality since all the equalities/inequalities are homogenous.I shall prove that the function is decreasing later .
To prove that the function is decreasing we differentiate twice we have :
$$f''(c)=-\frac{2a^2b}{(a+c)^3}-\frac{2ab^2}{(b+c)^3 }+2$$
With the assumptions above the function $f(c)$ is convex so the derivative is increasing .Remains to show the inequality at the equality case wich is not too hard i think .
From my first answer we have :
$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Remains to show :
$$P(a,b,c)\geq \sqrt{\frac{9}{4}+\frac{3}{2}\frac{\left(a-b\right)^2}{ab+bc+ca}}$$
With assumptions  $2c\geq a \geq c\geq b$ and $c+b\geq a $ and $a\geq 2c-2b$ .
We have :
Proof : we have the refinement of the inequality above :
$$\frac{9}{1}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}\geq \frac{9}{4}\frac{\left(a-b\right)^4}{(ab+bc+ca)^2}$$
Multiplying by the differents denominators and replacing $a$ by $c+b$ and $a$ by $2c-2b$ . Eliminating a variable because the inequality is homogenous we get a decreasing polynomial for the LHS and an increasing polynomial for the RHS wich are :
$$4((2-2x)(1+x)+x)^2(2+x)((2-2x)^4+1+x^4-(x^2+1)(2-2x)^2)((2-2x)^2+1+x^2-(x+1)(2-2x))\geq ((x+1)(x+2)(1+2x))^2(0.75(2+2x)+0.75((1+x)x)^{\frac{1}{3}})$$
With $0.33\leq x\leq 0.3675$ and $x=\frac{b}{c}$
Hope it helps you !


What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!

If $k$ and $n$ are not co-prime, you have to be very careful when dividing by $k\bmod n$ $-$ there may be more than one answer, as you saw. So $806/6\bmod 1000$ is not well-defined, because both $6\cdot 301$ and $6\cdot 801$ are equal to $806\bmod 1000$.
But in this case the problem is well-defined, because the expression $\dfrac{7^{2005}-1}{6}$ is an integer. So we can calculate it before reducing $\bmod 1000$. hgmath's answer shows one way to do this (and Raffaele's answer doesn't).


standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?

Determine a range of validity for this series.

Hint

Start with Taylor expansion $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+O\left(y^6\right)$$ Rise to the third power to obtain  $$\frac{1}{(1+y)^3}=1-3 y+6 y^2-10 y^3+15 y^4-21 y^5+O\left(y^6\right)$$ Now, replace $y$ by $\frac{4 x}{5}$ to get your result.

I am sure that you can take from here.

you have $(1+4/5x)^{-3}$=1-12/5x+64/25$x^2$.....
is the required exxpansion about x=0. for range, its true for |4/5x|<1


How do I solve $y'=\frac{x-y}{x+y}$? This is what I have so far:

$y'=\frac{x-y}{x+y}$

$ y'=\frac{x}{x+y}-\frac{y}{x+y}$

$ y'=\frac{1}{1+\frac{y}{x}}-\frac{1}{\frac{x}{y}+1} $

Make a substitution here:

$v=\frac{y}{x}, y=vx, y'=v'x+v$

$ v'x+v = \frac{1}{1+v}-\frac{1}{\frac{1}{v}+1}$

Do I just simplify the right side here, and then separate variables?  When I tried this, I got:

$\int \frac{1+v}{1-2v-v^2}dv = lnx+C_0$

I didn't know how to solve the integral here.

Any ideas?

You are complicating it. Sum $1$ to both sides to get $$y'+1=\frac{x-y}{x+y}+1=\frac{2x}{x+y}.$$ Now, because $(x+y)'=y'+1$, we get $$(x+y)'=\frac{2x}{x+y}.$$
Familiar, right? Write $u=y+x$, so $u'=2x/u \implies u^2/2=x^2+A$. We finally have $$y=-x\pm\sqrt{2x^2+B}.$$

You can solve this integral by using the partial fractions technique.
But tell me this. That simplification is correct? You got this result below:
$$\int \frac{1+v}{1-2v-v^2}dv$$

I've got this one:

$$\int \frac{1-v}{1-v}dv$$

That is 1 if v is different than 1, right?

Am I off the track here?

You got $$ \frac{1}{1-v}- \frac{1}{\frac{1}{v}-1} = K$$

$$\therefore  K = \frac{1}{1-v} -\frac{1}{\frac{1-v}{v}} $$

$$  \therefore K = \frac{1}{1-v} - \frac{v}{1-v}  $$

$$ \therefore K = \frac{1-v}{1-v}=1$$

So v can't be 0 and can't be 1 neither.


Homework Question. Joint Probability Distribution. Here is the question.
The joint PDF of X and Y is given by $f_{XY}(x,y) = {\frac 14} e^{-|x|-|y|}$. Find $P(X \le 1 ,and,  Y \le 0)$

Solving the problem I first found the marginal probabilities of X and Y. Can you please explain what I should do next.

To find the probability of some event $E$, you have to compute $$
  P(E) = \int_E \:dF
$$
where $F$ is your probability distribution. In your case, since you probability distribution has a density, you can also express that as $$
  P(E) = \int_E f \:d\lambda
$$
where $\lambda$ is the lebesgue measure on your probability space (which then is a subset of some $\mathbb{R}^d)$. The lebesgue measure is the ususal measure or length/area/volume/$\ldots$, so this is just a plain old integral. Additionally, in your case $E = (-\infty,1]\times(-\infty,0]$ is rectangular, which makes the integration especially simple. You get $$
  P(E) = \int_E f \:d\lambda = \int_{-\infty}^1\int_{-\infty}^0 f(x,y) \:dx\,dy \text{.}
$$

Solution using the mathStatica / Mathematica combo:

The joint pdf $f(x,y)$ is given as:

f = Exp[-Abs[x] - Abs[y]]/4;     domain[f] = {{x, -∞, ∞}, {y, -∞, ∞}};

You seek:

Prob[x < 1 && y < 0, f]

returns: $\frac 12 - \frac{1}{4 e}$

[ Even if this is homework, I guess your lecturer will want to see some workings, but this will give you something to aim at :) ]


A problem with planes, normal equations, lines and points. I was given the following problem and would like to know if my idea oh how to solve it is correct. I usually type my answers but due to my sketch I'll attach a picture with my interpretation of the problem and its possible solution

Question:
Find the normal of the lines $l$ that pass through a point $(1,0,2)$ and is orthogonal to a plane whose normal equation is $x+y-2z = 4$.

Basically my idea was the following, if I have a line orthogonal to a plane in $R^3$ then the direction vector of such plane will be parallel to my line. Then if I want to find the normal of the lines that pass through a point $P_o$ A line can be held with two vectors $(A,B,C)$ and $(a,b,c)$ whose dot product with the direction vector $(1,1,-2)$ is normal to the line and passes through $P_o$

Please, let me know if my work is correct or if I went wrong where did I fail to interpret the problem, what is wrong with my sketch. I genuinely appreciate any help thank you.

$(1,1,-2)$ is normal to your plane.  This is the direction vector of your line.

Vector equation: $L: (1,0,2) + (1,1,-2)t$

Parametric equations: $x = 1+t, y = 0+t, z = 2-2t$

Simultaneous equations: $\frac {x-1}{1} = \frac {y-0}{1}= \frac {z-2}{-2}$


Condition on a point on axis of the parabola so that $3$ distinct normals can be drawn from it to the parabola. The question is this:

  For the parabola: $$ (x-1)^2 + (y-1)^2 = \left(\dfrac{x+y} {\sqrt2}\right)^2 $$
  what is the condition on the point $(h,h)$ (which lies on the axis of
  the parabola) that $3$ distinct normals can be drawn from it to the parabola?

Now, I have this doubt: from any point (inside the axis) on the axis we can draw three normals to the parabola (no concrete proof, just visualisation), right? So isn't this question wrong?

After all the discussion in the comments part we have, perhaps, the solution to the edited question:

From a point on the parabola's axis it is possible to draw three different normals to the parabola iff the point is contained ("inside") the parabola but it is not the parabola's vertex (which could be considered not to be "inside" the parabola, anyway...but it's better, imo, to be thorough here and mention this).

Added: Thanks to one of the last comments of Henning below this answer, it is possible to see that the $\,y-$intercept of any normal to $\,y=x^2\,$ is $\,\frac12+a^2\;$ , from where it follows that no point on the parabola's axis and inside the parabola can have $\,y-$intercept $\,\le\frac12\;$ and have any hope of being one of the wanted points from which three normals to the parabola can be drawn!

This can be done in a more generic manner with all variables, in which case, from any point $P(h,k)$ there exist three normals to a parabola $y^2 = 4ax$ if $h> 2a$.


We throw two dices distinc. What is the probability that there will be at least a 6 given that the dice show different results? We throw two dices distinc.
What is the probability that there will be at least a 6 given that the dice show different results?

My work

Let

$A=$We have a six,

$B=$Different result

Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{6}$

This because: $P(B)=\frac{30}{36}$ and $P(A\cap B)=\frac{5}{36}$

Is good this?

Observe that the numbers on the two dice are distinct. They are also indistinguishable, so that any one of the six numbers has equal probability of showing. The probability that one of them is a six is therefore one third (two numbers out of six are showing).

I agree until your calculation for $P(A\cap B)=\frac{5}{36}$. Note that A or B could be the $6$, and that we want them to not both be 6. There's $12$ ways that A or B is $6$ and $1$ way that they are both $6$, so $P(A\cap B)=\frac{12 - 1}{36}=\frac{11}{36}$.

Substitute this back in and the final answer would be $\frac{11}{30}$


minimizing the norm of a curl over a domain According to my computations:

  The function which minimizes $$\int_\Omega \|\operatorname{curl} f\|^2\,dx$$ should satisfy $$\operatorname{curl}(\operatorname{curl}f) = 0$$ everywhere on $\Omega$, provided $\operatorname{curl} f = 0$ on $\partial \Omega$.

I followed the same kind of computation that the one demonstrating the the argmin of $\int_\Omega \|\nabla f\|^2~dx$ should satisfy $\Delta f = 0$. However, I am not sure whether my computations are right... May anyone check that please ?

1) We first start with a functional $$G(f) = \int_\Omega \|\operatorname{curl} f\|^2\,dx.$$

2) We compute
$$V(f,h) = \lim_{\epsilon\rightarrow 0} \frac{G(f+\epsilon h)-G(f)}{\epsilon} = 2\int_\Omega \operatorname{curl} f\cdot\operatorname{curl} h\,dx.$$

3) We use the identity :
$$\operatorname{div}(A\times B) = -A\cdot\operatorname{curl} B + B\cdot\operatorname{curl} A,$$ with $A=\operatorname{curl} f$ and $B=h$.

4) We obtain $$\int_\Omega \operatorname{curl} f\cdot \operatorname{curl} h\:dx = -\int_\Omega \operatorname{div}(\operatorname{curl} f\times h)\,dx + \int_\Omega h\cdot \operatorname{curl}(\operatorname{curl}f)\,dx.$$

5) We use the divergence theorem to obtain :
$$\int_\Omega \operatorname{curl} f\cdot \operatorname{curl} h\,dx = -\int_{\partial\Omega} \operatorname{curl} f\times h\,ds + \int_\Omega h\cdot\operatorname{curl}(\operatorname{curl}f)\,dx$$

6) We assumed $\operatorname{curl} f = 0$ on $\partial\Omega$, so the first term is zero.

7) $V(f,h)$ should equal zero for all $h$ for the function to be minimized, so $$\int_\Omega h\cdot\operatorname{curl}(\operatorname{curl}f) dx = 0\quad\forall h,$$ which implies $\operatorname{curl}(\operatorname{curl}f) = 0$ locally.

I guess this reasonning may be wrong in several places..... or is it right??
Thanks!

OP is mostly perfect until some later parts.

For the curl operator we have a similar integration by parts formula to the divergence theorem:
$$
\int_{\Omega}\nabla \cdot (u\nabla v) = \int_{\Omega}(u\Delta v + \nabla u\cdot \nabla v) = \int_{\partial \Omega} u\nabla v\cdot \nu \,dS,
$$
where $\nu$ is the outward unit normal to the boundary. The curl integration by parts formula is:
$$
\int_{\Omega}\nabla \cdot (\phi\times \psi) = \int_{\Omega}(\psi\cdot\nabla \times  \phi - \phi\cdot \nabla\times \psi) = \int_{\partial \Omega} \phi\times \psi\cdot \nu \,dS = \int_{\partial \Omega} \nu\times \phi\cdot \psi \,dS.\tag{1}
$$

Say now you wanna minimize (here I use $u$ and $v$ instead of $f$ and $h$):
$$
\mathcal{G}(v) = \int_{\Omega}|\nabla\times v|^2\,dx,
$$
if the minimizer is $u$ and the variational limit is also no problem:
$$
0= \frac{d}{d\epsilon}\mathcal{G}(u+\epsilon v)\Bigg|_{\epsilon= 0} = 2\int_{\Omega} \nabla \times u\cdot \nabla \times v.
$$

Now we use formula (1), letting $\phi = \nabla \times u$, and $\psi = v$:
$$
\int_{\Omega}(v\cdot\nabla \times (\nabla \times u)- \nabla \times u\cdot \nabla\times v) =\left\{\begin{aligned} \int_{\partial \Omega} \big(\nu\times (\nabla \times u)\big)\cdot v\,dS
\\
\\
 \int_{\partial \Omega} (v\times \nu) \cdot (\nabla \times u)\,dS
\end{aligned}\right.
$$
There are two choices of way of writing boundary terms to be zero:

You can set for the minimizer $\color{blue}{\nu\times (\nabla \times u)|_{\partial \Omega} = 0}$, and only tangential part suffices. $\nabla \times u|_{\partial \Omega = 0}$ is too strong.
You can set for the test function spaces $\color{red}{\nu\times v = 0}$, then $\nu \times u$ can be prescribed as any permissible boundary data $g$ (those who have 0 surface divergence).

Choosing blue, you have to modular a gradient in a proper function space $H$ to set up a well-defined problem:
$$
\left\{\begin{aligned}
\nabla\times (\nabla \times u) &=0 \quad \text{ in } \Omega,
\\
\nu \times (\nabla \times u) &= 0\quad \text{ on } \partial \Omega.
\end{aligned}\right.\tag{2}
$$
For $u \in H/\nabla p$ the quotient space, which means we don't tell the difference between $u $ and $u+\nabla p$, above problem has a unique solution. The only solution for (2) is zero, a.k.a. $u = 0$ or say $u = \nabla p$. Once we add a gauge condition $\nabla \cdot u = f$, then (2) is equivalent to the Neumann problem for $p$:
$$
\left\{\begin{aligned}
\Delta p &= f \quad \text{ in } \Omega,
\\
\nabla p\cdot \nu &= 0\quad \text{ on } \partial \Omega.
\end{aligned}\right.
$$

Choosing red, the problem is:
$$
\left\{\begin{aligned}
\nabla\times (\nabla \times u) &=0 \quad \text{ in } \Omega,
\\
\nu \times u &= g\quad \text{ on } \partial \Omega.
\end{aligned}\right.\tag{3}
$$
Again this problem has a unique solution in $H/\nabla p$, except this time $p$ is constant on the boundary which implies $\nabla p\times \nu =0$. If we set $g$ to be 0, then above problem again has a zero solution, which means $u = \nabla p$. Now a gauge condition $\nabla \cdot u = f$, then (3) can be transformed to a Dirichlet boundary problem for $p$:
$$
\left\{\begin{aligned}
\Delta p &= f \quad \text{ in } \Omega,
\\
p &= \mathrm{Constant}\quad \text{ on } \partial \Omega.
\end{aligned}\right.
$$

Summary: the minimizer $u = \mathrm{arg}\min\mathcal{G}(v)$ is a gradient field $\nabla p$, the boundary condition of $p$ relies on the boundary condition of $u$. Both problem (2) and (3) don't require the normal component of $\nabla \times u$ or $u$ to be imposed. If you prescribe $\nabla\times u = 0$ on the boundary, which implies $\nabla\times u\cdot \nu = 0$, the problem may be over-determined.

(This is supposed to be a shorter comment but I find myself powerless).

I haven't checked you reasoning, but I wanted to offer a simple (standard) proof. Some authors call this Dirichlet's principle.

Define $E(f) := \int_{\Omega} \| \nabla f \|^2 dx$.
It's clear that $E(f) \geq 0$.

Assume $f, g$ are equal on $\partial \Omega$
Let's see that, if $f$ is harmonic then $E(f) \leq E(g)$.

For that, let $u := f - g$. Now calculate $E(g) = E(f - u)$ using Green's first identity (here you'll use that $u = 0$ on $\partial \Omega$ and that $\Delta f = 0$).

You'll get immediately that $E(g) = E(f) + E(u)$ , and using that $E(u) \geq 0$ that $E(f) \leq E(g)$.


Find all solutions to a particular differential equation Find all solutions on ${R}$ of the differential equation $ y' = 3|y|^ \frac{2}{3} $

I believe I need to use separation of variables, but it will only work if the initial condition is nonzero. Therefore, not all solutions can be found this way.

Observe that: /y/^(2/3) = y^(2/3). So we solve: dy/dx = 3*y^(2/3).

Use separation of variables: y^(-2/3)dy = 3dx.

So: 3*y^(1/3) = 3x + C.

So: y = (x + C)^3 with y(0) = C^3.


$p^2$ misses 2 primitive roots When I Checked primitive roots of some primes P, I found this following phenomenon:

$14$ is a primitive root of prime $29$, but it's not primitive root of $29^2$

$18$ is a primitive root of prime $37$, but it's not primitive root of $37^2$

$19$ is a primitive root of prime $43$, but it's not primitive root of $43^2$

$11$ is a primitive root of prime $71$, but it's not primitive root of $71^2$
And they are all missing exactly one primitive root, which is P has one primitive root that cannot be found in primitive roots of $p^2$. My question is: What is the smallest prime P such that P has $2$ primitive roots that cannot be found in the primitive roots of $p^2$? ( Here I mean primitive roots between $0$ and $p-1$)

A small search with pari-gp shows that 367 is the smallest such prime, it misses the primitive roots 159 and 205. Then 653 misses four primitive roots 84,120,287 and 410.

A search up to 20000 shows that only 16631 misses 4, while several (1103, 6569, 13187, 14939, 15313, 16649 and 18587) misses 3 primitive roots.

By curiosity I have mesured how many primes have 0, 1, 2, .... primitive roots in the range 1, p-1 which are not primitive roots of $p^2$, it seems to behave like a Poisson distribution with $\lambda = -\log\log 2$. does somebody has an explanation?

Just to show how good is this estimate here is the count up to 409499 of primes with 0 to 6 primitive roots missing in mod $p^2$.

   Observed   Expected
0   23 949      23 962
1    8 695       8 782
2    1 696       1 609
3      210         197
4       19          18
5        0           1
6        1           0

So we should expect to have some prime with 7 or more after about 8 000 000 primes.

If you take a prime like 7, its square is 49, has 42 co-primes roots 0-49.  These can be represented as some x=g^n, where n ranges from 0 to 6*7-1 = 41.  When n is a multiple of p, then p divides its own period.

For example, 3 is a primitive root of 7.  3^5 gives 5 mod 7, and 3^35 gives 19 mod 49.  We now observe that 19 is a primitive root of 7, but in base 19, 7 divides its own period.

The number of small sevenites (ie 1 < n < p), is relatively small, none has gone as far as 20 in the search range, although 3511 has 10.  A run was done to find the 40 smallest sevenites for primes to something like 2,400,000.   This is done by insertion-sort using 60p as the upper banker.  From this table (it is about 2 cdrom's worth of zipped files), one extracts the list of small sevenites.  There are lots of bases which do not have small sevenites.

A compound sevenite is one where a higher power divides b^(p-1).  An example is the third-order 7³ divides 18^3-1, and the fourth-order 13^4 dividing 239^4-1.  There is only one small compound sevenite: 113³ in base 68.

No sevenite is known for the fibonacci series, but the sqrt(2) series 1,2,5,12,... has no fewer than 3 (13, 31 and 1546463), and the Heron triangle series as used to find the Messerine primes, and also integer area triangles of the form b-1, b, b+1, has one (103), these searches been taken to two millions (ie 2*120^4).

Sevenites are closely connected to the thing about 'x²=x, mod b^n'.  In practice, these are specific examples of the general case.  Where p is a sevenite of x, then x written in base p will agree to the last n places.

For example, 10 is a primitive root and sevenite of 487.  In base 487, the sevenite tail ending in 10 is ... 0, 10.  The 0 in column-one tells us that it's a small sevenite.

In base 13, the number 239 is written as 155.  The sevenite tail for 5, in 13 is 01550155.  Because the last four, but not five digits agree with 0000155, we see that 13^4 divides it, but 13^5 does not.

There is a fairly authoritive page on the wikipedia at `Euler's quotient'.

There's a list of sevenites http://z13.invisionfree.com/DozensOnline/index.php?showtopic=737 or the DozensOnline forum | mathematics | number theory | sevenites.