$24ml+1=k^2$ has no solution for all $l=1 \dots m$ Investigating solutions of $$24ml+1=k^2$$ for $l=1\dots m$
The question is to find the $m$-s for which the above equation has no solution for all $l=1..m$-s.
The first few $m$-s are:
$$3, 9, 24, 27, 81, 192, 243, 432, 729$$
Actually have found that the $m$ should be of form $2^a3^b$. Seems hang on the simple task, of finding general formula for this.

One interesting addition. Just tested the cases when the $$12ml+1=k^2$$ has no solution for all $$l=1..m$$ Seems it has no solution iff $m=3^a$ But has no proof yet.

Under $100$ there are $84$ $m$-values yield no integer integers $(k)$ for $l=1$.
To find them is simple. $\quad k\not\in\mathbb{N} \text { for }k=\sqrt{24 m + 1)}$.
For example, for any non-zero integer $(m)$,
$$\text{If }\space \sqrt{24 m + 1)}\ne
\big\lfloor \sqrt{24 m + 1)}\space\big\rfloor
\text{, then print } m \text{, endif }\tag{1}$$
This is the same as testing $\quad\sqrt{24 m \times(1) + 1)},\quad $
i.e. $l=1.\quad$ To find non-solution m-values for any other l-values, you must test the full $\quad 24m\times(l)+1, l\le m.\quad$ For examples,
\begin{align*}
\text{If }l=2,\quad& m\in\{ \emptyset \}\\
\text{If }l=3,\quad& m\in\{ 1,2,3\}\\
\text{If }l\in\{4,5,6,7,8\}\quad& m\in\{ \emptyset \}\\
\text{If }l=9,\quad& m\in\{ 1,2,3,4,5,6,7,8,9\}\\
\text{If }l\in\{10,11,12,\cdots, 23\}\quad& m\in\{ \emptyset \}\\
\mathbf{\vdots}
\end{align*}
Your derived list is correct but I cannot offer a better formula for finding these m-values.
$\textbf{Update}\\$
We can solve for $l$ and see which k-values yield an integer for a given m-value.
$$l = \frac{k^2 - 1}{24 m} \quad\text{and we know }m\ne0\tag{2}$$
We can find $k^2$ by taking multiples of $24m$ and adding 1 to see if any $\sqrt{k^2}$ yields an integer. Having found $k$ we plug it back into equation $(2)$ and find $l$. We have a valid $m$ if and only if the smallest $l$ generated is greater than $m$. The test values for $m$ can quickly be obtained from equation $(1)$ above.


Why is the unit disk closed? I read a question stating that if $z$ is complex, then $|z|\leq 1$ is a closed set. I think this is just saying that the unit disk is a closed set. Why is that so?

Look at the complement of the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$. The complement is given by $S^c = \{ z \in \mathbb{C}: |z| > 1 \}$. Consider a point $z_0 \in S^c$. $|z_0| > 1$ and hence let $|z_0| = 1 + r$ where $r > 0$. Consider the ball of radius $r$ centered at $z_0$ i.e. $B_r(z_0) = \{v \in \mathbb{C}:|v-z_0| < r \}$. Clearly, $B_r(z_0) \subset S^c$. This follows from triangle inequality since $$|z_0| = |z_0 - v + v| \leq |z_0 - v| + |v| \implies |v| \geq |z_0| - |z_0 - v| > (1+r) - r = 1$$
Hence, $|v|>1 \implies v \in S^c \implies  B_r(z_0) \subset S^c$. Hence, $S^c$ is open since given any point $z_0 \in S^c$ we can find a open neighborhood lying completely inside $S^c$ and hence $S$ is closed.

Equivalently, you can try to prove that the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$ contains all its accumulation points. The proof of this is again not hard. Look at a subsequence converging to an accumulation point and prove that if you have $|z_n| \leq 1$, then $\displaystyle \left| \lim_{n \rightarrow \infty} z_n \right| \leq 1$. (Hint: If not, what will happen?)

The unit disc $\bar D$ is the closure of the open unit disc $D$ defined as all complex numbers $z$ such that $|z|<1$. It just happens that with the standard topology of the plane, the closure of the open unit disc is the closed unit disc. So the set of all $z$ such that $|z| \leq 1$ is closed.

Note that in general the closure of the set is not the closed set. Indeed in the discrete topology this fails. What we need to insure that the closure of a ball coincide with the closed ball is the separation axiom called $T_1$. In topological spaces which do not have the $T_1$ axiom (discrete topology), this principle fails. Alternatively you can see it with the perspective Sivaram offers: check that the complement is open.


Question about variable and constant notation in some properties I am just starting to think about mathematical notation a lot more and some parts of it do not make as much sense to me as I would like. I am operating under the convention that beginning alphabet letters are constants and ending alphabet letters are variables.

I see this equation in algebra and calculus text books: $a(b+c)=ab+ac$
What this means to me is that it is true for constants, but says nothing about variables. But this clearly is true for variables as well, as you can do equations such as $x^2-2x=x(x-2)$.

Why is some math notation written this way, or rather, why isn't the property written as $x(y+z)=xy+xz$ since that would signify that it works for any unknown and not just constants. This goes for all the elementary algebra fraction equations as well. And then to make it even more confusing, as soon as exponents are involved, variables are now introduced, such as $x^mx^n=x^{m+n}$. Wouldn't a constant work just as well here? Am I just not understanding something basic about notation?

p.s. If there is a good resource to clarify mathematical notation I would love to see it.

Of course, they are all variables. Your perception is somewhat correct in the sense that customarily constants (or, "parameters") in the context of a specific problem are often denoted $\{a,b,c,\ldots\}$. But a symbol is a symbol is a symbol. Even the symbol $\pi$ isn't always used just to mean the number $3.1415926535...$

I don't quite get you. When you say $a(b+c) = ab +bc$ is true for 'constants',  does that mean it is true not only for $a=1$, $b=2$ and $c=3$, but also for $a=5$, $b=3$ and $c=3$, and so on and so forth. So there you are, it is varying in that sense. When I have an equation, say  $y=x²+c$ then only $x$ and $y$ are varying and $c$ is fixed, but $c$ can be any 'constant'. It's not varying in the sense in which $x$ and $y$ are varying. So depends on what the context is, I suppose.


The probability of choosing the third card Five cards are numbered as the following
$$5 , 5 , 5 , 4 ,4$$
Three cards are chosen randomly , whats the probability that the third card is numbered 4 ? Without replacement
My turn
Let the first two cards are numbered 5 and the third is4
So we get
$$3C1 \times 2C1 \times 2C1$$
Let the first is 5 and the second and the third are 4
So
$$3C1 \times 2C1 \times 1$$
 The same thing happens if the first and the third are 4 and the second is 5 so
The probability is $$\frac{ the sum }{5P3}$$
Is the solution correct ?

(Already answered in comments - there should be an Answer explaining carefully why this is trivial...)

The answer is obviously $2/5$; the first two cards chosen are irrelevant.

To be a little more formal: Let's give names to those five cards; say they're $C_1,\dots,C_5$ (so it happens that $C_1$, $C_2$ and $C_3$ are fives, while $C_4$ and $C_5$ are fours).

Say $X$ is the third card chosen. It's clear by symmetry that $$P(X=C_j)=P(X=C_k)$$for every $j$ and $k$; there's obviously no reason why the third card is more likely to be $C_2$ than $C_1$. Hence we have $$P(X=C_j)=1/5$$for every $j$, so $$P(X\text{ is a four})=P(X=C_4)+P(X=C_5)=2/5.$$

I have never liked calculating probability in terms of "binomial coefficients.  This is how I would do this problem:

There are four ways the first two cards can be drawn: 55, 54, 45, and 44.

The probability the first card is a 5 is 3/5.  If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s.  The probability of "55" is (3/5)(2/4)= 3/10.  In this case there are 3 cards left, one 5 and 2 4s, so the probability the third card drawn is a 4 is 2/3.  The probability of 554 is (3/10)(2/3)= 1/5.

The probability the first card is a 5 is 3/5.  If the first card is a 5 then there are 4 cards left, 2 5s and 2 4s. The probability of "54" is (3/5)(2/4)= 3/10. In this case there are 3 cards left, two 5s and one 4, so the probability the third card drawn is a 4 is 1/3.  The probability of "544" is (3/10)(1/3)= 1/10.

The probability the first card drawn is a 4 is 2/5.  If the first card is a 4 then there are 4 cards left, three 5s and one 4. The probability of "45" is (2/5)(3/4)= 3/5.  I this case there are 3 cards left, two 5s and one 4, so the probability the third card is a 4 is 1/3.  The probability of "454" is (3/5)(1/3)= 1/5.

Since there are only two 4s if the first two are "44" the third cannot be a 4.  The probability of "444" is 0.

Since these are independent events, the probability the third card is a 4 is 1/5+ 1/10+ 1/5= 2/10+ 1/10+ 2/10= 5/10= 1/2.

That is NOT the same as your $\frac{3C2+ 2C1+ 2C1}{5P3}= \frac{3+ 2+ 2}{60}= \frac{7}{60}$.


Quadrilateral with given angles We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!

If you are looking for the proof by elementary geometry, please see Hiroshi Saito(斉藤浩)'s great work about generalized Langley's problem
( https://www.gensu.co.jp/saito/challenge/pdf/3circumcenter_d20180609.pdf ). He introduced the amazing skill named "3 circumcenter method" invented by Ms. aerile_re(pen name), and you can find the solution of this problem in the article (Q1).

Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48
Multiply both denom and nom by 4Sin82
Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)
(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)
Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)
Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)
(2Sin24*Cos6 = Sin54 REPLACRMENT)
Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)
(1/4Sin54 = Sin18 REPLACEMENT)
Sinx / Sin(x + 64) = Sin18 / Sin82
X = 18
Y = 50


If $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$. So I basically have to prove what is on the title.

  Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.

One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.

I am having troubles proving the other inclusion. Any comments?

Thanks

The statement you are trying to prove is only necessarily true for commutative rings with $1$. In this case, you can argue that
$$I\cap J\subseteq (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$

A counterexample to the statement for general rngs is given by endowing the group $\mathbb Z$ with the zero product, that is defining $a\cdot b=0,\forall a,b\in\mathbb Z$. The ideals $(2)$ and $(3)$ are still comaximal, as for any $x\in\mathbb Z$ we can write $x=a\times 2+b\times 3$ for some $a,b\in\mathbb Z$, where $\times$ denotes regular multiplication, and $a\times 2=2+\cdots+2\in(2)$ where addition is performed $a$ times, and similarly $b\times 3\in (3)$. But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.

The direct proof (for unital rings) also can be given. Assume $x\in I\cap J$. Consider such elements $i,i'\in I$ and $j,j'\in J$ that $x=i+j$ and $1=i'+j'$. Note that $i=x-j$, $x\in J$ gives $i\in J$ and $j=x-i$, $x\in I$ gives $j\in I$.

We have $x=i\cdot1+j\cdot1=i(i'+j')+j(i'+j')=ii'+ij'+ji'+jj'$. Using the commutativity we get $x=i'i+ij'+i'j+jj'$. So there are such elements $r_1,...,r_4\in R$, $i_1,...,i_4\in I$, $j_1,...,j_4\in J$ that $x=r_1i_1j_1+...+r_4i_4j_4$ (put $r_1=...=r_4=1$). So $x\in IJ$.


Can I take the vector space generated by {$1 , x, x^2$} over the field $F_3$? Let $V$ be the vector space over the field $F_3$ (field with $3$ elements) of dimension $3$. Define an action of the symmetric group $S_3$ on $V$ by $\sigma.e_i$= $e_{\sigma (i)}$ where  {$e_1$, $e_2$, $e_3$} is a basis for $V$.

Can anyone please make me understand what this problem is saying?

I do not understand what is a vector space over the field $F_3$ (field with $3$ elements) ?
Can I  take the vector space generated by {$1 , x, x^2$} over the field $F_3$?

If the action is only on he basis elements then how would we find the operations like $ \sigma (e_1 + e_2)$?

If the action is only on he basis elements then how would we find the operations like $σ(e_1+e_2)$?

This seems to be the key to your question.

You are right: at the outset, the action is explicitly only defined on the set $\{e_1, e_2, e_3\}$. Given a group element $\sigma$, and one of these basis elements $b$, we know what $\sigma\cdot b$ is, and that is all.

But every mapping defined on the basis of a vector space gives rise to a unique linear transformation extending that mapping to the entire vector space spanned by the elements. Furthermore, if the mapping is a bijection and the vector space is finite dimensional, the extension is an isomorphism.

So what is happening is that by definition the action on an arbitrary element of the vector space is defined to be the linear extension: $\sigma\cdot (\sum\lambda_ie_i):=\sum\lambda_i(\sigma\cdot e_i)$.

You can check all the axioms to confirm that indeed, the originally defined group action on the basis has now been promoted to a group action of $S_3$ on the entire vector space spanned by the basis.

This "lifting" of the group action is what you were missing, I think.

There's another way to see how this happens. You can rephrase a group action of $G$ on a set $X$ to be a homomorphism from $G\to Sym(X)$.

If $X$ is the finite basis of an $n$ dimensional $F$ vector space, we have another group homomorphism from $Sym(X)\to GL(n,F)$. Composing these two homomorphisms we get a group homomorphism from $G\to GL(n,F)$, and that defines an action of $G$ on $span(X)$.

It says, essentially, that every element of $G$ acts like a linear automorphism on $span(X)$ (very nice elements of $Sym(span(X))$!)

$\mathbb F_3[x]/(p(x))$ will be just that vector space,  if $p(x)\in\mathbb F_3[x]$ is irreducible and $\operatorname{deg}p=3$.

So for instance,  take $p$ that doesn't have a root in $\mathbb F_3$.  Say, $p(x)=x^3+2x^2+1$.

To answer your last question,  use the property of group actions:   $\sigma (e_1+e_2)=\sigma (e_1)+\sigma (e_2)=e_{\sigma (1)}+e_{\sigma (2)}$, for $e_1,e_2\in V$.


Unitary Operators on Let $L^2_{\mathbb{P}}(\Omega)$ be a separable Hilbert space and $(\Omega,\Sigma,\mathbb{P})$ be a probability space.  Given two $f,g \in L^2_{\mathbb{P}}(\Omega)$ is there a unitary (or self-adjoint) operator $U$ on $L^2_{\mathbb{P}}(\Omega)$ satisfying
$$
U(f)=g\qquad \mathbb{P}-a.e?
$$

For a unitary operator you will need the condition $\|f\|=\|g\|$. W.l.o.g. assume $\|f\|=\|g\|=1$. Extend $f$ to an orthonormal basis $\{f_1,f_2,...\}$ with $f_1=f$ and extend $g$ to an orthonormal basis $\{g_1,g_2,...\}$ with $g_1=g$ and define $U(\sum a_n f_n)=\sum a_n g_n$. This gives  a unitary map with $Uf=g$.

Let $\phi:\Omega \rightarrow \Omega $ be a bijective measurable function preserving the $\mathbb{P}$-size of each set; ie:
$$
\mathbb{P}\left(
\phi^{-1}(B)
\right)=\mathbb{P}(B)\qquad (\forall B \in \Sigma).
$$
Then the composition operator $C_{\phi}:f\mapsto f\circ \phi \in L\left(L^2_{\mathbb{P}}(\Omega)\right)$ (where $L(L^2_{\mathbb{P}}(\Omega))$ is the set of continuous linear maps from $L^2_{\mathbb{P}}(\Omega)$ to itself) is unitary.

Well-defined

Using the push-forward property of measure
$$
\int_{\Omega}|C_{\phi}f|^2d\mathbb{P} =
\int_{\Omega}|f\circ \phi|^2d\mathbb{P}=
\int_{\phi(\Omega)}|f\circ \phi|^2d\phi_{\#}\mathbb{P}=
\int_{\Omega}|f\circ \phi|^2d\mathbb{P}<\infty;
$$
where $\phi_{\#}\mathbb{P}$ is the $\phi$-push-forward of $\mathbb{P}$.

Unitary:
Let $f,g \in L^2_{\mathbb{P}}(\Omega)$ and note that:
$$<C_{\phi}f,g> = \int_{\Omega}f\circ \phi g d\mathbb{P}=
\int_{\Omega}f g\circ \phi^{-1} d\mathbb{P}=
<f,C_{\phi^{-1}}g>;
$$
whence $C_{\phi}$ is unitary with adjoint $C_{\phi^{-1}}$!

Similarly, we compute that: $C_{\phi}\circ C_{\phi^{-1}}=C_{\phi^{-1}}\circ C_{\phi}=1_{L^2_{\mathbb{P}}(\Omega)}$.

A Fun Little Example...But and important One: The Fourier Transform

The Fourier Transform $f\mapsto \int_{\omega \in\Omega} e^{-2 \pi \lambda \omega} f(\omega) d\mathbb{P}(\omega)$ is a particularly important case.   I leave this exercise to you though

Then as Kavi stated, you just need to choose $\phi$ that maps $f$ into $g$.


Abstract Algebra: Polynomials are irreducible A) $x^4+x+1\in \Bbb Z_2[x]$ is irreducible

Proof:

  $x^4+x+1\in \Bbb Z_2[x]$ is primitive. The mod 2 reduction of $x^4+x+1\in \mathbb{Z}[x]$ is $f(x)=x^4+x+1\in \Bbb Z_2[x]$. Since $f(a) = 1 \neq 0$ for all $a\in\Bbb Z_2$ it follows that $f(x)$ has no linear factors. Suppose that $f(x)$ is reducible. Then it must be the product of quadratic factors. There are $3$ quadratic reducible polynomials in $\Bbb Z_2[x]$. The irreducible one is $x^2+x+1$ since this polynomial has no roots in $\Bbb Z_2$. Therefore $f(x)=(x^2+x+1)^2=x^4+x+1$ which is not the case.  Thus it is irreducible.

B) $\displaystyle\sum_{i=1}^{42}x^{i-1} \in \Bbb Z_2[x]$ is irreducible

I was able to prove problem A which I needed your guidance which on whether or not it was true.

But how to prove problem B?

Cyclotomic polynomials are irreducible. It is elaborated here:

http://en.wikipedia.org/wiki/Eisenstein's_criterion

Even I am interested in this question. Btw what do you mean by markup?

I think first one is right. But about Second I think it should be

Let

Q=$\sum_{i=1}^{42}$x$^{i-1}$

be an element of D[x], the polynomial ring with coefficients in D. Suppose there exists a prime ideal p of D such that
   x$^{i-1}$ ∈ p for each i ≠ 42,
Then Q cannot be written as a product of two non-constant polynomials in D[x]. If in addition Q is primitive (i.e., it has no non-trivial constant divisors), then it is irreducible in D[x]. If D is a unique factorization domain with field of fractions F, then by Gauss's lemma Q is irreducible in F[x], whether or not it is primitive (since constant factors are invertible in F[x]); in this case a possible choice of prime ideal is the principal ideal generated by any irreducible element of D.


Multivariable Calculus Divergence Theorem I have a cylinder where the top is given by $ z=-2$ the bottom is $z=2$ and the "lateral surface" is $x^2+y^2=1$. I have to find$\iint_{s}(4x+3y+z^2) \ dS$. I know this involves using the divergence theorem and then plugging in the answer into the volume of a cylinder but I have no idea how to set up the limits on the triple integral once I have found the partials and split up the integral for the divergence theorem. Thanks.

EDIT: The OP's desire to apply the divergence theorem is ambitious, but can be achieved. To apply the Divergence Theorem, we must have the surface integral in the form of $\iint_S \vec F\cdot\vec n dS$. So this problem gets sneaky. To this end:

Consider the vector field $\vec F_1=(4,3,(4x+3y)z/2)$. Since $\vec n = (x,y,0)$ on the lateral surface and $\vec n = (0,0,z/2)$ on the two disks, this accounts for $\iint_S (4x+3y)dS$ and gives us flux $\iiint_V \text{div}\vec F_1 dV =\iiint_V \frac12(4x+3y)dV = 0$, by symmetry. Now consider $\vec F_2 = z^2(x,y,z/2)$. By similar reasoning, its flux across $S$ is indeed $\iint_S z^2 dS$. So the flux of $\vec F_2$  is
$$\iiint_V  \text{div} F_2 dV = \iiint_V \frac72z^2\,dV =\frac72\int_0^{2\pi}\int_0^1\int_{-2}^2 z^2r dzdrd\theta = \frac{56}3\pi\,.$$
(Note that we do indeed get the same answer by the much simpler direct calculation.)

There is nothing wrong with this approach! The OP said "I know it involves using the divergence theorem.." but this does not mean that he is not interested in other approach.

Since the surface $S$ consists of three parts, then the integral can be evaluated as

  $$ \iint_{S} = \iint_{S_1} + \iint_{S_2} + \iint_{S_3} \longrightarrow (1) $$

First, we parametrize the lateral surface as

  $$ r(u,v)= \langle  \cos u, \sin u, v \rangle, \quad -2 \leq v \leq 2,\, 0\leq u \leq 2\pi. $$

Then, we have

$$\iint_{S} (4x+3y+z^2) dS = \int_{-2}^{2}\int_{0}^{2\pi} (4\cos u + 3\sin u +v^2)|\textbf{r}_u  \times \textbf{r}_v|dudv .$$

For the second and the third integrals in $(1)$, you do not need to parametrize, just use the technique. Now, you should be able to finish the problem.

Note: We used the following identity

  $$ \iint_{S} f \,dS = \iint_{T} f(\mathbf{x}(u, v)) \left\|{\partial \mathbf{x} \over \partial u}\times {\partial \mathbf{x} \over \partial v}\right\| du\, dv  $$


every $k$ cell is compact I did not understand the proof from the word "and some sets in this infinite sequence are from $\{G\}$....." to the end. Thanks for helping me to understand.

$k$ cell in $\mathbb{R}^k$ is defined as $\{x=(x_1,\dots, x_k):a_i\le x_i\le b_i\}=[a_1,b_1]\times\dots[a_k,b_k]$

It's not clear what the reasoning is. In my opinion the idea is that you have a sequence of subcells each covered by an infinite cover with no infinite subcover which is itself a subcover of $G$. This sequence converges to a point and this point is contained in an open set $U$ of $G$. But then $U$ must entirely contain a subcell and here it's the contradiction

In that prove we have infinite subcell all of it contain only one point and each of it can be covered by one open set of cover but still $G$ covered by infinite open cover, I think we must say if we can't cover $G$ by finite sub cover we split $G$ into $2^k$ of $k$ cells each of them covered by finite sub cover (one) if one of it can't cover take the same manner and split it (split up finite turn).


Finding point on parametric curve such that the tangent is parallel to line Let $C$ be a curve defined parametrically as $x=a\cos^3\theta$ and $y=a\sin^3\theta$, where $0\leq
\theta  \leq\pi/2$. Determine a point $P$ on $C$ such that the tangent at $P $ is parallel to the chord joining the points $(a,0)$ and $(0,a)$.

I have tried to show it by using mean value theorem. This function is continuous and differentiable on $0\leq
\theta  \leq\pi/2$. So by Lagrange mean value theorem, there exists $\theta_1; 0\leq
\theta_1  \leq\pi/2 $ such that $\left(\frac{dy}{dx}\right)_{ \theta_1}=\frac{y(\pi/2)-y(0)}{a-0}\implies -\tan\theta_1=\frac{a-0}{a-0}=1$. I don't know whether my approach is true or not.

As karakfa notes, you could simply see this by symmetry, but you seem to want a more formal approach.

The parametric derivative is in this case given as $$\frac{3a\cos \theta \sin^2 \theta}{-3a\cos^2 \theta \sin \theta}=-\tan\theta.$$

Since the line between your two points has slope $-1$, we find $\theta$ s.t.
$$-\tan\theta=-1 \implies\theta= \frac{\pi}{4}+\pi n, \quad n\in\mathbb{Z},$$

so the solution you are looking for is $$\theta=\frac{\pi}{4}.$$

From the given coordinates, the slope can be calculated to be = -1.

Now as in previous answers I can see that θ=π/4 has been calculated and well shown.

This value can be now be used in x=acos^3θ, where x=a/2.2^(1/2).Similarly, y(whenθ=π/4)= a/2.2^(1/2).

Those will be the answer to the requisite point on C.


prove that $p^{m/n}$, with $p$ prime and $\gcd(m, n)=1$, is irrational I need to prove that $p^{m/n}$, $m$ and $n$ naturals, $n > 1$, with $p$ prime and $\gcd(m, n)=1$, is irrational. It's suggested that this proof should be by contradiction or contraposition.

Suppose $\,p^{m/n}$ is rational. It's a  root of $\,x^n\! = p^m$ so by the Rational Root Test $\,x\,$ is an integer $\,a.$ Therefore, $ $ comparing powers of $\,p\,$ in $\,a^n = p^m$ $\Rightarrow$ $\,n\alpha = m\,$ so $\,n\mid m\,$ contra $\,n,m\,$ coprime.

Remark $ $ Without RRT: $\ $ if $\ p^{m/n}\! = a/b\ $ is rational then $\, p^m b^n = a^n\,$ hence comparing powers of $\,p\,$ on both sides of the prior equation yields  $\, m+n\beta = n\alpha\, $ so $\, n\mid m\, $ contra $\,n,m\,$ coprime

Assume the negation of the statement and set out for a contradiction. That is $$\frac{k}{l}^{\frac{n}{m}}=p$$

But the greatest common denominator is one so

$$\frac{k^{\frac{n}{m}}}{l^{\frac{n}{m}}}$$

Which must be a multiple of $l^{\frac{n}{m}}$. So $p=j^{\frac{n}{m}}$.

$p=({g^{\frac{1}{m}}})^n$ which contradicts the fact that $p$ was prime.


Indefinite integration and definite integration calculation Given the functions:

$A(x)=B(2x-1)-B(2x)$

B(x) is a function as x->infinity, limit of B(x)->0.

for $b>1$ find out:

$$\int_{b}^\infty A(x) dx - \frac12\int_{2b-1}^{2b} B(x) dx$$

The answer is $0$. But why?

I attempted to do:

$\int_{b}^\infty A(x) dx=\int_{b}^\infty B(2x-1)-B(2x) dx$

$=\int_{b}^\infty \sum_{k=2x-1}^{2x} B(k) dx$

And I think that it would be possible to swap the summation and integration, however got no further...

Not a full solution, but almost. Define
$$
I(b)=\int_{b}^\infty [B(2x-1)-B(2x)]dx - \frac12\int_{2b-1}^{2b} B(x) dx\ .
$$
For a generic $B(x)$ [not necessarily the one the OP wrote], we have
$$
\frac{d}{db}I(b)=0
$$
identically, using the fundamental theorem of calculus. Indeed
$$
\frac{d}{db}\int_b^\infty dx B(2x-1)=-B(2b-1)
$$
$$
\frac{d}{db}\int_b^\infty dx B(2x)=-B(2b)
$$
$$
\frac{d}{db}\int_{2b-1}^{2b}B(x)dx=2 B(2 b)-2 B(2 b-1)\ .
$$
Therefore, $I(b)$ is a constant (as a function of $b$). So, you only need to evaluate your $I(b)$ at a value of $b$ of your choice [pick a value for which the computation is easy] to establish your claim....

This is my answer (I modified the question as the B(x)= part is not quite useful):

let m=2x-1, n=2x

Then integral of A(x)=$\int_{2b-1}^\infty B(m) dm - \int_{2b}^\infty B(n) dn$

Then equals the integral at the right hand side.


Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$ without using Calculus techniques Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$.

It is easy if we apply Calculus techniques (e.g., derivatives, Lagrange multipliers, etc). However, the problem is assigned for students who have not yet learned Calculus. Is there a way to find the maximum without using Calculus techniques?

I have a few ideas:

Apply geometric means: $xy \leq \frac{x^2+y^2}{2} = -x + 3/2 $.
Change object function to $(xy)^2$, then we have $(xy)^2 = x^2(4-(x+1)^2)=x^2(-x^2 - 2x +3)$. That is, maximize  $x^2(-x^2 - 2x +3)$ on $(-3,1)$.

If you are deadset on a way without calculus, lets start with your idea $2$, i.e. how to maximize
$$p(x) = x^2y^2 = x^2\left(4-(x+1)^2 \right)= x^2(1-x)(3+x)$$

Now it is evident that this is a downward facing quartic with roots (with multiplicity) $-3, 0, 0, 1$, so the function is positive when $x \in (-3, 0)$ or $(0, 1)$.  It must have maxima in those intervals and one of them must be the global maximum we seek.

Case $1$: $x \in (-3, 0)$
$$p = (-x)(-x) (1-x)(3+x)$$
is a product of four positive terms, so by AM-GM, if the sum were a constant this would be maximized when all terms are equal.  Unfortunately the sum is not a constant, and the equality condition $(-x) = (1-x)$ is absurd.  Both these problems however can be addressed by appropriate scaling.  So consider maximizing for some real $\alpha, \beta > 0$,
$$\alpha \beta p = (-x)(-x)(\alpha - \alpha x)(3\beta + \beta x)$$
Now the sum is a constant if we have $2+\alpha = \beta$ and for equality we must have $-x = \alpha -\alpha x = 3\beta + \beta x \implies \alpha = \frac14(\sqrt{33}-3), \beta = \frac14(5+\sqrt{33}), x = -\frac14(3+\sqrt{33})$

As all the conditions for AM-GM get satisfied here, the maximum in $(-3,0)$ is when $x = -\frac14(3+\sqrt{33}) = x_1$, say.

Case $2$: $x \in (0, 1)$
Following the same logic, we now have
$$\alpha \beta p = (x)(x)(\alpha-\alpha x)(3\beta+\beta x)$$
and the conditions give
$2+\beta = \alpha, x = \alpha - \alpha x = 3\beta + \beta x \implies x_2 = \frac14(\sqrt{33}-3)$

All that is left is to see which among $x_1, x_2$ gives you the maximum, and it is easily seen $p(x_1) > p(x_2)$, so the maximum is $\sqrt{p(x_1)} = \frac14\sqrt{\frac32(69+11\sqrt{33})} \approx 3.52$.

By looking at the plot, the polynomial $$y=(4-(x+1)^2)x^2$$ has three extrema.

In the vicinity of an extrema, the curve looks like a parabola. We can bring the vertex of the parabola on the axis by translating the independent variable, $x\leftarrow x-m$, giving

$$(4-(x-m+1)^2)(x-m)^2=(3m^2+2m^3-m^4)+(-6m-6m^2+4m^3)x+(3+6m-6m^2)x^2+(-2+4m)x^3-x^4.$$

This is achieved when the first degree term vanishes, i.e. when

$$-6m-6m^2+4m^3=0$$ or

$$m=0,m=\frac{3\pm\sqrt{33}}4.$$

The approximating parabolas are

$$y=(3m^2+2m^3-m^4)+(3+6m-6m^2)(x+m)^2,$$ giving the extrema values $$3m^2+2m^3-m^4.$$


On $\phi(n) \sigma(n)$ being a square. I can't understand underlined statements:

The green one: For p an odd number, both p-1 and p+1 are even so all prime factors in $\prod (p-1)(p+1)$ must be belong to the set B, so the prime factors of $m_P$. So $\phi(n_P) \sigma(n_P) = mt_P^2$ holds for any collection of primes selected from set A. Where does $2^{\pi(y)-\pi((y+1)/2)}$ come from?

The red one: Why $m$ is same for $n_P$ and $n_Q$?

The blue one: How it is injective? How the identity element is empty? How $\Delta$ and $A$ make a group?

Orange/red: This is how the set $\mathcal{F}\subseteq P(\mathcal{A})$ is defined: every $\mathcal{P}\in\mathcal{F}$ has the same $m$.

Green: $2^{\pi(y)}$ and $2^{\pi((y+1)/2)}$ are the cardinality of $P(\mathcal{A})$ and $P(\mathcal{B})$, as stated in the second/third line in the paragraph.  Since $m$ is a product of elements from $\mathcal{B}$, there are at most $\lvert P(\mathcal{B})\rvert$ choices of $m$, but there are $\lvert P(\mathcal{A})\rvert$ choice of $\mathcal{P}$ so there is some $m$ which is hit by at least $\dfrac{\lvert P(\mathcal{A})\rvert}{\lvert P(\mathcal{B})\rvert}$ elements of $P(\mathcal{A})$.

Blue: You can recover $\mathcal{P}$ from $\mathcal{P}\triangle\mathcal{Q}$ by symmetric difference again with $\mathcal{Q}$.  Alternatively, you can view $2$ as the group of order $2$ and the power set $P(\mathcal{F})=2^{\mathcal{F}}$ as the direct product of $\lvert\mathcal{F}\rvert$ copies of $2$, then the group operation coincides with symmetric difference.

COMMENT.-Some questions that would imply   $\textbf{Proposition 8.7}$

Taking $n=pq$ with $p,q$ distinct primes one has $\phi(n)=(p-1)(q-1)$ and $\sigma (n)=1+p+q+pq=(p+1)(q+1)$ so
$$\phi(n)\sigma (n)=(p^2-1)(q^2-1)$$

For the primes $2$ and $3$ one has
 $$\phi(2p)\sigma (2p)=3(p^2-1)\\\phi(3p)\sigma (3p)=2^3(p^2-1)$$ Then in order to have   $\phi(2p)\sigma (2p)=\square$ we need primes $p$ such that $p^2-1=3x^2$. The two first are  $7$ and $97$.

Similarly for $\phi(3p)\sigma (3p)=2^3(p^2-1)$ we need primes $p$ such that $p^2-1=2x^2$. The two first are  $17$ and $577$.

►Are there infinitely many primes $p$ such that $p^2-1=3x^2$? or $p^2-1=2x^2$?

This question can be extended fixing all prime instead of $2$ and $3$ above but sometimes the coefficient $ a $ in $ ax ^ 2 $ can be large (for $ n = 61p $ you have $61^2-1 =2^3\cdot3\cdot11\cdot17$ for which $a=2\cdot3\cdot11\cdot17 = 930$ and surely  there are many more possible larger values. However there are many other primes like $19$ for which $\phi(19p)\sigma (19p)=360(p^2-1)$ so we need primes such that $p^2-1=10x^2$. Note that a single solution in an infinity of these cases would also prove the proposition. $$*****$$

We can verify that for $n=2p,7p$ and $97p$ we need primes such that $p^2-1=3x^2$ to have $\phi(2p)\sigma (2p),\phi(7p)\sigma (7p)$ and $\phi(97p)\sigma (97p)$ squares. In other words, a solution for one of three is also solution for the other two. Analogously for $n =3p$ and $n=17p$ we need to solve the same equation $p ^ 2-1 = 2x^2$.

►Are there infinitely many primes $p_i$ such that $p_i^2-1=by^2$ where the square-free part $b$ is the same for each $p_i$?

In this case just one solution for any of the $\phi(p_i\cdot p)\sigma (p_i\cdot p)=\square$ would prove the proposition.


What is the idea behind a projection operator? What does it do? I know what a projection operator is, but I am unable to explain it in words without using mathematical symbols. Can anyone help me?

I don't need examples or the definition - I want to know why and how its need arose, and what is the idea behind it?

This is how I used to imagine projections:

If  a mouse:

gets run over by a steamroller:

It will look like this:

Now if it gets run over by a steamroller another time, it will still look like this:

A projection map $f: A \to B$ is something you often need to simplify the space $A$. It is build so that differences which occur in the space $A$ will often no longer occur anymore in the space $B$. Then, if you know enough about the map $f$ and find out a bit about the space $B$ (which is easier as their are less differences) then you may find out interesting things about $A$.


Rotations of complex graphs Let $c_1 = -i$ and $c_2 = 3$. Let $z_0$ be an arbitrary complex number. We rotate $z_0$ around $c_1$ by $\pi/4$ counter-clockwise to get $z_1$. We then rotate $z_1$ around $c_2$ by $\pi/4$ counter-clockwise to get $z_2$.

There exists a complex number $c$ such that we can get $z_2$ from $z_0$ by rotating around $c$ by $\pi/2$ counter-clockwise. Find the sum of the real and imaginary parts of $c$.

I am having some trouble with this problem.  I have tried thinking this problem as if it were to be on the cartesian plane, but I still could not solve it.  Any help is appreciated.

Geometrically, this is simply stating that we wish to define a circle with center $c$ such that the points $z_0$ and $z_2$ lie on this circle.

In the complex plane, a point can be describe in polar coordinates by $re^{i\theta}+c$ for $0 \leq \theta < 2 \pi$ where $c$ is the center and $r$ is the radius.

The first point can then be described as $z_1 = z_0e^{i\frac{\pi}{4}} + c_1$ and the second point $z_2 = z_1e^{i\frac{\pi}{4}} + c_2$. We can then describe $z_2$ in terms of a circle based on $z_0$ as:

   $z_2 = \left (z_0 e^{i\frac{\pi}{4}} + c_1\right )e^{i\frac{\pi}{4}} + c_2= z_0e^{i\frac{\pi}{2}} + \underbrace{c_1e^{i\frac{\pi}{4}}+ c_2}_{\text{new center}, c} $

The center of this circle is $c = c_1e^{i\frac{\pi}{4}}+ c_2$.

From Euler's Formula: $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$. Using this, we can describe $c$ as:

   $$ c = c_1\cos(\frac{\pi}{4}) + ic_1\sin(\frac{\pi}{4}) + c_2 =  \frac{c_1}{\sqrt{2}} + i\frac{c_1}{\sqrt{2}} + c_2 $$

Note that in cartesian $c_1 = a_1+ b_1i$ and $c_2 = a_2+ b_2i$. Which implies $ic_1 = -b_1 + a_1 i$. Which leads to:

   $$ c =  \frac{a_1 + b_1i}{\sqrt{2}} + \frac{-b_1 + a_1 i}{\sqrt{2}} + \frac{\sqrt{2}a_2 + \sqrt{2}b_2i}{\sqrt{2}} = \frac{(a_1-b_1+\sqrt{2}a_2)}{\sqrt{2}} + \frac{(a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}i$$

Then we just need to find $\Re(c) + \Im(c)$:

   $$ \Re(c) + \Im(c) = \frac{(a_1-b_1+\sqrt{2}a_2) + (a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}= \frac{a_1+ \sqrt{2}{a_2}+\sqrt{2}b_2}{\sqrt{2}}$$

or

   $$ \Re(c) + \Im(c) = \frac{\Re(c_1) + \sqrt{2}\left ( \Re(c_2)+ \Im{c_2} \right ) }{\sqrt{2}}$$


Why is the coefficient of a fraction $\frac{1}{denominator}$ instead of simply 1? Why is the coefficient of a fraction $\frac{1}{denominator}$ instead of simply $1$?

Wouldn't the result be the same?

Depends of where you are taking the coefficients. For example Q is generated by 1, taking the coefficients in Q, but if you take them in Z you get just the integers and not all rationals. This becomes clearer when one studies modules and vector spaces.


Limits and Continuity: $\lim\limits_{n\to\infty}e^{\frac{1}{n}}$ Well, I'm doing this and I already find a resolution. I can write that limit as:
$e^{\lim\limits_{n\to\infty}\frac{1}{n}}$ and that results in $1$.

  My question is: Why can I do this?

A function $f(x)$, is continuous at a point, $a$ if:

$$\lim\limits_{x\to a} f(x) = f(a)$$

This is the definition of continuity. So if $n\to \infty$, then $\frac1n \to 0$, so:

$$\lim\limits_{n\to\infty} e^{1/n}=\lim\limits_{x\to 0} e^x=e^0=1$$

because $e^x$ is continuous at $0$.

You can do it because when $n$ gets really big $ \frac{1}{n}$ is going to get really close to zero, and anything to the zeroth power is always one. $e^{0} = 1$. I'm not sure if that answers your question though.

Jeez, constant downvotes. Anyone care to tell me what I did wrong?


sum of three inverse matrices The following Searle identity computes the sum of two inverses:

$A^{-1}+B^{-1} = A^{-1}(A+B)B^{-1}$.

Is there any generalisation of this for the sum of three inverses?

$A^{-1}+B^{-1}+C^{-1} = ? $

yes, we can generalise.

$$ A^{-1}+B^{-1}+C^{-1}= BB^{-1}CC^{-1}A^{-1}+AA^{-1}B^{-1}CC^{-1}+AA^{-1}BB^{-1}C^{-1}$$
$$ =A^{-1}B^{-1}C^{-1}(AB+BC+CA) $$
That is, sum of the n inverse matrices is equal to product of the n inverse matrices and sum of the original matrices taken (n-1) at a time.


In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?

I just dipped into a book, The Drunkard's Walk - How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3.

It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.

In a family with 2 children there are four possibilities:

1) the first child is a boy and the second child is a boy (bb)

2) the first child is a boy and the second child is a girl (bg)

3) the first child is a girl and the second child is a boy (gb)

4) the first child is a girl and the second child is a girl (gg)

Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.

There are many ways to explain this, but let me share with you how I do it in my class. I set $q=P(\mbox{You say that he has at least one daughter} \mid \mbox{He has $\{B,G\}$})$.

If you sample at family level, then you ask the father if he has at least a daughter or you check official records. In that case $q=1$.

If you sample at child level, by for instance meeting the father in the company of one child or otherwise happening to find out about one child, then $q=1/2$. In this case you will half the time miss this family because you happen to see the son first.

Using Bayes' formula, we get
$$
P(\{B,G\}) \mid \mbox{You say that he has at least one daughter})
= \frac{P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\})}
{P(\mbox{You say that he has at least one daughter})}.
$$

Since
$$P(\mbox{You say that he has at least one daughter}) = P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\}) + P(\mbox{You say that he has at least one daughter} \mid \{G, G\}) P(\{G, G\}),
$$

we get
$$
P(\{B,G\} \mid \mbox{You say that he has at least one daughter}) = (q \times 1/2)/(q \times 1/2 + 1 \times 1/4) = 2q/(1 + 2q),
$$

and it follows that
$$
P(\{G, G\} \mid \mbox{You say that he has at least one daughter})=1/(1 + 2q).
$$

If $q=1/2$, we get $P=1/2$, and $q=1$ gives $P=1/3$.


Is there any mathematician who felt guilty for one of his math discoveries ever? Quoted from Wikipedia:

  In 1888 Alfred Nobel's brother Ludvig died while visiting Cannes and a French newspaper erroneously published Alfred's obituary. It condemned him for his invention of dynamite and is said to have brought about his decision to leave a better legacy after his death. The obituary stated, Le marchand de la mort est mort ("The merchant of death is dead") and went on to say, "Dr. Alfred Nobel, who became rich by finding ways to kill more people faster than ever before, died yesterday." Alfred was disappointed with what he read and concerned with how he would be remembered.

Alfred Nobel felt guilty for invention of dynamite. He shares this emotion with many other scientists of the human history who became disappointed when they realized that their discoveries make the world a worse place to live. But what about mathematicians? We know almost all scientific discoveries are based on mathematical theorems and concepts but it seems feeling guilty is much more epidemic amongst physicists, chemists and biologists rather than mathematicians.

  Question: Is there any mathematician who felt guilty for one of his math discoveries ever? If yes, what was that particular theorem or concept? Please introduce historical references and other quotations which confirm existence of such an emotion.

Einstein, while not strictly a mathematician, expressed profound regrets later in his life when he pondered on the possibility of an atomic war. In 1939, he wrote to Roosevelt to urge the Americans to develop the atomic bomb before Germany, and Roosevelt took action as a direct result of this letter. In 1954, in an interview to The Reporter, Einstein said:

If I would be a young man again and had to decide how to make my
living, I would not try to become a scientist or scholar or teacher. I
would rather choose to be a plumber or a peddler in the hope to find
that modest degree of independence still available under present
circumstances.

(He is often credited to have said: "If only I had known, I should have become a watch-maker." But it appears that he never said those exact words.)

I've heard apocryphal stories that Newton was reticent to publish his calculus, concerned that the human race couldn't handle the power.  He only begin publicizing it when Leibniz started publishing prolifically, about 20 years later.  Perhaps somebody has a citation; I don't, hence I consider it apocryphal.


Prove that a subset $W$ of a vector space $V$ is a linear subspace of $V$ From Friedberg, 4th edition:

  Prove that a subset $W$ of a vector space $V$ is a subspace of $V$ if
  and only if $W \neq \emptyset$, and, whenever
   $a \in F$  and $x,y \in
 W$, then $ax \in W$ and $x + y \in W$.

I understand that in order to be a subspace, $W$ must contain the element $0$ such that for any element $x \in W$, it satisfies $x+0=x$. If $W$  is null, it cannot be a subspace, therefore its contrapositive statement must be true: If $W$ is a subspace it cannot be null.

I also understand that a subspace must be closed under addition and scalar multiplication. $W$ follows both these properties as stated in the question, but I am unsure as to how to 'prove' this. Aren't the conditions set in the question the definition of what it means to be a subspace? Is it sufficient to state that $W$ satisfies both these conditions and is therefore a subspace?

From the question, we know that $W$ is closed under addition and scalar multiplication. So all we have to do to prove that $W$ is a subspace is to show that $W$ contains the zero vector.
So, if $x$ is in $W$, and $ax$ is in $W$, then we can take $a=-1$. By this, we have -$x$ is in $W$ (Closure under scalar multiplication). Now, because we know that $W$ is closed under addition, we have $x+(-x)=x-x=0$, which is in $W$, thus showing that $W$ must contain the zero vector.

Is it okay to use 0,-1\in K as we never ask the existence of 0 or -1\in K in the definition of vector space


Orthogonal projection onto Hyperplane Let $Ax = y$ be a linear system with $A \in \Bbb R^{n,m}, n<m$. Consider $H_i := \{x \in \Bbb R^{m}: A_ix = y_i\}, i \in \{1,\dots,n\}, y_i \in \Bbb R$ where $A_i$ denotes the $i$-th row of $A$. Now, regard $\mathcal{P}_{H_i}z := z -  {A_i}^{\intercal} {y_i - A_i z  \over \|A_i\|^2} $. This is apparently an orthogonal projection onto the hyperplane $H_i$ but I fail to show it. Any help is appreciated.

$Px$ is the orthogonal projection onto $H$ if and only if $Px\in H$ and $\langle x-Px,z-Px\rangle = 0$ for all $z\in H$.

If $Px = x + \frac{y_i-A_ix}{\|A_i\|^2}A_i^T$ (there must be a plus!), then $A_iPx = A_ix + \frac{y_i-A_ix}{\|A_i\|^2}A_iA_i^T = A_ix + (y_i-A_ix) = y_i$, hence $Px\in H_i$. Also, if $z\in H_i$,
\begin{align}
\langle x-Px,y-Px\rangle
&= -\frac{y_i-A_ix}{\|A_i\|^2}\left\langle A_i^T,z-x-\frac{y_i-A_ix}{\|A_i\|^2}A_i^T\right\rangle\\
&= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - A_ix - \frac{y_i-A_ix}{\|A_i\|^2}A_iA_i^T\right)\\
&= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - A_ix - y_i + A_ix\right)\\
&= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - y_i\right) = 0,
\end{align}
since $A_iz=y_i$. So, $P$ is indeed the orthogonal projection onto $H_i$.

Hint: To show it, it's enough to prove $\mathcal P_{H_i}(z)$ is $z$ if $z\in H_i$ and is $0$ if $z\perp H_i$ (i.e. if $z=\lambda A_i^T$).


Fourier transform for dummies What is the Fourier transform? What does it do? Why is it useful (in math, in engineering, physics, etc)?

This question is based on the question of Kevin Lin, which didn't quite fit in Mathoverflow. Answers at any level of sophistication are welcome.

The ancient Greeks had a theory that the sun, the moon, and the planets move around the Earth in circles.  This was soon shown to be wrong.  The problem was that if you watch the planets carefully, sometimes they move backwards in the sky.  So Ptolemy came up with a new idea - the planets move around in one big circle, but then move around a little circle at the same time.  Think of holding out a long stick and spinning around, and at the same time on the end of the stick there's a wheel that's spinning.  The planet moves like a point on the edge of the wheel.

Well, once they started watching really closely, they realized that even this didn't work, so they put circles on circles on circles...

Eventually, they had a map of the solar system that looked like this:

This "epicycles" idea turns out to be a bad theory.  One reason it's bad is that we know now that planets orbit in ellipses around the sun.  (The ellipses are not perfect because they're perturbed by the influence of other gravitating bodies, and by relativistic effects.)

But it's wrong for an even worse reason that that, as illustrated in this  wonderful youtube video.

In the video, by adding up enough circles, they made a planet trace out Homer Simpson's face.  It turns out we can make any orbit at all by adding up enough circles, as long as we get to vary their size and speeds.

So the epicycle theory of planetary orbits is a bad one not because it's wrong, but because it doesn't say anything at all about orbits.  Claiming "planets move around in epicycles" is mathematically equivalent to saying "planets move around in two dimensions".  Well, that's not saying nothing, but it's not saying much, either!

A simple mathematical way to represent "moving around in a circle" is to say that positions in a plane are represented by complex numbers, so a point moving in the plane is represented by a complex function of time.  In that case, moving on a circle with radius $R$ and angular frequency $\omega$ is represented by the position

$$z(t) = Re^{i\omega t}$$

If you move around on two circles, one at the end of the other, your position is

$$z(t) = R_1e^{i\omega_1 t} + R_2 e^{i\omega_2 t}$$

We can then imagine three, four, or infinitely-many such circles being added.  If we allow the circles to have every possible angular frequency, we can now write

$$z(t) = \int_{-\infty}^{\infty}R(\omega) e^{i\omega t} \mathrm{d}\omega.$$

The function $R(\omega)$ is the Fourier transform of $z(t)$.  If you start by tracing any time-dependent path you want through two-dimensions, your path can be perfectly-emulated by infinitely many circles of different frequencies, all added up, and the radii of those circles is the Fourier transform of your path.  Caveat: we must allow the circles to have complex radii.  This isn't weird, though.  It's the same thing as saying the circles have real radii, but they do not all have to start at the same place.  At time zero, you can start however far you want around each circle.

If your path closes on itself, as it does in the video, the Fourier transform turns out to simplify to a Fourier series.  Most frequencies are no longer necessary, and we can write

$$z(t) = \sum_{k=-\infty}^\infty c_k e^{ik \omega_0 t}$$

where $\omega_0$ is the angular frequency associated with the entire thing repeating - the frequency of the slowest circle.  The only circles we need are the slowest circle, then one twice as fast as that, then one three times as fast as the slowest one, etc.  There are still infinitely-many circles if you want to reproduce a repeating path perfectly, but they are countably-infinite now.  If you take the first twenty or so and drop the rest, you should get close to your desired answer.  In this way, you can use Fourier analysis to create your own epicycle video of your favorite cartoon character.

That's what Fourier analysis says.  The questions that remain are how to do it, what it's for, and why it works.  I think I will mostly leave those alone.  How to do it - how to find $R(\omega)$ given $z(t)$ is found in any introductory treatment, and is fairly intuitive if you understand orthogonality.  Why it works is a rather deep question.  It's a consequence of the spectral theorem.

What it's for has a huge range.  It's useful in analyzing the response of linear physical systems to an external input, such as an electrical circuit responding to the signal it picks up with an antenna or a mass on a spring responding to being pushed.  It's useful in optics; the interference pattern from light scattering from a diffraction grating is the Fourier transform of the grating, and the image of a source at the focus of a lens is its Fourier transform.  It's useful in spectroscopy, and in the analysis of any sort of wave phenomena.  It converts between position and momentum representations of a wavefunction in quantum mechanics.  Check out this question on physics.stackexchange for more detailed examples.  Fourier techniques are useful in signal analysis, image processing, and other digital applications.  Finally, they are of course useful mathematically, as many other posts here describe.

Fast Fourier Transform is used in Engineering to reduce Computation time for solving Matrix Algebraic Equations and Matrix Difference Equations.


Find all the integer pairs $(r,s)$ that satisfy $s= (r^2 +3r +8) / (r^2 +r -2)$? I have been trying to solve this question but struggling to see where to start. Examples I've seen that works are the pairs: $(-3,2) , (4,2), (0,-4)$

Write the given equation as

\begin{align}
s&=1+2\left(\frac{r+5}{(r+2)(r-1)}\right)\\
 &=1+\frac{4}{r-1}-\frac{2}{r+2}.
\end{align}

With this, the only values of $r$ that I see satisfying the above equation leading to integer values of $s$ are $r=-5,-3,-1,0, 4$.


$K$-basis of cyclic $KG$-module determined by group elements Let $G$ be a finite group and $K$ a field with $\mathrm{char}(K) \nmid |G|$, and let $M$ be a cyclic $KG$-module.

For a cyclic generator $v \in M$ we can pick group elements $g_1, \dots, g_d \in G$ such that $\{ g_1v, \dots, g_d v \}$ is a $K$-basis of $M$. Does that automatically mean that $\{ g_1w, \dots, g_d w \}$ is a $K$-basis for all generators $w$ of $M$ as well? If not, can we choose those $g_i$ in a special way such that the answer becomes positive?

Thank you in advance!

What I know so far:
The set of all elements $w \in M$ such that $\{g_1 w, \dots, g_d w\}$ is a $K$-basis of $M$ is Zariski-open. To see this, we can assume w.l.o.g. $M = K^d$ and $G \leq \mathrm{GL}(d,K)$. Then the set of those elements $w$ coincides with the non-vanishing set of the polynomial $\det(g_1 X, \dots, g_d X)$, where $X = (X_1, \dots, X_d) \in K[X_1, \dots, X_d]^d$.

Therefore “almost all” generators $w$ form a $K$-basis of $M$ in this way. What about the remaining ones?

Let $G$ be the (dihedral of order $8$) subgroup of $\operatorname{GL}(2,\mathbb{R})$ generated by $g:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and  $h:=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

Then every non-zero element of $\mathbb{R}^2$ generates $\mathbb{R}^2$ as an $\mathbb{R}G$-module. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\{v,gv\}$ is a basis, but if $w=\begin{pmatrix}0\\1\end{pmatrix}$ then $gw=w$, so $\{w,gw\}$ is not a basis.

In this example $\{w,hw\}$ is a basis for every non-zero vector $w$, so you can choose the group elements so that your question has a positive answer. But if you take $G$ to be the group of rotational symmetries of a cube acting on $\mathbb{R}^3$ in the obvious way, then again every non-zero vector is a generator, but for any three distinct elements $g_1,g_2,g_3$, there is some non-zero $u\in\mathbb{R}^3$ fixed by $g_2g_1^{-1}$ and so for $w=g_1^{-1}u$, we have $g_1w=g_2w$ and so $\{g_1w,g_2w,g_3w\}$ is not a basis.

I think your trial will be not success. Because of the point of view "decomposition to the irreducible representations";

$K[G] \cong \oplus_{i=1}^h V_i$ as $K[G]$-module for some sub $K[G]$-modules $\{V_i\}_i$. (Giving $K[G]$-module and giving representation of $G$ are equivalent. And it is possible to take $\{V_i\}$ irreducible $K[G]$-modules, so it gives the decomposition to irreducible representation).

Of course $K[G]$ is a cyclic $K[G]$-algebra, but if you take element $w \in V_i$ for some $i$, $K[G]w \subset V_i$.


Prove that $i^i$ is a real number According to WolframAlpha, $i^i=e^{-\pi/2}$ but I don't know how I can prove it.

Here's a proof that I absolutely do not believe: take its complex conjugate, which is $\bigl({\bar i}\bigr)^{\bar i}=(1/i)^{-i}=i^i$. Since complex conjugation leaves it fixed, it’s real!

EDIT: In answer to @Isaac’s comment, I think that to justify the formula above, you have to go through exactly the same arguments that most of the other answerers did. For complex numbers $u$ and $v$, we define $u^v=\exp(v\log u)$. Now, the exponential and the logarithm are defined by series with all real coefficients; alternatively you can say that they are analytic, sending reals to reals. Thus $\overline{\exp u}=\exp(\bar u)$ and $\overline{\log(u)}=\log\bar u$. The result follows, always sweeping under the rug the fact that the logarithm is not well defined.

Let say: $f(z) = z^\theta $
We know Euler's formula: $e^{i \theta} = \cos(\theta) + i \sin(\theta)$
Using it we will get:
$$z^\theta = e^{\theta \ln(z)} = e^{\theta (\ln\|z\| + i\arg (z))} = e^{\theta \ln\|z\| }e^{ i \theta \arg (z))}= e^{\theta \ln\|z\| }{(\cos(\theta\arg (z)) + i \sin(\theta\arg (z)))} z \in C$$

So if $$z = i \wedge \theta = i \implies$$ $$ z^\theta = i^i = e^{i \ln(i)} = e^{i (\ln\|i\| + i\arg (i))} = e^{i \ln\|i\| }e^{ i i \arg (i))}= e^{i \ln(1) }e^{- \frac{\pi}{2}+2\pi k}= e^0 e^{- \frac{\pi}{2}+2\pi k}=e^{- \frac{\pi}{2}+2\pi k} $$ which is a bunch of Real numbers depending $k \in \mathbb Z$

So it is already proved that $i^i$ is a real number.


Is there a great mathematical example for a 12-year-old? I've just been working with my 12-year-old daughter on Cantor's diagonal argument, and countable and uncountable sets.

Why? Because the maths department at her school is outrageously good, and set her the task of researching a mathematician, and understanding some of the maths they did - the real thing.

So what else could we have done - thinking that we know our multiplication tables and fractions, but aren't yet completely confident with equations which have letters for unknown numbers?

I did think of proving that there are infinitely many primes - we can follow an argument - other suggestions welcome.

And incidentally, tell your local high school to do this ...

Six people at a dinner party is sufficient to ensure that there are either three mutual strangers or three mutual acquaintances. In fact, six is the smallest number that ensures this phenomenon. This is the diagonal Ramsey number $R(3,3)$, and the proof can be demonstrated with a couple pictures and just a dash of the pigeonhole principle. There are lots of directions she could go after understanding $R(3,3)$ (though most of it is not due to Ramsey).

Show her that 2 + 2 may not equal 4, but for example may equal 12, and explore the implications of that fact with her! Subtracting from each side 2 and then 2 again you'd get 0 = 8, and this would be a great intro to modular arithmetic.

Moreover, this is a great example of a situation in which dismissing what's known and allowing a seemingly wrong answer can open new doors. And it is also a way of showing that, in maths, there are no wrong ways (unless you contradict yourself, of course), but only roads that may lead nowhere (but that are worth exploring though).

An intro to quotients with simple and concrete examples would be mind-blowing as well, I believe. Let her come up with her own answer to what happens if, on the [0, 1] segment, we make 1 and 0 be the same element (we get a circle!). Then you can go on with more difficult examples and build a cylinder, a torus and a mobius string from a unit square! Funny enough, modular arithmetic is based upon quotients!
Show that the commutative property may not hold in infinite cases. Take the $\sum_{i=0}^\infty(-1)^n$ and ask her what she thinks it may sum up to. That sum waves up and down never approaching anything, but rearranging its terms it can be made divergent.

This is a great example of how things may not always be true in more difficult cases, even if they are in simpler ones.


If eigenvalues are positive, is the matrix positive definite? If the matrix is positive definite, then all its eigenvalues are strictly positive.

Is the converse also true?
That is, if the eigenvalues are strictly positive, then matrix is positive definite?
Can you give example of $2 \times 2$ matrix with $2$ positive eigenvalues but is not positive definite?

I think this is false. Let $A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle < 0$.
The point is that the matrix can have all its eigenvalues strictly positive, but it does not follow that it is positive definite.

For a matrix to be positive definite:

1) it must be symmetric
2) all eigenvalues must be positive
3) it must be non singular
4) all determinants (from the top left down the diagonal to the bottom right - not jut the one determinant for the whole matrix) must be positive.

If a 2x2 positive definite matrix is plotted it should look like a bowl. If the matrix is singular then it's a trough which follows the vector which takes the matrix to zero. If the sum of the cross terms is bigger than the trace then the plot is a saddle at zero.


A Tricky Math Question Along a road lie an odd number of stones placed at intervals of 10 metres.These stones have to be assembled around the middle stone. A person can carry only one stone  at a time. A man carried the job with one of the end stones by carrying them in  succession. In carrying all the stones he covered a distance of 3 km. Find the number of stones

Let us measure in units of $10$ meters, so he goes $300$ units.  If there are $n$ stones each side of center, he walks $n$ with the first stone, then $2(1+2+3+\dots (n-1))$ to collect all the stones on the side he started from, then $2(1+2+3+\dots n)$ to collect the ones on the other side.  $n+n(n-1)+n(n+1)=300, (2n+25)(n-12)=0, n=12$ so there are $25$ stones total.

its 301...3000meters/10 + 1(middle stone)


Condition for common roots of two Quadratic equations: $px^2+qx+r=0$ and $qx^2+rx+p=0$ The question is:

  Show that the equation $px^2+qx+r=0$ and $qx^2+rx+p=0$ will have a common root if $p+q+r=0$ or $p=q=r$.

How should I approach the problem? Should I assume three roots $\alpha$, $\beta$ and $\gamma$ (where $\alpha$ is the common root)? Or should I try combining these two and try to get a value for the Discriminant? Or should I do something else altogether?

The "if" part is clear, so we'll deal with the "only if". Moreover, we take the polynomials to be "true" quadratics ---that is, $p$ and $q$ are non-zero--- since otherwise the proposition is false (as @Winther mentions in a comment to the OP).

First, note that quadratics with a common root could have both roots in common, in which case they are equivalent. This means that multiplying-through by some $k$ turns one quadratic equation into the other, coefficient-wise:
$$q = k p, \quad r = k q, \quad p = k r \quad\to\quad p = k^3 p \quad\to\quad p(k^3-1) = 0 \quad\to\quad k^3 = 1$$

  Thus, since $k=1$, we have $p = q = r$.

If the quadratics don't (necessarily) have both roots in common, but do have root $s$ in common, then
$$\left.\begin{align}
p s^2 + q s + r &= 0 \quad\to\quad p s^3 + q s^2 + r s \phantom{+p\;} = 0 \\
q s^2 + r s + p &= 0 \quad\to\quad \phantom{p s^3 +\, } q s^2 + r s + p = 0
\end{align} \quad\right\}\quad\to\quad p(s^3-1) = 0 \quad\to\quad s^3 = 1$$

  Thus, since $s = 1$, substituting back into either polynomial gives $p+q+r=0$.

Easy-peasy!

But wait ...  The equations $k^3 = 1$ and $s^3 = 1$ have fully three solutions: namely, $\omega^{0}$, $\omega^{+1}$, $\omega^{-1}$, where $\omega = \exp(2i\pi/3) = (-1+i\sqrt{3})/2$. Nobody said coefficients $p$, $q$, $r$, or common root $s$, were real, did they?

If $k = \omega^n$, then we have in general that

  $$q = p \omega^n \qquad r = p \omega^{-n}$$

If $s = \omega^n$, then substituting back into either quadratic, and multiplying-through by an appropriate power of $\omega^{n}$ for balance, gives

  $$p\omega^{n} + q + r \omega^{-n} = 0$$

Just-slightly-less-easy-but-nonetheless-peasy!

If $p + q + r = 0 \to 1 $ is a root, otherwise both become one equation and so they surely have the same roots.


Inverse of matrix whose row sums is 1 Given a $10 $x$ 10$ invertible matrix with real entries, the sum of whose rows is $1$, prove that for the inverse, the sum of the elements in each row is $1$ ?  {Question from TIFR GS-2015}

"Sum of all rows equals $1$" means precisely that the vector $(1, \dotsc, 1)^T$ is a fixed point. Of course this vector is then a fixed point of the inverse, too.

Let $\textbf{1}$ be $10\times1$ vector with each component $1$ . Then $A(\textbf{1})=1(\textbf{1})$. So, $1$ is an eigenvalue of A, thus of $A^{-1}$ also. Thus, $1(\textbf{1})=A^{-1}(\textbf{1})$ (left multiply by $A^{-1}$), which implies that the sum of each row of the inverse matrix is also $1$.


Find the curvature of a curve given by it's implicit equation Please help me solve this:

Find the curvature of a curve given by it's implicit equation:

$ F(x,y) = 0 $

Now, I thought of solving this by converting it to a parametric equation, but I have no idea how to do so. Any tips?

Suppose this does define a smooth parametric curve $x = X(s),\; y = Y(s)$ parametrized by arc length $s$ in the neighbourhood of a point $(x_0, y_0) = (X(0), Y(0))$.  Differentiating the equation $F(X(s),Y(s))=0$ twice and using the chain rule (and using subscript notation for partial derivatives)

$$ \eqalign{F_1(X,Y) X' &+ F_2(X,Y) Y' = 0\cr
            F_{11}(X,Y) X'^2 &+ 2 F_{12}(X,Y) X' Y' + F_{22}(X,Y) Y'^2 + F_1(X,Y) X'' + F_2(X,Y) Y'' = 0\cr}$$
Since the parameter is arc length, the velocity vector $V =[X',Y']$ has length $1$ and is the unit tangent vector, i.e.
$$ X'^2 + Y'^2 = 1$$
and differentiating this gives us $$X' X'' + Y' Y'' = 0$$

The curvature is then $\kappa = |V'| = \sqrt{X''^2 + Y''^2}$.

Eliminating $X'$ and $Y'$ from the equations is rather tedious, but  should give us

$$ \kappa = \frac{\left| F_{11} F_2^2- 2 F_{12} F_1 F_2 + F_{22} F_1^2\right|}{(F_1^2 + F_2^2)^{3/2}}$$

WE can find  Implicit form of Curvature by implicit differentiation in the plane

Formula (3.1)


When does a variety have a point over a finite field for sufficiently large primes p? Let $X$ be an algebraic variety over the rational numbers. Suppose that $X$ has positive dimension. I would like to say that $X(\mathbb{F}_p)$ is non-empty for sufficiently large primes $p$. One idea is to use the Weil conjectures, but that seems like overkill. So my two questions are:

When can one deduce that a system of polynomial equations has a solution over $\mathbb{F}_p$ for sufficiently large primes $p$? I might have guessed that $\dim(X) > 0 $ is sufficient, but it seems like the affine variety $X^2+Y^2 = 0$ minus the point $(0,0)$ (which has no solutions for $p = -1 \mod 4$) suggests that extra conditions are necessary.
Is there an elementary proof of the (correct) version of part 1?

Using the Lang-Weil bound and maybe also the Chebotarev density theorem or something like that, I believe the correct statement is that if $\dim X \ge 1$ then $X(\mathbb{F}_p)$ is non-empty for all sufficiently large primes $p$ if and only if the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the set of irreducible components of $X_{\overline{\mathbb{Q}}}$ fixes at least one component of positive dimension. This condition doesn't hold in your example; there are two irreducible components over $\mathbb{Q}(i)$, given by the lines $X + iY = 0$ and $X - iY = 0$ (minus the origin), and the action of complex conjugation switches them.
I don't see any way to use a weaker result than the Lang-Weil bound.
I also want to point out that strictly speaking if $X$ is a variety over the rational numbers then it doesn't make sense to talk about $X(\mathbb{F}_p)$ because $\text{Spec } \mathbb{F}_p$ is not a variety (or scheme or anything) over the rational numbers; when people write expressions like this they are implicitly choosing a lift of $X$ to a scheme defined over some localization of $\mathbb{Z}$ defined by the denominators of some collection of polynomials cutting out $X$, and the behavior of $X(\mathbb{F}_p)$ for any fixed prime $p$ could a priori depend on this choice. I think it's true that the question of whether $X(\mathbb{F}_p)$ is non-empty for sufficiently large $p$ does not depend on this choice but I can't quote a theorem off the top of my head that guarantees this. In the Lang-Weil argument above let's say we're fixing a particular lift to keep things safe.

If $k$ is the field of rational numbers and $A$ is a finitely generated $k$-algebra with $X:=Spec(A)$, it follows $X(\mathbb{F}_p)$ is empty for any prime number $p>0$. If $\mathfrak{m}\subseteq A$ is a maximal ideal it follows $k \subseteq \kappa(\mathfrak{m})$ is a finite field extension of $k$, but $k$ has characteristic zero and hence this is impossible.
If $B:=\mathbb{Z}[x_1,..,x_n]/I$ where $\mathbb{Z}$ is the ring of integers and $Y:=Spec(B)$, we get a canonical map $\pi: Y \rightarrow S:=Spec(\mathbb{Z})$.
If you accept Zorns's Lemma/the "Axiom of Choice", it follows there is a maximal ideal $\mathfrak{m} \subseteq B$. The residue field $\kappa(\mathfrak{m})\cong \mathbb{F}_{q^r}$ is finite with characteristic $q>0$. Hence $\mathfrak{m}\cap \mathbb{Z}=(q)$, and it follows the fiber $\pi^{-1}(q)\subseteq Y$ is non-empty. Hence Zorn's Lemma  proves that $Y(\mathbb{F}_{q^r})$ is always non-empty for some $q,r$. The fact that the residue field $\kappa(\mathfrak{m})$ of a maximal ideal in $B$ is finite, is proved in the exercises in Atiyah-Macdonald's book "Commutative Algebra", hence this argument is in some sense "elementary". I have also seen simple online proofs that the residue field $\kappa(\mathfrak{m})$ is finite as well.


Finding multi dimensional, analytical relationship (curve fitting) for experimental data I have an unknown relationship like

$$
y = f(M, R, \sigma, P)
$$

I would like to find an analytical relationship that fits observed data the best.

As an example, if I keep $\sigma=0.3$ and $P=7$ constant, I get the following curves from the experiment:

I can fit each of these curves into equations of the form

$$
y = a (M + b)^n
$$

where $a$, $b$, $n$ are fitting parameters which I find via Least Squares.

However, how do I incorporate the relationship for $R$?

If I repeat the experiment but pick another $\sigma$ and $P$, the curves have the same shape (i.e., can  be described again by $y = a (M + b)^n$) but $a$, $b$, $n$ are all different. How can I incorporate $\sigma$ and $P$ to find the generic $f$ of a form $y = a (M + b)^n$?

There are two ways:

You know the science behind your experiment, so you can guess an analytic form. It looks like it's not the case for your problem.
You need to fit for many $R, \sigma, P$ parameters the curve and plot $a, b, n$ as a function of $R, \sigma, P$


Is the function T a linear transformation? Let the function $T$ be a mapping from $\mathbb{R^3} \rightarrow \mathbb{R^4}$ given by

$$T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix}1\\2\\-1\\-2\end{bmatrix}$$

Is this function a linear transformation?

Every linear transformation applies $$T(0)=0$$
The given transformation : $$T(0,0,0)=(1,2,-1,-2) \neq (0,0,0,0)$$
Hence, Not a linear transformation.

It might also be helpful to consider the proof of why T(0) must be 0 from the axioms of linearity

0 in a field with (second operation multiplication, which is what we consider here in this scalar multiplication) is defined to be the element at which $\forall f \in F$, we have   $0 + f = f + 0 = f$ (note that I made this definition two-sided, because in any ring [all fields are rings] the additive operation forms an abelian group). It can be shown using the distributive axiom of rings that $0 * f = f* 0 = 0$. This is also true for rings in general, not just fields, but we use fields with vector spaces for more useful axioms.
https://en.wikipedia.org/wiki/Proofs_of_elementary_ring_properties may also be helpful for general ring/field properties.

By linearity we have $T(0 * v) = 0 * T(v)$. This is not finished, but that tangent on fields was to really show that vector spaces have very similar structure, except that scalar multiplication is NOT binary operation. However, we can use the distributive property in similar manner as in Workaholic's answer. Vector spaces - Multiplying by zero scalar yields zero vector

Then we can see that this transformation will not be linear we the transformation applied to $\vec{0}$ is not also $\vec{0}$.


Column and Row Picture for Singular System of 100 Equations (Strang P55, 2.2.32) Start with 100 equations $\color{#8F00FF}{A}\mathbf{x} = \mathbf{0}$ for $\mathbf{x} = (x_1, ..., x_{1oo})$. Suppose elimination reduces the 100th equation to $0 = 0$, so the system is "singular".

  (a) Elimination takes linear combinations of the rows. So this singular system has
  the singular property: Some linear combination of the 100 rows is $\color{green}?$.

  (b) Singular systems $\color{purple}{A}\mathbf{x} = \mathbf{0}$ have infinitely many solutions. This means that some linear combination of the 100 columns is $\color{  #FF4F00}{??}$.

  (d) For a 100 by 100 singular matrix with no zero entries, describe in words the row picture and the column picture of
  $\color{  #9966CC}{A}\mathbf{x} = \mathbf{0}$. Not necessary to draw 100-dimensional space.

My Attempt for (a) and (b): The question postulates that for all $ 1 \leq r \leq 100, \sum_{c = 1}^{100}\color{#8F00FF}{a_{rc}}x_c=0.$ Pictorially:

Hitherto, my definition of "singular" is "dependent rows or columns" (cf P27). So $? = \color{green}{\vec{0}}$ and $?? = \color{  #FF4F00}{\vec{0}}$.

  Averred Answer for (d): The row picture has $100$ planes meeting along a common line $\color{#0073CF}{\text{through } \vec{0}}$. The
  column picture has $100$ vectors all in the same $\color{#0073CF}{99}$-dimensional hyperplane.

$\Large{1.}$ Why must the common line be $\color{#0073CF}{\text{through } \vec{0}}$? Why not any common line?

$\Large{2.}$ $A$ contains $100$ columns so $100$ column vectors. But $A$ also contains $100$ rows, so how and why $\color{#0073CF}{99}$-dimensional hyperplane?

Please mind that this question is from a germinal section of IoLA, 4th ed by Strang.

elimination takes linear combinations of the rows. So this singular system has singular property some linear combination of the 100 row is non singular
singular systems Ax=0 have infinitely many solutions This means that some linear combination of the 100 columns is non zero
when Ax=0 then linear combination of rows and columns will be same and donot have value greater than zero,


Knowing which theorem of calculus to use to prove number/nature of solutions How does one go about showing that the equation $$ arctanx = x^2  $$ has at least one solution and then in turn show that the equation has exactly one positive solution?

I figured for part a) "show at least one solution" it could be possible to rewrite
the equation as $$ x^2 - arctanx = 0 $$ and apply the intermediate value theorem. and for part b) "exactly one positive solution"  to examine the behaviour of the second derivative to get (concave up or down, could that give an indication of if the graph only has one positive solution?) but I'm not exactly sure if these are the correct approaches. Any suggestions as to the correct method would be much appreciated.

I am presuming that by $\arctan x$ you just mean one branch of that function that whose range is $[\pi,\pi]$ (There are more branches obtained by adding $2\pi k$ to that branch). Obviously $x=0$ is a solution. As $x$ grows from $0$ it obeys $\arctan x > x$ for small positive $x$. (this can easily be deduced from $y > \tan y$ from elementary trigonometry by taking $x=\tan y$). So f(x)=$x^2-\arctan x$ is negative for small positive values less than 1 since $x^2<x<\arctan x$ But this branch of $\arctan x$ is bounded above by $\pi$ and $x^2$ is unbounded so eventually they have to meet at the only positive solution $x_1$ of $f(x)=0$. For $x>x_1$ obviously $x^2>\arctan x$ and $f(x)$ is positive. This the only solution since the derivative of $x^2$ is $2x$ and is growing with $x$ so the function $x^2$ will always stay on one side of its tangent line at $x_1$, whereas the derivative of $\arctan x$ is $1/(1+x^2)$ is decreasing with increasing $x$ and thus its tangents will always stay on one side of its tangent at the point of intersection. We need to show that $2x>1/(1+x^2)$ This can be done by considering the function $2x^3+2x-1>0$ that is monotonously increasing. It has a zero at $x_0=0.42385$ where $f(x)<0$ so $x_0<x_1$ and the 2 functions stay away from each other for $x>x_1$
Note that if we consider more branches $\arctan x$ there will be more positive solutions to the right of $x_1$

Let $f(x)=x^2-arctanx$. Determine $f'(x)$. Then you should see that the equation $f'(x)=0$ has two solutions. Hence, by the mean value theorem, the equation $f(x)=0$ has at most three solutions !

Edit: this answer is not correct ! FRED


How to understand probability density function? I have a probability density function like this:

p(x) = $\begin{cases}1,&0\le x\le1 \\0,& \text{otherwise}   \end{cases}$

I want to know how to understand this PDF. Does this mean the probability of any value x $\in$ [0,1] is p(x)? I guess it's not. Then, if I want to know the probability that x=0.5, is P(0.5)=1 right? If it's right, $\int$p(x)dx=1 could be failed. So how to understand the PDF please?

The PDF gives you the relative probability of one value over another. In your example, we see that if $0 \le x_1 \le 1$ and $0 \le x_2 \le 1$, then $p(x_1)=1$ and $p(x_2)=1$, so the probabilities will be equal. Your PDF represents a uniform random variable over $[0,1]$. If you had something like this:

$$p(x) = \begin{cases}1,&0\le x\le\frac{1}{2} \\2,&\frac{1}{2}<x\le1\\0,& \text{otherwise}   \end{cases}$$

Then you would say that the probability of $P(X=1)$ is twice the probability of $P(X=0)$. The PDF does not say anything about the actual probabilities.


Constructing the natural numbers without set theory. As I understand it the idea of defining everything as sets is a relatively new idea in mathematics.  Does that mean there's a non-set theoretic definition of the natural numbers?  Could there be?

One way to construct the natural numbers (without sets) is type theory. We define an inductive type $\mathbb{N}$ with two constructors:

$$
1 : \mathbb{N}
$$
$$
S : \mathbb{N} \to \mathbb{N}
$$

Where $S$ "adds one" to a number. In this system $2$ is represented by $S(1)$, $3$ is represented by $S(S(1))$ etc. This is essentially a more direct implementation of the Peano axioms.

Yes, there is Peanos construction of them which is acctually my favourite method for constructing them.


Solve complex equation $|z|^5=z^5$ for $z$ I have a problem, this equation should be solved for $z$, with $z$ being complex. There should be $5$ solutions bc of the exponent. I already got the solutions from WolframAlpha, but dont know how to get there.

Clearly $|z|^5$ is non-negative and real so any solution to the equation $|z|^5 = z^5$ has that $z^5$ is non-negative and real. Furthermore, any non-negative and real number clearly satisfies the equation. So it is sufficient and necessary that $z^5$ is non-negative and real.

Zero is one solution. As for the remaining solutions, recall that in the complex plane, raising a nonzero complex number to the $5$th power multiplies the angle it forms with the positive real axis by 5. And it is sufficient and necessary that the resulting angle be 0. So it is sufficient and necessary that the original angle is a multiple of $\frac{2\pi}{5}$.

In other words, the solutions plotted in the complex plane are the rotations of the non-negative real axis by the 5 angles that are multiples of $\frac{2\pi}{5}$.

Hint: Left side is real non-negative, so the same is true for $z^5$. Use the $r,\phi$ notation of complex number and geometric interpretation of a product.


What's the meaning of this $(m,n) = 1$ I'm reading this pdf http://rutherglen.science.mq.edu.au/wchen/lndpnfolder/dpn01.pdf I understand some of the expression used in this but I don't understand the part $(m,n) = 1$

Is this a cartesian coordinate or some sort of operation?

The notation $\,(a,b) := \gcd(a,b)\,$ is widely used in number theory. Similarly, but less frequently, authors use $\,\ \ [a,b]\, := {\rm lcm}(a,b).\,$  Hence  $\,(a,b) = 1\,$ means $\,a,b\,$ are coprime: $\,c\mid a,b\,\Rightarrow\,c\mid 1.$

Here, as often, one can uniquely infer the meaning from its use. The first use  of the notation is in Theorem $1$ where it is claimed that $\,(a,b)=1\iff $ every positive divisor of $\,ab\,$ can be written  uniquely in the form $\,cd\,$ where $\,c\mid a,\ d\mid b,\,\ c,d\in\Bbb N.\,$ This implies that $\,a,b\,$ are coprime since, if  not, they have a common divisor $\,n> 1$ which has more than one such rep:  $\,c,d = n,1\,$ and $\,1,n.$

One of the reasons that the gcd tuple notation is so widely used is that it helps to emphasize analogies between gcds and ideals, which use the same tuple notation. They share many of the same arithmetical laws, e.g. associative, distributive, and $\,(a,b) = (a)\,$ if $\,a\mid b,\,$ so by using a common notation one can write proofs that work for both gcds and ideals. In PIDs like $\,\Bbb Z\,$ they are essentially arithmetically equivalent, since for ideals $\ (a,b) = (c) \iff c = \gcd(a,b),\,$ so one can read $\,(a,b)\,$ either as a gcd or an ideal. The analogies between ideals and gcds are clarified when one studies divisor theory (the modern version of Kronecker's alternative to Dedekind's ideal-theoretic approach to factorization in rings of algebraic integers). An introduction to divisor theory can be found in Borevich and I. R. Shafarevich, Number theory.  See also

Friedemann Lucius. Rings with a theory of greatest common divisors.
manuscripta math. 95, 117-36 (1998).

Olaf Neumann. Was sollen und was sind Divisoren?
(What are divisors and what are they good for?)  Math. Semesterber, 48, 2, 139-192 (2001).

Because there is little written in English on divisor theory, it would be very valuable to translate Neumann's very nice German exposition into English.  If anyone is interested in doing so please contact me.

$(m, n)$ is used by some to denote $\gcd(m, n),$ the greatest common divisor function of two integers, $m, n$.

So $(m, n) = 1$ means that the greatest common divisor of $m, n$ is $1$; i.e., $m, n$ are relatively prime.

Whether $(m, n)$ is used to mean $\gcd(m, n)$, or an ordered pair, depends on the context in which that notation is used.


Prove existence of Diagonalizable Matrix Suppose R, T $\in L(F^3)$ each have 2, 6, 7 as eigenvalues. Prove that there exists an invertible operator S $\in L(F^3)$ such that $R=S^{-1}TS$.

What I got so far is that since R and T have three eigenvalues each then R and T must be diagonalizable? I'm not really sure how to approach this question help?

Since both $R$ and $T$ have 3 distinct eigenvalues, they are diagonalizable i.e. there exists invertible matrices $P, Q$ such that $R = P D P^{-1}$ and $T = Q D Q^{-1}$, where $D = \textrm{diag}(2,3,5)$. Substituting one equation into the other, we get that
$$
R = P (Q^{-1} T Q)P^{-1} = (PQ^{-1}) T (PQ^{-1})^{-1}.
$$
Thus, $S = PQ^{-1}$ is the invertible matrix you're after.

enter image description here

Since R and T both have 3 distinct eigenvalues, R and T are both diagonalizable. Hence, there exists invertible operators A and B such that $R = ADA^{-1}$ and $T = BDB^{-1}.$ $$B^{-1}TB = B^{-1}BDB^{-1}B$$ $$AB^{-1}TBA^{-1} = ADA^{-1}$$ $$(AB^{-1})T(BA^{-1}) = R $$ $$S^{-1}TS = R$$


Let $X_{n}$ be the maximum score obtained after $n$ throws of a fair dice. Show that $(X_{n})_{n\in\mathbb{N} }$ is a Markov Chain. Let $(X_{n})_{n\in\mathbb{N} }$  be a sequence of randon variables defined by
$$X_{n}:=\mbox{maximum score obtained after } n \mbox{ throws of a fair dice.}$$

I need show that $(X_{n})_{n\in\mathbb{N} }$ is a Markov Chain. In this sense, we have to show that
$$P(X_{n+1}=i_{n+1}|X_{1}=i_{1},X_{2}=i_{2},\ldots,X_{n}=i_{n})=P(X_{n+1}=i_{n+1}|X_{n}=i_{n})\tag{1}$$
I know that if we consider  the random variable  $Y_{k}$ which are the result of $k$th throw of our fair dice, then $(Y_{k})_{k\in\mathbb{N}}$ is iid with uniform distribution over $\left\{1,2,3,4,5,6\right\}$, so, we have $X_{n}=\max\left\{Y_{1},\ldots,Y_{n}\right\}$. Furthermore, the cumulative distribution of $X_{n}$ is
$$F_{X_{n}}(k)=\left\{\begin{array}{ll}
0 & \mathrm{if}\: k<1 \\
\frac{i^{n}}{{6}^{n}} & \mathrm{if}\: i\leq k <i+1 \:\: \mathrm{for}\: i=1,2,3,4,5.\\
1 & \mathrm{if}\:\: k\geq 6
\end{array}\right.$$

Therefore, (1) is equivalent to
$$P(\max\left\{Y_{1},\ldots,Y_{n},Y_{n+1}\right\}=i_{n+1}|Y_{1}=i_{1},\max\left\{Y_{1},Y_{2}\right\}=i_{2},\ldots,\max\left\{Y_{1},\ldots,Y_{n}\right\}=i_{n})=P(\max\left\{Y_{1},\ldots,Y_{n},Y_{n+1}\right\}=i_{n+1}|\max\left\{Y_{1},\ldots,Y_{n}\right\}=i_{n})$$
I have not been able to prove this last, I need help in this matter.

Here is the Markov transition graph for your problem, where the states are labeled by the current maximum and $0$ represents the initial state:

All you need do is calculate the probabilities for the transitions and see that the final state (the sole attractor) is (in probability) $6$.  Each transition depends solely on the current state, not the sequence that led to the current state.

Markov Matrix for one-step transition probability would be

\begin{bmatrix}
\frac16 & \frac16 & \frac16 & \frac16 & \frac16 & \frac16 \\
0 & \frac26 & \frac16 & \frac16 & \frac16 & \frac16 \\
0 &  0 & \frac36  & \frac16 & \frac16 & \frac16 \\
0 &  0 & 0  & \frac46 & \frac16 & \frac16 \\
0 &  0 & 0 & 0  & \frac56 & \frac16 \\
0  &  0  &  0  &  0  &  0  &  1
\end{bmatrix}

The long term limiting probabilities would be as followed
\begin{align}
\pi_6 &= 1 & \pi_1 &= \pi_2 = \pi_3 = \pi_4 = \pi_5 = 0
\end{align}


Correlation between a random variable and its rank Let $X_1,\ldots,X_n$ be a random sample from $U(0,1)$ and $X_{(1)}<\ldots<X_{(n)}$ be the corresponding order statistics.

Define,
$$
R(X_1) = r\quad \text{if}\quad X_{(r)} = X_1;\quad r = 1(1)n
$$
i.e. $R(X_1)$ is the rank of $X_1$ in the ordered sample. Then what will be correlation between $X_1$ and $R(X_1)$ ?

A generalization of the problem statement is that $X_1, \dotsc,X_n$ is exchangeable. Then we will also get $R_1, \dotsc, R_n$ is exchangeable, and even the bivariate vectors
$$
\left( (\begin{smallmatrix} X_1\\R_1\end{smallmatrix}), \dotsc, (\begin{smallmatrix} X_n\\R_n\end{smallmatrix}) \right)
$$ is exchangeable. Let $\mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $\DeclareMathOperator{\E}{\mathbb{E}} \E X_1 R(X_1) =\E X_j R(X_j)$ for $j=1, \dotsc,n$. Then we find
$$
   \E X_1 R(X_1) =\frac1{n}\sum_{j=1}^n \E X_j R(X_j) = \\
\frac1{n}\sum_j X_{(j)} R(X_{(j)}) =\frac1{n}\sum_j \E X_{(j)} j =\\
\frac1{n} \E \sum_j j X_j=\frac1{n} \mu \sum_j j =\\
\frac{n(n+1)\mu}{2 n}=\frac{\mu (n+1)}{2}
$$
and then it follows fast that the correlation is zero, under much greater generality than in the question.

\begin{align}
E[X_1R(X_1)]=\frac1n\sum_{r=1}^nr\frac{\int_0^1xx^{r-1}(1-x)^{n-r}\mathrm dx}{\int_0^1x^{r-1}(1-x)^{n-r}\mathrm dx}=\frac1n\sum_{r=1}^n\frac{r^2}{n+1}=\frac{2n+1}6\;,
\end{align}

so with $E[X_1]=\frac12$ and $E[R(X_1)]=\frac{n+1}2$ we have

$$
\operatorname{Cov}(X_1,R(X_1))=\frac{2n+1}6-\frac12\cdot\frac{n+1}2=\frac{n-1}{12}\;.
$$

With $\operatorname{Var}(X_1)=\frac1{12}$ and $\operatorname{Var}(R(X_1))=\frac{n^2-1}{12}$ the correlation coefficient is

$$
\frac{n-1}{\sqrt{n^2-1}}=\sqrt{\frac{n-1}{n+1}}=\sqrt{1-\frac2{n+1}}\;.
$$


Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$

  Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?

For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down.

I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$

I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere.

Is there an algebraic proof/disproof of my above claim?

Thanks.

Assume $x\in\Bbb Z^+$.
If $x=m^2$ is a pefect square, then
$$m^2<x+\sqrt x=m^2+m<m^2+2m+1=(m+1)^2$$
an so $x+\sqrt x=R(x+\sqrt x)$ cannot be a perfect square.
Thus we need only consider the case that $x$ is not a perfect square, which makes $\lfloor x+\sqrt x\rfloor <\lceil x+\sqrt x\rceil$.

Let $n\in\Bbb Z^+$ be maximal with $n(n+1)<x$. Then $x=n^2+n+d$ with $1\le d\le (n+1)(n+2)-n(n+1)=2n+2$.
This makes
$$\lfloor x+\sqrt x\rfloor = n^2+n+d+n=(n+1)^2+d-1.$$
This is $\ge (n+1)^2$ and $\le n^2+4n+2<(n+2)^2 $.

Hence $ k^2=\lfloor x+\sqrt x\rfloor$
implies $k=n+1$, $x=n^2+n+1$. But $$(n+\tfrac12)^2=n^2+n+\tfrac14<x$$ implies that $x+\sqrt x$ should round up, not down.

Similarly, $k^2=\lceil x+\sqrt x\rceil = \lfloor x+\sqrt x\rfloor+1$ implies  $k=n+2$, $d=(n+2)^2+1-(n+1)^2=2n+2$, $x=n^2+3n+2$.
But $$ (n+\tfrac32)^2=n^2+3n+\tfrac 94>x$$
implies that $x+\sqrt x$ should be rounded down, not up.

We conclude that $R(x+\sqrt x)$ is never a perfect square for $x\in\Bbb Z^+$.

$n+\sqrt{n}$ never rounds to a square

By way of introduction, note first that $N^{2}+N$ can never be a
perfect square (when $N$ is a whole number) -- this is because perfect squares
of whole numbers must differ by (at least) $2N+1$.  If we let $n=N^{2}$ then
$N^{2}+N=n+\sqrt{n}$ and this can never be a perfect square; however, we can
ask if $n+\sqrt{n}$ can ever round to a perfect square -- i.e. is
there a perfect square which is within $\frac{1}{2}$ of it?

Let $G(n)=$ the nearest integer to $n+\sqrt{n}$, so that $G(n)$ lies within
$\frac{1}{2}$ of $n+\sqrt{n}$.$.$

Let $K$ be the nearest integer to $\sqrt{n}$, so

\begin{align}
K-\frac{1}{2}  & <\sqrt{n}<K+\frac{1}{2}\text{ and}\tag{A}\\
\left(  K-\frac{1}{2}\right)  ^{2}  & <n<\left(  K+\frac{1}{2}\right)
^{2}\text{ and}\nonumber\\
K^{2}-K+\frac{1}{4}  & <n<K^{2}+K+\frac{1}{4}\text{. Adding:}\tag{B}\\
K^{2}-\frac{1}{4}  & <n+\sqrt{n}<K^{2}+2K+\frac{3}{4}\text{ or}\nonumber
\end{align}

\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<\left(  K+1\right)  ^{2}-\frac{1}{4}\tag{C}%
\end{equation}

Thus: $n+\sqrt{n}$ lies between $K^{2}-\frac{1}{4}$ and $\left(  K+1\right)
^{2}-\frac{1}{4}$. It follows that $G(n)=K^{2}$ or $G(n)=\left(  K+1\right)
^{2}$ if $G(n)$ is to be a perfect square. We eliminate each case separately.

CASE 1: Suppose $G(n)=K^{2}$ Then $n+\sqrt{n}<K^{2}+\frac{1}{2}$ so (from line C):

\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<K^{2}+\frac{1}{2}\text{.}\tag{D}%
\end{equation}

Multiplying line (A) above by $-1$ and adding:

\begin{align*}
-K-\frac{1}{2}  & <-\sqrt{n}<-K+\frac{1}{2}\text{; adding to line (D):}\\
\left(  K^{2}-K\right)  -\frac{3}{4}  & <n<\left(  K^{2}-K\right)  +1\text{.}%
\end{align*}

Since $n$ and $K$ are whole numbers, we must have $n=K^{2}-K$. But from line
(B) above this gives $\ n+\frac{1}{4}<n$ or $\frac{1}{4}<0$: a contradiction.

CASE 2: Suppose $G(n)=\left(  K+1\right)  ^{2}$. Then

\begin{align*}
\left(  K+1\right)  ^{2}-\frac{1}{2}  & <n+\sqrt{n}<\left(  K+1\right)
^{2}+\frac{1}{2}\text{ and from line (A):}\\
-K-\frac{1}{2}  & <-\sqrt{n}<-K+\frac{1}{2}\text{. Adding:}\\
\left(  K+1\right)  ^{2}-K-1  & <n<\left(  K+1\right)  ^{2}-K+1\text{ or:}\\
\left(  K+1\right)  ^{2}-(K+1)  & <n<\left(  K+1\right)  ^{2}-\left(
K+1\right)  +2\text{.}%
\end{align*}

Since $n$ and $K$ are whole numbers and these are strict inequalities, we must
have $n=\left(  K+1\right)  ^{2}-(K+1)+1=(K+1)^{2}-K$. But from line (B) above:

\begin{align*}
\left(  K+1\right)  ^{2}-K  & <K^{2}+K+\frac{1}{4}\text{ or:}\\
K^{2}+2K+1-K  & <K^{2}+K+\frac{1}{4}\text{ or}\\
1  & <\frac{1}{4}\text{.}%
\end{align*}

This contradiction shows that $G(n)$ can never be a perfect (integer) square.


$p$-completion is pro-$p$ free Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $\widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, \mathbb{F}_p) \neq 0 \implies H^2(\widehat{G_p},\mathbb{F}_p) \neq 0$. There seems to be no need for a basis of $\widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $\widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.

This is false. All parafree groups have this property. Look at https://en.wikipedia.org/wiki/Parafree_group

One of the easiest example of a parafree group is

$\langle x,y,z: x^2y^2=z^3\rangle$.

This group is also residually-$p$ for all primes $p$.


Help to clarify proof of Euler's Theorem on homogenous equations Why is the last step (setting $\lambda = 1$ allowed?

I have trouble accepting this because if I set $\lambda =1 $ at the very start, then:

$f(\lambda x , \lambda y)=\lambda^r f(x,y)$

becomes

$f(x,y) = f(x,y)$

and

so I can't prove the theorem.

Why does setting $\lambda = 1$ at the end of the proof work?

EDIT

For example, let $f(x,y)=x^2y^2$

Therefore, $f(x,y)$ is homogeneous of degree 4.

Therefore, the second to last step will be:

$4x^2y^2=4\lambda^3x^2y^2$

The above equation is only true if $\lambda=1$

Why doesn't the theorem make a qualification that $\lambda$ must be equal to 1?

It seems to me that this theorem is saying that there is a special relationship between the derivatives of a homogenous function and its degree but this relationship holds only when $\lambda=1$. Please correct me if my observation is wrong.

You can't set $\lambda = 1$ in the line $f(\lambda x,\lambda y) = \lambda^r f(x,y)$ because the very next step is to differentiate with respect to $\lambda$ which makes no sense when the variable isn't present.

At the end of the proof, you are taking advantage of the fact that $\lambda$ is an arbitrary element of $\mathbb{R}$. So you pick an element that makes the calculations easy.

Variant of proof

Define the function $g : \mathbb{R} \rightarrow \mathbb{R}$ by $g(t) = f(tx,ty)$. Since $f$ is homogeneous, we can write $g(t) = t^r f(x,y)$. Find $g'(t)$.

Using $g(t) = t^r f(x,y)$, it is clear that $g'(t) = rt^{r-1} f(x,y)$.

Using $g(t) = f(tx,ty)$, we get that $g'(t) = \frac{\partial f}{\partial (tx)}\cdot\frac{d(tx)}{dt} + \frac{\partial f}{\partial (ty)}\cdot\frac{d(ty)}{dt} = x\frac{\partial f}{\partial (tx)}+y\frac{\partial f}{\partial (ty)}$.

So we have that for all $t$, $rt^{r-1} f(x,y) = x\frac{\partial f}{\partial (tx)} + y\frac{\partial f}{\partial (ty)}$. If we let $t=1$, then we have that $g(1) = f(x,y)$, our original function, and $rf(x,y) = x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}$, the desired result.

To address the specific example of $f(x,y) = x^2y^2$. You have the second to last step wrong on the RHS.

When doing the proof, we are working with $f(\lambda x, \lambda y)$ throughout. So when calculating $\dfrac{df}{d\lambda}$ we get:

$\begin{align}
\dfrac{df}{d\lambda} &= \dfrac{\partial f}{\partial (\lambda x)}\cdot \dfrac{d (\lambda x)}{d\lambda} + \dfrac{\partial f}{\partial (\lambda y)}\cdot \dfrac{d (\lambda y)}{d\lambda} \\
&= 2(\lambda x)(\lambda y)^2\cdot x + 2(\lambda x)^2(\lambda y)\cdot y \\
&= 4\lambda^3 x^2y^2 = \dfrac{d}{d\lambda}\left(\lambda^4 f(x,y)\right)
\end{align}$

So we could choose any $\lambda$ we want and it would still be a true equation. But then to get the desired result, we would have to divide that back out of both sides. Choosing $\lambda = 1$ saves a bit of algebra.

Since $f(x,y)=x^2y^2$, therefore, it can be written as
$$f(x,y)=x^2\left(\frac{y}{x}\right)\times x^2=x^4\left(\frac{y}{x}\right).$$
This shows that $f$ is a homogeneous function of degree $4$. Hence, by Euler's theorem, we have
$$x\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial x}=4f.$$


Mean Absolute Deviation for a Stable Distribution as a Function of the Tail Exponent Consider the standard Lévy-Stable (or Alpha Stable) distribution $S(\alpha,\beta, \mu, \sigma)$ where $\alpha$ is the tail exponent, $1 \leq \alpha \leq 2 $. Picking the symmetric case with $0$ mean (that is, $\beta =0$, $\mu =0$ the characteristic function becomes $C(t)= e^{-\left| t \sigma \right| ^{\alpha }}$.

We can numerically extract the mean absolute deviation by inverting the characteristic function. From the shape of the distribution shown below in the graph, and, where $X$ is the r.v., from the property that $E(|X|)_{\alpha=2}= \frac{2 }{\sqrt{\pi }} \sigma$ (the only explicit solution we get) and $E(|X|)_{\alpha=1}= \infty$ ( the Cauchy case), a good guess is:
$$E(|X|)= \left(\frac{a}{\log(\alpha)}\right)^{f(\alpha)} \sigma,$$
where $a=\frac{2 \log (2)}{\sqrt{\pi }}$ and $f ( \alpha)= 1$ when $\alpha =2$:

I wonder if someone knows of an explicit expression.

Added Comment: By tinkering I figured out we could track the function with $E(|X|)=\frac{a_1 \sqrt{\alpha }}{(a_2+\alpha)\, B\left(a_3 \alpha, a_4\right)}$,where $B$ is the Beta function, but the number of parameters is high.

Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$.
Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$,
$$ E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s $$

Note: Copied from question. Delete I guess if NNT answers his own question.

  Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$.
  Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$,
  $$ E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s $$