$\hat f_\lambda(x) = \hat f(x/\lambda)$ This is a small part of a larger problem I am trying to solve. This is stated as a basic property of the fourier transform. First we define for $f \in L^1(\mathbb{R}^d)$ and $\lambda \neq 0$,
$$ \hat f(x) := \int_{\mathbb{R}^d} f(t) e^{-2\pi i t x} dt
\quad\text{and}\quad
\hat f_\lambda(x) := \lambda f(\lambda x). $$

The property is that, $\hat f_\lambda(x) = \hat f(x/\lambda)$

I don't see how to prove this, not sure where to start really.

EDIT: The property I am trying to prove is the scaling property of this wikipedia article. http://en.wikipedia.org/wiki/Fourier_transform#Basic_properties

Following the notation of the wiki article. If $h(x)=f(ax)$, then $\hat{h}(\xi)=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Proof:

$\hat{h}(\xi)=\int h(x)e^{-2\pi i\xi x}\, dx=\int f(ax)e^{-2\pi i\xi x}\, dx=|a|^{-1}\int f(y)e^{-2\pi i\xi \left(\frac{y}{a}\right)}\, dy$.

This becomes: $|a|^{-1}\int f(y)e^{-2\pi i\left(\frac{\xi}{a}\right)y}\, dy=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Note this was for functions taking values in $\mathbf{R}$, in higher dimensions, say $n$,  the $|a|$ gets an exponent of $-n$.  I think that is the only change.

In this case
$$\hat f(\lambda t) = \int_{\mathbb{R}^d} f(x)e^{2\pi i \lambda x t}\,dx =
 \int_{\mathbb{R}^d} f(x/\lambda)e^{2\pi i  xt}{dx\over \lambda^d} $$
Note the $d$ power owing to the Jacobian of the transformation on $\mathbb{R}^d$.


Linear Algebra Practice Problems I am currently watching a lecture series on Youtube by Dr. Gilbert Strang, and it's been great so far. The only issue that I'm running into is that, without practicing the problems, I have no way to really gauge if I'm really understanding the material, and the concepts aren't sticking as well.

Specifically, we are currently covering the four major subspaces of linear algebra: the column space, the null space, the row space, and the left null space. One of the ideas in this chapter is to be able to find the basis for each of these spaces given a certain mxn matrix, and determining whether or not columns or rows are linearly independent.

I mostly understand the concepts themselves, but I really need to practice to be sure, and to get it to stick. I've tried searching google for practice problems, as well as past questions on this site, and so far I'm not really finding what I need. Would anyone happen to know a good source of practice problems for subspace bases and linear independence of vectors for matrices at the first year (undergraduate) linear algebra level? Thanks, I would appreciate it.

Here is a practical way to find the four subspaces you mention. By finding I mean: find a basis for them. (Because if we know a basis for a subspace, then we know that every other vector in the subspace is a linear combination of the basis vectors).

I am assuming you know that every linear function $f:{\mathbb R}^n \longrightarrow {\mathbb R}^m$ is given by an $m\times n$ matrix $A$ in the sense that $f(x)=Ax$ for every $x$ in ${\mathbb R}^n$. So assume we have such a matrix $A$.

For the column space: compute $A^T$ (the transpose), then do row operations until $A^T$ is in Row Echelon Form. Then the non-zero rows of the Row Echelon Form will be a basis for the Column Space (that is the same thing as the range of the linear function).

For the Null Space: put $A$ in Row Echelon Form. Remember that the leading entry of a row is the first non-zero entry of that row (if there is one). In the Row Echelon Form of $A$, each column that has no pivot(=no entry that is a leading entry of some row) corresponds to a free variable. If you find no free variables, the null space is the just the zero space (only has the zero vector in it, which means the function if one-to-one). If you find $k$ free variables, that means the Null Space has dimension $k$ (meaning: there will be $k$ vectors in a basis for it). To find the $k$ vectors for a specific basis, proceed as follows: set one free variable to $1$, and all others to zero, and solve the homogeneous system corresponding to the Row Echelon Form. The answer will be a vector in a basis for the Null Space. Now set another free variable to $1$ and all the others to $0$, and solve the linear system again, to get a second vector in a basis. So on, until you finish all the free variables, and get $k$ vectors for a basis for the null space.

As for the Row Space, it's just like the column space, but you work on $A$ instead of $A^T$.

The left Null Space is the same thing as the Null Space of $A^T$, so again switch $A$ with $A^T$.

To practice, try the three matrices below. Find a basis for the four subspaces in each case.

$A=\left(\begin{array}{ccc} -1 & 1 & 1 \\ 3 & -1 & 0 \\ 2 & -4 & -5 \end{array} \right)$.

$B=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{array} \right)$.

$C=\left(\begin{array}{ccc} -1 & 1 & 2 \\ -2 & 2 & 4 \\ 3 & -3 & -6 \end{array} \right)$.


Explain the Absorption Law I am currently in a Discrete Math class and reviewing some of my terminology and I don't really understand the Absorption Law. I am not looking for a proof or a truth table but an explanation in layman's terms.

Absorption Law
$$
A ∨ (A ∧ B) = A
$$
$$
A ∧ (A ∨ B) = A
$$

$$
A ∪ (A ∩ B) = A
$$
$$
A ∩ (A ∪ B) = A
$$

For the first one, either $A$ is true, or both $A$ AND $B$ are true.  In either case, $A$ is true.  On the other hand, if $A$ is true, then the first is true so the expression is true.

The others are similar.

As per logic perspective, we think values either as TRUE or FALSE, I.e: 0 or 1.

So here we can assume 0 or 1 only, for A and B respectively.

So consider A=1 and B=0,
Now put the values in asking law,
LHS=A.(A+B)
        =1.(1+0)
        =1.(1)
        =1=A   ...[because A=1]
            =RHS

Now you are thinking, why I cannot take value of B=1 and A=0,
Yes!!  You are independent to that,
If you take A=0 then multiply with adjacent terms in bracket, your whole answer will be zero, again, lhs will be equal to rhs.
So you will get answer as the value of A in every attempt.

Now for A+(A.B)=A
If A=1, B=0

Put,      1+(1.0)=1   [A=1]

Or if A=0, B=1
Then,
        A+(A.B)=A
        0+(0.1)=0     [A=0]

Hence you will get answer as A alwasy!!!!


Find an expression equalling 69 using the digits 1 9 9 2 in order A friend gave me a puzzle to find an expression for each of the integers $1$ to $100$ using the digits $1, 9, 9, 2$ such that they appear in that exact order. You're allowed any combination of the four basic operations, brackets, square roots, the factorial, exponentiation and also you can put digits together, e.g. we could have $23=19+(\sqrt{9})!-2$. I've managed every number relatively quickly except $69$ which I've been stuck on for ages and it's infuriating! Any ideas?

Edit: Apologies for my unclear explanation. You're only allowed to put the original digits together in the equation. The equation must make sense as a normal equation so you can't do things like combine $1$ with $\sqrt{9}$ to get $13$ because this can't be written down in a way which makes sense as a normal equation.

Also you aren't allowed to just use powers, unless you're using one of the four digits themselves, e.g. $2=1+(9/9)^2$ is okay because we've used the $2$ as a power but $69=(1-9)^2+\sqrt{9}+2$ isn't because there's now a superfluous $2$.

Since we can put digits together, one solution might be

$1+ ((\sqrt{9})!)((\sqrt{9})!) +2$

One expression which does not keep the exact order is

$9*9-12$ or $((\sqrt{9})!)((\sqrt{9}+1) +2$

I can't comment, therefore I am posting this as an answer and delete it later.

$(\sqrt{9})!9 = 69$, since $(\sqrt{9})!=6$, since you said, it is allowed to put digits together.

Getting rid of the 1 is not a problem $1\cdot n=n$.
But still I need to get rid of the 2.

edit:
$1 + \underbrace{(\sqrt{9})! (\sqrt{9})!}_{=66} + 2 = 69$


There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$. Prove that, if $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

Proposition: If $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

A (hypothesis): $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$.

B (conclusion): There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

My Work

B1: $(a + bi)(c + di) = 1$

$\therefore a + bi \not = 0$

B2: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

A1: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$.

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A2: There is a complex number $x + yi$ such that $x + yi = \dfrac{1}{a + bi}$.

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$

A3: $(a+bi)(x + yi) = (a + bi)(c + di)$ where $a + bi \not = 0$

$\implies x + yi = c + di$

$Q.E.D.$

EDIT: My Work #2

A1: $(a + bi)(c + di) = 1$

$\implies ac + adi + bic + bdi^2 = 1$

$\implies ac + adi + bic - bd = 1$

$\implies (ac - bd) + i(ad + bc) = 1$ is a complex number, where $(ac - bd)$ is the real part and $i(ad + bc)$ is the imaginary part.

A2: Let $ac - bd = 1$ and $ad + bc = 0$

$\therefore 1 + 0 = 1$

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A3: Let $(a + bi)(x + yi) = 1$

A4: $(a + bi)(c + di) = (a + bi)(x + yi)$ where ($a \lor b) \not = 0$.

$\implies c + di = x + yi$

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback. If there are any errors, please explain why and what the correct procedure is.

The aim of this exercise is to prove that the thing that you denote by $\frac1{a+bi}$ uniquely exists. Using this notation a priori implicitly assumes the statement we are about to prove.

Hint: Instead, as commented, try to explicitly find $c$ and $d$. For this, observe that $(a+bi)(a-bi)$ is real, and division is already known among reals.

For uniqueness of such $c+di$, we can take a general argument:
Suppose $w_1$ and $w_2$ are both inverses of $z$, i.e. $w_1z=zw_2=1$.
Then we have
$$w_1 = w_1\cdot 1 = w_1zw_2=1\cdot w_2=w_2$$

It seem that both proofs are strange or incomplete.

$\dfrac1{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i$

This is a unique complex number.


Find a continuous function $f:[1,\infty)\to\Bbb R $ such that $f(x) >0 $, $\int_1^\infty f(x)\,dx $ converges and $\int_1^\infty f(x)^2\,dx$ diverges I'm trying to find an example of a continuous function $f:[1,\infty) \to \Bbb R $ such that
$$f(x) >0 $$

$$\int_{1}^{\infty} f(x) \ dx \ \text{converges and } \int_{1} ^{\infty} f(x)^2 dx \ \text{diverges}.$$

Consider at $x=k\in{\mathbb N}_{\geq2}$ a trapezoidal spike of height $k$ with base $\left[k-{2\over k^3},\  k+{2\over k^3}\right]$ and top $\left[k-{1\over k^3},\  k+{1\over k^3}\right]$. Let $f: \>[1,\infty)\to{\mathbb R}$  be the function obtained by "concatenation" of these spikes. The area of the spike at $k$  is ${3\over k^2}$, so that
$$\int_1^\infty f(x)\>dx=3 \sum_{k=2}^\infty {1\over k^2}={\pi^2\over2}-3<\infty\ .$$
On the other hand each spike of $f^2$ contains a rectangle of width ${2\over k^3}$ and height $k^2$. It follows that
$$\int_1^\infty f^2(x)\>dx\geq\sum_{k=2}^\infty{2\over k}=\infty\ .$$
Add $e^{-x}$ to $f(x)$, as in Sangchul Lee's answer, to make $f$ positive everywhere.

A simpler example might be $f(x) = \frac{\sin{x}}{\sqrt{x}}$. Since $\frac{1}{\sqrt{x}}$ is continuously differentiable and monotonically decreasing to 0, and since $\sin{x}$ has a bounded and integrable anti-derivative, from Dirichlet's test $\int\limits_1^{\infty}\frac{\sin{x}}{\sqrt{x}}dx$ converges. Integration by parts shows that $\int\limits_1^{\infty}\frac{\sin^2{x}}{x}dx$ diverges.


Is $2^e$ irrational or rational? Please help me with the following question.

  Prove or disprove that $2^e$ is an irrational number, where $e$ is the Neipper number.

Thanks.

Determining whether $2^e$ is irrational or not is an open problem. And more generally, the task of determining whether $(\text{rational})^{\text{irrational}}$ is rational or not appears to be extremely difficult.

Read the comments to know more

just two lines:
(its all i could think and write)

let $$2^e=p$$
taking log of base 2 both sides

$$\log_22^e=\log_2p\implies e=\log_2p$$
and lets say that e is irrational

now assuming that $\log_2p=\frac ab, b\in N-\{1\}$ where $a$ is an irrational number and $N$is the set of all natural numbers and $2^a\ne n^mb$ where $ m$ being non-zero positive integers

so $$\log_2p=\frac ab\implies\log_2p^b=a\implies p^b=2^a\implies p=(2^a)^{\frac 1b}\ldots (1)$$

so from (1) it is clear that $p$ is always an irrational and from the very first statement we can say that a number power to an irrational number is an irrational number


How find this value of $x,y$ let $x,y\in R$, such

  $$\begin{cases}
\sqrt{1+(x+y)^2}=-y^6+2x^2y^3+4x^4\\
\sqrt{2x^2y^2-x^4y^4}\ge 4x^2y^3+5x^3
\end{cases}$$
  find the value of  $x,y$.

My try:
since

  $$|xy|\sqrt{2-x^2y^2}\ge 4x^2y^3+5x^3$$

then I can't,it is said this problem is Bulgaria National Olympiad(1979),But I can't find it .Thank you

Surely there's a whole family of solutions to this, if all you know is $x,y\in\mathbb{R}$.  How about trying $y=0$?  Then the first equation becomes
$$
\sqrt{1+x^2}=4x^4
$$
and the second becomes
$$
0 \ge 5x^3,
$$
or $x\le 0$.  The first equation has a solution for negative $x$ (about $-0.7474$), so that's one answer.  Is there something special about the particular solution you need?  (Or does "R" mean something different in Bulgaria?)


Is Fractal perimeter always infinite? Looking for information on fractals through google I have read several time that one characteristic of fractals is :

finite area
infinite perimeter

Although I can feel the area is finite (at least on the picture of fractal I used to see, but maybe it is not necessarly true ?), I am wondering if the perimeter of a fractal is always infinite ?

If you think about series with positive terms, one can find :

divergent series : harmonic series for example $\sum_0^\infty{\frac{1}{n}}$
convergent series :  $\sum_0^\infty{\frac{1}{2^n}}$

So why couldn't we imagine a fractal built the same way we build the Koch Snowflake but ensuring that at each iteration the new perimeter has grown less than $\frac{1}{2^n}$ or any term that make the whole series convergent ?

What in the definition of fractals allows or prevent to have an infinite perimeter ?

You can certainly define a fractal analogous to the Koch snowflake with finite perimeter.

The iterative rule for the standard Koch snowflake is to replace the middle third of each line segment with two line segments of equal length and meeting the original segment and each other at $60^\circ$ angles.  This applies the rule $x \mapsto 4x/3$ to the overall length of the curve, so applying it repeatedly produces a divergent sequence of lengths.

A modified rule might be to start with a red line segment and iteratively replace the middle half of each red line segment with the top of a trapezoid made of one red and two blue line segments (all of equal length and again making $60^\circ$ angles).  Then the overall red/blue lengths $(x,y)$ are subjected to the rule $(x,y)\mapsto(3x/4, y + x/2)$, which is convergent when iterated.  It is a nice exercise to show that this process, starting from a red segment of length $1$, produces an entirely blue curve of length $2$; and starting from an equilateral red triangle of perimeter $3$ produces a closed blue curve of perimeter $6$.  (Calculating the area inside the curve is another nice exercise.  Moreover, since the trapezoid can be constructed either red-blue-blue or blue-red-blue at each step, there are any number of different fractals with the same perimeter and area.)

If they have infinite sides, than they must have an infinite perimeter, especially if they are perfectly straight because the formula of perimeter of most shapes is adding up the amount of sides, and the fractal has infinite sides, then it should have an infinite perimeter.


Can someone explain why this is true? $\int f(x) g'(x)\,dx = \int f(x)\,dg$ I'm a software engineer (have been for 20+ years) and have over the past few years taken quite an interest in math. I would appreciate it if anyone is able to help me out with my question.

I understand the following:

$$\int g’(x)\,dx = \int dg$$

This makes sense to me.

$$\int g’(x)\,dx = g(x)+c$$
$$\int dg = g(x) + c$$

That being the case, $\displaystyle \int g'(x)\,dx = \int dg$.

As far as I know, this is also true:

$$\int f(x)g’(x)\,dx = \int f(x)\,dg$$

What I don't completely understand is why you can put $f(x)$ in each of these integrals and they continue to be equal. I mean, I understand that if $\int g’(x)\,dx = \int dg$, then I should be able to swap $g’(x)\,dx$ for $dg$ (or vice versa), but I'm trying to visualize how $f(x)$ doesn't somehow throw this off.

Just reading it, I am looking for the integral of $f(x)$ multiplied by $g’(x)$ with respect to $x$ and the integral of $f(x)$ with respect to $g$.

Can someone explain the logic of this to me and/or point me to an applicable proof? I've Googled around to see if I can find a good explanation, but I must not be describing my question in the search query well enough (hopefully I am doing an OK job describing it here).

What you should read is that, by definition, $dg=g'(x)\,dx$. This relation holds independently of integration.

For example, you can use it in the chain rule,

$$\frac{dg}{dt}=g'(x)\frac{dx}{dt}.$$

[If you want to be picky, $\dfrac{dg(x(t))}{dt}=g'(x(t))\dfrac{dx(t)}{dt}$.]

As Yves Daoust writes, this is from the simple fact:

$$\frac{dg}{dt} \cdot dt = /\text{ cancel dt }/=dg$$

This is independent of integration.


Solving differential equation $xdy + ydx = 0$ Solve following equation

$$xdy + ydx = 0$$

My process: group variables

$$xdy = -ydx \ \ \ ; \ \ \ \frac{x}{dx} = -\frac{y}{dy}$$

Integrate

$$\int{\frac{dx}{x}} = \int{\frac{-dy}{y}}$$
$$\ln |x| + K = - \ln |y| + K$$

Solution
$$\ln|xy| = K$$

However my textbook gives answer:

$$xy = K$$

Where am I going wrong?

Your approach is alright. Note that if $\ln A = k$, then $A = e^k$ and if $k$ is a constant, then so is $e^k$ and you could rename it to (for example) $K$.

Other approach (using the product rule):
$$xdy+ydx = 0 \Leftrightarrow d(xy) = 0 \implies xy = C$$

$$
Xdy + ydx =0
xdy = -- ydx
1/y dy = -- 1/x dx
$$

Integrating both sides;
$$
log |y| = -- log |x| + log |C| $$$$
log |y|  + log |x| = log |C| $$$$
log |xy| = log |C|
$$
Comparing respective parts;
$$xy = C$$$$
y = C/x
$$,
which is required general solution..


In the history of mathematics, has there ever been a mistake? I was just wondering whether or not there have been mistakes in mathematics. Not a conjecture that ended up being false, but a theorem which had a proof that was accepted for a nontrivial amount of time before someone found a hole in the argument. Does this happen anymore now that we have computers? I imagine not. But it seems totally possible that this could have happened back in the Enlightenment.

Feel free to interpret this how you wish!

In 1933, Kurt Gödel showed that the class called $\lbrack\exists^*\forall^2\exists^*, {\mathrm{all}}, (0)\rbrack$ was decidable.  These are the formulas that begin with $\exists a\exists b\ldots \exists m\forall n\forall p\exists q\ldots\exists z$, with exactly two $\forall$ quantifiers, with no intervening $\exists$s. These formulas may contain arbitrary relations amongst the variables, but no functions or constants, and no equality symbol. Gödel showed that there is a method which takes any formula in this form and decides whether it is satisfiable. (If there are three $\forall$s in a row, or an $\exists$ between the $\forall$s, there is no such method.)
In the final sentence of the same paper, Gödel added:

In conclusion, I would still like to remark that Theorem I can also be proved, by the same method, for formulas that contain the identity sign.

Mathematicians took Gödel's word for it, and proved results derived from this one, until the mid-1960s, when Stål Aanderaa  realized that Gödel had been mistaken, and the argument Gödel used would not work.  In 1983, Warren Goldfarb showed that not only was Gödel's argument invalid, but his claimed result was actually false, and the larger class was not decidable.
Gödel's original 1933 paper is Zum Entscheidungsproblem des logischen Funktionenkalküls (On the decision problem for the functional calculus of logic) which can be found on pages 306–327 of volume I of his Collected Works. (Oxford University Press, 1986.) There is an introductory note by Goldfarb on pages 226–231, of which pages 229–231 address Gödel's error specifically.

Yes, in the past, people used to do their math in Naive set theory. According to the Citizendium article Set theory, Gregg Cantor created a formal system for Naive set theory and then later, it was a surprize to discover that its formal system is inconsistent. After it was discovered to be inconsistent, mathematicians started working in the formal system of Zermelo-Fraenkel set theory and then again, shockingly, it was proven that there is no formal proof of the axiom of choice in Zermelo-Fraenkel set theory.


Isomorphism between $K_\mathbb{R}$ and $K\otimes_{\mathbb{Q}}\mathbb{R}$ I am reading Neukirch's Algebraic Number theory and I am a little bit confused about the part where he mentioned that $K_\mathbb{R}$ and $K\otimes_{\mathbb{Q}}\mathbb{R}$ are isomorphic via the map $\varphi: a\otimes x\mapsto(ja)x$, where $j:K\to K_\mathbb{R}$ is the map given by $a\mapsto (\tau a)$, where $\tau\in \operatorname{Hom}(K,\mathbb{C})$.

My questions are:
1) What type of isomorphism do we have here? Is it a $\mathbb{Q}$ or $\mathbb{R}$-vector space isomorphism?
2) And can someone show me how $\varphi$ is an isomorphism?

Thanks in advance!

Your question is a bit unclear.

  Neukrich probably said that with $\tau_1(a),\ldots,\tau_{r_1+r_2}(a)$ all the (pairs of) complex embeddings then we have an isomorphism $$K\otimes_{\Bbb{Q}}\Bbb{R}\to \Bbb{R}^{r_1}\times \Bbb{C}^{r_2}, a\otimes b\to (\tau_1(a)b,\ldots,\tau_{r_1+r_2}(a)b) \tag{1}$$

Given a complex embedding $\tau$ there is a corresponding absolute value $|a|_v=|\tau(a)|$.

$K_v$ is the completion of $K$ for $|.|_v$.

$v$ is the same for the complex conjugate embedding $\overline{\tau}$.

Either $\tau$ is a real embedding and $K_v\cong \Bbb{R}$, or it is a non-real embedding and $K_v \cong \Bbb{C}$.

On the other hand $K\cong \Bbb{Q}[x]/(f(x))$, $K\otimes_{\Bbb{Q}}\Bbb{R}\cong \Bbb{R}[x]/(f(x))$ and the latter is a field iff $f$ is irreducible over $\Bbb{R}$.

Conversely if  $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\dim_\Bbb{R}(K_v)$ then $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\deg(f)$ is $1$ or $2$ and if it is $2$ then $K_v=\Bbb{C}$ thus $K=\Bbb{Q}(\sqrt{-d})$ and $f$ is irreducible.

For any $\tau$ we have a surjective homomorphism $K\otimes_{\Bbb{Q}}\Bbb{R}\to K_v, a\otimes b\to \tau(a)b$.

$(1)$ follows from
$$\Bbb{R}[x]/(f(x)))=\Bbb{R}[x]/(\prod_j f_j(x))\cong \prod_j\Bbb{R}[x]/(f_j(x))$$


$\lim\limits_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2-x}$ Im trying to find:
$\lim\limits_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2-x}$

If I take the path $x=y$ that limit is $0$:

$\lim\limits_{(x,y)\to(0,0)}\frac{y^2}{y^2+y^2-y}=\lim\limits_{(x,y)\to(0,0)}\frac{y}{2y-1}=0$

If I take the path $x=y^2$ that limit is $1$:

$\lim\limits_{(x,y)\to(0,0)}\frac{y^4}{y^4+y^2-y^2}=\lim\limits_{(x,y)\to(0,0)}\frac{y^4}{y^4}=1$
So the limit doesn't exist.

But wolframalpha says that limit is $0$.
I tried polar coordinates and I get: $\lim\limits_{r\to0}\frac{rcos^2\alpha}{r-cos\alpha}$

If $cos\alpha \neq 0$ that limit is $0$.

If $cos\alpha = 0$ that means that $x = 0$. Then:

$\lim\limits_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2-x}$ = $\lim\limits_{(x,y)\to(0,0)}\frac{0}{y^2}=0$
So the limit is $0$.

Wich one is true? What is correct and what is wrong? (Excuse me for my english).

When you say, "If $\cos\alpha\neq0$", you are treating $\cos \alpha$ as a constant, which is equivalent to treating $y/x$ as a constant. So this means that if you approach $(x,y)=(0,0)$ along a path where $y/x$ is a constant, the limit is 0.

That is consistent with your finding in the first approach.

It sounds like asking to compute the limit of $x/y$ as both $x$ and $y$ tend to 0. The limit depends on which variable tends to 0 faster. The limit under a given path exists, but not the general limit, in the same way as a right-side limit and left-side limit for a univariate function may both exist, but differ.


find variable from polynomial when divisor and remainder is known I am asked the following question:

When ${x^4 - 3x^3 + px - 5}$ is divided by ${x - 3}$ the remainder is $16$, find $p$.

So my way of solving this was to use synthetic long division to divide ${x^4 - 3x^3 + px - 5}$ by $3$ that would leave me with a remainder expression that would have $p$.
$$
\begin{matrix}3&1&0&-3&p&-5\\
&&3&9&18&54+3p\\
&1&3&6&18+p&49+3p
\end{matrix}
%3| 1 |  0 | -3  |  p      |  -5
% |      3 |  9  |  18     |  54 + 3p
%
%  | 1 | 3 |  6  |  18 + p |  49 + 3p
$$

I get the remainder to be ${3p + 49}$

I would say then that ${3p + 49 = 16}$ and ${p = - 11}

But the textbook gives an answer of $7$.

Have I come to the wrong conclusion about how to work out ${p}$?

Hint: Note that $f(x)=(x-3)\cdot g(x)+r(x)$ implies $f(3)=r(3)$.


Subfield of a field F Question: Let $F$ be a field and let $K$ be a subset of $F$ with at least two elements. Prove that $K$ is a subfield of $F$, if for any $a,b \left ( b\neq 0 \right )$ in $K$, $a-b$ and $ab^{-1}$ belong to $K$.

This is an odd question but let's see.

It is necessary to investigate the case where $a=0$ and $a\neq 0$.

Let's proceed with the case where $a=0$ first.

Suppose $a=0, b\neq 0 \in K$. Let $K=\left \{ a,b \right \}$
Then, $a-b=0-b=-b$.
It doesn't seem as though I can proceed on with what I have here.
The additive inverse of $b$ is contained in $F$ by definition of a field but it is not necessary true that this additive inverse must be contained in $K$.

Any help is appreciated. Thanks in advance.

Let $x \in K.\,$ Then $x-x\in K$, hence $0 \in K$.

Let $x \in K.\,$ Then since $0 \in K$, we get $0 - x \in K$, hence $-x \in K$.

Let $x,y \in K.\,$ Then $-y \in K$, hence $x-(-y) \in K$, so $x+y \in K$.

Since $K$ has at least two elements, $K$ contains at least one nonzero element, $x$ say. Then $(x)(x^{-1}) \in K$, hence $1 \in K$.

Let $x \in K,\;x \ne 0$. Then since $1 \in K$, we get $(1)(x^{-1}) \in K$, hence $x^{-1} \in K$.

Let $x,y \in K$.

$\qquad$If $y \ne 0$, then $y^{-1} \in K$, so $(x)((y^{-1})^{-1}) \in K$, hence $xy \in K$.

$\qquad$If $y=0$, then $xy=0$, hence $xy \in K$.

Either way, $xy \in K$.

It follows that $K$ is a field.

It's ambiguous as to whether $a$ and $b$ must be distinct for those condition to hold.

If $a = 0$ then we have $b$ and $0-b = b$ and hmmm, if $b = -b$ but $b \ne b^{-1}$ then we have $0, b$ and that's it.  Not a field.

But if distinction is not implied then:

You have $a, b, a-b$ so you have $(a-b) -a = -b \in K$.  We have $a - (-b) = a+b$ so addition is closed.  And $b*b^{-1} = 1$ and and $1* b^{-1} = b^{-1}$ and $a(b^{-1})^{-1} = ab$ so multiplication is closed.  So a field.

But distinction can not be implied.


Determining Spot Rates A three-year, 4%, par-value bond with annual coupons sells for $990$, a two-year, $1000$, 3% bond with annual coupons sells for $988$, and a one-year, zero-coupon, $1000$ bond sells for $974$. Determine the spot rates $r_1$, $r_2$ and $r_3$.

This comes from Mathematical Interest Theory textbook section 8.3 #2. I understand how to compute similar problems, however I am unsure how to solve this given that the bonds have annual coupons(not zero coupon bonds). Any help would be appreciated thank you!

Normally you would start with:

$\frac {40}{1+r_1} +  \frac {40}{(1+r_2)^2} + \frac {1040}{(1+r_3)^3} = 990$

But that is a little messy so make this substitution.

Let $y_1,y_2, y_3 = \frac {1}{1+r_1}, \frac {1}{(1+r_2)^2},\frac {1}{(1+r_3)^3}$

$40y_1 +  40y_2 + 1040y_3= 990\\
30y_1 +  30y_2 + 1030y = 988\\
1000y_1 = 974$

Now you have a system of linear equations, that should be easy enough to solve for $y_1,y_2, y_3.$ From there you can calculate $r_1,r_2, r_3$


sphere and inscribed cube We can express ratio between the volume of a cube and volume of inscribed sphere  in this cube as $6/\pi$ . What is the ratio between these objects  when the cube is inscribed in a sphere?

Consider the cube within the sphere. It has sides $s$ and long diagonal $2a$, where $a$ is the radius of the sphere. Applying the Pythagorean theorem twice, we can show that $s=2a/\sqrt{3}$. It then follows that

$$\frac{V_{sphere}}{V_{cube}}=\frac{\frac{4}{3}\pi a^3}{\frac{8a^3}{3\sqrt{3}}}=\frac{\sqrt{3}\pi}{2}\approx 2.7207$$

I've calculated, sphere:incribed cube = 2.7206990463:1
Since its about ratio, we can make the diameter of the sphere anything, so I took 2. This makes the space diagonal of the cube=2. The sides of the cube are thus 2/root3. The volume of cube= 8root3/9. The volume of the sphere is 4(pi)r³/3 ad in this case, 4(pi)/3. If you take these two numbers (4(pi)/3 and 8root3/9), the ratio of sphere to inscribed cube is 2.7206990463:1


How do I calculate the following surface area? I have encountered this problem during my vector calculus studies which I cannot seem to setup correctly. This is a paraphrasing of the question:
Calculate the surface area of the surface $S$, a cone such that $z^2=x^2+y^2$, bounded above the disc
$x^2+y^2\le4$.

What formula would I use to compute this, or what is the process to getting the correct answer?

What I'd do is to plot the region the exercise is talking about (it's just a cone with the vertex in the bottom at (0,0,0) and a it's top at z = 2) and solve for the triple integral that gives you the volume. That is $\displaystyle\int_{\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2 \leq z^2 \leq 4\}}dxdydz$.

Or if you're used to cylindrical coordinates, try those. They're much easier in this case.

Here, given cone $$z = \sqrt{x^2 + y^2}$$ and as it is given that the required region is bounded above the disc $ \Rightarrow  z \in (0,2] $

You can solve the problem either by relying on the formula of lateral surface area $ S = \pi r l$ where $r =$ Radius, and $ l = $ slanted height.

Or you can use calculus.
Here I will use cylindrical coordinates.

My area element would be $dA = r d\theta dh \sec \frac{\pi}{4}$ ,Here $r = h$, so
$$ A = \sec \biggr (\frac{\pi}{4} \biggl )\int_0^2 \int^{2\pi}_0 h dr  d \theta $$

Your task is to figure out why there is $\sec \bigr (\frac{\pi}{4} \bigl )$ and $r = h$


Finding a closed form for the following integral: $\int_0^1\sqrt{1+x^k}dx$ Is there a nice closed form, for the following integral:
$$\int_0^1\sqrt{1+x^k}dx$$
And how can I derive it?
I have no idea how to get started, thanks for any help. This problem came up when I was working on a number theory problem in my university for some values of $k$.

If we expand $\sqrt{1+x^k}$ in series and then intergrate  then we get a close representation in terms of hypergeometric function
$$
\int_0^1 \sqrt{1+x^k}={}_2F_1\left(-\frac{1}{2},\frac{1}{k};\frac{k+1}{k};-1 \right)
$$


Time-dependent manifolds I'm studying Classical Mechanics, and in some cases the particles in the study are constrained to move in a certain manifold, which changes with time.

I've looked for bibliography about time-depending manifolds, or something similar I could apply to this study, but I haven't found anything yet,
so I'd greatly appreciate any bibliographical suggestion.

Any book dealing with non-autonomous differential equations (so pretty much any book) should fit the bill.  If you want better understanding of just manifolds Tu's Introduction to Manifolds is a reasonable choice.  For ODEs I would recommend Ordinary Differential Equations with Applications by Chicone (very theoretically rigorous and does most of its work in Banach Spaces).  For PDE, one text to consider is Taylor's Partial Differential Equations: basic theory (Springer).  If you want a treatment specifically in the language of manifolds, and taking a more differential geometry approach Pressley's Elementary Differential Geometry (Springer) is a pretty solid (and reasonably priced) introduction.


Must there be a sequence $(\epsilon_n)$ of signs such that $\sum\epsilon_nx_n$ and $\sum\epsilon_ny_n$ are both convergent? Let $(x_n)$ and $(y_n)$ be real sequences.

  (i) Suppose $x_n \rightarrow 0$ as $n \rightarrow \infty.$ Show that
  there is a sequence $(\epsilon_n)$ of signs (i.e., $\epsilon_n \in
 \{−1, +1\}$ for all $n$) such that $\sum \epsilon_nx_n$ is convergent.

  (ii) Suppose $x_n \rightarrow 0$ and $y_n \rightarrow 0.$ Must there
  be a sequence $(\epsilon_n)$ of signs such that $\sum\epsilon_nx_n$
  and $\sum\epsilon_ny_n$ are both convergent?

I'm struggling to come up with formal proofs, for (i) I've seen that we simply pick a limit and and then as soon as our sum passes the limit we set $\epsilon_n=-1$ until we pass it again and so on, oscillating about the limit but as $x_n \rightarrow 0$ we converge to it. for (ii) I don't think there must be such a sequence of $\epsilon_n$ but I can't construct a proof or counter example. So I would ask for a solution to (ii) and possibly a better way of constructing answers/tackling these problems in general.

Thank you

The answer to (ii) is yes, there is such a sequence of signs $\epsilon_n$. See Theorem 2.2.1 here, where this result (for any number of series, formulated for vector valued series) is referred to as the Dvoretzky-Hanani Theorem.

$\lim_\limits{n\to \infty} x_n = 0$

$\forall\epsilon>0,\exists N>0$ such that $n>N\implies |x_n|<\epsilon.$

I don't what to make poor epsilon pull double duty.  I am going to call the sequence of signs e.

$M_n = \sum_\limits {i=1}^n e_i x_i$
For the first N elements we can let $e_i$ be strictly positive.
When $N>n,$ we can choose the sign of $e_{n+1}$ such that if $M_n>M_N, e_{n+1}x_{n+1} < 0$ and if $M_n < M_N, e_{n+1}x_{n+1} > 0.$
With $e_{n+1}$ chosen in this way, since $|x_{n+1}| < \epsilon, |M_{n+1} - M_N| < \epsilon$

What about ii) my intuition says that it can't be done.

Suppose $\sum |y_n|$ and $\sum |x_n|$ diverge, but the sequences are going to zero.
for any sequence of signs $e_n$ such that $\sum e_n x_n$ converges, there exists a sequences of signs $f_n$ such that $e_n f_n y_n = |y_n|$  and $f_ny_n$ would still be a sequence that is going to zero.

For all convergent $\sum e_n x_n$ there exists a $y_n$ such that $y_n\to 0$ and $\sum e_n y_n$ is divergent.

But that is not quite the same thing as we have been asked to prove, is it?


Image of a maximal ideal Let $f:R\rightarrow S$ be a surjective homomorphism of commutative rings with unity. I want to prove that if $M$ is a maximal ideal then $f(M)$ is either $S$ or it is a maximal ideal of $S$. I get the feeling I should somehow use the correspondence theorem, but I just can't see how to exactly use it. Thank you in advance.

I also was wondering if the same statement holds for prime ideals?

If $f(M) \subseteq I \subseteq S$ is an ideal, then $M \subseteq f^{-1}(I) \subseteq R$. Since $M$ is maximal, we get $M=f^{-1}(I)$ or $f^{-1}(I)=R$, i.e. $f(M)=I$ or $I=S$. $\mathrm{QED}$

Let $f:R \to S$ be surjective, as above, and let $\mathfrak m$ be a maximal ideal. Then, since $f$ is surjective, the image of $\mathfrak m$ is an ideal also, which we denote by $f(\mathfrak m)$. We get an induced map on quotient rings: $f:R/\mathfrak m \to S/f(\mathfrak m)$. Now, the claim is equivalent to $S/f(\mathfrak m)$ being either a field or the zero ring.

If it is neither, it has a proper ideal, say $\mathfrak a$, whose inverse image is a proper ideal of $R/\mathfrak m$. But this is a contradiction, as $R/\mathfrak m$ is a field, thus having no proper ideals.


Derive a diffusion equation on the basis of the physical law $q=−D(n−n_0)^2n^2\frac{∂n}{∂x}+nv$ It is given that the mass flux density q of a substance in medium obeys the following physical law:

$$q=−D(n−n_0)^2n^2\frac{∂n}{∂x}+nv$$
where $n(x,t)$ is the concentration $([n]=ML−3)$ of the substance in the medium as a function of the space coordinate $x$ and time $t$, $D$ is a constant coefficient of diffusion, $n_0$ is a constant parameter of the problem and $v(x,t)$ is macroscopic velocity of the medium.

Derive a diffusion equation on the basis of the physical law.

Ive worked out that $D=M^{-5}L^{16}T$ but I'm really not confident with diffusion equations so any help towards the answer will be appreciated.

Dump this into the continuity equation,
$$\frac{\partial n}{\partial t}+\frac{\partial q}{\partial x}=0.$$
The PDE follows from calculating the derivatives.


Number of triangles sharing all vertices but no sides with a given octagon The number of triangles whose vertices are the the vertices of the vertices of an octagon but none of whose sides happen to come from the sides of the octagon.

My Attempt: Let $\{A,B,C,D,E,F,G,H\}$ be the vertices of an octagon. It is given that none of the side of octagon is the side of the triangle, so we do not take consecutive points.

So we take either $\{A,C,E,G\}$ OR $\{B,D,F,H\}$ points out of which we will take only three points, because we have form a triangle.

So This can be done by $\binom{4}{3}+\binom{4}{3} = 8$

But the only options given are 24, 52, 48, and 16.

Where have I made a mistake?

If two of the vertices are $A$ and $C$, what are the possible third vertex?  Look at the whole list $A,...,H$

See bro, what I did, from 8 vertices, three will make a triangle... So 8C3 = 56 but sides are not to be included and one side will be forming a side of triangle so deducting 8 from 56 we will get 48 which is the answer

I hope this concept it true... Thanks


Is it true that this function $f(n)=n^{13}$? Assume strictly monotone increasing function;
   such that $f:N^{+}\to N^{+}$, $h$ for all
  $n\in N^{+}$,
  $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

  Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$
$$f(f(m))=mf(m)$$
Put $n=m$,
$$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$
  What about following?

Let us denote by $f^{[k]}(n)$ the $k$th iterate of $f$. I cannot prove the claim, but I can prove that for all integers $n>1$ we have
$$
\lim_{k\to\infty}\frac{\log f^{[k+1]}(n)}{\log f^{[k]}(n)}=13.
$$
This is some kind of asymptotic evidence in favor of $f(n)=n^{13}$ being the only solution - alas, anything but conclusive.

This is seen as follows. We first prove that for all $k\ge3$ we have
$$
f^{[k]}(n)=f^{[2]}(n)^{A_k} f(n)^{B_k} n^{C_k}\qquad(*)
$$
for the sequence of vectors of positive integer determined by the recurrence relations
$$
\left(\begin{array}{r}
A_2\\B_2\\C_2\end{array}\right)=\left(\begin{array}{r}
1\\0\\0\end{array}\right),\qquad
\left(\begin{array}{r}
A_{k+1}\\B_{k+1}\\C_{k+1}\end{array}\right)=M\left(\begin{array}{r}
A_k\\B_k\\C_k\end{array}\right),
$$
where $M$ is the $3\times3$ matrix
$$
M=\left(\begin{array}{crr}
1&1&0\\1&0&1\\2015&0&0\end{array}\right).
$$
The proof follows from the given functional equation of $f$ by induction on $k$. The case $k=3$ is exactly the functional equation. The inductive step follows from the induction hypothesis by substituting $f(n)$ in place of $n$ and again applying the given functional equation.

The eigenvalues of $M$ are $\lambda_1=13$ and $\lambda_{2,3}=-6+i\sqrt{119}$.
The key is that of these $\lambda_1$ has the largest absolute value. Furthermore, if we write the vector
$$
(A_2,B_2,C_2)^T=x_1e_1+x_2e_2+x_3e_3
$$
in terms of unit eigenvectors $e_1,e_2,e_3$ belonging to the respective eigenvalues, we see that $x_1\neq0$.

For any $k\ge3$ we then have
$$
(A_k,B_k,C_k)^T=\lambda_1^{k-2}x_1e_1+\lambda_2^{k-2}x_2e_2+\lambda_3^{k-2}x_3e_3.
$$
For very large values of $k$ the first component dominates, and consequently
$$
\lim_{k\to\infty}\frac{A_{k+1}}{A_k}=\lim_{k\to\infty}\frac{B_{k+1}}{B_k}=\lim_{k\to\infty}\frac{C_{k+1}}{C_k}=13.
$$
It takes a while to see these limits if you calculate them. For some numerical support I fired up my Mathematica. The entrywise ratios of $M^{129}$ and $M^{128}$ are all in the interval $(12.9,13.2)$.

The claim follows from this by taking logarithms from $(*)$.

I don't know if this helps. It does seem to me that we should concentrate on large values of $n$ and asymptotics first. If only we could prove that $f$ must be a homomorphism of multiplicative monoids. Then it being strictly increasing would force it to a power function, and we know the exponent. If we know that $f$ is a power function the exponent can be determined without resorting to the above asymptotic gymnastics.

26.01.20

$\color{brown}{\textbf{The task analysis.}}$

From the issue conditions should
$$n\in\mathbb N,\quad f(n)\in \mathbb N,\quad f_2(n)=f(f(n)) \in \mathbb N,\quad f_3(n)=f(f(f(n)))\in \mathbb N,\tag1$$

$$n \le f(n)\le f_2(n)\le f_3(n).\tag2$$

Taking in account the issue equation, should
$$n^{2015}\,|\,f_3(n),\quad f(n)\,|\,f_3(n),\quad f_2(n)\,|\,f_3(n).\tag3$$

The issue equation can be presented in the form of
$$\dfrac{f_3(n)}{n^{2197}} = \dfrac{f_2(n)}{n^{169}}\, \dfrac{f(n)}{n^{13}}.\tag4$$

If $\ \exists(N)\forall(n\ge N):\ f(n) > n^{13},$ then
$\ \exists(N)\forall(n\ge N):$
\begin{align}
&\dfrac{f_2(n)}{n^{169}}\, \dfrac{f(n)}{n^{13}} = \dfrac{f(f_2(n))}{n^{2197}}
> \left(\dfrac{f_2(n)}{n^{169}}\right)^{13}, \\[4pt]
&\dfrac{f(n)}{n^{13}} > \left(\dfrac{f_2(n)}{n^{169}}\right)^{12}
=\left(\dfrac{f(f(n))}{n^{169}}\right)^{12}
> \left(\dfrac{f(n)}{n^{13}}\right)^{156},\\[4pt]
&f(n) < n^{13},
\end{align}

and this contradiction means that
$$\forall(N)\exists(n\in[N,f_2(N)]):\ f(n) \le n^{13}.\tag5$$

Similarly,
$$\forall(N)\exists(n\in[N,f_2(N)]):\ f(n) \ge n^{13}.\tag6$$

$\color{brown}{\textbf{Solutions building.}}$

From $(4)-(6)$ follows that
$$f(n) = n^{13}\tag{7}$$
is a solution and asymptotic case for all monotonic solutions.

Since different polynomials can not have infinity set of the common points, $n^{13}$ is the single polynomial, which satisfies $(5-6).$

Other monotonic solutions can be considered as the infinity series of deviations from $(7)$, based on the monotonity conditions in the form of
$$m^{14}\le (m+1)^{13}\tag8$$
or similiar ones.

Analysis shows,that required series are finite.

For example, the function
$$g_{14}(m) = \left\lceil14\,\dfrac{\ln m}{\ln(m+1)}\right\rceil\tag9$$
has the table of the values in the form of

On the other hand, every deviation near $n$ requires corresponding deviation near $f(n),$ and this lead to the secondary monotonity conditions, which can not be provided.

Similar table for the function
$$g_{169}(m) = \left\lceil169\,\dfrac{\ln m}{\ln(m+1)}\right\rceil\tag{10}$$

shows, that solutions are bounded with $m=45$ although $2^{13}=8192.$

Therefore, the solution $(7)$ is unique.


The stalk of the image sheaf on a normalization curve My question is based on an exercise (Ex II.6.9) in  Hartshorne's GTM52.

Suppose that $X$ is a projective curve over $k$ (an algebraically closed field). Let $\tilde{X}$ be its normalization and $\pi: \tilde{X}\rightarrow X$ be the projection map. For each point $P\in X$, let $\mathscr{O}_P$ be its local ring and $\tilde{\mathscr{O}_P}$ be the integral closure of $\mathscr{O}_P$. Then my question is:

  Is there a relationship between the stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ and the local ring $\tilde{\mathscr{O}_P}$? Are they equal? If it's true, how to prove it?

My attempt.

In the affine case, suppose $X=\mathrm{Spec}A$ for some domain $A$. I know that $\pi$ is just the ring extension $A\hookrightarrow\tilde{A}$, where $\tilde{A}$ is the integral closure of $A$ in its field of fractions. And $P$ corresponds to a prime ideal $\mathfrak{p}$ of $A$. Then $\tilde{\mathscr{O}_P}$ is the integral closure of $A_\mathfrak{p}$. The stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ can be finally expressed as a direct limit of local rings $\{\tilde{A}_f\}$, where $f$ is taken over all elements in $A\setminus\mathfrak{p}$. So the question is translated into a problem of commutative algebra. That is, is it true for  $\tilde{A_\mathfrak{p}}=\varinjlim\tilde{A}_f$?

Maybe we need something given by the assumption on $X$, i.e. $A$ is noetherian and of dimension $1$. But I still have no strategy to deal with it. Could anyone help me? Thanks in advance!

Yes, they are. The formation of integral closure commutes with localization. See Proposition 5.12 in Atiyah-MacDonald's Commutative Algebra book.

I think your attempt is right. $\tilde{A_\mathfrak{p}}=\varinjlim\tilde{A}_f$ is true because you are working in an integral domain(curve implies integral), so everything is included in the function field, and taking colimit is the same as taking the union.


How do I approach problems about the attributes of very large numbers? I have to solve a couple of problems involving very large numbers, ie. those that are too big to use in a calculator. I clearly need a general approach for this sort of problem. Any pointers would be appreciated, ie. no need to solve the problems completely, just give me some direction. (If I get desperate I'll ask for actual solutions :)

The two problems are:

1) Is $2222^{5555} + 5555^{2222}$ prime?

and

2) Can you demonstrate that the digital representation of $7^{35}$ contains at least one digit that occurs four times?

Based on a hint from Chas, I have an idea for (2).

For a number to contain one digit four times, it must have at least 31 digits. That is, a 30-digit number may not meet the criterion as it could have each of the ten digits 3 times. But as soon as we have 31-digits, a fourth occurrence of one of the digits must be present.

Therefore we meet the criterion if:

\begin{align}
7^{35} &> 10^{30} \\
log(7^{35}) &> log(10^{30}) \\
35 *log(7) &> 30 * log(10) \\
35 * 0.845 &> 30 \\
29.58 &> 30
\end{align}

The inequality is not true. Therefore I cannot prove the criterion can be met. Am I on the right track?

For Question 1, you can easily prove it is not prime by considering that both 5555 and 2222 are divisible by 1111

Continuing your proof on question 2:

You have shown correctly that $7^{35}$ has exactly 30 digits and that it is not enough to say there is at least one digit that repeats 4 times. By pigeon hole, the only time when this happens is when all the 10 digits repeat 3 times. So in this case, when we take the sum of the digits, it is a multiple of 3, and hence the number is divisible by 3.This is not possible has $7^{35}$ is not a multiple of 3. Thereby, we have proved that this case is not possible

Hint for the first question:

What is $2222^{5555} \mod 3$? What is $5555^{2222} \mod 3$?

Hint for the second question:

How many digits long is $7^{35}$?


Proving the integral of the Dirac delta function is 1 Was wondering if my solution is mathematically accurate enough:

The question in the book yields:

  Derive
  $$
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$
  From
  $$
f(x)=\int_{-\infty}^{\infty} f(x_i)\delta(x-x_i)\ dx_i
$$
  [Hint: let $f(x)=1$]

My method is:

  $$
f(x)=\int_{-\infty}^\infty f(x_i)\delta(x-x_i)\ dx_i
$$

  $$
f(x)=1
$$

  so
  $$
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$

I think this is not correct. Does someone know is this is correct, or how to do it better?

Cheers

The error is that in the definition mentioned, f(x) has to be compactly supported and making it constant equal 1 is not correct.

If f(x) is a function of x, then it cannot be removed from the integral. Therefore this proof only works if f(x) is a constant.


How to get $(n+1)!(n+2) = (n+2)!$ I mean it makes sense when I look at it that the two are equal, but I don't entirely understand how you get from one to the other - I presume there's some basic algebra involved - but I'm not sure what it is, how to figure it out, could someone explain this concept to me? I understand it somewhat relates to telescopic series, but I feel like if I don't understand this basic concept, I won't really get the rest of it.


Integral in spherical coordinates, $\Omega$ is the unit sphere, of $\iiint_\Omega 1/(2+z)^2dx\ dy\ dz$ $$\iiint_\Omega \frac{1}{(2+z)^2}dx\ dy\ dz$$

There is a VERY similar question How to integrate $\iiint\limits_\Omega \frac{1}{(1+z)^2} \, dx \, dy \, dz$ here

But this is different.

I like my spherical coordinates to have the angle in the x/z plane taken from "3 oclock" as normal, rather than from 12. So anyway, I got this:

$$\iiint_\Omega \frac{1}{(2+p\sin(\theta))^2}p^2 \cos(\theta)dp\ d\theta\ d\psi$$
Over
$$\Omega = \lbrace(p,\theta,\psi)|\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\psi\in\left[0,2\pi\right],p\in[0,1]\rbrace$$

I'm not sure what to do, I want to do a substitution but I've been explicitly told to use spherical coords, and it'd be a multivariate substitution (involving p and $\psi$)

Thanks.

Addendum:

I'm hoping you guys will make me feel silly and it'll be obvious, but I've been looking at it for a while, I can't see partial fractions helping,I can't get the p out of the denominator, I'm stumped, sadly.

Thoughts:
It actually seems quite easy in Cartesian. Also why can't I substitute, I'm going to have a go, treating it first as integral involving 2+something, then deal with the sin.

$$I=\int^1_0\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{\cos(\theta)p^2}{(2+p\sin(\theta))^2}\int^{2\pi}_0d\psi\ d\theta\ dp$$

$$=2\pi\int^1_0p\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{p\cos(\theta)}{(2+p\sin(\theta))^2}d\theta\ dp$$

Let $u=2+p\sin(\theta)$ then $\frac{du}{d\theta}=p\cos(\theta)$

Thus: $$du=p\cos(\theta)d\theta$$
$$\theta=\frac{\pi}{2}\rightarrow u=2+p$$
$$\theta=-\frac{\pi}{2}\rightarrow u=2-p$$

So:

$$I=2\pi\int^1_0p\int^{2+p}_{2-p}\frac{1}{u^2}du\ dp$$
$$=2\pi\int^1_0p\left[\frac{1}{u}\right]^{2-p}_{2+p}dp$$
$$=2\pi\int^1_0p\left(\frac{1}{2-p}-\frac{1}{2+p}\right)dp$$

Integrate this by parts (differentiate p, integrate the brackets)

Tidy up to get:

$$I=4\pi[\ln(3)-1]$$

Doing this in Cartesian would be hellish; the limits would be a nightmare.

It may be easier to write the integral using $u = \sin \theta$ instead.

The resulting integral is

$$2\pi \int_{-1}^1 \int_0^1 \frac{p^2}{(2+pu)^2} \, dp \, du$$

A table of integrals helps here.  Consulting wikipedia tells us that

$$\int \frac{x^2}{(ax+b)^2} \, dx = \frac{1}{a^3} \left( ax -2b \ln (ax + b) - \frac{b^2}{ax+b} \right) + C$$

That helps us take care of the radial integral:

$$\int_0^1 \frac{p^2}{(2+pu)^2} \, dp = \frac{1}{u^3} \left(u - 0 - 4 \ln (u+2) + 4 \ln 2 - \frac{4}{u+2} + \frac{4}{2} \right)$$

You can probably finish the problem from here.


Is infinite product of Z a group? Having the usual coordinate-wise addition, does infinite product of $\mathbb{Z}$ forms a group?

$a,b\in \prod ^\infty \mathbb{Z}$

$a=(a_1,a_2,...)$

$b=(b_1,b_2,...)$

$a\circ b=(a_1+b_1,a_2+b_2,...)$

Then, is $(\prod ^\infty \mathbb{Z},\circ)$ a group?

Thank you very much.

As Pedro said, yes.  It would be a good exercise to try and prove yourself that the axioms hold.


Showing that $ \int_0^1 \frac{t^{x}-1}{\ln(t)}dt$ exists for all $x$ How do I show that $\displaystyle  \int_0^1 \frac{t^{x}-1}{\ln(t)}\,dt $ exists for every $ x \in \mathbb{R} $?
I have tried bounding it but without success...

Notice that the function $f:t\mapsto \frac{t^x-1}{\ln t}$ can be extended by continuity on 1 since by the Hospital rule
$$
\lim_{t\to1}f(t)=\lim_{t\to1}\frac{xt^{x-1}}{1/t}=x,
$$
hence the integral $\int_{1/2}^1f(t)dt$ exists for all $x$. Now for the problem in 0 and by change of variable $u=-\ln t$ the problem becomes at $\infty$
$$\int_0^1 f(t)dt=\int_{0}^\infty\frac{e^{-u}-e^{-(x+1)u}}{u}du$$
so because the integral $\int_1^\infty \frac{e^{-u}}{u}du$ is convergent we see that the given integral converges iff $x+1>0$ i.e. $x>-1$.

Define $$f(x)=\int_0^1\frac{t^x-1}{\log t}\,dt.$$
Then, we have
\begin{align}
f'(x)&=\int_0^1\frac{\partial}{\partial x}\frac{t^x-1}{\log t}\,dt\\
f'(x)&=\int_0^1t^x\,dt\\
f'(x)&=\frac{1}{x+1}, \quad x>-1.\\
\end{align}
Therefore, the statement is wrong.


What's the projective limit of these polynomial rings ? Define an inverse system of polynomial rings over a commutative ring $k$ by the canonical projection $k[x_1,...,x_n] \to k[x_1,...,x_m]\;(m< n)$.

Question: What is the projective limit $\varprojlim_n k[x_1,...,x_n]$ ?

Note that we can write, as sets:

$$ k[x_1, x_2, \cdots, x_n] \cong k[x_1] \times x_2 k[x_1, x_2] \times \cdots \times x_n k[x_1, x_2, \cdots, x_n]$$

(all but the first component of the product are ideals) An element of the product is interpreted as the sum of all of its components. Furthermore, the canonical projection (assuming you mean to send $x_n$ to $0$) is just the projections onto the first $m$ components.

Therefore,

$$ \lim_n k[x_1, x_2, \cdots, x_n] \cong k[x_1] \times x_2 k[x_1, x_2] \times x_3 k[x_1, x_2, x_3] \times \cdots $$

What addition is should be straightforward. Working with multiplication should be similar in flavor to working in power series rings.

The elements should probably be thought of as infinite sums; given a finite set of variables, each such sum should only have finitely many monomials that involve only those variables.

I can answer your question. Call a linear combination of monomials $\prod_i x_i^{m_i}$ such that $m_i\ge 0, \sum_i m_i=m$ an $m$-form. Then as a subring
$$\varprojlim_{n \to \infty} k[x_1,...,x_n] = \lbrace f_1 + \cdots + f_m \mid m \ge 0,\;\; f_i\; i\text{-form}\rbrace \le k[[x_1,x_2,...]]$$

For the proof note that there are canonical projections $p_n: k[[x_1,x_2,...]] \to k[[x_1,...,x_n]]$ and the restriction to $S = \lbrace f_1 + \cdots + f_m \mid m \ge 0,\;\; f_i\; i\text{-form}\rbrace$ has image in $k[x_1,...,x_n]$. Hence $p= \prod_n p_n: S \to \prod_n k[x_1,...,x_n]$ is an embedding of rings whose image is exactly $\varprojlim_{n \to \infty} k[x_1,...,x_n]$.


balance scale problem for 13 (not 12) items The 12-item balance scale puzzle is very familiar.  The object is to find the lone non-standard item (if one exists) out of a group of 12 seemingly identical items, using a balance scale and a maximum of three weighings. Did you know that it is possible to accomplish the same outcome given a set of 13 seemingly identical items?  I've scoured the web for a discussion of this problem/solution and have not found one. Am I the only one that has solved this problem?

BTW: If you allow four (4) weighings on a balance scale, how many seemingly identical items can you analyze and be assured of finding the lone non-standard item within the group?

If there is sufficient interest, I will publish the answers in a future post.

The way to solve problems of this sort is to think about how much information you have, which is equivalent to how few possibilities remain.

Each weighing can have one of three possible outcomes: the left pan is heavier, ther right pan is heavier, or they weigh the same. You want each outcome to correspond to (roughly) the same number of remaining possibilities.

Initially, you have 26 or 27 possibilities: one item is heavier (and there are 13 choices for which that item that is), one item is lighter (out of 13), or maybe they're all the same. Since 27 = 3 * 3 * 3, you might hope that three weighings will suffice even if it's possible that they're all the same.

But for that to happen, the first weighing has to split the possibilities into three sets of exactly 9 each.

If on the first weighing you weigh 4 items against 4 others, the split is:

Left pan is heavier means one item in the left pan is heavier or one item in the right pan is lighter. That's 8 possibilities.
Right pan is heavier means one item in the left pan is lighter or one item in the right pan is heavier. Again 8 possibilities.
The pans balance. This must cover the 10 or 11 remaining possibilities. We cannot resolve these in only two more weighings. (Two weighings can have only 3 * 3 = 9 outcomes.)

So, four against four won't work. What about 5 against 5? This splits the 26 or 27 possibilities into sets of size 10, 10, and 6 or 7. Again, we cannot answer the question in only 2 more weighings.

More than 4 in each pan on the first weighing leaves too many possibilities for when the scales do not balance. Fewer than 5 leaves too many possibilities when they do balance.

The problem cannot be solved in only three weighings, even if you know there is an odd item. (Unless you know something else, like the color of the odd item, or that the odd item has a different density and there is some water we can immerse the apparatus in, or we can tie them together bolo-fashion, spin them, and observe how the center of mass moves. Or the scale has more than two pans.)

UPDATE: I just realized you only need to identify the odd item, not whether it's heavy or light. That means you start with only 13 possibilities. I'll be back.

I'M BACK...
Ignore the 13th item. In three weighings, you can tell if one of 12 items is odd, and if so whether it's light or heavy. I won't re-iterate this well-known solution. Re-interpret the "none of the 12 is odd" as "the so-far-ignored 13th item is odd".

You lose the ability to tell that there is an odd item, and if the odd item is the 13th you lose the ability to tell if it's light or heavy, but if you can assume there is an odd item you can always tell which it is, and sometimes (read usually) tell whether it's heavy or light.

The lone standard item will have weight greater or less than the remaining items.

Let us assume it has a greater weight than the rest. Though the solution will not change either way.

For 13 items:

Divide into 2 blocks of 4 and one block of 5.
Let them be A,B and C.
Put both A and B on the balance. If the defective piece is in either
A or B the balance will point it out. Assuming it is in B.
Then take the B set divide it into set of 2, weigh it again. That shall point
out which side has the defective item. Repeat this again. So overall
3 weighing in this case.
If both A and B are equal, then defective item lies in C block, which
has 5 items.
Divide C into 2,2 and 1 item. Repeat the weighing process with 2 weights on the balance.
If it is equal, then the item lying out side is the defective item.
If it is not equal, the balance will point out, which set has
defective item in it. Weigh again with 1 item on each side of the
balance to get the defective item.

Total weighing: 3 turns again.


Are there integer solutions for $y$ when $y^2 = x^3 +0x + 7$? Long story short I am curious if there is a way to find all the cube roots that will result into square root when I add the number $7$ to the cube root. Or with simple example:

$5^3$ is $125$ but when I add $7$ to $125,$ the result is $132$...which doesn't have a square root (not integer...). So if $5$ doesn't work, then what number once put to $3$rd $n$ will give number that once added by $7$ will give square root?

Here is the motivation behind my question: This one (Elliptic curve equation example...):

$y^2 = x^3 +0x + 7$

or better:

$y^2 = x^3 + 7$

Has a solution: $y = \sqrt{x^3 + 7}$

And if x = any integer number without remainder (1,2,3,4,5,6...), how can I find all the values of y where y is also integer?

In other words: the "Square root" of $x^3 + 7$ should be really a square root (e.q. $16,25,49...$) and not decimal?

I am looking for a simple solution, possibly with the help of modulo/primes since $7$ is a prime (and I'm also hoping to use this programically later in python).

Also if such integer exists at all it seems to be a very big number. I just tried to see if I can modulo $y$ to $2$ and it seems that no integer solutions can be found (well on wolframalpha...) even with $2$ as modulo.

Paste the following two lines to this website and press "Evaluate".

E = EllipticCurve([0, 7])
E.integral_points()

You'll get the result that there is no integral point on this curve (or better, for this integral model).

This also appears in GTM 106, "The Arithmetic of Elliptic Curves" by Silverman, page 266, Proposition 7.1.

You'll find the proof there.

Are there integer solutions to

$y^2 = x^3 + 7? \tag 1$

A challenging and engaging problem in the classical spirit of the school of Diophantus.

I don't have a solution for it.

Rather, one might say I have "half a solution".

We notice that if $(y, x)$ satisfies (1), then $y$ and $x$ are of opposite parity; that is,

$y \; \text{odd} \Longleftrightarrow y^2\; \text{odd} \Longleftrightarrow x^3 = y^2 - 7 \; \text{even} \Longleftrightarrow x \; \text{even}, \tag 2$

$y \; \text{even} \Longleftrightarrow y^2\; \text{even} \Longleftrightarrow x^3 = y^2 - 7 \; \text{odd} \Longleftrightarrow x \; \text{odd}.  \tag 3$

We may resolve the first case (2) as follows:

with $y$ odd and $x$ even, there exist $m, n \in \Bbb Z$ such that

$y = 2m + 1, \tag 4$

and

$x = 2n; \tag 5$

then (1) becomes

$4m^2 + 4m + 1 = 8n^3 + 7, \tag 6$

whence

$4m^2 + 4m - 8n^3 = 6 \Longrightarrow 4 \mid 6, \tag 7$

impossible.

Thus there is no such pair $(y, x)$.

The "second half" (3) appears to be much more difficult; indeed if we try

$y = 2m, \; x = 2n + 1, \tag 8$

we find

$4m^2 =  8n^3 + 12n^2 + 6n + 8$
$\Longrightarrow 4m^2 -  8n^3 - 12n^2 - 6n = 8, \tag 9$

which admits no such simple resolution.

As I said, I only have "half a solution"; but at least those seeking further need only consider the case (3).


Is $c \frac{\partial}{\partial x}=\frac{\partial c}{\partial x}=0$? Is $c \frac{\partial}{\partial x}=\frac{\partial c}{\partial x}=0$?

Or is one supposed to treat $c \frac{\partial}{\partial x}$ as "partially applied" operator (which expects a function).

I don't see how "linearity" of the derivative, that $\frac{\partial}{\partial x}(af(x)+ bg(x))= a\frac{\partial f}{\partial x}+ b\frac{\partial g}{\partial x}$, has anything to do with this. It is just a matter of notation- understanding what $c\frac{\partial}{\partial x}$ means as differential operator.

If c is a number , yeah thats right
We know that Differentiate of every number relative to x is equal to 0 so the second side is equal to 0 .
(differentiate of c and 1 relative to x is 0) so in the left side we have : [c(0)=0]
So yeah the equation is true.

Attention: my original language is not english and im not proffesional on it . so excuse me if i have mistake in my writing or in my words)


How to compute $\sum\limits_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$ I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!

Hint: There exists some $c_k$ independent of $n$ such that
$$
\frac1{(n-2)(n-1)n(n+1)(n+2)}=\sum_{k=-2}^2\frac{c_k}{n+k}.
$$
To find $c_k$, multiply both sides by $n+k$ and evaluate the result at $n=-k$. For example,
$$
c_{-2}=\left.\frac1{(n-1)n(n+1)(n+2)}\right|_{n=2}=\frac1{24}.
$$
Sanity check: $\sum\limits_{k=-2}^2c_k=0$ and every $c_k$ should be a multiple of $\frac1{24}$ with the sign of $(-1)^k$ and depending only on $|k|$.

Once this is done, note that the value $S$ of the series you are looking for is
$$
S=c_{-2}\cdot\left(\frac11+\frac12\right)+c_{-1}\cdot\left(\frac12\right)+c_{1}\cdot\left(-\frac13\right)+c_{2}\cdot\left(-\frac13-\frac14\right),
$$
which yields the value you got thanks to W|A, namely, $S=\dfrac1{96}$.

(n−3)!/(n+2)! = 1/[(n+2)(n+1)n(n-1)(n-2)] and you can easily solve by using fractional part.


Are elliptic curves algebraic varieties? I got a short question.

Are elliptic cubes also algebraic varietes?

Say we have $E:y^2=x^3+5x=:f(x)$

Then we can $f(x)=x(x^2+5)$

So it can't be an algebraic variety..

I feel like I am totally misunderstanding the definition of algebraic varieties.

Can someone help? Thanks

Elliptic curves are very famous examples of algebraic varieties. You seem to have forgotten that the function can have more than one variable; take the function $f(x,y)=y^2-x(x^2+5)$, then $E$ is just the variety $V(f)$.

You are messing up the notation and confusing yourself. You could write the polynomial
$$
f(x) = x^3 + 5x \in \mathbb{Q}[x] \text{ or } \mathbb{C}[x]
$$
or you could write the polynomial
$$
f(x,y) = y^2 - x^3 - 5x \in \mathbb{Q}[x,y]
$$
Once you factor $f(x)$ you find that it equals $x(x^2 + 5) = x(x+i\sqrt{5})(x-i\sqrt{5})$. This should give you the hint that it defines a single point in $\mathbb{Q}$ while giving you three distinct points in $\mathbb{C}$. The other polynomial does define an ellitpic curve as a variety over $\mathbb{Q}$.

Also, you could ramp up this discussion and frame it into a scheme theoretic language. Then, you should consider the scheme
$$
X = \text{Spec}(\mathbb{Z}[x,y]/(y^2 - x^3 - 5x))
$$
If you write this as the functor $X = \text{Hom}_{\textbf{CRing}}(\mathbb{Z}[x,y]/(y^2 - x^3 - 5x),-)$, then
\begin{align*}
X(\mathbb{F}_q) &= \{ (a,b) \in \mathbb{F}_q^2: b^2 - a^3-5a =0 \} \\
X(\mathbb{Q}_p) &= \{ (a,b) \in \mathbb{Q}_p^2: b^2 - a^3-5a =0 \} \\
X(\mathbb{Q}) &= \{ (a,b) \in \mathbb{Q}^2: b^2 - a^3-5a =0 \} \\
X(\mathbb{C}) &= \{ (a,b) \in \mathbb{C}^2: b^2 - a^3-5a =0 \}
\end{align*}
and so on...


Find $a^{b+c} b^{c+a} c^{a+b}$ if $\frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ If $\frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$, then what would be the value of $a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$?
I'm unable to proceed.

Let $ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k$

$\Rightarrow a=e^{k(b-c)}, b=e^{k(c-a)},c=e^{k(a-b)}$. Then the required value is $ e^{k(b^2-c^2)}. e^{k(c^2-a^2)}. e^{k(a^2-b^2)}=1$


Is there a better counter-example? (problem involving limit of composition of functions) This question arises from one of the Berkeley Problems in Mathematics, fall 1981. It asks me to find a counterexample to the following claim:

Suppose $\lim\limits_{x\rightarrow x_{0}}f(x)=a$, $\lim\limits_{t\rightarrow a}g(t)=b$, prove or find a counterexample for the claim $\lim\limits_{x\rightarrow x_{0}}g(f(x))=b$.

I feel this should not be true, for we can select functions with infinite 'singularities' near the origin like $x\sin[1/x]$. Then for each value $x=\frac{2}{k\pi}$ we should have the value to be equal to $x$. But this is not a counterexample. Looking up the book the authors give the following example:

$f=g=1$ when $t\not=0$, $f=g=0$ when $t=0$.

The authors then claimed that $\lim_{x\rightarrow 0}g(f(x))=1$ while $\lim_{t\rightarrow 0}g=0$. Unless I am confused with the definition of the limit, that there exists $\lambda$ such that for any $\epsilon$, if $|x-x_{0}|\le \lambda$, then $f(x)-A|\le \epsilon$, I do not see why $\lim_{t\rightarrow 0}g=0$ since $t=0$ is a removable singularity for $g$. So assuming I am wrong somewhere, I want to ask:

1): why the limit of $g$ at 0 is 0?

2): is there better counterexamples?

I think you just made a careless error. In the example provided, the limit of $g$ at 0 is not 0; it is 1, as you said. Here $x_0 = 0$, so that $a = 1$. So lim$_{t\rightarrow a}{g(t)}$ = lim$_{t\rightarrow 1}{g(t)} = 0$. And lim$_{x\rightarrow x_0}{g(f(x))} = $ lim$_{x\rightarrow 0}{g(f(x))} = 1$. This is probably as simple a counterexample as one could hope for.

Another counter-example that would be more familiar to a college or high-school calculus student would be the following.
consider $f,g: \mathbb{R} \to \mathbb{R}$ defined

$$
\begin{array}{ll}
g(x) = \frac{x^2 -1}{x-1} = \frac{(x+1)(x-1)}{x-1} \text{and}\\
f(x) = 1.
\end{array}
$$

Then
$$
\lim_{x \to 1} g(x) = \lim_{\epsilon \to 0} \frac{(1 + \epsilon +1)(1 + \epsilon-1)}{1+ \epsilon -1} = \lim_{\epsilon \to 0} \frac{(1 + \epsilon +1)(\epsilon)}{\epsilon} = 2.
$$

However
$$
\lim_{x \to 1}g(f(x)) = \lim_{x \to 1}g(1) = \frac{0}{0},
$$
which is undefined.

So we need that $f(x) \neq \lim_{x \to x_0} = a$ on a small interval $(x_0 -\epsilon, x_0 + \epsilon)$ containing $x_0$ to avoid problems (How to prove this limit composition theorem?).

The counter example in the text likely meant to say $f$ is identically 0 and $g(x)= 1$ for $x \neq 0$ and $g(0) = 0$. Hopefully you can see the parallel between this example and my own. When $f(x)$ is constant around $x_0$ then the limit of $g$ is forced to equal $g(f(x_0))$.

There may be a typo in your textbook. This is very common, and it is important to do a common sense check on math problems and "read" what the author meant to say rather than the exact words on the page.


Min value of $F(x) = x - \sqrt{1-x^2} $ Min value of $F(x) = x - \sqrt{1-x^2} $

But using derivatives  $F(x) = x - \sqrt{1-x^2} $

i stucked at $2\sqrt{1-x^2} + 2x = 0$

$\sqrt{1-x^2} > 0 $

I get $-1<x<1$

Please help me?
 is there another way to solve it btw?

I think there may be something wrong in this (that means, your derivative): $2\sqrt{1-x^2} + 2x = 0$

My answer:

  First let's notice that the function domain is $[-1,1]$ for that $1-x^2\geqslant 0$. Then it is easy to see $f'(x)=0$ when $f(x)$ reaches to minimum. So that
  $$
f'(x)=\frac{x}{\sqrt{1-x^2}}+1
$$
  set $f'(x)=0$ we get $x=-\frac{1}{\sqrt{2}}$, and $f(x)=-\sqrt{2}$ where it is the minimum. Rigorously speaking, you need to check the derivative on both side to judge whether it is the min value, but in this question it is obvious.

Rearrange

$$2\sqrt{1-x^2} + 2x = 0$$

to get

$$\sqrt{1-x^2} = -x$$

Square both sides

$$1-x^2=x^2$$

and solve for $x$.


(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...

Thus, using the expansion $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$, one sees that $$\log f(n)=n^{3/4}+c^{-1}n\log(1-cn^{-1/4})$$ is also $$\log f(n)=n^{3/4}+c^{-1}n\,(-cn^{-1/4}-\tfrac12c^2n^{-1/2}+o(n^{-1/2}))$$ that is, $$\log f(n)=-\tfrac12cn^{1/2}+o(n^{1/2})$$ In particular, $f(n)\to0$ for every $c>0$ and $f(n)\to\infty$ for every $c<0$.


Six real numbers so that product of any five is the sixth one One needs to choose six real numbers $x_1,x_2,\cdots,x_6$ such that the product of any five of them is equal to other number. The number of such choices is
A) $3$
B) $33$
C) $63$
D) $93$

I believe this is not so hard problem but I got no clue to proceed. The work I did till now.
Say the numbers be $a,b,c,d,e, abcde$. Then, $b\cdot c\cdot d\cdot e\cdot abcde=a$ hence $bcde= +-1$.
Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?

Here's an argument which extends to $n$ real numbers quite nicely! We can start by noting that
$$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
And by commutativity, we get $x_i^2 = x_j^2$ which implies that all the magnitudes are equal.

Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we conclude that $L=0$ or $1$.

Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s

If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
$$\frac{\prod_{i=1}^6 x_i}{1} = 1 \text{ and } \frac{\prod_{i=1}^6 x_i}{-1} = -1$$
Now we can count configurations. There will be
$$\sum_{i=0}^3 \binom{6}{2i} = 2^5$$
possibilities. And finally, we have $1+32 = 33$.

HINT:

It is obvious that you need to have the numbers absolute value equal to $1$ (besides one special case but we get to that later). The question is how to set the signs of these $1$s. The product will be negative if you have an odd number of negative signs in it and positive whenever you have an even number of negative signs. Since you are to determine signs for an even number of $1$s if you decide to give an even number of them a negative sign you have one of the following two situations

the $1$ that is NOT in the product has negative sign. In this case there are a $\text{even}-1=\text{odd}$ number of negative signs distributed along the other five $1$s which result in $-1$ so the condition is fullfilled
the $1$ that in NOT in the product has positive sign. In this case there are a $\text{even}-0=\text{even}$ number of negative signs distributed along the five other $1$s which result in $+1$ so the condition is once again fullfilled.

So your question translates to in how many ways can you distribute an even number of negative signs among the six $1$s. This is solved in the easiest way by considering the number of ways to distribute $0,2,4$ and $6$ negative signs.

Finally the one solution not having to deal with $1$s is the one dealing with all zeros.

Hope this helped


Permutations with Repetitions, how to select $n$ and $r$ I'm doing Permutations and Combinations and I'm facing difficulty with Permutations with repetitions. My book states that if there are $n$ objects chosen $r$ at a time, the total number of permutations is $n^r$.

I cannot understand how to decide what $n$ is and what $r$

In the following problem:

  In how many ways can 6 different rings be worn on 4 fingers?

While I can solve this using the Multiplication Principle, I keep thinking that $n$ is the number of rings (as in my textbook it says that there are '$n$' different things') and $r$ is the number of fingers.

It is reasonable to suppose that the order of rings on fingers matters. Blue then white then red on the ring finger looks better than red then blue then white.

To count the number of ways to choose the number of rings the various fingers will hold, use Stars and Bars. The number of choices is $\binom{9}{3}$.  For each choice, the rings can be permuted in $6!$ ways, for a total of $\binom{9}{3}6!$.

I would recommend that you think concretely instead of worrying about what is $n$ and $r$. Effectively that would mean that you derive the formula $n^r$ each time you solve the problem but in the long run that would be helpful.

You say that you can solve this using the multiplication principle, but the formula $n^r$ is only an abstract (special case) of the multiplication principle, so you are implicitly using the formula anyway. With experience the abstraction part would come naturally and forcing it is not good IMHO.

In this case just think about the rings one by one. The first ring has four choices (fingers) to go to, the second has four as well and so on, and that would give you the answer $4^6=4\times 4\times...\mbox{six times}$.


Numerical Analysis or Numerical Linear Algebra Book with Available Solutions? I am a beginning graduate student in Mathematics soon and I am planning to self-study Numerical Analysis and Numerical Linear Algebra. I know there are already reference-request questions about this, but I am looking for some more specific books.
I am looking for books in the following categories:

Books with difficult problems, and solutions in the back (ideal)
Books with difficult problems
Books with solutions in the back
Books with separate solution manuals
Problem books (ideally with solutions in the back, or solution manuals)

Could you please recommend me some textbooks, and tell me in which category they are? Thank you very much, any recommendations are immensely appreciated!

I was an algebra major student and later a postgraduate student in topological algebra. We had two main books with linear algebra problems (but, as far as I remember, non-Numerical). The books contained a lot of problems and, in the back, hints for some of (I guess, harder) problems. Maybe there are separate books containing full solutions for these problems.
[Kos] This is an English translation of the second edition of [Kos2]. This books contains relatively hard problems. One of its three chapters (about a hundred pages)  is devoted to linear algebra and geometry. Unfortunately, the second edition of the book contains
sufficiently many misprints, which are hopefully fixed in [Kos2].
[Pro] This books has about 350 pages and two thousands linear algebra problems.
References
[Kos] Exercises in algebra: a collection of exercises in algebra, linear algebra and geometry, ed. Kostrikin A.I., Gordon and Breach Publ., 1996.
[Kos2] A collection of linear algebra problems, 3-edn, corrected and extended, ed. Kostrikin A.I., Moscow, Phys.-mat. lit, 2001, in Russian (Сборник задач по линейной алгебре, под редакцией А.И. Кострикина, издание третье, исправленное и дополненное, Москва, Физико-математическая литература, 2001). I did not find an English translation.
[Pro] Proskuryakov I.V., A collection of linear algebra problems, in Russian (Проскуряков И.В. Сборник задач по линейной алгебре). I did not find an English translation.


Conditional expectation of first roll conditionally on sum of two rolls I have to find $E(X\mid Y)(y)$ where $X$ is the value of the first roll and $Y$ is the sum of the two dice.   I know that

$$E(X|Y)(y) = \sum_x{xP(X\mid Y)}=\frac{\sum_xxP(X=x, Y=y)}{P(Y=y)},$$

but this doesn't really get me anywhere.  From that summation, I get

$$E(X|Y)(y) = \frac{1\cdot P(1, Y=y)}{P(Y=y)} + \frac{2\cdot P(2, Y=y)}{P(Y=y)} + \dots + \frac{6\cdot P(6, Y=y)}{P(Y=y)}.$$

Here's where I'm stumped.  Can someone please answer this question for me.

Thanks!

EDIT:  To be clear, I need to find the expected value of the first roll given the sum of the two dice.

I get E(X∣Y=y)=y/2 for y≥2. Is that correct?

Use symmetry:

$(X,Y)$ and $(Y-X,Y)$ have the same distribution hence $$E(X\mid Y)=E(Y-X\mid Y).$$
$X+(Y-X)=Y$  hence $$E(X\mid Y)+E(Y-X\mid Y)=E(Y\mid Y)=Y.$$

Thus, $$E(X\mid Y)=\frac12Y.$$

See this.  Additionally/alternatively, Bayes' theorem would let you write $P(X \mid Y)$ in terms of $P(Y \mid X)$, which may make more sense to you.


Prove $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square Prove the number $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square.

I tried studying $a \mod b$ for different integers $b$, without succes.


Is there no such identity as $\csc^2+\sec^2=1$? $$\csc^2+\sec^2=1?$$

I thought I could just use reciprocal from the other formula $\sin^2+\cos^2=1$, can you explain what's wrong?

You can't just "use reciprocal", as you put it, because
$$\frac{1}{x^2+y^2}$$
is not the same thing as
$$\frac{1}{x^2}+\frac{1}{y^2}$$

We have $sin^{2}x + cos^{2}x = 1$ .Taking reciprocal we get $1 / sin^{2}x + cos^{2}x = 1$ ,then converting sin to cosec and cos to sec and then taking lcm of denominator we get $sec^{2}xcosec^{2}x/sec^{2}x + cosec^{2}x$ = 1


Conditional expectation of the sum of two random variables I've got some difficulties in calculating the conditional expectation of the sum of two RV.

I am not sure if I correctly formalized the scenario I am looking at. So I am trying to describe it first: I have two independent Poisson processes $P_1$ and $P_2$ with specific means $\lambda_x$ and $\lambda_y$. They are always executed one after the other. So $n$ runs of an experiment involve $n$ runs of both $P_1$ and $P_2$. I know that $E[N|T=t] = \lambda\,t$. What I'd like to calculate is the expectation that they sum of $n$ consecutive runs of the experiment given that the sum of one of the processes is $k$, i.e. $P_2=k$.
I've tried to formalize this into the following:

Let $X$ and $Y$ be two independent, exponentially distributed random variables with means $\lambda_x$ and $\lambda_y$. Let $Z$ be the sum of the two variables, i.e. $Z=X+Y$. So $Z$ should also be exponentially distributed, hence $Z\sim P(\lambda_x+\lambda_y)$.

I'd like to calculate the conditional expectation of $Z$, given that $Y=k$.

I have $P(Z=n| Y=k) = \frac{P(X+Y=n, Y=k)}{P(Y=k)} = \frac{P(X=n-k)P(Y=k)}{P(Y=k)}$. But this would cancel out the last term which doesn't seem to be right.

Any help is appreciated.

Thanks

The notation (or lack thereof) is confusing and what (I think) you are asking for makes no sense but... Let $\{N_1(t):t\ge 0\}, \{N_2(t);t \ge 0\}  $ be independent Poisson processes with rates $\lambda_1,\lambda_2.$ For a given time $t_0$ we observe values of $N_1(t_0), N_2(t_0)$ and then repeat $n$ times and sum the $n $ observed values for each process. That is the same as observing the values $N_1(nt_0), N_2(nt_0).$ The comments about one after another and consecutive runs do not matter.

Now let $t$ be any fixed time, possibly $t=nt_0,$ and for any $k\ge 0:$
$$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1t +k $$

The answer of Mr.Spot seems to bring me closer to the right answer, I believe. I again assume that $\lambda_1$ and $\lambda_2$ are the means and $n$ the number of repetitions. So the sum of the $n$ observed values would be $\lambda_1n$ and $\lambda_2n$.
Starting with what we had at the end:

$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1n + \lambda_2n $ and $k=\lambda_2n$.

We see that $n=\frac{k}{\lambda_2}$. Hence, we can write:

$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1 \frac{k}{\lambda_2} + k = k(\frac{\lambda_1}{\lambda_2}+1)$

However, filling in some values for $k, \lambda_1$ and $\lambda_2$ doesn't yield the expected results that I got from a simulation of the experiment. However,
$E[N_1(t)+N_2(t) \mid N_2(t)=k] = k(\frac{\lambda_2}{\lambda_1}+1)$ does work. Note its $\lambda_2$ over $\lambda_1$. But I don't know how to obtain it.

Thanks


Convergence in $L^1$ of defined step functions Question: Let $f \in L^1(m)$. For k=1,2,... let f_k be the step function defined by

$$
f_k(x)=k\int_{j/k}^{(j+1)/k} f(t)\,dt
$$
$$
\text{for } \frac{j}{k}<x\leq\frac{j+1}{k}, \quad j=...,-1,0,1,...
$$

Show that $f_k$ converges to f in $L^1$.

I guess I showed $f_k$ converges f a.e. by the Lebesgue Differentiation Theorem, since

$$\lim f_k(x)=\lim \dfrac{\int_{j/k}^{(j+1)/k} f(t)\,dt}{1/k}$$

If I can also show $|f_k|\leq |g|$ a.e. for some $g\in L^1$ then I can use the Lebesgue Dominated Convergence Theorem since $|f_k-f|\leq |g|+|f|$ and $f_k-f$ converges to $0$ a.e. My attemt was to show $|f_k|\leq |f|$, but I'm not even sure whether this inequality holds or not.

Any comments on my idea and suggestions?

Thanks!

You can prove this exactly as the standard result about approximate units in $L^1$.

Write $T_kf=f_k$. Then
$$T_kf(x)=\int_{\mathbb R} \Phi_k(x,t)f(t)\, dt,$$
for a certain "kernel" $\Phi_k$ given by
$$\Phi_k(x,t)=k\,\sum_{j\in\mathbb Z} \mathbf 1_{Q_{j,k}}(x,t)\, ,$$
where $Q_{j,k}$ is the square $(\frac{j}k,\frac{j+1}{k}]\times (\frac{j}k,\frac{j+1}{k}]\,$.

The kernel $\Phi_k$ is nonnegative, and a direct computation shows that $\int_{\mathbb R} \Phi_k(x,t)dx=1$ for all $t$. Using Fubini's theorem just as for convolutions, this implies that the operators $T_k$ are bounded on $L^1$ with $\Vert T_k\Vert\leq 1$.

We also have $\int_{\mathbb R} \Phi_k(x,t)dt=1$ for all $x$, and it is not hard to check
$$\lim_{k\to\infty} \int_{\{ t;\; \vert t-x\vert\geq\delta\}} \Phi_k(x,t)\, dt=0$$
for every fixed $\delta>0$, uniformly with respect to $x$. In fact, the above integral is equal to $0$ for all $x$ whenever $k>1/\delta$, since in this case $(x,t)$ does not belong to any $Q_{j,k}, j\in\mathbb Z$ if $\vert x-t\vert\geq \delta$. Using this, one shows in the "usual" way that $T_kf\to f$ in $L^1$ when $f$ is continuous with compact support. So the result holds for all $f\in L^1$ since the operators $T_k$ are uniformly bounded.


Why the inverse of this functions not working? If

$f(x) = (x+2)^2+2$

then we get 2 inverse functions for this function:

$f^{-1}(x) = -2+\sqrt{x-2}$
$f^{-1}(x) = -2-\sqrt{x-2}$

If we input functions and inverse functions in to each other we must get "x" back.
But why we don't get "x" when we input function $f(x) = (x+2)^2+2$ into second inverse function $f^{-1}(x) = -2-\sqrt{x-2}$

$f^-1(f(x)) = -2-\sqrt{f(x)-2}$

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

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

= $-2-(x+2)$

= $-2-x-2$

= $-4-x$ ≠ x

As a real-valued function on the reals $f$ is not an invertible function. So, treating some other function as if it were the inverse to $f$ is a recipe for disappointment.

A restriction of $f$, however, is an invertible function from $[-2, \infty)$ to $[2, \infty)$, and the first function you listed is the inverse.

Another restriction of $f$ is an invertible function from $(-\infty, -2]$ to $[2, \infty)$, and the second function you listed is the inverse.

In the second case, the error in your calculation of $f^{-1}(f(x))$ is that if $x \in (-\infty,- 2]$, then $\sqrt{(x+2)^2} \neq x+2$, since $\sqrt{}$ is being used to mean the positive square root, but $x+2$ is negative.

$x=(y+2)^2+2$ here I have interchanged $x$ and $y$.  Now solve for $y$ to get $f^{-1}$.

So, $y=\sqrt{x-2}-2=f^{-1}(x)$.

Remember,  however,  that since $f$ isn't $1-1$, it doesn't have an inverse.  In general,  it has to pass the horizontal line test to have an inverse.  The function I have found is, however,  the inverse to $f$ restricted to $[-2, \infty) $...


Zeno's Achilles & Tortoise - Where exactly is the proof wrong? (For those who don't know what this paradox is see Wikipedia  or the Stanford Encyclopedia of Philosophy.)
Let us define $a_i$ and $b_i$ recursively
$$
a_0 = 0\\
b_0 = 1\\
a_i = a_{i-1} + (b_{i-1} - a_{i-1})\\
b_i = b_{i-1} + (b_{i-1} - a_{i-1})/2
$$
It is easy to prove that $b_i>a_i\  \forall i$ using induction.
Thus while $|b_i-a_i|$ tends to $0$ , we will never have $a_i>b_i$.
We can now just replace $a_0$ as Achilles start position and $b_0$ as Tortoise start position. And then subsequent positions of Achilles is given by $a_i$s (Achilles new position is = Tortoise old position, which is the $1^{st}$ recursion). And Tortoise is assumed to move at half the speed of Achilles. Tortoise positions are represented by $b_i$s. (So, new position of Tortoise = Old Position + 1/2 the distance traveled by Achilles, which is the $2^{nd}$ recursion.)
Given, we have proven $b_i>a_i\  \forall i$, thus I claim Achilles will always be behind Tortoise (He will come closer and closer but will never overtake).
Obviously, I'm wrong but exactly where / which step of the proof above? (Please provide the exact mathematical step/argument where I went wrong.)
Some further discussion:
Basis the responses I got (which I'm unable to find fully convincing - and it maybe just me that I don't understand them well enough) I would like to add - In my opinion, the way I have defined $a_i$ and $b_i$ it is just a subset of positions that Achilles and Tortoise can take. In that subset what I have proved is correct i.e. Achilles cannot overtake Tortoise. But just in that subset <- And I think this is the key
Note that my $a_i$ and $b_i$ are all rational. I can embed infinite rationals between any 2 points on the real line. I think fundamentally the error in my proof is that I use induction on continuous variables. I'm not formally trained to express that mathematically in a precise way - Hence this question.
My question is not to challenge/discuss that Achilles will overtake or not etc or to come-up with another proof - My precise question is where exactly is my proof wrong.
Thanks

The issue is as follows. You have constructed an infinite sequence of times, at all of which Achilles is behind the tortoise. However, that doesn't mean that Achilles will always be behind the tortoise, because the set of times you have constructed is bounded.  Suppose Achilles has unit speed. Then they reach positions $a_1$ and $b_1$ at time $1$, $a_2$ and $b_2$ at time $3/2$, $a_3$ and $b_3$ at time $7/4$, and so on. It is easy to verify that all these times are less than $2$, so your argument only implies the tortoise is ahead for $t<2$. (Indeed, $t=2$ is exactly when Achilles overtakes the tortoise.)

There is nothing wrong with your proof… internal to the proof.
In respect of the universe of consideration, what is wrong with your proof is that it adopts a constrained frame of reference.  It declines to consider any value outside of the constraint that Achilles < tortoise.
In other words… as a proof… it proves that Achilles will always be behind the tortoise, given the assumption that Achilles will always be behind the tortoise.
Mathematically, the idea is to follow the approach in Sameer Baheti’s answer [noting that I have not checked the actual maths] — you allow yourself to use a different frame of reference — one that is not artificially constrained.


Can 2018 be written as a sum of two squares? If can, what is the expression? Can $2018$ be written as a sum of two squares?
If it can be, what are the numbers?

I know the first answer is it can be because $2018=2\times1009$ and $1009\equiv 1 \pmod{4}$. But I cannot find the numbers.

Yes, $2018 = 2\times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.

The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently.  At the end, one find $$1009 = 28^2+15^2 \quad\implies\quad 2018 = 43^2 + 13^2$$

A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.

You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169


Altitude in tetrahedron I was reading a book there , I saw the below question in the book  .

My doubt is that how they have written length of altitude is same as projection of AD on ABC  ?

Please explain if possible with a diagram

Hope the following picture can clarify ……

The idea is that volume of the tetrahedron equals $1/3$ base $\times$ height.

Or, Height = 3V / Area of the base.

But rather than working with tetrahedra and triangles, the math is a little nicer if we use parallelepeiped and parallelograms.

In which case, Height = Volume / Area.

Translate everything such that $A$ is at the origin.

The volume of the parallelepeiped formed by the vectors $AB, AC, AD$ equals the absolute value of the determinant of the matrix with those vectors as rows.

The area of parallelogram formed by $AB$ and $AC$ equals the absolute value of cross product $AB\times AC$


Having a cube, with a point at its center. What are the points that are equidistant from the center point to the cubes vertices? Having a cube, with a point at its center.

What shape do the points wich are equidistant between the center and the cubes vertices make?

The source of why I had this question is the following photo

What shape is resultant from this composition of equidistant points?

Thank you very much.

The answer to the question from screenshot is a truncated octahedron. If you consider center of the cube and a vertex, the locus of equidistant points is a perpendicular bisector plane. 8 of those planes create an octahedron. However, if you consider centers of other cubes, locus of equidistant points with them will give the initial cube itself. So the answer is the intersection of cube and octahedron, which is a truncated octahedron.

Within the pic of OP there is the quest for the volume of $D(0)$ as well. $D(0)$ was already being mentioned to be the truncated octahedron. Thus the searched for volume is $$V=8\sqrt 2\ s^3$$

where $s$ is the edge length of that polyhedron. That length in turn can be seen from that very pic to be (in units of the lattice constant $a$) $$s=\frac 12 \sqrt 2\ a$$

Thus the searched for total value would become $$V=4\ a^3$$


Prove all eigenvalues of $A^*A$ are non-negative Let $A$ be an $m\times n$ matrix. Prove that all eigenvalues of $A^*A$ are non-negative.

Hint: Pick an eigenvector $x$ of $A^*A$ and consider $x^*A^*Ax$.


Two six-sided dice are rolled. Find the probability distribution of $\max[X_1,X_2]$ I am not sure if I am understanding this problem correctly. It says;

  Two six-sided dice are rolled. Let X1 and X2
  denote the outcomes of the first and second die roll respectively.
  Define a new random variable Y such that:

  Y = max{X1 , X2}

Does this mean I am supposed to roll the two dice and find the largest roll?

From my own understanding, if I am supposed to find the largest roll from the two dice roll, is this example of mine correct?

ex: the outcomes are {2,3} then Y=3

Moving on to what is asked of me:

  Find the probability distribution of Y

For this part, do I need to find all the possible outcomes of the two dice? For example;

{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}

{2,1} {2,2} {2,3} {2,4} {2,5} {2,6}

{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}

etc....

And do I need to find the probability of the sum of two dice? For example;

Sum = 2 , Prob = 1/36

Sum = 3 , Prob = 2/36

Sum = 4 , Prob = 3/36

etc....

After this, what do I do from here to find the probability distribution of Y?

If I am totally off track, explanation from you guys will be great. Sorry if it looks like I'm asking too much. Please feel free to correct me on anything!

Notice that the mirror image of every L-shaped event corresponds to the outcome $Y$

As an example, consider the L-shaped event corresponding to the outcome $Y=3$

$$\begin{align}
                  \{1,3\} \\
                  \{2,3\} \\
\{3,1\}, \{3,2\}, \{3,3\}
\end{align}$$

We can easily see that $$\mathbb{P}(Y=3)=\frac{5}{36}$$

In general, you can generalize that $$\mathbb{P}(Y=y)=\frac{2y-1}{36}$$ where $y=1,2,3,4,5,6$

It may be useful to construct a table, with $X_1$ down the left, and $X_2$ along the top, and each cell gets the maximum of the two values.  The table below has been started for you; fill in the rest of the values.

$$
\begin{array}{|c|c|c|c|c|c|c|} \hline
  & \textbf{1} & \textbf{2} & \textbf{3}
  & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline
\textbf{1} & 1 & 2 & 3 & & & \\ \hline
\textbf{2} & 2 & 2 & 3 & & & \\ \hline
\textbf{3} & 3 & 3 & 3 & & & \\ \hline
\textbf{4} & & & & & & \\ \hline
\textbf{5} & & & & & &  \\ \hline
\textbf{6} & & & & & & \\ \hline
\end{array}
$$

The probability distribution for $Y = \max(X_1, X_2)$ can then be obtained by counting up the proportion of entries that are $1, 2, 3, \ldots$.


How many numbers must be selected from the set How many numbers must be selected from the set $\{1, 3, 5, 7, 9, 11, 13, 15 \}$ to guarantee that at least one pair of these numbers add up to $16$? Explain
your answer.

Just count the pairs that do add up to 16:

$$1+15 = 16, \quad3+13=16,\quad \dots$$

Then just ensure you choose enough numbers to guarantee that you have both numbers in one of these pairs.

solution pairs: (1,6) (2,5) (3,4)
    hole #1: for numbers 1 2 3
    hole #2: for numbers 4 5 6

111  1
---  ---
#1   #2

If you pick 4 numbers you are guaranteed to hit a solution


The value of an investment in Canada Savings Bonds is modelled by $A(t) = A_0 e^{0.0255t}$... Rest of question below. The value of an investment in Canada Savings Bonds is modeled by $$A(t) = A_0 e^{0.0255t}$$, where A is the amount the investment is worth after $t$ years, and $A_0$ is the initial amount invested. At what rate, correct to 3 decimal places, is the investment growing at the time when its value has doubled?

I understand that we have to find the derivative but how can we find it if we don't know an initial amount?

The question asks for rate when the investments value has been doubled or when

$$A(t)=2A_0 = A_0 e^{0.0255t}$$
So the initial value cancels and you have:
$$
2=e^{0.0255t}\Rightarrow \ln(2)=0.0255t\Rightarrow t=\frac{\ln(2)}{0.0255}
$$
Now, taking the derivative:
$$A'(t)=0.0255A_0e^{0.0255t}\Rightarrow A'(\frac{\ln(2)}{0.0255})=0.0255A_0e^{0.0255(\frac{\ln(2)}{0.0255})}\\
=.0255A_0e^{\ln(2)}=2(.0255)A_0=.051A_0$$
Which does depend on the principal, as it should.


How to linearise the constraint with a product term? I would like to linear the following constraint
$$y\cdot \sum_{i=1}^{N}\mu_ix_i \geq M$$
where $y$ is a positive integer variable, $x_i \in \{0,1\}, i =1,...,N$ are binary variables, $\mu_i, i =1,...,N$ and $M$ are positive constants.

Thank you.

If $D$ is a sufficiently large constant:

$$\begin{array}{l}
\sum_i \mu_i y_i \geq M \\
0\leq y_i \leq Dx_i \\
0\leq y-y_i\leq D(1-x_i)
\end{array}
$$

Then $x_i=0$ forces $y_i=0$ and $x_i=1$ forces $y_i=y$.

As long as the $x_i$ are not all zero, this inequality can always be satisfied by taking $y$ large enough.  So this is equivalent to $\sum_i x_i \geq 1$.


Question about homeomorphic surfaces I am wondering if there exist an example of two surfaces such that they are orientable (or not orientable) and have same Euler characteristic but they are NOT homeomorphic.

Thanks for your support.

Consider a punctured disc or the torus.

This cannot happen for closed surfaces by the classification theorem of surfaces.

No. By definition, both these surfaces would be homeomorphic to the same connect sum of tori, and so homeomorphic.


Examples of derivation of Lie algebras Let $A$ be an algebra over a field $F$. A derivation of $A$ is an $F$-linear map
$D : A\to A$ such that
$D(ab) = aD(b) + D(a)b$ for all $a, b \in A$.
The map $adx : L \to L$ is inner derivation. I'm looking for some examples non-inner derivations of Lie algebras.

The easiest example is perhaps to consider all derivations of the Heisenberg Lie Algebra $\mathfrak{h}_3(K)$, i.e., all linear maps $D\colon \mathfrak{h}_3(K) \rightarrow \mathfrak{h}_3(K)$ satisfying $D([x,y])=[D(x),y]+[x,D(y)]$ for all $x,y$. Here the brackets
are given by $[e_1,e_2]=-[e_2,e_1]=e_3$, where $(e_1,e_2,e_3)$ denotes a basis.
The inner derivations are of the form $ad (x)$, and are linear combinations of
$$
ad (e_1)=\begin{pmatrix} 0 & 0 & 0\cr
0 & 0 & 0\cr
0 & 1 & 0\cr
\end{pmatrix},\;
ad (e_2)=\begin{pmatrix} 0 & 0 & 0\cr
0 & 0 & 0\cr
-1 & 0 & 0\cr
\end{pmatrix},\, ad(e_3)=0.
$$
However, the Heisenberg Lie Algebra has many other derivations (outer derivations). In fact, all linear maps of the form
$$
D=\begin{pmatrix} d_1 & d_4 & 0\cr
d_2 & d_5 & 0\cr
d_3 & d_6 & d_1+d_5\cr
\end{pmatrix}
$$
are derivations of the Heisenberg Lie algebra.

I just the same calculations and i get

\begin{pmatrix} d_1 & d_4 & 0\cr
    d_2 & d_2 & 0\cr
    d_3 & d_5 & d_1 + d_2\cr
     \end{pmatrix}

for the derivations of  $\mathfrak{h}_3$.

I'm not sure who is right :P