Space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable Let's consider the space $\mathcal{L}^p(X, \Sigma, \mu)$ of all functions $f\colon X \to \mathbb{R}$ (or $\mathbb{C}$) for which:
$$
\int\limits_X|f|^p \mu(dx) < \infty.
$$
Here $X$ is a metric space, $\Sigma\subset 2^X$ — sigma-algebra on $X$ and $\mu$ — is the measure on $\Sigma$. As usual, functions that are equal almost everywhere considered to be equivalent, so technically we are dealing with classes $$[f]=\{g\in \mathcal{L}^\infty(X,\Sigma, \mu) \colon \mu \{x\in X\colon f(x)\neq g(x)\}=0\}.$$ This space is endowed with integral metric $$\rho_p(f,g)=\left(\int\limits_X|f-g|^p \mu(dx)\right)^{1/p}$$

The statement is that space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable, where $\forall A,B\in\Sigma$ $$\rho_\Delta(A,B)=\mu(A\Delta B)$$
Here, again, we consider $A\in \Sigma$ as a class $[A]=\{A'\in\Sigma\colon \mu(A\Delta A')=0\}$ to have the identity of indiscernibles.

How do one prove the statement?

As it has been pointed out by Ramiro, the assertion to be proved in the question is false. Only one of the implications is correct that is if $(\Sigma, \rho_{\Delta})$ is separable, then $L^p(X,\Sigma, \mu)$ is separable. This is what will be shown below.

Let $\chi_A$ be the characteristic function of the set $A$. Now observe that

$\| \chi_A - \chi_B\|_{L^p}^p = \int|\chi_A - \chi_B|^p = \int_{A\setminus B}1 + \int_{B\setminus A}1 = \mu(A\setminus B) + \mu(B\setminus A) = \rho_{\triangle}(A,B)$

Now consider the function $\varphi: \Sigma \to L^p(X, \Sigma,\mu )$ defined as $ A \mapsto \chi_A$.

This is an injective map from $(\Sigma, \rho_{\triangle})$ to $(L^p(X, \Sigma,\mu), \rho_p)$.

Observe that $\rho_{\triangle}(A_i,B) \rightarrow 0 \Leftrightarrow \|\chi_{A_i} - \chi_{B} \|_{L^p} \rightarrow 0$.

Consider a countable dense subset in $L^p$ say $f_i$. Now choose an element from each of the following sets $B(f_i,1/k)\cap \mathrm{Im}(\varphi)$ say $\chi_{A_{i,k}}$ where $i$ and $k$ vary over natural numbers. This will be a countable subset of $\mathrm{Im}(\varphi)$. For any given $\varepsilon$ and $\chi_A$ there exists $f_j$ such that $\| f_j - \chi_A \|_{L^p} < 1/k < \varepsilon/2$. Now let's choose $\chi_{A_{j,2k}}$, then $\| \chi_{A_{j,2k}} - \chi_A \|_{L^p} \leq \varepsilon $. Mistake in this argument was also pointed out by Glinka below.

Let $\lbrace A_i \rbrace$ to be a dense subset in $\Sigma$. Now consider the vector space spanned by $\mathbb{Q}\chi_{A_i}$. This is a countable set. This is dense in the vector space spanned by $\mathbb{R}\chi_{A_i}$ which is dense in $L^p$ since simple functions are dense in $L^p$. One can suitably modify the proof for complex valued functions.

The correct statement is that the space of summable functions (modulo a.e. equivalence) is separable if and only if there is a countable collection C of measurable sets of finite measure such that for each measurable set Q of FINITE measure and each integer N there is a set A in the collection C such that the measure of the symmetric difference between Q and A is less that 1/N. I have been unable to find a published proof of this statement, but the assertion is made in a book by Zaanen on measure theory published around 1967. However, the proof of the tricky direction made there is wrong. I know how to prove this result. If readers are interested I will provide a proof here.


How many different chess-board situations can occur? If you play a standard chess game on a normal $8 \cdot 8$ chess board with the usual rules: How many different "board representations" can exist?

Upper bound: Well, you have 16+16 = 32 chess pieces and 64 fields, so $\frac{64!}{32!} \approx 4.8 \cdot 10^{53}$ is an upper bound. This would mean that all chess pieces are distinguishable (e.g. white knights are not distinguishable) and you can distribut them freely accross the field (white pawns can never be in the first row).

Lower bound: If you look only at the pawns, you can imagine them as a number in base 3 (they can not move, move one in front or move two in front). You have 16 pawns, so $3^{17} \approx 1.3 \cdot 10^8$ is a lower bound.

I would also be happy with the number of different game situations (which is higher, as you might care of the possibility of castling, promotion or capture).

I guess the explanation will not be easy, so please provide a source.

$\left(56^{16}\times2\right)^2 =$ number of possible outcomes, right?

Each row of pawns can be in every place except the first row, so I thought of it as $4$ separate superimposed boards where the first two rows of two of the boards were removed [because pawns can't travel backward] then I multiplied the number of board-spaces to the 16th power because that's the number of states that a given board-square can be in at any given time, then I squared the answer to account for the fact that pieces can go missing overtime [because that's how you play].

These are my first thoughts however, it may also be the sum of $\left(56^8\times2\right)^2$ and/plus $\left(64^8\times2\right)^2$. I'm trying to approach it as an information theory problem (like an integrated circuit) but instead of bits-per-chip I'm trying (and probably failing miserably) to read it as board-square/board-space_states_per_board, it's particularly challenging because half of the pieces [plus horses and bishops (Bishops or, maybe castles? The "diagonal moving dudes")] are restricted to never being able to reach certain spaces, so if we could: subtract those from that piece's version of the board; think of the board as a binary data equation where each valid board-square was considered as being able to be in one of 3 states (yours the enemy's, or empty); add that data to equations that had those piece-states removed from them, and then subtract one from the total [because there will always be one piece left: therefore: meaning that the outcome where all spaces are empty is invalid]

-Hope this helps. I for one: would also like to solve this mystery.


A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$ Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.

My way:

Find the general term for both binomials separately
To find the independent term, consider the possible expansions by inspection then plug in the appropriate values into the general terms
Use calculator to sum up values (very lengthy process if there are lots of cases)


Power Series Example In my notes for complex analysis, there is an example asking the Radius of convergence for the series $\sum_{n=0}^\infty (az)^{2n}$

In the answers it says that the coefficient for $z_n$, $a_n$ is equal to $a^n$ if n is even and 0 if n is odd

Can someone please explain why this is?

Thank you very much

As $R^{-1}=\limsup\limits_n \,c_n^{1/n}$ by Hadamard's formula, we obtain at once that
$$R=\frac1{|\mkern1mu a\mkern1mu|}.$$


Finding inverse of polynomial in a field I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:

  In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.

My understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:

Proceeding with Euclid's algorithm:

$$ x^3 + 2x + 1  =(x^2 + 1)(x) + (x + 1) \\\\
  x^2 + 1 = (x + 1)(2 + x) + 2$$

We stop here because 2 is invertible in $\mathbb{Z}_3[x]$. We rewrite it using a congruence:

$$(x+1)(2+x) \equiv 2 \mod{(x^2+1)}$$

I don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?

Wikipedia has a page on this we a decent explanation, but it's still not clear in my mind.

Note that this question has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.

Thanks.

Write $f := x^3+2x+1$ and $g := x^2+1$. We want to find the inverse of $g$ in the field $\mathbb F_3[x]/(f)$ (I prefer to write $\mathbb F_3$ instead of $\mathbb Z_3$ to avoid confusion with the $3$-adic integers), i.e. we are looking for a polynomial $h$ such that $gh \equiv 1 \pmod f$, or equivalently $gh+kf=1$ for some $k\in \mathbb F_3[x]$. The Euclidean algorithm can be used to find $h$ and $k$:
\begin{align}
f &= x\cdot g+(x+1)\\
g &= (x+2)\cdot(x+1) + 2\\
(x+1) &= (2x)\cdot2 + 1
\end{align}
Working backwards, we find
\begin{align}
1 &= (x+1)-(2x)\cdot 2\\
&= (x+1)-(2x)(g-(x+2)(x+1))\\
&= (2x^2+x+1)(x+1)-(2x)g\\
&= (2x^2+x+1)(f-xg)-(2x)g\\
&= (2x^2+x+1)f- (x^3+2x^2)g\\
&= (2x^2+x+1)f - (2x^3+x^2)g\\
&= (2x^2+x+1)f + (x^3+2x^2)g.
\end{align}
So, the inverse of $g$ modulo $f$ is $h = x^3+2x^2 \pmod f = 2x^2+x+2 \pmod f$.


Function defined as the supremum of a sequence of functions - epsilon-delta proofs I'm really confused with working with sequences of functions and their sup.

Let A be a non-empty subset of $\mathbb{R}$. For all $n \in \mathbb{N}$ and let $f_n : A \to\mathbb{R}$ be a continuous function.

Suppose that for all $x \in A$ the sequence $f_1(x), f_2(x)...$ is bounded.

Define the function $g : A \to\mathbb{R}$ by: $g(x)=\sup\{f_1(x), f_2(x)...\}$

Then if we have some $a \in A$ and $\epsilon\gt0$, how would one go about proving that there is a $\delta\gt0$ such that, for all $x\in A$ with $|x-a|\lt\delta$ one has that $g(a)-g(x)\lt \epsilon$ ?

Additionally, how can we show whether the function $g$ is continuous?

I'm trying to do a standard epsilon-delta proof for the first part, but can't seem to introduce an $|x-a|$ term into the working.

Hint:
$$g(x) - g(a) \leqslant \sup g(x) - \inf g(a).$$
Use the fact:  $\sup A-\inf B=\sup(A-B).$


Where is $|z|^4 + z^2$ differentiable? Where is $|z|^4 + z^2$ differentiable?

well, I guess i suceeded at proving that its differentiable at $z_0=0$.
im stuck with the case where $z_o\neq0$. how do I do this by definition? (without using cauchy reimann's equations).
here is what I statrted and couldn't finish:

Edit: lets say im ignoring $z^2$ because its a ploynomial. here is what I get:

and yet, I still need to show that this limit doesn't exist.I can't see how.

As suggested in the comments, it suffices we consider $|z|^4$. Write $f(z) = |z|^4 = \bar z^2z^2$. Differentiating with respect to $\bar z$ gives $\partial_{\bar z}f(z) = 2\bar z z^2$ and this vanishes only when $z=0$. So $f$ is differentiable only at the origin.


help interpreting an abstract algebra test question This is a take-home test problem, and I don't want help solving it, just understanding what it's asking.  I've asked my prof a couple times, but she's either unwilling or unable to give me a straight answer, and I don't understand why.  The problem:

  Suppose that $f(x)=x^n+a_{n-1} x^{n-1}+\cdots+a_0\in \mathbb{Z}[x]$. If $r$ is rational and $x-r$ divides $f(x)$, prove that $r$ is an integer.

My question:  It's my understanding that when you say "$t$ divides $s$" you need to specify what domain you're in.  So they say that "$x-r$ divides $f(x)$", I'm confused about the domain, since $f(x)$ being an element of $\mathbb{Z}[x]$ is implying polynomials with integer coefficients, while $r$ being rational doesn't seem to fit with that.  Are they saying that $f(x) = (x-r)$ (another element of $\mathbb{Z}[x]$), or is it $f(x) = (x-r)$ (an element of $\mathbb{Q}[x]$)?

You are told that $f(x)$ is in $\mathbb{Z}[x]$, and that $f(x)$ is $x-r$ times an element of $\mathbb{Q}[x]$.  You are to deduce that $r$ must be an integer. The fact that $f(x)$ is monic (the lead coefficient is $1$) is crucial.

In a general Euclidean domain, given x,y in $R$, there are $q, r$ in $R$ with: $$a=bq+r $$ .

IF $r=0$ , then we say $b$ divides $a$. From Wiki's Euclidean Domain. $\mathbb Z[x]$ is a Euclidean domain. Notice that $\mathbb Z$ is embedded in $\mathbb Z[x]$ as the constant polynomials, so that integers are actually part of $\mathbb Z[x]$. IF $r$ is Rational; $r=p/q$ , you can always multiply through by $q$ to get an integer.


Polynomials $f$ and $f'$ with all roots distinct integers Edit 2. Since the question below appears to be open for degree seven and above, I have re-tagged appropriately, and also suggested this on MathOverflow (link) as a potential polymath project.

Edit 1. A re-phrasing thanks to a comment below:

  Is it true that, for all $n \in \mathbb{N}$, there exists a degree $n$ polynomial $f \in \mathbb{Z}[x]$ such that both $f$ and $f'$ have all of their roots being distinct integers? (If not, what is the minimal $n$ to serve as a counterexample?)

The worked example below for $n = 3$ uses $f$ with roots $\{-9, 0, 24\}$ and $f'$ with roots $\{-18, -4\}$.

(See also the note at the end, and the linked arXiv paper.)

  Question. For all $n \in \mathbb{N}$: Is it possible to find a polynomial in $\mathbb{Z}[x]$ with $n$ distinct $x$-intercepts, and all of its turning points, at lattice points?

This is clearly true when $n = 1$ and $n = 2$. A bit of investigation around $n = 3$ leads to, e.g., the polynomial defined by:

$$f(x) = x^3 + 33x^2 + 216x = x(x+9)(x+24)$$

which has $x$-intercepts at $(0,0)$, $(-9, 0)$, and $(-24, 0)$. Taking the derivative, we find that:

$$f'(x) = 3x^2 + 66x + 216 = 3(x+4)(x+18)$$

so that the turning points of $f$ occur at $(-4, -400)$ and $(-18, 972)$.

I am not even sure if this is true in the quartic${^1}$ case; nevertheless, this question concerns the more general setting. In particular, is the statement true for all $n \in \mathbb{N}$ and if not, then what is the minimal $n$ for which this is not possible?

$1$. Will Jagy kindly resolves $n=4$ since the monic quartic $f$ with integer roots $\{-7, -1, 1, 7\}$ leads to an $f'$ with roots $\{-5, 0, 5\}$. This example is also found as B5 in the paper here (PDF 22/24). The same paper has the cubic example above as B1, and includes a quintic example as B7:

$$f(x) = x(x-180)(x-285)(x-460)(x-780)$$

$$\text{ and }$$

$$f'(x) = 5(x-60)(x-230)(x-390)(x-684)$$

The linked arXiv (unpublished) manuscript seems to suggest that this problem is open.

The arXiv paper posted here (pdf) contains a list of references that have broached this problem in the past, and examples for $n=3$, $4$, and $5$ (which I have since incorporated into the OP).

However, even the case of $n = 6$ is listed as open$^\star$ (cf. Open Problem 6 on PDF 23/24) as of 2004. So, it appears the question asked here is open.

$\star$ (Edit): User i9Fn helpfully points to a 2015 paper containing the sextic polynomial

$$f(x) = (x − 3130)(x + 3130)(x − 3590)(x + 3590)(x − 10322)(x + 10322)$$

which leads to

$$f'(x) = 6x(x − 3366)(x + 3366)(x − 8650)(x + 8650)$$

thereby resolving the above open problem, and leading to an updated open question (cf. p. 363):

  Are there polynomials with the above-stated features and degree greater than six?

According to this latter paper's author, no such examples are known.

Try working backwards: find an integer polynomial $F$ of degree $n-1$ with all integer roots, such that its antiderivative has $n$ distinct roots. One way to check for this by looking for $n$ sign changes. Here's an example for $n-4$, I'll edit when I find a general solution.


Filling in a missing portion of a truth table I have the following truth table:
$$
\boxed{
\begin{array}{c|c|c|c}
a & b & c & x \\ \hline
F & F & F         & F \\
F & F & T         & F \\
F & T & F         & F \\
F & T & T         & F \\
T & F & F         & T \\
T & F & T         & T \\
T & T & F         & F \\
T & T & T         & T
\end{array}}
$$

I am supposed to use any logical operator: and, or, not. I tried like $10$ formulas already, and I am not able to think of what combination of $a,b,c$ will get that $x$ column right. Have I missed something really silly?

$x$ is only True in rows 5, 6, and 8, all of which have $a$ as True.  This means that we should look for:

$x$ = $a$ ∧ something

Whatever that "something" is, it should return True for all combinations of $b$ and $c$ except for $b$ True and $c$ False (which was row 7, where $x$ is False).

Hint:

   $$x\equiv a\land(\neg b\lor c).$$

How did I come up with this? Well, if $a$ is false, then any conjunction with $a\land(\text{something here})$ will be false as well. That takes care of the top half of the truth table. All that is left, then, is to figure out how to manipulate a combination of $b$ and $c$ to combine with $a$ to give the desired truth values. Thus, what you want is what is given above (grayed out in case you want to try to deduce it yourself).


Symbol for set all partitions of a set Is there a symbol for set all partitions of a set (like P for Power set)?

P.S. I need such symbol in writing some mathematical related notes.

See this answer to a similar question on MO. I suggest stealing notation from combinatorics and letting $\mathrm{B}_S$ denote the set of partitions of $S$, since the Bell number $\mathrm{B}_{|S|}$ is the number of partitions of $S$.

This is the commonly used symbol for the set of all subset of the set $E\:$ : $\mathscr{P}(E)$.


Solving systems of linear equations when matrix of coefficients is singular When attempting to solve a problem of type $AX=B$ where $A$ is the matrix of coefficients, $X$ contains the variables and $B$ is the right hand side, it turned out the $A$ was singular. As a result I could not premultiply both side by $A^{-1}$. In such a case, how would I solve for the variables without having to use augmented matrices?

Here is the question:

$$\begin{pmatrix} 1&2&1&1&1 \\ 2&1&2&1&1 \\ 1&2&3&1&1 \\ 2&2&1&1&3 \\3&3&5&2&2 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \\ x \\ y \end{pmatrix}=\begin{pmatrix} 6.3 \\ 6.7 \\ 7.7 \\ 9.8 \\10.9 \end{pmatrix}$$

Since $A$ is singular I cannot do $X=A^{-1}B$ which is the method I would typically use to solve for the unknowns. In later parts of the book I am studying they elude to a method with which one can use to solve for $X$, but do not further elaborate. Does anyone have an idea on how to solve this problem (or any problem in general where A is singular) without the use augmented matrices or is that the only way?

Even when the system of equations is singular, you can find a least-squares solution by solving the system $A^{T}Ax=A^{T}b$.

If the matrix is singular, the equation $Ax=b$ has no solution. This means that $b$ does not lie in the range of $A$. However it is possible to find a vector $x_0$ such that the image $Ax_0$ is ``closest'' to $b$ in a sense that can made precise. This requires a metric or norm on the target space.

One example of this is the Moore-Penrose pseudoinverse of $A$.


Hypothesis of Stone-Weierstrass Theorem Why do we need compactness in the space as hypothesis in Stone-Weierstrass Theorem?

  Theorem:
  Let $A \subset C(K)$ such that

  $A$ is a subalgebra that vanishes nowhere.
  For each $x, y \in K $ with $x \neq  y $, there exists $f \in A$ such that $f(x)\neq f(y)$.

  Then $ \overline A = C(K)$, where $C(K)$ is the space of continuous functions over the compact space $K$.

The Stone-Weierstrass theorem can be generalized in various ways, as discussed below (mostly based on General Topology by Willard, section 44). But presenting things in the greatest possible generality usually goes counter to the purpose of writing a textbook. Textbook authors are more concerned with presenting an insightful, digestible proof, and with proving the results that are actually used later.

Functions vanishing at infinity

The easiest generalization is to consider the algebra $C_0(X)$ of functions "vanishing at infinity" on a locally compact space $X$. This is the form that  Wikipedia presents. It's not really much of a generalization, since one can consider the one-point compactification of $X$, denoted $\widehat {X}=X\cup\{\infty\}$, and extend functions to $\infty$ by zero. By including the constant function in subalgebra $A$ one gets to the point where the compact case of Stone-Weierstrass can be used. Then one observes that to approximate a function that vanishes at $\infty$,  one does not need the constant functions after all.

Bounded functions

The space of all bounded real-valued continuous functions $C_b(X)$ still has the uniform norm, so one can hope to generalize the Stone-Weierstrass theorem verbatim. This does not actually work, though: let $X=[0,\infty)$ and denote by $A$ the algebra of all continuous functions $f\colon X\to\mathbb R$ such that $\lim_{x\to\infty} f(x)$ exists. This is a closed subalgebra which vanishes nowhere and separates points, but it does not coincide with $C_b(X)$; for example $\sin x\notin A$.

To state a version of Stone-Weierstrass theorem for this case, let's say that $Z\subset X$ is a zero set if there exists a continuous function $f:X\to\mathbb{R}$ such that $Z=f^{-1}(0)$. An algebra $A$ separates zero sets if for any two disjoint zero sets $Z_1$ and $Z_1$ there is $f\in A$ such that $\overline{f(Z_1)}\cap \overline{f(Z_2)}$ is empty.

Theorem. Suppose $X$ is a Tychonoff space (also known as $T_{3\frac12}$ space). If an algebra $A\subset C_b(X)$ separates zero sets, contains constant functions, and is closed in the uniform norm, then $A=C_b(X)$.

The proof still involves compactification of $X$, but this time one needs the Stone–Čech compactification.

Compact-open topology

If one considers $C(X)$, the set of all continuous functions on a non-compact space $X$, then that's no longer a normed space. While uniform convergence still makes sense, one can't hope to have uniform approximation here: for example, $f(x)=e^x$ cannot be uniformly approximated by polynomials on $\mathbb{R}$. A natural decision is to equip $C(X)$ with the compact-open topology; i.e., it's a locally convex space with topology generated by seminorms $\|f\|_K = \sup_K|f|$, for all compact subsets $K\subset X$.

Theorem. Suppose $X$ is a Tychonoff space. If $A\subset C(X)$ is an algebra that is closed in the compact-open topology, separates the points of $X$, and contains the constant functions, then $A=C(X)$.

For example, this implies that for every continuous function $f:\mathbb{R}\to\mathbb{R}$ there is a sequence of polynomials $p_n$ such that
$$
\forall M\ \lim_{n\to\infty}\sup_{|x|\le M}|f(x)-p_n(x)| =0
$$

Topology of uniform convergence

Willard presents two more forms of the Stone-Weierstrass theorem, using the topology of uniform convergence on $C(X)$ (which makes it a very disconnected topological space). They impose strong additional assumptions on the algebras of functions and are too involved to be stated here.

We need  the compactness of M because we are gonna cover it with a finite subcover of neightboors of a given point q arbitrary.

What a simple reasoning..


Does $\sum_{k\geq 1} e^{-\frac{k^{q}}{\sqrt{logk}}}<\infty$, where $0<q<1?$ Does $\sum_{k\geq 1} e^{-\frac{k^{q}}{\sqrt{2logk}}}<\infty$, where $0<q<1?$

Or that $\sum_{k\geq 1} (1-\frac{1}{\sqrt{2logk}k^{1-q}} )^{k} <\infty$?

But what I really want to bound is $\sum_{k\geq 1} (1-\frac{\sqrt{2logk}}{(2logk+1) k^{1-q}} )^{k} <\infty$, simply the above two are dominating for this one.

Attempt

1)Ratio and root test give limit equal to 1

2) Trying to get $\int_{1}^{\infty} e^{-\frac{x^{q}}{\sqrt{logx}}}dx$ into some known integral. A good candidate was the Gamma integral $\Gamma(\frac{1}{q})=\frac{1}{q}\int x^{1-\frac{1}{q}} e^{-x^{q}}dx =\int e^{-x^{q}}dx$. But $x^{q}>\frac{x^{q}}{\sqrt{logx}}$ and integration by parts gives

$$dy = (\frac{x^{q}}{\sqrt{logx}})'dx,$$
which derivative I can't find any write it in terms of y.

3)Say $q=\frac{1}{2}$. I tried to lower bound $\frac{x^{q}}{\sqrt{logx}}$, we have the error term from $e^{x}$'s Taylor expansion

$$\sqrt{\frac{x}{logx}}=(log(\sum_{k\neq 1}\frac{x^{k}}{k!}+1)+1)^{1/2}.$$

By lower bounding maybe we get a converging dominating series.

First of all the series must start from $2$ (you have a log at the denominator). Then note that for all sufficiently small $\epsilon>0$ holds, for a sufficiently large $k$, $k>N$ say, $$\frac{k^{q}}{\sqrt{\log\left(k\right)}}\geq k^{q-\epsilon}$$ since $k^{\epsilon}$ grows faster than $\sqrt{\log\left(k\right)}$. So $$\sum_{k>N}\exp\left(-\frac{k^{q}}{\sqrt{\log\left(k\right)}}\right)\leq\sum_{k>N}\exp\left(-k^{q-\epsilon}\right).$$ Now note that $$ e^{k^{a}}\geq k^{1+a}\Leftrightarrow k^{a}\geq\left(1+a\right)\log\left(k\right)$$ and this is true since every power of $k$ grow faster than $\log(k)$. So if we take $a=q-\epsilon$, for a sufficiently large $k$, we have $$\sum_{k>N}\exp\left(-k^{a}\right)\leq\sum_{k> N}k^{-\left(1+a\right)}\leq\sum_{k\geq1}k^{-\left(1+a\right)}=\zeta\left(1+a\right)$$ so the series is convergent.

small positive powers dominate large logarithmic powers, so k^q > (log(k)) ^ N for k large enough depending on q > 0, N arbitrarily big fixed, so your series is dominated by a series with e^-(log(k)^M) for arbitrarily big fixed M, and in particular for M=2, you get the series 1/(k^log(k)) which is obviously convergent as it is smaller than any zeta (N), N>2 say


Is $\mathrm{Cantor \, set} \times [0,1]$ self-similar? Consider the Cantor set $\times$ the interval $[0,1]$, i.e. Cantor sets put "one next to the other" as to "cover" the quare $[0,1]\times [0,1]$ is this set self-similar, i.e. the attractor of some homotheties?

Maybe $6$ maps ... map the square $[0,1]\times[0,1]$ onto each of these:
$$
\left[0,\frac{1}{3}\right] \times \left[0,\frac{1}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[0,\frac{1}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{2}{3},1\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{2}{3},1\right]
$$


Two Vectors and their Cross Product Let $v$ $=$ $\begin{pmatrix}0\\ c\\ 2c\end{pmatrix}$ and let $u$ $=$ $\begin{pmatrix}c^2-1\\ 2c\\ c-3\end{pmatrix}$. For what values of $c$ do the three vectors $v$, $u$ and the cross product of $v$ and $u$ form a basis of $R^3$.

I am mainly hoping to find a simpler way to do this.

I calculated the cross product to be:

$\begin{pmatrix}-3c^2-3c\\ 2c^3-2c\\ -c^3+c\end{pmatrix}$

Once I put all the vectors together, I tried to calculate the determinant of the matrix. But in the end, I got the following $6-degree$ characteristic polynomial:

$5c^6$ $-$ $c^4$ + $18c^3$ + $14c^2$

I know I am supposed to find the values of $c$ for which this polynomial does not equal $0$, but a polynomial of such large degree is very difficult to factorize. So i was just wondering if there is an easier way to do this problem. Is my current approach even correct?

Any help?

Excluding $c=0\implies\underline{u}=\underline{0}$:

As long as $\underline{u}$ and $\underline{v}$ are not parallel, then they together with $\underline{u}\times\underline{v}$ will form a basis for $\mathbb{R}^3$ since $\underline{u}\times\underline{v}$ is perpendicular to the plane containing $\underline{u}$ and $\underline{v}$.

Setting $\underline{u}=\lambda\underline{v}$ leads to the solution $c=-1$ only.

So $c$ can be any value apart from $0$ or $-1$.

Find which values of $c$ make $u$ and $v$ linearly independent (for instance you can row reduce a matrix with $u$ and $v$ as columns to see if it has full rank or not). Then $u\times v$ is necessarily orthogonal to both $u$ and $v$, and is therefore linearly independent with $u$ and $v$, so you get a basis for $\Bbb R^3$.


Odd and even number proofs Let $n$ be an integer. Prove that if $n^2+2n-1$ is even than $n$ is odd.

Proof: $$n^2+2n-1=2n$$ $$n^2-1=0$$ $$(n-1)(n+1)=0$$ $$n=-1,1$$

Which are odd.
Is this a complete proof? I feel like it only proves $n=-1,1$ not an odd number.

You are making a mistake as follows: Suppose that $n^2+2n-1$ is an even number. This means it is two times some integer. But what you seem to be doing, is equating it to $2n$, which makes it a specific integer, since $n$ is fixed (once we calculate $n^2+2n-1$).

On the other hand, every even integer is given by $2k$, where $k$ is an integer.

To do this question, suppose that $n^2+2n-1 = 2k$, then transposing, $n^2-1 = 2(k-n)$. Hence, $(n-1)(n+1) = 2(k-n)$.

At this point, it is enough to see that if $n$ was even, both $n-1$ and $n+1$ are odd.

The right hand side is even (divisible by $2$), whereas the left hand side, being a product of odd numbers, is odd. This is a contradiction.

Hence, $n$ must be odd.

Consider:

$EVEN = 2 \cdot ODD$ as you have stated or in more academic way:

$EVEN = 2 \cdot ODD + 0||2$

Can you see the difference?

By the way:

$ODD = 1 \cdot 1 + EVEN$


Can someone explain 4th dimensional objects? I'm not sure if I should ask this in mathematics or in physics.
From what I can tell, there are only 3 dimensions: X, Y, and Z. However, I have seen a lot of things about fourth and even fifth dimensional objects. I have tried for a year or two now to wrap my mind around the concept but I never have been able to do so. Can someone please enlighten me?

The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.

Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.

With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.

This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)

But I never understand I think we have to have a brain in 4D to understand.

You can also see that :

A conference really interesting Here

And a short video Here


Rolling dice possible outcomes You roll $10$ dice. How many possible outcomes include the numbers $1$
  to $6$?

It's essentially asking how many outcomes there are for 1 die to be 1, another to be 2, another to be 3, another to be 4, another to be 5, and another to be 6.

I know that we have $6 \cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 6!$ ways for 6 dice to contain 1-6, but I'm not sure about 10 dice. Thanks

Use Inclusion and Exclusion. All the ways to select at most 6 distinct faces, minus all the ways to select at most 5 of the 6, plus all the ways to select at most 4 four of the six, minus...

$$\dbinom 6 6~6^{10} - \dbinom 6 5~ 5^{10} + \dbinom 6 4~ 4^{10} - \dbinom 6 3~ 3^{10} + \dbinom 6 2~ 2^{10}-\dbinom 6 1 ~1^{10} + \dbinom 6 0 ~0^{10}$$

Based on the comments below, I try a different approach. I consider $1-$the number of cases where all numbers are not there.

Choose 5 out of 6 numbers. Each dice can be any number = $5^{10}$

So, the total number of ways is $6^{10} - 6*5^{10}$


$R = A \times B + C$, How retrieve A and C if C is negative? For A = 18,  B = 54 and $C = {-53, 53}$

$$R = A \times B + C$$

I can retrieve A and C from R with:
$$
A = \lfloor R \div B \rfloor
$$
$$
C={R} \pmod {B}
$$
example, A = 18, C = 53:
$$
R = 18 \times 54 + 53
$$
found A and C:
$$
A = \lfloor {1025} \div {54} \rfloor = {18}
$$
$$
C = {1025} \pmod {54} = {53}
$$
But if C is negative the result if false, A = 18, C = -53:
$$
R= 18 \times 54 - 53
$$
$$
A = \lfloor {919} \div {54} \rfloor = {17}
$$
$$
C = {919} \pmod {54} = {1}
$$
How can I retrive A =18 and C = -53 ?


Tangent plane to the graph of $f(x,y)=|xy|$ at $(0,0)$ Given the function $$f(x,y)=|xy|$$
how to know if the graph of f has a tangent plane at $(0,0)$?

I know that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ don't exist at $(0,0)$, is this enough to say that the tangent plane doesn't exist at that point?

My intuition says that maybe a tangent plane could exist since both axis are in the graph.

First, $$\frac {\partial f}{\partial x}\bigg|_{(x,y)}=\lim_{h\to\infty} \frac{f(x+h,y)-f(x,y)}{h}=\lim_{h\to\infty} \frac {| (x+h)(y)|- |xy|}{h}=\lim_{h\to\infty} |y| \frac{|x+h|-|x|}{h}$$
It follows that  $$\frac {\partial f}{\partial x}\bigg|_{(0,0)}=\lim_{h\to\infty} |0| \frac{|0+h|-|0|}{h}=\lim_{h\to\infty} |0| \frac{|h|}{h}=0$$

Also, $$\frac {\partial f}{\partial y}\bigg|_{(x,y)}=\lim_{h\to\infty} \frac{f(x,y+h)-f(x,y)}{h}=\lim_{h\to\infty} \frac {| (x)(y+h)|- |xy|}{h}=\lim_{h\to\infty} |x| \frac{|y+h|-|y|}{h}$$

so $$\frac {\partial f}{\partial y}\bigg|_{(0,0)}=\lim_{h\to\infty} |0| \frac{|0+h|-|0|}{h}=\lim_{h\to\infty} |0| \frac{|h|}{h}=0$$

This means that both $$\frac {\partial f}{\partial x}\bigg|_{(0,0)}$$ and $$\frac {\partial f}{\partial y}\bigg|_{(0,0)}$$ exist.

It remains to show that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\frac {\partial f}{\partial x}\bigg|_{(0,0)}(x-0)-\frac {\partial f}{\partial y}\bigg|_{(0,0)}(y-0)}{\|(x,y)\|}=0$$

But $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\frac {\partial f}{\partial x}\bigg|_{(0,0)}(x-0)-\frac {\partial f}{\partial y}\bigg|_{(0,0)}(y-0)}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{|xy|}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{|xy|}{\sqrt{x^2+y^2}}=0$$.

So with all this we can conclude that that $f$ has a tangent plane at $(0,0)$

By linearization we have that the tangent plane at $p = (a,b,f(a,b))$ is given by;

$$z = f(a,b) + (x-a)f_x(p) + (y-b) f_y(p)$$

i.e you have $\langle f_x(p), f_y(p), -1 \rangle \cdot \langle x-a,y-b,z-f(a,b)\rangle = 0$ i.e $\langle f_x(p),f_y(p), -1 \rangle$ is the normal vector for the plane at $p$.

$$\frac{\partial f}{\partial x}(0,0) = \lim_{t \to 0} \frac{f(t,0) - f(0,0)}{t} = 0$$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{t \to 0} \frac{f(0,t) - f(0,0)}{t} = 0$$


Difference between elementary submodel and elementary substructure This is a really "elementary" question, forgive the pun.

What is the difference between an elementary submodel and an elementary substructure (in first-order Logic)?

Sincere thanks for help.

Let M be a structure and T a theory that it models.
A substructure N of M need not model T.
if N models T, then we say that N is a submodel of M relative to T.
Otherwise, if N is a substructure of M that is not a submodel relative to T.
Thus the only distinction comes when a theory lurks in the background.

As an example, consider the signature S = (+, 0) where 0 is a unary relation symbol. then the naturals (N, +, 0) is an S-structure. Let A = {1, 2, 3, ...}. By restricting + to AxAxA and 0 to A, then A is naturally an S-structure. Since the inclusion map of A into N is an embedding (i.e. preserves atomics and negations of atomics over A), we get that A is a substructure of N. But relative to the full theory T of N, A is not a submodel of N since N models "exists x 0x" whereas M does not model "exists x 0x" as {0} intersect A = emptyset.

So there is indeed a difference between submodel and substructure, but it depends on the context.


Does there exist a circle, on the surface of a torus, that is not coplanar with the axis of revolution? I have got two closely related questions regarding Tori.

1.Does there exist a circle, on the surface of a torus, that is not co planar with the torus' axis of revolution?

2.Can the centre of a circle that is lying on the surface of a torus be the centre of a different circle also lying on the surface of the torus?

How does one proceed with proving (or disproving) them?

Both are true, so the easy way is to exhibit them:

Picture a standard torus with the axis of revolution along $z$ (i.e. rotated in the $x$-$y$ plane).  Horizontal cuts are not coplanar with the $z$ axis (indeed, are perpendicular to it.)  These are circles (and must be because they are parallel to the plane of rotation).  At all but the extreme top and bottom, there are two them, sharing the same center: on the $z$ axis.

For 1, in addition, there are two subtler circles through any point: http://mathworld.wolfram.com/VillarceauCircles.html, which will also not be coplanar with the $z$ axis.

Provided that your torus bows to 2r=R, then a pair of those Villarceau circles can be chosen to conform with your nr. 2.

--- rk


Prove $a/b+b/a$ for $a$ and $b$ natural is only natural for $a=b$ Is it possible to prove that for any natural $a,b$ the value of $a/b+b/a$ will not be natural with exception $a=b$?

$$\begin{align}
{a\over b}+{b\over a}=k&\implies a^2-kba+b^2=0\\
&\implies a=\left(k\pm\sqrt{k^2-4}\over2\right)b\\
&\implies\sqrt{k^2-4}\in\mathbb{N}\\
&\implies k=2\\
&\implies a=b
\end{align}$$

That's easy. $\frac{a}{b}+\frac{b}{a} = \frac{ a^2 + b^2 }{ab}$ So that $a/b+b/a$ is a natural number means $ a^2 +b^2 = k \cdot ab $ for some natural number k. We know the equation $x^2 -kx +1 =0$ has two integer roots if and only if $k=2$. To satisfy the condition, a should be equal to b.


How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:

$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$

$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$

where $b>0$, $T>0$.

Please kindly help?

Thanks to hits from Fabien, for the first one, let $t=\frac{T}{u^2+1}$:

$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t = \frac{2}{T}e^{-\frac{b^2}{2T}} \int_0^{+\infty}  e^{-\frac{b^2}{2T}u^2} \text{d}u  = \frac{\sqrt{2\pi}}{b\sqrt{T}} e^{-\frac{b^2}{2T}} $$

seems this tally with Mhenni Benghorbal's result.

For the second one, let $t=\frac{T}{u^2+1}$, $a^2=\frac{b^2}{2T}$:

$$I_2  =2 e^{-\frac{b^2}{2T}} \int_0^{+\infty} \frac{1}{1+u^2}  e^{-\frac{b^2}{2T}u^2} \text{d}u  =2 e^{-a^2}  \int_0^{+\infty} \frac{1}{1+u^2}  e^{-a^2u^2} \text{d}u := 2 e^{-a^2} I_3 $$

Then

$$I_3 = \int_0^\infty \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u = \int_0^\infty \text{d}u\, e^{-a^2u^2} \int_0^\infty e^{-x(1+u^2)} \text{d}x = \int_0^\infty \text{d}x\, e^{-x} \int_0^\infty e^{-(a^2+x)u^2} \text{d}u$$

So $$I_3 = \int_0^\infty \text{d}x e^{-x} \frac{1}{2} \sqrt{\frac{\pi}{a^2+x}} = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-x}}{\sqrt{a^2+x}} \text{d}x$$

Let $t=\sqrt{a^2+x}$, so $x=t^2-a^2$, $\text{d}x = 2t\text{d}t$,

$$I_3 = \sqrt{\pi} \int_a^\infty e^{-(t^2-a^2)} \text{d}t = \sqrt{\pi} e^{a^2} \frac{\sqrt{\pi}}{2}\, \text{erfc}(a)$$

So $$I_2 = 2e^{-a^2} I_3 =  \pi \, \text{erfc} \left(\frac{b}{\sqrt{2T}}\right)$$

Same as Mhenni Benghorbal's answer, $$I_2=  \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$

YEAH!

Hint :
Try the substitution $$t=\cfrac{T}{u^2+1}$$

The first integral has the shape of the gaussian.
The second one leads you to $$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1)  ) \,du$$

Considering $$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1)  ) \,du=\alpha\exp(-\beta)\cfrac{\sqrt{\pi}}{2\sqrt{\beta}} \\ I(0) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1}  \,du=\cfrac{\alpha\pi}{2}\end{cases}$$

with the substitution $\gamma^2 = \beta$ ends the problem.

Here are closed forms for the first and second  integrals respectively

  $$ I_1 = {
\frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}} $$

  $$I_2=  \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$

You can test them numerically.


Need help with $\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx$ I need help with this integral:
$$\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx\approx0.20597312051214...$$
Is it possible to evaluated it in a closed form?

A useful identity here is ${\rm arccosh\,} z = \ln( z + \sqrt{z^2 -1} )$.
Therefore $$\int_0^{\infty} e^{-x} \ln \ln( e^x + \sqrt{e^{2x}-1}) \;dx= \int_0^{\infty} e^{-x}\ln {\rm arccosh\,} e^x \;dx =\int_1^{\infty} \frac{\ln {\rm arccosh\,} y}{y^2} \;dy\, $$
Change integration variable $z={\rm arccosh\,}y$,
$$= \int_0^{\infty} \frac{\ln z \sinh z}{ \cosh^2 z} \;dz\,.$$
Integrate by parts and use the definition of the Euler's $\gamma$ constant and $\Gamma$ function see here,
$$= -\gamma + \ln\left[\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \right]\,. $$
We can further simplify this using $\Gamma(1-z)\Gamma(z)=\pi/\sin \pi z$  and $\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ for $z=3/4$ and $\sin (3\pi/4)=1/\sqrt{2}$. This gives $$=-\gamma - 3\ln 2 - 2 \ln \pi + 4 \ln \Gamma(1/4)\,. $$

This is a partial answer only.

Let $x=\log u$; then $dx=du/u$, and
$$
I=\int_{1}^{\infty} \frac{1}{u^2}\log\log(u+\sqrt{u^2-1})du.
$$
Next let $u=\cosh v$; then $du=\sinh v dv$, and $u+\sqrt{u^2-1}=\cosh v + \sinh v=e^v$, so
$$
I=\int_{0}^{\infty}\frac{\sinh v \log v}{\cosh^2 v} dv=\int_{0}^{a}\log{v}\cdot d\left(1-\frac{1}{\cosh v}\right) - \int_{a}^{\infty}\log v \cdot d\left(\frac{1}{\cosh v}\right),
$$
where we can choose $a\in(0,\infty)$.  Now integrate by parts to get
$$
I=\log{a}\left(1-\frac{1}{\cosh a}\right)-\int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv +\frac{\log a}{\cosh a} +\int_{a}^{\infty}\frac{dv}{v\cosh v}= \\\log a - \int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv+\int_{a}^{\infty}\frac{dv}{v \cosh v}.
$$
If we let $a\rightarrow 0$, this eliminates the middle term:
$$
I=\lim_{a\rightarrow 0+}\left[\log{a} + \int_{a}^{\infty}\frac{dv}{v\cosh v}\right].$$
Perhaps someone can take it from here?


What is the solution of $\lim_{n\rightarrow +\infty}(2+3^n)^{\frac{1}{2n}}$? Here's what I did: $$\lim_{n\rightarrow +\infty}(2+3^n)^{\frac{1}{2n}}=\lim_{n\rightarrow +\infty}e^{\frac{1}{2n}\ln(2+3^n)}$$
What should I do next in order to solve it?


Simple Logarithms Equation $$3^x = 3 - x$$

I have to prove that only one solution exists, and then find that one solution.

My approach has been the following:

$$\log 3^x = \log (3 - x)$$

$$x\log 3 = \log (3 - x)$$

$$\log 3 = \frac{\log (3 - x)}{x}$$

And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.

You may consider the function given by
$$
f(x)=3^x-(3-x),\quad x \in \mathbb{R}.
$$ We have
$$
f'(x)=3^x \cdot\ln3+1>0,\quad x \in \mathbb{R},
$$ Thus the function is strictly increasing on $\mathbb{R}$.

We have
$$
\begin{align}
f(0)&=1-(3-0)=-2<0\\\\
f(1)&=3-(3-1)=1>0
\end{align}
$$ then the unique solution $x_0$ is such that $x_0 \in (0,1)$.

You may observe that
$$
3^x=3-x
$$ is equivalent to
$$
(3-x)\ln 3 \times e^{(3-x)\ln 3}=3^3 \ln 3
$$ then, using a solution of $Xe^X=3^3 \ln 3$ in terms of the Lambert function, we get

  $$
x_0=3-\frac{W(27\ln 3)}{\ln 3}=\color{red}{0.741551813...}
$$

Edited version

After $x \cdot \log 3 = \log (3 – x)$ is $x = \frac {\log (3 – x)}{\log 3} $

which is equivalent to $y = x$ and $y = \frac {\log (3 – x)}{\log 3} $

Plotting the above TWO curves on the same graph, you will find they intersect at only one point (i.e. one solution).

Note also that:- When x > 3, the 'y' of the $y = x$ is much higher than the 'y' from $y = \frac {\log |3 – x|}{log 3}$. That means they will never meet when $x > 3$.


Function positively homogeneous of degree $1$ Let $f:U \to \mathbb{R}$ be a differentiable, positively homogeneous of degree $1$ in an open $U \subset \mathbb{R}^m$ containing $0$. Show that $f$ is a restriction to $U$ of a linear transformation from $\mathbb{R}^m$ to $\mathbb{R}$. Conclude that the function $f:\mathbb{R}^2 \to \mathbb{R}$ given by$$
f(x,y) = \begin{cases}
\dfrac{x^3}{x^2 + y^2}; & (x, y) \neq (0, 0)\\
0; & (x, y) = (0, 0)
\end{cases}
$$
  is not differentiable in $0$.

I do not know how to start this question. In previous questions, I proved:

Euler's relations for positively homogeneous functions of degree $k$.
For all $k \in \mathbb{R}$ there exists a function $f \in C^{\infty}(\mathbb{R}^{m}\setminus \{0\})$ positively homogeneous of degree $k$ such that $f(x)>0$ for all $x \in \mathbb{R}^m$ but is not a polynomial.

But, I do not know if it helps. I would like some hints. Thanks in advance.

$\def\vec{\boldsymbol}$For $\vec{x} = (x_1, \cdots, x_m)^{\mathrm{T}} \in U$, define$$
g(\vec{x}) = f(\vec{x}) - \sum_{k = 1}^m x_k \frac{\partial f}{\partial x_k}(\vec{0}),
$$
then $g$ is also positively homogeneous of degree $1$. Note that $f(\vec{0}) = 0$ and $f$ is differentiable at $\vec{x} = \vec{0}$, then$$
\lim_{\vec{y} \to \vec{0}} \frac{g(\vec{y})}{\| \vec{y} \|} = 0. \tag{1}
$$
Now suppose $B(\vec{0}, δ) \subseteq U$. For any $\vec{x} \in U\setminus \{0\}$, note that $t \vec{x} \in U$ and $g(\vec{x}) = \dfrac{g(t \vec{x})}{t}$ for any $0 < t < \dfrac{δ}{\| \vec{x} \|}$, then from (1) there is$$
g(\vec{x}) = \lim_{t \to 0^+} \frac{g(t \vec{x})}{t} = \| \vec{x} \| · \lim_{t \to 0^+} \frac{g(t \vec{x})}{\| t \vec{x} \|} = 0.
$$
Also, $g(\vec{0}) = f(\vec{0}) = 0$. Therefore $g = 0$, i.e.$$
f(\vec{x}) = \sum_{k = 1}^m x_k \frac{\partial f}{\partial x_k}(\vec{0}). \quad \forall \vec{x} \in U
$$

Proof assuming that $Df$ is continuous at the origin: we have $\frac {||f(x+h)-f(x)-Df(x)(h)||} {||h||} \to 0$ as $||h|| \to 0$. Replace $x$ by $tv$ and $h$ by $tu$ where $u$ and $v$ are fixed. Using homogeinity we get $\frac {||f(v+u)-f(v)-Df(tv)(u)||} {||u||} \to 0$ as $t \to 0$. Assuming continuity of $Df$ at $0$ this gives $f(v+u)-f(v)=Df(0) (u)$ for all $u$ and $v$. Put $v=0$ to get $f(u)=f(0)+Df(0)(u)$ for all $u$. Homogeinity implies that $f(0)=0$ so $f$ is the linear map $Df(0)$.


Properly drawing a Penrose tiling using the pentagrid method As part of my work, I create tools for artists to make various types of patterns for artistic purposes. I am trying to make a tool to create a Penrose tiling and I would like to use the pentagrid method of generating it, as it seems like the easiest way to allow the user to arbitrarily scale and translate the plane and still generate a nice random aperiodic tiling that easily stretches to infinity in any direction. Whereas using something like inflation or deflation becomes problematic when the user decides to scale the tile size down or translate by a large amount in any direction.

Given that, I've found several references on the how to generate a tiling from a pentagrid, such as these:

Pentagrids and Penrose
The Empire Problem in Penrose Tilings
Penrose Tiling
Penrose Tilings Tied Up in Ribbons

They've been incredibly helpful, but I feel like I'm missing a step. I'm able to generate the pentagrid:

But when I attempt to generate the tiles, I end up with tiles that are separated:

I'm working in Swift and I'm generating the vertexes of each tile by doing the following at each intersection point:

var vertexes: [CGPoint] = []
for i in 0..<testPts.count {
    var nextVertex = intersectionPt
    for gridIndex in 0..<numGrids {
        let normal = e [ gridIndex ]
        let k = ceil((testPts [ i ] • normal) - gamma [ gridIndex ])
        nextVertex += k * normal
    }
    vertexes.append(nextVertex)
}

The testPts array contains 4 points - one in each face of the pentagrid that has a vertex at the intersection point. The e array contains the normals for each direction of the pentagrid. The gamma array contains the offsets of each grid in the pentagrid. Values in gamma are between -1.0 and 1.0.

Is this the expected result? Some of the references I've read have made vague statements about needing to move the tiles together (without suggesting how to do that), while others have seemed to indicate that they'll all fall into the correct positions.

If they do need to be moved together, is there a particular algorithm to do that?

If they don't need to be moved together, then what have I missed or misunderstood?

Well I think I figured it out. The key is to keep track of which intersections of which grid lines are associated with each tile. Choose a starting tile and say that it is aligned. Then take each of the 4 tiles that were generated from the previous and next intersections of the 2 intersecting grid lines and move them towards the current tile until they touch. Then do the same for each tile you just moved, etc., until you've aligned all the tiles. In code it looks something like this (after adding the first tile to the queue):

    while (!queue.isEmpty) {
        // get the front element of the queue
        let startingTile = queue [ 0 ]
        queue.remove(at: 0)

        // find the (up to) 4 tiles that are adjacent to it
        let adjacentTiles = startingTile.unalignedAdjacentTiles()

        // for each tile we found
        for adjacentTile in adjacentTiles {
            // move it to touch this tile
            let offset = startingTile.minimumOffsetTo(adjacentTile)
            adjacentTile.realign(by: offset)

            // mark it as aligned
            adjacentTile.aligned = true

            // add it to the queue
            queue.append(adjacentTile)
        }
    }

Doing that, I get results like this:

Somehow I think it's strange that you have to move the tiles.
For instance, in this Desmos demonstration, I've generated the tiles for the associated intersections in the pentagrid without having to move them.
https://www.desmos.com/calculator/g4wcmzmn3s


In context of graph search,what is mini-max principle after doing some Google search i am not able to find an effective and understandable explanation of mini-max principle,anyone please explain it to me.

MINIMAX PRINCIPLE
Minimax is a method in decision theory for minimizing the xpected maximum loss. It is  applied in two players games such as tic-tac-toe, or chess where the two players take alternate moves. It has also been extended to more complex games whic require general decision making in the presence of increased uncertainty. All these games have a common property that they are logic games. This means that these games can be described by a set of rules and premises. So it is possible to know at a given point of time, what are the next available moves.


Decreasing sequence on a measure space I have the following problem:
Let $\mu$ be a measure on the measure space $(X,\mathbf{A})$. Let $\{A_{n}\}_{n=1}^{\infty}$ be a decreasing sequence in $\mathbf{A}$ so that $A_{n} \in \mathbf{A}$ for all $n\geq1$ and $A_{1} \supseteq A_{2} \supseteq A_{3}...,$
Let $A = \cap_{n=1}^{\infty}A_{n}$
(a) Assume that $\mu(A_{n})\leq \frac{1}{n}$ for all $n\geq 1$. Is it true that $\mu(A)=0$. (Provide a proof or counter example.)
I don't know how to approach such a problem and I have more of them so I would be really thankful for a general way to solve these types of problems.
Anyone with some knowledge of this?

More generally we have a theorem that states the following:
Let $\mathcal{A}$ be an algebra and let $\mu:\mathcal{A} \to [0,\infty]$ be a set function for which $\mu(\emptyset)=0$. If we have a decreasing sequence of sets $A_n\subseteq A_{n-1}$ in $\mathcal{A}$ such that $\mu(A_n)<\infty$ for some $n\in \mathbb{N}$, then:
$$\mu\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim_{n\to \infty}\mu(A_n)$$
We say that the measure(or set function more generally) is continuous from above in the set $\bigcap_{n=1}^{\infty} A_n$. If you are not familiar with algebras or more general measure theory then just know that every $\sigma$-algebra is an algebra and since it is given that $\mu$ is a measure it will satisfy the conditions of this theorem.

Hint
$\mu(A) \leq \mu(A_n)$ for every $n\in \mathbb{N}.$


Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.

I begin with rewriting the complex equation to $a+bi$ format.

1 $z^3 = i = 0 + i$

2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$

3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.

4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.

5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$

6 $k = 0$,  $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$

7 $k = 1$,  $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$

8 $k = 2$,  $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$

So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6},  \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?

Thanks!

Way easier way;

$$z^3=i \\
\iff z^3-i=0 \\
\stackrel{-i=i^3}{\iff}z^3+i^3=0 \\
\iff (z+i)(z^2-iz-1) = 0 \\
\iff z_1=-i,\;  z_2=\frac12 (i-\sqrt 3), \; z_3=\frac12 (i+\sqrt 3)$$

Disregard this answer if your exercises restrict you to trigonometric/polar form.

Use polar coordinates.

$z^{3}=i=e^{i(\frac{\pi}{2}+2k\pi)}$, $k\in \mathbb{Z}$

And from here it is much simpler

EDIT: what I mean by much simpler...

The OP kind of used polar form, but not really. You should stock with polar form until the very end. The OP introduces $\theta$, $cosinus$ and $sinus$ functions etc. You mix everything up and forget that $(\frac{\pi}{2}+2\pi)$, divided by $3$, gives $\frac{5\pi}{6}$...


What is the set of all probability distributions? I have a little problem which is in part due to my lack in english skills.
I am currently reading a book and I dont understand the following chapter (RW stands for random walk):
part1
part2
The problem is i dont understand what the author means with
$\lambda \in P(\mathbb{R})$ (the set of all probability distributions). What is a probability distribution? THe wikipedia article is not quite clear. Im not sure if $\lambda$ is a distribution function or a probability measure on R.
Now up to this point The F and its n-th fold have been distribution functions. But the proof that a standard model always exists would imply that the F's and the $\lambda$ are probability measures now.

thanks in advance

A probability distribution is a function F(x) defined on the real line with the following properties: $F(x)$ is monotone increasing, $\lim_{x \to -\infty}F(x)=0,\ \lim_{x \to \infty}F(x)=1$.


Find GCD $(10^6\times 6^2 \times 5 ; 6 \times 15 \times 3^7)$ Find GCD $(10^6\times 6^2 \times 5 ; 6 \times 15 \times 3^7)$

I have no idea.

Write
$$
10^6\cdot 6^2\cdot 5 = 2^8\cdot 3^2\cdot 5^7
$$
$$
6\cdot 15\cdot 3^7 = 2\cdot 3^9\cdot 5
$$
Does that help?

My bad. I was wrong in counting number of factors for 2nd number. Edited the correct answer

Your expression is same like
$$(2^9 \times 3^3 \times 5^7 , 2 \times 3^9 \times 5)$$
which can be divided by the largest number
$$(2 \times 3^3 \times 5)$$

So the GCD is $$(2 \times 3^3 \times 5)$$ which is equal to $90$.


$c\mid a,b\iff c\mid\gcd(a,b)$ [GCD Universal Property] If $c$ is a common divisor of $a$ and $b$ then $c$ divides the greatest common divisor of $a$ and $b$.  What can we use to prove this?

The general definition of a gcd $G$ is that it is a common divisor that is divisibly greatest, i.e. if $d$ is any common divisor then $\,d\mid G,\,$ so $\, d\le G,\,$ thus $G$ is  a greatest common divisor.  Combining both directions we obtain the following handy bidirectional form of the general definition of a gcd
$$g\,\text{ is a gcd of }\,a,b\,\text{ in }R\ \ \text{ if }\ \  \bbox[5px,border:1px solid #c00]{d\mid a,b\iff d\mid g}\ \text{ holds for all}\ \ d\in R\qquad\qquad\ \ \ \ \ \ \ \ $$
Indeed putting $\,d=g\,$ in $(\Leftarrow)$ yields $\,g\mid a,b,\,$ so $\,g\,$ is a common divisor of $\,a,b,\,$  and necessarily  divisibly greatest since direction $(\Rightarrow)$ shows every common divisor $\,d\,$ divides $\,g.$
Below is a proof of the "divisibly greatest" form of the gcd in $\Bbb Z,\,$ via Bezout.
Theorem $\ \ \ \   d\mid a,b\iff d\mid (a,b)\ \ \ $ [GCD Universal Property]
${\bf Proof}\ \ (\Rightarrow)\ \ \ d\mid a,b\,\Rightarrow\, d\mid (a,b) = i\,a+j\,b,\, $ some $\, i,j\in\Bbb Z,\,$ by Bezout.
$(\Leftarrow)\ \ \ \ d\mid (a,b)\mid a,b\,\Rightarrow\, d\mid a,b\ $ by transitivity of $ $ "divides".
Note that the proof shows that a linear common divisor is always greatest, i.e. any common divisor $ \:d\:$ of $ \:a,b\:$ that is an integral linear combination of them $ \:d = i\,a + j\,b\:$ is necessarily the greatest common divisor, since, as above, every common divisor $ \:c\:$ divides $ \:d,\:$ hence $ \:c\le d.\,$ If we inline the linked Bezout proof here we obtain a direct proof by Euclidean descent (induction).
Remark  $\ $ Dually we have the universal property of LCM
Lemma $\ \ \ a,b\mid m\iff [a,b]\mid m\ \ \ $ [LCM Universal Property]
In more general UFD's such as $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ there generally is not a linear representation (Bezout equation) for gcds but we can instead use prime factorizations to prove the above properties (then these properties boil down to the universal propery of min & max on the exponents of primes, e.g. see here).

Or we can prove the GCD Universal Property directly by induction on $\,\color{#90f}{{\rm size} := a\!+\!b}.\,$ It's clearly true if $\,a\!=\!0\,$ or $\,b\!=\!0,\,$ or if $\,a\! =\! b\!:\ c\mid a,a\!\iff\! c\mid (a,a)=a.\,$ Else $\,a\!\neq\! b\!\neq\!0.\,$ By symmetry, wlog $\,a>b.\,$  so $\, c\mid a,b\!\iff\! \color{#0a0}{c\mid a\!-\!b,b}\!\iff\! c\mid(a\!-\!b,b)=(a,b)\,$ since  $\,\color{#0a0}{\rm green}\,$ instance has smaller $\,\color{#90f}{{\rm size}} = (a\!-\!b)+b = a < \color{#90f}{a+b},\,$ so $\rm\color{}{induction}\,$ applies.

This can be shown based on what we are given by showing $a$ and $b$ in terms of a product of $c$. If $c$ is a common divisor of $a$ and $b$, that means a and b are each the product of at least one $c$ and another natural number: $a = c^np$ and $b=c^nq$, with $a,b,c,n,p,q \in \mathbb N$. There are then two possibilities:

Either $c$, or a power of $c$ $c^n$, is the GCD of $a$ and $b$. Trivially, $c$ divides any $c^n$ where $n>0$.
Neither $c$  nor any $c^n$ is the GCD of $a$ and $b$; in that case, $p$ and $q$ have their own GCD $m > 1$, and the GCD of $a$ and $b$ is $c^nm$; again, by inspection, we see that $c$ divides any $c^nm$ to produce $c^{n-1}m$.


Integral Definition of Center of Mass: Rope in the Plane If the placement of a rope is defined by an even function $y (x) $ from $x=-a $ to $x=a $, what is the the center of mass of the rope?)

I imagined it would be the average value of the function over the interval, but that is not the case. Or if it is I am making an error in a larger problem I'm working on.

Can anyone explain more about the integral definition of center of mass? The Wikipedia page isn't very helpful.

If $\rho$ is the density mass and $S$ is the surface of the rope, the mass $M$ is
$$
M=\iint_S\rho\,\mathrm dA
$$
and the moments are
$$
M_x=\iint_S\rho\,y\,\mathrm dA\qquad M_y=\iint_S\rho\,x\,\mathrm dA
$$
Hence the center of mass coordinates are
$$
\begin{align}
\bar x&=\frac{M_y}{M}=\frac{\iint_S x\,\mathrm dA}{\iint_S\mathrm dA}\\
\bar y&=\frac{M_x}{M}=\frac{\iint_S y\,\mathrm dA}{\iint_S\mathrm dA}
\end{align}
$$
If the rope is between $f(x)$ and $g(x)$ and between $a$ and $b$
the center of mass coordinates are
$$
\begin{align}
\bar x&=\frac{M_y}{M}=\frac{\int_{a}^b \int_{g(x)}^{f(x)} x\,\mathrm dy\,\mathrm dx}{\int_{a}^b \int_{g(x)}^{f(x)} \mathrm dy\,\mathrm dx}=\frac{\int_{a}^b x\left[f(x)-g(x)\right]\mathrm dx}{\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx}=\frac{1}{S}\int_{a}^b x\left[f(x)-g(x)\right]\mathrm dx\\
\\
\bar y&=\frac{M_x}{M}=\frac{\int_{a}^b \int_{g(x)}^{f(x)} y\,\mathrm dy\,\mathrm dx}{\int_{a}^b \int_{g(x)}^{f(x)} \mathrm dy\,\mathrm dx}=\frac{\int_{a}^b \frac{1}{2}\left[f^2(x)-g^2(x)\right]\mathrm dx}{\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx}=\frac{1}{S}\int_{a}^b \frac{1}{2}\left[f^2(x)-g^2(x)\right]\mathrm dx
\end{align}
$$
where $S=\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx$ is the area of the rope.

If the rope is the surface between $y=f(x)$ even, $y=0$, $x=-a$ and $x=a$
we have
$$
\begin{align}
\bar x&=\frac{M_y}{M}=\frac{\int_{-a}^a xf(x)\mathrm dx}{\int_{-a}^a f(x)\mathrm dx}=\frac{\int_{-a}^a xf(x)\mathrm dx}{2\int_{0}^a f(x)\mathrm dx}=0\\
\\
\bar y&=\frac{M_x}{M}=\frac{\int_{-a}^a \frac{1}{2}f^2(x)\mathrm dx}{\int_{-a}^a f(x)\mathrm dx}=\frac{1}{2}\frac{\int_{0}^a f^2(x)\mathrm dx}{\int_{0}^a f(x)\mathrm dx}
\end{align}
$$

You'll need line integrals:
$$ y_c = \frac{\int y(x)\,d\ell}{\int d\ell} = \frac{\int y(x)\sqrt{(y'(x))^2+1}\,dx}{\int\sqrt{(y'(x))^2+1}\,dx}, $$
where $d\ell=\sqrt{dx^2+dy^2}$ is the little portion of length (and thus mass).


Conversion from binary fraction to octal Can someone verify my results?
    Binary to Octal

1) 10101.11 is 25.75

2) 0.01101 is 0.6875

3) 10110110.001 is 266.125

To convert quickly, group into chunks of three, starting at the decimal point.

Say I have the binary number:  $$1001010110100000101.10010101$$  Lets write this as

$$001 \ 001 \ 010 \ 110 \ 100 \ 000 \ 101 \ .100 \ 101 \ 010$$  Notice I added some zeros, which don't change the number, to make sure I have chunks of three.  Now just write each of these chunks of three as a unique number in $0,1,2,3,4,5,6,7$ by converting from binary as you normally would.

So for my number above, $001=1$, $010=2$, $110=6$, $100=4$, $000=0$, $101=5$, $100=4$, $101=5$, and $010=2.$  Hence in octal we have that $$1001010110100000101.10010101=126405.452$$

Now try using this method to convert the numbers you have above.  (I won't do it for you) The decimal parts are not correct at the moment.

Why does this work?  The reason this works is because $8$ is a power of $2$.  In particular $8=2^3$, which is why we split into blocks of $3$.  Each block of $3$ in binary corresponds to a unique number in octal, and vice-versa.  Say I have a binary number $$x=a_n a_{n-1} \cdots a_2 a_1 a_0.a_{-1}a_{-2}\cdots a_{-m}.$$  Each $a_i$ here represents a digit that is either $0$ or $1$.  Then $$x=a_n 2^n +a_{n-1}2^{n-1}+\cdots +2a_1 +a_0+a_{-1} 2^{-1}+\cdots +a_{-m}2^{-m}.$$  We now want to write this in base 8.  Lets group terms:

$$x=\cdots +8^1\left(4a_5+2a_4+a_3\right)+ 8^0\left(4a_2+2a_1 +a_0\right)+8^{-1}\left(4a_{-1}+2a_{-2}+a_{-3}\right)+8^{-2}\left(\cdots \right)\cdots .$$
Then we see that in octal, we just want to group triplets of the binary numbers.

In a very similar way, we can convert from Hexadecimal to binary, and binary to hexadecimal, by grouping into chunks of $4$.  (Hexadecimal is base $16=2^4$)

$$(0.011011101)_2$$

Start selecting three Bits of numbers from left side. For example, first bit will be $011$, second bit will be $011$ and the third bit will be $101$. We started selected three Bits from left side. And after selecting all the bits convert that all the bits into octal number. So the answer of the question given above will be:

$$(0.335)_8.$$


Dinner invitation problem combinatorics A man has eight friends $A$, $B$, $C$, $D$, $E$, $F$, $G$ and $H$. He would like to invite three of them for dinner. If $A$ and $B$ must be together, i.e. he cannot just invite one of them, but $C$ must not come with $D$, i.e. he cannot invite both of them, then how many different invitations can he make?

I was thinking that the original combination should be $ ^{8} C_{2}$ , and then I should progress somehow using the addition rule, but in the solutions I was given it says $ ^{6} C_{1}$ instead. Could someone explain why?

Add up the following:

Number of combinations with A and B: $\binom{8-2}{3-2}=6$
Number of combinations with C but without A or B or D: $\binom{8-4}{3-1}=6$
Number of combinations with D but without A or B or C: $\binom{8-4}{3-1}=6$
Number of combinations without A or B or C or D: $\binom{8-4}{3-0}=4$

The answer is therefore $6+6+6+4=22$.

I would think of this as a "subsets" problem: the friends he invites to dinner form a subset of the set of all friends.  In general, if a set contains n elements then it has $2^n$ subsets.  To include the constraints, "A and B must be together" and "C must not come with D", think of "AB" as a single element and "CD" as a single element.  Then "the set of all friends" contains 6 elements, "AB", "CD", "E", "F", "G", and "H", so has $2^6= 64$ subsets.  But where "C" and "D" are concerned we can invite either C or D but not both so that every subset the contains "CD" can be treated as either "C" or "D".  That multiplies the number of possible subsets by 2.  There are 128 different ways to invite friends.  (That includes the empty set.  If you do not want to include "invite no one" as a "dinner", there are 127 ways.)


Equality-constrained QCQP — what to do next? I am studying the following problem. I want to obtain all the local minima of
$$
\min_{x\in D} x^Tx \quad \quad \quad \quad \quad \quad (P.1)
$$
where $x\in\mathbb{R}^n$ and $D = \{x: x^TA_ix = 1, \forall i=1,\dots,m\}$ and $A_1,\dots,A_m$ are a collection of $m<n$ positive definite $n\times n$ matrices. According to the Lagrange multiplier theorem ("Nonlinear programing" by Dimitry Bertsekas):

Proposition 3.1.1 (Lagrange Multiplier Theorem - Necesary Conditions) Let $x^*$ be a local minimum of $f$ subject to $h(x)=(h_1(x),\dots,h_m(x))=0$ and assume that the gradients $\nabla h_1(x^*),\dots,\nabla h_m(x^*)$ are linearly independent. Then, there exists a unique vector $\lambda^*=(\lambda_1^*,\dots,\lambda_m^*)$ called Lagrange multiplier vector such that
$$
\nabla f(x^*) + \sum_{i=1}^m\lambda_i^*\nabla h_i(x^*)  =0 \quad \quad \quad \quad \quad \quad (3.3)
$$

Thus, the set of ALL local minima can be divided in two, the ones which have $\{\nabla h_i\}_{i=1}^m$ linearly independent (regular points), and such points may be found using (3.3), and other minima which have $\{\nabla h_i\}_{i=1}^m$ linearly dependent (nonregular points).
My question is: I already know how to find the regular points. My interest is regarding the nonregular ones. Can those nonregular points be studied (obtained) in my case? For reference, note that $h_i(x) = x^TA_ix-1$ and $\nabla{h_i}(x) = 2A_ix$.
From this point, let me explain what I have found up to now:
Suppose $\hat{x}$ is one of such points. I can partition the set of vectors $\mathcal{A}=\{A_i\hat{x}\}_{i=1}^m$ into two disjoint sets $\mathcal{B}=\{B_i\hat{x}\}_{i=1}^s\subset \mathcal{A}$ and $\mathcal{C}=\{C_i\hat{x}\}_{i=1}^r\subset \mathcal{A}$ with $r+s=m$ such that the vector in $\mathcal{B}$ are linearly independent and the ones in $\mathcal{C}$ are a linear combination of the ones in $\mathcal{B}$.
Then, there exists real constants $\{\alpha_{ij}\}_{j=1}^m$ such that
$$
C_i\hat{x} = \sum_{j=1}^s \alpha_{ij}B_j\hat{x}
$$
And for this point to lie in $D$ I require that $\hat{x}^TC_i\hat{x} = 1$ but also $\hat{x}^TB_j\hat{x}, j=1\dots,s$. Hence, it must be the case that $\sum_{j=1}^s \alpha_{ij}=1$ for $i=1,\dots,r$ and $j=1,\dots,s$. (is this useful?).
My original intention was to cast this as another (reduced) nonlinear program in which the gradients $\{\nabla h_i\}$ are linearly independent. My attempt was to try to show that $\hat{x}$ will be a local minimum of $(P.1)$ ignoring the "linearly dependent components". This is, if I solve (P.1)  for all combinations of $\{A_i\}_{i=1}^m$, for example building $D$ only with the $A_i$ for separate, then by pairs, etc... then $\hat{x}$ would show up as a solution of one of those, in which the corresponding $\{\nabla h_i\}$ are linearly dependent. Is this true? The more I think about it the more I convince myself it is not the case. But what do you think? As an example of what I mean here, is the following. Consider $m=2$, $n=3$ and $A_1 = \text{diag}(1,1,0.1)$, $A_2 = \text{diag}(1,1,0.5)$. The constraints $x^TA_1x=1$, $x^TA_2x=1$ look like the red and blue surfaces here:

Their intersection is the yellow circle. Note that points in such intersection will have linearly dependent gradients, basically since $A_1x=A_2x$ for $x$ in such interection. However, if I study (P.1) just by taking into account $A_1$, I could still find those points. Of course this is just an example that motivated my idea, but It seems not to generalize.
My other attempt was to try to find explicitly all points that have $\{A_i\hat{x}\}_{i=1}^m$ linearly dependent. This is, find coeficients $\alpha_1,\dots,\alpha_m$ such that the matrix $H(\alpha)=\sum_{i=1}^m\alpha_iA_i$ is singular, and then look for $\hat{x}$ in the kernel of $H(\alpha)$ and see if such set of points intersect the problem constraints, etc... However, I haven't been able to obtain anything new from here.
What do you suggest me to do next? Do you think is possible to obtain such points? Actually any suggestion, advice, comment would be helpful for me.

$$\begin{array}{ll} \underset{\mathrm x \in \Bbb R^n}{\text{minimize}} & \mathrm x^\top \mathrm x\\ \text{subject to} & \mathrm x^\top \mathrm A_1 \mathrm x = 1\\ & \qquad\vdots\\ & \mathrm x^\top \mathrm A_m \mathrm x = 1\end{array}$$
where $m < n$. We define the Lagrangian
$$\mathcal L ({\rm x}, \mu) := \mathrm x^\top \mathrm x + \mu_1 \left( \mathrm x^\top \mathrm A_1 \mathrm x - 1 \right) + \cdots + \mu_m \left( \mathrm x^\top \mathrm A_m \mathrm x - 1 \right)$$
Taking the partial derivatives and finding where they vanish,
$$\begin{aligned} \left( \mathrm I_n + \mu_1 \mathrm A_1 + \cdots + \mu_m \mathrm A_m \right) \mathrm x &= 0_n \\ \mathrm x^\top \mathrm A_1 \mathrm x &= 1 \\ &\vdots\\ \mathrm x^\top \mathrm A_m \mathrm x &= 1 \end{aligned}$$
I have no idea how to solve this system of $n+m$ equations in $n+m$ unknowns. Since we are not looking for the solution ${\rm x} = 0_n$, from the first $n$ equations, we have
$$\det \left( \mathrm I_n + \mu_1 \mathrm A_1 + \cdots + \mu_m \mathrm A_m \right) = 0$$
but I do not know what to do with this. Suggestions are most welcome.


Is this number irrational? Is the following (decimal) number irrational?

0.10100100010000100000100000010000000100... etc.

My intuition tells me it is irrational. My informal "proof" is simply that it doesn't contain a repeating set of digits.

For irrationality, is it both a necessary and sufficient condition that the digits never revert to a repeating sequence?
Is there a more formal proof for this case?

A formal justification of your informal proof can be achieved by noting that in the process of long division, the fact that you have a finite number of possible remainders guarantees that eventually a remainder will be repeated. That is, for any rational number its decimal expansion becomes periodic.

The number you gave when you asked the question is indeed an irrational number because it goes on without repeating or ending.  I've used to learn that.  It keeps adding zeroes on and on and on beside the ones.  This never goes into a repeating sequence because it keeps adding on the zeroes.  I don't know if there's a more formal proof for this kind of case, but remember that an irrational number can never be written as a fraction.


Rotate an area around a diagonal line. I know how to find the volume of the figure formed when you rotate a $2$-dimensional area around a horizontal or vertical line, but what if it were a diagonal line instead?

For example:

Rotate the area between the curve $y=x^2$ and $y=x$ around the line $y=x$.

That should create a diagonal "football" shape, starting at the origin.

Anyways, the way I thought to go about the problem was to rotate the equation $45°$ to the right, and then solve the problem as if it were rotated around the x-axis. So I decided to convert $y=x^2$ to a polar equation, add $\frac{pi}{4}$ to $\theta$, and then turn it back into a rectangular equation. That worked, and it gave me this equation, which gives a diagonal parabola opening toward the first quadrant:

$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2}xy(x^2+y^2)$

Only problem is that I need to isolate $y$, such that I can solve this problem like an ordinary rotation problem, and that seems impossible. What should I do to solve the problem? I don't know much linear algebra, but could that be used to solve this instead?

The line through $(t,t)$ which is perpendicular to $y=x$ is given by $y=-x+2t$,

so it intersects $y=x^2$ where $\displaystyle x=\frac{\sqrt{1+8t}-1}{2}$ and $\displaystyle y=-x+2t=\frac{4t+1-\sqrt{1+8t}}{2}$.

The distance from $(x,y)$ to the line $y=x$ is given by $\displaystyle R=\frac{|y-x|}{\sqrt{2}}=\frac{|2t+1-\sqrt{1+8t}|}{\sqrt{2}}$,

so $\displaystyle V=\int_0^1\pi R^2 ds=\int_0^1\pi\big(2t^2+6t+1-(2t+1)\sqrt{1+8t}\big)\sqrt{2}dt$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\left[\frac{2}{3}t^3+3t^2+t\right]_0^1-\int_0^1(2t+1)\sqrt{1+8t}dt\bigg)$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\frac{14}{3}-\int_1^3\big(\frac{1}{4}(u^2-1)+1\big)u\cdot\frac{1}{4}udu\bigg)=\pi\sqrt{2}\bigg(\frac{14}{3}-\frac{1}{16}\big[\frac{u^5}{5}+u^3\big]_1^3\bigg)$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\frac{14}{3}-\frac{1}{16}\cdot\frac{372}{5}\bigg)=\frac{\pi\sqrt{2}}{60}$

Convert $y=x^2$ to a polar equation

$$r\sin\theta = r^2\cos^2 \theta$$
$$r\sin\theta = r^2\cos^2 \theta \rightarrow \hbox{ (add }\frac{\pi}{4}\hbox{ to }\theta\hbox{)}$$
$$r\sin (\theta + \frac{\pi}{4}) = r^2 \cos^2(\theta + \frac{\pi}{4})$$
$$\frac{\sqrt{2}}{2}r\left(\sin \theta + \cos \theta\right) = \frac{1}{2} \left(r\cos\theta -r\sin \theta\right)^2 \rightarrow\hbox{ (convert back) }$$
$$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2} (x-y)^2$$
which is not $\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ anyway.
Thus, it must be a mistake somewhere :)
$\rho((x^2,x),y-x=0)$ can be done much simpler: $\rho=|x^2-x|\cdot \frac{\sqrt{2}}{2}$, so you just have to $\pi\int \rho^2(t) dt$ where $t=\frac{\sqrt{2}}{2}(x+y)=\frac{\sqrt{2}}{2}(x^2+x)$.
Edit:$\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ itself leads to an irreducible cubic.


Proof of the extreme value theorem without using subsequences I am preparing a lecture on the Weierstrass theorem (probably best known as the Extreme Value Theorem in english-speaking countries), and I would propose a proof that does not use the extraction of converging subsequences. I did not explain subsequences in my calculus course, and I must choose between skipping the proof of the theorem and finding some proof which works only for functions $\mathbb{R} \to \mathbb{R}$.

I remember I once read a proof based on some bisection technique, but I can't find a reference right now.
I would be grateful for any reference to books, papers, web sites about this alternative proof.

Edit: since somebody modified my question, I will write down the precise theorem I want to prove.

Theorem. Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. Then $f$ has at least a maximum and a minimum point.

Note: This answers the question as it was originally asked, which unfortunately wasn’t what was intended. I’ve added a sketch of an answer to the intended question below it.

I don’t have a reference at hand, but the proof is straightforward. Suppose that $f$ is continuous on $[a_0,b_0]$ with $f(a_0)<0<f(b_0)$. Recursively construct a sequence of intervals $[a_k,b_k]$ as follows.

If $f\left(\frac{a_k+b_k}2\right)=0$, stop: you’ve found the desired point.
If $f\left(\frac{a_k+b_k}2\right)<0$, let $a_{k+1}=\dfrac{a_k+b_k}2$ and $b_{k+1}=b_k$.
If $f\left(\frac{a_k+b_k}2\right)>0$, let $a_{k+1}=a_k$ and $b_{k+1}=\dfrac{a_k+b_k}2$.

Then $\langle a_k:k\in\Bbb N\rangle$ is non-decreasing, $\langle b_k:k\in\Bbb N\rangle$ is non-increasing, $a_k<b_m$ for all $k,m\in\Bbb N$, and $b_k-a_k=(b-a)2^{-k}$ for all $k\in\Bbb N$, so $\sup_ka_k=\inf_kb_k$, and by continuity the function must be $0$ at this point.

Obviously you’d want to fill in a lot of detail and show how it generalizes to the full intermediate value theorem, but that’s it in a nutshell.

To use bisection to prove the extreme value theorem, let $f$ be a continuous real-valued function on $[a,b]$. First use bisection to show that $f$ is bounded above: if not, at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)\ge k$, and otherwise choose the right half and a point $x_k$ in it such that $f(x_k)\ge k$. Then $a_k\le x_k\le b_k$ for all $k\in\Bbb N$, so $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f$ cannot be defined at $\sup_ka_k$.

Now let $y=\sup_{x\in[a,b]}f(x)$, and repeat the bisection process, but this time at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)>y-2^{-k}$; if not, the right half contains such a point $x_k$, and we choose the right half. Then $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f(\sup_ka_k)=y$.

Here's an outline of a "bisection" proof.

Suppose $\sup_{x \in [a,b]} f = \alpha$.  Define a sequence of intervals $I_k = [a_k,b_k]$ recursively as follows: $I_0 = [a,b]$ and for $k > 0$

$
I_k =
\begin{cases} [a_{k-1}, \frac{a_{k-1} + b_{k-1}}{2}] \quad \text{if }
\sup_{x \in [a_{k-1},\frac{a_{k-1} + b_{k-1}}{2}]} = \alpha \\
[\frac{a_{k-1} + b_{k-1}}{2},b_{k-1}] \quad \text{otherwise}.
\end{cases}
$

Then there exists $x \in \cap_{k} I_k$ (though this requires some care to show, and is a theorem by itself), and you can show $f(x) = \alpha$.


What is $\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+...}}}$ $$\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+...}}}=?$$
Where $i$ is the imaginary unit, $i=\sqrt{-1}$.
I just want to know whether this is possible or not.

I prefer the limit notation in writing these expressions. Nevertheless, if I have interpreted your notation correctly, then you want something like this:

$L=\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}}$

$\implies L^2-1=\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}$

$\implies (L^2-1)^2-i=\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}$

$\implies ((L^2-1)^2-i)^2+1=\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}$

$\implies (((L^2-1)^2-i)^2+1)^2+i=\sqrt{i^4+\sqrt{i^5+...}}=L$  (Recall that $i^4=1, i^5=i$ and so on)

At least in principle, you can get the answer by solving the equation.

However, remember that complex square roots are multivalued and there is no such thing as a 'positive square root' for complex numbers as used in your notation.

Hence, there still remains some ambiguity in the question and we get multiple answers by this method. https://www.wolframalpha.com/input/?i=solve+%28%28%28L%5E2-1%29%5E2-i%29%5E2%2B1%29%5E2%2Bi%3DL

Hope this helps.

Above equation shown below:

$L=\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}}$

I just realized my previous answer does not add up.

Since nested radical's have solution's which are real numbers,

hence we consider the value of "L" as real.

@tatan gave equation for "L"

$\ (((L^2-1)^2-i)^2+1)^2+i=L$

Solving for $(i)$ we get:

$(i)=[(L^2-1)^8-4(L^2-1)-L]/[(4(L^2-1)^6-1]$  ---(1)

Since right hand side of equation (1) is real &

the left hand side is imaginary we have a contradiction.

Hence the reply is negative to "OP" question, if a

solution is possible to his given equation.


There exist infinitely many integers $q$ such that $3^{q-1}\cdot q\cdot p$ is cubic number？ Let $p$ is give postive integers, show that:there exist infinitely many integers $q$ such that $3^{q-1}\cdot q\cdot p$ is cubic number?

Alternatively, I considered the form $q=p^2\cdot l^3$?But How to prove there exist infinitely $l$ such $p^2\cdot l^3-1\equiv0\pmod 3?$I can't figure it out

Hint.

Assume first that $3\not\mid p$.

Set $q=3k$. (If $3\mid p$ but $9\not \mid p$, set $q=3k+1$, while if $9\mid p$ but $27\not \mid p$ set $q=3k+2$, etc). Then
$$
N=3^{q-1}qp=(3^{k})^3kq
$$
It thus suffices to show that there exist infinitely many $k$, such that $pk$ is a cubic number.


Find function $f(x)$ satisfying $\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$ I am looking for a non-trivial function $f(x)\in L_2(0,\infty)$ independent of the parameter $n$ (a natural number) satisfying the following integral equation: $$\displaystyle\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$$
or prove that $f(x)=0$ is the only solution.

The similar question is here but there are no parameters in the integral and the answer is based upon hit and trial method.

I want to know what would be the nice approach to tackle this problem.

EDIT: Such a function may exist, here is an example due to Stieltjes. A function $f(x) = \exp(-x^{1/4}) \sin x^{1/4}$ satisfies $\int_0^{\infty} f(x) x^n dx = 0$ for all integers $n \ge 0$.

We use the substitution $x=u^4$ to write
$I_n = \int_0^{\infty} f(x) x^n dx = 4 \int_0^{\infty} e^{-u} \sin(u) u^{4n+3} du$;
then integrate by parts four times (differentiating the power of $u$, and integrating the rest) to show that $I_n$ is proportional to $I_{n-1}$,
and finally check that $I_0=0$.(the Edit copied from here)

I'll prove a partial result that hopefully someone can complete to showing that $f(x) \equiv 0$ is the unique continuous and $L^2(\mathbb{R}_{>0})$ solution. Beyond that, perhaps one can use that continuous functions are dense in $L^2$ to get the requested result.  The main tool I want to contribute is the following lemma:

$\mathbf{Lemma}:$ Consider $f \in C(\mathbb{R}_{>0}) \cap L^2(\mathbb{R}_{>0})$. Then for any choice of $0<x_0<y_0 \leq \infty$, we have

$$f(x_0) = \lim_{n \to \infty} ne^{nx_0}\int_{x_0}^{y_0}\frac{f(x)}{1+e^{nx}}dx $$

$\mathbf{Proof}:$ Fix $\epsilon>0$ and, by continuity, choose $\delta>0$ such that $|x-x_0| < \delta \implies|f(x)-f(x_0)| < \epsilon$. Note that

$$\lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{1}{1+e^{nx}}dx = 1$$

and

\begin{align}
\lim_{n \to \infty} \left | n e^{nx_0} \int_{x_0+\delta}^{y_0}\frac{f(x)}{1+e^{nx}} dx \right |
& \leq  \lim_{n \to \infty} n e^{nx_0} \int_{x_0+\delta}^{y_0} e^{-nx} |f(x)| dx \\
& \leq  \lim_{n \to \infty} n e^{nx_0} e^{-(n-1)(x_0+\delta)} \int_{x_0+\delta}^{y_0} e^{-x} |f(x)| dx \\
& \leq  \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}f(x)\|_1 \\
& \leq \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}\|_2 \cdot \|f\|_2  \\
& = 0
\end{align}

by Holder's inequality (neither of these depended on the choice of $\delta$-- they're just straightforward computation/comparison). Given these results, we see that

\begin{align}
\lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{y_0} \frac{f(x)}{1+e^{nx}}dx - f(x_0) \right |
& = \lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{f(x)-f(x_0)}{1+e^{nx}}dx \right | \\
& \leq \lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{|f(x)-f(x_0)|}{1+e^{nx}}dx \\
& \leq \epsilon
\end{align}

Since $\epsilon>0$ was arbitrary, this proves the lemma.
$$\tag*{$\blacksquare$}$$

This lemma gives us some insight into the behavior of functions satisfying your condition. In particular, if $f$ satisfies these hypotheses and the condition under consideration, we must have for any $x_0 > 0$ that

$$0 = \lim_{n \to \infty} ne^{nx_0} \int_{0}^\infty \frac{f(x)}{1+e^{nx}}dx = f(x_0) + \lim_{n \to \infty} ne^{nx_0} \int_{0}^{x_0} \frac{f(x)}{1+e^{nx}}dx$$
Or

$$f(x_0) = -\lim_{n \to \infty} ne^{nx_0} \int_0^{x_0} \frac{f(x)}{1+e^{nx}}dx $$

From this result, we have $\forall$ $0<x_0<y_0<\infty$ that

$$\lim_{n \to \infty} ne^{nx_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx = \lim_{n \to \infty} e^{-n(y_0-x_0)} \left [ ne^{ny_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx \right ] = 0 \cdot (-f(y_0)) = 0 $$

and similarly $\forall$ $0 < w_0 < x_0$ with $f(w_0) \neq 0$, the expression

$$ne^{nx_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx = e^{n(x_0-w_0)} \left [ ne^{nw_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx \right ]  $$
diverges to $-\text{sgn}(f(w_0)) \cdot \infty$ as $n \to \infty$, since the term in brackets approaches $-f(w_0)$.

Now, suppose $f$ is not identically $0$, so the set

$$\{t >0 | f \equiv 0 \text{ on } (0,t) \}$$

is bounded above; define $T_1$ as the supremum of this set if it is nonempty and $0$ if it is empty. Suppose $\exists T_2>T_1$ such that
$$ T_1 < t < T_2 \implies f(t) \neq 0$$
Then $f$ is nonzero on $(T_1,T_2)$ by definition, so by the Intermediate Value Theorem $f$ cannot change sign in this interval. Thus either $f(x) \geq 0$ or $-f(x) \geq 0$ on $(0,T_2)$-- WLOG assume the former, so that for any $x_0 \in (T_1,T_2)$ the first identity after the lemma gives that $f(x_0) \leq 0$, showing $f(x_0)=0$, a contradiction. We therefore conclude that no such $T_2$ exists, i.e. $\forall t>T_1$ $\exists s \in (T_1,t)$ such that $f(s)=0$. That is to say, on any interval containing $0$ on which $f$ is not identically zero, $f$ must have infinitely many roots greater than $T_1$. In particular, the suggestion $f(x)=e^{-x^{1/4}} \sin(x^{1/4})$ cannot be a solution.

Note that if one can use the above results to show that $T_1 = 0$, it would establish that no continuous solution with compact support on $\mathbb{R}_{>0}$ exists (relevant because such functions are dense in $L^2(\mathbb{R}_{>0})$).

There is no one continuous function that will satisfy that for all n.

By trivial solution I assume you mean $f(x)=0$. One potential (although also fairly trivial, or at least boring) answer is

$$f(x)=(1+e^{nx})g(x)$$

where

$$\int_0^{\infty} g(x) \, dx=0$$

say maybe $\sin(x)$ or indeed any periodic function that acts in this way. However, this of course is not strictly defined as our $\epsilon-\delta$ definition will not hold for convergence, i.e.

$$\lim_{x\to\infty} \int_0^n \sin(x) \, dx$$

is not defined, so that is not an option either.

Let us now  consider to look at this graphically. Imagine, if you can, the graph of such a function, say defined from $0$ to $\infty$ such that our integral is zero. Without a loss of generality, let us say that this function is positive to start with, so at some point, say $x_0$ this function will have to cross the $x$ axis and become negative.

This means that our function has to be periodic and increasing at exactly the same rate and time as $1+e^{nx}$, thus resulting in only the trivial solution as above, or that the finitely bounded area before this critical value $x_0$ has identical area to our unbounded area beyond $x_0$.

This means that our function $f(x)$ satisfies $$\int_0^{x_0}f(x) \, dx=-\int_{x_0}^\infty f(x) \, dx$$

As the function must satisfy this for all n this answer should suffice to show that the function $f(x)$ must be trivial, as if $f(x)$ satisfies this for some n, the the denominator will grow at a different rate for a different n, thus showing that there is no such $f(x)$.


Taylor's theorem: $f'' + f = 0, f(0) = f'(0) = 0$. I am having a hard time coming up with a solution to this problem.

Suppose that $f$ is twice differentiable and that $f'' + f = 0$. If $f(0) = f'(0) = 0$, use Taylor's theorem to show that $f = 0$.

The definition of Taylor's theorem we were given uses the Lagrange remainder $f^{(n+1)}(c) \frac{(x-a)^{n+1}}{(n+1)!}$. Technically we don't have to use Taylor's theorem, but my professor said he doesn't know of a way to do it without it.

The obvious use of Taylor's theorem is that for all $x$ we have $f(x) = f''(c) \cdot x^2/2$ for some $|c| < |x|$. And then this means $f(x) = -f(c) \cdot x^2/2$. Then I don't know how to proceed. I can't think of a way except to show that $f(x) \ne 0$ leads to a contradiction. But what shall I contradict? Here are some of the promising ideas I've had that don't seem to lead anywhere: (suppose WLOG that $x>0$ in the following)

I can construct a sequence $(c_n)$ where $f(x) = (-1)^n f(c_n)\cdot x^{2n}/(2n)!$ since it is easy to show that $f = -f'' = f^{(4)} = \cdots$. So I have all these nonzero $f(c_n)$. But where will that take me?
How about another sequence? There is a $c_1$ such that $0 < c_1 < x$ and $f(c_1) = -f(x) \cdot 2/x^2 \ne 0$. Then there's a $c_2$ with $0<c_2<c_1$ and $f(c_2) = -f(c_1) \cdot 2/{c_1}^2 \ne 0$, and so on. This leads to another sequence $(c_n)$ with nonzero image. It converges (since it's strictly decreasing and bounded below by 0), but does anything useful happen there?
Since each of the $c_i$ are distinct, and $f(c_i)$ has opposite sign to $f(c_{i-1})$ for all $i$, with the IVT I can show that $f$ has infinitely many zeros in the interval $(0, x)$. I don't see how this can be manipulated to a contradiction though.

How can I proceed?

From $f''=-f $ we get that $f $ is infinitely differentiable.

Let us first work on an interval $[-m,m]$. As $f$ is continuous, there exists $M>0$ with $|f(t)|\leq M$ for all $t\in[-m,m]$.

Now we write the degree-$n$ Taylor expansion:
$$
f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}2+\cdots+f^{(n-1)}(0)\frac{x^{n-1}}{(n-1)!}+f^{(n)}(c)\,\frac{x^n}{n!},
$$
where $c$ is between $0$ and $x$. The condition $f''=-f$, together with $f(0)=f'(0)=0$, guarantees that $f^{(k)}(0)=0$ for all $k$. Then
$$
f(x)=\frac{f^{(n)}(c)}{n!},
$$
and so for $n $ even we have $|f^{(n)}|=|f| $, and then
$$|f(x)|\leq\frac{M}{n!}.$$
As we can get this estimate for all even $n$, we conclude that $f(x)=0$ for all $x\in [-m,m]$. As the choice of $m$ was arbitrary, $f(x)=0$ for all $x$.

Unless I am grossly mistaken, we don't need Taylor's theorem.  This is a second order differential equation with constant coefficients.  The solution is $c_1 \cos(x) + c_2 \sin(x)$. The initial conditions give that $c_1 = c_2 = 0$.  Your Professor doesn't know of a way to solve this without Taylor's theorem?


Term for similarity transformation which is not a translation What's the best (i.e. most concise) term to refer to an orientation-preserving similarity transformation which is not a translation? Here are some descriptions I could think of, but all of them feel rather bulky. I hope they are as equivalent to one another as I think they are, and I hope there is something simpler equivalent to all of them.

a direct similarity transformation which is not a translation (by a non-zero displacement)
a homothety (possibly with factor $1$) followed by rotation (with possibly zero angle)
an orientation-preserving similarity with at least one fixed point
a direct similitude with a well-defined similitude center, or the identity
what $z\mapsto a(z-c)+c$ describes in the complex plane, for some fixed $a,c\in\mathbb C$ (with $a\neq 0$)

As a native German, I tend to think about this using the German term “Drehstreckung” which literally translates to “rotation-dilation”. I'm somewhat surprised by the difficulty I have in finding an exact English translation for this concept.

I'm pretty sure your literal translation of Drehstreckung, "rotation-dilation", is the term that my high-school textbook used for this kind of transformation. Sure enough, I see the exact same terminology in lecture notes for a course in Linear Algebra with Probability at Harvard and in a "MATLAB help page" for linear-algebra students at Johns Hopkins.

Not entirely sure, but there are some elements of Mobius Transformation.


Proof that $B \land ( B \lor C) = B$? In my logic design exam today I was given this question:

  Show that:
  $$ B  \land ( B \lor C) = B $$

It's asking for a proof for this expression. Could someone please explain how such expression can be proven? I'm not that good at Boolean algebra but I believe that it's in the simplest form.

A truth table will show it

B    C    B or C    B and (B or C)

T    T      T            T
T    F      T            T
F    T      T            F
F    F      F            F

Distribute out expression, and you're left with B and B = B or B and C. so that should imply B.

$B \wedge (B \lor C) = (B\wedge B) \lor (B\wedge C )= B \lor (B\wedge C ) = B$


Reasonably-Good Bounding Sphere of an Affine-Transformed Sphere I have a unit sphere transformed by a given $4\times 4$ affine transform (it is a 3D transform; using homogenous coordinates with $w=1$ allows representing translation).

I now need to calculate a bounding sphere of this transformed sphere.  How can I do this?

For completeness, I should mention I've thought of some bad ways to do it.

For example, one could take the sphere's axis-aligned bounding box, transform that, and then compute the minimal bounding sphere of those 8 points.  This is unsatisfactory because the bounding sphere will be very coarse.  E.g. if the transform were the identity, we can see that the new bounding sphere will have $3 \sqrt{3} ~\times$ the volume.

So as additional requirements: the preferred method will produce a fairly tight bounding sphere and also be reasonably simple ($O(1)$ time, small constant).

Okay, well I thought about it some more, and it's actually pretty easy.  Here are some algorithms, all of which produce the optimal bounding sphere:

In your example, don't consider the axis-aligned box's corners.  Consider its face-centers.
As a refinement of this, forward transform the origin as well as the points $<1,0,0>, <0,1,0>, <0,0,1>$.  The transformed origin is the center of the new bounding sphere, and the longest vector from the transformed origin to one of those transformed points is the radius.
Decompose matrix into position/rotation/scale.  As a commenter mentioned one can use SVD, but because it's affine, there's also an easier way.  Then the radius of the bounding sphere is the max scaling along any axis, and the position is the translation (or you can find it by forward-transforming the origin).


Problem related to remainder A polynomial in $x$ leaves a remainder $2$ and $3$ when divided by $x-1$ and $x+1$. What is the remainder, when divided by $x^2-1$ ?

Hint.  The remainder will be a linear polynomial, say $ax+b$, and you need to determine $a$ and $b$.  The division can be written as
$$f(x)=(x^2-1)q(x)+ax+b\ .$$
From the given information you have
$$f(x)=(x-1)r(x)+2\ .$$
Now substituting $x=1$ in these two equations will give some information about $a$ and $b$, and you can use the other given remainder in a similar way to get a second equation.  You can then solve for $a$ and $b$.

Let $\displaystyle f(x)=(x^2-1)g(x)+A(x-1)+B(x+1)$ where $g(x)$ is a polynomial and $A,B$ are constants

Using Polynomial remainder theorem, $\displaystyle2=f(1)=B\cdot(1+1)\iff B=1$

Similarly,  $\displaystyle3=f(-1)=A(-1-1)\iff A=?$


How to solve inequalities with absolute values on both sides? If you have an inequality that has two absolute value bars like $|4x+1|<|3x|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think you do $4x+1<-3x$. Is this correct?

You could also square everything
$$
|f(x)| < |g(x)| \Leftrightarrow |f(x)|^2 < |g(x)|^2 \\
\Leftrightarrow f(x)^2 < g(x)^2 \\
\Leftrightarrow 0< g(x)^2-f(x)^2 \\
\Leftrightarrow 0< (g(x)-f(x))(g(x)+f(x)), \\
$$
which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign.

I just solve the equalities without absolute values: LHS = RHS and LHS = -RHS. Then I use a number line test to check values in each interval to determine the intervals of the solution set. This led me to all values between -1 and -1/7, not including engpoints, since the original inequality was strict.


Prove a given group is free From Hungerford's "Algebra":

What type of tools does one have to tackle a problem like this? I seem at a loss at how to show a group is free at all. One can consider each group as the homomorphic image of a free group, but how does one work in the other direction?

The group $F/H$ is (isomorphic to) the free group on $X \setminus Y$, since it has the correct universal property: If $G$ is a group with underlying set $|G|$, then there are natural bijections $$\hom(F/H,G) \cong \{\phi \in \hom(F,G) : H \subseteq \ker(\phi)\}$$
$$= \{\phi \in \hom(F,G) : Y \subseteq \ker(\phi)\} \cong \{f \in \hom(X,|G|) : f|_Y = 1_G\}$$
$$ \cong \hom(X \setminus Y,|G|).$$

Take any function $X \setminus Y \to G$ where $G$ Is an arbitrary group. Then extend it is any way you want to a function $X \to G$. This induces a unique group homomorphism $F \to G$ which in turn induces a group homomorphism $F/H \to G$. A bit of work shows that this extension from $X \setminus Y$ to $F/H$ is unique, so the universal property of the free group is satisfied. So $F/H$ is free on $X \setminus Y$.

This all makes sense since we $X \setminus Y$ is (morally speaking) its own image in $F/H$.


Transformation of positive semi-definite matrices Let $a,b,c,d,e$ be positive reals such that the following matrix is positive semi-definite:
$$
\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\
a+3b+3c+d & a+2b+c & a+b  \\
a+2b+c & a+b & a \\ \end{pmatrix}
$$

Does it follow that also the following matrix is positive semi-definite?
$$
\begin{pmatrix} e & d & c \\
d & c & b  \\
c & b & a \\ \end{pmatrix}
$$

[By iterated subtraction of rows and columns, the two matrices have the same determinant; but do these operations preserve also this stronger property?]

Let

$$
A = \begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a + 2b+c \\a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\end{pmatrix}, \quad
B = \begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \end{pmatrix}.$$

Then $A=UBU^T$ and $B=VAV^T$, where

$$U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad
V = U^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix} $$

Therefore $A$ and $B$ represent the same quadratic form, but in different bases,  hence $A$ is positive semi-definite if and only if $B$ is positive semi-definite.

Added to answer: $x^TBx = (V^Tx)^TA(V^Tx) \geq 0$, hence $B$ is positive semi-definite.

Edit: I make a rectification of an erroneous extension to semi-definite matrices (following remarks of @Paolo Leonetti).

I will consider only the particular case of positive definite matrices according to Sylvester criterion (see https://en.wikipedia.org/wiki/Positive-definite_matrix).
This criteria is applied by checking the positivity of  determinants of "russian dolls matrices" beginning

by the North-West corner $M(1,1), M(1:2,1:2)$ and $M(1:3,1:3)=M$, or
by the South-East corner $M(3,3), M(2:3,2:3)$ and $M(1:3,1:3)=M$.

It is this latter criteria that we will consider.

$$A=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\
a+3b+3c+d & a+2b+c & a+b  \\
a+2b+c & a+b & a \\ \end{pmatrix} \ \ \text{and} \ \ B=\begin{pmatrix} e & d & c \\
d & c & b  \\
c & b & a \\ \end{pmatrix}$$

Let $M_{23}=M(2:3,2:3)$ (we keep in a matrix $M$ the elements which are in column 2 and 3, i.e., we suppress line 1 and column 1).

Thus it suffices to check the equivalence:

$$det(A)>0 \ \& \ det(A_{23})> 0 \ \& \ a > 0  \ \ \Longleftrightarrow \ \ det(B)\geq0 \ \& \ det(B_{23})\geq 0 \ \& \ a \geq 0 \ \ \ (1)$$

But $det(A)=det(B)$ (you say it in your text), and one can verify that:

$$det(A_{23})=det(B_{23})=ac-b^2$$

Thus (1) is established.

Remark: coefficients $1,2,1$, then $1,3,3,1$, then $1,4,6,4,1$ that appear in matrix $A$ come from Pascal triangle. This remark could open the way to a new theorem and new proof, generalizable to larger matrices of this form.