Is this derivation of implicit differentiation using chain rule is technically valid? In my text book I found a derivation of implicit differentiation using the chain rule to get this formula:

$$\frac{dy}{dx}=\frac{-\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

Which: $x=x,y=g(x),F=F(x,y)$

I tried to derive same formula as following:

$$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}\frac{\partial x}{\partial x}$$
$$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}$$
$$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+2\frac{\partial y}{\partial x}$$

then substracting $\frac{\partial y}{\partial x}$ from each side:
$$0=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial x}$$
$$\frac{\partial y}{\partial x}=-\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}$$
$$\frac{\partial y}{\partial x}=\frac{-\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

I think something is technically wrong in the step of subtracting but I'm not sure. Is my derivation valid ?

Thanks

Your notation is terribly confusing, so let me do things properly for you.

You have a point $(a,b)$ such that $F(a,b) = 0$. You also have that $\frac {\partial F} {\partial y} (a,b) \ne 0$. The implicit function theorem guarantees the existence of a differentiable function $g$ such that $y = g(x)$ in some neighbourhood of $a$ and $F(x,g(x)) = 0$ on it. You want to find $g'$ on this neighbourhood.

Begin by writing that $0 = F(x,g(x))$, and derive with respect to $x$:

$$0 = (0)' = F(x,g(x))' = \frac {\partial F} {\partial x} + \frac {\partial F} {\partial y} g'(x) ,$$

whence it follows that

$$g'(x) = - \frac {\frac {\partial F} {\partial x}} {\frac {\partial F} {\partial y}} .$$

Look at your first line, in the case where $g(x)-3x, F(x,y) = x^2y^3$.  It does not appear to be true.


Find the derivative of $\frac{(2x−1)e^{−2x}}{(1−x)^2}$ I need to find the derivative of $$\frac{(2x−1)e^{−2x}}{(1−x)^2}$$

I seems very complex to me so I'm wondering if there is a rule or formula I should be using? I attempted it using the chain rule first for the numerator (since I have $ ( 2 x- 1)$ multiplied by $e^{- 2 x}$ as my numerator) and then my plan was to use this rule: $(\frac{u}{v})′=\frac{vu′-uv′}{v^2}$.

It gets messy and complicated. Could someone please explain how you'd attempt this problem?

If you want to differentiate the product of $3$ functions, the product rule is still available:

$$
(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'
$$

(and if you had decided to group (gh) or (fh) at the outset, you can check you'll still get the same outcome)

So, if you take $f = 2x -1 $, $g = e^{-2x}$ and $h = (1-x)^{-2}$, you're set (notice I changed the exponent on $h$ to make this a product rule; you could similarly formulate things in terms of the quotient rule if you prefer)


Trial solution for modulational instability I have a simple linear differential equation of the following form:

$\frac{\partial a}{\partial z} + i\beta_2\frac{\partial^2 a}{\partial t^2} = i\gamma P(a+a^*)$

I seek the solution to $a(z,t)$, and $a^*$ is the complex conjugate. $\beta_2$, $\gamma$ and $P$ are all constants, and $i$ is the imaginary unit.

Apparently, the following trial solution is valid:

$a = c_1\exp[i(kz-\omega t)] + c_1^*\exp[-i(kz-\omega t)]$

if the following dispersion relation is satisfied:

$k = \pm \sqrt{\beta_2^2\omega^4 + 2\gamma P\beta_2\omega^2}$

However, I am unable to validate this. When I plug the trial solution into the differential equation, there is no way that a $k^2$ pops out, which would eventually let me get a dispersion relation of the form shown (in a square root).

This problem is related to modulation instabilities (MI) in nonlinear optical systems. The wiki page on MI makes the claim that this solution is valid, but does not provide a proof, and the book referenced by that page does the same thing, without proof. Any ideas?

Substiuting trial solution in the differential equation. Compare +ve and -ve exponential part. I got 2 equation(1 & 2), multiply them and find k^2. To match with dispersion relation, subtract these two equations. After that, you can figure out by yourself.


Geometric representation of all the points for $\cos(z) = 2$, $z \in \mathbb C$ So I am having trouble visualizing the solutions to
$$\cos(z) = 2, z \in \mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $\cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?

If $z=x+iy$ then

$$ \cos(x+iy) = \cos(x)\cos(iy) - \sin(x)\sin(iy) = \cos(x)\cosh(y) - i\sin(x)\sinh(y) $$

Therefore

\begin{align} \cos(x)\cosh(y) &= 2 \\ \sin(x)\sinh(y) &= 0 \end{align}

Obviously, no real solution exists, so $y\ne 0 \implies \sin x = 0$.

But $\cosh (y) > 0, \forall y \implies \cos x > 0 \implies \cos(x) = 1 \implies x = 2n\pi$, which leaves

$$ \cosh(y) = 2 \implies y = \pm \cosh^{-1}(2) = \pm \ln (2 + \sqrt{3}) $$

So the solution is

$$ x = 2n\pi \pm i\ln(2 + \sqrt{3}) $$

For a visualization, think of the solutions as points on the 2 lines $y = \pm \ln(2+\sqrt{3})$, spaced out evenly by $2n\pi$

Have you thought about considering the polar form of the equation? It might help you think geometrically. $\cos z = \text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $\ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.


Odds of specific scenario after $4$ dice rolls I have started studying for a statistics competition but seem to be a bit rusty and unsure about the answer (which I can't check).

"A fair die is rolled $4$ times. What's the probability of each of the numbers $1, 2$ and $3$ appearing at least once each?"

I tried thinking that I can get the numbers $1, 2$ and $3$ then a fourth roll providing $6$ different numbers. That would leave me $6$ possibilities of getting $1, 2$ and $3$ on the $3$ first rolls, in that order.

Since I don't need the $1,2$ and $3$ necessarily on the $3$ first rolls, I thought they can actually be placed in $C_{4,3} =4$ "positions" of rolls.

And also, they can come in $3!=6$ possible orders.

Adding all that up equals to ${\rm{6}}{\rm{.4}}{\rm{.6 = 144}}$ possibilities of getting $1, 2$ and $3$ at least once, from a total of ${6^4} = 1296$ different outcomes.

$$P = {{144} \over {1296}} \approx 0.11$$

It is a multiple choice and the answer is one from: $${5 \over {36}},{1 \over {12}},{5 \over {12}},{1 \over {24}}$$

And none of the above is equivalent to 0.11, unfortunately. Can anyone help figuring out which step I got wrong?

Let's just count the number of ways it could happen.

If they each appear once each, so the last number is a 4,5,6:

* pick if the last number is 4,5,6 (3)

* order everything (4!)

If one of them appears twice:

* pick which appears twice (3)

* order everything, but unorder the numbers that appear twice (4!/2!)

Add those products and then divide by 6^4.


Gaussians going towards delta "functions" We have a sequence of random variables $x_1, x_2, x_3,...$ that are independent and are $N(0, 1/n)$ random variables. We want to show that $(x_1)^2 + (x_2)^2 + (x_3)^2 +...$ converges in probability to 1.

I have tried using Borel-Cantelli Lemma, but I am unsuccessful. Then, a lightbulb clicked, and I thought maybe the $x_n$ are converging to a delta function, so this may be the reason why. However, I cannot prove this rigorously either. This is mildly frustration; how does one approach these problems, and how do you show this?

Let $X_k$ be i.i.d. normal random variables with zero mean and variance $\sigma^2 = \frac{1}{n}$. Define $Y_n = \sum_{k=1}^n X_k^2 = \frac{1}{n} \sum_{k=1}^n Z_k^2$, where $Z_k$ are i.i.d. standard normal variables. (Already from this form, $Y_n \to \mathbb{E}(Y_n)$ with probability 1 by the law of large numbers).

Clearly:
$$
   \mathbb{E}(Y_n) = \frac{1}{n} \sum_{k=1}^n \mathbb{E}(Z_k^2) = \frac{1}{n} \sum_{k=1}^n 1 = 1
$$
By the law of total variance:
$$
   \mathbb{Var}(Y_n) = \frac{1}{n^2} \sum_{k=1}^n \mathbb{Var}(Z_k^2) = \frac{1}{n^2} \sum_{k=1}^n \mathbb{E}((Z_k^2-1)^2) = \frac{1}{n^2} \sum_{k=1}^n \left(\underbrace{\mathbb{E}(Z_k^4)}_{=3} - 2 \underbrace{\mathbb{E}(Z_k^2)}_{=1} + 1\right) = \frac{2}{n}
$$
Now, using Chebyshev's inequality:
$$
   \mathbb{P}\left( |Y_n - 1| > \epsilon \right) < \frac{\mathbb{Var}(Y_n)}{\epsilon^2}
$$

Hence for arbitrary $\epsilon$, and $\delta > 0$ there exists $m \in \mathbb{N}$, such that for all $n > m$, $\mathbb{P}( | Y_n -1| > \epsilon) < \delta$, i,e. $Y_n$ converges in probability to 1.

To the revised question if $X$ is $N(0,1)$ then $X^2$ is chi-square with 1 degree of freedom.  So $(i X_i)^2$ is chi-square with 1 df.  It has mean 1 and variance 2.  So $X_i^2$ has mean $1/i$ and variance $2/i^2$.  Now the mean of the sum of the $X_i^2$ is the partial sum of a divergent series and hence the sum cannot converge.  This assumes that $N(0,1/n)$ has standard deviation $1/n$ not the variance.  If the variance is $1/n$ then $iX_i^2$ is chi square 1.  So the mean of $X_i^2$ is $1/i^2$ and variance $2/i^4$ and the mean of the sum of the $X_i^2$ is a convergent series but does not converge to $1$. It converges to $π^2/6$. The sum of the random variables has the $n$th term approaching $0$ but the sum is a non-degenerate random variable. Note that the assumption here was the Xi is distributed N(0,1/i) as it was originally posed.


Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$ Let $x$, $y$ and $z$ be real numbers. Prove that:
  $$4(x^6+y^6+z^6)+5(x^5y+y^5z+z^5x)\geq\frac{(x^2+y^2+z^2+xy+xz+yz)^3}{8}$$

I think cyclic homogeneous polynomial sixth degree inequalities with three variables  are interesting enough because it's still open.

By the way, the case of fifth degree and less is very easy and we can kill it by $uvw$.

My trying by $uvw$.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that
$$\sum_{cyc}(32x^6+40x^5y)\geq(9u^2-3v^2)^3$$ or
$$\sum_{cyc}(32x^6+20x^5y+20x^5z)-27(3u^2-v^2)^3\geq20\sum_{cyc}(x^5z-x^5y)$$ or
$$4w^6+516u^3w^3-244uv^2w^3+2592u^6-4644u^4v^2+1872u^2v^4-72v^6-3(3u^2-v^2)^3\geq$$
$$\geq\frac{20}{9}(x-y)(y-z)(z-x)(27u^3-18uv^2+w^3).$$
We can show that the left side is non-negative and after squaring of the both sides

we'll get a fourth degree inequality of $w^3$, which is nothing I think.

Any hint would be desirable.

Thank you!


When does $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ = $\lambda$($A$) hold? I was working on a proof of a homework assignment when I stumbled on an issue.

Here is the problem I was working on: Suppose that if $f$ is a nonnegative $\mathcal M$-measurable (Lebesgue measurable)  function and that $E\subset \mathcal M$.
Let $A$ = {${x\in E}$: $f$($x$) $>$ $0$}.  Show that if $\int \limits_{\Bbb E} f d\lambda$ = $0$, then $\lambda (A)$ = $0$.

So far, this is what I have:

Note that $0$ $<$ $f\chi_A$ $\le$ $f\chi_E$.  So, we have,

$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$.

By our assumption that $\int \limits_{\Bbb E} f d\lambda$ = $0$, we further see

$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$ = $0$.

This is where I got stuck.
I want to say that $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ = $\lambda$($A$).  However, I was told that this does not hold in general.

So my question is: Why does it not hold in general?  Under what conditions does it hold?

$\int f\chi_A d\lambda = \lambda (A)$ fails in most cases. Just take A to be an interval and play with some f. If the equation holds for some f do you expect it to hold for $nf$ for all positive integers n? No, right? Here is the correct proof: let $A_n =\{x :f(x)>1/n\}$. Then $\int f\chi_{A_n} \leq \int f\chi_A =0$. On the other hand $f\chi_{A_n} \geq (1/n) \chi_{A_n}$. Hence $\int (1/n) \chi_{A_n} =0$ which menas $\lambda (A_n)=0$. This is true for each n and hence $\lambda (\cup A_n)=0$. now verify that $\cup A_n$ is nothing but A.

$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$ = $0$

$$\iff \int \limits_{\Bbb R} f \chi_{A} \,d\lambda = 0$$

$$\to f \chi_{A} = 0 \ \lambda-\text{a.s.} \tag{1}$$

$$\to \chi_{A} = 0 \ \lambda-\text{a.s.} \tag{2}$$

$$\to \lambda(A) = 0$$

$(1)$ Did you already prove this in class?

If not consider $A_n = \{f(x) > \frac1n \}$

$(2)$ Gasp! What if $f=0$?


Proof of Fermat's Theorem for Extrema I'm trying to follow a proof of Fermat's theorem for extrema -- that if some function $f$ has a local minimum at $(a,b)$ and its first partial derivatives exist, then they're equal to zero. The proof in  the textbook I'm using doesn't quite make sense to me (or, at least, it's very first step).

It first defines a function $g(x) = f(x,b)$. I don't quite understand this step, as it seems we would want to work with a function of $x$ and $y$, and since we're using $(a,b)$ as a point, these are effectively constants. It also seems that we're defining a single variable function, $g$, as a two-variable function, $f$.

If we accept this step for just the moment, it argues that $f$ having a max or min at $(a,b)$, then $g$ has a max or min at $a$. Since the first step is still quite confusing, I don't quite understand how this translates, though it seems that the authors simply took $x = a$. From here, they deduce that $g$ have a max or min at $a$ implies $g'(a) = 0$, $g'(a) = f_x (a,b)$, and thus $f_x (a,b) = 0$. They then use the same argument to show that $f_y (a,b) = 0$.

I think my main confusion is in the first step, but I also can't quite see how a max or min at $(a,b)$ translates into a max or min at $a$.

Thanks in advance for any insights on this.

The proof is correct. We are assuming that $f$ has a local minimum at $(a,b)$. There's nothing wrong with defining $g(x)=f(x,b)$, right?! The function $g$ is a function of a single variable $x$ and, since when $(x,y)$ is close to $(a,b)$, $f(x,y)\geqslant f(a,b)$, then, in particular, when $x$ is close to $a$, $g(x)\geqslant g(a)$ (since $(x,b)$ is close to $(a,b)$). So, the single variable function $g$ has a minimum at $a$. Since it is differentiable, its derivative at $a$ is $0$. But, by definition of partial derivative, $g'(a)=f_x(a,b)$.


Prove that $13^n -4^n$ is divisible by $9$ for every integer $\geq $ Prove that $13^n - 4^n$ is divisible by $9$ for every integer $n$ is greater than or equal to $1$.

I have written down my answer but I don't know if it is correct.

$    13(13^k) - 4^{(k+1)}$
$=    9(13^k) + 4(13^k) - 4{(4^k)}$
$=    9(13^k) + 4(13^k - 4^k)$
$=    9(13^k) + 4(9a)$ where $9a = 13^k -4^k$
$=    9(13^k + 4a)$
$=    9b$ where $b=13^k + 4a$

Just use $a^n -b^n=(a-b)(a^{n-1} + ... + b^{n-1})$

Another way is using induction by n as you did it (more or less) in the OP.

Take base case $ n=1$, $9$ is divisible by $9$. Assume it to be true for $ n>=1$, then replace $n$ by $n+1$.

$13.13^n -4.4^n $ can be written as $9.13^n \,  + 4.(13^n-4^n)$ and both parts of it are divisible by 9.


Finding number with prime number of divisors? The problem is to find the number of numbers in [l,r] that have prime number of divisors.
Example for $l=1,r=10$
The answer is $6$
since $2,3,4,5,7,9$ are the numbers having prime number of divisors

constraints are $l,r\le10^{12}$

and $r-l\le10^6$.
Can someone help me with the fastest solution?

Here is my approach

I stored primes  3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 in an array.
Loop through i=1 to 1000000
   if i is prime
       for j in primes
           calculate val=i^(j-1)
              add the val to answer list
for each query [l,r], the answer is all primes in [l,r]+
                     the numbers in the list in range [l,r]

but finding all primes in [l,r] takes too much time.
Any sugesstion?

While prime factorization doesn't exactly happen instantly for large numbers, you can determine the number of divisors of a number using the exponents from the prime factorization.

For every distinct power in a number's prime factorization, there is an integral exponent. Add 1 to every one of these powers, multiply the resulting numbers, and you will obtain the number of divisors that divide even into the original number.

Some examples are called for to demonstrate the power of this fact (of which I can't recall a rigorous proof).

Example 1: 60. The prime factorization of 60 is $2^23^15^1$. Just looking at each of the exponents, we have 2,1,1. Add one to each of them to get 3,2,2 and multiply them to get 3*2*2 = 12. This means that 60 has 12 divisors (including 1 and 60 itself). Indeed, the positive divisors of 60 are, in ascending order, 1,2,3,4,5,6,10,12,15,20,30,60.

Example 2: 17. 17 is prime, so its prime factorization is (explicitly with exponents) 17^1. Take the exponent 1, add one to it to get 2, and now you have the number of positive divisors of 17: 1,17.

Example 3: $p$, where $p$ is any prime. Needless to say, the prime factorization is $p^1$, and the positive divisors of $p$ are 1,$p$. Since the number of divisors is always 2 (which is prime), we can conclude that all primes have a prime number of divisors.

Now we can consider composites. Every natural number $n\geq 1$ must have 1 and $n$ at least in its prime divisor list. For every other divisor $d$ of $n$, there is a corresponding divisor $d/n$ that keeps the number of divisors even. The only special case is when $d=\frac{n}{d} \implies d^2 = n$. But then $d$ is just the square root of $n$ (!) and will only occur once in the list of $n$'s divisors, thus producing an odd number of divisors. So the only composite numbers with an odd number of divisors are perfect squares. There is no need to check any non-square composite number because they will have an even number of at least 4 divisors.

Computational Pro-Tip: when searching for all numbers on the interval $[a,b]$, "simply" list the primes in the interval. Then assign $m = \lfloor \sqrt{a} \rfloor$, and $n = \lceil \sqrt{b} \rceil$. Find the prime factorization for every natural number on the interval $[m,n]$, double each of the exponents in said prime factorization, add one to every exponent, multiply all of them as before and determine if this product is prime. If so, just append the square of corresponding integer to the list that keeps track of the numbers with a prime number of positive divisors.

Your answer is simply the count of prime numbers plus some extra numbers $p^{2^k},k\geq 1,$ where $p$ is prime, or more generally, all the numbers of the form $p^{2^k},k\geq 0$ and $p$ prime.

See this for more details.

By the way, I am also trying this problem of codechef...but not AC yet :(


Big $\mathscr{O}$ notation - division operation Consider that the following constant $h$ (time-step size) and $p$ as the order of convergence of some numerical method. On a source I read

$$\dfrac{\mathscr{O}(h^{p+1})}{\mathscr{O}\left(\frac{h}{2}^{p+1}\right)}=\mathscr{O}\left(h\right)\tag{1}$$

Why does the above result hold true? As far as I am concerned, if I remember correctly, it holds true that $$\mathscr{O}\left(\frac{h}{2}^{p+1}\right)=\frac{1}{2}^{p+1}\cdot\mathscr{O}\left(h^{p+1}\right)=\mathscr{O}\left(h^{p+1}\right)$$
However, I cannot realize the reason why equation $(1)$ holds true. Could you please help me understand that?

For definition
$$O(g) =  \left\lbrace f:\exists C > 0, \exists U_\delta(x_0), \forall x \in U_\delta(x_0) \left(|f(x)| \leqslant C \cdot g(n) \right) \right\rbrace$$
is used notation $f(x)=O(g(x)), x\to x_0$ where under $x_0$ we understand also some $\infty$, respectively we understand neigbourhood for $x_0$. Obviously, mathematically correct record should be $f \in O(g)$, though often is used "$=$".
For simplicity let's consider positive functions. It's well known property of $O$, that for constant with respect to $x$ we can write
$$ C \cdot O(g) = O(C \cdot g) = O(g) $$
where all equalities we can understand as set equalities, i.e. holds in both directions.
Also it's well know properties
$$ O(f) \cdot O(g) = O(f \cdot g) $$
$$g \neq 0  \Rightarrow \dfrac{O(f)}{g} = O\left( \dfrac{f}{g}\right)$$
In your example, equation (1) $\dfrac{\mathscr{O}(h^{p+1})}{\mathscr{O}\left(\frac{h}{2}^{p+1}\right)}=\mathscr{O}\left(h\right)\tag{1}$ I see following possibility: if we have $\frac{h^{p+1}}{2}$ in place of $O\left( \left( \frac{h}{2} \right)^{p+1} \right)$ and, so, on right hand then you have $O(h) = O(1)$
But in this case you second equation is not true, because now you have both functions, so
$$\mathscr{O}\left(\frac{h}{2}^{p+1}\right)=\frac{1}{2}^{p+1}\cdot\mathscr{O}\left(h^{p+1}\right) \ne \mathscr{O}\left(h^{p+1}\right)$$ .


Proof of uniform continuity of the convolution. Assume throughtout that $1 \leq p < \infty$ and $1< q \leq \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$ and $m$ is the lebesge measure. I have to prove the following theorem:

  Theorem 2: If $f \in L^p(\mathbb{R}^n)$ and $g \in L^q(\mathbb{R}^n)$, then $f \star g$ is uniformly continuous.

We saw the following theorem in the lecture with proof:

  Theorem 1: If $1 \leq p < \infty$ and $1< q \leq \infty$ and $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is locally $p$-integrable, then $$\lim_{\rho \rightarrow 0} \int_K |f(x+\rho) -f(x)|^pdm = 0$$
  for any Lebesgue measurable subset $K \in \mathbb{R}^n$.

My attempt to prove uniform continuity is the following:

Let $\epsilon > 0$. Since I can use Theorem 1, there exists a $\delta > 0$ such that for all $|\rho| < \delta$ we have $$\int_{\mathbb{R}^n}|f(y+\rho)-f(y)|^pdm < (\frac{\epsilon}{||g||_{L^q}+1})^p$$

Now let $x,\rho \in \mathbb{R}^n$ with $|\rho|<\delta$. The convolution $(f \star g)(x)$ is defined as an integral $\int f(x-y)g(y)dm(y)$ outside of a set $A$, and as $0$ inside $A$, where A is the set where $f(x-y)g(y)$ is not Lebesgue integrable with respect to $y$. If $x+\rho$ and $x$ are in $A^c$, then I can estimate
\begin{align*}
|(f \star g)(x + \rho)-(f \star g)(x)| &= |\int_{\mathbb{R}^n}(f(x + \rho -y)-f(x-y))g(y)|dm(y) \\
&\leq ||f(x+\rho - \cdot)-f(x- \cdot)||_{L^p}||g||_{L^q} \\
&=(\int_{\mathbb{R}^n}|f(y+\rho)-f(y)|^pdm(y))^{1/p}||g||_{L^q} \\
&< \frac{\epsilon}{||g||_{L^q}+1}||g||_{L^q} \\
&<\epsilon.
\end{align*} .
This estimate works now fine if I assume that $x+\rho$ and $x$ are in $A^c$ where I can write the convolution as an Integral. And if $x+\rho$ and $x$ are in $A$, then the convolution is equal to $0$ and thus $$|(f \star g)(x + \rho)-(f \star g)(x)|=0 < \epsilon$$.

But if I have the cases

$x+\rho \in A$ and $x \in A^c$
$x+\rho \in A^c$ and $x \in A$

then I can no longer write the convolution $f \star g$ as an integral. How do I save the proof? Or do I have to use a completely different argument to include those cases?

The proof is actually correct.
Holder inequality gives that: $\ \ \forall x\in R^n,\ \ \int_{R^n}{\left|f(x-y)g(y)\right|\le\left(\int_{R^n}\left|f\right|^p\right)^\frac{1}{p}}\left(\int_{R^n}\left|g\right|^q\right)^\frac{1}{q}<+\infty$
This shows that A is empty, and your proof is correct.


Abelian groups and Sylow-subgroups For a finite group $G$, if the number of all Sylow p-Subgroups is one, then, $G$ is abelian, right? My question is, whether the other direction is also true.

Well, like Bernard pointed out even the first direction is false but if you put nilpotent instead of abelian then it is true in both direction.


How to check if a point is inside a rectangle? There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle?

$M$ of coordinates $(x,y)$ is inside the rectangle iff

$$(0<\textbf{AM}\cdot \textbf{AB}<\textbf{AB}\cdot \textbf{AB}) \land (0<\textbf{AM}\cdot \textbf{AD}<\textbf{AD}\cdot \textbf{AD})$$
(scalar product of vectors)

find the equations of the lines and equate the distance from the point P TO THE GIVEN TWO Pairs of lines .if the point is inside the rectangle,the distances to the pairs will be less than the distances between the line in the pair


How can we calculate the mathematical expectation of a Dirac delta? The expected value of a function $g$ of a random variable $X$ is defined as the Stieljes integral:

$$E[g(x)] = \int_{-\infty}^\infty g(x) \,dF(x),$$

where $F(x)$ is the cumulative distribution function of $X$. In practice, we also have:

$$E[g(x)] = \int_{-\infty}^\infty g(x) f(x) \,dx,$$

where $f(x)$ is the density function. I am working with the function $g(x) = a \, \text{sech}^2(a \, x)$. From an engineer's point of view, this function resembles a Dirac delta (multiplied by 2) for large enough $a$. Consequently, I tried to calculate $E[\delta(x)]$ rather than $E[a \, \text{sech}^2(a \, x)]$ for very large $a$. On the one hand, I got:

$$E[\delta(x)] = \int_{-\infty}^\infty \delta(x) f(x)\, dx = f(0).$$

On the other hand, using the first formula,

$$E[\delta(x)] = \int_{-\infty}^\infty \delta(x)\, dF(x).$$

Since $\delta$ is equal to zero almost everywhere, the set $\{x:\delta(x) \neq 0\} = \{ 0 \}$ is countable and hence has a Lebesgue measure of zero. I have read that if the Lebesgue measure is zero, the integral is zero even if the value of the function is infinite. Hence, it follows that

$$E[\delta(x)] = 0.$$

Obviously, there is a flaw in my calculations. Actually I am familiar with Riemann integration and have little understanding of Lebesgue integration.

P.S.: Computer simulations give one result or the other depending on the ratio between $a$ and the sample size.

The dirac delta is what is called a generalised function.

Basically, the dirac delta is defined to work as a useful exception to the usual rule of a discrete atom having zero Lebesgue measure.

As an exception, naturally it doesn't follow the rule.   So, if it is useful to do so then we might extend the notion of expectation to say:

$$\mathsf E(\delta(X-c)) = f(c)$$


How to calculate coordinates of point P on an ellipse that has a tangent line at 45 degrees I have the equation of an ellipse: $\frac{x^2}{4}+\frac{y^2}{9}=1$

In the upper left quadrant, how do I find the coordinates of tangent point P whose tangent line is at an angle of 45 degrees with the x-axis (slope of 1)?

I haven't done this kind of math in nearly a decade, so I haven't been able to figure it out.

A way to do this without derivative:

Let the tangent line be $y=x+c$. Substitute into equation of the ellipse: $$\frac{x^2}{4}+\frac{(x+c)^2}{9}=1$$ $$\implies13x^2+8cx+(4c^2-36)=0\tag{*}$$

Since the tangent touches the ellipse (at exactly one point), the discriminant of $(*)$ has to be zero: $$\Delta=-144(c^2-13)=0\implies c^2=13\implies c=\pm\sqrt{13}.$$ $c=\sqrt{13}$ is obvious.

Back to $(*)$, $13x^2+8\sqrt{13}x+(4\sqrt{13}^2-36)=0\implies x=\dfrac{-4}{\sqrt{13}}$, a repeated root.

The ellipse has the parametric equation $x=2\cos t, y=3\sin t$ (with $t\in[0,2\pi])$. The tangent vector at $t$ is $(-2\sin t,3\cos t)$. It makes an angle $45^\circ$ with the horizontal if $-2\sin t=3\cos t$, that is $\tan t=-\frac{3}{2}$. Given that you want the point to the in the upper left quadrant, that means $t=\pi-\arctan\frac32$. Then the point on the ellipse has coordinates

$$x=2\cos(\pi-\arctan\frac32)=-2\cos\arctan\frac32$$
$$y=3\sin(\pi-\arctan\frac32)=3\sin\arctan\frac32$$

Let $\alpha=\arctan\frac32\in]0,\pi/2[$ (so that $\cos\alpha$ and $\sin\alpha$ are both positive). From $\cos^2\alpha+\sin^2\alpha=1$, you get $1+\tan^2\alpha=\dfrac1{\cos^2\alpha}$ then

$$\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{2}{\sqrt{13}}$$
$$\sin\alpha=\tan\alpha\cos\alpha=\frac{3}{\sqrt{13}}$$

Hence $x=-\dfrac{4}{\sqrt{13}}$ and $y=\dfrac{9}{\sqrt{13}}$.


Number of particular terms in a product I hope this question is not too stupid, but I know very little about combinatorics and i can't find an answer..

So fix $r\in \mathbb{N}$. Let $x_1,\ldots, x_r$ be variables, let the product between them be commutative and consider the quantity $(x_1+\ldots+x_r)^{2r-2}$. Of this i want to consider only the terms of the form $x_1^2x_2^2\ldots x_{i-1}^2 x_{i+1}^2\ldots x_r^2$, i.e. the terms in which the variable $x_i$ doesn't appear and all the other variables $x_j$ appear with power exactly 2. Of course, because the product is commutative, consider all the permutations.

Fixed $i$ between $1$ and $r$, how many terms of this said form $x_1^2x_2^2\ldots x_{i-1}^2 x_{i+1}^2\ldots x_r^2$ are there? the answer should be $\frac{(2r-2)!}{2^{r-1}}$, but i can't figure out why, would you enlighten me?

Here's a very useful and general fact. Consider a product:

$$(a_1+a_2+...)(b_1+b_2+...)...$$

There are a finite number of factors, and a finite number of terms within each factor, but I'll write it as above to keep the notation trimmer. Let's say there are $n$ factors. I'll use the term $n$-product to refer to a product of $n$ arbitrary numbers. Then the above product is equal to the sum over all possible $n$-products that can be formed by taking one term from inside each set of parentheses. For example:

$$(a+b)(c+d+e)=ac+ad+ae\quad\leftarrow\text{terms where we chose $a$ from the first factor}\\
+bc+bd+be\quad\leftarrow\text{terms where we chose $b$}$$


How to evaluate the limit $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}$ First I tried direct substitution, which resulted in the indeterminate form. Then because of the square roots, I tried rationalizing (both numerator and denominator), but still get the indeterminate form. I can't use L'Hopital's rule because we haven't learned that yet (not sure if it would work anyway). Don't really know what else to try algebraically... Any hints?

I took a course on limits a few years ago. There some limits which converge, but many diverge (exercise). Limits which approach 0 are very special; there is a general theorem about these type of limits. The theorem says that $lim_{x \to 0} f{x} = f(0)$ under some special conditions. In your case, the function does not satisfy these conditions (exercise). You must look at the asymptotic behavior. By taylor expansion, near 0, $\sqrt{bx+a}=\sqrt{a}(1+\frac{bx}{2a})+O(x^2)$ Ok WE ARE GOOD TO GO NOW!!! Let's look at your limit (exercise) your limit reduces to -8/4=-2.(exercise)


You’ve got a bit of a problem here since the reduced matrix clearly has full rank, so its only null vector is $0$. Indeed, the original matrix $A$ is nonsingular: its determinant is equal to $-3.755$, so it’s not even close. Also, if you compute $Ax$, you’ll find that its elements, though small, aren’t particularly close to zero. Their first three places after the decimal point are zero, though, so at the limit of precision that you’ve got in the original data, it could be considered a null vector.

When working with inexact data like this, even if you know theoretically that the matrix should be singular, the actual matrix that you work with might turn out not be so in practice. On top of that, Gaussian elimination isn’t a particularly stable algorithm, so it can introduce its own errors as well. A common way to deal with this is to approximate a null vector of the matrix by computing its SVD (singular value decomposition) and then taking the right singular vector that corresponds to the smallest singular value. It appears that this is what Octave did. Computing the SVD of $A$ in Mathematica yields $(0.408695,0.208933,-0.881299,0.112373)$ for this, which jibes with the vector $x$ in your question. The corresponding singular value is $0.000478394$, which, since the matrix is symmetric, has the same absolute value as its least-modulus eigenvalue. This is zero to three places, which is about as good as the input data that you have.

In Matlab, you can compute the SVD of a matrix by using the function svd, not surprisingly. See the documentation for how to interpret its results. Since you’re looking for a right null vector, you’ll need to extract a singular vector from the “$V$” matrix—the right-hand one.

Will Jagy’s observation that your matrix $A$ appears to be an integer matrix plus a multiple of the identity leads me to suspect that the underlying problem that you’re trying to solve is finding eigenvectors of the integer matrix $$B=\begin{bmatrix}23&30&20&14\\30&49&32&49\\20&32&20&32\\14&49&32&103\end{bmatrix}$$ and indeed, to three decimal places one of its eigenvalues is $-0.941$. However, this is not a very well-conditioned matrix and the one you obtained by adding $0.941\,I$ to it is even worse. What this means is that it’s going to be very sensitive to truncation and other sources of error. Try this experiment: see what happens to the value of the determinant of $B-\lambda I$ as you add more and more significant digits to the approximation of this eigenvalue: carrying more precision in the approximation to the eigenvalue gets us closer and closer to singularity. Unfortunately, this also makes the matrix more and more ill-conditioned, which is bad news for Gaussian elimination. Indeed, when doing this same experiment with row-reduction instead of determinants, Mathematica eventually starts throwing warnings that the result of RowReduce may contain significant numerical errors.

From the reduced matrix in your question it looks like you’re actually carrying more precision in your calculations than you have in the matrix $A$ in your question, but if the digits after the $23$ in the reduced matrix are the more precise value that you got for this eigenvalue, then it’s off by quite a bit. That’s also going to throw off your calculations, so you might want to double-check how you’re computing eigenvalues.


Find the ratio of curved surface area of frustum to the cone. In the figure, there is a cone which is being cut and extracted in three segments having heights $h_1,h_2$ and $h_3$ and the radius of their bases $1$ cm, $2$cm and $3cm$, then
  The ratio of the curved surface area of the second largest segment to that of the full cone.

$\color{green}{a.)2:9}\\
b.)4:9\\
c.)\text{cannot be determined }\\
d.) \text{none of these}\\$

I found that $h_1=h_2=h_3\\$

and

$ \dfrac{A_{\text{2nd segment}}}{A_{\text{full cone}}}=\dfrac{\pi\times (1+2)\times \sqrt{h_1^2+1} }{\pi\times 3\times \sqrt{(3h_1)^2+3^2} }=\dfrac13$

But book is giving option $a.)$

Area is proportional to the square of linear dimension. So the area of the full cone is $k(3^2)$ for some $k$. The area of the second largest segment is $k(2^2) - k(1^2)$, so the ratio is $3:9 = 1:3$. You are right, and the book is wrong.

The lateral heights for all the segments will be the same and so will the heights.

The curved surface area of a cone $= \pi$(radius)(lateral height)

For whole cone = $\pi(3)(3L)$ (where $L$ is the lateral height of each segment)

For 2nd largest segment = $\pi(2)(L)$ (since segment $2$ has lateral height of $L$)

So the ratio will give = $\dfrac{\pi 2L}{\pi9L} = \dfrac 29$


Show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$. Let $x_n \ge 0$ for all $n \in \mathbf{N}$ and $x>0$, show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$.

My textbook does the following proof:

Let $\epsilon >0$, we must find an $N$ such that $n \ge N$ implies $|\sqrt{x_n} - \sqrt{x}|< \epsilon$ for all $n \ge N$.

\begin{align}
|\sqrt{x_n} - \sqrt{x}| &= |\sqrt{x_n} - \sqrt{x}|\left(\frac{\sqrt{x_n} + \sqrt{x}}{\sqrt{x_n} + \sqrt{x}}\right) \\
& = \frac{|x_n - x|}{\sqrt{x_n} + \sqrt{x}} \\
& \le \frac{|x_n - x|}{\sqrt{x}} \ \ \ \ \ \ \cdots \ \ \ \ \ \ (1)
\end{align}

Since $(x_n) \rightarrow x$ and $x>0$, we can choose $N$ such that $|x_n - x| < \epsilon\sqrt{x}$ whenever $n \ge N$ and so for all $n \ge N$, $|\sqrt{x_n} - \sqrt{x}| < \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon$.

What I'm wondering is, in Eqn.$(1)$, why is $\sqrt{x}$ kept in the denominator? Couldn't one just have simply $\le |x_n - x|$ and choose $N$ such that $|x_n - x| < \epsilon$ for $n \ge N$ and the rest follows?

They keep the denominator for the case that $x < 1$. If you assume $x\ge 1$ you can use your assumption.

Because $\sqrt{x_n}+\sqrt{x}\geq \sqrt x$ and thus $$\frac{1}{\sqrt{x_n}+\sqrt x}\leq \frac{1}{\sqrt x}.$$
When you'll see continuous function, such a proof is easier using the continuity of $x\longmapsto \sqrt x$.


Finding Extrema of $f(x,y)=x^4+y^4-4xy$ Let $f(x,y)=x^4+y^4-4xy$

How do I find all the relative extrema and saddle points of $f$ which lie within the open square ${(x,y) | -2<x<2,-2<y<2}$. And also if $f$ was in the closed region ${(x,y) | -2\le x\le 2,-2\le y\le 2}$ what would be the absolute maxima and minima of $f$ on this closed region and where do these extreme values occur?

What I have Tried:

I first set $f_1=4(x^3-y)$ and $f_2=4(y^3-x)$ I don't know what to do next though.

Hint: Look at the corresponding closed region $R$, so that the extrema will either be critical points in int$R$ or they will lie on the boundary of $R$. For the saddle points, use the second derivative test.

Second hint:

Step 1:

Solve the following equations and use the points in the second derivative test to decide if they are local extrema or saddle points. Suggestion: let $z=y/x$

$f_{x}=4x^{3}-4y=0\\f_{y}=4y^{3}-4x=0$

Step 2:

Now consider each piece of the boundary, making the appropriate substitution and extremize the function (of a single variable) you get in each case. For example, if we look at the top side of the square, where $y=2$ and $-2<x<2$, we get

$f(x,2)=x^{4}-8x+16$, from which $f'(x,2)=4x^{3}-8=0$ gives $x=2^{1/3}$ as a critical point. Now determine what type of critical point it is, substitute into $f(x,y)$ and compare it with what you got in step 1. Then check $x=2, x=-2$ in $f(x,2)$ and write down the result.

repeat for the other sides of the square.

All told, you get a table of values from which you can read off the maxima and minima.

you can use Lagrangians ( Constrained Optimization). I give you an example here :

http://mat.gsia.cmu.edu/classes/QUANT/NOTES/chap4.pdf


Simplifying $ \log_{e^2}(e^{4a}+ae^{4a}) $ Problem

Simplify logaritm:

$$ \log_{e^2}(e^{4a}+ae^{4a}) $$

preferably in a way that end result contains only natural logarithm.

 Attempt to solve

I know few computational rules about logarithms:

$$ \log_a(xy) = \log_a(x)+\log_a(y) $$

$$ \log_a(\frac{x}{y})=\log_a{x}-\log_a(y) $$

$$ \log_a(x^n)=n\log_a(x) $$

And formula for change of basis :

$$ \log_a(x)=\frac{\log_b(x)}{\log_b(a)} $$

There is sum inside this logarithm and it appears we don't have formula for this. On wikipedia i found two formulas about summation / substraction inside logarithm:

$$ \log_b(a+c)=\log_b a+ \log_b (1+\frac{c}{a}) $$

$$ \log_b(a-c)=\log_b a+ \log_b (1-\frac{c}{a}) $$

I have no former experience / knowledge of these formulas but for now the assumption is these are correct.

I would try to first try  change of basis to $\log_{e^2}()\rightarrow \log_e() = \ln()$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{\log_e(e^{4a}+ae^{4a})}{\log_e(e
^2)} $$

$$ \log_{e}(e^{4a}+ae^{4a})=\log_{e^2}(e^{4a}+ae^{4a})\cdot \log_e({e^2}) $$

$$ \log_{e}(e^{4a}+ae^{4a})=2 \cdot \log_{e^2}(e^{4a}+ae^{4a})$$

We get the original logarithm with base $e$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{1}{2}\ln(e^{4a}+ae^{4a}) $$

Now the sum:

$$ \frac{1}{2}\ln((a+1)e^{4a})
 $$

$$4a \frac{1}{2} \ln ((a+1)e) $$
$$ 4a \cdot \frac{1}{2} \ln (a+1) + \ln(e) $$

$$ \frac{4a}{2} \ln {(a+1)} + 1 $$

$$ 2a \ln (a+1) + 1 $$

If there is an error let me know.

You were doing fine down to $$\frac 12\ln((a+1)e^{4a})$$ but then you cannot pull out the $4a$ because it is not an exponent on $a+1$.  Instead you should do
$$\frac 12\ln((a+1)e^{4a})=\frac 12\left(\ln (a+1)+\ln(e^{4a})\right)\\=\frac 12\ln(a+1)+2a$$


Increasing, continuous, concave downward function normalised between 0.5 and 1 What would be a good function which is increasing, continuous, concave downward with

$$\lim_{x \to 0} f(x) = 0.5$$

and

$$\lim_{x \to \infty} f(x) = 1.$$

It should be concave downward whose concavity can be parametrically controlled.

How do you like $f_k(x) = \dfrac{1}{1+e^{-kx}}$, where $k\in(0,\infty)$ is a parameter?

Consider the function $$f (t)=1-\frac {1}{2} e^{-t}, \forall x\in \mathbb {R} $$ which is continuous, increasing, concave downward and satisfies the limits


For a group homomorphism, does there exist a subgroup of the domain isomorphic to the image? Let $\phi : G \to H$ be a group homomorphism. Is is it true that, there exists a subgroup $K$ of $G$ such that $K$ is isomorphic to the image of $\phi$?

Edit: How about for finite $G$?

HINT: $\Bbb Z$ has finite homomorphic images of all sizes; does it have finite subgroups of all sizes?

Yeah. If there is a homomorphism $\phi$ from $G$ to $H$, then there is a isomorphism from $G/\ker {\phi}$ to $\phi(G)$
(namely, from $a\ker\phi$ to $\phi(a)$ is the isomorphism). As $\ker{\phi}$ is a normal subgroup of $G$ it is evident that its factor group is a subgroup.


Evaluating a double integral with monte carlo integration? I am trying to find an equation that estimates the following integral:

$$\int_0^1 \int_0^1 e^{(x+y)^2} \,dx\,dy$$

where I am given a list of different uniform RV's $m_1,m_2,\ldots$ and $n_1,n_2,\ldots$.

I'm not quite sure how to go about evaluating this double integral?

I know the integral for a single integral over that interval is a summation over n, but how does it work for two integrals where the function is already given?

Any insight would be great!

thank you

The measure $dx\,dy$ is is the uniform distribution on $[0,1]\times[0,1].$ So (assuming the uniform random variables are independent—and you should have said something about their joint distribution) the sample mean
$$
\frac 1 k \sum_{i=1}^k e^{(m_i+n_i)^2}
$$
should approximate the value of the integral.

i want to give you the answer only, which I found from a website. I am giving you the link. Hope it will help you. Actually, I don't know how to solve this problem.
https://www.wolframalpha.com/input/?i=%E2%88%AB%280%2C1%29%E2%88%AB%280%2C1%29+e%5E%28%28x%2By%29%5E2%29+dxdy


How many 4 digit number can be formed by 0,1,2,3,4,5 divisible by 4 How many 4 digit number can be formed by 0,1,2,3,4,5 divisible by 4 with repetition

My Approach:

Last two digits can be 00,04,12,20,24,32,40,44,52 that is 9 possibilities for last two digits.

For the hundredth place digit all 6 possibilities exist.For the thousand place we have 5 options (1,2,3,4,5 ).

Hence the solution is 5*6*9=270

Is my solution correct? Please correct me if I am wrong.

Your ans of $270$ is correct, but you might find this way simpler:

Since a number can't start with zero, there will be $5\cdot6^3$ valid numbers,
of which exactly $1/4$ will be divisible by $4$

[Think of a number system to base 6 to understand the rationale]

Best way is to identify the first(1000) and last 4 digit numbers(5552) divisible by 4. Then they will be in arithmetic progression with d=4. Then you can identify n=number of terms[(5552-1000)/4+1=1139 numbers can be formed], which gives you exact answer for any such problems with repetition.


±1-random walk from 5 until 20 or broke You play a game where a fair coin is flipped. You win 1 if it shows heads and lose 1 if it shows tails. You start with 5 and decide to play until you either have 20 or go broke. What is the probability that you will go broke?

You can use symmetry here - Starting at $5$, it is equally likely to get to $0$ first or to $10$ first. Now, if you get to $10$ first, then it is equally likely to get to $0$ first or to $20$ first.

What does that mean for the probability of getting to $0$ before getting to $20$?

All answers so far are great but some readers seem to feel they lack an intuitive explanation - and perhaps the maths to back it up.

Consider that you have an equal chance of moving up or down along your two paths.
$$P_{u} = P_{d} = \frac{1}{2}$$
And your path lengths up and down are:

$$L_{u} = |20 - 5| = 15 $$
$$L_{d} = |0- 5| = 5 $$
$$L_{u} = 3 L_{d}$$

Now we know we have to end up, whatever the route, having moved a total distance of $L_{u}$ or $L_{d}$.

We know the probability of getting to $0$ plus the probability of getting to $20$ has to be 1 with the same proportionality constant. The probability that you complete one path (ie reach $0$ or $20$) before the other one is completed depends on the pathlength of the competing process (ie the longer it takes for one outcome to occur the better the chances for the alternative). As such:

$$P_{0} \propto L_{u}$$

So, canceling the constant of proportionality, we get the probability of going broke as:

$$\frac{P_{0}}{P_{total}} = \frac{L_{u}}{L_{u}+L_{d}} = \frac{3L_{d}}{4L_{d}}=\frac{3}{4} $$
Where $P_{total} = 1$ so $P_{0} = \frac{3}{4}$


Find a basis of a subspace constructively Let $V\subset \mathbb{R}^n$ be a subspace of the standard euclidean space.  Is there a constructive method to find a basis of $V$?

Obviously, if we can realize $V$ as the image or kernel of an explicit linear transformation, then there are methods to do this, but the only way I know to find such a linear transformation requires you already have a basis (or at least a spanning set of vectors).

If $V$ is subset of standard Euclidian space than $V$ can have dimensions 0,1,2 or 3.
If $V$ is an euclidean plane then it has dim $2$ and the base is any of two vectors that are nonparallel and lay on that plane.
If $V$ is a line, then you can take any vector that is parallel with that line for a base.
If $V$ is a point then dim is $0$.
In any other cases you have provide additional information about $V$, for example linear operator that describes it and from there you can get vector base.


Help to prove the existance of a function Let $f:X \rightarrow Y$ be a function. Prove that there exists a function $g:Y \rightarrow X$ such that $f \circ g = I_Y$ if and only if $f$ is a surjection.

I need help on proving the following:

$f$ is a surjection $\Rightarrow  \exists g:Y \rightarrow X, f \circ g = I_y$

What I've tried:

$f$ is a surjection $\Rightarrow \forall y \in Y, \exists x \in X, f(x)=y$, definition of a surjection

From here, I thought of constructing a function, $g:Y \rightarrow X$, since all $y$s have some preimage, $x \in X$ . But this will not be possible if $f$ is not injective. In that case, the codomain of $g$ will need to be appropriately restricted, so that $y_1=y_2 \Rightarrow g(y_1)=g(y_2)$. But I can't restrict the codomain of $g$ since the conclusion requires the codomain of $g$ to be exactly $X$. I don't know what else to do. Also, is my attempt anywhere near a solution or am I thinking about this problem in a wrong way?

This is a consequence of AXIOM OF CHOICE


Asymptotic Expansion, Regular Perturbation Regular perturbation. Find the first two terms in an asymptotic expansion of the small parameter $ϵ$ of the solution of
  $$
xy'+y=ϵy^{1/2},\quad x>0,\quad y(1)=1.
$$
  Explain why the expansion is not valid as $x\to\infty$. What form of rescaling would be necessary to examine behaviour for large $x$?

I've been learning about how to construct asymptotic solutions to regularly perturbed DEs, but I'm unsure of how to treat the $$y^{1/2}$$ on the right hand side.

So I divide through by $x$, and set an expansion for $y$ in terms of $y_i$, but how do I do the square root of a series?

Thank you for any help given!

The lowest order solution is obviously $y=x^{-1}$ leading to the first order equation
$$
(xy_1)'=x^{-1/2}, \; y_1(1)=0
\implies xy_1=2(x^{1/2}-1)
\implies y_1=2(x^{-1/2}-x^{-1})
$$
In the next approximation one gets
$$
(x(y_0+ϵy_1+ϵ^2y_2))'=ϵy_0^{1/2}(1+\tfrac12ϵy_0^{-1}y_1)
\\
(xy_2)'=1-x^{-1/2}, y_2(1)=0
\\
y_2=1-2x^{-1/2}+x^{-1}
$$
Thus the first three terms of the asymptotic expansion are
$$
y(x)=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(1-2x^{-1/2}+x^{-1})
$$

For the exact solution you get
$$
(\sqrt{xy})'=\frac12\frac{(xy)'}{\sqrt{xy}}=\frac12ϵx^{-1/2}
\implies
\sqrt{xy}=ϵx^{1/2}+1-ϵ
\\
\implies y=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(x^{-1}-2x^{-1/2}+1)
$$

So the above three term expansion is already the exact solution.

With regard to the later part of the question, the closed form of the solution by @LutzL doesn't seem to have issue when ${x \to ∞}$. Nevertheless, I think that what it may be getting at is that this expansion is asymptotic, so $$\epsilon (\frac{1}{√x}-\frac{1}{x}) = o(\frac{1}{x})$$ as ${\epsilon \to 0} $. However, for $x=O(\frac{1}{\epsilon^2})$ this is not the case, so the expansion is invalid as ${x \to ∞}$. Thus mutliscale perturbation theory is required, which is enough to say for the question.

I'm not sure if this is right, but this is my answer for when I hand this in
(DEs II has got lousy).


Evaluate for a double integral Evaluate $$\int_{0}^{1} \int_{e^y}^{e} \frac{x}{\ln x}\ dx\ dy$$

I'm new to double integral. Can anyone guide me for this question? Thanks in advance

Hint. This double integral is improper because the integrand is not bounded along the line $x=1$. Note that
$$\int_{y=0}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy
=\lim_{\epsilon\to 0} \int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy.$$
By changing the order of integration (Fubini's Theorem), we get
$$\int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy=
\int_{x=e^{\epsilon}}^{e} \left(\int_{y=\epsilon}^{\ln(x)} \frac{x}{\ln x} dy\right) dx=\int_{x=e^{\epsilon}}^{e}  \frac{x}{\ln x}(\ln(x)-\epsilon) dx.$$
Can you take it from here?

P.S. In order to avoid the limit you should use a version of Fubini's Theorem for improper integrals.

The idea of multiple integrals is to treat them as iterated integrals. For example, the integral provided can be rewritten as $$\int_{y=0}^{y=1}\left(\int_{x = e^y}^{x=e}\frac{x}{\ln x} dx\right)dy$$
It should be clear that we must evaluate the inner integral first, and the inner integral must resolve to a function of $y$. Assuming you could integrate the inner integral (let's call it $F(x)$), the double integral becomes $$\int_0^1(F(e)-F(e^y))dy$$ All that remains is to evaluate this final integral.


Differential calculus vs Integral calculus I have never done integration in my life and I am in first year of university. Is it harder than taking the derivative? I've heard its just going backwards. Also, my high school taught me only differentiation, I don't know why we never touched on integration. I'm going to be starting it next week and I want to know what I'm facing. Is it generally considered harder than differentiation? Thank you in advance.

If you were fine with derivatives, you will be fine with integrals in 1st year calc. It never hurts to pay attention in class (which kind of implies attending class) and to do your homework.  In fact, if you have trouble with a problem, you should do more of the same kind as soon as you know the answer. As to difficulty:

Integrals start out harder than derivatives and wind up easier.  The reason derivatives are easier is that if a function has a derivative you can compute what it is.  There is an algorithm for doing so.  Sometimes the computation may be long and complicated.  But theoretically anyway, anyone can do this.

With the integral, you will be given a lot of problems to solve, but there is no algorithm.  The kind of problems you get in first year calculus will be solvable if you learn enough tricks. (They are chosen to be solvable).  There are hundreds of tricks because over the course of many years lots and lots of smart mathematicians have worked them out.  You'll probably learn 3 - 5 tricks in your first year class.

Integrals which are not subject to the tricks (that is, most of them) are evaluated with numerical or approximating methods and for practical purposes that works very well.  In fact, with the availability of computers, many of the old tricks have fallen into disuse, because it is faster to just do the numerical computation.

Integrals and derivatives are the reverse of one another in the same sense that addition and subtraction are.  As you know, taking an operation in one direction is often easier than reversing it.  Thus multiplication is easier than division,and raising things to powers is easier than the reverse: finding roots.

One expression of the connection between derivatives and integrals is the Fundamental Theorem of the Calculus, which you will probably be taught.  However, the two subjects are more intertwined than the Fundamental Theorem suggests. Indeed, I find the myriad entanglements fascinating.

Now why does the difficulty turn around later?  Derivatives are about differences and division which make a function less smooth; and the smoother a function is, the easier it is to work with.  For example, if f is differentiable, f' may only be continuous; and in fact it can be discontinuous.  You've taken a relatively smooth function and degraded it.

Integrals are about addition, that is to say averaging, which always smooths things out.  You can integrate discontinuous functions and they will come out continuous.  Integrate again and they will be differentiable, which is even smoother than continuous. We like smooth.

Integration is much harder than differentiation, especially trig substitution, but I'm a fifth grader and you're a high schooler, so you may understand it better. Here's a handy rule:
Antiderivative of x to the n power: x to the n+1 power/n+1


Asymptotic behaviour of the number of sign changes in the sequence $\cos n\alpha$ Let $0$ $\leq$ $\alpha$ $\leq$ $\pi$. Denote by $V_n$$(\alpha)$ the number of sign changes in the sequence ${u_n}$ where $u_n$ $=$ $\displaystyle \cos n\alpha$. Then find the limit of the sequence $\frac{V_n(\alpha)}{n}$ as $n$ $\to$ $\infty$.

I am not getting any idea as to how to proceed because the sign changes is irregular. Any help is appreciated.

The limit indeed exists and is
$$
\lim_{n\to\infty}\frac{V_n(\alpha)}n=\frac{\alpha}\pi.
$$
Assume first that $\alpha/\pi$ is rational, thus $\alpha=\pi p/q$ with $0\leqslant p\leqslant q$ and $q\geqslant1$. After $2q$ steps one is back at the angle $0$ on the circle and one did exactly $p$ full circles hence one experienced $2p$ changes of sign and the result holds.

(Here the number of changes of sign is the maximal length of a subsequence $(u_{n_k})$ such that $u_{n_k}u_{n_{k+1}}\lt0$ for every $k$.)

When $a$ is irrational, the rotation of angle $\alpha$ is ergodic with invariant measure the uniform measure on the circle hence the proportion of times one observes a change of sign converges to the combined measure of the portions $(\frac\pi2-\alpha,\frac\pi2)$ and $(-\frac\pi2-\alpha,-\frac\pi2)$, that is, to $2\alpha/(2\pi)$.

The number of signs is either even ($V_n = 2n$) or odd ($V_n = 2n+1$), and in both cases the answer is $2$.


Computing the angle between two vectors using the inner product Question is: Consider $\mathbb{R}^3$ with the inner product $\langle u1,u2\rangle=x_1x_2+3y_1y_2+z_1z_2$ where $u_1=(x_1,y_1,z_1)$ and $u_2=(x_2,y_2,z_2)$ are two vectors in $\mathbb{R}^3$. What is the cosine of the angle between the vectors $v=(2,1,3)$ and $w=(-1,2,-1)$ with respect to the above inner product?

I don't even understand what exactly the inner product is,I tried using the formula
 $\cos(\theta)=\frac{\langle v,w\rangle}{|v||w|}$ but I don't understand how to use the inner product in this case. Any help would be much appreciated. I don't understand what to even substitute in the formula.

Hint: $$\cos\theta=\dfrac{\langle u_1,u_2\rangle}{|u_1||u_2|}\\
\implies \cos\theta=\dfrac{-2+6-3}{\sqrt{4+3+9}\sqrt{1+12+1}}$$
For the meaning of the inner product, see the images and the information here.


Examples of algebras that have a bounded approximate identity We note  that  $L^{1}(\mathbb R)$ is an algebra with respect to convolution without identity element.
However, regarded as a Banach algebra, $(L^{1}(\mathbb R), \ast) $ has a bounded approximate identity with respect to convolution, that is, there is a set $\{e_{r}\}_{r>0}\subset L^{1}$ such that $\|e_{r}\|_{L^{1}}\leq C$ for all $r>0$ and $C$ is some constant and $\|e_{r}\ast f- f\|_{L^{1}} \to 0$ as $r\to 0$ for $f\in L^{1}.$

My Questions are: (1) What are other examples of Banach algebras (preferably function spaces) that have a bounded approximate identity?
(2) Is $L^{1}$ the only convolution algebra which has a bounded approximate identity?

Well, of course if $G$ is any locally compact abelian group then $L^1(G)$ is a Banach algebra under convolution. If $G$ is not discrete then $L^1(G)$ has no identity, while if $G$ is first-countable then there is a bounded approximate identity. (If $G$ is not first countable there's still a net that gives a bounded approximate identity, but perhaps not a sequence.)

Say $K$ is a locally compact Hausdorff space, $K$ is not compact, but $K$ is a countable union of compact sets. Let $A=C_0(K)$, the space of functions that vanish at infinity. Then $A$ is a Banach algebra (with pointwise multiplication) with no identity but with a bounded approximate identity. (Again, if you settle for a net instead of a sequence you don't need to assume that $K$ is a countable union of compact sets.)

I am interested in the Sobolev algebra W^{1,2} of absolutely continuous functions f on the non negative half line such that both f and f' are square integrable. This is a Banach algebra without a unit under pointwise multiplication, and it does have an approximate identity.

However, there is no bounded approximate identity in W^{1,2}.

As a matter of fact, if A is a Banach algebra without a unit, that is reflexive regarded as a Banach space, and whose product is separately weakly continuous, then  every approximate identity in A is unbounded. The reason is that for a bounded approximate identity (e_n) there would be a weakly convergent subsequence, still an approximate identity, that would converge to a unit, a contradiction.


derivative of a gradient just a very simple question:

I have a continuous and differentiable function $\phi$ defined on $R^3$ with its gradient $\nabla\phi$.

Could you please tell me if it makes sense to take a derivative of this gradient with respect to $\phi$?

So:

$\frac{\partial}{\partial{\phi}} \nabla \phi = ?$

Would it make sense to say $\frac{\partial}{\partial{\phi}} \nabla \phi = \frac{\partial}{\partial{\phi}}(\phi_{x}, \phi_{y},\phi_{z}) = (\phi_{xx}, \phi_{yy},\phi_{zz}) $?

Thanks for your time,
Matt

Well from my point of view, you are thinking something but your notation of expression is describing other thing !!!! Like, what does it mean by $\nabla \phi $ ? We know in multi-variable calculus we can not express the derivative of a multi-variable function with just one scaler valued function.we need the help of vector valued function or expression .and here the $\nabla \phi$ comes into action.so derivative of $\phi$ is not a single expression (scalar valued function). So, by the derivative of $\phi$ you are just meaning $\nabla \phi $.

Now coming to your question. here, what does it mean by $\partial \phi $ ?
It is not a appropriate notation here,because here you wanting to say how $\nabla \phi$ changes as I change $\phi$ a little bit.Now,the problem is,this $\phi$ is not a single variable rather a scalar valued function.So you have to work on variable after variable of $\phi$ and component after component of $\nabla \phi$ (As,now it is a vector valued function) you can't express the whole expression as
$$\color{red}{\frac{\partial}{\partial{\phi}} \nabla \phi = \frac{\partial}{\partial{\phi}}(\phi_{x}, \phi_{y},\phi_{z}) = (\phi_{xx}, \phi_{yy},\phi_{zz})}$$
rather.if  $$\nabla \phi=
    \begin{bmatrix}
    \phi_x \\
    \phi_y \\
    \phi_z \\
    \end{bmatrix}
$$
then you should compute $\frac{\partial \phi}{\partial{x}}$ and $\frac{\partial}{\partial{x}}\nabla \phi$ individually.
And finally after dividing them then we may  say something like that you want. But that would yields,
$$\frac{\partial}{\partial{\phi}} \nabla \phi=\frac{1}{\phi_x} \begin{bmatrix}
    \phi_{xx} \\
    \phi_{yx} \\
    \phi_{zx} \\
    \end{bmatrix}\text{(differentiating partially with respect to $x$)}$$

But it is not a completed answer though.you have to follow the procedure in respect of y and z also,to get a compact answer.It will be somehow a rigorous computation.


clarification asked for 'difference between convolution and crosscorrelation?' I don't understand answer formulated in ways like this "Thus, $p\ast q$ is the distribution of $X+Y$.
The cross-correlation $p\circ q$ is the distribution $c=(c_n)_n$ defined by $c_n=\sum\limits_kp_kq_{n+k}=P[Y-X=n]$ for every $n$. Thus, $p\circ q$ is the distribution of $Y-X$."

Can someone explain this in an easier way?
See: What's the difference between convolution and crosscorrelation?

PS: By convolution I meant the type of convolution that is used in image and signal processing. Stuff like this: http://www.songho.ca/dsp/convolution/convolution.html#convolution_2d

You can think of convolution as "flip and shift". Order doesn't matter - convolving A with B is the same as convolving B with A.

Correlation is convolution but without the "flip". Order in this case is important - correlating A with B is not the same as correlating B with A


Probability of choosing any real number in the range $[0,1]$? Basically I want to give a counter argument (if possible) to the statement:

"It is unlikely to randomly pick any real number in the interval $[0,1]$"

The statement implies that there is a small probability. So if you could either prove that the probability is $0$, or if you can prove that you cant define the probability in the first place, then that should be enough to make a counter argument?

I've done a bunch of reading and there seems to be multiple ways to tackle this, like measure theory, logical possibility, conceptual possibility etc.... Some guy said that the probability of choosing one specific number on a uniform distribution is $0$, but here we are talking about infinitely many numbers in the range $[0,1]$.

I'm assuming there is someone smarter than me here who knows this stuff, any help is appreciated.

If i understood correctly, then you are asking wheter there exists a uniform probability distribution $P$ over $\mathbb{R}$ and if so, what is $P([0,1])$.

The answer is no. Suppose however towards a contradiction, that such a measure exists.
Now by basic probability axioms we can write
$$P(\mathbb{R}) = \sum_{k\in \mathbb{Z}}P([k,k+1))$$
since $\{[k,k+1) \: : \: k\in \mathbb{Z}\}$ is a (countable) partition of $\mathbb{R}$. If the distribution is uniform, then $P([k,k+1))=P([n,n+1))$ for all $k,n \in \mathbb{Z}$, so we have two cases to cover:

$$P([k,k+1))=\alpha > 0  \Rightarrow P(\mathbb{R})= \sum_{k\in \mathbb{Z}} \alpha = \infty$$
and
$$P([k,k+1))= 0  \Rightarrow P(\mathbb{R})= \sum_{k\in \mathbb{Z}} 0 = 0$$
but both cases give a contradiction, since we assumed $P$ to be a probability measure.

It's simple. A single real number is a single point. [0,1] has lenght 1. [0,1/2] has lenght 1/2 so the probability of picking a number out of that interval is 1/2/1=1/2. A point has no (measurable) lenght. So the probability of picking it is 0 /1=0. Also picking out any rational number is zero, regardless of the fact that there is an infinite number of them, chances of landing on a rational number on a real line are 0. Because each single rational has on its own no measurable lenght. Sets like all rationals in  [0,1], have no defined lenght because they have discontinuities, lots of them, they are missing all the irrationals like pi, square of 2 etc. Therefore only reals are smooth and complete enough to have a meaningful definition of lenght defined on their intervals.


Let $f:[a,b]\to\Bbb R$ be a continuous function and$g:[a,b]\to\Bbb R, g(x)=\sup\{f(t):a\le t\le x\}.$ Show $g$ is a continuous function over $[a,b]$ I am a 31-year-old college student trying to study from some practice exam materials that my professor has given me. I will include an example of one of the problems I am struggling with.

Let $f: [a,b] \rightarrow  \Bbb R$ be a continuous function and define $$g: [a,b] \rightarrow \Bbb R, \quad g(x)= \sup\{f(t): a\leq t\leq x\}.$$

I am supposed to show that $g$ is a continuous function over $[a,b]$. I recall doing this by showing that $g$ is a monotone function or something along those lines but I am not quite sure how to proceed with this.
I am stuck with what it is exactly that I need to show to prove that g is continuous.

We need for all $\epsilon$ a $\delta$
$$|g(x)-g(x+h)|<\epsilon, |\delta|<h$$
$$\Leftrightarrow |\sup\{f(t): a\leq t\leq x\}-\sup\{f(t): a\leq t\leq x+h\}|<\epsilon, h<|\delta|$$
Immediately by continuity we have a $\delta$ such that $$|f(x)-f(x+h)|<\epsilon/2, h<|\delta|$$
$$...\leq|\sup\{f(t): a\leq t\leq x\}-\sup\{f(t): a\leq t\leq x\}|+|\epsilon/2|<\epsilon, h<|\delta|$$


Show $\frac{\sin x_1\sin x_2\cdots\sin x_n}{\sin(x_1+x_2)\sin(x_2+x_3)\cdots\sin(x_n+x_1)}\le\frac{\sin^n(\pi/n)}{\sin^n(2\pi/n)}$, for $\sum x_i=\pi$ Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that
$$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$
Show that
$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n
$$

This problem also post  MO,Until now No one solve it,I think there might be a solution here, because I 've heard that there are a lot of people here who are good at and like inequality, so the possibility of solving this inequality is very high, and I really look forward to them.

Update 1

1) I found that for $n=4, 5$, although there are infinitely many stationary points,
each stationary point is the global maximizer. In other words, the objective function is constant for all stationary points. Is this true for $n > 5$?

Take $n=4$ for example.
Let
$$g(x_1, x_2, x_3) = \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2} ,$$
\begin{align}
f(x_1, x_2, x_3) = \ln g(x_1, x_2, x_3)  &= \ln \sin x_1 + \ln \sin x_2 + \ln \sin x_3 + \ln \sin (x_1 + x_2 + x_3)\nonumber\\
&\qquad - 2\ln \sin (x_1+x_2) - 2\ln \sin (x_2 + x_3).
\end{align}
The stationary points are those feasible points with $\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0.$
I found that $$\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0 \Longrightarrow
g(x_1, x_2, x_3) = \frac{1}{4}.$$

In detail, we have
\begin{align}
\cot x_1 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) &= 0, \qquad (1)\\
\cot x_2 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) - 2\cot (x_2 + x_3) &= 0,\\
\cot x_3 + \cot (x_1 + x_2 + x_3) - 2\cot (x_2 + x_3) &= 0.
\end{align}
By letting $u_1 =\cot x_1, \ u_2 = \cot x_2, \ u_3 = \cot x_3$,
we have $(1) \Longrightarrow u_1u_2 + u_2u_3 + u_3u_1 - u_2^2 - 2 = 0$.
On the other hand, $g(x_1, x_2, x_3) - \frac{1}{4} = \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2} = 0.$

2) We can see this from another view.

Case $n=4$: Let $u_i = \cot x_i, \ i=1,2,3$. We have (noting that $x_4 = \pi - x_1-x_2-x_3$)
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)}\nonumber\\
=\ & \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2}\nonumber\\
=\ & \frac{(u_1u_2 + u_2u_3+u_3u_1-1)(1+u_2^2)}{(u_1+u_2)^2(u_2+u_3)^2}\nonumber\\
=\ & \frac{1}{4} - \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2}.
\end{align}

Case $n=5$: Let $u_i = \cot x_i, \ i=1,2,3, 4$. We have (noting that $x_5 = \pi - x_1-x_2-x_3-x_4$)
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_5)\sin (x_5+x_1)}\nonumber\\
=\ & \frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin (x_1+x_2+x_3+x_4)}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_1+x_2+x_3)\sin (x_2+x_3+x_4)}\nonumber\\
=\ & \frac{(u_1u_2u_3 + u_1u_2u_4 + u_1u_3u_4 + u_2u_3u_4 - u_1-u_2-u_3-u_4)(1+u_2^2)(1+u_3^2)}
{(u_1+u_2)(u_2+u_3)(u_3+u_4)(u_1u_2+u_2u_3+u_3u_1-1)(u_2u_3+u_3u_4+u_4u_2-1)}.\quad (2)
\end{align}
Denote (2) as $f(u_1, u_2, u_3, u_4)$.
It follows from $\frac{\partial f}{\partial u_4} = 0$
that $u_1 = g(u_2, u_3, u_4)$. Let $h(u_2, u_3, u_4) = f(g(u_2, u_3, u_4), u_2, u_3, u_4)$.
It follows from $\frac{\partial h}{\partial u_4} = 0$ that $u_2 = F(u_3, u_4)$.
Then $h(F(u_3, u_4), u_3, u_4) = \frac{5\sqrt{5}-11}{2}.$
Remark: Here $g, h, F$ are some rational functions whose expressions are not given, for the sake of simplicity.

Update

Proof of $n=4$:

We need to prove that
$$\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} \le \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4.$$
It suffices to prove that
$$(\sin(x_1+x_2))^2(\sin(x_2+x_3))^2 - 4\sin x_1 \sin x_2 \sin x_3 \sin(x_1+x_2+x_3) \ge 0.$$
Using substitutions $$\cos x_1 = \frac{1-w_1^2}{1+w_1^2}, \ \sin x_1 = \frac{2w_1}{1+w_1^2}, \ \cos x_2 = \frac{1-w_2^2}{1+w_2^2}, \ \sin x_2 = \frac{2w_2}{1+w_2^2}, \\
\cos x_3 = \frac{1-w_3^2}{1+w_3^2}, \ \sin x_3 = \frac{2w_3}{1+w_3^2},$$ the inequality becomes
$$\frac{16 Q^2}{(w_1^2+1)^2 (w_2^2+1)^4 (w_3^2+1)^2}\ge 0$$
where
\begin{align}
Q &= w_1^2 w_2^3 w_3+w_1^2 w_2^2 w_3^2-w_1 w_2^4 w_3+w_1 w_2^3 w_3^2-w_1^2 w_2^2-w_1^2 w_2 w_3-w_1 w_2^3-6 w_1 w_2^2 w_3\nonumber\\
&\qquad -w_1 w_2 w_3^2-w_2^3 w_3-w_2^2 w_3^2+w_1 w_2-w_1 w_3+w_2^2+w_2 w_3.
\end{align}
We are done.
Remark: We can prove $n=4$ without using above substitutions. For $n=5$, it is not so simple.

Previously written

This is not an answer. I want to point out that for $n=4, 5$, there exist infinitely many feasible points such that equality occurs. In other words, if the inequality holds, there exist infinitely many global maximizers.

1) $n=4$.

Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
(\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2 = 0.
\end{align}
Remark: We may solve $x_1$ from (1), that is, $x_1 = \mathrm{arccot}\frac{\sqrt{2}-\cos x_2}{\sin x_2}, \ x_2 \in (0, \frac{\pi}{2}).$

Let $x_3 = x_1,\ x_4 = \pi - x_1 - x_2 - x_3.$ We have $x_1, x_2, x_3, x_4 > 0; \ x_1 + x_2 + x_3 + x_4 = \pi$ and
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} - \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4\\
=\ & \frac{(\sin x_1)^2\sin x_2 \sin (2x_1 + x_2)}{(\sin (x_1+x_2))^4} - \frac{1}{4}\\
=\ & - \frac{((\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2)^2}{4(\cot x_1 + \cot x_2)^4}\\
=\ &0.
\end{align}

2) $n = 5$.

Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
-4(\cot x_2)^2(\cot x_1)^2 + (-2(\cot x_2)^3 + 6\cot x_2)\cot x_1 + (\cot x_2)^4  + 4(\cot x_2)^2 - 1 = 0.
\end{align}
Let $y_1 = \cot x_1, \ y_2 = \cot x_2$. We have $-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1
+y_2^4+4y_2^2-1 = 0$ which results in $\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3 = 0$
since $x_1, x_2 \in (0, \frac{\pi}{2}).$

Let $x_3 = x_2, \ x_4 = x_1, \ x_5 = \pi - x_1 - x_2 - x_3 - x_4.$
We have $x_1, x_2, x_3, x_4, x_5 > 0; \ x_1 + x_2+x_3+x_4+x_5=\pi.$
Note that $\big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\big)^5 = \frac{5\sqrt{5}-11}{2}$. We have
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}
{\sin (x_1+x_2) \sin (x_2 + x_3) \sin (x_3 + x_4) \sin (x_4+x_5) \sin (x_5+x_1)}
- \Big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\Big)^5\\
=\ &\frac{(\sin x_1)^2(\sin x_2)^2\sin (2x_1 + 2x_2)}{(\sin (x_1+x_2))^2\sin 2x_2 (\sin (x_1 + 2x_2))^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &-\frac{5\sqrt{5}-11}{16}\frac{
(\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\sqrt{5}+1)(\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3)^2}
{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\
=\ &0.
\end{align}

This is just too long for a comment .

I can prove a particular case of the general case of a variant using Jensen's inequality .

    Let $x_i>0$ with $1\leq i\leq n$ such that $\sum_{i=1}^{n}x_i=\pi$ with $|x_i-x_{i+1}|=\epsilon$  and $n=2q+1$ where $q$ is a natural number then we have:
    $$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(\frac{2\pi}{n}+n\epsilon)}}\le\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n
$$
    Where $\frac{\pi}{n}-q\epsilon>0$

Proof we can rewrite the inequality as :

$$
\dfrac{\sin{(\frac{\pi}{n}-q\epsilon)}\sin{(\frac{\pi}{n}-(q-1)\epsilon)}\cdots\sin{(\frac{\pi}{n})}\cdots\sin{(\frac{\pi}{n}+q\epsilon})}{\sin{((\frac{2\pi}{n}-(2q-1)\epsilon))}\sin{((\frac{2\pi}{n}-(2q-3)\epsilon))}\cdots \sin(\frac{2\pi}{n}+n\epsilon)}\le\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n
$$

We make the function $$f(x)=\frac{\sin{(\frac{\pi}{n}-q\epsilon+x)}}{\sin{(\frac{2\pi}{n}-(2q-1)\epsilon+2x)}}$$

The function $$g(x)=\ln(f(x))$$ is concave on the  interval $[0,2q\epsilon]$ provided that $\sin{(\frac{\pi}{n}-q\epsilon+x)}\leq \sin{(\frac{2\pi}{n}-(2q-1)\epsilon+2x)}$  remains to apply the Jensen's inequality to the function we get :

$$g(0)+g(\epsilon)+g(2\epsilon)+\cdots+g(2q\epsilon)\leq \ln\left(\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n\right)$$

So we have showed :

$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le \frac{\sin{(\frac{2\pi}{n}+n\epsilon)}}{\sin{(\frac{2\pi}{n})}}
\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n $$

Furthermore I have checked numerically that we have :

$\exists \,\epsilon>0$ such that :

$$ \frac{\sin{(\frac{2\pi}{n}+\epsilon)}}{\sin{(\frac{2\pi}{n})}}
\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n \le \left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n})}}\right)^n$$

If we swap the variables we get $n!$ functions of the kind $f(x)$ wich is a difficulty .

Another way

I recall the Levinson's inequality (Theorem B):

    Let $f:I=[a,b]\to \mathbf{R}$ be a $3$-convex function , and $a_n$ , $b_n$ $\in [a,b]$ for $n=1,2,\cdots,n$ such that :
    $$\operatorname{max}(a_1\cdots a_n)\leq \operatorname{min}(b_1\cdots b_n), a_1+b_1=a_2+b_2=\cdots=a_n+b_n$$
    Then :$$\frac{\sum_{i=1}^{n}f(b_i)}{n}-f\Big(\frac{\sum_{i=1}^{n}b_i}{n}\Big)\geq \frac{\sum_{i=1}^{n}f(a_i)}{n}-f\Big(\frac{\sum_{i=1}^{n}a_i}{n}\Big)$$

A part of the initial inequality is an application of the Levinson Theorem since we have :

$f(x)=\ln(\sin(x))$  and $f'''(x)\geq 0$ on $]0,\frac{\pi}{2}]$

And $0<b_i=x_i+x_{i+1}\leq \frac{\pi}{2}$ and $0<a_i=x_i<\frac{\pi}{2}$

With the condition :$$\operatorname{max}(a_1\cdots a_n)\leq \operatorname{min}(b_1\cdots b_n), a_1+b_1=a_2+b_2=\cdots=a_n+b_n$$


Suppose $\theta \neq \frac pq * \pi$. Show ${\{e^{in\theta} : n \ \epsilon \ N}\}$ is dense in $S^1=\{x + iy: x^2 + y^2 = 1\} \subseteq C$ Essentially, I need to show that e^inx is dense in the complex unit circle for irrational angle x. I'm not sure how to go about this but I think I need to show that between any two points on the unit circle, we can find some e^inx.

Its so simple.$$exp(jx)=cos(x)+jsin(x)$$. Since $$sin^2(x)+cos^2(x)=1$$ this complex number always is on the circumference of a circle with radius 1


A question on multinomial theorem using binomial theorem $(3x^2+2x+c)^{12}=\sum A_r x^r$

and

$\frac{A_{19}}{A_5}=\frac 1 {2^7}$

Find $c$. I really have no idea what to do with this. This was on a test. I have studied only binomial theorem. So, please use it only(combinatorics approach is always welcome).


Numbers with no finite representation on paper It occurred to me that there must be a lot of numbers without any form of finite representation on paper. Is there a name for these numbers?

For example...

Integers and rationals have a very simple representation e.g. 3/4

Irrational numbers obviously can also have a finite representation:
1.41421356... can be written as "the solution to the equation x^2 = 2"

Transcendental numbers can also have a finite representation:
e can be written as "the limit of (1 + 1/n)^n as n approaches infinity"

In other words, with a finite amount of effort one can give the reader enough information to calculate the value of the specified number exactly (to any degree of accuracy the reader chooses)

However, there must be a lot of numbers where this simply is not possible.

Consider the number 1.2736358762987349862379358... where this is just a string of (genuinely) random digits. There is no way to provide a finite definition that will specify this number to an arbitrary degree of accuracy. Similarly, there is no equation to which this number is a solution (I think, although I don't know how one would prove this).

Does this mean there are "gaps" in the real numbers. The number above is definitely somewhere between 1.2 and 1.3 but there is no way I can specify the value of this number (without writing an infinite number of digits). The number exists on the number line but I will never be able to do anything with it.

Is there a name for these numbers? Can anyone point me to some interesting resources on this topic?

I'm only asking as an interested hobbyist so apologies if this question isn't very scientific.

That depends on what you mean by "representation." One way to cash this out is to talk about the definable numbers. These are more general than the computable numbers, but they are still countable because there are still only countably many possible descriptions of a number in a language over a finite alphabet.

I think what you need is this Completeness Theorem:

Each contracting sequence of intervals has a common point; that is, the intersection of all the intervals of the family is not empty.

The basic idea is if the real number $c$ in the proof (below) is the number that you want a finite representation for (your randomly defined number for example), the construction given in the proof represents a finite way of defining the contracting sequence of intervals such that $c$ is in the intersection of all of them.

Here's the proof given in First Concepts of Topology, by Chinn and Steenrod.

Definition. an infinite sequence of closed intervals of real numbers $I_0, I_1, ..., I_n$ is called a contracting sequence if each interval contains the next and hence all intervals that follow: $$I_0 \supset I_1 \supset I_1 \supset ... \supset I_n \supset ...$$
A contracting sequence is called a regularly contracting sequence if $I_0 = [m, m+ 1]$ is the interval from an integer $m$ to the next integer $m + 1,$ and, for each $n >= 1$ $I_n$ is one of the intervals obtained by partitioning $I_{n-1}$ into ten equal parts. Thus $I_1$ is a tenth part of $I_0,$ $I_2$ a tenth part of $I_1$ and so on. The length of $I_n$ is $10^{-n}.$

Consider first the case of a regularly contracting sequence. For each $n = 0, 1, 2, ...$ let $a_n$ denote the left-hand endpoint of $I_n.$ Then $a_0 = m$ is an integer. Since $I_n$ has length $10^{-n}$, $I_n = [a_n, a_{n+10^{-n}}].$ The points that divide $I_{n-1}$ into ten equal parts are

$$a_{n-1}, a_{n-1} + \frac{1}{10^n}, a_{n-1} + \frac{2}{10^n}, ..., a_{n-1}+\frac{10}{10^n}.$$

Since $I_n$ is one of these ten subintervals, its left endpoint must be $$a_n = a_{n-1} + \frac{k_n}{10^n},$$ where $k_n$ is one of the digits $0, 1, 2, ..., 9.$ In this way the sequence of intervals determines an integer $m$ and an infinite sequence of digits $k_1, k_2, ..., k_n, ....$ Let $c$ denote the real number
$$c = m + \frac{k_1}{10} + \frac{k_2}{10^2} + ... + \frac{k_n}{10^n} + ....$$
(that is, the decimal expansion of $c$ is $m.k_1k_2,...$).

Since $$a_n = m + \frac{k_1}{10} + \frac{k_2}{10^2} + ... + \frac{k_n}{10^n},$$ it is clear that $a_n <= c$. Now the decimal expansions of $a_n + \frac{1}{10^n}$ and of $c$ agree out to the n'th digits, but the n'th digit of $c$ is $k_n$ while that of $a_n + \frac{1}{10^n}$ is $k_n + 1$. It follows that $$a_n <= c <= a_n + \frac{1}{10^n}.$$

These inequalities assert that $c \in I_n$ and since they hold for every integer $n,$ it follows that $c$ lies in each interval of the sequence, hence in the intersection of all of them. This proves the theorem in case the sequence contracts regularly.

The case where the sequence does not contract regularly is similar, and the same book gives a proof. I am not going to enter it here right now, but would be happy to put the second part in also.


Any I.i.d sequence of random variables are uniformly integrable This is an exam question. My teacher says it is true. But I'm thinking it might be false for random variables such as Cauchy. Can anyone confirm that this is false in general?

I think Cauchy distribution would be fine because it is bounded. In general I think it is true, you can take the maximum point of the distribution to use with your delta in your epsilon-delta proof.

Also, given that all the distributions are the same, you can of course bound the absolute integrals. The absolute integrals are the same as the usual integral as the pdfs are positive functions.


Modulus of inverse of a number In the Chinese Remainder Thm., $y_i$ is the modular inverse of $\frac{P}{p_i}$, where $p_i$ is the $i$th modulus in a set of $n$ congruencies and $P$ is $\prod_{i=1}^np_i$ (right?). So if I calculated $P$ to be 420 and $p_1=3$, why is $y_1\equiv140^{-1}\mod3 = 2$? Is $140^{-1}$ in this case not $\frac{1}{140}$? I would think $140^{-1}\mod 3$ would be a non-integer, namely $\frac{1}{140}$.

The notation $a^{-1}$ just means the multiplicative inverse, meaning a number such that if I multiply it by $a$ I will get the multiplicative identity. So in the real numbers $a^{-1}$ just happens to be $1/a$ so when I multiply $a$ with $1/a$ then I get 1. When doing modular arithmetic, the definition of $a^{-1}$ is the same but writing it as a fraction doesn't make sense. So working modulo 3, first note that $140=2$ so $140^{-1}=2^{-1}$ and which number multiplied by 2 gives you 1 mod 3? It is 2 because 2 is its own inverse mod 3.

So $140^{-1}=2^{-1}=2$ mod 3. Some people choose to give a meaningful interpretation to fraction in modular arithmetic too so you can say 1/2 mod 3 but remember that it really means $2^{-1}$. This is what Don Antonio is doing.


If one egg is found to be good, then what is the probability that other is also good? A basket contains 10 eggs out of which 3 are rotten. Two eggs are taken out together at random. If one egg is found to be good, then what is the Probability that other is also good?

I applied conditional probability. It says that one of them is good, so the probability of the other one being good can be found in the 9 eggs left out of which 6 are good, so Probability = $6/9$

Am I right with my understanding?

Am I right with my understanding?

You're overlooking a subtle point here, one that is often overlooked. You do get the right result, but due to a bit of luck.

The problem is you're overlooking a choice — in addition to having selected a pair of eggs from the basket, you are also choosing to assign the labels "one" and "other" to the two eggs.

Correctly modeling how that choice is made extremely important.

Fortunately, the problem surely means for the choice to be made in the easier-to-understand fashion: either one of the following two equivalent models was used

The labels were assigned randomly, with each of the two choices having equal probability
The eggs were selected not as a pair, but one at a time, and the first egg selected being called "one" and the second egg selected being called "other"

The reason that this is surely right is that the problem appears to indicate either of the following typical procedures, which directly correspond to the two bullet points above:

Take a pair of eggs from the basket
Pick one of the two eggs
Check whether the chosen egg is good or rotten
Guess whether the other is good or rotten

or

Take one egg from the basket
Check whether it is good or rotten
Take another egg from the basket
Guess whether it is good or rotten

An example of a different procedure requiring a different model, incidentally, could be described as follows:

Take a pair of eggs from the basket
Your colleague checks whether they are good or rotten
If both are rotten, things stop here. Otherwise, your colleague tells you one is good
You ask your colleague to point to a good egg.
You now guess whether the other egg is good or rotten.

(the actual problem being modeled doesn't have to include a colleague; that character can just be a fictional entity that enables the analysis to actually pick out a good egg)

I think it's $\frac{2}{7}$ if you pick them together at the same time and not put them back in the basket.


Matrix transformation $f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ I am self-learning Linear Algebra (mainly through Khan Academy and Youtube). I encountered this linear transformation, and was wondering how would this be solved.
$f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$
$f \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $  = $ \begin{pmatrix}
4 & 1 & 3\\
2 &-1 & 3\\
2 & 2 & 0
\end{pmatrix} $ $\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $
Establish $x, y$ and $z$ such that,
$f\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $ = $\begin{pmatrix}
4\\
5\\
-1
\end{pmatrix} $
Do I just need to multiply the values of $f$ for the $3\times3$ matrix? What does this mean overall?

You need to solve the non-homogeneous system
$$ \begin{pmatrix}
4 & 1 & 3\\
2 &-1 & 3\\
2 & 2 & 0
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =  \begin{pmatrix}
4\\
5\\
-1
\end{pmatrix}$$
Form then the augmented coefficients matrix and reduce it by rows:
$$ \begin{pmatrix}
4 & 1 & 3&:&4\\
2 &-1 & 3&:&5\\
2 & 2 & 0&:&-1
\end{pmatrix}\stackrel{R_1\leftrightarrow R_3\cdot\left(\frac12\right)}\longrightarrow\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
2 &-1 & 3&:&5\\
4 & 1 & 3&:&4\\
 \end{pmatrix}\stackrel{R_2-2R_1\\R_3-4R_1}\longrightarrow$$
$$\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
0 &-3 & 3&:&6\\
0 & -3 & 3&:&6\\
 \end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
0 &-3 & 3&:&6\\
0 & 0 & 0&:&0\\
 \end{pmatrix}$$
Well, there are infinite solutions to the above system: begin with row $3$ , and remember the columns represent $\;x, y, z\;$ from left to right, so:
$$R_2:\;\;-3y+3z=6\implies y=z-2\\R_1:\;\;x+y=-\frac12\implies x=-\frac12-y=-\frac12-z+2=\frac32-z$$
Now just choose a nice $\;z\;$ , say $\;z=0\implies y=-2\,,\,\,x=\cfrac32\;$ , and one solution to your problem is
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
3/2\\
-2\\
0
\end{pmatrix}$$
as you can easily check.

You are given that $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4 & 1 & 3 \\ 2 & -1 & 3 \\ 2 & 2 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}$.
Of course, $\begin{pmatrix}4 & 1 & 3 \\ 2 & -1 & 3 \\ 2 & 2 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.
So we have $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.
That gives us three equations:
$$x= 4x+ y+ 3z\quad\text{which reduces to}\quad 3x+ y+ 3z= 0.$$
$$y= 2x- y+ 3z\quad\text{which reduces to}\quad 2x- 2y+ 3z= 0.$$
$$z= 2x+ 2y\quad\text{which reduces to}\quad 2x+ 2y- z= 0.$$
An obvious solution to that is $x= y= z= 0.$
Notice that the determinant is
$$\begin{align*}
\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & -2 & 3 \\ 2 & 2 & -1 \end{array}\right|&= 3\left|\begin{array}{cc} -2 & 3 \\ 2 & -1\end{array}\right|- \left|\begin{array}{cc}2 & 3 \\ 2 & -1\end{array}\right|+ 3\left|\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right|\\[5pt]
&= 3(2- 6)- (-2- 6)+ 3(4+ 4)\\[5pt]
&= -12+ 8+ 24\\[5pt]
&= 20\\[5pt]
\end{align*}$$
Since that is not $0, x= y= z= 0$  is the only solution.


Proving that the closure of a closed formula is the formula itself Closed Formula. A formula (in a first order language) is said to be closed if it has no free variables.
Closure of a Formula. Let $A[x_0, . . . , x_{n−1}]$ be a formula (in a first order language) whose free variables are among
$x_0, . . . , x_{n−1}$ and $x_{n−1}$ is free in $A$, where $x_0, . . . , x_{n−1}$ are the first $n$ variables in alphabetical order. We call
$\forall x_{n-1}\ldots\forall x_1\forall x_{0}A$
the closure of $A$.

Question.
Using these definitions, how can I show that if $A$ is closed, it is its own closure?

I thought that I will begin by assuming that $Q$ is the closure of $P$. But I don't understand what exactly it is that I need to show. Do I need to show that $P\leftrightarrow Q$ or that they are syntactically exactly same formula or something else? In any case I don't how to proceed. Can anyone help?

According to the definition, if the formula $A$ is closed, it has no free variables.

Thus, there is no $n$ such that:

  the free variables of $A$ are among $x_0, \ldots,x_{n−1}$ and $x_{n−1}$ is free in $A$.

Thus, the closure of $A$ has no "added" universal quantifiers upfront, which means that it is $A$ itself.

If we have a formula $A$, like $x_2=c_1$, the definition gives us a "recipe" to produce its closure (in this case we have $n=3$):

  $∀x_2∀x_1∀x_0(x_2=c_1)$

because we have to prenex the universal quantifiers for the first $n$ variables in alphabetical order.

If instead $A$ is $∃x_2(x_2=c_1)$, and we apply the definition:

  "let $A[x_0,\ldots,x_{n−1}]$ be a formula whose free variables are among $x_0,\ldots,x_{n−1}$ and $x_{n−1}$ is free",

we get no $n$, because neither $x_2$ nor $x_1$ nor $x_0$ are free.

Thus, we have to "prenex" nothing and we get: $∃x_2(x_2=c_1)$ which is $A$ itself.

Note : in this case, it is not correct to say that $∀x_2∀x_1∀x_0(∃x_2(x_2=c_1))$ is the closure of $∃x_2(x_2=c_1)$.

Of course, the two are equivalent, because we can prove that:

  if $x$ does not occur free in $\alpha$, then $\vdash \alpha \leftrightarrow ∀x \alpha$.

You need to show that the formulae are syntactically equal.

The closure of a formula $\phi$ is the formula $\forall x_1 \ldots \forall x_n.\phi$ where $fv(\phi) = \{ x_1, \ldots, x_n \}$. If $\fv(\phi) = \emptyset$, then we of course will not prefix $\phi$ with any quantifiers, and the result is obvious just $\phi$.


Inequality Problem of examination Show that if $n > 2$, then $(n!)^2 > n^n$.

I cannot find any way out to these.Please help.

I tried break $n!$ and then argue but failed.

For $n=k, (k!)^2>k^k$

For $n=k+1,\{(k+)!\}^2=(k+1)^2\cdot(k!)^2>(k+1)^2\cdot k^k$

It is sufficient to prove $(k+1)^2\cdot k^k>(k+1)^{k+1}\iff k+1>\left(1+\frac1k\right)^k$

But $\left(1+\frac1k\right)^k<3$ See this and this

$$(n!)^2=1(n)\times2(n-1)\times3(n-2)\times \cdots n(1)$$
But
$$n\ge a+1 \implies a(n)\ge a(a+1)\implies (a+1)(n-a)\ge n$$
And equality is achieved iff $a=0$ or $a=n-1$. Since for $n>2$ we have $a=1\neq n-1$ in our product, we have the strict inequality $(n!)^2>n^n$.


Traveling Salesman with paths instead of points I have a rectangular area filled with vector paths (an SVG document, to be precise). Starting at the origin, I need to visit every part of every path.

For an open path, like a line or an arc, I would most often walk to one end of the path, walk the length of the path, then walk to another path.

For a closed path, like a circle or a square, I would most often walk to a point on the path near my previous stop, walk the perimeter of the path, then walk to another path.

I say "most often" above because it would also be valid to visit a path, not necessarily at an endpoint, walk along part of it, then leave to another path, later coming back to finish the abandoned walk.

The purpose of this problem is to minimize total travel distance for a laser cutter or CNC mill that needs to cut along all the paths.

Are there any existing algorithms (or open source tools) for solving this problem?

EDIT: Two paths never cross. Anywhere that might happen will have been altered from 4 vertices and 2 paths to 5 vertices and 4 paths by a previous step in my workflow.

EDIT2: The set of paths is disjoint. Some steps in the solution will be between pieces of disconnected paths.

As far as I see, taking all crosspoints as vertices and all parts of pathes as edges we come to the Chinese postman problem. Unlike the Travelling salesman problem the Chinese postman problem is polynomially solvable. You can find $O(n^3)$ solution here.


This is regarding a verbal math problem A Shopkeeper buys sweets at 11 for a rupee. He bought an equal number of sweets at 10 for a rupee. he sold all the sweets at 8 for a rupee. find loss or gain %

Please solve this for me. I am not able to understand this statement. Can you also explain the meaning of this problem statement ?

My previous answer was totally wrong and I apologize. Let me start again.

Per unit, the shopkeeper pays $\frac{1}{11}$ for the first kind of sweets and $\frac{1}{10}$ for the second kind of sweets. This makes an average value of $\frac{21}{220}$ when he buys. When he sells the sweets, he charges $\frac{1}{8}$ per unit. The difference is then $\frac{1}{8}-\frac{21}{220}=\frac{13}{440}$ and this is the profit per unit. Then, dividing the profit per unit by the price paid per unit, this gives $\frac{13}{42}$ which is almost $31$ percents.

He bought each sweet for 100 paisa/10 = 10 paisa;
he sold each sweet for 100 paisa/8 = 12.5 paisa.

So the gain% = ( (buying price - selling price) * 100 ) / selling price.


How does chaos arise in Hamiltonian systems? I have a question about how chaos arises in Hamiltonian systems. I've been taking a upper year undergrad class on dynamics and it has focused on chaos a lot. So far however we have only seen the more standard signatures of chaos. We've studied the Lorenz system and the damped driven pendulum which are both dissipative. As a result they have strange attractors and a negative sum to their Lyaponov exponents.

All this falls apart for me when I think about Hamiltonian systems and more specifically Louisville's Theorem. If volumes in phase space are conserved for Hamiltonian systems, how does chaos arise? The solutions cannot fall into a strange attractors and I expect the Lyaponov spectrum to be symmetric.

I have read answers that make references to KAM theory, but I don't understand how that is relevant. KAM theory seems to deal with when perturbations preserve toriodal phase spaces, but if the perturbed solutions become chaotic there still lies the question as to whether or not that is in fact a solution. Or am I missing something that has to do with perturbation theory?

Let me not claim that any system that can transit into Chaos must be "Hamiltonian". But there is a very large class of such systems, to which you probably intuitivley refer, and those are systems that can be represented at microscopic level by Fokker Planck equations. Many classical cases of Chaos are of this family, such as in fluid dynamics, pattern formation, chemical reaction, mechanics, laser theory...

The Fokker Planck equations, roughly said, resemble somehow partition functions encapsulating the Hamiltonian. This is when you could identify, for instance, a potential.

Let me suggest you two books that will help you to learn this aspects in detail:

Stochastic Methods: A Handbook for the Natural and Social Sciences

Synergetics: Introduction and Advanced Topics

While the first will help you to learn about the Fokker-Planck equations on microscopic level the second helps you to understand the connect to how Chaos arises in such Hamiltonian systems via phase transition. The second book however not only describes Hamiltonian systems.

In general to your question, how chaos arises in Hamiltonian systems, there ara several factors some of the we collected in the following discussions:

>>>here

and >>>here

What is very basic requirement, is that your non-linear system has at least 3 instable eigen-modes while all other eigen-modes stable. Once you vary the parameters (eigenvalues) of the instable eigen-modes phase transition and bifurcations of the stable solutions of the system can occur in some cases where system behaves then chaotic. That is if for instance a system before phase transition has two instable eigen-modes and after the phase transition one more parameter (eigenvalue) gets instable and the system changes to a phase where 3 eigen-modes are instable. However this is just a scetch, and you need to rigorousely read into the stuff.


Language of a grammar vs regular expression vs nfa I read some note about Automaton Course. i see this note, that following all is the same. but i think the L(g) is not equal to NFA and regular expression. anyone could help me with defining the language of this figures (nfa, regular expression and grammar):

Note that $(a \mid b)^{*}$ means $\{\varepsilon, a, b, aa, bb, ab, ba, \dots \}$, so $(bb \mid b)^{*} = \{\varepsilon, b, bb, bbb, \dots \} = b^{*}$. That leads to

$$R = (a \mid b) (bb \mid b)^{*} = (a \mid b) b^{*}$$

So the language created by $R$ is $L(R) = \{ a^lb^k \mid k \in \mathbb{N}_0, 0 \leq l \leq 1\}$.

Converting your NFA to an equivalent DFA we get

Now we can see that $M$ accepts all words of the type $xy$ where $x$ can either be $a$ or $b$ (but not nothing), and $y = b^{*}$ (including $y = \varepsilon$). That means $L(R) = L(M)$.

Finally, let's take a look at $G$:

\begin{align*}
S &\rightarrow aAB \mid bAb \\
A &\rightarrow bbA \mid b \mid \varepsilon \\
B &\rightarrow bB \mid bAB \mid \varepsilon
\end{align*}

Case 1:
$S \rightarrow aAB$: We can derive the $A$-part with either nothing, one $b$ or as much $b$'s as we like. We can derive the $B$-part with either nothing, as many $b$'s as we like or go the $A$-part.

Case 2: $S\rightarrow bAb$: In the $A$-part we can derive nothing, one $b$, or as many $b$'s as we like. This, how you pointed out, leads to the fact the the word $b$ wouldn't be accepted.

So $G$ is equivalent to:

\begin{align*}
S &\rightarrow aT \mid bTb \\
T &\rightarrow bT \mid  \varepsilon
\end{align*}

which leads to $L(G) = \{a^l b^kb \mid k \in \mathbb{N}, 0 \leq l \leq 1\} \neq L(M) = L(R)$.

The regular expression tells us that the language accepted is either an 'a' and then any number of 'b's or a non-zero number of 'b's. It's not hard to see that any such string will be accepted by the NFA (after 'ab' or a single 'a' we have states 2 and 3 active in the NFA and then the bs just cycle them around.) Additionally, the empty string isn't accepted. After the first letter we can never again be in state 1, and no other state has an outgoing 'a' edge. This checks out.

As for the CFG: 'A' can clearly become any number of 'b's. It is the smallest set containing 0, 1 and 2 + any number it contains. B can similarly be any number of 'b's. But, it seems you are right. S can become 'a' followed by any number of 'b's or 2 + any number of 'b's. We want only one 'b' to be required. Perhaps this is a typo?