How to parametrise a circle in a plane I've been studying for an exam and in one of the exams it asks me the parametrise the unit circle in $R^3$ which lines in the plane $$x+y=0$$.

I know how to parametrise a normal circle, but I don't really have any idea how to parametrise it in a plane like that. Any help would be appreciated, thanks

The unit sphere has equation $x^2+y^2+z^2=1$, its intersection with the plane is the circle you are looking for. So you have $$y=-x$$ and $$2x^2+z^2=1$$. You can now use polar coordinates so put $x=-y={1 \over \sqrt 2} \cos t$ and $z=\sin t$ so that  $$2{1 \over  2} \cos^2 t+ \sin^2 t=1$$ is identically verified.

Here is what I think. The way to derive a circle is to set up the identity of distance, i.e. the square of the distance of a point on the circle to its center (1 for unit circle) is equal to the sum of squares(Pythagorean theorem). If you are given the points on $x+y=0$ plane, then the setting of such points would be limited to $(x,-x,z)$. Do the same thing. Since now you are guaranteed the points in a plane, it is for sure a circle, not a sphere.


How should I go about obtaining the explicit solution to this logistic first-order nonlinear ordinary differential equation? I have to find the explicit solution to this harvesting problem in a population model where $\frac{dN}{dt}=rN(1-\frac{N}{K})-H(N)$ such that $H=qEN$, subject to initial condition $N(0)=N_0$. Here $N=N(t)$ and $r$,$K$,$q$ and $E$ are positive constants. I also have to deduce the two limiting behaviors as $t$ approaches infinity. So far I have broken down the equation to $\frac{1}{N}\frac{dN}{dt}+\frac{r}{k}N=(r-qE)$. Any hints on how to do the problem would be very much appreciated. I'm having trouble understanding what they mean by finding the explicit solution and deducing two limiting behaviors.

First of all, "finding the explicit solution" means finding an equation of the form $N=N(t)$ that satisfies your differential equation. "Deducing the limiting  behaviors" (I think) means determining what happens to the population/harvest if the initial amount of plants (in your case) is below vs. above the "carrying capacity".

The logistic differential equation is separable, so you just separate the variables, use partial fraction decomposition, and solve.


Find the determinant of $n\times n$ matrix Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
      -x & a_2&a_3&a_4&\cdots &a_n\\
      a_{1} & -x & a_3&a_4&\cdots &a_n\\
      a_1&a_{2} & -x &a_4&\cdots &a_n\\
      \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
      a_1&a_{2} & a_3&a_4&\cdots & -x\\
      \end{array}\end{bmatrix}$, then how to find the $\det (M)$?
Proof: First I started by taking the $a_i$ from each  $i$-th columns, then$ |M|=\prod_{i=1}^{n}{a_i} \begin{vmatrix}\begin{array}{ccccccc}
      \frac{-x}{a_1} & 1&1&1&\cdots &1\\
      {1} & \frac{-x}{a_2} & 1&1&\cdots &1\\
      1&1 & \frac{-x}{a_3} &1&\cdots &1\\
      \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
      1&1 & 1&1&\cdots & \frac{-x}{a_n}\\
      \end{array}\end{vmatrix}$. After this is there any easiest way to find the determinant.

Subtract the first row from each of the other rows.  Most of the terms are now zero, and you can expand across the first row.  Each product misses one of the $(x+a_i)$ factors, replaced by $a_i$.  So the determinant is
$$(-1)^n\prod_i(x+a_i)\left[\frac x{x+a_1}-\frac {a_2}{x+a_2}-\frac {a_3}{x+a_3}...\right]\\
=(-1)^n\prod_i(x+a_i)\left[1-\sum_i\frac{a_i}{x+a_i}\right]$$

Hint: $\det(M)$ is a polynomial in $x$ and $M$ is clearly singular if $x=-a_k$ for some $k$


Circle theorem/triange angle question I am doing practise papers and there is one question I cannot understand even with the mark scheme. I have added the pictures below:

Question (with added annotations):

Mark scheme:

The question is to find angle x (23 degrees, apparently).

At first I thought that angle OCD might be 69 degrees because of the alternate angle theorem, but I realised that it wasn't on a tangent so that wouldn't work. Other than that I have no idea - could someone please explain (in detail if possible, maths isn't my strength!) the steps to find it? Thanks in advance!

Steps hinted (justify/prove each step in the following):

The triangle $\;\Delta OCD\;$ is isosceles (why?), and thus $\;\angle OCD=\angle ODC\;$, and since $\;\angle DOC\;$ is the other angle , it must be that

$$111^\circ-x^\circ=\angle DOC=180^\circ-2\angle OCD$$

But $\;\angle OCD\;$ is an exterior one to $\;\Delta OCE\;$ and thus $\;\angle OCD=2x^\circ\;$ (why?), so we get

$$111^\circ-x^\circ=180-4x^\circ\implies3x^\circ=69^\circ\implies x=23^\circ$$

To get the $2x$:
$180-(180-2x)$ is obviously $2x$ so $OCD = 2x$ and $CDO = 2x$ because angles in the same segment are equal (the segment being the semcircle).


How to prove that $O(n) \cdot O(n^2) = O(n^3)$? Would the solution be something like

let $fn \in O(n)$ so $ fn \le c1 \cdot n$  and  $ gn \in O(n^2) $ so $ gn \le c2 \cdot n^2$ where $c1,c2$ constants

so $ fn*gn \le C \cdot n^3$ so $ fn*gn \in O(n^3)$ so $O(n) \cdot O(n^2) = O(n^3)$

Yes, you are exactly right.

Maybe even $$fn \cdot gn \leq c_1 n \cdot c_2 n^2 \leq C n^3 = O(n^3)$$ for more clarity.

note: When speaking of complexity classes, the equality sign is often used insted of $\in$ which would be formally more correct.

If you also need to prove $O(n^3) \subseteq O(n) \cdot O(n^2)$, the argument would go the same similarly (see comments below for details).

Recall that

$$O(n)=\omega_1(n)\cdot n \quad \limsup_{n\to \infty} |\omega_1 (n)|=c_1\in \mathbb{R}$$

$$O(n^2)=\omega_2(n)\cdot n^2 \quad \limsup_{n\to \infty} |\omega_2 (n)|=c_2\in \mathbb{R}$$

therefore

$$O(n) \cdot O(n^2)=\omega_1 (n)\cdot\omega_2(n)\cdot n^3=\omega_3(n)\cdot n_3=O(n^3)$$

indeed

$$\limsup\left|\omega_3(n)\right|\le\limsup\left|\omega_1 (n)\right|\cdot \limsup\left|\omega_2(n)\right|=c_1\cdot c_2=c_3\in \mathbb{R}$$


What is the statistical equilibrium for this simulation of happy bubbles? Happy Bubbles

I hope this is not too specific or practical, but I just made a simulation of sorts and seem to have hit quite close to an equilibrium (by accident). Now I am wondering if and how you would calculate the exact function or exact value for this situations statistical "sweetspot".

The Problem

I was playing around with D3.js and created a canvas with randomly appearing bubbles with a 5% chance every iteration. Also every iteration bubbles touching any border will disappear. This can happen because the bubbles move around every iteration.

The Parameters

Starting with a single bubble
Every bubble has a x- and a y-coordinate which represents the center of the bubble
Bubble Size: 20px radius
Iterationlength: 100 ms
Canvas Size: 600px by 600px
Bubble Speed: -5px to 5px on X-Axis and Y-Axis

(Math.random() * 10) - 5 (the random number is a floating point number between 0 (including) and 1 (excluding))

Starting position per bubble: Math.random() * 600 for x and y coordinate (seperately)
Bubble pops when it gets within 20px of any outer edge (again using the center as reference)

The Equilibrium

With these Parameters new bubbles and disappearing bubbles seem to cancel each other out. The number of bubbles ranges from 100 to 130 but has not breached those limits in more then an hour.

The Question

Do these Parameters ensure such an equilibirum or is it nothing special, that the number seems to be rather stable? How would one calculate this?

Caveats

One I can think of is that the bubble could move from 20.000001 to 15.000001 in one iteration. But I guess that is a small detail. Maybe this isn't even relevant.

Edit:

Here is the jsFiddle! (Be carefull with less then 100 ms iteration time, it might stress the browser)

This is a simple model to estimate $\mathcal{N}$, the expected number of bubbles after some sort of equilibrium is reached. In this model, the bubble will be randomly displaced at each iteration.

The first claim is $\mathcal{N}$ is proportional to $\mathcal{T}$, the expected life time of a bubble . After we start the system and wait long enough for the system to stabilize, there are two ways to estimate $\mathcal{N}$.

Observe the system for $N \gg 1 $ iterations. Sum the number of bubbles observed in these iterations and then divide the final result by $N$.
Follow all bubbles created and destroyed during these $N$ iterations. Sum over their life time and then divide by $N$.

In general, these two ways of counting will differ by a number which is of the order of $\mathcal{T}$. If $\mathcal{T}$ is finite.
This two method will produce the same estimation of $\mathcal{N}$ as $N \to \infty$.

If on average, $\alpha$ bubble is created during each iteration, one will have

$$\mathcal{N} = \alpha \mathcal{T}$$

In this problem $\alpha = 0.05$. If the bubble is created within $20$px from the edge, the bubble will be destroyed in the next iteration.
The same thing happens to the bubble at the end of its life. All this process
will add $1$ to the expected life time $\mathcal{T}$.

For a probability $\left(\frac{560}{600}\right)^2$, the bubble will be created
within the safe zone $[20,580] \times [20,580]$.
Let $L = 580-20 = 560$ and $t$ be the number of iteration since the bubble
is created. Introduce random variables $X, Y, \eta_x(t), \eta_y(t)$ and we will
update the position of the bubble by following recipe

$$\begin{cases}
x(0) &= 20+X\\
y(0) &= 20+Y
\end{cases}
\quad\text{ and }\quad
\begin{cases}
x(t) &= x(t-1) + \eta_x(t)\\
y(t) &= y(t-1) + \eta_y(t)\\
\end{cases}
\quad\text{ for }\quad t > 0
\tag{*1}
$$

Subject to the constraint the bubble is created in the safe-zone, the variables $X, Y$ and $\eta_x(t), \eta_y(t)$ are uniform random variables taking values over $[0,L]$ and $[-5,5]$ respectively.

Let $T$ be the random variable corresponds to the life time of a bubble (excluding the last iteration discussed above).
Introduce a function
$$u(x,y) = {\bf E}\big[ T \big| (X,Y) = (x,y) \big]\quad\text{ for }\quad (x,y) \in [0,L]\times[0,L]$$
If $u(x,y)$ can be approximated by a continuous function, then the
$\mathcal{T}$ we want will be given by a integral

$$\mathcal{T} = 1 + \left(\frac{560}{600}\right)^2 {\bf E}[T]
\quad\text{ with }\quad
{\bf E}[T] =
\frac{1}{L^2} \int_0^L\int_0^L u(x,y) dx dy
$$

We can view the random update in $(*1)$ as a Brownian motion. If equilibrium has been reached, we should have

$$u(x,y) = {\bf E}\big[ u(x + \eta_x, y + \eta_y) \big] + 1$$
This implies $u(x,y)$ should approximately satisfies a PDE

$$u(x,y) = u(x,y)
+  \frac{{\bf E}[\eta_x^2]}{2} \frac{\partial^2 u(x,y)}{\partial x^2}
+  \frac{{\bf E}[\eta_y^2]}{2} \frac{\partial^2 u(x,y)}{\partial y^2}
+ 1
$$
Let $\sigma^2 = {\bf E}[\eta_x^2] = \frac{5^2}{3}$, this has the form of a Possion
equation
$$\nabla^2 u(x,y) = -\frac{2}{\sigma^2}$$
and the appropriate condition for this problem is $u(x,y) = 0$ on the boundaries.

If I'm not mistaken, the solution of these boundary value problem is given by

$$u(x,y) = \frac{32L^2}{\sigma^2\pi^4} \sum_{n=1,\text{odd}}^\infty\sum_{m=1,\text{odd}}^{\infty}
\frac{1}{n m (n^2+m^2)}
\sin\left(\frac{\pi n x}{L}\right)
\sin\left(\frac{\pi m y}{L}\right)
$$
Integrate this, we get

$$\mathcal{T} = 1 + \left(\frac{560}{600}\right)^2\frac{128L^2}{\sigma^2\pi^6} K
\quad\text{ where }\quad
K = \sum_{n=1,\text{odd}}^\infty\sum_{m=1,\text{odd}}^{\infty}
\frac{1}{n^2 m^2 (n^2+m^2)}
\approx 0.52973
$$
Substitute this into the formula $\mathcal{N} = \alpha T$, we get $$\mathcal{N}_{600} \approx 115.65$$
which now falls within the observed range of OP's numerical result.
If one repeat the same argument to other canvas size $s$, we have

$$\mathcal{N}_s = 1 + \left(1 - \frac{40}{s}\right)^2
\frac{384(s - 40)^2}{25\pi^6} K
$$
Following is a table for $\mathcal{N}_s$ for other canvas sizes $s$:

$$
\begin{array}{r:r}
s & \mathcal{N}_s\\
\hline
300 &  21.54\\
400 &  44.47\\
500 &  75.84\\
600 & 115.65\\
700 & 163.92\\
800 & 220.64\\
900 & 285.82
\end{array}
$$

I fitted the average path lengths I obtained.
$x$ axis: Size of square canvas $\times 100$ pixels.
$y$ axis: Average number of iterations per bubble.


Area of a quadrilateral proof Prove that the area of a quadrilateral is one half the product of the lengths of its diagonals and the sine of the angle between the diagonals.

Hint:

Let $S$ be the area of a triangle, $a,b$ be the length of two edges and $\theta$ be the angle between them, you have the following formula
$$S=\frac{1}{2}ab\sin\theta$$

On more hint: $\sin\theta=\sin(\pi-\theta)$.

!**enter image description here
**
**The proof:

Area of a quadrilateral =
 area of the first triangle + area of the second triangle + area of the third triangle + area of the fourth triangle

= [0.5 .d . b . sin θ ] + [0.5 . b . c . sin (180 - θ) ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin (180 - θ) ]

As, sin (180 - θ) = sin θ

Area = [0.5 . d . b . sin θ ] + [0.5 . b . c . sin θ ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin θ ]

Area = 0.5 . sin θ . [ (d . b) + (b . c) + (a . c) + (d . a) ]

As, (d . b) + (b . c) = b . (c + d)
and  (a . c) + (d . a) = a . (c + d)

Area = 0.5 . sin θ . [ b . (c + d) + a . ( c + d) ]

Area = 0.5 . sin θ . (c + d) . [b + a]

As, (b + a) = p , (c + d) = q

Area = 0.5 . p . q . sin θ


Simplify $\sin \arctan x$, assuming $x>0$ Now, naturally I understand that $$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$$ and therefore I should have just $$\sin \arctan x = \frac{x}{\sqrt{x^2 + 1}}.$$ However, I am interested as to the effect of restraining the domain to $\{x > 0\}$, as I believe the function may simplify further. Could someone point me in the right direction?

It cannot be simplified further. There is nothing simpler than  $ x/ \sqrt{ 1 -x^2} $ next in simplicity only to $ \tan^{-1}x $. Draw a triangle to look at the sides $ (\sqrt{1+x^2}, 1,x )$ for yourself. Domain is $ (-\infty, \infty) $.Choose only the positive part if you need only positive argument. Also there is atan2 function or equivalent to make a proper quadrant choice for numerical work for several CAS.


Help finding inverse of $f(x)=\frac{e^x-e^{-x}}{2}$ I'm trying to find the inverse of $f(x)=\frac{e^x-e^{-x}}{2}$. My textbook says $f^{-1}(x)=\ln(x+\sqrt{x^2+1})$, but I haven't been able to get that answer. Switching $x$ and $y$, I tried solving for $y$ a few different ways, with limited success. Initially, I tried:

$$x=\frac{e^y-e^{-y}}{2}\\
2x=e^y-e^{-y}\\
2x=e^{-y}(e^{2y}-1)\\
\ln(2x)=\ln(e^{-y}(e^{2y}-1))\\
\ln(2x)=\ln(e^{-y})+\ln(e^{2y}-1)\\
\ln(2x)=-y+\ln(e^{2y}-1)\\
\ln(2x)=-y+\ln((e^y-1)(e^y+1))\\
\ln(2x)=-y+\ln(e^y-1)+\ln(e^y+1)$$

I can't see anything else I could do with that. I also tried squaring both sides back on that first line, but it lead me down another dead-end. The only thing that seemed to yield an answer was:

$$x=\frac{e^y-e^{-y}}{2}\\
2x=e^y-e^{-y}\\
2x=e^y-\frac{1}{e^y}\\
2x=\frac{e^{2y}-1}{e^y}\\
2xe^y=e^{2y}-1\\
0=e^{2y}-2xe^y-1\\
(e^y)^2-2x(e^y)-1=0$$
Then, the quadratic equation yields,
$$e^y=\frac{2x\pm \sqrt{4+4}}{2}=\frac{2x\pm 2\sqrt{2}}{2}\\
e^y=x\pm \sqrt{2}\\
y=\ln(x\pm \sqrt{2})$$

I'm not sure what to do about the $\pm$. I imagine one would be eliminated to make the domain of $f^{-1}$ match the range of $f$, but I'm not sure how to determine which (actually, the textbook says the domains and ranges of both $f$ and $f^{-1}$ are $(-\infty ,\infty)$. More importantly, the book's answer, $f^{-1}(x)=\ln(x+\sqrt{x^2+1})$ doesn't match mine! Help! :)

You have
$$
(e^y)^2 - 2x(e^y) - 1 = 0
$$
From this you get
$$
e^y = \frac{2x\pm \sqrt{{\color{red}{4x}^2} + 4}}{2}= x\pm \sqrt{x^2 + 1}
$$
Here you only get the $+$ in the $\pm$ because otherwise the numeerator is negative and you have $e^y>0$ (assuming here that we are just talking about real numbers). So
$$
y = \ln(e^y) = \dots
$$

enter image description here](https://i.stack.imgur.com/IxpIA.jpg)
Some parts of ans has been done directly without detailed steps


How do we say that two logically equivalent definitions are not semantically equivalent? This is a question of terminology (and also relates to philosophy).

Very often we have a certain term, such as "differentiable function", and then we have a definition:

  (1) def. A differentiable function is a function such that ...

Then we have a theorem stating

  (2) theorem. A function is differentiable iff ...

If this theorem has a correct proof, then we could also have made (2) into a definition, and state that (1) and (2) are "equivalent" definitions.

However, we cannot simply assume that they are equivalent. This has to be proven.

Now my question is one of terminology: Is there an accepted simple term or phrase that captures the fact that "these two definitions are logically equivalent, but they are not semantically the same definition so that their logical equivalence requires a proof"?

For example, I'm thinking of something like "these two semantically independent definitions are equivalent".

A comparison with analytical philosophy might clarify: Frege's concept of "sense" and "reference": it may be that the Evening star and the Morning star are in fact the same particular planet Venus (they have the same reference), but as they are perceived by us, they have different identities (they have different senses). It then requires (physical) proof that they have the same reference. Similarly, our definitions may have the same reference (they refer to the same subset of functions, i.e. are equivalent), but they do not have the same sense (the definitions are different in terms of the meaning they convey when we read them). Hence it requires (mathematical) proof that the two definitions with different 'senses' (meanings), in fact are logically equivalent (have the same 'reference').

You could have said that they are equivalent definition, but that may not as strict as only one of them would be a definition and the other being a theorem.

If we want to be strict the most concise way is to decompose the definition and theorem first. They are of the form:

$$\phi\overset{def}\equiv\psi_{def}$$
$$\phi\Leftrightarrow\psi_{thm}$$

what one would do is to start with a theorem that says that $\psi_{def}\equiv\psi_{thm}$, if needed there can be a lot of statements that are equivalent. At this point we are only talking about equivalent statements. After this we introduce the definition (which can use any chosen statement).

What I've seen is that the parts of a definition as above has it's separate name. One call $\phi$ the definiendum and $\phi_{def}$ the differentiae. In similar way we would call $\phi$ and $\psi_{thm}$ the sides of the equivalence (of the theorem). That way we can say that the sides of the theorem are equivalent to the differentiae of the definition (or isolated that $\psi_{thm}$ is equivalent with the differentiae). However I find that somewhat clumsy and the term is not that usually used in mathematics anyway.

Let's test your idea:

Theorem: A function is differentiable iff  it is not not differentiable.

So then, is  "not not differentiable" a good definition of differentiable?


How would you describe calculus in simple terms? I keep hearing about this weird type of math called calculus. I only have experience with geometry and algebra. Can you try to explain what it is to me?

There came a time in mathematics when people encountered situations where they had to deal with really, really, really small things.

Not just small like 0.01; but small as in infinitesimally small.  Think of "the smallest positive number that is still greater than zero" and you'll realize what sort of problems mathematicians began encountering.

Soon, this problem became more than just theoretical or abstract.  It became very, very real.

For example, velocity.  We know that average velocity is the change in position per change in time (i.e., 5 miles per hour).  But what about velocity at a point in time?  What does it mean to be going 5 mph at this moment?

One solution someone came up with was to say "it's the change in position divided by the change in time, where the change in time is an infinitesimally small amount of time".  But how would you handle/calculate that?

Another problem came about trying to find the area under a curve.  The current accepted solution was to divide the curve into rectangles, and add together the area of the rectangles.  However, in order to find the exact area under the curve, you'd need to divide it into rectangles that were infinitesimally tiny, and, therefore, add up an infinite amount of tiny rectangles -- to something that was finite (area).

Calculus came about as the system of math dedicated to studying these infinitesimally small changes.  In fact, I do believe some people describe calculus as "the study of continuous changes".

This is the simplest explanation.

http://www.math-prof.com/Calculus_1/Calc_Ch_01.asp

ie...its simply a way of dealing with an equation where the denominator has become 0. Instead of leaving it as insoluble...you have a method of proceeding.


Idea for primality testing based on a trigonometric product This is an idea that I had about 3 months ago. I tried some college professors, they didn't care. I tried to solve, but with no luck. I ask for help to find the closed form of the following product series. I know it would be safer only to put the product, but I just would like to see this solved, and if I put why I think this is important, the chance of someone helping would be higher.

My idea is:

  If $\prod_{k=2}^{X-1}{\sin(\pi\frac{X}k)}≠0$, than the number $X$ is prime. Else, $X$ is a non-prime.

Why:

$\sin(\pi\frac{X}k)$ tests if a number k divides X. If the value returned is $0$, than k divides X. Because for a number to divide another, the result must be an integer, and $\sin(\pi\alpha)$ is equal to $0$ if and only if $\alpha$ is an integer. So, to test if a number, for example, k=3 divides 9, we could try that like $\sin(\pi\frac{9}3)=\sin(3\pi)=0$. Now let's try to see if 2 divides 11 this way: $\sin(\pi\frac{11}2)=\sin(5.5\pi)=-1≠0$
Using logic, we could extend this to the product above. We only need one sine being $0$ for the product to be $0$. That means that with only one divisor from range $2$ to $X-1$, the number X is a non-prime. And the only way for the product to be $0$ is if one or more of the sines is $0$: You can't multiply numbers $≠0$ and get $0$.

For example, 7 is prime if: $$\sin(\pi\frac{7}2)\times \sin(\pi\frac{7}3)\times \sin(\pi\frac{7}4)\times \sin(\pi\frac{7}5)\times \sin(\pi\frac{7}6)\ne 0$$ and that's the case.

I know I could test only until $\sqrt{7}$ rounded up, but if someone can get the closed form of the product, that wouldn't matter, and we wouldn't have to calculate the square root.

Thank you for reading this.

Assuming $X$ is an integer $\geq3$ your test is fine from a logical point of view. But it is completely useless if you want to find out whether $X:=63\,176\,591$ is prime. You would need millions of terms of the sine series in order to compute a  single factor of your product, say $\sin\bigl(\pi{X\over 31}\bigr)$. But even if you computed all these factors to $10^6$ places, in a numerically unlucky situation you would not be able to decide definitively that the product is $\ne0$.

This test is useless because this product tends to zero for big numbers. Even the result for 1273 (prime) is $0$ for the computer, because it gets too small. The sine values varies between 0 and 1, so the product gets smaller and smaller... I even did an approximation for that product (possibly wrong), which gave me a complex answer but also returned $0$ for big numbers.

I'm trying to think now of a function that returns $f(integer)=0$, but gives $|f(non-integer)|\ge1$

Since I already wasted some time trying to do this approximation I'll share it. But again, I don't know if this is correct:

$P(X)=\prod_{k=2}^{X-1}{\sin(\pi\frac{X}k)}$

$P(X)=(2i)^{2-x}\prod_{k=2}^{X-1}{(e^{i\pi X/k}-e^{-i\pi X/k})}$

$P(X)=(2i)^{2-x}\prod_{k=2}^{X-1}{e^{i\pi X/k}}\times\prod_{k=2}^{X-1}{(1-e^{-2i\pi X/k})}$

$$P(X)=(2i)^{2-x}\prod_{k=2}^{X-1}{e^{i\pi X/k}}\times(-1)^{X-2}\times\prod_{k=2}^{X-1}{(e^{-2i\pi X/k}-1)}$$

$\prod_{k=2}^{X-1}{e^{i\pi X/k}}=exp(\sum_{2}^{X-1}{\frac{i\pi X}{k}})=exp(i\pi X\times\sum_{2}^{X-1}{\frac{1}{k}})$

$\prod_{k=2}^{X-1}{e^{i\pi X/k}}\approx exp[i\pi X(lnX+0.57722+\frac{1}{2X}-1)]$

$$\prod_{k=2}^{X-1}{e^{i\pi X/k}}\approx exp[i\pi X(lnX-0.4228+\frac{1}{2X})]$$

Now using this: Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$

$\prod_{k=2}^{X-1}{(e^{-2i\pi X/k}-1)}=(-1)^{X-2}\times\prod_{k=2}^{X-1}{(1-e^{-2i\pi X/k})}$

$=(-1)^{X-2}\times\prod_{k=2}^{X-1}{(1-\frac{1}{exp(2i\pi X/k)}})=$

$=(-1)^{X-2}\times\prod_{k=2}^{X-1}{(1-\frac{1}{exp(i\pi X/k)})(1+\frac{1}{exp(i\pi X/k)}})=$

$=(-1)^{X-2}\times\prod_{k=2}^{X-1}{(\frac{exp(i\pi X/k)-1}{exp(i\pi X/k)}\times\frac{exp(i\pi X/k)+1}{exp(i\pi X/k)})}$

Where $\frac{exp(i\pi X/k)+1}{exp(i\pi X/k)}=a_k$ and $\frac{exp(i\pi X/k)-1}{exp(i\pi X/k)}=\frac{1}{a_{k-1}}$

$$\prod_{k=2}^{X-1}{(e^{-2i\pi X/k}-1)}=(-1)^{X-2}\times(\frac{1+exp(\frac{i\pi X}{X-1})}{exp(\frac{i\pi X}{X-1})})\times(\frac{exp(i\pi X)}{1+exp(i\pi X)})$$

So, the answer would be:

$$P(X)\approx(2i)^{2-X}\times exp[i\pi X(lnX-0.4228+\frac{1}{2X})] \times (\frac{1+exp(\frac{i\pi X}{X-1})}{exp(\frac{i\pi X}{X-1})})\times(\frac{exp(i\pi X)}{1+exp(i\pi X)})$$


How do I find the subspace of $P_{2}$ from the following subsets? One of the following is a subspace of $P_{2}$.
A. $W= \{ax^2+bx+c\mid a+b+c=1\}$
B. $W= \{ax^2+bx+c\mid a+b+c=0\}$
C. $W= \{ax^2+bx+c\mid a \geq 0\}$
D. $W= \{ax^2+bx+c \mid b,c \geq 0\}$
So it's either $B, C$, or $D$ because the zero vector condition won't work on $A$. What I don't understand is the difference between $C$ and $D$. How does having 'a' greater than or equal zero differ from having '$b$' and '$c$' greater than or equal zero? Can someone please explain how it works? This concept is a bit new to me and I couldn't find any examples of this particular case online.

How does having 'a' greater than or equal zero differ from having 'b' and 'c' greater than or equal zero? Can someone please explain how it works?

So, for example, consider the polynomials
$$
p_1(x) = 2x^2 + 3x + 4,\\
p_2(x) = -3x^2 + 4x + 5,\\
p_3(x) = 0x^2 - 5x + 6,\\
p_4(x) = -5x^2 + 6x - 7.
$$
The polynomials $p_1$ and $p_3$ are elements of the set described in C because the $x^2$ coefficient (the "$a$" coefficient) is non-negative. The polynomials $p_1$ and $p_2$ are in the set described in D because the other coefficients (the "$b$" and "$c$" coefficients) are non-negative. $p_4$ is in neither set because it is not true that $a \geq 0$, and it is also not true that we have both $b \geq 0$ and $c \geq 0$.

Hint to answer the title question:
A subspace must be closed under multiplication by a scalar such as $−1$.


Can anyone gives me an example of $V$ such that there exists at least one pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) > V(ac)$ $D$ is an Integral Domain.

$V$ only satisfies the first Euclidean property, i.e. for all $\,a,b\in D\,$ if $\,b\neq 0\,$ then there are $\,q,r\in D\,$ such that $\, a = qb +r\,$ with either $r=0$ or $\, V(r) < V(b),\,$ where $V$ maps $D$ into  (well-ordered)  $\,\Bbb N$.
?
Can anyone gives me an example of $V$ such that there exists at least one  pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) > V(ac)$?

I have been really trying hard to construct one. But I can't.

Let $D=\Bbb Q[X]$ and
$$ V(f)=\begin{cases}
2\deg f&\forall x\gg 0\colon f(x)>0 \\
2\deg f+1&\forall x\gg 0\colon f(x)<0 \\
\text{whatever}&f=0\end{cases}$$

For $a,b\ne0$, we can find by polynomial division $q,r$ with $a=qb+r$ and either $r=0$ or $\deg r<\deg b$, in which case also have $V(r)<V(b)$.
Thus the desired conditins hold.
Now let $a=c=-1$ and note that $V(a)=1$, $V(ac)=0$.


About differential of functions... Ok, probably it is a silly question, but I'm studying calculus for the first time, and I still can't "see clearly" the definition of differential of a function.

We define $dy$ - or equivalently $df(x)$ - as $dy = f'(x)\cdot\Delta x$. Now, if we consider the function $y=f(x)=x$, it follows that $df(x)=dx=f'(x)\cdot\Delta x=1\cdot\Delta x=\Delta x$; hence we can rewrite the previous relation as $dy=f'(x)\cdot dx$; and consequently $f'(x)=\frac{dy}{dx}$.

The question is: is there any good reason why I should substitute $\Delta x$ with its differential $dx$ in that relation?

I know it might sound weird, but I can't intuitively "feel" the fact that $\Delta x = dx$, I think it just makes things more confused...

Source (lang: italian): angeloangeletti.it/UNICAM_RECANATI/APPUNTI/04_lez.pdf

Well, to understand this, you have to understand limits. What a derivative is doing is it is taking the slope of a polynomial function - f(x). What you are doing is you're taking two points exactly $\Delta$x apart. The points would then have x-coordinates x and x+$\Delta$x. When you take the slope of the line between those two points (Click this to help), you get a number, and when $\Delta$x is infinitely close to, but not equal to 0, we call it dx, and the slope of the line is the derivative at that point. Since we are getting numbers for points, we can plot the derivatives at all points and match them into a function. This function is the first derivative of your original polynomial function, f'(x). Taking the derivative of that function is the second derivative, f''(x).

This is very useful because you can find, easily, rates of change, volume from outlines, and so on. For example, the derivative of a graph of the distance a car has traveled at a certain time is equal to the graph of its velocity at a certain time.

The volume of a cone/pyramid, and therefore the volume of a sphere, are all derived by calculus.

P.S. In the image, h = $\Delta$x


Is there a specific symbol for denoting a linear subspace like $\subseteq$ for denoting a subset? I was under the impression that this was a proper notation, but I was just corrected on this but did not get a proper explanation of what it should be. I get that it is definitely not the same, but what is the right symbol then?
So I am specifically asking for a linear subspace. Or does this not make any difference?

Some authors denote that as $\leq$. The inclusion $\subseteq$ of set $X$ form a partial order of the power set $\mathcal{P}(X)$ of a set $X$ and if $X$ is vector space then it's subspaces form the suborder as a subset of $\mathcal{P}(X)$. In order theory, there are many orders that use different symbols $\leq,\ \subseteq,\ \lesssim,\ \preceq,\ \unlhd$, etc.
Usually the $\subset$ symbol is used for sets. However, any symbol can be used as long as it is predefined or can be distinguished from the context. There are a myriad of orders, so there is no need to create other symbols for them.

Indeed a linear subspace is a subset of a given linear space for which the linearity properties are preserved but there is not a specific symbol for that.
In a compact form we could use $$U=\operatorname{span}\{v_1,\ldots,v_k\}\subseteq V$$ beeing $\{v_1,\ldots,v_k\} \subseteq V$ a basis for $U$ or more in general any subset of the linear space $V$ depending on the context we are dealing with.


Prove $\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$. Given $a,b,c,d\ge 0$ and $a^2+b^2+c^2+d^2=4$ show the following holds :
$$\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$$
Now I tried to fully expand but that becomes too ugly. I want a non-expand proof of this fact. For some motivation I solved a three variable version which I assumed to be true and it came out to be true!
That one is :
$a,b,c\ge 0$ and $a^2+b^2+c^2=3$ show that :
$\displaystyle \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\le 1$. The second one is easier and I DO NOT NEED a solution of the second one. Please help me out on the first one because it is similar but not solvable by a similar method.

You can treat the function
$$
f(a,b,c,d) = \frac{a}{b+3} = \frac{b}{c+3} + \frac{c}{d+3} + \frac{d}{a+3}
$$
as a 4-d surface, and use calculus to find any critical points.  Finding the partial derivatives of $f$ with respect to $a,b,c,d$ end up looking very symmetrical, and you end up finding that (1,1,1,1) is a critical point.  Now, you can check that this satisfies your quadratic condition, and you can either calculate the second derivatives (inadvisable), or do a local test (not a complete proof, but easier) to confirm that this point is indeed a local maximum.
Since at this local maximum, your inequality is satisfied, and you can check that there are no other critical points, ( $(-1,-1,-1,-1)$ technically works, but this is outside the given constraints), you can conclude that this inequality holds.

Addendum:  techincally speaking you should check the boundaries before you can make the conclusion.  The boundary conditions actually devolve into variants of the three term case for which you have already solved.  So everything should work out peachy there as well.


How does a cellular automaton "know" when to halt? I came across the (proven) claim that the cellular atomaton "rule $110$" is universal, meaning that it can do every calculation in principle. But to calculate a result, the automaton must "halt" somehow.

  How does the automaton "know" when it has to halt, in other words, how is "halting" simulated in this automaton ?

A turing machine has the "halt"-state but "rule $110$" apparently only can manipulate the symbols.

The 110 never halts. The machine will, for instance, continue forever toward the left filling in $1$'s. So the only stable configuration with finitely many $1$'s (or infinitely many, for that matter) is the configuration with no $1$'s at all.

So my guess is you have to look at a subset of the tape for a specific repeating pattern or something.


Ratio of sums vs sum of ratio Is anyone aware of any general (or perhaps not so general) relationship (inequality for instance) relating

$A(x,y)= \frac{\sum_z f(x,y,z)}{\sum_z g(y,z)}$

and

$B(x,y)= \sum_z\left(\frac{f(x,y,z)}{g(y,z)}\right)$

?

Specific context (for what I'm dealing with, but not necessarily the question) is that $\sum_{x,y,x}f(x,y,z)=1$ and $\sum_{y,x}g(y,z)=1$ and $f(x,y,z)\geq 0$ and $g(y,z)\geq 0\quad \forall x,y,z$. I.e. probabilities (or more generally, I guess, measures).

It seems like it could 'vaguely' be related to log sum inequalities (when transformed) or Jensen's inequality perhaps?

If you assume that both $f$ and $g$ are nonnegative, you have $A(x, y) \leq B(x, y)$.

Proof:

$ \frac{f(x, y, z)}{\sum_{z}g(y, z)} \leq \frac{f(x,y,z)}{g(y,z)} $, since $g(y, z) \leq \sum_{z}g(y, z)$.

So
$A(x,y) = \frac{\sum_z f(x,y,z)}{\sum_z g(y,z)} = \sum_z \frac{f(x, y, z)}{\sum_z g(y, z)} \leq \sum_z \frac{f(x, y, z)}{g(y, z)} = B(x,y)$.

One example would be
Jensen's inequality:

https://en.wikipedia.org/wiki/Jensen%27s_inequality


Finding set of vectors that spans the solution set Question: Find a set of vectors $\{u,v\}$ in $\mathbb{R}^4$ that spans the solution set of the equations:

$$\begin{align}w - x + y + z = 0 \\
5w + 2x - y + z = 0\end{align}$$

Reducing these I get:

When I reduce these, I'm getting

$x = -2z - 6w$

$y = -3z - 7w$

$z = u$

$w = v$

I am getting this as my answer:

$$(x,y,z,w) = u(-2,-3,1,0) + v(-6,-7,0,1)$$

But this is wrong for some reason and I don't understand what I did wrong. Could anyone help me solve this problem correctly?

EDIT: So apparently, the y and z values are correct, however I have my x and w values wrong somehow. According to my instructor, the first entry for u should be -1/7, however I'm not sure how to get that.

I'll advance based on your calculations.

$$ (x,y,z,w) = ( -2u-6v, -3u-7v,u,v ) = ( -2u,-3u,u,0 )+ ( -6v,-7v,0,v )$$

  $$\implies (x,y,z,w) = u( -2,-3,1,0 ) + v( -6,-7,0,1 ). $$

Can you see your basis now?


Characteristic of a field and basis of a dual space. Let $K$ be a field. $E=K_n[x]=\{p(x)\in K[x]|\text{degree}(p(x))\leq n\}$

For a $p(x)\in E$ we define $p^{(0)}(x)=p(x)$ and for every positive integer $i$, $p^{(i)}(x)=(p^{(i-1)})'(x)$. For $i\in \{0,...,n\}$ consider $w_i:E\longrightarrow K$ given by $w_i(p(x))=p^{(i)}(1)$. Prove that $B=(w_0,...,w_n)$ is basis of $E^*$ if and only if $K$ has characteristic $0$ or characteristic $p>n$.

I can't solve this problem because I don't even understand the problem statement.

First of all, what is $p^{(i)}(x)$? Of course I understand that $p(x)=a_0+a_1x+...+a_nx^n$ so $p^{(3)}(x)$ would be $((a_0+a_1x+...+a_nx^n)^2)'$?

Second of all, what is the characteristic of $K$?

Thank you for your time and excuse me if I made any grammar mistake, my english is not too good.

$p^{(3)}$ would be $((a_0+a_1x+...+a_nx^n)^2)'$.

No, you have to build it up so

$$
(a_0+a_1x+...+a_nx^n)^{(3)}=
(a_0+a_1x+...+a_nx^n)^{(2)}{'}=
(a_0+a_1x+...+a_nx^n)^{(1)}{''}=
(a_0+a_1x+...+a_nx^n)^{(0)}{'''}=
(a_0+a_1x+...+a_nx^n){'''}
$$
and you now just have to differentiate the polynomial three times.

The characteristic of a field $\mathbb{F}$ is just the least positive $k$ such that $k\cdot 1=0$; or if there are none, $0$. In fact such a $k$ will be a prime number $p$.

That answers your questions, good luck with the calculation to show the $w_i$ are linearly independent exactly when the characteristic is $0$ or greater than $n$.

You can take derivatives formally.

Every finite field with $p^n$ elements has characteristic $p$ which is the smallest nonzero integer $k$ such that $1+1+\cdots+1=k\times 1=0.$ So you'd have modulo $k$ operations.

$R$ and $C$ have characteristic zero, sums of the additive identity are never zero.


Given a real $x$ and an integer $N \gt 1$, prove that there exist integers $h$ and $k$ with $0 \lt k \le N$ such that $|kx-h|\lt 1/N$. Given a real $x$ and an integer $N \gt 1$, prove that there exist integers $h$ and $k$ with $0 \lt k \le N$ such that $|kx-h|\lt 1/N$. Hint. Consider the $N+1$ numbers $tx-[tx]$ for $t=0,1,2,\dots, N$ and show that some pair differs by at most $1/N$.

I'm trying to prove this hint. Let $a_t=tx-[tx]$ for $t=0,1,2,\dots, N$, and assume that for every different pair $|a_i-a_j| \ge 1/N$.

I can't lead to a contradiction from this. I would greatly appreciate any help.

PLEASE delete, I misread the (conditions of the) OP.


Ring structure on the Galois group of a finite field Let $F$ be a finite field. There is an isomorphism of topological groups $\left(\mathrm{Gal}(\overline{F}/F),\circ\right) \cong (\widehat{\mathbb{Z}},+)$. It follows that the Galois group carries the structure of a topological ring isomorphic to $\widehat{\mathbb{Z}}$. What does the multiplication $*$ look like? If $\sigma$ is the Frobenius, we have $\sigma^n * \sigma^m = \sigma^{n*m}$, and this describes $*$ completely. Is there any way to give an explicit and natural formula for $\alpha * \beta$ if $\alpha,\beta$ are $F$-automorphisms of $\overline{F}$? Also, is there any more conceptual reason why the Galois group carries the structure of a topological ring (without computing the Galois group)?

Maybe the following is a more precise version of the question: Consider the Galois category $\mathcal{C}$ of finite étale $F$-algebras together with the fiber functor to $\mathsf{FinSet}$. The automorphism group is exactly $\pi_1(\mathrm{Spec}(F))=\widehat{\mathbb{Z}}$. Which additional structure on the Galois category $\mathcal{C}$ is responsible for the ring structure on its automorphism group?

Yes, there is a natural way of describing the ring structure on $Gal(\overline{F}\backslash F)$, just using the Galois theoretic fact that Galois groups of infinite Galois extensions can be considered as inverse limits of systems of Galois groups of finite extensions.

For simplicity, suppose that $F=\mathbb{F}_p$.

For each $n\in\mathbb{N}$, $Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$ is cyclic of order $n$, generated by the Frobenius $\sigma(x)=x^p$, so there are natural group isomorphisms $\iota_n:\frac{\mathbb{Z}}{n\mathbb{Z}}\cong Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p), r+n\mathbb{Z}\mapsto \sigma^r$.

Therefore $Gal(\mathbb{F}_{p^m}\backslash\mathbb{F}_p)$ carries a natural ring structure given by $\sigma^r*\sigma^s=\sigma^{rs}$, and this makes the maps $\iota_n$ ring isomorphisms.

There are natural surjective transition maps $\nu_{n,m}:Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)\to Gal(\mathbb{F}_{p^m}\backslash\mathbb{F}_p), \alpha\to\alpha|_{\mathbb{F}_{p^m}}$ whenever $m\mid n$, making this collection of Galois groups into an inverse system, and it is a fact in Galois theory that $Gal(\overline{F}\backslash\mathbb{F}_p)$ is the limit of this system.

Therefore, given general $F$-automorphisms $\alpha,\beta\in Gal(\overline{F}\backslash\mathbb{F}_p)$, we can write $\alpha=(\alpha_n)_{n\in\mathbb{N}}$, $\beta=(\beta_n)_{n\in\mathbb{N}}$, where $\alpha_n,\beta_n\in Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$, $\nu_{n,m}(\alpha_n)=\alpha_m$, $\nu_{n,m}(\beta_n)=\beta_m$, and multiplication in $Gal(\overline{F}\backslash\mathbb{F}_p)$ is given by $\alpha*\beta=(\alpha_n*\beta_n)_{n\in\mathbb{N}}$, where multiplication in $Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$ was defined above.

This is the natural way of describing the ring structure on $Gal(\overline{F}\backslash\mathbb{F}_p)$, and since $\widehat{\mathbb{Z}}$ is identivcally defined as the inverse limit of the system of cyclic groups $\frac{\mathbb{Z}}{n\mathbb{Z}}$, it is equivalent to the ring structure on $\widehat{\mathbb{Z}}$.


Very *mathematical* general physics book I am searching for a book to study physics. So far, I've been suggested

  Resnick, Halliday, Krane, Physics,

but it doesn't seem to be very suited for a math major. Can you suggest some more mathematical books?

By mathematical I mean:

rigorous, with complete good proofs;
at the same time, intuitive and throughout in explanations.

There are now quite a few excellent ones, but most of these are pitched at fairly sophisticated readers-graduate students or professional mathematicans. The thinking according to such textbooks, of course,is that the readers are very far along in thier mathematical training and are ready to use that mathematics to learn physics at a very high level. Whether or not that's true is debatable. In any event, here's a list to get you started:

Lectures on Quantum Mechanics for Mathematics Students by L. D. Faddeev and O. A. Yakubovskii: Really the only one for undergraduates-a beautiful and justly famous Russian treatment now in English. But even this beginning text requires a good knowledge of differential equations, real and complex analysis, linear algebra and group theory.

Quantum Mechanics for Mathematicians by Leon A. Takhtajan : One of several such texts for second year graduate students, most of which need at least first year graduate courses in analysis,differential geometry and topology. Excellently written,though.

Physics for Mathematicians, Mechanics I by Michael Spivak :  Awesome book by the master. Perfect follow up to his 5 volume opus on graduate differential geometry and shows in depth what all that beautiful manifold theory was good for. Let's hope he finishes the projected other 3 volumes.

Quantum Theory for Mathematicians by Brian C. Hall  A recent addition to the list I haven't seen. But if it's as good as his Lie groups book, it'll be well worth checking out.

Mathematical Methods in Quantum Mechanics by Gerald Teschl : This book is really more about the methods of quantum theory then the science itself. That being said,it's by a master and presents the material beautifully for second year graduate students.

Quantum Field Theory by Gerald B. Folland: Another terrific textbook by Folland, covering everything about QFT that can be made rigorous at this point,which sadly isn't as much as we'd like. You better have your graduate analysis chops on before tackling this one.

Semi-Riemannian Geometry With Applications to Relativity by Barrett O'Neill : Exactly what the title says it is. A classic,one of the best books on relativity theory for mathematicians. The book to read after Spivak.

Modern Geometric Structures and Fields by S. P. Novikov and I. A. Taimanov: An incredible and completely unorthodox beginning graduate course in differential geometry by one of the premier mathematicians in the world that completely interweaves the careful theory with  virtually the entire modern structure of physics-from basic mechanics through relativity theory through the beginnings of string theory and quantum field theory.A good working knowledge of rigorous calculus of one and several variables and linear algebra is all you need to tackle this one. An absolute must have if you're interested in the relationship between mathematics and physics. I DO wish it had more exercises.

UPDATE: I've added a few more in response to some of the other posts here:

Foundations of Mechanics 2nd edition by Ralph Abraham and Jerrold E. Marsden: A remarkable and careful text for both physicists and mathematicians giving one of the first detailed presentations of classical mechanics from the modern differential geometric point of view. Very advanced, really for strong graduate students in mathematics or very strong graduate students in physics. A good follow up to Spivak's mechanics book and a first year graduate mathematics course sequence.

Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems  by Jerrold E. Marsden and Tudor S. RatiuL A very mathematical first year graduate course in classical mechanics,again from the moder geometric point of view and emphasizing symmetry. Can act as the preliminary text for the text immediately preceding this one. About the same level as Spivak,but it's not as rigorous and focuses more on advanced physics then geometry. Very good collateral reading to that text.

Symplectic Techniques in Physics by Victor Guillemin and Shlomo Sternberg: I've never seen this book, but I've heard good things about it. It's by 2 masters,so that by itself makes it worth checking out.

Ordinary Differential Equations by V.I. Arnold:  A terrific intermediate level course emphasizing the role of linear transformations and manifolds in the study of linear ODEs. A strong course in linear algebra and geometry is needed; this beautiful text is a must for students interested in the geometry of differential equations and physics.

Geometrical Methods in the Theory of Ordinary Differential Equations by V.I. Arnold: Graduate level follow up to the preceding text. An introduction to ordinary differential equations and dynamical systems emphasizing thier role in chaos theory and physics. A very difficult and sometimes mystifying book, but the sheer richness of the ideas and the interplay of math and physics makes it worth the effort.

I'll also mention in passing at the end the wonderful physics textbooks by Walter Griener. Not only are they concise,wide ranging and extremely clear, they have more solved examples then any physics book I've ever seen. They're written by a master. If you can get them in the original German and can read those,that would be even better as some errors creeped in in translation.

Hope that gets you started. Good luck!

Try Principles of Physics by Resnick, Halliday and Walker


Simple algorithm to get the square of an integer using only addition? This problem was mentioned in passing in a reading and it piqued my curiosity.

I'm not sure where to start. Any pointers? (perhaps square root was meant?)

An algorithm to do this for integers would be (in pseudo-code):

 input a
 b = abs a
 s = 0
 while b > 0 do
      s = s + a
      b = b - 1
 return s

$$a^2= 2((a - 1) + (a - 2) + ... + 1) + a$$

Imagine a square of side 5 units
(Going clockwise) Add one side (5), then another (4) on and on...

It will give you this:
$$2\times(4 + 3 + 2 + 1) + 5 = 25 = 5^2$$


Solution of Linear Differential Equations If $[A, [A, B]] = 0 = [B, [A, B]]$, where $[A,B]=AB-BA$, then

$$e^{tA} e^{tB}=e^{t(A+B)+(t^2/2)[A,B]}$$

The book suggests proving that $e^{t(A+B)+(t^2/2)[A,B]}$ is solution of
$$\dot{X}=AX+XB, \qquad X(0)=I$$
if and only if
$$\frac{d}{dt}( e^{-(t(A+B)+(t^2/2)[A,B])}Ae^{t(A+B)+(t^2/2)[A,B]} - t[A,B])=0$$

I derived the function $t \mapsto e^{t(A+B)+(t^2/2)[A,B]}$, but I couldn't show that is a solution of the initial value problem or use the hint.

This is exercise 2.52 in the book of Differential Equations from Chicone.

One can show by induction that for all $n\in\mathbb{N}^*$, $$[A,B^n]=nB^{n-1}[A,B]$$
Thus
$$ \forall n\in\mathbb{N}^*, A\frac{B^n}{n!}-\frac{B^n}{n!}A=\frac{B^{n-1}}{(n-1)!}[A,B] $$
Suming this gives
$$ [A,\exp(B)]=\exp(B)[A,B] $$
Let $\varphi(t)=\exp\left(t(A+B)+\frac{t^2}{2}[A,B]\right)$ then $\varphi'(t)=(A+B+t[A,B])\varphi(t)$ and $\varphi(0)=I_n$. Now let $\psi(t)=\exp(tA)\exp(tB)$, then
$$ \psi'(t)=\exp(tA)(A\exp(tB)+\exp(tB)B) $$
and $[tA,\exp(tB)]=\exp(tB)[tA,tB]$ that is to say $A\exp(tB)-\exp(tB)A=t\exp(tB)[A,B]$ so we have
$$ \psi'(t)=(A+B+t[A,B])\psi'(t) $$
Since $\psi(0)=I_n$, according to Cauchy-Lipschitz theorem we have $\psi=\varphi$ that is to say
$$ \exp(tA)\exp(tB)=\exp\left(tA+tB+\frac{t^2}{2}[A,B]\right) $$


Why do we use $\frac{\pi}{180}$ to convert from degrees to radians? If I want to convert from degrees to radians, I can use the function that takes degree value as an input, multiplies it with $\frac{\pi}{180}$ and returns the radian value: $\operatorname{DtoR}(d)=d \times \frac{\pi}{180}$.

And if I want to go from radians to degrees I need to only go backwards and  divide radian value with $\frac{\pi}{180}$ (e.g. multiply it with $\frac{180}{\pi}$): $\operatorname{RtoD}(r)=r \times \frac{180}{\pi}$.

My question is this: Why does multiplying/dividing with $\frac{\pi}{180}$ converts degrees into radians/radians into degrees? Why exactly that number, not some other? Also, does this work only for unit circle, or for any circle?

A full circle is $360^\circ$ and also $2\pi$ radians. Thus $360^\circ = 2\pi\text{ rad}$. We simplify by dividing by two.

Since circumference =2pier if we take half circle it would be pier in which rotation of angle is 180 degree and  arc length would be pie*r then we get theta =pie =180 degree


Can I have a logical explanation for why this number is so ridiculously close to a whole number? $$e^ {\pi\sqrt{163}}=262537412640768743.9999999999992\cdots$$

Why does this number run so incredibly close to a whole number? Can I have a logical explanation for why this finding? I know how to calculate it, but I want an answer that would explain it to me by intuition only. Thank you very much for any help. It is not the same as the other similar problem because I am looking for a logical reason and the other is wondering whether or not it is concidence.

Here's an explanation of these numbers. Say you want $e^{\pi \cdot n}$ to be within a positive error called $\epsilon$ of a natural number $L$, but you want $n$ to be natural as well. In other words...

$$L-\epsilon \lt e^{\pi \cdot n} \lt L+\epsilon$$

You can "solve" for n. Take the natural log and divide by $\pi$ to get...

$${1 \over {\pi}} \cdot \ln{(L-\epsilon)} \lt n \lt {1 \over {\pi}} \cdot \ln{(L+\epsilon)}$$

Now it should be clear that if $L$ ranges through $1$ to $\infty$ that the boundary for $n$ will range $0$ to $\infty$.

Since $L$ must be an natural number, there is little reason to think the values $n$ can take will be any more likely to not contain a natural number.

So now you'd use the above and pick an error tolerance. Values that allowed for both a natural $n$ and $L$ will be within $\epsilon$ of $L$. I'm guessing these are the so called Heegner numbers, you'd just allow for $n$ which are the square root of other numbers.


For every $x \in [\frac{\pi}{2},\pi]$, $\sin(x)+\cos(x)\geq 1$. Prove rigorously by contradiction. For every $x \in [o,\frac{\pi}{2}]$, $\sin(x)+\cos(x)\geq 1$.
How do you prove this rigorously by contradiction?

I understand you start by assuming that $\sin(x)+ \cos(x)<1$ and prove this is a false statement. I can see from drawing the graphs this is false but how do I show it algebraically?

Both $\cos(x)$ and $-\sin(x)$ are convex functions on the interval $(\pi/2,\pi)$. Since in the endpoints of such interval se have $\cos(x)-\sin(x)=-1$, it follows that $\cos(x)-\sin(x)\leq -1$ all over $[\pi/2,\pi]$. The last inequality implies that $\sin(x)+\cos(x)$ is a decreasing function over $[\pi/2,\pi]$, so, for any $x\in(\pi/2,\pi]$,
$$\sin(x)+\cos(x)<\sin(\pi/2)+\cos(\pi/2)=1.$$

We start with the assumption $$\sin x+\cos x<1$$ and rearrange, $$\sin x<1-\cos x$$ square both sides $$\sin ^2 x<1-2\cos x+\cos^2x$$ and manipulate further. $$\sin^2x-\cos^2x<1-2\cos x$$ Since $\sin^2x+\cos^2x=1$ and $\sin^2x+\cos^2x-2\cos^2x=\sin^2x-\cos^2x$, $$1-2\cos^2x<1-2\cos x$$ $$2\cos^2x>2\cos x$$ $$\cos x>1$$ which is false since the codomain of $\cos x$ is $-1\le x\le 1$.


Matrix decomposition definition Wikipedia says "In the mathematical discipline of linear algebra, a matrix decomposition or matrix factorization is a factorization of a matrix into a product of matrices. There are many different matrix decompositions; each finds use among a particular class of problems."

But in my opinion decomposition term should be used to represent breaking a matrix in different sub-matrices or some new matrices created after some operation on original matrix which if used together and passed through some algorithm(not necessarily product), shall reproduce the original matrix.

Is there some different terminology to represent what I am expecting to say?

"But in my opinion decomposition term should be used to represent breaking a matrix in different sub-matrices or some new matrices created after some operation on original matrix which if used together and passed through some algorithm(not necessarily product), shall reproduce the original matrix."

To give one of many examples, the $LU$ decomposition will

represent breaking a matrix into two new matrices, namely $L$ and $U$.  These matrices are
created after some operation on (the) original matrix, the operation being Gaussian elimination.  Then
if used together, that is, the matrices $L$ and $U$ are used together, and
passed through some operation, the operation being matrix multiplication, they
reproduce the original matrix, because $LU=A$.

I think this fits perfectly into what you have said.  As far as I know there is no other commonly used term for this kind of thing.

I have tried answering this question, It was long so I am putting it as link. I am not a mathematician but programmer, please provide some inputs.


Finding the equations of the tangents where a quadratic equation cuts the $x$-axis and the angle between the tangents (differentiation involved) Calculate the equations of the tangent where $y=x^2-5x-24$ cuts the $x$-axis.

$(x-8)(x+3)$ factorising

$x=8,  x=-3 $

$y'(x)=2x-5$

$y'(8)=11$

$y'(-3)=-11$

$y=11x+c$

$0=11(8)+c$

And then I find, $c$, and repeat for the other tangent equation which gives:

$y=11x-88$ and $y=-11x-33$

The second question is what is the angle between the tangents and I don't know how to find it. I know it has something to do with tan-theta.
Could someone also check if my arithmetic is correct for the first part.

Notice, the slope of tangent at general point of the curve $y=x^2-5x-24$ $$\frac{dy}{dx}=\frac{d}{dx}(x^2-5x-24)=2x-5$$ Now, the point where curve  $y=x^2-5x-24$ cuts the x-axis has $y=0$ thus we have $$0=x^2-5x-24$$ Solving the quadratic equation for the values of $x$ as follows  $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-24)}}{2(1)}$$ $$=\frac{5\pm 11}{2} \iff x=8, -3$$ Hence, we get two points $(8, 0)$ & $(-3, 0)$ where curve intersects x-axis

Now, the slope of the tangent at $(8, 0)$ is $$=\frac{dy}{dx}|_{x=8}=2\times 8-5=11$$ Hence its equation $$y-0=11(x-8)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=11x-88}}$$

Similarly, the slope of the tangent at $(-3, 0)$ is $$=\frac{dy}{dx}|_{x=-3}=2\times (-3)-5=-11$$ Hence its equation $$y-0=-11(x-(-3))$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=-11x-33}}$$
Hence, the angle between the tangents is given as $$\tan \theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ now, setting $m_1=11$ & $m_2=-11$, we get
$$\tan \theta=\left|\frac{11-(-11)}{1+11(-11)}\right|$$
$$=\left|\frac{11}{-60}\right|=\frac{11}{60}$$  Hence, $$\theta=\tan^{-1}\left(\frac{11}{60}\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{acute angle:}\ \theta=\tan^{-1}\left(\frac{11}{60}\right)\approx10.39^\circ}}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{obtuse angle:}\ \theta=\pi-\tan^{-1}\left(\frac{11}{60}\right)\approx 169.61^\circ}}$$

Basically the slope or the coeffecient of x in the tangent equation represents tan(theta)

So what you need to do is get artan(-11)= -84.81 then arctan(11)=84.81 and find the difference =169.62 deg you might want to take use the acute angel 10.39 deg


Find orthogonal matrix missing values I was solving this problem, where I need to find the value x, which is missed in the orthogonal matrix A.
$$
A =
\begin{pmatrix}
x&0.5&-0.5&-0.5\\
x&0.5&0.5&0.5\\
x&-0.5&-0.5&0.5\\
x&-0.5&0.5&-0.5\\
\end{pmatrix}
$$
One of the properties of orthogonal matrix is that the dot product of orthogonal matrix and its transposed version is the identity matrix:
$$I=A.A^T$$
$$
A^T =
\begin{pmatrix}
x&x&x&x\\
0.5&0.5&-0.5&-0.5\\
-0.5&0.5&-0.5&0.5\\
-0.5&0.5&0.5&-0.5\\
\end{pmatrix}
$$
So I figured that $$x=0.5$$
That would result in identity matrix I:
$$
I =
\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{pmatrix}
$$
As the result I had to provide the sum of missing values, which $0.5\times4 = 2$. The grader (program) returned the error. Is there something I am missing (there most likely is, I guess)?
Thank you in advance!

$A$ is orthogonal $ \iff x= \pm \frac{1}{2}.$

Thank you everyone, indeed there are two possible answers $$x = 1/2$$ $$x = -1/2$$, therefore there can't be a definitive answer.


Whats wrong with this proof? (trying to prove a function is surjective) Let $f : A → B$ and $g : A' → B'$ be functions that are onto, and $h : A × A' → B × B'$
be the function $h(x, y) = (f(x), g(y))$.

Part1. Dr. Bob tries to prove that $h$ is onto with the following argument. Let
$A = B = \{1, 2\}$, $A' = B' = \{a, b\}$, $f$ be given by $f(1) = 2$ and $f(2) = 1$, and $g$ be
given by $g(a) = a$ and $g(b) = b$. Then clearly $h$ is onto, because $h(2, a) = (1, a)$,
$h(1, a) = (2, a)$, $h(2, b) = (1, b)$, and $h(1, b) = (2, b)$. What is wrong with Dr. Bob’s
proof?

I can't figure out what is wrong, bob has the outputs for $h$ as $(1,a)$, $(2,a)$, $(1,b)$, and $(2,b)$. Isn't that every element that is possible for the output? Isn't that enough to make it onto?

The second part of this problem is to correctly prove that $h$ is onto, but my proof would like bob's proof, which is incorrect.

What if $A, B$ are different sets than $\{1, 2\}$? What if $A = \mathbb{Q}$ and $B = \mathbb{N}$?

You can't proof a general statement by specializing to a convenient case.

You need to get back to basics and definition.

Take $\forall z\in A'\times B'$, $z=(z_x,z_y)$, with $z_x\in A'$ and $z_y\in B'$.

Since $z_x\in A'$, $\exists x\in A$ such that $f(x)=z_x$ (by definition of $f$)

Similarly $\exists y\in B$ such that $g(y)=z_y$

Thus

$\forall z=(z_x,z_y)\in A' \times B'$,  $\exists (x,y)\in A \times B$ such that $h(x,y)=(f(x),g(y))=(z_x,z_y)$


Show that an entire function $f$ s.t. $|f(z)|>1$ for $|z|>1$ is a polynomial I have been struggling on the following problem.

Suppose $f$ is an entire analytic function such that $|f(z)|>1$ if $|z|>1$. Show that $f$ is a polynomial.

My idea is as followed: all zeros of $|f(z)|$ lie inside $|z|\leq 1$. Applying Argument Principle, we can show that number of zeros of $f$ is bounded. So we can assume

$f(z)=(z-z_1)...(z-z_M)g(z)$

where g is entire analytic without any zeros.
Then I would like to apply Liouville's Theorem: the point is that it isn't too clear to me why $|\dfrac{1}{g(z)}|$ is a bounded function.

If the Taylor series about 0 does not terminate, $f(1/z)$ has an essential singularity at $0$ (why?)

Then from the Casorati–Weierstrass theorem (have you learnt this?) you know $f(1/z)$ cannot be bounded from below near 0. Contradiction!

The number of zeros of $f$ can be finite as the set of zeros of $f$ will have a accumulation point and $f$ will be equivalent to $0$ (not possible) then
$$g=\frac{f}{(z-a)\cdots(z-x)}$$
where $a$ to $x$ are the finite number of zeros of $f$ then $g$ is analytic consider $h=1/g$ see that $h(z)$ is never zero as we have deleted the zeros of $f$ and also $f$ tends to infinity as $z$ tends to infinity (by the given condition) which implies
$$|h(z)| \le a+|z|^n,$$
where $n$ is the number of zeros of $f$, but by fundamental theorem of algebra we get $h(z)=c$ (constant) thus
$$f=\frac{(z-a)\cdots(z-x)}{k}.$$


Zero to the zero power – is $0^0=1$? Could someone provide me with a good explanation of why $0^0=1$?

My train of thought:

$x>0$

$0^x=0^{x-0}=0^x/0^0$, so

$0^0=0^x/0^x=\,?$

Possible answers:

$0^0\cdot0^x=1\cdot0^0$, so $0^0=1$
$0^0=0^x/0^x=0/0$, which is undefined

PS. I've read the explanation on mathforum.org, but it isn't clear to me.

In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined.

Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$?

Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined.

However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period.

Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory).

So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we  make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$).

Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$
is the number of ways of making no selections from no choices.

Coda. In the end, it is a matter of definition and utility. In Calculus and  algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions.

We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way.

We must note that the term $0^0$ is not a normal number since there is no arithmetic calculation that leads to form the $0^0$. We have two options:

A) We set $0^0=0$ and demand that all arithmetic operations are subjected to number $0$ and not the term $0^0$, e.g. writing $0^0=0^{(1-1)}$ is forbidden. This definition is nice since adding zero doesn't change any equation.

$\space$

B) We set $0^0=1$ and demand that all arithmetic operations are subjected to number $1$ and not the term $0^0$, e.g. writing $0^0=0^{(1-1)}$ is forbidden. This definition is nice since the limit $x^0 \rightarrow 1$ as $x \rightarrow 0$ is consistent with this definition.

Summa summarum: if we want to define the number $0^0$, then the most useful choice is $1$.


Is $f$ a continuous function? I seek to understand whether a certain discontinuous function on the dyadic rationals can be recast using the Cantor set as a continuous function.

Let $X$ be the dyadic rationals in the interval $(\frac12,1]$

Let $2^{\nu_2(x)}$ be the highest power of $2$ that divides $x$.

Let $f:X\to X$ be given by

$f(x)=\begin{cases}
\frac14(3x+2^{\nu_2(x)})& \text{ if} & x>\frac23 \\
\frac12(3x+2^{\nu_2(x)})& \text{ if} & x<\frac23\\
\end{cases}$

Now cut the interval at $2/3$ and glue the two parts at $1/2=1$ and thereby, $f$ will run from $(\frac23,\frac76)$.

I think at the moment this is NOT a continuous function (under the normal metric) because e.g. $\lim_{n\to\infty}f(7/8-2^{-n})=f(7/8)-1/8$.

  The question is this:  Can reinterpreting the binary expansion of $x$ as its Cantor set by sending $2^{-m}\mapsto2\cdot3^{-m}$ make this a continuous function via the devil's staircase?

Let $\phi(x):\sum_{n=1}^\infty 2^ni:i\in\{0,1\}\mapsto\sum_{n=1}^\infty 3^ni:i\in\{0,2\}$

Then is $\phi f\phi^{-1}(x)$ continuous?

As a little background - I'm aware I am asking the xy question.  My objective is to make $f$ a continuous function from a segment of $\Bbb R$ to itself. To do so would prove there are no nontrivial cycles in the Collatz conjecture via Sharkovskii's theorem.

For this to be successful, the "recasting" of the function as a continuous function on a segment of $\Bbb R$ must only have cycles of the same order as any cycles of the original $f$.

I'm pretty sure there's a way to do this but sadly I'm also pretty sure it's not exactly what I'm doing here.  But to see what I'm doing wrong or how this is impossible will be as much help to me as a successful answer.

I suppose there will be mechanisms to relate discrete and continuous mathematics, but if your question is destined to the absence of cycles in the Collatz conjecture, you could try to put your question in the  graph theory, for example,

is it a forest, or equivalently, is a disjoint union of trees?

Is a directed tree to the case where the edges are all directed towards a particular vertex, or all directed away from a particular vertex?

is a connected graph? is a labeled graph?

the numbers n=1(mod 3) are the nodes of the union of disjoint trees?


Prove "Any open subset of $\Bbb R$ is an at most countable union of pairwise disjoint open intervals”. Let $E$ be a nonempty open subset of $\Bbb R$. There is a collection $A$ of subsets of $\Bbb R$ satisfying the following:

$(i)$ Each element of $A$ is an open interval (possibly infinite) and any two distinct members of $A$ are disjoint;
$(ii)$ $A$ is at most countable;
$(iii)$ $E =\bigcup\limits_{G\in A}G$.

Prove that ”Any open subset of $\Bbb R$ is an at most countable union of pairwise disjoint open intervals”.

I don't know any structure theorems. I only have done a course on real analysis. Which dealt with compact sets and all.

Let $\mathcal{I}$ be the set of the intervals in $\mathbb{R}$ with rational endpoints and $E \subset \mathbb{R}$ an open set.

For all $x \in E$, prove there exist a rational $r_x > 0$ and a rational $q_x$ such that $x \in\, ]r_x-q_x, q_x + r_x[$ and $]r_x-q_x, q_x + r_x[ \subset E$. Hint: use the density of $\mathbb{Q}$ in $\mathbb{R}$.
Deduce that $E$ can be written as a union of elements of $\mathcal{I}$.
Prove that $\mathcal{I}$ is countable. Hint: use the fact that $\mathbb{Q}\times \mathbb{Q}$ is countable.
Deduce you can write $E$ as a countable union of elements of $\mathcal{I}$.
Let $E = \cup_{\alpha \in A} I_{\alpha}$ be such a union. For each $\alpha \in A$, prove there exists a largest open interval $J_{\alpha}$ such that $I_{\alpha} \subset J_{\alpha} \subset E$. This $J_{\alpha}$ is not necessary in $\mathcal{I}$.
Conclude you can write $E$ as a countable union of pairwise disjoint intervals.


Combinatorics using a geometric diagram How can I do this without trial-and-error?

It has something to do with a triangle and summing the next row?

After seeing JorleArbake's comment linking to http://abelkonkurransen.no/problems/abel_1314_r1_sol_nb.pdf it includes a solution described here:

Their argument is to recursively find the number of paths from the starting cell by finding the number of paths starting from cells in the middle of the grid instead.

For all of the triangles on the bottom row, there is only one path starting from it and leading to the end (namely it itself).  From there, notice that picking a starting square on the next level up, you may continue downwards to any of the adjacent cells, thus inheriting all of their (possibly multiple) different paths having started from each of them.  Continuing this process working upwards, you find that you can find the total number of paths from some parent cell is equal to the sum of the total number of paths from each adjacent child.  Filling in the grid from bottom up leads eventually to $56+41+56 = 153$

-Edit- Below is my original answer to this question, which still has some merit in leaving up since it shows a method to at least bound the size on the answer using elementary arguments.

In a very common question similar to this one, the definition of "adjacent" would have been such that at every step we have only three moves possible (straight down, down-left, or down-right), in which case the answer would be $3^4=81$ seen quickly from the multiplication principle.

With the definition of "adjacency" given here as two cells are adjacent if their respective borders share at least one point in common (i.e., share a corner vertex or edge), you can quickly see that there will be at least 3 choices available at every step (and in particular at some steps will have 4 or even 5 choices available), and that the available paths from the previously mentioned problem will all be valid paths for this one.  As such we know that the answer will be strictly greater than 81.  This rules out the possibility of $(a)$ as being the correct answer.

Notice then also that there will be at most 5 choices available at each step, so the number of possibilities must be strictly less than $5^4=625$, thus ruling out the possibility of $(e)$ being the answer.

To proceed further, intuitively one might expect that instead of looking at $3^4$ and $5^4$, you could look at the average number of choices raised to the fourth power (through some argument relating to geometric mean).  The number of choices from each triangle in order is: 3,4,3,4,4,3,5,3,4,4,3,5,3,5,3,4.  So, the average seems slightly under 4.  Thus we would expect the total number of possibilities to be less than $4^4 = 256$, ruling out $(d)$ as being a correct answer.

This leaves us with two possibilities left, either $(b)$ or $(c)$.  I do not see at the moment a good argument to choose between these two.

Just want to point out that the average number of choices is $3.75$, so the answer must be lower than $197.75$. The answer must be b) $153$ (from choices provided).


Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.

Remember, that for any nonzero $x$, one has

$$
x+\frac{1}{x} \geq 2 \tag{1}
$$

We apply this inequality repeatedly. Put

$$
T=\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab}
$$

We have
$$
T=\frac{a^2+b^2}{abc}+\frac{c}{ab} =\frac{\sqrt{a^2+b^2}}{ab} \bigg(\frac{\sqrt{a^2+b^2}}{c}+\frac{c}{\sqrt{a^2+b^2}
}\bigg) \geq \frac{2(\sqrt{a^2+b^2})}{ab},
$$

by using (1) with $x=\frac{\sqrt{a^2+b^2}}{c}$. Then,

$$
T \geq \frac{2}{\sqrt{ab}} \sqrt{\frac{a}{b}+\frac{b}{a}} \geq \frac{2\sqrt{2}}{\sqrt{ab}}
$$
by using (1) with $x=\frac{a}{b}$. So $T \geq \sqrt{\frac{8}{ab}}$.

We see that if $ab \leq 8$, we are done. So we may assume $ab \geq 8$. By symmetry, we may also assume
$ac \geq 8, bc \geq 8$. But then

$$
1 \leq \dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \leq \frac{1}{8}+\frac{1}{8}+\frac{1}{8} =\frac{3}{8},
$$

which is impossible.

multiply abc

$a+b+c\geq abc $

$a^2+b^2+c^2\geq abc $

$a^2+b^2+c^2-(a+b+c)$

$⇔(a-\frac12)^2+(b-\frac12)^2+(c-\frac12)^2\geq \frac34$

therefore

$a^2+b^2+c^2\geq \frac34+abc$


Video Lectures for Linear Algebra Are there any video lectures for linear algebra other than Strang's?. I'm looking for something more pure. Ideally, from a professor who follows a book similar to Hoffman and Kunze or Axler.

Thanks!

I have made a new series of videos to accompany Linear Algebra Done Right. These videos are available here: http://www.linear.axler.net/LADRvideos.html

http://www.khanacademy.org/math/linear-algebra

Here is one place to look


Probability of guessing a PIN-code A friend and I recently talked about this problem:

Say my friend feels a little adventurous and tells me that exactly three of four digits of his PIN-code are the same, what is the probability that I will guess it in three tries?

At first I thought this shouldn't be too difficult to count, but the digit restriction threw me off. Essentially I want to count how many possible PIN-codes there are with the restriction that $3$ of $4$ digits are the same. I tried thinking in terms of using sums, but I got stuck. I actually ended up making a quick MATLAB-script that computed the number of possible PIN-codes using a brute force method. Assuming that my script is correct there are $360$ codes that abide by this restriction out of a total of $10^4=10\hspace{4 px}000$ possible PIN-codes. Using this it is easy to calculate the rest, but I am now wondering how one might go about this in a more elegant way.

A PIN-code is a $4$-digit number where the possible digits are $0,1,2,...,9$. So for my question two examples of possible codes are $3383$ and $2999$. Let's assume that there are no further restrictions, although in reality there likely are, and that each digit is equally likely. It is important to note that I do not know if it is $0,1,...,8$, or $9$ that appears three times.

This question is not homework or anything, it is really just for curiosity. Thanks for any help!

(By the way I saw this question: Combinatorics and Probability Problem but it did not help me.)
EDIT: I made an error in my script. Updated.

Suppose $n$ is repeated. There are 9 other numbers that can occur. And the other digit can occur in 4 possible positions giving $36$ possibilities.

There are $10$ possibilities for $n$ so the total number of combinations with exactly three digits the same is $360$.

10000 guesses that would be including double and triple and four same number
0000,0001,0002, 1000 ,1111,2245 ,5001  ,8768 ,9990.  .eg
So the chance of get a four number pin is 1 in 10, 000 that's is 0000 is included


Implicit function theorem. Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^1$ function,and $a\in \mathbb{R}^n$ such that $\frac{\partial f}{\partial x_i}(a)\ne 0$ for all $i$ and $f(a)=0$.
a) Prove that there is a neighbour of $a$ such that the equation $f=0$ fix $n$ functions:
$x_1(x_2,...,x_n),...,x_n(x_1,x_2,...,x_{n-1})$
b) Find $\Pi^n_{i=1}\frac{\partial x_i}{\partial x_{i+1}}$, where the indices are $mod(n)$ and all of the relevant derivatives are at $a$.

I have absolutely no idea how to begin...any ideas?

Il order not to mix everything let us denote $X_i$ the new functions, more precisely as $\partial f\over \partial x_i$$ \not =0$, there  exists near $(a_1,....a_{i-1},a_{i+1},...a_n)$  a function $X_i (x_1,...,x_{i-1},x_{i+1},...x_n)$ such that $f(x_1,...,x_{i-1},X_i, x_{i+1},...x_n)=0$ and $X_i (a_1,...,a_{i-1},a_{i+1},...a_n)=a_i$.

Deriving the equation $X_i (x_1,...,x_{i-1},x_{i+1},...x_n)$ by $x_{i+1}$ we get :

${\partial f \over \partial x_i} (a).{\partial X_{i+1} \over \partial x_i} +{\partial f \over \partial x_{i+1}} (a)=0 .$

Then ${\partial X_{i+1} \over \partial x_i}= -{ {\partial f \over \partial x_{i+1}} (a)\over {\partial f \over \partial x_{i}} (a)}$

We deduce  the product :

$\Pi_1^n{\partial X_{i+1} \over \partial x_i}= (-1)^n$

Hint:

For part one, Take $i$ and  Apply implicit function theorem on equation $f(x_1 x_2,...,x_n)=0$ at point $x=a$ to represent $x_i$ as function of $(x_1,..,x_{i-1}, x_{i+1}..,x_n)$ i.e., $x_i(x_1,..,x_{i-1}, x_{i+1}..,x_n)$. Note that you know $\frac{\partial f}{\partial x_i}(a)\ne 0$.

For part b) use chain rule .

Note that $\Pi^n_{i=1}\frac{\partial x_i}{\partial x_{i+1}} =\frac{\partial x_1}{\partial x_{1}} =1   $


Please explain the intuition behind the dual problem in optimization. I've studied convex optimization pretty carefully, but don't feel that I have yet "grokked" the dual problem.  Here are some questions I would like to understand more deeply/clearly/simply:

How would somebody think of the dual problem?  What thought process would lead someone to consider the dual problem and to recognize that it's valuable/interesting?
In the case of a convex optimization problem, is there any obvious reason to expect that strong duality should (usually) hold?
It often happens that the dual of the dual problem is the primal problem.  However, this seems like a complete surprise to me.  Is there any intuitive reason to expect that this should happen?
Does the use of the word "dual" or "duality" in optimization have anything to do with the dual space in linear algebra?  Or are they just different concepts that go by the same name.  What about the use of the word "dual" in projective geometry — is there a connection there?
You can define the dual problem and prove theorems about strong duality without ever mentioning the Fenchel conjugate.  For example, Boyd and Vandenberghe prove a strong duality theorem without mentioning the Fenchel conjugate in their proof.  And yet, people often talk as if the Fenchel conjugate is somehow the "essence" of duality, and make it sound as if the whole theory of duality is based on the Fenchel conjugate.  Why is the Fenchel conjugate considered to have such fundamental importance?

Note: I will now describe my current level of understanding of the intuition behind the dual problem.  Please tell me if you think I might be missing any basic insights.
I have read the excellent notes about convex optimization by Guilherme Freitas, and in particular the part about "penalty intuition".  When we are trying to solve
\begin{align*}
\text{minimize} &\quad f(x) \\
\text{such that} & \quad h(x) \leq 0
\end{align*}
one might try to eliminate the constraints by introducing a penalty when constraints are violated.  This gives us the new unconstrained problem
\begin{equation}
\text{minimize} \quad f(x) + \langle \lambda ,h(x) \rangle
\end{equation}
where $\lambda \geq 0$. It's not hard to see that for a given $\lambda \geq 0$, the optimal value of this unconstrained problem is less than or equal to the optimal value for the constrained problem.  This gives us a new problem — find $\lambda$ so that the optimal value for the unconstrained problem is as large as possible.  That is one way to imagine how somebody might have thought of the dual problem.  Is this the best intuition for where the dual problem comes from?
Another viewpoint: the KKT conditions can be derived using what Freitas calls the "geometric intuition".  Then, if we knew the value of the multipliers $\lambda$, it would be (often) much easier to find $x$.  So, a new problem is to find $\lambda$.  And if we can somehow recognize that $\lambda$ is a maximizer for the dual problem, then this suggests that we might try solving the dual problem.
Please explain or give references to any intuition that you think I might find interesting, even if it's not directly related to what I asked.

Here's what's really going on with the dual problem.  (This is my attempt to answer my own question, over a year after originally asking it.)

(A very nice presentation of this material is given in Ekeland and Temam.  These ideas are also in Rockafellar.)

Let $V$ be a finite dimensional normed vector space over $\mathbb R$.  (Working in an inner product space or just in $\mathbb R^N$ risks concealing the fundamental role that the dual space plays in duality in convex optimization.)

The basic idea behind duality in convex analysis is to think of a convex set in terms of its supporting hyperplanes.  (A closed
convex set $\Omega$ can be "recovered" from its supporting hyperplanes
by taking the intersection of all closed half spaces containing $\Omega$.
The set of all supporting hyperplanes to $\Omega$ is sort of a
"dual representation" of $\Omega$.)

For a convex function $f$ (whose epigraph is a convex set), this strategy leads
us to think about $f$ in terms of affine functions $\langle m^*, x \rangle - \alpha$
which are majorized by $f$.  (Here $m^* \in V^*$ and we are using the notation $\langle m^*, x \rangle = m^*(x)$.)

For a given slope $m^* \in V^*$, we only need to consider the "best" choice of $\alpha$ -- the other affine minorants with slope $m^*$ can be disregarded.
\begin{align*}
& f(x) \geq \langle m^*,x \rangle - \alpha \quad \forall x \in V \\
\iff & \alpha \geq  \langle m^*, x \rangle - f(x) \quad \forall x \in V \\
\iff & \alpha \geq \sup_{x \in V} \quad \langle m^*, x \rangle - f(x)
\end{align*}
so the best choice of $\alpha$ is
\begin{equation}
f^*(m^*) = \sup_{x \in V} \quad \langle m^*, x \rangle - f(x).
\end{equation}
If this supremum is finite, then
$\langle m^*,x \rangle - f^*(m^*)$ is the best affine minorant of
$f$ with slope $m^*$.
If $f^*(m^*) = \infty$, then there is no affine minorant of $f$ with slope $m^*$.

The function $f^*$ is called the "conjugate" of $f$.  The definition and basic facts about $f^*$ are all highly intuitive.  For example, if $f$ is a proper closed convex function then $f$ can be recovered from $f^*$, because any closed convex set (in this case the epigraph of $f$) is the intersection of all the closed half spaces containing it.  (I still think the fact that the "inversion formula" $f = f^{**}$ is so simple is a surprising and mathematically beautiful fact, but not hard to derive or prove with this intuition.)

Because $f^*$ is defined on the dual space, we see already the fundamental role played by the dual space in duality in convex optimization.

Given an optimization problem, we don't obtain a dual problem until we specify how to perturb the optimization problem.  This is why equivalent formulations of an optimization problem can lead to different dual problems.  By reformulating it we have in fact specified a different way to perturb it.

As is typical in math, the ideas become clear when we work at an appropriate level of generality.  Assume that our optimization problem is
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,0).
\end{equation*}
Here $\phi:X \times Y \to \bar{\mathbb R}$ is convex.  Standard convex optimization problems can be written in this form with an appropriate choice of $\phi$.  The perturbed problems are
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,y)
\end{equation*}
for nonzero values of $y \in Y$.

Let $h(y) = \inf_x \phi(x,y)$.  Our optimization problem is simply to evaluate $h(0)$.

From our knowledge of conjugate functions, we know that
\begin{equation*}
h(0) \geq h^{**}(0)
\end{equation*}
and that typically we have equality.  For example, if $h$ is subdifferentiable at $0$ (which is typical for a convex function) then $h(0) = h^{**}(0)$.

The dual problem is simply to evaluate $h^{**}(0)$.

In other words, the dual problem is:
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - h^*(y^*).
\end{equation*}
We see again the fundamental role that the dual space plays here.

It is enlightening to express the dual problem in terms of $\phi$.  It's easy to show that the dual problem is
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - \phi^*(0,y^*).
\end{equation*}

So the primal problem is
\begin{equation*}
\operatorname*{minimize}_{x \in X} \quad \phi(x,0)
\end{equation*}
and the dual problem (slightly restated) is
\begin{equation*}
\operatorname*{minimize}_{y^* \in Y^*} \quad \phi^*(0,y^*).
\end{equation*}
The similarity between these two problems is mathematically beautiful, and we can see that if we perturb the dual problem in the obvious way, then the dual of the dual problem will be the primal problem (assuming $\phi = \phi^{**}$).  The natural isomorphism between $V$ and $V^{**}$ is of fundamental importance here.

The key facts about the dual problem -- strong duality, the optimality conditions, and the sensitivity interpretation of the optimal dual variables -- all become intuitively clear and even "obvious" from this viewpoint.

An optimization problem in the form
\begin{align*}
\operatorname*{minimize}_x & \quad f(x) \\
\text{subject to} & \quad g(x) \leq 0,
\end{align*}
can be perturbed as follows:
\begin{align*}
\operatorname*{minimize}_x & \quad f(x) \\
\text{subject to} & \quad g(x) + y \leq 0.
\end{align*}

This perturbed problem has the form given above with
\begin{equation*}
\phi(x,y) =
\begin{cases}
f(x) \quad \text{if } g(x) + y \leq 0 \\
\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}
To find the dual problem, we need to evaluate $-\phi^*(0,y^*)$, which is a relatively straightforward calculation.
\begin{align*}
-\phi^*(0,y^*) &= -\sup_{g(x) + y \leq 0} \quad \langle y^*,y \rangle - f(x) \\
&= -\sup_{\substack{ x \\ q \geq 0 }} \quad \langle y^*, -g(x) - q \rangle - f(x) \\
&= \inf_{\substack{ x \\ q \geq 0 }} \quad f(x) + \langle y^*, g(x) \rangle + \langle y^*, q \rangle.
\end{align*}
We can minimize first with respect to $q$, and we will get $-\infty$ unless $\langle y^*, q \rangle \geq 0$ for all $q \geq 0$.  In other words, we will get $-\infty$ unless $y^* \geq 0$.

The dual function is
\begin{equation*}
-\phi^*(0,y^*) =
\begin{cases}
\inf_x \quad f(x) + \langle y^*, g(x) \rangle \quad \text{if } y^* \geq 0 \\
-\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}

This is the expected result.

Here are some counter-examples to help you understand KKT conditions and strong duality. The answer is from my other post: https://math.stackexchange.com/a/4154563/273731
${\bf counter-example 1}$
If one drops the convexity condition on objective function, then strong duality could fails even with relative interior condition.
The counter-example is the same as the following one.
${\bf counter-example 2}$
For non-convex problem where strong duality does not hold, primal-dual optimal pairs may not satisfy KKT condition.
Consider the optimization problem
\begin{align}
 \operatorname{minimize} & \quad e^{-x_1x_2} \\
 \text{subject to} & \quad x_1\le 0.
\end{align}
The domain for the problem is $D = \{ (x_1,x_2) \ge 0 \}$. The problem is not convex by calculating the Hessian matrix. Clearly, any $x_1 = 0, x_2 \in\mathbb R_+$ is a primal optimal solution with primal optimal value $1$ .
The Lagrangian is
$$
L(x_1,x_2,\lambda) = e^{-x_1x_2} + \lambda  x_1.
$$
The dual function is
\begin{align}
 G(\lambda) &= \inf L(x_1,x_2,\lambda) =
 \begin{cases}
  0&  \lambda\ge 0\\
  -\infty& \lambda < 0
 \end{cases}
\end{align}
Thus, $\lambda \geq 0$ is dual optimal solution with dual optimal value $0$, so dual gap is $1$, strong duality fails. As for the KKT conditions, remember the domain is $D = \{ (x_1,x_2) \ge 0 \}$
\begin{align*}
 &\lambda-x_2e^{-x_1x_2}=0\\
 &x_1\le 0\\
 &\lambda\ge 0\\
 &\lambda x_1=0\\
\end{align*}
Pick any primal-dual pair satisfying $x_1 = 0, x_2\ge 0, \lambda\ge0, \lambda\ne x_2$, the KKT conditions fail.
${\bf counter-example 3}$
For a non-convex problem, even strong duality holds, solutions for KKT conditions may not be primal-dual optimal solution.
Consider the optimization problem on $\mathbb R$
\begin{align}
 \operatorname{minimize} & \quad x^3 \\
 \text{subject to} & -x^3-1\le 0.
\end{align}
The objective function is not convex by calculating the Hessian matrix. Clearly, $x=-1$ is the unique primal optimal solution with primal optimal value $-1$.
The Lagrangian is
$$
L(x,\lambda) = x^3 - \lambda  (x^3+1).
$$
The dual function is
\begin{align}
 G(\lambda) &= \inf L(x,\lambda) =
 \begin{cases}
  -1&  \lambda=1\\
  -\infty& otherwise
 \end{cases}
\end{align}
Thus, $\lambda = 1 $ is dual optimal solution with dual optimal value $-1$, so dual gap is $0$, strong duality holds. While the KKT conditions are
\begin{align*}
 &3x^2(1-\lambda)=0\\
 &-x^3-1\le 0\\
 &\lambda\ge 0\\
 &\lambda (-x^3-1)=0\\
\end{align*}
Solutions for KKT conditions are $x=-1, \lambda=1$ or $x=0,\lambda=0$. Notice that $x=0,\lambda=0$ satisfies KKT conditions but has nothing to do with primal-dual optimal solutions.
The discussion indicates for non-convex problem, KKT conditions may be neither necessary nor sufficient conditions for primal-dual optimal solutions.
${\bf counter-example4}$
For a convex problem, even strong duality holds, there could be no solution for the KKT condition, thus no solution for Lagrangian multipliers.
Consider the optimization problem on domain $\mathbb R$
\begin{align}
 \operatorname{minimize} & \quad x \\
 \text{subject to} & \quad x^2\le 0.
\end{align}
Obviously, the problem is convex with unique primal optimal solution $x=0$ and optimal value $0$; Feasible set is $\{0\}$, therefore Slater's condition fails.
The Lagrangian is
$$
L(x,\lambda) = x + \lambda  x^2.
$$
The dual function is
\begin{align}
 G(\lambda) &= \inf L(x,\lambda) =
 \begin{cases}
  -\infty&  \lambda\le 0\\
  -\frac{1}{4\lambda} &\lambda >0
 \end{cases}
\end{align}
Thus, dual optimal value is $0$, so dual gap is $0$, strong duality holds. However, there are no solution for dual optimal solution because the optimal value is attained as $\lambda\rightarrow \infty$. As for the KKT conditions
\begin{align*}
 &1+2\lambda x=0\\
 &x^2\le 0\\
 &\lambda\ge 0\\
 &\lambda x^2=0\\
\end{align*}
No solution for KKT conditions.
This is the convex problem where the dual problem has no feasible solution and KKT conditions have no solution but the primal problem is simple to solve.
${\bf counter-example 5}$
For a differentiable convex problem, there could be no solution for the KKT conditions, even if the primal-dual pair exists. In that case, the strong duality fails.
Consider the optimization problem on domain $D:=\{(x,y):y>0\}$
\begin{align}
 \operatorname{minimize} & \quad e^{-x} \\
 \text{subject to} & \quad \frac{x^2}{y}\le 0.
\end{align}
The problem can be proved to be convex with primal optimal solution $x=0, y>0$ and optimal value $1$; Feasible set is $\{(0,y): y > 0\}$, therefore Slater's condition fails.
The Lagrangian is
$$
L(x,y,\lambda) = e^{-x} + \lambda  \frac{x^2}{y}.
$$
After some careful calculation, the dual function is
\begin{align}
 G(\lambda) &= \inf L(x,y,\lambda) =
 \begin{cases}
  0&  \lambda\ge 0\\
  -\infty &\lambda <0
 \end{cases}
\end{align}
Thus, dual optimal value is $0$, so dual gap is $1$, strong duality fails. We can pick primal-dual pair to be $x=0, y=1, \lambda=2$. As for the KKT conditions
\begin{align*}
 &-e^{-x}+\frac{2\lambda x}{y}=0\\
 &\frac{x^2}{y}\le0\\
 &\lambda\ge 0\\
 &\lambda \frac{x^2}{y}=0\\
\end{align*}
with $y>0$, thus no solution for KKT conditions.
This counter-example warns us that we have to be careful about the strong duality condition even for differentiable convex problems.