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# Blind Signal Separation for Sparse Signals

Assume we have $N$ measurements $z_1, z_2, \dots, z_N \in \mathbb{R}^{n_z}$ that are generated by

$$z_i = M v_i + e_i$$

where $v_i \in \mathbb{R}^{n_v}$, $n_v < n_z$ and $e_i \in \mathbb{R}^{n_z}$ is an error sampled from normal distribution $\mathcal{N}(0, \sigma^2 I)$. Unknown (tall) matrix $M \in \mathbb R^{n_z \times n_v}$ should have full column rank.

Given only the measurements $z_1, z_2, \dots, z_N$ and no information about $M$, $v_i$ and $e_i$, my task is to find a matrix $B \in \mathbb R^{n_v \times n_z}$ that estimates

$$\hat v_i \approx B \hat z_i$$

for new samples $(\hat v_i, \hat z_i)$ generated by same distributions as before. This is referred to as blind signal separation (BSS). I don't know the distribution of $v_i$ at all, but I have the information that the entries $v_{i,j}$ will always be around $0$ for all indices $j=1,\dots,n_v$ except a small number of indices, up to $k$ many (it's around $\frac{k}{n_v} \leq 0.2$). So, $v_i$ is always "approximately sparse".

1. Is there a good way to use this information to improve certain BSS algorithms to deliver a $B$ that will more probably deliver solutions $\hat v_i$ which tend to have the above sparsity property?

2. Or are there certain algorithms for BSS that are especially well suited for this special case?

3. Is there maybe a clever way to reformulate the BSS problem to contain the sparsity property?

To explain a bit about my application. I want to build software that is filled with piano music and should extract the notes being played. I take several MP3 files and generate samples $z_i$ from wavelet-transforming the sound signal at different time stamps. So a $z_i$ is basically the vector of amplitudes to different frequencies at a certain time. Then the $v_i$ is the vector of the intensities of the different notes being played (obviously unknown resp. not practicable to look up). If you play one note $v_{i,j}=\delta_{ij}$ you get all its harmonic frequencies in $z_i$ which makes it difficult to filter out.

## migrated from mathoverflow.netAug 4 at 11:52

This question came from our site for professional mathematicians.

• It looks similar to Blind Deconvolution problem just without the assumption of $M$ being a Deconvolution operator. – Royi Aug 6 at 9:54


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You are reading an older version of this FlexBook® textbook: CK-12 Algebra II with Trigonometry Concepts Go to the latest version.

6.9: Long Division of Polynomials

Difficulty Level: At Grade Created by: CK-12

The area of a rectangle is $6x^3 - 12x^2 + 4x - 8$ . The width of the rectangle is $2x - 4$ . What is the length?

Guidance

Even though it does not seem like it, factoring is a form of division. Each factor goes into the larger polynomial evenly, without a remainder. For example, take the polynomial $2x^3-3x^2-8x+12$ . If we use factoring by grouping, we find that the factors are $(2x - 3)(x - 2)(x + 2)$ . If we multiply these three factors together, we will get the original polynomial. So, if we divide by $2x - 3$ , we should get $x^2 - 4$ .

$& \overset{\quad \ \ \ \quad {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{2x-3 \Big ) 2x^3-3x^2-8x+12}$

How many times does $2x$ go into $2x^3$ ? $x^2$ times.

$& \overset{\qquad \ \ \ \overset{{\color{blue}x^2}}{\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}} {{\color{red}2x}-3 \big ) {\color{red}2x^3}-3x^2-8x+12}\\& \quad \quad \quad \underline{2x^3-3x^2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;}\\& \qquad \qquad \ \ 0$

Place $x^2$ above the $x^2$ term in the polynomial.

Multiply $x^2$ by both terms in the divisor ( $2x$ and -3) and place them until their like terms. Subtract from the dividend $(2x^3-3x^2-8x+12)$ . Pull down the next two terms and repeat.

$& \overset{\quad \quad \ \ \overset{\qquad \quad \ x^2 \qquad \ \ -4} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}2x}-3 \Big ) 2x^3-3x^2-8x+12}\\& \quad \quad \quad \ \underline{2x^3-3x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad \ \quad \ \ {\color{red}-8x}+12\\& \quad \quad \quad \quad \quad \quad \quad \quad \underline{-8x+12 \;\;}$

$2x$ goes into $-8x$ 4 times.

After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtracting we notice that everything cancels out. Therefore, just like we thought, $x^2 - 4$ is a factor.

When dividing polynomials, not every divisor will go in evenly to the dividend. If there is a remainder, write it as a fraction over the divisor.

Example A

$(2x^3-6x^2+5x-20) \div (x^2-5)$

Solution: Set up the problem using a long division bar.

$& \overset{\quad \ \ \ \quad {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2-5 \Big ) 2x^3-6x^2+5x-20}$

How many times does $x^2$ go into $2x^3$ ? $2x$ times.

$& \overset{\quad \ \ \ \quad \overset{\qquad \quad \ \ {\color{red}2x}} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}x^2}-5 \Big ) {\color{red}2x^3}-6x^2+5x-20}\\& \quad \quad \quad \ \underline{2x^3-10x^2 \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad \quad 4x^2+5x-20$

Multiply $2x$ by the divisor. Subtract that from the dividend.

Repeat the previous steps. Now, how many times does $x^2$ go into $4x^2$ ? 4 times.

$& \overset{\quad \ \ \ \quad \overset{\qquad \qquad \ 2x+{\color{red}4}} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}x^2}-5 \Big ) 2x^3-6x^2+5x-20}\\& \quad \quad \quad \ \underline{2x^3-10x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad \quad {\color{red}4x^2}+5x-20\\& \quad \quad \quad \quad \quad \quad \quad \underline{4x^2 \qquad \ -20 \;}\\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5x$

At this point, we are done. $x^2$ cannot go into $5x$ because it has a higher degree. Therefore, $5x$ is a remainder. The complete answer would be $2x+4+\frac{5x}{x^2-5}$ .

Example B

$(3x^4+x^3-17x^2+19x-6) \div (x^2-2x+1)$ . Determine if $x^2-2x+1$ goes evenly into $3x^4+x^3-17x^2+19x-6$ . If so, try to factor the divisor and quotient further.

Solution: First, do the long division. If $x^2-2x+1$ goes in evenly, then the remainder will be zero.

$& \overset{\qquad \qquad \quad \overset{\qquad \qquad 3x^2+7x-6} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2-2x+1 \Big ) 3x^4+x^3-17x^2+19x-6}\\& \qquad \qquad \quad \ \ \underline{3x^4-6x^3+3x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \qquad \qquad \quad \ 7x^3-20x^2+19x\\& \quad \quad \quad \qquad \qquad \quad \ \underline{7x^3-14x^2+7x\;\;\;\;\;\;\;\;}\\&\quad \quad \quad \qquad \qquad \qquad \quad -6x^2+12x-6\\&\quad \quad \quad \qquad \qquad \qquad \quad \ \underline{-6x^2+12x-6 \;\;\;}\\&\quad \quad \quad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0$

This means that $x^2-2x+1$ and $3x^2+7x-6$ both go evenly into $3x^4+x^3-17x^2+19x-6$ . Let’s see if we can factor either $x^2-2x+1$ or $3x^2+7x-6$ further.

$x^2-2x+1=(x-1)(x-1)$ and $3x^2+7x-6=(3x-2)(x+3)$ .

Therefore, $3x^4+x^3-17x^2+19x-6=(x-1)(x-1)(x+3)(3x-2)$ . You can multiply these to check the work. A binomial with a degree of one is a factor of a larger polynomial, $f(x)$ , if it goes evenly into it. In this example, $(x-1)(x-1)(x+3)$ and $(3x-2)$ are all factors of $3x^4+x^3-17x^2+19x-6$ . We can also say that 1, 1, -3, and $\frac{2}{3}$ are all solutions of $3x^4+x^3-17x^2+19x-6$ .

Factor Theorem: A polynomial, $f(x)$ , has a factor, $(x - k)$ , if and only if $f(k) = 0$ .

In other words, if $k$ is a solution or a zero , then the factor, $(x - k)$ divides evenly into $f(x)$ .

Example C

Determine if 5 is a solution of $x^3+6x^2-8x+15$ .

Solution: To see if 5 is a solution, we need to divide the factor into $x^3+6x^2-8x+15$ . The factor that corresponds with 5 is $(x - 5)$ .

$& \overset{\qquad \ \overset{\qquad \quad x^2+11x+5} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x-5 \Big ) x^3+6x^2-50x+15}\\& \quad \quad \quad \ \underline{x^3-5x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad 11x^2-50x\\& \quad \quad \quad \quad \quad \quad \underline{11x^2-55x \;\;\;\;\;\;}\\&\quad \quad \quad \quad \quad \quad \quad \quad \quad 5x+15\\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \underline{5x-25 \;}\\&\quad \quad \quad \quad \quad \quad \quad \quad \qquad \quad \ 40$

Since there is a remainder, 5 is not a solution.

Intro Problem Revisit

First, do the long division.

$& \overset{\qquad \qquad \quad \overset{\qquad \qquad 3x^2 + 2} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{2x -4 \Big ) 6x^3-12x^2+4x-8}\\& \qquad \qquad \quad \ \ \underline{6x^3-12x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \qquad \qquad \quad \ 4x - 8\\& \quad \quad \quad \qquad \qquad \quad \ \underline{4x - 8\;\;\;\;\;\;\;\;}\\&\quad \quad \quad \qquad \qquad \quad \ \ 0$

This means that $2x - 4$ and $3x^2+2$ both go evenly into $6x^3-12x^2+4x-8$ .

$3x^2+2$ can't be factored further, so it is the rectangle's length.

Guided Practice

1. $(5x^4+6x^3-12x^2-3) \div (x^2+3)$

2. Is $(x+4)$ a factor of $x^3-2x^2-51x-108$ ? If so, find any other factors.

3. What are the real-number solutions to #2?

4. Determine if 6 is a solution to $2x^3-9x^2-12x-24$ .

1. Make sure to put a placeholder in for the $x-$ term.

$& \overset{\qquad \ \ \ \ \overset{\qquad \qquad 5x^2+6x-27} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2+3 \Big ) 5x^4+6x^3-12x^2+{\color{red}0x}-3}\\& \quad \quad \quad \ \underline{5x^4 \qquad \ +15x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad 6x^3-27x^2+{\color{red}0x}-3\\& \quad \quad \quad \quad \quad \quad \underline{6x^3 \quad \quad \quad \ +18x \;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \qquad \ \ -27x^2-18x-3\\& \quad \quad \quad \quad \quad \qquad \quad \underline{-27x^2 \quad \quad \quad -81 \;\;}\\& \quad \quad \quad \quad \quad \quad \qquad \qquad \quad -18x+78$

The final answer is $5x^2+6x-27-\frac{18x-78}{x^2+3}$ .

2. Divide $(x + 4)$ into $x^3-2x^2-51x-108$ and if the remainder is zero, it is a factor.

$& \overset{\qquad \ \ \ \overset{\qquad \quad x^2-6x-27} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x+4 \Big ) x^3-2x^2-51x-108}\\& \quad \quad \ \ \underline{x^3+4x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \ -6x^2-51x\\& \quad \quad \quad \quad \ \ \underline{-6x^2-24x \;\;\;\;\;\;\;\;\;\;}\\&\quad \quad \quad \quad \quad \quad \quad -27x-108\\&\quad \quad \quad \quad \quad \quad \quad \ \underline{-27x-108 \;}\\&\quad \quad \quad \quad \quad \quad \qquad \qquad \qquad 0$

$x + 4$ is a factor. Let’s see if $x^2 - 6x -27$ factors further. Yes, the factors of -27 that add up to -6 are -9 and 3. Therefore, the factors of $x^3-2x^2-51x-108$ are $(x + 4), (x - 9)$ , and $(x + 3)$ .

3. The solutions would be -4, 9, and 3; the opposite sign of each factor.

4. To see if 6 is a solution, we need to divide $(x - 6)$ into $2x^3-9x^2-12x-24$ .

$& \overset{\qquad \ \ \overset{\qquad \quad \ 2x^2+3x+6} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x-6 \Big ) 2x^3-9x^2-12x-24}\\& \quad \quad \ \ \underline{2x^3-12x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \quad \quad \quad \quad \quad \quad 3x^2-12x\\& \quad \quad \quad \quad \quad \quad \underline{3x^2-18x \;\;\;\;\;\;\;\;\;\;}\\&\quad \quad \quad \quad \quad \quad \quad \quad 6x-24\\&\quad \quad \quad \quad \quad \quad \quad \quad \underline{6x-36 \;\;\;\;\;\;}\\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ 12$

Because the remainder is not zero, 6 is not a solution.

Vocabulary

Long division (of polynomial)
The process of dividing polynomials where the divisor has two or more terms.
Divisor
The polynomial that divides into another polynomial.
Dividend
The polynomial that the divisor goes into. The polynomial under the division bar.
Quotient
The answer to a division problem.

Practice

Divide the following polynomials using long division.

1. $(2x^3+5x^2-7x-6) \div (x+1)$
2. $(x^4-10x^3+15x-30) \div (x-5)$
3. $(2x^4-8x^3+4x^2-11x-1) \div (x^2-1)$
4. $(3x^3-4x^2+5x-2) \div (3x+2)$
5. $(3x^4-5x^3-21x^2-30x+8) \div (x-4)$
6. $(2x^5-5x^3+6x^2-15x+20) \div (2x^2+3)$

Determine all the real-number solutions to the following polynomials, given one factor.

1. $x^3-9x^2+27x-15; (x+5)$
2. $x^3+4x^2-9x-36; (x+4)$
3. $2x^3+7x^2-7x-30; (x-2)$
4. Determine all the real number solutions to the following polynomials, given one zero.
5. $6x^3-37x^2+5x+6; 6$
6. $6x^3-41x^2+58x+15; 5$
7. $x^3+x^2-16x-16; 4$
8. Find the equation of a polynomial with the given zeros.
9. 4, -2, and $\frac{3}{2}$
10. 1, 0, and 3
11. -5, -1, and $\frac{3}{4}$
12. Challenge Find two polynomials with the zeros 8, 5, 1, and -1.

Vocabulary Language: English

Long division (of polynomial)

Long division (of polynomial)

Long division of polynomials is the process of dividing polynomials when the divisor has two or more terms.
Quotient

Quotient

The quotient is the result after two amounts have been divided.

Mar 12, 2013

Apr 23, 2015

We need you!

At the moment, we do not have exercises for Long Division of Polynomials.

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# MTI RADAR, DSP algorithm question

Im new to RADAR signal processing.
I using this doc for ref: https://www.mathworks.com/help/phased/examples/ground-clutter-mitigation-with-moving-target-indication-mti-radar.html

I trying to rebuild the code there, I dont need all the definitions of the antennas and signals because I already have file with I/Q pulses example and I need to process them.

I stuck on the part where I should do [Matched Filter] -> [3-Pulse Canceller]
The way I do it, which i told is wrong, but the result just look better for me is:

1. calculating Matched Filter for every pulse (assuming I working with 64 pulses) and I arrange the filtered data on Matrix like this:

1. take the matched filter result row by row and entering the 3PC filter like this:

and than continue the processing.. the matlab code:

%% Matched Filter
mf_buffer_Q = filter(coeff,1,buffer_Q,[],2);
mf_buffer_I = filter(coeff,1,buffer_I,[],2);

%% Three-Pulse Canceller
h = [1 -2 1];
TriPC_buffer_Q = filter(h,1,mf_buffer_Q,[],2);
TriPC_buffer_I = filter(h,1,mf_buffer_I,[],2);

the plot result:

the other way to do it is what i think is wrong but I told is the right way is:

%% Matched Filter
mf_buffer_Q = filter(coeff,1,buffer_Q,[],1);
mf_buffer_I = filter(coeff,1,buffer_I,[],1);

%% Three-Pulse Canceller
h = [1 -2 1];
TriPC_buffer_Q = filter(h,1,mf_buffer_Q,[],2);
TriPC_buffer_I = filter(h,1,mf_buffer_I,[],2);

that means:

the plot result for this:

what output seems better? what is the right way to do this 2 filters? one horizontal and the second vertical
or I just should do both filters horizontal.

thanks.


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# Maximum height after collision in a bowl

1 vote
61 views

Consider two masses $m, M$ that fall from the top of a hemispherical bowl (from rest) and slide down without any friction.

We want to find how high each of the masses will go after the collision assuming the collision is elastic.

This is how far I can go. It seems to me there are too many unknowns and I am missing some key simplification.

edited Oct 8, 2018
Instead of using the conservation of energy you can use the Law of Restitution, which says that for an elastic collision the relative velocity of approach equals the relative velocity of separation. You still have to use the conservation of momentum also.

You are making the solution harder than it needs to be.  There are 3 steps :

1. Calculate the velocities at collision, which occurs at the bottom of the circle and is $\sqrt{2gh}$ for each .

2. Calculate the velocities after collision. Think of it as a linear collision.

3. The final heights are again given by $v^2=2gh$  but using the velocities after collision.
How do I know the collision occurs at the bottom of the bowl? Is there a mathematical/physical proof that that is true? I can see that they would hit each other at the bottom if they had equal masses.
For the same reason that all bodies accelerate at the same rate $g$ due to gravity.

You can think of the hemispherical bowl as an inclined plane with a variable angle of inclination $\theta$. The acceleration down the incline is $g\sin\theta$ which is independent of mass. Both blocks start from rest at the same height and accelerate at the same rate, so they have the same speed at every height. They reach the bottom at the same time and with the same speed.

If the two blocks have the same mass and the collision is elastic (KE conserved) then they will both return to the height from which they started, because of conservation of energy.

If the masses are different but the collision is still elastic, the smaller mass will rise higher than it started and the larger mass will rise lower than it started. When it reaches the top of the hemisphere the smaller mass will keep going straight up. This is actually quite important because if the hemisphere was instead a sphere then the smaller block would continue moving in a circle. Unless it has enough KE to "loop-the-loop" then it wil lose contact with the sphere at some point and become a projectile with a horizontal component of velocity.  As a result it would not climb as high as when it moves straight up, because it is not stationary at its highest point.

If the two blocks were attached to a smooth circular track then after the collision the following relation holds, unless the smaller block does a "loop-the-loop". (It would be very interesting to calculate (i) what minimum ratio of $\mu=M/m$ is required for the smaller mass to "loop-the-loop", and (ii) how many collisions there are for each value of $\mu$ - there will be a collision each time the smaller block reverses direction and "loops-the-loop" .)

Using conservation of energy (ie gravitational PE), and assuming the blocks are stationary at their maximum heights, the maximum heights after the collision will be related by $$mgh+Mgh=mgh'+Mgh''$$  I was tempted to suggest that the COM will remain in the same position horizontally throughout, but this is not true. The COM starts on the right of the centre line (assuming $M \gt m$) but it is on the centre line when the blocks meet at the lowest point. So the COM does move, because there are external forces acting horizontally on the two blocks - namely the normal contact forces from the hemisphere. (If the hemisphere were resting on a frictionless surface, the COM of the hemisphere and blocks would stay in the same position horizontally, because there are no external forces acting horizontally on this system.)

Another nuance which might affect the result is that the blocks as drawn would actually rotate - ie change their orientation in space. This *might* affect the speed with which they slide down the hemisphere, and also the position at which they collide. It would not affect the above relation, because all kinetic energy (rotational as well as translational) is zero when the blocks come to rest.
Outstanding explanation.
I shall leave you to obtain the solution to the problem. In a few days I will post a solution.

1 vote

The two blocks reach the bottom of the bowl at the same time with the same speed $u=U=\sqrt{2gh}$, as discussed in the comments. These are the initial speeds in the collision.

Momentum and kinetic energy are both conserved during the elastic collision. Instead of applying conservation of KE it is simpler to apply the Law of Restitution with a coefficient $e=1$. This implies that the relative velocity of separation equals the relative velocity of approach.

If the final speeds are $v, V$ for small and large blocks respectively, taking the +ve direction as left for the smaller block and right for the larger, then conservation of linear momentum and the law of restitution give $$(M-m)u=mv-MV$$ $$2u=v+V$$ From these we get $$v=\frac{3M-m}{M+m}u$$ $$V=\frac{3m-M}{m+M}u$$ Note that if $M \gt 3m$ then the larger block does not rebound ($V\lt 0$); it continues moving in its initial direction, chasing the smaller block. The smaller block always rebounds because $3M \gt m$ if $M \gt m$ hence $v \gt 0$ always.

The heights reached after the collision by the smaller and larger blocks are respectively $$h'=\frac{v^2}{2g}=(\frac{3M-m}{m+M})^2h$$ $$H'=\frac{V^2}{2g}=(\frac{3m-M}{M+m})^2h$$

answered Oct 9, 2018 by (26,660 points)
selected Oct 9, 2018
1 vote

This is my solution. It is brute force algebra so not as elegant as the solution above.

There is also energy conservation for each body after the collision

Linear momentum conservation means that

and because the collision happens at the bottom of the bowl $v{My}'=v{my}'=0$.

Multiplying this equation by itself gives

and substituting in the equations for energy conservation for each mass after the collision we get the heights they reach after the collision

answered Oct 9, 2018 by (270 points)
edited Oct 10, 2018


text:
# if ::iterator == 0 causes error:

## Recommended Posts

gimp    142
I have a handle class. The handle class hold an iterator in to a map.
typename typedef std::map<std::string, std::pair<ResourceType*, size_t> >::iterator ResourceItr;

ResourceItr	m_Resource;

On construction my code sets the Resource handle to 0. Later I test for the validity of the handle :
template <class ResourceType, typename IOType>
inline bool CHandle<ResourceType, IOType>::IsNull(void) const
{
return (m_Resource == 0); <<<<<<<<<<<<<<<<<<<<<
}

For some reason since upgrading to VS 2005 this test of wether a iterator is null is no longer valid. Can anyone think of why this may no longer be valid?

##### Share on other sites
smart_idiot    1298
I don't think you're allowed to do that with iterators.

##### Share on other sites
Assassin    246
don't test for null, test
if (it == container.end())

##### Share on other sites
gimp    142
hmmmm... I guess I could just add a bool to the hanlde class to track it that way. I can't use .end as the handle is hel by object that exist before the container exists and needs to be assigned later.

I'm quite suprised that I shouln't do that. I wonder where I picked that up from...

##### Share on other sites
Guest Anonymous Poster
you want to test if the ResourceType* is NULL right?

well the map stores its data as a std::pair,
so to test for ResourceType*==0 you would have to do:

inline bool CHandle::IsNull(void) const
{
return (m_Resource->second.first == 0); <<<<
}

i have no idea what the overloaded == operator does in the map::iterator though... but it should never have worked...

##### Share on other sites
petewood    819
If the iterator is actually just a pointer (which is quite possible) then it could evaluate correctly. I wonder whether the newer STL libraries have debug behaviour which detects misuse. Just a thought, no factual basis.


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# 1/110 Scale Shuttle model is a success!

### Help Support The Rocketry Forum:

#### georgegassaway

Update - See message # 9, where I have spilled the beans, this is a prototype for a KIT!

Recently a long thread in TRF drifted into some discussion of the various shuttle models Ive done:

One of those was a 1/110 shuttle stack I built in 1979. The thread got me to thinking about doing another 1/110 shuttle, which would be a better model. Such as using the Guillows foam orbiter, and vac-formed parts of my own.

https://www.rocketryforum.com/showpost.php?p=64966&postcount=135

But rather than stretch that thread any farther, Ive decided to start a new one here.

Last Monday (Dec 14th) , I flew a boilerplate for it, using a D12-3. It flew nicely, about 300 feet, perhaps more, with a good straight liftoff and weathercocking into the wind (5-8 mph) as a very stable non-shuttle model of the same weight would tend to do. I had anticipated I'd need to do some tweaking of the aerodynamics but it flew so well, I could not determine anything about the flight to tweak.

Flew it a second time, another good flight. And the third time, on an E9-4. Another good flight.

Steward Jones shot video of all of those flights, and put them onto YouTube. Here is the link for the high quality version:

And here is the link for a low-res version, for those whose computers balk at playing hi-quality smoothly:

At the BRB December launch yesterday, I flew it three more times. It was windier, at 10 mph or more. It weathercocked more, but still in a manner as a typical model, ejection at a good altitude. All were D12-3 flights, as an E9 flight might have put the orbiter into the woods, and I only have this one foam orbiter at the moment.

Now that the boilerplate has proven itself out, Ill be ordering some parts (like some good 3.0 tube for the ET, in place of the 3.125 Mailing Tube the boilerplate has). And will be getting some wooden molds made up for the ET Nose, ET Aft Dome, and some other parts (for vac-forming better copies than this boilerplate used).

I have put together a web page showing the origins of the 1/110 model and this past weeks flight photos, at:

https://homepage.mac.com/georgegassaway/GRP/Scale/Shuttle2009.htm

The idea of this model is not a detailed scale model. But a relatively easy scale shuttle model to fly for fun.

- George Gassaway

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#### GregGleason

##### Well-Known Member
Very nice, George.

I am not a BG expert but it looked like the shuttle performed well from my untrained eye. In fact, I am surprised at how much it looks like the real thing when it's gliding.

Is this your early Christmas present?

:happydeer:

Greg

#### georgegassaway

I am not a BG expert but it looked like the shuttle performed well from my untrained eye. In fact, I am surprised at how much it looks like the real thing when it's gliding.

Is this your early Christmas present?
This is the beauty of that Guillow’s foam orbiter. The only glide trimming it needed was to add a little bit of noseweight, and to flatten out the plug-in rudder that has a slight natural bend to it . No messing around with elevon angles, and the molded-in elevons do not move (part of the trick with this model is to get it to not react on boost to the slight upward angle of the elevons which are JUST RIGHT for glide once the correct glide CG is achieved).

As for Christmas, well, the “present” part was the fact that the test flying of the boilerplate went so smoothly.

I’ll report a bit later on what this model is about.

- George Gassaway

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#### Gus

##### Well-Known Member
George,

Great idea and thank Steward for the great videos.

Would you mind posting a closeup pic of your orbiter attachment mechanism?

Steve

#### powderburner

##### Well-Known Member
Very nice work, George! Glide mode looked very good

Are my old eyeballs tricking me or is that wing just a teeny bit "non-scale" with a slightly stretched span?

#### georgegassaway

Gus, included are some pics of the attachments.

Since this is a boilerplate, it didn’t matter how pretty it looks, but that it be functional. Also it suffered some “still building it at 3 AM-itis”. And at the field, I found out the long 1/4” plastic tube I was using for a lug just was not sliding smooth enough on the 3/16” rod, so on the field I had to glue on a short 3/16” lug and a short 1/4” lug (did not have a second 3/16” lug in my range box)

The attachments are basically 1/8” dowels on the ET and 1/8” launch lugs on the orbiter.

Originally I glued the two aft-most lugs on the orbiter in the wrong place, about 1/2” forward of where they belonged. Rather than remove them, I kept them and used them for alignment when I glued on two 1/2” lugs behind them (I had used tape to build up the diameter of some scrap 1/8” dowel and slid that thru the old and new lugs to keep the proper alignment as the new lugs were glued on).

When I do the later models, mechanically the attachment method will be the same. But physically they will be better. I plan to work up some cast parts for the ET to orbiter attachments (no plywood), and some sort of cast part for the orbiter to help align those two aft lugs.

Powderburner - yes the span is a bit oversized. That was what Guillows did to improve the glide performance. Also the fuselage width is a hair skinny. By span, it would be 1/94. By fuselage length, it would be about 1/116 scale. So, it works out nicely for a 1/110 scale model, and the oversized wings help with the glide. Actually I think the Estes orbiter also had oversized wings, as did Luther Hux’s 1/72 and 1/48 R/C model airplane orbiters of the early 1980’s. I also used oversized wings on some of my orbiters to improve the glide (except for the ones aimed at potential use for an eventual scale contest model instead of sport flying models).

- George Gassaway

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#### Pantherjon

##### Well-Known Member
Very nicely done George! Where can one get those Guillows gliders?

#### Gus

##### Well-Known Member
George,

Thanks.

Does the underside of the orbiter attach parallel to the wall of the ET, or is the nose of the orbiter canted slightly outward? The photos seem to suggest the nose is canted but it might just be the angle of the photo.

Pantherjon,

They're between $4 and$5 but I picked up a bunch of them at Michaels last year on the discount rack for \$1 apiece.

#### georgegassaway

OK, here is what this 1/110 shuttle boilerplate has been about. And I am a little bit surprised nobody figured it out.

A proof-of-concept prototype for a kit that I plan to produce in the next 2 months or so.

I wanted to be sure it flew well before saying anything, or starting to arrange for obtaining raw parts (like tubes) and molds for vac-forming many parts and RTV and casting resin for casting a few other parts.

The kit will use vac-formed parts for the ET nose and aft dome. Vac-formed parts for the SRB noses, aft skirts and SRB nozzles. There MAY be a vac-formed wrap for the intertank, depending on whether I can get the master pattern for that laser-cut (I did a test vac-forming with a similar intertank type of pattern and it worked out well).

Orbiter will be the same, the Guillow’s foam orbiter.

The idea for this model is to be easy to assemble, and for it to fly well. It is not going to have lots of scale details. The idea will be to focus more on flying it than looking at it on a shelf.

The cast parts will not be aimed at being scale detail parts. Instead, they will be aimed at making the model easier to assemble. Such as aligning and gluing the SRB’s to the ET. I have the intention for the ET tube and SRB tubes to be pre-drilled with holes, probably 1/16”. The cast attachment parts will have very short alignment pins (1/16&#8221 that will fit into those pre-drilled holes. So, it will be like gluing together a plastic model that has alignment pins that fit into holes. For the ET to Orbiter attachments, same sort of thing. The ones on the ET will be cast and make use of alignment pins to fit into holes in the ET tube.

There will be very little (possibly zero) use of any paper marking guides for assembly. Reference lines will be pre-marked onto the tubes. I will be making up jigs to assist me with drilling all the various holes at the right places and marking the lines. I *think* I will be able to work up a marking guide for the gray leading edges of the orbiter, for those who would want to have it look more accurate, so they’d have pre-drawn reference lines on the foam orbiter to assist with where to paint it gray.

There will be custom decals. I already have some that I drew up for my 1/72 shuttle, which can be resized (And tweaked since the Guillow’s orbiter cockpit window area is shaped a little differently). Some other aspects of the decals will have to be made up new, since I used some other brand of decals on my 1/72 contest shuttle in addition to my custom ones.

Assuming I do this, it will be done in three phases.

One, I'll be working up the prototype kits at home, and send a couple of them to some people I know for some feedback. That will be the most time-consuming portion since this means making up the master parts for the cast pieces, the jigs, marking jigs, decals, instructions, and even building and flying a boilerplate 2.0 to check out how it does with all the “real” parts (for example the 1.0 prototype uses a mailing tube 3.12” in diameter, not 3.0”, and no intertank wrap other than a printed one). And all of that has to happen before I send out a couple of pre-production prototypes for feedback.

Two, I’ll produce a limited production run of about 10 kits for sale. The money from those first 10 kits will help to recoup the cost of the molds , supplies, and provide some seed money for the next step. I already have two local BRB club members who are pretty much insisting they get to buy kit #1 and kit #2 (both wanted #1), so I do not have too much concern about selling the other 8.

Three, well, if the first 10 sell well, I will buy a lot more supplies and go into unlimited production of the kit. Also I’d like to get feedback from everyone who buys those first 10 kits, so if there was a need to tweak the kit, or instructions, or whatever, I can do that before churning out a lot more kits.

Oh, and four, I may also get into selling a few other things, rather than just sell the shuttle as my only product. Will have to see how it goes.

- George Gassaway

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#### Meatball 1

##### Well-Known Member
Cool stuff, George. Good luck with the kit making, et al. Gotta love a shuttle model, detailed or not.

#### RoyAtl

##### Well-Known Member
One of those was a 1/110 shuttle stack I built in 1979. The thread got me to thinking about doing another 1/110 shuttle, which would be a better model. Such as using the Guillows foam orbiter, and vac-formed parts of my own.
Sam Boden of SoAR did a semi-scale shuttle stack with that orbiter, and it gave an excellent flight at our November launch.

#### rokitflite

##### Well-Known Member
Oh, and four, I may also get into selling a few other things, rather than just sell the shuttle as my only product. Will have to see how it goes.

- George Gassaway
Collectible Plates???!!!

#### zog139

##### Well-Known Member
Neat idea George. I had 2 of the gliders laying around for a long time and decided to put them in flight. If you look at this thread

( go down to post # 19 ) I took a Bertha and added two gliders and it flew real nice with a D12. I actually added led's to the bottom of the wings and flew it at our night launch in Sept with both gliders sporting running lights along the bottom of the wings. Thougth I had some pics somewhere but seemed to have misplaced them !

:tree:

#### Brent

##### Well-Known Member
Very cool George. Put me down for a kit.

#### rockets2000

##### Well-Known Member
Nice rocket George. I'll take one too.

#### Rocketcrab

##### Well-Known Member
I'm in! :santa-smile:

#### wilsotr

##### Well-Known Member
Nice, George! I'm interested too ....

#### Saluki

##### Well-Known Member
TRF Supporter
I'm interested in getting my name on a list for this kit. Put me down for a kit.

#### Delta-IV

##### Well-Known Member
George,

Its great to see some other scale kits being developed. I'll definitely want to get one.

#### Fred22

##### Well-Known Member
Put me down for one please
Cheer
fred

#### mjennings

##### Well-Known Member
cool work George, looks like you'll have no problem selling the first 10. Any idea on an approximate price, or are you holding off figuring that out until things are a little further developed?

#### nitroracer1

##### Member
put me down to purchase one

Kevin

#### ryrego

##### Member
George - wow cool to see someone had the same idea i did !

I modeled mine also after the legacy Estes kit, but I didn't much like the fin layout. I also wanted to experiment with detached boosters, so of course the shuttle is the perfect platform. This is my first post ... i will try to attach some photos. I am currently in Orlando, but moving soon to Huntsville !

My shuttle also used the foam preformed glider. My 'inspiration' was a clear 1-liter bottle which i thought would make a perfect round bottom for the external tank. The rest is just paper! Yes, the nose cones are just paper. So i planned on putting a big (D/E) engine in the tank and smaller (B/C) in the boosters. Boosters and shuttle were held on with straws and dowels.

The first flight was terrible and the whole stack blew apart. I adjusted the weight and the second flight went straight up, but the boosters still came off early and the tank and shuttle went majorly unstable. The next couple flights I used all (3) C6-3's and it went a little better. This link is to the first shot in a really nice sequence. Use the arrow to go right and see the sequence.

Enjoy!
Ryan

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#### ryrego

##### Member
More Pics after the virgin flight

#### dlazarus6660

##### Well-Known Member
ryrego,

Did you post a build for this? Got notes? I'm planning a build like this and it would be easier to copy/clone yours.

Daniel

#### ryrego

##### Member
no, and no. i am more of a 'on-the-fly' or 'prototype' builder - i should start a journal though to keep my notes.

Anyway, check this out - it is the model I used for the external tank nose. For the other nose/nozzles I used Roger's transition tool - they were more basic/linear pieces.

https://jleslie48.com/shuttle20/th/

Everything else was based on the Estes model like the dowels/straws that connect the shuttle to the tank. The 3 motors and the fin placement made this build quite unstable. I would like to improve that. The back of the tank has two fins on it - not sure that provides any useful control.... I had to put a large amount of weight at the top/rear of the external tank. And that barely keeps it under control.

Also, each booster is held on by one dowel at the top, and two at the bottom. The straws that go around the dowels are not tight, so this may be another reason for instability. I suggest a more snug fitting joint for those if you want to have them detachable.

See photo of the boosters separating. Note they are still firing - and the shuttle and tank started doing flips not long after.... I think everything hit the ground still on fire...

Conclusion: just 'wing' it and it will turn out just as good, if not better

#### georgegassaway

Not much progress on the shuttle kit lately. But I hope to get it going again soon.

I did do a bit of testing with some potential decals. In this case I just printed them onto normal paper and used double-sided tape to hold them on. Just to see what theyd look like.

Flew the boilerplate again 2 weeks ago at the February BRB Launch. Flew it twice on a D12. Again two good flights. This makes 8 flights so far.

Photos below.

- George Gassaway

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#### GregGleason

##### Well-Known Member
Nice George!

The 3rd pic looks like the bottom of the shuttle got a little scorched, or is it something else?

I like pic 4 with the moon in the background. Sweet!

Greg

#### georgegassaway

The 3rd pic looks like the bottom of the shuttle got a little scorched, or is it something else?
Yep. I had lost the ignitor plug for the D12. I made the mistake of using some wadding and masking tape to hold the ignitor in place. The way I applied the masking tape was across the end, which left two sides open. So when the D12 ignited, it shot a bit of flame out sideways, into the orbiter.

I have seen this making tape induced sideways ignition flame with other models too, and should have known better. Well, I knew better, I needed to remember better for a model like this.

So, I will mention in the instructions not to do that. When using the Estes ignitor plugs, theres been no problem at all.

- George

#### georgegassaway

An update on the shuttle model kit.

Not much progress on it lately. But , but I did finally do a cosmetic upgrade to the boilerplate. I painted the ET, and added paper wraps to the SRB’s. I simply taped them on, with the seams facing into the ET. I am still trying to decide the most practical way to do the SRB wraps as far as cost, ease of application, and so forth.

I will say that I used the easy way of painting of the ET without masking the SRB’s, and applied the SRB wraps without trying to work on the seams facing the ET. It was a lot easier to cut them off, paint the ET, do the wraps, then glue the SRB’s back on. That’s the sequence the kit will use, paint ET, complete SRBs, and then glue the SRB’s to the ET.

Flew the model 4 times on D12-3 power, at the March 20th BRB Launch. All flights were good. The paint and wraps had added some extra mass to it, but it still flew fine (just a little lower of course). The boilerplate has now flown 12 times, all good flights.

Pics below.

Please check out the kits I'm currently auctioning on eBay. I need to pay some bills and hopefully have enough left to help provide the seed money to produce this kit. Link to the auctions: https://tinyurl.com/yzamjlg

- George Gassaway

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# zbMATH — the first resource for mathematics

Metrics with positive topological entropy on $$S^ 2$$. (Métriques à entropie topologique positive sur $$S^ 2$$.) (French) Zbl 0778.58035
Sémin. Théor. Spectrale Géom., Chambéry-Grenoble 10, Année 1991-1992, 97-107 (1992).
It is believed that geodesic flows on compact manifolds with negative curvatures have positive entropy and that they are ergodic. However, Donnay (1989) and Burns and Gerber (1989) proved that to build ergodic metrics with positive entropy on $$S^ 2$$ is topologically not possible and that the properties can be obtained for negative curvature cases like their examples. In this paper it is proved that it is not the problem of the curvature and it is possible to construct metrics with positive curvature and positive entropy on $$S^ 2$$. However, the problem is still open for Liouville entropy and ergodicity.
Reviewer: Y.Kozai (Tokyo)
##### MSC:
 37A99 Ergodic theory 37D40 Dynamical systems of geometric origin and hyperbolicity (geodesic and horocycle flows, etc.) 53D25 Geodesic flows in symplectic geometry and contact geometry 54C70 Entropy in general topology
##### Keywords:
ergodic metrics with positive entropy on $$S^ 2$$
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text: The number of triangles is always two less than the number of sides. Count the number of sides in each of the polygons featured in this batch of worksheets for 6th grade and 7th grade students. The other part of the formula, − is a way to determine how many triangles the polygon can be divided into. Sum of Interior Angles of a Polygon. Let's Review To determine the total sum of the interior angles, you need to multiply the number of triangles that form the shape by 180°. Figure 1 Triangulation of a seven‐sided polygon to find the interior angle sum.. Theorem 39: If a convex polygon has n sides, then its interior angle sum is given by the following equation: S = ( n −2) × 180°. Question 1057870: The sum of the interior angles of a polygon is twice the sum of its exterior angles. In order to find the measure of a single interior angle of a regular polygon (a polygon with sides of equal length and angles of equal measure) with n sides, we calculate the sum interior anglesor $$(\red n-2) \cdot 180$$ and then divide that sum by the number of sides or $$\red n$$. A plane figure having a minimum of three sides and angles is called a polygon. Sum of interior angles of n-sided polygon = n x 180 ° - 360 ° = (n-2) x 180 ° Method 4 . Given an integer N, the task is to find the sum of interior angles of an N-sided polygon. Set up the formula for finding the sum of the interior angles. Interior Angle = Sum of the interior angles of a polygon / n. Where “n” is the number of polygon sides. The formula is = (−) ×, where is the sum of the interior angles of the polygon, and equals the number of sides in the polygon.. Add the interior angles, set the sum equal to 720, and solve for x: About the Book Author. 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This post is about irrotational and solenoidal behavior of Electric field and Magnetic field respectively which is also explain Maxwell’s second and fourth equation. Closed line integral (MAXWELL’S SECOND EQUATION) $\oint \vec{E} \cdot \vec{dl}=0$ Potential difference can exit between two different and distinct points but with the same point, it is always zero. Potential is […]

### Nonlinearities in Structure

When the force-displacement relation of the system is linear, then that particular type of system is named as “linear system.” The simplest example of such a system is mass-spring mechanical assembly. In this mechanical system, the stiffness of a spring remains constant. But in real-world problems, the stiffness of a system is not linear; however, it may be idealized […]

### Basics Of Electromagnetic Theory

This post is about basic idea of Electro-magnetic theory. This post cover what is Electric field, magnetic field , how it propagate and what is strength of Electric field and magnetic field which is explained by Gauss Law and Amperes Law respectively. It also includes some idea about Gauss divergence theorem, Stocks theorem, permittivity and […]

### Carving on Acrylic sheet using CNC-machine

Carving is a subtractive manufacturing/machining process in which the tool scrapes out the material from the workpiece to generate the desired shape object. The process has been here for centuries; sculptures made of stone or paintings inside the caves are examples. Advancement in Computer-Aided Manufacturing (CAM) has modernized the traditional carving process. Nowadays, Computerized Numerical […]

### Norm and condition number of a matrix

In a numerical analysis with a vector involvement, norms are essential to predict the various errors involved in the numerical analysis. A norm is a function ||.|| in a vector space V. If A is a n*n matrix, then its norm is a real number and denoted by ||A||. A norm satisfies the following properties: […]


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# Eigenvalues

I assume you meant "eigenvalues", not "eigenvectors". Hint: Consider the action of $AA^{-1}$ on an eigenvector of A.


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Article

# Gradient Estimates for the Perfect and Insulated Conductivity Problems with Multiple Inclusions

• ##### Biao Yin
Communications in Partial Differential Equations (Impact Factor: 1.19). 09/2009; DOI: 10.1080/03605300903564000
Source: arXiv

ABSTRACT In this paper, we study the perfect and the insulated conductivity problems with multiple inclusions imbedded in a bounded domain in $\mathbb{R}^n, n\ge 2$. For these two extreme cases of the conductivity problems, the gradients of their solutions may blow up as two inclusions approach each other. We establish the gradient estimates for the perfect conductivity problems and an upper bound of the gradients for the insulated conductivity problems in terms of the distances between any two closely spaced inclusions.

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• Source
##### Article: Optimal estimates and asymptotics for the stress concentration between closely located stiff inclusions
[Hide abstract]
ABSTRACT: If stiff inclusions are closely located, then the stress, which is the gradient of the solution, may become arbitrarily large as the distance between two inclusions tends to zero. In this paper we investigate the asymptotic behavior of the stress concentration factor, which is the normalized magnitude of the stress concentration, as the distance between two inclusions tends to zero. For that purpose we show that the gradient of the solution to the case when two inclusions are touching decays exponentially fast near the touching point. We also prove a similar result when two inclusions are closely located and there is no potential difference on boundaries of two inclusions. We then use these facts to show that the stress concentration factor converges to a certain integral of the solution to the touching case as the distance between two inclusions tends to zero. We then present an efficient way to compute this integral.
12/2013;
• Source
##### Article: Singular Behavior of Electric Field of High Contrast Concentrated Composites
[Hide abstract]
ABSTRACT: A heterogeneous medium of constituents with vastly different mechanical properties, whose inhomogeneities are in close proximity to each other, is considered. The gradient of the solution to the corresponding problem exhibits singular behavior (blow up) with respect to the distance between inhomogeneities. This paper introduces a concise procedure for capturing the leading term of gradient's asymptotics precisely. This procedure is based on a thorough study of the system's energy. The developed methodology allows for straightforward generalization to heterogeneous media with a nonlinear constitutive description.
01/2014;
• Source
##### Article: Characterization of the gradient blow-up of the solution to the conductivity equation in the presence of adjacent circular inclusions
[Hide abstract]
ABSTRACT: We consider the conductivity problem in the presence of adjacent circular inclusions having arbitrary constant conductivity. When two inclusions get closer and their conductivities degenerate to zero or infinity, the gradient of the solution can be arbitrary large. We characterize the gradient blow-up by deriving an explicit formula for the singular term of the solution in terms of the Lerch transcendent function. This derivation is valid for inclusions having arbitrary constant conductivity. We illustrate our results with numerical calculations.
12/2013;


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# Commuting invariants and duals of C_p vector spaces

Let $K$ be a field complete with respect to some discrete valuation, with perfect residue field of characteristic $p$. Let $\mathbb{C}_p$ be the completion of an algebraic closure of $K$, and set $G_K := \text{Gal}(\mathbb{C}_p/K)$.

Let $V$ be a $\mathbb{C}_p$ vector space. Is it true in general that $\text{Hom}_K(V^{G_K}, K) \simeq \text{Hom}(V, \mathbb{C}_p)^{G_K}$?

Now let $T_p(X)$ be the Tate module of an abelian variety $X/K$. Then it is true that

$$\text{Hom}_K((T_p(X) \otimes_{\mathbb{Z}_p} \mathbb{C}_p)^{G_K}, K) \simeq \text{Hom}_{\mathbb{Z}_p}(T_p(X), \mathbb{C}_p)^{G_K}.$$

I'm looking for a proof of this which is somewhat elementary: I'm okay with using vanishing theorems for the Galois cohomology of $\mathbb{C}_p(i)$, but preferably not much more than that. Does anybody have such a simple way to see this?

Thanks!

-
Concerning your first question: a $G_K$-equivariant map from V to $C_p$ exists, for example, whenever $V$ is of Hodge-Tate type with one weight equal to $0$. This happens for lots of repns for which $V^{G_K} = 0$. – Laurent Berger Dec 17 '12 at 8:41
I'm afraid that I'm not familiar enough with $C_p$-representations to see why such a thing exists. If $V$ is Hodge-Tate with a weight of $0$, wouldn't there be a corresponding copy of $K$ in the invariants? – Tony Jan 15 '13 at 16:54


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# Homework Help: Problem: particle in infinitely deep potential

1. Oct 15, 2011

### paly4

1. The problem statement, all variables and given/known data

A square well: 0≤x≤L, V=0, for others x V=∞
ψ=N[x(x-L)+Ax2(x-L)+Bx(x-L)2]
H=-[STRIKE]h[/STRIKE]/2m*d2/dx2+V(x)

a) Determine N
b) Use the Rayleigh-Ritz variational method to obtain the <ψ|H|ψ> as a function of A and B, and minimize the energy with respect to A and B
c) Discuss how many distinct values of the pair (A, B) that you obtain and do you get more than one energy? If yes, what are they?
d) Compare you results to the exact particle in the box and discuss the error.
a') Consider the same system, but now V(x)=x for 0≤x≤L
Compute the first order perturbation correction to the ground state energy
b') Compute the first order perturbation correction to the ground state wave function

3. The attempt at a solution

a) I found with SAGE mathematical software that N2=$\frac{\sqrt{210}}{\sqrt{(2B2-3AB+2A2)L7+7(A-B)L6+7L5}}$
b) I'm not sure in that. But I've calculated <E>=-[STRIKE]h[/STRIKE]/2m*∫ψ(x)*ψ''(x)*dx from 0 to L, it's another value depending from A, B and L.
I don't understand, what does "minimize the energy with respect to A and B" means. It already depends only from A, B and L.
c) It looks like that there must be some border values for energy, so it will be possible to determine A and B.
d) Exact particle is E=[STRIKE]h[/STRIKE]2/8mL2, just need to know A and B in the value from b) to compare.

a') <E>=-[STRIKE]h[/STRIKE]/2m*∫ψ(x)*(d2/dx2+x)[ψ(x)]*dx from 0 to L. Am I right in that solution? Or it is just -[STRIKE]h[/STRIKE]/2m*∫ψ(x)*(x)*ψ(x)*dx
b') have no idea, how to calculate correction for the ground state of the wave function.

Last edited: Oct 15, 2011
2. Oct 16, 2011

### vela

Staff Emeritus
The variational principle gives you an upper bound on the energy of the ground state; therefore, to get the best estimate, you want to find the least upper bound. In this case, your estimate will vary with A and B, so you want to find the value or values of A and B that minimize the function.

3. Oct 16, 2011

### paly4

Thank you.
I've got 3 values of the <E>=5, 7 and 21 *([STRIKE]h[/STRIKE]2/mL2) after the finding derivatives of <E> with respect to A, B and AB.
So, the minimum of <E> is 5([STRIKE]h[/STRIKE]2/mL2).

But I'm still stuck with a') and b').

4. Oct 17, 2011

### vela

Staff Emeritus
5. Oct 27, 2011

### paly4

Vela, thanks for helping me!


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## ganeshie8 one year ago The starting population of a certain kind of migratory birds in a forest is $$10000$$. $$3\%$$ of the population die each year. Reproduction rate is $$4\%$$ per year and $$500$$ birds immigrate to the forest each year. Estimate the population of birds at the end of $$30$$ th year

1. anonymous

let $$P(n)$$ denote the population after $$n$$ years. we're given that $$P(0)=10000$$ and the recurrence relation $$P(n)=0.04P(n-1)-0.03P(n-1)+500\\P(n)=0.01P(n-1)+500$$

2. anonymous

so $$P(n)-0.01P(n-1)=500$$ignoring our initial condition, a constant particular solution $$P(n)=k$$ gives $$k-0.01k=500\implies 0.99k=500\implies k=500/0.99=\frac{50000}{99}$$

3. ganeshie8

Ahh that looks much better than what im doing.. i ended up with a geometric series after writing out the first few terms

4. anonymous

now consider the homogeneous recurrence relation $$P(n)-0.01P(n-1)=0$$suppose that we had a geometric sequence $$P(n)=r^n$$ which gives $$r^n-0.01r^{n-1}=0\\r^{n-1}\left(r-0.01\right)=0\implies r=0.01$$ so our homogeneous solution looks like $$P(n)=C(0.01)^n$$ and thus the general solution to our original recurrence is $$P(n)=C(0.01)^n+\frac{50000}{99}$$

5. anonymous

now we want to solve for $$C$$ using our initial condition: $$P(0)=10000\\C(0.01)^0+\frac{50000}{99}=10000\\C=10000(1-5/99)\approx 9494.95$$

6. anonymous

oops, I messed up the recurrence -- it should've actually read

7. ganeshie8

$$P(n)=C(1.01)^n+\frac{50000}{99}$$ ?

8. anonymous

$$P(n)=P(n-1)+0.04P(n-1)-0.03P(n-1)+500\\P(n)=1.01P(n-1)+500$$ in which case we'd get $$r=1.01$$ and our particular solution would be different as well

9. ganeshie8

Ohk gotcha!

10. anonymous

$$k=1.01k+500\\-0.01k=500\\C=-50000$$ so the general solution is $$P(n)=C(1.01)^n-50000$$

11. dan815

|dw:1437282863276:dw|

12. dan815

would that work?

13. anonymous

now applying our initial condition: $$10000=C(1.01)^0-50000\\60000=C$$ so our model is $$P(n)=60000(1.01)^n-50000$$

14. anonymous

now we plug in $$n=30$$ so $$P(30)\approx 30870.93$$

15. anonymous

in general, the theory of linear recurrence relations with constant coefficients $$x_n+c_1x_{n-1}+c_2x_{n-2}+\dots+c_{n-1}=g(n)$$ mirrors that of hte analogous differential equations in that we can first find a particular solution and also investigate the general homogeneous solution of the homogeneous case $$g=0$$ using the ansatz $$x_n=r^n$$

16. anonymous

the method of undetermined coefficients works here, too, only with slightly different 'rules' since we're no longer dealing with differentiation, and even variation of parameters carries over :-)

17. dan815

wolfram gives 30,991.5, with DE method http://www.wolframalpha.com/input/?i=da%2Fdt%3D0.01*a%2B500%2C+a%280%29%3D10000%2C+t%3D30

18. anonymous

yep, @dan815 your approach will overshoot since this is essentially the same as dealing with continuous compounding

19. ganeshie8

I see what you're doing to get a particular solution, but I'm not so sure why making the function a constant, $$P(n)=k$$ is working

20. dan815

yeah i think

21. anonymous

@ganeshie8 we pick a constant because we want $$P(n)-1.01P(n-1)$$ to be a constant; if there were non-constant terms then they would not be entirely annihilated -- consider $$P(n)=An+B$$ which gives $$An+B-1.01(A(n-1)+B)=-0.01An+(1.01A-0.01B)$$

22. anonymous

we'd want $$-0.01An$$ to be identically zero, which would make $$A=0$$ so this reduces to a constant solution $$P(n)=B$$. it's the same logic that is used in the method of undetermined coefficients with linear differential equations

23. ganeshie8

so a partiuclar solution need not be a function of independent variable ($$n$$) ? oh nvm i think i get it, we just want "some" particular solution and the constant funciton is good enough i tihnk xD

24. dan815

|dw:1437283645284:dw|

25. anonymous

yeah, the particular solution is just any $$P_p$$ that satisfies the recurrence regardless of initial conditions. its purpose is that we can then consider solutions of the equation in $$P_p+P_c$$; since our equation is linear, this reduces to: $$(P_p+P_c)(n)-1.01(P_p+P_c)(n)=500\$$P_p(n)-1.01P_p(n))+(P_c(n)-1.01P_c(n))=500\\500+(P_c(n)-1.01P_c(n))=500\\\quad\implies P_c(n)-1.01P_c(n)=0so finding *any* particular solution reduces the problem to solving the homogeneous case 26. ganeshie8 yeah i did solve it by expanding like that, i think oldrin's method looks more elegant and general.. 27. dan815 |dw:1437283962537:dw| 28. anonymous it's the same thing in linear algebra with \(Ax=b$$ -- you find a particular solution $$x_p$$ so you can then look at $$x=x_p+x_c$$ so $$Ax=A(x_p+x_c)=Ax_p+Ax_c=b+Ax_c$$ and thus $$b+Ax_c=b\implies Ax_c=0$$; a particular solution is the origin vector that 'steps' us into the affine space of solutions 29. dan815 ah ok 30. anonymous$$P(n)=500+1.01P(n-1)$$is an arithmetico-geometric sequence which permits a very easy trick to solve it:$$P(n)=500+1.01(500+P(n-2))\\\quad\quad=500(1+1.01+\dots+1.01^{k-1})+1.01^kP(n-k)$$so$$P(n)=500\cdot\frac{1.01^n-1}{1.01-1}+1.01^nP(0)$$31. anonymous so then you just $$\dfrac{1.01^n-1}{1.01-1}=100(1.01^n-1)=100\cdot1.01^n-100$$ so$$P(n)=50000\cdot1.01^n-50000+10000\cdot1.01^n=60000\cdot1.01^n-50000


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$$\require{cancel}$$

3.5: Energy Landscapes

In the previous section we proved that the total energy is conserved. In the section before that, we looked at potential energies. Typically, the potential energy is a function of your position in space. When we plot it as a function of spatial coordinates, we get an energy landscape, measuring an amount of energy on the vertical axis. Of course we can also plot the total energy of the system - and since that is conserved, it is the same everywhere, and thus becomes a horizontal line or plane. Because kinetic energy cannot be negative, any point where the potential energy is higher than the total energy is not allowed: the system cannot reach this point. When the potential energy equals the total energy, the kinetic energy (and thus the speed) has to be zero. Whenever the potential energy is lower than the total energy, there is a positive kinetic energy and thus a positive speed.

Probably the simplest energy landscape is that of the harmonic oscillator (mass on a spring) - it’s a simple parabola. The point at which the horizontal line representing the total energy crosses the parabola corresponds to the extrema of the oscillation: these are its turning points. The bottom of the parabola is its mid-point, and you can immediately see that that’s where the kinetic energy (and thus the speed) will be highest.

Of course you can have more complex energy landscapes than that. In particular, you can have a landscape with multiple extrema, see for example Figure $$\PageIndex{1}$$ A particle that is being acted upon by forces described by this potential energy, follows a trajectory in this landscape, which can be visualized as a ball rolling over the hills and valleys of the landscape. Think back to the harmonic oscillator example. If we let go of a ball in a parabolic vase at some point on the slope, the ball will roll down and pick up speed, then roll up the opposite slope and lose speed, until it reaches the same height where its speed will again be zero. The same is true in more complicated landscapes. Particularly interesting are local maxima. If you put a ball exactly on top of one of them, it will stay there - it is a fixed point, but an unstable one, as any arbitrarily small perturbation will push it down. If you let go of a ball at a level above a local maximum, it may hop over it to the next minimum,but if your initial position (your initial energy) was too low, your ball can get stuck oscillating about a local minimum - a metastable point.

Energy landscapes are even useful when the total energy is not conserved - for example because of friction terms. Friction causes energy to dissipate from the system, which is equivalent to having your ball move in the landscape with friction. For low friction, your ball will oscillate, but get less high every time, until it comes to rest at the minimum. For high friction, it won’t even oscillate, but just get to the minimum - exactly what an over damped system in real life does.

Example $$\PageIndex{1}$$: Worked example: the Lennard-Jones Potential

The Lennard-Jones potential energy is a commonly used model to describe the interactions between uncharged atoms and molecules. This potential energy can be written in two equivalent ways:

$U_{\mathrm{LJ}}(r)=\frac{A}{r^{12}}-\frac{B}{r^{6}}=4 \varepsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^{6}\right] \nonumber$

where $$r$$ is the distance between the atoms or molecules, and A, B, $$\varepsilon$$, and $$\sigma$$ are positive constants.

1. Find the dimensions of A, B, $$\varepsilon$$, and $$\sigma$$.
2. Express $$\varepsilon$$ and $$\sigma$$ in A and B.
3. Sketch the potential (in its second form) as a function of $$\frac{r}{\sigma}$$, and use this sketch to give a physical interpretation of $$\varepsilon$$ and $$\sigma$$.
4. Does the Lennard-Jones potential lead to attractive or repulsive forces at short distances? And what about long distances?
5. Find all equilibrium points of this potential energy, and determine their stability.

Solution

1. $\begin{array}{l}{[U]=\text { Energy } \Longrightarrow[U]=M \times \frac{L}{T^{2}} \times L=\frac{M L^{2}}{T^{2}}} \\ {[A]=\text { Energy } \times \text { Length }^{12} \Longrightarrow[A]=\frac{M L^{2}}{T^{2}}} \\ {[B]=\text { Energy } \times \text { Length }^{6} \Longrightarrow[A]=\frac{M L^{8}}{T^{2}}}\end{array}\nonumber$ Because the powers of the terms $$\left(\left(\frac{\sigma}{r}\right)^{12}\right) \text { and }\left(\left(\frac{\sigma}{r}\right)^{6}\right)$$ are different, while we add them together, they haveto be dimensionless, so $[\sigma]=L \text { and }[\varepsilon]=[U]=\frac{M L^{2}}{T^{2}}\nonumber$.
2. $\begin{array}{l}{4 \epsilon \sigma^{12}=A \text { and } 4 \varepsilon \sigma^{6}=B} \\ {\frac{A}{B}=\sigma^{6} \Longrightarrow \sigma=\left(\frac{A}{B}\right)^{1 / 6}}\end{array}\nonumber$ By substituting $$\sigma$$ in the expressions for either A or B we can derive an expression for $$\varepsilon$$: $\begin{array}{l}{4 \sigma^{6} \varepsilon=B} \\ {4 \varepsilon A=B^{2} \Longrightarrow \varepsilon=\frac{B^{2}}{4 A}}\end{array}\nonumber$
3. (c) Figure $$\PageIndex{2}$$. Interpretation:≤is a measure for the depth of the potential well.æsets the length scale andtherefore the position of the equilibrium point.
4. Method 1: We calculate the force as minus the derivative of the potential energy: $F=-\frac{\partial U}{\partial r}=4 \varepsilon\left(\frac{12 \sigma^{12}}{r^{13}}-\frac{6 \sigma^{6}}{r^{7}}\right)\nonumber$ For small r we have $$r^{-13} \gt \gt r^{-7}$$, so F is positive and therefore repulsive. Conversely, for larger we have $$r^{-13} \lt \lt r^{-7}$$, so F is negative and therefore attractive. Method 2: Use the sketch in (c) to to see that the slope of the potential is negative for small r, which implies a repulsive force, and the slope of the potential is positive for larger, which implies an attractive force.
5. For an equilibrium point we have $0=\frac{\partial U}{\partial r}=4 \varepsilon\left(\frac{12 \sigma^{12}}{r^{13}}-\frac{6 \sigma^{6}}{r^{7}}\right)=24 \frac{\varepsilon \sigma^{6}}{r^{7}}\left(\frac{2 \sigma^{6}}{r^{6}}-1\right)\nonumber$ so there is only one equilibrium point, at $r_{\mathrm{eq}}=2^{1 / 6} \sigma\nonumber$ To determine the stability at this point, we consider the second derivative of U(r): $\left.\frac{\partial^{2} U}{\partial r^{2}}\right|_{r=r_{\mathrm{eq}}}=4 \varepsilon\left.\left(42 \frac{\sigma^{6}}{r^{8}}-156 \frac{\sigma^{12}}{r^{14}}\right)\right|_{r=r_{\mathrm{eq}}}=4 \varepsilon\left(\frac{42}{2^{4 / 3} \sigma^{2}}-\frac{156}{2^{7 / 3} \sigma^{2}}\right)=-36 \cdot 2^{2 / 3} \frac{\varepsilon}{\sigma^{2}}<0\nonumber$ which means that the equilibrium point is stable. Alternatively, we could have determined the stability by considering the graph drawn at (c), from which we can see that the equilibrium point corresponds toa global minimum of the potential energy and hence is stable.


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# Queen - Live at LIVE AID 1985

Does it get any better?

# DreamHost is awesome!

There are many crappy hosting providers. In fact, almost all of them are crappy.

Almost.

DreamHost is awesome!

In their standard shared-hosting plan they provide:

• unlimited webspace
• unlimited MySQL databases
• unlimited email accounts
• unlimited user accounts
• with separate web spaces
• with SSH access
• free LetsEncrypt SSL certificates for every domain and subdomain
• up-to-date PHP and MySQL versions
• excellent and friendly customer support

(to be continued…)

They do offer domain registration as well, but I prefer using Hover for this. In cases where everything else is on DreamHost, I just switch the domain's name servers to the DreamHost name servers. And it just works.

By now, I recommend DreamHost to everyone who asks, especially to all my clients.

# List comprehensions for physicists

A list comprehension provides a compact way of mapping a list to another list by applying a function to each of the elements of the list.

From "Dive Into Python 3"

Yeah. Well. Talk English to me! Please.

Until recently I had no clue what list comprehensions do or how to use them or what they're good for. Reading Dive Into Python 3 about two years ago, without much prior programming knowledge, maybe wasn't the best idea I ever had.

Today I know quite a bit more about programming in general and about Python in particular. Still, list comprehensions kept eluding me until I read From List Comprehensions to Generator Expressions by the inventor of Python himself, Guido van Rossum.

In this article, he starts by showing how lists are stated in mathematics. As a physicist, an expression like

$$\{ x \in \mathbf{N}\ |\ x < 10 \}$$

is perfectly clear to me. It's the set of all natural numbers that are smaller than 10, i.e.:

$$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$$

depending on your preference of zero being a natural number or not.

A python list of these integers would be:

[0,1,2,3,4,5,6,7,8,9]

Und suddenly, everything became clear to me.

[f(x) for x in S if P(x)]

is the list of the function values $f(x)$ of all $x$ in the set $\mathbf{S}$ that fulfill the condition $\mathbf{P}$.

Using a Python list comprehension, the first mathematical expression above reads:

[x for x in N if x < 10]

Of course, this list is not possible, because $\mathbf{N}$ is infinite.

The function value $f(x)$ is very handy as well. The set of the squares of all integers between 1 and 4:

$$\{ y\ |\ y = x^{2}; x \in \mathbf{N}; 0 < x < 5 \}$$

expressed via a list comprehension is:

[x**2 for x in [1,2,3,4]]

Looking at list comprehensions in mathematical terms helped me understand them. It's a great example of one and the same idea expressed differently in different fields and thus more or less understandable by people in that field.

# Run Selenium from PyCharm

In order to run the Selenium WebDriver from within PyCharm, you need to explicitly add the path of the browser driver executable.

It is in fact not necessary to add the PyCharm Selenium plugin.

My PyTest fixture for the Selenium browser now looks like this:

from selenium import webdriver
import pytest

@pytest.fixture(scope="session")
def browser():
browser = webdriver.PhantomJS(executable_path="/usr/local/bin/phantomjs",
desired_capabilities={
})
browser.implicitly_wait(3)

yield browser

browser.quit()

I'm using PhantomJS as a headless browser because it is way faster than Firefox or Chrome. I wish I could disable all loading of CSS files.

### Update 2017-01-18

Putting the explicit path to the browser driver executable into the test fixture causes problems when I run the same test on Snap CI. A better solution is to add the directory of the (PhantomJS) executable into PyCharm itself.

In my PyCharm Community Edition 2016.2 I go to Run >> Edit Configurations…, select the run configuration that runs my functional test and put

PATH=/usr/local/bin/

into Environment variabls. That's all. Now PyCharm finds the executable of PhantomJS in that directory and I can remove the executable_path line from the fixture.

I do not know why PyCharm doesn't just import the \$PATH from my shell.

# Disable CSRF validation when unit-testing a Flask app!

In my current Flask project, I wanted to test a view method that includes a form with a POST request. But it wouldn't work.

The view method looks something like this:

# views.py

if form.validate_on_submit():
t = Tournament(name=form.name.data)
db.session.commit()

flash('A tournament was added to the database.')
return redirect(url_for('tennis.index'))

form=form)

Flask-Testing provides a TestCase class with several useful assert methods. So the test would look like this:

# test_views.py

class TestTennisViews(TestCase):
[…]

# WHEN submitting correct form data to the add_tournament view
data=dict(
name='Chicharito Open',
))

# THEN it redirects to the tennis overview page
self.assertRedirects(r,'/tennis/')
# and it shows a message
self.assertMessageFlashed('A tournament was added to the database.')
# and there is a tournament in the database
assert Tournament.query.count() > 0
# and it has the correct tournament
t = Tournament.query.first()
assert t.name == 'Chicharito Open'

assertRedirects checks if the status code is 301 or 302. In my tests, it was always 200, and the test failed. Of course, nothing was added to the database, either.

I googled a lot, without success. I tried all combinations of the following keywords, among others:

flask
unit test
wtforms
POST request
testing

The solution hit me when taking a break (it works!). It lies in the form.validate_on_submit() part of the view function. validate_on_submit() does two things:

1. It validates the form fields according to the validators specified in the form definition.
2. It validates a CSRF token.

Guess what: there is no CSRF token generated when executing a POST request directly in the test.

The solution is easy, as FLASK-WTF provides a configuration option to disable CSRF:

# config.py

[…]

class TestingConfig(Config):
TESTING = True
WTF_CSRF_ENABLED = False

[…]

class FunctionalTestingConfig(TestingConfig)
WTF_CSRF_ENABLED = False

Of course, only disable CSRF in your test configuration! I created a separate configuration for my functional tests, so that CSRF is enabled in those. When using the browser with Selenium, the CSRF token is provided by the WTForms form on the web page.

# Testing Bootstrap select dropdown field with Selenium (Python)

I am building a Flask app that uses the Bootstrap CSS framework. Forms fields are generated with WTForms via the Flask-WTF extension.

One of my forms includes a select field:

Surface

<select class="form-control" id="surface" name="surface">
<option value="clay">Clay</option>
<option value="hard">Hard</option>
<option value="grass">Grass</option>
</select>

My functional tests with Selenium went fine until I changed the desired test value from the default ("Clay") to a different option ("Hard"). (duh!!) So I knew I did something wrong and looked for the correct way to choose an option from a select field.

All the tests use the Selenium WebDriver.

from selenium import webdriver

browser = webdriver.Firefox()

### How it doesn't work

Via Google and Stackoverflow I found several possible solutions, none of which worked for me.
 My favorite one uses the Selenium Webdriver Select class:

from selenium.webdriver.support.ui import Select

select = Select(browser.find_element_by_id('surface'))
select.select_by_visible_text("Hard")
select.click()

Then there are solutions using the find_element_by_xpath method:

option = browser.find_element_by_xpath("//select[@id='surface']/option[@value='hard']")
option.click()

or find_element_by_css_selector:

option = browser.find_element_by_css_selector("select#surface > option[value='hard']")
option.click()

and one other looking for the option fields:

dropdown = browser.find_element_by_id('surface')
for option in dropdown.find_elements_by_tag_name('option'):
if option.text == 'Hard':
option.click()
break

Apparently, Bootstrap does not play with Selenium. Or whatever.

### How it does works

I don't remember where I got the idea to my solution, but it's simple and obvious. It selects the dropdown menu, and then types in the text that I want to click. Then hits enter and voilá, the desired option is selected.

from selenium.webdriver.common.keys import Keys

dropdown = browser.find_element_by_id('surface')
dropdown.send_keys('Hard')
dropdown.send_keys(Keys.ENTER)

If you have a better way to select an option in a Bootstrap select field with the Selenium WebDriver, then please tell me!

# serving society

Dude. I service society by rocking! I'm out there on the front lines, liberating people with my music. Rockin' ain't no walk in the park, lady!

Jack Black in "School of Rock"

# Tolle Dinge in Deutschland

Im Juli waren wir auf Deutschlandbesuch. Dort sind mir natürlich viele gute Dinge aufgefallen. Gute Dinge, die es teilweise nur dort gibt, und sicher nicht in Mexiko:

• Sauberes und leckeres Wasser direkt aus der Leitung
• Lecker Brötchen
• Filterkaffee
• Lecker Bier
• Döner Kebap
• Stille
• Waldwege
• Sofortwarme Duschen
• Lecker Brot
• Lecker Kuchen
• Bier auf der Straße trinken
• Öffentliche Mülleimer
• Zwiebelmett

Was es nicht gibt:

• Essensstände im Park

# Too Good To Be Good

This New York Times article makes some very good observations hat are not only true for the "upscale Polanco neighboorhood":

Waiters will grab a plate from under your fork.

[…]

30 servers searched for someone to serve among the restaurant’s 15 diners.

[…]

In other words: Behind every obsequious waiter is an arrogant customer.

Some interesting information I didn't know before but, judging from my own observations, have no doubt are true:

The World Economic Forum says Mexico has the second highest level of income inequality among 35 nations in the Organization for Economic Cooperation and Development. The 10 percent of Mexico’s population with the highest income use more than one-third of the nation’s resources, while the bottom 10 percent use less than 2 percent.

That's one crazy country.


text:
zbMATH — the first resource for mathematics

Heuristics for the multi-vehicle covering tour problem. (English) Zbl 0973.90019
Summary: The multi-vehicle covering tour problem is defined on a graph $$G=(V\cup W,E)$$, where $$W$$ is a set of vertices that must collectively be covered by up to $$m$$ vehicles. The problem consists of determining a set of total minimum length vehicle routes on a subset of $$V$$, subject to side constraints, such that every vertex of $$W$$ is within a prespecified distance from a route. Three heuristics are developed for this problem and tested on randomly generated and real data.

MSC:
 90B20 Traffic problems in operations research
Full Text:


text:
# Complexity of $\mathbb{Z}^n$ tilings

Let $\mathcal{T} \subset \mathbb{Z}^n$ be a finite set. Let $\Lambda \subset \mathbb{Z}^n$ be a full rank lattice. We say that $\mathcal{T}$ is a $\Lambda$-tile for $\mathbb{Z}^n$ if the following two conditions are satisfied:

(i) $(\mathcal{T} + \lambda_1) \cap (\mathcal{T} + \lambda_2) = \emptyset$ for all $\lambda_1,\lambda_2 \in \Lambda$ with $\lambda_1 \neq \lambda_2$.

(ii) $\bigcup_{\lambda \in \Lambda}{(T+\lambda)} = \mathbb{Z}^n$.

There are many interesting open problems concerning tiles. For instance, a 40 year old conjecture of Golomb and Welch states that the $\ell_1$ ball (in $\mathbb{Z}^n$) of radius $r$ does not tile $\mathbb{Z}^n$ unless $r = 1$ or $n = 2$.

My question is what is the complexity of the following problem: Given $n$ and $\mathcal{T}$, decide whether there exists $\Lambda$ such that $\mathcal{T}$ is a $\Lambda$-tile for $\mathbb{Z}^n$. Is there a reference for that?

• A recent paper addressing this in $\mathbb Z^2$ is arxiv.org/abs/1602.05738 (it is shown that tileability is decidable in 2 dimensions, by showing that if $\mathcal T$ tiles, then it does so periodically). Mar 14 '16 at 18:02
• Thank you for the reference. The paper you mentioned deals with the non-lattice case, which is way more complicated... Mar 15 '16 at 9:01
• So I'm trying to understand which is your question: 1) Given $\mathcal T$ and $\Lambda$, is it NP-complete to decide if $\mathcal T$ is a $\Lambda$-tile? or 2) Given $\mathcal T$, is it NP-complete to decide if $\mathcal T$ is a $\Lambda$-tile for some lattice $\Lambda$? I think the answer to 1) is that it's P. Mar 15 '16 at 12:07
• Thanks, I changed the text. It is more like 2). Can you elaborate on 1)? Mar 16 '16 at 0:10
• So here's my attempt at 1): Given $\Lambda$, form a matrix whose columns are a basis for $\Lambda$. I believe the necessary and sufficient condition for $\mathcal T$ to be a $\Lambda$-tile is (a) $|\det A|=|\mathcal T|$ and (b)$A^{-1}(\mathbf u-\mathbf v)\not\in\mathbb Z^n$ for all distinct $\mathbf u$ and $\mathbf v$ in $\mathcal T$. Mar 16 '16 at 4:24