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# Homework Help: Why do distant objects appear to move more slowly?

1. Apr 16, 2014

1. The problem statement, all variables and given/known data
This is not really a homework question and I am not sure if this should be in the Maths forum or Physics forum.

See the title.
For example,a plane with the same length and speed is moving at constant speed near and far away from me.
From my experience I know that the far plane will appear to move more slowly.Why?

3. The attempt at a solution
I have seen many people saying that as the distance from the eye increases,the distance the plane have to move across the field of view increases.That's why it is moving slowly.
This does not make much sense to me.
I don't have a proof based understanding of this.

One idea that came to my mind is that as the distance from the eye increases,as the plane goes from a point A to B(Eye is C),the angle ACB is less than the angle ACB it makes when it moves nearer to me.
So as the angle is less, the distance it moves on the image is less.
(I can't explain this so well but I hope you understand what I mean)

My explanation seems too childish.If it's right,can you make it make sense Mathematically?

2. Apr 16, 2014

### PhysicoRaj

Your idea already seems mathematical for me. Math doesn't always mean equations and operations. It is also logic. You have used (knowingly or unknowingly) the math that is needed. Still, this could be connected to mathematical equations as $S=rΘ.$

3. Apr 16, 2014

What is $S=rΘ$?

4. Apr 16, 2014

### PhysicoRaj

Actually it is the consequence of the definition of the plane angle, how we made it.
Any plane angle, Θ is defined as the ratio of the arc length of a circle which subtends the angle Θ, to the radius of the circle.
If S is the arc length, r the radius then
Θ=S/r
S=rΘ

Note that Θ is in radians.

5. Apr 16, 2014

But if light is being refracted by the lens,why not use the refraction formula?
$$n_1\sin(i)=n_2\sin(r)$$

What's the use of S=rΘ here?

6. Apr 16, 2014

### PhysicoRaj

I'm not sure if I got you right.. by refraction formula, how do you prove that farther objects appear to move slowly?

7. Apr 16, 2014

No.Not the refraction formula,sorry.I'll use a ray diagram.

Then I will do a little trigonometry to prove it.

#### Attached Files:

• ###### gyg.png
File size:
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8. Apr 16, 2014

### PhysicoRaj

So your object moves from A to B while the image moves from A' to B' and A'B' is smaller than AB hence it appears to move slowly, right?
If you do the trigonometry, you'll end up with something similar to S=rΘ only that the angle Θ will be inside a trig function.

9. Apr 16, 2014

No it's not like that.

The AB is closer than AB(Note the colour difference)
So as the plane move from A to B,the image moves from A' to B'.
When the plane move from A to B,the image move from A' to B' and the distance A'B' is greater than A'B'.So the far plane takes a longer time to move across the retina than the near plane.The rest is handled by the brain

#### Attached Files:

• ###### gyg.png
File size:
6.5 KB
Views:
133
10. Apr 16, 2014

### PhysicoRaj

I get it. This is nothing but the use of $S=rθ$ here. If you would look into the derivation for the magnification of a simple microscope (the lens here) you can observe that the angle subtended by the object at the eye, and at the lens is considered. Not getting into details, we can deduce that relation to meet our needs, and it clearly yields $\tan θ = \frac{h}{D}$ where $h$ is equivalent to $S$ and $D$ to $r$. $\tan θ$ can be approximated to $θ$ for small angles. Fetching $θ=\frac{S}{r}$

Since $S$ is constant, $\theta$ is inversely proportional to $r$. Hence as distance, $r$ increases, $\theta$ decreases. It's 'angular velocity' decreases.

Last edited: Apr 16, 2014
11. Apr 17, 2014


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# How to output results in a specific format？

i get some results using Table,codes are as this

nu = 8.163265306122447;
Table[FindRoot[(I*m[k])*Tanh[I*m[k]] - nu, {m[k], k*\[Pi],
k*\[Pi] - \[Pi]/2, k*\[Pi] + \[Pi]/2}], {k, 1, 10}]

and it outputs

{{m[1] -> 1.78621}, {m[2] -> 5.28713}, {m[3] -> 8.66944}, {m[4] ->
11.9678}, {m[5] -> 15.2155}, {m[6] -> 18.4326}, {m[7] ->
21.6303}, {m[8] -> 24.8149}, {m[9] -> 27.9906}, {m[10] -> 31.1597}}

i just wan to output the results in this form and in one cell

m[1]=1.78621;
m[2]=5.28713;
m[3]=8.66944;
...

so how to deal with it? iam sorry i have no trying codes as i don't konw how to try.

nu = 8.163265306122447;
s1 = Table[
FindRoot[(I*m[k])*Tanh[I*m[k]] - nu, {m[k], k*\[Pi],
k*\[Pi] - \[Pi]/2, k*\[Pi] + \[Pi]/2}], {k, 1, 10}]

s2 = s1 /. {"->" -> "="}

however it doesn't work

• You're on the wrong way. Please press F1 and read the document of ReplaceAll (/.) carefully. ReplaceAll is a function you can't circumvent as long as you're using Mathematica. – xzczd Nov 6 '19 at 13:33
• i don't konw how to repalceall about the [,-> and some specific characters – dcydhb Nov 6 '19 at 13:48
• Now read the document of Rule and Equal, and read this Chinese tutorial: note.youdao.com/ynoteshare1/… especially the 函数和它的马甲 section. – xzczd Nov 6 '19 at 13:50
• nu = 8.163265306122447; Table[FindRoot[(I*m[k])*Tanh[I*m[k]] - nu, {m[k], k*\[Pi], k*\[Pi] - \[Pi]/2, k*\[Pi] + \[Pi]/2}][[1]], {k, 1, 10}] /. Rule :> Equal // Column – Bob Hanlon Nov 6 '19 at 14:05
• @Bob Hanlon,it seems that i want to use =and delete [ and ] – dcydhb Nov 6 '19 at 14:14

The code depends on what are you going to achieve. If your aim is simply to assign the values to m[i] the most straightforward way is the following. Let res be your result:

res = {{m[1] -> 1.78621}, {m[2] -> 5.28713}, {m[3] ->
8.66944}, {m[4] -> 11.9678}, {m[5] -> 15.2155}, {m[6] ->
18.4326}, {m[7] -> 21.6303}, {m[8] -> 24.8149}, {m[9] ->
27.9906}, {m[10] -> 31.1597}};

Let us first remove the internal curly braces:

res1 = Flatten[res, 1]

(*  {m[1] -> 1.78621, m[2] -> 5.28713, m[3] -> 8.66944, m[4] -> 11.9678,
m[5] -> 15.2155, m[6] -> 18.4326, m[7] -> 21.6303, m[8] -> 24.8149,
m[9] -> 27.9906, m[10] -> 31.1597}  *)

And now you easily get the value of any of your m-s. Say, for m[5] you should only evaluate

m[5] /. res1

(* 15.2155 *)

This construct you can substitute in any your code, where you need to substitute m[i].

You can also assign the values for all m[i] once forever:

MapAt[(Apply[Set, #] &), res1, All]

(*  {1.78621, 5.28713, 8.66944, 11.9678, 15.2155, 18.4326, 21.6303, \
24.8149, 27.9906, 31.1597}   *)

Now, let us check, say, the value of m[2]:

m[2]

(*  5.28713  *)

Finally, may it by that you only wish to represent them in a form that is convenient for a visual inspection? In that case do the following:

Column[MapAt[(Apply[Equal, #] &), res1, All]]

Have fun!

• it seems that i can't use your code,res = {{m[1] -> 1.78621}, {m[2] -> 5.28713}, {m[3] -> 8.66944}, {m[4] -> 11.9678}, {m[5] -> 15.2155}, {m[6] -> 18.4326}, {m[7] -> 21.6303}, {m[8] -> 24.8149}, {m[9] -> 27.9906}, {m[10] -> 31.1597}}; res1 = Flatten[res, 1] Column[MapAt[(Apply[Equal, #] &), res1, All]] and it runs out MapAt::psl: Position specification All in MapAt[Equal@@#1&,{m[1]->1.78621,m[2]->5.28713,m[3]->8.66944,m[4]->11.9678,m[5]->15.2155,m[6]->18.4326,m[7]->21.6303,m[8]->24.8149,m[9]->27.9906,m[10]->31.1597},All] is not an integer or a list of integers. >> – dcydhb Nov 7 '19 at 2:05
• and the most important is that when i cpoy this result cell to MS word or txt or matlab`,the form changed and you can't use it. – dcydhb Nov 7 '19 at 2:07
• That was my question from the very beginning. You should precisely specify what do you need this result for. It appears that I answered not to the question you had in mind. Where do you want to export it and for what purpose? What are you going to do with it in Word? Or in Editor? Sorry, I do not know matlab at all. – Alexei Boulbitch Nov 7 '19 at 9:43
• To address your claim that error message shows up. This code works for me (Mma 12, Win. 10). The reason that it made some problems for you may be that on your machine Mma "remembered" some of the used values. Try to apply Clear to all variables used in this code. – Alexei Boulbitch Nov 7 '19 at 9:45
• @xzczd That is quite possible. However, from the discussion I got an impression that the OP did not clearly formulated his final goal. It seems now that it is about export, rather than representation in Mma. It is quite probable that the solution OP needs is completely different, nothing to do with what I offered. But I propose that OP first explains his aims more clearly. – Alexei Boulbitch Nov 7 '19 at 13:46

format: plain


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# 2. Maclaurin Series

By M. Bourne

In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a.

We now take a particular case of Taylor Series, in the region near x = 0. Such a polynomial is called the Maclaurin Series.

The infinite series expansion for f(x) about x = 0 becomes:

f(x)~~f(0)+f’(0)x +(f’’(0))/(2!)x^2 +(f’’’(0))/(3!)x^3 +(f^("iv")(0))/(4!)x^4 +...

f’(0) is the first derivative evaluated at x = 0, f’’(0) is the second derivative evaluated at x = 0, and so on.

[Note: Some textbooks call the series on this page "Taylor Series" (which they are, too), or "series expansion" or "power series".]

### Example 1: Expanding sin x

Find the Maclaurin Series expansion for f(x) = sin x.

f(x)=x-1/6x^3 +1/120x^5 -1/5040x^7+...

to see if the polynomial is a good approximation to f(x) = sin x.

We observe that our polynomial (in grey) is a good approximation to f(x) = sin x (in green) near x = 0. In fact, it is quite good between -π ≤ x ≤ π.

Don't miss the Taylor and Maclaurin Series interactive applet where you can explore this concept and the other examples on this page further.

### Example 2: Expanding e^x

Find the Maclaurin Series expansion of f(x) = e^x.

Here's a graph of the answer we just obtained, and the original f(x) = e^x graph:

We can see the approximation is very close to the original f(x)=e^x in the region -2 < x < 2.5.

## Exercise

Find the Maclaurin Series expansion of cos x.

Here's a graph showing the answer we just obtained, and the original f(x) = cos(x) graph:

Once again, we observe that our polynomial (in grey) is a good approximation to f(x) = cos x (in green) between -π ≤ x ≤ π.

Continues below

## Finding Pi Using Infinite Series

### Example 3

In the 17th century, Leibniz used the series expansion of arctan x to find an approximation of π.

d/(dx)arctan\ x=1/(1+x^2)

The value of this derivative when x = 0 is 1. That is, f’(0)=1.

Similarly for the subsequent derivatives:

d^2/(dx^2)arctan\ x=-(2x)/((1+x^2)^2

f’’(0)=0

d^3/(dx^3)arctan\ x=2(3x^2-1)/((1+x^2)^3

f’’’(0)=-2

d^4/(dx^4)arctan\ x=-24x(x^2-1)/((1+x^2)^4

f^("iv")(0)=0

d^5/(dx^5)arctan\ x=24(5x^4-10x^2+1)/((1+x^2)^5

f^("v")(0)=24

Now we can substitute into the Maclaurin Series formula:

arctan\ x ~~0+(1)x+0/2x^2-2/(3!)x^3 +0/(4!)x^4+24/(5!)x^5+...

=x-x^3/3+x^5/5-x^7/7+...

Considering that (see the 45-45 triangle)

arctan 1=pi/4

we can substitute x = 1 into the above expression and get the following expansion for π

pi=4(1-1/3+1/5-1/7+...)

All very well, but it was not a good way to find the value of π because this expansion converges very slowly.

Even after adding 1000 terms, we don't have 3 decimal place accuracy.

4sum_(n=0)^1000((-1)^n)/(2n+1)=3.142\ 591\ 654...

(We noe know that π = 3.141 592 653 5..., and we know many other more efficient ways to find pi.)

Here's the graph of y = arctan x (in green) compared to the polynomial we just found (in grey).

### Example 4

Let's now find the value of pi using the same expansion, but this time we substitute x=1/sqrt(3). We'll use the fact tan (pi/6) = 1/sqrt(3).

Now arctan\ x ~~x-x^3/3+x^5/5-x^7/7+...

 ~~1/sqrt(3)-(1/sqrt(3))^3/3+(1/sqrt(3))^5/5-(1/sqrt(3))^7/7 +...

The value of pi will be 6 times the above value.

Using this expansion, we only need to use 18 steps to get 10 decimal place accuracy for pi. That is:

6sum_(n=0)^18((-1)^n)(1//sqrt(3))^(2n+1)/(2n+1) =6 xx 0.523\ 598\ 775\ 61 =3.141\ 592\ 653\ 6...

#### Why is it so much more accurate here?

The approximating curve in this case has what's called a radius of convergence (limited domain) where it "works" (that is, converges). Outside that domain, the approximation diverges.

We'll see another example of this situation in the Taylor and Maclaurin Series interactive applet.

top

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

### Calculus Lessons on DVD

Easy to understand calculus lessons on DVD. See samples before you commit.


text:
# Writing

Members Public

## Integer Tiles: Unveiling the Rectangle's Inner Workings

Here's a simple-sounding challenge. Problem. Imagine that you have a number of rectangles, each of which has at least one integer side. If they come together to tile a larger rectangle, must it also have one integer side? It looks like high school combinatorics, but at the 1985 Summer Meeting

Members Public

## Square One: The Chess Principles You Need to Start Winning

Chess is seen as a game of strategy. A mental battle between two genius minds. A reflection of one's thinking ability. But if you ever played it yourself, you'd know that chess has nothing to do with IQ. There are $$10^{10^{50}}$$ ways a game can play out. Each

Members Public

## Exam Theory: How to Learn and Revise for Perfect Grades

There's a good chance you prepare for exams poorly. I say this because most students don't learn how to learn. They default to what feels natural, to what their peers are doing, or to what their teachers tell them to do. But it's possible to do so much better. Imagine

Members Public

## Harvard Bound: What It Takes to Get Into an Elite College

I believe that anyone can get into an elite college. That any student, no matter their grades or extracurriculars, can forge a path into the red-brick halls of Harvard and other top universities. I know this is controversial. I know that an air of uncertainty and impossibility surrounds the topic


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### Just Touching

Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

### Biggest Bendy

Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic.

# Calculating with Cosines

##### Stage: 4 and 5 Challenge Level:

Try to make sense of each method one line at a time. The following mathematical topics may be useful:

Pythagoras' theorem

Cosines of angles greater than $90^{\circ}$

Distance between two points on a coordinate grid

Trigonometry in right-angled triangles


text:
# Indexing over all possible functions in better than linear time

Given two sets X and Y, the number of functions mapping X to Y is $\vert Y\vert^{\vert X \vert}$.

In particular I am interested in binary strings of relatively small length, e.g. 8. There are $2^8$ (comprising set $B$) possible bit strings and $2^{2048}$ functions that map onto the same set (let's call the set of these $F$ ). So I need 2048 bits to represent these functions uniquely.

My goal is to find a function, that given an arbitrary an arbitrary 2048 length bit string, will return the appropriate function from the $F$. I can imagine doing this with space on order $2^8$, and average time linear in $|F|$, using a hash table mapping $B\rightarrow B$ and some enumeration over mappings.

The question is, can I do better than this, especially with regard to the linear time (without of course just storing all functions)?

• What do you mean by returning a function? How do you represent a function? – Tsuyoshi Ito Mar 12 '12 at 14:15
• Why $2^{2048}$? ($|X|=|Y|=2^8$). – Marzio De Biasi Mar 12 '12 at 14:33
• @Vor: Yeah, 8×256 is surprisingly large. :) – Tsuyoshi Ito Mar 12 '12 at 15:47
• It is easier to just think about the natural numbers instead of the binary strings. Every function is just a ordered k tuple of integers from 0 to n-1. Now you can consider it as integers in base n. All you need is to implement all the simple bijections, which is nothing more than base conversion. – Chao Xu Mar 12 '12 at 23:44
• Just interpret the 2048-bit string as the 8-bit values of the function, listed in order (i.e., $f(0) \oplus f(1) \oplus \dots \oplus f(255)$, where $\oplus$ is concatenation). To find $f(x)$, read off the 8 bits beginning at bit $8x$. – mjqxxxx Mar 13 '12 at 4:24


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## samnatha Group Title pleasssse helpp me ! find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2 - 2x + 2y - 15 = 0 one year ago one year ago

1. malevolence19

You need implicit differentiation $2x+\frac{dy}{dx}2y-2+2 \frac{dy}{dx}=0 \implies x-2+\frac{dy}{dx}(y+2)=0 \implies \frac{dy}{dx}=\frac{2-x}{y+2}$ $y-y_0=\left. \frac{dy}{dx} \right\|_{x_0,y_0}(x-x_0)$ So you need a point. Then you can find the equation then identify K

2. samnatha

see i can't do it that way as we have not done differentiation so it confuses me i know there is another way like you can rearrange the first equation so that x = .... and then sub it into the other one but after i do that i'm no sure what to do

3. malevolence19

If you just solve the first one for x: $x=-4y-K$ And plug it in you get: $(4y+K)^2+y^2+2(4y+K)+2y=0$ From this you can solve for K in terms of y. As long as you know the slope and everything is write for the original line this will give you the value which allows it to cross tangential. But the slope changes every small increment you move so I don't feel convinced that the slope (in general) can just be -1/4...

4. samnatha

how do i get the slope from that ?

5. samnatha

do i multiply out the equations so i get this $16y ^{2} + 8yk + k ^{2} + y ^{2} + 8y + 2k + 2y - 15$


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# Sum of the digits of 1729!

Find the sum of digits of the decimal representation of $$1\times2\times\cdots \times1729$$.

×

Problem Loading...

Note Loading...

Set Loading...


text:
# Linear search on list or tuples in Python

PythonServer Side ProgrammingProgramming

#### Beyond Basic Programming - Intermediate Python

Most Popular

36 Lectures 3 hours

#### Practical Machine Learning using Python

Best Seller

91 Lectures 23.5 hours

#### Practical Data Science using Python

22 Lectures 6 hours

In this article, we are going to learn how to apply a linear search on lists and tuples.

A linear search starts searching from the first element and goes till the end of the list or tuple. It stops checking whenever it finds the required element.

## Linear Search - Lists & Tuples

Follow the below steps to implement linear search on lists and tuples.

• Initialize the list or tuple and an element.
• Iterate over the list or tuple and check for the element.
• Break the loop whenever you find the element and mark a flag.

## Example

Let's see the code.

# function for linear search
def linear_search(iterable, element):
# flag for marking
is_found = False
# iterating over the iterable
for i in range(len(iterable)):
# checking the element
if iterable[i] == element:
# marking the flag and returning respective message
is_found = True
return f"{element} found"

# checking the existence of element
if not is_found:

# initializing the list
numbers_list = [1, 2, 3, 4, 5, 6]
numbers_tuple = (1, 2, 3, 4, 5, 6)
print("List:", linear_search(numbers_list, 3))
print("List:", linear_search(numbers_list, 7))
print("Tuple:", linear_search(numbers_tuple, 3))
print("Tuple:", linear_search(numbers_tuple, 7))

If you run the above code, then you will get the following result.

## Output

List: 3 found
Tuple: 7 not found


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How do you find the volume bounded by x-8y=0 & the lines x+2y revolved about the y-axis?

May 21, 2018

Volume of revolution =$\frac{188}{3} \pi {a}^{3}$

Explanation:

$x - 8 y = 0$, PASSES through the origin
$x + 2 y = a$ has the intercepts,
$x = a , \mathmr{and} y = \frac{a}{2}$
The ihtersection of the two lines happen to be,
$8 y + 2 y = a$
$y = \frac{a}{10}$
$x = 8 \frac{a}{10}$

^3Considering y axis to represent the height of the cone,
Area of cross section is given by
$A = \pi {\left(r\right)}^{2}$
where,
r is the x coordinate.
For x-8y=0, from y=0 to y=a/10,
x=8y
A=pi(x)^2
A=pi(8y)^2
$A = 64 \pi {y}^{2}$
Volume1 = $\int A \mathrm{dy} = {\int}_{0}^{\frac{a}{10}} 64 \pi {y}^{2} \mathrm{dy} = 64 \frac{\pi}{3} {y}^{3} = 64 \frac{\pi}{3} \left({\left(\frac{a}{10}\right)}^{3} - {0}^{3}\right)$
$V o l u m e 1 = \frac{64}{1000} \frac{\pi}{3} {a}^{3}$
For x=2y=1 from y=a/10 to y=a/2
$x = a - 2 y$
A=pi(x)^2
$A = \pi {\left(a - 2 y\right)}^{2}$
Volume2=$\int A \mathrm{dy} = {\int}_{\frac{a}{10}}^{\frac{a}{2}} \pi {\left(a - 2 y\right)}^{2} \mathrm{dy}$
=-pi/2(a-2y)^3/3=-pi/6((a-2xx(a/2)^3-(a-2xxa/10)^3)

$= - \frac{\pi}{6} \left(a - \frac{2}{8} {a}^{3} - a + \frac{2}{1000} {a}^{3}\right)$
$= - \frac{\pi}{6} \left(- \frac{2}{8} {a}^{3} + \frac{2}{1000} {a}^{3}\right)$

$= 2 \left(\frac{\pi}{6}\right) \left(\frac{1}{8} - \frac{1}{1000}\right) {a}^{3}$
$\frac{1}{8} = \frac{125}{1000}$
$= \frac{\pi}{3} \left(\frac{125}{1000} - \frac{1}{1000}\right) {a}^{3}$

Volume2=$124 \frac{\pi}{1000} {a}^{3}$

Volume of revolution = Volume1+Volume2
$= \frac{64}{1000} \left(\frac{\pi}{3}\right) {a}^{3} + \frac{124}{1000} \left(\frac{\pi}{3}\right) {a}^{3}$
$64 + 124 = 188$
Volume of revolution =$\frac{188}{3} \pi {a}^{3}$


text:
Thread: Problem related to Population Growth

1. Problem related to Population Growth

I am having a test the day right after spring break and throughout this time i don't know where to look for help, so i hope someone can help me with this problem

Suppose that a certain population is modeled by the following Differential Equation: dP/dt = 2P(P/120-2)
For what values of P is decreasing? For what values of P is increasing?

I tried to solve the D.E. by setting all the P's on one side but that didnt seem to work at all

2. You don't actually have to find P(t), just see where dP/dt is negative (and where positive).

3. so do I just have to plug and chuck? or is there a easier way to find the exact solution?

the solution I got is: P is descreasing when 0<P<240
P is increasing when P>240

4. Correct

Not sure what your worry is - if you didn't do it by solving dP/dt = 0 = 2P(P/120-2) or else p^2 / 60 - 4p then try that


text:
Other Open Access

Architecture of Virtual Power Plant for Ancillary Services

Ali Jibran; Massucco Stefano; Silvestro Federico

Dublin Core Export

<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Ali Jibran</dc:creator>
<dc:creator>Massucco Stefano</dc:creator>
<dc:creator>Silvestro Federico</dc:creator>
<dc:date>2018-06-08</dc:date>
<dc:description>The increased penetration of distributed energy resources opens up issues in power system as a whole. This creates markets opportunities for ancillary services; particularly TSO deals with the issues of congestion management, reserves, reactive power control etc. Literature suggests different techniques where TSO and DSO interact with each other, and in this way, DSO can offer flexibility to TSO in terms of provision of ancillary services. The paper discusses the issues that the current power system face due to the profound effects of new generating resources, and then examines in detail the way these issues are resolved in a conventional manner. Then, the paper discusses some literature proposals for the interaction between TSO-DSO for solving the issues in an efficient manner, and finally presents the architecture where a Virtual Power Plant (VPP) is developed to facilitate DSO with a platform for the provision of ancillary services.</dc:description>
<dc:identifier>https://zenodo.org/record/3891603</dc:identifier>
<dc:identifier>10.5281/zenodo.3891603</dc:identifier>
<dc:identifier>oai:zenodo.org:3891603</dc:identifier>
<dc:relation>info:eu-repo/grantAgreement/EC/H2020/676042/</dc:relation>
<dc:relation>doi:10.5281/zenodo.3891602</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:title>Architecture of Virtual Power Plant for Ancillary Services</dc:title>
<dc:type>info:eu-repo/semantics/other</dc:type>
<dc:type>publication-other</dc:type>
</oai_dc:dc>

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views


text:
### Wrong

#### Score: 0 / 1

A horizontal force F = 5.0 N acts on the block. The block slides over the trolley. What is the acceleration of the trolley?

Your choiceExpected choiceAnswerComment

5.0 m s–2

1.0 m s–2

0.75 m s–2

0.60 m s–2

### Wrong

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Δp in a time Δt. What is the impulse acting on the object during this change?

Your choiceExpected choiceAnswerComment

Δp

ΔΔt

(Deltap)/(Deltat)

zero

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

The object must be moving at constant speed.

The velocity of the object in any direction must be zero.

The forces acting horizontally on the object must equal the forces acting vertically on the object.

The resultant force acting on the object must be zero.

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

(v^2sin^2theta)/(2g)

(v^2cos^2theta)/(2g)

(vsintheta)/(g)

(vcostheta)/(g)

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

T<W/2

T=W/2

W/2<T<=W

T>W

### Wrong

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Vand car Y is VY.

Which of the following vectors represent the velocity of Y relative to X?

Your choiceExpected choiceAnswerComment

### Wrong

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What is the average acceleration of the ball between X and Y?

Your choiceExpected choiceAnswerComment

2000 m s–2

4000 m s–2

5000 m s–2

8000 m s–2

### Wrong

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8 m s–1.  What is the total upward force acting on the parachutist?

Your choiceExpected choiceAnswerComment

0.N

70 N

560 N

700 N

### Wrong

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Your choiceExpected choiceAnswerComment

the upward push of the chair on the student.

the downward force on the student.

the downward push of the chair on Earth.

the upward force on Earth.

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

10 kg m s–2

30 kg m s–2

40 kg m s–2

50 kg m s–2

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

206/4V

V

206/210V

4/206V

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

2.5m

5.0m

10m

14m

### Wrong

#### Score: 0 / 1

20 m s–1.

What is the order of magnitude of the force acting on the stopper?

Your choiceExpected choiceAnswerComment

10–3 N

100 N

101 N

103 N

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

Ff.

F.

F+f.

f.

### Wrong

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What is the acceleration of the train?

Your choiceExpected choiceAnswerComment

sin θ

cos θ

tan θ

g/(tan theta)

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

decrease

increase

increase then remain constant.

remain constant.

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

Only the magnitude of the acceleration at a given instant

Both the velocity and the acceleration at a given instant

Only the distance travelled in a given time

Both the distance travelled in a given time and the magnitude of the acceleration at a given instant

### Wrong

#### Score: 0 / 1

2/3.

How much energy is dissipated in pulling the truck up the plane?

Your choiceExpected choiceAnswerComment

16 kJ

24 kJ

32 kJ

64 kJ

### Wrong

#### Score: 0 / 1

Ignoring air resistance what is the range RR of the football?

Your choiceExpected choiceAnswerComment

ut

ut cos θ

ut sin θ

ut tan θ

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

No forces act on the object.

No resultant force acts on the object.

The momentum of the object is zero.

No work is being done on the object.

### Wrong

#### Score: 0 / 1

What is the time taken for the stone to hit the surface of the lake?

Your choiceExpected choiceAnswerComment

(vsintheta)/g

(vcostheta)/g

(2vsintheta)/g

(2vcostheta)/g

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

3

2

radius

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

The object continues to move with a changing velocity.

The object continues to move with a constant velocity.

The object continues to move with a changing direction.

The object continues to move in the same direction.

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

The object must be stationary.

The object must move with a constant speed.

The resultant force acting on the object must be zero.

No forces must act on the object.

### Wrong

#### Score: 0 / 1

cm s–1 to a speed of 15 cm s–1. What is the impulse acting on the car?

Your choiceExpected choiceAnswerComment

7.5 mN s

37.5 mN s

0.75 N s

3.75 N s

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

The object must be stationary.

The object must be moving with constant acceleration.

The resultant force acting on the object must be zero.

There must be no external forces acting on the object.

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

m only

v only

m and v

neither m nor v

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

0.50 ms−1

2.0 ms−1

4.5 ms−1

8.0 ms−1

### Wrong

#### Score: 0 / 1

Vx in the x-direction and the component of its velocity Vy in the y-direction, vary with time t

xx-direction component

y-direction component

The particle is moving

Your choiceExpected choiceAnswerComment

in a parabola.

with simple harmonic motion.

with constant velocity in a straight line.

in a circle.

### Wrong

#### Score: 0 / 1

v. Air resistance is negligible.

A second projectile has double the mass of the first projectile and is launched with the same velocity. Air resistance is still negligible. Which of the following paths best represents the path of the projectile? (The original path is shown as a dotted line)

Your choiceExpected choiceAnswerComment

### Wrong

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Which of the following correctly shows the magnitude of the average velocity and the magnitude of the average speed?

Your choiceExpected choiceAnswerComment

### Wrong

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Your choiceExpected choiceAnswerComment

Zero

Increasing

Decreasing

Constant

### Wrong

#### Score: 0 / 1

What is the magnitude of the mean net force on the ball during the collision?

Your choiceExpected choiceAnswerComment

(m(u-v))/((t_2+t_1)

(m(u-v))/((t_2-t_1)

(m(u+v))/((t_2+t_1)

(m(u+v))/((t_2-t_1)

### Wrong

#### Score: 0 / 1

Which of the following correctly compares the magnitudes of the net force and of the drag (resistance) force on the two balls?

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

Product of force and distance

Product of force and distance moved in the direction of the force

Product of power and time

Product of force and displacement

### Wrong

#### Score: 0 / 1

What is the change in speed of the object shown by the graph?

Your choiceExpected choiceAnswerComment

0.5 m s–1

2.0 m s–1

36 m s–1

72 m s–1

### Wrong

#### Score: 0 / 1

What is the speed of the object just after it leaves the spring?

Your choiceExpected choiceAnswerComment

ysqrt(k/m

ysqrt(m/k

yk/m

ym/k`

### Wrong

#### Score: 0 / 1

Which of the following statements is correct whilst the stone is in motion?

Your choiceExpected choiceAnswerComment

The vertical component of the stone’s displacement is constant.

The horizontal component of the stone’s displacement is constant.

The vertical component of the stone’s velocity is constant.

The horizontal component of the stone’s velocity is constant.

### Wrong

#### Score: 0 / 1

Ignoring air resistance, which statement is true?

Your choiceExpected choiceAnswerComment

Y will hit the ground before X.

Y will hit the ground after X.

Y will hit the ground at the same time as X.

The outcome can only be determined if the initial speed of X is known.

### Wrong

#### Score: 0 / 1

Which quantity is zero when the ball is at the highest point Y of the path?

Your choiceExpected choiceAnswerComment

The horizontal component of the ball’s acceleration

The vertical component of the ball’s acceleration

The horizontal component of the ball’s velocity

The kinetic energy of the ball

### Wrong

#### Score: 0 / 1

A change in which of the following will affect the horizontal distance travelled?

Your choiceExpected choiceAnswerComment

m only

u only

both m and u

neither m nor u

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

vh and the vertical component vv of velocity of an object are shown on the graphs. Air resistance is negligible.

These graphs could represent the motion of an object fired from a cliff

Your choiceExpected choiceAnswerComment

vertically upwards.

at an angle above the horizontal.

horizontally.

at an angle below the horizontal.

### Wrong

#### Score: 0 / 1

What is the speed of the object after 1.0 s?

Your choiceExpected choiceAnswerComment

0.50ms–1

1.0ms–1

1.5ms–1

2.0ms–1

### Wrong

#### Score: 0 / 1

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

The horizontal distance travelled by the ball to the point where it lands on the floor depends on

Your choiceExpected choiceAnswerComment

m and h only.

m and v only.

h and v only.

mh and v.

### Wrong

#### Score: 0 / 1

The magnitude of the force exerted on the skier by the tow rope must be

I. greater than the magnitude of the total resistive force acting on the skier
II. equal to the magnitude of the force exerted on the tow rope by the skier
III. equal to the magnitude of the force causing the boat to accelerate.

Which of the above factors is/are correct?

Your choiceExpected choiceAnswerComment

I and II only

I and III only

II only

III only

### Wrong

#### Score: 0 / 1

What can the student deduce from this graph only, and what quantity from the graph is used to make this deduction?

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

Which of the following correctly describes the changes, if any, to the ball’s vertical speed and horizontal speed?

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

Which graph shows how the vertical component of the velocity v of the ball varies with t as it falls to the sea?

Your choiceExpected choiceAnswerComment

### Wrong

#### Score: 0 / 1

How much kinetic energy is gained by the girl during the period represented on the graph? Frictional forces are negligible.

Your choiceExpected choiceAnswerComment

200 J

400 J

600 J

1200 J

### Wrong

#### Score: 0 / 1

The object is then projected with the same initial conditions but air resistance is taken into account. Which of the following is the trajectory when air resistance is taken into account? The original trajectory is shown as a dotted line.

Your choiceExpected choiceAnswerComment


text:
# Past Events

Date/Time: Thursday, November 23, 2017 - 09:30 to 10:30
Venue: SMS seminar room
Speaker: Ramprasad Kale, 4th Year student, SMS, NISER
Title: Some Topics in Sequential Estimation

Confidence intervals constructed from a fixed-sample size procedure may sometimes be too wide to be of any practical use. Hence, to avoid such a problem, it is desirable to construct confidence intervals with fixed-width. G. B. Dantzig (1940) showed that there did not exist any fixed sample size...

Date/Time: Wednesday, November 22, 2017 - 15:30
Venue: Seminar Hall
Speaker: Anantadulal Paul, NISER
Title: Counting complex curves with singularity in CP^2

The study of of degree d curves in $\mathbb{CP}^2$ is one of the interesting topic which arises in the context of Enumerative Geometry. The specific problem, which I will talk about, can be stated as follows:

Let $\mathbb{CP}^2$ be a compact complex surface and $L \rightarrow \mathbb{CP}^... Date/Time: Wednesday, November 22, 2017 - 14:00 to 15:00 Venue: Conference Room, School of Mathematical Sciences Speaker: Sabyasachi Mukhopadhyay, University of Hohenheim Title: Modelling spatio-temporal variation in rainfall using a hierarchical Bayesian regression model Date/Time: Wednesday, November 22, 2017 - 11:00 to 12:00 Venue: conference Room, School of Mathematical Sciences Speaker: Ananta Kumar Majee, University of Tübingen Title: Rate of convergence of a semi-discrete finite difference scheme for stochastic balance laws driven by Levy noise Date/Time: Tuesday, November 21, 2017 - 11:30 Venue: Seminar Hall, SMS Speaker: Abhrojyoti Sen, SMS, NISER Title: Vanishing Pressure Limit For a Non-strictly Hyperbolic System In this talk we will be discussing about vanishing pressure limit for the equation u_t + (u^2/2)_x = 0 ρ_t + (ρu)_x = 0, with initial data u(x, 0) = u_0(x), ρ(x, 0) = ρ_0(x), where u is the velocity component and ρ is the density component. This equation... Date/Time: Monday, November 20, 2017 - 15:30 Venue: Seminar hall Speaker: Atibur Rahaman , NISER Title: Partial dual construction for C*-Quantum groups Given a Hopf algebra H in a braided category \mathcal{C} and a projection H\longrightarrow A to a Hopf subalgebra, one can construct a Hopf algebra r_{A}(H), called the partial dualization of H , with a projection to Hopf algebra dual to A. A non-degenerate Hopf pairing \omega :A\otimes B \... Date/Time: Monday, November 20, 2017 - 11:30 to 12:30 Venue: SMS, Seminar Room Speaker: Prof. Raj Srinivasan, University of Saskatchewan, Canada Title: Non Homogenous Poisson Process and Its Applications Abstract: Traditionally, Poisson process with constant rate have been extensively used to model arrival and potential service processes arising in queueing and queueing networks models under the Markov regime. Generalization to non Markovian models, usually has taken... Date/Time: Thursday, November 16, 2017 - 16:00 to 17:15 Venue: SMS SEMINAR HALL Speaker: Dr. Kunal Krishna Mukherjee , IIT Madras Title: On dynamical system preserving weights The canonical unitary representation of a locally compact separable group arising from an ergodic action of the group on a von Neumann algebra with separable predual preserving a faithful normal semifinite (infinite) weight is weak mixing. On the contrary, there exists a non... Date/Time: Monday, November 13, 2017 - 11:30 to 12:30 Venue: Seminar Room, School of Mathematical Sciences Speaker: Suprio Bhar, TIFR Bangalore Title: Stochastic PDEs in the space of Tempered distributions In this talk, we will discuss some recent results on the existence and uniqueness of strong solutions of certain classes of stochastic PDEs in the space of Tempered distributions. We show that these solutions can be constructed from the solutions of "related" finite dimensional stochastic... Date/Time: Friday, November 10, 2017 - 15:45 to 16:45 Venue: Math Conference Room Speaker: Sudeep Stephen, The University of Newcastle, Australia Title: Zero forcing in graphs Abstract: For a two-colouring of the vertex set of a simple graph$G=(V,E)$, consider the following colour-change rule: a red vertex is converted to blue if it is the only red neighbour of some blue vertex. A vertex set$S\subseteq V\$ is called \emph{zero-forcing} if,...

## Contact us

School of Mathematical Sciences

NISERPO- Bhimpur-PadanpurVia- Jatni, District- Khurda, Odisha, India, PIN- 752050

Tel: +91-674-249-4081

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text:
Answer: What was the name of the utopia discovered in “Lost Horizon”?

The Question: What was the name of the utopia discovered in “Lost Horizon”?

Camelot
Shangri-La


text:
#### Archived

This topic is now archived and is closed to further replies.

# Finding the angle between two sprites

This topic is 5817 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

hello there. I want my enemy to chase down the player by finding the angle between the two and setting its vector to that angle. Here's what I am doing:

float opp = ( p_playerY - enemyY );
float adj = ( p_playerX - enemyX  );

float angle = atan( ( opp / adj ) ) * 180.0f / 3.14f;

This should give me the angle right? well, the enemy doesn't move toward the player like he should. Can someone help or give another solution to this problem? Thanks. [edited by - randomDecay on August 15, 2002 2:06:46 PM] [edited by - randomDecay on August 15, 2002 2:07:12 PM]

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Off the top of my head I''d say it''s something to with operator precedence. Try using this

  float opp = ( p_playerY - enemyY );float adj = ( p_playerX - enemyX  );float angle = (atan( ( opp / adj ) ) * 180.0f) / 3.14f;

##### Share on other sites
damn, didn''t help...

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atan2( y, x ) yields the angle.
Alternativelly, you can use do this:

dir_x = ai_x - player_x
dir_y = ai_y - player_y;

dir_x & dir_y are now the direction vectors the ai should travel along to reach the player.

ie ai_x + dir_x equals player_x

To move the ai one step towards the player over time ( dt ), then do something like

ai_x += dir_x * dt;
ai_y += dir_y * dt;

These operations would all be easier if you use a vector library.

##### Share on other sites
maxsteel -- your second solution will work thanks, but I have one question for you about it...

let''s say enemy XY = (96, 96) player XY = ( 416, 288 )

dirx = 96 - 416 = -320
diry = 96 - 288 = -192

you said enemyX + dirX = playerX which is not true :
96 + -320 = -224

On a computer screen the enemy should be going SOUTH EAST with the coordinates above. but if you work out your solution, ie --

enemyX += ( dirX * dt ) (let''s say dt = 1/2 )

enemyX = 96 + ( -320 * .5 ) = -64 ?? Shouldn''t the enemy be going EAST in the x direction?

Thanks and please tell me if I understand what I am doing

##### Share on other sites
Try this:

dir_x = player_x - ai_x;
dir_y = player_y - ai_y;

(Flip the operands in the subtraction)

##### Share on other sites
sweet, it works perfectly! One thign I am concerned about though: all my movement for enemies is INDEPENDENT of the frame rate. Is this method dependent on the framerate? I can''t seem to get that atan2 method working, so I guess I am going to have to use this.

Thanks maxsteel and everyone else who helped!

##### Share on other sites
if you divide dir_x and dir_y by sqrt(dir_x^2 + dir_y^2), you will get the same values for the x and y offset that you would have gotten if your atan2 method was used, and then used sin and cos on the angle to generate an x and y offset, if that answers your question.

As for being frame rate dependent, that depends on if you base movement on the system clock (frame rate dependent), or plan to set a strict amount of movement each frame (not frame rate dependent).

if you want your movement to be independent of frame rate, you will always multiply your x and y offset values by the same value for movement.

Make sense?

[edited by - Waverider on August 15, 2002 3:43:57 PM]

##### Share on other sites
so if I always use the same dt, it will be the same offset even on two different computers with different framerates?

But the way I see it is with a higher framerate, the PursuePlayer() function will be called more more often than a lower framerate. So the enemy will move faster on some machines with higher framerates.

[edited by - randomDecay on August 15, 2002 3:49:20 PM]

##### Share on other sites
Any game that moves a set amount each frame will move faster on a higher end machine if the frame rate itself isn't controlled somehow. (EVERYTHING will move faster)

If you want to, say, limit the frame rate to 24, then you could keep the object positions from updating until 1/24 second in time passes, then allow the movement routine to execute on all the objects.

Otherwise, you can update the object positions every frame based on how much time has actually elapsed since the last frame, and the objects will move the same speed no matter what machine they are running on.

To answer your question, if you use the SAME dt every frame, a faster machine will run the movement routine more often, causing everything to go faster. So if you base your dt on how much time elapses instead of setting it the same value every frame, you will get the same movement speed on every machine, even thought the frame rate will be higher on the faster machines (IF you don't want to control the frame rate).

[edited by - Waverider on August 15, 2002 4:13:02 PM]

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text:
Some quick code snippets for using an Ant build script from within a Gradle build:

## Plain and simple

If the Ant script uses optional tasks, these tasks must explicitly be defined, and the implementing libraries must be added as dependencies. In this example, I am using JUnit 4.8.2 with the Ant-provided junit and junitreport tasks:


text:
## Dennett on free will

I just finished reading Daniel Dennett's Freedom Evolves. It's about free will and freedom. His overall point is that over time, over evolution, we have acquired more or more powerful free choices.

## "Stop that crow!"

Thruout the book, Dennett seems to worry that his ideas on free will will dispel some useful illusion. He often uses the phrase "Stop that crow!", which he borrows from the Dumbo movie, where the elephant can fly as long as he's holding what he thinks is a magic feather.

## Free will

Now, I consider "free will" to be a chimerical concept in its common usage. The term bundles several very different things together:

• Subjectively making choices
• Lack of physical determinability. As opposed to observing a person's brain in complete detail and predicting the "free" choice he makes later that day.
• Status as automaton or not.
• Moral responsibility, as opposed to "you can't blame me because somebody or something made' me do it"

Dennett never dissects the strands of meaning in this way. But in chapters 2 and 3, he demonstrates that there is no essential connection between free will and lack of physical determinability. He also refutes the (silly IMO) position that quantum indeterminability somehow makes for free will.

He motivates the non-connection between choices and physical determinability with the example of Conway's game of Life.

Although Life is "really" a pixelated rectangle with every pixel on equal footing, Life experimenters distinguish and name higher-level patterns - "entities" within the game, one might say. Experimenters also make design choices that often include avoiding harm to their creations, eg by placing blocks where gliders might smash into them. Avoiding is fundamentally a choice. Entities within the game itself could theoretically be designed to react to incoming disruptions by making avoidance choices,

Now, Life is actually Turing Complete - given a large space, you can actually build a Universal Turing Machine in it. And so Life entities avoidance choices could theoretically reach any specified standard of intelligence.

And of course Life is fully pre-determined. So entities in Life could make real choices in a fully pre-determined world.

## I blogged about chapter 3 without knowing it

Chapter 3 (Thinking about Determinism) basically repeats Who's Still Afraid of Determinism? Rethinking Causes and Possibilities, a paper by him and Christopher Taylor on causality. By co-incidence, I blogged about it earlier1.

## The Evolution of Moral Agency

The chapter I liked best was chapter 7, the Evolution of Moral Agency. In it he advances a theory, new to me but pulled variously from Robert Frank and George Ainsley, that we have moral emotions in order to resist temptation. And why resist temptation? It's actually in our long term interests (Frank). And why not just directly act on our long-term interests? Because temptation increases hyperbolically as it gets nearer (Ainsley), and we can counter it best with a competing stimulus, that being moral emotions (Frank), and that works best if it's not under conscious control (Dennett himself).

This theory has the ring of truth to it.

## The Future of Human Freedom

I found his final chapter weak (as Dennett goes). He's concerned with "Holding the line against creeping exculpation". Ie, as we learn more and more about why people make the choices they do, must that mean we can less and less

Given Dennett's past writings, I was surprised that he didn't bring his previous ideas to bear more. Instead, he writes a woolly final chapter, littered with poorly chosen real world examples that he distances himself from.

### What I would have said

Rather than merely criticize, I will offer my own answer. I would have said, leaning on Dennett's own earlier ideas on design, that we are blameworthy or praiseworthy exactly for occurences that we designed.

Important points:

• I did not say "declaratively designed". Don't imagine an inventor with a blueprint and a contraption run amuck, or anything of the like. I mean "design" in the sense that Dennett has used it previously, the sense in which design permeates life. In particular:
• Lack of declarative faculty is no bar to responsibility. My dog can't describe or reflect on her choices, but she can properly be praised or blamed. A little. To a doggy standard of responsibility, not a human one.
• Self-deception is no bar to responsibility.
• An important but subtle distinction: we are responsible for the design that was realized, not for the realized outcome. That means that "wiggle room" between design and realized outcome is no bar to responsibility. In particular, I contemplate:
• Probabilistic tradeoffs, regardless which outcome is realized. A drunk driver, even thought she happened not to hit anyone, has done a blameworthy thing.
• Contingent outcomes. So bureaucrats, when one says "I just recommended X" and the other "I just followed the recommendation", don't avoid responsibility. They each partly designed the outcome, contingent on the another's action.

The virtues of this definition:

• It defeats the creeping exculpation that concerns Dennett.
• It doesn't bar exculpation where there's really no mens rea`.
• It's fuzzy on just the right things:
• When it's fuzzy whether an occurence is the realization of some design.
• (The flip side of the previous) When it's fuzzy whether a design was realized. (again, distinct from whether an outcome was realized)
• It allows multiple responsibility for the same occurence in situations, like the example that Dennett gives on page 74.
• It gives the right answer when a designer is a surprising entity such as a group, an organization, or a system.
• We don't have to know the responsible person or group's inner workings in order to assign responsibility, we can treat him/her/it as a black box that effectuations of designs come out of. Black-boxing generally makes analysis easier.
• We can understand "attempts" and "incitement" and similar qualified sins in this framework.
• We can understand lack of responsibility due to being deceived in this framework.
• It distinguishes our responsibility from that of non-choosers (fire) and weak choosers (animals) in a principled, non-question-begging way.

## Footnotes:

1 Short summary of my position: When trying to define causality, we should really be looking at quality of explanation.


text:
# American Institute of Mathematical Sciences

May  2017, 11(2): 267-282. doi: 10.3934/amc.2017018

## Rank equivalent and rank degenerate skew cyclic codes

 Department of Mathematical Sciences, Aalborg University, Fredrik Bajers Vej 7G, 9220, Denmark

Received  February 2016 Revised  March 2016 Published  May 2017

Fund Project: The author is supported by The Danish Council for Independent Research (Grant No. DFF-4002-00367).

Two skew cyclic codes can be equivalent for the Hamming metric only if they have the same length, and only the zero code is degenerate. The situation is completely different for the rank metric. We study rank equivalences between skew cyclic codes of different lengths and, with the aim of finding the skew cyclic code of smallest length that is rank equivalent to a given one, we define different types of length for a given skew cyclic code, relate them and compute them in most cases. We give different characterizations of rank degenerate skew cyclic codes using conventional polynomials and linearized polynomials. Some known results on the rank weight hierarchy of cyclic codes for some lengths are obtained as particular cases and extended to all lengths and to all skew cyclic codes. Finally, we prove that the smallest length of a linear code that is rank equivalent to a given skew cyclic code can be attained by a pseudo-skew cyclic code.

Citation: Umberto Martínez-Peñas. Rank equivalent and rank degenerate skew cyclic codes. Advances in Mathematics of Communications, 2017, 11 (2) : 267-282. doi: 10.3934/amc.2017018
##### References:
 [1] D. Boucher, W. Geiselmann and F. Ulmer, Skew-cyclic codes, Appl. Algebra Engin. Commun. Comp., 18 (2007), 379-389.  doi: 10.1007/s00200-007-0043-z.  Google Scholar [2] D. Boucher and F. Ulmer, Coding with skew polynomial rings, J. Symb. Comput., 44 (2009), 1644-1656.  doi: 10.1016/j.jsc.2007.11.008.  Google Scholar [3] J. Ducoat, Generalized rank weights: a duality statement, in Topics in Finite Fields, (2015), 101-109.  doi: 10.1090/conm/632/12622.  Google Scholar [4] J. Ducoat and F. Oggier, Rank weight hierarchy of some classes of cyclic codes, in 2014 IEEE Inf. Theory Workshop, (2014), 142-146.   Google Scholar [5] E. M. Gabidulin, Theory of codes with maximum rank distance, Problemy Peredachi Informatsii, 21 (1985), 3-16.   Google Scholar [6] E. M. Gabidulin, Rank q-cyclic and pseudo-q-cyclic codes, in IEEE Int. Symp. Inf. Theory, (2009), 2799-2802.   Google Scholar [7] W. C. Huffman and V. Pless, Fundamentals of Error-Correcting Codes Cambridge Univ. Press, Cambridge, 2003. doi: 10.1017/CBO9780511807077.  Google Scholar [8] J. Kurihara, R. Matsumoto and T. Uyematsu, Relative generalized rank weight of linear codes and its applications to network coding, IEEE Trans. Inf. Theory, 61 (2015), 3912-3936.  doi: 10.1109/TIT.2015.2429713.  Google Scholar [9] R. Lidl and H. Niederreiter, Finite Fields Addison-Wesley, Amsterdam, 1983.  Google Scholar [10] U. Martínez-Peñas, On the roots and minimum rank distance of skew cyclic codes, Des. Codes Crypt., 83 (2017), 639-660.  doi: 10.1007/s10623-016-0262-z.  Google Scholar [11] U. Martínez-Peñas, On the similarities between generalized rank and Hamming weights and their applications to network coding, IEEE Trans. Inf. Theory, 62 (2016), 4081-4095.  doi: 10.1109/TIT.2016.2570238.  Google Scholar [12] O. Ore, On a special class of polynomials, Trans. Amer. Math. Soc., 35 (1933), 559-584.  doi: 10.2307/1989849.  Google Scholar [13] O. Ore, Theory of non-commutative polynomials, Ann. Math., 34 (1933), 480-508.  doi: 10.2307/1968173.  Google Scholar [14] D. Silva and F. R. Kschischang, On metrics for error correction in network coding, IEEE Trans. Inf. Theory, 55 (2009), 5479-5490.  doi: 10.1109/TIT.2009.2032817.  Google Scholar [15] H. Stichtenoth, On the dimension of subfield subcodes, IEEE Trans. Inf. Theory, 36 (1990), 90-93.  doi: 10.1109/18.50376.  Google Scholar

show all references

##### References:
 [1] D. Boucher, W. Geiselmann and F. Ulmer, Skew-cyclic codes, Appl. Algebra Engin. Commun. Comp., 18 (2007), 379-389.  doi: 10.1007/s00200-007-0043-z.  Google Scholar [2] D. Boucher and F. Ulmer, Coding with skew polynomial rings, J. Symb. Comput., 44 (2009), 1644-1656.  doi: 10.1016/j.jsc.2007.11.008.  Google Scholar [3] J. Ducoat, Generalized rank weights: a duality statement, in Topics in Finite Fields, (2015), 101-109.  doi: 10.1090/conm/632/12622.  Google Scholar [4] J. Ducoat and F. Oggier, Rank weight hierarchy of some classes of cyclic codes, in 2014 IEEE Inf. Theory Workshop, (2014), 142-146.   Google Scholar [5] E. M. Gabidulin, Theory of codes with maximum rank distance, Problemy Peredachi Informatsii, 21 (1985), 3-16.   Google Scholar [6] E. M. Gabidulin, Rank q-cyclic and pseudo-q-cyclic codes, in IEEE Int. Symp. Inf. Theory, (2009), 2799-2802.   Google Scholar [7] W. C. Huffman and V. Pless, Fundamentals of Error-Correcting Codes Cambridge Univ. Press, Cambridge, 2003. doi: 10.1017/CBO9780511807077.  Google Scholar [8] J. Kurihara, R. Matsumoto and T. Uyematsu, Relative generalized rank weight of linear codes and its applications to network coding, IEEE Trans. Inf. Theory, 61 (2015), 3912-3936.  doi: 10.1109/TIT.2015.2429713.  Google Scholar [9] R. Lidl and H. Niederreiter, Finite Fields Addison-Wesley, Amsterdam, 1983.  Google Scholar [10] U. Martínez-Peñas, On the roots and minimum rank distance of skew cyclic codes, Des. Codes Crypt., 83 (2017), 639-660.  doi: 10.1007/s10623-016-0262-z.  Google Scholar [11] U. Martínez-Peñas, On the similarities between generalized rank and Hamming weights and their applications to network coding, IEEE Trans. Inf. Theory, 62 (2016), 4081-4095.  doi: 10.1109/TIT.2016.2570238.  Google Scholar [12] O. Ore, On a special class of polynomials, Trans. Amer. Math. Soc., 35 (1933), 559-584.  doi: 10.2307/1989849.  Google Scholar [13] O. Ore, Theory of non-commutative polynomials, Ann. Math., 34 (1933), 480-508.  doi: 10.2307/1968173.  Google Scholar [14] D. Silva and F. R. Kschischang, On metrics for error correction in network coding, IEEE Trans. Inf. Theory, 55 (2009), 5479-5490.  doi: 10.1109/TIT.2009.2032817.  Google Scholar [15] H. Stichtenoth, On the dimension of subfield subcodes, IEEE Trans. Inf. Theory, 36 (1990), 90-93.  doi: 10.1109/18.50376.  Google Scholar
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2019 Impact Factor: 0.734


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# set_coordinates¶

dolfin.cpp.fem.set_coordinates()

Sets mesh coordinates from function Mesh connectivities d-0, d-1, …, d-r are built on function mesh (where d is topological dimension of the mesh and r is maximal dimension of entity associated with any coordinate node). Consider clearing unneeded connectivities when finished.

Parameters: & geometry (MeshGeometry) – (MeshGeometry ) Mesh geometry to be set Function & position (const) – (Function ) Vectorial Lagrange function with matching degree and mesh void


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# Mentally Traumatizing Movies

#### Jennywocky

##### guud languager
It won't be any better the second time around... the CG are a bit dated now and it's tough not to pick on them a little bit. Maybe give the commentary a go and then try re-watching it. It won't be any scarier but maybe the insight gained from the commentary might make it a modicum more interesting at least.
I ended up watching it again last Friday.

Yeah, I'm still harshing on it. It's not much more than a cobbled pastiche of better scifi/horror flicks. I can see all the influences. The character arcs aren't really handled well. Considering it only clocks in at 1:25 or 1:30, it seems like a bunch of stuff was tossed in the edit room too.

Plus, I didn't really find it scary. Just dumb people doing dumb things or having random awful things happen to them without much coherence. Sometimes that can work (like Nightmare on Elm Street sequence) if the film is self-aware enough to not take itself seriously and/or focus on the dark humor instead, but Event Horizon isn't that film.

##### Push to Start
I ended up watching it again last Friday.

Yeah, I'm still harshing on it. It's not much more than a cobbled pastiche of better scifi/horror flicks. I can see all the influences. The character arcs aren't really handled well. Considering it only clocks in at 1:25 or 1:30, it seems like a bunch of stuff was tossed in the edit room too.

Plus, I didn't really find it scary. Just dumb people doing dumb things or having random awful things happen to them without much coherence. Sometimes that can work (like Nightmare on Elm Street sequence) if the film is self-aware enough to not take itself seriously and/or focus on the dark humor instead, but Event Horizon isn't that film.
Ahhh. Jenny, I'm sorry you won't get that time back.

#### Jennywocky

##### guud languager
Ahhh. Jenny, I'm sorry you won't get that time back.
I chalked it up as "suffering for the art."

I rented "Don't Breathe" tonight on DVD for $1.50. It's gotta be better. #### Bad Itch ##### Push to Start I chalked it up as "suffering for the art." I rented "Don't Breathe" tonight on DVD for$1.50. It's gotta be better.
*cackle* Good luck

#### Creeping Death

##### Consigliere
Was any else a little bothered by that scene in the Green Mile when MCD grabbed Tom Hanks' junk?

##### Push to Start
Never saw the green mile but now I want to write to Tom Hanks.

"Dude when that dude grabbed your junk in that movie... what was your first thought? Was it Brando? Butter?"

#### TAC

##### Inspectorist
Jacobs Ladder is a good one to traumatize. I find psychological horror does the trick much better.

#### Reluctantly

##### Resident disMember
Was any else a little bothered by that scene in the Green Mile when MCD grabbed Tom Hanks' junk?
I was just disappointed he didn't get to butter his bread.

#### Rixus

##### I introverted think. Therefore, I am.
OK. I've heard people raving about The Conjuring 2. First one was just kind of, "Meh." About to give it a try. Drunken review coming in about 2 hours.

#### WhatWasThat

##### Member
It has some comedic moments as well but "A Clockwork Orange" is pretty disturbing at times. It's one of my favorite films though.

Since someone already mentioned Requiem for a Dream, Pi (by the same director) deserves a mention as well.

Eternal Sunshine of the Spotless Mind
Whatever Happened to Baby Jane?
Angel Baby

Jacobs Ladder is a good one to traumatize. I find psychological horror does the trick much better.
Yep most "normal" horror movies are more funny than scary (bad acting, obvious twists/jump scares, gore for no reason, same old story every time, etc)

Jacob's Ladder was great though in a sort of "I probably don't ever want to see this again but it's an amazing movie" sort of way. All of the movies I listed basically deal with people losing their mind, distortions of reality, etc.

#### Jennywocky

##### guud languager
OK. I've heard people raving about The Conjuring 2. First one was just kind of, "Meh." About to give it a try. Drunken review coming in about 2 hours.
The Conjuring (1) was better in my opinion.

The second one was just more of the same + some over the top demon stuff.

the main reason I find it interesting is that I already knew of the Warrens long before this. I spent some years online trying to find and collect books about demonology. I had the bio book of them called "THe Demonologist" which even had a black and white photo section including pics of the raggedy anne doll named Annabelle. ("Annabelle" the movie kind of sucks too, btw. Some unsettling stuff I guess, but it's all kind of trope and not emotionally engaging.)

---

I saw CLockwork Orange back in my film class in college. I should rewatch it. I just remember the juxtaposition of positive/classical music with disturbing psychopathic violence, which considering I wasn't heavily into film at the time, was really novel to me as a technique.

....well, I remember the milk bar as well. I grew up in quaint little religious rural america. That was surreal to me at the time.

#### Rixus

##### I introverted think. Therefore, I am.
Unfortunately, I didn't get around to watching much of it. I kind of passed out drunk halfway through and don't remember it. I will watch it sober tonight.

I did see the start, and it looked OK. If the same as the first. I didn't like the nun - too much and a bit OTT. I'd rather not see what's scaring us, really. The shouting old guy was a good touch, but I don't know what that dog thing was.

#### Jennywocky

##### guud languager
Unfortunately, I didn't get around to watching much of it. I kind of passed out drunk halfway through and don't remember it....
hehehe

#### zerkalo

##### .
which considering I wasn't heavily into film at the time
same...i remember finding the film really underwhelming
same goes for all other stanley kubrick movies ive seen

#### Rixus

##### I introverted think. Therefore, I am.
The Conjuring (1) was better in my opinion.

The second one was just more of the same + some over the top demon stuff.

the main reason I find it interesting is that I already knew of the Warrens long before this. I spent some years online trying to find and collect books about demonology. I had the bio book of them called "THe Demonologist" which even had a black and white photo section including pics of the raggedy anne doll named Annabelle. ("Annabelle" the movie kind of sucks too, btw. Some unsettling stuff I guess, but it's all kind of trope and not emotionally engaging.)
I just finished Conjuring 2 (totally sober this time.) I actually enjoyed it. I would have rather the nun be more extreme and not seen properly - perhaps unable to see her face would have creeped be out more. But it had some nice haunting moments and had a good atmosphere. Could have tormented the little boy a bit more.

The pictures in the credits were a nice touch - it adds something to it.

I've been wondering about Anabelle. I thought they'd covered the doll enough in the first one, really. My friend says it's really terrifying, but she is a complete wuss with films. When we went to see Lights Out, she was clinging to my arm so tightly her nails dug in and actually drew blood.

#### Jennywocky

##### guud languager
I just finished Conjuring 2 (totally sober this time.) I actually enjoyed it. I would have rather the nun be more extreme and not seen properly - perhaps unable to see her face would have creeped be out more. But it had some nice haunting moments and had a good atmosphere. Could have tormented the little boy a bit more.

The pictures in the credits were a nice touch - it adds something to it.

I've been wondering about Anabelle. I thought they'd covered the doll enough in the first one, really. My friend says it's really terrifying, but she is a complete wuss with films. When we went to see Lights Out, she was clinging to my arm so tightly her nails dug in and actually drew blood.
Basically it has some jump scares, but the characterization / human drama / character behavior is pretty dumb, which for me is a dealbreaker.

I like the pictures in the credits as well. I didn't say Conjuring 2 was a bad movie... just that if you've seen the first, there's not a ton of surprises in the second, and it felt like it was trying to outdo the first movie. Although the nun character is pretty unsettling.

#### lucky12

##### walking on air
Come and see.

Russian movie; follows the life of a young boy who gets caught up in the war. The people in his area have to fight off the SS.

Its got a lot of drawn out scenes that make you feel uncomfortable. The director manifests fear and discomfort in a lot of ways throughout the movie.

Sent from my SM-G925W8 using Tapatalk

#### dutchdisease

Forgot I had created this thread, surprised to see it is still thriving.

#### Rixus

##### I introverted think. Therefore, I am.
I usually reactivate it on a Friday when trying to find something worth watching - and then Jenny comes along and gives some always useful advice. It's becoming a tradition.

#### nanook

##### a scream in a vortex
didn't bother to watch c2 all the way to the end because ... yawn

but that scene wherein the kid is pulled at the legs, in bed, southwards was slightly unnerving because this is one of the things that happen to me in pseudo lucid dreams all the fucking time. sometimes i just witness it with aloof curiosity, sometimes i am actually scared. either way, i don't need a movie to give me bad dreams.

#### zerkalo

##### .
Come and see.

Russian movie; follows the life of a young boy who gets caught up in the war. The people in his area have to fight off the SS.

Its got a lot of drawn out scenes that make you feel uncomfortable. The director manifests fear and discomfort in a lot of ways throughout the movie.

Sent from my SM-G925W8 using Tapatalk

----------------
saw andrzej zulawski's posession with my friend...all the yelling was hilarious at first...but then i had enough of it and all the violence and gore....then came the scene where....(you'll know it when you see it) and i was like "okay......................................................................"
i still have no idea wtf that abomination was about

#### Jennywocky

##### guud languager
"Come and See" seemed to have pretty decent reviews. I'll be adding that and "Evolution" to my view list.

Saw Rogue One today. Can't say it was mentally traumatizing, especially since we have a good idea how it ends ahead of time. But damn, is it bittersweet... and the Alliance actually is grayer "morality wise" than Lucas ever permitted.

#### crippli

##### disturbed
I don't think films have mentally traumatized me with a permanent notable signature.. Maybe with the exception of Two girls One cup. I think it's the soundtrack.

Films like Never Let Me Go or Dancer in the Dark made me uneasy, as I thought the makers captured a form of essence of human nature.

#### Jennywocky

##### guud languager
I don't think films have mentally traumatized me with a permanent notable signature.. Maybe with the exception of Two girls One cup. I think it's the soundtrack.

Films like Never Let Me Go or Dancer in the Dark made me uneasy, as I thought the makers captured a form of essence of human nature.
I have seen Never Let Me Go; yes, I wouldn't call it "traumatizing" in terms of impact on audience, but it is a heavy movie, I hurt after watching it. it really makes you think and has some relevancy to today's world. (I have to get around to Dancer in the Dark, but I've seen a number of von Trier movies... Breaking the Waves is the one that comes to mind now, another heavy movie. All of his movies generally create some degree of discomfort.)

#### zerkalo

##### .
you guys....what did you like about never let me go?

#### Jennywocky

##### guud languager
you guys....what did you like about never let me go?
Was there something you didn't like about it?

it was a little slow in spots (I've never read the book), but the acting was good, there was pathos, and there's already ambiguity over property ownership with the advent of new medical procedures in today's world. It has a bit of an unsettling undercurrent but is mostly profoundly sad.

#### zerkalo

##### .
Was there something you didn't like about it?
not really...the film/book were just a little too underwhelming so im surprised you found it so powerful
but then again the appreciation i could have had for the story was probably killed by the book(read it before watching the film).....it dragged and dragged....the movie had less unnecessary filler in it but i already knew the story so the magic was lost i guess

i did really like the ruth(i hope im not mixing up names here? its been years)(its the keira kneightley character) character. preferred her to the main girl
in the book it always felt like ruth, although selfish and what not, was always more transparent, freespirited and involuntarily honest...unlike the main girl who was always repressed and annoyingly "mature"

i also love the song scene and the song itself ^_^

#### Jennywocky

##### guud languager
not really...the film/book were just a little too underwhelming so im surprised you found it so powerful
but then again the appreciation i could have had for the story was probably killed by the book(read it before watching the film).....it dragged and dragged....the movie had less unnecessary filler in it but i already knew the story so the magic was lost i guess
Yes, it felt a little different going into it cold. It was definitely a "cool" movie in terms of emotional heat -- the emotions for me were triggered mainly by the slow burn of watch them trying to escape their fate over time and slowly watching hope crumble. I felt like I knew Kathy very well by the end and could understand her... how she keeps everything under wraps and is so controlling of her own surface-level expression and helps people by doing things for them / being with them. She also missed out on some opportunities by keeping so much inside and pretending it did not exist.

i did really like the ruth(i hope im not mixing up names here? its been years)(its the keira kneightley character) character. preferred her to the main girl in the book it always felt like ruth, although selfish and what not, was always more transparent and involuntarily honest...unlike the main girl who was always repressed and psuedo mature
Well, I wouldn't call it "pseudo mature" in that her motivations were good, but yes it was a choice on her part about how to express herself and what she chose to do with her time. All types tend to pick the way they feel most comfortable in terms of expressing themselves, based on their values, and all the characters were children and then teens, all of them immature. But I agree about Ruth -- she might have been blatantly selfish/immature at times but she was also very upfront and you knew exactly where she stood, and she just was what she was. Tommy probably was the least fleshed out; he had anger issues which tracked from what was happening to him but I didn't know a ton else about him...

i also love the song scene and the song itself ^_^
I still can feel the "mood" of the movie as I sit here and recall it. Just melancholy. Not necessarily in a bad way, but... basically watching people come to grips that their lives are not their own due to a fluke they could not control and there will be no reprieve, and how is that fair with some lives having more importance than others... although isn't our world already like this?

BTW, Alex Garland (the guy who adapted the book into a script) is the director/writer of Ex Machina, the screenwriter for Sunshine and 28 Days Later, etc.

#### zerkalo

##### .
at first i disliked ruth, but the more the story progressed, the more i warmed up to her
i felt bad for her, she always wore her heart on her sleeve and naively thought that kathy liked her, when all this time kathy secretly found her company insufferable
i kept getting more and more restless reading kathy's monologues as she literally lives her entire life in her head, never confronting people about things that upset her, never doing anything about anything, looking down on people like ruth...and in the end what good did any of that do? nothing. it was ruth who really lived her life

the film was better in that there was more emphasis on this...umm....lost time sentiment...and the hallowing regret i guess
the book focused on...i dunno man...other things

#### zerkalo

##### .
by the way...it just came back to me as i typed my last post
but have you seen the adaptation of/read the authour's other book? i love love loved the remains of the day
it has very similar themes, but was 10 times better
also the film has emma thompson <3

#### Jennywocky

##### guud languager
by the way...it just came back to me as i typed my last post
but have you seen the adaptation of/read the authour's other book? i love love loved the remains of the day
it has very similar themes, but was 10 times better
also the film has emma thompson <3
yes!

It has been a long long time since i have seen it so I do not remember many details at this point (aside from the ending) -- Anthony Hopkins, Emma Thompson, and Christopher Reeve before he was paralyzed. I have been planning to rewatch it for years but just never got around to it; yet the feelings aroused by it still linger.

#### zerkalo

##### .
saw in a year with 13 moons yesterday
the slaughterhouse scene was impossible to watch
rest of the film is ridiculously depressing too, Elvira has to be the most pitiful character ever made

also ive been listening to a song for europe non stop since finishing the film...not so christmas-y


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# Strings in C++ : Finding the first non-repeated character in a string

We are given a string, and told to find the first non-repeated character in it. Let’s look at an example: let’s say our string is “alligator”. The first non-repeated character in this string is “i”.

Now, how do we go about finding this first non-repeated character programmatically?

APPROACH 1

There is a very direct algorithm which we can employ — Take the first character of the given string (which is ‘a’ in our example string “alligator”). Compare it with each of the other characters in the string. If that character does not appear anywhere else, return that character and terminate the algorithm. Else, if the character does appear somewhere else in the string (as in the case of the ‘a’ in “alligator”), go to the 2nd character of the string, and repeat the same process (i.e. check if the 2nd character of the string gets repeated anywhere). Continue this process for each character of the string (i.e. in the k th iteration, we would compare the k th character of the string with each character in the string to determine if it appears anywhere else). If, at the end of the algorithm, we find that all characters get repeated somewhere, i.e. there is no non-repeating character, print a message to that effect.

———

The above algorithm is simple to understand and implement. It will run in $O(n^2)$ time where $n$ is the length of the input string.

——

In pseudocode, the algorithm would look as follows:

——-

Input: String s of length n.

Output: First non-repeated character in s.

Algorithm:

For i going from 0 upto n-1:

flag = true; // flag is a boolean variable initialized to true.

For each j going from 0 upto n-1:

If ( s[i] == s[j] )

flag = false;

End For. //end the inner For loop.

If ( flag == true ) //i.e. if flag is true when we come out of the inner For loop

return s[i];  //character s[i] is the 1st non-repeated character in s.

End For. // end the outer For loop.

//If the algorithm reaches this statement, it means that there is no non-repeated character in s.

// Hence we return a default character, which would not otherwise be returned.

// We choose the null character — NUL for this purpose.

return NUL;

———

So that was the pseudocode for the algorithm. We can translate the above code into C++ as follows:

———

#include <iostream.h>

// function f takes a string s, and returns the first non-repeated character in s.

// In case there is no non-repeated character, it returns the null character ” (having ASCII value 0)

char f (char *s)

{

//we first check if s is the null string.

if ( !s)

return ”; // return the null character is s is the null string.

int i = 0, j;

bool flag;

while ( s[i] ! = ”)

{

flag = true;

j = 0;

while ( s[j] ! = ”)

{

if ( s[i] == s[j] )

flag = false;

j ++;

}

if (flag)

return s[i];

i++;

}

return ”; //i.e. no non-repeated character was found.

}

————


text:
HW4 Solution

1) a) Source should have area <= area of core and divergence angle <= 2 alpha_m
So maximum diameter = 8 mu m
Divergence angle = 2 alpha_m => NA_{source} = 0.18

b) lambda = 1.55 mu m,  w_o = 8/2 (mu m) = 4 ( mu m )
theta = 2 lambda_o / {pi w_o} = 2 1.55 / {pi 4 } = 0.2467 (rad) = 14.13^o
Check alpha_m: alpha_m = sin^{-1} (NA) = 10.37^o > 14.13^o/2
Yes, it is optimum.

c) If the source area = core area, the distance should be zero since beam will expand with distance. In sub-optimal situation, the distance should not be larger z_o = pi w_o^2/lambda = pi 4^2/1.55 = 32.43 (mu m)

d) No, this small beam size will lead to increase in theta such that theta > 2 alpha_m.
Check: theta ={2 times 1.55}/ { pi times 4/2} = 2 times 14.13^o >= 2 times 10.37^o
Even if you can make the distance between source and fiber to zero, light will leak since it is not totally internal reflection the cladding and core interface inside the fiber!

2) a = 25 mu m, n_1 = 1.485, n_2=1.465, lambda = 850 nm, lambda_0 = 1360 nm, S_0 = 0.09 {ps}/{km-nm^2}, Delta lambda = 2nm
a) V = {2 pi a}/ lambda NA = {2 pi times 25}/ 0.85 sqrt{1.485^2 - 1.465^2} = 44.89
b) Since V > 2.405, this fiber operates at multi-mode regime. Number of modes can be found as
M = Int ( V^2 / 2 )= 1007
c) To have single mode operation for 1900nm to 2100nm, we need to have cut-off at the short wavelength and long wavelength will satisfy the single mode condition automatically. Therefore, lambda_c = 1900 nm
2.405 = {2 pi a} / lambda_c NA => NA = 2.405 lambda_c / {2 pi a} =2.909 times 10^{-2}
=> sqrt{n_1^2 - n_2^2} = 2.909 times 10^{-2} => n_1^2 = n_2^2 + ( 2.909 times 10^{-2})^2 => n_1 = 1.46529

3) a = 30 mu m, n_2 = 1.46, D_{modal} = 10 {ns} / km, L = 10km, n_1 approx n_{1g} and n_1 approx n_2
D_{modal}= n_1 Delta / c => Delta = c D_{modal}/ n_2 ={3 times 10^8 times 10^{-8} times 10^{-3}} / 1.46 => Delta = 2.055 times 10^{-3}
NA approx n sqrt {2 Delta} = 1.46 times sqrt {2 times 2.069 times 10^{-3}} = 0.093594
Also NA = n_a sin alpha_m where n_a is the refractive index outside the fiber, i.e. n_a = 1 for air outside the fiber.
=> alpha_m = 5.37^o
Using textbook's formula, R_b = 1 /{2 Delta tau} = 1 / {2 times 10 times 20} ({Gb}/s) = 5 ({Mb}/s)
Using conservative estimate, R_b = 1 /{4 Delta tau} = 2.5 ({Mb}/s)

4) a = 4 mu m, n_1 = 1.467, n_2 = 1.464, lambda = 1.3 mu m
a) V = {2 pi 4} / 1.3 sqrt{1.467^2 - 1.464^2} = 1.813
b) V < 2.405 => single mode regime
c) Single mode of 800nm to 1100nm => lambda_c = 800 nm
2.405 = {2 pi a} / lambda_c NA => a = {2.405 lambda_c} / {2 pi NA} = {2.405 times 0.8} / {2 pi sqrt{1.467^2 - 1.464^2}}
=> a = 3.266 ( mu m ), i.e. diameter <= 6.532 (mu m)
Notes: Core radius needs to be reduced for shorter lambda_c.

5) a= 5 mu m, Delta = 0.002. Take typical n = 1.45 for glass. Single mode cutoff wavelength lambda_c occurs at V=2.405.
2.405 = {2 pi a} / lambda_c NA approx 2.405 = {2 pi a} / lambda_c n sqrt{2 Delta}
lambda_c = {2 pi a NA} / {2.405} = { 2 pi a n sqrt{2 Delta}} / 2.405
lambda_c approx {2 pi times 5 times 1.45 times sqrt{2 times 0.002}} / 2.405 = 1.1979 ( mu m)

6)


text:
# Linear functionals on a vector space

Given $f$ and $g$, linear functionals on the same vector space such that $[x,f]=0$ whenever $[x,g]=0$, where $[x,f]=f(x).$ I am supposed to show that there exists a scalar $\alpha$ such that $f=\alpha g$. What I have done is below. Please see if it is right.

Fix $x$ in V. If $[x,g]\ne0$, we define $\alpha=[x,f]/[x,g].$ Then, $[x,f]=\alpha[x,g].$ i.e $f(x)=\alpha g(x)$, which follows that $f=\alpha g$

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There are a couple of reason why the above is not correct: first, your $\alpha$ depends on $x$: you have not proven that $f=\alpha g$, just that, for that particular $x$ you have $f(x)=\alpha g(x)$. And you don't say what to do if $[x,g]=0$ (it's clear what to do, but you should say it). Really, the first part is the big problem: you would need to show that if $x$ and $y$ are such that $[x,g]\neq 0$ and $[y,g]\neq 0$, then $[x,f]/[x,g] = [y,f]/[y,g]$. Otherwise, your $\alpha$ will change with $x$. –  Arturo Magidin Sep 13 '11 at 3:24
P.S. You can edit your question and modify your attempt; there are a couple of ways to finish off your argument... perhaps the Rank-Nullity Theorem can be of some help? –  Arturo Magidin Sep 13 '11 at 3:41
okay. I'm a bit confused as to what to do next. what exactly am I to show. –  smanoos Sep 13 '11 at 3:42

Assuming that you can find an $x$ such that $[x,g]\neq 0$, you have defined a scalar $\alpha$ as $\alpha = [x,f]/[x,g]$. This implies that the specific scalar $f(x)$ is equal to the scalar $\alpha g(x)$. From this, we can conclude that $f$ and $\alpha g$ agree at $x$; but this is not enough. In order to show that $f=\alpha g$ as functions, you need to show that $f$ and $\alpha g$ agree on all of $V$. So you would still need to show that for every $y\in V$ you have $f(y)=\alpha g(y)$; that is, that $$f(y) = \left(\frac{[x,f]}{[x,g]}\right)g(y).$$ You have not done so: you have only shown that this equation holds for $y=x$, not for every $y\in V$.

If you can prove that this is indeed true (that the displayed equation does hold), then you will be almost done, but not quite: you would still need to consider what happens if you cannot find an $x$ with $[x,g]\neq 0$ (that is, what happens if $g(x)=0$ for all $x\in V$?).

Now, going back to what you still need to prove, there are a number of ways of doing this.

One way would be to show that for every $x$ and $y$, if $g(x)\neq 0$ and $g(y)\neq 0$, then $$\alpha = \frac{[x,f]}{[x,g]} = \frac{[y,f]}{[y,g]}.$$ One possibility for doing this would be to try to come up with some linear combination of $x$ and $y$ which is sent to $0$ by $g$; then use the hypothesis that everything sent to $0$ by $g$ is sent to $0$ by $f$, and see if you can deduce the above equality.

Another possibility is to think about the dimensions of the kernels of $f$ and $g$, and what that tells you about $x$ and $y$.

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Thank you for your help –  smanoos Sep 13 '11 at 12:46
Can you please give me a hint on the linear combination of x and y? The following is what I assumed and it works only if $g(y)=0$: $u=(\alpha y+x)-x$ –  smanoos Sep 13 '11 at 14:58
@smanoos: Suppose $g(x)=a$, and $g(y)=b$. How much is $g(bx-ay)$? (Note that his makes sense, because $a$ and $b$ are scalars, so $bx-ay$ is a vector). –  Arturo Magidin Sep 13 '11 at 15:54
Thank you so much. I think this one works perfectly. You are a genius. –  smanoos Sep 13 '11 at 23:33


text:
Common fixed point for three pairs of self-maps satisfying common (E.A) property in generalized metric spaces.(English)Zbl 1273.54056

Summary: Using the concept of common (E.A) property, we prove a common fixed point theorem for three pairs of weakly compatible self-maps satisfying a new contractive condition in the framework of a generalized metric space. Our results do not rely on any commuting or continuity condition of mappings. An example is provided to support our result. The results obtained in this paper differ from the recent relative results in the literature.

MSC:

 54H25 Fixed-point and coincidence theorems (topological aspects) 47H10 Fixed-point theorems
Full Text:

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This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.