text:
## Compassionate one year ago What is the quadratic formula and how do I use it? @Zale101

1. Compassionate

Patiently waits

2. dan815

|dw:1444022758804:dw|

3. dan815

it originates from completing the square method

4. Compassionate

What type of expressions would I need to use this one, is it limited to binomials?

5. dan815

all equations of the form ax^2+bx+ c= 0 a,b,c can be any real number including 0

6. Zale101

The quadratic formula is what i believe a short cut way to solve for x. You can factor to solve for x, but when you have unfavorable quadratics, then the quadratic formula comes to handy. From where did it came from? Well, consider the standard quadratic equation: $$\large ax^2+bx+c=0$$ if you tried to solve for x, you will come up with the quadratic formula. Let us try it! It $$\large ax^2+bx+c=0$$ Divide everything by a $$\large x^2+\frac{b}{a}x+\frac{c}{a}=0$$ $$\large x^2+\frac{b}{a}x+\frac{b^2}{4a}-\frac{b^2}{4a}+\frac{c}{a}=0$$ $$\large (x+\frac{b}{2a})^2-\frac{b^2}{4a}+\frac{c}{a}=0$$ $$\large (x+\frac{b}{2a})^2=\frac{b^2}{4a}-\frac{c}{a}$$ $$\large (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$$ square root both sides $$\large x+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}}$$ $$\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

7. dan815

a=/= 0

8. Zale101

This is he proof for the quadratic formula. YOU don't need to do it but you can always memorize it!

9. Zale101

Sorry for keeping you waiting...

10. Compassionate

It's okay. So what is all this talk I hear about binomials in quadratic formulas? What's the point?

11. Zale101

binomials in the quadratic? If you have two binomials multiplying together (x-2)(x+1), you can solve for x without needing the quadratic formula. You can simply set the two binomials to zero and solve for x. However, both methods can give you the value of x.

12. Compassionate

Why is it if I have (x-2)(x+1) the solutions really become x = 2, x = -1

13. Compassionate

$s(t) = –g \times t^2 + vt + h$ i think this would be easier to remember.

14. zepdrix

When you have $$\large\rm f(x)=(x-2)(x+1)$$, you're asking why this gives us "solutions" of x=2 and x=-1?

15. Compassionate

yes

16. zepdrix

By "solutions", we mean zeroes, or where the function intercepts the x-axis. This happens when y=0.$\large\rm 0=(x-2)(x+1)$From here you need to think back to basic arithmetic. If you have $$\large\rm 0=\color{royalblue}{A}\color{orangered}{B}$$, then one of the numbers, A or B, has to be 0. That's the only way you can get 0 in multiplication. Applying this to our polynomial, it tells us that the factor (x-2) or (x+1) has to be 0.$\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}$So either $$\large\rm 0=x-2$$ (adding 2 to each side gives us x=2) or $$\large\rm 0=x+1$$ (subtracting 1 from each side gives us x=-1)

17. zepdrix

$\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}$I hope that makes some sense. We can only get zero if we have this:$\large\rm 0=\color{royalblue}{0}\color{orangered}{(x+1)}$or this$\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{0}$

18. Compassionate

I get it. That makes sense. So what about applying the gravity formula. Let me give you an example You throw a ball at the speed of 10 m/s standing on the ground. It reaches a maximum height of 50 feet. When will it reach this height? (Time) I know there's something with a vertex, or there's a vertex formula? But this seems so confusing, because sometimes I'll need to find the height, sometimes I need to find the speed -- can you shed some light on this?

19. Compassionate

Scratch that. Lets say I'm given y = 3x2 + x – 2 and I am asking to not only find the vertex, but graph it as well

20. zepdrix

Ummm, well you have a basic understanding of what a parabola looks like, yes? If I give you $$\large\rm f=(x)^2$$, it's a bowl shape opening upward. If I give you $$\large\rm g=-(x)^2$$, it's the same shape but opening downward. When we throw other stuff into the mix,$\large\rm h=-(x-2)^2+3$We have the exact same shape, we're only moving the "base of that bowl" or vertex as we call it. Having a coefficient on the squared term though,$\large\rm h=-3(x-2)^2+3$will affect the growth rate and change the shape a bit.

21. zepdrix

So for your example: $$\large\rm y=3x^2+x-2$$ Hmm this might be a tricky one to start with if you're not familiar with completing the square. Maybe I can use a simpler example? :D

22. Compassionate

Sure. Pick an example. I understand that the quadratic formula finds both x-intercepts, usually two whole positive and negative numbers, and sometime it's just one positive or negative intercept, but say I wanted to find the vertex

23. zepdrix

Just a note about that factoring thing earlier, and the "solutions" that we were guaranteed. Most polynomials can not be factored nicely. Especially anything dealing with physics like your motion equation. So it's usually not very easy to turn it into something like this: $$\rm 0=(x-2)(x+1)$$ That's why the Quadratic Function is amazing. It's always works. If you find that more intuitive to use rather than these weird algebraic shortcuts, then by all means. That was just a side note :) Ok back on topic. Yes I have an example.

24. zepdrix

I'm thinking of a function like this:$\large\rm f(x)=\color{orangered}{x^2+8x}+5$I highlighted the x's in orange because we want to complete the square on them. We would like to turn f(x) into something like this:$\large\rm f(x)=\color{orangered}{(x+number)^2}+other~number$Lemme use letters instead of words there..$\large\rm f(x)=\color{orangered}{(x-h)^2}+k$When it is in this form, we call it vertex form. And the vertex of the function is located at $$\large\rm (h,k)$$.

25. zepdrix

For our function, Think of it like this:$\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5$I've split the middle term into two equal parts. I want to show you the square method. It's not very practical, but it gives a lot of intuition into this weird method.

26. zepdrix

27. zepdrix

|dw:1444025568081:dw|I'm going to let this length across be x and the other part be 4.

28. zepdrix

Since it's a square, it has equal dimension on the other side,|dw:1444025623649:dw|

29. zepdrix

So now we'll fill in these smaller squares. Notice the upper left square is an $$\large\rm x~by~x$$ square.

30. zepdrix

|dw:1444025674721:dw|And the other rectangular shapes have side lengths 4 and x.

31. zepdrix

So in this example, to complete the square, fill it in the rest of the way, we need that 4 by 4 box filled. So we need 16, ya?

32. zepdrix

|dw:1444025797639:dw|Once we have that 16, this square can be simplified down into $$\large\rm (x+4)(x+4)$$ those are the side lengths. Which we would further simplify to $$\large\rm (x+4)^2$$

33. zepdrix

Maybe I'm going into too much weird detail here :) lol

34. zepdrix

So umm

35. zepdrix

$\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5$We'll add 16 to complete our square because that's the number we needed.$\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}+5$But in the land of math, we're not allowed to just add 16 willy nilly like that. It's changes the function. So, in order to keep the same function, we need to balance it. A way we can do this is by subtracting 16 at the same time.$\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}-16+5$

36. Compassionate

I've seen thje example, and I kind of get it, but it's a little complex for me. I'm just familiar with the quadratic formula and the vertex = -b/2a thing. I really do appreciate your time and wanting to help me, it's just some things I don't get

37. zepdrix

Ok if you're more comfortable with the shortcuts, we can skip right to the meat then :) $\large\rm f(x)=\color{orangered}{x^2+8x}+5$$\large\rm b=8,\qquad \frac{b}{2}=4,\qquad\left(\frac{b}{2}\right)^2=16$So we'll add 16 to complete our square, and subtract 16 at the same time to balance it.$\large\rm f(x)=\color{orangered}{x^2+8x+16}-16+5$And our orange part simplifies down to: $$\large\rm \left(x+\frac{b}{2}\right)^2$$ That's why I listed all three of the b parts, we'll need all of them.

38. zepdrix

$\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-16+5$

39. Compassionate

Question: in most quadratic problem involving motion, will I generally be asked to find the time? Say, if I'm given something like 16t^2 + 96t + 200, and I solve the vertex, that will give me the time the ball reaches it's maxiumum height, and solving for the positive x-intercept will give me the time it touches the ground. In High School Math, are we just generally finding the time of objects?

40. zepdrix

Good good. Yes. In motion problems those are the three features you'll spend a lot of time on. Maximum height, Starting and end point.

41. Compassionate

In my textbooks, we never learned about completing the square. Just basic formulas and how to find them. I haven't gotten to that part in my books. I mean, I'm familiar with factoring polynomials but not completing their squares

42. zepdrix

So in this example we're working on:$\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-11$Our vertex is located at: $$\large\rm (-4,-11)$$

43. zepdrix

So that would give you your base point.|dw:1444026482133:dw|

44. zepdrix

But knowing exactly how it grows is a little tricky to graph. That's where those "solutions" come into play, ya?|dw:1444026562732:dw|Wherever it is that they might be

45. zepdrix

Oh, also notice... when you put it into vertex form, you'll avoid using the quadratic function altogether. You really only want to use the quadratic function from standard form.

46. zepdrix

$\large\rm f(x)=\left(x+4\right)^2-11$$\large\rm 0=\left(x+4\right)^2-11$It's much easier to solve for x explicitly from here. Add 11 to each side,$\large\rm 11=(x+4)^2$Square root,$\large\rm \pm\sqrt{11}=x+4$Subtract 4,$\large\rm x=-4\pm\sqrt{11}$

47. zepdrix

quadratic formula* not function.. sorry typo :(

48. Compassionate

What's the need to complete the square, whjy can't we just plug it into the quadratic formula, find the x-intercepts, then find hte vertex?

49. Compassionate

I understand you're knowledgeable of the subject, but I don't know how to complete squares. You're confusing me

50. zepdrix

Hmm. It's pretty straight forward finding the x-coordinate of your vertex. We were given this $$\large\rm x^2+8x$$ for the x's. The vertex is located at $$\large\rm x=-\frac{b}{2}$$. But finding the y-coordinate is a lot trickier. We don't have a nice shortcut like the quadratic formula. We simply have to put it into vertex form. If you search hard enough, maybe you can find a weird shortcut formula that allows you to skip completing the square. But I'm not familiar with any such method D:

51. Compassionate

I apprecaite your help, but it's 1:30am. Maybe we can pickup tomorrow.

52. zepdrix

Haha I was thinking the same thing XD Sorry for rambling on so long!

53. zepdrix

I get carried away sometimes hehe

54. Compassionate

It's great, actually, you're enthusiastic. Never be ashamed of it. TTYL

55. zepdrix

\c:


text:
Forces and Kinematics Problem - Am I right?

Can someone tell me if I'm right? 0.98 seems a bit high for the coefficient of friction?

A 25kg sled is pulled at a steady 1.3 m/s by a rope inclined 40 degrees above horizontal. If the pulling force is 26.7 N, what is the coefficient of friction between the sled runners and the snow?

m = 25 kg, v = 1.3 m/s, Fp = 26.7 N, mgcos40 = Fn = 187.7 N,

Ftotal = ma (Ftotal = 0 because of no acceleration).
Fp + mgsin40 - Ff = 0,
26.7 + 157.5 = Ff, Ff = 184.2.

u = Ff/Fn = 184.2/187.7 = 0.98.

Related Introductory Physics Homework Help News on Phys.org
Hootenanny
Staff Emeritus
Gold Member
Using g as 9.8 I got 0.697(3sf)

Hootenanny
Staff Emeritus
Gold Member
$$F= \mu R \\ \mu R + 25gcos50=26.7 \\ \mu 25gsin50=26.7 - 25gcos50 \\ \mu = \frac{26.7-25gcos50}{25gsin50} \\ \mu = 0.697$$
Sorry it took my a while to sought out the LaTex. lol I just wanted you to know I was workin on it.

Last edited:
I don't understand that very much. what's muRmuR? like...lol

Sorry, I'm having a lot of trouble and that whole long equation just confused me more.

Hootenanny
Staff Emeritus
Gold Member
it shouldn't come out like that. I'm jus correcting it. I Hate Latex

ok cause i dunno where you got all the numbers from

Hootenanny
Staff Emeritus
Gold Member
$$F= \mu R$$
$$\mu R + 25gcos50=26.7$$
$$\mu 25gsin50=26.7 - 25gcos50$$
$$\mu = \frac{26.7-25gcos50}{25gsin50}$$
$$\mu = 0.697$$
I've tried to lay it out better

R is the normal force?

by the way, thank you for helping me along so much so far :)

Hootenanny
Staff Emeritus
Gold Member
No worries. Yeah, R is the normal reaction. mu is the co-effiecent of friction.

you used 50 degrees as the angle. it's 40 though? :\

Hootenanny
Staff Emeritus
Gold Member
cos50=sin40.

Hootenanny
Staff Emeritus
Gold Member
$$F= \mu R$$ is the equation for maximal frictional force.

numerically, that's correct. but i don't understand that concept at all. can you demonstrate it with 40 degrees? i apologize for being so lost :(

F = uR is, yes.

Hootenanny
Staff Emeritus
Gold Member
Okay. Have you drawn a diagram with the slope and all the forces acting?

F(total) = ma

F(total) = 0

F(normal) = (m)(g)(cos40) = (25)(9.8)(cos40) = 187.7

F(pulling) - F(friction) + 25(9.8)(sin40) = 0

26.7 + 25(9.8)(sin40) = F(friction)

26.7 + 157.5 = F(friction)

184.2 = F(friction)

F(friction) = F(normal) * mu

mu = F(friction)/F(normal) = 184.2/187.7 = 0.98

The coefficient of friction equals 0.98.

Yes, I drew a free body diagram.

Hootenanny
Staff Emeritus
Gold Member
blockitoff said:
F(total) = ma
F(pulling) - F(friction) + 25(9.8)(sin40) = 0
Third line down. it should read: F(pulling)-F(Friction)-25(9.8)(sin40)=0 because the component of the weight it acting in the opposite direction to the pulling force.

oh i see! let me work it out real quick :)

Hootenanny
Staff Emeritus
Gold Member
Sorry I didnt spot it earlier!

F(total) = ma

F(total) = 0

F(normal) = (m)(g)(cos40) = (25)(9.8)(cos40) = 187.7

F(pulling) - F(friction) - 25(9.8)(sin40) = 0

26.7 - 25(9.8)(sin40) = F(friction)

26.7 - 157.5 = F(friction)

-130.8 = F(friction)

F(friction) = F(normal) * mu

mu = F(friction)/F(normal) = -130.8/187.7 = -0.697

Ok I got it! MY ONLY QUESTION that remains is how did I know to just make -0.697 positive to be 0.697?

Hootenanny
Staff Emeritus

format: plain


text:
# Repulsive Force between two Electromagnets

This question might seem a bit too basic(or advanced!), but it is the one to which I'm really desperate to find a simple enough explanation and solution.

Ok, So here's the problem,

I have two identical electromagnets with a field strength $B$ each. Each of radius $r$ and length $l$ with $N$ turns say, how do I calculate the net repulsive/attractive force between them if

1. they attract each other and are pulled apart or
2. they repel each other and are pushed against each other

I want to levitate one magnet carrying a weight (so its stabilized) above the other electromagnet. Even if the solution is for different fields it is no problem.

• Wikipedia has a long list of examples of such calculations. – Winther Sep 10 '15 at 18:53
• Sebastian Riese gave you a good pointer there, but unfortunately it's not going to work for scenarios with ferromagnetic materials, which you will need, if you want to make this work. You could use a simulation program like MaxFem for the calculations, but there is a rather steep learning curve associated with that. As a practical pointer: start with permanent magnets and figure out a way to use a weak electromagnet just for stability control. Do not try to create the load bearing magnetic field with an electromagnet, cheap electromagnets don't have the power density of permanent magnets. – CuriousOne Sep 10 '15 at 20:20
• @Winther: My first site to visit was Wikipedia but it did not specify for electromagnets. – Gouthamm4G Sep 11 '15 at 13:48
• @CuriousOne: Since I already have the materials for making an electromagnet, I didn't want to use permenant magnets. Besides, if I use electromagnet I could easily vary the magnetic field in real time for future changes. – Gouthamm4G Sep 11 '15 at 13:51
• And you will need those material to control the levitation. All I am saying is that you will find it much easier to do if you start with permanent magnets to create >90% of the required forces. You don't have to do it this way, it just turns out to be the by far better approach. – CuriousOne Sep 11 '15 at 13:57

$$\vec F = I \oint_{\gamma_2} d\vec r \times \vec B(\vec r).$$
Where $\vec B(\vec r)$ is the field generated by coil 1 and $I$ is the current in coil 2 and the integral is along the curve $\gamma$ which is described by the wires of the coil 2. Note, that there might also be a torque.
Knowing the maximal or homogeneous field $B$ in the interior of the coils does not help much.
The field of the coil is given by the Biot-Savart law: $$\vec B(\vec r) = \frac{\mu_0 I}{4\pi} \int_{\gamma_1} \frac{d\vec r' \times (\vec r - \vec r')}{\left| \vec r - \vec r' \right|^3}.$$


text:
# User talk:Ronewolf

I hope you enjoy editing here and being a Wikipedian! Please sign your messages on discussion pages using four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question and then place {{helpme}} before the question on your talk page. Again, welcome! RJFJR 14:55, 16 November 2007 (UTC)
Personally, I think that the most natural units are just like Planck units except they would normalize 4πG and ε0 rather than normalizing G and 4πε0 as Planck units do. Then you will see those factors of 4π disappear from the equations of physical law. c and ħ would remain 1 as in Planck units. These are sorta like Lorentz–Heaviside units and it turns out that the elementary charge, measured in these most natural units comes out to be ${\displaystyle \scriptstyle {\sqrt {4\pi \alpha }}}$ = 0.30282212 . I've been told in QED or QCD theory, that this number is also confirmed as the natural measure of charge of electrons. You can think of the fine-structure constant, α, as taking on the value it does because of the inherent amount of charge, measured in natural units, that nature has bestowed upon the electron and the quarks with charge. 74.104.160.199 (talk) 16:51, 13 September 2009 (UTC)


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# A simultaneous Wielandt positivity theorem

A simultaneous Wielandt positivity theorem We consider matrix semigroups $$\mathcal {S}$$ S which are closed under multiplication by complex scalars, and whose norm closure contains no zero-divisors. We show that when every non-zero $$S$$ S in $$\mathcal {S}$$ S is indecomposable and the spectral radius of $$S$$ S is equal to the spectral radius of $$|S|$$ | S | for all $$S$$ S in $$\mathcal {S}$$ S , then $$\mathcal {S}$$ S is effectively positive, in the sense that there exists a diagonal unitary matrix $$D$$ D so that for each $$S$$ S in $$\mathcal {S}$$ S , $$S=\alpha _S D |S| D^{-1}$$ S = α S D | S | D - 1 for some $$\alpha _S \in \mathbb {T}$$ α S ∈ T . We also show the same conclusion holds even if individual indecomposability is weakened to indecomposability of the semigroup as a whole, as long as the semigroup is convex. We give examples showing that all hypotheses are required. We also extend some of these results to compact operators, under additional conditions. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Positivity Springer Journals

# A simultaneous Wielandt positivity theorem

, Volume 19 (1) – May 25, 2014
12 pages

/lp/springer_journal/a-simultaneous-wielandt-positivity-theorem-3fL4vzf4jW
Publisher
Springer Journals
Subject
Mathematics; Fourier Analysis; Operator Theory; Potential Theory; Calculus of Variations and Optimal Control; Optimization; Econometrics
ISSN
1385-1292
eISSN
1572-9281
D.O.I.
10.1007/s11117-014-0289-1
Publisher site
See Article on Publisher Site

### Abstract

We consider matrix semigroups $$\mathcal {S}$$ S which are closed under multiplication by complex scalars, and whose norm closure contains no zero-divisors. We show that when every non-zero $$S$$ S in $$\mathcal {S}$$ S is indecomposable and the spectral radius of $$S$$ S is equal to the spectral radius of $$|S|$$ | S | for all $$S$$ S in $$\mathcal {S}$$ S , then $$\mathcal {S}$$ S is effectively positive, in the sense that there exists a diagonal unitary matrix $$D$$ D so that for each $$S$$ S in $$\mathcal {S}$$ S , $$S=\alpha _S D |S| D^{-1}$$ S = α S D | S | D - 1 for some $$\alpha _S \in \mathbb {T}$$ α S ∈ T . We also show the same conclusion holds even if individual indecomposability is weakened to indecomposability of the semigroup as a whole, as long as the semigroup is convex. We give examples showing that all hypotheses are required. We also extend some of these results to compact operators, under additional conditions.

### Journal

PositivitySpringer Journals

Published: May 25, 2014

## You’re reading a free preview. Subscribe to read the entire article.

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text:
# Generate thumbnail image in addition to main image? (command line render)

I'd like to set up a command line script that renders out each frame to the desired output image type (EXR, PNG, TGA, etc.) at full resolution, but then also generates and saves a small thumbnail JPG image at a specified width (say 400px for example).

I'd like to generate this thumbnail from the command line or with a script instead of using the file output node in the compositor. Makes it easier to replicate for multiple render jobs.

Note1: I do not want to render twice. I want to render once and then scale and save out a second smaller file. I don't want to do this from inside the blender file directly, but with an external script.

Note2: I would be open to a second blender instance that renders by only using the compositor to input the large image rendered in the first instance with the new output resolution. That way it's only processing flat images and not doing any scene rendering.

Here's a second option, which I now believe is the best solution, as it is the most elegant and versatile that I've found so far.

The solution is essentially an extension of the Compositing nodes Scale and FileOutput option. However the critical difference is the addition of an Image node being mixed with the scaled Render Output image before being piped into the FileOutput node. The image node uses a dummy UV image created at the required Thumbnail output dimensions. When mixed with this dummy image, the FileOutput node will output the new dimensions without having to change the Scene Render Dimensions.

This option is better than my original solution because it is simpler to implement, shorter code, as well as not needing to rely on external file paths.

Command line usage (last argument pipes in thumbnail width):

blender -b "file.blend" -o "//out/" -P thumbnail.py -- 500

and the script thumbnail.py:

import bpy
import sys

class Thumbnail:

tWidthArg = int(sys.argv[9])

scn = bpy.context.scene
resX = scn.render.resolution_x
resY = scn.render.resolution_y
thumbWidth = 400

if tWidthArg > 0:
thumbWidth = tWidthArg

thumbHeight = ((thumbWidth/resX)*resY)

#create new internal dummy image the size we want for thumbnail
bpy.ops.image.new(name="ThumbNailDummyImage", width=thumbWidth, height=thumbHeight, color=(0.0, 0.0, 0.0, 0.0), alpha=False, generated_type='BLANK', float=False)

#set up compositing nodes
nTree = scn.node_tree
Src = nTree.nodes["Render Layers"]
Dst = nTree.nodes["Composite"]

#image input node - use dummy uv image
imageNode = nTree.nodes.new('CompositorNodeImage')
imageNode.name = "INPUTIMAGE"
imageNode.image = bpy.data.images['ThumbNailDummyImage']
imageNode.location = (Src.location.x+300, Src.location.y-600)

#scale node to resize thumbnail to correct size
scaleNode = nTree.nodes.new('CompositorNodeScale')
scaleNode.space = "ABSOLUTE"
scaleNode.location = (imageNode.location.x, imageNode.location.y+200)
scaleNode.inputs['X'].default_value = thumbWidth
scaleNode.inputs['Y'].default_value = thumbHeight

#mix the scale render output with the dummy uv image to force new output size
mixNode = nTree.nodes.new('CompositorNodeMixRGB')
mixNode.location = (imageNode.location.x+300, imageNode.location.y)

#setup output filenode
fileOutNode  = nTree.nodes.new('CompositorNodeOutputFile')
fileOutNode.location = (mixNode.location.x+300, mixNode.location.y)
fileOutNode.layer_slots[0].name='Thumb'
fileOutNode.file_slots[0].path='Thumb'
fileOutNode.format.file_format= 'JPEG'
fileOutNode.format.color_mode = 'RGB'
fileOutNode.base_path = '//out/'

#connect nodes

Here is a very basic script, that adds an app handler to receive callbacks whenever a frame has been rendered, and prints the frame number and path to rendered image file to system console:

import bpy

def render_post_callback(scene):
frame = scene.frame_current
filepath = scene.render.frame_path()
print("Frame: %i, Filepath: %s" % (frame, filepath))

bpy.app.handlers.render_post.append(render_post_callback)

Note: It works for the render settings output options, but not for File Output nodes. As of now, there is no equivalent of frame_path() for File Output nodes.

It's easy to retrieve the base paths (output folders) from File Output nodes:

base_paths = [node.base_path for node in bpy.context.scene.node_tree.nodes if node.type == 'OUTPUT_FILE']

You could let an external script check these paths and generate thumbnails for all rendered images, unless there is already a thumbnail (should be as easy as to name the thumbnails like the original images in a different folder and do a name check). Should work as long as there aren't any unrelated image files in the rendered images folders.

You could write a simple python script, and then call it from the command line. See this related post:

Another way is to use blender's python API, and run a python script in blender from the command line (see this post), e.g.:

blender --background /path/to/file.blend --python /path/to/script.py --render-anim or

blender -b /path/to/file.blend -P /path/to/script.py -a

The render resolution can be set with bpy.context.scene.render.resolution_x and bpy.context.scene.render.resolution_y.

There are several ways you could go about creating a second thumbnail image (related post):

• Render twice using a python script, at different resolutions.

• Create some compositing nodes from python to scale the image in the compositor.

For example, this:

import bpy

#define thumbnail dimensions and name:
thumb_x = 400
thumb_y = 400
thumb_name = "thumbnail"

# function to find nodes connected to a node
for n_inputs in node_in.inputs:

scene = bpy.context.scene
scene.use_nodes = True # make sure composite nodes are enabled
for node in scene.node_tree.nodes:
if node.type == 'COMPOSITE':
tree = bpy.context.scene.node_tree

scale_node = tree.nodes.new(type="CompositorNodeScale")
scale_node.space = 'ABSOLUTE'
scale_node.location.x = 200
scale_node.inputs["X"].default_value = thumb_x
scale_node.inputs["Y"].default_value = thumb_y

node = tree.nodes.new(type="CompositorNodeOutputFile")
node.format.file_format = 'JPEG'
node.location.x = 400
node.file_slots[0].path = thumb_name

bpy.ops.render.render(animation=True) # render animation

Will create an absolute scale node and connect whichever node is connected to the Composite node to it. It will also create a File Output node:

• Scale using an external command line image processing program such as ImageMagick.

• Scale using a python image manipulation library (see this post).

• I don't want to render twice, and I'd rather avoid an external image processing program/library if possible. I just want to replicate the process whereby Blender takes the temporary EXR file it renders and then converts it to the output file format and saves. I want to replicate that process with a smaller file and not get hit with the cost of a second render. Surely this is possible? Jan 31, 2014 at 19:00
• @ToddMcIntosh The best option is probably creating some compositing nodes with python (it will handle animations), it's not actually that difficult. I'll try and write something out in a bit..
– gandalf3
Jan 31, 2014 at 21:07
• Yeah, the only problem with the Scale/FileOutput solution, is that it has to be in a different blend file as the scale dimensions only scale the image then overlay it on a large black background the size of the Render resolution. I'm working on a solution using the bpy.app.handlers.render_complete event to add a callback that creates a second scene with separate Render resolution. I'll post it once I get it working. Feb 1, 2014 at 0:41
• @ToddMcIntosh Your right.. That's annoying :( I thought I found a solution, but it just outputs transparent (black since jpg doesn't support transparency) images. Here it is in case you want to look at it.
– gandalf3
Feb 1, 2014 at 2:59
• I found a trick that makes the Scale/FileOutput option work. You have to mix the render image with another image that's at the output dimensions you want, and then it works. I've added a new solution that does this dynamically by creating a dummy UV image. Feb 2, 2014 at 6:32

Ok, I've written a script that solves the question in my own specific context. The following script works when the render job is rendering one frame in the sequence per Blender instance. That works fine for me since the context I'm using it in is a render farm (Brenda with AWS) which handles each frame in the animation queue as an individual render.

CAVEAT: This does not work when trying to render the whole frame range from one instance of Blender as the Event Handler only fires after the last frame is rendered.

thumbnail.py (to be called as part of the command line job launch)

import bpy

class Thumbnail:

doneThumbRender = bpy.props.BoolProperty()
doneThumbRender = False
lastFullFramePath = bpy.props.StringProperty()
bpy.context.scene.frame_current = bpy.context.scene.frame_start
outThumbnailWidth = bpy.props.IntProperty()

def callbackFunc(context):

lastFullFramePath = bpy.context.scene.render.frame_path(frame=bpy.context.scene.frame_current)

if bpy.data.scenes.find("THUMB") < 0:

#Set up Thumbnail Scene
bpy.ops.scene.new()
thumbScn = bpy.data.scenes[1]
thumbScn.name = "THUMB"
bpy.data.screens['Default'].scene = bpy.data.scenes['THUMB']
bpy.data.scenes['THUMB'].render.engine = "CYCLES"

#Setup Composite nodes to resize source image
bpy.context.scene.use_nodes = True
inputImageLoc = lastFullFramePath
Scene = bpy.context.scene
Tree = Scene.node_tree
Src = Tree.nodes["Render Layers"]
Dst = Tree.nodes["Composite"]

#Create input image node for large source image
imageNode = Tree.nodes.new('CompositorNodeImage')
imageNode.name = "INPUTIMAGE"

#Scale Node
scaleNode = Tree.nodes.new('CompositorNodeScale')
scaleNode.space = "RENDER_SIZE"

inputImageXSize = Scene.node_tree.nodes['INPUTIMAGE'].image.size[0]
inputImageYSize = Scene.node_tree.nodes['INPUTIMAGE'].image.size[1]

#Calc new render height based on new thumbnail width as aspect ratio of source image
newRenderHeight = ((outThumbnailWidth/inputImageXSize)*inputImageYSize)

#Set Thumb Scene Render settings
Scene.render.resolution_percentage = 100
Scene.render.resolution_x = outThumbnailWidth
Scene.render.resolution_y = newRenderHeight
Scene.render.filepath = "//out/thumb"
Scene.render.image_settings.file_format ="JPEG"
Scene.render.image_settings.color_mode = "RGB"
Scene.cycles.samples = 1
Scene.frame_start = bpy.data.scenes['Scene'].frame_current
Scene.frame_end = Scene.frame_start
bpy.data.scenes['THUMB'].frame_current = Scene.frame_start

#Check for camera in Thumb Scene and add if missing
numCams = 0

for ob in Scene.objects:
if ob.type == 'CAMERA':
bpy.context.scene.camera = ob
numCams + numCams + 1
break

if numCams==0:
for ob in Scene.objects:
if ob.type == 'CAMERA':
bpy.context.scene.camera = ob
break

#Do Thumb Render
if Thumbnail.doneThumbRender == False:

Scene = bpy.context.scene
Tree = Scene.node_tree
getImageNode = Tree.nodes["INPUTIMAGE"]


text:
### Random-access

In [1]

A typical illustration of this distinction is to compare an ancient scroll (sequential; all material prior to the data needed must be unrolled) and the book (direct: can be immediately flipped open to any arbitrary page). A more modern example is a cassette tape (sequential—one must fast forward through earlier songs to get to later ones) and a CD (direct access—one can skip to the track wanted, knowing that that would be the one retrieved).

In data structures, direct access implies the ability to access any entry in a list in constant (independent of its position in the list and of list's size, i.e. $O\left(1\right)$) time. Very few data structures can guarantee this, other than arrays (and related structures like dynamic arrays). Direct access is required, or at least valuable, in many algorithms such as binary search, integer sorting or certain versions of sieve of Eratosthenes.[3]

Other data structures, such as linked lists, sacrifice direct access to permit efficient inserts, deletes, or reordering of data. Self-balancing binary search trees may provide an acceptable compromise, where access time is equal for any member of a collection and only grows logarithmically with its size.


text:
# Uniform Distribution

1. Jan 27, 2006

### circa415

If X ~ U(a, b) then f(x) = 1/(b-a)

but what if b-a is less than 1

for instance if X ~ (.5,1) then f(x) = 2???

I'm a bit confused. Any help would be appreciated.

2. Jan 27, 2006

### StatusX

A probability distribution p(x) may take on values greater than one. It doesn't mean something has a probability of more than one of happening because, remember, p(x) isn't itself a probability. The probability that X lies in the range x->x+dx is given by p(x)dx, and as long as the integral of this over any range is not greater than one, which is assured by normalization and the fact that p(x) must be non-negative, the actual probability will never be greater than one.

Last edited: Jan 27, 2006


text:
# The Unapologetic Mathematician

## The Order on the Real Numbers

We’ve defined the real numbers $\mathbb{R}$ as a topological field by completing the rational numbers $\mathbb{Q}$ as a uniform space, and then extending the field operations to the new points by continuity. Now we extend the order on the rational numbers to make $\mathbb{R}$ into an ordered field.

First off, we can simplify our work greatly by recognizing that we just need to determine the subset $\mathbb{R}^+$ of positive real numbers — those $x\in\mathbb{R}$ with $x\geq0$. Then we can say $x\geq y$ if $x-y\geq0$. Now, each real number is represented by a Cauchy sequence of rational numbers, and so we say $x\geq0$ if $x$ has a representative sequence $x_n$ with each point $x_n\geq 0$.

What we need to check is that the positive numbers are closed under both addition and multiplication. But clearly if we pick $x_n$ and $y_n$ to be nonnegative Cauchy sequences representing $x$ and $y$, respectively, then $x+y$ is represented by $x_n+y_n$ and $xy$ is represented by $x_ny_n$, and these will be nonnegative since $\mathbb{Q}$ is an ordered field.

Now for each $x$, $x-x=0\geq0$, so $x\geq x$. Also, if $x\geq y$ and $y\geq z$, then $x-y\geq0$ and $y-z\geq0$, so $x-z=(x-y)+(y-z)\geq0$, and so $x\geq z$. These show that $\geq$ defines a preorder on $\mathbb{R}$, since it is reflexive and transitive. Further, if $x\geq y$ and $y\geq x$ then $x-y\geq0$ and $y-x\geq0$, so $x-y=0$ and thus $x=y$. This shows that $\geq$ is a partial order. Clearly this order is total because any real number either has a nonnegative representative or it doesn’t.

One thing is a little hazy here. We asserted that if a number and its negative are both greater than or equal to zero, then it must be zero itself. Why is this? Well if $x_n$ is a nonnegative Cauchy sequence representing $x$ then $-x_n$ represents $-x$. Now can we find a nonnegative Cauchy sequence $y_n$ equivalent to $-x_n$? The lowest rational number that $y_n$ can be is, of course, zero, and so $\left|y_n-(-x_n)\right|\geq x_n$. But for $-x_n$ and $y_n$ to be equivalent we must have for each positive rational $r$ an $N$ so that $r\geq\left|y_n-(-x_n)\right|\geq x_n$ for $n\geq N$. But this just says that $x_n$ converges to ${0}$!

So $\mathbb{R}$ is an ordered field, so what does this tell us? First off, we get an absolute value $\left|x\right|$ just like we did for the rationals. Secondly, we’ll get a uniform structure as we do for any ordered group. This uniform topology has a subbase consisting of all the half-infinite intervals $(x,\infty)$ and $(-\infty,x)$ for all real $x$. But this is also a subbase for the metric we got from completing the rationals, and so the two topologies coincide!

One more very important thing holds for all ordered fields. As a field $\mathbb{F}$ is a kind of a ring with unit, and like any ring with unit there is a unique ring homomorphism $\mathbb{Z}\rightarrow\mathbb{F}$. Now since $1gt;0$ in any ordered field, we have $2=1+1>0$, and $3=2+1>0$, and so on, to show that no nonzero integer can become zero under this map. Since we have an injective homomorphism of rings, the universal property of the field of fractions gives us a unique field homomorphism $\mathbb{Q}\rightarrow\mathbb{F}$ extending the ring homomorphism from the integers.

Now if $\mathbb{F}$ is complete in the uniform structure defined by its order, this homomorphism will be uniformly complete. Therefore by the universal property of uniform completions, we will find a unique extension $\mathbb{R}\rightarrow\mathbb{F}$. That is, given any (uniformly) complete ordered field there is a unique uniformly continuous homomorphism of fields from the real numbers to the field in question. Thus $\mathbb{R}$ is the universal such field, which characterizes it uniquely up to isomorphism!

So we can unambiguously speak of “the” real numbers, even if we use a different method of constructing them, or even no method at all. We can work out the rest of the theory of real numbers from these properties (though for the first few we might fall back on our construction) just as we could work out the theory of natural numbers from the Peano axioms.

December 4, 2007 -


text:
# Rotation angle

1. Sep 28, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
My book says that $$\epsilon \vec{n} \times \vec{q}$$ represents a small rotation about q.
n is an normal vectors
Obviously the cross product is orthogonal to both n and q, but I did not know it corresponded to an angle?
BY the way, how do you make a cross product in tex?
2. Relevant equations

3. The attempt at a solution

Last edited: Sep 28, 2007
2. Sep 28, 2007

### mjsd

cross product in tex? \times will be fine

the epsilon correspond to the small amplitude of your infinitestimal rotation I think

3. Sep 28, 2007

### ehrenfest

But why does this represents an angle?

4. Sep 28, 2007

### Dick

It doesn't represent an angle. It represents an infinitesimal displacement. A small displacement from a vector q rotated around an axis along n should be perpendicular to both. Hence nxq.


text:
# Difference between discretization and quantization in physics

I am just trying to understand the fundamental difference between these two concepts in physics:

• From discreteness of some quantity: one usually interprets it as a quantity being only able to take on distinct set of values, and not all the possible values as would be possible for a quantity with a continuous spectrum, e.g. the position. On the other hand, all computational problems in physics (e.g. numerical ones) are performed via a discretization process, where the continuous spectrum of values is mapped to a discrete one by choosing a spacing (e.g. in integration techniques). But on the other hand, one also uses the term discrete when referring to the values e.g. a spin $+\frac{1}{2} /-\frac{1}{2}$ that the system can take, namely spin up and down, but we do not say that spin is quantized. (or do we?)

• From quantization: Most common is the idea of quanta of energy, that comes in discrete packets of energy $\hbar \omega,$ e.g. when solving the set of possible eigenvalues for the hydrogen atom, and hence discovering a discrete set of energy values that difference always by an integral amount of a certain quanta, so one refers to the energy here as being quantized. On the other hand, quantization is also a term used in QFT, when describing on a fundamental level what defines the particle picture "first quantization" and the wave description "second quantization".

My questions are:

1. Is it true to say that quantization, or more specifically a quantity in physics being quantized, implies discreteness? Or is this conceptually completely a nonsensical comparison?

2. Formulated differently, why would it be wrong to call the energy spectrum of the hydrogen atom (say) as "discretized" instead of "quantized"?

3. Do these concepts ever become interchangeable, or are always fundamentally different?

4. In a nutshell, the question would be: Why call it "Quantum Mechanics" and not "Discrete Mechanics"?

I apologize for the vague character of this post, my aim is solely to understand the fundamental difference of these two concepts in physics.

Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once". Observable quantities become certain operators acting on this vector space of states.

As you can see, this doesn't have anything, on the face of it, to do with discretisation. It turns out, however, that a lot of the operators we are interested in have discrete eigenvalues, and this implies that the corresponding physical values are discrete. Position, however, has a continuous spectrum, as do many other quantum observables.

There are plenty of sources explaining exactly how one goes from classical sets of states and numerical observables to quantum states (vector spaces - Hilbert spaces in particular) and quantum observables (operators); I won't cover that. All I shall say is that quantisation is a big mathematical process replacing a load of classical things with quantum things, and this sometimes leads to certain physical quantities being discretised.

• Could you please expand on your first paragraph a bit further please? Because I feel it is touching on the essence of what quantization is. – user098876 Sep 22 '15 at 14:30
• Unbound states in quantum mechanics are still quantised states but the energy spectrum isn't descrete, for example. – gented Jun 23 '17 at 19:52
• @GennaroTedesco Hi, but what does "quantised" in such case entail then? Is there a simple way of fleshing out what's behind it? – user929304 Jun 29 '17 at 14:01
• @user929304 Quantised means everything that follows from a quantisation procedure. In "classical" quantum mechanics the quantisation procedure constists of introducing a Hilbert space and postulating that states are elements thereof, with their time evolution ruled by self-adjoint operators (the Hamiltonian) plus or minus some other observables. – gented Jun 29 '17 at 14:48

Forget about language issues for a moment. The crucial feature of quantum mechanics is quantum-mechanical interference of probability amplitudes, which most people understand as a wave phenomenon (but may be described more elegantly and conceptually cleanly in terms of noncommutative matrix operations.) A mere consequence of that is that the spectra (sets of eigenvalues) of operators representing physical quantities (observables) are often, but not always, discrete. (Think back on wave equation spectra, etc.) The quantized energy spectra were the original puzzle that led physicists in the early 20th century to grip onto the discreteness, as classical mechanics for such quantities lacked it. That's why the name "quantum" was used, but other people working in the Schroedinger representation called it "wave mechanics".

By contrast, discretizing classical problems by putting them on a notional approximating lattice will, in general, not produce QM interference phenomena. Discretization is just a schematic numerical analysis approximation of continuous systems.

1. Quantization often relies on discreteness, so matrix methods of discrete mathematics are quite useful there, but its salient feature is QM interference of amplitudes, not just discrete structures. After all, even classical drum eigenvalue equations involve such discrete mathematical structures and quantized eigenvalues (and even interfering modes, but no probability waves). It is the bizarre probabilistic framework that makes QM.

2. It is fine to call the discrete part of a system's energy spectrum "discrete", but its continuous, scattering, part of course should be called "continuous", even though , it too is part of the quantum spectrum.

3. It is a terrible idea to mix these concepts. You'd invite stares, incomprehension, and discomfort.

4. In a nutshell, because it is not always discrete, and would not reflect the salient feature this answer starts with.

Aside : Angular momentum and spin are quantized observables in QM, and we do call them that. (Their quantization is not a convenience issue of numerical analysis: they behave in genuinely peculiar non-commutative ways.)

In quantum mechanics, bound states (i.e. energy eigenstates whose energy is less than the limiting value of the potential as $|{\bf x}| \to \infty$, whose wavefunctions therefore decay exponentially at large ${\bf x}$) typically have discrete energy levels, while scattering states (whose wavefunctions typically approach $e^{ikr}/r^\frac{d-1}{2}$ in $d$ spatial dimensions) have continuous energy spectra. So quantization doesn't necessarily imply discreteness.

• This is very interesting. Would you please be so kind to expand the two points you made in the parentheses? – user929304 Jun 29 '17 at 15:13
• @user929304 The mathematical distinction between bound and scattering states is discussed at physics.stackexchange.com/questions/9773/…. – tparker Jun 29 '17 at 20:07

They are very different concept. There are observables that are continous and observables that are quantized. A quantized quantity can take only discrete values. On the other hand, all measures are performed only with finite precision. So each observable can be measured only with finite precision, indipendently of its continuous or quantized character.

Let $A$ be an observable. There are four cases:

• $A$ is quantized and the accuracy is such that we can recognize this;
• $A$ is quantized but the accuracy is bad and experimental data agree with the hypotesis "$A$ is continuous";
• $A$ is continuous but the accuracy is bad and it appears as quantized;
• $A$ is continuous and the accuracy is good enough to see it.

In each case, $A$ is discrete in the sense that all test can be made only with finite precision. Tests with higher and higher precision allow to to perceive the actual nature of $A$, i.e. the one that best fits into the theoretical model for the physical system.

Note that not necessary a bad accuracy leads to conjecture $A$ is quantized or continuous. For example, dealing with a list of experimental values of $A$, it could happen that no evidence appears that allows to suppose a quantized nature for $A$.

For example, that energy levels of the hydrogen atom are quantized is an experimental fact, encoded into the theoretical model. Theory is developed such that the Hamiltonian of the hydrogen atom allows for quantized eigenvalues. This result is hidden if the accuracy is poor and verified if the accuracy is sufficiently good.

Hence dscreteness and quantization are unrealated as a matter of principle but intertwined as a matter of fact (i.e, in experimental practise).

• "A quantized quantity can take only discrete values." Wrong. – user50229 Aug 2 at 15:57
• No, I'd say that's right. 'Quantized' means 'integer multiple of a fundamental quantity'. Eletric charge, for instance, is a quantizied quantity. As a consequence, it takes only discrete values. Maybe you were referring to the fact that operators appearing in QM may have continuous spectrum. That's a different matter. – Federico Aug 4 at 19:51
• Energy of light is quantised and comes in multiples of $\hbar$, namely $\hbar \omega$. $\omega$ is continuous. – user50229 Aug 5 at 11:35

Quantisation does not imply discreteness. In fact the opposite is true, they are very different concepts. Part of the answer is in the definition of "discrete". Practical examples:

Quantisation: Consider the photon as discrete packets of energy ℏω as mentioned in the question. Using this equation and the fixed speed of light, a "red" photon has 1.97 eVolts of energy, oscillates 4.76*10^14 times per second and travels 630 nanometers in one oscillation. An "orange" photon has 2.03 evolts of energy, oscillates 4.91*10^14 times per second and travels 611 nanometers in one oscillation. A "yellow" photon has 2.09 eVolts of energy, oscillates 5.05*10^14 times per second and travels 593 nanometers in one oscillation.

You will never see a 1.97eV photon oscillating at a different rate. High energy photons oscillate fast, low energy photons oscillate slow. Although each photon is indeed "discrete" in the sense that it is a single unit, it is NOT "discrete" in the sense that it can take on "any value" in the continuous range of energy from mini-evolts to mega-evolts.

Discreteness: Consider the hydrogen molecule. Due to the structure of the hydrogen molecule, the electron is bound to the proton at very specific energy levels. When you break or re-adjust these bonds, you may see a 10.2eV photon emitted, or a 1.89eV photon emitted, or a 12.1eV photon emitted. These are very "discrete" energy levels and cannot take on "any value". You would never see say a 5.123eV photon emitted from hydrogen due to the breaking of an electron bond.

Also, first quantisation deals with properties of single photons, while the second quantisation deals with the properties of groups of photons travelling together.

I think the term "discretisation" in numerical analysis is completely different from "quantisation". In the first case the quantity is actually continuous(e.g. a function of space) but we divide into small parts just to get an practical result. Given enough time and ample computational strength the parts can be as small as one wishes.

Whereas, "Quantisation" is completely different. It says that a function(better word would be "eigenvalue of an operator function") can only take some given values, say integer ones. Thus "discretised energy spectrum" would mean that a continuous spectrum is being looked at different energies. But "Quantised" means there is no existencce of energy level in between two of them.

Although in QM people use the words "discrete" and "quantised" in identical sense. But this discrete is not the one used in numerical simulations.

And to answer "why we use the name quantum mechanics instead of discrete mechanics?" its probably because Germans discovered it..

• German has the word "diskret" which means discrete (and discreet for that matter). – Sebastian Riese Sep 13 '15 at 14:59

## protected by Qmechanic♦Sep 13 '15 at 16:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


text:
Introduction

In a standard* (e.g. not 2020) tax year, estimated payments are due quarterly according to the schedule outlined in Publication 505.

Payment Period Due Date
January 1 – March 31 April 15
April 1 – May 31 June 15
June 1 – August 31 September 15
September 1 – December 31 January 15

* Saturday, Sunday, holiday rule, fiscal year taxpayers, and January payment exceptions can be ignored for the purpose of this question; use the standard table provided.

Challenge

Input:

A date/timestamp, either provided as input to your solution (function parameter or stdin) as a string (at a minimum specifying a specific year, month, an day of month) or native Date object (or equivalent) or the current (dynamically updated) date/time based on the clock of the executing machine.

Output:

The most specific payment period, one of Q1, Q2, Q3, or Q4 as a string, either printed to stdout/console or programmatically returned from the function, to which a timely payment made on the input date would be applied.

Input Output
January 16 - April 15 Q1
April 16 - June 15 Q2
June 16 - September 15 Q3
September 16 - January 15 Q4

A day shall commence at 00:00 (midnight) and cover time periods up to but not including 00:00 (midnight) of the following day. The last moment of a given day, to a ms granularity is 23:59:59.999.

• can it take day of the year, as in the last day of the year is day 365? Jun 17 '21 at 0:26
• Publication 505 is 48 pages. Mind stating which page you got your table from so I don't have to pour over 48 pages to find the specific reference you're using for your tables? Jun 17 '21 at 0:53
• ALSO, this is for estimated taxes not general tax payments. Update your question to be specific about that, because estimated taxes don't actually fall within that fiscal quarter. I'm not sure you can golf this based on that. Jun 17 '21 at 0:55
• @ThomasWard Page 26, but the relevant bit is copied into the problem. It's an exercise in golfing, not tax code, hence ignoring the edge cases. The publication is just for flavor :) Jun 17 '21 at 2:59
• Coule we take three arguments day, month and year?
– user100690
Jun 17 '21 at 5:18

JavaScript (ES6), 53 bytes

Expects a string in "DD-MM-YYYY" format.

s=>"Q"+"4243131"[[d,m]=s.split-,(m*24|d>15)*9%23%7]

Try it online!

39 bytes

If we can just take the month and the day.

m=>d=>"Q"+"4243131"[(m*24|d>15)*9%23%7]

Try it online!

35 bytes

As suggested by @ovs, using a larger lookup table with a much simpler formula is actually shorter.

m=>d=>"Q"+"_4111223334444"[m+=d>15]

Try it online!

• Can you explain the thought process that goes around making formulas like these? Is it mainly brute force? what is the starting point? I am always amazed by how people find "magic formulas" for things
– user100752
Jun 17 '21 at 8:07
• @EliteDaMyth That's mainly brute force on several hard-coded templates and with hard-coded ranges for each variable. I was actually pretty sure that a more standard formula would work for that one, but I couldn't find any that was shorter than the hash. Jun 17 '21 at 10:18
• Does this work? It produces the same output for your extensive test-suite.
– ovs
Jun 17 '21 at 13:42
• @ovs I think that works indeed. Jun 17 '21 at 13:53

Retina 0.8.2, 72 63 bytes

\d+
$* 1+-(1{1,5}(1)*)-1{1,15}(1)* 11$3$2$1
(111)*1+
Q$#1 0|5 4  Try it online! Link includes test cases. Takes input in YYYY-MM-DD format. Explanation: \d+$*

Convert to unary.

1+-(1{1,5}(1)*)-1{1,15}(1)*
11$3$2$1  Add 2, 3 or 4 to the month depending on whether the month is greater than 5 or the day is greater than 15 or not. (111)*1+ Q$#1

Subtract 1 from the month, then integer divide it by 3 and prefix Q to the result. (This is to avoid zero width matches.)

0|5
4

Change Q0 and Q5 to Q4.

• @Arnauld Bah, I'd totally misread the table...
– Neil
Jun 17 '21 at 11:42

R, 65 62 bytes

function(d)paste0("Q",1+sum(c(3,5,8)<el(format(d-15,"%m"):1)))

Try it online!

Expects input in R's native Date class. Subtract 15 days from the input, extract the month (as a string such as "04"), and convert to numeric with el(...:1) (see this tip). We then get the result by adding 1 for each of the values 3, 5, and 8 which are smaller than the result.

• I like the lexicographic comparison, but I think numeric can be shorter. Jun 17 '21 at 14:16
• Ha! You edited while I was typing, and yours is better! Well done! Jun 17 '21 at 14:16

Charcoal, 28 bytes

≔Ｉ⪪Ｓ-θ§12344÷⁻§θ¹⁺›¹⁶⊟θ›⁶⊟θ³

Try it online! Link is to verbose version of code. Takes input in the form YYYY-MM-DD. Explanation:

≔Ｉ⪪Ｓ-θ

Input the date, split it on - and cast to integer.

§12344÷⁻§θ¹⁺›¹⁶⊟θ›⁶⊟θ³

Take the month, subtract 1 each if the day is less than 16 or the month is less than 6, divide by 3, then cyclically index into the string 12344, thus casting -1 and 4 to 4.

Python 3.8 (pre-release), 109 bytes

lambda n,i=int:'Q'+['?111223334444'[i(k:=(S:=n.split('-'))[1])],'1234'[i(k)//3-1]][k in'1469'and(i(S[0])<16)]

Try it online!

Way too long.

The OP has confirmed very kindly that this is the only solution of mine which is valid.

• You're allowed to ignore useless arguments -there's a meta post about it. Jun 17 '21 at 20:06


text:
# Permutations Formula

Permutations Formula

Have you heard of the famous Arab mathematician, Al-Khalil? He wrote a book titled, "Book of cryptographic messages," and that book contained the first use of Permutations and Combinations.

Permutations are extremely important in Math and they are also essential in various fields of Science such as Physics and Biology.

In this lesson, you will be introduced to the formulas of permutations. You will recall about permutations and learn how to apply the formula to solve problems related to permutations. This lesson includes several simple examples to help you understand the concept clearly. Let's now start this lesson by learning the definition of permutation.

## Lesson Plan

 1 What Is Permutation? 2 Important Notes on Permutation Formula 3 Solved Examples on Permutation Formula 4 Interactive Questions on Permutation Formula

## What Is Permutation?

Before discussing the permutation formula, we should understand the concept of “Permutation.”

Defining Permutation: A permutation is an arrangement of a certain number of objects in a specific order.

In fact, permutation is nothing but counting the number of ways in which the given objects can be arranged.

Permutation can be defined as the different ways in which a collection of objects or alphabets can be arranged.

When dealing with permutation, remember that position is important.

Given below are the different formulas of permutation.

1. This formula is used to find the number of possible ways in which anything can be arranged while ensuring that repetition is not allowed.

 P = \begin{align}\frac{n!}{(n-r)!}\end{align}

2. This formula is used to find the number of possible ways in which anything can be arranged while allowing repetition.

 $$P\left ( n,r \right )$$ = $$n^{r}$$

$$n$$ = Total number of elements in a set.

$$r$$ = Number of elements selected.

Did you know that we use the combination formula when the order does not matter? This formula is given as:

 $$C(n,r)$$= \begin{align}\frac{n!}{(n-r)!r!}\end{align}

## What Is the General Formula of Permutation?

Formula for Permutation:  $$P$$ = $$\dfrac{n!}{(n-r)!}$$ ; $$0 \leq r \leq n$$

Here “!” represents the product of the integers less than or equal to $$n$$, but it should be greater than or equal to one.

Some common examples are:

$$1!$$ = $$1\times 1$$ = $$1$$

$$2!$$ = $$2\times 1$$ = $$2$$

$$3!$$ = $$3\times 2\times 1$$ = $$6$$

$$4!$$ = $$4\times 3\times 2\times 1$$ = $$24$$

Did you notice that all the numbers are the factors of the result (product)?

 $$P$$ = $$\dfrac{n!}{(n-r)!}$$ ; $$0 \leq r \leq n$$

## How to Solve Word Problems Using Permutation?

To solve a word problem using permutation, we will use the general formula based on the information given.

Let us understand factorials using an example.

Represent the factorial for number 4.

Solution: Always remember, factorials are represented with a “!” sign which means 4 will be written as 4!

Now, to find out the value of 4! we will solve it as follows:

$$4!$$ = $$4\times 3\times 2\times 1$$ = 24

Now let us solve a problem using the permutation formula.

Example: Emily is having 4 chairs and she wants to place her three dolls on those chairs. In how many possible ways can Emily place her 3 dolls on 4 chairs?

Solution: Given n = 4 and r = 3

On applying the permutation formula

P = \begin{align}\frac{n!}{(n-r)!}\end{align}

We get,

\begin{align}P &= \dfrac{n!}{(n-r)!}\\ & = \dfrac{4!}{(4-3)!}\\ &= \dfrac{4!}{1!}\\ &= \dfrac{4\times 3\times 2\times 1}{1}\\ &= 24\end{align}

She can place 3 dolls in 24 ways on the 4 chairs.

Important Notes
• If you have five vegetables such as a potato, an onion, a broccoli, a tomato, and an okra that are to be added while cooking a certain dish, but you changed the order of adding these vegetables, it will not make a difference. The dish will taste the same irrespective of the order in which the vegetables were added as long as the vegetables are not replaced by different varieties. Here, "what you add or position" is important and that forms an important aspect of combination.
• If your bag has a lock code, say 345, and you try opening it using 453 or 534, that will not work. You will have to use the numbers 345, and in the same order. Here, "what u choose" is important and that's permutation.

## Solved Examples

 Example 1

There are 7 books on a bookshelf.

Can you find out in how many different ways will you be able to rearrange them such that no book can be placed in a spot more than once (repetition is not allowed)?

Solution

Since we have seven books, we will have:

Seven choices for the $$1^{st}$$ book; Six choices for the $$2^{nd}$$ book; Five choices for the $$3^{rd}$$ book; Four choices for the $$4^{th}$$ book; Three choices for the $$5^{th}$$ book; Two choices for the $$6^{th}$$ book; One choice for the $$7^{th}$$ book

Now, multiply the choices: $$7\times 6\times 5\times 4\times 3\times 2\times 1$$ = $$5,040$$

 $$\therefore$$ 5,040 ways are possible
 Example 2

There are 5 pencils in a pencil box.

Help Andrew figure out all the possible ways in which the pencils can be rearranged such that a pencil cannot be placed twice in the same position.

Solution

We have 5 pencils. We know that these pencils cannot be placed twice in the same position.

There will be 5 spot choices for the $$1^{st}$$ pencil; $$4$$ spot choices for the $$2^{nd}$$ pencil; $$3$$ spot choices for the $$3^{rd}$$ pencil; $$2$$ spot choices for the $$4^{th}$$ pencil; $$1$$ spot for the $$5^{th}$$ pencil

Therefore, when we multiply the choices, we get: $$5\times 4\times 3\times 2\times 1$$ = $$120$$

Thus, there are 120 possible arrangements.

 $$\therefore$$ There are 120 possible arrangements.
 Example 3

In how many ways can you rearrange the word “SMILE” so that it may or may not make a valid word?

Note that you cannot place any alphabet more than once at a spot.

Solution

The word “SMILE” has 5 letters.

Hence, the spot choices for these letters will be the same as that in the previous example about 5 pencils.

Therefore, to find out all the possible rearrangements, multiply the choices: $$5\times 4\times 3\times 2\times 1$$ = 120

 $$\therefore$$ 120 ways are possible.
 Example 4

How many 5-digit codes can be created using the digits from 0 to 9 if repetition is allowed?

Solution

Given n = 10 and r = 5

\begin{align}P\left ( n,r \right ) &= n^{r}\\ & =10^{5}\\ &=100,000\end{align}

 $$\therefore$$ 100,000 five-digit codes can be created
 Example 5

Consider a set of 5 alphabets p, q, r, s, t.

In how many ways can 4 alphabets be selected without repetition?

Solution

The set of alphabets are p, q, r, s, t.

4 alphabets are to be selected.

$$\therefore$$  $$p_{r}^{n}$$ = $$p_{4}^{5}$$ = $$\dfrac{n!}{(n-r)!}$$

= $$\dfrac{5!}{(5-4)!}$$

= $$5\times 4\times 3\times 2\times 1$$

= 120

 $$\therefore$$ 120 ways

Challenging Questions
1. Find the sum of all the possible 4-digit numbers formed using the digits 3, 4, 5, and 6 using each digit once?

## Interactive Questions

Here are a few activities for you to practice.

## Let's Summarize

We hope you enjoyed learning about permutation formula with the examples and practice questions. Now you will be able to easily solve problems on permutation formula with repetition, permutation formula example, and combination formula.

At Cuemath, our team of math experts are dedicated to make learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it problems, online classes, videos, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.

## 1. What is permutation in math?

When it comes to math, permutation can be defined as the process of changing the order or sequence of an already ordered set of objects.

## 2. What is the difference between permutation and combination?

The most basic difference between permutation and combination is that you have to consider the order in case of permutation, but not when it comes to combination.

The order matters in permutation, but doesn’t matter in the case of a combination.

For example, a mobile phone unlocks with a password pin, which is $$3428$$. If you type $$2438$$, it will not get unlocked because the order of number matters here. That is called a permutation.

If it were a combination, then your phone would unlock by any sequence you make with the numbers, 2, 3, 4 and 8

## 3. What is “r” in the permutation formula?

The exact formula for permutation is $$\dfrac{n!}{(n-r)!}$$ ; 0 ≤ r ≤ n

Here, while n denotes the total number of items in a set, $$r$$ denotes all the items taken for permutation.

More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus


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# How to secure different types of CloudFront origins

## Background

When CloudFront goes to the origin to fetch a content it behaves like any other third-party service. In effect, by default, whatever CloudFront has access to, everyone else has too.

Usually, this is a problem.

How to protect the origin so that only CloudFront has access and can not be accessed directly?

It depends on the origin itself.

### Protecting S3 buckets

This is the most straightforward case, and it is the most likely to pop up when you search for serving private content. You have a private bucket, but you want CloudFront to have access to it.

This is the only usage of the Origin Access Identity, which is a principal that you can reference in the bucket’s resource policy.

This is how to make it work: You create an OAI, assign it to the distribution, then update the bucket policy. This gives access to that distribution while also keeping the bucket private.

### Protecting API Gateways

This is when you have an API and you want to restrict access only to the distribution.

API Gateway supports restricting access by API-keys, which means there is a header that must be sent along with the request.

CloudFront can set a custom header when it accesses the origin so that it can send the x-api-key with the secret value.

The API is kept private, as it requires the secret, but when accessed through CloudFront the request goes through.

### Protecting VPCs

#### The approach that does not work

VPC security is usually done by security groups, and you can configure them to allow only CloudFront IPs. There is an official list of IPs that the service use and also a notification so that you can handle the future changes to the list.

Actually, there is an official script to manage the settings of a security group to only allow CloudFront IPs.

It seems good, but unfortunately, this is not a good solution.

It makes sure that only CloudFront can access the instances, but does not limit it to your distribution. In effect, any CloudFront distribution can access it, even outside your account.

Which means, anybody can create a new distribution, point it to your instance, and have access.

#### The kinda expensive approach

If you have an ALB, then you can configure WAF to require a custom header to be present in order to allow the request. Then add that header to the distribution, and you are all set.

The downside? Cost.

WAF has a fixed monthly cost, and for small applications, it could make the majority of the bill.

See here for a guide on how to set it up.

If you have instances that you can configure, you can implement an API-key-like functionality. See the next section for the details.

### Protecting a custom backend

When you have a custom backend, be it Apache, Nginx, or a NodeJs server that you can configure freely, then the easiest way is to implement the API-key-like functionality. Set a custom header on the CloudFront distribution with a secret value, then configure your backend to only allow requests with that header set.


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The greatest number theorist in mathematical universe, Leonhard Euler had discovered some formulas and relations in number theory, which were based on practices and were correct to limited extent but still stun the mathematicians. The prime generating equation by Euler is a very specific binomial equation on prime numbers and yields more primes than any other relations out there in number theory. Euler told that the equation $f(x)=x^2+x+k$ yields many prime numbers with the values of x being input from x=0 to $x=k-2$ ; k being a prime.

Let’s see how many primes we can get by using different values of $k$ and $x$:

 Serial Number Value of k (prime) Value of x (from x=0 to x=k-2) Value of f(x)=(x^2+x)+k Not a Prime? 1 2 0 2 2 3 0 3 3 1 5 4 5 0 5 5 1 7 6 2 11 7 3 17 8 7 0 7 9 1 9 No 10 2 13 11 3 19 12 4 27 No 13 5 37 14 11 0 11 15 1 13 16 2 17 17 3 23 18 4 31 19 5 41 20 6 53 21 7 67 22 8 83 23 9 101 24 13 0 13 25 1 15 No 26 2 19 27 3 25 No 28 4 33 No 29 5 43 30 6 55 No 31 7 69 No 32 8 85 NO 33 9 103 34 10 123 No 35 11 145 No 36 17 0 17 37 1 19 38 2 23 39 3 29 40 4 37 41 5 47 42 6 59 43 7 73 44 8 89 45 9 107 46 10 127 47 11 149 48 12 173 49 13 199 50 14 227 51 15 257 52 19 0 19 53 1 21 54 2 25 No 55 3 31 56 4 39 No 57 5 49 58 6 61 59 7 75 No 60 8 91 No 61 9 109 62 10 129 No 63 11 151 64 12 175 No 65 13 201 No 66 14 229 67 15 259 68 16 291 69 17 325 No 70 23 0 23 71 1 25 No 72 2 29 73 3 35 No 74 4 43 75 5 53 76 6 65 No 77 7 79 78 8 95 No 79 9 113 80 10. 133 81 11 155 No 82 12 179 83 13 205 No 84 14 233 85 15 263 86 16 295 No 87 17 329 88 18 365 No 89 19 403 90 20 443 91 21 485 No

The above table yields many prime numbers, which again can be put at the place of k and so on the table can be progressed.

According to Euler, 41 was the most appropriate value of k yielding more prime numbers than any other k. In the list below, each value of f(x) is a prime for k=41:

 k= 41 x= 0 f(x)= 41 1 43 2 47 3 53 4 61 5 71 6 83 7 97 8 113 9 131 10 151 11 173 12 197 13 223 14 251 15 281 16 313 17 347 18 383 19 421 20 461 21 503 22 547 23 593 24 641 25 691 26 743 27 797 28 853 29 911 30 971 31 1033 32 1097 33 1163 34 1231 35 1301 36 1373 37 1447 38 1523 39 1601

So, the Euler’s Prime Generating Equation can be written as
$f(x) = x^2+x+41$ ; where $x$ is an integer ranging from 0 to 39.

What if we increase the value of x beyond the limit of 39? What will we get?

The next values of f(x) in this series would be 1681, 1763, 1847, 1933, 2021, 2111, 2203, 2297, 2393, … .
Are all these prime numbers too? The answer is no. 1681 is not a prime number, neither are 1763 and 2021. Though all others (in above sequence at-least) are prime numbers.

1. Huen Yeong Kong says:

I believe I have a formula far better than Euler’s formulae. It will generate consecutive primes from 2 to 7993. Here is the
formula: factor(product(2*i,i,1,8000)); written in Maxima software. Try it. What do you think?

2. Huen Yeong Kong says:

Here is a shortened version tor pagination problem:
Numerical example:

## Fox – Rabbit Chase Problems

Part I: A fox chases a rabbit. Both run at the same speed $v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is…

## Solving Integral Equations (4) : Integral Equations into Differential Equations

Introduction In earlier parts we discussed about the basics of integral equations and how they can be derived from ordinary differential equations. In second part, we also solved a linear integral equation using trial method. Now we are in a situation from where main job of solving Integral Equations can be started. But before we go ahead to that mission, it will be…

## Calendar Formula: Finding the Week-days

This is the last month of the glorious prime year 2011. We are all set to welcome upcoming 2012, which is not a prime but a leap year. Calendars have very decent stories and since this blog is based on mathematical approach, let we talk about the mathematical aspects of calendars. The international calendar we use is called Gregorian Calendar,…

## Abel Prize for 2014 to Yakov Sinai

Mathematical Physicist Yakov Gregory Sinai, (b. 21st September 1935, 78 years old) has been awarded the prestigious Abel Prize for 2014 by the Norwegian Academy of Science and Letters (NASL). The President of the NASL, Nils C. Stenseth, announced the winner of the 2014 Abel Prize at the Academy in Oslo on 26 March 2014. The Abel Prize recognizes contributions of…


text:
Lesson 9

When Are They the Same?

Lesson Narrative

In this lesson students apply their knowledge of solving equations by considering two real world situations: two tanks where one is filling and the other is emptying and two elevators traveling above and below ground level. Using the given expressions for each situation, students are asked to determine when the amount of water in the tanks or the travel time of elevators will be the same. It is the work of the student to recognize that they can set the two expressions equal and solve the equation for the unknown and this work sets up the concept of substitution for the coming section on systems of linear equations.

Learning Goals

Teacher Facing

• Create an equation in one variable to represent a situation in which two conditions are equal.
• Interpret the solution of an equation in one variable in context.

Student Facing

Let’s use equations to think about situations.

Student Facing

• I can use an expression to find when two things, like height, are the same in a real-world situation.

Building Towards

Glossary Entries

• coefficient

A coefficient is a number that is multiplied by a variable.

For example, in the expression $$3x+5$$, the coefficient of $$x$$ is 3. In the expression $$y+5$$, the coefficient of $$y$$ is 1, because $$y=1 \boldcdot y$$.

• constant term

In an expression like $$5x+2$$, the number 2 is called the constant term because it doesn’t change when $$x$$ changes.

In the expression $$7x+9$$, 9 is the constant term.
In the expression $$5x+(\text-8)$$, -8 is the constant term.
In the expression $$12-4x$$, 12 is the constant term.

• term

A term is a part of an expression. It can be a single number, a variable, or a number and a variable that are multiplied together. For example, the expression $$5x + 18$$ has two terms. The first term is $$5x$$ and the second term is 18.


text:
# Strongly decreasing sequence of sets

• November 14th 2010, 04:50 PM
hjortur
Strongly decreasing sequence of sets
This was a homework question few weeks ago, and the teacher didn't give
out the solution to this problem (it doesn't directly concern complex analysis), so I was wondering weather someone knows the answer to this one.

Let $A\subset \mathbb{C}$, and $D$ be such that
$D(A)$ is the set $A$ with all isolated points of $A$ removed. So the question is, is there a set $A\subset \mathbb{C}$
such that the sequence, $(D^{n}(A))_{n\in\mathbb{N}}$ is strongly decreasing?
That is $D(A)\subset A$, $D^{2}(A)\subset D(A)$, but not equal?

My first thoughts went something like this: Let $M:=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}$. Then $D(M)=\{0\}$ and
$D^{k}(M)=\emptyset$ if $k\geq 2$, so this wont do.

Now for every $x\in M$ $x\ne0$ we can make $\varepsilon$-balls such that $\varepsilon_{x_1}\cap\varepsilon_{x_2}=\emptyset$ if $x_1\ne x_2$ and neither is 0. Then we can construct a sequence
for every $x\in M$ such that its limit is $x$ and it is entirely contained in $\varepsilon_x$. Lets call the set of those sequences $S_M$.
Now let $M_1=M\cup S_M$.
$M_1$ is such that $D(M_1)=M$ so now we got 4 strictly decreasing items in the sequence.

For every given $n\in\mathbb{N}$ we can continue this construction
such that the sequence is strictly decreasing for the first $n$ terms.
But that is not the same as the infinite sequence is strictly decreasing.

I am starting to think that this is impossible, but I am not sure.
Does anyone have any ideas?
• November 14th 2010, 05:50 PM
Drexel28
Quote:

Originally Posted by hjortur
This was a homework question few weeks ago, and the teacher didn't give
out the solution to this problem (it doesn't directly concern complex analysis), so I was wondering weather someone knows the answer to this one.

Let $A\subset \mathbb{C}$, and $D$ be such that
$D(A)$ is the set $A$ with all isolated points of $A$ removed. So the question is, is there a set $A\subset \mathbb{C}$
such that the sequence, $(D^{n}(A))_{n\in\mathbb{N}}$ is strongly decreasing?
That is $D(A)\subset A$, $D^{2}(A)\subset D(A)$, but not equal?

My first thoughts went something like this: Let $M:=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}$. Then $D(M)=\{0\}$ and
$D^{k}(M)=\emptyset$ if $k\geq 2$, so this wont do.

Now for every $x\in M$ $x\ne0$ we can make $\varepsilon$-balls such that $\varepsilon_{x_1}\cap\varepsilon_{x_2}=\emptyset$ if $x_1\ne x_2$ and neither is 0. Then we can construct a sequence
for every $x\in M$ such that its limit is $x$ and it is entirely contained in $\varepsilon_x$. Lets call the set of those sequences $S_M$.
Now let $M_1=M\cup S_M$.
$M_1$ is such that $D(M_1)=M$ so now we got 4 strictly decreasing items in the sequence.

For every given $n\in\mathbb{N}$ we can continue this construction
such that the sequence is strictly decreasing for the first $n$ terms.
But that is not the same as the infinite sequence is strictly decreasing.

I am starting to think that this is impossible, but I am not sure.
Does anyone have any ideas?

So, you're asking does there exist a set $A\subseteq\mathbb{C}$ so that if $D(A)$ is the set of limit points of $A$ then

$A\supset D(A)\supset D\left(D\left(A\right)\right)\supset D\left(\left(\left(A\right)\right)\right)\supset\c dots$

What makes sense to me is to go a route similar to what you did. So, let's think of, just for now, a subset of the real numbers which survives two applications of $D$. Something like $D=[-5,-4]\cup\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\rig ht\}$. What about a set which survives three? What if we took now the subset of the complex numbers which has the real line, appended to the vertical line $\left\{\frac{i}{n}:n\in\mathbb{N}\}$ and then for each $n_0\in\mathbb{N}$ we considered $\frac{i}{n_0}+\frac{1}{m}$ for all $m\in\mathbb{N}$. Then, I feel that $D\left(A\right)=\mathbb{R}\cup\left\{\frac{i}{n}:n \in\mathbb{N}\right\}$ and $D\left(D\left(A\right)\right)=\mathbb{R}$. What if you added to each point of the $\frac{i}{n_0}+\frac{1}{m}$ a sequence converging to it, and to each one of those points append a, etc.

I don't feel this is correct though. Let me think about it.


text:
Address 115 S 2nd Ave W, Marsing, ID 83639 (208) 899-8037

how to calculate compressibility error Homedale, Idaho

Close × Code %% True Airspeed from Indicated Airspeed Calculation % % This model shows how to compute true airspeed from indicated airspeed using the Ideal Airspeed Correction % block. It constitutes a fragment of a complete aerodynamics problem, including only measurement and calibration.Airspeed Correction ModelsTo view the airspeed correction models, enter the following at the MATLAB® command line:aeroblk_indicated aeroblk_calibrated aeroblk_indicated Or more technically, Bernoulli's equation assumes an incompressible flow, but as air is in fact compressible using the Bernoulli equation to deduce airspeed requires that compressibility be taken into account. Hot Network Questions How can you tell if the engine is not brand new?

Please try the request again. Image from Mechanical Measurements and Metrology - Measurement of Fluid Velocity by S. What is the value of the true air speed (TAS), the Equivalent Air Speed (EAS) and the Calibrated Air Speed (CAS) of your airplane ?

Solution and explanation The formula for As the air is brought to rest inside the pitot tube, Bernoulli tells us that 100% of the dynamic pressure is perfectly converted to static pressure, which the sensors then measure.

I'm sorry, but my English is not perfect yet... Calibrated airspeed is equivalent airspeed modified with compressibility effects of air which affect the airspeed indicator. At SL ISA, for M < 0,4, EAS is almost equal to CAS. 5. So for a small twin piston (MEP, multi engine piston) flying low and slow, the compressibility error is negligible.

For slow speeds, the data required are static air temperature, pressure altitude and IAS (or CAS for more precision). Best Regards, Tony Pike Give Sergei back his dignity......Simples! A pitot-static airspeed sensor contains a restricted volume of air. The Calculate CAS subsystem uses the flap setting and indicated airspeed to calculate the calibrated airspeed.

Compressibility error = square root[1 + M2/4 + M4/40 + M6/1600] Relationship between CAS and EAS : CAS = EAS x square root[1 + M2/4 + M4/40 + M6/1600] So as ISBN 978-0-07-295046-5 ^ Houghton, E.L. Last edited by hans37; 03-12-2010, 10:06. A shock wave is an extreme demonstration of air being compressible, but air is actually compressing and decompressing over a much wider range of speeds, without forming shocks.

Execution of Batch class Should zero be followed by units? Password Register Forgot Password? It measures the lower dynamic pressure due to the velocity of the fluid in the tube (Bernoulli's principle). The above formula is accurate within 1% up to Mach 1.2 and useful with acceptable error up to Mach 1.5.

To start viewing messages, select the forum that you want to visit from the selection below. Travelling to Iceland and UK How to handle a senior developer diva who seems unaware that his skills are obsolete? Baz Comment Post Cancel hans37 Registered Member Join Date: Oct 2010 Posts: 6 #12 27-10-2010, 18:13 Re: compressibility factor,new type of questions Thank You Baz for explanation! The transonic region is the range of speeds from the speed at which the air flow first becomes supersonic over one part of the aircraft, to the speed at which all

We found that TAS = 503 kt. Comment Post Cancel Previous Next - Themes --- Black Red -- vBulletin Default Theme --- Blue Green --- Blue Yellow --- Cool Blue --- Denim --- Gradient --- Grey Stripes --- In new CQB-15 there are a lot of new questions that aren't in cqb14/GenNav. We are still using CQB14 and our present syllabus caters for that.

Thus the pitot tube measures a higher impact pressure than you would get by adding dynamic + static pressures. For this reason, TAS cannot be measured directly. BTW: Could You confirm it, that your CAA switch for CQB-15 with beginning of next year? The freestream air has dynamic pressure and static pressure.

The aeroblk_calibrated model implements a conversion to true airspeed.Each model consists of two main components:COESA Atmosphere Model Block calculates the change in atmospheric conditions with changing altitude.Ideal Airspeed Correction Block transforms I appreciate the convenience of showing stall speed in EAS, but if I said, for example that an aircaft at 400kts and sea level would handle the same and be subject Best regards! I get that the ASI will read slower because the ram air is compressing against the air already in the pitot tube.

Within the airspeed indicator, % there is a certain amount of trapped air. Licence agreement and Code of conduct About SKYbrary Disclaimers ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection to All they have is a simple scientific calculator and a CRP-5 or similar 'whizz wheel' and absolute precision is pretty near impossible on a CRP-5. Last edited by Oktas8; 5th Sep 2012 at 13:31. « Previous Thread | Next Thread » Thread Tools Show Printable Version Search this Thread Advanced Search Posting Rules You may

When flying at high altitudes and higher % airspeeds, calibrated airspeed (CAS) is always higher than equivalent airspeed. % Calibrated airspeed is equivalent airspeed modified with compressibility effects of air % My question is when does this happen? Thank you all for help! Using this calibration table, the indicated airspeed (IAS) is determined from calibrated airspeed by modifying it with calibration error of the airspeed indicator.

Published with MATLABÂ® R2016b MATLAB and Simulink are registered trademarks of The MathWorks, Inc. Generated Mon, 17 Oct 2016 16:00:58 GMT by s_ac15 (squid/3.5.20) Modern aircraft instrumentation use an Air Data Computer to perform this calculation in real time and display the TAS reading directly on the EFIS. As the speed increases above this value, the term $\frac{1}{2}\rho v^{2}$ no longer gives the difference between the Total and static pressure.

However for a jet, flying fast and high, it is another story and as you can see on the first graph, during the climb at a constant CAS, the difference between Since these shock waves form as a function of the speed of sound, people specify Mach number in this context. If you climb at constant dynamic pressure, which means at constant EAS, then CAS is increasing and TAS as well. 6.


text:
## WeBWorK Problems

### Scoring a Multiple Choice Problem

by Louis Zulli -
Number of replies: 5
A colleague would like WW to score his multiple choice problems in the following manner: Correct answer = 5pts, Incorrect answer = 0pts, No Answer=1pt.

How can a problem be written to achieve this effect?

Thanks.

### Re: Scoring a Multiple Choice Problem

by Davide Cervone -
There are several approaches to this. One would be to write a custom grader, but it can also be done with a custom answer evaluator. I assume you are planning to use radio buttons? It's been a while since I looked at them, but it should be possible to use a Parser-based string answer checker that allows blank answers and that uses a custom checker routine that returns a score of 1, .2 or 0, depending on the answer. The professor would need to assign the problem a value of 5 points. It is pretty easy to provide a custom checker for the Parser checkers, either with the "checker=>" option to the string's cmp routine, or with the custom_cmp() function from the answerCustom.pl file in /pg/macros.

I usually write the example for you, but haven't had the chance. If I can, I'll do it this weekend, but that's only if I get my grading done. (Or can't take it any more.)

Davide

### Re: Scoring a Multiple Choice Problem

by Louis Zulli -
Thanks Davide.

Your reply reminded me of another issue. We created a draft of a multiple choice quiz that uses radio buttons. Each problem uses new_multiple_choice().

However, we appear to have a difficulty. Once a radio button is selected, there seems to be no way to deselect it. Another button can be selected to replace it, but it seems impossible to leave all buttons deselected once one is selected.

I suppose we can resolve this issue and the one in the original post by adding a button for "I choose not to respond." Or we could abandon radio buttons and just use answer boxes. But are there better solutions?

Louis

### Re: Scoring a Multiple Choice Problem

by Davide Cervone -
That is how the browsers have implemented radio buttons.

It would be possible to use JavaScript to make it possible to deselect the radio buttons. For example, there could be a "No Answer" button that would uncheck all the radio buttons. It might even be possible to arrange for CTRL-click to uncheck a button. But these would require that the student have JavaScript enabled in his or her browser.

I'm not sure if you are allowing multiple submissions of the problem, or how you are planning to deal with the answer results at the top of the page, but if the student can do multiple submissions, it would be possible to do the "No Answer" button without JavaScript. But you are getting pretty close to just having a "I choose not to respond" choice in the radio buttons.

(BTW, if I remember correctly, it is possible to prevent the answer results section from showing up at all if you don't want to be telling them about correct answers.)

Davide

### Re: Scoring a Multiple Choice Problem

by Davide Cervone -
Louis:

I have added the ability to uncheck the radio buttons, either by clicking them a second time, or shift-clicking them (the problem author specifies which). You need to set an additional option for the RadioButtons() function: either uncheckable=>1 or uncheckable=>"shift".

Note that this requires the user to have JavaScript active. It is possible for students to have JavaScript turned off, and so unchecking would not be possible in that case. Currently no message is issued, but it would be possible to have a message added to the problem to warn the user that they would be able to uncheck the buttons if they turned on JavaScript.

Davide

### Re: Scoring a Multiple Choice Problem

by Davide Cervone -
OK, I've just committed a new file (parserRadioButtons.pl) to the CVS repository for pg/macros. This implements a Parser-based method for handling radio buttons. There is some documentation at the top of the file, but here is a simple example:
    DOCUMENT();

"PG.pl",
"PGbasicmacros.pl",
"Parser.pl",
);

TEXT(beginproblem);

$radio = RadioButtons(["Choice 1", "Choice 2"],1); BEGIN_TEXT Choose one:$BR
\{$radio->buttons\} END_TEXT ANS($radio->cmp);
$showPartialCorrectAnswers = 1; ENDDOCUMENT();  This sets up a "RadioButtons" object by giving the choices and the index of the correct answer. There are a number of additional options that can be supplied (see the parserRadioButtons.pl file for details), which control things like what to use as the answers when the students select the various buttons (these can be numbers, letters, or the button strings themselves, or any other strings you set), and what to use between the entries (spaces,$BR, $PAR, etc). To handle your special point arrangement, you can supply a custom answer checker that returns partial credit according to your model above. It turns out that the standard Parser objects don't provide a method for returning partial credit (I'll have to fix that), but we can use the List checker, which does. Here is an example:  DOCUMENT(); loadMacros( "PG.pl", "PGbasicmacros.pl", "PGanswermacros.pl", "Parser.pl", "parserRadioButtons.pl", ); TEXT(beginproblem);$radio = RadioButtons(["Choice 1", "Choice 2"],1);

BEGIN_TEXT
Choose one:$BR \{$radio->buttons\}
END_TEXT

Context($radio->{context});$cmp = List($radio)->cmp(list_checker=> sub { my ($correct,$student,$ans) = @_;
return .2 if $student->[0] eq ""; return ($student->[0] eq $correct->[0] ? 1 : 0); } );$cmp->install_pre_filter('erase');  # allow blank answers

ANS($cmp);$showPartialCorrectAnswers = 1;

ENDDOCUMENT();

Here we have to work a little bit to make everything work out. First, we must set the context to the one from the $radio object, so that the list will have access to the string that were set up for the$radio. Then we make a new answer checker based on the list checker with a custom list checker. The custom checker returns .2 if the student didn't answer anything, and returns 1 if correct, or 0 if not. Finally, we need to allow blank answers, so we remove the prefilter that WeBWorK installs to avoid checking blanks.

You will need to assign the problem a score of 5 (in the homework set editor) in order to make the totals come out the way you want.

Hope that does what you need.

Davide


text:
Precalculus (6th Edition)

$[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$
Given: $x^{3}+i=0$ or $x^{3}=-i$ $-i$ can be written in trigonometric form as: $-i=0-i=(\cos 90^{\circ}-\sin 90^{\circ})$ Absolute value of third root is given as $\sqrt[3] 1=1$ Now, the arguments can be given as: $k=0,1,2$ Roots: $(\cos 30^{\circ}+\sin 30^{\circ})$,$(\cos 150^{\circ}+\sin 150^{\circ})$,$(\cos 270^{\circ}+\sin 270^{\circ})$ Solution set of the equation can be written as: $[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$


text:
## Duke Mathematical Journal

### Irreducible symplectic $4$-folds and Eisenbud-Popescu-Walter sextics

#### Abstract

Eisenbud, Popescu, and Walter [4] have constructed certain special sextic hypersurfaces in ${\mathbb P}^5$ as Lagrangian degeneracy loci. We prove that the natural double cover of a generic Eisenbud-Popescu-Walter (EPW) sextic is a deformation of the Hilbert square of a $K3$-surface $(K3)^{[2]}$ and that the family of such varieties is locally complete for deformations that keep the hyperplane class of type $(1,1)$; thus we get an example similar to that (discovered by Beauville and Donagi [2]) of the Fano variety of lines on a cubic $4$-fold. Conversely, suppose that $X$ is a numerical $(K3)^{[2]}$, suppose that $H$ is an ample divisor on $X$ of square $2$ for Beauville's quadratic form, and suppose that the map $X\dashrightarrow|H|^{\vee}$ is the composition of the quotient $X\to Y$ for an antisymplectic involution on $X$ followed by an immersion $Y\hookrightarrow|H|^{\vee}$; then $Y$ is an EPW sextic, and $X\to Y$ is the natural double cover

#### Article information

Source
Duke Math. J., Volume 134, Number 1 (2006), 99-137.

Dates
First available in Project Euclid: 4 July 2006

https://projecteuclid.org/euclid.dmj/1152018505

Digital Object Identifier
doi:10.1215/S0012-7094-06-13413-0

Mathematical Reviews number (MathSciNet)
MR2239344

Zentralblatt MATH identifier
1105.14051

#### Citation

O'Grady, Kieran G. Irreducible symplectic $4$ -folds and Eisenbud-Popescu-Walter sextics. Duke Math. J. 134 (2006), no. 1, 99--137. doi:10.1215/S0012-7094-06-13413-0. https://projecteuclid.org/euclid.dmj/1152018505

#### References

• A. Beauville, Variétés Kähleriennes dont la première classe de Chern est nulle, J. Differential Geom. 18 (1983), 755--782.
• A. Beauville and R. Donagi, La variété des droites d'une hypersurface cubique de dimension $4$, C. R. Acad. Sci. Paris Sér. I Math. 301 (1985), 703--706.
• G. Casnati and F. Catanese, Even sets of nodes are bundle symmetric, J. Differential Geom. 47 (1997), 237--256.
• D. Eisenbud, S. Popescu, and C. Walter, Lagrangian subbundles and codimension $3$ subcanonical subschemes, Duke Math. J. 107 (2001), 427--467.
• W. Fulton and P. Pragacz, Schubert Varieties and Degeneracy Loci, Lecture Notes in Math. 1689, Springer, Berlin, 1998.
• F. J. Gallego and B. P. Purnaprajna, Very ampleness and higher syzygies for Calabi-Yau threefolds, Math. Ann. 312 (1998), 133--149.
• D. Gieseker, Geometric invariant theory and applications to moduli problems'' in Invariant Theory (Montecatini, Italy, 1982), Lecture Notes in Math. 996, Springer, Berlin, 1983, 45--73.
• D. Huybrechts, Compact hyper-Kähler manifolds: Basic results, Invent. Math. 135 (1999), 63--113.; Erratum, Invent. Math. 152 (2003), 209--212.
• A. Iliev and K. Ranestad, Addendum to ʽʽ K$3$ surfaces of genus $8$ and varieties of sums of powers of cubic fourfolds,ʼʼ preprint, 2002.
• V. A. Iskovskih, Fano threefolds, I, Math. USSR-Izv. 11 (1977), no. 3, 485--527.
• S. Mukai, Moduli of vector bundles on $K3$ surfaces and symplectic manifolds, Sugaku Expositions 1 (1988), 139--174.
• D. Mumford, J. Fogarty, and F. Kirwan, Geometric Invariant Theory, 3rd ed., Ergeb. Math. Grenzgeb. (2) 34, Springer, Berlin, 1994.
• K. G. O'Grady, Involutions and linear systems on holomorphic symplectic manifolds, Geom. Funct. Anal. 15 (2005), 1223--1274.
• —, Irreducible symplectic $4$-folds numerically equivalent to $(K3)$, preprint.
• C. Okonek, M. Schneider, and H. Spindler, Vector Bundles on Complex Projective Spaces, Progr. Math. 3, Birkhäuser, Boston, 1980.


text:
## Stream: maths

### Topic: Non-positive-definite inner product spaces

#### Eric Wieser (Feb 01 2021 at 09:59):

Currently, docs#inner_product_space requires positive definite-ness.
nLab suggests that this is not a necessary condition for something to be an "inner product space", while wikipedia suggests that it is a necessary condition.

Without wanting to argue about whether the mathlib definition is correct, what would be a suitable name for the version whose inner product is not-necessarily-positive definite?

#### Kevin Buzzard (Feb 01 2021 at 10:00):

"space equipped with a symmetric bilinear form"?

#### Kevin Buzzard (Feb 01 2021 at 10:02):

I've never seen the nLab's convention used anywhere else BTW

#### Andrea Berlingieri (Feb 01 2021 at 10:33):

Maybe it's a silly question, and slightly off-topic w.r.t. your question, but isn't Wikipedia's definition "problematic" when the inner product goes to $\mathbb{C}$ , since it is not linearly ordered and thus the positive definiteness condition doesn't make sense in that case?

#### Kevin Buzzard (Feb 01 2021 at 10:33):

in the C case you assume sesquilinearity so (x,x) is always real-valued.

#### Andrea Berlingieri (Feb 01 2021 at 10:34):

Oh, I see. Thanks.

#### Eric Wieser (Feb 01 2021 at 10:51):

To motivate my question - this comes up when wanting to work in minkowski spacetime, for instance, where one unit basis vector squares to -1. It would be nice to still get norm and inner notation(and maybe dist), and have at least the bits of inner_product_space that show they are consistent with each other

#### Yury G. Kudryashov (Feb 01 2021 at 14:44):

You can't have norm in this case because norm is a real-valued function, and in Minkowski spacetime some squared norms are negative.

#### Eric Wieser (Feb 01 2021 at 16:02):

Oh, that's a good point - and I guess dist also has a square-root in it that makes it undesirable there. So I guess I don't have a reason to care about lemmas for norm and dist

#### Jean-Philippe Burelle (Feb 04 2021 at 14:57):

Many useful lemmas actually only involve the squared norm, and can be useful in indefinite signature. For instance, the "Pythagorean theorem" works as-is. Cauchy-Schwarz can also be adapted, but there are different cases depending on the sign of the "squared norm" of the vectors.

Last updated: May 10 2021 at 07:15 UTC