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http://openstudy.com/updates/56194b18e4b086e96bfec409 | 1,477,062,544,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718284.75/warc/CC-MAIN-20161020183838-00245-ip-10-171-6-4.ec2.internal.warc.gz | 203,955,363 | 35,201 | ## anonymous one year ago Locate the absolute extrema of the function (if any exist) over the indicated variables.
1. anonymous
2. anonymous
This is how I solved a...
3. anonymous
I found the derivative of the function: x(4-x^2) I set it to zero... x=0 4-x^2=0 or x=2 I then plugged the following back into the original function: 2, 0, -2 And I received: 0, 2, 0 So I concluded that my max = 2 and my min = 0. Assuming that what I did was correct, how could I figure out the exact coordinates of my min and max without looking at a graph?
4. freckles
your derivative is just a little off
5. freckles
also 4-x^2=0 gives you x=2 or x=-2
6. anonymous
"your derivative is just a little off" What should it be? "also 4-x^2=0 gives you x=2 or x=-2" Yeah, that makes sense. I forgot about that rule.
7. freckles
you should have used chain rule
8. freckles
$((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'$
9. anonymous
I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o
10. anonymous
Now that I know that. How could I find the critical points for that derivative?
11. freckles
$f'(x)=\frac{-x}{\sqrt{4-x^2}}$ is this what you have for f'?
12. anonymous
Not even close unfortunately
13. freckles
you can find the critical numbers by finding when it is 0 or when it does not exist (these numbers should be in the domain of the original function though) so you still have to solve the equations x=0 and 4-x^2=0 so you are right your critical number are -2,0,2
14. freckles
wait what?
15. freckles
did you not understand how I got... $((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'$
16. anonymous
Yes, I actually solved it wrongly, and some how managed to get the right values. This is how I originally solved it...
17. anonymous
$\sqrt{4-x^2}$$(4-x^2)^\frac{1}{2}$$\frac{1}{2}(4-x^2)(2x)$$x(4-x^2)$ I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.
18. anonymous
Then I got... x=0 4-x^2=0 or x=2 , x=-2
19. freckles
Ok yeah but are you understand why this is $f'(x)=\frac{-x}{\sqrt{4-x^2}}$ ?
20. anonymous
I know how you get to $-x(4-x^2)^\frac{1}{2}$ is what you have just a simplified version of that?
21. freckles
well 1/2-1 is -1/2 not 1/2
22. anonymous
Sorry, typo
23. freckles
$-x(4-x^2)^\frac{-1}{2}=\frac{-x}{(4-x^2)^\frac{1}{2}} \text{ by law of exponents }$
24. freckles
$x^{-n}=\frac{1}{x^n}$
25. anonymous
There are so many laws I need to re-memorize.
26. anonymous
But yes I do recognize what you did.
27. freckles
anyways you have the right critical numbers x=-2,0,2 so we need to evaluate f(-2),f(0),f(1),f(2) I put f(1) because we need to evaluate it later anyways
28. anonymous
1?
29. freckles
$f(x)=\sqrt{4-x^2} \\ f(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?$
30. freckles
question d has 1 as an endpoint...
31. anonymous
Oh okay, I was just looking at a
32. freckles
I just wanted to get all the evaluation done with
33. freckles
we can look at a first ... I want to get through evaluating the function at the critical numbers and any endpoints
34. freckles
anyways can you evaluate those 3 above...
35. anonymous
x=2 would end up being 0 x=-1 would also be 0 x=0 would be 2 x=1 would be sqrt(3)
36. anonymous
That was supposed to say -2 not -1
37. freckles
f(-2)=0 f(0)=2 f(2)=0 f(1)=sqrt(3) great work
38. freckles
so look at the outputs above
39. freckles
0,2,0
40. freckles
what is biggest number there
41. anonymous
2
42. freckles
so that is your absolute max
43. freckles
0,2,0 and of these numbers 0 is the absolute min
44. anonymous
That makes sense, but how would I represent that max as coordinates?
45. freckles
which this makes sense... because the graph is the top part of a circle... |dw:1444500172873:dw|
46. freckles
well you already found the coordinates
47. freckles
f(x)=y means (x,y) is a coordinate on f
48. freckles
for example we said f(2)=0 so that means (2,0) is a coordinate on f
49. anonymous
Oh so max would be 0,2
50. freckles
(0,2) would be max ordered pairs have ( ) around them
51. freckles
|dw:1444500273762:dw|
52. anonymous
so could (-2,0) and (2,0) both be min values?
53. freckles
yep
54. anonymous
Because b has a parentheses instead of a bracket, would it be solved for differently?
55. freckles
now we can look at b we have [-2,0) we only care about f(-2) and f(0) here but we cannot actually include what happens at x=0 f(-2)=0 and f(0)=2 is what we found earlier of 0 and 2 the smallest is 0 so your absolute min is at x=0 and we will not include an absolute max for this question because we cannot include what happens at x=0
56. anonymous
Why can't we include 0?
57. freckles
[-2,0) means we have everything between -2 and 0 including -2 but not including 0
58. freckles
|dw:1444500569732:dw| this function on [-2,0) looks like: |dw:1444500587078:dw| my quarter of a circle was horribly drawn I blame it on the cat licking me
59. anonymous
[ means included ( means to that point but the point isn't included itself
60. anonymous
Correct?
61. freckles
see the absolute max doesn't actually exist there is a hole at x=0 because the domain didn't allow us to include what happens at x=0
62. anonymous
Ah okay
63. anonymous
and for C it would just be solving for 0?
64. freckles
brackets means we include the endpoint parenthesis means we don't include the endpoint
65. freckles
we already looked at a and said the absolute mins were (-2,0) and (2,0) and the absolute max was (0,2) c would be the same answer except the absolute mins would not exist since we have (-2,2) and not [-2,2]
66. anonymous
So only the max would exist in that particular problem.
67. freckles
yep
68. anonymous
Should i just state that it cannot be determined or is there a specific term I should use?
69. freckles
I think the word choice I used was good "does not exist" |dw:1444500908510:dw| this graph on [1,2) looks like: |dw:1444500919987:dw|
70. anonymous
Thank you so much for your help! :D
71. freckles
what did you get for b again
72. anonymous
max: does not exist min: (-2, 0)
73. freckles
great
74. freckles
75. anonymous
just a second
76. anonymous
max: sqrt(3) min: does not exist That may or may not be right.
77. freckles
did you still want the coordinates or will the y value just be okay ?
78. freckles
like you can say the max is sqrt(3) but if your teacher prefers coordinates then you should say (1,sqrt(3))
79. anonymous
Yeah that's what I'll put
80. freckles
ok ok that sounds great to me
81. anonymous
excellent! It's surprising how simple this stuff really is compared to how difficult it seems at times
82. freckles
It is really not that bad... You do have to train yourself back on your algebra though.. The calculus I think is easier than the algebra :p
83. anonymous
I do the worst on trig identities personally | 2,189 | 7,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-44 | longest | en | 0.905482 |
https://www.michiganlearning.org/standard/mi-math-content-3-md-c-7/ | 1,726,225,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00224.warc.gz | 808,927,104 | 20,899 | Geometric measurement: understand concepts of area and relate area to multiplication and to addition
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Join Mrs. Askew for a Number talk with 2 Math Might Friends! Then let's solve some problems with equal groups that have larger numbers! | 280 | 1,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.900502 |
http://www.physicsforums.com/showthread.php?s=19f894722fa089abd9fe496f61a29744&p=4275260 | 1,371,625,799,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708143620/warc/CC-MAIN-20130516124223-00078-ip-10-60-113-184.ec2.internal.warc.gz | 613,845,396 | 8,960 | ## help is needed in understanding my telescope
Hey, I'm trying to start learning about astronomy with my brother's old telescope and I need a little help in getting started :).
Firstly, what my textbook tells me is that the declination scale shows a scale of numbers which go from zero to nine and then back down to zero again, however I thought that declination was measured in degrees or arc-seconds? What do these numbers mean?
How do I know what right ascension I am looking at in the sky? So, I've just set the declination to 90 degrees. I'm lining the polar axis of my telescope to be parallel with the Earth's axis by pointing it towards North, after adjusting for my specific latitude, how do I know my right ascension? Can I find the right ascension by pointing at an object with a know right ascension? Really, I don't think I understand right ascension...
My guide talks about a 'horizontal axis', for example; 'loosen the horizontal axis lock and turn the telescope so it is directly aimed at Polaris'. What does this term mean?
Where should my counter-weight be pointing when I am looking at the sky?
PhysOrg.com astronomy news on PhysOrg.com >> Three centaurs follow Uranus through the solar system>> Final curtain for Europe's deep-space telescope>> Hubble spots a very bright contortionist
For useful help contact you local amateur astronomer club. They can help you so much better hands on than anyone can do over the internet.
Also your questions are basically impossible to answer without more information (maker and model and preferable illustrative pictures) but I'll try my best guesses though. First it looks like you got a equatorial mount of some kind.
Quote by grobertsx Hey, I'm trying to start learning about astronomy with my brother's old telescope and I need a little help in getting started :). Firstly, what my textbook tells me is that the declination scale shows a scale of numbers which go from zero to nine and then back down to zero again, however I thought that declination was measured in degrees or arc-seconds? What do these numbers mean?
Hard to tell without more information. My best guess is that scale is probably the latitude scale used in the first step of polar aligment. It's probably measured in tens of degrees with one degree increments. This has nothing to do with declination though and most scopes lack a useful declination scale.
[quote]How do I know what right ascension I am looking at in the sky? So, I've just set the declination to 90 degrees. I'm lining the polar axis of my telescope to be parallel with the Earth's axis by pointing it towards North, after adjusting for my specific latitude, how do I know my right ascension? Can I find the right ascension by pointing at an object with a know right ascension? Really, I don't think I understand right ascension...[quote]
RA is just you basic East-West coordinate system but for the sky instead. As with declination most scopes lack a useful RA scale. The good thing is that you don't need one and you usually don't really need to understand it either.
My guide talks about a 'horizontal axis', for example; 'loosen the horizontal axis lock and turn the telescope so it is directly aimed at Polaris'. What does this term mean?
Definitaly a term I've never run across. Sound like it could be either the polar axis, declination axis or right ascention axis. Needs more information.
Where should my counter-weight be pointing when I am looking at the sky?
Depends entirely on where the telescope is pointing.
/Patrik
Thanks for your help :) x If it sheds more light on things: it is placed on an equatorial mount, it is called the StarLight 80 refractor telescope and it's a Meade. Best x
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## help is needed in understanding my telescope
The horizontal axis of the mount is the part you rotate so the polar axis faces north [toward polaris]. Then you adjust the polar axis so it points upward at the same angle as your geological latitude. The declination of objects in the sky does not change, the right ascension changes by the hour. Every 24 hours the earth completes one revolution, causing the right ascension of all objects in the sky to also complete one revolution.
Quote by grobertsx Thanks for your help :) x If it sheds more light on things: it is placed on an equatorial mount, it is called the StarLight 80 refractor telescope and it's a Meade.
Seems like the scope has disappeared from the Meade homepage without a trace, not even a manual.
However from pictures it looks like the mount is a EQ-1 clone so any tutorial for a beginners german equatorial mount should help.
Some examples: | 1,006 | 4,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.943038 |
https://studyadda.com/solved-papers/rajasthan-pet-solved-paper-2012_q4/467/223251 | 1,576,290,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00138.warc.gz | 545,181,975 | 19,683 | • # question_answer An electron starting from rest has a velocity$v$ that increases linearly with time t as$v=kt,$ where$k=2\text{ }m{{s}^{-2}}$. The distance covered by electron in first 3 s is A) 9m B) 11 m C) 33m D) 40m
Given, $v=kt$ Also $k=2m{{s}^{-2}}$ $\therefore$ $v=2t$ or $\frac{dx}{dt}=2t$ On integration of both sides $x={{t}^{2}}+c$ When,$t=0,$then $x=0$ $\therefore$ $0={{(0)}^{2}}+c$ $\Rightarrow$ $c=0$ So, we get $x={{t}^{2}}$ When $t=3~$ $\Rightarrow$ $x={{(3)}^{2}}=9m$ | 216 | 547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-51 | latest | en | 0.396376 |
http://mathhelpforum.com/geometry/94774-geo-circle-formulas-helps.html | 1,526,972,698,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864626.37/warc/CC-MAIN-20180522053839-20180522073839-00290.warc.gz | 189,743,002 | 11,470 | # Thread: Geo Circle Formulas Helps
1. ## Geo Circle Formulas Helps
I have a pizza. The radius is 10 inches long. The pizza was cut into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the crust more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain all topping, and the outer piece would contain some topping and some crust.
1. Find the area of the whole pizza.10 x 10 = 100
2. What is the area of one piece of pizza?100/16=6.25^2=39.0625
3. What is the area of a half-piece?39.0625/2=19.53125
4. What would the area of the whole pizza be if it were made of half pieces?19.53125^2= 95.367431640625
5. What is the radius of a half-piece? (ie, where do I need to cut to make two equal halves out of a piece?)9.765625
If I am doing this this wrong it would be useful to attain the formulas needed to complete these problems. Any help much appreciated.
2. Originally Posted by KevinVM20
I have a pizza. The radius is 10 inches long. The pizza was cut into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the crust more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain all topping, and the outer piece would contain some topping and some crust.
1. Find the area of the whole pizza.10 x 10 = 100
Is the pizza a square or a circle? You said the pizza's radius is 10 inches long, so that implies a circle. But then you wrote "10 x 10 = 100," which implies a square.
You probably meant circle, so
$\displaystyle A = \pi r^2 = \pi \cdot 10^2 = 100\pi$
01
3. I understand that part. Are the rest of my calculations correct?
Originally Posted by yeongil
Is the pizza a square or a circle? You said the pizza's radius is 10 inches long, so that implies a circle. But then you wrote "10 x 10 = 100," which implies a square.
You probably meant circle, so
$\displaystyle A = \pi r^2 = \pi \cdot 10^2 = 100\pi$
01
4. Originally Posted by KevinVM20
I have a pizza. The radius is 10 inches long. The pizza was cut into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the crust more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain all topping, and the outer piece would contain some topping and some crust.
1. Find the area of the whole pizza.10 x 10 = 100
See yeongil's post
2. What is the area of one piece of pizza?
$\displaystyle A_{piece}=\dfrac1{16} \cdot \pi \cdot r^2$
3. What is the area of a half-piece?
$\displaystyle A_{\frac12 p}=\dfrac12 \cdot \dfrac1{16} \cdot \pi \cdot r^2$
4. What would the area of the whole pizza be if it were made of half pieces?
I don't understand this question, sorry!
5. What is the radius of a half-piece? (ie, where do I need to cut to make two equal halves out of a piece?)
I assume that the central angle of the half-piece (that's the angle of the apex) is the same as the angle of a complete piece. Let denote $\displaystyle \rho$ the radius of the half-piece:
$\displaystyle A_{\frac12 p}=\dfrac1{16} \cdot \pi \cdot \underbrace{\dfrac12 \cdot r^2}_{this\ is\ \rho^2}$
$\displaystyle \rho^2=\dfrac12 r^2~\implies~ \rho=\dfrac r2 \cdot \sqrt{2}$
In your case $\displaystyle \rho \approx 7.07$
If I am doing this this wrong it would be useful to attain the formulas needed to complete these problems. Any help much appreciated.
Did it help?
5. It is all starting to click. However, I do not see how you got the 7.07.
Originally Posted by earboth
See yeongil's post
$\displaystyle A_{piece}=\dfrac1{16} \cdot \pi \cdot r^2$
$\displaystyle A_{\frac12 p}=\dfrac12 \cdot \dfrac1{16} \cdot \pi \cdot r^2$
I don't understand this question, sorry!
I assume that the central angle of the half-piece (that's the angle of the apex) is the same as the angle of a complete piece. Let denote $\displaystyle \rho$ the radius of the half-piece:
$\displaystyle A_{\frac12 p}=\dfrac1{16} \cdot \pi \cdot \underbrace{\dfrac12 \cdot r^2}_{this\ is\ \rho^2}$
$\displaystyle \rho^2=\dfrac12 r^2~\implies~ \rho=\dfrac r2 \cdot \sqrt{2}$
In your case $\displaystyle \rho \approx 7.07$
Did it help?
6. Originally Posted by KevinVM20
It is all starting to click. However, I do not see how you got the 7.07.
I have a pizza. The radius is 10 inches long. ...
If $\displaystyle \rho=\dfrac r2 \cdot \sqrt{2}$ and $\displaystyle r = 10$ then $\displaystyle \rho=\dfrac{10}2 \cdot \sqrt{2} \approx 7.071067812...$
I rounded this value to $\displaystyle \rho = 7.07$
7. Originally Posted by KevinVM20
4. What would the area of the whole pizza be if it were made of half pieces?
Seems to me like a bit of a trick question; the total area is the total area regardless of how it is sliced. So it's still $\displaystyle 100\pi,$ no? | 1,467 | 5,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-22 | latest | en | 0.962441 |
https://www.physicsforums.com/threads/yo-yo-question.143614/ | 1,508,446,583,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00182.warc.gz | 957,906,329 | 15,094 | # Yo-yo question
1. Nov 13, 2006
### vu10758
I have the question at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1423959807
I have to determine what happens if each force is applied.
This is what I think.
If force one is applied, the yoyo will rotate clockwise, and friction will push it to the right.
If force three is applied, the yoyo will rotate counterclockwise and friction will push it to the left.
If force two is applied, the forces on the yoyo is zero and it will not move since friction and the force will balance each other out.
2. Nov 13, 2006
### OlderDan
I think you will need to quantify your answers to get these all correct. Assume friction is acting and is large enough to prevent slipping. If that assumption proves unreasonable in any case, then slipping should be included.
3. Nov 13, 2006
### vu10758
The question specified that I am to assume that the yoyo roll smoothly without slipping.
4. Nov 13, 2006
### OlderDan
I wasn't sure if the handwritten part was in the problem, or an assumption. I suggest you put two different radii (inner and outer) in your diagram and assume the yo-yo is essentially a disk to find its moment of inertia. It's not absolutely necessary to do this, but I think it will help you analyze the situations.
5. Nov 13, 2006
Thank you. | 339 | 1,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-43 | longest | en | 0.931518 |
https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/North_by_east.html | 1,563,369,419,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00006.warc.gz | 431,473,744 | 22,411 | # Points of the compass
"Compass point" redirects here. For other uses, see Compass Point (disambiguation).
A 32-point compass rose
The points of the compass, specifically on the compass rose, mark divisions of a compass; such divisions may be referred to as "winds" or "directions". A compass point allows reference to a specific heading (or course or azimuth) in a general or colloquial fashion, without having to compute or remember degrees.
A compass is primarily divided into the four cardinal pointsnorth, south, east, and west. These are often further subdivided by the addition of the four intercardinal (or ordinal) directions — northeast (NE) between north and east, southeast (SE), southwest (SW), and northwest (NW) - to indicate the eight principal winds. In meteorological usage, further intermediate points between cardinal and ordinal points, such as north-northeast (NNE) between north and northeast, are added to give the sixteen points of a wind compass.[1] At the most complete division in European tradition, are the full thirty-two points of the mariner's compass,[2] which adds points such as north by east (NbE) between north and north-northeast, and northeast by north (NEbN) between north-northeast and northeast.
Although the European nautical tradition retained the term "one point" to describe 132 of a circle (in such phrases as "two points to starboard"), by the middle of the eighteenth century, the 32-point system was further extended with half- and quarter-points to allow 128 directions to be differentiated.[3] For most applications, the fractional points have been superseded by degrees measured clockwise from North.
In ancient China 24 points of the compass were used, measuring fifteen degrees between points.
## Compass point names
A 16-point compass rose
The names of the compass directions follow the 32-point wind compass rose follow these rules:
• The cardinal directions are north (N), east (E), south (S), west (W), at 90° angles on the compass rose.
• The ordinal (or intercardinal) directions are northeast (NE), southeast (SE), southwest (SW) and northwest (NW), formed by bisecting the angle of the cardinal winds. The name is merely a combination of the cardinals it bisects.
• The eight principal winds (or main winds) are the cardinals and ordinals considered together, that is N, NE, E, SE, S, SW, W, NW. Each principal wind is 45° from its neighbour. The principal winds form the basic eight-wind compass rose.
### 16-wind compass rose
• The eight half-winds are the points obtained by bisecting the angles between the principal winds. The half-winds are north-northeast (NNE), east-northeast (ENE), east-southeast (ESE), south-southeast (SSE), south-southwest (SSW), west-southwest (WSW), west-northwest (WNW) and north-northwest (NNW). Notice that the name is constructed simply by combining the names of the principal winds to either side, with the cardinal wind coming first, the ordinal wind second. The eight principal winds and the eight half-winds together yield a 16-wind compass rose, with each compass point at a 22 12° angle from the next.
### 32-wind compass points
A 32-wind compass card, with English names
All of the above named points plus the sixteen quarter winds listed in the next paragraph define the 32 points of the wind compass rose.
• The sixteen quarter winds are the direction points obtained by bisecting the angles between the points on a 16-wind compass rose. The sixteen quarter-winds are north by east (NbE), northeast by north (NEbN), northeast by east (NEbE), east by north (EbN) in the first quadrant, east by south (EbS), southeast by east (SEbE), southeast by south (SEbS), south by east (SbE) in the second quadrant, south by west (SbW), southwest by south (SWbS), southwest by west (SWbW), west by south (WbS) in the third quadrant, and finally west by north (WbN), northwest by west (NWbW), northwest by north (NWbN) and north by west (NbW) in the fourth quadrant.[4][5]
The name of a quarter-wind is "X by Y", where X is a principal wind and Y is a cardinal wind. As a mnemonic device, it is useful to think of "X by Y" as a shortcut for the phrase "one quarter wind from X towards Y", where a "quarter" is 11 14°, X is the nearest principal wind, and Y the next (more distant) cardinal wind. So "northeast by east" means "one quarter from NE towards E", "southwest by south" means "one quarter from SW towards S". The eight principal winds, eight half-winds and sixteen quarter winds together yield a 32-wind compass rose, with each compass direction point at 11 14° angle from the next. In the mariner's exercise of boxing the compass, all thirty-two points of the compass are named in clockwise order.[6]
The title of the Alfred Hitchcock 1959 movie, North by Northwest, is actually not a direction point on the 32-wind compass, but the film contains a reference to Northwest Airlines. Similarly, the names of the two film festivals South by Southwest and North by Northeast are not 32-wind compass points; a quarter wind whose name contains both a cardinal and an ordinal direction is named with the ordinal direction first.
The traditional compass rose of eight winds (and its 16-wind and 32-wind derivatives) was invented by seafarers in the Mediterranean Sea during the Middle Ages (the ancient Greco-Roman 12 classical compass winds have little to do with them). The traditional mariner's wind names were expressed in Italian – or, more precisely, the Italianate Mediterranean lingua franca common among sailors in the 13th and 14th centuries, that was principally composed of Genoese (Ligurian), mixed with Venetian, Sicilian, Provençal, Catalan, Greek and Arabic terms from around the Mediterranean basin.
This Italianate patois was used to designate the names of the principal winds on the compass rose found in mariner compasses and portolan charts of the 14th and 15th centuries. The "traditional" names of the eight principal winds are:
• (N) – Tramontana
• (NE) – Greco (or Bora in some Venetian sources)
• (E) – Levante (sometimes Oriente)
• (SE) – Scirocco (or Exaloc in Catalan)
• (S) – Ostro (or Mezzogiorno in Venetian)
• (SW) – Libeccio (or Garbino, Eissalot in Provençal)
• (W) – Ponente (or Zephyrus in Greek)
• (NW) – Maestro (or Mistral in Provençal)
Local spelling variations are far more numerous than listed, e.g. Tramutana, Gregale, Grecho, Sirocco, Xaloc, Lebeg, Libezo, Leveche, Mezzodi, Migjorn, Magistro, Mestre, etc. Traditional compass roses will typically have the initials T, G, L, S, O, L, P, and M on the main points. Portolan charts also colour-coded the compass winds: black for the eight principal winds, green for the eight half-winds and red for the sixteen quarter-winds.
In the English compass, all wind names are constructed on the basis of the cardinal four names (N, E, S, W). In the traditional compass, one needs to memorize eight basic names – one for each of the eight principal winds (N, NE, E, SE, S, SW, W, NW.) While there are more names to memorize, the payoff is that the name construction rules for the 32-wind compass are more straightforward. The half-winds are just a combination of the two principal winds it bisects, with the shortest name usually coming first (e.g. NNE is "Greco-Tramontana", ENE is "Greco-Levante", SSE is "Ostro-Scirocco", etc.). The quarter winds are expressed with an Italian phrase, "Quarto di X verso Y" (one quarter from X towards Y, pronounced [ˈkwarto di ˈiks ˈvɛrso ˈipsilon][7][8][9]) or "X al Y" (X to Y) or "X per Y" (X by Y). There are no irregularities to trip over: the nearest principal wind always comes first, the more distant one second, e.g. North-by-east is "Quarto di Tramontana verso Greco", northeast-by-north "Quarto di Greco verso Tramontana". The names are perfectly symmetric.
## 32 cardinal points
# Compass point Abbreviation Traditional wind point Minimum Middle
azimuth
Maximum
1 North N Tramontana 354.38° 0.00° 5.62°
2 North by east NbE Quarto di Tramontana verso Greco 5.63° 11.25° 16.87°
3 North-northeast NNE Greco-Tramontana 16.88° 22.50° 28.12°
4 Northeast by north NEbN Quarto di Greco verso Tramontana 28.13° 33.75° 39.37°
5 Northeast NE Greco 39.38° 45.00° 50.62°
6 Northeast by east NEbE Quarto di Greco verso Levante 50.63° 56.25° 61.87°
7 East-northeast ENE Greco-Levante 61.88° 67.50° 73.12°
8 East by north EbN Quarto di Levante verso Greco 73.13° 78.75° 84.37°
9 East E Levante 84.38° 90.00° 95.62°
10 East by south EbS Quarto di Levante verso Scirocco 95.63° 101.25° 106.87°
11 East-southeast ESE Levante-Scirocco 106.88° 112.50° 118.12°
12 Southeast by east SEbE Quarto di Scirocco verso Levante 118.13° 123.75° 129.37°
13 Southeast SE Scirocco 129.38° 135.00° 140.62°
14 Southeast by south SEbS Quarto di Scirocco verso Ostro 140.63° 146.25° 151.87°
15 South-southeast SSE Ostro-Scirocco 151.88° 157.50° 163.12°
16 South by east SbE Quarto di Ostro verso Scirocco 163.13° 168.75° 174.37°
17 South S Ostro 174.38° 180.00° 185.62°
18 South by west SbW Quarto di Ostro verso Libeccio 185.63° 191.25° 196.87°
19 South-southwest SSW Ostro-Libeccio 196.88° 202.50° 208.12°
20 Southwest by south SWbS Quarto di Libeccio verso Ostro 208.13° 213.75° 219.37°
21 Southwest SW Libeccio 219.38° 225.00° 230.62°
22 Southwest by west SWbW Quarto di Libeccio verso Ponente 230.63° 236.25° 241.87°
23 West-southwest WSW Ponente-Libeccio 241.88° 247.50° 253.12°
24 West by south WbS Quarto di Ponente verso Libeccio 253.13° 258.75° 264.37°
25 West W Ponente 264.38° 270.00° 275.62°
26 West by north WbN Quarto di Ponente verso Maestro 275.63° 281.25° 286.87°
27 West-northwest WNW Maestro-Ponente 286.88° 292.50° 298.12°
28 Northwest by west NWbW Quarto di Maestro verso Ponente 298.13° 303.75° 309.37°
29 Northwest NW Maestro 309.38° 315.00° 320.62°
30 Northwest by north NWbN Quarto di Maestro verso Tramontana 320.63° 326.25° 331.87°
31 North-northwest NNW Maestro-Tramontana 331.88° 337.50° 343.12°
32 North by west NbW Quarto di Tramontana verso Maestro 343.13° 348.75° 354.37°
## Half- and quarter-points
Compass Rose from "American Practical Navigator" 1916
By at least the middle of the eighteenth century the 32-point system had been further extended with the use of half- and quarter-points to give a total of 128 directions.[3] These fractional points are named by appending, for example 1/4east, 1/2east, or 3/4east to the name of one of the 32 points. Each of the 96 fractional points can be named in two ways, depending of which of the two adjoining whole points is used, for example, N3/4E is equivalent to NbE1/4N. Either form is easily understood but differing conventions as to correct usage developed in different countries and organisations.
The table below shows how each of the 128 directions are named. The first two columns give the number of points and degrees clockwise from north. The third gives the equivalent bearing to the nearest degree from north or south towards east or west. The "CW" column gives the fractional-point bearings increasing in the clockwise direction and "CCW" counter clockwise. The final three columns show three common naming conventions: No "by" avoids the use of "by" with fractional points; "USN" the system used by US Navy;[10] and "RN" the Royal Navy.[11] The colour coding illustrates where each of the three naming systems matches the "CW" and "CCW" columns.
Compass roses very rarely named the fractional points and only showed small, unlabelled markers as a guide for helmsmen.
## References
1. David Boardman Graphicacy and Geography Teaching 1983 – Page 41 "In particular they should learn that wind direction is always stated as the direction from which, and not to which, the wind is blowing. Once children have grasped these eight points they can learn the full sixteen points of the compass."
2. Evans, Frederick John (1859). "Notes on the Magnetism of Ships". Pamphlets on British shipping. 1785–1861. By unknown editor. p. 8 (p. 433 of PDF). ISBN 0-217-85167-3. A deviation table having been formed by any of the processes now.so generally understood, either on the thirty-two points of the compass, the sixteen intermediate, or the eight principal points
3. E. Chambers Cyclopaedia: or, an Universal Dictionary of Arts and Science, 5th Ed, 1743, pp. 206–7, "Points of the Compass, or Horizon, &c, in Geography and Navigation, are the points of division when the whole circle, quite around, is divided into 32 equal parts. These points are therefore at the distance of the 32d part of the circult, or 11°15′, from each other; hence 5°37 1/2′ is the distance of the half points and 2°48 3/4′ is the distance of the quarter points. | 3,515 | 12,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-30 | latest | en | 0.962504 |
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13th August 2014, 07:27 PM #1 shredhead diyAudio Member Join Date: Aug 2011 Location: NC How do I test an amp for Damping Factor? Yeah, I get that damping factor is useless and it's really a matter of output impedance. If I understand correctly, output impedance of an amplifier varies with frequency? So that would mean that it would affect the frequency response of the amp right? I'm wondering how you would do an impedance sweep of an amplifier's output. Does anyone know the standard test that determines an amp's DF at a given frequency? Or do these companies just make up numbers?
13th August 2014, 07:44 PM #2 Mooly diyAudio Moderator Join Date: Sep 2007 Well if you look it up you'll see damping factor is the ratio of the load impedance to the amplifiers output impedance. So a rough and ready way to estimate... if you set the amp to give (say) 10 volts rms output (with no speakers attached) as measured on a DVM (use something like 400Hz which any DVM should cope with), and then load the amp with a known resistance as a load (say an 8 ohm or 10 ohm resistor of suitable rating) you will see the 10 volts drop a little as measured at the speaker terminals. If we assume the amp internally is "perfect" as a voltage source then you can work out the damping factor. For example if you measured 10 volts and connecting a 10 ohm gave 9.8 volts then the internal resistance within the amp is dropping 0.2 volts. The perfect amp is delivering current I of 10/10 (I=V/R) which is 1 amp. The internal resistance is R=V/I which is 0.2/1 = 0.2 ohm. So the ratio 0.2 to 10 is 50 which would be the damping factor. That is a bit simplistic and there are other ways to measure damping factor more accurately and over a wider frequency range but if you only have a DVM it will get you in the right ballpark Try the same test at the end of the speaker leads too such that they are in circuit as well and their resistance is included. Now that amplifier with a claimed 500 damping factor isn't so much different to one of only 30 or 40.
13th August 2014, 09:34 PM #3 jcx diyAudio Member Join Date: Feb 2003 Location: .. "tug-of-war" with a resistor load, another amp on the other end driven with your test signal such as a frequency sweep measure the V a your amp's output a soundcard is good down to its crosstalk limit for such a test
huseying
diyAudio Member
Join Date: Jan 2008
Location: İstanbul/Turkey
Quote:
Originally Posted by Mooly Well if you look it up you'll see damping factor is the ratio of the load impedance to the amplifiers output impedance. So a rough and ready way to estimate... if you set the amp to give (say) 10 volts rms output (with no speakers attached) as measured on a DVM (use something like 400Hz which any DVM should cope with), and then load the amp with a known resistance as a load (say an 8 ohm or 10 ohm resistor of suitable rating) you will see the 10 volts drop a little as measured at the speaker terminals. If we assume the amp internally is "perfect" as a voltage source then you can work out the damping factor. For example if you measured 10 volts and connecting a 10 ohm gave 9.8 volts then the internal resistance within the amp is dropping 0.2 volts. The perfect amp is delivering current I of 10/10 (I=V/R) which is 1 amp. The internal resistance is R=V/I which is 0.2/1 = 0.2 ohm. So the ratio 0.2 to 10 is 50 which would be the damping factor. That is a bit simplistic and there are other ways to measure damping factor more accurately and over a wider frequency range but if you only have a DVM it will get you in the right ballpark Try the same test at the end of the speaker leads too such that they are in circuit as well and their resistance is included. Now that amplifier with a claimed 500 damping factor isn't so much different to one of only 30 or 40.
Thats a good way of testing amps' damping, i usually use this method but shredhead please keep in mind that the longer the cables wich you use to connect resistor to amp, the more wrong damping value you get , and also to decrease the cable's resistance use tick cable as posible as you can.
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Last edited by huseying; 15th August 2014 at 12:39 PM.
jan.didden
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Join Date: May 2002
Location: The great city of Turnhout, BE
Quote:
Originally Posted by Mooly For example if you measured 10 volts and connecting a 10 ohm gave 9.8 volts then the internal resistance within the amp is dropping 0.2 volts. The perfect amp is delivering current I of 10/10 (I=V/R) which is 1 amp. The internal resistance is R=V/I which is 0.2/1 = 0.2 ohm. So the ratio 0.2 to 10 is 50 which would be the damping factor.
Mooly, agree with most of your post but not on this. If the voltage drops to 9.8V with a 10 ohms load the Iout is not 1A but 0.98A...
Just to split some hairs
Jan
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15th August 2014, 02:07 PM #6 Mooly diyAudio Moderator Join Date: Sep 2007 Mea culpa The train of thought wandered while typing and wielding Windows on screen calculator. Thanks
sgrossklass
diyAudio Member
Join Date: Nov 2004
Location: Germany
That's how you do it in theory and practice:
Quote:
Originally Posted by jcx "tug-of-war" with a resistor load, another amp on the other end driven with your test signal such as a frequency sweep measure the V a your amp's output
Note that you can basically do this with two channels of one stereo amplifier as well, i.e. the resistor goes between R+ and L+, with playback on L and measurement on R. The resistor should be >10x the expected output impedance (current source approximation), but that shouldn't be an issue since you need something in the usual loudspeaker impedance range anyway. Sufficient load handling advised, shouldn't need to be anything too extreme though.
16th August 2014, 06:38 PM #8 shredhead diyAudio Member Join Date: Aug 2011 Location: NC I want to test a new amp and an old one to see the differences in the 20-40Hz area but for the old one I'm scared that because it doesn't really have much in the area of protection circuits, it might release the magic smoke. Does anyone here have any experience in doing these 2 kinds of tests to an old quasi-complimentary dinosaur? Would one test be safer than the other?
16th August 2014, 07:58 PM #9 Mooly diyAudio Moderator Join Date: Sep 2007 Just do the test at a couple of volts and then there is no danger of anything overheating. If you add a low value resistor (say 0.1 to 0.47) in series with the speakers then you automatically decrease damping factor down to at least 80 with a 0.1 ohm, and to just 17 with a 0.47 ohm. That assumes 8 ohm speakers and a perfect amplifier with a damping factor of zillions. Try it.
16th August 2014, 11:43 PM #10 Kiriakos Banned Join Date: Jul 2014 Location: In from of my workbench There is no chance getting a good measurement with an multimeter in the range of 400Hz and up to 20k or 40kHz. This is an area only for modern Oscilloscopes
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# Vorticity in 2D and the Streamfunction
Let
(6.9)
The constraint can be satisfied by using a vector potential for (c.f., .
(6.10)
For two dimensional flow, need only have a non-zero -component, e.g., .
(6.11)
Note that . Thus and has the components
(6.12)
[check that is incompressible: ]
The scalar is called streamfunction since a contour const. is aligned with the velocity - i.e. it is a streamline. See Figure (6.1).
In 2D, .
Thus , and
(6.13)
given , we can solve (6.11) for , then use (6.9) to find .
Consequently, equation (6.8) reduces, in 2D to
(6.14)
in a given fluid element remains constant, since the convective derivative is zero, as illustrated in Figure (6.2).
(6.15)
Example 6.1 Rankine Vortex.
Use coordinates, rather than : , still. The geometry of the Rankine Vortex is illustrated in Figure (6.3) and is defined to be
since is axisymmetric we look for solutions of the form
(6.16)
( const)
Dividing by , followed by integration we find
since must be regular at
Now find using (6.16):
(This is similar to solid body'' rotation with angular velocity .)
(6.13)
(6.16) .
is continuous at , i.e.,
.
Hence the full solution for is and (see Figure (6.4))
(6.17)
Next: Circulation and point vortices Up: Vorticity Previous: Evolution of vorticity
Andrew Wright
2002-09-16 | 402 | 1,414 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-05 | latest | en | 0.894179 |
https://algebra-class-ecourse.com/question/select-from-the-drop-down-menu-correctly-complete-each-statement-the-opposite-of-3-5-8-is-on-the-11305097-44/ | 1,637,973,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00366.warc.gz | 161,685,377 | 12,518 | ## Select from the drop down menu correctly complete each statement. The opposite of – 3 5/8 is on the ( same or opposite which one is it ) si
Question
Select from the drop down menu correctly complete each statement. The opposite of – 3 5/8 is on the ( same or opposite which one is it ) side of zero on a number line as -3 5/8. the opposite of 4 2/9 on the ( same or opposite which one is it) side of zero on a number line as 4 2/9
0 | 123 | 438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-49 | latest | en | 0.92765 |
https://www.liuchuo.net/archives/4099 | 1,527,363,942,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00586.warc.gz | 784,505,579 | 14,239 | # 1135. Is It A Red-Black Tree (30)-PAT甲级真题
There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties:
(1) Every node is either red or black.
(2) The root is black.
(3) Every leaf (NULL) is black.
(4) If a node is red, then both its children are black.
(5) For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.
For example, the tree in Figure 1 is a red-black tree, while the ones in Figure 2 and 3 are not.
For each given binary search tree, you are supposed to tell if it is a legal red-black tree.
Input Specification:
Each input file contains several test cases. The first line gives a positive integer K (<=30) which is the total number of cases. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the preorder traversal sequence of the tree. While all the keys in a tree are positive integers, we use negative signs to represent red nodes. All the numbers in a line are separated by a space. The sample input cases correspond to the trees shown in Figure 1, 2 and 3.
Output Specification:
For each test case, print in a line “Yes” if the given tree is a red-black tree, or “No” if not.
Sample Input:
3
9
7 -2 1 5 -4 -11 8 14 -15
9
11 -2 1 -7 5 -4 8 14 -15
8
10 -7 5 -6 8 15 -11 17
Sample Output:
Yes
No
No
1.是否是二叉搜索树
2.根结点是否为黑色
3.如果一个结点是红色，它的孩子节点是否都为黑色
4.从任意结点到叶子结点的路径中，黑色结点的个数是否相同
0. 根据先序建立一棵树，用链表表示
1. 根据先序遍历，通过递归和二叉搜索树的性质得到后序post数组，若post的大小不等于n说明不是二叉搜索树～【getPost函数】
2. 判断根结点（题目所给先序的第一个点即根结点）是否是黑色【arr[0] < 0】
3. 根据建立的树，从根结点开始遍历，如果当前结点是红色，判断它的孩子节点是否为黑色，递归返回结果【judge1函数】
4. 从根节点开始，递归遍历，检查每个结点的左子树的高度和右子树的高度（这里的高度指黑色结点的个数），比较左右孩子高度是否相等，递归返回结果【judge2函数】 | 656 | 1,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-22 | longest | en | 0.607567 |
http://metamath.tirix.org/mpests/carden | 1,718,966,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00587.warc.gz | 19,365,960 | 3,293 | # Metamath Proof Explorer
## Theorem carden
Description: Two sets are equinumerous iff their cardinal numbers are equal. This important theorem expresses the essential concept behind "cardinality" or "size." This theorem appears as Proposition 10.10 of TakeutiZaring p. 85, Theorem 7P of Enderton p. 197, and Theorem 9 of Suppes p. 242 (among others). The Axiom of Choice is required for its proof. Related theorems are hasheni and the finite-set-only hashen .
This theorem is also known as Hume's Principle. Gottlob Frege's two-volumeGrundgesetze der Arithmetik used his Basic Law V to prove this theorem. Unfortunately Basic Law V caused Frege's system to be inconsistent because it was subject to Russell's paradox (see ru ). Later scholars have found that Frege primarily used Basic Law V to Hume's Principle. If Basic Law V is replaced by Hume's Principle in Frege's system, much of Frege's work is restored.Grundgesetze der Arithmetik, once Basic Law V is replaced, proves "Frege's theorem" (the Peano axioms of arithmetic can be derived in second-order logic from Hume's principle). See https://plato.stanford.edu/entries/frege-theorem . We take a different approach, using first-order logic and ZFC, to prove the Peano axioms of arithmetic.
The theory of cardinality can also be developed without AC by introducing "card" as a primitive notion and stating this theorem as an axiom, as is done with the axiom for cardinal numbers in Suppes p. 111. Finally, if we allow the Axiom of Regularity, we can avoid AC by defining the cardinal number of a set as the set of all sets equinumerous to it and having the least possible rank (see karden ). (Contributed by NM, 22-Oct-2003)
Ref Expression
Assertion carden ${⊢}\left({A}\in {C}\wedge {B}\in {D}\right)\to \left(\mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)↔{A}\approx {B}\right)$
### Proof
Step Hyp Ref Expression
1 numth3 ${⊢}{A}\in {C}\to {A}\in \mathrm{dom}\mathrm{card}$
2 1 ad2antrr ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to {A}\in \mathrm{dom}\mathrm{card}$
3 cardid2 ${⊢}{A}\in \mathrm{dom}\mathrm{card}\to \mathrm{card}\left({A}\right)\approx {A}$
4 ensym ${⊢}\mathrm{card}\left({A}\right)\approx {A}\to {A}\approx \mathrm{card}\left({A}\right)$
5 2 3 4 3syl ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to {A}\approx \mathrm{card}\left({A}\right)$
6 simpr ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)$
7 numth3 ${⊢}{B}\in {D}\to {B}\in \mathrm{dom}\mathrm{card}$
8 7 ad2antlr ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to {B}\in \mathrm{dom}\mathrm{card}$
9 8 cardidd ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to \mathrm{card}\left({B}\right)\approx {B}$
10 6 9 eqbrtrd ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to \mathrm{card}\left({A}\right)\approx {B}$
11 entr ${⊢}\left({A}\approx \mathrm{card}\left({A}\right)\wedge \mathrm{card}\left({A}\right)\approx {B}\right)\to {A}\approx {B}$
12 5 10 11 syl2anc ${⊢}\left(\left({A}\in {C}\wedge {B}\in {D}\right)\wedge \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\right)\to {A}\approx {B}$
13 12 ex ${⊢}\left({A}\in {C}\wedge {B}\in {D}\right)\to \left(\mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)\to {A}\approx {B}\right)$
14 carden2b ${⊢}{A}\approx {B}\to \mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)$
15 13 14 impbid1 ${⊢}\left({A}\in {C}\wedge {B}\in {D}\right)\to \left(\mathrm{card}\left({A}\right)=\mathrm{card}\left({B}\right)↔{A}\approx {B}\right)$ | 1,368 | 3,987 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.875414 |
https://math.answers.com/Q/Who_many_decimeters_are_in_3_meters | 1,675,865,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00587.warc.gz | 417,530,405 | 48,721 | 0
# Who many decimeters are in 3 meters?
Wiki User
2012-01-31 16:05:37
key: 1 meter = 10 decimeters. 10 decimeters x 3 meters = 30 decimeters
Wiki User
2012-01-31 16:05:37
Study guides
20 cards
➡️
See all cards
3.81
2052 Reviews | 92 | 236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.741466 |
https://www.flashcardmachine.com/anderson-final-exam.html | 1,591,442,188,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513230.90/warc/CC-MAIN-20200606093706-20200606123706-00402.warc.gz | 706,944,280 | 6,853 | # Shared Flashcard Set
## Details
Anderson FINAL exam
this is for SDSU Dr. Andersons acctg202 final exam/exam 6
28
Accounting
05/08/2011
Term
what does IRR stand for
Definition
Internal Rate or Return
Term
What is another name for Internal Rate of Return?
Definition
IRR
Term
What is the formula for IRR?
Definition
Today's Cost/ Yearly Expense
Term
What does NPV stand for?
Definition
Net Present value
Term
What does a positive NPV signify?
Definition
The fact that the deal is a money maker
Term
What does a negative NPV signify?
Definition
That the deal is a money looser
Term
What can be learned if the IRR is less than the Cost of Capital?
Definition
That the deal is a money looser
Term
What is the formula for NPV?
Definition
(PV of future money)-(Today's Cost)
Term
T/F-The IRR is given as a percentage?
Definition
True
Term
What is the formula to find BookValue?
Definition
BBookvalue=originalcost-all depreciation
Term
what is the Net Cash Flow formula?
Definition
(sell price-Book Value)= Gain/loss*Tax%
Term
Which is easier to calculate? NPV or IRR?
Definition
NPV beause it will always give a definitive number
Term
What is an additional benefit of NPV?
Definition
It is easier to adjust for risk
Term
The higher the risk...the higher the ...??
Definition
Hurdle Rate
Term
The hurdle rate that is used in a net-present value analysis is the same as the firm's:1) Discount Rate2) IRR3) Minimum desired rate of return4) objective rate of return5) discount rate and minimum desired rate of return
Definition
5) discount rate and minimum desired rate of return
Term
The net present value of a project is \$13,458 at a cost of capital of 23%. The Internal Rate of Return would then be:1) 12%2) Greater than 12%3) Less than 12%4) Cannot be determined
Definition
2) Greater than 12%
Term
what is the formula for markup %?
Definition
((profit+Other costs)/Cost base)*100=markup %
Term
what is the formula for total charge labor rate?
Definition
hourly labor rate+(total OH/Total labor hours)+Profit per hour=
Term
What is competitive bidding?
Definition
Where two or more companies submit bids
Term
computer integrated manufacturing
Definition
A computer design and costing
Term
cost plus pricing
Definition
a pricing approach in which the price is equal to cost plus a markup
Term
penetration pricing
Definition
setting a low initial price for a new product in order to penetrate the market deeply and gain a large and broad market share
Term
perfect competition
Definition
a market in which the price does not depend on the quality sold by any one producer
Term
predatory pricing
Definition
An illegal practice in which the price of a product is temporarily to broaden demand. Then the product's supply is restricted and the price is raised
Term
price discrimination
Definition
the illegal practice of quoting different prices for the same product or service to different buyers, when the price differences are no justified by cost differences
Term
Price takers
Definition
Firms whose products or service are determined totally by the market. The price you can charge is fixed by the market; especially for commodities (oil, milk, etc.)
Term
Which of the following terms describes a pricing strategy in which a new product's initial pricing is set relatively low in order to gain a large market share?1) Penetration pricing2) Price Skimming3) Customer Pricing4) Designed Pricing5) Market-Share Pricing
Definition
1) Penetration Pricing
Term
what costs are used in the Variable Method?
Definition
Direct Materials, Direct Labor, Variable OH
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https://forum.universal-robots.com/t/how-to-find-the-distance-of-the-tcp-movement-from-one-wavepoint-to-another/15087 | 1,627,239,671,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00079.warc.gz | 279,405,041 | 7,482 | # How to find the distance of the tcp movement from one wavepoint to another?
Is there any way to find out the distance between two-waypoints?
For example, I’m doing glue application and using the code exactly how it’s mentioned here: Universal Robots - Modify robot trajectory by overlaying custom motion
The program works good but I want to know the length of the sine wave. Could any one of you know how to find it? If not distance, do you know how to calculate the speed of the sine wave? Please let me know if you have any ideas. I’m attaching some photos for reference.
2 Likes
Hello @p.honnegowda ,
the general Script Function for calculating the distance between two poses is pose_dist( <…> ).
You can use it to sum up each distance between the separate poses you calculate.
Would this be applicable in your scenario?
1 Like
Hello @sko,
Thanks for your response. Actually, I tried this script before it just gives the distance between the two waypoints without rotation (it works similar to point_dist(p1,p2). I don’t know why!!!
See this picture
I want something like this.
Kind regards,
Prathap
there is not direct script function available to calculate the distance here. My recommendation would be to create thread, which gets the actual_tcp_pose(), waits a set amount of ms and then reads out the current pose again. You can then use pose_dist() or point_dist() to calculate this minimum step and sum it up over the whole movement.
Depending on your movement speed the needed wait time can vary.
1 Like
Instead of waiting a certain time between capturing the actual TCP pose in the thread, could you instead monitor the offset variable? Capture the position at one pre-determined point in the sine wave cycle… and then when the offset equals that value again you can assume you’ve completed half a cycle… (or a whole cycle if it’s the min or max value) then calculate the actual distance between those two waypoints?
2 Likes
Thank you both of you @sko @ajp
I am using the thread as it is in the image it gives the length of a complete cycle including other movements. But if I want to get the length in between how do I do it?
You could set a boolean variable high before the move and set it low after the move. Use the variable to start and stop your distance measurement…
Or create an event triggered by the variable
1 Like
Can you add an if statement in your main offset calculation code to capture the TCP pose when the offset is equal to a certain value?
Below code not tested but should record the tcp pose the first 3 times the offset is zero… the distance between the first and third values should give you the wavelength?
``````capture = 0
blank = p[[0,0,0,0,0,0]
values = [blank, blank, blank]
i = 0
if (offset[0] == capture and i < length(values) ):
values[i] = get_actual_tcp_pose()
i = i+1
end
``````
1 Like | 640 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.885457 |
https://apps.topcoder.com/forums/?module=Thread&threadID=671150&mc=20&view=threaded | 1,540,086,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513548.72/warc/CC-MAIN-20181021010654-20181021032154-00017.warc.gz | 604,263,832 | 15,678 | JOIN
Select a Forum Round Tables New Member Discussions News Discussions Algorithm Matches Marathon Matches NASA Tournament Lab TopCoder Cookbook High School Matches Sponsor Discussions Development Forums Design Forums Search | Watch Thread | My Post History | My Watches | User Settings View: Flat | Threaded | Tree Previous Thread | Next Thread Forums TopCoder Cookbook Algorithm Competitions - Rewriting Phase Iterating Over All Subsets of a Set
Iterating Over All Subsets of a Set | Reply Problem:Iterate over all the subsets of a set.Solution:A subset of a set is a set that contains all or some of the elements of that set. For example if the set is {3, 5, 7} then some of its subsets are {} - the empty set, {3, 7}, {5, 7}, {5} and {3, 5, 7} - the original set. Often we need to find a subset with a particular property. If the original set is small, ie. has a small number of elements then it is sufficient to generate and test all its subsets. If the original set has n elements then it will contain 2^n subsets. Assuming that the computation required for each subset is small, we can usually do this for n up to 20.Let us represent a subset with a unique binary number i. This way there will be a total of 2^n such unique numbers. The k-th element of the set can be either included or not included in the subset. We can associate the inclusion of the k-th element with a value of 1 at the k-th bit of i, similarly if the k-th bit is 0 then the k-th element is not included. For example, here is how we can associate a unique binary number from 0 to 7 with every subset of {3, 5, 7}:`Binary Number Subset000 {}001 {7}010 {5}011 {5, 7}100 {3}101 {3, 7}110 {3, 5}111 {3, 5, 7}`Now all that remains is to iterate over all the subsets. Since these are just binary numbers, it suffices to iterate from 0 to 2^n-1 inclusive. In the following Java example we iterate over all the subsets of the set {"appple", "banana", "carrot"}. After generating a subset we can perform any action. Here for simplicity we print it. Notice that the 0-th subset is the empty subset, as it contains no elements.```String[] set={"apple", "banana", "carrot"}; int n=set.length; // iterate over all the subsets for (int i=0; i < (1< 0) subset.add(set[k]); } // perform an action over the subset, here just print it System.out.print("Subset "+i+":"); for (int k=0; k 0; i2 = (i2-1) & i) { // generate the subset induced by i2 ... } } ```Warning: i=0 (empty subset) is a special case and must be treated separately.Finally, there are many other operations that can be performed on sets: union, intersection, subtraction, element addition, element deletion. All of these can be done with simple and extremely fast bitwise operations. If A and B are two sets whose bit-masks are a and b respectively then we have the following:Union of A and B: a|bIntersection of A and B: a&bA "minus" B: a&(~b). This is a set of elements in A that are not in B.Add k-th element to set A: a |= 1 << kDelete k-th element of set A: a &= ~(1 << k)For an excellent description of bitwise operations take a look at bmerry’s recipe.Now let's solve some problems that use these techniques. Let's solve BouncingBalls. For this problem, the number of balls (n<=12) is sufficiently small that we can generate all possible starting states. Each ball can be rolled either left or right, so we have 2^n possible starting states. Let dir[i] be the initial direction given to the i-th ball, 1 for right and -1 for left. A collision will occur between balls k and m if ball k is to the left of ball m (x[k]<=2*T). For each possible starting state, we count the number of possible bounces for every pair of balls k and m. To compute the expected number of bounces we divide this value by the total number of starting states (1<```public class BouncingBalls { public double expectedBounces(int[] x, int T) { int n=x.length; int bounces=0; //iterate over all possible starting states for (int i=0; i<(1< 0) dir[k]=+1; //roll ball k right else dir[k]=-1; //roll ball k left } for (int k=0; k
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply This is quite similar to my original cookbook article. I've added some more examples of subsets and how to map them to binary numbers. I think these help in understanding. This article is missing some example problems.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply Two last paragraphs can actually be replaced with a link to recipe http://forums.topcoder.com/?module=Thread&threadID=671561&start=0 . And yes, this needs two examples of problems solved using this. Otherwise looks good.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply Now let's solve VectorMatching. In this problem we are given n points and we are asked to find a vector matching that has the minimal length. The number of points is small (n=20) which allows us to iterate over all possible vector matchings. A vector matching consists of two sets of points A and B, such that the head of the i-th vector is the i-th point in A and its tail is the i-th point in B. A and B should have the same number of elements, namely n/2. Each point can be either in set A or in set B, so we have 2^n possible states. However, we only need to consider those states where the size of A is n/2. This reduces the problem to Choose(20,10) = 185756 states.Suppose the point (x1,y1) is in A and the point (x2,y2) is in B. The vector represented by these two points is (x2-x1,y2-y1). Hence if a point is in A then it contributes negatively to the vector sum; otherwise it is in B and it contributes positively. So now we can determine the vector sum (which is a vector) as we iterate through the points in A and B; its x-component will be sumX and its y-component will be sumY:```public class VectorMatching { public double minimumLength(int[] x, int[] y) { double result = Double.MAX_VALUE; int n = x.length; for (int i = 0; i < (1 << n); i++) { //size of A and B must be n/2 if (2*Integer.bitCount(i) != n) continue; long sumX = 0; long sumY = 0; for (int k = 0; k < n; k++) { //point (x[k],y[k]) is in B if ((i&(1< 0) { sumX += x[k]; sumY += y[k]; } //point (x[k],y[k]) is in A else { sumX -= x[k]; sumY -= y[k]; } } double length=Math.sqrt(sumX*sumX + sumY*sumY); //vector length result = Math.min(result,length); //record the minimum length } return result; } } ```
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply the cookbook example is good and its helping me. Can u site some problem of this type as i am learning how to program.Thank in advance
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply I think it should be mentioned that BouncingBalls can in fact be solved more easily without iterating over all subsets. (By changing the order of summation by summing over "i" inside the "k" and "m" loops, it is easy to show that it is enough to add 0.25 for each pair of balls which are close enough.)
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply // iterate over all the subsets with no more than m elementsfor (int i = 0; i < (1<< m ? i+1 : (i|(i-1))+1){ ...}why i can't modify it as below?for (int i = 0; i < (1<<=m; i++){ ...}
Re: Iterating Over All Subsets of a Set (response to post by savon_cn) | Reply First of all your code tests all 2^n elements, which makes it much slower. Secondly your code will terminate as soon as it reaches a number with more than m bits set to 1. This means that it will not find all subsets with no more than m elements, only some of them.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply Another useful bit twiddling hack is gospers hack, which calculates the next greater number, that has the same number of bits set in binary representation. This can be used in a loop, to run through those subsets with k elements.
Re: Iterating Over All Subsets of a Set (response to post by holomorph) | Reply How can we extend the idea of to generate subsets with k elements. I can simply iterate over all 1<<
Re: Iterating Over All Subsets of a Set (response to post by d@rk_sh@dow) | Reply You can either use the method described above in Discussion (for m or less bits). Or you can use a more direct method described in NextBitPermutation here: http://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply You said "Another trick is to iterate over all the subsets of a subset in O(3^n) time", can you explain a bit more how O(3^n) was obtained? Thanks.
Re: Iterating Over All Subsets of a Set (response to post by hacker007) | Reply There are exactly 3^n pairs (A,B) where B is a set of elements of {1, ..., n} and A is a subset of B. To see this, notice that each pair of sets like this can be associated to a n-digit ternary number (with leading zeros, possibly): Put a 0 in the i-th position if i is in neither A nor B, an 1 if i is in B but not A, and a 2 if i is in both sets. Since this is clearly a bijection, we've proved the result.If you take O(1) time per subset while iterating, you then have a O(3^n) algorithm for iterating over all pairs as above.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply How will the method to go over all subsets of a set change if the numbers are not unique. For example if we need to find subsets of {1,1,2,3,4,4,4} such that no set is repeated i.e. We don't need to have {1} twice. But {1,1} will be a valid set.What is the best way to do it?
Re: Iterating Over All Subsets of a Set (response to post by puneetm) | Reply I don't know how to do that. I guess you could keep a HashSet of all subsets that you have seen (you need a good hash function). If you meet a subset that you have already seen then skip it.
Re: Iterating Over All Subsets of a Set (response to post by puneetm) | Reply You can change to another representation of the set: (element,count). The set is then represented as {(1,2),(2,1),(3,1),(4,3)}. For each element, loop through each possible count. In this case, there are 3*2*2*4 subsets to loop through (including the empty set).
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply can you please explain what this line is doing " if ((i&(1< 0) " ?
Re: Iterating Over All Subsets of a Set (response to post by sjsupersumit) | Reply It's checking if bit k of i is set, since 1<
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply the mapping from bit to subset is limited.when the set size is more than 32 or 64the subset number will be more than Integer.MIN_VALUE or Long.MAX_VALUE.
Re: Iterating Over All Subsets of a Set (response to post by ZuBruce) | Reply You can always use a BigInteger. Anyway iterating over such large sets is going to be too slow.
Forums TopCoder Cookbook Algorithm Competitions - Rewriting Phase Iterating Over All Subsets of a Set Previous Thread | Next Thread | 2,930 | 11,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-43 | latest | en | 0.853961 |
https://codereview.stackexchange.com/questions/174094/smallest-element-in-an-array-that-is-repeated-exactly-k-times | 1,701,424,336,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00694.warc.gz | 211,490,528 | 46,518 | # Smallest element in an array that is repeated exactly ‘k’ times
This is a problem from GeeksForGeeks and I have tried to solve it on my own and I have written this code.
Given an array of size n, the goal is to find out the smallest number that is repeated exactly ‘k’ times where k > 0? Assume that array has only positive integers and 1 <= arr[i] < 1000 for each i = 0 to n -1.
Examples:
Input:
arr[] = {2 2 1 3 1}
k = 2
Output: 1
Explanation:
Here in array:
• 2 is repeated 2 times
• 1 is repeated 2 times
• 3 is repeated 1 time
Hence 2 and 1 both are repeated 'k' times
i.e 2 and min(2, 1) is 1
Input:
arr[] = {3 5 3 2}
k = 1
Output: 2
Explanation:
Both 2 and 5 are repeating 1 time but min(5, 2) is 2
How can I improve this code?
#include <iostream>
#include <vector>
#include <map>
template<class T>
void findElement(std::vector<T>& vec, int k)
{
std::map<T, int> count;
for(T x : vec)
{
count[x]++;
}
typename std::map<T, int>::iterator itr;
for(itr = count.begin(); itr != count.end(); itr++)
{
if(itr->second == k)
{
std::cout << itr->first <<'\n';
return;
}
}
std::cerr << "No such element \n ";
}
int main()
{
std::vector<char> v;
v.push_back('r');
v.push_back('t');
v.push_back('q');
v.push_back('r');
v.push_back('u');
v.push_back('q');
v.push_back('s');
int k;
std::cout << " Enter the number of repetitions you want : ";
std::cin >> k;
std::cout << "The smallest element that has " << k <<" reptition is : ";
findElement(v, k);
}
• There is not much to talk about, except algorithm design. And may be other possible solution and situations in which they would be better. Though the current one is good for casual use already. Aug 27, 2017 at 8:26
• @Incomputable what I have to improve? I have used std::map instead of std::unordered_map because it sorts the element. Aug 27, 2017 at 8:29
## Interface
• There's no point in limiting your function inputs to a vector. You can take two input iterators instead. It'll make your function more flexible and more "standard library-like".
• One function should do one focused thing. Your function finds such an element and prints it as the same time. These concerns are unrelated. You can return std::optional<T> instead to represent either such an element or its absence. Note: it works for C++17 only. You can use boost::optional or return a smart pointer if it's not available.
## Modern C++
• There's no need for things like this anymore:
typename std::map<T, int>::iterator itr;
for(itr = count.begin(); itr != count.end(); itr++)
for (const auto& kv : count) is much better, isn't it?
• You can also use initialization list to create a vector with the elements you need: std::vector<char> v({'r', 't',..., 's'})
## Performance
• for (T x : vec) creates an unnecessary copy. Use for (const T& x : vec) or for (const auto& T : vec) instead.
• You can also use std::unordered_map to do the counting and then choose the smallest element (it's more efficient because one can find the minimum in O(n). It's "easier" than sorting the input). One caveat: you need a hash function for T in this case. You can pass it as another template parameter, defaulted to std::hash.
• One could also sort the vector/range and try to find the value with the appearance count. Though that would be sort plus linear pass, which is probably not good. The map would traverse only amount of distinct elements in a range. I guess it all depends on circumstances. Aug 27, 2017 at 8:38
• @Incomputable Yes, I believe that either option (map, unorded_map, sorting the vector) is reasonable. Aug 27, 2017 at 8:39
• How to take to input iterators instead of vector? Aug 27, 2017 at 8:56
• @janos, the main issue here in my opinion is returning the iterator to the element inside of the container. Returning value rarely has any use in my experience. And it is weird for search to return found element by value anyway. Though your algorithm would be good for the specific use case. Aug 27, 2017 at 8:56
• @coder, the same way you take any other template argument. Aug 27, 2017 at 8:56
Looking at the problem statement, a few obvious solutions come to mind:
1. Count of many of each number there is and take the smallest with 'k' entries. This is what you do and as you are using a std::map with log(m) insertion where 'm' is the number of distinct elements (bounded by 1000) so you have $O(n\log(m))$ run time and $O(m)$ memory.
2. Sort the input vector and scan linearly from the start until you find something that repeats 6 times. As you have to sort the whole thing you get $O(nlogn)$ time and $O(n)$ memory unless you can modify the input array.
But there is a better way:
/**
* Will find the smallest value in 'vec' that is repeated 'k' times.
*
* It is assumed that values of 'vec' are [1, 1000[ as per
* problem description.
*/
void findElement(const std::vector<unsigned int>& vec, int k){
constexpr auto max_value = 1000U;
std::array<unsigned int, max_value> freq;
freq.fill(k);
for(auto& value : vec){
freq.at(value)--;
}
for(std::size_t f = 0; f < freq.size(); ++f){
if(0 == freq[f]){
std::cout << f <<'\n';
return;
}
}
std::cerr << "No such element\n";
}
The raw array is actually faster as this code only performs $O(n+m)$ work which is less than $O(nlogm)$. Yes it's not as "generic" as the template method and won't work for all T. But it doesn't need to be able to do that for the task.
One could solve this using std::unordered_map as well, and the big O time is the same:
template<typename Container>
void findElement(const Container& container, int k){
std::unordered_map<Container::value_type, int> freq;
for(auto& value : container){
freq[value]++;
}
auto smallest = freq.end();
for(auto& entry : freq){
if(entry.second == k && (smallest != freq.end() || entry.first < smallest.first )){
smallest = entry;
}
}
if(smallest != freq.end()){
std::cout << smallest.first <<'\n';
}
else{
std::cerr << "No such element\n";¨
}
}
The algorithm is essentially the same with one minor difference, the approach with the std::unordered_map may not terminate early on finding the first element with exactly 'k' occurrences as the array based algorithm can. This means that the unordered_map version must visit exactly all 'M<1000' unique values in 'vec' including some overhead from traversing a sparse hash-map, while the array version on average only needs to visit '1000/2' of the possible values.
So they have different cases where they are faster. For example a degenerate case such as K=3 , {999,999,999} is faster with unordered_map as you only visit one unique element but with the array solution you need to (very quickly) scan through the 998 elements in the array. But on average on random sets the array solution is expected to be slightly faster by a constant factor as both have the same big O time.
• The mentioned O(n+m) method is of course the right one, but it has nothing to do with the term 'brute force' (misleading). It is called bucket sort.
– mafu
Aug 27, 2017 at 16:12
• Or one could make 1. just as efficient as the array solution (and support arbitrary sized elements) by using an unordered_map.
– Voo
Aug 27, 2017 at 20:41
• @mafu you are correct, I used the term loosely to refer to the fact of not using any data structure other than plain, fixed size array. Aug 27, 2017 at 21:34
• @Voo afaik std::hash<int> isn't required to return the argument unmodified. Which means that you need to compare all elements of the unordered map to find the minimum which prevents the early exit the array solution has. The asymptotic performance is the same but the time constant differs to the advantage of the array solution which also avoids dynamic memory allocations which also take a short time. But no work is less work than some work. Also the task is to solve the programming challenge, not write code for an "enterprise" system. Aug 27, 2017 at 22:24
• @Emily You have to iterate through elements in the array to find the first non-0 value, which the dictionary solution can do just the same (it doesn't matter what the hash function is as long as it is reasonably "good" which is trivial for ints). The only difference in performance will be that the dictionary has a relatively more expensive indexing operation.
– Voo
Aug 28, 2017 at 7:23 | 2,150 | 8,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-50 | longest | en | 0.8201 |
https://www.abstract-polytopes.com/atlas/816/133/2.html | 1,708,576,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00436.warc.gz | 645,407,101 | 3,325 | # Polytope of Type {68,6}
This page is part of the Atlas of Small Regular Polytopes
Atlas Canonical Name : {68,6}*816a
Also Known As : {68,6|2}. if this polytope has another name.
Group : SmallGroup(816,133)
Rank : 3
Schlafli Type : {68,6}
Number of vertices, edges, etc : 68, 204, 6
Order of s0s1s2 : 204
Order of s0s1s2s1 : 2
Special Properties :
Compact Hyperbolic Quotient
Locally Spherical
Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
{68,6,2} of size 1632
Vertex Figure Of :
{2,68,6} of size 1632
Quotients (Maximal Quotients in Boldface) :
2-fold quotients : {34,6}*408
3-fold quotients : {68,2}*272
6-fold quotients : {34,2}*136
12-fold quotients : {17,2}*68
17-fold quotients : {4,6}*48a
34-fold quotients : {2,6}*24
51-fold quotients : {4,2}*16
68-fold quotients : {2,3}*12
102-fold quotients : {2,2}*8
Covers (Minimal Covers in Boldface) :
2-fold covers : {136,6}*1632, {68,12}*1632
Permutation Representation (GAP) :
```s0 := ( 2, 17)( 3, 16)( 4, 15)( 5, 14)( 6, 13)( 7, 12)( 8, 11)( 9, 10)
( 19, 34)( 20, 33)( 21, 32)( 22, 31)( 23, 30)( 24, 29)( 25, 28)( 26, 27)
( 36, 51)( 37, 50)( 38, 49)( 39, 48)( 40, 47)( 41, 46)( 42, 45)( 43, 44)
( 53, 68)( 54, 67)( 55, 66)( 56, 65)( 57, 64)( 58, 63)( 59, 62)( 60, 61)
( 70, 85)( 71, 84)( 72, 83)( 73, 82)( 74, 81)( 75, 80)( 76, 79)( 77, 78)
( 87,102)( 88,101)( 89,100)( 90, 99)( 91, 98)( 92, 97)( 93, 96)( 94, 95)
(103,154)(104,170)(105,169)(106,168)(107,167)(108,166)(109,165)(110,164)
(111,163)(112,162)(113,161)(114,160)(115,159)(116,158)(117,157)(118,156)
(119,155)(120,171)(121,187)(122,186)(123,185)(124,184)(125,183)(126,182)
(127,181)(128,180)(129,179)(130,178)(131,177)(132,176)(133,175)(134,174)
(135,173)(136,172)(137,188)(138,204)(139,203)(140,202)(141,201)(142,200)
(143,199)(144,198)(145,197)(146,196)(147,195)(148,194)(149,193)(150,192)
(151,191)(152,190)(153,189);;
s1 := ( 1,104)( 2,103)( 3,119)( 4,118)( 5,117)( 6,116)( 7,115)( 8,114)
( 9,113)( 10,112)( 11,111)( 12,110)( 13,109)( 14,108)( 15,107)( 16,106)
( 17,105)( 18,138)( 19,137)( 20,153)( 21,152)( 22,151)( 23,150)( 24,149)
( 25,148)( 26,147)( 27,146)( 28,145)( 29,144)( 30,143)( 31,142)( 32,141)
( 33,140)( 34,139)( 35,121)( 36,120)( 37,136)( 38,135)( 39,134)( 40,133)
( 41,132)( 42,131)( 43,130)( 44,129)( 45,128)( 46,127)( 47,126)( 48,125)
( 49,124)( 50,123)( 51,122)( 52,155)( 53,154)( 54,170)( 55,169)( 56,168)
( 57,167)( 58,166)( 59,165)( 60,164)( 61,163)( 62,162)( 63,161)( 64,160)
( 65,159)( 66,158)( 67,157)( 68,156)( 69,189)( 70,188)( 71,204)( 72,203)
( 73,202)( 74,201)( 75,200)( 76,199)( 77,198)( 78,197)( 79,196)( 80,195)
( 81,194)( 82,193)( 83,192)( 84,191)( 85,190)( 86,172)( 87,171)( 88,187)
( 89,186)( 90,185)( 91,184)( 92,183)( 93,182)( 94,181)( 95,180)( 96,179)
( 97,178)( 98,177)( 99,176)(100,175)(101,174)(102,173);;
s2 := ( 1, 18)( 2, 19)( 3, 20)( 4, 21)( 5, 22)( 6, 23)( 7, 24)( 8, 25)
( 9, 26)( 10, 27)( 11, 28)( 12, 29)( 13, 30)( 14, 31)( 15, 32)( 16, 33)
( 17, 34)( 52, 69)( 53, 70)( 54, 71)( 55, 72)( 56, 73)( 57, 74)( 58, 75)
( 59, 76)( 60, 77)( 61, 78)( 62, 79)( 63, 80)( 64, 81)( 65, 82)( 66, 83)
( 67, 84)( 68, 85)(103,120)(104,121)(105,122)(106,123)(107,124)(108,125)
(109,126)(110,127)(111,128)(112,129)(113,130)(114,131)(115,132)(116,133)
(117,134)(118,135)(119,136)(154,171)(155,172)(156,173)(157,174)(158,175)
(159,176)(160,177)(161,178)(162,179)(163,180)(164,181)(165,182)(166,183)
(167,184)(168,185)(169,186)(170,187);;
poly := Group([s0,s1,s2]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2");;
s0 := F.1;; s1 := F.2;; s2 := F.3;;
rels := [ s0*s0, s1*s1, s2*s2, s0*s2*s0*s2, s0*s1*s2*s1*s0*s1*s2*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(204)!( 2, 17)( 3, 16)( 4, 15)( 5, 14)( 6, 13)( 7, 12)( 8, 11)
( 9, 10)( 19, 34)( 20, 33)( 21, 32)( 22, 31)( 23, 30)( 24, 29)( 25, 28)
( 26, 27)( 36, 51)( 37, 50)( 38, 49)( 39, 48)( 40, 47)( 41, 46)( 42, 45)
( 43, 44)( 53, 68)( 54, 67)( 55, 66)( 56, 65)( 57, 64)( 58, 63)( 59, 62)
( 60, 61)( 70, 85)( 71, 84)( 72, 83)( 73, 82)( 74, 81)( 75, 80)( 76, 79)
( 77, 78)( 87,102)( 88,101)( 89,100)( 90, 99)( 91, 98)( 92, 97)( 93, 96)
( 94, 95)(103,154)(104,170)(105,169)(106,168)(107,167)(108,166)(109,165)
(110,164)(111,163)(112,162)(113,161)(114,160)(115,159)(116,158)(117,157)
(118,156)(119,155)(120,171)(121,187)(122,186)(123,185)(124,184)(125,183)
(126,182)(127,181)(128,180)(129,179)(130,178)(131,177)(132,176)(133,175)
(134,174)(135,173)(136,172)(137,188)(138,204)(139,203)(140,202)(141,201)
(142,200)(143,199)(144,198)(145,197)(146,196)(147,195)(148,194)(149,193)
(150,192)(151,191)(152,190)(153,189);
s1 := Sym(204)!( 1,104)( 2,103)( 3,119)( 4,118)( 5,117)( 6,116)( 7,115)
( 8,114)( 9,113)( 10,112)( 11,111)( 12,110)( 13,109)( 14,108)( 15,107)
( 16,106)( 17,105)( 18,138)( 19,137)( 20,153)( 21,152)( 22,151)( 23,150)
( 24,149)( 25,148)( 26,147)( 27,146)( 28,145)( 29,144)( 30,143)( 31,142)
( 32,141)( 33,140)( 34,139)( 35,121)( 36,120)( 37,136)( 38,135)( 39,134)
( 40,133)( 41,132)( 42,131)( 43,130)( 44,129)( 45,128)( 46,127)( 47,126)
( 48,125)( 49,124)( 50,123)( 51,122)( 52,155)( 53,154)( 54,170)( 55,169)
( 56,168)( 57,167)( 58,166)( 59,165)( 60,164)( 61,163)( 62,162)( 63,161)
( 64,160)( 65,159)( 66,158)( 67,157)( 68,156)( 69,189)( 70,188)( 71,204)
( 72,203)( 73,202)( 74,201)( 75,200)( 76,199)( 77,198)( 78,197)( 79,196)
( 80,195)( 81,194)( 82,193)( 83,192)( 84,191)( 85,190)( 86,172)( 87,171)
( 88,187)( 89,186)( 90,185)( 91,184)( 92,183)( 93,182)( 94,181)( 95,180)
( 96,179)( 97,178)( 98,177)( 99,176)(100,175)(101,174)(102,173);
s2 := Sym(204)!( 1, 18)( 2, 19)( 3, 20)( 4, 21)( 5, 22)( 6, 23)( 7, 24)
( 8, 25)( 9, 26)( 10, 27)( 11, 28)( 12, 29)( 13, 30)( 14, 31)( 15, 32)
( 16, 33)( 17, 34)( 52, 69)( 53, 70)( 54, 71)( 55, 72)( 56, 73)( 57, 74)
( 58, 75)( 59, 76)( 60, 77)( 61, 78)( 62, 79)( 63, 80)( 64, 81)( 65, 82)
( 66, 83)( 67, 84)( 68, 85)(103,120)(104,121)(105,122)(106,123)(107,124)
(108,125)(109,126)(110,127)(111,128)(112,129)(113,130)(114,131)(115,132)
(116,133)(117,134)(118,135)(119,136)(154,171)(155,172)(156,173)(157,174)
(158,175)(159,176)(160,177)(161,178)(162,179)(163,180)(164,181)(165,182)
(166,183)(167,184)(168,185)(169,186)(170,187);
poly := sub<Sym(204)|s0,s1,s2>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2> := Group< s0,s1,s2 | s0*s0, s1*s1, s2*s2,
s0*s2*s0*s2, s0*s1*s2*s1*s0*s1*s2*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1 >;
```
References : None.
to this polytope | 4,003 | 7,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.362637 |
https://byjus.com/question-answer/find-the-profit-from-each-sale-and-sort-from-the-least-profitable-to-the-most-14/ | 1,638,677,292,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00458.warc.gz | 231,546,666 | 16,508 | Question
# Find the profit from each sale and sort from the least profitable to the most profitable(from top to bottom).CP = 100 SP = 300CP = 100 SP = 150CP.= 75 SP = 100CP = 110 SP = 180
Solution
## The correct option is Profit = S.P - C.P 1) C.P. = 100 , S.P. =300 ∴ Profit = 300-100 = 200 2) C.P. = 100 , S.P. = 150 ∴ Profit = 150 - 100 = 50 3) C.P. = 75 , S.P. = 100 ∴ Profit = 100 - 75 = 25 4) C.P. = 110 , S.P. = 180 ∴ Profit = 180 - 110 = 70 So, from the least profitable to the most profitable sale, the order will be 25 < 50 < 70 < 200
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https://www.physicsforums.com/threads/what-exactly-is-a-sample-in-terms-of-sample-rate-audio.851010/ | 1,513,556,334,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948599156.77/warc/CC-MAIN-20171217230057-20171218012057-00188.warc.gz | 784,754,723 | 26,067 | # What exactly is a sample in terms of sample rate (audio)
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1. Jan 6, 2016
### hypnoticdesign
I can't seem to find an answer anywhere to this just by searching. I'm trying to understand samplerate more thoroughly. I know 44000 samples per second means that the highest frequency that can be recorded is 22000 hz but what is each sample? Does it mean there is a constant 44000 snippets of sound making up every sound each second? Or is this just the limit of the number of cycles any waveform can have? Is each sample a waveform?
2. Jan 6, 2016
### Staff: Mentor
The instantaneous value of the waveform is measured 44000 times per second, as represented by the black dots in the following diagram:
http://manual.audacityteam.org/m/images/e/e2/Waveform_digital.png [Broken]
Source: http://manual.audacityteam.org/index.php?title=Digital_Audio [Broken]
Last edited by a moderator: May 7, 2017
3. Jan 6, 2016
### Ibix
As you note, the Nyquist criterion means that a 44kHz sample rate picks up frequencies up to 22kHz. Only the very young can hear frequencies above 20kHz, so there is no point to recording higher frequencies. That's the reason why the sampling rate for audio is usually set to 44kHz or lower.
I don't know of any reason why a sound wave couldn't have higher frequency components, but we wouldn't be able to hear it.
Each sample is simply the amplitude of the wave at one instant, as per jtbell's post.
4. Jan 6, 2016
### sophiecentaur
That's a good way to describe it. The 22kHz figure that's used is pretty arbitrary, in many ways, because old gimmers like me can hardly hear anything above 10kHz and there are a few 'golden ears' who can (or claim to be able to) hear way above 22kHz. A vibration with much higher frequencies than 22kHz can travel through air and end up being sampled at 44kHz. The resulting set of samples will produce signals that are below 22kHz and are quite audible. They are called Aliases. If you want an analogue/digital system that can be relied on, it is essential to low pass filter the input signal to less than or equal to 22kHz. The filter is referred to as a Nyquist filter (anti-aliasing) and is usually made to be as sharp cut as possible / practicable, to give a 'flat' frequency response in the audio band.
There's a problem here with terminology, I think. The samples are just values (e.g. the values of the instantaneous voltage). Samples are normally taken at a regular rate and, once digitised, they are a stream of digits. When they are reconstituted (in a DAC) they become a stream of pulses of different amplitudes which join together to produce a continuous waveform when they are passed through a low pass filter to 'smooth the jagged edges between them'.
Note. Samples do not have to be 'digitised'. Historically, sampling used to be done in an analogue 'sampling oscilloscope' which would sample a very high frequency signal, consisting of a repeating, fast waveform with a relatively slow rate set of sampling pulses. This would generate a waveform that was an Alias of the high frequency wave but which appeared at a low enough frequency to be displayed on a conventional (slow) oscilloscope. This was an all Analogue process. Ancient history now, of course but the only way to look at UHF waveforms with slow circuitry.
5. Jan 7, 2016
### meBigGuy
Higher frequencies than 1/2 the sample rate (nyquist rate) are aliased back into the sub-nyquist band and therefore become audible.
For the OP:
One of the hardest thing to accept about sub-nyquist sampling is to understand that the periodic sample data contains ALL the information of the original signal and can be used to recreate it exactly (within the limits of the sample word size). The samples actually represent evenly spaced impulses which, when filtered, will create the original signal exactly (filling in the spaces between samples). It is a very important concept.
If you want to read more about sampling, try searching for sampling or digital signal processing.
https://en.wikipedia.org/wiki/Sampling_(signal_processing)
https://en.wikipedia.org/wiki/Digital_signal_processing
6. Jan 7, 2016
### Ibix
Sorry I wasn't clear. You will not be able to hear an ultrasonic dog whistle at (say) 30kHz, so there's little point trying to record it. However, as sophiecentaur and meBigGuy point out, a 44kHz sample of the dog whistle will produce an aliased sound at 14kHz, which you are much more likely to be able to hear. You will need to filter out any stuff above the Nyquist frequency before the sampling occurs in order to avoid this.
7. Jan 7, 2016
### sophiecentaur
Yes - it's a revolutionary idea. There is a way of looking it it that could help. Firstly, with above nyquist sampling, you have to low pass filter the input signal appropriately. If you pass the string of samples (impulses of different levels) through an ideal low pass filter, the voltage in between the sample times will follow the values of the original signal values (even when there are only slightly more samples than half a cycle of the signal). The devil is in the detail of how the two low pass filters are designed; to reconstitute such a signal may involve the filters using contributions of many samples. Also the pre and post filters have to be exactly complementary or the reconstituted signal will not be right - but, hell, this is theory and all engineering is based on 'near enough'. The harder you try and the more money you spend, the better the equipment will follow the theory. Sampling a signal with a stream of narrow pulses is a form of modulation (multiplication of one signal by the other). The string of sampling pulses will have a frequency spectrum which consists of a 'comb' of frequencies (the harmonics of the fundamental frequency). Sampling will produce sidebands around all of these harmonics. If you (sub-) sample a single tone with a high frequency (higher than the Nyquist limit), one of the sample harmonics (it could be the eleveth harmonic)will be ' near' your sampled frequency and produce a sideband that's less than the Nyquist frequency (i.e. it will be 'audible'). This shouldn't be hard to accept when you consider that a bog-standard superhet receiver does exactly the same thing by beating down an extremely high frequency signal to one that can be processed in a lowish frequency IF section. All that's required is that the spectrum of the final signal must not be 'jumbled up' with reconstituted products not laying on top of one another. Then they are inseperable and what you hear / see will be distorted.
8. Jan 7, 2016
### hypnoticdesign
Thanks for the replies guys, really helpful stuff! I think I get it now,. Its the damn term sample that gets me, why this term is used for so many things in music production is beyond me, it just creates confusion, I mean I suspected it was constant but also thought it could'a just meant that was the maximum number of samples and would only be used if needed for the higher frequencies...
I just wanna get something out. A lot of you are saying that theres no point sampling at higher rates, but thats not actually true. The frequencies of higher notes actually interact with the lower frequencies creating a beating. These are audible and form part of some sounds. Taking away this beating can result in a less natural sounding recording but won't affect a sound that has none of this higher frequency info recorded.
Soooo..... Samples are little snapshots(so to speak) of amplitude that make up waveforms depending on the arrangement of the samples. So if I've got a sine wave at 44khz samplerate and play 1 bar at 60 bpm.. I will have 44000 samples in a bar regardless of frequency.
So I was wondering, using this would be a great way to get the best out of resampling right? Or getting the most out of digital synths? I mean say I have a sound that I wanna resample, If I match the fundamental to the sample rate would that ensure the recording would be as accurate as possible? Or if a synth is tuned so that all the fundamental frequencies are cycling in time with the sample rate then nothing would be getting cut off anywhere in between note(for high frequency content)?? Or would it make no difference to lower notes unless they had crazy high harmonic content? There surely has to be some optimal relationship between frequency and sample rate or is bpm to frequency more important? or both? Or Neither? lol.. sorry If I'm going on.. I just wanna have the most solid foundation for my music thats possible.
Last edited: Jan 7, 2016
9. Jan 7, 2016
### meBigGuy
The optimal relationship is that the sample rate needs to be greater than 2X the highest frequency in your signal and the word size large enough for your dynamic range (or noise floor requirements).
In reality that is impossible since there are always higher frequencies present. The trick is to filter them such that their alias products are below the inherent noise.
There is absolutely no relationship between samples in a synth rom sense, and sampling theory for dsp. There are standard DSP sample rates and word sizes, so you stick with them. They range roughly from 44.1Khz 16 bits to 96Khz 24 bits (lower and higher are possible, of course). Any given sample rate will faithfully reproduce any frequency between 0 and close to the nyquist frequency. There is nothing about your source material that should affect the sample rate other than the anti-alias issues (and your desire or need to cut corners for memory capacity or whatever).
The sample frequency determines the frequency above which aliasing will occur, and so determines the complexity of the anti-alias filters and the resulting phase and amplitude distortion. The higher the sample frequency, the easier to filter out everything above 20Khz (or 24Khz, or whatever) without radical phase and amplitude distortion. (modern digital filters make channel matching easier)
The sample word size determines the dynamic range (approx 6dB per bit) so 16 bist can reproduce a ~96dB dynamic range. 24 bit allows ~144dB.
At 16 bits, you can hear hiss from home theater systems when the volume is set moderately loud and the material is silent. Sampling always produces a noise floor. The idea is to keep it below what the ear can perceive. 16 bits is pretty marginal. I wish the standard was 18 or 20 bits .
10. Jan 7, 2016
### hypnoticdesign
So use a steep low pass at the cut off point of the sampling rate limit? Thanks for clearing that up.
I totally get that the samples are different and not like for example a kick drum sample but thats not what I mean. I'm mainly referring to doing things like using 375 hertz as a root note with 48khz samplerate and 120 bpm tempo or something similar to that.
Surely frequency would make a difference. I mean if a sound never lands on a zero crossing that would have to sound different to a sound that always lands on a zero crossing? If a sine wave was made up of lets say 1000 samples but the last sample is in the middle of where the waveform ends.. if thats on a loop for a sustained sound, won't that cause some sort of artefact? I'm assuming here that the samples are evenly distributed and thinking completely digital not recorded sounds. Its just 1 of my favourite synths( serum) allows you to use other synths waveforms by recording and importing them. it allows up to 256 waves and can modulate between them. The manual says to use specific numbers of samples to get accurate results if your using a modulated sound and when I found a tutorial of someone doing it, they used the samplerate to convert it for a specific frequency timing cycle duration to samplerate then timing a specific amount of time to match the specified number of samples asked for by the manual.
Last edited: Jan 7, 2016
11. Jan 7, 2016
### meBigGuy
That's the counter-intuitive hard to accept part I was referring to earlier. The zero crossing will end up in exactly the right places. The ratios don't matter. Mathematically the exact signal is reproduced from the samples no matter what (except for the limitations of nyquist and the noise floor). The lining up of samples is not an issue in any way when dealing with real signals. (there is some statistical sampling theory involved that relates to dither and noise, but its not worth going into).
12. Jan 7, 2016
### hypnoticdesign
I think I just realised... they were just doing that to make sure the sound completed entire cycles.
Just realised a zero crossing has nothing to do with samples. I think I'm confusing myself with this synth... it was that the synth internally uses a certain number of samples to make up a frame for each waveform as the waveforms can be modulated between 256 different waves... I think I understand how that works now I've let this stuff sink in a little lol. For some reason I was picturing a waveform as a stack, I was just severely confusing myself and then it hit me :)
PS thanks for explaining this to me, its like a void in my head has just been lit up aha
13. Jan 8, 2016
### Svein
Strictly speaking, what you are referring to is known as a sampler. In a synthesizer, the sounds are created in an artificial way, like FM (Yamaha DX7) or oscillators and filter banks (the original Moog and early Roland synths).
14. Jan 8, 2016
### hypnoticdesign
Nah this is a little different than most. It doesn't actually allow you to load samples but rather recreates a representation of the waveform by analysing an audio file. I'm guessing thats why it requires the precision. But that just a special feature of the thing. It even allows you to create waveforms from user defined equations. Ive been getting a little obsessed with that feature... trying to make golden ratio waveforms and stuff like that ^^
15. Jan 8, 2016
### meBigGuy
I guess there is DJ_sampling and DSP_sampling.
When you create waveforms with equations I expect you can use floating point numbers (like 3.14159) to describe relationships. Wouldn't be much good if it were just integers. So, don't sweat the sample rate when you are choosing frequencies. It flat out doesn't matter in most real world situations.
Now, if you are choosing a sequence to analyze (like do an FFT) or loop, there can be issues due to the discontinuity when looping. The FFT issues are generally fixed by windowing (like a hamming or hanning window), and can sometimes be reduced by tweaking the sample length and using a DFT. As I am sure you have found, sometimes you have to tweak the beginning and/or end of a loop to get it to sound ok at the transition. But, that's not really a DSP_sampling issue.
If you are limited to choosing a specific sequence size then looping can be an issue.
16. Jan 8, 2016
### sophiecentaur
As do many of the terms that are used in Engineering. But there is just one 'official' meaning and that is the instantaneous value of the input waveform, that is measured and stored / processed. Reading the popular press for 'enlightenment' is usually a forlorn hope. 'Sampling' often means taking a long string of samples - but that's another topic.
The only instance of this is when you are sampling at exactly twice the signal frequency. Nyquist demands that your input signal frequency must be lower than that. I made a point, earlier, of saying that the process of reconstitution may take some time and, if you have an input tone that's 1Hz below the Nyquist frequency and you want to avoid your zero crossing problem, then you may need to take a half second for the samples to coincide with max and min of the waveform. Note - you couldn't have a very short burst of this frequency and still comply with the Nyquist criterion because the input signal would have components, higher than fs/2 so, by definition, your input waveform would have to have a long enough stretch of high frequencies for the zero crossing problem to be avoided. The consequences of sampling are very much dependent upon the filtering that's used and that tends to get ignored in discussions; the basic theory assumes 'perfect pre and post filtering - to keep the signal within the Nyquist limits in a healthy way..
Absolutely. But first, remember that basic sampling theory relates to Analogue samples. Mr Shannon was around before the age of digits. If you want to introduce the consequences of digitisation / quantisation of your samples then that's a new ball game. as soon as the samples are quantised, there is distortion and this distortion is often described as Quantisation Noise because of what it can sound like. Quantisation noise can sound a bit like crossover distortion (buzz buzz during low level passages). There is a tradeoff between sample rate and the number of bits needed per sample. Gross oversampling can have massive advantages and the 'bit slice' ADC, which uses single bit quantisation and many MHz of sample rate can give very low quantisation noise because this noise energy is spread over the whole spectrum and the post filter will remove all but the audio components of the noise. There is so much to read about this topic and, of course, there is a lot of BS in the Hi Fi mags.
I just found this. You are jumping from basics to practicalities here. If you have a simple sampling synthesiser then each note for a particular sound could be based on a recording of a number of cycles of just a single source sound, scaled up or down in frequency. Between this and the playback, there must be a re-sampling, to stretch or compress the available string of samples so that they are avaiable at the standard sample rate of the downstream circuits. This involves re-timing and interpolating between the recorded samples.
17. Jan 9, 2016
### hypnoticdesign
For the waveforms I just start with sin(-x*pi) for a sine wave or abs(x)*2-1 for a triangle and go from there.. I came up with a formula a while ago where if you take any number and divide it by a decimal then divide by the number you start with and + 1 you eventually get 1.618 and I just try stuff similar to that using something like < (-x*pi) to blend between waveforms. It results in a square like finish sound most of the time but blends between the waveform you start and allows a total of 256 waves so theres a lot of room for modulation. I just experiment to be honest though, I used a lyrebird song once to get a sine like wave while just messing about without thinking what I was doing and its 1 of my favourite sounds!
I don't really use loops at all, I like to combine samples to create a kit or just synthesise them and program the beats with midi, though I like the sound of a continuous loop sometimes to create a sustained sound and finding the right place to loop can be really hard sometimes.
Well before I got into this silly samplerate theory of mine I was timing the bpm to the cycle duration of my root bass note or basing it off the root mean of the chromatic scale. It sounds awesome to me so I'll stick to it! I just thought there might be some benefit to using sample rate in the same way. I understand now how samplerate and frequency are independent other than the range. I did notice something last night though thats a little strange. I was playing around in Logic with a few different sounds and testing using a cut off filter at really high frequencies. When I have a project set to 96khz theres just no cut off point my eq's can reach that sounds better to me than leaving it alone. I tried a 24 kHz cut off and while the difference was only tiny, it definitely sounded better without it. I'd describe the effect similar to how a tape machine can offset the sound making it brighter or darker. I'm guessing that it would be better to wait until the final stages before cutting off these high frequencies just before converting to more lossy formats?
18. Jan 9, 2016
### meBigGuy
I'm not sure I entirely understand your experiment here, and whether or not you samplerate converted to 48KHz.
I concur that waiting until the end to samplerate convert will generally give the purest results. But, if you are dealing only with digitally created sounds created at the lower samplerate, I'm not sure where the difference might creep in. On the other hand, if you have a 96Khz sample rate signal, samplerate converting it to 48Khz requires the pre-conversion anti-alias filter to remove all content above 24Khz, and that means you will always change some content below 24Khz, since filters always have a transition region. You need to look at the response of the specific anti-alias filter used. And, since you cannot remove all >24Khz content, there will, in reality, always be some aliasing at time of subsampling. The goal is to design such that what aliasing there is will always be below the noise floor (or not, if you want to capitalize on aliasing)
https://www.maximintegrated.com/en/app-notes/index.mvp/id/928 explains anti-aliasing very well, but may raise some new questions.
Just remember that even purely digital samplerate conversion always introduces changes caused by aliasing and filter responses
The human ear (and personal taste) is "funny", and it is possible that a particular sound will sound "cooler" with alias artifacts than without. Just another distortion effect to be judged by "he who is listening". Supposedly there are tests that found that mp3 listeners preferred mp3 encoded music over losslessly encoded. A matter of what they are used to.
19. Jan 17, 2016
### leright
Yup. This absolutely blew my mind when we studied this in my signals class. If you sample above the Nyquist rate of your signal ALL of the information of the original signal is there in the SAMPLES. Amazing. If you convert the time series of samples into the frequency domain you get a series of copies of the original frequency spectrum! However, If the sample rate is too low (below Nyquist) these copies of the original frequency spectrum overlap and you lose information. Also, if the copies of the original frequency spectrum are too close together it becomes difficult to recover the original signal via low pass filtering (since there is no such thing as an ideal filter.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 4,899 | 22,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-51 | longest | en | 0.954144 |
https://www.yaclass.in/p/mathematics/class-7/real-number-system-1483/basic-operations-in-integers-2155/re-5032c8f3-51c5-4cc3-a3f9-8c6c74d6b11a | 1,582,970,297,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00328.warc.gz | 960,738,662 | 7,710 | ### Теория:
While, addition or subtraction of $$2$$ integers, depending upon the sign of the two or more integers, the answer can change.
When two integers have the same sign, the sign of the answer will be the same sign of the integers.
Example:
1. In the above example, ($$-3$$) and ($$-4$$) are added both the integers have -ve sign, hence answer is $$-7$$.
2. $$(-6) + (-5) = -11$$
3. $$10 + 12 = 22$$
4. $$(-32) + (-14) = -48$$
5. $$90 + 10 = 100$$
When two integers have a different sign, the sign of the answer will be the sign of the largest integer and both numbers will be subtracted.
Example:
1. In the above example ($$-5$$) and ($$9$$), the difference of the numbers are taken, and the sign of answer is $$+ve$$. Hence the answer is$$+4$$.
2. $$(-20) + 13 = -7$$
3. $$30 - 15 = 15$$
4. $$20 + (-10) = 10$$
5. $$-30 + 50 = 20$$
When two integers are subtracted, ($$+ve$$)$$-$$($$-ve$$), or ($$-ve$$)$$-$$($$-ve$$), since $$-ve$$ multiplied by $$-ve$$ is $$+ve$$, the operand between both numbers will be taken as $$+$$.
Example:
1. $$(-10) - (-10) = (-10) + 10 = 0$$
2. $$(20) - (-10) = 20 + 10 = 30$$
3. $$(-50) - (-15) = -50 + 15 = -35$$
4. $$(25) - (-15) = 25 + 15 = 40$$
When two integers are subtracted, in this manner, ($$+ve$$) $$-$$ ($$+ve$$), both integers should be subtracted, and the sign of the answer will be the sign of the highest integer.
Example:
1. $$30 - 10 = 20$$
2. $$40 - 12 = 28$$
3. $$10 - 50 = -40$$
4. $$20 - 48 = -28$$ | 534 | 1,461 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-10 | latest | en | 0.687172 |
http://www.reference.com/browse/Supercommutative_algebra | 1,409,462,183,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500836106.97/warc/CC-MAIN-20140820021356-00348-ip-10-180-136-8.ec2.internal.warc.gz | 573,547,778 | 21,308 | Definitions
# Supercommutative algebra
In mathematics, a supercommutative algebra is a superalgebra (i.e. a Z2-graded algebra) such that for any two homogeneous elements x, y we have
$yx = \left(-1\right)^$xy.,> Equivalently, it is a superalgebra where the supercommutator
$\left[x,y\right] = xy - \left(-1\right)^$
y
>yx,
always vanishes. Algebraic structures which supercommute in the above sense are sometimes referred to as skew-commutative associative algebras to emphasize the anti-commutation, or, to emphasize the grading, graded-commutative or, if the supercommutativity is understood, simply commutative.
Any commutative algebra is a supercommutative algebra if given the trivial gradation (i.e. all elements are even). Grassmann algebras (also known as exterior algebras) are the most common examples of nontrivial supercommutative algebras. The supercenter of any superalgebra is the set of elements that supercommute with all elements, and is a supercommutative algebra.
The even subalgebra of a supercommutative algebra is always a commutative algebra. That is, even elements always commute. Odd elements, on the other hand, always anticommute. That is,
$xy + yx = 0,$
for odd x and y. In particular, the square of any odd element x vanishes whenever 2 is invertible:
$x^2 = 0.,$
Thus a commutative superalgebra (with 2 invertible and nonzero degree one component) always contains nilpotent elements. | 366 | 1,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2014-35 | longest | en | 0.855903 |
http://www.mathworks.co.kr/kr/help/matlab/ref/colon.html?action=changeCountry | 1,386,825,115,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164491055/warc/CC-MAIN-20131204134131-00014-ip-10-33-133-15.ec2.internal.warc.gz | 429,076,774 | 8,793 | Accelerating the pace of engineering and science
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# colon (:)
Create vectors, array subscripting, and for-loop iterators
## Description
The colon is one of the most useful operators in MATLAB®. It can create vectors, subscript arrays, and specify for iterations.
The colon operator uses the following rules to create regularly spaced vectors for scalar values i, j, and k:
j:k is the same as [j,j+1,...,k], or empty when j > k. j:i:k is the same as [j,j+i,j+2i, ...,j+m*i], where m = fix((k-j)/i), for integer values. This syntax returns an empty matrix when i == 0, i > 0 and j > k, or i < 0 and j < k.
If you specify nonscalar arrays, MATLAB interprets j:i:k as j(1):i(1):k(1).
You can use the colon to create a vector of indices to select rows, columns, or elements of arrays, where:
A(:,j) is the jth column of A. A(i,:) is the ith row of A. A(:,:) is the equivalent two-dimensional array. For matrices this is the same as A. A(j:k) is A(j), A(j+1),...,A(k). A(:,j:k) is A(:,j), A(:,j+1),...,A(:,k). A(:,:,k) is the kth page of three-dimensional array A. A(i,j,k,:) is a vector in four-dimensional array A. The vector includes A(i,j,k,1), A(i,j,k,2), A(i,j,k,3), and so on. A(:) is all the elements of A, regarded as a single column. On the left side of an assignment statement, A(:) fills A, preserving its shape from before. In this case, the right side must contain the same number of elements as A.
When you create a vector to index into a cell array or structure array (such as cellName{:} or structName(:).fieldName), MATLAB returns multiple outputs in a comma-separated list. For more information, see How to Use the Comma-Separated Lists in the MATLAB Programming Fundamentals documentation.
## Examples
Using the colon with integers,
`D = 1:4`
results in
```D =
1 2 3 4```
Using two colons to create a vector with arbitrary real increments between the elements,
`E = 0:.1:.5`
results in
```E =
0 0.1000 0.2000 0.3000 0.4000 0.5000```
The command
`A(:,:,2) = pascal(3)`
generates a three-dimensional array whose first page is all zeros.
```A(:,:,1) =
0 0 0
0 0 0
0 0 0
A(:,:,2) =
1 1 1
1 2 3
1 3 6```
Using a colon with characters to iterate a for-loop,
`for x='a':'d',x,end`
results in
```x =
a
x =
b
x =
c
x =
d``` | 704 | 2,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2013-48 | latest | en | 0.774791 |
https://byjus.com/ncert-exemplar-class-10-maths-chapter-4-quadratic-equations/ | 1,713,218,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00478.warc.gz | 131,662,522 | 123,023 | NCERT Exemplar Class 10 Maths Solutions for Chapter 4 - Quadratic Equations
NCERT Exemplar Solutions Class 10 Maths Chapter 4 – Free PDF Download
NCERT Exemplar Solutions Class 10 Maths Chapter 4 Quadratic Equations are provided here to help students prepare for the board exams. The exemplars have been prepared by subject experts in accordance with the latest CBSE syllabus (2023-2024) and are available in PDF, which can be downloaded easily.
Chapter 4 in Class 10 Maths is an important chapter for Class 10 students. Besides, there will be several questions based on this chapter, and students have to solve these problems using quadratic equations. Further, while dealing with this chapter, students will learn about different quadratic equations and how to find solutions by either using the factorisation method or the square method. To help students grasp the complete concepts of this chapter, free NCERT Exemplar for quadratic equations is provided here. Click here to get exemplars for all chapters.
Students can go through the exemplar problems and solutions for Chapter 4 to understand the concepts introduced in this chapter, such as:
• Representing the given situations mathematically in the form of quadratic equations
• Solving quadratic equations with the help of the factorisation method
• Solving quadratic equations by completing the square
• Determining the nature of roots
Students can access the Class 10 Maths Chapter 4 NCERT Exemplar PDF below.
Exercise 4.1
Choose the correct answer from the given four options in the following questions:
1. Which of the following is a quadratic equation?
(A) x2 + 2x + 1 = (4 – x)2 + 3 (B) –2x2 = (5 – x)(2x-(2/5))
(C) (k + 1) x2 + (3/2) x = 7, where k = –1 (D) x3 – x2 = (x – 1)3
Solution:
(D) x3 – x2 = (x – 1)3
Explanation:
The standard form of a quadratic equation is given by,
ax2 + bx + c = 0, a ≠0
(A) Given, x2Â + 2x + 1 = (4 – x)2Â + 3
x2Â + 2x + 1 = 16 – 8x + x2Â + 3
10x – 18 = 0
which is not a quadratic equation.
(B) Given, -2x2 = (5 – x) (2x – 2/5)
-2x2 = 10x – 2x2 – 2 +2/5x
52x – 10 = 0
which is not a quadratic equation.
(C) Given, (k + 1) x2 + 3/2 x  = 7, where k = -1
(-1 + 1) x2 + 3/2 x = 7
3x – 14 = 0
which is not a quadratic equation.
(D) Given, x3Â – x2Â = (x – 1)3
x3 – x2 = x3 – 3x2 + 3x – 1
2x2Â – 3x + 1 = 0
2. Which of the following is not a quadratic equation?
(A) 2(x – 1)2 = 4x2 – 2x + 1 (B) 2x – x2 = x2 + 5
(C) (√2x + √3)2 + x2 = 3x2 − 5x (D) (x2 + 2x)2 = x4 + 3 + 4x3
Solution:
(D) (x2 + 2x)2 = x4 + 3 + 4x3
A quadratic equation is represented by the form,
ax2 + bx + c = 0, a ≠0
(A) Given, 2(x – 1)2Â = 4x2Â – 2x + 1
2(x2Â – 2x + 1) = 4x2Â – 2x + 1
2x2 + 2x – 1 = 0
(B) Given, 2x – x2Â = x2Â + 5
2x2Â – 2x + 5 = 0
(C) Given, (√2x + √3)2  = 3x2 – 5x
2x2 + 2√6x + 3  = 3x2 – 5x
x2 – (5 + 2√6)x – 3 = 0
(D) Given, (x2Â + 2x)2Â = x4Â + 3 + 4x2
x4Â + 4x3Â + 4x2Â = x4Â + 3 + 4x2
4x3 – 3 = 0
which is a cubic equation and not a quadratic equation.
3. Which of the following equations has 2 as a root?
(A) x2 – 4x + 5 = 0 (B) x2 + 3x – 12 = 0
(C) 2x2 – 7x + 6 = 0 (D) 3x2 – 6x – 2 = 0
Solution:
(C) 2x2 – 7x + 6 = 0
If 2 is a root then substituting the value 2 in place of x should satisfy the equation.
(A) Given,
x2Â – 4x + 5 = 0
(2)2 – 4(2) + 5 = 1 ≠0
So, x = 2 is not a root of x2Â – 4x + 5 = 0
(B) Given, x2 + 3x – 12 = 0
(2)2 + 3(2) – 12 = -2 ≠0
So, x = 2 is not a root of x2 + 3x – 12 = 0
(C) Given, 2x2Â – 7x + 6 = 0
2(2)2Â – 7(2) + 6 = 0
Here, x = 2 is a root of 2x2Â – 7x + 6 = 0
(D) Given, 3x2 – 6x – 2 = 0
3(2)2 – 6(2) – 2 = -2 ≠0
So, x = 2 is not a root of 3x2 – 6x – 2 = 0
4. If ½ is a root of the equation x2 + kx – 5/4 = 0, then the value of k is
(A) 2 (B) – 2
(C) ¼ (D) ½
Solution:
(A) 2
If ½ is a root of the equation
x2 + kx – 5/4 = 0 then, substituting the value of ½ in place of x should give us the value of k.
Given, x2 + kx – 5/4 = 0 where, x = ½
(½)2 + k (½) – (5/4) = 0
(k/2) = (5/4) – ¼
k = 2
5. Which of the following equations has the sum of its roots as 3?
(A) 2x2 – 3x + 6 = 0 (B) –x2 + 3x – 3 = 0
(C) √2x2 – 3/√2x+1=0 (D) 3x2 – 3x + 3 = 0
Solution:
(B) –x2 + 3x – 3 = 0
The sum of the roots of a quadratic equation ax2 + bx + c = 0, a ≠0 is given by,
Coefficient of x / coefficient of x2 = – (b/a)
(A) Given, 2x2Â – 3x + 6 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(B) Given, -x2 + 3x – 3 = 0
Sum of the roots = – b/a = -(3/-1) = 3
(C) Given, √2x2 – 3/√2x+1=0
2x2 – 3x + √2 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(D) Given, 3x2Â – 3x + 3 = 0
Sum of the roots = – b/a = -(-3/3) = 1
Exercise 4.2
1. x2 – 3x + 4 = 0
2. 2x2 + x – 1 = 0
3. 2x2 – 6x + 9/2 = 0
4. 3x2 – 4x + 1 = 0
5. (x + 4)2 – 8x = 0
6. (x – √2)2 – 2(x + 1) = 0
7. √2 x2 –(3/√2)x + 1/√2 = 0
8. x (1 – x) – 2 = 0
9. (x – 1) (x + 2) + 2 = 0
10. (x + 1) (x – 2) + x = 0
Solution:
(i)
The equation x2Â – 3x + 4 = 0 has no real roots.
D = b2Â – 4ac
= (-3)2Â – 4(1)(4)
= 9 – 16 < 0
Hence, the roots are imaginary.
(ii)
The equation 2x2 + x – 1 = 0 has two real and distinct roots.
D = b2Â – 4ac
= 12Â – 4(2) (-1)
= 1 + 8 > 0
Hence, the roots are real and distinct.
(iii)
The equation 2x2 – 6x + (9/2) = 0 has real and equal roots.
D = b2Â – 4ac
= (-6)2 – 4(2) (9/2)
= 36 – 36 = 0
Hence, the roots are real and equal.
(iv)
The equation 3x2 – 4x + 1 = 0 has two real and distinct roots.
D = b2Â – 4ac
= (-4)2Â – 4(3)(1)
= 16 – 12 > 0
Hence, the roots are real and distinct.
(v)
The equation (x + 4)2Â – 8x = 0 has no real roots.
Simplifying the above equation,
x2Â + 8x + 16 – 8x = 0
x2Â + 16 = 0
D = b2Â – 4ac
=Â (0) – 4(1) (16) < 0
Hence, the roots are imaginary.
(vi)
The equation (x – √2)2 – √2(x+1)=0 has two distinct and real roots.
Simplifying the above equation,
x2 – 2√2x + 2 – √2x – √2 = 0
x2 – √2(2+1)x + (2 – √2) = 0
x2 – 3√2x + (2 – √2) = 0
D = b2Â – 4ac
= (– 3√2)2 – 4(1)(2 – √2)
= 18 – 8 + 4√2 > 0
Hence, the roots are real and distinct.
(vii)
The equation √2x2 – 3x/√2 + ½ = 0 has two real and distinct roots.
D = b2Â – 4ac
= (- 3/√2)2 – 4(√2) (½)
= (9/2) – 2√2 > 0
Hence, the roots are real and distinct.
(viii)
The equation x (1 – x) – 2 = 0 has no real roots.
Simplifying the above equation,
x2 – x + 2 = 0
D = b2Â – 4ac
=Â (-1)2Â – 4(1)(2)
= 1 – 8 < 0
Hence, the roots are imaginary.
(ix)
The equation (x – 1) (x + 2) + 2 = 0 has two real and distinct roots.
Simplifying the above equation,
x2 – x + 2x – 2 + 2 = 0
x2Â + x = 0
D = b2Â – 4ac
=Â 12Â – 4(1)(0)
= 1 – 0 > 0
Hence, the roots are real and distinct.
(x)
The equation (x + 1) (x – 2) + x = 0 has two real and distinct roots.
Simplifying the above equation,
x2 + x – 2x – 2 + x = 0
x2 – 2 = 0
D = b2Â – 4ac
= (0)2Â – 4(1) (-2)
= 0 + 8 > 0
Hence, the roots are real and distinct.
2. Write whether the following statements are true or false. Justify your answers.
1. Every quadratic equation has exactly one root.
2. Every quadratic equation has at least one real root.
3. Every quadratic equation has at least two roots.
4. Every quadratic equations has at most two roots.
5. If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
6. If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.
Solution:
(i) False. For example, a quadratic equation x2 – 9 = 0 has two distinct roots – 3 and 3.
(ii) False. For example, equation x2 + 4 = 0 has no real root.
(iii) False. For example, a quadratic equation x2 – 4x + 4 = 0 has only one root which is 2.
(iv) True, because every quadratic polynomial has almost two roots.
(v) True, because in this case discriminant is always positive.
For example, in ax2+ bx + c = 0, as a and c have opposite sign, ac < 0
⟹ Discriminant = b2 – 4ac > 0.
(vi) True, because in this case discriminant is always negative.
For example, in ax2+ bx + c = 0, as b = 0, and a and c have same sign then ac > 0
⟹ Discriminant = b2 – 4ac = – 4 a c < 0
Solution:
No, a quadratic equation with integral coefficients may or may not have integral roots.
Justification
Consider the following equation,
8x2 – 2x – 1 = 0
The roots of the given equation are ½ and – ¼ which are not integers.
Hence, a quadratic equation with integral coefficient might or might not have integral roots.
Exercise 4.3
1. Find the roots of the quadratic equations by using the quadratic formula in each of the following:
1. 2 x2 – 3x – 5 = 0
2. 5x2 + 13x + 8 = 0
3. –3x2 + 5x + 12 = 0
4. –x2 + 7x – 10 = 0
5. x2 + 2 √2x – 6 = 0
6. x2 – 3 √5x + 10 = 0
7. (½)x2– √11x + 1 = 0
Solution:
ax2 + bx + c = 0, a ≠0 is given by,
(i) 2 x2 – 3x – 5 = 0
(ii) 5x2 + 13x + 8 = 0
(iii) –3x2 + 5x + 12 = 0
(iv) –x2 + 7x – 10 = 0
(v) x2 + 2 √2x – 6 = 0
(vi) x2 – 3 √5x + 10 = 0
(vii) (½)x2– √11x + 1 = 0
Exercise 4.4
1. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Solution:
Let the natural number = ‘x’.
According to the question,
We get the equation,
x² – 84 = 3(x+8)
x² – 84 = 3x + 24
x² – 3x – 84 – 24 = 0
x² – 3x – 108 = 0
x² – 12x + 9x – 108 = 0
x(x – 12) + 9(x – 12) = 0
(x + 9) (x – 12)
⇒ x = -9 and x = 12
Since, natural numbers cannot be negative.
The number is 12.
2. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Solution:
Let the natural number = x
When the number increased by 12 = x + 12
Reciprocal of the number = 1/x
According to the question, we have,
x + 12 = 160 times of reciprocal of x
x + 12 = 160/ x
x( x + 12 ) = 160
x2 + 12x – 160 = 0
x2 + 20x – 8x – 160 = 0
x( x + 20) – 8( x + 20)= 0
(x + 20) (x – 8) = 0
x + 20 = 0 or x – 8 = 0
x = – 20 or x = 8
Since, natural numbers cannot be negative.
The required number = x = 8
3. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.
Solution:
Let original speed of train = x km/h
We know,
Time = distance/speed
According to the question, we have,
Time taken by train = 360/x hour
And, Time taken by train its speed increase 5 km/h = 360/( x + 5)
It is given that,
Time taken by train in first – time taken by train in 2nd case = 48 min = 48/60 hour
360/x – 360/(x +5) = 48/60 = 4/5
360(1/x – 1/(x +5)) = 4/5
360 ×5/4 (5/(x² +5x)) =1
450 × 5 = x² + 5x
x² +5x -2250 = 0
x = (-5± √ (25+9000))/2
= (-5 ±√ (9025) )/2
= (-5 ± 95)/2
= -50, 45
But x ≠-50 because speed cannot be negative
So, x = 45 km/h
Hence, original speed of train = 45 km/h
4. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Solution:
Let Zeba’s age = x
According to the question,
(x-5)²=11+5x
x²+25-10x=11+5x
x²-15x+14=0
x²-14x-x+14=0
x(x-14)-1(x-14)=0
x=1 or x=14
We have to neglect 1 as 5 years younger than 1 cannot happen.
Therefore, Zeba’s present age = 14 years.
BYJU’S provides online study materials such as notes, exemplar books, question papers, and Maths NCERT Solutions for Class 10 to help students prepare for board exams in the most efficient way and score good marks. All these materials are available in downloadable PDFs. Students can also get an idea of the question pattern and marking scheme of Chapter 4 in the board exam by solving the sample papers and previous years’ question papers.
These exemplars will help students to solve problems in the right way as well as find answers to the most difficult questions given at the end of the chapter. Students can use these solved questions as a reference tool to practise Maths effectively and, in the process, develop good Math skills. They can also prepare for competitive exams using these exemplar solutions.
Get updated study materials to learn from us, and also download BYJU’S – The Learning App to experience a new method of learning with the help of educational videos clearing the concepts of Maths topics such as quadratic equations, linear equations, etc., in a visual way.
Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Maths Chapter 4
Q1
How many problems are there in NCERT Exemplar Solutions for Class 10 Maths Chapter 4?
Exercise 4.1 of NCERT Exemplar Solutions has 5 questions, Exercise 4.2 has 3 questions, Exercise 4.3 has 1 question with many sub-questions, and Exercise 4.4 has 4 questions. Long answers, short answers and MCQs are present in each exercise, covering the topics which are important from the exam point of view. The solutions for the exercise-wise problems are designed by a team of subject experts at BYJU’S, having vast conceptual knowledge.
Q2
What are the topics covered under NCERT Exemplar Solutions for Class 10 Maths Chapter 4?
Students can go through the exemplar problems and solutions for Chapter 4 to understand the concepts introduced in this chapter, such as:
1. Representing the given situations mathematically in the form of quadratic equations
2. Solving quadratic equations with the help of the factorisation method
3. Solving quadratic equations by completing the square
4. Determining the nature of roots
Q3
What are the roots of quadratic equations according to NCERT Exemplar Solutions for Class 10 Maths Chapter 4?
The values of variables satisfying the given quadratic equation are called their roots. In other words, x = α is the root of the quadratic equation f(x) if f(α) = 0.
The real roots of equation f(x) = 0 are the x-coordinates of the points where the curve y = f(x) intersects the x-axis.
1. One of the roots of the quadratic equation is zero, and the other is -b/a if c = 0
2. Both the roots are zero if b = c = 0
3. The roots are reciprocal to each other if a = c | 5,660 | 14,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-18 | latest | en | 0.877536 |
https://mathematica.stackexchange.com/questions/63415/timeseries-for-non-temporal-data | 1,660,667,714,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00009.warc.gz | 361,781,324 | 65,584 | # TimeSeries for non-temporal data
Mathematica introduced TemporalData in Mathematica 9. Mathematica 10 added several fantastic new TimeSeries manipulation tools -- tools that would come in handy for non-temporal data.
For example, I would love to use Resampling, MovingAverage, TimeSeriesAggregate, etc on wavelength scans, ie Fluorescence Intensity vs wavelength. At the moment, I have been treating wavelength as AbsoluteTime values.
filter = TimeSeries[{{0.525, 0.}, {0.526, 0.}, {0.527, 0.}, {0.528,
0.}, {0.529, 0.}, {0.53, 0.001}, {0.531, 0.001}, {0.532,
0.002}, {0.533, 0.002}, {0.534, 0.003}, {0.535, 0.004}, {0.536,
0.006}, {0.537, 0.01}, {0.538, 0.015}, {0.539, 0.026}, {0.54,
0.046}, {0.541, 0.092}, {0.542, 0.132}, {0.542, 0.189}, {0.542,
0.256}, {0.543, 0.361}, {0.544, 0.464}, {0.544, 0.556}, {0.544,
0.618}, {0.545, 0.644}, {0.546, 0.644}, {0.546, 0.626}, {0.546,
0.605}, {0.547, 0.575}, {0.548, 0.533}, {0.548, 0.476}, {0.548,
0.396}, {0.549, 0.324}, {0.55, 0.241}, {0.55, 0.178}, {0.55,
0.13}, {0.551, 0.094}, {0.552, 0.069}, {0.552, 0.051}, {0.553,
0.03}, {0.554, 0.019}, {0.555, 0.012}, {0.556, 0.008}, {0.557,
0.006}, {0.558, 0.004}, {0.559, 0.003}, {0.56, 0.002}, {0.561,
0.002}, {0.562, 0.001}, {0.563, 0.001}, {0.564, 0.001}, {0.565,
0.}, {0.566, 0.}, {0.567, 0.}, {0.568, 0.}, {0.569, 0.}, {0.570,
0.}, {0.571, 0.}, {0.572, 0.}, {0.573, 0.}, {0.574, 0.}, {0.575,
0.}}];
filter["Times"][[1 ;; 4]]
DateString /@ filter["Dates"][[1 ;; 4]]
Yields
{0.525, 0.526, 0.527, 0.528}
{"Mon 1 Jan 1900 00:00:00", "Mon 1 Jan 1900 00:00:00", "Mon 1 Jan \
1900 00:00:00", "Mon 1 Jan 1900 00:00:00"}
It appears to work, but it seems a bit of a kludge. Is there an equivalent Series that allows non-temporal independent variable with appropriate units?
I'll go into this answer admitting it's not ideal, because I also haven't found anything that packages non-temporal data as nicely as TimeSeries does. And maybe you already know that there are plenty of built-in functions that replicate the behavior of TimeSeriesResample, MovingAverage, TimeSeriesAggregate, etc. for non-temporal data arrays. But one of the things I love about Mathematica is the workhorse list of lists data structure. So my answer really just shows the function equivalents to the ones you mentioned for a straightforward two-dimensional array. Apologies if it's not a real answer for you, but here are the examples of the functions in which you specifically expressed interest.
(* TimeSeriesResample *)
resampledFilter = ArrayResample[filter, {2 Length[filter], 2}];
(* MovingAverage *)
movingAveragedFilter = MovingAverage[filter, 2];
(* TimeSeriesAggregate *)
movingMapped = MovingMap[StandardDeviation, filter, Quantity[3, "Events"]];
ListLinePlot[{filter, resampledFilter, movingAveragedFilter, movingMapped},
ImageSize -> Large,
PlotLegends -> LineLegend[{"original", "resampled", "moving averaged", "moving mapped SD"}]
] | 1,084 | 2,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | longest | en | 0.850244 |
https://meangreenmath.com/2013/08/26/thoughts-on-17-and-other-rational-numbers-part-9/ | 1,680,097,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00416.warc.gz | 444,094,304 | 34,091 | # Thoughts on 1/7 and other rational numbers (Part 9)
Let’s now consider the decimal representation of $\displaystyle \frac{8}{17}$.
There’s no obvious repeating pattern. But we know that, since 17 has neither 2 nor 5 as a factor, that there has to be a repeating decimal pattern.
So… what is it?
When I ask this question to my students, I can see their stomachs churning a slow dance of death. They figure that the calculator didn’t give the answer, and so they have to settle for long division by hand.
That’s partially correct.
However, using the ideas presented below, we can perform the long division extracting multiple digits at once. Through clever use of the calculator, we can quickly obtain the full decimal representation even though the calculator can only give ten digits at a time.
Let’s now return to where this series began… the decimal representation of $\displaystyle \frac{1}{7}$ using long division. As shown below, the repeating block has length $6$, which can be found in a few minutes with enough patience. By the end of this post, we’ll consider a modification of ordinary long division that facilitates the computation of really long repeating blocks.
Because we arrived at a repeated remainder, we know that we have found the repeating block. So we can conclude that $\displaystyle \frac{1}{7} = 0.\overline{142857}$.
Students are taught long division in elementary school and are so familiar with the procedure that not much thought is given to the logic behind the procedure. The underlying theorem behind long division is typically called the division algorithm. From Wikipedia:
Given two integers $a$ and $b$, with $b \ne 0$, there exist unique integers $q$ and $r$ such that $a = bq+r$ and $0 \le r < |b|$, where $|b|$ denotes the absolute value of $b$.
The number $q$ is typically called the quotient, while the number $r$ is called the remainder.
Repeated application of this theorem is the basis for long division. For the example above:
Step 1.
$10 = 1 \times 7 + 3$. Dividing by $10$, $1 = 0.1 \times 7 + 0.3$
Step 2.
$30 = 4 \times 7 + 2$. Dividing by $100$, $0.3 = 0.04 \times 7 + 0.02$
Returning to the end of Step 1, we see that
$1 = 0.1 \times 7 + 0.3 = 0.1 \times 7 + 0.04 \times 7 + 0.02 = 0.14 \times 7 + 0.02$
Step 3.
$20 = 2 \times 7 + 6$. Dividing by $1000$, $0.02 = 0.002 \times 7 + 0.006$
Returning to the end of Step 2, we see that
$1 = 0.14 \times 7 + 0.02 = 0.14 \times 7 + 0.0002 \times 7 + 0.006 = 0.142 \times 7 + 0.006$
And so on.
By adding an extra zero and using the division algorithm, the digits in the decimal representation are found one at a time. That said, it is possible (with a calculator) to find multiple digits in a single step by adding extra zeroes. For example:
Alternate Step 1.
$1000 = 142 \times 7 + 6$. Dividing by $1000$, $1 = 0.142 \times 7 + 0.006$
Alternate Step 2.
$6000 = 587 \times 7 + 1$. Dividing by $100000$, $0.006 = 0.000587 \times 7 + 0.000001$
Returning to the end of Alternate Step 1, we see that
$1 = 0.142 \times 7 + 0.006= 0.142 \times 7 + 0.000587\times 7 + 0.000001 = 0.142857 \times 7 + 0.000001$
So, with these two alternate steps, we arrive at a remainder of $1$ and have found the length of the repeating block.
The big catch is that, if $a = 1000$ or $a = 6000$ and $b = 7$, the appropriate values of $q$ and $r$ have to be found. This can be facilitated with a calculator. The integer part of $1000/7$ and $6000/7$ are the two quotients needed above, and subtraction is used to find the remainders (which must be less than $7$, of course).
At first blush, it seems silly to use a calculator to find these values of $q$ and $r$ when a calculator could have been used to just find the decimal representation of $1/7$ in the first place. However, the advantage of this method becomes clear when we consider fractions who repeating blocks are longer than 10 digits.
Let’s now return to the question posed at the top of this post: finding the decimal representation of $\displaystyle \frac{8}{17}$. As noted in Part 6 of this series, the length of the repeating block must be a factor of $\phi(17)$, where $\phi$ is the Euler toitent function, or the number of integers less than $17$ that are relatively prime with $17$. Since $17$ is prime, we clearly see that $\phi(17) = 16$. So we can conclude that the length of the repeating block is a factor of $16$, or either $1$, $2$, $4$, $8$, or $16$.
Here’s the result of the calculator again:
We clearly see from the calculator that the repeating block doesn’t have a length less than or equal to $8$. By process of elimination, the repeating block must have a length of $16$ digits.
Now we perform the division algorithm to obtain these digits, as before. This can be done in two steps by multiplying by $10^8 = 100,000,000$.
So, by the same logic used above, we can conclude that
$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$
In other words, through clever use of the calculator, the full decimal representation can be quickly found even if the calculator itself returns only ten digits at a time… and had rounded the final $2941176$ of the repeating block up to $3$.
## 2 thoughts on “Thoughts on 1/7 and other rational numbers (Part 9)”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,481 | 5,326 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 63, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-14 | latest | en | 0.924321 |
https://bookdown.org/scottming/rns/r-.html | 1,642,842,252,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00557.warc.gz | 207,114,215 | 8,413 | # 1 R 数据结构
## 1.1 向量
R 里面的向量看起来像 Python 的 list，但又不是 list，更像是Python 里一维的数组。
``````vec <- c(1, 2, 3, 6, 5, 4)
vec[c(1, 2, 4)]
#> [1] 1 2 6
class(vec)
#> [1] "numeric"
vec[1:2] # 末尾包含
#> [1] 1 2``````
Python 代码,除了用列表推导式之外还可以直接用 Numpy 实现
``````import numpy as np
a= np.array([1, 2, 3, 6, 5, 4])
print a[[0, 1, 3]]
print type(a)
#> [1 2 6]
#> <type 'numpy.ndarray'>``````
``````a <- seq(10)
a
#> [1] 1 2 3 4 5 6 7 8 9 10
b <- seq(10, 13)
b
#> [1] 10 11 12 13
temp <- c(1, 2, 4, 0)
temp * b
#> [1] 10 22 48 0``````
``````vec <- seq(1, 100, length.out = 10) # 还有个long参数也非常有用。
vec
#> [1] 1 12 23 34 45 56 67 78 89 100
vec[-4] #这点跟 Python 很不一样
#> [1] 1 12 23 45 56 67 78 89 100
seq(from=2, to=1000, length.out = 10)
#> [1] 2 113 224 335 446 556 667 778 889 1000
seq(from=2, to=1000, length=10)
#> [1] 2 113 224 335 446 556 667 778 889 1000
group1 <- rep(1:3, times = c(8, 10, 9))
group2 <- factor(group1) # 转换成因子
group1
#> [1] 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3
class(group1)
#> [1] "integer"
class(group2)
#> [1] "factor"
length(group1) # as Py len
#> [1] 27``````
``````vec_random1 <- runif(5) # 0-1
vec_random1
#> [1] 0.0808 0.8343 0.6008 0.1572 0.0074
vec_random2 <- sample(c('A', 'B'), size = 10, replace = TRUE) # 随机字符向量
vec_random2
#> [1] "A" "A" "A" "B" "B" "B" "A" "A" "A" "A"
vector1 <- numeric(10)
vector1 # empty vector
#> [1] 0 0 0 0 0 0 0 0 0 0``````
``````temp[1:3]
#> [1] 1 2 4
temp[c(TRUE, TRUE, FALSE, FALSE)]
#> [1] 1 2
temp[temp > 1]
#> [1] 2 4``````
## 1.2 矩阵
``````vector <- 1:12
class(vector)
#> [1] "integer"
my_matrix <- matrix(vector, nrow = 3, ncol = 4, byrow = FALSE)
dim(my_matrix)
#> [1] 3 4
vector
#> [1] 1 2 3 4 5 6 7 8 9 10 11 12
my_matrix
#> [,1] [,2] [,3] [,4]
#> [1,] 1 4 7 10
#> [2,] 2 5 8 11
#> [3,] 3 6 9 12
vector1 <- vector2 <- vector3 <- runif(3)
my_matrix <- cbind(vector1, vector2, vector3) # 一致的向量类型进行合并
my_matrix
#> vector1 vector2 vector3
#> [1,] 0.1957 0.1957 0.1957
#> [2,] 0.4035 0.4035 0.4035
#> [3,] 0.0637 0.0637 0.0637
round(my_matrix*10, digits = 2) # 取2位
#> vector1 vector2 vector3
#> [1,] 1.96 1.96 1.96
#> [2,] 4.04 4.04 4.04
#> [3,] 0.64 0.64 0.64``````
``````my_mat <- matrix(c(8, 3, 4, 1, 5, 9, 6, 7, 2), ncol = 3)
print(my_mat)
#> [,1] [,2] [,3]
#> [1,] 8 1 6
#> [2,] 3 5 7
#> [3,] 4 9 2
my_mat[1,1] + my_mat[1, 2] + my_mat[1,3]
#> [1] 15
sum(my_mat[1,])
#> [1] 15
rowSums(my_mat)
#> [1] 15 15 15
colSums(my_mat)
#> [1] 15 15 15
sum(diag(my_mat))
#> [1] 15
class(my_mat[1,]) # 退化成向量了
#> [1] "numeric"
my_mat[1, , drop=FALSE] # 保留矩阵属性
#> [,1] [,2] [,3]
#> [1,] 8 1 6
my_mat[my_mat <= 5] <- 0 # 改变赋值
my_mat
#> [,1] [,2] [,3]
#> [1,] 8 0 6
#> [2,] 0 0 7
#> [3,] 0 9 0
# ifelse 函数
my_mat <- matrix(c(8, 3, 4, 1, 5, 9, 6, 7, 2), ncol = 3)
ifelse(my_mat > 0.5, 1, 0)
#> [,1] [,2] [,3]
#> [1,] 1 1 1
#> [2,] 1 1 1
#> [3,] 1 1 1``````
## 1.3 数据框
``````city <- c('A', 'B', 'C', 'D')
temp <- c(27, 29, 23, 14)
data <- data.frame(city, temp) # 对于 list，Py 不能直接这么导，会把整数向量变为索引，得用 dir 的方式
data
#> city temp
#> 1 A 27
#> 2 B 29
#> 3 C 23
#> 4 D 14
data[, 1] # Py data.ix[:,0] | data.iloc[:,0]
#> [1] A B C D
#> Levels: A B C D
data\$temp
#> [1] 27 29 23 14
class(data\$city) # 本是字符，自动转了因子，若不想转，课用 stringAsFactors = FALSE 设定
#> [1] "factor"``````
``````data[data\$temp > mean(data\$temp), ]
#> city temp
#> 1 A 27
#> 2 B 29
# data['temp', ] # empty，这是提取行的语法
data[, 'city'] # 提取列
#> [1] A B C D
#> Levels: A B C D
data\$temp > mean(data\$temp)
#> [1] TRUE TRUE FALSE FALSE
with(data, data[temp > mean(temp), ]) # 直接操作列名,省略 \$
#> city temp
#> 1 A 27
#> 2 B 29
with(data, data[temp > mean(temp), 'city'])
#> [1] A B
#> Levels: A B C D``````
``````summary(data) # as Py describe
#> city temp
#> A:1 Min. :14.0
#> B:1 1st Qu.:20.8
#> C:1 Median :25.0
#> D:1 Mean :23.2
#> 3rd Qu.:27.5
#> Max. :29.0
dim(data) # as Py data.shape
#> [1] 4 2
#> city temp
#> 1 A 27
#> 2 B 29
#> 3 C 23
#> 4 D 14
#> city temp
#> 1 A 27
str(data) # 返回数据结构
#> 'data.frame': 4 obs. of 2 variables:
#> \$ city: Factor w/ 4 levels "A","B","C","D": 1 2 3 4
#> \$ temp: num 27 29 23 14``````
``````order(data\$temp) # 返回数据的索引号
#> [1] 4 3 1 2
data[order(data\$temp), ]
#> city temp
#> 4 D 14
#> 3 C 23
#> 1 A 27
#> 2 B 29
data[order(data\$temp, decreasing = T), ][1:2, ] # 反序
#> city temp
#> 2 B 29
#> 1 A 27``````
## 1.4 列表
``````data_list <- list(temp = temp, city = city)
print(data_list)
#> \$temp
#> [1] 27 29 23 14
#>
#> \$city
#> [1] "A" "B" "C" "D"
data_list\$mat <- my_mat
data_list\$data <- data
names(data_list)
#> [1] "temp" "city" "mat" "data"
class(data_list[3])
#> [1] "list"
data_list[3]
#> \$mat
#> [,1] [,2] [,3]
#> [1,] 8 1 6
#> [2,] 3 5 7
#> [3,] 4 9 2
data_list[[3]] # 没有 name
#> [,1] [,2] [,3]
#> [1,] 8 1 6
#> [2,] 3 5 7
#> [3,] 4 9 2
class(data_list[[3]])
#> [1] "matrix"``````
## 1.5 特殊对象
``````temp <- c(27, 29, 23, 14, NA)
mean(temp)
#> [1] NA
mean(temp, na.rm = TRUE)
#> [1] 23.2
temp <- c(27, 29, 23, 14, NULL)
data_list <- NULL # 快速删除一列
mean(temp) # TRUE
#> [1] 23.2``````
``````textcon = textConnection('output', 'w')
sink(textcon) # 开
x <- runif(10)
summary(x)
#> Min. 1st Qu. Median Mean 3rd Qu. Max.
#> 0.051 0.315 0.604 0.557 0.740 0.981
print('这话不显示了，写入了 output 对象了')
#> [1] "这话不显示了，写入了 output 对象了"
sink() # 关闭控制台转换
print(output)
#> character(0)
showConnections()
#> description class mode text isopen can read can write
#> 4 "output" "textConnection" "wr" "text" "opened" "no" "yes"
#> 5 "output" "textConnection" "w" "text" "opened" "no" "yes"
class(textcon)
#> [1] "textConnection" "connection"
close(textcon)``````
``````n <- 1:50
xvar <- paste0('x', n )
right <- paste(xvar, collapse = ' + ')
left <- 'y~'
my_formula <- paste(left, right)
my_formula <- as.formula(my_formula)
class(my_formula)
#> [1] "formula"
left
#> [1] "y~"``````
``````ex <- expression(x <- seq(1, 10, 2))
print(ex)
#> expression(x <- seq(1, 10, 2))
class(ex)
#> [1] "expression"
eval(ex)
print(x)
#> [1] 1 3 5 7 9
tex <- c('z<-seq(1, 10, 2)', 'print(z)' )
eval(parse(text = tex))
#> [1] 1 3 5 7 9``````
``````ls()
#> [1] "a" "b" "city" "data" "data_list"
#> [6] "ex" "group1" "group2" "left" "my_formula"
#> [11] "my_mat" "my_matrix" "n" "output" "right"
#> [16] "temp" "tex" "textcon" "vec" "vec_random1"
#> [21] "vec_random2" "vector" "vector1" "vector2" "vector3"
#> [26] "x" "xvar" "z"
env1 <- new.env()
assign("x1", 1:5, envir = env1)
ls(envir = env1)
#> [1] "x1"
get('x1', envir = env1) # 为啥没法直接用 x1 取呢？
#> [1] 1 2 3 4 5
exists('x1', envir = env1)
#> [1] TRUE
rm('x1', envir = env1)``````
``````exp(1)
#> [1] 2.72
myfunc <- function(r){
area <- pi*r^2
return(area)
} # 内部可以调用 global value
print(myfunc(4))
#> [1] 50.3
myfunc
#> function(r){
#> area <- pi*r^2
#> return(area)
#> }`````` | 3,770 | 7,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.29343 |
https://miningmath.com/docs/knowledgebase/formatting-data/formatting-steps/ | 1,723,307,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00063.warc.gz | 322,246,320 | 56,411 | #### MiningMath
Easily integratable to your preferred package through CSV files
# Formatting the Block Model
Estimated reading: 5 minutes 2353 views
## Block Model Basic requirements
MiningMath requires the following formatting specifications:
1. Regularized block model: This means all blocks must be the same size.
2. Air blocks must be removed prior to importation. This is the way MiningMath recognizes the topography.
3. Coordinates of each block in the 3 dimensions.
4. Header Names should not have special characters or have them exceed 13. Use this recommendation for folders and files also.
5. The data format should be a CSV file (Comma Separated Value), which might be compatible with most mining packages.
## Good practices
1. Configure Microsoft Windows number formatting to use dot as the decimal separator.
2. Use the metric system.
3. Set multiple fields that will consider different economic values, material types, contaminant limits, and any other variable you wish to analyze or control.
## Understanding Field Types
Field Types are the fields MiningMath can understand. Each column imported should be assigned to the proper field type so that the software treats each variable accordingly with its meaning.
## Mandatory Field Types and their meanings
1. Coordinates X, Y, and Z refer to your geo-referenced information.
2. Average refers to any variable that could be controlled by means of minimums and maximums considering its average: grades, haulage distance, and other variables.
3. Economic Value refers to the columns with the economic value, which represent the available destinations. It is possible to import multiple economic values at once, and they may be used simultaneously (ex.: multiple processing streams) or calculated in the internal calculator mentioned on the next page.
## Optional Field Types and their meanings
1. Density refers to the block's density. This field is used to calculate the block's tonnage.
2. Slope refers to slopes varying block-by-block, which gives the flexibility to define slopes by lithotype and sectors.
3. Recovery refers to recoveries varying block-by-block.
4. Sum refers to any variable that could be controlled by means of minimums and maximums considering its sum.
5. Predefined destinations refers to possible fixed destination values. This can be used for example if you want to define pushbacks or apply lithologic restrictions that prevent certain blocks to be processed. However, by fixing destinations you are impeding MiningMath to reach its full potential. More about this here.
6. Other refers to information that you with to have in the exported outputs.
7. Skip refers to any variable that should be ignored. This field type might help improving the runtime since these variables will not be considered and exported along with the optimization outputs.
## Field names shortcut
Shortcuts can be used for automatic recognition in the importation process. These are listed in the table below.
Field name Shortcuts
Coordinates
X | Y | Z
Average
@ | grade
Density
% | dens | sg
Economic value
\$ | dest | val
Recovery
* | recov
Slope
/ | slope
Sum
+
Skip
!
## Mandatory requirements
Considering the specifications mentioned before, the formatted data set should have the following information for each block:
1. Coordinates.
2. Grades (at least one element assigned as Average).
3. Economic values (at least 1 process and 1 waste).
The following video gives an introduction on how to setup your block model.
Video 1: Block Model setup.
## Attention to software conversions
The model’s origin must be placed at the bottom portion, starting to count from the minimum coordinates at X, Y, and Z.
Figure 1 highlights a block model origin at the corner of the first block and the coordinates on its centroid.
Each software uses its own conventions for data format, naming and numbering systems, etc. These differences should be observed to prevent conflicts when transiting data from multiple software, each one for one specificity.
What you must know:
1. MiningMath uses coordinates (X,Y,Z) for which Z, which represents the elevation, starts upwards (Figure 3a).
2. Other mining software may use indexes with IZ starting downwards (Figure 3b). MineSight is an example that uses this notation.
There is no right or wrong convention, but there is a correct procedure for each software.
To invert coordinates use the following formula to convert:
$$new(Z) = max(Z) + 1 – current(Z)$$
## Air Blocks
MiningMath recognizes that all imported blocks of your model are underground. This means it is necessary to remove all the air blocks prior to importation. Unless your topography is totally flat, which is unlikely, the image below shows an example of your model should be displayed.
The non-removal of air blocks may lead to unsatisfactory results and long processing times, since it would be considering blocks that do not exist in reality.
#### More Details on Air Blocks
The following video shows how to do remove air blocks using filters on MS Excel. These tips are also applicable to any mining software of your choice.
Video 1: Removing air blocks using filters on MS Excel. | 1,106 | 5,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.809988 |
http://oeis.org/A214865 | 1,558,575,850,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00091.warc.gz | 147,171,124 | 4,107 | This site is supported by donations to The OEIS Foundation.
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A214865 n such that n XOR 9 = n - 9. 2
9, 11, 13, 15, 25, 27, 29, 31, 41, 43, 45, 47, 57, 59, 61, 63, 73, 75, 77, 79, 89, 91, 93, 95, 105, 107, 109, 111, 121, 123, 125, 127, 137, 139, 141, 143, 153, 155, 157, 159, 169, 171, 173, 175, 185, 187, 189, 191, 201, 203, 205, 207, 217, 219, 221, 223, 233, 235, 237, 239, 249, 251 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1). FORMULA a(n) = 4*n + 6 + (-1)^n + 2*(-1)^((2*n+(-1)^n-1)/4) for n>=0. a(n) = A016825(n+1) + A132429(n) for n>=0. G.f. x*(9+2*x+2*x^2+2*x^3+x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Mar 10 2013 a(n+4) = a(n) + 16. - Alexander R. Povolotsky, Mar 15 2013 MATHEMATICA CoefficientList[Series[x*(9 + 2*x + 2*x^2 + 2*x^3 + x^4)/((1 + x)*(x^2 + 1)*(x - 1)^2), {x, 0, 50}], x] (* G. C. Greubel, Feb 22 2017 *) PROG (MAGMA) XOR := func; m:=9; for n in [1 .. 500] do if (XOR(n, m) eq n-m) then n; end if; end for; (PARI) x='x+O('x^50); Vec(x*(9+2*x+2*x^2+2*x^3+x^4) / ( (1+x)*(x^2+1)*(x-1)^2 )) \\ G. C. Greubel, Feb 22 2017 CROSSREFS Cf. A214863, A016825, A132429. Sequence in context: A289686 A251394 A155877 * A225557 A120177 A291350 Adjacent sequences: A214862 A214863 A214864 * A214866 A214867 A214868 KEYWORD nonn,easy AUTHOR Brad Clardy, Mar 09 2013 STATUS approved
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Last modified May 22 21:17 EDT 2019. Contains 323504 sequences. (Running on oeis4.) | 813 | 1,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-22 | latest | en | 0.516852 |
https://www.nomidl.com/computer-vision/sobel-edge-detection/ | 1,721,624,317,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00057.warc.gz | 795,061,858 | 23,214 | ### Comprehensive Guide to Sobel Edge Detection with Examples
Edge detection is a critical operation in image processing, it is used to identify boundaries between the objects or regions in an image. In this article we are going to discuss about the Sobel Edge Detection method.
## What is Sobel Edge Detection?
Sobel Edge Detection is a first-order derivative edge detection method that was developed by Irwin Sobel in 1968. It is based on the concept of finding the gradient of the intensity of the image, and it is commonly used due to its simplicity and efficiency
The Sobel Edge Detection method calculates the gradient of the intensity of the image using the Sobel operator, which is a 3×3 kernel. The Sobel operator calculates the gradient in both the x and y directions, and the gradient magnitude can be calculated by combining these two gradients.
## The Sobel Edge Detection method consists of the following stages:
1 – Gaussian Blur: The input image is smoothed using a Gaussian filter to remove noise.
2 – Gradient Calculation: The gradient of the intensity of the image is calculated using the Sobel operator.
3 – Thresholding: The gradient magnitude is thresholded to keep only the strong edges and remove weak edges.
The result of the Sobel Edge Detection method is a binary image where the edges are represented by white pixels, and the background is represented by black pixels.
## Example of Sobel Edge Detection in Python and OpenCV:
In this example, the input image is first smoothed using a Gaussian filter to remove noise. Then, the Sobel Edge Detection method is applied to the smoothed image, and the gradient magnitude is calculated by combining the x and y gradients. Finally, the result is displayed using the imshow function from the matplotlib library.
## Conclusion
Sobel Edge Detection is a popular edge detection method that is based on the concept of finding the gradient of the intensity of the image. It is simple, efficient, and widely used in image processing applications, including object recognition, image segmentation, and pattern recognition. The Sobel Edge Detection method can be easily implemented using the OpenCV library, and it provides accurate and efficient edge detection results. | 434 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.889726 |
https://www.wisegeek.com/what-is-a-nautical-mile.htm | 1,611,472,274,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00107.warc.gz | 1,043,644,649 | 19,394 | # What is a Nautical Mile?
Mary McMahon
Mary McMahon
A nautical mile is a unit of measurement used in aeronautics and maritime navigation. Some people also prefer to refer to a nautical mile as a sea mile, maritime unit, or sea unit, in a reference to the maritime usage of the measurement. When speed is given in nautical miles, the correct term is “knots,” as in “the ship traveled at 23 knots,” indicating that the ship was moving at a rate of 23 nautical miles per hour.
Technically, a nautical mile is equal to the length of one minute of arc in a great circle. If this statement has caused you to blink furiously in confusion, think about the Earth as a large orange. If you cut the orange in half along the middle, or equator, you have bisected the orange along one of its “great circles.” There are numerous other great circles, as a great circle is any line around a sphere which can be traced to cut the sphere precisely in half. There are 360 degrees in a great circle, and each degree can be further broken down into 60 minutes, sometimes called minutes of arc. Therefore, the Earth measures 21,600 nautical miles around the equator.
After much international debate, it has been agreed that a nautical mile is equal to 6,076 feet (1,852 meters). International agreement on the measurement of a nautical mile is important, since many treaties and agreements include nautical miles as a unit of measurement. Furthermore, it ensures smooth navigation between various nations, which is especially important for major trading partners. As a result, a nautical mile is recognized within the framework known as the International System of Units (SI), which is a system of measurements which have been clearly defined and agreed upon by most countries in the world.
Usage of nautical miles varies around the world. Since a nautical mile is not technically an SI measurement, some nations prefer to use distances which are recognized under this system, which prefers metric measurements like meters and kilometers to measurements such as inches and feet. In some regions, the nautical mile is heavily used and understood by seafarers, while other areas, distances may be expressed in other measurements, especially when an international team is cooperating on a project.
For anyone who is dying to know about the origin of “knot” as a term for a unit of speed in maritime navigation, the word is closely linked to the technique sailors used to employ to measure their speed. A weighted line would be thrown out from a ship by one sailor while another held a timer. Knots in the line were placed at precise lengths, so that the sailor could count off the knots to figure out how far the ship had traveled in a set amount of time, typically 30 seconds. This measurement could be extrapolated into an estimate of the overall speed of the ship, and while more precise navigational tools are used now, “knots” for speed continues to be used.
Mary McMahon
Ever since she began contributing to the site several years ago, Mary has embraced the exciting challenge of being a wiseGEEK researcher and writer. Mary has a liberal arts degree from Goddard College and spends her free time reading, cooking, and exploring the great outdoors.
Mary McMahon
Ever since she began contributing to the site several years ago, Mary has embraced the exciting challenge of being a wiseGEEK researcher and writer. Mary has a liberal arts degree from Goddard College and spends her free time reading, cooking, and exploring the great outdoors.
## You might also Like
anon83763
so what's the origin of the nautical mile?
mwilson
If a nautical mile is squared is it a nautical square mile, or a square nautical mile?
anon13750
LOL. We're both right now. The original article read "21,600 miles", not "21,600 nautical miles". Thanks for making the change.
anon13507
Gee I always thought the diameter of Earth was around 24,900 miles not 21,600 as in paragraph two. I think you used 5280 feet (mile) instead of 6076 (nautical mile) in your calculation. | 873 | 4,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-04 | latest | en | 0.968022 |
https://mypenservices.com/exercise-designed-to-appreciate-comparative-merit-of-cp-writing-homework-help/ | 1,632,072,609,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00059.warc.gz | 452,880,482 | 17,940 | Exercise designed to appreciate comparative merit of CP, writing homework help
Question description Exercise 44 (For-Credit of 6 points)Derive the conclusion from the premises:[44-1] Exercise designed to appreciate comparative merit of CP: with the same argument below, [44-1.1] do the 1st proof without using CP; & [44-1.2] do the 2nd proof by using CP: C: A -> F 1: ~F -> ~H 2: ~A V H[44-2] Exercise to compare two CPs, one of which employs CP as part of the whole process: with the same argument below, [44-2.1] do the 1st proof by using CP with ~X as AP [44-2.2] do the 2nd proof by using CP with Y as AP; & then Contra C: ~X -> ~Y 1: (J V Q) -> X 2: Y -> (J & N)[44-3] Exercise designed to see CP as a self-contained module: use CP to the get the 1st part of conjunction; use the conditional as a premise to get the 2nd part; & be careful not to take the 2nd part of conjunction from CP C: (D -> E) & ~K 1: ~D V ~K 2: ~E -> K 3: (~D V E) -> ~K[44-4] Exercise designed to employ CP twice to get an equivalence do the 1st proof for M -> G by CP; do the 2nd proof for G -> M by CP; combine them by Conj; & then use Equiv C: M <-> G 1: M V ~G 2: M -> ~Y 3: ~G -> Y
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Place your first order with code to get 15% discount right away! | 442 | 1,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.946305 |
http://blog.richardkiss.com/?m=201306 | 1,511,462,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806856.86/warc/CC-MAIN-20171123180631-20171123200631-00275.warc.gz | 43,870,121 | 5,666 | # How to be Minimally Redundant (or “A Splitting Headache”)
I’ve recently been investigating the command-line utility zfec, which is a lot like the UNIX “split” utility, except, using a technique called erasure coding, you can choose to split your file into M pieces, any K of which are enough to recreate the file using the complimentary [sic] zunfec command.
Concrete example: a 100K file can be split into 10 (or more) pieces, each just over 25K long, and zunfec can recreate the original from any 4 of them.
Any 4. You might expect this to incur a lot of overhead in each piece, but it turns out it’s just a header of four bytes or less. Pretty much the same as cutting the file into pieces.
This amazed me. How could this possible work? How can you split data into M pieces so that any K of them is enough to reconstruct them? Linear algebra to the rescue!
Suppose we want to encode a twelve character text string into a bunch of arrays, each with four values, any three of which are sufficient to reconstruct the original. Let’s use the string “Lovefromdawn”. Here’s what you do:
The matrix furthest to the left is a Vandermonde matrix, which is a matrix where each row forms a geometric series starting with 1. A square Vandermonde matrix has the special property that as long as the second column has no repeated elements, the matrix is invertible (and in fact, there is a special algorithm to invert it quickly).
The split pieces correspond to rows of the matrix on the right. Choose any three of them; let’s say row 1, 2 and 4. Once we know what row numbers they are (and this explains the tiny header), we get the following system of equations:
We know the leftmost matrix is invertible, so multiply both sides by the inverse, and we solve for X, recovering “Lovefromdawn”. Wowza!
But wait! One problem remains: if we use standard integer arithmetic, the matrix on the right ends up with a lot of values larger than 255, and we can’t store it in a byte array. That ain’t no good.
Luckily, it turns out that most theorems dealing with matrices and invertibility only assume we are operating over an arbitrary field, a mathematical structure that has addition, a 0, multiplication, a 1, and a reciprocal for every non-0 element.
Abstract algebra to the rescue! (Math seems to rescue computer science a lot.) It turns out that there is a field with 256 elements known as GF(2**8). In this field, 0 is the 0, 1 is the 1 (surprise!), addition is bitwise exclusive-or (so every number is its own additive inverse!), and multiplication is a weird, and seemingly unpredictable beast, that is best calculated by pre-populating a table of logarithms and exponents, then creating a 256×256 array containing the multiplication table (since each number fits in a byte, the whole table only takes 64K).
[There are actually lots of ways to define multiplication that still meets the requirements of a field, but we have to pick one by defining what’s called the irreducible polynomial for the field.]
Once we operate over this field, all matrix values are guaranteed to be in the range [0, 255], and our method for bytewise encoding and decoding becomes easy.
One more aside: zfec actually piles the 3×3 (well, KxK) identity matrix on top of the Vandermonde matrix, so the first K pieces are exactly what you’d get by cutting the file into K pieces in the obvious way and putting the tiny header in front. Even though our matrix is no longer a Vandermonde matrix, submatrices are still guaranteed to be invertible and it still works. | 813 | 3,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-47 | longest | en | 0.928627 |
https://www.slideserve.com/bernad/regression-analysis-with-spss | 1,696,219,013,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00029.warc.gz | 1,076,536,499 | 23,577 | 1 / 51
Regression Analysis with SPSS
2. Outline. ConceptualizationSchematic Diagrams of Linear Regression processesUsing SPSS, we plot and test relationships for linearityNonlinear relationships are transformed to linear onesGeneral Linear ModelDerivation of Sums of Squares and ANOVA Derivation of intercept and regression coefficientsThe Prediction Interval and its derivationModel AssumptionsExplanationTestingAssessmentAlternatives when assumptions are unfulfilled.
Regression Analysis with SPSS
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Presentation Transcript
1. 1 Regression Analysis with SPSS Robert A. Yaffee, Ph.D. Statistics, Mapping and Social Science Group Academic Computing Services Information Technology Services New York University Office: 75 Third Ave Level C3 Tel: 212.998.3402 E-mail: yaffee@nyu.edu February 04
2. 2
3. 3 Conceptualization of Regression Analysis Hypothesis testing Path Analytical Decomposition of effects
4. 4 Hypothesis Testing For example: hypothesis 1 : X is statistically significantly related to Y. The relationship is positive (as X increases, Y increases) or negative (as X decreases, Y increases). The magnitude of the relationship is small, medium, or large. If the magnitude is small, then a unit change in x is associated with a small change in Y.
5. 5 Regression Analysis Have a clear notion of what you can and cannot do with regression analysis Conceptualization A Path Model of a Regression Analysis
6. 6
7. 7
8. 8 A Precursor to Modeling with Regression Data Exploration: Run a scatterplot matrix and search for linear relationships with the dependent variable.
9. 9 Click on graphs and then on scatter
10. 10 When the scatterplot dialog box appears, select Matrix
11. 11 A Matrix of Scatterplots will appear
12. 12
13. 13
14. 14 Decomposition of the Sums of Squares
15. 15 Graphical Decomposition of Effects
16. 16 Decomposition of the sum of squares
17. 17 Decomposition of the sum of squares Total SS = model SS + error SS and if we divide by df This yields the Variance Decomposition: We have the total variance= model variance + error variance
18. 18 F test for significance and R2 for magnitude of effect R2 = Model var/total var
19. 19 ANOVA tests the significance of the Regression Model
20. 20 The Multiple Regression Equation We proceed to the derivation of its components: The intercept: a The regression parameters, b1 and b2
21. 21 Derivation of the Intercept
22. 22 Derivation of the Regression Coefficient
23. 23 If we recall that the formula for the correlation coefficient can be expressed as follows:
24. 24
25. 25
26. 26
27. 27 Significance Tests for the Regression Coefficients We find the significance of the parameter estimates by using the F or t test. The R2 is the proportion of variance explained.
28. 28 F and T tests for significance for overall model
29. 29 Significance tests If we are using a type II sum of squares, we are dealing with the ballantine. DV Variance explained = a + b
30. 30 Significance tests T tests for statistical significance
31. 31 Significance tests Standard Error of intercept
32. 32 Programming Protocol
33. 33 Select a Data Set (we choose employee.sav) and click on open
34. 34 We open the data set
35. 35 To inspect the variable formats, click on variable view on the lower left
36. 36 Because gender is a string variable, we need to recode gender into a numeric format
37. 37 We autorecode gender by clicking on transform and then autorecode
38. 38 We select gender and move it into the variable box on the right
39. 39 Give the variable a new name and click on add new name
40. 40 Click on ok and the numeric variable sex is created
41. 41 To invoke Regression analysis, Click on Analyze
42. 42 Click on Regression and then linear
43. 43 Select the dependent variable: Current Salary
44. 44 Enter it in the dependent variable box
45. 45 Entering independent variables These variables are entered in blocks. First the potentially confounding covariates that have to entered. We enter time on job, beginning salary, and previous experience.
46. 46 After entering the covariates, we click on next
47. 47 We now enter the hypotheses we wish to test We are testing for minority or sex differences in salary after controlling for the time on job, previous experience, and beginning salary. We enter minority and numeric gender (sex)
48. 48 After entering these variables, click on statistics
49. 49 We select the following statistics from the dialog box and click on continue
50. 50 Click on plots to obtain the plots dialog box
51. 51 We click on OK to run the regression analysis
52. 52 Navigation window (left) and output window(right)
53. 53 Variables Entered and Model Summary
54. 54 Omnibus ANOVA
55. 55 Full Model Coefficients
56. 56 We omit insignificant variables and rerun the analysis to obtain trimmed model coefficients
57. 57 Beta weights These are standardized regression coefficients used to compare the contribution to the explanation of the variance of the dependent variable within the model.
58. 58 T tests and signif. These are the tests of significance for each parameter estimate. The significance levels have to be less than .05 for the parameter to be statistically significant.
59. 59 Assumptions of the Linear Regression Model Linear Functional form Fixed independent variables Independent observations Representative sample and proper specification of the model (no omitted variables) Normality of the residuals or errors Equality of variance of the errors (homogeneity of residual variance) No multicollinearity No autocorrelation of the errors No outlier distortion
60. 60 Explanation of the Assumptions 1. Linear Functional form Does not detect curvilinear relationships Independent observations Representative samples Autocorrelation inflates the t and r and f statistics and warps the significance tests Normality of the residuals Permits proper significance testing Equality of variance Heteroskedasticity precludes generalization and external validity This also warps the significance tests Multicollinearity prevents proper parameter estimation. It may also preclude computation of the parameter estimates completely if it is serious enough. Outlier distortion may bias the results: If outliers have high influence and the sample is not large enough, then they may serious bias the parameter estimates
61. 61 Diagnostic Tests for the Regression Assumptions Linearity tests: Regression curve fitting No level shifts: One regime Independence of observations: Runs test Normality of the residuals: Shapiro-Wilks or Kolmogorov-Smirnov Test Homogeneity of variance if the residuals: White’s General Specification test No autocorrelation of residuals: Durbin Watson or ACF or PACF of residuals Multicollinearity: Correlation matrix of independent variables.. Condition index or condition number No serious outlier influence: tests of additive outliers: Pulse dummies. Plot residuals and look for high leverage of residuals Lists of Standardized residuals Lists of Studentized residuals Cook’s distance or leverage statistics
62. 62 Explanation of Diagnostics Plots show linearity or nonlinearity of relationship Correlation matrix shows whether the independent variables are collinear and correlated. Representative sample is done with probability sampling
63. 63 Explanation of Diagnostics Tests for Normality of the residuals. The residuals are saved and then subjected to either of: Kolmogorov-Smirnov Test: Tests the limit of the theoretical cumulative normal distribution against your residual distribution. Nonparametric Tests 1 sample K-S test
64. 64 Collinearity Diagnostics
65. 65 More Collinearity Diagnostics condition numbers = maximum eigenvalue/minimum eigenvalue. If condition numbers are between 100 and 1000, there is moderate to strong collinearity
66. 66 Outlier Diagnostics Residuals. The predicted value minus the actual value. This is otherwise known as the error. Studentized Residuals the residuals divided by their standard errors without the ith observation Leverage, called the Hat diag This is the measure of influence of each observation Cook’s Distance: the change in the statistics that results from deleting the observation. Watch this if it is much greater than 1.0.
67. 67 Outlier detection Outlier detection involves the determination whether the residual (error = predicted – actual) is an extreme negative or positive value. We may plot the residual versus the fitted plot to determine which errors are large, after running the regression.
68. 68 Create Standardized Residuals A standardized residual is one divided by its standard deviation.
69. 69 Limits of Standardized Residuals If the standardized residuals have values in excess of 3.5 and -3.5, they are outliers. If the absolute values are less than 3.5, as these are, then there are no outliers While outliers by themselves only distort mean prediction when the sample size is small enough, it is important to gauge the influence of outliers.
70. 70 Outlier Influence Suppose we had a different data set with two outliers. We tabulate the standardized residuals and obtain the following output:
71. 71 Outlier a does not distort and outlier b does.
72. 72 Studentized Residuals Alternatively, we could form studentized residuals. These are distributed as a t distribution with df=n-p-1, though they are not quite independent. Therefore, we can approximately determine if they are statistically significant or not. Belsley et al. (1980) recommended the use of studentized residuals.
73. 73 Studentized Residual
74. 74 Influence of Outliers Leverage is measured by the diagonal components of the hat matrix. The hat matrix comes from the formula for the regression of Y.
75. 75 Leverage and the Hat matrix The hat matrix transforms Y into the predicted scores. The diagonals of the hat matrix indicate which values will be outliers or not. The diagonals are therefore measures of leverage. Leverage is bounded by two limits: 1/n and 1. The closer the leverage is to unity, the more leverage the value has. The trace of the hat matrix = the number of variables in the model. When the leverage > 2p/n then there is high leverage according to Belsley et al. (1980) cited in Long, J.F. Modern Methods of Data Analysis (p.262). For smaller samples, Vellman and Welsch (1981) suggested that 3p/n is the criterion.
76. 76 Cook’s D Another measure of influence. This is a popular one. The formula for it is:
77. 77 Using Cook’s D in SPSS Cook is the option /R Finding the influential outliers List cook, if cook > 4/n Belsley suggests 4/(n-k-1) as a cutoff
78. 78 DFbeta One can use the DFbetas to ascertain the magnitude of influence that an observation has on a particular parameter estimate if that observation is deleted.
79. 79 Programming Diagnostic Tests Testing homoskedasiticity Select histogram, normal probability plot, and insert *zresid in Y and *zpred in X
80. 80 Click on Save to obtain the Save dialog box
81. 81 We select the following
82. 82 Check for linear Functional Form Run a matrix plot of the dependent variable against each independent variable to be sure that the relationship is linear.
83. 83 Move the variables to be graphed into the box on the upper right, and click on OK
84. 84 Residual Autocorrelation check
85. 85
86. 86
87. 87
88. 88
89. 89
90. 90
91. 91
92. 92
93. 93
94. 94 Alternatives to Violations of Assumptions 1. Nonlinearity: Transform to linearity if there is nonlinearity or run a nonlinear regression 2. Nonnormality: Run a least absolute deviations regression or a median regression (available in other packages or generalized linear models [ SPLUS glm, STATA glm, or SAS Proc MODEL or PROC GENMOD)]. 3. Heteroskedasticity: weighted least squares regression (SPSS) or white estimator (SAS, Stata, SPLUS). One can use a robust regression procedure (SAS, STATA, or SPLUS) to obtain downweighted outlier effect in the estimation. 4. Autocorrelation: Run AREG in SPSS Trends module or either Prais or Newey-West procedure in STATA. 4. Multicollinearity: components regression or ridge regression or proxy variables. 2sls in SPSS or ivreg in stata or SAS proc model or proc syslin.
95. 95 Model Building Strategies Specific to General: Cohen and Cohen General to Specific: Hendry and Richard Extreme Bounds analysis: E. Leamer.
96. 96 Nonparametric Alternatives If there is nonlinearity, transform to linearity first. If there is heteroskedasticity, use robust standard errors with STATA or SAS or SPLUS. If there is non-normality, use quantile regression with bootstrapped standard errors in STATA or SPLUS. If there is autocorrelation of residuals, use Newey-West autoregression or First order autocorrelation correction with Areg. If there is higher order autocorrelation, use Box Jenkins ARIMA modeling.
More Related | 2,892 | 12,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-40 | latest | en | 0.797914 |
https://coderanch.com/t/409314/java/Negative-int | 1,521,749,182,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00638.warc.gz | 551,011,216 | 8,204 | Win a copy of Java EE 8 High Performance this week in the Java/Jakarta EE forum!
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# Negative int
Ranch Hand
Posts: 122
Hi,
Given an int, how to determine whether it is positive or negative. For e.g. int x = -1; How to determine x is negative?
Thanks,
Visu Nekk
Ranch Hand
Posts: 433
look at the bit pattern of x..!!
if right most bit is 1, this means it is negative..!!!
x = -1 ===> 11111111 11111111 11111111 11111111
Wanderer
Sheriff
Posts: 18671
I'm not sure the original poster was thinking of bit patterns. The answer might be as simple as
Deepak Chopra
Ranch Hand
Posts: 433
if he was asking this..!! he could have tested it first..before posting!!
Jim Yingst
Wanderer
Sheriff
Posts: 18671
This is the beginner section. Maybe he doesn't know how to use <, or write an if statement, or print the result. Who knows?
Sheriff
Posts: 23286
46
And also, if the rightmost bit of a binary integer is 1, that means it's an odd integer, not that it's a negative integer.
Marshal
Posts: 57482
175
Originally posted by Paul Clapham:
And also, if the rightmost bit of a binary integer is 1, that means it's an odd integer, not that it's a negative integer.
You mean leftmost, surely.
BTW: If you get a binary String with the Integer.toBinaryString() methods, it excludes leading 0s, so it always has 1 as the leftmost displayed digit. You will have to use a for loop and the insert method of StringBuilder (or similar) to pack it with 0s from the left.
For always has 1 read "almost always has 1" The value of 0 wouldn't have a 1 in![/edit]
[ January 30, 2008: Message edited by: Campbell Ritchie ]
Jim Yingst
Wanderer
Sheriff
Posts: 18671
[PC]: And also, if the rightmost bit of a binary integer is 1, that means it's an odd integer, not that it's a negative integer.
[CR]: You mean leftmost, surely.
I'm pretty sure he meant rightmost. Leftmost is what Sunny should have said, but rightmost is what he did say. Paul was commenting on that.
Deepak Chopra
Ranch Hand
Posts: 433
Ohh I am sorry, What I meant was left most bit..!!
Being a left handed person i often got confusion regarding that..! as My left is actually my right hand..but for other it is left hand..!!
Campbell Ritchie
Marshal
Posts: 57482
175
Yes, you are right; it was Sunny Jain who ought to have said leftmost.
Ranch Hand
Posts: 3389
Originally posted by Sunny Jain:
Ohh I am sorry, What I meant was left most bit..!!
Being a left handed person i often got confusion regarding that..! as My left is actually my right hand..but for other it is left hand..!!
Good that the confusions are resolved earlier here. Also i believe it does not make any confusions wiht L-R , R-L associativities
The City calls upon her steadfast protectors. Now for a tiny ad: The WEB SERVICES and JAX-RS Course https://coderanch.com/t/690789/WEB-SERVICES-JAX-RS | 837 | 3,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-13 | latest | en | 0.904412 |
https://cheapeaglesjerseys.com/qa/question-what-does-the-overall-heat-transfer-coefficient-depend-on.html | 1,632,107,959,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00444.warc.gz | 213,912,017 | 7,106 | # Question: What Does The Overall Heat Transfer Coefficient Depend On?
## What affects overall heat transfer coefficient?
The factors affecting overall heat transfer coefficient are : Physiochemical properties of fluids ( both cold and hot ) such a viscosity , density, specific heat, thermal conductivity.
Geometry of the exchanger ( equivalent length and heat exchanging area ) Velocity of flowing fluids..
## Can the overall heat transfer coefficient be negative?
In case of constant wall temperature, using adiabatic wall temperature as reference temperature can result in negative heat transfer coefficient, which means the heat flux has a different direction with the defined driving temperature difference.
## What is the difference between heat transfer coefficient and overall heat transfer coefficient?
Overall heat transfer coefficient R = Resistance(s) to heat flow in pipe wall (K/W) Other parameters are as above. The heat transfer coefficient is the heat transferred per unit area per kelvin. … The areas for each flow will be different as they represent the contact area for each fluid side.
## What does a high overall heat transfer coefficient mean?
The larger the coefficient, the easier heat is transferred from its source to the product being heated. In a heat exchanger, the relationship between the overall heat transfer coefficient (U) and the heat transfer rate (Q) can be demonstrated by the following equation: where. Q = heat transfer rate, W=J/s [btu/hr]
## How do you increase the coefficient of convective heat transfer?
To improve convective heat flow, increasing the area of contact either between the athlete or the apparel and the flow, as well as the speed of the flow is important. For cooling, materials selection and apparel designs can focus on allowing air to flow between the body and the apparel.
## Is heat transfer coefficient same as thermal conductivity?
Thermal conductivity is a property of the material that directly relates the rate of heat transfer to the thermal gradient while heat transfer coefficient is an empirical function that correlates the effective heat transfer across the boundary to the difference in bulk temperatures measured at the interfaces.
## What does heat transfer coefficient depend on?
Heat transfer coefficient depends on both the thermal properties of a medium, the hydrodynamic characteristics of its flow, and the hydrodynamic and thermal boundary conditions.
## Does heat transfer coefficient depend on temperature?
The value of the surface heat transfer coefficient depends strongly on temperature.
## Does heat transfer increase with flow rate?
On the other hand, an increase in the heat transfer coefficient is observed if the area is maintained constant. Doubling the mass flow rate will result in a 92% increase in the heat transfer coefficient. However, there is a concomitant increase in the pressure drop, proportional to the mass flow rate raised to 0.95.
## What are the 4 types of heat transfer?
Various heat transfer mechanisms exist, including convection, conduction, thermal radiation, and evaporative cooling.
## What is the heat transfer coefficient of air?
Convective Heat Transfer Coefficient for Air The convective heat transfer coefficient for air flow can be approximated to. hc = 10.45 – v + 10 v1/2 (2) where. hc = heat transfer coefficient (kCal/m2h°C) v = relative speed between object surface and air (m/s) | 657 | 3,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-39 | latest | en | 0.91721 |
https://www.jiskha.com/display.cgi?id=1266455462 | 1,503,374,471,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00212.warc.gz | 917,431,611 | 5,303 | # AP Chemistry
posted by .
A 1.2M NaOCl solution is prepared by dissolving solid NaOCL in distilled water at 298K. The hydrolysis reaction
OCl-(aq) + H2O(l) --> HOCl(aq) + OH-(aq) occurs.
a. Calculate the value of the equilibrium constant at 298K for the hydrolysis reaction.
b. Calculate the value of [OH-] in the 1.2M NaOCl solution at 298K.
---
So far I have the expression:
K= {[HOCl][OH-]}/[OCl-]
...but I don't understand how to find K if you're only given the concentration of NaOCl. Help???
Thanks!
• AP Chemistry -
You can derive this equation with a little work but the equilibrium constant is, in reality, the hydrolysis constant for OCl. That is Kh = Kw/Ka where
Ka is for HOCl.
• AP Chemistry -
Okay, then. I think I know what you're getting at, but is it ok if you point out where I can start? (Not a little sure on deriving equations...)
• AP Chemistry -
To be honest about it, all of the derivations I've seen were done in books used 30 years ago when we called that constant the hydrolysis constant and had to derive that it was Kw/Kb. In modern texts, the word hydrolysis constant is never used. They simply write the equation for the hydrolysis, as you have done, and say OCl is acting as a base (note that it pulls a hydrogen away from HOH to make HOCl so it is a Bronsted-Lowry base). Then these same modern texts note that Ka*Kb = Kw, so obviously, Kb = Kw/Ka. Voila! And that's as close as the modern texts get to deriving it BECAUSE modern students have no clue as to what a hydrolysis constant is. It's simply not used in the lingo. WHY? Mostly because the idea is that OCl is acting as a base and everyone KNOWS that KaKb=Kw so we can find Kb by Kb = Kw/Ka.Said another way, Kb IS the hydrolysis constant.
But I can show you how to do it if you wish.
• AP Chemistry -
Yeah, that would be nice. Thanks much!
• AP Chemistry -
I have tried several times to post this but the site has been having some problems. Maybe this time.
We take a salt NaA and throw it into some water. The Na^+ doesn't hydrolyze but the A^- does. I will omit all of the charges to make this simpler to type.
A + HOH ==> HA + OH
Kh = (HA)(OH)/(A)
Now multiply that expression by (H)/(H) which gives
(H)*(HA)(OH)
--------------- = Kh
(H)*(A)
Note that the numerator contains (H)(OH) which is Kw. What is left?
(HA)
-----
(H)(A)
but that is just 1/Ka; therefore,
Kh = Kw/Ka
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More Similar Questions | 1,269 | 4,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-34 | latest | en | 0.960648 |
https://www.lessons99.com/data-structures-part-3.html | 1,591,517,454,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00430.warc.gz | 774,452,494 | 14,096 | # Data Structures Part 3
Data Structures Part 3
Stacks:
A stack is a homogeneous collection of items of any one type, arranged linearly with access at one end only called top. This means the data can be added or removed only from top. Formally this type of stack is called as Last in First out (LIFO). Data is added to the stack using push operation, and removed using pop operation. Different implementations are possible; although the concept of a stack is unique.
A stack is a dynamic structure. It changes as elements are added to and removed from it. A stack can be implemented as a constrained version of a linked list. A stack is referenced via a pointer to the top element of the stack. The link member in the last node of the stack is set to NULL to indicate the bottom of the stack.
Stacks and linked lists are represented identically. The difference is that insertions and deletions occur anywhere in a linked list, but only at the top of a stack.
• Function push creates a new node and places it on top of the stack.
• Function pop removes a node from the top of the stack, frees the memory that was allocated to the popped node, and returns the popped value.
Stack Operations:
Implementation of a simple stack of integers
struct stackNode
{
int data;
struct stackNode *nextPtr;
};
/* Insert a node at the top of the stack */
void push(struct stackNode **topPtr, int info)
{
struct stackNode *newPtr;
newPtr = (struct stackNode *)
malloc(sizeof (struct stackNode));
if (newPtr != NULL)
{
newPtr ->data = info;
newPtr->nextPtr = *topPtr;
*topPtr = newPtr;
}
else
printf (“%d not inserted. No memory”
“available.n”, info);
}
/*Remove a node from the stack top */
int pop (struct stackNode **topPtr)
{
struct stackNode *tempPtr;
int popValue;
tempPtr = *topPtr;
popValue = (*topPtr) ->data;
*topPtr = (*topPtr) ->nextPtr;
free (tempPtr);
return popValue;
}
/*Is the stack empty? */
int isEmpty(struct stackNode *topPtr)
{
}
Evaluation of arithmetic expressions:
• Notation can be infix, postfix or prefix.
Infix: operator is between operands
A + B
Postfix: operator follows operands
AB+
Prefix: operator precedes operands
+AB
• Operators in a postfix expression are in correct evaluation order.
Postfix Expressions
Infix Postfix
a + b * c abc*+
(precedence of * is higher than of +)
a + b * c / d abc*d/+
(precedence of * and / are same and they are left associative)
• Parentheses override the precedence rules:
(a + b) * c
ab+c*
• More examples
Infix Postfix
(a + b) * (c – d) ab+cd-*
a – b / (c + d * e) abcde*+/-
((a + b) * c – (d – e))/ (f + g) ab+c*de – – fg+/
Order of precedence for 5 binary operators:
power (^)
multiplication (*) and division (/)
The association is assumed to be left to right except in the case of power where the association is assumed from right to left.
I.e. a + b + c = (a+b) +c = ab+c+
a^b^c = a^ (b^c) = abc^^
Conversion of Infix expression to postfix notation
Suppose Q is an arithmetic expression in Infix notation. This algorithm finds the equivalent postfix expression P. Scan Q from left to right and repeat steps 3 to 6 for each element of Q until Stack is empty.
• If an operand is encountered, add it to P.
• If a left parenthesis is encountered, push it onto the Stack.
• If an operator is encountered, then repeatedly pop from Stack and add P each operator (on the top of Stack) which has the same precedence as or higher precedence than current symbol. i. Add the current symbol on to the stack.
• If a right parenthesis is encountered, then i. repeatedly pop from Stack and add to P each operator (on the top of Stack. until a left parenthesis is encountered).
ii. Remove the left parenthesis.
• Exit
Evaluating postfix expression:
This algorithm evaluates postfix expression P using Stack.
• Scan P from left to right and repeat step 3 and 4 for each element of P until the end of the string is encountered.
• If an operand is encountered, put it on Stack.
• If an operator (*) is encountered, then
i. Remove the two top elements of Stack, where A is the top element and B is the next to the top element.
ii. Evaluate B (*) A.
iii. Place the result of (ii) back to Stack.
• Set VALUE equal to the top element on Stack.
• Exit C Programming & Data Struct.
Implementation of a stack as an array:
#define maxstack 100
struct stack {
int items[maxstack];
int top;
};
int isEmpty(struct stack s){
return (s.top < 0);
}
int isFull(struct stack s){
return (s.top >= maxstack-1);
}
void push (struct stack *s, int x){
if (s->top >= maxstack-1)
printf (“The stack is full.n”);
else {
s->top = s->top +1;
s->items[s->top] = x;
}
}
int pop (struct stack *s){
int x;
if (s->top < 0)
printf (“Stack is empty.n”);
else {
x = s->items[s->top];
s->top = s->top –1;
return x; }
}
int main()
{
struct stack S;
int c, i;
S.top = -1;
while ((c=getchar() )!=’n’)
push (&S, c);
while (!isEmpty(S))
printf (“%c”, pop(&S));
printf (“n”);
}
Queues:
A queue is a list with the restriction that insertions only occur at the back, and deletions only occur at the front. A queue is FIFO data structure. The fundamental operations, in addition to size, clear, and empty, are enqueue (insert at back) and dequeue (delete and return the element at the front). This type of frequently used list is known as queue. We have two pointers to access the queue. They are
• Front (used for deletion)
• Rear (Used for insertion)
Insertion:
if rear>n queue overflow
else
increment the rear pointer and insert the value in the rear position.
Deletion:
if front =0 then queue underflow
else
increment the front pointer and return the front-1 value.
Some of the practical applications of a queue are:
• An operating system always makes use of a queue (ready queue, waiting queue) for scheduling processes or jobs.
• All the input output calls made by the disk to and from the memory are handled by a queue.
• We have many processors like 80386 and queue plays an active part in there architecture.
• Queues can be commercially used in online business applications for processing customer requests in first in first serve manner.
• When a resource like CPU is shared between among multiple consumers, a queue solves the hatchet.
• In all the basic algorithms for searching or sorting, every one of them makes use of queue.
• Queues are very efficient when we want to take out elements in the same order we have put in inside.
Implementation of a queue as an array:
#include<stdio.h>
#include<conio.h>
#define SIZE 5
int i, rear, front, item, s[SIZE];
void insert(int item, int s[]);
void del(int s[]);
void display(int s[]);
void main()
{
int ch;
clrscr ();
front=0;
rear=-1;
do
{
printf (“nn 1.INSERTION n 2.DELETION n 3.EXIT n”);
printf (“n ENTER YOUR CHOICE : “);
scanf (“%d”, &ch);
switch (ch)
{
case 1:
printf (“nt INSERTION n”);
if (rear>=SIZE-1)
{
printf (“tn QUEUE IS FULLn”);
}
else
{
printf (“nENTER AN ELEMENT : “);
scanf (“%d”, &item);
insert (item, s);
}
display(s);
break;
case 2:
printf (“nt DELETION n”);
if (front>rear)
{
printf (“tn QUEUE IS EMPTYn”);
}
else
{
del(s);
}
display(s);
break;
}
}
while (ch!=3);
getch ();
}
void insert(int item, int s[])
{
if (rear<SIZE)
{
rear=rear+1;
s [rear]=item;
}
}
void del(int s[])
{
int i;
item=s [front];
for (i=0;i<=rear; i++)
s [i]=s[i+1];
rear–;
printf (“n DELETED ELEMENT IS %dnn”, item);
}
void display(int s[])
{
printf (“n”);
for (i=front; i<=rear; i++)
{
printf (” t %d”, s[i]);
}
}
Inserting a node in required position:
Inserting a node at the Rear end:
• Allocate memory for the new node.
• Assign the value to the data field of the new node.
• Set the link field of the new node to NULL.
• If the queue is empty, then set FRONT and REAR pointer points to the new node and exit.
• Set the link field of the node pointed by REAR to new node. A
• Adjust the REAR pointer points to the new node.
Deleting a node from required position:
Deleting a node at the Front end:
• If queue is empty, then display message “Queue is empty – no deletion is possible” and exit.
• Otherwise, move the FRONT pointer points to the second node.
• Free the first node.
0 | 2,147 | 8,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-24 | latest | en | 0.822764 |
http://www.allthetests.com/quiz33/quiz/1461626498/Integers-Quiz | 1,521,443,042,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646602.39/warc/CC-MAIN-20180319062143-20180319082143-00744.warc.gz | 329,555,321 | 15,089 | # Integers Quiz
10 Questions - Developed by: Ms.Gordon/Tytianna - Developed on: - 1.398 taken
## Pick the right ones.
• 1
6 x 2 - 6 + 6 =
• 2
12 - 6 - 3 + 5 =
• 3
6 - 6 + 12 +18 -10 x 2 =
• 4
16 - 8 =
• 5
14 - 14 + 3 - 6
• 6
18+6=
• 7
14-5+6-7=
• 8
85-9-62-7=
• 9
89-9=
• 10
245-65+5= | 159 | 287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-13 | latest | en | 0.365155 |
https://gyroplacecl.com/gyroscopic-representation-understanding-the-mechanics-and-applications/ | 1,718,346,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00167.warc.gz | 269,623,703 | 22,450 | Gyroscopic Representation: Understanding the Mechanics and Applications
Gyroscopic representation refers to the use of mathematical models or physical devices, called gyroscopes, to depict and analyze rotational motion. These representations help understand phenomena like precession, nutation, and wobble in various fields such as physics, engineering, and navigation.
1) What is Gyroscopic Representation? An Introduction to the Concept
Gyroscopic Representation is a concept that deals with the visual representation of dynamic systems using gyroscopes. This unique approach combines principles from both physics and animation to create stunning visuals that accurately depict complex movements.
To better understand Gyroscopic Representation, let’s start by exploring what gyroscopes are. A gyroscope is essentially a spinning wheel or disk in which the axis of rotation maintains its orientation regardless of any external forces acting upon it. This property, known as rigidity in space, allows gyroscopes to be used for various purposes like stabilization and navigation.
Now imagine harnessing this rigid characteristic of a gyroscope and applying it to represent an object’s movement simultaneously across all three dimensions – X, Y, and Z axes! That’s exactly what Gyroscopic Representation does; it uses several interconnected gyroscopes strategically placed within a system to visually capture every intricate motion occurring within.
The brilliance behind this concept lies in how these interconnected gyros work together harmoniously. Each individual gyroscope represents one dimension – think yaw (rotation around the vertical axis), pitch (tilting up or down), and roll (tilting left or right). By combining these separate representations through precise calibration techniques, we achieve an accurate depiction displaying fluidity across multiple dimensions simultaneously.
But why should we bother with such complexity when there are already established methods for visually representing objects’ movements? The answer lies in our pursuit of achieving fidelity between reality and simulation while captivating audiences with mesmerizing visuals!
Traditional approaches often make use of simplified 2D animations where rotational dynamics aren’t adequately conveyed due to limitations imposed by two-dimensional rendering. In contrast,cally replicates real-world behavior giving incredibly realistic results.T using sophisticated algorithms benefitng t utilizea combinationt clever programmingto truly emulateof experts workingssforces Directorateathom paireddefense simulations., they can confidently visualizeg spacecraft docking maneuversrgy making sure unmanned missionsat run smoothly withoutntions risk.nd danger.s
MoreoverFurthermore, Gyroscopic Representation challenges our creative limits by introducing intricate levels of aesthetics to the animation world. With its abilityo convey motion withextremem accuracyoss alldimensionalitiesns,s it opens up new opportunities for artistic exploration and storytelling. complex movements can be beautifully animatedgiving life too still objectsobjects,to become emotive characters inanimated films or captivating visualizationsngage viewers on a deeper levelvel.
So whether you’re an enthusiast wantingnting to explore the cinematic possibilitiespossibilities,or ae scientist lookingforging innovative solutionslutions fromire defense simulations,lations,Gyroscopicgy Representation is sureto revolutionize your perceptionon of movement representation.presen Existing animations will pale inn comparison,and as weworkes continuehence tonduit pushingbordersoundaries offeriexploring even more mesmerizing visualsthat pushddhe boundariesnand challenge our understandingf movemenvement itselflf.for good!
2) How does Gyroscopic Representation Work? Exploring the Mechanics
How does Gyroscopic Representation Work? Exploring the Mechanics
Gyroscopes are fascinating devices that defy our intuition and push the boundaries of physics. These rotating mechanisms have been puzzling scientists and engineers for centuries, leading to numerous applications in various fields such as navigation systems, space technology, robotics, and even toys like yo-yos.
At its core, gyroscopic representation is all about harnessing angular momentum to stabilize an object’s rotation axis. This mechanical phenomenon enables gyroscopes to maintain their orientation regardless of external forces acting upon them. To delve deeper into this mysterious concept, let’s explore the mechanics behind how gyroscopes work.
Firstly, it is important to understand angular momentum – a property possessed by any spinning or rotating object. Angular momentum depends on two primary factors: moment of inertia (a measure of an object’s resistance to changes in rotation) and angular velocity (how quickly it spins). The product obtained by multiplying these two factors gives us the magnitude of total angular momentum.
Now comes gyroscope’s central element – a rapidly spinning wheel called rotor or spin axis – which exhibits both moment of inertia and high rotational speed simultaneously. As we increase either parameter alone—either through increasing mass distribution away from its center or accelerating its rotation—the resulting effect significantly magnifies overall angular momentum experienced by the system; hence enhancing its stability against applied outside disturbances.
As per Newton’s first law—an object tends not change spontaneously unless acted upon —this augmentedinertia means more effort would be requiredto modifygyroscope’s initial state.Not only would additional force/energy needsto bemustered butalso subsequently oriented alongthe desired directionofchangewhen attempting totilt,stabilizeor alterthespinning body.Secondarily,becauseangularmomentumconservespace-timegeometry,the spinner automatically provides counterbalancing torque–which prevents deviationfromthepre-established course.These twoworkhand-in-handtosteer thenavigationalproperties of a gyroscope.
To wrap our minds further around the mechanism, let’s consider an example: imagine holding a bicycle wheel by its axle and spinning it rapidly. As you tilt the axis vertically or horizontally, do you notice how resisting external forces come into action? This resistance is gyroscopic stability in motion!
In simplistic term,Rotational Inertiatells us to resistchanges;whileconservationofAnguluar Momentum ensuresthingsspinalong theirpresentaxis.The outcomeoffusingsuchphysical lawsformafewfascinatingpodium tricksandsomepracticalapplications -frommaintaining steady flightinthin airusingAirplanes’HorizonReferenceUnit(HRUs)to stellar navigationthroughhighlymethodicalMagneticTorquersusedoncomsatstoaplaceonyour shelffordecorativegyroscopesandlevitatingtops,toevenunleashingentoysouvenir trivial pursuitsspreadingdelightpackagedasYo-Yos forchildren—motion experience Diehard-Poise&PhysicsLeaningTeller-Wobble Jane!
Furthermore,guidancetracking,satelliteattitude controlorunderwaterrobotics demonstratehowimprovedholisticcounter engineering often harnessthiswondrous effectanti-nuclear weapons’ guidance systems.So,youseneednotjustgawk atentertainment providedbyAsymmetricalRotatingSound/Dance Exhibitsbutsteerdaily captureshowswhenround steer-tiresworksynchronouslywiththeCar’sangularmomentum–reboots/ undisturbedcommuting journey daily elegancelifestyleitimpliessharpercorner turning thrills.Gravity-defyingprofessional stuntslikeEasySittingBikingKartWheeliegivesacademicians ideasovertailiedtwister-Bud-Smasher-yappopping futureacademic discoveriespromisemoreefficientgalactic travel.Sate-wriggleHidalgotravel coursesandicorn-painter-linkedsuperfast-meridinasalspinning givingOpen-endUSAEducation.
In conclusion, gyroscopic representation stems from understanding the principles of angular momentum and rotational inertia. By exploiting these fundamental aspects of physics, we can harness the power of a rapidly spinning wheel to achieve incredible stability in various applications ranging from navigation systems to toys. The mechanics behind how gyroscopes work continues to captivate scientists, engineers,and enthusiasts alike as they uncover new ways to utilize this mesmerizing phenomenon for advancements in technology and beyond!
3) Step-by-Step Guide: Mastering the Art of Gyroscopic Representation
Title: Unveiling the Secrets of Gyroscopic Representation: A Step-by-Step Guide
Introduction:
Welcome aboard, fellow adventurers! Today, we embark on a captivating journey that will unravel the enigmatic world of gyroscopic representation. In this comprehensive step-by-step guide, we’ll equip you with essential knowledge and techniques to master this intriguing art form like a true connoisseur.
1) Laying Down the Foundation:
Just as every masterpiece needs a sturdy canvas, our first step is to understand the fundamental principles behind gyroscopic representation. By definition, it refers to drawing or creating an illusion of rotating objects using various artistic mediums such as pencils or digital tools.
2) Gathering Supplies – The Toolkit:
Now that you grasp the essence of what awaits us ahead let’s assemble our toolkit for conquest. Begin by gathering essentials such as quality paper or digital sketching software (based on your preference), fine-tipped pens/pencils in varying thicknesses for maximum control over details and shading effects. Keep handy erasers for precise fixes while ensuring not to lose them into “eraser-hungry” voids!
3) Embracing Perspective – Be One with Motion
Perspective serves as one vital element in achieving mesmerizing gyroscope illustrations. Start off by familiarizing yourself with vanishing points – those magical spots where parallel lines converge towards infinity—introducing depth and realism within your artwork.
4) Framing Imagination – Outlining Your Vision
Imagine standing at an amusement park ride booth; now envision translating its excitement onto paper! Sketch out basic shapes representing roller coaster tracks or Ferris wheels suiting your inspiration—a blank slate ready to be transformed into kinetic beauty closely resembling reality itself!
5) Pencil Rhapsody – Shading Techniques Galore
Shadings are crucial elements when encapsulating motion seamlessly within static-based drawings—the heartthrob moments? Use impeccable hatching/crosshatching techniques to breathe life into your artwork, imbuing it with shades and depth never witnessed before. Experiment! Explore!! Elevate!!!
6) Color-Whirl Extravaganza:
Ready to splash vivid hues onto the canvas? Select a vibrant color palette that complements your vision while keeping in mind contrasting colors for highlighting directional movements of various elements within gyroscope illustrations. Remember, color psychology plays its hidden role!
7) Tackling Technology – Digital Marvels Unveiled
Though traditional methods are timeless treasures, venturing into digital art brings forth versatile opportunities to manifest extraordinary gyroscopic wonders without breaking pencils or shedding tears over accidental smudges! Harness software wizardry like Adobe Illustrator or Procreate—let’s digitize those whirlwinds!
8) Practicing Patience – A Virtue Worth Gathering
Last but far from least: Rome wasn’t built in a day; similarly mastering the art of gyroscopic representation demands time and patience molded together splendidly. With each stroke perfected gradually through practice sessions (accompanied by small doses of caffeine), you’ll soon witness bewitching creations emerging at every turn.
Conclusion:
Congratulations on immersing yourself deeply into this captivating guide towards unveiling the secrets behind mesmerizing mechanical motion illustrations – Gyroscopic Representation. Equipped with knowledge spanning fundamental principles, artistic technique mastery blended harmoniously with technological triumphs—the sky is not just merely a limit anymore—it becomes an infinite playground waiting for creative souls like yours so let loose those reins bound only by imagination dance!!!
Are you curious about gyroscopic representation? Do terms like precession, nutation and roll axis leave you scratching your head? Well, fret no more! In this blog post, we are going to dive deep into the world of gyroscopes and answer some frequently asked questions that will help unravel their mysteries. So hold on tight as we take a spin through the fascinating realm of gyroscopic representation!
1) What is Gyroscopic Representation?
Gyroscopic representation refers to the way in which a gyroscope behaves when subjected to external forces or torques. A gyroscope is essentially a spinning wheel or disc that maintains its orientation regardless of any movements or disturbances it encounters. This unique property makes them an invaluable tool in various fields such as navigation systems, aerospace engineering, robotics and even toys like tops.
2) How does Precession work?
Precession can be quite perplexing at first glance but fear not – understanding it isn’t rocket science! When an external force is applied perpendicular to the rotational axis of a spinning gyroscope (known as torque), rather than tilting over immediately in response, it actually follows suit by rotating around another orthogonal axis called precession. This interesting phenomenon gives rise to mind-bending effects often seen with these devices.
Think about riding a bicycle – have you ever noticed how turning while pedaling causes your bike to lean towards one side instead of toppling right away? That’s because your front wheel acts just like a miniature gyroscope where leaning corresponds to precessional motion!
3) What on Earth is Nutation then?
Nutation sounds peculiar indeed – almost fantastical! But rest assured; there’s nothing mystical about it! When disturbed from equilibrium during rotation (such as applying varying torques), gyroscopes undergo what scientists refer whimsically as “nutation.” It manifests itself by causing small periodic wobbling motions superimposed upon larger rotations known collectively as preevolved figures. These peculiar oscillations are the hallmark of gyroscopic representation.
Think of a gyroscope wobbling on its axis as if it were performing an intricate dance routine just to maintain balance amidst external forces acting upon it. So, next time you see those subtle twirls and gyrations, think nutation!
4) What’s all this about the Roll Axis?
Ah yes, let us now embark on our final voyage through the realm of gyroscopes – into none other than their roll axis! When referring to a gyroscope’s roll axis, we’re discussing how it rotates around its primary rotational or spin-axis while undergoing precession. Essentially, imagine your favorite thrill ride at an amusement park – often called “The Rotor” or something equally thrilling! The riders stick against the wall as they’re spun insanely fast within a large drum-like chamber that gradually tilts away from vertical orientation – analogous with understanding and visualizing rotation around this elusive yet captivating roll axis!
So there you have it folks – some frequently asked questions regarding gyroscopic representation answered in detail above for your enlightenment and entertainment. Now armed with these insights into precession, nutation and even their beloved roll axes; consider yourself well-informed when encountering discussions swirling around anything related to these magnificent spinning wonders known as gyroscopes!
5) Unleashing Creativity with Gyroscopic Representations: A Deep Dive
Unleashing Creativity with Gyroscopic Representations: A Deep Dive
Are you ready to take a deep dive into the world of gyroscopic representations and unlock your true creative potential? In this blog post, we will explore how these fascinating visual tools can revolutionize your approach to problem-solving and discover innovative solutions. So grab yourself a cup of coffee, sit back, and get ready for an eye-opening journey!
Gyroscopic representations are unique visualizations that use rotating three-dimensional objects to represent complex ideas or concepts. These mesmerizing visuals not only capture attention but also stimulate the mind in ways traditional two-dimensional diagrams cannot. By incorporating motion into our thinking process, gyroscopes challenge our brains to think outside the box and break free from conventional thought patterns.
One remarkable aspect of using gyroscopes as creativity enhancers is their ability to engage both hemispheres of the brain effectively. The left hemisphere controls logical thinking while the right hemisphere drives imagination and artistic expression. With gyroscope-inspired approaches, we tap into both sides simultaneously — allowing for a harmonious fusion between analytical problem-solving skills and imaginative flair.
Imagine tackling a tricky marketing puzzle where you need ideas on improving brand recognition across multiple platforms.Projecting this challenge onto spinning spheres within your mind’s eye allows new connections between different elements involved in branding activities . As each sphere rotates independently – one representing social media outreach strategies; another symbolizing print advertising campaigns- sparks fly as links form effortlessly before us! This dynamic visualization prompts unexpected insights by engaging disparate parts together visually , enabling breakthroughs previously unimagined!
Moreover,to deepen our understanding further its worth mentioning that human beings innately respond more strongly when information comes through interactive channels.Our brains crave novelty,and tracking movement via intricate rotations gets neurons buzzing.Additionally,giving shape,presence nad animation solidifies comprehension leading facilitating fluid communication which indeed paves way fr impactful presentations among team members.It effectively sets collaborative meetings abuzz stimulating productive conversations and driving diverse perspectives ensuring a more comprehensive solution.
The benefits of using gyroscopic representations extend beyond the realms of business or marketing. Fields such as education, science, engineering, art, among others can also leverage this powerful tool to foster deeper understanding and creativity.A physics teacher employing spinning models illustrating complex concepts like planetary motion captivates students’ attention.Sequences showcasing atoms morphing into molecules – with each transformation intricately spun- not only take learning to another level but also make it an enjoyable experience for young minds.Nurturingfantasy,daring imagination channelizing them constructively drive revolutionary advancements in fields that fuel our world!
In conclusion,given the power within gyroscopic representations to unlock hidden dimensions,promote cross-pollination between different areas,and engage both logical analysis alongside creative thinking; we cannot ignore their potential impact on unleashing human ingenuity. So whether you’re seeking fresh ideas at work,reinventing your educational approach or just want a burst inspiration during moments of artistic lull consider embracing the mesmerising magic lying behind twirling spatial arrangements!Embrace swirls& spins,bend norms welcome fluidity,& prepare embark stirring journey filled endless possibilities photography technology animation industrial design ,this is where transformative,cross-disciplinary revolutions thrive.Enjoy riding wave toward new horizons propelled by magnificent motions swayed unruly tides whisps immerse awash brilliance wonders life creatively surges before open mind & inspired soul!!!
6) Enhancing Visualization Skills through Gyroscopic Representation Techniques
Title: Elevating Visualization Skills through Gyroscopic Representation Techniques
Introduction:
Visualization skills are vital in various professional domains, enabling individuals to comprehend complex concepts and data effortlessly. Representing information visually allows us to absorb it more readily, leading to clearer understanding and improved decision-making. In this blog post, we will delve into the intriguing realm of gyroscopic representation techniques – an innovative approach that can revolutionize how we enhance our visualization abilities.
1) Understanding Gyroscopic Representation Techniques:
Gyroscopic representation techniques involve harnessing the power of motion-based visuals inspired by gyroscope principles. By employing three-dimensional movements, rotations, and orientations as a means of visual communication or demonstration toolsets for diverse topics/artifacts/phenomena/macrosystems/microsystems/nano systems/etc., these methods offer dynamic ways to illustrate intricate concepts vividly.
2) Leveraging Animation for Enhanced Engagement
One key advantage offered by gyroscopic representation lies in its ability to create engaging animations that captivate audiences across industries. The utilization of rotational motions imparts a sense of dynamism within otherwise static depictions; thus helping learners grasp abstract ideas with greater ease due to their enhanced engagement levels.
The artistry behind constructing animated representations using gyrations supports retention rates since movement stimulates both hemispheres while aiding cognitive assimilation processes significantly—a testament to the potent synergy between creativity and scientific thinking enabled via these unique strategies.
3) Embracing Interactive Learning Opportunities:
Collaborative digital platforms increasingly integrate interactive elements powered by gyroscopy technology—allowing users not only observe but also actively participate in 360-degree explorations remote immersions/experiences/practices/demonstrations/lab works/books readings demos(the word depends on your context). Such approaches foster active learning experiences where students/participants/consumers interact tactilely with virtual presentations/models/simulations(ditto*) enhancing receptiveness exponentially(*replace excessively complicated words).
For instance,Illustratively(in let’s talk about two interdisciplinary subjects of interest here:) simulated molecular structures within the realm of organic chemistry become tangible, providing aspiring chemists/biologists/energists/architects with a more palpable understanding(notice how ‘tangible’ and ‘more palpable are both convenient adjectives). These immersive experiences serve as catalysts in accelerating subject comprehension rates by bridging-the gap between abstract ideas(theory)and physical reality(application).
4) Empowering Data Visualization:
Gyroscopic representation techniques have also emerged as powerful allies for data scientists/analysts/researchers seeking to present complex information compellingly. Dynamic visualizations can transform datasets into captivating narratives that effortlessly convey critical insights.
By harnessing rotational movements aligned intuitively with dataset variables, scholars experts/professionals(choose whichever suits your context better), or even enthusiasts(cityplanners/materialscientist/historians/market analysts/metaphysician*, etc.) gain deeper understandings by observing dynamic relationships unfolding before their eyes(minds?). It becomes easier(*better option: significantly facilitates streamlined comprehension due/)to detect patterns/trends/recurrences/spatial correlations(time associations)/anomalies hidden within voluminous/datatracts(database/logsanoithebest?*)—as they come alive through fluid gyroscopic renderings on interactive platforms(pressures-mapped-journeys/dynamic-topographic-flows.dot*)
*(Choose one field [or don’t do it at all]; depends on what you find fitting)
5) Bridging Communication Gaps:
Beyond educational contexts/application areas(*switch this likely phrase around depending on context liking*), incorporating gyroscopic representations fills an often-overlooked void in effective communication across disciplines and industries . This methodology revolutionizes presenting multi-dimensional concepts/theories/narrative11ideas/artwork(feastronamics’/storidystopian trove[deeptech/social-implications]/immaterial world’s revealing approachability; neuroscience) to non-experts or peers from diverse fields of expertise/professions.
Gyroscopic representation aligns, unifies and homogenizes dissimilar approaches and elements through intuitive visualizers facilitating a shared understanding/collective reception telling stories/data insights spiritedly leaving lasting impressions(a morphing jigsaw transcending stereotypical heights*) – promoting collaborative efforts beyond borders, fostering novel perspectives on various subjects domains(strangling boundaries ethic bondage!). And yes(*crosswords*, playfulness is also key).
Conclusion:
In this fast-paced world where information overload often impedes comprehension, gyroscopic representation techniques offer a refreshing approach in enhancing our visualization abilities. From educational settings to data analysis realms/cross-disciplinary collaborations(handVScryingearthemblems*), these mesmerizing techniques enable us to grasp complex concepts effortlessly while bridging communication gaps with their immersive power.
By embracing innovation derived from the principles of gyroscope physics/motion actions/demonstrations(estimation?), we open new channels for precise expression captivating storytelling—ultimately empowering individuals across professions/disciplines(self-consciousness plus**). Improve your visualization skills today by exploring these transformative practices that propel knowledge sharing into uncharted territories.(*Either define hand [artists/sculptors/handbooks] or do not mention anything after collaboration)
**(choose one here: performance mindfulness/phpi phenomenon awareness/incredibly powerful ways!/[blackmaker stuff])
Rate author | 4,636 | 26,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.911682 |
https://rcogenasia.com/electricity-generation/how-does-ac-electricity-work.html | 1,638,238,498,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00013.warc.gz | 570,477,600 | 17,749 | # How does AC electricity work?
Contents
## How does AC current flow in a wire?
In AC, a changing magnetic field creates a changing electric field, and a changing electric field creates a changing magnetic field. When AC flows in a wire, a changing magnetic field is created in the wire. This induces its own current in the wire in the opposite direction to the original.
## How does AC power work in a house?
The electrical current in your house is AC. This comes from power plants that are operated by the electric company and is carried through power lines. The direction of current is switching back and forth 60 times each second (60 Hertz) in the United States. … The light bulb does not care if it is using DC or AC current.
## How does electricity flow through a wire?
Electric current (electricity) is a flow or movement of electrical charge. The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not move. … The wire is “full” of atoms and free electrons and the electrons move among the atoms.
## How does AC current move forward?
In alternating current, the electrons don’t move in only one direction. Instead, they hop from atom to atom in one direction for a while, and then turn around and hop from atom to atom in the opposite direction. … In alternating current, the electrons don’t move steadily forward. Instead, they just move back and forth.
THIS IS INTERESTING: How does an electric guitar sound without being plugged in?
## Why AC current is used in homes?
Because high voltages are more efficient for sending electricity over long distances, AC has an advantage over DC. This is because the high voltages from the power plant for home use can be easily reduced to a safer voltage. The voltage is changed with a transformer.
## What is the working principle of AC generator?
AC generators work on the principle of Faraday’s law of electromagnetic induction. This law states that electro motive force is generated in a current carrying loop that is placed in a uniform magnetic field. | 430 | 2,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.928365 |
https://reference.wolfram.com/language/ref/Factorial.html | 1,723,066,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00728.warc.gz | 383,983,765 | 20,730 | # Factorial
n!
gives the factorial of n.
# Details
• Mathematical function, suitable for both symbolic and numerical manipulation.
• For noninteger n, the numerical value of n! is given by Gamma[1+n].
• For integers and half integers, Factorial automatically evaluates to exact values.
• Factorial can be evaluated to arbitrary numerical precision.
• Factorial automatically threads over lists.
• Factorial can be used with Interval and CenteredInterval objects. »
# Background & Context
• Factorial represents the factorial function. In particular, Factorial[n] returns the factorial of a given number , which, for positive integers, is defined as . For n1,2,, the first few values are therefore 1,2,6,24,120,720,. The special case is defined as 1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects. For a general complex number , , where the Gamma function is defined by for all complex values of except when is a negative integer (in which case is complex infinity). Factorials of half integers are given by rational multiples of .
• Factorials are best known for counting fixed orderings of the elements of a list, known as permutations, which can be generated using Permutations. There are permutations of a list of (distinct) elements, a fact that follows from there being spots to place the first element, spots to place the second element once the first is placed, spots to place the third element once the first two elements are placed and so on until a single spot remains in which to place the last element. There are therefore permutations of , namely , , , , and .
• More generally, for an -element multiset having distinct elements with copies of the distinct element (so ), the number of permutations equals the multinomial coefficient , given by Multinomial. The multinomial coefficient also counts the ways to partition an -element set into labeled subsets of sizes n1,,nk. Hence the binomial coefficient , given by Binomial and defined to count the -element subsets of an -element set, satisfies .
• The factorial function satisfies the recurrences and . It grows faster than any exponential function, as shown by Stirling's approximation . Factorials also appear in fundamental results in number theory and analysis. Wilson's theorem states that if and only if is prime. If is an infinitely differentiable scalar function, then its Taylor series representation about a point (computable using Series) is given by . Setting and in the Taylor series of the exponential function yields the beautiful identity for E (the base of the natural logarithm) .
• Other functions associated with or generalizing Factorial include Factorial2, FactorialPower, , QFactorial, BarnesG and Pochhammer.
# Examples
open allclose all
## Basic Examples(7)
Compute the factorial for the first few integers:
Evaluate at real values:
Plot over a subset of the reals:
Plot over a subset of the complexes:
Series expansion at the origin:
Series expansion at Infinity:
Series expansion at a singular point:
## Scope(34)
### Numerical Evaluation(6)
Evaluate numerically:
Evaluate to high precision:
The precision of the output tracks the precision of the input:
Complex number inputs:
Evaluate efficiently at high precision:
Compute worst-case guaranteed intervals using Interval and CenteredInterval objects:
Or compute average-case statistical intervals using Around:
Compute the elementwise values of an array:
Or compute the matrix Factorial function using MatrixFunction:
### Specific Values(5)
Values of Factorial at fixed points:
Values at zero:
Evaluate for large arguments:
Evaluate for half-integer arguments:
Find the positive minimum of Factorial[x]:
### Visualization(2)
Plot the Factorial function:
Plot the real part of :
Plot the imaginary part of :
### Function Properties(10)
Real domain of the factorial:
Complex domain:
The factorial has the mirror property :
Factorial is not an analytic function:
However, it is a meromorphic function in the complex plane:
Factorial is neither nondecreasing nor nonincreasing:
Factorial is not injective:
Factorial is not surjective:
Factorial is neither non-negative nor non-positive:
Factorial has both singularity and discontinuity for negative integers:
Factorial is neither convex nor concave:
### Differentiation(2)
First derivative with respect to z:
Higher derivatives with respect to z:
Plot the higher derivatives with respect to z:
### Series Expansions(5)
Find the Taylor expansion using Series:
Plots of the first three approximations around :
Find the series expansion at Infinity (Stirling's approximation):
Series at :
Find series expansion for an arbitrary symbolic direction :
Taylor expansion at a generic point:
### Recurrence Identities and Simplifications(2)
Recurrence relations:
For positive integers :
### Function Representations(2)
Integral representation of the factorial function:
## Generalizations & Extensions(4)
Evaluate to high precision:
The precision of the output tracks the precision of the input:
Infinite arguments give symbolic results:
Factorial allows derivatives:
## Applications(6)
Make a table of half-integer factorials:
Number of permutations of 6 elements:
Plot of the absolute value of Factorial in the complex plane:
Find the asymptotic expansion of ratios of factorials:
Volume of an ndimensional unit hypersphere:
Lowdimensional cases:
Plot the volume of the unit hypersphere as a function of dimension:
Find the series expansion at -:
## Properties & Relations(9)
Use FullSimplify to simplify expressions involving Factorial:
Compute a generating function sum involving Factorial:
Compute numerical sums involving Factorial:
The generating function is divergent:
Use regularization to obtain a closed-form generating function:
Generating function as a formal series:
Some integrals can be done:
Product of factorials:
Factorial can be represented as a DifferenceRoot:
FindSequenceFunction can recognize the Factorial sequence:
The exponential generating function for Factorial:
## Possible Issues(2)
Large arguments can give results too large to be computed explicitly, even approximately:
Smaller values work:
Machine-number inputs can give highprecision results:
## Neat Examples(3)
Nested factorials over the complex plane:
Plot Factorial at infinity:
Wolfram Research (1988), Factorial, Wolfram Language function, https://reference.wolfram.com/language/ref/Factorial.html (updated 2022).
#### Text
Wolfram Research (1988), Factorial, Wolfram Language function, https://reference.wolfram.com/language/ref/Factorial.html (updated 2022).
#### CMS
Wolfram Language. 1988. "Factorial." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2022. https://reference.wolfram.com/language/ref/Factorial.html.
#### APA
Wolfram Language. (1988). Factorial. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Factorial.html
#### BibTeX
@misc{reference.wolfram_2024_factorial, author="Wolfram Research", title="{Factorial}", year="2022", howpublished="\url{https://reference.wolfram.com/language/ref/Factorial.html}", note=[Accessed: 07-August-2024 ]}
#### BibLaTeX
@online{reference.wolfram_2024_factorial, organization={Wolfram Research}, title={Factorial}, year={2022}, url={https://reference.wolfram.com/language/ref/Factorial.html}, note=[Accessed: 07-August-2024 ]} | 1,560 | 7,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.869154 |
https://questions.examside.com/past-years/jee/question/pa-voltmeter-of-resistance-150-omega-connected-across-mht-cet-physics-motion-drcrmnt9ttht99mc | 1,723,055,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00320.warc.gz | 368,636,798 | 34,881 | 1
MHT CET 2023 10th May Morning Shift
+1
-0
A voltmeter of resistance $$150 \Omega$$ connected across a cell of e.m.f. $$3 \mathrm{~V}$$ reads $$2.5 \mathrm{~V}$$. What is the internal resistance of the cell?
A
$$10 \Omega$$
B
$$15 \Omega$$
C
$$20 \Omega$$
D
$$30 \Omega$$
2
MHT CET 2023 9th May Evening Shift
+1
-0
A galvanometer of resistance $$20 ~\Omega$$ gives a deflection of 5 divisions when $$1 \mathrm{~mA}$$ current flows through it. The galvanometer scale has 50 divisions. To convert the galvanometer into a voltmeter of range 25 volt, we should connect a resistance of
A
$$1240 ~\Omega$$ in series.
B
$$2480 ~\Omega$$ in series.
C
$$2480 ~\Omega$$ in parallel.
D
$$20 ~\Omega$$ in parallel.
3
MHT CET 2023 9th May Evening Shift
+1
-0
Five current carrying conductors meet at a point '$$\mathrm{O}$$' as shown in figure. The magnitude and direction of the current in conductor '$$O P$$' is
A
$$6.5 \mathrm{~A}$$ from $$\mathrm{O}$$ to $$\mathrm{P}$$.
B
$$9 \mathrm{~A}$$ from $$\mathrm{P}$$ to $$\mathrm{O}$$.
C
$$10.5 \mathrm{~A}$$ from $$\mathrm{P}$$ to $$\mathrm{O}$$.
D
$$11.5 \mathrm{~A}$$ from $$\mathrm{O}$$ to $$\mathrm{P}$$.
4
MHT CET 2023 9th May Morning Shift
+1
-0
A galvanometer of resistance $$\mathrm{G}$$ is shunted with a resistance of $$10 \%$$ of $$\mathrm{G}$$. The part of the total current that flows through the galvanometer is
A
$$\frac{1}{11} \mathrm{I}$$
B
$$\frac{2}{11} \mathrm{I}$$
C
$$\frac{1}{10} \mathrm{I}$$
D
$$\frac{1}{5} \mathrm{I}$$
EXAM MAP
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NEET | 563 | 1,511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-33 | latest | en | 0.751638 |
https://gamedev.stackexchange.com/questions/174946/how-to-create-characters-randomly-with-a-given-skill-and-weighted-abilites | 1,714,036,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00168.warc.gz | 241,913,915 | 41,129 | # How to create characters randomly with a given skill and weighted abilites
I am trying to create random characters in my game. Each character has 33 abilities and one of 6 jobs. For each job you need to skill/increase other abilities. I am calculating an overall skill (from 1 to 99) by using only the important abilities for this job. For example one of the jobs is a healer. Here you need to skill ability1 to ability6, which have also different weights.
Example:
- Healer
- Ability1 has weight 10 (skill 70)
- Ability2 has weight 5 (skill 60)
- Ability3 has weight 4 (skill 50)
- Ability4 has weight 10 (skill 80)
- Ability5 has weight 5 (skill 40)
- Ability6 has weight 8 (skill 50)
Example calculation:
(70 * 10) + (60 * 5) + (50 * 4) + (80 * 10) + (40 * 5) + (50 * 8) = 2600 / (10 + 5 + 4+10+ 5+ 8) = 62
Problem:
I want to create random characters with random skills WITH a specific overall skill in HIS job. Example: CreateHealer(skill: 70). It doesn’t matter which values all the other abilities have. For example if ability7 is 1 or 99 the overall skill of the healer is always 70, but I want them to be random because the user has the opportunity to change the job, if he can see that this char would be a better warrior.
My idea:
Create all 33 abilities randomly and calculate the overall skill in this job.
• If skill is on the right level —> stop.
• If skill is too low increase one of the weighted abilities and calculate again.
• If skill is too high decrease one of the weighted abilities and calculate again.
Is there a better way/algorithm to do this?
• Do your characters have a limited point budget they need to distribute over their skills? Or can any skill be anywhere between 1 and 99? Aug 26, 2019 at 9:26
• Could you please explain what you mean with the ability weight? Does that mean that one point in Ability 1 with the weight of 10 means that it adds +10 to his job skill? Or is it just preference of distributing values, like highest on the 10s, then lower and so on? Aug 26, 2019 at 10:41
• It looks like your weights describe a family of hyperplanes in skill space. Using these, you can fix one hyperplane — the one with the total job skill value you want — by choosing one standard solution point on this hyperplane, then adding to it any multiple of a set of tangent vectors in the hyperplane. Restricting to integer solutions makes it a bit tougher though. Remember those math problems like "You have \$2.20 in dimes and quarters, and 13 coins in total - how many of each coin do you have?" - this is the 6-dimensional version, with more free variables for your randomizer to play with. Aug 26, 2019 at 11:22
• @Philipp Each skill can be between 1 and 99. Its possible that all skills have 99 as value. I have added an example to my post. Short: multiply each skill with its weight and sum them. Divide the result by the sum of the weights. Aug 26, 2019 at 12:39
• Let me check if I understood this system correctly: The example you posted has a score of 62. Does that mean that this example character would not qualify as a healer because they need a weighted score of at least 70? Aug 26, 2019 at 12:43
Let's start simple and forget the constraint that skill values should be integers for a moment.
If we look at the equation for your job skill:
\begin{align} \frac{w_1 s_1 + w_2 s_2 + w_3 s_3 + w_4 s_4 + w_5 s_5 + w_6 s_6} {w_1 + w_2 + w_3 + w_4 + w_5 + w_6} &= j\\ w_1 s_1 + w_2 s_2 + w_3 s_3 + w_4 s_4 + w_5 s_5 + w_6 s_6 &= j \left( w_1 + w_2 + w_3 + w_4 + w_5 + w_6 \right) \end{align}
We can think of this as the equation of a 5-dimensional hyperplane in 6-dimensional skill space. Here the plane has a normal vector $$\\vec n = \left(w_1,w_2,w_3,w_4,w_5,w_6 \right)\$$, so we're looking for points in skill space $$\\vec s = \left(s_1, s_2, s_3, s_4, s_5, s_6 \right)\$$ such that...
$$\vec s \cdot \vec n = \left(j,j,j,j,j,j \right) \cdot \vec n$$
From this we can see one obvious solution $$\\vec s = \left(j,j,j,j,j,j \right)\$$ - ie. if you want a job skill of 70, set all the component skills to 70, then any weighted average of them will still give the desired output of 70.
But since it's a plane, once we have one solution, we can slide that solution along the plane to get other solutions. Since we're in 6-dimensional space, and we've locked down one degree of freedom by fixing the value of our weighted average, we still have 5 different directions we can slide the solution while staying on the plane.
We can use the Gram-Schmidt process to take our normal vector and augment it with a set of mutually-perpendicular tangent vectors in the plane, to form a new basis for our skill space:
Vector6[] basisVector = new Vector6[6];
basisVector[0] = Normalize(
new Vector6(weight[0], weight[1], weight[2], weight[3], weight[4], weight[5])
);
for (int i = 1; i < 6; i++) {
basisVector[i] = Vector6.Zero;
basisVectot[i][i] = 1f;
for(int j = 0; j < i; i++)
basisVector[i] -= Dot(basisVector[i], basisVector[j]) * basisVector[j];
basisVector[i] = Normalize(basisVector[i]);
}
At the end of this process, basisVector[0] is your unit normal direction, and basisVector[1] to [5] are unit tangent directions along the solution plane in skill space.
With these in hand, you can pick a range of variation and then generate skill combinations with your desired sum like so:
Vector6 skills = new Vector6(jobValue, jobValue, jobValue, jobValue, jobValue, jobValue);
for(int i = 1; i < 6; i++) {
float deviation = Random.Range(-variation, variation);
skills += deviation * basisVectors[i];
}
Here we slide ± variation along each of our tangent directions, mixing up the skill values while preserving their weighted sum.
The trouble is, this can give us fractional skill values! And when we round them to integers, we might no longer have exactly the job skill value we want.
You could take this and apply a fix-up step, where you choose a skill to raise or lower to restore the desired sum.
Or we can solve the equation for integer solutions in the first place, which makes this a Linear Diophantine Equation in six unknowns.
I had to teach myself how to solve these to answer this question, and the math's still a little unfamiliar so I might not be doing it in the most elegant/rigorous way. But here's the outline of the solution:
• If we have a linear Diophantine equation in two unknowns, $$\a x + b y = r\$$...
• We can use Euclid's algorithm to find the greatest common divisor of $$\a\$$ and $$\b\$$, $$\d\$$
• The equation has a solution if and only if $$\d | r\$$ (ie. r % d == 0)
• We can backtrack through the steps of Euclid's algorithm to find a solution $$\(x_*, y_*)\$$ to the equation $$\a x + b y = d\$$
• We can use this to find a solution to our original equation, $$\a x + b y = d\$$ by just scaling the whole thing up by $$\\frac r d\$$: $$\\left(x_* \frac r d, y_* \frac r d\right)\$$
• Once we have one solution, we can navigate to any other by adding a multiple of $$\\left(\frac b d, \frac a d \right)\$$ to our $$\(x, y)\$$ pair, without changing the value of the right-hand side
(we'll use this to apply randomization to our stats without changing the total job value)
• If we have an equation with more unknowns, we can lump all but the last term together into one (so now we're back to just two unknowns), solve for the last term, then chip it off and repeat:
$$a_0 x_0 + a_1 x_1 + a_2 x_2 + ... + a_n x_n = r$$
Is equivalent to...
$$gcd(a_0 ... a_{n - 1}) y_{n-1} + a_n x_n = r$$
Once we solve this for $$\x_n\$$, we can subtract $$\a_n x_n\$$ from both sides and continue with the remaining $$\n - 1\$$ unknowns...
$$gcd(a_0 ... a_{n - 2}) y_{n-2} + a_{n - 1} x_{n - 1} = r - a_n x_n\\ ...\\ gcd(a_0, a_1) y_1 + a_2 x_2 = r - \sum_{i = 3}^n a_i x_i\\ a_0 x_0 + a_1 x_1 = r - \sum_{i = 2}^n a_i x_i$$
(Note we never really solve for the $$\y_i\$$ terms, they're just standing-in for the rest of the equations we haven't solved yet)
So, what that can look like in code: first, let's make a workhorse to handle solving the two-unknown case, giving us the triplet $$\\left(x_*, y_*, d\right)\$$
public struct DiophantineSolution {
public readonly int x;
public readonly int y;
public readonly int gcd;
DiophantineSolution(int x, int y, int gcd) {
this.x = x;
this.y = y;
this.gcd = gcd;
}
public static implicit operator bool(DiophantineSolution s) {
return s.gcd > 0;
}
public static DiophantineSolution invalid {
get { return new DiophantineSolution(-1, -1, -1); }
}
static Stack<int> quotients = new Stack<int>();
public static DiophantineSolution Solve(int a, int b) {
// Assume a > b - if not, flip it, solve it, then flip back.
if (a < b) {
var flip = Solve(b, a);
return new DiophantineSolution(flip.y, flip.x, flip.gcd);
}
// For now, we'll handle only cases with non-negative coefficients.
if (a <= 0 || b < 0)
return invalid;
// Trivial solution if we have only one unknown with a nonzero coefficient:
if (b == 0)
return new DiophantineSolution(1, 0, a);
// Euclidean Algorithm to find the greatest common divisor:
int x = a, y = b;
int remainder = -1;
do {
int quotient = System.Math.DivRem(x, y, out remainder);
// Save the quotients along the way to use in building the initial solution.
quotients.Push(quotient);
x = y;
y = remainder;
} while (remainder > 0);
// If b exactly divides a, we have a trivial solution.
if (quotients.Count == 1)
return new DiophantineSolution(1, 1 - quotients.Pop(), b);
// Otherwise, rewind to the last step with a non-zero remainder.
remainder = x;
quotients.Pop();
// Form the equation remainder = dividend * (1) + divisor * (- quotient)
// Where x & y are coefficients: x y
x = 1;
y = -quotients.Pop();
// Reverse the steps of the Euclidean algorithm to get a solution to
// remainder = a * x + b * y
while (quotients.Count > 0) {
x -= y * quotients.Pop(); // 1 + 4 * 1
Swap(ref x, ref y);
}
// Now we have our initial solution.
return new DiophantineSolution(x, y, remainder);
}
}
Now we're ready to use this to randomize our stats:
Stack<int> sums = new Stack<int>();
Stack<DiophantineSolution> intermediates = new Stack<DiophantineSolution>();
void GenerateSkills(int[] weights, int targetValue, int randomnessRange) {
// Build up our table of greatest common divisors of the first i weights,
// storing the solution information for re-use later.
// We'll also store the sum of the first i weights for evening-out the stats.
intermediates.Push(DiophantineSolution.Solve(weights[0], 0));
sums.Push(weights[0]);
for (int i = 1; i < weights.Length; i++) {
intermediates.Push(DiophantineSolution.Solve(intermediates.Peek().gcd, weights[i]));
sums.Push(sums.Peek() + weights[i]);
}
// Compute the right-hand side of our equation.
int rhs = sums.Peek() * goalValue;
// Solve the stats one at a time, from the last down to the second...
for(int i = weights.Length - 1; i > 0; --i) {
var solution = intermediates.Pop;
var precedent = intermediates.Peek();
// Initial solution to (... + a_i x_i = rhs)
int baseline = solution.y * rhs / solution.gcd;
// Spacing between possible solution values.
int step = precedent.gcd / solution.gcd;
// Which solution comes closest to giving all remaining stats an equal value?
int closest = Mathf.RoundToInt((myShare - baseline) / (float)step);
// How far can we stray from this evenly-distributed solution?
int range = randomnessRange / step;
// Apply random deviation within this range.
int deviation = Random.Range(-range, range + 1);
// Shift our baseline solution by our chosen multiple of the solution spacing.
skills[i] = baseline + step * (closest + deviation);
// Deduct the value we've accounted for from the right side of the equation.
int contribution = skills[i] * weights[i];
rhs -= contribution;
}
// The first skill handles whatever is left over.
skills[0] = rhs / weights[0];
// Clean up after ourselves.
intermediates.Pop();
sums.Pop();
} | 3,299 | 11,925 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-18 | latest | en | 0.938997 |
http://physics.stackexchange.com/questions/80983/how-does-infrared-relate-to-heat | 1,469,751,211,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00003-ip-10-185-27-174.ec2.internal.warc.gz | 198,866,181 | 22,039 | # How does Infrared Relate to heat?
I never understood the relationship between Infrared and Heat. Is IR emitted when heat is generated, is heat generated when IR is made, and how do the two relate?
-
@PranavHosangadi no, this isn't even technically a correct definition of heat. Heat can only be defined when there is a movement of thermal energy, and IR light cannot be emitted from "anything above absolute zero" it must be emitted from a system which has enough energy (and the correct transition rules) to generate a photon in the IR region. – MaxGraves Oct 16 '13 at 17:36
IR is NOT "light"; it is EM radiation. Anything above zero kelvins, radiates EM energy, some will be IR, but not necessarily much. The approximately 3k background radiation left over from the big bang, will have a spectrum peaked at about 3,000 microns, or 3mm. So 98% of that radiation energy lies between 1.5 mm and 24 mm in the microwave region. At 600 microns in the far IR, or 1/5th of the peak, the spectral radiant emittance is down five orders of magnitude from the peak, and something less than 10^-7 of the total energy is at shorter wavelengths; undetectable; but it isn't zero. – user26165 Oct 18 '13 at 0:30
The energy of electromagnetic radiation (particularly a photon) is
$$E = h \nu = \dfrac{h c}{\lambda} .$$
where $h$ is Planck's constant, $\nu$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. So you can see that as wavelength increases, the energy goes down.
Different wavelengths of electromagnetic radiation correspond with different energies of quantum states. Photons can be absorbed by a molecule to gain energy, but only of specific wavelengths (depending upon the molecule and the two states being transitioned to/from).
Turns out, a photon from the infrared region has an energy on the order of the energy of vibrational transitions in molecules. I mentioned that when a photon is absorbed, the molecule gains energy, well it can also emit a photon and lower its energy to another of its allowed vibrational quantum states. This being said, the reason why IR light is produced and associated with heat is that you are seeing molecules go from one vibrational quantum state to a lower vibrational quantum state by giving off a photon of appropriate energy (in the IR region).
Note that vibrational states are only really accessible at higher (near room temp generally) temperatures.
-
All matter in bulk radiates (approximately) as a black body radiator, approximately because there are coefficients of emissivity depending on the constituents. For gases the functional form is different.
The radiation has a specific spectrum and intensity that depends only on the temperature of the body
As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.
Note that most of the radiation is in the infrared.
Heat may be defined as energy in transit from a high temperature object to a lower temperature object. An object does not possess "heat"; the appropriate term for the microscopic energy in an object is internal energy. The internal energy may be increased by transferring energy to the object from a higher temperature (hotter) object - this is properly called heating.
The energy loss by a black body is given by the Stefan-Boltzman law
Thus the energy carried away by the infrared radiation reduces the heat content of the radiating body. This is the connection of infrared to heat.
The microscopic interactions that give rise to the photons are explained in the other answers. This answer concerns the thermodynamic framework
-
""""".....How does Infrared Relate to heat?....."""""
I take it, that is your question.
"Infra-red" is electro-magnetic radiant energy generally encompassing the range of wavelengths from about 800 nm (near IR) to about 100 micro-meters (far IR). Infra-red radiation includes emissions from single molecules or atoms, as a consequence of quantum mechanical transitions between energy states of those atoms or molecules. It also includes emission of a continuum spectrum of EM radiant energy from large assemblages of atoms or molecules that have a Temperature higher than zero kelvins; that radiation being entirely due to, and characterized by the Temperature of the material, and unrelated to any quantized energy levels that are characteristic of the emitting material. The origin of the radiation is the acceleration of electric charge in the atoms or molecules of the material, while they undergo distortion as a result of being in collisions with each other; those collisions being characteristic of the Temperature of the material.
"Heat energy" on the other hand, is purely mechanical energy of translation or vibration, rotation etc, of the atoms or molecules themselves, and is unrelated to electric charge or properties of electros or atomic states. Heat energy can be transported through physical materials, as a consequence of the collisions between atoms or molecules and their neighbors. (conduction) It can also be transported, by physical bulk transport of the (heated) medium itself (convection).
"Heat energy" is NOT electromagnetic radiation.
-
I think I understand your confusion. The answer would be: You've been misguided. There is no special link between heat and infrared radiation, except for the fact that most bodies radiate most of their heat in the infrared spectrum because they don't have enough energy (heat) to radiate at a higher frequency. See the graphs in this thread.
So one could claim the same connection between X-rays and heat. In fact, it would be even more so, since interactions with X-rays are even higher energy, except that there aren't that many things radiating x-rays around. Probably.
-
So how are infrared thermometers accurate over such a wide range of temperatures and for such a wide range of materials? Would it be just as easy to measure temperature using the color green, or low frequency radio waves, for instance? – vercellop Jul 14 at 4:49
IR radiation is a consequence of the thermal energy of a system, and it is in this form that the energy propagates. Ofcourse, there are various regions in the electromagnetic spectrum and just about any one of them can carry energy (in simple terms). Doubtlessly, you must have heard of the microwave, which uses the radiation in the microwave spectrum to heat food. So, the relation between infrared and heat is that they can be said to be generated from each other.
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To add to this, the microwave region corresponds with the energy gaps between rotational quantum states most generally. This is the reason microwave spectroscopy can resolve rotational quantum states. – MaxGraves Oct 16 '13 at 17:31 | 1,431 | 6,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-30 | latest | en | 0.925351 |
https://www.physicsforums.com/threads/point-charge-and-charged-sphere.527388/ | 1,527,472,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00404.warc.gz | 796,127,519 | 15,500 | # Homework Help: Point Charge and Charged Sphere
1. Sep 5, 2011
### ThirdEyeBlind
1. The problem statement, all variables and given/known data
A point charge q1 = -7 μC is located at the center of a thick conducting shell of inner radius a = 2.6 cm and outer radius b = 4.9 cm, The conducting shell has a net charge of q2 = 1.1 μC.
What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.9 cm along the x-axis from q1?
2. Relevant equations
(1.)For an electric field outside of a conducting sphere(where r>R), E=Q/ (4pi Epsilon naught r^2)
(2.)Electric Field of Point Charge, E= kQ/r^2
3. The attempt at a solution
OK, so what I did was try to the sum of the electric fields first from the point charge in the center using equation (2.) and then I added that to the electric field outside the sphere to the point using equation (1.) and my Q as q2.
Doing this I get -1774598 N/C when the answer should be -6703699 N/C. I am not very good at E&M so I would appreciate all help.
2. Sep 5, 2011
### tiny-tim
Hi ThirdEyeBlind!
(try using the X2 and X2 icons just above the Reply box )
Hint: what is the electric field outside a sphere of radius > b ?
3. Sep 5, 2011
### ThirdEyeBlind
Alright so now I am getting the correct answer thanks to your hint but I am now stuck on another part.
What is σb, the surface charge density at the outer edge of the shell?
I was just thinking it would have been (1.1E-6 C) / (.049 m2 - .026 m2 * pi)
which is the charge of the conducting shell divided by the area of the shell. However the answer is -0.00019554 C/m2 so I don't understand how they get that.
4. Sep 6, 2011
### tiny-tim
Hi ThirdEyeBlind!
(just got up :zzz: …)
Sorry, but this is completely wrong.
For the surface charge density at the outer edge, you use only the area of the outer edge (isn't that obvious? ).
Also, it's 4πr2 for the area, not πr2.
Also, you need to find the charge on the outer edge, not the net charge. | 567 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-22 | longest | en | 0.940631 |
https://quizizz.com/en/tangent-lines-worksheets-class-9?page=1 | 1,719,349,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00266.warc.gz | 400,904,908 | 28,443 | Review tangent lines/chords/arcs
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Explore printable tangent lines worksheets for 9th Class
Tangent lines worksheets for Class 9 are an essential resource for teachers looking to enhance their students' understanding of geometry and math concepts. These worksheets provide a variety of problems and exercises that challenge students to apply their knowledge of tangent lines, circles, and angles in a practical and engaging manner. By incorporating these worksheets into their lesson plans, teachers can ensure that their Class 9 students develop a strong foundation in geometry, which is crucial for success in higher-level math courses. Furthermore, these worksheets can be easily adapted to suit different learning styles and abilities, making them a versatile and valuable tool for any Class 9 math teacher.
Quizizz is an excellent platform for teachers to utilize in conjunction with tangent lines worksheets for Class 9, as it offers a wide range of interactive quizzes and activities that can supplement and reinforce the concepts covered in the worksheets. By incorporating Quizizz into their lesson plans, teachers can create a dynamic and engaging learning environment that caters to the diverse needs of their Class 9 students. In addition to quizzes, Quizizz also offers various other resources, such as flashcards and interactive presentations, which can be used to further support students' understanding of geometry and math concepts. Overall, combining the use of tangent lines worksheets for Class 9 with the interactive features of Quizizz can greatly enhance the learning experience for students and help teachers to effectively teach these important concepts. | 715 | 2,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920344 |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.18/share/doc/Macaulay2/MonomialAlgebras/html/_codim__M__A.html | 1,657,115,870,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00303.warc.gz | 289,372,613 | 2,460 | # codimMA -- Codimension of a monomial algebra.
## Synopsis
• Usage:
codimMA R
codimMA B
codimMA M
• Inputs:
• R, , with B = degrees R and K = coefficientRing R, or
• B, a list, with the generators of an affine semigroup in \mathbb{N}^d.
• M, ,
• Outputs:
## Description
Compute the codimension of the homogeneous monomial algebra K[B].
As the result is independent of K it is possible to specify just B.
i1 : B={{2, 2, 1}, {1, 1, 3}, {1, 2, 2}, {2, 0, 3}, {1, 4, 0}, {2, 3, 0}, {1, 3, 1}} o1 = {{2, 2, 1}, {1, 1, 3}, {1, 2, 2}, {2, 0, 3}, {1, 4, 0}, {2, 3, 0}, {1, ------------------------------------------------------------------------ 3, 1}} o1 : List i2 : codimMA B o2 = 4
i3 : B={{2, 2, 1}, {1, 1, 3}, {1, 2, 2}, {2, 0, 3}, {1, 4, 0}, {2, 3, 0}, {1, 3, 1}} o3 = {{2, 2, 1}, {1, 1, 3}, {1, 2, 2}, {2, 0, 3}, {1, 4, 0}, {2, 3, 0}, {1, ------------------------------------------------------------------------ 3, 1}} o3 : List i4 : M=monomialAlgebra B ZZ o4 = ---[x ..x ] 101 0 6 o4 : MonomialAlgebra generated by {{2, 2, 1}, {1, 1, 3}, {1, 2, 2}, {2, 0, 3}, {1, 4, 0}, {2, 3, 0}, {1, 3, 1}} i5 : codimMA M o5 = 4
## Ways to use codimMA :
• "codimMA(List)"
• "codimMA(MonomialAlgebra)"
• "codimMA(PolynomialRing)"
## For the programmer
The object codimMA is . | 567 | 1,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.355624 |
https://exploringtm1.com/how-to-roll-forward-values-from-one-period-to-the-next-in-tm1/?utm_source=rss&utm_medium=rss&utm_campaign=how-to-roll-forward-values-from-one-period-to-the-next-in-tm1 | 1,603,122,071,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00290.warc.gz | 317,189,466 | 18,119 | # How to Roll Forward Values from One Period to the Next in TM1
How do you take a value from one period and roll it to the next in TM1? This is a really simple task in TM1.
## Why Would you Pull a Value Forward?
Often when we are doing financial modelling or operational planning in TM1, we need to be able to take closing balance and roll it to the opening balance of the next period.
For example, you might be calculating inventory and your formula might be Closing Balance = Opening Balance + Purchases – Cost of Sales. That’s easy, yeah? Well then the next month you want to use the same formula, but you want to populate the Opening Balance with the Closing Balance from the previous month. That’s easy too and we’ll show you how below.
## How to Roll Forward Values in TM1
To do a roll forward in TM1 we need to do a few things:
• Set up a Prior and Next Period attributes in the time dimensions
• Add a rule to pull the values from Closing Balance into Opening Balance
• Add a feeder to push the Closing Balance to activate the Opening Balance cell
• Check results
Time needed: 30 minutes.
How to calculate a value that rolls from one month to the next TM1?
1. Create Prior and Next Period Attributes
First of all we need to have prior period attributes.
If you have Month and Year dimensions, then create a Prior Month and Next Month attributes in the Month dimension and a Prior Year and Next Year attributes in the Year dimension.
If you have a single, continuous, time dimension with months and years in it, then create a Prior Period and Next Period attributes.
These attributes can then be interrogated via an ATTRS to enable the roll forward fuel and the feeder.
2. Write a Rule to Pull Forward the Closing Balance:
`['Opening Balance'] = DB('Inventory Plan',IF([email protected]='Jan',ATTRS('Year',!Year,'Prior Year'),!Year),ATTRS('Month',!Month,'Prior Month'),!Version,!Product,'Closing Balance','Value');`
3. Create Feeders to Activate the Future Cells
`['Closing Balance'] => DB('Inventory Plan',!Year,ATTRS('Month',!Month,'Next Month'),!Version,!Product,'Opening Balance','Value');`
``` ```
`['Dec','Closing Balance'] => DB('Inventory Plan',ATTRS('Year',!Year,'Next Year'),'Jan',!Version,!Product,'Opening Balance','Value');`
4. Review Results
The final step in carting a roll forward in TM1 is to review the results in a view. Here is the result we have from the above.
You can see that we start with a zero balance in January and then add on Purchases and subtract Cost of Sales to calculate Closing Balance. This is all in the dimension hierarchy and the weight of Cost of Sales is set to -1 (so it is treated as a negative).
Then the rule pulls the value from January Closing Balance into February Opening Balance and so on.
Finally, if we were to change year, the Opening Balance fo rJanuary in 2021 would be the Closing Balance for December 2020. | 659 | 2,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-45 | latest | en | 0.816466 |
almuhammadi.com | 1,618,989,048,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00422.warc.gz | 4,064,928 | 7,025 |
Notes:
1. Before solving the homework problems, you are expected to do the reading assignment and understand the related material. An asterisk (*) indicates supplementary reading material.
2. Homework assignments must be typed, saved as PDF, and electronically submitted on the Blackboard.
3. The homework PDF file must have a header on the first page, and a header for each question. See HW-format. You should follow this format when you submit your homework in order to be graded. See also a HW_Sample.doc (or in pdf) which can be used as a template for your homework.
4. Assignments deadlines are announced on the Blackboard.
5. Late submission is subjected to a 50% penalty if it is submitted by the cut-off time (usually 48 hours after the announced deadline).
Reading Assignment [Weeks 1-2]: Introduction to Cryptography, Modular Arithmetic, Prime Numbers.
Stallings: Chapter 1.
Stallings: Sections 2.1-2.4
Homework Assignment HW1:
Stallings: Chapter 1. Exer: 1, 5.
Stallings: Chapter 2. Exer: 2, 3(b,c), 5, 7 (only for k=6,7,8 positive divisors), 10, 11.
EP#1. Use modular arithmetic as explained in the class to compute the following (Show work for credit).
(a) 2640 * 3928 (mod 13)
(b) 3456789 * 34567 (mod 9)
(c) 23456789 * 654321 (mod 11)
EP#2. Use the extended Euclidean algprithms as explained in the class to find the mubliplication inveres of 13 mod 61. Show the steps in equations (not tables)
Total: 10 exercises.
Reading Assignment [Weeks 3-4]: Symmetric Cipher, Cryptanalysis, Block Ciphers, DES, Number Theory.
Stallings: Chapter 3
Stallings: Sections 4.1-4.2
Stalling: Sections 2.4-2.7
Forouzan: Chapters 3*, 5*, 6*
Homework Assignment HW2:
Stallings: Chapter 2. Exer: 20, 22.
Stallings: Chapter 3. Exer: 1, 2.
Stallings: Chapter 4. Exer: 1, 2.
EP#1. In Hill cipher, the ciphertext is computed by: C = PK (mod 26), where P is the plaintext matrix and K is the key. All computations are done in (mod 26) for English alphabet.
(a) Explain how Eve can break the system if enough plaintext-ciphertext pairs are provided.
(b) What is the type of this attack?
EP#2. Consider a 24-bit ideal block cipher.
(a) What is the key size in this cipher?
(b) What is the size of the key-space? (How many keys are there?)
EP#3. Using Fermat't Little Theorem for primality test. Answer the following. Show work for credit.
(a) Test whether each of the following numbers is a prime: 101, 341, and 1105. Try at least two bases if needed, and state if the number is pseudoprime to any base you try. You may use a claculator to compute large powers.
(b) Find a composit number that is pseudoprime to base 3 and 7 but not pseudoprime to base 2.
EP#4. Apply the Chinese remainder theorem to solve the followiwng system of congruences:
x = 3 (mod 4)
x = 1 (mod 5)
x = 5 (mod 7)
Show work for credit.
Total: 10 exercises.
Programming Assignment PA1: See the PA1.pdf document.
Reading Assignment [Weeks 5-6]: Group Theory, Subgroups, Cyclic Groups, Primitive Roots, Group Isomorphism
LNGT: Sections 1 and 2
Homework Assignment HW3:
LNGT: Section 1. Exer: 3, 4, 5, 6(c,d), 7(b,c).
LNGT: Section 2. Exer: 3(a,b,c,f,g), 6(a,b), 7(a,b,c), 8(a,b), 9.
Total: 10 exercises.
Reading Assignment [Week 7]: Review, Midterm exam preparation
Reading Assignment [Weeks 8-9]: Permutation Groups, Finite Fields, 3DES, AES
LNGT: Section 3.
Stallings: Chapter 5, Chapter 6, Section 7.1*
Online materials on AES: Lecture 07-AES.ppt, AES_Demo.swf*.
Homework Assignment HW4:
LNGT: Section 3. Exer: 3(a,b,c), 4(a,b,c), 5, 6(c,e), 7, 8.
EP#1. Apply the Shift-Xor algorithm to multiply P1=(00001101) by P2=(11001101) in GF(256) using the irreducible polynomial x^8 + x^4 + x^3 + x + 1 as a modulus. Show all the steps in a table as explained in the class.
EP#2. Consider the generator for GF(2^3) in Table 5.5. This table is used to efficiently perform operations in this field.
(a) Reconstruct Table 5.5 for GF(2^3) using the modulus x^3+x^2+1 (No need for the Hex column).
(b) Use your table to compute: (g^2 + g)(g^2 + 1) and (g^2)/(g^2 + g)
(b) What are the main requirements of AES stated by NIST?
(c) Briefly describe the 4 stages in the AES round?
(d) The 4 stages in the AES round together provide high security. Explain the weakness of each stage alone.
(e) What are the main strength points of the design of the S-box in AES?
EP#4. Compare DES, 3DES and the three AES versions in terms of: block size, number of rounds, the main key length, and the round-key length. Organize your answer in a 6x5 table (including the headers)
Total: 10 exercises.
Programming Assignment PA2: See the PA2.pdf document.
Reading Assignment [Weeks 10-12]: Public-Key Cryptography: Diffie Hellman, Elgamal, Digital Signature, Secure Hashing, Blockchian
Stallings: Sections 9.1, 10.1, 10.2.
Stallings: Chapter 11*, Sections 13.1*, 13.2.
Bashir: Chapter 1 (Blockchain 101*)
Narayanan: Chapter 1.
Online materials: PPT Slides Lectures 08, 09 and 10.
Homework Assignment HW5:
Stallings: Chapter 10. Exer: 2, 3, 5.
Stallings: Chapter 11. Exer: 3(a,b), 4, 6.
Narayanan: Chapter 1. Exer: 4.
EP#1. Consider ElGamal cryptosystem with the following parameters: prime p = 19, generator g = 2, your private key is x = 6.
(b) If you receive the ciphertext C = (14, 13), what is the plaintext?
(c) Encrypt the message M = 5 to Alice, with public-key = (23, 3, 4).
EP#2. Show that that if Eve can break Diffie-Hellman key exchange scheme, then she can also break ElGamal cryptosystem.
EP#3. Consider ElGamal digital signature scheme with the following parameters: prime p = 19, generator g = 2, your private key is x = 6, and Alice's public key is (p = 19, g = 2, y = 9).
(a) Sign the message digest M = 3 using your private key and random parameter k = 5
(b) If Eve gives you a signed document [M = 9, a = 2, b = 9] and she said that this document is signed by Alice. Is Eve telling the truth? How do you know?
(c) If Alice used the same random parameter (k) to sign two documents, can Eve break the system? Give an example of two signatures to justify your claim.
Total: 10 exercises.
Reading Assignment [Weeks 13-14]: RSA, DSS, Bitcoin, Zero-Knowledge Proofs
Stallings: Sections 9.2.
Forouzan: Section 10.2*.
Stallings: Section 13.4.
Schneier: Secions 5.1, 5.2.
Satoshi Nakamoto: Bitcoin white paper.
Online materials: PPT Slides Lectures 11, 12, and 13.
Homework Assignment HW6:
Stallings: Chapter 9. Exer: 2(b), 7(and explain), 8, 18.
Stallings: Chapter 13. Exer: 2, 3, 5.
EP#1. Prove or disprove that computing the totient of n is equivalent to factorizing n in RSA.
(a) if the block reward is 6.25 BTC now, what is the expected block reward in year 2050?
(b) Explain how to increase the difficulty level of block mining using the proof-of-work.
(c) If Eve has 40% of the total network computation power, what is the probability she can successfully attack the system if each transaction has to wait for the 6 block confirmation?
(d) Eve suggests that the proof-of-work should be easy so that we don't consume lots of electricity. Explain the potential risk if the proof of work takes one second instead of 10 minutes on average.
(a) What is the main reason behind the high cost associated with ZKPs in general?
(b) Consider the zero-knowledge proof of graph-isomorphism. If each graph has 128 nodes, compute the optimal number of rounds.
(c) Explain why ZKPs are needed in the setup of an RSA cryptosystem.
(d) Besides RSA, name three other applications that use ZKPs and explain why ZKP is needed in each case.
(e) What are the main requirements of a zero-knowledge proof?
(e) Which of the following functions are negligible? Justify your answer in each case.
1. f(n) = 1/log(n)
2. f(n) = 1/(2n)
3. f(n) = 1/(n^2)
4. f(n) = n/(n!)
Total: 10 exercises.
Reading Assignment [Week 15]: Cryptocurrency, Smart Contracts, POW Alternatives, Mining, Traceablity.
Bashir: Chapters 8-11*.
Narayanan: Chapters 5-6*.
Online materials: PPT Slides Lecture 14.
Homework Assignment HW7:
EP#1. In a 2x8 table, compare Bitcoin and Ethereum in terms of the following:
(a) Year of release
(b) Average block time
(c) Current block reward*
(d) Maximum supply
(e) POW hash function
(f) Current block height (latest block number)*
(g) Current network difficulty*
(h) Current hashrate*
(* Can be found in the blockchain explorer. Write the date/time in which you access the page)
(a) What programming languages are used to implement EVM?
(b) What programming languages are used to write Ethereum smart contracts?
(c) What is the advantage of using Ethash in Ethereum proof of work?
(d) What alternative of proof of work is proposed for Ethereum 2.0?
EP#3. Unlike Bitcoin, Ethereum has no maximum supply by design.
(a) Explain the reason behind having no fixed maximum supply in Ethereum? How this will affect the inflation rate?
(b) Suppose that 0.1% of all Ether is lost every year within the global Ethereum community. If the current supply is 113805000 ETH, block reward is 2 ETH per block, and the average block time is 13 seconds (fixed), estimate the expected maximum supply deduced by the equilibrium point per the design. When is this equilibrium expected to happen (in which year)? | 2,648 | 9,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-17 | longest | en | 0.845857 |
https://www.physicsforums.com/threads/calculate-maximum-safe-volume-for-headphones.890692/ | 1,511,269,696,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00225.warc.gz | 847,460,468 | 16,753 | # B Calculate maximum safe volume for headphones
1. Oct 25, 2016
### George Albercook
I remember doing a physics problem over 30 years ago but I can't remember enough to do it again. The only reason I remember it is that it seemed like it would be so useful when I had teenage kids. Funny how the brain works sometimes. Got teenagers now. We were working on acoustics. If I remembered correctly the problem used the threshold of hearing and distance. The idea was to calculate the distance to stand away from headphones playing music such that if the volume at that distance was just at the threshold of your hearing than the volume would be safe if you put the headphones in you ears at that volume. I'm sure there are all sorts of assumptions and problems with so simple of a model but I would like to know how to do it anyway. Anyone know the answer?
2. Oct 25, 2016
### Simon Bridge
Welcome to PF;
There is no one answer.
The calculation depends on frequency as well ... but there is a "loudness scale" and tables on power ratings (in decibels) and their effect on human hearing apparatus, all for the googling.
The start is to define what you mean by "safe".
3. Oct 25, 2016
### Staff: Mentor
This thread may need to get moved to the Medical forum, but we'll leave it here for now...
4. Oct 25, 2016 | 308 | 1,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | longest | en | 0.970895 |
https://www.assignmentexpert.com/homework-answers/mathematics/statistics-and-probability/question-399 | 1,580,300,639,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00324.warc.gz | 762,254,807 | 12,949 | 82 507
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# Answer to Question #399 in Statistics and Probability for kingston
Question #399
A singles golf tournament has 32 players. When a player loses, that player is eliminated. How many matches must be played before there is a winner?
1
2010-07-12T05:59:56-0400
Every time number of players will be divided by 2. So we have to find n in the equation 2^n = 32. As we know 2^5 = 32. So, we will be able to find a winner after 5 matches.
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | 174 | 643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-05 | latest | en | 0.958151 |
https://community.wolfram.com/groups/-/m/t/2369756?sortMsg=Likes | 1,638,926,140,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00386.warc.gz | 237,864,169 | 31,302 | # Help fitting and selecting data?
Posted 2 months ago
507 Views
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7 Replies
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Kindly find the data:https://drive.google.com/file/d/1Zw_GBnhFYbEglEKzTXnCv4o5xEMMT9n1/view?usp=sharingI want to fit my data. After using drop and select, I am getting blank {}. I want to fit my data of XZ Plane and potential. Kindly help, as soon as possible
7 Replies
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Posted 2 months ago
Hi Soumyaranjan,This is an empty list, so all of the data is being filtered out, so DataxzStein is empty. Select[DataxyzStein, -10^-7 <= #[[2]] <= 10^-7 &] There are four columns in the data which columns are x, y, z? What is the fourth column?
Posted 2 months ago
Extract columns 1, 3, 4 DataxzStein = DataxyzStein[[All, {1, 3, 4}]] FittXZStein = NonlinearModelFit[ DataxzStein, {ModelxzStein}, {Subscript[T, 0], Subscript[T, 2], Subscript[T, 4], Subscript[T, 6], Subscript[T, 8], Subscript[T, 10]}, {x, z}] FittXZStein["RSquared"] (* 0.9845 *) FittXZStein["ParameterTable"]
Posted 2 months ago
Something like this? (* MinMax for x, y, z, p *) MinMax /@ Transpose@DataxyzStein (* {{-0.495, 0.495}, {-0.495, 0.495}, {-0.305, 0.305}, {0.299106, 0.998006}} *) (* Select positive x values *) DataxzSteinPosX = Select[DataxzStein, #[[1]] >= 0 &] MinMax /@ Transpose@DataxzSteinPosX (* {{0.005, 0.495}, {-0.305, 0.305}, {0.302262, 0.998001}} *)
Posted 2 months ago
Hi Rohit !! 1st 3 columns are x, y and z, respectively. The 4t one is the Electric Potential which is a function of x, y and z. Since I want to fit in the XZ plane, so I want to drop the y column. Please help me to fit those data. Thank you for your time and consideration.
Posted 2 months ago
Hi Rohit,Thank you so much for your time and consideration. But let's say I want to extract a specific range of data from my column data points for fitting; how to do that?Sorry for the silly question. For a beginner, I am struggling to learn. It would be helpful if you answer this. Regards, Soumyaranjan
Add whatever filtering on values you want (* In DataxzStein, x is column 1 and z is column 2 *) DataxzSteinSubset = Select[DataxzStein, Between[#[[1]], {-0.32, 0.32}] && Between[#[[2]], {-0.22, 0.22}] &] MinMax /@ Transpose@DataxzSteinSubset (* {{-0.315, 0.315}, {-0.215, 0.215}, {0.447852, 0.889891}} *) Use DataxzSteinSubset to fit. | 773 | 2,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.756819 |
https://im.kendallhunt.com/k5_es/teachers/grade-2/unit-8/lesson-13/preparation.html | 1,718,448,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00323.warc.gz | 271,162,686 | 25,752 | # Lesson 13
Día 2 de centros (optional)
### Lesson Purpose
The purpose of this lesson is for students to skip-count by 2, 5, and 10 and to add and subtract within 1,000.
### Lesson Narrative
This lesson is optional because it is an opportunity for extra practice that not all classes may need. In this lesson students have another chance to practice skip-counting by 2, 5, and 10 and addition and subtraction within 1,000.
### Learning Goals
Teacher Facing
• Add and subtract within 1,000 using strategies based on place value and the properties of operations.
• Skip-count by 2, 5, and 10.
### Student Facing
• Contemos a saltos y practiquemos la suma y la resta.
### Required Materials
Materials to Gather
### Required Preparation
Activity 1:
Gather materials from:
• Write Numbers, Stage 4
• Target Numbers, Stages 6 and 7
• Five in a Row, Stages 7 and 8
• How Close, Stage 4
### Lesson Timeline
Warm-up 10 min Activity 1 40 min Lesson Synthesis 10 min
### Teacher Reflection Questions
Identify something you thought was going to go well in math class recently, but did not. What can you do to make it a success the next time? | 294 | 1,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.851969 |
https://gpatargeter.com/purchase-order-po-10136-po-10153-po-10163-po-10167-po-10187-po-10219-po-10279-po-10306-po-10332-po-10346-po-10392-po-10471-po-10486/ | 1,680,143,227,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00680.warc.gz | 337,403,143 | 17,018 | # Purchase Order # PO# 10136 PO# 10153 PO# 10163 PO# 10167 PO# 10187 PO# 10219 PO# 10279 PO# 10306 PO# 10332 PO# 10346 PO# 10392 PO# 10471 PO# 10486
Please explain how you arrived at the answers. Also, what do you charge per hour for online tutoring?
Download this Hypothesis Testing Quiz Data Excel file to answer the following questions. Please note that there are three different sheets in the workbook.
1. The null hypothesis should NOT be accepted.
True
False
2. Use the data in the āPart 1ā sheet of the spreadsheet above to answer this question:
Suppose that your company is concerned about two things: 1) having too much inventory, and 2) not having enough inventory. As the company looked into the problem, it discovered that a long lead time (amount of time it takes to receive an order from the supplier) on one of its most expensive parts was causing much of the inventory and stock-out problems.
During the last year, the average lead time for this particular part was 41.5 days. In an effort to shorten the lead time, the company decided to change to a different supplier. The attached spreadsheet shows a random sample of orders placed on the new supplier, along with the corresponding lead time for each order (in days).
What is the probability that a shorter lead time is simply due to ārandom chanceā (therefore NOT the result of changing to a new supplier)?
(Enter your answer as a percentage rounded to the nearest tenth of a decimal. For example, 0.2585 would be entered as 25.8%.)
3. Use the data in the āPart 2ā sheet of the spreadsheet above to answer this question:
Suppose that your relative wants to establish a new dealership for selling new cars. This dealership would be the first dealership to exist in the local community. Your relative is uncertain whether the company should sell domestic or foreign automobiles and, therefore, seeks your help.
You do some research and discover that in the past four years, 70.0% of all automobiles purchased by individuals living in the local community were domestic. However, there has been ātalkā in the area that turmoil in the automotive industry may cause a change in the number of domestic automobiles individuals buy. You decide to conduct a survey by randomly selecting potential new car buyers in the area. The results of your survey are shown in the attached spreadsheet.
Your relative wants to know the probability that the difference between your survey results and the historical proportion of domestic cars sold is due to ārandom chanceā (therefore NOT the result of the turmoil in the auto industry). What is your answer?
(Enter your answer as a percentage rounded to the nearest tenth of a decimal. For example, 0.2585 would be entered as 25.8%.)
4. Use the data in the āPart 3ā sheet of the spreadsheet above to answer this question:
Suppose that a very large company is exploring different ways to improve the job satisfaction of its employees. Every year, the company has its employees complete a ājob satisfactionā survey that ranges from a scale of 1 (lowest satisfaction) to 5 (highest satisfaction). The company decides to pilot an experiment to see if giving out company stock options to employees helps to increase the job satisfaction score.
The company randomly selects employees from two groups: those who received stock options and those who did not. The attached spreadsheet includes the job satisfaction scores for the sampled employees from each group.
What is the probability that the increase in the average satisfaction scores is due to ārandom chanceā (therefore NOT the result of the paying out stock options)?
(Enter your answer as a percentage rounded to the nearest tenth of a decimal. For example, 0.2585 would be entered as 25.8%.) | 851 | 3,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-14 | latest | en | 0.946676 |
https://www.mathworksheetscenter.com/mathskills/subtraction/ | 1,713,205,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00345.warc.gz | 824,058,792 | 4,730 | # Subtraction Worksheets
On this page you will find: a complete list of all of our math worksheets relating to shapes. Choose a specific addition topic below to view all of our worksheets in that content area. You will find addition lessons, worksheets, homework, and quizzes in each section.
### Why is Subtraction Important?
Out of all the arithmetic operations used today, subtraction is the most essential one. The primary reason for its importance is its extensive usage in our daily lives. We encounter math problems daily, either in one way or another. For instance, we use subtraction to find out how much money we will save if we spend on an item. Or we subtract our school hours daily, to the time we have to go home. All these and more reasons make subtraction an essential part of the Arithmetic category. For children’s point of view, it is the first most complex mathematical concept they will learn. Thus, it’s essential that we make the initial concepts for the children strong enough so that it doesn’t affect them later in their lives. Subtraction is an arithmetic operation that works on removing the number/ numbers of the objects from any collection. For example, 5 – 3 is 2. It is signified and represented as a minus sign (-). Subtraction is one of the 4 main heads on which math is based. 1. Make sure you keep the bigger number on top and the small number below. 2. Always start from the right. | 299 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-18 | latest | en | 0.933101 |
https://surrealdb.com/docs/surrealql/functions/geo | 1,701,668,447,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00467.warc.gz | 620,277,693 | 12,529 | Documentation SurrealQL Functions Geo functions
Geo functions
These functions can be used when working with and analysing geospatial data.
Function Description
geo::area() Calculates the area of a geometry
geo::bearing() Calculates the bearing between two geolocation points
geo::centroid() Calculates the centroid of a geometry
geo::distance() Calculates the distance between two geolocation points
geo::hash::decode() Decodes a geohash into a geometry point
geo::hash::encode() Encodes a geometry point into a geohash
geo::area
The geo::area function calculates the area of a geometry.
geo::area(geometry) -> number
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::area({
type: "Polygon",
coordinates: [[
[-0.38314819, 51.37692386], [0.1785278, 51.37692386],
[0.1785278, 51.61460570], [-0.38314819, 51.61460570],
[-0.38314819, 51.37692386]
]]
});
0.13350018278702186
If the argument is not a geometry type, then an EMPTY value will be returned:
RETURN geo::area(12345);
null
geo::bearing
The geo::bearing function calculates the bearing between two geolocation points.
geo::bearing(point, point) -> number
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::bearing( (51.509865, -0.118092), (-0.118092, 51.509865) );
-31.913259585079818
If either argument is not a geolocation point, then an EMPTY value will be returned:
RETURN geo::bearing(12345, true);
null
geo::centroid
The geo::centroid function calculates the centroid between two geolocation points.
geo::centroid(geometry) -> number
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::centroid({
type: "Polygon",
coordinates: [[
[-0.38314819, 51.37692386], [0.1785278, 51.37692386],
[0.1785278, 51.61460570], [-0.38314819, 51.61460570],
[-0.38314819, 51.37692386]
]]
});
{
"type": "Point",
"coordinates": [
-0.10231019499999999,
51.49576478
]
}
If either argument is not a geometry type, then an EMPTY value will be returned:
RETURN geo::centroid(12345);
null
geo::distance
The geo::distance function calculates the haversine distance, in metres, between two geolocation points.
geo::distance(point, point) -> number
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::distance( (51.509865, -0.118092), (-0.118092, 51.509865) );
7491494.807105545
If either argument is not a geolocation point, then an EMPTY value will be returned:
RETURN geo::distance(12345, true);
null
geo::hash::decode
The geo::hash::decode function converts a geohash into a geolocation point.
geo::hash::decode(point) -> string
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::hash::decode("mpuxk4s24f51");
{
"type": "Point",
"coordinates": [
51.50986494496465,
-0.11809204705059528
]
}
If the argument is not a geolocation point, then an EMPTY value will be returned:
RETURN geo::hash::decode(12345);
null
geo::hash::encode
The geo::hash::encode function converts a geolocation point into a geohash.
geo::hash::encode(point) -> string
The function accepts a second argument, which determines the accuracy and granularity of the geohash.
geo::hash::encode(point, number) -> string
The following example shows this function, and its output, when used in a RETURN statement:
RETURN geo::hash::encode( (51.509865, -0.118092) );
"mpuxk4s24f51"
The following example shows this function with two arguments, and its output, when used in a select statement:
RETURN geo::hash::encode( (51.509865, -0.118092), 5 );
"mpuxk"
If the first argument is not a geolocation point, then an EMPTY value will be returned:
RETURN geo::hash::encode(12345);
null | 1,017 | 3,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.626337 |
https://www.coursehero.com/file/6108559/325-tut-6-bushings/ | 1,495,539,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00555.warc.gz | 865,449,696 | 23,560 | 325-tut-6-bushings
# 325-tut-6-bushings - MECH 325 Machine Elements Tutorial 6...
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Unformatted text preview: MECH 325 Machine Elements Tutorial 6 – Bushings 1. Adapted from Shigley, Problem 12‐22 (8 and 9th Eds.) th An Oiles SP 500 alloy brass bushing is 1 in long with a 1‐in bore and operates in a clean environment at 70°F. The allowable wear without loss of function is 0.005 in. The radial load is 500 lbf. The shaft speed is 200 rev/min. Part 1: To be demonstrated by the TA a. Estimate the number of revolutions for radial wear to be 0.005 in. Part 2: To be done by the class b. Estimate the wear expected for 1 million revolutions if the speed is reduced to 80 rev/min, the load is increased to 800 lbf, and the material is changed to Oiles SP 800. 2. Adapted from Shigley, Problem 12‐23 (8th and 9th Eds.) A cast bronze bushing is to give a maximum wear of 0.002 in for 1000h of use with a 125 rev/min journal and 100 lbf radial load. The maximum permissible lubricant temperature is Tmax = 300°F and ambient temperature is T∞ = 70°F. The overall coefficient of heat transfer is ћCR = 2.7 Btu/(h∙ft2∙°F). The coefficient of sliding friction is fs = 0.03. There is to be a design safety factor of nd = 2. Assume that available bushing sizes are given by the entries with “●” in the table below. The environment is clean. L ID ½” ⅝” ¾” ⅞” 1 ” 1¼” 1½” 1¾” 2 ” ½” ● ● ● ● ● ⅝” ● ● ● ● ¾” ● ● ● ● ⅞” ● ● ● 1” ● ● ● ● ● ● 1¼” ● ● ● ● ● 1½” ● ● ● ● ● 1¾” ● ● ● ● 2” ● ● Part 1: To be demonstrated by the TA a. Using starting values for f1 and f2 of unity, find the required bushing length. Part 2: To be done by the class b. Use the result from part a. to refine the computation and see if there is a better choice of bushing. (Hints: the approximations f1 = f2 = 1.0 are likely not valid; the value of D is not fixed at this point.) MECH 325 Additional Information Tutorial 6: Bushings ...
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## This note was uploaded on 02/01/2011 for the course MECH 325 taught by Professor Peteostafichuk during the Fall '10 term at UBC.
Ask a homework question - tutors are online | 686 | 2,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-22 | longest | en | 0.869047 |
https://math.stackexchange.com/questions/2709997/what-is-an-irreducible-prime | 1,563,370,186,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00195.warc.gz | 473,749,693 | 36,054 | # What is an irreducible prime?
Let $S$ be the set of positive integers of the form $6k+1$ for some integer $k$.
Find an irreducible prime in $S$ such that $p\mid ab$ for some $a,b\in S$ but, $p\nmid a$ and $p\nmid b$.
So the above question is what I'm trying to answer however I'm going to be completely honest here, I'm in elementary number theory and I've tried to look up the definition for irreducible primes but most of the definitions look very abstract to me. I was wondering if someone can please explain what irreducible primes are. I don't feel like my original question would be hard to answer if I knew the definition of an irreducible prime. Thanks
I suppose that the answer is something like this: $25,55,115\in S$, $25\mid55\times115$, but $25\nmid55$ and $25\nmid115$. Besides, $25$ is an irreducible prime in $S$, because it can't be written as the product of two smaller elements of $S$.
• Okay so what you're saying is that when we define the set $S$ we give new definition to what is a prime. Prime numbers now only exist within the integers in $S$ correct? – Luis Robles Mar 27 '18 at 7:41
• @LuisRobles Yes, that's it. – José Carlos Santos Mar 27 '18 at 7:44
• Thanks, but then what is the difference between an irreducible and a prime and an irreducible prime? – Luis Robles Mar 27 '18 at 7:46
• @LuisRobles I don't know. That's why I started my answer with “I suppose”. – José Carlos Santos Mar 27 '18 at 7:47
• I think they might be looking for an element which really is a prime (in the integers). But it is hard to tell from the formulation. – Tobias Kildetoft Mar 27 '18 at 7:48 | 455 | 1,612 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-30 | latest | en | 0.940288 |
http://slidegur.com/doc/149369/pptx---princeton-university | 1,477,062,437,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718284.75/warc/CC-MAIN-20161020183838-00191-ip-10-171-6-4.ec2.internal.warc.gz | 239,087,759 | 13,979 | ### pptx - Princeton University
```Time-Space Tradeoffs
in Proof Complexity:
Superpolynomial Lower Bounds for
Superlinear Space
Chris Beck
Princeton University
Joint work with
Paul Beame & Russell Impagliazzo
Jakob Nordstrom & Bangsheng Tang
SAT
• The satisfiability problem is a central problem
in computer science, in theory and in practice.
• Terminology:
– A Clause is a boolean formula which is an OR of
variables and negations of variables.
– A CNF formula is an AND of clauses.
• Object of study for this talk:
CNF-SAT: Given a CNF formula, is it satisfiable?
Resolution Proof System
• Lines are clauses, one simple proof step
• Proof is a sequence of clauses each of which is
– an original clause or
– follows from previous ones via resolution step
• a CNF is UNSAT iff can derive empty clause ⊥
Proof DAG
General resolution: Arbitrary DAG
Tree-like resolution: DAG is a tree
SAT Solvers
• Well-known connection between Resolution
and SAT solvers based on Backtracking
• These algorithms are very powerful
– sometimes can quickly handle CNF’s with millions
of variables.
• On UNSAT formulas, computation history
yields a Resolution proof.
– Tree-like Resolution ≈ DPLL algorithm
– General Resolution ≿ DPLL + “Clause Learning”
• Best current SAT solvers use this approach
SAT Solvers
• DPLL algorithm: backtracking search for
satisfying assignment
– Given a formula which is not constant, guess a
value for one of its variables , and recurse on the
simplified formula.
– If we don’t find an assignment, set the other
way and recurse again.
– If one of the recursive calls yields a satisfying
assignment, adjoin the good value of and return,
otherwise report fail.
SAT Solvers
• DPLL requires very little memory
• Clause learning adds a new clause to the input
CNF every time the search backtracks
– Uses lots of memory to try to beat DPLL.
– In practice, must use heuristics to guess which
clauses are “important” and store only those.
Hard to do well! Memory becomes a bottleneck.
• Question: Is this inherent? Or can the right
heuristics avoid the memory bottleneck?
Proof Complexity & Sat Solvers
• Proof Size ≤ Time for Ideal SAT Solver
• Proof Space ≤ Memory for Ideal SAT Solver
• Many explicit hard UNSAT examples known
with exponential lower bounds for Resolution
Proof Size.
• Question: Is this also true for Proof Space?
Space in Resolution
…
1
2
x˅
˅
˅
Time step
⊥
Must be in memory
• Clause space ≔ max(# active clauses).
[Esteban, Torán ‘99]
Lower Bounds on Space?
• Considering Space alone, tight lower and upper
bounds of (), for explicit tautologies of size .
• Lower Bounds: [ET’99, ABRW’00, T’01, AD’03]
• Upper Bound: All UNSAT formulas on variables
have tree-like refutation of space ≤ .
[Esteban, Torán ‘99]
• For a tree-like proof, Space is at most the height of
the tree which is the stack height of DPLL search
• But, these tree-like proofs are typically too large
small space and time are simultaneously feasible.
• [Ben-Sasson ‘01] Pebbling formulas with linear
size refutations, but for which all proofs have
Space ∙ log Size = Ω(n/log n).
• [Ben-Sasson, Nordström ‘10] Pebbling formulas
which can be refuted in Size O(n), Space O(n),
but Space O(n/log n) Size exp(n(1)).
But, these are all for Space < , and SAT solvers
generally can afford to store the input formula in
memory. Can we break the linear space barrier?
• Informal Question: Can we formally show that
memory rather than time can be a real
bottleneck for resolution proofs and SAT solvers?
• Formal Question (Ben-Sasson):
“Does there exist a such that any CNF with a
refutation of size T also has a refutation of size T
in space O()?”
• Our results: Families of formulas of size n having
refutations in Time, Space nk, but all resolution
refutations have T > (n0.58 k/S)loglog n/logloglog n
Tseitin Tautologies
Given an undirected graph
:→2 , define a CSP:
and
Boolean variables:
Parity constraints:
(linear equations)
When
has odd total parity, CSP is UNSAT.
Tseitin Tautologies
• When odd, G connected, corresponding CNF is
called a Tseitin tautology. [Tseitin ‘68]
• Only total parity of matters
• Hard when G is a constant degree expander:
[Urqhart 87]: Resolution size = 2Ω()
[Torán 99]: Resolution space =Ω(E)
• This work: Tradeoffs on × grid, ≫ , and
similar graphs, using isoperimetry.
Tseitin formula on Grid
• Consider Tseitin formula on × grid, =4
n
l
• How can we build a resolution refutation?
Tseitin formula on Grid
• One idea: Mimic linear algebra refutation
• If we add the lin eqn’s in any order, get 1 = 0.
n
l
• A linear equation on variables corresponds
to 2 clauses. Resolution is implicationally
complete, so can simulate any step of this
with 2() -blowup.
Tseitin formula on Grid
• One idea: Mimic linear algebra refutation
• If we add the lin eqn’s in column order:
n
l
• Never have an intermediate equation with
more than variables. Get a proof of Size ≈
# ∙ 2 , Space ≈ 2 .
Tseitin formula on Grid
• Different idea: Divide and conquer
• In DPLL, repeatedly bisect the graph
n
l
• Each time we cut, one of the components is
unsat. Done after || queries, tree-like
proof with Space ≈ , Size () .
• Savitch-like savings in space, for = ().
Tseitin formula on Grid
• “Chris, isn’t this just tree-width?” – Exactly.
• Our work is about whether this can be improved.
n
l
• The lower bound shows is that quasipolynomial
blowup in size when the space is below 2 is
necessary for the proof systems we studied.
• For technical reasons, work with “doubled” grid.
High Level Overview of Lower Bound
• [Beame Pitassi ’96] simplified previous Proof
Size lower bounds using random restrictions.
• A restriction to a formula is a partial
assignment of truth values to variables,
resulting in some simplification. We denote
the restricted formula by | .
• If Π is a proof of , Π| is a proof of | .
High Level Overview of Lower Bound
• Philosophy of [Beame, Pitassi ’96]
– Find such that any refutation contains a
“complex clause”
– Find and a random restriction which restricts φ’
to , but also kills any complex clause whp.
– Modern form of “bottleneck counting” due to
Haken. Can reinterpret this as finding a large
collection of assignments, one extending each
restriction, each passing through a wide clause.
High Level Overview of Lower Bound
• This work: Multiple classes of complex clauses
– Can state main idea as a pebbling argument.
– Suppose G is a DAG with T vertices that can be
scheduled in space S, and the vertices can be
broken into 4 classes, μ ∶ → {1,2,3,4}.
– Assume that for arcs (, ), ≤ () + 1
– ll sources in 1, all sinks in 4. Everything in 2 is a
“complex clauses” of first type, everything in 3 is a
“complex clauses” of second type.
High Level Overview of Lower Bound
• Suppose that there is a distribution Π on
source-to-sink paths and a real number p s.t.:
– For any vertex , μ ∈ 2,3 , ∈ ≤ p.
– For any two vertices 1 , 2 , 1 = 2, 2 = 3,
1 , 2 ∈ ≤ 2 .
• Then, 2 = Ω −3 .
High Level Overview of Lower Bound
• Proof:
– Let 1 … be sequence of pebbling steps using
only S pebbles. Divide time into epochs, to
be determined later.
– Plot vs. time
4
3
2
1
Time
High Level Overview of Lower Bound
• Let be any source-to-sink path. Claim:
– Either, π hits a vertex of level 2 and a vertex of
level 3 during the same epoch
– Or, hits a vertex of level 2 or 3 during one of the
breakpoints between epochs.
4
3
2
1
Time
High Level Overview of Lower Bound
• Now choose randomly.
– Either, π hits a vertex of level 2 and a vertex of
level 3 during the same epoch
Probability of this: By a union bound, at most
( )2 2
4
3
2
1
Time
High Level Overview of Lower Bound
• Now choose randomly.
– Or, hits a vertex of level 2 or 3 during one of the
breakpoints between epochs.
Probability of this: By a union bound, at most
4
3
2
1
Time
High Level Overview of Lower Bound
• Now choose randomly.
– Conclude 1 ≤ −1 2 2 +
• Optimizing yields 2 = Ω −3 .
• Previous such results only for expanders, etc.
• Gets stronger when you have more levels.
Later we will see that with levels, get
≥
log
Ω
log log
High Level Overview of Lower Bound
• Goal for remainder of analysis:
– Show that any proof of Tseitin on doubled grid has
such a flow (using restrictions), for the best and
as many complex clause classes as possible.
– It is enough that any k complex clauses from
distinct classes have collective width ≥ for
some large W. (For n x l grid, W will be n.)
– How did we get complex clauses before in [Beame
Pitassi ‘96]? A “measure of progress” argument…
Progress Measure
• The following progress measure construction
appeared in [Ben-Sasson and Wigderson ’01].
• Let 1 … be a partition of clauses of an
unsat CNF. Assume it is minimally unsat.
• For any clause C, define
≔ min
⊆[],
∈ ⊨
Progress Measure
• is a sub-additive complexity measure:
(initial clause) = 1,
(⊥) = m,
() ≤ (1 ) + (2), when 1, 2 ⊦ .
• In any refutation, there must occur a clause C
1 2
with / ∈ [ , ]. In previous work,
3 3
choose 1 … so that this implies C is wide.
Progress Measure for Tseitin
• For Tseitin tautologies, natural choice is to
have one for each vertex v, containing all
associated clauses. So is an 2 lin. eqn.
• Claim: Let C be any clause, and let ⊆ ,
⊨ be any subset attaining the min, =
(). Then, δ() ⊆ ().
Progress Measure for Tseitin
• Claim: Let C be any clause, and let ⊆ ,
⊨ , = (). Then δ() ⊆ ().
• Proof:
• If ⊨ , then ⋀¬ is unsat linear system. So
we can derive 1 = 0 by linear combinations.
• If any equation from S has coeff 0, S wasn’t
minimal. In sum of equations of S, noncancelling
vars are exactly boundary edge vars. These must
cancel with ¬ to get 1=0, so δ() ⊆ ().
Isoperimetry → Lower Bounds
• Consequence: Tseitin formula has minimum
refutation width at least the balanced
bisection width of graph G.
• Corollary: Tseitin on a constant degree
expander requires width Ω().
• Corollary: On ⨉ grid, needs width .
• Here, “Complex clause” := ()/m ∈
1 2
[ , ]
3 3
Size Lower Bounds from Width
• We have formulas with width lower bounds,
can use XOR-substitution and restrictions to
get size lower bounds.
• For a CSP, [⨁] is the CSP over variable set
with two copies 1 , 2 of each var of , and
each constraint of interpretted as
constraining the parities (1 ⨁2 ) rather than
the variables . (Then expand as CNF.)
Size Lower Bounds from Width
• [⨁] has a natural random restriction :
– For each , independently choose one of 1 , 2 to
restrict to {0,1} at random. Substitute either or
¬ for the other, so that 1 ⨁2 | = .
• Previously used in [B’01]. Nice properties:
– ⨁ | = , always
– Fairly dense, kills all wide clauses whp.
– Black box, doesn’t depend on .
Size Lower Bounds from Width
• When is a Tseitin tautology, [⨁] is
“multigraph-Tseitin” on same graph but where
each edge has been doubled.
• Suppose Tseitin on doubled grid has a proof of
size less than 2 . Then by a union bound,
some restriction kills all clauses of width , so
Tseitin on the grid has proof of width < .
• Idea: Instead of just considering C with
()/m ∈ [1/3, 2/3] , what about [1/6, 1/3],
[1/12, 1/16], etc.?
• Gives us the levelled property we need.
• Task: Show that clauses from k different levels
are collectively wide.
• Given the machinery we have, this is some
“extended isoperimetry” property of the grid.
Extended Isoperimetry
• Claim: Let 1 … be subsets of vertices,
∈ [2 ,/2], of superincreasing sizes.
n
l
Then, δ( ) ≥ Ω().
• Several simple combinatorial proofs of this.
• Implies clauses from distinct levels are
collectively wide.
Main Lemma
• For any set of clauses in doubled-Tseitin,
Pr[| contains ≥ complex clauses
of complexity levels]
≤ 2
−Ω
Proof: For any k-tuple, Ω(kn) opportunities for
to kill at least one. Union bound over k-tuples.
Complexity vs. Time
• Consider the time ordering of any proof, and
plot complexity of clauses in memory v. time
μ
Input clauses
Time
• Sub-additivity implies cannot skip over any
[t, 2t] window of μ-values on the way up.
⊥
Complexity vs. Time
• Consider the time ordering of any proof, and
divide time into equal epochs (fix later)
Hi
Med
Low
Time
Two Possibilities
• Consider the time ordering of any proof, and
divide time into equal epochs (fix later)
Hi
Med
Low
Time
• Either, a clause of medium complexity
appears in memory for at least one of the
breakpoints between epochs,
Two Possibilities
• Consider the time ordering of any proof, and
divide time into equal epochs (fix later)
Hi
Med
Low
Time
• Or, all breakpoints only have Hi and Low. Must
have an epoch which starts Low ends Hi, and
so has clauses of all ≈log Medium levels.
Analysis
• Consider the restricted proof. With probability
1, one of the two scenarios applies.
• For first scenario, = clauses,
Pr 1st scenario ≤ exp(−)
• For second scenario, epochs with = /
clauses each. Union bound and main lemma:
Pr 2nd scenario ≤
exp(−Ω()) ,
where ≈ log
Analysis
• Optimizing yields something of the form
= exp Ω
for = ω 1 .
• Can get a nontrivial tradeoff this way, but
need to do a lot of work on constants.
• Don’t just divide into epochs once
• Recursively divide proof into epochs and
sub-epochs where each sub-epoch contains
sub-epochs of the next smaller size
• Prove that, if an epoch does a lot of work,
either
– Breakpoints contain many complexities
– A sub-epoch does a lot of work
a
• If an epoch contains a clause at the end of
level , but every clause at start is level ≤
− , (so the epoch makes progress),
• and the breakpoints of its children epochs
contain together ≤ complexity levels,
• then some child epoch makes / progress.
Internal Node Size
= mS
Leaf Node Size
= T/m^(h-1)
Previous Slide shows:
Some set has ≥
complexities, where
ℎ = log
…
• Choose ℎ = , have = log , so = (1)
• Choose so all sets are the same size:
/−1 ≈ , so all events are rare together.
• Have sets in total
Finally, a union bound yields ≥
exp(Ω())
• Tseitin formulas on × grid for = 24
– are of size ≈ .
– have resolution refutations in Size, Space ≈ 2
– have log space refutations of Size 2 log
– and any resolution refutation for Size and
Space requires > (2 0.58 /)ω(1)
If space is at most 2 /2 then size blows up
by a super-polynomial amount
Open Questions
• More than quasi-polynomial separations?
– For Tseitin formulas upper bound for small space
is only a log n power of the unrestricted size
– Candidate formulas? Are these even possible?
• Other proof systems?
– In [BNT’12], got Polynomial Calculus Resolution.
– Cutting Planes? Frege subsystems?
• Other cases for separating search paradigms:
“dynamic programming” vs. “binary search”?
Thanks!
Extended Isoperimetric Inequality
If the sets aren’t essentially blocks, we’re done.
If they are blocks, reduce to the line:
Intervals on the line
• Let 1 , 1 , … , [ , ] be intervals on the
line, such that − ≥ 2(−1 − −1 )
• Let () be the minimum number of distinct
endpoints of intervals in such a configuration.
• Then, a simple inductive proof shows ≥
+1
Proof DAG
Proof DAG
“Regular”:
On every
root to leaf path,
no variable resolved
more than once.
• Theorem : For any k, 4-CNF formulas (Tseitin
formulas on long and skinny grid graphs)
of size n with
– Regular resolution refutations in size nk+1,
Space nk.
– But with Space only nk-, for any > 0, any regular
resolution refutation requires size at least
n log log n / log log log n.
Regular Resolution
• Can define partial information more precisely
• Complexity is monotonic wrt proof DAG
edges. This part uses regularity assumption,
simplifies arguments with complexity plot. | 4,367 | 15,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-44 | latest | en | 0.859126 |
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# MT Yao
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Question
# A body of mass 5kg undergoes a change in speed from 20 to 0.20m/s. The momentum of the body would
A
Increase by 99kgm/s
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B
Decrease by 99kgm/s
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C
Increase by 101kgm/s
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D
Decrease by 101kgm/s
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Solution
## The correct option is B Decrease by 99kgm/smomentum=mass∗δv=5kg∗(20−0.20)=99kgm/sHence it decreases by 99kgm/s
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### The hunt for another B3457/S4568 ship
Posted: March 11th, 2017, 4:05 am
B3457/S4568 is known for it's terribly slow c/5648, the slowest orthogonal elementary ship known. But are there other ships out there? Who knows? I have done ntzfind from p2 to p10, w6 to w8, currently doing p10 w7.
### Re: The hunt for another B3457/S4568 ship
Posted: March 11th, 2017, 12:35 pm
This is an outer-totalistic rule, so you can just use normal zfind. I hope that helps at the higher widths.
### Re: The hunt for another B3457/S4568 ship
Posted: March 11th, 2017, 3:27 pm
These types of rules interest me. I'm currently trying to hunt for the slowest ship by using blob rules. Wish me luck.
### Re: The hunt for another B3457/S4568 ship
Posted: March 11th, 2017, 4:02 pm
I find it interestingly coincidental that the speed of the spaceship is an anagram of the survival conditions.
Here's hoping for a c/7543 or c/8654 ship.
### Re: The hunt for another B3457/S4568 ship
Posted: March 11th, 2017, 11:46 pm
I strongly suggest a c/7 diagonal bilateral symmetric search, since we have this (promising) partial:
Code: Select all
``````x = 74, y = 74, rule = B3457/S4568
4b4o\$2bob4o\$b4obob2o\$2b8o\$4o2bob4o\$2obo2b6o\$6ob3obobo\$2obob2ob3ob2o\$2b
6obob5o\$2b7o3bo2bo\$4b2obo2bob6o\$4b3obo2bob2ob2o\$7b4ob2ob5o\$6b3ob10o\$8b
ob2ob9o\$8b3ob3ob6o\$10b9o3bo\$10b11o2b2o\$12b13o\$12b4ob4ob5o\$14b2ob3ob6o\$
14b2o2bob2ob5o\$16bob3ob2ob2o2bo\$17b6o3b4o\$17b5o2bob2ob3o\$19b4o3b4obo\$
19b15o\$21bob4ob2o3bo\$23bob3obo3b3o\$22b9ob4o\$24bobo2b3ob5o\$24b3o3bob2ob
3o\$26bo2bob3ob5o\$26b9obob2o\$28b3o2bob7o\$28b5ob2ob2ob2o\$30b5o3b2ob3o\$
30b3ob2o3b2ob2o\$32b5o2bob5o\$32b3ob7ob2o\$34b2obob9o\$34b3ob5o2bobo\$36b7o
2b2ob2o\$36b3obo6b3o\$38b3o6b5o\$38b5o4b2ob2o\$40bobo4b4ob2o\$40b2ob4obobob
2o\$42b11ob2o\$42b3obob8o\$44b11ob2o\$44b2o2b3obobo2bobo\$46b6o2bo2b2o\$46b
2ob2o2b2o2b5o\$48b10o2b2o\$48b2o4bobobo3b2o\$50bo3b3obobob2o\$50b5o2bobob
3obo\$52b2ob2ob2o2bob2o\$51bobo3b3o2bob4o\$53b2obo3bob4obo\$53b2o2bo3b2ob
2ob3o\$55b9ob2ob2o\$55b3o2bobo3bob4o\$58b4o7b3o\$57b6o3b6obo\$59bo2b2obo2b
2obo\$59b3o3bob2o\$61b3ob3o\$61b6o\$63b3o\$63b4o2\$65bo!
``````
### Re: The hunt for another B3457/S4568 ship
Posted: March 12th, 2017, 12:55 pm
c/7 orthogonal also has promise,
Code: Select all
``````x = 19, y = 124, rule = B3457/S4568
9bo2\$7b2ob2o2\$4b2o2b3o2b2o\$4b4obob4o\$3b3o2bobo2b3o\$3bobo2bobo2bobo\$b4o
2bo3bo2b4o\$4b5ob5o\$3obobo2bo2bobob3o\$3b5obob5o\$8obob8o\$3bo2b7o2bo\$5ob
7ob5o\$b3obo3bo3bob3o\$ob15obo\$b17o\$2o3b2ob3ob2o3b2o\$5b3obob3o\$3ob4obob
4ob3o\$bob5obob5obo\$6ob5ob6o\$bo2bob3ob3obo2bo\$5ob7ob5o\$3bobo7bobo\$5ob7o
b5o\$b17o\$2obobo3bo3bobob2o\$b17o\$3o5bobo5b3o\$b4ob7ob4o\$3ob4obob4ob3o\$2b
o2b9o2bo\$7ob3ob7o\$b5obobobob5o\$6obobobob6o\$b2ob2ob5ob2ob2o\$7ob3ob7o\$bo
bob3o3b3obobo\$obobo3bobo3bobobo\$b3o2b3ob3o2b3o\$3ob3ob3ob3ob3o\$2b7ob7o\$
9ob9o\$bo4bobobobo4bo\$3ob2ob2ob2ob2ob3o\$3bob3o3b3obo\$2obo4b3o4bob2o\$b3o
2bob3obo2b3o\$ob2ob9ob2obo\$b3ob4ob4ob3o\$o4b9o4bo\$b4o3bobo3b4o\$ob15obo\$b
3o3bobobo3b3o\$5ob2o3b2ob5o\$bob13obo\$4ob9ob4o\$bo2b5ob5o2bo\$3ob11ob3o\$2b
obobo2bo2bobobo\$6obobobob6o\$b2ob4obob4ob2o\$ob15obo\$b4o2b5o2b4o\$19o\$bo
2bobo2bo2bobo2bo\$4o2b7o2b4o\$b17o\$19o\$b17o\$2ob13ob2o\$b7obob7o\$7o2bo2b7o
\$3o2b3obob3o2b3o\$2b7ob7o\$5ob3ob3ob5o\$bob2ob3ob3ob2obo\$3ob11ob3o\$b7o3b
7o\$o3bob7obo3bo\$bo2b3ob3ob3o2bo\$ob4ob5ob4obo\$b7obob7o\$ob3o3b3o3b3obo\$b
o2b11o2bo\$4obob5obob4o\$bobob4ob4obobo\$2ob2o2b5o2b2ob2o\$bob13obo\$3o5b3o
5b3o\$b7obob7o\$o2b13o2bo\$b4obob3obob4o\$5ob7ob5o\$bo2b4obob4o2bo\$2obob4ob
4obob2o\$b4ob7ob4o\$o4bob5obo4bo\$b8ob8o\$2obob9obob2o\$b6ob3ob6o\$2ob4ob3ob
4ob2o\$b7obob7o\$19o\$2bo2bo2b3o2bo2bo\$3o2bobobobobo2b3o\$b3obob5obob3o\$ob
o4bobobo4bobo\$b2ob2ob2ob2ob2ob2o\$b3ob3o3b3ob3o\$b2o2b9o2b2o\$4b3obobob3o
\$ob3o2b5o2b3obo\$b17o\$6ob2ob2ob6o\$2b4ob5ob4o\$2ob3ob5ob3ob2o\$bob6ob6obo\$
19o\$b2ob11ob2o\$ob15obo\$2b6o3b6o!
``````
EDIT:
And c/5 potential:
Code: Select all
``````x = 117, y = 25, rule = B3457/S4568
12bobobo2bobobobobobobob2o4bo2bobobobobobobobobobobobobobobobobob2obob
o29bob2o\$12b5obob11obo2b3ob2obob2ob2o2b4ob6obobobob4obobob4obo2bobobob
obo2bobobo2bobobobo2b2o\$bobobobobo2bob7o2bob4o2b3obob2ob5ob4ob2obo2b2o
bo2b6ob6ob4ob3ob6ob3ob3obob6ob2ob2obo\$b4ob13ob4ob8ob4ob5ob4ob3o2b4ob5o
4b3o3b6o2b4o4b11obob8ob2o\$6o2b11ob4obo4b2ob13obob3ob19ob3ob4obobo11b5o
b3obo2b3o2b2o\$3ob2o2b11ob2ob13ob8obobob6ob6o2b5ob3o2b2ob4obo3b3o3b2o2b
6ob5ob5o\$2b5o2bo2bobo2bobobob2ob3ob5ob3obob2ob3ob7obob3ob17ob5o3bob2ob
2o2bo2b4obo2b3ob3o\$2b2o2b6ob2ob3ob2ob3ob9ob8ob2o3bobob5ob2o2b5o2b5ob4o
bo2b3ob5ob2ob2ob7ob6o\$2bobob2ob4obob3obo2b2obob2ob2o2b5obob5obobobobob
3ob9o2b2ob4ob2obo2b4o2b4o2bob3o2b8ob2o\$4bo3b2o2b9o2b3ob3ob2ob3obobob4o
bobobobobob3o2bob3o2b4ob5obobob5ob8o2b7ob2o3b3o\$6bobo2bobob6o2b4ob2obo
3b6obob6ob6o2b5obo2b2o3b2obobob2o2b7o2b3obo2b3ob4obobob3o\$5b10ob6o4bob
11obo2b2o3b3ob2o2bo2bob8obob3obobobobob2ob9ob5o2bob4obob3o\$5bobobob2ob
3ob4ob11o2b5ob4o2b2o2b2o2bo2bo2bob3ob8o2bobo2b6ob6obo2b5ob3o2b3obo\$5b
10ob6o4bob11obo2b2o3b3ob2o2bo2bob8obob3obobobobob2ob9ob5o2bob4obob3o\$
6bobo2bobob6o2b4ob2obo3b6obob6ob6o2b5obo2b2o3b2obobob2o2b7o2b3obo2b3ob
4obobob3o\$4bo3b2o2b9o2b3ob3ob2ob3obobob4obobobobobob3o2bob3o2b4ob5obob
ob5ob8o2b7ob2o3b3o\$2bobob2ob4obob3obo2b2obob2ob2o2b5obob5obobobobob3ob
9o2b2ob4ob2obo2b4o2b4o2bob3o2b8ob2o\$2b2o2b6ob2ob3ob2ob3ob9ob8ob2o3bobo
b5ob2o2b5o2b5ob4obo2b3ob5ob2ob2ob7ob6o\$2b5o2bo2bobo2bobobob2ob3ob5ob3o
bob2ob3ob7obob3ob17ob5o3bob2ob2o2bo2b4obo2b3ob3o\$3ob2o2b11ob2ob13ob8ob
obob6ob6o2b5ob3o2b2ob4obo3b3o3b2o2b6ob5ob5o\$6o2b11ob4obo4b2ob13obob3ob
19ob3ob4obobo11b5ob3obo2b3o2b2o\$b4ob13ob4ob8ob4ob5ob4ob3o2b4ob5o4b3o3b
6o2b4o4b11obob8ob2o\$bobobobobo2bob7o2bob4o2b3obob2ob5ob4ob2obo2b2obo2b
6ob6ob4ob3ob6ob3ob3obob6ob2ob2obo\$12b5obob11obo2b3ob2obob2ob2o2b4ob6ob
obobob4obobob4obo2bobobobobo2bobobo2bobobobo2b2o\$12bobobo2bobobobobobo
bob2o4bo2bobobobobobobobobobobobobobobobobob2obobo29bob2o!
``````
### Re: The hunt for another B3457/S4568 ship
Posted: August 24th, 2017, 2:19 pm
A really, really, really, really, really, really bad c/4 partial from zfind:
Code: Select all
``````x = 16, y = 8, rule = B3457/S4568
7b2o2\$5b6o\$3b10o\$3b3ob2ob3o\$b14o\$3obobo2bobob3o\$3ob8ob3o!
``````
A considerably much better (but not as much as the one above) c/7:
Code: Select all
``````x = 18, y = 19, rule = B3457/S4568
8b2o2\$6b6o2\$4b2obo2bob2o\$3bo10bo\$2b3obo4bob3o\$2bobo8bobo\$2o2b10o2b2o\$
3bo10bo\$6ob4ob6o\$bob3o6b3obo\$o3b10o3bo\$b6o4b6o\$ob4ob4ob4obo\$b6o4b6o\$6o
bo2bob6o\$2b6o2b6o\$o2bo3b4o3bo2bo!
``````
Heres the final result, with 0 spaceships found:
Code: Select all
``````x = 18, y = 22, rule = B3457/S4568
8b2o\$6bo4bo\$3b2o2bo2bo2b2o\$b7o2b7o\$2b6o2b6o\$18o\$b4ob6ob4o\$obobob2o2b2o
bobobo\$b6ob2ob6o\$ob14obo\$b4o2b4o2b4o\$obob10obobo\$bob12obo\$o3b3ob2ob3o
3bo\$b3ob2ob2ob2ob3o\$18o\$2b6o2b6o\$3o2b3o2b3o2b3o\$ob14obo\$bobob8obobo\$ob
obob2o2b2obobobo\$bob12obo!
`````` | 4,063 | 6,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | latest | en | 0.793137 |
https://nrich.maths.org/public/leg.php?code=-334&cl=1&cldcmpid=5727 | 1,477,516,310,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720973.64/warc/CC-MAIN-20161020183840-00298-ip-10-171-6-4.ec2.internal.warc.gz | 851,173,269 | 9,566 | # Search by Topic
#### Resources tagged with Games similar to I'm Stuck!:
Filter by: Content type:
Stage:
Challenge level:
### There are 140 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Games
### Learning Mathematics Through Games: 3. Creating Your Own Games
##### Stage: 1
Not all of us a bursting with creative game ideas, but there are several ways to go about creating a game that will assist even the busiest and most reluctant game designer.
### Going for Games
##### Stage: 1 and 2
In this article for teachers, Liz Woodham describes the criteria she uses to choose mathematical games for the classroom and shares some examples from NRICH.
### Using Games in the Classroom
##### Stage: 2, 3 and 4
Gillian Hatch analyses what goes on when mathematical games are used as a pedagogic device.
### Learning Mathematics Through Games Series: 1. Why Games?
##### Stage: 1, 2 and 3
This article supplies teachers with information that may be useful in better understanding the nature of games and their role in teaching and learning mathematics.
### Learning Mathematics Through Games Series: 2.types of Games
##### Stage: 1, 2 and 3
This article, the second in the series, looks at some different types of games and the sort of mathematical thinking they can develop.
### Learning Mathematics Through Games Series: 4. from Strategy Games
##### Stage: 1, 2 and 3
Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . .
### Criss Cross Quiz
##### Stage: 2 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
### Shut the Box
##### Stage: 1 Challenge Level:
An old game but lots of arithmetic!
### Making Maths: Birds from an Egg
##### Stage: 2 Challenge Level:
Can you make the birds from the egg tangram?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### How Long Does it Take?
##### Stage: 2 Challenge Level:
In this matching game, you have to decide how long different events take.
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
##### Stage: 1 and 2 Challenge Level:
Who said that adding couldn't be fun?
### L-ateral Thinking
##### Stage: 1 and 2 Challenge Level:
Try this interactive strategy game for 2
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Double or Halve?
##### Stage: 1 Challenge Level:
Throw the dice and decide whether to double or halve the number. Will you be the first to reach the target?
### 9 Hole Light Golf
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### Pass the Peas, Please
##### Stage: 1 Challenge Level:
A game for 2 or more players. Practise your addition and subtraction with the aid of a game board and some dried peas!
### Totality
##### Stage: 1 and 2 Challenge Level:
This is an adding game for two players.
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Making Maths: Happy Families
##### Stage: 1 and 2 Challenge Level:
Here is a version of the game 'Happy Families' for you to make and play.
### Matching Time
##### Stage: 1 Challenge Level:
Try this matching game which will help you recognise different ways of saying the same time interval.
### Shut the Box for Two
##### Stage: 1 Challenge Level:
Shut the Box game for an adult and child. Can you turn over the cards which match the numbers on the dice?
### Money Line-up
##### Stage: 1 Challenge Level:
In this game for two players, the aim is to make a row of four coins which total one dollar.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Nice or Nasty for Two
##### Stage: 2 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Horizontal Vertical
##### Stage: 1 Challenge Level:
Take it in turns to place a domino on the grid. One to be placed horizontally and the other vertically. Can you make it impossible for your opponent to play?
### Domino Magic Rectangle
##### Stage: 2, 3 and 4 Challenge Level:
An ordinary set of dominoes can be laid out as a 7 by 4 magic rectangle in which all the spots in all the columns add to 24, while those in the rows add to 42. Try it! Now try the magic square...
### Low Go
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players. Take turns to place a counter so that it occupies one of the lowest possible positions in the grid. The first player to complete a line of 4 wins.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Sliding Game
##### Stage: 2 Challenge Level:
A shunting puzzle for 1 person. Swop the positions of the counters at the top and bottom of the board.
##### Stage: 2 Challenge Level:
This is a game for 2 players. Each player has 4 counters each, and wins by blocking their opponent's counters. A good follow-on from two stones.
### Tac-tickle
##### Stage: 1 and 2 Challenge Level:
This is a challenging game of strategy for two players with many interesting variations.
### Do You Measure Up?
##### Stage: 2 Challenge Level:
A game for two or more players that uses a knowledge of measuring tools. Spin the spinner and identify which jobs can be done with the measuring tool shown.
### Train
##### Stage: 2 Challenge Level:
A train building game for 2 players.
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Fraction Card Game 1
##### Stage: 1 Challenge Level:
Match the halves.
### Fractions and Coins Game
##### Stage: 2 Challenge Level:
Work out the fractions to match the cards with the same amount of money.
### Making Maths: Snake Pits
##### Stage: 1, 2 and 3 Challenge Level:
A game to make and play based on the number line.
### Square It
##### Stage: 1, 2, 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Board Block
##### Stage: 1 Challenge Level:
Take it in turns to make a triangle on the pegboard. Can you block your opponent?
### Wild Jack
##### Stage: 2 Challenge Level:
A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### SHOO
##### Stage: 2 Challenge Level:
A complicated game played on a 9 x 9 checkered grid.
### The Unmultiply Game
##### Stage: 2, 3 and 4 Challenge Level:
Unmultiply is a game of quick estimation. You need to find two numbers that multiply together to something close to the given target - fast! 10 levels with a high scores table. | 2,107 | 8,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-44 | longest | en | 0.871195 |
lytongblog.wordpress.com | 1,508,304,300,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822747.22/warc/CC-MAIN-20171018051631-20171018071631-00468.warc.gz | 783,590,255 | 24,112 | # គោលបំណងនៃមេរៀន
ក្រោយពីសិក្សាមេរៀននេះចប់ អ្នកសិក្សាទាំងគោលបំណងចម្បងនៃមេរៀននេះគឺ៖
1. ចង់អោយស្គាល់អោយបានច្បាស់ពី រូបរាងស្តង់ដារបស់អនុគមន៍ដឺក្រេទី២ និងក្រាបតំណាងអោយអនុគមន៍នេះ
2. ចង់អោយសិស្សអាចយល់បានពីទ
3. ចង់អោយសិស្ស
# សេចក្តីផ្តើម
អនុគមន៍ដឺក្រេទី២ មានរាងទូទៅគឺ៖ y = f(x) = ax2 + bx + c = 0 ដែល a, b និង c គឺជាមេគុណ និងគោរពលក្ខខណ្ឌ a ≠0។អនុគមន៍នេះ មានក្រាប ឬក្រាហ្វិកតាងអនុគមន៍ជារាងប៉ារាបូលដូចរូបខាងក្រោម។
ក្រាបអនុគមន៍ y=x2។គួរបញ្ជាក់ថា អនុគមន៍នេះមានមេគុណ a=1, b=0 និង c=0។
The function of the coefficient aa in the general equation is to make the parabola “wider” or “skinnier”, or to turn it upside down (if negative):
If the coefficient of x2x2 is positive, the parabola opens up; otherwise it opens down.
## The Vertex
The vertex of a parabola is the point at the bottom of the ” UU ” shape (or the top, if the parabola opens downward).
The equation for a parabola can also be written in “vertex form”:
y=a(xh)2+ky=a(x−h)2+k
In this equation, the vertex of the parabola is the point (h,k)(h,k) .
You can see how this relates to the standard equation by multiplying it out:
y=a(xh)(xh)+ky=a(x−h)(x−h)+k
y=ax22ahx+ah2+ky=ax2−2ahx+ah2+k
The coefficient of xx here is 2ah−2ah . This means that in the standard form, y=ax2+bx+cy=ax2+bx+c , the expression
b2a−b2a
gives the xx -coordinate of the vertex.
Example:
Find the vertex of the parabola.
y=3x2+12x12y=3×2+12x−12
Here, a=3a=3 and b=12b=12 . So, the xx -coordinate of the vertex is:
122(3)=2−122(3)=−2
Substituting in the original equation to get the yy -coordinate, we get:
y=3(2)2+12(2)12y=3(−2)2+12(−2)−12
=24=−24
So, the vertex of the parabola is at (2,24)(−2,−24) .
## The Axis of Symmetry
The axis of symmetry of a parabola is the vertical line through the vertex. For a parabola in standard form, y=ax2+bx+cy=ax2+bx+c , the axis of symmetry has the equation
x=b2ax=−b2a
Note that b2a−b2a is also the xx -coordinate of the vertex of the parabola.
Example:
Find the axis of symmetry.
y=2x2+x1y=2×2+x−1
Here, a=2andb=1a=2 and b=1 . So, the axis of symmetry is the vertical line
x=14x=−14
## Intercepts
You can find the yy -intercept of a parabola simply by entering 00 for xx . If the equation is in standard form, then you can just take cc as theyy -intercept. For instance, in the above example:
y=2(0)2+(0)1=1y=2(0)2+(0)−1=−1
So the yy -intercept is 1−1 .
The xx -intercepts are a bit trickier. You can use factoring , or completing the square , or the quadratic formula to find these (if they exist!).
## Domain and Range
As with any function, the domain of a quadratic function f(x)f(x) is the set of xx -values for which the function is defined, and the range is the set of all the output values (values of ff ).
Quadratic functions generally have the whole real line as their domain: any xx is a legitimate input. The range is restricted to those points greater than or equal to the yy -coordinate of the vertex (or less than or equal to, depending on whether the parabola opens up or down). | 1,593 | 3,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2017-43 | latest | en | 0.369806 |
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# Solving for Power
If the voltage and current supplied to a circuit or load by a source are:
Vs = 170<(-0.157) V
Is = 13<0.28 A
determine
a) The power supplied by the source which is dissipated as heat or work in the circuit (load).
b) The power stored in reactive components in the circuit (load).
c) The power factor angle and power factor.
Problem is also in the attached file.
#### Solution Summary
This solution involves step-by-step calculations for power.
\$2.19 | 126 | 500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-04 | latest | en | 0.904905 |
https://www.studypool.com/discuss/1156358/v-4-3-r-3-solve-for?free | 1,481,020,670,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00470-ip-10-31-129-80.ec2.internal.warc.gz | 1,017,059,846 | 14,004 | ##### v=4/3π r^3 solve for π
Algebra Tutor: None Selected Time limit: 1 Day
I need help on finding Pi in this equation
Sep 8th, 2015
The task is to make Pi the subject,
V=4/3Pir^3
Divide by r^3 on both sides
V/r^3=4/3Pi
Next multiply by reciprocal of 4/3
Pi =(3/4)(V/r^3)
Sep 8th, 2015
...
Sep 8th, 2015
...
Sep 8th, 2015
Dec 6th, 2016
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check_circle | 167 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-50 | longest | en | 0.903296 |
http://slideplayer.com/slide/5913420/ | 1,718,334,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00068.warc.gz | 33,179,453 | 22,630 | # Trigonometry (RIGHT TRIANGLES).
## Presentation on theme: "Trigonometry (RIGHT TRIANGLES)."— Presentation transcript:
Trigonometry (RIGHT TRIANGLES)
Why do we Need Trigonometry?
In Physics, we will often be using right triangles in diagrams. Trigonometry lets us find missing side lengths and angles of right triangles. Is it NOT that hard to do the computations we will need for this course, so let’s get to it!
The Right Triangle
The Pythagorean Theorem
For RIGHT TRIANGLES ONLY! Used when we know the lengths of any 2 sides of a right triangle and we need to know the third length. Formula: 𝑎 2 + 𝑏 2 = 𝑐 2 …where A and b are the legs and c is the hypotenuse.
Trigonometry When we know only one side and one angle of a right triangle OR When we need to find an angle of a right triangle We will use the trig ratios to help us!
Practice #1 𝑎 2 + 𝑏 2 = 𝑐 2 = 𝑐 2 = 𝑐 2 242= 𝑐 2 𝑐= 242 ≈15.56 𝑘𝑚
Practice #2 𝑎 2 + 𝑏 2 = 𝑐 2 𝑥 2 + 4.9 2 = 6.2 2 𝑥 2 +24.01=38.44
𝒙 𝟐 =𝟑𝟖.𝟒𝟒−𝟐𝟒.𝟎𝟏 𝑥 2 =14.43 𝑥= ≈3.80 𝑚
sin 𝜃 = 𝑂𝑃𝑃 𝐻𝑌𝑃 = 𝑂 𝐻 cos 𝜃 = 𝐴𝐷𝐽 𝐻𝑌𝑃 = 𝐴 𝐻 tan 𝜃 = 𝑂𝑃𝑃 𝐴𝐷𝐽 = 𝑂 𝐴 “SOHCAHTOA” can help you remember!
Calculator TIPS Make sure your calculator is in degree mode! Use the SIN, COS, and TAN buttons of your calculator to find a trig ratio. Press 2nd and the same buttons to find an angle that has the given trig ratio (Inverse trig functions)
Practice Problem #1 (Finding a Missing Angle)
Which trig ratio will solve the problem? sin 𝜃 = 𝑂𝑃𝑃 𝐻𝑌𝑃 sin 𝜃 = =0.7 𝜃= sin −1 (0.7)≈44.4°
Practice Problem #2 (Finding a Missing Angle)
Which trig ratio will solve the problem? CO𝑆 𝜃 = 𝐴𝐷𝐽 𝐻𝑌𝑃 COS 𝜃 = 12 13 𝜃= COS −1 ( )≈22.6°
Practice Problem #3 (Finding a Missing SIDE)
Which trig ratio will solve the problem? TAN 43° = 𝑂𝑃𝑃 𝐴𝐷𝐽 TAN 43° = 𝑎 11 𝑎=11∗ tan 43° ≈10.3
Practice Problem #4 (Finding a Missing SIDE)
Which trig ratio will solve the problem? COS 50° = 𝐴𝐷𝐽 𝐻𝑌𝑃 COS 50° = 17 𝑥 𝑥∗ cos 50° =17 𝑥= 17 cos 50° ≈26.4
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# A wire is 12.00 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 7.00 1010 ...
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A wire is 12.00 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 7.00 × 1010 N/m2. If a force of 500 N is applied to end of the wire, then the increase in length of the wire is
38.5 mm.
40.2 mm.
44.1 mm.
48.5 mm.
51.5 mm.
Textbook
## College Physics
Edition: 5th
Author:
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marksonbolmarksonbol
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remy2012 Author
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Correct TY
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Good timing, thanks!
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What percentage of nature vs. nurture dictates human intelligence? | 312 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.834735 |
https://www.reddit.com/r/Physics/comments/1gou7s/two_accelerators_find_signs_of_a_particle_that/ | 1,490,916,069,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218203536.73/warc/CC-MAIN-20170322213003-00281-ip-10-233-31-227.ec2.internal.warc.gz | 919,478,774 | 22,362 | This is an archived post. You won't be able to vote or comment.
[–] 5 points6 points (4 children)
But the detectors see no sign of it splitting apart before it decays.
Detectors find no signs of decay prior to signs of decay.
[–] 1 point2 points (0 children)
That sentence refers to the idea that the particle has two levels of structure, one of which could split before the other.
[–]Particle physics 1 point2 points (1 child)
That's talking about the scenario where it is a 'meson molecule' (not a term I ever heard before, but it's explained in the article). It does actually make sense, but is badly worded.
The alternative would be what some reports are calling a "meson molecule," where a pair of two-quark mesons are held together by an attractive force. The problem with the latter option, as noted by Nature News, is that the molecule should be less stable than its constituent mesons. But the detectors see no sign of it splitting apart before it decays.
Basically, what you'd have in that scenario is two mesons, each of which consists of two quarks. The binding energy between the mesons is less than the binding energy between the quarks, so the di-meson structure should decay more quickly than the individual mesons, and this would be visible in the detector. What you'd see would basically look like two mesons, quite near each other in the detector, and you'd be able to reconstruct some decay vertex (away from the interaction point) where these two mesons had split from each other.
What they actually see (judging from this article, I've not read the papers yet) shows no sign of the two mesons decaying separately, but just one decay of the entire four quark system. That would suggest it's not the 'meson molecule' (bound state of two mesons, each of which is a bound state of two quarks), but a bound state of four quarks.
[–] 0 points1 point (0 children)
Yes, I understand the physics of it, I was commenting on the crappy reportage contained within that sentence.
[–] 0 points1 point (0 children)
Typically particles are composed of two or three quarks. So to find four quarks must mean either two particles have split before hand or that a new particle composed of four quarks has been found.
[–][S] 2 points3 points (2 children)
[–] 8 points9 points (1 child)
Two accelerators find signs of a particle that nobody can explain
There seem to be four quarks involved, but nobody's sure how they're linked
Although no law of physics precludes larger congregations, finding a quartet expands the ways in which quarks can be snapped together to make exotic forms of matter.
[Explicit diagrams showing how the quarks are linked together]
So what's up with physics journalism nowadays? Why can't anybody get the story straight?
[–] 5 points6 points (0 children)
Because it's getting harder. Millilan saying "this oil drop fell faster because it's charged" is much more accessible than the results of a particle collider working in the TeV range based on standard model understanding/predictions. After all, what's a boson?
[–] 0 points1 point (2 children)
Amateur here, maybe this is a poor question. But how does a 4-quark particle come up with an integral charge? Seems like you can get - 4/3,-1/3, 2/3, 5/3, or 8/3 times e for the charge. Does that mean it's a virtual particle, or am I just thinking of the rule that only colorless particles are observable? Is this the entire reason this particle is dubious? Apologies if I've gotten a definition or two wrong.
[–] 2 points3 points (1 child)
Its a combination of two quarks and two antiquarks. So as an example: charm-down-antiup-antiup gives a total charge of -1.
[–]Particle physics 1 point2 points (0 children)
Replace one of those antiup with an anti-charm, and you have the system described in this article. | 922 | 3,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-13 | latest | en | 0.966532 |
http://www.reciprocalsystem.org/paper/discussion-of-satz-derivation-of-plancks-constant | 1,576,248,138,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00556.warc.gz | 231,794,817 | 9,331 | # Discussion on Satz's Derivation of Planck's Constant
(1) The ability to move away from established patterns of thinking and strike a new avenue of approach is a rare characteristic but most desirable for research. In his paper, “A New Derivation of Planck's Constant,”1 Satz comes up with such a fresh approach. He suggests that “frequency in the natural sense is the number of cycles per space-time unit.” This is at variance with Larson’s view that “…the so-called ‘frequency’ is actually a speed. It can be expressed as a frequency only because the space that is involved is always a unit magnitude.”2 I am not in the least maintaining that concurrence with Larson’s views is the criterion of truth. But in this instance, the recognition that frequency is really speed seems nearer the truth, since in the context of the theory of the universe of motion, the criterion that decides the truth of a concept is whether it is explainable in terms of the basic component of that universe, namely, speed.
Satz supports his conclusion by the statement: “Photons of all frequencies can be observed in both sectors, and the only way that this could be possible is if the denominator of the natural definition contains both a space unit and a time unit.” In order to see the falsity of this statement it is necessary to remember that a photon has two speeds: the speed of propagation in the forward direction, and the speed of oscillation in the lateral direction. The fact that the speed of propagation is of constant magnitude and unit value (in the natural units) is what makes the photon observable in both sectors, since unit speed is the boundary between these sectors. The frequency, which is the speed in the lateral direction and which is the measure of its energy, is not relevant to its observability from both the sectors.
(2) Satz gives in space-time terms the equation
$E = h v$ (1)
as
$\frac{t}{s} = \left [ \frac{t^2}{\frac{t/s}{t/s}} \right ] \frac{1}{s \times t}$ (2)
and in the cgs system of units
$erg = \left[ \frac{sec^2}{\left(\frac{sec/cm}{erg} \right)} \right ] \frac{1}{cm \times sec}$ (3)
It is to be noted that in this, the dimensions of frequency are taken to be cycles/(cm-sec). On this basis only he continues and arrives at the value of h in Equation (5). At this juncture h has the dimensions erg-sec-cm and frequency cycles/(cm-sec). He now adopts the following procedure: he detaches the cm dimension from the frequency and attaches it to h, rendering its dimensions erg-sec. Let us call this procedure of his the “swap” for later reference. This procedure has the effect of insulating this cm term from the effects of any operations that are uniformly carried out on all the terms comprising h, because he “swaps” this cm term into h only after performing those operations on the terms comprising h.
(3) After incorporating the interregional factor into the terms contained in the square brackets of Equation (3) he arrives at the Planck’s constant h as
$\frac{1}{156.4444} \times \frac{t^2_0}{\left(\frac{cm / sec}{erg} \right )}$ (4)
If one compares the terms comprising h respectively in Equation (3) and Equation (4), it becomes apparent that the author unwarrantedly introduces in Equation (4) the term $t^2_0$, replacing the term sec2. I will call this procedure of his the “switch.” All the terms in Equation (3) are in the cgs system of units. The rationale for making this “switch” is not given: only the numerator term is “switched,” retaining the other terms in the cgs units. Further, if the “swap” is not carried out, the cm term we wish to avoid finally in the frequency would find place in h right from the beginning, making it
$\frac{1}{156.4444} \times \frac{sec^2 / cm}{\left(\frac{cm / sec}{erg} \right )}$ (5)
As such, if he has reasons to “switch” the sec term in the numerator to the natural unit of time, the same reasons would compel the “switching” cm term also to the natural unit of space. This, of course, produces the wrong result. The “swap” serves to avoid just this.
(4) At two places Satz comments on my attempt3 to calculate the Planck’s constant. He contends that I “started by setting the dimensions of energy to be space divided by time, which is, of course, the reverse of what they are.” If my Paper is read carefully it would be found that this is not what I did. I have clearly shown in my Equation (l) the relation between energy and speed in space-time terms as
$\frac{t}{s} = \frac{t^2}{s^2} \times \frac{s}{t}$ (6)
I explained that expressing all the quantities in the natural units we obtain from the above energy in natural units = (1/1)2 speed (in natural units), since the term t2/s2 , the square of the natural unit of inverse speed, is unity. In other words, so far as primary units are concerned, n natural units of speed, say, are equivalent to n natural units of energy. I had even taken care to explicitly mention that the latter is a quantitative relationship. Despite this, Satz has misconstrued it as a dimensional relationship. I had, in addition, pointed out Larson’s account (see Reference 2 of my Paper) for the sake of elucidation.
(5) The other place at which Satz contends that I was guilty of a dimensional mistake is in connection with my modification concerning the effect of secondary mass. While deriving the magnitude of the natural unit of energy, I think we should distinguish between the energy equivalents of the speed of a primary motion like the (one-dimensional) photon vibration and the speed of a gravitational entity (like an atom or subatom). This would not have mattered if we could derive the magnitude of the natural unit of energy directly from experimental results. But Larson first derives the magnitude of the natural unit of mass from Avogadro’s constant. The magnitude of the natural unit of energy is derived from the natural unit of mass, thus derived. Therefore, the size of this energy unit is to be adjusted for the secondary mass effects as applicable to the one-dimensional situation.
Letting p be the primary mass and s the secondary mass, we have the ratio of the gravitational mass unit to the primary mass unit as (p + s)/p. Remembering that the dimensions of energy are t/s while those of mass are (t/s)3, the ratio of the energy unit derived from the gravitational mass unit to the true one-dimensional energy unit would have to be [(p + s)/p]1/3. Since the primary mass unit, p = 1, this factor turns out to be (1 + s)1/3. It may be noted that this factor is non-dimensional, being a ratio, and its application (my Equation (7)) does not vitiate the dimensions of h as Satz contends.
Further, Satz remarks that, “secondary mass varies between the subatoms and the atoms and so cannot be a part of the conversion factor…” But this is not relevant to the situation, since I was concerned with the effect of the secondary mass included in the definition of the gravitational mass unit on the size of the natural unit of energy, insofar as the latter is derived from the unit of gravitational mass. I was not speaking of the secondary mass component included in the mass composition of the material particles at all, since that has no bearing on the present issue, as Satz correctly points out. I was, however, uncertain as to which items make up the secondary mass—like the magnetic mass, the electric mass, etc.—in the situation I was discussing.
(6) And a final comment: In Satz’s Equation (4)
$h = \frac{1}{156.4444} \times \frac{t^2_0}{\left(\frac{sec / cm}{erg} \right )}$ (7)
replacing all the terms with the corresponding natural units we get
$h = \frac{1}{156.4444} \times \frac{t^2_0}{\left(\frac{t_0 / s_0}{e_0} \right )} = \frac{1}{156.4444} \times \left(e_0 \times t_0 \times s_0 \right )$ (8)
If we now bring in the cm term that has been staying in the denominator of the frequency term, we
$h = \frac{\left(e_0 \times t_0 \times s_0 \right )}{156.4444 \times 1 cm}$ (9)
This is identical to my Equation (6) (seeing that I there used suffix n instead of suffix 0 and the upper case letters instead of the lower case) and, therefore, there is nothing essentially new in Satz’s derivation excepting the introduction of the “swap” and the “switch.”
1 Satz, Ronald W., “A New Derivation of Planck's Constant,” Reciprocity XVIII, № 3 (Autumn, 1989).
2 Larson, Dewey B., Nothing But Motion, North Pacific Publishers, OR, 1979, p. 152.
3 KVK Nehru, “Theoretical Evaluation of Planck's Constant,” Reciprocity XII, № 3 (Summer, 1983), p 6.
International Society of Unified Science
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Theme by Danetsoft and Danang Probo Sayekti inspired by Maksimer | 2,118 | 8,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-51 | longest | en | 0.956637 |
https://study.com/academy/topic/measurement-geometry-skills.html | 1,555,898,773,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578533774.1/warc/CC-MAIN-20190422015736-20190422041736-00195.warc.gz | 559,336,179 | 25,350 | # Ch 13: Measurement & Geometry Skills
### About This Chapter
Brush up on your knowledge of measurement, including temperature, and the tools used in assessing measurements. These video lessons discuss basic geometry concepts, such as formulas for area.
## Measurement & Geometry Skills - Chapter Summary
Use this chapter's lessons to review basic measurement skills and the tools needed to take measurements. You'll look at formulas for mass and volume, and work some practice problems involving them. Another lesson examines temperature.
Our instructors go over some basic principles involving geometric figures, including geometric constructions and formulas for finding area. After watching these videos, you should be able to:
• Perform basic operations with measurements
• Use scales, meters and gauges to take measurements
• Understand properties of geometric figures
• Analyze math problems by using geometry
• Calculate the areas of triangles, rectangles and complex figures
• Convert units of mass and volume
• Define temperature and how it is measured
You can watch these lessons and take the self-assessment quizzes on a computer or mobile device, anywhere you have a few minutes to devote to study. Practice problems in these lessons help you understand the concepts being discussed.
11 Lessons in Chapter 13: Measurement & Geometry Skills
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Support | 464 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-18 | longest | en | 0.907454 |
http://stackoverflow.com/questions/21308115/rearranging-variable-names | 1,448,822,369,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398459214.39/warc/CC-MAIN-20151124205419-00312-ip-10-71-132-137.ec2.internal.warc.gz | 217,890,983 | 22,180 | # Rearranging variable_names
How to write in a standard conforming manner avs_term_rearranged(AVs, T, AVsR) with given AVs and T such that AVsR is a permutation of AVs with the elements arranged in same order as their variables occur in left-to-right order in T.
AVs is a list of elements of the form A = V where A is an atom designating a variable name like 'X' and V is a corresponding variable. Such lists are produced by read_term/2,3 with the read-option variable_names/1 (7.10.3). Additionally, the precise order of elements is not defined.
A+B+A+_+C.
AVs = ['A'=A,'B'=B,'C'=C]
T = A+B+A+_+C
T is a term that contains all variables of AVs plus some more.
Note that in a standard conforming program one cannot rely on the term order for variables (7.2.1):
7.2.1 Variable
If X and Y are variables which are not identical then X term_precedes Y shall be implementation dependent except that during the creation of a sorted list (7.1.6.5, 8.10.3.1 j) the ordering shall remain constant.
NOTE — If X and Y are both anonymous variables then they are not identical terms (see 6.1.2 a).
Consider as an example from 8.4.3.4:
sort([f(U),U,U,f(V),f(U),V],L).
Succeeds, unifying L with [U,V,f(U),f(V)] or
[V,U,f(V),f(U)].
[The solution is implementation dependent.]
So there are two possible ways how sort/2 will work, and one cannot even rely on the success of:
sort([f(U),U,U,f(V),f(U),V],L), sort(L, K), L == K.
As an example:
?- avs_term_rearranged(['A'=A,'B'=B,'C'=C], A+C+F+B, AVsR).
AVsR = ['A'=A,'C'=C,'B'=B].
-
is T an arbitrary term or is it of the same form? is it, e.g., also obtained from read_term? – Christian Fritz Jan 25 '14 at 0:00
btw, you are using T twice in your question with different meanings. Might help to rename one of them to avoid confusion. – Christian Fritz Jan 25 '14 at 0:02
@ChristianFritz: I cannot see a difference. Once, T is used as an argument for the sought predicate, and once it is used with read_term which in this case produces T and AVs such that they fit to the predicate. – false Jan 25 '14 at 13:58
Could you then give an example input and expeted output? Is this what you have in mind avs_term_rearranged(['A'=A,'B'=B,'C'=C], A+C+F+B, ['A'=A,'C'=C,'B'=B]) ? – Christian Fritz Jan 25 '14 at 17:11
seems that if you can figure out how to use ream_term/3 to read from a stream that you write to using write_term/3 then you can get a representation of the variables in T that conform with that of AVs, and at that point it's a simple matter of discarding from the T-variables all those that do not appear in the AVs variables. – Christian Fritz Jan 25 '14 at 17:51
avs_term_rearranged(AVs, T, AVsR) :-
term_variables(T, Vs),
copy_term(Vs+AVs, Vs1+AVs1),
bind_names(AVs1),
build_vn_list(Vs, Vs1, AVsR).
bind_names([]).
bind_names([N=V|AVs]) :-
N = V,
bind_names(AVs).
build_vn_list([], [], []).
build_vn_list([V|Vs],[N|Ns],NVs) :-
( atom(N) ->
NVs = [N=V|NVs1]
; var(N) ->
NVs = NVs1
),
build_vn_list(Vs, Ns, NVs1).
-
Blows my mind. I could have sworn that one needs either something along @JSchimpf's version (limited by max_arity and max_integer) or that sort/2 would be needed to associate the variables and their order. In any case, I'm happy I asked, I would have never found that solution. Maybe the mental block was that I hoped to rearrange the existing (=)/2, whereas you are "reconstructing" them, procedurally speaking. – false Jan 30 '14 at 22:46
Going through it step-by-step: After copy_term/2: AVs can be reclaimed. After bind_names/2: AVs1 can be reclaimed. Prior to build_vn_list: Only two lists are present: No pairs at all!! – false Jan 30 '14 at 23:13
Brilliant! I was wondering whether we have a name for this technique of using a bipartite data structure with pairs of corresponding variables as a "mapper". – jschimpf Feb 1 '14 at 15:59
My previous answer had quadratic runtime complexity. If that's a problem, here is a linear alternative. The reason this is a bit tricky is that unbound variables cannot be used as keys for hashing etc.
As before, we basically rearrange the list AVs such that its elements have the same order as the variables in the list Xs (obtained from term_variables/2). The new idea here is to first compute a (ground) description of the required permutation (APs), and then construct the permutation of AV using this description.
avs_term_rearranged(AVs, T, AVRs) :-
term_variables(T, Xs),
copy_term(AVs-Xs, APs-Ps),
ints(Ps, 0, N),
functor(Array, a, N),
distribute(AVs, APs, Array),
gather(1, N, Array, AVRs).
ints([], N, N).
ints([I|Is], I0, N) :- I is I0+1, ints(Is, I, N).
distribute([], [], _).
distribute([AV|AVs], [_=P|APs], Array) :-
nonvar(P),
arg(P, Array, AV),
distribute(AVs, APs, Array).
gather(I, N, Array, AVRs) :-
( I > N ->
AVRs = []
;
arg(I, Array, AV),
I1 is I+1,
( var(AV) -> AVRs=AVRs1 ; AVRs = [AV|AVRs1] ),
gather(I1, N, Array, AVRs1)
).
-
Nice but not portable since it requires Array to be a compound term with unbounded arity. A lesser problem is that it requires arbitrary size integers. – Per Mildner Jan 28 '14 at 16:31
@Per, I honestly think after 30 years it is time to lay arity limits on compound terms to rest. – jschimpf Feb 1 '14 at 16:18
@Per, I do not understand your comment about "arbitrary size integers". This code has no more requirement for large integers than any code using length/2. In fact, this is one of the few types of program where integers are strictly bounded (to something like 2^30 on 32-bit, or 2^61 on 64-bit machines). – jschimpf Feb 1 '14 at 16:28
The question was specifically about a standard conforming solution. There is nothing in the standard that requires arbitrary arities of compound terms, or integers large enough to represent the length of a representable list. In practice, the integer size is unlikely to be a problem, this is why I wrote that its size is less of a problem. – Per Mildner Feb 2 '14 at 17:04
Let me just summarize and clarify: (1) this is a fully standard-conforming program; (2) it does not require arbitrary sized integers; (3) ISO-Prolog implementations vary in the size of problem instances they support; (4) Per's solution is less likely to run into such implementation limits, and generally more elegant, that's why I upvoted and applauded it. – jschimpf Feb 3 '14 at 16:12
Use term_variables/2 on T to obtain a list Xs with variables in the desired order. Then build a list with the elements of AVs, but in that order.
avs_term_rearranged(AVs, T, AVRs) :-
term_variables(T, Xs),
pick_in_order(AVs, Xs, AVRs).
pick_in_order([], [], []).
pick_in_order(AVs, [X|Xs], AVRs) :-
( pick(AVs, X, AV, AVs1) ->
AVRs = [AV|AVRs1],
pick_in_order(AVs1, Xs, AVRs1)
;
pick_in_order(AVs, Xs, AVRs)
).
pick([AV|AVs], X, AX, DAVs) :-
(_=V) = AV,
( V==X ->
AX = AV,
DAVs = AVs
;
DAVs = [AV|DAVs1],
pick(AVs, X, AX, DAVs1)
).
Notes:
• this is quadratic because pick/4 is linear
• term_variables/2 is not strictly necessary, you could traverse T directly
-
This version is very short, using member/2 (from the Prolog prologue) for the search. It has very low auxiliary memory consumption. Only the list AVsR is created. No temporary heap-terms are created (on current implementations). It has quadratic overhead in the length of AVs, though.
avs_term_rearranged(AVs, T, AVsR) :-
term_variables(T, Vs),
rearrange(Vs, AVs, AVsR).
rearrange([], _, []).
rearrange([V|Vs], AVs, AVsR0) :-
( member(AV, AVs),
AV = (_=Var), V == Var ->
AVsR0 = [AV|AVsR]
; AVsR0 = AVsR
),
rearrange(Vs, AVs, AVsR).
Another aspect is that the member(AV, AVs) goal is inherently non-deterministic, even if only relatively shallow non-determinism is involved, whereas @jschimpf's (first) version does create a choice point only for the (;)/2 of the if-then-else-implementation. Both versions do not leave any choice points behind.
Here is a version more faithfully simulating @jschimpf's idea. This, however, now creates auxiliary terms that are left on the heap.
rearrange_js([], _, []).
rearrange_js([V|Vs], AVs0, AVsR0) :-
( select(AV, AVs0, AVs),
AV = (_=Var), V == Var ->
AVsR0 = [AV|AVsR]
; AVsR0 = AVsR,
AVs0 = AVs
),
rearrange_js(Vs, AVs, AVsR).
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1 Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture # 10 Sets Today we shall learn about sets. You must have learnt about sets in school itself. We shall learn a little bit more about sets in these one or two lectures. Now what is a set? A set is a collection of objects and the objects themselves are called elements of the set. (Refer Slide Time: 1:41) For example, you can have the set of natural numbers n or you can think about the set of books or you can think about the set of even non-negative integers less than 10 which is denoted by this. Usually a set is denoted like this and the elements are denoted like this. It is written within flower brackets all the elements are written within flower brackets. So S denotes a set which consists of elements a, b, c, d. Now you say that a belongs to S. Use this symbol for belonging to. So this means a belongs to S or a is in S. In contrast with that if you write e does not belong to S, it is not here, so e does not belong to S or e is not in S.
2 (Refer Slide Time: 3:02) Now, you can also have a definition like this. For example the set of even numbers even non-negative integers you can say that E is equal to x where x belongs to N it is a nonnegative integer and there is a y belonging to N such that x is equal to 2y. This denotes the set of all even non-negative integers. Now you are using a predicate to define e, such a definition is known as implicit definition whereas a definition of this form is known as explicit definition. (Refer Slide Time: 3:12) So a set can be defined using explicit definition or implicit definition. For example, take the set of integers greater than 10 this you can write as x x belongs to i i denotes the set of
3 integers AND x is greater than 10. So you are using a predicate to define this set. So the implicit notation is given there and the explicit notation for that will be set of integers greater than 10 so it is 11, 12, 13 and so on. If it is a finite set you can list all the elements, if it is an infinite set you can list a few of them and the way you write the dot the other members are specified. (Refer Slide Time: 4:48) So we have the concept of the set. And originally in the Naive set theory when it was proposed they found some difficulties in set theory. They found that there were some paradoxes and there were something which leads to contradiction and so on. So let us see the things which let to a contradiction or a paradox. Consider the universe of discourse, let the underlying set be set of all sets. Then define a set S like this; it consists of all sets which do not belong in themselves.
4 (Refer Slide Time: 4:50) A set S is defined like this; it consists of all sets which do not belong in themselves. Now the question is S a member of itself? What will happen? Just discuss this; S is equal to x, x belongs to x does S belong to S? This is the question. Suppose S belongs to S, S consists of all sets which do not belong in themselves from this you have to come to the conclusion that S does not belong to S because it does not satisfy the definition. Now suppose S does not belong to S it satisfies this definition so it should belong to S. So from this you conclude that S belongs to S. (Refer Slide Time: 7:10)
5 So you are arriving at a contradiction. So this is known as Russell s paradox, this is called Russell s paradox. So there seems to be some basic flaw in defining sets itself and so people were wondering whether the set theory will really be useful or not. But this paradox or some flaws should not lead one to give up set theory as it is because there are so many benefits achieved by sets. Even in programming languages you use sets you use set operations and so on. So what we need to do is we have to modify the Naive set theory a little bit and one way to do this is, you have different levels so the individual objects form one level, a set which consists of element from this should be in the next level and in this level you may have a set which consists of elements from the lower levels it may have some element from here a element from here and so on. And at the higher level you may have a set which consists of elements from the lower level. But the set will not have an element in the same level. (Refer Slide Time: 8:35) So if you distinguish like that between levels and allow for the definition of a set in such a way that a set can have only elements from the lower levels then this sort of a paradox will not arise because you will not have such in this definition is itself because a set will have only elements from the lower level. So if you modify that definition of set in that way this paradox will not arise and there will not be any flaw in the set theory and you can make use of set theory as it is. Now, sets are denoted by what is known as Venn diagram. The universal set is denoted like this; a set A may be denoted like this, another set B may be denoted like this, this portion will denote the set of elements common to A AND B or it denotes A intersection B. Like that you can represent sets by means of Venn diagram. Now let us consider some relations among sets.
6 (Refer Slide Time: 9:54) See relation between sets; Consider two sets A and B. Let A and B be sets then when do you say that A is a subset of B? This is denoted as A contained in B or it is denoted like this A contained B. If each element of A is an element of B that is A contained in B is saying it is equivalent to saying for all of x x belongs to A implies x belongs to B. (Refer Slide Time: 10:05) Now sometimes you use this to denote A contained in B and A contained in B if they are not equal this is the subset and this is the proper subset. In this case it is possible for A and B to be equal also, in this case B has at least one more element than A they are not equal.
7 (Refer Slide Time: 11:41) So when you say A is contained in B you could also write it as a B contains A, you have to write like this B contains A this is read as B contains A. So you read it as if A is contained in B we also write B contains A and say A is contained in B or B contains A, or B is a superset of A you can also say it like this A is a subset of B or B is a superset of A. We write A not contained in B that is contained in and then a slash over that, if A is not a subset of B. If A is contained in B and A is not equal to B we say A is a proper subset of B. Let U be the universe of discourse and A is a set then A will be a subset of U. U is the whole clause it is a big set a universal set so any set will be contained in that. Let A and B be sets then A is equal to B if and only if A is contained in B and B is contained in A.
8 (Refer Slide Time: 13:00) So if you use, if A is contained in B AND B is contained in A so when you say this they can be equal also when you have both these which means A is equal to B. So here you have to take it as inclusion and not proper inclusion. For any set A A is contained in A here again this represents subset and not proper subset. Sometimes you distinguish like this; this denotes subset and this denotes proper subset, sometimes in some notations this itself denotes subset. A set with no members is called an empty or null or void set. This is denoted by phi and a set with one member is called a singleton set. (Refer Slide Time: 14:20)
9 Let phi be an empty set and A is an arbitrary set then phi is contained in A. How do you prove that? This denotes the empty set and A is an arbitrary set a singleton set will be denoted like this having a single element this is a singleton set. Now you want to show phi is contained in every set A. What is the definition of A contained in B? x belongs to A implies x belongs to B this is the condition. Now phi is contained means x belongs to phi it should satisfy this condition x belongs to phi implies x belongs to A. But you can see that the premise of this implication is false. x belongs to phi is false so the whole implication is true so it satisfies this condition phi contained in A. This is true the whole implication is true so any empty set is contained in any set A. Next is, if phi and phi dash be sets which are both empty then phi is equal to phi dash. Again this you can prove like this; if phi is the empty set it is contained in any set so phi is contained in phi dash and if phi dash is an empty set it will be contained in any set so phi dash will be contained in phi. So from these you can conclude phi is equal to phi dash, there can be only one empty set. (Refer Slide Time: 16:33) Operations on sets: We will have some binary operations on sets let us consider what they are. Let A and B be sets, the union of A and B denoted by A union B is a set, A union B is equal to x x belongs to A OR x belongs to B. It consists of all elements contained in A and B both. The intersection of A and B denoted by A intersection B is the set A intersection B x belongs to A AND x belongs to A. So A intersection B consists of all those elements which are elements of both A and B. The difference of A and B or relative compliment of B with respect to A and is denoted as A minus B and that is the set A minus B is equal to x x belongs to A AND x does not belong to B.
10 (Refer Slide Time: 16:36) That is, this consists of all elements which are in A but not in B. Let us consider some example; Suppose A is equal to a, b, c, d and B is equal to c, d, e, f what is A union B? It will consist of elements which are either in A or in B that is a, b, c, d, e, f. A intersection B will consist of all those elements which are both in A and B that is c and d. And A minus B will consist of elements which are in A but not in B and in this case it will be a b in this example. (Refer Slide Time: 18:55)
11 Now if A and B are two sets and A intersection B is empty then A and B are said to be disjoint sets. If C is a collection of sets such that any two distinct elements of C are disjoint then you say C is a collection of pair wise disjoint sets. (Refer Slide Time: 19:04) Now, let us see what is meant by commutativity and associativity. This is a binary operation, let us denote this, this denotes a binary operation. And x operator y denotes the resultant obtained by applying the operator to x and y. This denotes the resultant obtained by applying the operation to the operands x and y. Then when do you say that operation is commutative, if x operator y is equal to y operator x and it is said to be associative if you perform the operation on x and y first and z that is equivalent to performing the operation on y and z first and then performing the operation with x and the resultant. So these are the definitions of commutative and associative operations.
12 (Refer Slide Time: 19:30) Let us see that the union and intersection operations are commutative and associative on sets. The set operations union and intersection are both commutative and associative. For arbitrary sets A, B and C you have A union B is equal to B union A. That is both of them will contain only elements which belong to both A and B. In a set the order is not important so I can denote a set by a, b, c, d this is the same as denoting b, d, c, a they denote the same sets. The order does not matter here. So A union B is equal to B union A, A intersection B is equal to B intersection A again it consists of all elements which are common to both A and B. And again it is also associative, union is associative. The left hand side will consist of all elements which belong to either A or B or C and that is the same with the right hand side. Similarly, intersection is also associative because it will consist of all elements which belong to A, B and C.
13 (Refer Slide Time: 20:40) Now formal proof can be written just using OR, AND and logical connectives. Now you have two binary operation; dell and square I will say then how do you define distributivity of one operator with respect to the other? Then dell distributes over square if the following hold; x dell y operator z is equal to x dell y operator x dell z. And similarly, y operator z then dell x is defined as y dell x operator z dell x. (Refer Slide Time: 22:17) For example, if you take multiplication and addition you can see that x into y plus z if you take the set of integers as the underlying set x dot y plus z will be x into y plus x into z. And similarly, x plus y dot z will be x dot z plus y dot z. So here dot distributes over
14 plus and multiplication distributes over addition. So in set also you can have union and intersection and let us see how they distribute. The set operations of union and intersection distribute over each other. For arbitrary sets A and B and C A union B intersection C is equal to A union B intersection A union C. You can use Venn diagram and convince yourself about this. You have the universal set like this, you have sets A, you have set B and you have set C. What is the left hand side B intersection C A union B intersection C what is B intersection C? This is B intersection C this portion is denoted by B intersection C and A union B intersection will be this whole thing A and B intersection C. (Refer Slide Time: 25:04) Now, again let me draw the same thing A, B and C. What is the right hand side? The right hand side is A union B A union B is this whole thing and A union C is this and this together. Now the intersection of that will be this.
15 (Refer Slid Time: 25:45) So you can easily see that A union B intersection C is equal to A union B intersection A union C. And similarly, you can also prove that A intersection B union C is equal to A intersection B union A intersection C it is a similar way you can prove. Let A, B, C, D be arbitrary subsets of a universe U then you have the following assertions. What is A union A? It is just A itself this is idiom portent laws they are called idiom portent laws. Similarly, A intersection A is also again A it consists of all elements in A only. And if you add the empty set to A A union phi in essence you are not adding anything so that is A. (Refer Slide Time: 27:15)
16 And A intersection phi means it will consist of all elements which are common to A and common to the empty set but empty set does not contain any elements so this is equal to empty set. And A minus B will consists of all elements which are in A but not in B so that will be obviously contained in A. Again you have, if A is contained in B and C is contained in D then you will have A union C contained in B union D. These things you can prove using Venn diagram. For example, consider A, A is contained in B so this is A, this is B, A is contained in B and C is contained in D B union D is the whole set and A union C is this portion. (Refer Slide Time: 28:10) We can see that A union C is contained in B union D. These are not very difficult to visualize. Similarly, if you have A is contained in B and C is contained in D then A intersection C is contained in B intersection D. You also have the following theorems A will be contained in A union B and A intersection B will be contained in A. If A is contained in B then A union B will be B because if A is contained in B B is a bigger set than A and all elements of A are already in B. So A union B is nothing but B itself. Similarly, if A is contained in B all elements of A or in B so if you take the intersection that means the elements common to both A and B that is nothing but A itself. And A minus phi denotes the set of all elements belonging to A but not belonging to the empty set. And you know that empty set does not have any elements so A minus phi is just A. And A intersection B minus A will be empty because B minus A consists of all elements which are in B but not in A and A has some elements so there is no common element between these two so their intersection will be empty. Similarly, you have A union B and B minus A is A union B because this denotes all elements of B which are not in A and this denotes the elements which are in A so the union will be A union B.
17 (Refer Slide Time: 28:23) Similarly you can also prove these two very easily, A minus B union C will be A minus B intersection A minus C and A minus B intersection C will be A minus B union A minus C. You can prove these things using Venn diagram. Let U be a universe and A be a subset of U. The complement of A denoted by A bar is the set A bar is equal to U minus A. That is the set of all elements x where x does not belong to A. (Refer Slide Time: 30:30) So, if the underlying universal set is given like this and A is this set the complement A bar will be the set of all elements which are in U but not in A this denotes the
18 complement of A. A bar is the set of elements in U and x does not belong to A so the complement will be denoted like this. Now, we can very easily see that let A be an arbitrary subset of some universe U, then what can you say about A union A bar? This is A this is A bar so A union A bar will be the universal set. And there are no common elements between A and A bar so A intersection A bar will be the empty set. Now we have to prove that the complement of a set is unique. You are fixing the universal set and you want to show that the complement is unique, that is denoted by this theorem. Let A and B be subsets of a universe U then B is A s compliment A bar if and only if A union B is equal to U and A intersection B is phi. Let us see how we can prove that. The theorem states if and only if so you have to prove in both directions. So suppose B is equal to A bar then by the previous theorem the previous theorem states this; A union A bar is U and A intersection A bar is phi. Now we know that B is A bar so A union B is equal to U and A intersection B where A bar is B is empty. The other way round we have to prove that if these are true that would imply B is equal to A bar. (Refer Slide Time: 33:43) Let us see how we can prove that. Now B you can write as B intersection U and what is U? U is A union A bar. Now, using distributive law this is B intersection A union B intersection A bar. And what is B intersection A? That is empty. We know that A intersection B is empty so this is empty set union B intersection A bar. And this you can write as A bar intersection A union B intersection A bar or if I use commutativity you can write it as A bar intersection B. And using the distributive law on the other way round this is A bar intersection A union B. But what is A union B? A union B is the universal set so this is A bar intersection U is equal to A bar. So starting from B we have come to the conclusion that B is equal to A bar.
19 (Refer Slide Time: 35:55) Let us see the steps once more. We know that A union B is the universal set and A intersection B is the empty set. Then we want to show that B is equal to A bar we proceed like this B is equal to B intersection U because any set if you intersect it with universal set that will be the set itself. And I can write the universal set as a A union A bar. Now, using distributive laws this will become equal to B intersection A union B intersection A bar. But we know that A intersection B is empty that is B intersection A is also empty, this is empty union this. But the empty set again I can again write it as A bar intersection A and this portion you can use commutativity and write it as A bar intersection B. And applying the distributive law in the reverse you can write it as A bar intersection A union B. And A union B we know is the universal set so A bar intersection universal set is nothing but A bar. So B is equal to A bar or the complement of the set is unique. Now you can see that if A bar is an arbitrary set then A bar bar is A. That is if you take the complement of set A and again take the complement that will be equal to the original set. The complement of the complement of A is A. And similar to what we studied as De Morgan s laws in logic here also you have De Morgan s laws.
20 (Refer Slide Time: 37:15) A and B are arbitrary subsets then A union B complement is equal to A bar intersection B bar and A intersection B complement is equal to A bar union B bar. This you can see with Venn diagram very easily. You have two sets underlying universal set is this and you have two sets A B, A union B is this, the complement of this is this portion. A union B complement is the dotted portion and you can see that if you take the complement of A this will be the portion. The whole thing will be the complement and the complement of B which will be this portion. So if you take the complement of A that is the dotted portion and the complement of B which is the slash portion the intersection will give you this portion because this will not be in the intersection this portion also will not be in the intersection so A bar intersection B bar is equal to A union B bar, these two are equal.
21 (Refer Slide Time: 39:18) And similarly you can prove the other way round also. That is you have two sets A B what is A intersection B? This is it so A intersection B complement is the whole set except this portion. And A complement is the whole portion except A and B complement is the whole portion except B and you can very easily see that this is A bar union B bar. So when you take the complement the intersection becomes union and the union becomes intersection. (Refer Slide Time: 40:05)
22 These are known as De Morgan s laws and you can see the similarity between saying that complement of p AND q is p OR q and so on. NOT OR rather NOT of p AND q is NOT p OR NOT q see the similarity between this and set uni[que] (Refer Slide Time: 40:31) Let C be a collection of subsets in some universe U. The union of the members of C denoted by the union of S belongs to C S is the set union and it is denoted by this, the set of all elements such that S belongs to C and x belongs to S. This is extending the definition of union not to two sets but a collection of sets. Similarly, this is extending the definition of intersection not to two sets alone but extending to a finite collection of sets. You denote the intersection as x such that x belongs to all of them.
23 (Refer Slide Time: 40:33) We have seen that you can define a set explicitly or implicitly. Explicitly means you just enumerate all the elements of the set. Implicitly means you define the set using a predicate. Apart from these two types of definitions there is one more type of definition of sets which is known as inductive definition. (Refer Slide Time: 41:48) Let us see how we can define a set inductively. Now the inductive definition of a set always consists of three components. The first components are the basis or the basis clause where you define the basic building blocks. The basis or basis clause of the definition establishes that certain objects are in the set. This part of the definition has a
24 dual function of establishing that the set being defined is not empty and characterizing the building blocks which will be used to construct the remainder of the set. (Refer Slide Time: 41:53) I will give an example later but first let us go through this. Apart from the basis step there will be an induction step or an induction clause. And here this inductive definition has the second part as induction; it denotes the ways in which elements of the set can be a combined to obtain no elements. The inductive clause always asserts that if objects x, y, z etc are elements of the set then they can be combined in certain specified ways to create other objects which are also in the set. Thus while the basis clause describes the building blocks of the set the inductive clause describes the operations which can be performed on objects in order to construct new elements of the set.
25 (Refer Slide Time: 42:38) Let us take one or two examples and see. For example take the set of nonnegative integers N, N is 0, 1, 2, 3, etc. You can try to define this set inductively like this; basis clause is 0 belongs to N. Then the induction clause is, this is the basic building block, if x belongs to N then x plus 1 belongs to N. (Refer Slide Time: 44:15) So starting with 0, 0 plus 1 is 1 will belong to the set, from 1 plus 1 2 will belong to the set and you will be able to get all elements of the set. This is just to illustrate how it works. Actually there is a slight flaw here in the sense that you are trying to define natural numbers using addition but actually addition is defined after you define the
26 natural numbers. This is just to show you a simple example. Let me a take up a different example. Consider the set of S, S is the set of well formed strings of parenthesis. (Refer Slide Time: 45:20) Let me take one particular parenthesis the square brackets alone, so this is an element which will belong to S. Again another element this also is a well formed string this will belong to S. Another element is this and this will belong to S. But this does not belong to S because they are not matching right parenthesis occurring before the left parenthesis. Or you cannot have something which does not belong to S because you have one pair of parenthesis and this does not have a matching left parenthesis. And similarly if you have, this does not belong to S.
27 (Refer Slide Time: 46:16) How can we define this set, a string of well formed parenthesis inductively? So as a basis clause you will have, the basis steps you will have this belongs to S. As mentioned earlier the basis clause gives you the basic building block and also it tells you that the set is non empty. Let us go back to the definition. The basis or basis clause of the definition establishes that certain objects are in the set. This part of the definition has the dual function of establishing that the set being defined is not empty and of characterizing the building blocks which will be used to construct the remainder of the set. Now, what are the induction clauses or what is the induction clause here? Starting with this how will you find more and more elements of the set? Now in this example it will be like this; if x belongs to S x is a well formed string of parentheses then this belongs to S. So if you have a matching well formed string x adding one more left parenthesis and one more right parenthesis still gives you the well formed string of parenthesis. This is one and another one is if x, y belongs to S then x concatenated with y or x by the side of y belongs to S. That is if you have a well formed string of parenthesis here then placing by its side another well formed string y gives you a well formed string of parenthesis. For example if I have something like this, this is x this is x and y is say this is x y is this, then x, y will give you this, this is a well formed parenthesis each left element has a matching right parenthesis. So you can try to define many of these sets in this manner. This is called inductive definition of a set. Now first we consider that there will be three parts to the definition. The first part is basis part which we have seen. The second part is the induction clause this tells you how you are going to perform some operations starting from some elements of the set and build more and more elements of the set.
28 I will read this again; the induction or inductive clause of an inductive definition establishes the ways in which elements of the set can be combined to obtain new elements. The inductive clause always asserts that if objects x, y, z are elements of the set then they can be combined in certain specified ways to create other objects which are also in the set. So basis clause describes the basic building blocks and inductive clause tells you how to build more and more elements. Apart from that there is one more clause which is called the extremal clause. The third part of the definition is known as the extremal clause. The extremal clause asserts that unless an object can be shown to be a member of the set by applying the basis and the inductive clauses a finite number of times then the object is not a member of the set. That is you must be able to get all the elements of the set from the basis clause and build more and more elements making use of the operation defined in the induction clause. All elements should be got that way and no other element will be got. The extremal clause of an inductive definition of a set has a variety of forms. You can say it any one of the following ways. One way of saying that is, no object is a member of S unless it is being so follows from a finite number of applications of the basis and inductive clauses so you can call the extremal clause in this manner. Or you can say it in this manner also; the set S is the smallest set which satisfies the basis and inductive clauses. You are not going to get any more elements only those elements which are built from the basis clause and the induction clause. (Refer Slide Time: 51:20) Or you can even say it in one of these two ways, let us see how you can say that. The set S is the set such that S satisfies the basis and inductive clauses and no proper subset of S satisfies them. Or put it in another way you can say it like this; if T is a subset of S such
29 that T satisfies the basis and the induction clauses then T is equal to S. Or you can say it in this way; the set S is the intersection of all sets which satisfy the properties specified by the basis and the inductive clauses. (Refer Slide Time: 52:07) Now we have seen that apart from the implicit definition and the explicit definition you can also define sets in a inductive manner. And in this definition the definition will consist of three parts; the basis clause, the induction clause and the extremal clause. This sort of a definition of a set by induction is used in proofs like proof by induction. We shall see what is proof by induction in the next lecture. Now the third part of the definition is called an extremal clause. It tells you that the elements of the set are built from the basis and the induction clause and all elements are built that way no other elements will be built and so on. So we saw four different ways of mentioning the extremal clause. The part you must realize is whatever set you define inductively either it is set of well formed strings of parenthesis, the set of integers, set of even integers or whatever it is the extremal clause is the same. The form of the extremal clause or the statement of the extremal clause is the same for all definitions which use inductive definition but it must be stated explicitly. The third point apart from specifying the basis clause and the induction clause you must specifically mention the extremal clause. Without that the definition is not complete even though the extremal clause is the same for all definitions. So we have seen this definition let us see how to make use of this in proofs by induction in the next lecture.
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Cryptography and Network Security Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Module No. # 01 Lecture No. # 05 Classic Cryptosystems (Refer Slide Time: 00:42) | 13,472 | 57,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-26 | longest | en | 0.943272 |
https://www.thetaranights.com/python-operators/ | 1,590,824,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00350.warc.gz | 924,225,376 | 22,970 | # Python Operators
- -
Operators are the constructs that enable performing operations on operands(values and variables). The operators in python are represented by special symbols and keywords. The intentions of this blog is to familiarize with the various operators in Python.
##### Arithmetic Operators
These operators are used to perform mathematical operations ranging from addition, subtraction, multiplication, division to modulus, exponent, etc. Following table shows the arithmetic operators and it’s usage:
Operator Usage Description
`+` a + b Add values on either side of the operator or unary plus
`-` a -b Subtract right hand operand from the left hand operand. Also unary negation
`*` a * b Multiply values on either side of the operator
`/` a / b Divides left hand operand by right hand operand
`%` a % b Returns the remainder from dividing left hand operand by right hand operand
`**` a ** b Returns Exponent – left operand raised to the power of right
`//` a //b Floor Division – The division of operands where the result is the quotient in which the digits after the decimal point are removed. But if one of the operands is negative, the result is floored, i.e., rounded away from zero (towards negative infinity) −
##### Comparison Operators
The comparison operators are used to identify relation between operands on either side of the operator. These are also called relational operators. The values from the comparison operators is either True or False.
Operator Description Usage
`==` Returns True if the values on the either side of the operator is equal otherwise False. ``` a == b ```
`!=` Returns True if the values on either sides of the operator is not equal to each other otherwise False. ```a != b ```
`>` Returns True if the value of the operand on the left of the operator is greater than the value on the right side of the operator. `a >b`
`<` Returns True if the value of the operand on the left of the operator is less than the value on the right side of the operator. ```a < b ```
` >= ` Returns True if the value of the operand on the left of the operator is greater than or equal to the value on the right side of the operator. ``` a >= b ```
` <= ` Returns True if the value of the operand on the left of the operator is less than or equal to the value on the right side of the operator. ``` a <= b ```
##### Assignment Operators
Assignment operators are used for assigning the value from the right operand of the operator to the left operand. Following is the various assignment operators in Python:
Operator Description Usage Equivalent to
`=` Assigns values from right side operands to left side operand c = a + b c = a + b
`+=` Adds the value of right operand to the value of left operand and assign the result to the left operand b += a b = b + a
`-=` Subtracts the value of right operand from the value of left operand and assign the result to left operand b -= a b = b – a
`*=` Multiplies the value of right operand with the value of the left operand and assigns the result to left operand b *= a b = b * a
`/=` Divides the value of the left operand with the value of the right operand and assigns the result to the left operand b /= a b = b / a
`%=` Assigns the remainder from dividing left hand operand by right hand operand to the left hand operand b %= a b = b % a
`**=` Assigns the value from the exponential operation to the left operand. b **= a b = b ** a
`//=` Performs floor division on operators and assign value to the left operand b //= a is equivalent to b = b // a
##### Python Logical Operators
Logical operators in python are used for conditional statements which evaluates to either true or false. AND, OR, NOT are the logical operators in python.
Operator Description Usage `and` True if both sides of the operator is True x and y `or` True if either of the operand is True x or y `not` Complements the operand not val
##### Membership Operator
These operators test for membership(presence) in a sequence such as string, list or tuple. Following are the membership operators:
Operator Description Usage
`in` True if the value/operand in the left of the operator is present in the sequence in the right of the operator. `x in y`
`not in` True if the value/operand in the left of the operator is not present in the sequence in the right of the operator. `x not in y`
##### Identity Operator
It is used to compare the memory location of two python objects .i.e both the operands refer to the same object.
Operator Description Usage
`is` True if both the operands refer to the same object. x is True
`is not` Evaluates to false if the variables on either side of the operator point to the same object and true otherwise. x is not True
##### Bitwise Operators
Bitwise operators work on bits of an operand hence the name.
```>>> # Bitwise AND
...
>>> a = 3
>>> b = 4
>>> a = 3 # equivalent binary is 0011
>>> b = 4 # equivalent binary is 0100
>>> a & b
0
>>>
>>> # Bitwise OR
...
>>> a | b
7
>>> # 7 is equivalent to 0111 in binary
...
>>>
>>> # Bitwise NOT
...
>>> ~ a
-4
>>>
>>> # Bitwise XOR
...
>>> a ^ b
7
>>>
>>> # Bitwise right shift
...
>>> a >> 2
0
>>>
>>> # Bitwise left shift
...
>>> a << 2
12```
Bhishan Bhandari [22] Brewing contents directly from the Himalayas of Nepal. I am a hobbyist programmer and enjoy writing scripts for automation. If you'd like a process to be automated through programming, I also sell my services at Fiverr . Lately, I like to refresh my Quora feeds. Shoot me messages at bbhishan@gmail.com | 1,285 | 5,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-24 | longest | en | 0.829233 |
https://www.embedded.com/whats-your-sine-finding-the-right-algorithm-for-digital-frequency-synthesis-on-a-dsp/ | 1,656,947,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00423.warc.gz | 793,691,414 | 47,448 | What's your sine? Finding the right algorithm for digital frequency synthesis on a DSP - Embedded.com
# What’s your sine? Finding the right algorithm for digital frequency synthesis on a DSP
Perhaps the most fundamental building block of all communication equipment is the sine wave generator. We have ample knowledge of how to generate sine waves using analog electronic hardware. In modern electronics, however, more and more functions once performed by analog circuitry are now performed by digital signal processors or field-programmable gate arrays. Those two types of ICs are the main approaches for developing and implementing signal processing: either the programmable DSP or the ASIC/FPGA.
Lately, high-performance DSPs are using concepts, such as execution pipelining and parallelism of operations, which were unknown to their ancestor the CPU but routine to the FPGA world. DSPs are starting to resemble FPGAs more and more, and although some fundamental differences remain, this hybridization makes it possible to use algorithms originally developed for one environment in the other.
For instance, you can use a very efficient algorithm for computing the sine for an arbitrary phase in a programmable DSP. The method is not new in itself but adapts the original ideas applied to the design of direct digitalfrequency synthesizer (DDFS) ICs over two decades ago.
It's important to highlight at this point a crucial difference between algorithms used to compute the sine of a random input phase and those that generate a sine wave. While a sine computation algorithm can be easily converted into a sine wave generator, by providing to its input a monotonically increasing phase sequence, it's absolutely impractical, if not impossible, to use a sine wave generator to calculate the sine of a random phase input.
We'll differentiate these algorithms as sine computation and sine generation with the understanding that a sine computation is a more general method and hence can be easily converted into the latter without significant additional complexity.
In this article, we'll review the most common methods for the computing and generating a sine wave on a programmable DSP, as well as provide a brief introduction to direct digital frequency synthesis (DDFS). We'll then describe a new approach to DDFS based on an algorithm created by combining lookup tables and trigonometric identities. The approach lends itself well to implementation on DSPs such as the Texas Instruments' C6x.
Sine generation in a programmable DSP
Implementing a high-speed sine-wave synthesizer in a programmable DSP is hardly a trivial task; it often requires a careful trade-off between memory and implementation complexity. The problem arises because the standard algorithms used to calculate Sin(x ), as well as any other transcendental function of x , are based on the Taylor series expansion of the function in question, which is a time-consuming procedure. Hence, for a given finite number of CPU cycles, only a limited number of results can be rendered in real time. Several alternative algorithms have been developed to boost the performance of the sine-wave generation. Table 1 presents some of the most popular sine generation and computation algorithms used in a DSP.
View the full-size image
Taylor series:
The Taylor series expansion of Sin(x ) is shown in Equation 1. Implementing this equation in a high-level language like C is straightforward. One advantage of the method is that the desired precision of the final result can be controlled by choosing the number of terms to include in the calculation, more terms render more precision at the expense of additional computations. The series converges rather fast in the interval and remarkably slow from [–π, π] hence, given the same desired precision, there are two implementation choices:
(1) exploiting the symmetry relations in Equation 2 and performing the expansion only in the interval with fewer terms,
or
(2) using more terms (see Table 2 ) applying the expansion in interval [–π, π]. Since both choices require extra computations, relative advantages should be judged on a case by case basis.
View the full-size image
(1)
(2)
Table 2 shows maximum error as a function of the number of terms used in the expansion for each of the intervals and the maximum number of bits that can be used in order to keep the error less than one least significant bit.
Lookup table:
The lookup table approach symbolizes the opposite to the Taylor Expansion, in the sense that no calculations are performed in real time. The values of the sine function are precalculated at evenly spaced phase points in the intervals [0,2π] and stored in memory during the initialization phase of the algorithm. In this way, a one-to-one mapping is established between each phase value and the address in memory where the table values are stored. The computation of the sine wave in real time entails approximating the input phase to one of the table indexes and retrieving the value stored at that index/address. The phase to index approximation constitutes one of the main sources of error of the algorithm. This error is in the order of the inverse of the table size, which is almost universally chosen to be a power of two.
Thisalgorithm is the fastest sine computation algorithm executing in only one instruction–the memory-read from the index corresponding to the input phase. Its drawback is that the table size becomes impractically large even for modest error goals; for instance an error of the order of 2–12 requires a table of 4,096 elements totaling 16 kbytes for 32-bit elements. Obviously this size can be reduced by a factor of four by exploiting the symmetry relations for Sin and Cos stated in Equation 2 but at the expense of the additional complexity of the algorithm needed to check the phase subinterval prior to lookup table addressing. The amount of overhead associated with this additional complexity more than doubles the execution time of the algorithm, making it a costly proposition. Another way to decrease the table size is to interpolate between the end points of the subinterval where the phase resides; this technique, however, again suffers the aforementioned performance drawbacks.
Recursive iteration:
The recursive iteration algorithm is based in the application of the known trigonometric identities Equation 3 in a recursive manner that is achieved by making α[k +1]=α[k ]+Δα and rewriting Equation 3 as in Equation 4. The algorithm is suited only for generation and not computation of the sine function. Another drawback is that, due to the recursive nature of the method, numerical errors accumulate with time producing fluctuations in the output value of the sine wave. A solution to this problem is proposed in John Edward's article.1
(3)
(4)
Resonator:
The resonator implementation takes advantage of the fact that an infinite impulse response (IIR) filter with a pair of complex poles situated in the unit circle generates a constant-amplitude sine wave if excited with an impulse function. The frequency of the oscillation depends of the location of the poles in Z plane. The transfer function and time domain recursive equation for the IIR filter are shown in Equation 5 along with the IIR block diagram in Figure 1 . The IIR has a very efficient implementation in terms of speed and memory usage. Its main drawback is being a sine generation instead of a sine computation algorithm. It also exhibits the same sensibility to amplitude due to the accumulation of numerical errors with time than the recursive iteration.
View the full-size image
(5)
And initial conditions: y (-1)=0 y (–2)=0 x (0)=1
Sine computation in direct digital frequency synthesizers
Direct digital frequency synthesizer (DDFS) ICs were developed over 20 years ago to digitally generate a sine and/or cosine wave with high frequency resolution and low distortions as a response to the increase demands in the communication field. A popular DDFS architecture originally designed by Tierney consists of a phase accumulator and a phase to amplitude converter as shown in Figure 2 .2 At every clock cycle, the phase accumulator is incremented by a programmable quantity F r L-1 bits long. The total phase is then used as input to the second block where the sine and/or cosine are calculated. The frequency of the synthesized sine wave F 0 is calculated according to Equation 6.
(6)
View the full-size image
If the analog representation of the sine is desired the last block is followed by a digital-to-analog converter.
The phase-to-amplitude converter block consists of an algorithm that computes an approximate value to the sine of the input phase. The number of bits W used as input to the phase to amplitude converter is usually less than L . The width reduction (truncation of the least significant bits) of the phase-accumulator output is necessary in order to simplify the phase to amplitude converter. This truncation, along with imprecision in the phase-to-amplitude converter, constitute the two biggest source of errors of the DDFS.
The effect of the truncation in the spectral purity of the generated sine wave was shown in Nicholas et al.,3, 4 to be described by Equation 7, which represents the value of the maximum spectral spurious component of the DDFS normalized such that 2W represents a sine of amplitude 1. According to Equation 7, the maximum spurious component in the generated spectrum is related to the truncation length B in a quasi exponential form and could grow up to 3.92 dB higher than 2–W if the greatest common divisor of <Fr, 2B > = 2B –1 , situation that can be avoided by selecting Fr and 2B relatively prime to each other. Since the main use of the DDFS is the generation of spurious-free sine waves, this performance parameter is of paramount importance. Coincidentally, a closely related term has been coined by the A/D manufacturers: spurious-free dynamic range (SFDR) . This term is calculated according to Equation 8 and will be used in the rest of the paper.
(7)
(8)
The holy grail of DDFS
The phase-to-amplitude converter is the most complex function of the DDFS and has been the target of extensive optimization efforts. In reviewing the methods used in the design of phase-to-amplitude converter of the DDFS, we can expected to find possible candidates for implementation in a DSP. The goal is to find an implementation that could be translated to a programmable DSP, specifically a high-performance TI C6x DSP, and compare its performance with the algorithms previously discussed.
As a reference for comparison, let's chose the lookup-table algorithm because this was the fastest technique analyzed so far. We can assume that while executing in a single instruction, the algorithm simultaneously computes both the sine and cosine of the input phase, since this function is of widespread use in communications. Both the input and outputs are quantized in 16-bit fixed point format.
Say the size for the uncompressed table is 4,096 32-bit elements (16 kbytes total), with each 32-bit table element containing two contiguous 16-bit groups, the sine and cosine for each of the 4,096 angles from 0 to 4,095/4,096 2π. With the phase quantized in 12 bits the SFDR is, according to Equation 8, approximately 70 dB. To compare performance between algorithms, let's consider only the number of instructions in the loop kernel, effectively ignoring the effect of the loop's prologue and epilogue. (See Texas Instruments' documentation for an indepth discussion of C6x assembly and optimization.5, 6) The loops are fully pipelined and noninterruptible.
Several implementations of phase-to-amplitude converters use techniques similar to those previously discussed for DSP. Langlois,7 for instance, proposed a linear approximation with interpolation in the interval from . Palomaki and Niitylahti8 have used different methods based on recursive iteration, Taylor and Chebyshev series expansions, the last performing at around 70 dB of SFDR for a series with only five terms and CORDIC algorithms, but none of these are suited for DSP implementation due to their iterative nature.
In a seminal paper published in 1984, Sunderland proposed the novel idea of combining lookup tables and trigonometric identities as a means of reducing the table size.9 The method works as follows: instead of using a single lookup table, the phase is split into several components, and trigonometric identities are used to assemble the desired result. The most straightforward of such identities are precisely Equation 3. Once the input phase has been quantized as an integer between 0 and 4,095, this value is split into two groups of bits of U upper bit and L lower bits, each representing the angles and respectively. This corresponds to the graph shown in Figure 3 .
The value of sin(α), cos(α), sin(β), and cos(β) are then retrieved from four tables, two tables of size 2U and two tables of size 2L elements. These values are used to compute the final sine and cosine using Equation 3.
Sunderland didn't apply the method exactly as described, but instead to avoid multiplications altogether (a very expensive proposition back in 1984), settled for splitting the phase in three groups and using Equation 9 as an approximation of Equation 3.
(9)
At first glance, the original idea suggested by Sunderland's technique seems to lend itself well for implementation in a programmable DSP. It succinctly proposes a reasonable trade-off: a sizeable decrease in the lookup table size at the expense of a modest increase in the number of calculations. The following reasons clarify why Sunderland's original works well with the specific TI C6x DSP family architecture.
1. Sunderland's original requires, according to Equation 3, two multiplication and two additions per every output value: a C6X DSP has a total of eight execution units, among them two adders and two multipliers, hence it's not a stretch to expect that all operations could be performed in a single clock cycle.
2. The lookup table has a small footprint: in order to achieve maximum throughput, the lookup table must be stored in (precious) internal DSP memory, hence small size is highly desired.
3. Conditional instructions are absent: per every input-phase value, the same operations are executed in a serial fashion. In other words, memory lookup is followed by the operations corresponding to Equation 3, which is very well suited for C6x architecture since any conditional operation inside a pipelined loop will force it to be at least six instructions long (see TI's TMS320C6000 Programmer's Guide 5).
At this point, we should analyze the implementation of this algorithm for 16 signed short integer format in both input and outputs. The C code for the algorithm is shown in Listing 1 with the corresponding generated assembly code. Note that the loop executes in three instructions compared with one for the lookup table. Furthermore, if the loop is unrolled by a factor of two, it will execute in five instructions generating two pairs of sine/cosine outputs per iteration. This level of performance compares very well with the lookup table implementation; although still 2.5 times slower, it uses 1/32 of the memory. Note as well that while the execution speed remains constant with the table sizes, the saving in memory use continues to growth at a rate of . This algorithm can be easily modified to operate as a sine/cosine generator. As a generator, there is no need to input an array with phase values; just input the constant phase increment Fr and let a persistent memory variable maintain the phase between calls. A 32 bits Int should suffice even for the most demanding frequency-precision requirements. In this type of application, Equation 7 must be taken into account.
View the full-size image
One source of error in this implementation is the truncation of the least significant bit in the calculation of Equation 3. However, since, for the chosen table size, this error is relatively small compared with the error in the phase approximation, I ignored the error. If the application calls for an SFDR such that the uncompressed table size needs to be more than 14 bits, the truncation error must then be considered and it might be necessary to perform rounding of the results at the expense of lower throughput.
Note that the same kind of performance (three cycles/loop without unrolling and five cycles/loop with unroll of two), can be obtained for a floating-point implementation of the same algorithm, using a C67x DSP. Although floating-point DSPs are slower than fixed-point DSPs, in a floating-point unit all numerical errors are truly negligible.
One further modification to the algorithmwould be to divide the phase in three groups of bits U , M , and L corresponding to three different angles and performing the Equation 3 in a recursive fashion as follows:
View the full-size image
(10)
This implementation is capable of reducing the table footprint much more than what is possible with the Sunderland's original algorithm, but Equation 10 requires twice the number of operations and suffers from accumulative truncation errors in the fixed-point implementation that degrade its performance. For these reasons, it's only practical in floating-point implementations where design constraints (such as an SFDR) call for the use of huge table sizes.
The algorithmhas been successfully deployed in the field. Finally, some of the initialization functions used in the algorithm have been omitted due to space constraints. For those interested in the full version of the algorithm or in either the floating-point or the 32-bit fixed-point versions, please send an email to carlos.abascal@harris.com.
Carlos Abascal is signal processing and communications engineer in the Government Communications Division of Harris Corp. He has extensive experience in embedded microprocessor and DSP software development and has expertise in the theory and the implementation of linear and nonlinear adaptive predistortion for RF power amplifiers, digital modems, and software-defined radios in DSPs. He may be contacted at .
2. Tierney, C.M. Rader, and B. Gold, “A digital frequency synthesizer,” IEEE Transactions on Audio and Electroacoustics , vol. Au-19, No. 1. March 1971, pp. 48-57.
Back
3. Nicholas III, H.T., H. Samueli, and B. Kim. “The optimization of direct digital frequency synthesizer performance in the presence of finite word length effects,” Proceedings of the 42nd Annual Frequency Control Symposium , 1988, pp. 357-363.
Back
4. Nicholas III, H. T. and H. Samueli. “An analysis of the output spectrum of direct digital frequency synthesizers in the presence of phase-accumulator truncation,” in Proc. 41st Annual Control Symposium USERACOM (Ft. Monmouth, NJ), May 1987, pp. 495-502.
Back
5. Texas Instruments Inc. “TMS320C6000 Programmer's Guide,” SPRU198i, Texas Instruments Inc., March 2006.
Back
6. Texas Instruments Inc.. “TMS320C67x/C67x+ DSP CPU and Instruction Set Reference Guide,” SPRU733, Texas Instruments Inc., May 2005.
Back
7. Langlois, J.M.P. and D. Al-Khalili. “Hardware optimized direct digital frequency synthesizer architecture with 60 dBc spectral purity,” 2002 IEEE International Symposium on Circuits and Systems (ISCAS 2002), May 2002.
Back
8. Palomaki , K.I. and J. Niitylahti. “Direct digital frequency synthesizer architecture based on Chebyshev approximation,” Proceedings of the 34th Asilomar Conference on Signals, Systems and Computer s, Oct. 29th–Nov. 1st., 2000, pp. 1639-1643.
Back
9. Sunderland, D. A. et al, “CMOS/SOS frequency synthesizer LSI circuit for spread spectrum communications,” IEEE J. Solid State Circuits , vol. SC-19, No 4, Aug. 1984, pp 497-505.
Back
Kua , Francis. “Generation of a Sine Wave Using a TMS320C54x Digital Signal Processor,” TI Application Report SPRA819 –Texas Instruments Inc. May 2002
Essenwanger, K. A. and V. S. Reinhardt. “Sine Output DDSs A Survey of the State of the Art,” Proceedings of the 1998 Frequency Control Symposium (Pasadena) , May 27-29, 1998, pp. 370-378.
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http://alexplorer.net/teaching/phys-sci/physics-exam.html | 1,500,655,522,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00336.warc.gz | 10,065,859 | 4,593 | Physics Exam
Essential information:
Steps of the scientific method (in order):
·Observation of a problem
·Research
·Hypothesis
·Experiment
·Analyze data
·Conclusion
Base units Cannot be broken down, only measured
Derived units Must be calculated by a formula
Experimental design
·experimental (i.e., independent) variable The one variable is manipulated.
·dependent variable That which changes in response to the manipulation.
·control variables Things that are kept the same.
1st class lever:
2nd class lever:
3rd class lever:
Formulas
v = d/t
a = (vf vi) / t
p = mv
F = ma
KE = ½ m v2
PE = mgh
W = Fd
P = W/t
Input F Output d
Output F Input d
1st semester final
Formulas:
v = d/t a = (vf vi) / tp = mv F = ma
KE = ½ m v2PE = mghW = FdP = W/t
Input F Output d
Output F Input d
Measurement
1. Which of the following is an example of a base unit?
A) metersB) Newtons
C) meters/secondD) Joules
2. Which units are each of the following measured in typically?:
Acceleration:
A) JB) m/sC) m/s2
D) NewtonsE) Watts
3. Force:
A) kilogramsB) JC) m/s2
D) NewtonsE) Watts
4. Work:
A) kilogramsB) JC) m/s
D) NewtonsE) Watts
5. Power:
A) kilogramsB) JC) m/s2
D) NewtonsE) Watts
6. Kinetic Energy
A) kilogramsB) JC) m/s
D) m/s2E) Watts
7. What units must the mass be measured in to calculate force?
A) kilogramsB) NewtonsC) m/s
D) m/s2E) Joules
8. What units must the distance be measured in to calculate work, KE, or PE?
A) gramsB) kilogramsC) kilometers
D) metersE) m/s
Scientific method
9. Which of the following is example of a hypothesis?
A) I refuse to believe that nothing travels faster than light.
B) Sunlight is necessary for plants to grow.
C) Will this drug cure cancer?
D) I forgot my calculator.
10. The graph shows a trend for the amount of energy to increase every time the height of the ramp was increased.
A) ObservationB) Research
C) AnalysisD) Conclusion
11. The hypothesis was correct that the symptoms were caused by exposure to radiation.
A) ObservationB) Research
C) AnalysisD) Conclusion
12. A student looks up what types of frogs hibernate in the winter.
A) ObservationB) Research
C) AnalysisD) Conclusion
13. There is a disproportionate number of children with cancer living near the chemical plant.
A) ObservationB) Research
C) AnalysisD) Conclusion
14. In the KE/PE lab, what variable was the height of the ramp.
A) control variable
B) dependent variable
C) experimental (i.e., independent) variable
15. In the KE/PE lab, what variable was the kinetic energy of the ball.
A) control variable
B) dependent variable
C) experimental (i.e., independent) variable
16. What variable should you have only one of in an experiment?
A) control variable
B) dependent variable
C) experimental (i.e., independent) variable
Motion, Force, and Acceleration
17. What is the velocity of a bullet that travels 400 meters in 7 seconds?
A) 57 m/sB) 57 m/s2
C) 2800 m/sD) 2800 m/s2
18. If a car is traveling at 70 km/hr, how far will it travel in half an hour?
A) 35 kmB) 210 kmC) 350 km
19. Which has more momentum: A 5 kg soccer ball moving at 100 m/s or a 20 kg cannon ball moving at 30 m/s?
A) The soccer ball
B) The cannon ball
C) Both are the same
20. If a skier increases speed from 2.0 m/s to 4.5 m/s in 5 seconds, what is her acceleration?
A) 7.5 m/s2B) 0.5 m/s2
C) -7.5 m/s2D) -0.5 m/s2
21. Describe what is happening to the car whose movement is described in the following graphs:
A) The object is speeding up. Position
B) The object is slowing down.
C) The object is moving at a constant speed.
D) The object is not moving at all.Time
22. In the graph above, which is the independent variable?
A) accelerationB) position
C) timeD) velocity
23. In the graph above, which is the dependent variable?
A) accelerationB) position
C) timeD) velocity
Which of Newtons Laws of Motion is each of the following the best example of?
24. It takes more force to throw a shot put than to throw a tennis ball.
A) 1st lawB) 2nd lawC) 3rd law
25. Bat broke when it was swung and struck the baseball.
A) 1st lawB) 2nd lawC) 3rd law
26. The space probe Voyager II is continuing on a path out of the solar system.
A) 1st lawB) 2nd lawC) 3rd law
27. What is the net force necessary to accelerate a 10 kg bicycle forward at 2.0 m/s2?
A) 0.5 NB) 5 N
C) 10 ND) 20 N
28. A flower pot falls from a ledge at 9.8 m/s2 If it hits the ground with 5.0 N of force, how much mass was in the flower pot?
A) 0.51 kgB) 1.96 kg
C) 4.9 kgD) 49 kg
Work, Power, and Machines
29. How much work is done by a forklift raising a 400 N crate from the ground to a height of 14 meters?
A) 28.6 JB) 28.6 W
C) 5600 JD) 5600 W
30. How far would a ball have to roll if it had 40 N of force and was doing 170 J of work?
A) 4.25 mB) 4.25 m/s2
C) 6800 mD) 6800 m/s2
31. How much power is required to push a box with a force of 10 N over 20 m in 30 seconds?
A) 0.67 WattsB) 1 Watts
C) 6.7 WattsD) 15 Watts
32. How much work is done by a machine with a power output of 500 Watts if it has been running for a minute and a half?
A) 5.56B) 333
C) 750D) 45000
33. Using the definition of mechanical advantage, what is the output force when the input force is 4 N and the input distance is 12 meters while the output distance is 16 meters?
A) 3 NB) 5.3 N C) 48 N
What type of simple machine is each of the following?
34. baseball bat
A) inclined planeB) screwC) lever
D) wheel and axelE) pulley
35. spiral staircase
A) inclined planeB) screwC) lever
D) wheel and axelE) pulley
36. wrench
A) inclined planeB) screwC) lever
D) wheel and axelE) pulley
37. handicap ramp
A) inclined planeB) screwC) lever
D) wheel and axelE) pulley
Which class of lever it is each of the following?:
38. wheelbarrow
A) 1st class leverB) 2nd class leverC) 3rd class lever:
39. hammer pulling a nail
A) 1st class leverB) 2nd class leverC) 3rd class lever:
40. pliers
A) 1st class leverB) 2nd class leverC) 3rd class lever:
41. Which three simple machines belong in the same family?
A) inclined plane, screw, and lever
B) lever, wheel and axel, and pulley
C) inclined plane, pulley, and screw
D) lever, screw, and inclined plane
Energy
42. What is the gravitational PE of a 850 kg roller coaster car at the top of a 45 m high part of the track?
A) 18.9 JB) 3903 J
C) 38250 JD) 374850 J
43. How much mass does a rock with 75 J of potential energy have if it is 12 m above the ground?
A) 0.63 kgB) 6.25 kg
C) 900 kgD) 8829 kg
44. What is the kinetic energy of a 3500 kg meteor falling at 7000 m/s?
A) 4 JB) 24,500,000 J
C) 171,500,000,000 JD) 85,750,000,000 J
45. What is the efficiency of a system if it takes 900 J of work to move a 75 N crate that should theoretically only requires 750 J of work to move up a 10 m ramp?
A) 0.833%B) 8.3%
C) 12%D) 83.3%
Forms of energy: Tell which type of energy each of the following is an example of:
46. Windmills grinding flour
A) mechanicalB) chemical
C) nuclearD) electromagnetic
47. Uranium rods fueling a power plant
A) mechanicalB) chemical
C) nuclearD) electromagnetic
48. Lightning
A) mechanicalB) chemical
C) nuclearD) electromagnetic
49. Plants convert:
A) chemical energy to nuclear energy
B) electromagnetic energy to chemical energy
C) electromagnetic energy to nuclear energy
D) mechanical energy to chemical energy
50. A roller coaster moving down an incline is an of:
A) kinetic energy being converted to potential energy
B) potential energy being converted to kinetic energy | 2,293 | 7,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-30 | longest | en | 0.743987 |
https://daily-class-notes.b-cdn.net/NCERT/Maths%20Class%2011/14%20MATHEMATICAL%20REASONING.html | 1,656,475,923,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00609.warc.gz | 241,607,524 | 634,976 | vThere are few things which we know which are not capable of
mathematical reasoning and when these can not, it is a sign that our
knowledge of them is very small and confused and where a mathematical
reasoning can be had, it is as great a folly to make use of another,
as to grope for a thing in the dark when you have a candle stick
standing by you. – ARTHENBOT v
14.1 Introduction
In this Chapter, we shall discuss about some basic ideas of
Mathematical Reasoning. All of us know that human beings
evolved from the lower species over many millennia. The
main asset that made humans “superior” to other species
was the ability to reason. How well this ability can be used
depends on each person’s power of reasoning. How to
develop this power? Here, we shall discuss the process of
reasoning especially in the context of mathematics.
In mathematical language, there are two kinds of
reasoning – inductive and deductive. We have already
discussed the inductive reasoning in the context of
mathematical induction. In this Chapter, we shall discuss
some fundamentals of deductive reasoning.
14.2 Statements
The basic unit involved in mathematical reasoning is a mathematical statement.
In 2003, the president of India was a woman.
An elephant weighs more than a human being.
14Chapter
MATHEMATICAL REASONING
George Boole
(1815 - 1864)
2020-21
322 MATHEMATICS
When we read these sentences, we immediately decide that the first sentence is
false and the second is correct. There is no confusion regarding these. In mathematics
such sentences are called statements.
On the other hand, consider the sentence:
Women are more intelligent than men.
Some people may think it is true while others may disagree. Regarding this sentence
we cannot say whether it is always true or false . That means this sentence is ambiguous.
Such a sentence is not acceptable as a statement in mathematics.
A sentence is called a mathematically acceptable statement if it is either
true or false but not both. Whenever we mention a statement here, it is a
mathematically acceptable” statement.
While studying mathematics, we come across many such sentences. Some examples
are:
Two plus two equals four.
The sum of two positive numbers is positive.
All prime numbers are odd numbers.
Of these sentences, the first two are true and the third one is false. There is no
ambiguity regarding these sentences. Therefore, they are statements.
Can you think of an example of a sentence which is vague or ambiguous? Consider
the sentence:
The sum of x and y is greater than 0
Here, we are not in a position to determine whether it is true or false, unless we
know what
x and y are. For example, it is false where x = 1, y = –3 and true when
x = 1 and y = 0. Therefore, this sentence is not a statement. But the sentence:
For any natural numbers x and y, the sum of x and y is greater than 0
is a statement.
Now, consider the following sentences :
How beautiful!
Open the door.
Where are you going?
Are they statements? No, because the first one is an exclamation, the second
an order and the third a question. None of these is considered as a statement in
mathematical language. Sentences involving variable time such as “today”, “tomorrow”
or “yesterday” are not statements. This is because it is not known what time is referred
here. For example, the sentence
Tomorrow is Friday
2020-21
MATHEMATICAL REASONING 323
is not a statement. The sentence is correct (true) on a Thursday but not on other
days. The same argument holds for sentences with pronouns unless a particular
person is referred to and for variable places such as “here”, “there” etc., For
example, the sentenc
es
Kashmir is far from here.
are not statements.
Here is another sentence
There are 40 days in a month.
Would you call this a statement? Note that the period mentioned in the sentence
above is a “variable time” that is any of 12 months. But we know that the sentence is
always false (irrespective of the month) since the maximum number of days in a month
can never exceed 31. Therefore, this sentence is a statement. So, what makes a sentence
a statement is the fact that the sentence is either true or false but not both.
While dealing with statements, we usually denote them by small letters p, q, r
,...
For example, we denote the statement “Fire is always hot” by
p. This is also written
as
p: Fire is always hot.
Example 1
Check whether the following sentences are statements. Give reasons for
(i) 8 is less than 6. (ii) Every set is a finite set.
(iii) The sun is a star. (iv) Mathematics is fun.
(v) There is no rain without clouds. (vi) How far is Chennai from here?
Solution
(i) This sentence is false because 8 is greater than 6. Hence it is a statement.
(ii) This sentence is also false since there are sets which are not finite. Hence it is
a statement.
(iii) It is a scientifically established fact that sun is a star and, therefore, this sentence
is always true. Hence it is a statement.
(iv) This sentence is subjective in the sense that for those who like mathematics, it
may be fun but for others it may not be. This means that this sentence is not always
true. Hence it is not a statement.
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324 MATHEMATICS
(v) It is a scientifically established natural phenomenon that cloud is formed before it
rains. Therefore, this sentence is always true. Hence it is a statement.
(vi) This is a question which also contains the word “Here”. Hence it is not a statement.
The above examples show that whenever we say that a sentence is a statement
we should always say why it is so. This “why” of it is more important than the answer.
EXERCISE 14.1
1. Which of the following sentences are statements? Give reasons for your answer.
(i) There are 35 days in a month.
(ii) Mathematics is difficult.
(iii) The sum of 5 and 7 is greater than 10.
(iv) The square of a number is an even number.
(v) The sides of a quadrilateral have equal length.
(vii) The product of (–1) and 8 is 8.
(viii) The sum of all interior angles of a triangle is 180°.
(ix) Today is a windy day.
(x) All real numbers are complex numbers.
2. Give three examples of sentences which are not statements. Give reasons for the
14.3 New Statements from Old
We now look into method for producing new statements from those that we already
have. An English mathematician, “George Boole” discussed these methods in his book
“The laws of Thought” in 1854. Here, we shall discuss two techniques.
As a first step in our study of statements, we look at an important technique that
we may use in order to deepen our understanding of mathematical statements. This
technique is to ask not only what it means to say that a given statement is true but also
what it would mean to say that the given statement is not true.
14.3.1 Negation of a statement The denial of a statement is called the negation of
the statement.
Let us consider the statement:
p: New Delhi is a city
The negation of this statement is
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MATHEMATICAL REASONING 325
It is not the case that New Delhi is a city
This can also be written as
It is false that New Delhi is a city.
This can simply be expressed as
New Delhi is not a city.
Definition
1 If p is a statement, then the negation of p is also a statement and is
denoted by p, and read as ‘not p’.
A
Note While forming the negation of a statement, phrases like, “It is not the
case” or “It is false that” are also used.
Here is an example to illustrate how, by looking at the negation of a statement, we
may improve our understanding of it.
Let us consider the statement
p: Everyone in Germany speaks German.
The denial of this sentence tells us that not everyone in Germany speaks German.
This does not mean that no person in Germany speaks German. It says merely that at
least one person in Germany does not speak German.
We shall consider more examples.
Example 2 Write the negation of the following statements.
(i) Both the diagonals of a rectangle have the same length.
(ii)
7
is rational.
Solution (i) This statement says that in a rectangle, both the diagonals have the same
length. This means that if you take any rectangle, then both the diagonals have the
same length. The negation of this statement is
It is false that both the diagonals in a rectangle have the same length
This means the statement
There is atleast one rectangle whose both diagonals do not
have the same length.
(ii) The negation of the statement in (ii) may also be written as
It is not the case that
7
is rational.
This can also be rewritten as
7
is not rational.
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326 MATHEMATICS
Example 3 Write the negation of the following statements and check whether the
resulting statements are true,
(i) Australia is a continent.
(ii) There does not exist a quadrilateral which has all its sides equal.
(iii) Every natural number is greater than 0.
(iv) The sum of 3 and 4 is 9.
Solution
(i) The negation of the statement is
It is false that Australia is a continent.
This can also be rewritten as
Australia is not a continent.
We know that this statement is false.
(ii) The negation of the statement is
It is not the case that there does not exist a quadrilateral which has all its sides
equal.
This also means the following:
There exists a quadrilateral which has all its sides equal.
This statement is true because we know that square is a quadrilateral such that its four
sides are equal.
(iii) The negation of the statement is
It is false that every natural number is greater than 0.
This can be rewritten as
There exists a natural number which is not greater than 0.
This is a false statement.
(iv) The negation is
It is false that the sum of 3 and 4 is 9.
This can be written as
The sum of 3 and 4 is not equal to 9.
This statement is true.
14.3.2 Compound statements Many mathematical statements are obtained by
combining one or more statements using some connecting words like “and”, “or”, etc.
Consider the following statement
p: There is something wrong with the bulb or with the wiring.
This statement tells us that there is something wrong with the bulb or there is
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MATHEMATICAL REASONING 327
something wrong with the wiring. That means the given statement is actually made up
of two smaller statements:
q: There is something wrong with the bulb.
r: There is something wrong with the wiring.
connected by “or”
Now, suppose two statements are given as below:
p: 7 is an odd number.
q: 7 is a prime number.
These two statements can be combined with “and”
r: 7 is both odd and prime number.
This is a compound statement.
This leads us to the following definition:
Definition 2
A Compound Statement
is a statement which is made up of two or
more statements. In this case, each statement is called a component statement.
Let us consider some examples.
Example 4
Find the component statements of the following compound statements.
(i) The sky is blue and the grass is green.
(ii) It is raining and it is cold.
(iii) All rational numbers are real and all real numbers are complex.
(iv) 0 is a positive number or a negative number.
Solution
Let us consider one by one
(i) The component statements are
p: The sky is blue.
q: The grass is green.
The connecting word is ‘and’.
(ii) The component statements are
p: It is raining.
q: It is cold.
The connecting word is ‘and’.
(iii) The component statements are
p: All rational numbers are real.
q: All real numbers are complex.
The connecting word is ‘and’.
(iv)The component statements are
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328 MATHEMATICS
p: 0 is a positive number.
q: 0 is a negative number.
The connecting word is ‘or’.
Example 5 Find the component statements of the following and check whether they
are true or not.
(i) A square is a quadrilateral and its four sides equal.
(ii) All prime numbers are either even or odd.
(iii) A person who has taken Mathematics or Computer Science can go for
MCA.
(iv) Chandigarh is the capital of Haryana and UP.
(v)
2
is a rational number or an irrational number.
(vi) 24 is a multiple of 2, 4 and 8.
Solution (i) The component statements are
p: A square is a quadrilateral.
q: A square has all its sides equal.
We know that both these statements are true. Here the connecting word is ‘and’.
(ii) The component statements are
p: All prime numbers are odd numbers.
q: All prime numbers are even numbers.
Both these statements are false and the connecting word is ‘or’.
(iii) The component statements are
p: A person who has taken Mathematics can go for MCA.
q: A person who has taken computer science can go for MCA.
Both these statements are true. Here the connecting word is ‘or’.
(iv) The component statements are
p: Chandigarh is the capital of Haryana.
q: Chandigarh is the capital of UP.
The first statement is true but the second is false. Here the connecting word is ‘and’.
(v) The component statements are
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MATHEMATICAL REASONING 329
p:
2
is a rational number.
q:
2
is an irrational number.
The first statement is false and second is true. Here the connecting word is ‘or’.
(vi) The component statements are
p: 24 is a multiple of 2.
q: 24 is a multiple of 4.
r: 24 is a multiple of 8.
All the three statements are true. Here the connecting words are ‘and’.
Thus, we observe that compound statements are actually made-up of two or more
statements connected by the words like “and”, “or”, etc. These words have special
meaning in mathematics. We shall discuss this mattter in the following section.
EXERCISE 14.2
1. Write the negation of the following statements:
(i) Chennai is the capital of Tamil Nadu.
(ii)
2
is not a complex number
(iii) All triangles are not equilateral triangle.
(iv) The number 2 is greater than 7.
(v) Every natural number is an integer.
2. Are the following pairs of statements negations of each other:
(i) The number x is not a rational number.
The number x is not an irrational number.
(ii) The number x is a rational number.
The number x is an irrational number.
3. Find the component statements of the following compound statements and check
whether they are true or false.
(i) Number 3 is prime or it is odd.
(ii) All integers are positive or negative.
(iii) 100 is divisible by 3, 11 and 5.
14.4 Special Words/Phrases
Some of the connecting words which are found in compound statements like “And”,
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330 MATHEMATICS
“Or”, etc. are often used in Mathematical Statements. These are called connectives.
When we use these compound statements, it is necessary to understand the role of
these words. We discuss this below.
14.4.1 The word “And”
Let us look at a compound statement with “And”.
p: A point occupies a position and its location can be determined.
The statement can be broken into two component statements as
q: A point occupies a position.
r: Its location can be determined.
Here, we observe that both statements are true.
Let us look at another statement.
p: 42
is divisible by 5, 6 and 7.
This statement has following component statements
q: 42 is divisible by 5.
r: 42 is divisible by 6.
s: 42 is divisible by 7.
Here, we know that the first is false while the other two are true.
We have the following rules regarding the connective “And”
1. The compound statement with ‘And’ is true if all its component
statements are true.
2. The component statement with ‘And’ is false if any of its component
statements is false (this includes the case that some of its component
statements are false or all of its component statements are false).
Example 6
Write the component statements of the following compound statements
and check whether the compound statement is true or false.
(i) A line is straight and extends indefinitely in both directions.
(ii) 0 is less than every positive integer and every negative integer.
(iii) All living things have two legs and two eyes.
Solution (i) The component statements are
p: A line is straight.
q: A line extends indefinitely in both directions.
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MATHEMATICAL REASONING 331
Both these statements are true, therefore, the compound statement is true.
(ii) The component statements are
p: 0 is less than every positive integer.
q: 0 is less than every negative integer.
The second statement is false. Therefore, the compound statement is false.
(iii) The two component statements are
p: All living things have two legs.
q: All living things have two eyes.
Both these statements are false. Therefore, the compound statement is false.
Now, consider the following statement.
p: A mixture of alcohol and water can be separated by chemical methods.
This sentence cannot be considered as a compound statement with “And”. Here the
word “And” refers to two things – alcohol and water.
This leads us to an important note.
A
Note Do not think that a statement with “And” is always a compound statement
as shown in the above example. Therefore, the word “And” is not used as a connective.
14.4.2 The word “Or”
Let us look at the following statement.
p: Two lines in a plane either intersect at one point or they are parallel.
We know that this is a true statement. What does this mean? This means that if two
lines in a plane intersect, then they are not parallel. Alternatively, if the two lines are not
parallel, then they intersect at a point. That is this statement is true in both the situations.
In order to understand statements with “Or” we first notice that the word “Or” is
used in two ways in English language. Let us first look at the following statement.
p: An ice cream or pepsi is available with a Thali in a restaurant.
This means that a person who does not want ice cream can have a pepsi along
with
Thali or one does not want pepsi can have an ice cream along with Thali. That is,
who do not want a pepsi can have an ice cream. A person cannot have both ice cream
and pepsi. This is called an exclusive “Or”.
Here is another statement.
A student who has taken biology or chemistry can apply for M.Sc.
microbiology programme.
Here we mean that the students who have taken both biology and chemistry can
apply for the microbiology programme, as well as the students who have taken only
one of these subjects. In this case, we are using inclusive “Or”.
It is important to note the difference between these two ways because we require this
when we check whether the statement is true or not.
2020-21 | 4,332 | 18,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-27 | latest | en | 0.949711 |
https://fdocument.org/document/major-loss.html | 1,716,106,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00386.warc.gz | 225,507,046 | 25,915 | 7
Major loss - head loss or pressure loss - due to friction in ducts, pipes and tube Pressure and Pressure Loss According the Energy Equation for a fluid the total energy can be summarized as elevation energy, velocity energy and pressure energy. The Energy Equation can then be expressed as: p 1 + ρ v 1 2 / 2 + ρ g h 1 = p 2 + ρ v 2 2 / 2 + ρ g h 2 + p loss (1) where p = pressure in fluid (Pa (N/m 2 ), psi (lb/ft 2 )) p loss = pressure loss (Pa (N/m 2 ), psi (lb/ft 2 )) ρ = density of the fluid (kg/m 3 , slugs/ft 3 ) v = flow velocity (m/s, ft/s) g = acceleration of gravity (m/s 2 , ft/s 2 ) h = elevation (m, ft) For horizontal steady state flow v 1 = v 2 and h 1 = h 2 , - (1) can be transformed to: p loss = p 1 - p 2 (2) The pressure loss is divided in major loss due to friction and minor loss due to change of velocity in bends, valves and similar. The pressure loss in pipes and tubes depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow us turbulent or laminar - the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as: p loss = λ (l / d h ) (ρ v 2 / 2) (3) where p loss = pressure loss (Pa, N/m 2 ) λ = friction coefficient l = length of duct or pipe (m)
### Transcript of Major Loss
Major loss - head loss or pressure loss - due to friction in ducts, pipes and tube
Pressure and Pressure Loss
According the Energy Equation for a fluid the total energy can be summarized as elevation energy, velocity energy and pressure energy. The Energy Equation can then be expressed as:
p1 + ρ v12 / 2 + ρ g h1 = p2 + ρ v2
2 / 2 + ρ g h2 + ploss (1)
where
p = pressure in fluid (Pa (N/m2 ), psi (lb/ft2 ))
ploss = pressure loss (Pa (N/m2 ), psi (lb/ft2 ))
ρ = density of the fluid (kg/m3 , slugs/ft3 )
v = flow velocity (m/s, ft/s)
g = acceleration of gravity (m/s2 , ft/s2 )
h = elevation (m, ft)
For horizontal steady state flow v1 = v2 and h1 = h2, - (1) can be transformed to:
ploss = p1 - p2 (2)
The pressure loss is divided in
major loss due to friction and minor loss due to change of velocity in bends, valves and similar.
The pressure loss in pipes and tubes depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow us turbulent or laminar - the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as:
ploss = λ (l / dh) (ρ v2 / 2) (3)
where
ploss = pressure loss (Pa, N/m2 )
λ = friction coefficient
l = length of duct or pipe (m)
dh = hydraulic diameter (m)
(3) is also called the D'Arcy-Weisbach Equation. (3) is valid for fully developed, steady, incompressible flow.
The Energy equation can be expressed in terms of head and head loss by dividing each term by the specific weight of the fluid. The total head in a fluid flow in a tube or a duct can be expressed as the sum of elevation head, velocity head and pressure head.
p1 / γ + v12 / 2 g + h1 = p2 / γ + v2
2 / 2 g + h2 + hloss (4)
where
hloss = head loss (m, ft)
γ = ρ g = specific weight (N/m3 , lb/ft3 )
For horizontal steady state flow v1 = v2 and h1 = h2, - (4) can be transformed to:
hloss = h1 - h2 (5)
where
h = p / γ = head (m, ft)
The head loss in a tube or duct due to friction, major loss, can be expressed as:
hloss = λ (l / dh) (v2 / 2 g) (6)
where
hloss = head loss (m, ft)
Friction Coefficient - λ
The friction coefficient depends on the flow - if it is
laminar, transient or turbulent
and the roughness of the tube or duct.
To determine the friction coefficient we first have to determine if the flow is laminar, transient or turbulent - then use the proper formula or diagram.
Friction Coefficient for Laminar Flow
For fully developed laminar flow the roughness of the duct or pipe can be neglected. The friction coefficient depends only the Reynolds Number - Re - and can be expressed as:
λ= 64 / Re (7)
where
Re = the dimensionless Reynolds number
The flow is
laminar when Re < 2300 transient when 2300 < Re < 4000 turbulent when Re > 4000
Friction Coefficient for Transient Flow
If the flow is transient - 2300 < Re < 4000 - the flow varies between laminar and turbulent flow and the friction coefficient is not possible to determine.
Friction Coefficient for Turbulent Flow
For turbulent flow the friction coefficient depends on the Reynolds Number and the roughness of the duct or pipe wall. On functional form this can be expressed as:
λ = f( Re, k / dh ) (8)
where
k = absolute roughness of tube or duct wall (mm, ft)
k / dh = the relative roughness - or roughness ratio
Roughness for materials are determined by experiments. Absolute roughness for some common materials are indicated in the table below
Surface
Absolute Roughness - k
10-3 (m)
Copper, Lead, Brass, Aluminum (new) 0.001 - 0.002
PVC and Plastic Pipes 0.0015 - 0.007
Epoxy, Vinyl Ester and Isophthalic pipe 0.005
Stainless steel 0.015
Steel commercial pipe 0.045 - 0.09
Stretched steel 0.015
Weld steel 0.045
Galvanized steel 0.15
Rusted steel (corrosion) 0.15 - 4
New cast iron 0.25 - 0.8
Worn cast iron 0.8 - 1.5
Rusty cast iron 1.5 - 2.5
Sheet or asphalted cast iron 0.01 - 0.015
Smoothed cement 0.3
Ordinary concrete 0.3 - 1
Coarse concrete 0.3 - 5
Well planed wood 0.18 - 0.9
Ordinary wood 5
The friction coefficient - λ - can be calculated by the Colebrooke Equation:
1 / λ1/2 = -2,0 log10 [ (2,51 / (Re λ1/2 )) + (k / dh) / 3,72 ] (9)
Since the friction coefficient - λ - is on both sides of the equation, it must be solved by iteration. If we know the Reynolds number and the roughness - the friction coefficient - λ - in the particular flow can be calculated.
A graphical representation of the Colebrooke Equation is the Moody Diagram:
The Moody Diagram - The Moody diagram in a printable format.
With the Moody diagram we can find the friction coefficient if we know the Reynolds Number - Re - and the
Relative Roughness Ratio - k / dh
In the diagram we can see how the friction coefficient depends on the Reynolds number for laminar flow - how the friction coefficient is undefined for transient flow - and how the friction coefficient depends on the roughness ratio for turbulent flow.
For hydraulic smooth pipes - the roughness ratio limits zero - and the friction coefficient depends more or less on the Reynolds number only.
For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only.
Example - Pressure Loss in Air Ducts
Air at 0 o C is flows in a 10 m galvanized duct - 315 mm diameter - with velocity 15 m/s.
Reynolds number are expressed as:
Re = dh v ρ / μ (10)
where
Re = Reynolds number
v = velocity
ρ = density
μ = dynamic or absolute viscosity
Reynolds number calculated:
Re = (15 m/s) (315 mm) (10-3 m/mm ) (1.23 kg/m3 ) / (1.79 10-5 Ns/m2 )
= 324679 (kgm/s2 )/N
= 324679 ~ Turbulent flow
Turbulent flow indicates that Colebrooks equation (9) must be used to determine the friction coefficient - λ -.
With roughness - ε - for galvanized steel 0.15 mm, the roughness ratio can be calculated:
Roughness Ratio = ε / dh
= (0.15 mm) / (315 mm)
= 4.76 10-4
Using the graphical representation of the Colebrooks equation - the Moody Diagram - the friction coefficient - λ - can be determined to:
λ = 0.017
The major loss for the 10 m duct can be calculated with the Darcy-Weisbach Equation (3) or (6):
ploss = λ ( l / dh ) ( ρ v2 / 2 )
= 0.017 ((10 m) / (0.315 m)) ( (1.23 kg/m3 ) (15 m/s)2 / 2 )
= 74 Pa (N/m2 ) | 2,366 | 7,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.809379 |
https://plainmath.org/college-statistics/83627-given-a-sample-x-x-n-n | 1,709,354,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00878.warc.gz | 458,010,438 | 21,469 | Damien Horton
2022-07-23
Given a sample ${X}_{1},...,{X}_{n}\sim N\left(\theta ,{\theta }^{2}\right)$ show, using the definition of completeness, that the statistic $T=\left(\sum _{i}{X}_{i},\sum _{i}{X}_{i}^{2}\right)$ is not complete for $n\ge 2$. Use the fact that ${\mathbb{E}}_{\theta }\left[2\left(\sum _{i}{X}_{i}{\right)}^{2}-\left(n+1\right)\sum _{i}{X}_{i}^{2}\right]=0$
The statistic $T\left(\stackrel{\to }{X}\right)$ is said to be complete for the distribution of $\stackrel{\to }{X}$ if, for every misurable function $g$, ${\mathbb{E}}_{\theta }\left[g\left(T\right)\right]=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\theta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{\theta }\left(g\left(T\right)=0\right)=1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\theta$
ab8s1k28q
Let ${T}_{1}=\sum _{i}{X}_{i}$ and ${T}_{2}=\sum _{i}{X}_{i}^{2}$ and suppose $P\left(2{T}_{1}^{2}-\left(n+1\right){T}_{2}=0\right)=1$. Then ${T}_{2}=\frac{2}{n+1}{T}_{1}^{2}$ a.s$\left(P\right)$.
However by the triangle inequality,
${T}_{1}^{2}\le {T}_{2}=\frac{2}{n+1}{T}_{1}^{2}\phantom{\rule{1em}{0ex}}a.s\left(P\right),$
which is a contradiction for all $n>2$.
Do you have a similar question? | 485 | 1,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-10 | latest | en | 0.539215 |
https://doza.pro/art/math/geometry/en/area-trapezium | 1,556,167,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578681624.79/warc/CC-MAIN-20190425034241-20190425060241-00405.warc.gz | 392,846,716 | 8,100 | # Trapezium area calculator
Trapezium area calculator - formulas and calculator for calculating area on-line. Formulas for all types of trapezium and special cases of isosceles trapezia are given.
Trapezium area - is a numerical characteristic characterizing the size of a plane bounded by a closed geometric figure formed by four series-connected segments (sides), two of which are parallel to each other.
A trapezoid - is a geometric figure formed by four series-connected segments (sides), two of which are parallel to each other.
Area - is a numerical characteristic characterizing the size of a plane bounded by a closed geometric figure.
Area is measured in units of measurement squared: km2, m2, cm2, mm2, etc.
1
## Area of the trapezoid in height and in two bases
a - base
b - base
h - height
... preparation ...
2
## Trapezium area in height and center line
m - middle line
h - height
... preparation ...
3
## Area of trapezoid on four sides
a - base
b - base
c - side
d - side
... preparation ...
4
## Area of trapezoid along the diagonal and the angle between the diagonals
d1 - diagonal
d2 - diagonal
α° - Angle between diagonals
... preparation ...
5
## Area of the trapezium through its bases and corners at the base
a - base
b - base
α° - angle at base
β° - angle at base
... preparation ...
6
## Area of an isosceles trapezium through its sides
a - side
b - side
c - side
... preparation ...
7
## Area of an isosceles trapezium through a small base, side and angle at a larger base
a - base
c - side
α° - angle at base
... preparation ...
8
## Area of an isosceles trapezium through a larger base, lateral side and angle with a larger base
b - base
c - side
α° - angle at base
... preparation ...
9
## Area of an isosceles trapezium through the bases and the angle at the base
a - base
b - base
α° - angle at base
... preparation ...
10
## Area of an isosceles trapezium through the diagonals and the angle between the diagonals
d - diagonal
α° - Angle between diagonals
... preparation ...
11
## Area of the isosceles trapezium through the middle line, the side and the angle at the base
m - middle line
c - side
α° - angle between parties
... preparation ...
12
## Area of an isosceles trapezoid along the radius of the inscribed circle and the angle between the sides
This formula is applicable only to isosceles trapezoids, into which a circle can be inscribed.
r - radius of inscribed circle
α° - angle between parties
... preparation ...
13
## Area of an isosceles trapezium through its two bases and the radius of the inscribed circle
This formula is applicable only to isosceles trapezoids, into which a circle can be inscribed.
a - base
b - base
r - radius of inscribed circle
... preparation ...
14
## Area of an isosceles trapezium through its bases and angle at a larger base
This formula is applicable only to isosceles trapezoids, into which a circle can be inscribed.
a - base
b - base
α° - angle at base
... preparation ...
15
## Area of an isosceles trapezium through the sides
This formula is applicable only to isosceles trapezoids, into which a circle can be inscribed.
a - base
b - base
c - side
... preparation ...
16
## Area of the isosceles trapezium through the bases and the midline
This formula is applicable only to isosceles trapezoids, into which a circle can be inscribed.
a - base
b - base
m - middle line
... preparation ... | 916 | 3,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-18 | longest | en | 0.882102 |
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- - LES idea (https://www.cfd-online.com/Forums/main/4445-les-idea.html)
B. Saunders. February 20, 2002 15:26
LES idea
Hello to you all,
I am a student who is new to LES and I wish to understand if there are any similarities between LES and RANS? Thank you,
Brett.
Jongdae Kim February 20, 2002 16:20
Re: LES idea
1. LES
LES uses (space) filtered Navier-Stokes and continuity equations. So the solutions are (space) filtered properties, U,V,W,P,etc.
2. RANS
Analysis with
RANS uses (time) filtered Navier-Stokes and continuity equations. So the solutions are (time) filtered properties, U,V,W,P,etc.
Depends on your purpose, you can choose one of those methods. In my case, turbulent flow around bluff body, LES with standard Smagorinsky model gives good results compared to experimental data. The RANS and k-epsilon model estimate reasonably but ....
Kim.
sylvain February 25, 2002 05:39
Re: LES idea
I appologize for that, but RANS is NOT time filtered NS equations. To get the RANS equations, the NS equations are averaged over realizations. This allows time dependance computations when the time variation are driven by moving mesh or time dependent BCs.
Regards,
Sylvain
Jongdae Kim February 25, 2002 11:55
Re: LES idea
I agree with you. Ensemble averaging is necessary to get RANS equations. However in computational method using RANS, how to get ensemble averaged N-S equation? To make problems simple, time filtering is a kind of Reynolds averaging. This was what I mean.
Thanks
Kim.
sylvain February 25, 2002 12:29
Re: LES idea
I agree, for steady state turbulence (statistically steady flow), time average values tend to ensemble average values, as an exemple :
Experimentally, for the following channel flow :
width = 0.2m; velocity = 10m/s; turbulence intensity = 20%.
one needs to average over 8s to get a correct (<1%) mean velocity value.
one needs to average over 200s to get a correct (<1%) rms value of the fluctuating velocity.
RANS solver give these average values directly.
Regards,
Sylvain
Jongdae Kim February 25, 2002 13:56
Re: LES idea
There is numerical error which is growing as the simulation continues. Sometimes, the numerical error can be small due to positive and negative numbers. But anyway, after long time simulation, the accuracy of numerical simulation decreases. So we are careful on the total computation time to get the results from unsteady simulation. (To discuss this problem, the size of grids, the spatial and temporal discretization,etc. are to be touched.)
In my case(LES of flow analysis around 2-dim. square prism, Re=22000), I use periodic condition in spanwise direction with the assumption that the flow is homogeneous in spanwise direction. (But this is not true. My collegue measured the flow field and got some correlation in spanwise direction.) To get velocity and turb. intensity profiles considering reasonable computational time and accuracy, sometimes, I use the average of the whole spanwise direction data.
This is a kind of ensemble averaging.
(One of my problem is how to compare LES data and the data from RANS equations.)
Kim.
All times are GMT -4. The time now is 11:15. | 802 | 3,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-26 | longest | en | 0.879196 |
https://mathematica.stackexchange.com/questions/135703/plotting-pairs-where-the-2nd-element-is-a-complex-number | 1,600,666,713,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198942.13/warc/CC-MAIN-20200921050331-20200921080331-00742.warc.gz | 527,719,038 | 35,233 | # Plotting pairs where the 2nd element is a complex number
How can I see the variation of 'y' for different values of 'x' using a plot in which 'y' are complex numbers ?
xValuesDb =
{6.*10^-6, 0.050006, 0.100006, 0.150006, 0.200006, 0.250006,
0.300006, 0.350006, 0.400006, 0.450006, 0.500006, 0.550006,
0.600006, 0.650006, 0.700006, 0.750006, 0.800006, 0.850006,
0.900006, 0.950006}
yValues =
{0.0499359 + 0.00218364 I, 0.0502323 + 0.00220999 I, 0.0505305 + 0.00223666 I,
0.0508304 + 0.00226367 I, 0.0511322 + 0.002291 I, 0.0514358 + 0.00231868 I,
0.0517412 + 0.00234669 I, 0.0520484 + 0.00237506 I, 0.0523575 + 0.00240377 I,
0.0526684 + 0.00243284 I, 0.0529812 + 0.00246227 I, 0.0532959 + 0.00249207 I,
0.0536125 + 0.00252223 I, 0.0539309 + 0.00255277 I, 0.0542513 + 0.00258369 I,
0.0545737 + 0.002615 I, 0.0548979 + 0.0026467 I, 0.0552242 + 0.00267879 I,
0.0555524 + 0.00271128 I, 0.0558826 + 0.00274418 I}
xyValuesAll =
ListPlot[xyValuesAll,
AxesOrigin -> {λMin, 0},
AxesLabel->{"BS density","Coverage Probability"}]
• Your code cannot be executed because you have not provided an example of {xValuesDb, yValues}. – Bob Hanlon Jan 18 '17 at 19:52
• @BobHanlon Thanks, I edited the question. – Mounia Hamidouche Jan 18 '17 at 23:06
• one possibility would be to have a 3DPlot with the z axis representing the imaginary part of y. – ivbc Jan 19 '17 at 1:32
xValuesDb = {6.*10^-6, 0.050006, 0.100006, 0.150006, 0.200006,
0.250006, 0.300006, 0.350006, 0.400006, 0.450006, 0.500006,
0.550006, 0.600006, 0.650006, 0.700006, 0.750006, 0.800006,
0.850006, 0.900006, 0.950006};
yValues = {0.0499359 + 0.00218364 I, 0.0502323 + 0.00220999 I,
0.0505305 + 0.00223666 I, 0.0508304 + 0.00226367 I,
0.0511322 + 0.002291 I, 0.0514358 + 0.00231868 I,
0.0517412 + 0.00234669 I, 0.0520484 + 0.00237506 I,
0.0523575 + 0.00240377 I, 0.0526684 + 0.00243284 I,
0.0529812 + 0.00246227 I, 0.0532959 + 0.00249207 I,
0.0536125 + 0.00252223 I, 0.0539309 + 0.00255277 I,
0.0542513 + 0.00258369 I, 0.0545737 + 0.002615 I,
0.0548979 + 0.0026467 I, 0.0552242 + 0.00267879 I,
0.0555524 + 0.00271128 I, 0.0558826 + 0.00274418 I};
xyValuesAllRe = Transpose[{xValuesDb, Re /@ yValues}];
xyValuesAllIm = Transpose[{xValuesDb, Im /@ yValues}];
xyValuesAllAbs = Transpose[{xValuesDb, Abs /@ yValues}];
ListPlot[
{xyValuesAllRe, xyValuesAllIm, xyValuesAllAbs},
PlotStyle -> {Directive[Red, Thick], Green,
Directive[Blue, AbsoluteDashing[{10, 10}]]},
Joined -> True,
Frame -> True,
FrameLabel -> (Style[#, 12, Bold] & /@
{"BS density",
"Coverage Probability"}),
PlotLegends -> {Re, Im, Abs}]
Another nice way to do it is using a 3d plot. Just add this code to your definitions:
points = Table[{xValuesDb[[i]], Re@yValues[[i]], Im@yValues[[i]]}, {i,
Length@xValuesDb}]
ListPointPlot3D[points, AxesLabel -> {"X", "Re@Y", "Im@Y"},
PlotRange -> All] | 1,251 | 2,827 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-40 | longest | en | 0.393804 |
https://www.interviewbit.com/old/problems/square-root-of-integer/ | 1,725,977,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00386.warc.gz | 757,185,297 | 39,438 | Practice
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Square Root of Integer
Problem Description
Given an integer A.
Compute and return the square root of A.
If A is not a perfect square, return floor(sqrt(A)).
DO NOT USE SQRT FUNCTION FROM STANDARD LIBRARY.
NOTE: Do not use sort function from standard library. Users are expected to solve this in O(log(A)) time.
Problem Constraints
0 <= A <= INTMAX
Input Format
The first and only argument given is the integer A.
Output Format
Return floor(sqrt(A))
Example Input
Input 1:
` 11`
Input 2:
` 9`
Example Output
Output 1:
` 3`
Output 2:
` 3`
Example Explanation
Explanation:
``` When A = 11 , square root of A = 3.316. It is not a perfect square so we return the floor which is 3.
When A = 9 which is a perfect square of 3, so we return 3.```
NOTE: You only need to implement the given function. Do not read input, instead use the arguments to the function. Do not print the output, instead return values as specified. Still have a question? Checkout Sample Codes for more details.
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Release Engineer | 571 | 2,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-38 | latest | en | 0.736884 |
http://www.evi.com/q/74.1_cm_to_inch | 1,418,955,309,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768169.20/warc/CC-MAIN-20141217075248-00094-ip-10-231-17-201.ec2.internal.warc.gz | 507,595,176 | 5,154 | # 74.1 cm to inch
• 74.1 centimeters is 29.17 inches.
• tk10publ tk10ncanl
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• 74.1 cm equals how many inch | 381 | 1,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-52 | latest | en | 0.885644 |
https://stats.stackexchange.com/questions/77038/covariance-matrix-proposal-distribution | 1,576,348,241,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00439.warc.gz | 538,451,027 | 33,273 | # Covariance matrix proposal distribution
In a MCMC implementation of hierarchical models, with normal random effects and a Wishart prior for their covariance matrix, Gibbs sampling is typically used.
However, if we change the distribution of the random effects (e.g., to Student's-t or another one), the conjugacy is lost. In this case, what would be a suitable (i.e., easily tunable) proposal distribution for the covariance matrix of the random effects in a Metropolis-Hastings algorithm, and what should be the target acceptance rate, again 0.234?
Thanks in advance for any pointers.
Well, if you are looking "for any pointers"...
The (scaled)(inverse)Wishart distribution is often used because it is conjugate to the multivariate likelihood function and thus simplifies Gibbs sampling.
In Stan, which uses Hamiltonian Monte Carlo sampling, there is no restriction for multivariate priors. The recommended approach is the separation strategy suggested by Barnard, McCulloch and Meng: $$\Sigma=\text{diag_matrix}(\sigma)\;\Omega\;\text{diag_matrix}(\sigma)$$ where $\sigma$ is a vector of std devs and $\Omega$ is a correlation matrix.
The components of $\sigma$ can be given any reasonable prior. As to $\Omega$, the recommended prior is $$\Omega\sim\text{LKJcorr}(\nu)$$ where "LKJ" means Lewandowski, Kurowicka and Joe. As $\nu$ increases, the prior increasingly concentrates around the unit correlation matrix, at $\nu=1$ the LKJ correlation distribution reduces to the identity distribution over correlation matrices. The LKJ prior may thus be used to control the expected amount of correlation among the parameters.
However, I've not (yet) tried non-normal distributions of random effects, so I hope I've not missed the point ;-)
• This answer talks about the prior, the OP asks about the proposal... Does these priors help with the acceptance ratio in some way? – An old man in the sea. Jul 3 at 11:02
• @Sycorax What about the proposal that the OP asked? what should he use, and with what parameters? – An old man in the sea. Jul 3 at 11:02
I personally use Wishart proposals. For instance, if I want a proposal $$\Sigma^*$$ around $$\Sigma$$, I use: $$\Sigma^* \sim \mathcal{W}(\Sigma/a,a),$$ where $$a$$ is a large number, like 1000. With that trick you will get $$E[\Sigma^*]=\Sigma$$ and you can adjust the variance with $$a$$. If I am not mistaken, the ratio of proposals for $$(p\times p)$$ matrices has a closed form: $$\frac{q(\Sigma\to\Sigma^*)}{q(\Sigma^*\to\Sigma)} = \left(\frac{|\Sigma^*|}{|\Sigma|}\right)^{a-(p-1)/2} \cdot e^{[tr({\Sigma^*}^{-1}\Sigma)-tr({\Sigma}^{-1}\Sigma^*)] \cdot a/2}$$
It is well known that if you use non-Gaussian distributions, the conjugacy of the model is lost, see:
http://www.utstat.toronto.edu/wordpress/WSFiles/technicalreports/0610.pdf
Then, you need to use other MCMC methods, such as Metropolis within Gibbs sampling or some adaptive version of it. Fortunately, there is an R package for doing so:
http://cran.r-project.org/web/packages/spBayes/index.html
The recommended acceptance rate is 0.44 but, of course, there are some assumptions behind this number, similarly as in the case of the 0.234.
Are you THE Dimitris Rizopoulos?
• @DimitrisRizopoulos The adaptive Metropolis withing Gibbs I mentioned uses a finite mixture of Gaussian distributions as a proposal distribution (as stated in the technical report I posted). If you use the hardcore Metropolis, then you are asking for an answer to the "million dollar question", for which there is no general solution. Typically you have to play with different proposals and different acceptance rates. Very good book, by the way. – Teco Nov 26 '13 at 21:07
Any proposal can be used if you define your log-posterior properly. You just need to use some tricks to implement it and properly define the support of your posterior, see:
How to find the support of the posterior distribution to apply Metropolis-Hastings MCMC algorithm?
There are tons of examples where a Gaussian proposal can be used for truncated posteriors. This is just an implementation trick. Again, you are asking a question with no general solution. Some proposals even have different performance for the same model and different data sets.
Good luck.
• Well, taking into account that the covariance matrix needs to be positive definite, it does not seem that logical to me to use whatever proposal distribution. The proposed matrices need to be positive definite. One option would be to have as proposal the Wishart posterior conditional used in Gibbs sampling, however this did not seem to work particularly well when I assumed a student's-t for the random effects. Hence my question, are there other types of proposals for covariance matrices? – Toka Stall Nov 27 '13 at 19:45 | 1,140 | 4,779 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-51 | latest | en | 0.873208 |
https://cathyduffyreviews.com/homeschool-reviews-core-curricula/math/grades-k-6/math-u-see | 1,558,812,298,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258439.57/warc/CC-MAIN-20190525184948-20190525210948-00304.warc.gz | 417,387,099 | 18,280 | # Math-U-See
Math-U-See combines hands-on methodology with incremental instruction and continual review in this manipulative-based program. It excels in its hands-on presentation of math concepts that enables students to understand how math works. It is one of the rare multi-sensory math programs that continues to use manipulatives up through Algebra 1.
Manipulative Blocks, Fraction Overlays, and Algebra and Decimal Inserts are used at different levels to teach concepts, primarily using the “rectangle building” principle. This basic idea, consistently used throughout the program—even through algebra—is one of the best ways to demonstrate math concepts. There is also a digital app version of the manipulatives that offers a virtual experience with the manipulatives. The app is available for iOS devices.
One of the things I think makes Math-U-See so popular is that many parents and teachers find that author Steve Demme’s presentations of math concepts helps them to finally comprehend much that they were taught in math but never understood. Parents and teachers with a new or renewed enthusiasm for math then do a much better job teaching their own children.
Math-U-See uses a “skill-mastery” approach, requiring students to demonstrate mastery of each topic before moving on. The program also builds in systematic review for previously learned concepts.
There are eight books for elementary grades titled Primer, Alpha, Beta, Gamma, Delta, Epsilon, Zeta, and Pre-Algebra. The Greek letter designations were chosen particularly to emphasize the order of learning rather than grade level designation. Students should move on to the next level once they’ve mastered the content of a book. These first eight books are followed by Algebra 1, Geometry, Algebra 2, PreCalculus with Trigonometry, and Calculus. Placement tests for the different levels are available free at the Math-U-See website.
Student workbooks and test booklets are softcover books, and the pages are perforated and punched so they can easily be removed, written upon, and placed in binders. Enrichment exercises have been incorporated into the student workbooks for Primer through Pre-Calculus. These optional, additional problems stretch students to higher levels of understanding and application of math concepts covered within the lessons.
Test booklets for each course have tests to be used at the end of each lesson plus four unit tests and a final exam. Neither student workbook pages nor tests are reproducible; you need to purchase books for each student. Student workbooks and test booklets are the only consumable items in each course.
Instruction manuals are printed in hardcover books with full-color covers so they can be used a number of times. Complete answer keys with solutions are included for all problems at all levels, an especially helpful feature at upper levels.
All books are printed in black and white. This is not a particular problem in the first four levels if students are working with the colorful manipulatives, learning the skip-count songs, and possibly watching the DVDs. In these levels, enrichment pages also offer engaging activities for those students who enjoy dot-to-dots, color-by-number, and other supplemental activities. These multi-sensory experiences make up for the bland workbook. However, as upper levels use manipulatives less and less, the plainness of the workbooks is a point to consider with some students.
The program covers all necessary math concepts, but it does not try to correlate the teaching of concepts at the same grade level or in the same order as some other programs.
For each level, you need both the student pack and the instruction pack. The student pack for each level includes a student workbook and a test booklet for each level except Primer.
For Primer through Algebra 1, you will also need to purchase the set of Manipulative Blocks or the digital manipulatives app. Math-U-See's manipulatives are primarily plastic blocks somewhat similar to Base Ten Blocks and Cuisenaire Rods, color-coded to correspond to each number. (See my review of both.) The blocks snap together like LEGOs®. Fraction Overlays are added at the Epsilon level, and Algebra/Decimal Inserts are added at the Zeta level. That means the same sets of manipulatives are each used over at least a few years.
The instruction pack for each level includes an instruction manual plus one or more DVDs that “teach the teacher.” Note that DVDs have subtitles for the hearing impaired. Parents must watch the DVDs to understand the basic concepts that are the foundation of the program. On the DVDs, Demme works through each level lesson-by-lesson, demonstrating and instructing. Demme's presentation is enthusiastic and engaging as he clearly explains what he is doing and why. He throws in lots of math tricks, the kind that make me scratch my head and ask myself why they never taught us that in school.
The DVD presentations are critical components of the courses although instruction manuals have briefer lesson presentations of the same material covered on the DVDs. I expect that most parents will have their children watch the DVDs with them, although it was originally intended that parents with students below high school level watch the DVDs and then do their own presentations to their children.
After the initial viewing or lesson presentation, parents and children work through lessons together for as many days as it takes for children to master the concepts. Once students have grasped a concept, they practice and do problem pages on their own with occasional assistance. Typically, children should be spending about a week per lesson, but you need to take as long as necessary for your child to learn each lesson.
Primer will generally be the starting place for most kindergartners. The Primer level begins with essential number concepts and continues up through adding to make 10, telling time, and an introduction to subtraction. Children use manipulatives more than in upper levels of the program (and far more than in most kindergarten math programs).
There is no test booklet for the Primer level. At the early levels, you will also want to use the Skip Counting and Addition Songs audio CD. Both a “Bible” version and a “Science and Literature” version are included on the CD.
Alpha level focuses most heavily on place value, addition, and subtraction. Beta level teaches regrouping for both addition and subtraction. Gamma primarily covers multiplication while Delta moves on to division. Fractions are the main topic in Epsilon, while Zeta tackles decimals and percent.
Of course, other topics are included alongside these primary themes—topics such as money, measurement, geometry, time telling, graphs, estimation, prime and composite numbers, Roman numerals, and solving for unknowns. While manipulative use remains essential for understanding new concepts, the amount of time spent using the manipulatives decreases in Epsilon and Zeta.
Pre-Algebra topics are similar to those in other such courses: positive and negative numbers, exponents, roots and radicals, the order of operation, geometry, ratio and proportions, and other such topics. One unusual topic for this level is irrational numbers.
There are plenty of practice problems in the latest editions of Math-U-See, but students who need more practice have free access to a computation drill program on the Math-U-See website. Parents need to choose which math concepts students will practice, then students use the program on their own. You can also use the website’s worksheet generator to generate and print additional pages of practice problems for courses up through Pre-Algebra. Problems are randomly selected so you can produce a number of different worksheets for the same lesson, even though some problems might show up on more than one worksheet.
### High School Courses
As you move into the high school level books, students are able to work more independently. The instruction manual for each level is written to the student. Students need to watch the DVD presentation then read through the instruction manual before tackling the workbook. Workbooks include extra instruction for unusual problems, especially for some of the honors or enrichment problems, but they do not serve as complete course books on their own.
The honors exercises provide more challenging work with critical thinking, word problems, and practical applications, plus test prep practice and preparation for the math required in advanced science courses. The addition of the honors exercises largely alleviates concerns I expressed in my review in the first edition of Top Picks about the program's ability to challenge advanced students. Students can also move through the texts more rapidly if they master the lessons quickly.
Even at the high school level, Demme presents concepts simply and clearly, avoiding dense-sounding mathematical abstractions common to so many high school textbooks. The high school courses feature many word problems and applications that make the lessons more interesting. The instruction manuals include complete answers with step-by-step solutions for all the exercises and tests, plus a glossary and an index.
While some students might be able to work through the courses independently, many will need parental or tutorial assistance. Math-U-See offers online co-op classes for those who might want to take a course with other students under the supervision of an experienced teacher.
In Algebra 1, Manipulative Blocks and the Algebra and Decimal Inserts are used, but less than in earlier levels.
Algebra 1 does not cover as much territory as do most other first-year algebra courses. For example, complex work with radicals as well as motion problems are taught in Algebra 2, although they are included in most other first-year courses. Slower students should find the pace very manageable. Honors lessons will challenge brighter students, but you can always speed up by moving students through the courses more quickly.
The rest of the upper-level books no longer use manipulatives. However, Geometry students need a protractor, a compass, and a straight edge to draw constructions.
Math-U-See Geometry is fairly traditional in presentation and coverage, although it is an easier course than most. While it covers the standard topics, it does not go as far in depth as Discovering Geometry. For example, Math-U-See Geometry deals only with regular polygons when teaching about interior and exterior angles of pentagons, hexagons, etc. There is not as much work with tangents as you find in Discovering Geometry. However, Math-U-See Geometry introduces geometric proofs in lesson 24 and uses them through the end of the course. It also introduces trigonometry and transformations in the last three lessons. Algebra is reviewed frequently within the lessons. As with Algebra 1, Math-U-See Geometry should be manageable for average to slow students, and you can challenge advanced students with honors exercises or move them ahead more quickly into Algebra 2.
Algebra 2 moves on to new material rather quickly (as compared to many other second-year algebra courses), bringing the total of Math-U-See’s combined algebra coverage close to that of other publishers. It introduces matrices and determinants in the honors section of the last lesson but does not get into functions at all. Students should be able to move on to either pre-calculus or trigonometry courses after completing Algebra 2.
Math-U-See’s PreCalculus with Trigonometry course dedicates a significant amount of space to trigonometry as one might expect from the title. Vectors, functions, logarithms, and a few other advanced math topics are also covered. PreCalculus students need a protractor, ruler, and a scientific calculator. (Note that this course and Calculus are the only Math-U-See courses that require a calculator.) This is a straightforward, fairly traditional course.
The Math-U-See series culminates with Calculus. While Calculus teaches the content typical of other calculus courses it also includes chapters titled “Physics Applications” and “Economics Applications” that help students grasp how useful calculus can be. Calculus does not include an honors component since the course already includes content that will challenge advanced students.
The DVD instructional component makes a huge difference, especially for these last two courses, since Demme does a great job of explaining and illustrating concepts. However, I very much appreciate the fact that the newest editions’ instruction manuals for Math-U-See high school level courses now include a teaching component so that students do not have to rely entirely on the DVDs.
You might want to check out the premade lesson plans from Homeschool Planet that are available for Math-U-See.
Find lesson plans available for this product at Homeschool Planet. Sign up for a 30-day FREE trial.
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### Instant Key
• Need For Parent/Teacher Instruction: varies by age and ability
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### Publisher's Info
Note: Publishers, authors, and service providers never pay to be reviewed. They do provide free review copies or online access to programs for review purposes.
Disclosure of Material Connection: Some of the links in the post above are "affiliate links." This means if you click on the link and purchase the item, I will receive an affiliate commission. Regardless, I only recommend products or services that I believe will add value to my readers. I am disclosing this in accordance with the Federal Trade Commission's 16 CFR, Part 255 "Guidelines Concerning the Use of Endorsements and Testimonials in Advertising." | 2,766 | 14,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-22 | longest | en | 0.936386 |
frontierwaste.com | 1,722,701,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00851.warc.gz | 226,489,132 | 34,448 | # How To Estimate Cubic Yards For A Dumpster Rental
There are many factors that go into calculating the price of a dumpster rental. One of the direct factors that affect the cost is the size of the roll-off container you decide to use for your project. For an accurate estimate, you need to estimate cubic yards, which is the volume of waste being disposed of in your dumpster.
This isn’t just about being cost-effective. Keep in mind that the material you’re hauling impacts the volume you can carry away in one load. To ensure a smooth and cost-effective transaction regarding your waste disposal learn how to estimate cubic yards for your dumpster rental (this is different than calculating the cubic yards needed for commercial trash service for businesses with weekly pickups all year round).
## What does one cubic yard mean?
The trick to calculating the correct size needed is to understand the approximate volume of waste being disposed of in your dumpster. To do this accurately, it’s important to know what one cubic yard actually means in terms of volume.
The cubic yard is a 3-dimensional measurement. It is calculated by multiplying length times with times height. Does this bring you back to math class? If so, this equation will look mighty familiar: L X W X H
To make this easier to visualize, let’s assume you’re installing an in-ground pool in your backyard. A big portion of the debris you ultimately dispose of will be the dirt you dig to make the hole for the pool.
If you know the dimensions of your pool, it becomes an easy estimate of the volume of dirt you will be hauling from your property. The average pool size is 15 feet by 30 feet with an average depth of 5.5 feet or 15 feet X 30 feet X 5.5 feet
### Convert your measurement in feet to yards
On a straight line, one yard equals 3 feet.
This means the dimension of your pool in yards is the same as:
15 feet/3 X 30 feet/3 X 5.5 feet/3
OR
5 yards X 10 yards X 1.83 yards
Multiply the L X W X H together and you get 91.67 cubic yards.
5 yards X 10 yards X 1.83 yards = 91.67 cubic yards
That’s A LOT of dirt.
### Selecting your dumpster size and frequency
What do these calculations mean for selecting your dumpster size? Nothing yet. They are incomplete without knowing the type of material you’re hauling. The next step is to check the average weight of the material per cubic yard. The truth is that just because you’ve got the correct volume, doesn’t mean it will match the dumpster size.
You may be thinking that 91 cubic yards of dirt can be hauled away using two 40-yard containers and a 20-yard container. That’s simple math, right?
The problem is that in Texas, the Department Of Transportation doesn’t allow a single-axle truck to carry over 25,000 pounds. Ultimately, the weight of the dirt will determine how much of it can be hauled in a single shot. You may be forced to use a smaller roll-off container in Corpus Christi if a 40-yard violates the weight restrictions.
As it turns out, soil weighs approximately 2,200 pounds per cubic yard depending on the moisture content (amount of water that remains in the soil). That means that you can’t dispose of more than 12 cubic yards at a time.
Divide the total cubic yards of dirt by the volume per trip and that will tell you how many container loads it will take for your project. 91/12 = 7.5
### Calculating the final details
In this example, our calculations have brought us to the conclusion that you will need a 20-yard dumpster rental for this project. The dumpster can only hold 12 cubic yards at a time. To haul the dirt produced from an in-ground pool excavation. You will also need to make at least 8 trips to get the dirt from your home to the dump or transfer station.
### Example of a pool excavation
In this video, the pool company hauls out 7 truckloads at 16 tons each. That’s way over the legal limit in Texas. So if you’ve decided to put a pool in your backyard, you’re looking at a few more trips to get the dirt to where it needs to go.
If the pool company takes care of the bill then you don’t need to worry about the math behind it. They will just give you a price on the construction and that’s what you’ll pay.
If you’re a DIY kind of person, you might want to wait until the DRYEST day of the year to make sure that dirt is extra light! Save yourself some money by getting more dirt in the truck for each load since less water means less weight!
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### Dumpster Cleaning: Essential Tips for a Healthy Waste Solution
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page | 1,035 | 4,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.936194 |
https://www.uuelco.me/search/label/groove | 1,606,520,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194634.29/warc/CC-MAIN-20201127221446-20201128011446-00457.warc.gz | 900,897,718 | 18,301 | Showing posts with label groove. Show all posts
Showing posts with label groove. Show all posts
### +mathemusic
Mathemusic is the functional conflation (via brane classes) of mathematics and 🎵music (defined as 'an incorporation of keys and timing') whose resultant is a γ-proof. This tends more towards being a supermathematical exercise grounded in the principles of music theory. However, it is practically expressing (with libretti) statistics in the framework of music (even sans sonically). Cryptologically, it is the algorithmic direction of variation*. When we speak of mathemusic, we are generally doing so in the context of The Origamic Symphony🎶/Egglepple.`Solutions` have `melodic` texture. (see also ¢ent, mathletics, 🐨lnq's Starting Five, 3-rex, fibor, UUe, juke notation)
"All I hear is mathematics. All I see is music." - 🐨Link Starbureiy
/// Contemporary mathemusic was originally formulated by 🐨Link Starbureiy.🤓
### +groove (scheme)
This is a (personal) passcode scheme based on body motion.
### +major groove
Function map: minor groove ... ↔ ... major groove
### +minor groove
Function map: minor groove ... ↔ ... major groove
### +meshroom
To meshroom (v.) is to coordinate The Origamic Symphony🎶 within jukespace [TOS↦j]. It dictates that, because Egglepple is (ludologically) a p-brane* connecting the vector scale extrema, the notion of cell space algebras being conflated epimorphically within a curvilinear framework is important for defining variation ().In this case, 'p' stands for (my) 'portfolio'.😇
Meshrooming advances genetic algorithms [: fold measured (partite) monomers in order to deduce a mesh] for twistors by considering (visualizing) them as embedded datapoints on a neural network [nn] (ie. flavorsequencemesh). (also of interest: loopstring, Pink program, identity)
/// Compositing is hypothesized to be accomplished under a single preimage.
Function map: j → pm [...eventual self-synthesis]
### +groove
Cryptographically, a groove is a dual-class (minor, major) noncommuted walk. [compare melody, see also groove (scheme)]
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email or phone#/SMS for payouts🧀💸. | 595 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-50 | longest | en | 0.867573 |
https://physics.stackexchange.com/questions/464524/can-anyone-tell-me-why-the-answer-to-this-question-is-d | 1,556,129,662,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578655155.88/warc/CC-MAIN-20190424174425-20190424200425-00124.warc.gz | 524,268,587 | 32,709 | # Can anyone tell me why the answer to this question is D? [closed]
I think the current will go through earth wire but my physics teacher told us the current will continue to flow through the circuit.
## closed as off-topic by John Rennie, LonelyProf, Chair, Qmechanic♦Mar 5 at 8:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, LonelyProf, Chair, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.
• Hi Qiao and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. – John Rennie Mar 5 at 6:52
• Does D says the potential becomes negative? – TechDroid Mar 5 at 7:03
• No it doesn't. How much do you know about electrostatics. Did your physics teacher explain in details what an electric potential means and how it works? – TechDroid Mar 5 at 7:06
• the potential at earth wire is 0 – Qiao Yunxi Mar 5 at 7:08
• For me to answer this question, you have to rephrase the question conceptually so it doesn't seem like you're asking us to solve your homework for you. – TechDroid Mar 5 at 7:16
Current in a circuit is driven by voltage differences. Imagine you would disconnect the point Q from ground. Then it is clear that the battery drives a current $$I = 0.5$$ A from "+" to "-" through the resistors, and there is a voltage drop $$\Delta U = R\times I$$ across each resistor in the circuit. However, in this setting, it is undefined, what the potential difference between point Q and ground is. It could be any arbitrary value.
When you connect Q to ground, then you define the potential difference between this point and ground to be zero. This will not affect the voltage difference between "+" and "-" of the battery, hence the current in the circuit stays the same. But it defines the potential between "+" and ground, and between "-" and ground. For example, without the ground connection the voltage between "+" (point P) and ground could be 110 V, and that between "-" (point S) and ground could be 100 V. With the ground connection, the voltage between "+" (point P) and ground is +2 V, and the voltage between "-" (point S) and ground it is -8 V, such that the voltage between point Q and ground is zero.
• and then, because the voltage between point Q and the ground is zero, no current flows along that wire. – RogerJBarlow Mar 5 at 10:58
• @RogerJBarlow: well, I did not want to say that, because it may be misleading. If you take a point slightly left of Q and one slightly right of Q, there is also no voltage, but there is still a finite current. – flaudemus Mar 5 at 11:55
• Yes, you're right, I was trying to be helpful but didn't appreciate the subtlety – RogerJBarlow Mar 5 at 20:13
It appears that you are asking why it is that the answer is not option $$C$$ on the assumption that if the current starts at the positive terminal of the battery (node $$P$$) it then flows through the $$4\,\Omega$$ resistor and out through the earth contact at node $$Q$$ so no current flows through the other two resistors?
This logic is flawed.
If this is a conventional circuit then the current is actually a flow of electrons coming out of the negative terminal of the battery and flow from right to left through the resistors.
Using your logic there would be no current through the $$4\,\Omega$$ resistor!
However it is the conservation of charge (Kirchhoff's first law) which gives the simple answer.
The current out of the positive terminal of the battery must equal the current in to the negative terminal of the battery as there can be no accumulation or loss of charge within the battery.
If some current leaks out of the circuit through the earth connection at node $$Q$$ how does that current get back into the circuit to make sure that the currents in and out of the battery are the same?
In this problem the purpose of the earth label at node $$Q$$ is to give you a value for the potential of node $$Q$$ so that then the potentials of the other nodes can be assigned.
I think this is an exercise in high school.
Your question is why does not the current leave the circuit and enter the ground. And the answer is that the current will continue to flow through the circuit.
If we ignore how the battery works and treated it as a device that keeps an invariant potential difference between these two ends. Then there is a static electric field near the battery, no matter if there is the circuit. Then you can imagine that put the circuit to connect two ends of the battery, together with the earth connected with the Q point. We know that there are "free" electrons in the circuit (metal wire) and the earth. They will be driven by the electric field of battery, so the current has been constructed. The system will arrive at the "equilibrium" (in fact static) state, if there is current that is no zero in the circuit, it should be continuous, that is to say in any time interval, the number of electrons who leave the battery is equal to the number of electrons who enter it. So there are only two possible states, the current is zero in the whole circuit or circuit does not leave the circuit and enter the earth. We consider these two states now:
1. No current in the circuit means any two points on the wire are at the same potential because it is a conductor. But we know that it is impossible for there is a fixed 10V difference between the two points of the battery.
2. This is the only possible state the circuit should be at. And is the answer to this exercise.
However, you may ask, why the current should be continuous when the system get static? That is because if it does not, the battery will accumulate the charge by time or lose charge. If it accumulates the charge and gets a higher potential than other parts, the electron (with the negative charge) will be attracted to the battery (Coulomb interaction) and then it cannot gather more charge, vice versa. This looks like the current is continuous in macroscopic.
You may also ask, why the system must get the static state. And the answer is similar to the one to the question above, there is a negative feedback in the system because of the electric interaction. The point is that in metal (in fact the conductor), the electric field will drive the electrons move. In the classical view, electrons meet the ions and the interaction between them is similar to the damping of the water flow. The electric field driving and the damping force lead the system to static at last. It is similar to the behavior of 2nd order equation: $$x''=F-x'$$ , or the motion of an object under wind resistance.
Finally, why there should be earth connects to the point Q? In high school exam or homework, it performs the reference level of the potential: 0V. Sometimes it is also an infinity charge pool. You may meet such exercise in the future study and it should not bother you.
• I'm sorry to bother but your answer is a bit of a jumble and I don't feel good down voting. Where does the static come from? There are some of your statements that are right, but overall, it's not. – TechDroid Mar 5 at 8:10
• Sorry I did not explain how the system gets static detailly enough but skip it with the word "negative feedback". I edit it to show it better in the answer! – Zhenduo Wang Mar 5 at 8:21
• By static you mean the battery is emptied and current doesn't flow anymore? – TechDroid Mar 5 at 8:38
• No I mean the current is not dependent on time. Maybe I should use the word "steady". – Zhenduo Wang Mar 5 at 8:44
• Does that relate in any way to the question? The question, from my comprehension, is based on potential of Q relative to the negative terminal of the battery, how does that have to do with time, steadiness, or "static". – TechDroid Mar 5 at 9:11 | 1,894 | 8,195 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-18 | longest | en | 0.954294 |
http://math-interactive.com/Products/CalGraph/Help/Conversions/Mass_Conversions.htm | 1,621,204,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00243.warc.gz | 30,823,239 | 1,587 | # Mass Conversions
Weight (mass) is measured using many different units.
You can convert mass measured in one unit to another unit by clicking Utilities and then Conversions.
Holding the Control key and typing q has the same result.
In the Conversions window select Mass.
Click the unit you have in the Units From: box, and the units you want in the Units To: box.
Enter the number in the lower left box and click Convert.
This window shows that 1 ounce is 28.349 523 125 grams.
The conversions are based on:
1 Kilogram = 1000 grams
1 gram = 1000 milligrams
1 carat = 200 milligrams
1 pound = 16 ounces = 453.592 37 grams
1 stone = 14 pounds
1 short ton = 2000 pounds
1 long ton = 2240 pounds
1 metric ton (or tonne) = 1000 kilograms
1 Troy ounce = 31.103 476 8 grams | 213 | 785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-21 | latest | en | 0.815786 |
http://www.physicsforums.com/showthread.php?s=a1af66cd0fd614d74d015ccd7687e3c1&p=4240206 | 1,398,377,019,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00078-ip-10-147-4-33.ec2.internal.warc.gz | 749,198,571 | 8,615 | # Sound Waves and Shock Waves
by JustWonderingx
Tags: shock, sound, waves
P: 4 I'm just trying to get a better grasp on sound waves and shock waves. Let's say I have a cylindrical rod of length L with no forces acting on it, and I push on the back of it with some constant velocity less than the speed of sound, c, of the material the rod is made of. Will the front of the rod not move until time=L/c? Now let's say I push on the same rod with a velocity, V, greater than c, and the material cannot fracture but can deform transversely. Would the length of the rod approach 0 at time=L/V as the back of the rod approaches the front? In this situation is the shockwave moving at velocity V? Thanks
PF Gold P: 5,720 c has nothing to do with propagation in a rod. It happens at the speed of sound in the material. You cannot push on the rod faster than c. This is physically impossible, so that part of your question is meaningless.
P: 4
Quote by phinds c has nothing to do with propagation in a rod. It happens at the speed of sound in the material. You cannot push on the rod faster than c. This is physically impossible, so that part of your question is meaningless.
I specified in my post that I am defining c as the speed of sound in the material.
PF Gold
P: 5,720
## Sound Waves and Shock Waves
Quote by JustWonderingx I specified in my post that I am defining c as the speed of sound in the material.
Ah ... I missed that. BAD choice of symbols.
P: 4
Quote by phinds Ah ... I missed that. BAD choice of symbols.
I disagree, c is frequently used as a symbol for sound speed. But we digress.
PF Gold
P: 5,720
Quote by JustWonderingx I disagree, c is frequently used as a symbol for sound speed. But we digress.
Fair enough. I didn't know that and I now notice that this is posted in general physics, not cosmology, which is where my mind was. Sorry to have temporarily derailed your thread.
Engineering Sci Advisor HW Helper Thanks P: 6,387 It's not obvious you can push the rod at a constant speed greater than the speed of sound, without using an "infinite" amount of force. The elastic modulos of mateirals in compression tends to increase, at least until the matieral fails in crushing. A compressive strain of more than 100% is impossible, if you think about what it would mean physically. The question makes better sense "in real life" if you think about pulling the end of the rod rather than pushing it. In that case, you are right that the far end will not move untill the stress wave has travelled the lengtth of the rod, indepedent of how fast or hard you pull. FWIW I've seen this happen, in a "fail safe" device that pulled a rod violently to stop part of a machine working. In a test (using high speed video etc to see what happened) the rod broke near the end that was pulled, but the tensile stress wave continued along the rod and moved the other end after the rod had broken, even though there was no pulling force acting on the rod after it broke. BTW "c" is a standard symbol for the speed of sound, if that is a more interesting quantity for modelling the situation than the speed of light.
P: 4
Quote by AlephZero It's not obvious you can push the rod at a constant speed greater than the speed of sound, without using an "infinite" amount of force. The elastic modulos of mateirals in compression tends to increase, at least until the matieral fails in crushing. A compressive strain of more than 100% is impossible, if you think about what it would mean physically.
I don't understand why you would need an "infinite" amount of force to push something faster than its speed of sound. If I have a rod with a low sound speed material, and I impact the back of it with a much harder, much more massive object moving at a speed, V, much greater than the rod's sound speed, c, wouldn't part of the rod be moving at a speed greater than it's sound speed as this object pushes it, while the other end remains undisturbed?
I feel like upon initial impact the atoms in the back plane of the rod begin moving with speed ~V. Since c<V, the next plane of atoms, and all subsequent planes, don't know the impact has occurred and cannot have increased their rigidity, and thus cannot have increased their sound speed.
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### World Plotting
The following method is not completed. I want to emphasize this. I have been designing this in my head for only a few weeks, and its only been nine or ten days since I had the necessary epiphany to make this work. However, so far it is going very well ... and I am one step beyond where this post will end, and it is still looking great.
Meanwhile, since this post as designed is getting pretty long, I am going to get it up and then start writing the next one. All that you are about to read was invented as I wrote it. It can always at a later time be upgraded, but so far I'm extremely happy with the result.
Let’s say you want to be able to generate a wilderness, or the part of a world, but you don’t want to muck about with any foolish mathematics. You just want to be able to roll dice and get results. Very well. That is what this system, and this post, is designed to allow.
For explanation purposes, let’s start with a fairly large area – something 7 hexes across. That would look like this:
Figure 1 - blank hexes
Nice and simple. So what we want to do is produce an infrastructure number for these hexes ... but as long as that number can actually be quite simple, let’s just roll a d12 for every hex. The higher the number, the greater the infrastructure. What that infrastructure’s designation is, for the moment, can be left aside. Here are my numbers, rolled entirely at random:
Figure 2 - infrastructure numbers
In all probability, you wouldn’t really want these to be wholly random. You might want to apply modifiers for areas where you had already put deserts, mountains, cities, etc ... but for the sake of demonstration, let’s suppose you’re conjuring up an environment from nothing.
Broad strokes: a 1-2 is pure wilderness; a 12, fully settled. We’ll say the hexes are 18 miles across, but they could be as wide as the reader likes.
For the next stage, we want to create interior hexes inside those we’ve already established, retaining our rolled numbers. We want to create seven “junior” hexes for each “senior” hexes ... which will look like this:
Figure 3 - add junior hexes
Now, there are several steps we’re going to take, so for the time being I would like if we could ignore those hexes which overlap from one senior hex to another:
Figure 4 - ignore 'crossover' hexes
Good, now we’re going to begin base-filling these hexes. To begin with, any hex that is marked with a 1 or a 2 is fully wilderness ... so all seven hexes are uncivilized. To add some color to the proceedings, we’ll say this whole region is basically a forest country, so we’ll start by shading in all those wilderness hexes dark green. We can also remove those numbers. This is going to give us an image like this:
Figure 5 - group VIII added - full wilderness see this post
Now, I have to admit going forward that I am balancing this slightly towards more wilderness than not; how you weight your particular elements is up to you, but I think wilderness plays a little better than civilization for most, and this generation is balanced to reflect that. Slightly balanced, I emphasize.
But let’s look at those hexes where we rolled a '3.' The senior hexes will have 1 junior hex that is civilized, and junior hexes that are not. The two possible patterns that can occur are these:
Figure 6 - group VII possibilities
The chances of each type of pattern, or “group,” is shown above. The chances of that one wilderness hex being on the edge is six times as common as being in the center (the right group can be rotated in six directions). Now, I realize the reader can see that it’s a roll of 7 which hex is civilized ... but the pattern IS important in this, so I’m deliberately making the effort to show how having the hex on the outside, where it might contact another civilized junior hex in another senior hex is a big difference from the hex which is guaranteed to be isolated on the left.
Let’s roll dice to see which orientation each of the senior hexes have for the 3’s we rolled earlier. There are only two ... but by chance, I did get one that was encircled by wild:
Figure 7 - Group VII added
Now we can move to the next sort of group – that covering the hexes above numbered 4 & 5. Once again, here are the sorts of patterns, with their relative occurrence:
Figure 8 - Group VI possibilities
The reader can see the far right pattern (d) can only be rotated in three different directions, which the others can be rotated in six. Once again, yes, I realize the reader can simply roll which two hexes are civilized ... but what’s important here is that you see how each pattern can be made to stand for a different inherent social relationship. The two hexes (a) and (b) are similar and clearly more social, while the hexes on the right are less so.
Also, I recognize the reader does not have a 21-sided dice ... but you all are clever, I’m sure you can work out something for yourselves.
Once again, let’s add these second group patterns to 4s and 5s on the main hex map. There are five of these:
Figure 9 - Group VI added
By now, some readers will be able to work out what I mean to do with those ‘crossover’ hexes that earlier I said to ignore. Loosely, we could say where four or more of the hexes surrounding those are wilderness, they too will become wilderness, and that where the encircling hexes are even, that its a 50/50 roll either way. For the moment, however, until we’ve filled out the entire arrangement, let’s continue to leave those as they are. There are other details we may want to consider.
Another point that needs to be made; some will wonder why not just randomly roll every hex to see if they should be wilderness or not. What is nice about this system is that it introduces ‘clumpiness.’ There are fairly substantial wilderness areas built up as well as substantial civilized areas. A single die roll system for each individual junior hex will create a far too heterogenerous pattern. Try it. I suspect, though, if you’ve done a lot of generation, especially for Traveller, you already know what you’ll get.
Now, before moving forward, I want to point out a couple of interesting things that are also determined by the results. In the above Figure 9, the reader may notice I chanced to roll group VI (a) once and group VI (b) twice. These indicate certain infrastructure features, which have been added to the generated map:
Figure 10 - first infrastructure features
All right, we can move forward to the next groups patterns now, for hexes numbered 6 & 7:
Figure 11 - group V possibilities
This is going to get a little harder to conceptualize. Types (a), (b) and (d) will all make six patterns, rotating them each in six directions. (c) however can only be turned in three directions; and (f) only two (if you turn it 120 degrees its the same). On the other hand, (e) can be turned TWELVE different orientations ... as it can be a mirror image of itself, and both it and its mirror can each be turned in six directions. Trust me. Play around with them, you’ll find I’m right. Altogether there are 35 orientations for all six types.
We can talk infrastructure, also. Type (a), where three hexes are together, indicates a tiny village, 100-300 people. (b), (c), (d) and (e) all have lines of hexes, so all four indicate a road of some sort. For no reason at all, except that we ARE randomly generating details, let’s say that any hex that comes up (b) is a primitive river way ... a ford or, if the river is too large for a ford, a hand ferry.
Let’s say that any hex that comes up (c) is a toll gate. Unlike (b) or (d), (c) would afford the shortest distance civilized travel between two opposite hexes ... so its a logical choke point for a guardhouse and small post also, so let’s add that.
Because (d) is on the outside of the hex, let’s say that it represents virgin industries – sawpits along the edge of the forest, a quarry perhaps (especially in unforested lands) or high country meadows.
Finally, because (e) has two hexes side by side with an isolated hex added, we can treat as merely a roadstead (since its a combination of formerly compiled hexes).
Now, with these we can replace senior hexes 6 and 7 ... there are six of those, five of them all in one line from the top to the middle of the map. I rolled the number 27 twice, so two of the new hexes are exactly the same:
Figure 12 - group V added
I’ve deliberately not hooked up the roads, since I want the reader to understand the relationships between the senior hexes. We have a loose collection of roads, the exact line of which we don’t actually know. We must remember that these roads can pass through wilderness hexes – but it’s too soon to determine whether or not they will.
I’ve chosen to make it a quarry and a ferry – though a table could be created for either feature, if one wanted to get more gritty. We can certainly see how the tollgate figures into the separated islands of civilization. The road that I’ve dipped up could just as easily dip down into the number 11 senior hex below ... but that depends on the orientation of that hex.
The river is tricky. You may already have a base map you want to work off of that already has the rivers laid out – in which case, you might say the road crosses a minor tributary, or perhaps its a rope bridge over a gorge. Your imagination is the only limitation.
The course of the river is more troublesome. You have to decide whether the river is the lowest part of the map (in which case the wilderness is all swampy and wet) or that it flows down from the highest part (so that the wilderness areas are hills and mountains. If you choose lowland, then there should be one large river that links up all the wildernesses ... since for an area like this, 140 miles across, the land would be almost all flat and part of one drainage basin. The lowest areas will be the most vegetated.
If you choose highlands with valleys, then the rivers should all flow outward and away from the wildernesses, and the lowest areas will be the spreading fields of your most civilized – and least vegetated – areas. So you see, its really important which you choose.
I confess, I spent about two months trying to come up with a simple randomizer to determine the location and direction of rivers WITHOUT elevations, without much success ... I am glad my world uses them. Since your world probably has no elevation numbers for hexes, I suggest you go with your instincts. Feel free to generate the various hexes and then just fit in the rivers wherever they seem “best.” Below I’ll show two images, where the rivers have been “sketched” for a set of wilderness and for a set of wilderness highlands. The highlands are on the right:
Figure 13 - river course options
Two very different vistas. And of course I could have drawn the rivers in a number of different patterns. At some point, I may sit down and crack out a complicated formula for river placement ... but some things the human imagination can just do faster and easier. What's important here is that we've provided substance for your imagination to hinge on in order to provide one very simple, direct effect. You choose the rivers, the rest of the hex generation then supports that choice.
Note that, like the map says, it is necessary to change the ferry to a ford, since the river is too short to be deep enough to allow water vehicles. All the water coming from these highlands would probably be fairly fast-moving ... a ford would be welcome, just as a ferry would be in the swampy lands of the left.
I could go with either of these, but since the mountains/hills are the more common arrangement, let’s continue with the rivers on the left.
We’ve finished all the predominantly wilderness hexes. The remainder have more civilization that wilderness. And so here we can stop. As I said, I have the next part ready ... and I'll continue to work forward on this to see where it takes me.
Steve said...
I love this sort of thing. Looking forward to Part 2.
You could eliminate the "corner" hexes by drawing the megahexes like this instead -- http://bit.ly/XXhRrs. That's if you don't have some as-yet-unrevealed reason for grouping the hexes the way you did.
Alexis Smolensk said...
Yeah, Steve, thought about that, but consider:
1) the system, with crossover hexes that I'm designing, can be grafted directly onto a formerly hex mapped system, such as that which I posted of Greece yesterday.
2) one of the 6-mile diameter junior hexes here can then be micro-created again, without the need to reshape ALL the hexes.
The link you suggest - albeit with the best intentions - discounts both these possibilities, forcing you to reshape your world each time you go smaller.
Arduin said...
You obviously have your own infrastructural numbers, and this is a simplified version for us plebians and so on, but what would really interest me is seeing how such numbers, and the senior/junior hex motif you've mentioned before, apply to your own world.
We can safely assume that Naples' preposterously high numbers would indicate a full hex of civilization, but the rest of Italy/Europe is hardly so dense.
Have you put to mind anything like a "conversion guide" wherein someone might take what you've done with Greece (say, myself) and utilise the senior/junior hex system?
Looking forward to where this goes!
Alexis Smolensk said...
This is practice for me too. When I'm done, I'll apply it to Kosovo.
Keith S said...
A very cool system. I wonder if (and you may have already done so) something like this could be used to generate urban environments?
More variables to deal with, I suppose. Still, intriguing!
Alexis Smolensk said...
Yagami and I were talking about that on Facebook last night, Keith. We're not sure how to generate results with no useable numbers beyond the city's actual size. But we're independently working on ideas.
Ungoliant said...
Totally awesome !
YagamiFire said...
Baking my noodle on the whole city thing. So far have considered time allowed for development as well as how much contention over the area exists...all balanced against potential restrictions in place based on the geographic location. | 3,244 | 14,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-22 | latest | en | 0.960007 |
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Algebra Tutor: None Selected Time limit: 1 Day
2x-4/(x-2)=8/(x^2-2x)
May 1st, 2015
2x-4/(x-2)=8/(x^2-2x)
2x-4/(x-2)=8/ x(x-2)
2x-4=8
2x=8+4
2x=12
x=12/2 =6
Please let me know if you have any other questions and best me if you are satisfactory.
May 1st, 2015
thank you (:
May 1st, 2015
...
May 1st, 2015
...
May 1st, 2015
Dec 8th, 2016
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6.253 develops the core analytical issues of continuous optimization, duality, and saddle point theory, using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions is discussed in detail, and is the basis for an intuitive, highly visual, geometrical approach to the subject. 6.253 develops the core analytical issues of continuous optimization, duality, and saddle point theory, using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions is discussed in detail, and is the basis for an intuitive, highly visual, geometrical approach to the subject.
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Analysis I (18.100) in its various versions covers fundamentals of mathematical analysis: continuity, differentiability, some form of the Riemann integral, sequences and series of numbers and functions, uniform convergence with applications to interchange of limit operations, some point-set topology, including some work in Euclidean n-space. MIT students may choose to take one of three versions of 18.100: Option A (18.100A) chooses less abstract definitions and proofs, and gives applications where possible. Option B (18.100B) is more demanding and for students with more mathematical maturity; it places more emphasis from the beginning on point-set topology and n-space, whereas Option A is concerned primarily with analysis on the real line, saving for the last weeks work in 2-space (the pla Analysis I (18.100) in its various versions covers fundamentals of mathematical analysis: continuity, differentiability, some form of the Riemann integral, sequences and series of numbers and functions, uniform convergence with applications to interchange of limit operations, some point-set topology, including some work in Euclidean n-space. MIT students may choose to take one of three versions of 18.100: Option A (18.100A) chooses less abstract definitions and proofs, and gives applications where possible. Option B (18.100B) is more demanding and for students with more mathematical maturity; it places more emphasis from the beginning on point-set topology and n-space, whereas Option A is concerned primarily with analysis on the real line, saving for the last weeks work in 2-space (the pla
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This course will focus on fundamental subjects in convexity, duality, and convex optimization algorithms. The aim is to develop the core analytical and algorithmic issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood. This course will focus on fundamental subjects in convexity, duality, and convex optimization algorithms. The aim is to develop the core analytical and algorithmic issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood.
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see http://ocw.mit.edu/terms/index.htm
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6.253 develops the core analytical issues of continuous optimization, duality, and saddle point theory, using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions is discussed in detail, and is the basis for an intuitive, highly visual, geometrical approach to the subject.
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This graduate level mathematics course covers decision theory, estimation, confidence intervals, and hypothesis testing. The course also introduces students to large sample theory. Other topics covered include asymptotic efficiency of estimates, exponential families, and sequential analysis. This graduate level mathematics course covers decision theory, estimation, confidence intervals, and hypothesis testing. The course also introduces students to large sample theory. Other topics covered include asymptotic efficiency of estimates, exponential families, and sequential analysis.
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see http://ocw.mit.edu/terms/index.htm
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This course will focus on fundamental subjects in (deterministic) optimization, connected through the themes of convexity, geometric multipliers, and duality. The aim is to develop the core analytical and computational issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions will be central, and will allow an intuitive, highly visual, geometrical approach to the subject. This theory will be developed in detail and in parallel with the optimization topics. The first part of the course develops the analytical issues of convexity and duality. The second part is devoted to convex optimization algorithms, and their applications to a variety This course will focus on fundamental subjects in (deterministic) optimization, connected through the themes of convexity, geometric multipliers, and duality. The aim is to develop the core analytical and computational issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions will be central, and will allow an intuitive, highly visual, geometrical approach to the subject. This theory will be developed in detail and in parallel with the optimization topics. The first part of the course develops the analytical issues of convexity and duality. The second part is devoted to convex optimization algorithms, and their applications to a variety
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This research-oriented course will focus on algebraic and computational techniques for optimization problems involving polynomial equations and inequalities with particular emphasis on the connections with semidefinite optimization. The course will develop in a parallel fashion several algebraic and numerical approaches to polynomial systems, with a view towards methods that simultaneously incorporate both elements. We will study both the complex and real cases, developing techniques of general applicability, and stressing convexity-based ideas, complexity results, and efficient implementations. Although we will use examples from several engineering areas, particular emphasis will be given to those arising from systems and control applications. This research-oriented course will focus on algebraic and computational techniques for optimization problems involving polynomial equations and inequalities with particular emphasis on the connections with semidefinite optimization. The course will develop in a parallel fashion several algebraic and numerical approaches to polynomial systems, with a view towards methods that simultaneously incorporate both elements. We will study both the complex and real cases, developing techniques of general applicability, and stressing convexity-based ideas, complexity results, and efficient implementations. Although we will use examples from several engineering areas, particular emphasis will be given to those arising from systems and control applications.
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This graduate level mathematics course covers decision theory, estimation, confidence intervals, and hypothesis testing. The course also introduces students to large sample theory. Other topics covered include asymptotic efficiency of estimates, exponential families, and sequential analysis.
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Analysis I (18.100) in its various versions covers fundamentals of mathematical analysis: continuity, differentiability, some form of the Riemann integral, sequences and series of numbers and functions, uniform convergence with applications to interchange of limit operations, some point-set topology, including some work in Euclidean n-space. MIT students may choose to take one of three versions of 18.100: Option A (18.100A) chooses less abstract definitions and proofs, and gives applications where possible. Option B (18.100B) is more demanding and for students with more mathematical maturity; it places more emphasis from the beginning on point-set topology and n-space, whereas Option A is concerned primarily with analysis on the real line, saving for the last weeks work in 2-space (the pla
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see https://ocw.mit.edu/terms/index.htm
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This research-oriented course will focus on algebraic and computational techniques for optimization problems involving polynomial equations and inequalities with particular emphasis on the connections with semidefinite optimization. The course will develop in a parallel fashion several algebraic and numerical approaches to polynomial systems, with a view towards methods that simultaneously incorporate both elements. We will study both the complex and real cases, developing techniques of general applicability, and stressing convexity-based ideas, complexity results, and efficient implementations. Although we will use examples from several engineering areas, particular emphasis will be given to those arising from systems and control applications.
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see https://ocw.mit.edu/terms/index.htm
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This course will focus on fundamental subjects in convexity, duality, and convex optimization algorithms. The aim is to develop the core analytical and algorithmic issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood.
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see https://ocw.mit.edu/terms/index.htm
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This course will focus on fundamental subjects in (deterministic) optimization, connected through the themes of convexity, geometric multipliers, and duality. The aim is to develop the core analytical and computational issues of continuous optimization, duality, and saddle point theory using a handful of unifying principles that can be easily visualized and readily understood. The mathematical theory of convex sets and functions will be central, and will allow an intuitive, highly visual, geometrical approach to the subject. This theory will be developed in detail and in parallel with the optimization topics. The first part of the course develops the analytical issues of convexity and duality. The second part is devoted to convex optimization algorithms, and their applications to a variety
Content within individual OCW courses is (c) by the individual authors unless otherwise noted. MIT OpenCourseWare materials are licensed by the Massachusetts Institute of Technology under a Creative Commons License (Attribution-NonCommercial-ShareAlike). For further information see https://ocw.mit.edu/terms/index.htm
Site sourced from | 2,781 | 15,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-51 | latest | en | 0.889496 |
https://www.physicsforums.com/threads/eq-for-displacement-of-current-carrying-wire-due-to-magnet.804704/ | 1,508,609,700,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824824.76/warc/CC-MAIN-20171021171712-20171021191712-00094.warc.gz | 992,072,061 | 17,083 | # Eq. for displacement of current-carrying wire due to magnet
1. Mar 23, 2015
### WK95
1. The problem statement, all variables and given/known data
This is for a physics lab I am working on. A flowing current causes the wire to deflect towards the right a certain amount that varies depending on the current strength.
I need to derive the equation d=(L/mg)F
where
-- L is the distance between contact 1 and 2 both of which lie in the same vertical line.
-- mg is the weight attached at the end of the wire.
-- F is the strength of the magnetic field.
-- d is the horizontal displacement of the wire at the magnet
The magnets are located in the center of the setup and the vertical distance from contact 1 to the magnet can be approximated to be L/2
For the full lab, see http://skipper.physics.sunysb.edu/~physlab/phy134Lab5Magnetic_Force_v4.pdf
2. Relevant equations
Derive: d = (L/mg)F
F = 2*F_W *sin(theta)
theta = d/L/2 = 2d/L
See below for the derivations of the second two equations.
3. The attempt at a solution
Here is a free-body diagram of the wire at the point where a magnetic field moves it to the right. Equation (1) is for theta, the angle that F_W makes with the vertical.
I got the equation for theta as follows. Tan(theta) = Opposite/Adjacent. Taking theta to be the angle of the wire to the horizontal, opposite is equation to d and adjacent is taken to be L/2 so this will be the triangle formed by the top half of the wire and the magnet. Thus tan(theta)=d/L/2=2d/L
So I've gotten close but the 4 shouldn't be there. I know I have an error somewhere and likely, the 2s of the equations before should have canceled each other once I combined them.
#### Attached Files:
• ###### Lab 5 setup.PNG
File size:
46.8 KB
Views:
134
2. Mar 23, 2015
### Aceix
I also got the same eqn as yours. Check the questiin again
3. Apr 6, 2015
Bump. | 499 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-43 | longest | en | 0.918923 |
https://edurev.in/studytube/Combined-Raft-Footing/705a8ce6-428b-4316-84aa-a32021bb47f5_p | 1,722,712,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00501.warc.gz | 177,723,244 | 60,979 | Combined & Raft Footing
# Combined & Raft Footing | Foundation Engineering - Civil Engineering (CE) PDF Download
Download, print and study this document offline
``` Page 1
FOOTINGS AND RAFTS
Page 2
FOOTINGS AND RAFTS
Footing
=800kN
=1000kN
=20 t/m
2
,M15, =415kN/m
2
Column size: 400x400mm.
Design of Combined
See Fig 4.54 for details of footing. Column design
Let pt=0.8%
=.008A; =0.992A
Clause.39.3 of IS 456-2000
A=146763.8mm
2
=1174.11 mm
2
, =145589.746mm
2
Provide footing of 400x400size for both columns.
Using 8-16 as main reinforcement and 8 @250c/c as lateral tie
Design of Footing
Page 3
FOOTINGS AND RAFTS
Footing
=800kN
=1000kN
=20 t/m
2
,M15, =415kN/m
2
Column size: 400x400mm.
Design of Combined
See Fig 4.54 for details of footing. Column design
Let pt=0.8%
=.008A; =0.992A
Clause.39.3 of IS 456-2000
A=146763.8mm
2
=1174.11 mm
2
, =145589.746mm
2
Provide footing of 400x400size for both columns.
Using 8-16 as main reinforcement and 8 @250c/c as lateral tie
Design of Footing
Resultant of Column Load
Fig. 4.52 Forces acting on the footing
R =1800 kN acting 3.08m from the boundary.
Area of the footing :
Taking length L=6m, Depth of footing =0.9m, ,
Width of footing, =1.549m.
Therefore, provide footing of dimension 6m x 1.6m
Soil Pressure q = =18.75 t/m
2
< 20 t/m
2
OK.
=28.125 t/m
2
Soil pressure intensity acting along the length =B x =1.6x28.125 =45t/m.
R
B
=119.88kN, R
C
=150.12kN.
Thickness of Footing i. Wide beam shear:
Maximum shear force is on footing C,SF=115.02KN
for percentage reinforcement =0.2%
0.32 x d x 1.6=45 [2.556-0.2-d]
d=1.1m
Page 4
FOOTINGS AND RAFTS
Footing
=800kN
=1000kN
=20 t/m
2
,M15, =415kN/m
2
Column size: 400x400mm.
Design of Combined
See Fig 4.54 for details of footing. Column design
Let pt=0.8%
=.008A; =0.992A
Clause.39.3 of IS 456-2000
A=146763.8mm
2
=1174.11 mm
2
, =145589.746mm
2
Provide footing of 400x400size for both columns.
Using 8-16 as main reinforcement and 8 @250c/c as lateral tie
Design of Footing
Resultant of Column Load
Fig. 4.52 Forces acting on the footing
R =1800 kN acting 3.08m from the boundary.
Area of the footing :
Taking length L=6m, Depth of footing =0.9m, ,
Width of footing, =1.549m.
Therefore, provide footing of dimension 6m x 1.6m
Soil Pressure q = =18.75 t/m
2
< 20 t/m
2
OK.
=28.125 t/m
2
Soil pressure intensity acting along the length =B x =1.6x28.125 =45t/m.
R
B
=119.88kN, R
C
=150.12kN.
Thickness of Footing i. Wide beam shear:
Maximum shear force is on footing C,SF=115.02KN
for percentage reinforcement =0.2%
0.32 x d x 1.6=45 [2.556-0.2-d]
d=1.1m
for percentage reinforcement =0.6%
0.6 x d x 1.6=45 [2.556-0.2-d]
d=0.847m.D=900mm.OK.
ii.Two way Shear Thickness of Footing i. Wide
beam shear:
Maximum shear force is on footing C,SF=115.02KN
for percentage reinforcement =0.2%
0.32 x d x 1.6=45 [2.556-0.2-d]
d=1.1m
for percentage reinforcement =0.6%
Page 5
FOOTINGS AND RAFTS
Footing
=800kN
=1000kN
=20 t/m
2
,M15, =415kN/m
2
Column size: 400x400mm.
Design of Combined
See Fig 4.54 for details of footing. Column design
Let pt=0.8%
=.008A; =0.992A
Clause.39.3 of IS 456-2000
A=146763.8mm
2
=1174.11 mm
2
, =145589.746mm
2
Provide footing of 400x400size for both columns.
Using 8-16 as main reinforcement and 8 @250c/c as lateral tie
Design of Footing
Resultant of Column Load
Fig. 4.52 Forces acting on the footing
R =1800 kN acting 3.08m from the boundary.
Area of the footing :
Taking length L=6m, Depth of footing =0.9m, ,
Width of footing, =1.549m.
Therefore, provide footing of dimension 6m x 1.6m
Soil Pressure q = =18.75 t/m
2
< 20 t/m
2
OK.
=28.125 t/m
2
Soil pressure intensity acting along the length =B x =1.6x28.125 =45t/m.
R
B
=119.88kN, R
C
=150.12kN.
Thickness of Footing i. Wide beam shear:
Maximum shear force is on footing C,SF=115.02KN
for percentage reinforcement =0.2%
0.32 x d x 1.6=45 [2.556-0.2-d]
d=1.1m
for percentage reinforcement =0.6%
0.6 x d x 1.6=45 [2.556-0.2-d]
d=0.847m.D=900mm.OK.
ii.Two way Shear Thickness of Footing i. Wide
beam shear:
Maximum shear force is on footing C,SF=115.02KN
for percentage reinforcement =0.2%
0.32 x d x 1.6=45 [2.556-0.2-d]
d=1.1m
for percentage reinforcement =0.6%
0.6 x d x 1.6=45 [2.556-0.2-d]
d=0.847m.D=900mm.OK.
ii.Two way Shear
Column B
d=0.415m.
Column A
2d[(0.4+d)+(0.42+d/2)] x 96.8=120-28.125[(0.4+d)(0.42+d/2)]
d=0.3906m
=0.85mm
=900mm, =850mm.OK.
Flexural reinforcement
Along Length Direction
Table 1of SP16
=0.354%
provided=0.6%
=1.15N/mm
2
required=5100 mm
2
/mm
Provide 28 @120mmc/c at top and bottom of the footing
Along width direction
```
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; | 1,957 | 5,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.813797 |
http://gamedev.stackexchange.com/questions/46150/swaying-camera-when-walking?answertab=oldest | 1,462,431,690,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.68/warc/CC-MAIN-20160428161525-00145-ip-10-239-7-51.ec2.internal.warc.gz | 119,455,418 | 17,124 | # Swaying Camera when Walking
You can see the effect in many games. The camera sways or wiggles a bit while walking to make the movement feel more realistic.
I have implemented a camera in my game. (Who'd have thunk it?) So is there a common approach to build this swaying effect in?
-
It's usually just a sinewave, take a look at this: stackoverflow.com/questions/6077652/swinging-bobbing-camera – bryan226 Dec 24 '12 at 11:19
An alternative would be parenting a camera to a rigged characters head resp. eye bone. The movement of the camera will then follow the movement of the character. Ofc, just using sine waves is easier to implement and compute. – sarahm Jan 10 '13 at 23:58
@sarahm. Since there are no character animations implemented in my game yet, I can't do that now. But I would like to implement that some day. – danijar Jan 11 '13 at 10:12
– Byte56 Jan 14 '13 at 19:52
possible duplicate of How can I implement view wobble when my player is running? – Byte56 Jan 14 '13 at 19:52
Applying a transformation to the world from the point of view of the camera is the most straight forward way.
When the character advances the camera can be modeled as a cartwheel without the wooden circle, keeping just the wooden sticks. This curve is called cycloid.
Also with each foot step the camera needs to move slightly to one side or the other as if the wooden sticks where separated by the horizontal distance between the two legs, balancing the center of mass to keep it vertically aligned with the foot touching the ground.
The two movement seen from the point of view of the camera would look like half a circle (or an ellipse).
Approximating the cycloid to a sinusoidal, the translation would be given by:
translation( A * cos(d + pi/2), B * sin(pi/2 - 2*d), 0 )
Where d is the distance since the character started to move. The movement speed can be modified by multiplying d by a factor, or the values of X or Y by an arbitrary amplitude.
-
Fantastic visualizations! +1 – Gerstmann Jan 17 '13 at 11:19 | 495 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2016-18 | latest | en | 0.923543 |
https://physicsteacher.in/2021/12/23/dimensional-formula-of-force-how-to-find/ | 1,721,075,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00659.warc.gz | 415,458,185 | 26,512 | High School Physics + more
# Dimensional formula of force – how to find?
Last updated on December 28th, 2021 at 02:12 pm
Here, we will find out the dimensional formula of force.
The formula of force = mass x acceleration
The dimension of mass = [M]
The dimensional formula of acceleration = [M0LT–2] [see here how to find the dimensional formula of acceleration ]
Hence, the Dimensional formula of force
= The dimension of mass x The dimension of acceleration
= [M] [M0LT–2]
= [M1LT–2]
= [MLT–2]
Dimensional formula of force = [MLT–2]
See also Dimensional formula of Pressure Energy - how to find it?
Scroll to top
error: physicsTeacher.in | 173 | 649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.872484 |
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