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https://www.mentorforbankexams.com/2018/04/rrb-general-intelligence-mcqs-set-1.html | 1,580,153,665,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251705142.94/warc/CC-MAIN-20200127174507-20200127204507-00496.warc.gz | 983,091,968 | 57,758 | RRB General Intelligence MCQ’s (Set – 1)
Dear Aspirants,
Welcome to Mentor for Bank Exams. Railway exam will be conducted in the month of April/May 2018 as per the official notification.Here we are giving the important General Intelligence MCQ’s for upcomming RRB ALP/Technician and RRB Group D in the exam oriented new pattern, these will help you to score good in Reasoning section.
1. Select the missing number from the given responses.
456 22 434
268 29 239
194 ? 121
a) 32
b) 54
c) 73
d) 86
Explanation: 456 - 434 = 22
268 - 239 = 29
194 - 121 = 73.
2. Pointing towards a lady in a photograph a man said "she is the only daughter-in-law of my Mother's husband". How that lady relates to that man?
a) Aunt
b) Mother
c) Wife
d) Sister
Explanation:
3. A word is represented by only one set of number is given in any one of the alternatives. The sets of number given in the alternatives are represented by two classes of alphabets as in two matrices given below. The column and rows of matrix I are numbered from 0 to 4 and that of matrix II are numbered from 5 to 9.A letter from these matrices can be represented first by its row and next by its column, e.g; J can be represented by 01,34,etc and O can be represented by 43,10,etc,similarly,you have to identify the set for the word JOLT.
a) 01 10 66 99
b) 20 43 78 04
c) 13 32 85 59
d) 34 22 96 76
4. Arrange the following words as per the order in dictionary.
1. Inappropriate
2. Inappeasable
3. Inaptitude
4. Inapplicable
5. Inapprehensible
a) 25431
b) 13425
c) 24513
d) 52341
5. If "-" shows "+","+" shows " x ","x" shows "÷","÷" shows "-" then 36-14÷9+3 is equal to
a) 19
b) 20
c) 23
d) 31
Explanation: 36 – 14 ÷ 9 + 3 is changed to 36 + 14 - 9 X 3as given in the question.
36 + 14 -27
50 - 27 = 23.
6. Which set of letters when sequentially placed at the gaps in the given letter series shall complete it?
a b _ _ a _ c _ a _
a) cbdab
b) cdbbd
c) cdbdd
d) cdbdb
Explanation: The correct sequence is abcdabcdab.
7. If PRIVATE is coded as 1234567 and RISK is coded as 2398, how is RIVETS coded?
a) 232679
b) 232679
c) 234769
d) 234976
Explanation: PRIVATE = 1234567
RISK = 2398
By using the values which are assigned to the given letters, we can code
RIVETS = 234769
8. A defective compass points towards the North when it should point towards the South-west. If a person is holding that compass facing North-east, than in which direction that defective compass will indicate?
a) South
b) East
c) South-west
d) North-west
Explanation:
By comparing the two diagrams it is clear that, the pointer which was showing North-east is directed towards the South, by damaged compass.
9. Statement: If both the parents have type B blood, their child can have only type B blood.
Conclusion I: Only people with type B blood can have children with type B blood.
Conclusion II: All children with type B blood come from parents who also have type B blood.
a) Only conclusion I is correct
b) Only conclusion II is correct
c) Both conclusions I and II are correct
d) Neither conclusion I nor II is correct
Explanation: The statement suggests that two people with type B blood, cannot have a child with any other type of blood. This does not imply that only people with type B blood can beget children with type B blood, nor does it imply that every child with type B blood comes from parents who share the same blood type.
10. The following numbers/words/letters follow a particular pattern. Select the related word/letter/number from the given alternatives.
62: 40 :: 53: ?
a) 28
b) 34
c) 42
d) 52
Explanation: The analogy is 6^2+2^2=36+4=40
Similarly: 5^2+ 3^2= 25+9=34
11. Six persons P, Q, R, S, T and U not necessarily in the same order were seated in a row facing the north. T was 2nd to the left of U who was at a gap of 2 from R. S was 2nd to the left of R and was neighbouring P. R was not seated at either of the extreme ends. Who was seated to the immediate left of Q?
a) T
b) R
c) U
d) S
Explanation: S P R T Q U
12. Find the odd one out:
a) 934
b) 732
c) 622
d) 914
Explanation: (7-3)/2=2
(6-2)/2=2
(9-1)/2=4
(9-3)/2=3 and not 4.
13. Complete the series:
BD, CF, EI, ?
a) IN
b) IM
c) HM
d) HN
Explanation: There are 2 series:
Series 1: B C _ E _ _ H
Series 2: D _ F _ _ I _ _ _ M.
14. Select the related word/letter/number from the given alternatives.
FK: UP :: IL:?
a) RN
b) SO
c) RM
d) RO
Explanation: The position values of the alphabets are used where A=1, B=2, C=3 and so on and the reverse values are taken where A=26, B=25, C=24 and so on
Forward value of F(6)= Reverse value of U(6)
Forward value of K(11)= Reverse value of P(11)
Forward value of I(9)= Reverse value of R(9)
Forward value of L(9)=Reverse value of O(12)
15. Select the related word/letter/number from the given alternatives.
BE14: HK38:: CF18: ?
a) IJ38
b) KL46
c) IL42
d) JM46 | 1,508 | 4,838 | {"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-2020-05 | latest | en | 0.888731 |
https://scienceeureka.com/define-youngs-modulus-and-bulk-modulus/ | 1,686,187,303,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00376.warc.gz | 535,291,945 | 14,752 | Define Young’s modulus and Bulk modulus
Young’s modulus:
Within the elastic limit, longitudinal stress divided by longitudinal strain is called Young’s modulus.
Young’s modulus (Y) = longitudinal stress / longitudinal strain
If longitudinal stress tends to infinity and longitudinal strain tends to zero. Young’s modulus tends to infinity. For a perfectly rigid body, the longitudinal stress is infinite and the longitudinal strain zero. So, Young’s modulus Y is infinite. on the other hand, if any external force is applied to a plastic body, its longitudinal stress is zero and Young’s modulus (Y) to becomes zero.
Bulk modulus:
Within the elastic limit, volume stress divided by volume strain is called the bulk modulus of elasticity.
Bulk moudulus (K) = volume stress / volume strain
Every substance – solid, liquid or gas has some volume and hence the bulk modulus is meaningful for all substances. The bulk moduli of a perfectly rigid body and a perfectly plastic body are infinite and zero respectively.
Since liquid and gaseous substances undergo only volume strain, the bulk modulus is the only elastic modulus for them. Among the different bulk moduli of gas very useful is its isothermal and adiabatic bulk moduli. The isothermal bulk modulus of air = 1.01 x 105Pa and the adiabatic bulk modulus of air = 1.42 x 105Pa. | 281 | 1,338 | {"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-2023-23 | longest | en | 0.789429 |
https://www.physicsforums.com/threads/symmetry-factor-of-a-general-feynman-diagram.696572/ | 1,519,027,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00774.warc.gz | 935,247,735 | 15,119 | # Symmetry factor of a general feynman diagram
1. Jun 12, 2013
### omephy
I am studying QFT from Srednicki's book. Let me ask a question about symmetry factor from this book.
Let, for specific values of $V$ and $P$ from eqn (9.11) we get some terms. One of them is a disconnected diagram consisted of two connected diagrams $C_1$ and $C_2$. The disconnected diagrams symmetry factor is, say, S; that is the term for disconnected diagram has a numerical coefficient: $\frac{1}{S}$. Now we write the term for disconnected diagram according to the eqn (9.12): $D = \frac{1}{S_D} \prod_I (C_I)^{n_I}$. In this case is this true: $S=\frac{1}{n_1 !} \times \frac{1}{n_2!} \times C_1$'s symmetry factor $\times C_2$'s symmetry factor? Here, $$S_D = \prod_I n_I !$$
Last edited: Jun 12, 2013
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Draft saved Draft deleted | 285 | 963 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-09 | longest | en | 0.883794 |
https://mikesmathpage.wordpress.com/2017/04/14/sharing-kelsey-houston-edwardss-video-about-pi-and-e-with-kids/ | 1,659,887,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00016.warc.gz | 392,720,119 | 26,091 | # Sharing Kelsey Houston-Edwards’s video about Pi and e with kids
Yesterday I a new video from Kelsey Houston-Edwards that just blew me away. At this point I don’t have the words to describe how much I admire her work. What she is doing to make challenging, high level math both accessible and fun for everyone is amazing.
If I exchange Infinitely many digits of Pi and E are the two resulting numbers transendental?
Before showing the boys Houston-Edwards’s video, I wanted to see what they thought about the question. So, we just dove in:
Next, I took a great warm up idea from Houston-Edwards’s video and asked the boys if they could find *any* two irrational numbers that you could use to swap digits and produce a rational number.
Now, with that little bit of prep work, we watched the new video:
After the video we talked about what we learned. I think just tiny bit of prep work we did really helped the boys get a lot more out of the video.
One of the fun little challenge questions from the video was to show that (assuming $\pi$ and $e$ differ in infinitely many digits, then you will produce uncountably many different numbers by swapping different digits. I didn’t expect that the boys would be able to construct this proof, so I gave them a sketch of how I thought about it (and hopefully my idea was right . . . . )
I think that kids will find the ideas in Houston-Edward’s new video to be fascinating. It is so fun (and sadly so rare) to be able to share ideas that are genuinely interesting to professional mathematicians with kids. As always, I can’t wait for next week’s PBS Infinite series video! | 361 | 1,623 | {"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": 2, "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} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.963629 |
https://www.gnu.org/software/c-intro-and-ref/manual/html_node/Complex-Data-Types.html | 1,721,461,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00687.warc.gz | 706,993,652 | 3,023 | Next: , Previous: , Up: Primitive Types [Contents][Index]
### 11.3 Complex Data Types
Complex numbers can include both a real part and an imaginary part. The numeric constants covered above have real-numbered values. An imaginary-valued constant is an ordinary real-valued constant followed by ‘i’.
To declare numeric variables as complex, use the `_Complex` keyword.4 The standard C complex data types are floating point,
```_Complex float foo;
_Complex double bar;
_Complex long double quux;
```
but GNU C supports integer complex types as well.
Since `_Complex` is a keyword just like `float` and `double` and `long`, the keywords can appear in any order, but the order shown above seems most logical.
GNU C supports constants for complex values; for instance, ```4.0 + 3.0i``` has the value 4 + 3i as type `_Complex double`. See Imaginary Constants.
To pull the real and imaginary parts of the number back out, GNU C provides the keywords `__real__` and `__imag__`:
```_Complex double foo = 4.0 + 3.0i;
double a = __real__ foo; /* `a` is now 4.0. */
double b = __imag__ foo; /* `b` is now 3.0. */
```
Standard C does not include these keywords, and instead relies on functions defined in `complex.h` for accessing the real and imaginary parts of a complex number: `crealf`, `creal`, and `creall` extract the real part of a float, double, or long double complex number, respectively; `cimagf`, `cimag`, and `cimagl` extract the imaginary part.
GNU C also defines ‘~’ as an operator for complex conjugation, which means negating the imaginary part of a complex number:
```_Complex double foo = 4.0 + 3.0i;
_Complex double bar = ~foo; /* `bar` is now 4 - 3i. */
```
For standard C compatibility, you can use the appropriate library function: `conjf`, `conj`, or `confl`.
#### Footnotes
##### (4)
For compatibility with older versions of GNU C, the keyword `__complex__` is also allowed. Going forward, however, use the new `_Complex` keyword as defined in ISO C11.
Next: , Previous: , Up: Primitive Types [Contents][Index] | 528 | 2,045 | {"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} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.796247 |
http://forums.wolfram.com/mathgroup/archive/2002/Jun/msg00145.html | 1,529,322,886,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125550-00013.warc.gz | 114,835,898 | 8,456 | Re: Use of ShowProgress Option output
• To: mathgroup at smc.vnet.net
• Subject: [mg34805] Re: Use of ShowProgress Option output
• From: "Carl K. Woll" <carlw at u.washington.edu>
• Date: Sat, 8 Jun 2002 05:21:11 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Andre,
Here is a perhaps frowned on possibility. Add a new definition to Print,
such as
In[4]:=
Unprotect[Print];
Print[a__/;printflag]:=
Module[{},
printflag=False;
If[ValueQ[iterlist],iterlist={iterlist,{a}},iterlist={a}];
Print[a];
printflag=True;]
Protect[Print];
Now, load the nonlinearfit package and run it:
In[7]:=
<< Statistics`NonlinearFit`
In[8]:=
data = {{1.0, 1.0, .126}, {2.0, 1.0, .219},
{1.0, 2.0, .076}, {2.0, 2.0, .126}, {.1, .0, .186}};
In[9]:=
printflag=True;
Clear[iterlist]
NonlinearFit[data,
theta1 theta3 x1 / (1 + theta1 x1 + theta2 x2),
{x1, x2}, {theta1, theta2,
theta3},RegressionReport->BestFitParameters,ShowProgress->True
]
printflag=False;
Iteration:1 ChiSquared:0.236342 Parameters:{1., 1., 1.}
Iteration:2 ChiSquared:0.0448893 Parameters:{2.69113, 3.34008, 0.173414}
Iteration:3 ChiSquared:0.00265359 Parameters:{5.00694, 9.2595, 0.425435}
Iteration:4 ChiSquared:0.000156222 Parameters:{4.45296, 15.2544, 0.588897}
Iteration:5 ChiSquared:0.0000809401 Parameters:{4.27179, 15.2961, 0.620383}
Iteration:6 ChiSquared:0.0000647486 Parameters:{3.88567, 15.3226, 0.662373}
Iteration:7 ChiSquared:0.0000553597 Parameters:{3.76195, 15.3201, 0.681971}
Iteration:8 ChiSquared:0.0000493747 Parameters:{3.51402, 15.2834, 0.715091}
Iteration:9 ChiSquared:0.0000454058 Parameters:{3.33655, 15.2364, 0.743497}
Iteration:10 ChiSquared:0.0000444845 Parameters:{3.13323, 15.1582, 0.777529}
Iteration:11 ChiSquared:0.0000435527 Parameters:{3.13156, 15.1594, 0.780051}
Out[11]=
2.44277 x1
---------------------------
1 + 3.13151 x1 + 15.1594 x2
Now, let's check to see if iterlist has the list you want:
In[13]:=
iterlist
Out[13]=
{{{{{{{{{{{Iteration:1 ChiSquared:0.236342 Parameters:{1., 1., 1.}},
{Iteration:2 ChiSquared:0.0448893 Parameters:{2.69113, 3.34008,
0.173414}}},
{Iteration:3 ChiSquared:0.00265359 Parameters:{5.00694, 9.2595,
0.425435}}},
{Iteration:4 ChiSquared:0.000156222 Parameters:{4.45296, 15.2544,
0.588897}}},
{Iteration:5 ChiSquared:0.0000809401 Parameters:{4.27179, 15.2961,
0.620383}}},
{Iteration:6 ChiSquared:0.0000647486 Parameters:{3.88567, 15.3226,
0.662373}}},
{Iteration:7 ChiSquared:0.0000553597 Parameters:{3.76195, 15.3201,
0.681971}}},
{Iteration:8 ChiSquared:0.0000493747 Parameters:{3.51402, 15.2834,
0.715091}}},
{Iteration:9 ChiSquared:0.0000454058 Parameters:{3.33655, 15.2364,
0.743497}}},
{Iteration:10 ChiSquared:0.0000444845 Parameters:{3.13323, 15.1582,
0.777529}}},
{Iteration:11 ChiSquared:0.0000435527 Parameters:{3.13156, 15.1594,
0.780051}}}
Yup, there it is. If you are worried about changing Print, you could write a
function (AndreFit?) which takes care of changing the Print definition,
setting printflag and clearing iterlist, and then running NonlinearFit, and
then restoring everything back to normal.
Carl Woll
Physics Dept
U of Washington
----- Original Message -----
From: "Andre Heinemann" <heinemann at hmi.de>
To: mathgroup at smc.vnet.net
Subject: [mg34805] Use of ShowProgress Option output
> Hi,
>
> ... may be my question was not clear in detail, but I realy can't find a
> solution.
> I will show you an example from the help browswer to ilustrate my problem.
>
>
> data={{1.0,1.0,.126},{2.0,1.0,.219},{1.0,2.0,.076},{2.0,
> 2.0,.126},{.1,.0,.186}};
>
> NonlinearRegress[data,
> theta1 theta3 x1/(1+theta1 x1+theta2 x2),{x1,x2},{theta1,theta2,theta3},
> RegressionReport->BestFitParameters,ShowProgress->True]
>
> I get:
>
> From In[396]:=
> Iteration:1 ChiSquared:0.23634239210284666` Parameters:{1.,1.,1.}
> .
> .(snip snap) 8><
> .
> From In[396]:=
> Iteration:11 ChiSquared:0.000043552662305849804` \
> Parameters:{3.13156,15.1594,0.780051}
>
> Out[396]=
> {BestFitParameters\[Rule]{theta1\[Rule]3.13151,theta2\[Rule]15.1594,
> theta3\[Rule]0.780062}}
>
> My aim is to get !automaticaly! a list like this:
>
>
{{Iteration:1,{1.,1.,1.}},...,{Iteration:11,{3.13156,15.1594,0.780051}},...,
}
>
> May be this problem is very easy to solve, but I found only the "Hand
> of copy and paste - this is not the way I want to use a computer (-;
>
> So if you have any idea how to solve this problem, please give me a hint.
>
> Thanks,
> Andre
>
>
```
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• Previous by thread: Re: Use of ShowProgress Option output
• Next by thread: PlotVectorField | 1,678 | 4,712 | {"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-2018-26 | latest | en | 0.438232 |
https://sage-answers.com/what-temperature-do-water-melt/ | 1,695,986,219,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00365.warc.gz | 535,833,928 | 12,298 | # What temperature do water melt?
## What temperature do water melt?
32°F
At temperatures above 32°F (0°C), pure water ice melts and changes state from a solid to a liquid (water); 32°F (0°C) is the melting point. For most substances, the melting and freezing points are about the same temperature.
## What temperature does water melt and boil?
Increasing the temperature above the boiling point, 212°F (100°C), causes water to change from liquid to gas (water vapor). The melting/freezing and boiling points change with pressure. The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower.
Is the melting point of water 0?
The universal constant for the freezing/melting point of water is 0 degrees Celsius, equivalent to 32 degrees Fahrenheit. Salt makes water freeze at a lower temperature, however, which is why it is applied to roads to prevent ice formation.
### At what temperature water becomes ice?
32 degrees Fahrenheit
Water, like all types of matter, freezes at a specific temperature. The freezing point for water is 0 degrees Celsius (32 degrees Fahrenheit). When the temperature of water falls to 0 degrees Celsius and below, it begins to change to ice. As it freezes, it releases heat to its surroundings.
### How cold can liquid water get?
How low can you go? For water, the answer is -55 degrees Fahrenheit (-48 degrees C; 225 Kelvin). University of Utah researchers found that is the lowest temperature liquid water can reach before it becomes ice.
Does water completely melt?
The melting point depends slightly on pressure, so there is not a single temperature that can be considered to be the melting point of water. However, for practical purposes, the melting point of pure water ice at 1 atmosphere of pressure is very nearly 0 °C, which is 32 °F or 273.15 K.
#### What is the melting point of gold?
1,948°F (1,064°C)
Gold/Melting point
#### What is water’s freezing point?
32°F (0°C)
Water/Melting point
What is diamonds melting point?
The ultimate melting point of diamond is about 4,027° Celsius (7,280° Fahrenheit).
## What is frozen water called?
Ice is the common name for frozen water. Liquid water becomes solid ice when it is very cold. The freezing point is 0° Celsius (32° Fahrenheit or 273 kelvin). Ice is commonly made in a home refrigerator or freezer.
## What is the coldest place on Earth?
Oymyakon is the coldest permanently-inhabited place on Earth and is found in the Arctic Circle’s Northern Pole of Cold.
How cold is dry ice?
-109° F
Unlike regular ice, dry ice doesn’t melt into a liquid as it warms up. Instead, it converts directly back into its gaseous form in a process known as sublimation. At -109° F, dry ice is also significantly colder than the 32° F surface temperature of regular ice.
### What temperature is hotter than the boiling point of water?
Steam occurs when water goes above 212 degrees Fahrenheit, which is hotter than water when it is at its stable point. While water boils at 212 F, steam is at a much higher temperature as water turns to vapor. While it is possible for water to remain in a liquid form after it has reached 212 F, it becomes an unstable liquid.
### Does water freeze and melt at the same temperature?
At temperatures above 32°F (0°C), pure water ice melts and changes state from a solid to a liquid (water); 32°F (0°C) is the melting point. For most substances, the melting and freezing points are about the same temperature.
What is the temperature of water as ice melts?
– At what temperature does fresh water turn to ice? 32°F (0°C). – Based on their experiments melting the crushed ice, at what temperature does ice melt to make water – what is the melting point of ice? 32°F (0°C). – How can the melting and freezing temperatures be the same? – Why is the ice/salt mixture colder?
#### What is the initial temperature of the water?
to find the initial temperature (t0) in a specific heat problem. In fact, water has one of the highest specific heats of any “common” substance: It’s 4.186 joule/gram °C. That’s why water is so useful in moderating the temperature of machinery, human bodies and even the planet. | 948 | 4,202 | {"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} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.872574 |
http://www.maths.date/cn/pe/solve/203 | 1,553,543,699,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00233.warc.gz | 324,683,273 | 5,167 | ### Squarefree Binomial Coefficients题号：203 难度： 25 中英对照
The binomial coefficients nCk can be arranged in triangular form, Pascal's triangle, like this:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1
.........
### Solution
### Code
import java.math.BigInteger;
import java.util.*;
public class p203 {
private static int N = 51;
public static void main(String args[]) {
long start = System.nanoTime();
long result = run();
long end = System.nanoTime();
System.out.println(result);
System.out.println((end - start) / 1000000 + "ms");
}
private static long run() {
ArrayList<Long> list = new ArrayList<>(0);
long line[] = {1};
// for循环生成杨辉三角,保存在list中
for (int i = 0; i < N; i++) {
for (int j = 0; j < line.length; j++) {
}
line = nextLine(line);
}
// list 中元素排序
Collections.sort(list);
// 移除list 中重复元素
removeDups(list);
BigInteger count = new BigInteger("0");
for (int i = 0; i < list.size(); i++) {
if (isSquareFree(list.get(i))) {
count = count.add(new BigInteger("" + list.get(i)));
}
}
return count.longValue();
}
private static void removeDups(ArrayList<Long> arr) {
for (int i = 0; i < arr.size(); i++) {
long curr = arr.get(i);
int j = i + 1;
while (j < arr.size() && curr == arr.get(j)) {
arr.remove(j);
}
}
}
private static long[] nextLine(long[] n) {
long[] next = new long[n.length + 1];
for (int i = 0; i < next.length; i++) {
if (i == 0 || i == next.length - 1) {
next[i] = 1;
} else
next[i] = n[i - 1] + n[i];
}
return next;
}
// 遍历判断是否是 square free number
public static boolean isSquareFree(long n) {
for (long i = 2; i <= n; i++) {
int count = 0;
while (n % i == 0) {
n /= i;
count++;
if (count == 2) return false;
}
}
return true;
}
}
34029210557338
3ms | 616 | 1,780 | {"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} | 3.71875 | 4 | CC-MAIN-2019-13 | latest | en | 0.206884 |
http://docplayer.net/9879256-Ppf-s-of-germany-and-france.html | 1,576,288,354,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00156.warc.gz | 42,948,427 | 28,516 | # PPF's of Germany and France
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1 Economics 165 Winter 2 Problem Set #1 Problem 1: Let Germany and France have respective labor forces of 8 and 6. Suppose both countries produce wine and cares according to the following unit labor requirements: a w G = 2; a c G = 8; a w F = 1; a c F =. Further, suppose that the representative German and Frenchman both have utility given by U i (Q w, Q c ) = Q w Q c. a) Depict the PPF for both countries and label the interceps as well as the slope(include their exact values, pls). Which country has the comparative advantage in producing wine/cars? If we plot cars on the x-axis and wine on the y, we see that the slope of the German PPF is 4 and that of the French is. Since the opportunity cost of a car is 4 bottles of wine in Germany and bottles in France, we see that the Germans have a comparative advantage in car production and the French have a comparative advantage in the production of wine. PPF's of Germany and France Wine Cars Germany France b) Suppose each country wanted to go it alone. What would be the quantities produced (=consumed) and the utility levels obtained under autarky? (A little help: recall from micro that in the consumption optimum MRS = MU c /MU w = p c /p w.) For each country, MRS = MU c /MU w = W/C. The price ratio will be the slope of the PPF in each country. In Germany this is 4. Hence, W = 4C. The point on Germany s PPF where W = 4C is when C = 5 and W =. Hence, this is what Germany produces. Since Germany has a population of 8, each consumer will get 1/16 units of car and 1/4 of a unit of wine. Hence, the utility of the representative German is 1/64. In France, the price ratio is, so W = C. The only point on the French PPF satisfying this condition is at C = 3 and W = 3. This is France s production choice. Since the population is 6, each worker will consume 1/ cars and 1/2 wine. Consequently, the utility of the representative Frenchman is 1/4. c) Now consider free trade. Depict the world s relative supply function (again, pls label and provide numbers). What is the world s free trade equilibrium relative price? What quantities does each country produce, how much do they import/export? What are the utility levels obtained? How high are their respective wages? Consider what each country will do under free trade. Given a set of prices, each country will want to produce at the point on the PPF that will generate the most wealth. Then given those same prices and that level of wealth, each will chose to consume at the point that maximizes utility. Start with Germany. If the price of cars is very low relative to the price of wine, Germany will want to produce only wine. For example, if p c = 2 and p w = 1, then if Germany produces only wine, it will have a GDP of 4(1) = 4 and if it produces only cars its GDP will be (2) =. As a result, Germany will produce only wine. Since you always need to reduce wine production by 4 units if you increase car production by 1 unit, Germany will never produce cars as long as p c /p w < 4. At the point where p c /p w = 4, Germany will be indifferent to producing cars or wine. That is,
2 any point on the PPF will generate the same GDP for the country. This corresponds to the flat part of the relative supply graph. Finally if p c /p w >4, Germany will produce only cars. France goes through a similar though process, but since the slope of its PPF is -, the relative price at which France is indifferent between producing cars and wine is p c /p w =. For a lower price ratio, France will specialize completely in wine production and for a higher ratio it will specialize completely in car production. Relative Supply and Demand 25 P(c)/P(w) 15 5 Supply Demand Q(c)/Q(w) The relative demand curve can be found by aggregating consumer demand. This task is greatly simplified since all consumers have the same utility function. Since the utility function is Cobb-Douglas, we know that all consumers will have constant expenditure shares. (If you don t recall this fact about Cobb-Douglas utility, just solve the consumer s utility max problem.) This means that every individual has demand functions given by W = M/2p w ; and C = M/2p c, where M is wealth and the 2 in the denominator comes from the result that all workers spend half their income on C and half on W. If we divide C by W we find that the relative demand function is: Q C /Q w = p w /p c. The only point on the supply curve that satisfies this equation is where Q C /Q w = 1/6 and p c /p w = 6. This is where the market will be in equilibrium. When p c /p w = 6, we know that Germany will produce only cars ( of them) and France will produce only wine (6 units). The value of Germany s cars is 6 and it will spend half its wealth on cars and half on wine (because of the representative agent s utility function). Hence Germany will consume 3/6 = 5 cars and 3/1 = 3 units of wine. France s 6 bottles of wine are worth 6. Since France has the same preferences as Germany and the same wealth, it follows that facing the same prices (one world price) France will consume the same bundle. Hence 5 cars and 3 units of wine. This means that Germany exports 5 cars and imports 3 units of wine, while the opposite is true in France. Because Germany has a population of 8, each person there consumes 1/16 units of cars and 3/8 units of wine. In France (population 6) each person consumes 1/12 units of cars and 1/2 units of wine. Since Germans consume 3/4 the amount that French people consume, their wages are 3/4 as high. That is w G = (3/4)w F. Finally, plugging the consumption amount into the utility functions reveals that the representative consumer in Germany receives utility of (1/16)(3/8) = 3/128, and the representative consumer in France receives a utility of (1/12)(1/2) = 1/24. d) Consider a third country: Spain, which has a labor force of 45 and unit labor requirements a w S = 1 and a c S = 15. Repeat c) for all three countries. Does Spain gain from trade? What about the other two? Start by noting that the Spanish PPF will have a slope of 15. This means that for them in autarky, p c /p w = 15. At these prices, consumers will maximize utility by consuming in the aggregate where W/C = 15, or W = 15C. The only point on Spain s PPF satisfying this condition is W = 22.5 and C = 1.5. At this level of production each of the 45 workers will receive 1/2 units of wine and 1/3 units of cars. This generates a utility of 1/6 for the consumers. Since Spanish people have the same utility function as the French and Germans, the relative demand curve is unchanged. However, the relative supply curve will change to reflect the addition of Spain s production. We see from part c that there are plateaus on the supply curve when the relative price is 4 and. These represent the prices at which Germany and France, respectively, are indifferent to where they produce on their PPFs. Spain is indifferent at a price ratio of 15, so now we add a new plateau. The resulting graph is...
3 Relative Supply and Demand P(c)/P(w) Q(c)/Q(w) Supply Demand The price ratio that clears the world market is p c /p w =. (We will also see this when we solve for the allocation.) At this price ratio, Germany specializes in car production (producing cars), Spain specializes in wine production (producing 45 units of wine) and France is indifferent amongst al the production points on its PPF (all will give France the same GDP). When Germany produces cars, its GDP is () =. When Spain produces 45 units of wine, its GDP is 45(1) = 45. Regardless of where France produces on its PPF, it will have GDP = 6. (It could produce 6 units of wine worth 6, units of cars worth 6 or any combination in between, but regardless, the value of Frances production will be 6.) When we know the wealth of every country, we know what it will consume. Once again, recall that due to the utility function, each consumer will spend half his wealth on cars and half on wine. So the Germans will be able to afford 5/ = 5 cars and 5/1 = 5 units of wine. The French will buy 3/ = 3 cars and 3/1 = 3 units of wine and finally the Spanish will consume 22.5/ = 2.25 cars and 22.5/1 = 22.5 units of wine. If we add up all the consumptions of cars, we see that aggregate car consumption is C = =.25; and aggregate wine production is W = = 2.5. Since Germany is supplying cars and aggregate car demand is.25, it must be the case that France produces.25 cars. Since Spain supplies 45 units of wine and 2.5 are consumed in the aggregate, France is producing 57.5 units of wine. Sure enough, this production point is on the French PPF, so the relative price of is supportable in equilibrium. As a result, Germany exports 5 cars and imports 5 wine, Spain exports 22.5 wine and imports 2.25 cars, and France exports 27.5 wine and imports 2.75 cars. When we divide country consumption by population we find that a German worker consumes 1/16 cars and 5/8 wine; a French worker consumes 1/ cars and 1/2 wine; and a Spaniard will consume (2.25)/45 = 1/ cars and 1/2 wine. Since the Frenchman and the Spaniard consume the same bundle, they must have the same wage. The German consumes 25% more. Hence w F = w S = (.8)w G. Plugging consumption bundles into the utility function reveals that U F = U S = 1/4 and U G = 5/128. Comparing these values to autarky, we see that Germany and Spain gain from trade and France neither gains nor loses. Problem 2: a) List the four main results of the Hechscher-Ohlin model and state each in your own words.. i) Heckscher-Ohlin Result: If two countries differ only in their relative factor endowments and they are opened up to trade freely, each country will export the good whose production process uses relatively more intensively that factor in which the country is relatively more abundant. ii) iii) Rybczynski Effect: If a country experiences an increase in its endowment of one of the factors, production of the good which utilizes that factor relatively more intensively will increase more than proportionately to the increase in the factor endowment, while production of the other good will actually decrease. Stolper-Samuelson Effect: If the market price of one of the goods increases, the return to the factor utilized relatively more intensively in the production of that good will increase more than proportionately to the increase in the output price.
5 PPFs for Hightechnia and Bananarepublic Computers Bananarepublic Hightechnia Bananas b) How does the move to free trade affect the real incomes in terms of bananas and computers of U, S, and L in Hightechnia? What can you say about consumption possibilities of the country as a whole? We know that when we move to free trade the price of computers in Hightechnia will increase and the price of bananas will decrease. Without loss of generality, we can renormalize prices in such a way that the price of computers increases while the price of bananas remains constant. When the price of computers increases, the nominal wages of all factors involved in computer production will initially increase proportionately. But this would mean that Unskilled workers in the computer industry earned more than in the banana industry. As a result more labor will be attracted to the computer industry and wages in the computer industry would fall part of the way back down while those in the banana industry rose. In the end, the wages of all unskilled laborer will have risen less than proportionately to the increase in computer prices. The reduction of wages in the computer industry resulting from the movement in U from B to C will further increase wages for S. Since S has seen its return increase more than proportionately to p C, the purchasing power of S over C has increased. Also since the price of bananas is unchanged S has increased purchasing power over B as well. Hence the real income to S has been unambiguously increased. We saw earlier that nominal wages to U rise less than proportionately to the price increase of C. Hence real wages have fallen in terms of C (unskilled workers can t afford as much in computers as they could in autarky). However, since nominal wages rose and the price of bananas is unchanged, real wages have risen in terms of bananas. The total effect on U is ambiguous. If the representative consumption basket is weighted heavily toward computers then U is worse off under trade. If consumption is weighted toward bananas, U is better off. Nominal returns to L have fallen since the price of bananas is unchanged but the wage to labor has increased. This will unambiguously decrease the real wage of L. The consumption possibilities of the country as a whole have increased, since the country will now be consuming at a point above the PPF. c) Repeat b) for Bananarepublic. The story here will be just the opposite. Now the price of bananas will rise and the price of computers will fall. Without loss of generality, we will renormalize prices so that only the price of bananas changes (it increases). Here real wages for L will be unambiguously increased, real wages for S will be unambiguously decreased and there will be an ambiguous outcome for U. If U s consumption is weighted toward computers, U will be better off, and if consumption is weighted toward bananas U will be worse off. This country too benefits in the aggregate as consumption will lie above the PPF.
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### Economics 181: International Trade Homework # 4 Solutions
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, Fall 2007 Version A Special Codes 00000 PLEDGE: I have neither given nor received unauthorized help on this exam. SIGNED: PRINT NAME: Econ 202 Final Exam 1. On average over the past 50 years, the U.S. | 9,564 | 40,126 | {"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} | 3.71875 | 4 | CC-MAIN-2019-51 | latest | en | 0.957125 |
https://dsp.stackexchange.com/questions/59703/why-wiener-filter-not-iir | 1,675,644,295,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00378.warc.gz | 235,666,131 | 37,993 | Why Wiener filter not IIR
I would like to limit discussion to discrete time causal Wiener filter.
By definition it is an FIR filter, and it also has optimality under Mean Square Error criterion within the class of linear filters.
How do we know that there isn't a better IIR filter (something with a feedback loop), which is also linear?
You're right, there usually is a better IIR filter (if you have enough data). The discrete-time Wiener filter is not "by definition" FIR. It is common to constrain the filter to the FIR case because it's often more straightforward to implement, and because such a filter can be made adaptive more easily. Also, in practice you often want to consider only a finite data window, which makes the optimum Wiener filter FIR.
However, you don't need to impose the constraint that the filter length be finite. The most general causal discrete-time Wiener filter for data available over the infinite past is an IIR filter.
A derivation of the FIR and the IIR cases can be found (among others) in Digital Signal Processing by Proakis and Manolakis.
EDIT: If we only have a finite amount of data available, the optimum Wiener filter has a finite impulse response. Assume we have $$N$$ data points (one current and $$N-1$$ consecutive past data points), then the best we can do with a linear time-invariant filter is to linearly combine those $$N$$ data points, by multiplying each data point by a coefficient and adding them up:
$$y[n]=\sum_{k=0}^{N-1}h[k]x[n-k]\tag{1}$$
The coefficients $$h[k]$$ are optimized such that the mean square error between the filter output $$y[n]$$ and some desired signal is minimized. The $$N$$ coefficients $$h[k]$$ are the impulse response of the corresponding FIR Wiener filter.
• @CowboyTrader: For the IIR Wiener filter the data sequence is assumed to be infinite. Jul 23, 2019 at 12:59
• @CowboyTrader: The book I mentioned is generally a good book on DSP, not only about discrete-time filters. But it does have a lot of good stuff on filters. Jul 23, 2019 at 13:00
• @CowboyTrader: As an IIR filter it's both, so it takes infinitely many past values into account, but it can be expressed recursively. Jul 23, 2019 at 13:12
• @CowboyTrader: If you only have $N$ data points, you can only linearly combine them using $N$ coefficients, so the optimal filter must be FIR. This doesn't say anything about implementation, but as a filter it must have a finite impulse response. Jul 23, 2019 at 13:27
• @CowboyTrader: Using recursion you will either get infinite memory - no use for finite data - or, in the degenerate case, finite memory, which could also be implemented without recursion. Jul 23, 2019 at 13:37 | 649 | 2,682 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-06 | latest | en | 0.935882 |
https://www.justintools.com/unit-conversion/length.php?k1=picometers&k2=centimeters | 1,726,321,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00851.warc.gz | 769,719,217 | 27,659 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# LENGTH Units Conversionpicometers to centimeters
1 Picometers
= 1.0E-10 Centimeters
Category: length
Conversion: Picometers to Centimeters
The base unit for length is meters (SI Unit)
[Picometers] symbol/abbrevation: (pm)
[Centimeters] symbol/abbrevation: (cm)
How to convert Picometers to Centimeters (pm to cm)?
1 pm = 1.0E-10 cm.
1 x 1.0E-10 cm = 1.0E-10 Centimeters.
Always check the results; rounding errors may occur.
Definition:
The picometer (symbol: pm) or picometre (International spelling) is a unit of length in the metric system, equal to 1×10−12 m, or one trillionth (1/1000000000 ..more definition+
Centimeter (symbol: cm), or centimetre (British spelling) is a unit of length in the metric system, equal to one hundredth of a meter, centi being the SI prefix for a ..more definition+
In relation to the base unit of [length] => (meters), 1 Picometers (pm) is equal to 1.0E-12 meters, while 1 Centimeters (cm) = 0.01 meters.
1 Picometers to common length units
1 pm = 1.0E-12 meters (m)
1 pm = 1.0E-15 kilometers (km)
1 pm = 1.0E-10 centimeters (cm)
1 pm = 3.2808398950131E-12 feet (ft)
1 pm = 3.9370078740157E-11 inches (in)
1 pm = 1.0936132983377E-12 yards (yd)
1 pm = 6.2137119223733E-16 miles (mi)
1 pm = 1.056970721911E-28 light years (ly)
1 pm = 3.7795280352161E-9 pixels (PX)
1 pm = 6.25E+22 planck length (pl)
Picometersto Centimeters (table conversion)
1 pm = 1.0E-10 cm
2 pm = 2.0E-10 cm
3 pm = 3.0E-10 cm
4 pm = 4.0E-10 cm
5 pm = 5.0E-10 cm
6 pm = 6.0E-10 cm
7 pm = 7.0E-10 cm
8 pm = 8.0E-10 cm
9 pm = 9.0E-10 cm
10 pm = 1.0E-9 cm
20 pm = 2.0E-9 cm
30 pm = 3.0E-9 cm
40 pm = 4.0E-9 cm
50 pm = 5.0E-9 cm
60 pm = 6.0E-9 cm
70 pm = 7.0E-9 cm
80 pm = 8.0E-9 cm
90 pm = 9.0E-9 cm
100 pm = 1.0E-8 cm
200 pm = 2.0E-8 cm
300 pm = 3.0E-8 cm
400 pm = 4.0E-8 cm
500 pm = 5.0E-8 cm
600 pm = 6.0E-8 cm
700 pm = 7.0E-8 cm
800 pm = 8.0E-8 cm
900 pm = 9.0E-8 cm
1000 pm = 1.0E-7 cm
2000 pm = 2.0E-7 cm
4000 pm = 4.0E-7 cm
5000 pm = 5.0E-7 cm
7500 pm = 7.5E-7 cm
10000 pm = 1.0E-6 cm
25000 pm = 2.5E-6 cm
50000 pm = 5.0E-6 cm
100000 pm = 1.0E-5 cm
1000000 pm = 0.0001 cm
1000000000 pm = 0.1 cm
(Picometers) to (Centimeters) conversions | 1,018 | 2,508 | {"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} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.789026 |
https://www.jakkalsvallei.co.za/44852_kolkata_power_calculation_for_rotary_crusher.html | 1,718,871,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00600.warc.gz | 750,593,911 | 9,852 | (PDF) Static analysis and strength calculation of drive shaft …
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Rock Crushing Rule of Thumb
1. The total cost of a jaw crusher installation underground may exceed six times the cost of the crusher itself (purchased new), while on surface the factor is usually between three and four. Source:...
2. With a typical 6:1 reduction ratio, the power consumption of a large jaw crusher (48 by 60) is approximately 1.8 tons per horsepower-hour (2.2 t/kWh). Source: Arthur Taggart
1. The total cost of a jaw crusher installation underground may exceed six times the cost of the crusher itself (purchased new), while on surface the factor is usually between three and four. Source:...
2. With a typical 6:1 reduction ratio, the power consumption of a large jaw crusher (48 by 60) is approximately 1.8 tons per horsepower-hour (2.2 t/kWh). Source: Arthur Taggart
3. The power consumption of a 42-inch gyratory crusher is approximately 2.4 tons per horsepower-hour (2.9 t/kWh). Source: Arthur Taggart
4. Power consumption of a jaw crusher when idling is about 50% of full load, for a gyratory it is approximately 30%. Source: Richard Taggart
See more
Study of the Energy-Power Parameters of the Crushing …
Using the methodology described above, it was found that the calculated power of the crusher drive DMRiE 14.5×13 at crushing of the fluxed limestone [ 26] is 395 kW, which is 1.2 times less than the power of the motor DAZO 450U-8UI (500 kW) in operation at the moment.
rotary crushers for coal mine
rotary coal crushers australia mining equipment - pbcollege.in. Mining Products, Inc. Astec Mining Group (AMG) products have a rich history that comes from decades of a complete line of comminution and sizing equipment, as well as underground mining Rotary Coal Breakers . work in conjunction with a primary crusher.
Power Calculation For Rotary Crusher
Calculated Power In Jaw Crusher Crusher Mills, Cone. hammer crusher power calculations Grinding Mill China with conventional hammer mill and rotary drum dryer …
Design, Fabrication and Testing of a Double Roll Crusher
The actual capacity of roll crushers is only about 25% of the theoretical value due to voids between particles and loss of speed in gripping the feed particle [4], [5]. Thus the actual capacity, Q = .356 t/hr 2.1.9 Crushing Power The required crushing power is a function of Bond work index, capacity and the reduction ratio.
Drive power calculation of a crushing machine …
Fig. 1. Scheme of a rotary-cone crusher: a – sectional view of the crusher; b – section A-A. The power spent on material crushing is consumed to overcome the moment of force M: М = Р .l, (2) where P – magnitude of the force required to carry out crushing; l= …
Calculated Power In Jaw Crusher
How the the rock crusher motor power is calculated ? – . Rock crushers power P ( kW ) can be calculated as : P = mR2n3jk1/108 × 103η. …. Rock Crusher; Jaw Crusher; Drying Hammer Crusher; Ring Hammer Crusher.
Power Calculation For Rotary Crusher
power calculation for rotary crusher adani group of mines. Calculating Power Requirement For Crushing. 2 hammer mill Wanting to know the way crusher people calculate power demand is like asking NASA for lift off thrust demand"Supercritical steam power plants meet notably the requirements for high The devices most commonly used …
Crushing Machine
Roller Crusher Product Price: Rs 5 Lakh / Unit Get Best Price Product Details: For meeting the versatile requirements of our valuable clients, we are engaged in offering superior quality range of Roller Crusher. This range consists of a pair of cast manganese steel smooth roller which is mounted on a heavy cast iron base frame.
kolkata dynamic load calculation of crusher
Jan 15, 2016 · Refrigeration Load Calculation Software. Load Calculations. Commercial, Industrial. Last Software Update: 30 June 2010. Last Entry Update: 25 August 2018. Ratings . 0. ... CAEPIPE performs linear and non-linear, static and dynamic pipe stress calculations for piping systems of anyplexity, in any industry. Load Calculations. Code ... | 2,421 | 10,875 | {"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-2024-26 | latest | en | 0.835074 |
https://rdrr.io/bioc/qvalue/src/R/lfdr.R | 1,544,852,508,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00140.warc.gz | 740,407,031 | 14,411 | # R/lfdr.R In qvalue: Q-value estimation for false discovery rate control
#### Documented in lfdr
#' @title Estimate local False Discovery Rate (FDR)
#'
#' @description
#' Estimate the local FDR values from p-values.
#'
#' @param p A vector of p-values (only necessary input).
#' @param pi0 Estimated proportion of true null p-values. If NULL, then \code{\link{pi0est}} is called.
#' @param trunc If TRUE, local FDR values >1 are set to 1. Default is TRUE.
#' @param monotone If TRUE, local FDR values are non-decreasing with increasing p-values. Default is TRUE; this is recommended.
#' @param transf Either a "probit" or "logit" transformation is applied to the p-values so that a local FDR estimate can be formed that
#' does not involve edge effects of the [0,1] interval in which the p-values lie.
#' @param adj Numeric value that is applied as a multiple of the smoothing bandwidth used in the density estimation. Default is \code{adj=1.0}.
#' @param eps Numeric value that is threshold for the tails of the empirical p-value distribution. Default is 10^-8.
#'
#' @details It is assumed that null p-values follow a Uniform(0,1) distribution.
#' The estimated proportion of true null hypotheses \eqn{\hat{\pi}_0}{pi_0} is either
#' a user-provided value or the value calculated via \code{\link{pi0est}}.
#' This function works by forming an estimate of the marginal density of the
#' observed p-values, say \eqn{\hat{f}(p)}{f(p)}. Then the local FDR is estimated as
#' \eqn{{\rm lFDR}(p) = \hat{\pi}_0/\hat{f}(p)}{lFDR(p) = pi_0/f(p)}, with
#' adjustments for monotonicity and to guarantee that \eqn{{\rm lFDR}(p) \leq
#' 1}{lFDR(p) <= 1}. See the Storey (2011) reference below for a concise
#' mathematical definition of local FDR.
#'
#' @return
#' A vector of estimated local FDR values, with each entry corresponding to the entries of the input p-value vector \code{p}.
#'
#' @references
#' Efron B, Tibshirani R, Storey JD, and Tisher V. (2001) Empirical Bayes analysis
#' of a microarray experiment. Journal of the American Statistical Association, 96: 1151-1160. \cr
#' \url{http://www.tandfonline.com/doi/abs/10.1198/016214501753382129}
#'
#' Storey JD. (2003) The positive false discovery rate: A Bayesian
#' interpretation and the q-value. Annals of Statistics, 31: 2013-2035. \cr
#' \url{http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.aos/1074290335}
#'
#' Storey JD. (2011) False discovery rates. In \emph{International Encyclopedia of Statistical Science}. \cr
#' \url{http://genomine.org/papers/Storey_FDR_2011.pdf} \cr
#' \url{http://www.springer.com/statistics/book/978-3-642-04897-5}
#'
#' @examples
#' # import data
#' data(hedenfalk)
#' p <- hedenfalk$p #' lfdrVals <- lfdr(p) #' #' # plot local FDR values #' qobj = qvalue(p) #' hist(qobj) #' #' @author John D. Storey #' @seealso \code{\link{qvalue}}, \code{\link{pi0est}}, \code{\link{hist.qvalue}} #' @aliases lfdr #' @keywords local False Discovery Rate, lfdr #' @export lfdr <- function(p, pi0 = NULL, trunc = TRUE, monotone = TRUE, transf = c("probit", "logit"), adj = 1.5, eps = 10 ^ -8, ...) { # Check inputs lfdr_out <- p rm_na <- !is.na(p) p <- p[rm_na] if (min(p) < 0 || max(p) > 1) { stop("P-values not in valid range [0,1].") } else if (is.null(pi0)) { pi0 <- pi0est(p, ...)$pi0
}
n <- length(p)
transf <- match.arg(transf)
# Local FDR method for both probit and logit transformations
if (transf == "probit") {
p <- pmax(p, eps)
p <- pmin(p, 1 - eps)
x <- qnorm(p)
mys <- smooth.spline(x = myd$x, y = myd$y)
y <- predict(mys, x)$y lfdr <- pi0 * dnorm(x) / y } else { x <- log((p + eps) / (1 - p + eps)) myd <- density(x, adjust = adj) mys <- smooth.spline(x = myd$x, y = myd$y) y <- predict(mys, x)$y
dx <- exp(x) / (1 + exp(x)) ^ 2
lfdr <- (pi0 * dx) / y
}
if (trunc) {
lfdr[lfdr > 1] <- 1
}
if (monotone) {
o <- order(p, decreasing = FALSE)
ro <- order(o)
lfdr <- cummax(lfdr[o])[ro]
}
lfdr_out[rm_na] <- lfdr
return(lfdr_out)
}
## Try the qvalue package in your browser
Any scripts or data that you put into this service are public.
qvalue documentation built on Nov. 1, 2018, 4:44 a.m. | 1,330 | 4,143 | {"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} | 2.796875 | 3 | CC-MAIN-2018-51 | latest | en | 0.580496 |
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SAMPLE FINANCIAL RATIOS QUESTIONS AND ANSWERS
Financial Ratios Quiz and Test | AccountingCoach
For multiple-choice and true/false questions, simply press or click on what you think is the correct answer. For fill-in-the-blank questions press or click on the blank space provided. If you have difficulty answering the following questions, learn more about this topic by reading our Financial Ratios
Financial Ratio Multiple Choice Questions | My Accounting
Test your knowledge of the financial ratios with multiple choice questions and quizzes. Test your knowledge of the financial ratios with multiple choice questions and quizzes. Skip to content. Menu. Accounting Topics. Home » Financial Ratio Analysis » Financial Ratios. Correct! Correct. Close me! Incorrect! Incorrect.
Financial Statement Ratios - Practice Test Questions
Financial Statement Ratios Chapter Exam Instructions. Choose your answers to the questions and click 'Next' to see the next set of questions. You can skip questions if you would like and come back
Ratio Analysis: Problems and Solutions | Accounting
The following is the Balance Sheet of a company as on 31st MarchFrom the following particulars found in the Trading, Profit and Loss Account of A Company Ltd., The following is the summarised Profit and Loss Account of Taj Products Ltd. for the year ended From the following Balance Sheet and additional information, you are required to calculate: (i) See all full list on accountingnotes[PDF]
Financial Analysis Question Paper, Answers and Examiners
Question Paper, Answers and Examiners Comments Financial Analysis Questions, Answers and Examiners’ Comments LEVEL 5 DIPLOMA IN CREDIT MANAGEMENT JANUARY 2013 Instructions to candidates Answer some of the basic matters they learned in Level 3 such as ratio analysis.
Quiz & Worksheet - Financial Statement Analysis | Study
Quiz & Worksheet - Financial Statement Analysis Quiz; Financial statement ratios are calculated by using two or more line items from a financial statement and performing a mathematical[PDF]
Chapter 2 Financial Statement and Ratio Analysis
answers to these and other questions. Firm managers use accounting information to help Financial Statement and Ratio Analysis LO1 The Financial Statements 1.2 The Income Statement Financial Statement and Ratio Analysis LO1 The Financial Statements 1.3 Statement of Cash Flows
Top 20 Financial Modeling Interview Questions (With Answers)
What is financial modeling? Why is it useful? Is it only confined to company’s financial affairs? This How do you build a Financial Model? Go through this Financial Modeling in Excel Training to build a What is working capital and how do you forecast it? This is a basic question of finance included in What are the design principles of a good financial model? Another easy Financial Modeling See all full list on wallstreetmojo
Finance Interview Questions - Most Common Questions & Answers
Check out CFI's interview guides with the most common questions and best answers for any corporate finance job position. Interview questions and answer for finance, accounting, investment banking, equity research, commercial banking, FP&A, more! Free guides and practice to ace your interview.[PDF]
Financial Analysis Question Paper, Answers and Examiners
Question Paper, Answers and Examiners Comments Level 5 Diploma www Financial Analysis Questions, Answers and Examiner’s Comments LEVEL 5 DIPLOMA IN CREDIT MANAGEMENT JUNE 2012 Instructions to candidates 3. Using appropriate ratios, comment on the viability of obtaining the rent of £800,000 per
Related searches for sample financial ratios questions and | 718 | 3,684 | {"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-2019-43 | latest | en | 0.865292 |
http://fmwww.bc.edu/repec/bocode/l/lppinv.sthlp | 1,696,331,973,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00683.warc.gz | 23,092,368 | 4,513 | {smcl} {* *! version 1.1.2 30sep2022}{...} {vieweralsosee "[TS] arima" "mansection TS arima"}{...} {vieweralsosee "" "--"}{...} {vieweralsosee "[TS] arima postestimation" "help arima postestimation"}{...} {vieweralsosee "" "--"}{...} {viewerjumpto "Syntax" "lppinv##syntax"}{...} {viewerjumpto "Description" "lppinv##description"}{...} {viewerjumpto "Methods and formulas" "lppinv##methods"}{...} {viewerjumpto "Examples" "lppinv##examples"}{...} {viewerjumpto "Remarks" "lppinv##remarks"}{...} {viewerjumpto "References" "lppinv##references"}{...} {p2colset 1 15 17 2}{...} {title:Title} {phang} {bf:lppinv} {hline 2} solve an under-, over- and identified linear problem without an objective function (a "hybrid" LP-LS problem) with the help of the Moore-Penrose pseudoinverse and singular value decomposition (SVD) and test the normalized RMSE {marker syntax}{...} {title:Syntax} {p 8 17 2} {cmdab:lppinv} {help varlist|matname:{it:varlist|matname}} {ifin} [{cmd:,} {it:options}] {synoptset 35 tabbed}{...} {synopthdr} {synoptline} {syntab:"Hybrid" LS-LP problem type} {synopt:{opt cols}}non-typical constrained OLS ({bf:cOLS}) (see {help lppinv##description:Description}) {p_end} {synopt:{opt tm}}Transaction Matrix ({bf:TM}), options {opt cols} and {opt tm} are mutually exclusive (not specifying any of them equals {bf:custom}) {p_end} {syntab:Constructing the LHS} {synopt:{opth m:odel(varlist|matname)}}the MODEL part of {bf:a} (see {help lppinv##methods:Methods and formulas}) {p_end} {synopt:{opth c:onstraints(varlist|matname)}}the CONSTRAINTS part of {bf:a} {p_end} {synopt:{opth s:lackvars(varlist|matname)}}the SLACK VARIABLES part of {bf:a} {p_end} {synopt:{opt zerod:iagonal}}set all the diagonal elements of {bf:a} to 0 {p_end} {syntab:SVD-based estimation} {synopt:{opth tol:erance(real)}}{helpb [M-1] tolerance:roundoff error}, a number to determine when a number is small enough to be considered zero (optional, not specifying {it:tol} is equivalent to specifying {it:tol}=0){p_end} {synopt:{opth l:evel(#)}} confidence level (by default: {helpb clevel:c(level)}) {syntab:Monte-Carlo-based t-test} {synopt:{opth seed(#)}}random-number seed, # is any number between 0 and 2^31-1 (or 2,147,483,647) (by default:{helpb set_seed: c(rngseed_mt64s)}) {p_end} {synopt:{opth iter:ate(#)}}number of iterations, # must be divisible by 50 (by default: {bf:500}){p_end} {synopt:{opth dist:ribution(string)}}random-variable generating function, name of an earlier declared {helpb m2_ftof:Mata object} returning a {bf:real matrix (r x c)} with two arguments, real scalars {bf:r} and {bf:c} (by default: {bf:lppinv_runiform}, see {help lppinv##examples:Examples} on how to pass {bf:rnormal()} to {cmd:lppinv}, the full list of built-in functions is available {help mf_runiform:here}){p_end} {synopt:{opt nomc}}skip the Monte Carlo-based t-test{p_end} {synopt:{opt notrace}}hide any output with the exception of dots{p_end} {synoptline} {p2colreset}{...} {p 4 6 2} {it:depvar} and {it:indepvars} may contain time-series operators if data are {helpb tsset}; see {help tsvarlist}. {p_end} {p 4 6 2} {opt by}, {opt collect}, {opt fp}, {opt rolling}, {opt statsby}, and {cmd:xi} are allowed; see {help prefix}.{p_end} {marker weight}{...} {p 4 6 2} {opt weight}s are not allowed; see {help weights}. {p_end} {p 4 6 2} See {manhelp regress_postestimation R:regress postestimation} for features available after estimation.{p_end} {marker description}{...} {title:Description} {pstd} The algorithm solves "hybrid" linear programming-least squares (LP-LS) problems with the help of the Moore-Penrose inverse (pseudoinverse), calculated using {help mf_svsolve:singular value decomposition (SVD)}, with emphasis on estimation of non-typical constrained OLS ({bf:cOLS}), Transaction Matrix ({bf:TM}), and {bf:custom} (user-defined) cases. The pseudoinverse offers a unique solution and may be the best linear unbiased estimator (BLUE) for a group of problems under certain conditions, see Albert (1972). Over- and identified problems are accompanied by {helpb regress:regression} analysis, which is feasible in their case. For such and especially all remaining cases, a Monte-Carlo-based {helpb ttest:t-test} of mean {bf:NRMSE} (normalized by standard deviation of the RHS) is performed, the sample being drawn from a uniform or user-provided distribution (via a {help m2_ftof:Mata function}). {pstd} Non-typical constrained OLS ({bf:cOLS}) is based on constraints in model and/or data but not in parameters. Typically, such models are of size ≤ {bf:2N} where {bf:N} is the number of observations (Bolotov, 2014). Furthermore, the number of their parameters may vary in the LHS from row to row (e.g. level vs derivative). {pstd} {bf:Example of a non-typical cOLS problem:} {break}{it:Estimate the trend and the cyclical component of a country's GDP} {it:given the textbook or any other definition of its peaks, troughs, and} {it:saddles.} {pstd} Transaction Matrix ({bf:TM}) of size ({bf:M x N}) is a formal model of interaction between {bf:M} and {bf:N} elements in a system (Bolotov, 2015). For example, {break}{bind: • }an input-output table (IOT) is a type of {bf:TM} where {bf:M = N} and the elements are industries; {break}{bind: • }a matrix of trade/investment/etc. is a type of {bf:TM} where {bf:M = N} and the elements are countries or (macro)regions in which diagonal elements must, in some cases, be equal to zero; {break}{bind: • } a matrix of country/product structure where {bf:M ≠ N} and some elements are known; {break}{bind: }... {pstd} {bf:Example of an TM problem:} {break}{it:Estimate the input-output table or a matrix of trade/investment}, {it:the technical coefficients or (country) shares of which are unknown.} {pstd} {cmd:lppinv} returns matrix {bf:r(solution)}, scalar {bf:r(nrmse)}, and {helpb ttest:t-test} results. In addition, matrix {bf:r(a)} is available with the help of the command: {break}{cmd:. return list, all}. {marker methods}{...} {title:Methods and formulas} {pstd} The problem is written as a matrix equation {bf:a @ x = b} where {bf:a} consists of coefficients for CONSTRAINTS and for SLACK VARIABLES (the upper part) as well as for MODEL (the lower part) as illustrated in Figure 1. Each part of {bf:a} can be omitted to accommodate a special case: {break}{bind: • }{bf:cOLS} problems require no case-specific CONSTRAINTS; {break}{bind: • }{bf:TM} problems require case-specific CONSTRAINTS, no problem CONSTRAINTS, and an optional MODEL; {break}{bind: • }SLACK VARIABLES are non-zero only for inequality constraints and are omitted if problems don't include any; {break}{bind: }... {pstd} {break}{bf:Figure 1: Matrix equation a @ x = b} {break} {bind: }a{bind: } |{bind: }b {break}+–––––––––––––––––––––––––––––––––––––––+–––––––––––––––––+–––––––––––––+ {break}| CONSTRAINTS (PROBLEM + CASE-SPECIFIC) | SLACK VARIABLES | CONSTRAINTS | {break}+–––––––––––––––––––––––––––––––––––––––+–––––––––––––––––+–––––––––––––+ {break}|{bind: }MODEL{bind: } |{bind: }MODEL{bind: }| {break}+–––––––––––––––––––––––––––––––––––––––––––––––––––––––––+–––––––––––––+ {break}Source: self-prepared {pstd} The solution of the equation, {bf:x = pinv(a) @ b}, is estimated with the help of {help mf_svsolve:SVD} and is a {bf:minimum-norm least-squares} {bf:generalized solution} if rank of {bf:a} is not full. To check if {bf:a} is within computational limits, its (maximum) dimensions can be calculated using the formulas: {break}{bind: • }{bf:(2 * N) x (K + K*)}{bind: }{bf:cOLS} without slack variables; {break}{bind: • }{bf:(2 * N) x (K + K* + 1)}{bind: }{bf:cOLS} with slack variables; {break}{bind: • }{bf:(M * N) x (M * N)}{bind: }{bf:TM} without slack variables; {break}{bind: • }{bf:(M * N) x (M * N + 1)}{bind: }{bf:TM} with slack variables; {break}{bind: • }{bf:M x N}{bind: }{bf:custom} without slack variables; {break}{bind: • }{bf:M x (N + 1)}{bind: }{bf:custom} with slack variables; {pstd} where, in {bf:cOLS} problems, {bf:K} is the number of independent variables in the model (including the constant), {bf:K*} ({bf:K*} \not \in {bf:K}) is the number of extra variables in CONSTRAINTS, and {bf:N} is the number of observations; in {bf:TM} problems, {bf:M} and {bf:N} are the dimensions of the transaction matrix; and in custom cases, {bf:M} and {bf:N} or {bf:M x (N + 1)} are the dimensions of {bf:a} (fully user-defined). {marker remarks}{...} {title:Remarks} {pstd} For Python-savy users there is a Python version of {cmd:lppinv} {browse "https://pypi.org/project/lppinv/"} with similar functionality. {marker examples}{...} {title:Examples} cOLS problem: {cmd:. sysuse gnp96.dta, clear} {cmd:. gen correction = runiform()} {cmd:. lppinv gnp96, cols m(time) c(d.gnp96) s(correction)} TM problem (with Monte Carlo t-test based on uniform distribution): {cmd:. clear} {cmd:. set obs 30} {cmd:. gen rowsum = rnormal(15, 100)} {cmd:. gen colsum = rnormal(12, 196)} {cmd:. lppinv rowsum colsum, tm level(90)} {cmd:. lppinv rowsum colsum, tm zerod level(90)} {cmd:. matlist r(solution)} TM problem (with Monte Carlo t-test based on normal distribution): ... {cmd:. mata: function lppinv_normal(r, c) return(rnormal(r,c, 0, 1))} {cmd:. lppinv rowsum colsum, tm level(90) dist(lppinv_normal)} {cmd:. matlist r(solution)} {marker references}{...} {title:References} {phang} Albert, A., 1972. {it:Regression And The Moore-Penrose Pseudoinverse.} New York: Academic Press. {phang} Bolotov, I. 2014. {it:Modelling of Time Series Cyclical Component on a Defined} {it:Set of Stationary Points and its Application on the US Business} {it:Cycle}. [Paper presentation]. The 8th International Days of Statistics and Economics: Prague. {browse "https://msed.vse.cz/msed_2014/article/348-Bolotov-Ilya-paper.pdf"} {phang} Bolotov, I. 2015. {it:Modeling Bilateral Flows in Economics by Means of Exact} {it:Mathematical Methods.} [Paper presentation]. The 9th International Days of Statistics and Economics: Prague. {browse "https://msed.vse.cz/msed_2015/article/111-Bolotov-Ilya-paper.pdf"} | 3,116 | 10,007 | {"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} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.540173 |
https://cs.stackexchange.com/questions/7014/proof-for-p-complete-is-not-closed-under-intersection | 1,716,245,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00250.warc.gz | 163,692,250 | 42,249 | # Proof for P-complete is not closed under intersection
Unfortunately I have no idea how to show this:
Show that the set of ${\sf P}$-complete languages is not closed under intersection.
As far as I understand my lecture notes, ${\sf P}$-completeness is defined as follows:
• $A \subset \Sigma^{*}$ is complete for ${\sf P}$ iff $A \in \text{P}$ and $\forall B \in {\sf P}, B \le_L A$
• $\le_L$ is ${\sf LOGSPACE}$-reduction: for $A,B \subset \Sigma^{*}$, the relation $A \le_{L} B$ is defined by $$A \le_{L} B \quad\text{iff}\quad \exists f \in {\sf FLOGSPACE}, (x \in A \Leftrightarrow f(x) \in B)$$
• This is the reason for quotation. I've searched in the literature (Papadimitriou, Bovet Crescenzi) and also on the internet, but I didn't find something useful. That's why I'm asking for help. Nov 29, 2012 at 12:56
• What reductions are you using? Might they be log-space many-to-one reductions (the reduction model favoured e.g. in Papadimitriou's text)? Nov 29, 2012 at 13:53
• Please accept my apologies for not providing the definition. I thought the definition is uniform. Here it is: $A \subset \Sigma^{*}$ is complete for $\text{P}$ if 1. $A \in \text{P}$ 2. $B \le_L A$ for all $B \in \text{P}$ Nov 29, 2012 at 14:14
• It all depends on the definition of $\le_L$. How is the reduction defined? Nov 29, 2012 at 17:07
• @A.Schulz: These lecture notes are a bit unstructured, but I think the definition above refers to the definition of the $\text{LOGSPACE}$-reduction: Let $A,B \subset \Sigma^{*}. A \le_{L} B: \Leftrightarrow \exists f \in \text{FLOGSPACE}$ with $(x \in A \Leftrightarrow f(x) \in B).$ Nov 29, 2012 at 17:22
Let $A$ be any P-complete problem (say circuit evaluation). Here are two other P-complete problems: $A_0 = \{0x : x \in A \}$ and $A_1 = \{1x : x \in A\}$. The problem $A_0 \cap A_1 = \emptyset$, while definitely in P, isn't P-complete. The latter is true even if we allow Turing reductions, assuming $L \neq P$.
• @A.Schulz: Concerning this answer I have to questions to you or Yuval: 1.) What could be an example for $x$? 2.) I understand why $A_0 \cap A_1 = \emptyset$, but how do I see that this is in $P$? Dec 4, 2012 at 15:40
I think this questions exploits only a technicality. However, you can follow this path
1. The class of ${\sf P}$-complete languages does not contain $\Sigma^*$ and $\emptyset$. This is true for all classes and not only ${\sf P}$, since in order to reduce a language $L_1\neq\emptyset,\Sigma^*$ to another language $L_2$, there has to be at least one element in $L_2$ and at least one element in $\bar L_2$
2. Show that there are two disjoint ${\sf P}$-complete languages (use Yuval's answer).
• This depends on what reduction-class is being considered. Because $\mathsf P$-completeness is somewhat trivial under polynomial-time many-to-one reductions, I would assume for that very reason that he was considering something like log-space many-to-one reductions, in which case if is not known whether $\mathsf L$ is a subset of the $\mathsf P$-complete problems. Nov 29, 2012 at 13:52
• @NieldeBeaudrap: You are right. However, if somebody talks about reduction and the class ${\sf P}$ I usually would expect polytime many-one reductions as the standard reduction. My answer, on the other hand, holds for all many-one reductions, since $\emptyset$ can never be (many-one) complete due to technical reasons. Nov 29, 2012 at 14:10
• As we don't know whether $\mathsf L \ne \mathsf P$, I would say that precisely if $\mathsf P$ is involved, you should ask about whether a stronger notion of reduction is being used. Anyhow, the second part of your answer certainly is true; I just think that you should soften the first part to note that, whether or not "$\mathsf P$-complete = $\mathsf P \smallsetminus \{\Sigma^\ast, \varnothing\}$", it is certainly true that $\varnothing$ is not $\mathsf P$-complete (as this is actually all that your answer requires). Nov 29, 2012 at 15:56
• @A.Schulz: Thank you for your answer, Professor Schulz. Concerning your first aspect: I've provided our definition of $\text{P}$-complete above and as far as I can tell, although it excludes $\Sigma^*$, it doesn't exclude $\emptyset$. So our definition is wrong, isn't it? Concerning your second aspect: I think the answer provided later on by Yuval is that what you meant? Nov 29, 2012 at 15:59
• @Uriel: I extended my answer. Dec 4, 2012 at 12:04 | 1,306 | 4,400 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-22 | latest | en | 0.860926 |
https://timewithai.com/2019/07/18/probability-rules/ | 1,582,600,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00120.warc.gz | 584,714,908 | 58,941 | # Probability Rules
### Basic Probability Rules
In the previous section we considered situations in which all the possible outcomes of a random experiment are equally likely, and learned a simple way to find the probability of any event in this special case. We are now moving on to learn how to find the probability of events in the general case (when the possible outcomes are not necessarily equally likely), using five basic probability rules. Fortunately, these basic rules of probability are very intuitive, and as long as they are applied systematically, they will let us solve more complicated problems; in particular, those problems for which our intuition might be inadequate.
#### Rule 1
For any event A, 0 ≤ P(A) ≤ 1.
This first rule simply reminds us of the basic property of probability that we’ve already learned. The probability of an event, which informs us of the likelihood of it occurring, can range anywhere from 0 (indicating that the event will never occur) to 1 (indicating that the event is certain). One practical use of this rule is that is can be used to identify any probability calculation that comes out to be more than 1 as wrong.
#### Rule 2
P(S) = 1; that is, the sum of the probabilities of all possible outcomes is 1.
# Complement Rule
Let’s move on to rule 3. In probability and in its applications, we are frequently interested in finding out the probability that a certain event will not occur. An important point to understand here is that “event A does not occur” is a separate event that consists of all the outcomes in the sample space S that are not in A. It is for this reason that the event “event A does not occur” is called “the complement event of A,” since it compares event A to the whole sample space. Notation: we will write “not A” to denote the event that A does not occur. Here is a visual representation of how event A and its complement event “not A” together represent the whole sample space.
### Comment
Such a visual display is called a “Venn diagram.” A Venn diagram is a simple way to visualize events and the relationships between them using rectangles and circles. We will use Venn diagrams throughout this module.
Rule 3 deals with the relationship between the probability of an event and the probability of its complement event. Given that event A and event “not A” together make up the whole sample space S, and since rule 2 tells us that P(S) = 1, the following rule should be quite intuitive:
#### Rule 3: The Complement Rule
P(not A) = 1 – P(A); that is, the probability that an event does not occur is 1 minus the probability that it does occur.
We are now moving to rule 4, which deals with another situation of frequent interest, finding P(A or B), the probability of one event or another occurring. Before we get to the actual rule, however, we need some clarifications and definitions.
When a parent says to his or her child in a toy store “Do you want toy A or toy B?”, this means that the child is going to get only one toy and he or she has to choose between them. Getting both toys is usually not an option.
In contrast,
In probability, “OR” means either one or the other or both.
and so,
P(A or B) = P(event A occurs or event B occurs or both occur)
Having said that, it should be noted that there are some cases where it is simply impossible for the two events to both occur at the same time, in which case we don’t have to worry about the possibility that both occur when we try to find P(A or B). The distinction between events that can happen together and those that cannot is an important one.
Consider the following two events:
A—a randomly chosen person has blood type A
B—a randomly chosen person is a woman.
In this case, it is possible for events A and B to occur together.
Definition: Two events that cannot occur at the same time are called disjoint or mutually exclusive. (We will use disjoint.)
We can therefore say that in the first example events A and B are disjoint, and in the second example they are not disjoint. Using Venn diagrams, we can visualize two events that are disjoint and compare them to two events that are not:
The Venn diagrams suggest that another way to think about disjoint versus not disjoint events is that disjoint events do not overlap. They do not share any of the possible outcomes, and therefore cannot happen together. On the other hand, events that are not disjoint are overlapping in the sense that they share some of the possible outcomes and therefore can occur at the same time.
The purpose of the following activity is to strengthen your intuition and understanding about disjoint versus not disjoint events.
# Addition Rule for Disjoint Events
Now that we understand the idea of disjoint events, we can finally get to rule 4. Rule 4 actually has two versions, one for finding P(A or B) in the special case when events A and B are disjoint, and a more general version for when the events are not necessarily disjoint. We will first present the version of rule 4 that is restricted to disjoint events, and later in the section (after rule 5) we will revisit rule 4 and present the more general version.
#### Rule 4: The Addition Rule for Disjoint Events
The Addition Rule for Disjoint Events: If A and B are disjoint events, then P(A or B) = P(A) + P(B).
### Comment
When dealing with probabilities, the word “or” will always be associated with the operation of addition; hence the name of this rule, “The Addition Rule.”
## Example
Recall the blood type example:
Blood TypeOABAB
Probability0.440.420.100.04
* A person with type A can donate blood to a person with type A or AB.
* A person with type B can donate blood to a person with type B or AB.
* A person with type AB can donate blood to a person with type AB only.
* A person with type O blood can donate to anyone.
What is the probability that a randomly chosen person is a potential donor for a person with blood type A?
From the information given, we know that being a potential donor for a person with blood type A means having blood type A or O. We therefore need to find P(A or O). Since the events A and O are disjoint, we can use the addition rule for disjoint events to get: P(A or O) = P(A) + P(O) = 0.42 + 0.44 = 0.86. It is easy to see why adding the probability actually makes sense. If 42% of the population has blood type A and 44% of the population has blood type O, then 42% + 44% = 86% of the population has either blood type A or O, and thus are potential donors to a person with blood type A. This reasoning about why the addition rule makes sense can be visualized using the pie chart below:
# P(A and B) for Independent Events
Rule 4, the addition rule, deals with finding P(A or B). We are now moving on to rule 5, which deals with yet another situation of frequent interest, finding P(A and B), the probability that both events A and B occur. In other words,
P(A and B) = P(event A occurs and event B occurs)
For example, we might be interested in the probability that if two people are chosen at random, both the first has blood type O and the second has blood type O. Since a person with blood type O can donate blood to anyone, this probability might be of particular interest in this context.
Using a Venn diagram, we can visualize “A and B,” which is represented by the overlap between events A and B:
### Comment
There is one special case for which we know what P(A and B) equals without applying any rule.
So, if events A and B are disjoint, then (by definition) P(A and B)= 0. But what if the events are not disjoint?
Recall that rule 4, the Addition Rule, has two versions. One is restricted to disjoint events, which we’ve already covered, and we’ll deal with the more general version later in this module. The same is true of rule 5. Rule 5 has two versions. The version we’ll present here is restricted to a special case that we’ll now discuss, and there is a more general version that we’ll present in the next module.
The version of rule 5 that will be presented here applies to the special case in which the two events are independent of each other.independent(definition)
Two events A and B are said to be independent if the fact that one event has occurred does not affect the probability that the other event will occur. If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.
### Comment
It is quite common for students to initially get confused about the distinction between the idea of disjoint events and the idea of independent events. The purpose of this comment (and the activity that follows it) is to help students develop more understanding about these very different ideas.
The idea of disjoint events is about whether or not it is possible for the events to occur at the same time (see the examples on page 3 of the Probability Rules section).
The idea of independent events is about whether or not the events affect each other in the sense that the occurrence of one event affects the probability of the occurrence of the other (see the examples above).
The following activity deals with the distinction between these concepts.
The purpose of this activity is to help you strengthen your understanding about the concepts of disjoint events and independent events, and the distinction between them.
In each of the following questions, you are presented with a random experiment and two events related to it. You are asked to decide whether the events are disjoint or not, and whether the events are independent or not.
# Multiplication Rule for Independent Events
Now that we understand the idea of independent events, we can finally get to rule 5. As mentioned before, Rule 5 actually has two versions, one for finding P(A and B) in the special case in which the events A and B are independent, and a more general version for use when the events are not necessarily independent. We will first present the version of rule 5 that is restricted to independent events, and in the next section we will revisit Rule 5 and present the more general version.
#### Rule 5: The Multiplication Rule for Independent Events
If A and B are two independent events, then P(A and B) = P(A) * P(B).
### Comment
When dealing with probabilities, the word “and” will always be associated with the operation of multiplication; hence the name of this rule, “The Multiplication Rule.” | 2,333 | 10,458 | {"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} | 4.6875 | 5 | CC-MAIN-2020-10 | latest | en | 0.947322 |
https://semesterpapers.com/2021/04/07/mat-121-college-algebra-written-assignment-5/ | 1,624,369,993,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00061.warc.gz | 468,329,014 | 10,977 | # Mat 121 college algebra written assignment 5
MAT-121: COLLEGE ALGEBRA
Written Assignment 5
2.5 points each
6.1
Algebraic
For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.
1.
2.
For the following exercises, find the formula for an exponential function that passes through the two points given. Round results to 3 decimal places.
1. (0, 21) and (2, 38)
2. (−1, 5) and (4, 2)
For the following exercises, use the compound interest formula, .
1. After a certain number of years, the value of an investment account is represented by the equation . What is the value of the account?
2. What was the initial deposit made to the account in the previous exercise?
3. How many years had the account from the previous exercise been accumulating interest?
Real-World Applications
1. In the year 1980, the median price for a house in New Hampshire was \$95,000. By the year 2000, the median house value had appreciated to \$133,300. What was the annual growth rate between 1980 and 2000 (state as a percent to one decimal place)? Assume that the value continued to grow by the same percentage. What will the median price for a house in New Hampshire in the year 2020? Round to the nearest hundred dollars.
6.2
Graphical
For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept.
For the following exercises, graph and its transformation on the same axes. Give the horizontal asymptote, the domain, and the range.
1.
For the following exercises, start with the graph of . Then write a function that results from the given transformation.
1. Shift f (x) 3 units downward and right 2 units.
2. Shift f (x) 2.25 units up and left 1.5 units.
3. Shift f (x) 4 units right and reflected about the x-axis.
Numeric
For the following exercises, evaluate the exponential functions for the indicated value of x. Round your result to 3 decimal places.
1. for g (5).
6.3
Algebraic
For the following exercises, rewrite each equation in exponential form.
1.
For the following exercises, rewrite each equation in logarithmic form.
1.
For the following exercises, solve for x by converting the logarithmic equation to exponential form.
1.
For the following exercises, use the definition of common and natural logarithms to simplify. Round answer to 4 decimal places
1.
Numeric
For the following exercises, evaluate the base b logarithmic expression without using a calculator.
1.
For the following exercises, evaluate the common logarithmic expression without using a calculator.
1.
For the following exercises, evaluate the natural logarithmic expression without using a calculator.
1.
Technology
For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.
1.
Extensions
1. Is the following true: ? Verify the result.
Real-World Applications
1. The intensity levels of two earthquakes measured on a seismograph can be compared by the formula where M is the magnitude given by the Richter Scale. On April 7th, 2018, an earthquake of magnitude 4.6 hit near Perry OK, USA. On the same day an earthquake of magnitude 6.3 hit near Porgera Papua New Guinea. How many times greater was the intensity of the Papua New Guinea earthquake than the OK earthquake? Round to the nearest whole number. NOTE: Remember that the value of a log is an exponent. So represents the exponent for the base of the common log.
6.4
For the following exercises, state the domain, range, the vertical asymptote, x– and y-intercepts (if they exist, if they do not exist write
DNE), and end behavior of the function.
For the following exercises, sketch the graph of the indicated function.
For the following exercises, use a graphing calculator to find approximate solutions to each equation. If you do not have a graphing calculator then use the following site’s online graphing calculator. You will want to use the “Intersection” function. Round results to 4 decimal places
https://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
Graphical
For the following problem, start with the graph of . Then write a function, g(x), that results from the given transformation. Graph both f(x) and its transformation on the same axes.
1. Shift f (x) 1 unit up, right 2 units, and reflected about the x axis.
6.5
Algebraic
For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
26.
For the following exercises, condense to a single logarithm if possible.
27.
For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
28.
For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.
29. to base 10
For the following exercises, suppose and . Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of m and n. Show the steps for solving.
30.
Numeric
For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.
31.
Extensions
32. Use the product rule for logarithms to find all x values such that . Show the steps for solving.
6.6
Algebraic
For the following exercises, use like bases to solve the exponential equation.
33.
For the following exercises, use logarithms to solve. Round your results to 4 decimal places.
34.
For the following exercises, use the definition of a logarithm to solve the equation.
35.
For the following exercises, use the one-to-one property of logarithms to solve. Give an exact value for the result.
36.
For the following exercises, solve each equation for x.
37.
For the following exercises, solve for the indicated value.
38. An account with an initial deposit of \$10,500 earns 4.25% annual interest, compounded continuously. How long will it take until the account is worth twice the initial deposit? Round your answer to the nearest tenth. Use the formula
6.7
Real-World Applications
1. A doctor prescribes 165 milligrams of a therapeutic drug that decays by about 36% each hour. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours and use the model determine the half-life of the drug, round to the nearest tenth of an hour.
2. The half-life of Plutonium 238 is 87.75 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.
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If you think your paper could be improved, you can request a review. In this case, your paper will be checked by the writer or assigned to an editor. You can use this option as many times as you see fit. This is free because we want you to be completely satisfied with the service offered. | 2,207 | 10,082 | {"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-2021-25 | latest | en | 0.901308 |
https://www.unitconverter.dev/volume/imperialpint/imperialtablespoon/1/ | 1,713,080,946,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00864.warc.gz | 970,397,460 | 18,594 | # 🧮 Volume
## Imperial Pint to Imperial Tablespoon
The volume conversion of 1 imperial pint is 32 imperial tablespoon.
Imperial Pint
to
Imperial Tablespoon
Imperial Pint Imperial Tablespoon
0.01 0.32
0.05 1.6
0.1 3.2
0.25 8
1 32
5 160
10 320
20 640
50 1600
100 3200
### Volume
Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre. The volume of a container is generally understood to be the capacity of the container; i. e., the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. Volumes of complicated shapes can be calculated with integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space. | 299 | 1,235 | {"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} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.849954 |
https://metanumbers.com/1859898000000 | 1,642,597,166,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00645.warc.gz | 467,655,978 | 7,826 | # 1859898000000 (number)
1,859,898,000,000 (one trillion eight hundred fifty-nine billion eight hundred ninety-eight million) is an even thirteen-digits composite number following 1859897999999 and preceding 1859898000001. In scientific notation, it is written as 1.859898 × 1012. The sum of its digits is 48. It has a total of 16 prime factors and 448 positive divisors. There are 493,516,800,000 positive integers (up to 1859898000000) that are relatively prime to 1859898000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 48
• Digital Root 3
## Name
Short name 1 trillion 859 billion 898 million one trillion eight hundred fifty-nine billion eight hundred ninety-eight million
## Notation
Scientific notation 1.859898 × 1012 1.859898 × 1012
## Prime Factorization of 1859898000000
Prime Factorization 27 × 3 × 56 × 239 × 1297
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 9299490 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,859,898,000,000 is 27 × 3 × 56 × 239 × 1297. Since it has a total of 16 prime factors, 1,859,898,000,000 is a composite number.
## Divisors of 1859898000000
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32, 40, 48, 50, 60, 64, 75, 80, 96, 100, 120, 125, 128, 150, 160, 192, 200, 240, 250, 300, 320, 384, 400, 480, 500, 600, 640, 800, 960, 1200, 1600, 1920, 2400, 3200, 4800, 9600 Load all the 448 divisors
448 divisors
Even divisors 392 56 28 28
Total Divisors Sum of Divisors Aliquot Sum τ(n) 448 Total number of the positive divisors of n σ(n) 6.20598e+12 Sum of all the positive divisors of n s(n) 4.34609e+12 Sum of the proper positive divisors of n A(n) 1.38526e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.36378e+06 Returns the nth root of the product of n divisors H(n) 134.263 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,859,898,000,000 can be divided by 448 positive divisors (out of which 392 are even, and 56 are odd). The sum of these divisors (counting 1,859,898,000,000) is 6,205,983,062,400, the average is 138,526,407,64.,285.
## Other Arithmetic Functions (n = 1859898000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 493516800000 Total number of positive integers not greater than n that are coprime to n λ(n) 1927800000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 68321022996 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 493,516,800,000 positive integers (less than 1,859,898,000,000) that are coprime with 1,859,898,000,000. And there are approximately 68,321,022,996 prime numbers less than or equal to 1,859,898,000,000.
## Divisibility of 1859898000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 5 0 3
The number 1,859,898,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1859898000000)
Base System Value
2 Binary 11011000100001010100011110100001010000000
3 Ternary 20120210201111012211112010
4 Quaternary 123010022203310022000
5 Quinary 220433032342000000
6 Senary 3542231512532520
8 Octal 33041243641200
10 Decimal 1859898000000
12 Duodecimal 2605640b1140
20 Vigesimal 3cd0i2a000
36 Base36 nqfb8xhc
## Basic calculations (n = 1859898000000)
### Multiplication
n×y
n×2 3719796000000 5579694000000 7439592000000 9299490000000
### Division
n÷y
n÷2 9.29949e+11 6.19966e+11 4.64974e+11 3.7198e+11
### Exponentiation
ny
n2 3459220570404000000000000 6433797420453258792000000000000000000 11966206954706175120723216000000000000000000000000 22255924382644105694682867991968000000000000000000000000000000
### Nth Root
y√n
2√n 1.36378e+06 12297.9 1167.81 284.379
## 1859898000000 as geometric shapes
### Circle
Diameter 3.7198e+12 1.16861e+13 1.08675e+25
### Sphere
Volume 2.69498e+37 4.34698e+25 1.16861e+13
### Square
Length = n
Perimeter 7.43959e+12 3.45922e+24 2.63029e+12
### Cube
Length = n
Surface area 2.07553e+25 6.4338e+36 3.22144e+12
### Equilateral Triangle
Length = n
Perimeter 5.57969e+12 1.49789e+24 1.61072e+12
### Triangular Pyramid
Length = n
Surface area 5.99155e+24 7.5823e+35 1.5186e+12
## Cryptographic Hash Functions
md5 66dbd884aff5751044dbf396c6c9a772 f0b587ccd4d252bf96ced4922f0c5838fe742e8a f8be42a1aa0c152cab0381e552e8b7d516652b407f08691f24e24a728d794f54 ed1caba5575a144a3a6bb98d21994beba389a49c7a291c701b1cc5e1b1152777c62f7626b1a0a345d250dc0c4cc4dafb4b6d7cc05a94f3e139ed71888a734c24 5000cd482022c63e94e8ef84bcc2b24a629bf105 | 1,919 | 5,067 | {"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-2022-05 | latest | en | 0.748227 |
http://www.slideserve.com/binh/geometry-semester-2-model-problems-ca-essential-standards-part-2 | 1,506,043,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688103.0/warc/CC-MAIN-20170922003402-20170922023402-00533.warc.gz | 580,857,025 | 18,351 | 1 / 36
# Geometry Semester 2 Model Problems (CA Essential Standards) Part 2 - PowerPoint PPT Presentation
Geometry Semester 2 Model Problems (CA Essential Standards) Part 2. PowerPoint by Kenneth C Utt John Kimball High -- TUSD. Model Problem #33. In the diagram shown, m CBD = 95˚. What is the measure of CDB ?. 44 o + 95 o + x o = 180 o 139 o + x o = 180 o x o = 180 o – 139 o
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### Geometry Semester 2 Model Problems (CA Essential Standards) Part 2
PowerPoint by
Kenneth C Utt
John Kimball High -- TUSD
Model Problem #33 Part 2
In the diagram shown, mCBD = 95˚. What is the measure of CDB ?
44o + 95o + xo = 180o
139o + xo = 180o
xo = 180o – 139o
xo = 41o
95o
Model Problem #34 Part 2
In the diagram shown, P is a point on ML. What is the measure of the angle marked X?
180 – (90 + 31) = MPN
180 – 121 = 59o
180 – (59 + 55) = LPQ
180 – 114 = 66o
180 – (90 + 66) = PQL
180 – 156 = 24o
180 – 24 = x = 156o
59o
66o
24o
Model Problem #35 Part 2
If ABC is rotated 90˚clockwise about the origin to form A’B’C’, what would be the coordinates of A’?
A’ =(-1,-3)
B’ = (-4,-4)
C’ = (-3,-2)
Model Problem #36 Part 2
The coordinates of the vertices of ∆JKL, are J(-2,-1), K(1,3), L(4,-3). If ∆JKL is translated 2 units down and 4 units to the right to create , what are the coordinates of the vertices of ∆J’K’L’?
J’ = (-2 + 4,-1 – 2) = (2,-3)
K’ = (1 + 4, 3 – 2) = (5, 1)
L’ = (4 + 4, -3 – 2) = (8,-5)
Model Problem #37 Part 2
If quadrilateral DEFG is reflected across the y-axis, it would create quadrilateral D’E’F’G’. What are the coordinates of point G’ ?
G’ = (6,2)
Model Problem #38 Part 2
A diagram from a proof of the Pythagorean Theorem is shown. Write an equation that represents the area of the entire square in two ways. On the left side, express the area as the product of the length and the width. On the right, represent the sum of the areas of the triangles and the smaller square. Then use the equation to prove the theorem.
(a + b)2
c2 + 4(a ∙ b) ÷ 2
Model Problem #39 Part 2
A right triangle’s hypotenuse has length 11. If one leg has length 6, what is the length of the other leg?
112 = 62 + b2
121 = 36 + b2
85 = b2
√85 = b
Model Problem #40 Part 2
In a basketball game, a player from the home team threw the ball from corner C to a player standing at point E. (E is the midpoint of AD). Then the player at point E threw the ball to a player at corner B. If the court was 80 feet long and 50 feet wide, how far was the ball thrown? (Leave in simplified radical form)
√4100 =
10√41 =
2 ∙10 √41
20√41
502 + 402 = h2
2500 + 1600 = h2
√4100 = h
Model Problem #41 Part 2
In the figure shown, sin A 0.4, cos A 0.5, and tan A 0.9. What is the approximate length of BC?
Sin A = 0.4 = BC/40
(40)(0.4) = BC
16 = BC
Model Problem #42 Part 2
In the figure below, if tan A = 4/3, what are sin A and cos A?
5
4
32 + 42 = h2
9 + 16 = h2
√25 = 5 = h
sin A = 4/5
cos A = 3/5
3
Model Problem #43 Part 2
A ladder is leaned against a wall at an angle of 65° to the ground. How far off the ground does the ladder touch the wall?
sin 65o = ?/30’
0.9 = x/30’
27’ = x
Model Problem #44 Part 2
Triangle JKL is shown in the diagram. Which equation should be used to find the length of LJ?
A sin 28o = LJ/54
B sin 28o = 54/LJ
C cos 28o = LJ/54
D cos 28o = 54/LJ
sin
cos
Model Problem #45 Part 2
On a swing set, on engineer used a support bar that was 20 feet long. If the support bar forms a 70° angle to the ground, how far apart will the support bars be at the base?
sin 70o = 0.94
cos 70o = 0.34
tan 70o = 2.75
sin 70o = 0.94
cos 70o = 0.34
tan 70o = 2.75
cos 70o = Adj/20
6.8 ∙ 2 = 13.6
Model Problem #46 Part 2
In the diagram, mB = 75o and AC = 11.9 in. Which equation could be used to find BC?
A x = 11.9(tan 75o)
B x = 11.9(sin 75o)
C x = 11.9/tan 75o
C x = 11.9/sin 75o
tan 75o = 11.9 / x
tan 75o (x) = 11.9
x = 11.9 / tan 75o
Model Problem #47 Part 2
The right triangle in the diagram has one side with a length of 5√3. What is the length of the side marked x?
A 5
B 15
C 5√6
D 10√3
x
5√3
60o
Model Problem #48 Part 2
In the circle shown, the measure of BC= 60o, and the measure of ABD = 62o. What is the measure of CD?
Arc BD = 124o
124o – 60o = 64o
Model Problem #49 Part 2
In the circle shown, DF and CE are chords intersecting at G. If DG = 9, FG = 4, and EG = 12, what is the length of CG?
9
4 ∙ 9 = x ∙ 12
36 = 12x
3 = x
12
x
4
Model Problem #50 Part 2
In the circle shown, what is the measure of angle 1?
1 = ½(104o – 38o)
½(66o)
33o
Model Problem #51 Part 2
LM is tangent to a circle, whose center is C, at point M. MQ is a diameter. If mQNP = 65˚ and mNPM = 50˚, what is mPMR?
180o -65o = 115o
180o – (115o + 50o)
180o – 165o
15o
90o – 15o
75o
15o
115o
Model Problem #52 Part 2
A square is circumscribed about a circle. What is the ratio of the circumference of the circle to the perimeter of the square?
A ¼
B ½
C 2/
D /4
Square = 4 ∙10 = 40
Circle = 10 = 10
10/40
1/4
10
Model Problem #53 Part 2
A cylinder rolls across a table top for 10 complete revolutions. If the diameter of the base is 6 inches, how far did the cylinder travel? (Leave the answer in terms of ).
10 ∙ ∙ 6 =
60 ins.
Model Problem #54 Part 2
The prism shown has a base in the shape of a right triangle. What is the lateral surface area of the prism?
3 x 6 = 18
4 x 6 = 24
32 + 42 = h2
25 = h2
5 x 6 = 30
18 + 24 + 30 = 72 cm2
Model Problem #55 Part 2
What is the volume of the prism shown?
4 ( 8 + 6 ) ÷ 2 = 28 = Base
28 x 10 = 280 mm3
Model Problem #56 Part 2
A target for a yard game is made with areas that are alternately painted white and gray, as shown in the diagram. The inner circle is white and has a radius of 1 inch. Each of the other three rings has a radius 1 inch more than the ring before it. What is the area of the white portion of the target?
32 - 22 + 12
9 - 4 + 1
6 or 18.84
Model Problem #57 Part 2
A rectangle that is 12 feet wide has a perimeter of 40 feet. What is the area of the rectangle?
40 – (12 + 12) = 16
16 ÷ 2 = 8
8 ∙ 12 = 96 ft2
Model Problem #58 Part 2
Each side of a triangle measures 4 m. What is the area of the triangle? (Leave the answer in simplified radical form).
22 + b2 = 42
4 + b2 = 16
b2 = 12
b = √4 ∙ 3 or 2√3
4 ∙ 2√3 ÷ 2
4√3 m2
Model Problem #59 Part 2
Quadrilateral ABCD is a rhombus. If AC = 10 inches and BD = 8 inches, what is the area of ABCD?
4
10 ÷ 2 = 5 & 8 ÷ 2 = 4
4 ∙ (5 ∙ 4) ÷ 2 =
2 ∙ 20 = 40 in2
5
Model Problem #60 Part 2
The diagram shows a trapezoid with a height of 4 cm. What is the area of the trapezoid?
4 (3 + 9) ÷ 2
2 (3 + 9)
2 ∙ 12 = 24 cm2
Model Problem #61 Part 2
The volume of a right rectangular prism is calculated to be 18 cubic centimeters. If the length, the width, and the height of the prism are all doubled, what would be the volume of the new prism?
2 x 3 x 3 = 18
4 x 6 x 6 = 144 cm3
Model Problem #62 Part 2
The cylinder shown has a height of 4 cm and the diameter of the base is 10 cm. What is the volume of the cylinder?
V = Bh
V = 52 ∙ 4
V = 25 ∙ 4
V = 100
V = 314 cm3
Model Problem #63 Part 2
The pyramid shown has a square base that measures 8 cm on each side. The slant height of the pyramid is 6 cm. What is the surface area of the prism?
8 x 8 = 64
4 x 8 x 6 ÷ 2 = + 96
160 cm2
Model Problem #64 Part 2
Given: AB. What is the first step in constructing the perpendicular bisector to AB?
a. Draw a line segment connecting points E and F.
b. From point C, draw an arc that intersects the line at points A and B.
c. Draw a line segment connecting points A and B.
d. From points A and B, draw equal arcs that intersect at points E and F.
Model Problem #65 Part 2
Darla is constructing an equilateral triangle. Which of the following could be her first step?
A B
C D
Model Problem #66 Part 2
Marsha is using a straightedge and compass to do the construction shown. Which statement best describes the construction Martha is doing?
a. a line through P parallel to line l by constructing two lines perpendicular to the same line
b. a line through P parallel to line l by copying an angle
c. a line through P perpendicular to line l
d. a line through P congruent to line l
Model Problem #67 Part 2
Amina is bisecting an angle. Which of the construction diagrams shown below best represents the beginning of Amina’s construction?
A B
C D | 3,191 | 9,230 | {"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} | 4.0625 | 4 | CC-MAIN-2017-39 | latest | en | 0.774552 |
https://artofproblemsolving.com/wiki/index.php?title=2006_iTest_Problems/Problem_26&oldid=99216 | 1,702,334,659,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00524.warc.gz | 121,704,004 | 11,994 | # 2006 iTest Problems/Problem 26
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A rectangle has area $A$ and perimeter $P$. The largest possible value of $\tfrac A{P^2}$ can be expressed as $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.
## Solution
Let $l$ and $w$ be the length and width of the rectangle, respectively. The area of the rectangle is $lw$, and the perimeter of the rectangle is $2l+2w$. We wish to maximize $\frac{lw}{(2l+2w)^2}$.
By the AM-GM Inequality, $\frac{l+w}{2} \ge \sqrt{lw}$, with equality occurring when $l = w$. Multiply both sides by 4 to get $2l+2w \ge 4\sqrt{lw}$.
Because the length and width of a rectangle is positive, $2l+2w \ge 4\sqrt{lw} > 0$. Thus, squaring both sides would not affect the inequality sign, so $(2l+2w)^2 \ge 16lw$. Finally, since $(2l+2w)^2$ is positive, we can divide both sides by $(2l+2w)^2$ and $16$ to get $\frac{lw}{(2l+2w)^2} \le \frac{1}{16}$. The equality case $l = w$ satisfies the equality statement, so the highest possible ratio is $\frac{1}{16}$. Therefore, $m+n = \boxed{17}$. | 367 | 1,132 | {"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": 24, "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} | 4.5625 | 5 | CC-MAIN-2023-50 | latest | en | 0.846738 |
https://solvedlib.com/n/i-3-7-your-3-x27-y-da-answer-an-ysec-x-0-1-value-lt-y-lt,21055268 | 1,723,121,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00415.warc.gz | 425,504,751 | 18,198 | I 3 7 your 3 'Y)dA answer an ysec?(x), 0 1 value <y< S,osxs4y
Question:
I 3 7 your 3 'Y)dA answer an ysec?(x), 0 1 value <y< S,osxs4y
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Pant BWhat is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at noma incidence? Express your answer in centimeters_ Use 1.33 for the index of refraction of water:View Available Hint(s)AZd
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Figure 1: Problem TLC plate: Notice that the tail only appears to be streaking form the boltom spot:
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[-/2 Points]DETAILSZILLDIFFEQ9 2.6.007.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERUse a numerical solver and Euler's method to obtain four-decimal approximation of the indicated value_ First use h =0.1 and then use h 0.05 y)? , y(0) = 0.4; Y(0.5) y(0.5) (h = 0.1) y(0.5) (h = 0.05)
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 How many years will it take for $510 to grow to$1,014.79if it's invested at 8 percent compounded annually?The number of years it will take for $510 to grow to$1,014.79at 8percent compounded annually is ___ years
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The combined cost of Car X and Car Y is 7250$ post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Intern Joined: 01 May 2017 Posts: 18 Location: United States (IL) Concentration: Technology, Strategy GPA: 3.95 The combined cost of Car X and Car Y is 7250$ [#permalink]
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Updated on: 10 Mar 2018, 11:40
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The combined cost of Car X and Car Y is 7250$. Had the Car X been 25% cheaper, and the Car Y been 20% costlier, Combined cost would have been 6,450$. How much Car X is costlier than Car Y?
A : 500
B : 800
C : 1000
D : 2750
E : 2250
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This discussion does not meet community quality standards. It has been retired.
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Originally posted by prady2231 on 10 Mar 2018, 01:25.
Last edited by prady2231 on 10 Mar 2018, 11:40, edited 2 times in total.
Intern
Joined: 01 May 2017
Posts: 18
Location: United States (IL)
Concentration: Technology, Strategy
GPA: 3.95
Show Tags
10 Mar 2018, 08:21
The combined cost of Car X and Car Y is 7250$. Had the Car X been 25% cheaper, and the Car Y been 20% costlier, Combined cost would have been 6,450$. How much Car X is costlier than Car Y?
A : 500
B : 600
C : 700
D : 800
E : 1000
I am getting $2750 as answer. if car X is 25% cheaper - it's price doesn't become 1/4th -- but rather 3/4th of original price. first equation : $$X+Y = 7250$$ second equation : $$0.75X + 1.20 Y = 6450$$. Solving both for X, Y we have : Y = 6750/3 using first equation and solution of Y to find X - Y ( ANswe r to prompt question of " how much is X costlier than Y) We get $$X - Y = 2750$$ Please let me know what you guy's think. Hope this helped.... _________________ Regards, Gladi “Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back) Senior SC Moderator Joined: 22 May 2016 Posts: 2361 The combined cost of Car X and Car Y is 7250$ [#permalink]
Show Tags
10 Mar 2018, 11:03
[quote="prady2231"]The combined cost of Car X and Car Y is 7250$. Had the Car X been 25% cheaper, and the Car Y been 20% costlier, Combined cost would have been 6,450$. How much Car X is costlier than Car Y?
A : 500
B : 600
C : 700
D : 800
E : 1000[/quote]
[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Gladiator59][b]Gladiator59[/b][/url] , I got $2,750 as well. [EDIT] X + Y = 7250 Y = 7250 - X .75X + 1.2Y = 6450 .75X + 1.2(7250 - X) = 6450 .75X + 8700 - 1.2X = 6450 -0.45X = -$2250
X = 5,000
Y = 7250 - 5,000 = 2250
X - Y = (5,000 - 2,250) = $2,750 Answer E [url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=prady2231][b]prady2231[/b][/url] , Would you please check the question's wording and/or the answers? EDITED: Corrected arithmetic at line (X - Y) Intern Joined: 01 May 2017 Posts: 18 Location: United States (IL) Concentration: Technology, Strategy GPA: 3.95 Re: The combined cost of Car X and Car Y is 7250$ [#permalink]
Show Tags
10 Mar 2018, 11:46
1
Corrected the options. Correct answer is 2750.
VP
Joined: 09 Mar 2016
Posts: 1287
The combined cost of Car X and Car Y is 7250$[#permalink] Show Tags 10 Mar 2018, 12:10 prady2231 wrote: The combined cost of Car X and Car Y is 7250$. Had the Car X been 25% cheaper, and the Car Y been 20% costlier, Combined cost would have been 6,450$. How much Car X is costlier than Car Y? A : 500 B : 800 C : 1000 D : 2750 E : 2250 $$x+y = 7250$$ --> $$y = 7250 -x$$ $$x-0.25x + 1.2y = 6450$$ $$x-0.25x+1.2(7250 -x) = 6450$$ $$x-0.25x+8700-1.2x = 6450$$ $$x-0.25x-1.2x=6450-8700$$ $$-0.45x=-2250$$ $$x = 5000$$ $$7250-5000= 2250$$ so its E looks like the OA was incorrect Bunuel can you pay attention to this discussion shouldn't the correct answer be 2,250 ? niks18 ?:) --== Message from the GMAT Club Team ==-- THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please re-post it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you. The combined cost of Car X and Car Y is 7250$ &nbs [#permalink] 10 Mar 2018, 12:10
Display posts from previous: Sort by | 1,875 | 5,972 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-04 | latest | en | 0.903941 |
https://www.puzzles-world.com/2019/04/kavya-and-sana-with-dolls-puzzle.html | 1,555,786,378,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00437.warc.gz | 825,754,518 | 17,610 | # Kavya and Sana with Dolls puzzle
## Kavya and Sana with Dolls puzzle
While playing with dolls,
Kavya realised that if she had 15 more,
she would own a quarter as many dolls as Sana.
But if Sana owned 20 fewer,
she would still have five times more dolls than Kavya.
How many dolls does Kavya own?
Explanation :
Dolls with Kavya - x
Dolls with Sana - y
x+15=y/4 => y=4x+60
y-20=5x => y=5x+20
=> 4x+60=5x+20
=> x=40 and y=220
So Kavya owns 40 Dolls
and Sana owns 220 Dolls | 171 | 477 | {"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} | 4.03125 | 4 | CC-MAIN-2019-18 | longest | en | 0.970416 |
http://datatreker.com/category/r | 1,560,861,440,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998724.57/warc/CC-MAIN-20190618123355-20190618145355-00001.warc.gz | 47,160,402 | 36,786 | ## Donsker’s invariance principle
Donsker’s invariance principle is a classical extension of the central limit theorem. It says that the path of a discrete random walk converges, if properly rescaled, in distribution to a standard Brownian motion. More details can be found in your favorite text book or on wikipedia. Here just comes an animation:
library(dplyr)
library(gganimate)
library(transformr)
set.seed(42)
n <- 1000 # maximal steps of random walk
n_seq <- seq(from=10, 100, by = 10) # starting phase
n_seq <- c(n_seq, seq(from=100, n, by = 100))
number_iteration <-length(n_seq)
X <- sample(c(-1,1), n, replace=TRUE, prob=c(0.5, 0.5))
S <- cumsum(X) # random walk S_n
W <- numeric(number_iteration*n)
id <- numeric(number_iteration*n)
time <- numeric(number_iteration*n)
for (i in 1:number_iteration){
for (j in (1:n)){
index <- floor((j-1)/n * n_seq[i])
if (index==0) {
W[j+(i-1)*n] <-((((j-1)/n * n_seq[i]) -(index)) *( S[index+1]))/ sqrt(n_seq[i])
}
else {
W[j+(i-1)*n] <- (S[index] + (((j-1)/n * n_seq[i]) -(index)) *( S[index+1]-S[index]))/ sqrt(n_seq[i])
}
}
id[(1+n*(i-1)):(n*i)] <- rep(n_seq[i],n)
time[(1+n*(i-1)):(n*i)] <- (1:n)/n
}
rw <- data.frame(x = time,
y = W,
id =id)
p <- ggplot(rw, aes(x=x, y=y)) +
geom_line()+
ylab(expression(W^{(n)}~ (x)))+
transition_states(
id,
transition_length = 2,
state_length = 1) +
exit_shrink() +
ease_aes('sine-in-out')+
labs(title = "Donsker's invariance principle. n = {closest_state}")
animate(p, nframes=number_iteration*2)
## Brownian motion via Hilbert basis
Brownian motion is a central object of probability theory. One perspective to look at its construction uses Hilbert spaces.
Let be a Hilbert basis equipped with standard scalar product . Hence, every can be written in a unique way as
where and . Let be an i.i.d. sequeence of standard gaussian random variables and consider
Using the independence of we obtain
Brownian motion is then given by (a continuous modification of) , . Hence, “all” we need to do is to find a Hilbert basis. The most known is the following.
#### Haar basis
Here we refer to http://datatreker.com/haar-basis for an introduction.
haar_mother <- function(x){
(x >0 & x <= 0.5) - (x >0.5 & x <= 1)
}
haar <- function(x,j,k){
(2^(j/2) * haar_mother(2^j*x-k))
}
set.seed(42)
j_max <- 10
n_max <- 11 # maximal resolution
delta <- 2^{-n_max}
x <- (0: 2^n_max)/2^n_max
xi_0 <- rnorm(1) # xi corresponding to constant function
xi <- list() # list of random variables xi
for (j in 0:j_max){
xi[[j+1]] <- rep(0, 2^j)
for (k in 0:(2^j-1))
{
xi[[j+1]][k+1] <-rnorm(1)
}
}
# data.frame containing approximation of the Brownian motion
df <- data.frame(x=numeric(),
y=numeric(),
id=numeric())
for (i in x){
y <- c(1,rep(1, i*2^n_max), rep(0, 2^n_max-i*2^n_max))
alpha <- list() # wavelet coefficients for y
bm <- sum(y)/(length(y)+1) * xi_0
for (j in 0:j_max){
alpha[[j+1]] <- rep(0, 2^j)
for (k in 0:(2^j-1))
{
alpha[[j+1]][k+1] <-sum(haar(x,j,k)*y)*delta
bm <- bm + alpha[[j+1]][k+1]*xi[[j+1]][k+1]
}
df_new <- data.frame(x=i,
y=bm,
id=j)
df <- bind_rows(df, df_new)
}
}
#### Animation using gganimate
ggplot(df, aes(x=x, y=y)) +
geom_line()+
transition_states(
id,
transition_length = 2,
state_length = 1
) +
labs(title = "Construction of Brownian motion using Haar basis. Step: {closest_state}") +
ease_aes('sine-in-out')
The result looks the same as with Levy’s construction which is not astonishing. The implementation above is not optimal and is much slower than the previous one.
It might be worth to consider a different Hilbert basis.
#### Trigonometric Hilbert basis
The trigonometric basis is defined as
together with .
set.seed(42)
N_max <- 200 # number of trigonometric basis functions used
n_max <- 11 # maximal resolution
delta <- 2^{-n_max}
x <- (0: 2^n_max)/2^n_max
xi <- rnorm(N_max+1) # list of random variables xi
# data.frame containing approximation of the Brownian motion
df <- data.frame(x=numeric(),
y=numeric(),
id=numeric())
for (i in x){
y <- c(1,rep(1, i*2^n_max), rep(0, 2^n_max-i*2^n_max))
alpha <- numeric() # wavelet coefficients for y
alpha[1] <- mean(y)
bm <- alpha[1] * xi[1]
for (n in 1:N_max){
alpha[n] <-sum(sqrt(2)*cos(x*pi*n)*y)*delta
bm <- bm + alpha[n]*xi[n]
df_new <- data.frame(x=i,
y=bm,
id=n)
df <- bind_rows(df, df_new)}
}
#### Animation using gganimate
p <- ggplot(df, aes(x=x, y=y)) +
geom_line()+
transition_states(
id,
transition_length = 2,
state_length = 1
) +
labs(title = "Construction of Brownian motion using trigonometric basis. Step: {closest_state}") +
ease_aes('sine-in-out')
animate(p, nframes=1000)
The above convergence is very slow. The irregularity properties of the Brownian motion are not yet very visible.
## Haar basis
Hilbert spaces play a prominent role in various fields of mathematics. An orthonormal basis of such a space is called a Hilbert basis. The purpose of this blog is to illustrate a very clasic and basic Hilbert basis – the Haar basis.
Let be a Hilbert basis of equipped with the standard scalar product . Hence, every can be written in a unique way as
where and .
A classic Hilbert basis consists of Haar functions that are supported on . They are defined using the Haar wavelet:
The Haar basis consists then of rescaled (by ) versions of shifted by ,
together with the constant function . (Note: the constant function has to be added since we consider the interval and not ). In R this looks like:
haar_mother <- function(x){
(x >0 & x <= 0.5) - (x >0.5 & x <= 1)
}
haar <- function(x,j,k){
2^(j/2) * haar_mother(2^j*x-k)
}
#### Animation of the Haar basis
j_max <- 3 # maximal depth
n_max <- 10 # resolution of grid
df <- data.frame(x=numeric(),
y=numeric(),
id=numeric())
x <- (1: 2^n_max)/2^n_max
id<-1
for (j in 0:j_max){
for (k in 0:(2^j-1))
{
df_new<- data.frame(x=x,
y=haar(x,j,k),
id=id)
df <- bind_rows(df, df_new)
id <- id+1
}
}
ggplot(df, aes(x=x, y=y)) +
geom_step()+
transition_states(
id,
transition_length = 2,
state_length = 5
) +
labs(title = 'Illustration of Haar basis') +
ease_aes('sine-in-out')
#### Approximation via Haar basis
Now, every function can be written as
The coefficients are called the Haar Wavelet coefficients. Let us calculate them in the discrete setting. We give us a mesh and values and some function :
j_max <- 12
n_max <- 13 # maximal resolution
delta <- 2^{-n_max}
x <- (1: 2^n_max)/2^n_max
y <- x* sin(1/x)
alpha <- list() # list of Haar coefficients
for (j in 0:j_max){
alpha[[j+1]] <- rep(0, 2^j)
for (k in 0:(2^j-1))
{
alpha[[j+1]][k+1] <-sum(haar(x,j,k)*y)*delta # approximation of scalar product
}
}
y_approx <- rep(0, length(y)) # approximated values of y
#### Calculating the approximations for different values of j
df <- data.frame(x=numeric(),
y=numeric(),
id=numeric())
y_approx <- mean(y) # this is the contribution of the constant function
for (j in 0:j_max){
for (k in 0:(2^j-1))
{
y_approx <- y_approx + alpha[[j+1]][k+1]*haar(x,j,k)
}
df_new <- data.frame(x=x,
y=y_approx,
id=j)
df <- bind_rows(df, df_new)
}
## Animation of convergence
ggplot(df, aes(x=x, y=y)) +
geom_step()+
transition_states(
id,
transition_length = 2,
state_length = 1
) +
labs(title = "Haar basis approximation of x sin(1/x). Depth: {closest_state}") +
ease_aes('sine-in-out')
## Lévy’s construction of Brownian motion
Brownian motion is a central object of probability theory. The idea of Lévy’s construction is to construct the Brownian motion step by step on finite sets
of dyadic points. As is dense in the Brownian motion is then obtained as the uniform limit of linear interpolation on .
It is pretty easy to illustrate this construction using R and the package gganimate. We use the same notation as in the proof of Wiener’s theorem given on page 23 in “Brownian motion” by Peter Mörters and Yuval Peres.
library(dplyr)
library(gganimate)
library(transformr)
set.seed(42)
n_max <- 14 # maximal number of steps
D <- (0: 2^n_max)/2^n_max # this is in fact D_n_max
B <- list()
Z <- rnorm(1)
B[[1]] <- c(0, Z/2 + rnorm(1), Z)
for (n in 2:n_max){
B[[n]] <- rep(0, 2^n+1)
index_known <- seq(1,2^n+1, by=2) # indices where values are known from previous steps
B[[n]][index_known] <- B[[n-1]]
index_unknown <- seq(2, 2^n, by=2) # indices where values are not yet defined
for (d in index_unknown){
B[[n]][d] <- 0.5*(B[[n]][d-1]+B[[n]][d+1])+ rnorm(1)*2^(-(n+1)/2)
}
}
## interpolation and transformation in a data.frame
df <- data.frame(time=numeric(),
value=numeric(),
id=numeric())
for (n in 1:n_max){
D_n<-(0: 2^n)/2^n
B_interpol<- approx(D_n, B[[n]], xout = D)\$y # interpolation
df_new <- data.frame(time=D,
value=B_interpol,
id=n)
df <- bind_rows(df, df_new)
}
## animation
ggplot(df, aes(x=time, y=value)) +
geom_line()+
transition_states(
id,
transition_length = 2,
state_length = 1
) +
labs(title = 'Levys construction of Brownian motion. Step: {closest_state}', x = 'time', y = 'position') +
ease_aes('sine-in-out')`
## What are typical football results II?
We continue What are typical football results? The notion of weaker and stronger has not been made precise. Is is true that a team that has say 5 Elo points more than another team is really stronger? What might be an appropriate threshold? A glance at the current Elo ranking might give an indication that teams in within 50 points may be considered as equally strong. But is this true? At which threshold the probabilities of win, draw, lose will change?
## What are typical football results?
In this post we continue our investigation of Who wins the 2018 FIFA World Cup™? and take a first look on historical data of FIFA football matches. These are obtained from the site www.eloratings.net using the wayback machine and some copy-and-paste. Unfortunately, our data set obtained in this way is not complete and we did not obtain data on all FIFA matches in this millennium. However, we were able to retrieve all matches of the FIFA World cup 2018 participants plus the matches of Italy, the Netherlands, and Austria. Yes, Italy and the Netherlands are not qualified, but we still are convinced that these two teams are amongst the strongest teams in the world. We added Austria to pay homage to the country where we spent a lot of quality time.
We will try to answer questions like:
• What is the most probable outcome of a game? [->]
• What is the probability to have a win, a draw or a lose? [->]
• What is the probability that the stronger team wins? [->] And with what result? [->]
The answers to these questions can be found following the links after the questions. Detailed answers can be found below.
## Who wins the 2018 FIFA World Cup™?
In just a few weeks the 2018 FIFA World Cup™ starts. People already discuss passionately who is going to win and how the chances for their teams are. Almost everybody has an intuition, opinion, idea, feeling or whatsoever about the performances of the different nations. There might be a consensus among football experts and fans on the top favorites, e.g. Brazil, Germany, Spain, but more debate on possible underdogs. However, most of these predictions rely on subjective opinions and are very hard if not impossible to quantify. An additional difficulty is the complexity of the tournament, with billions of different outcomes, making it very difficult to obtain accurate guesses of the probabilities of certain events.
How can we make reasonable, objective and quantitative estimates of the outcomes? For example, what is the probability that Brazil, Germany or Spain will win the cup? What are the chances that England will make it to the Round of 16? What are the chances that Brazil beats Germany in the semifinals 7:1?
In this and the following posts, we give quantitative answers to all kind of these questions. This post will start with what we can learn by studying previous matches and tournaments. Once we found some appropriate data we will investigate which models are out there to model an event like the FIFA World Cup. | 3,407 | 11,977 | {"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} | 3.3125 | 3 | CC-MAIN-2019-26 | longest | en | 0.657859 |
http://www.docstoc.com/docs/20247221/Quantum-Mechanics-Example-1-Quantum-mechanics | 1,371,690,354,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709947846/warc/CC-MAIN-20130516131227-00082-ip-10-60-113-184.ec2.internal.warc.gz | 415,160,962 | 15,219 | # Quantum Mechanics Example 1 Quantum mechanics
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``` UNIVERSITY OF CALIFORNIA, SANTA CRUZ
BOARD OF STUDIES IN COMPUTER ENGINEERING
CMPE 240: INTRODUCTION TO LINEAR DYNAMICAL SYSTEMS
Gabriel Hugh Elkaim Fall 2005
Quantum Mechanics Example
• wave function and Schrodinger equation
• discretization
• preservation of probability
• eigenvalues & eigenstates
• example
1 Quantum mechanics
• single particle in interval [0, 1], mass m
• potential V : [0, 1] → R
Ψ : [0, 1] × R+ → C is (complex-valued) wave function
interpretation: |Ψ(x, t)|2 is probability density of particle at position x, time t
1
(so |Ψ(x, t)|2 dx = 1 for all t)
0
evolution of Ψ governed by Schrodinger equation:
h2
¯
h˙
i¯ Ψ = V − 2
x Ψ = HΨ
2m
√
where H is Hamiltonian operator, i = −1
1
2 Discretization
let’s discretize position x into N discrete points, k/N , k = 1, . . . , N
wave function is approximated as vector Ψ(t) ∈ CN
2
x operator is approximated as matrix
−2 1
1 −2 1
2 2
=N 1 −2 1
.. .. ..
. . .
1 −2
2
so w = v means
(vk+1 − vk )/(1/N ) − (vk − vk−1 )(1/N )
wk =
1/N
(which approximates w = ∂ 2 v/∂x2 )
discretized Schrodinger equation is (complex) linear dynamical system
˙
Ψ = (−i/¯ )(V − (¯ /2m)
h h 2
h
)Ψ = (−i/¯ )HΨ
where V is a diagonal matrix with Vkk = V (k/N )
hence we analyze using linear dynamical system theory (with complex vectors & matrices):
˙ h
Ψ = (−i/¯ )HΨ
h
solution of Shrodinger equation: Ψ(t) = e(−i/¯ )tH Ψ(0)
h
(−i/¯ )tH
matrix e propogates wave function forward in time t seconds (backward if t < 0)
3 Preservation of probability
d 2 d ∗
Ψ = ΨΨ
dt dt
= ˙ ˙
Ψ∗ Ψ + Ψ ∗ Ψ
= h h
((−i/¯ )HΨ)∗ Ψ + Ψ∗ ((−i/¯ )HΨ)
= h
(i/¯ )Ψ HΨ + (−i/¯ )Ψ HΨ
∗
h ∗
= 0
(using H = H T ∈ RN ×N )
hence, Ψ(t) 2 is constant; our discretization preserves probability exactly
h
U = e−(i/¯ )tH is unitary, meaning U ∗ U = I
unitary is extension of orthogonal for complex matrix: if U ∈ CN ×N is unitary and
z ∈ CN , then
U z 2 = (U z)∗ (U z) = z ∗ U ∗ U z = z ∗ z = z 2
2
4 Eigenvalues & eigenstates
H is symmetric, so
• its eigenvalues λ1 , . . . , λN are real (λ1 ≤ · · · ≤ λN )
• its eigenvectors v1 , . . . , vN can be chosen to be orthogonal (and real)
h h
from Hv = λv ⇔ (−i/¯ )Hv = (−i/¯ )λv we see:
h
• eigenvectors of (−i/¯ )H are same as eigenvectors of H, i.e., v1 , . . . , vN
h h h
• eigenvalues of (−i/¯ )H are (−i/¯ )λ1 , . . . , (−i/¯ )λN (which are pure imaginary)
• eigenvectors vk are called eigenstates of system
• eigenvalue λk is energy of eigenstate vk
h
• for mode Ψ(t) = e(−i/¯ )λk t vk , probability density
2
h
|Ψm (t)|2 = e(−i/¯ )λk t vk = |vmk |2
doesn’t change with time (vmk is mth entry of vk )
5 Example
Potential Function V (x)
1000
900
800
700
600
500
V
400
300
200
PSfrag replacements
100
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
• potential bump in middle of infinite potential well
¯
• (for this example, we set h = 1, m = 1 . . . )
3
lowest energy eigenfunctions
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
PSfrag replacements
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
• potential V shown as dotted line (scaled to fit plot)
• four eigenstates with lowest energy shown (i.e., v1 , v2 , v3 , v4 )
now let’s look at a trajectory of Ψ, with initial wave function Ψ(0)
• particle near x = 0.2
• with momentum to right (can’t see in plot of |Ψ|2 )
• (expected) kinetic energy half potential bump height
4
0.08
0.06
0.04
0.02
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
0.15
PSfrag replacements
0.1
0.05
0
0 10 20 30 40 50 60 70 80 90 100
eigenstate
• top plot shows initial probability density |Ψ(0)|2
• bottom plot shows |vk Ψ(0)|2 , i.e., resolution of Ψ(0) into eigenstates
∗
time evolution, for t = 0, 40, 80, . . . , 320:
|Ψ(t)|2
PSfrag replacements
x
cf. classical solution:
5
• particle rolls half way up potential bump, stops, then rolls back down
• reverses velocity when it hits the wall at left
(perfectly elastic collision)
• then repeats
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
PSfrag replacements 0.1
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
4
x 10
t
N/2
plot shows probability that particle is in left half of well, i.e., |Ψk (t)|2 , versus time t
k=1
6 MATLAB code
% quantum mechanics example for cmpe240
% -----------------------------------
%
% Symmetric Potential
% define matrix that is discretized
% version of Laplacian, ie, del squared operator
% ----------------------------------------------
n=100; deltax=1/n; x=0:deltax:1;
delsq = toeplitz([-2 1 zeros(1,n-1)]);
%not including 1/deltax^2 term
6
% define potential function
% -------------------------
Vmax=.1; v1=0*ones(1,40); v2=Vmax*sin((0:20)/20*pi).^2;
v3=0*ones(1,40); v=[v1,v2,v3];
do=1; if do
% hamiltonian operator
% --------------------
hbar=1; m=1; H = -hbar^2/(2*m)*delsq + diag(v); A = 1/(i*hbar)*H;
% finding energies of eigenstates
[V,D]=eig(H);
[E,index]=sort(diag(D)); %sort energies
V=V(:,index); %sort eigenfunctions
for m=1:(n+1); %normalizing eigenfunctions
V(:,m)=V(:,m)/norm(V(:,m));
end;
% initial wave fct
% ----------------
% stationary wave packet;
% Psi0 = exp(-(x-0.2).^2/0.1^2);
% moving wave packet;
k=2*pi/0.2;
Psi0=exp(-(x-0.2).^2/0.1^2).*exp(i*k*x); Psi0=conj(Psi0’);
Psi0 = Psi0/norm(Psi0); meanE=Psi0’*H*Psi0; Alpha=V’*Psi0;
absAlpha=abs(Alpha).^2;
fprintf(’Mean energy is %f\n’, meanE);
fprintf(’Vmax is %f \n\n’, Vmax);
% time evolution
% --------------
deltat=40; M=expm(deltat*A); PsiMat=Psi0; Psi=Psi0;
nsteps=500; for z=1:nsteps; Psi=M*Psi; PsiMat=[PsiMat Psi]; end;
PDense=abs(PsiMat).^2; % Probability Density.
PDensel=PDense(1:50,:); Pl=sum(PDensel);
%deltat=200; M=expm(deltat*A);
7
%PsiMat=Psi0; Psi=Psi0;
%nsteps=10;
%for z=1:nsteps; Psi=M*Psi; PsiMat=[PsiMat Psi]; end;
%PDense2=abs(PsiMat).^2; % Probability Density.
end;
% some plots
% ----------
close all; figure(1); plot(x,v);
xlabel(’x’); ylabel(’V’); title(’potential function’);
grid on;
if do
figure(2);
m=4;
for l=1:m
subplot(m,1,l);
plot(x, real(V(:,l)),x,v/Vmax*.3,’:’);
axis([0 1 -.3 .3]);
end;
subplot(m,1,1); title(’ Low energy eigenfunctions’);
subplot(m,1,m); xlabel(’x’);
figure(3);
subplot(2,1,1);
plot(x,PDense(:,1)); grid on;
xlabel(’x’); ylabel(’|Psi0|^2’);
title(’Wavefunction at t=0’);
subplot(2,1,2);
plot(0:100,absAlpha,’o’,0:100,absAlpha);
grid on;
xlabel(’n’);
ylabel(’|Alphasubn|^2’);
title(’|Alphasubn|^2 vs. n’);
figure(4);
m=8;
for l=1:m;
subplot(m,1,l);
plot(x,PDense(:,l),x,v,’:’);
8
axis off;
end;
subplot(m,1,1); title(’Psi at intervals of 40 t.u.’);
subplot(m,1,m); xlabel(’x’);
figure(5);
m=8;
for l=1:m;
subplot(m,1,l);
plot(x,PDense(:,l+8),x,v,’:’);
axis off;
end;
subplot(m,1,1); title(’Psi at intervals of 40 t.u.’);
subplot(m,1,m); xlabel(’x’);
figure(6); plot(0:nsteps,Pl)
grid on; zoom on;
axis([0 nsteps 0 1]);
xlabel(’time’); ylabel(’Pleft’);
title(’Pleft vs. t’);
% momentum operator
% -----------------
%P= hbar/(2*deltax)*toeplitz([0, i,zeros(1,n-1)]);
end;
courtesy of Stephen Boyd @ Stanford University
9
```
Related docs | 3,124 | 7,998 | {"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} | 2.984375 | 3 | CC-MAIN-2013-20 | longest | en | 0.734341 |
https://www.cheatography.com/spoopyy/cheat-sheets/linear-algebra/ | 1,566,696,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00024.warc.gz | 761,439,411 | 15,880 | # Linear Algebra Cheat Sheet by spoopyy
### Basis
A set S is a basis for V if 1. S spans V 2. S is LI.
If S is a basis for V then every vector in V can be written in one and only one way as a linear combo of vectors in S and every set containing more than n vectors is LD.
### Basis Test
1. If S is a LI set of vectors in V, then S is a basis for V 2. If S spans V, then S is a basis for V
### Change of Basis
P[x]_B' = [x]_B [x]_B' = P-1 [x]_B [B B'] -> [ I P-1 ] [B' B ] -> [ I P ]
### Cross Product
if u = u1i + u2j + u3k AND v = v1i + v2j + v3k THEN u x v = (u2v3 - u3v2)i - (u1v3 -u3v1)j + (u1v2 - u2v1)k
### Definition of a Vector Space
u + v is within V u+v = v+u u+(v+w) = (u+v)+w u+0 = u u-u = 0 cu is within V c(u+v) = cu+cv (c+d)u = cu+du c(du) = (cd)u 1*u = u
### Diagonalizable Matrices
A is diagonalizable when A is similar to a diagonal matrix. That is, A is diagonalizable when there exists an invertible matrix P such that P-1AP is a diagonal matrix
### Dot Products Etc.
length/norm ||v|| = sqrt(v_12 +...+ v_n2 ||cv|| = |c| ||v|| v / ||v|| is the unit vector distance d(u,v) = ||u-v|| Dot product u•v = (u_1v_1 +...+ u_nv_n) n cos(theta) = u•v / (||u|| ||v||) u&v are orthagonal when dot(u,v) = 0
### Eigenshit
The scalar lambda(Y) is called an Eigenvalue of A when there is a nonzero vector x such that Ax = Yx. Vector x is an Eigenvector of A corresponding to Y. The set of all eigenvectors with the zero vector is a subspace of Rn called the Eigenspace of Y. 1. Find Eigenvalues: det(YI - A) = 0 2. Find Eigenvectors: (YI - A)x = 0 If A is a triangular matrix then its eigenvalues are on its main diagonal
### Gram-Schmidt Orthonormalization
1. B = {v1, v2, ..., vn} 2. B' = {w1, w2, ..., wn}: w1 = v1 w2 = v2 - projw1v2 w3 = v3 - projw1v3 - projw2v3 wn = vn - ... 3. B'' = {u1, u2, ..., un}: ui = wi/||wi|| B'' is an orthonormal basis for V span(B) = span(B'')
### Important Vector Spaces
Rn C(-inf, +inf) C[a, b] P P_n M_m,n
### Inner Products
||u|| = sqrt d(u,v) = ||u-v|| cos(theta) = / (||u|| ||v||) u&v are orthagonal when = 0 proj_v u = * v
### Kernal
For T:V->W The set of all vectors v in V that satisfies T(v)=0 is the kernal of T. ker(T) is a subspace of v. For T:Rn ->Rm by T(x)=Ax ker(T) = solution space of Ax=0 & Cspace(A) = range(T)
### Linear Combo
v is a linear combo of u_1 ... u_n .
### Linear Independence
a set of vectors S is LI if c1v1 +...+ ckvk = 0 has only the trivial solution. If there are other solutions S is LD. A set S is LI iff one of its vectors can be written as a combo of other S vectors.
### Linear Transformation
V & W are Vspaces. T:V->W is a linear transformation of V into W if: 1. T(u+v) = T(u) = T(v) 2. T(cu) = cT(u)
### Non-Homogeny
If xp is a solution to Ax = b then every solution to the system can be written as x = xp
### Nullity
Nullspace(A) = {x ε Rn : Ax = 0 Nullity(A) = dim(Nullspace(A)) = n - rank(A)
### Orthogonal Sets
Set S in V is orthogonal when every pair of vectors in S is orthogonal. If each vector is a unit vector, then S is orthonormal
### One-to-One and Onto
T is one-to-one iff ker(T) = {0} T is onto iff rank(T) = dim(W) If dim(T) = dim(W) then T is one-to-one iff it is onto
### Rank and Nullity of T
nullity(T) = dim(kernal) rank(T) = dim(range) range(T) + nullity(T) = n (in m_x n) dim(domain) = dim(range) + dim(kernal)
### Rank of a Matrix
Rank(A) = dim(Rspace) = dim(Cspace)
### Similar Matrices
For square matrices A and A' of order n, A' is similar to A when there exits an invertible matrix P such that A' = P-1 AP
### Spanning Sets
S = {v1...vk} is a subset of vector space V. S spans V if every vector in v can be written as a linear combo of vectors in S.
### Test for Subspace
1. u+v are in W 2. cu is in w
2 Pages
//media.cheatography.com/storage/thumb/spoopyy_linear-algebra.750.jpg
PDF (recommended) | 1,422 | 3,924 | {"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} | 3.90625 | 4 | CC-MAIN-2019-35 | longest | en | 0.768735 |
https://xianblog.wordpress.com/tag/forecasting/ | 1,685,951,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00446.warc.gz | 1,142,005,869 | 26,423 | ## understanding elections through statistics [book review]
Posted in Books, Kids, R, Statistics, Travel with tags , , , , , , , , , , , , , , , , , , , , , , , , on October 12, 2020 by xi'an
A book to read most urgently if hoping to take an informed decision by 03 November! Written by a political scientist cum statistician, Ole Forsberg. (If you were thinking of another political scientist cum statistician, he wrote red state blue state a while ago! And is currently forecasting the outcome of the November election for The Economist.)
“I believe [omitting educational level] was the main reason the [Brexit] polls were wrong.”
The first part of the book is about the statistical analysis of opinion polls (assuming their outcome is given, rather than designing them in the first place). And starting with the Scottish independence referendum of 2014. The first chapter covering the cartoon case of simple sampling from a population, with or without replacement, Bayes and non-Bayes. In somewhat too much detail imho given that this is an unrealistic description of poll outcomes. The second chapter expands to stratified sampling (with confusing title [Polling 399] and entry, since it discusses repeated polls that are not processed in said chapter). Mentioning the famous New York Times experiment where five groups of pollsters analysed the same data, making different decisions in adjusting the sample and identifying likely voters, and coming out with a range of five points in the percentage. Starting to get a wee bit more advanced when designing priors for the population proportions. But still studying a weighted average of the voting intentions for each category. Chapter three reaches the challenging task of combining polls, with a 2017 (South) Korea presidential election as an illustration, involving five polls. It includes a solution to handling older polls by proposing a simple linear regression against time. Chapter 4 sums up the challenges of real-life polling by examining the disastrous 2016 Brexit referendum in the UK. Exposing for instance the complicated biases resulting from polling by phone or on-line. The part that weights polling institutes according to quality does not provide any quantitative detail. (And also a weird averaging between the levels of “support for Brexit” and “maybe-support for Brexit”, see Fig. 4.5!) Concluding as quoted above that missing the educational stratification was the cause for missing the shock wave of referendum day is a possible explanation, but the massive difference in turnover between the age groups, itself possibly induced by the reassuring figures of the published polls and predictions, certainly played a role in missing the (terrible) outcome.
“The fabricated results conformed to Benford’s law on first digits, but failed to obey Benford’s law on second digits.” Wikipedia
The second part of this 200 page book is about election analysis, towards testing for fraud. Hence involving the ubiquitous Benford law. Although applied to the leading digit which I do not think should necessarily follow Benford law due to both the varying sizes and the non-uniform political inclinations of the voting districts (of which there are 39 for the 2009 presidential Afghan election illustration, although the book sticks at 34 (p.106)). My impression was that instead lesser digits should be tested. Chapter 4 actually supports the use of the generalised Benford distribution that accounts for differences in turnouts between the electoral districts. But it cannot come up with a real-life election where the B test points out a discrepancy (and hence a potential fraud). Concluding with the author’s doubt [repeated from his PhD thesis] that these Benford tests “are specious at best”, which makes me wonder why spending 20 pages on the topic. The following chapter thus considers other methods, checking for differential [i.e., not-at-random] invalidation by linear and generalised linear regression on the supporting rate in the district. Once again concluding at no evidence of such fraud when analysing the 2010 Côte d’Ivoire elections (that led to civil war). With an extension in Chapter 7 to an account for spatial correlation. The book concludes with an analysis of the Sri Lankan presidential elections between 1994 and 2019, with conclusions of significant differential invalidation in almost every election (even those not including Tamil provinces from the North).
R code is provided and discussed within the text. Some simple mathematical derivations are found, albeit with a huge dose of warnings (“math-heavy”, “harsh beauty”) and excuses (“feel free to skim”, “the math is entirely optional”). Often, one wonders at the relevance of said derivations for the intended audience and the overall purpose of the book. Nonetheless, it provides an interesting entry on (relatively simple) models applied to election data and could certainly be used as an original textbook on modelling aggregated count data, in particular as it should spark the interest of (some) students.
[Disclaimer about potential self-plagiarism: this post or an edited version will eventually appear in my Books Review section in CHANCE.]
## ABC forecasts
Posted in Books, pictures, Statistics with tags , , , , , , , , on January 9, 2018 by xi'an
My friends and co-authors David Frazier, Gael Martin, Brendan McCabe, and Worapree Maneesoonthorn arXived a paper on ABC forecasting at the turn of the year. ABC prediction is a natural extension of ABC inference in that, provided the full conditional of a future observation given past data and parameters is available but the posterior is not, ABC simulations of the parameters induce an approximation of the predictive. The paper thus considers the impact of this extension on the precision of the predictions. And argues that it is possible that this approximation is preferable to running MCMC in some settings. A first interesting result is that using ABC and hence conditioning on an insufficient summary statistic has no asymptotic impact on the resulting prediction, provided Bayesian concentration of the corresponding posterior takes place as in our convergence paper under revision.
“…conditioning inference about θ on η(y) rather than y makes no difference to the probabilistic statements made about [future observations]”
The above result holds both in terms of convergence in total variation and for proper scoring rules. Even though there is always a loss in accuracy in using ABC. Now, one may think this is a direct consequence of our (and others) earlier convergence results, but numerical experiments on standard time series show the distinct feature that, while the [MCMC] posterior and ABC posterior distributions on the parameters clearly differ, the predictives are more or less identical! With a potential speed gain in using ABC, although comparing parallel ABC versus non-parallel MCMC is rather delicate. For instance, a preliminary parallel ABC could be run as a burnin’ step for parallel MCMC, since all chains would then be roughly in the stationary regime. Another interesting outcome of these experiments is a case when the summary statistics produces a non-consistent ABC posterior, but still leads to a very similar predictive, as shown on this graph.This unexpected accuracy in prediction may further be exploited in state space models, towards producing particle algorithms that are greatly accelerated. Of course, an easy objection to this acceleration is that the impact of the approximation is unknown and un-assessed. However, such an acceleration leaves room for multiple implementations, possibly with different sets of summaries, to check for consistency over replicates. | 1,517 | 7,735 | {"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} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.943294 |
https://twiki.esc.auckland.ac.nz/do/view/OpsRes/DVDSupplyChain | 1,721,277,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00144.warc.gz | 523,861,590 | 8,769 | # Case Study: The DVD Supply Chain Problem
## Problem Description
This problem has been adapted from Supply Chain Network Economics: Dynamics of Prices, Flows and Profits by Anna Nagurney (2006)
DVD players are produced in Asia and bought by DVDs-4-U for sale throughout New Zealand.
The cost of producing DVD players comes from the setup cost and the production cost. To set up the manufacturing process costs $5 per player being produced. To produce players using this manufacturing process costs dollars. The cost of performing DVDs-4-U's quality assurance test and then shipping the players to New Zealand is given by for players. The storage of players costs DVDs-4-U . DVDs-4-U estimate it costs customers dollars in petrol to visit their nearest DVDs-4-U retail store and buy DVD players. This gives a unit transaction cost of dollars per player. Their market research shows that the demand curve for the DVD players is given by where is the effective price per player for the customers, i.e., retail price + unit transaction cost. ## Student Tasks 1. The manufacturers in Asia are requesting$45 for each DVD player they send to DVDs-4-U. Given this price, solve the Manufacturer's Problem to find the amount of DVD players they want to ship to DVDs-4-U. Write a management summary of your solution.
What to hand in Your AMPL files for solving the Manufacturer's Problem. Your management summary.
2. Given the $45 cost for importing the DVD players from Task 1, DVDs-4-U have decided to sell them for$85. Given these prices, solve the Retailer's Problem to find the amount of DVD players they want to receive from Asia and sell in New Zealand. Write a management summary of your solution.
What to hand in Your AMPL files for solving the Retailer's Problem Your management summary.
3. Since DVDs-4-U have decided to sell the DVD players for $85, they have estimated the effective price for consumers in New Zealand to be$88. Given this price, solve the Consumer's Problem to find the amount of DVD players that are purchased in New Zealand. Write a management summary of your solution.
What to hand in Your AMPL files for solving the Consumer's Problem. Your management summary.
4. Given your solutions to Tasks 1, 2 and 3, what do you conclude about the prices in this supply chain?
Variational inequalities suggest prices of $43.80 for sending a DVD player from Asia to DVDs-4-U,$80.60 for each player sold in New Zealand and an effective price of \$90.80. How do these prices "fit" in this supply chain? | 544 | 2,523 | {"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} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.956701 |
http://conceptmap.cfapps.io/wikipage?lang=en&name=DLVO_theory | 1,553,440,996,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00150.warc.gz | 39,961,440 | 19,903 | # DLVO theory
The DLVO theory (named after Boris Derjaguin and Lev Landau, Evert Verwey and Theodoor Overbeek) explains the aggregation of aqueous dispersions quantitatively and describes the force between charged surfaces interacting through a liquid medium. It combines the effects of the van der Waals attraction and the electrostatic repulsion due to the so-called double layer of counterions. The electrostatic part of the DLVO interaction is computed in the mean field approximation in the limit of low surface potentials - that is when the potential energy of an elementary charge on the surface is much smaller than the thermal energy scale, ${\displaystyle k_{B}T}$. For two spheres of radius ${\displaystyle a}$ each having a charge ${\displaystyle Z}$ (expressed in units of the elementary charge) separated by a center-to-center distance ${\displaystyle r}$ in a fluid of dielectric constant ${\displaystyle \epsilon _{r}}$ containing a concentration ${\displaystyle n}$ of monovalent ions, the electrostatic potential takes the form of a screened-Coulomb or Yukawa potential,
${\displaystyle \beta U(r)=Z^{2}\lambda _{B}\,\left({\frac {e^{\kappa a}}{1+\kappa a}}\right)^{2}\,{\frac {e^{-\kappa r}}{r}},}$
where ${\displaystyle \lambda _{B}}$ is the Bjerrum length, ${\displaystyle \kappa ^{-1}}$ is the Debye–Hückel screening length, which is given by ${\displaystyle \kappa ^{2}=4\pi \lambda _{B}n}$, and ${\displaystyle \beta ^{-1}=k_{B}T}$ is the thermal energy scale at absolute temperature ${\displaystyle T}$.
## Overview
DLVO theory is a dispersion stabilizing theory in which zeta potential is used to explain that as two particles approach one another their ionic atmospheres begin to overlap and a repulsion force is developed.[1] In this theory, two forces are considered to impact of colloidal stability: Van der Waals forces and Coulombic (Entropic) forces.
The total potential energy is described as the sum of the attraction potential and the repulsion potential. When two particles approach each other, electrostatic repulsion increases and the interference between their electrical double layers increases. Meanwhile, the Van der Waals attraction increases as they get closer. At each distance, the net potential energy of the smaller value is subtracted from the larger value.[2]
The combination of these forces results in a deep attractive well, which is referred to as the primary minimum. At larger distances, the energy profile goes through a maximum energy barrier, and subsequently passes through a shallow minimum, which is referred to as the secondary minimum.[3]
At maximum energy barrier, repulsion is greater than attraction. Particles rebound after interparticle contact, and remain dispersed throughout the medium. The maximum energy needs to be greater than the thermal energy. Otherwise, particles will aggregate due to the attraction potential.[3] The height of the barrier indicates how stable the system is. Since particles have to overcome this barrier in order to aggregate, two particles on a collision course must have sufficient kinetic energy due to their velocity and mass.[2] If the barrier is cleared, then the net interaction is all attractive, and as a result the particles aggregate. This inner region is often referred to as an energy trap since the colloids can be considered to be trapped together by Van der Waals forces.[2]
For a colloidal system, the thermodynamic equilibrium state may be reached when the particles are in deep primary minimum. At primary minimum, attractive forces overpower the repulsive forces at low molecular distances. Particles coagulate and this process is not reversible.[4] However, when the maximum energy barrier is too high to overcome, the colloid particles may stay in the secondary minimum, where particles are held together weaker than the primary minimum.[5] Particles form weak attractions but are easily redispersed. Thus, the adhesion at secondary minimum can be reversible.[6]
## History
In 1923, Debye and Hückel reported the first successful theory for the distribution of charges in ionic solutions.[7] The framework of linearized Debye–Hückel theory subsequently was applied to colloidal dispersions by Levine and Dube[8][9] who found that charged colloidal particles should experience a strong medium-range repulsion and a weaker long-range attraction. This theory did not explain the observed instability of colloidal dispersions against irreversible aggregation in solutions of high ionic strength. In 1941, Derjaguin and Landau introduced a theory for the stability of colloidal dispersions that invoked a fundamental instability driven by strong but short-ranged van der Waals attractions countered by the stabilizing influence of electrostatic repulsions.[10] Seven years later, Verwey and Overbeek independently arrived at the same result.[11] This so-called DLVO theory resolved the failure of the Levine–Dube theory to account for the dependence of colloidal dispersions' stability on the ionic strength of the electrolyte.[12]
## Derivation
DLVO theory is the combined effect of van der Waals and double layer force. For the derivation, different conditions must be taken into account and different equations can be obtained.[13] But some useful assumptions can effectively simplify the process, which are suitable for ordinary conditions. The simplified way to derive it is to add the two parts together.
### van der Waals attraction
van der Waals force is actually the total name of dipole-dipole force, dipole-induced dipole force and dispersion forces,[14] in which dispersion forces are the most important part because they are always present. Assume that the pair potential between two atoms or small molecules is purely attractive and of the form w = -C/rn, where C is a constant for interaction energy, decided by the molecule's property and n = 6 for van der Waals attraction.[15] With another assumption of additivity, the net interaction energy between a molecule and planar surface made up of like molecules will be the sum of the interaction energy between the molecule and every molecule in the surface body.[14] So the net interaction energy for a molecule at a distance D away from the surface will therefore be
${\displaystyle w(D)=-2\pi \,C\rho _{1}\,\int _{z=D}^{z=\infty \,}dz\int _{x=0}^{x=\infty \,}{\frac {xdx}{(z^{2}+x^{2})^{3}}}={\frac {2\pi C\rho _{1}}{4}}\int _{D}^{\infty }{\frac {dz}{z^{4}}}=-{\frac {\pi C\rho _{1}}{6D^{3}}}}$
where
• w(r) is the interaction energy between the molecule and the surface
• ${\displaystyle \rho _{1}}$ is the number density of the surface.
• z is the axis perpendicular with the surface and passes across the molecule. z = D at the point where the molecule is and z = 0 at the surface.
• x is the axis perpendicular with z axis, where x = 0 at the intersection.
Then the interaction energy of a large sphere of radius R and a flat surface can be calculated as
${\displaystyle W(D)=-{\frac {2\pi C\rho _{1}\rho _{2}}{12}}\int _{z=0}^{z=2R}{\frac {(2R-z)zdz}{(D+z)^{3}}}\approx -{\frac {\pi ^{2}C\rho _{1}\rho _{2}R}{6D}}}$
where
• W(D) is the interaction energy between the sphere and the surface.
• ${\displaystyle \rho _{2}}$ is the number density of the sphere
For convenience, Hamaker constant A is given as
${\displaystyle A=\pi ^{2}C\rho _{1}\rho _{2}}$
and the equation will become
${\displaystyle W(D)=-{\frac {AR}{6D}}}$
With a similar method and according to Derjaguin approximation,[16] the van der Waals interaction energy between particles with different shapes can be calculated, such as energy between
two spheres: ${\displaystyle W(D)=-{\frac {A}{6D}}{\frac {R_{1}R_{2}}{(R_{1}+R_{2})}}}$
sphere-surface: ${\displaystyle W(D)=-{\frac {AR}{6D}}}$
Two surfaces: ${\displaystyle W(D)=-{\frac {A}{12\pi D^{2}}}}$ per unit area
### Double layer force
A surface in a liquid may be charged by dissociation of surface groups (e.g. silanol groups for glass or silica surfaces[17]) or by adsorption of charged molecules such as polyelectrolyte from the surrounding solution. This results in the development of a wall surface potential which will attract counterions from the surrounding solution and repel co-ions. In equilibrium, the surface charge is balanced by oppositely charged counterions in solution. The region near the surface of enhanced counterion concentration is called the electrical double layer (EDL). The EDL can be approximated by a sub-division into two regions. Ions in the region closest to the charged wall surface are strongly bound to the surface. This immobile layer is called the Stern or Helmholtz layer. The region adjacent to the Stern layer is called the diffuse layer and contains loosely associated ions that are comparatively mobile. The total electrical double layer due to the formation of the counterion layers results in electrostatic screening of the wall charge and minimizes the Gibbs free energy of EDL formation.
The thickness of the diffuse electric double layer is known as the Debye screening length ${\displaystyle 1/\kappa }$ . At a distance of two Debye screening lengths the electrical potential energy is reduced to 2 percent of the value at the surface wall.
${\displaystyle \kappa ={\sqrt {\sum _{i}{\frac {\rho _{\infty i}e^{2}z_{i}^{2}}{\epsilon _{r}\epsilon _{0}k_{B}T}}}}}$
with unit of m−1 where
• ${\displaystyle \rho _{\infty i}}$ is the number density of ion i in the bulk solution.
• z is the valency of the ion. For example, H+ has a valency of +1 and Ca2+ has a valency of +2.
• ${\displaystyle \varepsilon _{0}}$ is the vacuum permittivity, ${\displaystyle \epsilon _{r}}$ is the relative static permittivity.
• kB is the Boltzmann constant.
The repulsive free energy per unit area between two planar surfaces is shown as
${\displaystyle W={\frac {64k_{B}T\rho _{\infty }\gamma ^{2}}{\kappa }}e^{-\kappa D}}$
where
• ${\displaystyle \gamma }$ is the reduced surface potential
${\displaystyle \gamma =\tanh \left({\frac {ze\psi _{0}}{4kT}}\right)}$
• ${\displaystyle \psi _{0}}$ is the potential on the surface.
The interaction free energy between two spheres of radius R is
${\displaystyle W={\frac {64\pi k_{B}TR\rho _{\infty }\gamma ^{2}}{\kappa ^{2}}}e^{-\kappa D}}$ [18]
Combining the van der Waals interaction energy and the double layer interaction energy, the interaction between two particles or two surfaces in a liquid can be expressed as:
${\displaystyle W\left(D\right)=W(D)_{A}+W(D)_{R}\,}$
where W(D)R is the repulsive interaction energy due to electric repulsion and W(D)A is the attractive interaction energy due to van der Waals interaction.
## Application
Since the 1940s, the DLVO theory has been used to explain phenomena found in colloidal science, adsorption and many other fields. Due to the more recent popularity of nanoparticle research, DLVO theory has become even more popular because it can be used to explain behavior of both material nanoparticles such as fullerene particles and microorganisms.
## Shortcomings
Additional forces beyond the DLVO construct have been reported to also play a major role in determining colloid stability.[19] DLVO theory is not effective in describing ordering processes such as the evolution of colloidal crystals in dilute dispersions with low salt concentrations. It also can not explain the relation between the formation of colloidal crystals and salt concentrations.[20]
## References
1. ^ Jan W. Gooch (2007). Encyclopedic Dictionary of Polymers. p. 318. ISBN 978-1-4419-6246-1.
2. ^ a b c
3. ^ a b
4. ^ "Laboratory of Colloid and Surface Chemistry (LCSC)". www.colloid.ch. Retrieved 2015-12-04.
5. ^ Boström, Deniz; Franks, Ninham. "Extended DLVO theory: Electrostatic and non-electrostatic forces in oxide suspensions". Advances in Colloid and Interface Science. 123 (26).
6. ^ "DLVO Theory - folio". folio.brighton.ac.uk. Retrieved 2015-12-04.
7. ^ Debye, P.; Hückel, E. (1923), "The theory of electrolytes. I. Lowering of freezing point and related phenomena", Physikalische Zeitschrift, 24: 185–206.
8. ^ Levine, S. (1939), "Problems of stability in hydrophobic colloidal solutions I. On the interaction of two colloidal metallic particles. General discussion and applications", Proceedings of the Royal Society of London A, 170 (145): 165, Bibcode:1939RSPSA.170..165L, doi:10.1098/rspa.1939.0024.
9. ^ Levine, S.; Dube, G. P. (1940), "Interaction between two hydrophobic colloidal particles, using the approximate Debye-Huckel theory. I. General properties", Transactions of the Faraday Society, 35: 1125–1141, doi:10.1039/tf9393501125.
10. ^ Derjaguin, B.; Landau, L. (1941), "Theory of the stability of strongly charged lyophobic sols and of the adhesion of strongly charged particles in solutions of electrolytes", Acta Physico Chemica URSS, 14: 633.
11. ^ Verwey, E. J. W.; Overbeek, J. Th. G. (1948), Theory of the stability of lyophobic colloids, Amsterdam: Elsevier.
12. ^ Russel, W. B.; Saville, D. A.; Schowalter, W. R. (1989), Colloidal Dispersions, New York: Cambridge University Press.
13. ^ M. Elimelech, J. Gregory, X. Jia, R. A. Williams, Particle Deposition and Aggregation Measurement: Modelling and Simulation (Boston: 1995).
14. ^ a b Jacob N. Israelacvili, Intermolecular and Surface Forces (London 2007).
15. ^ London, F. (1937), Trans Faraday Soc, 33, 8–26.
16. ^ Derjaguin B. V. (1934)Kolloid Zeits 69, 155–164.
17. ^ Behrens, S. H. and Grier, D. G., "The charge on glass and silica surfaces," Journal of Chemical Physics 115, 6716–6721 (2001)
18. ^ Bhattacharjee, S.; Elimelech, M.; Borkovec, Michal (1998), "DLVO interaction between colloidal particles: Beyond Derjaguins approximation", Croatica Chimca Acta, 71: 883–903.
19. ^ Grasso, D., Subramaniam, K., Butkus,M., K Strevett, Bergendahl, J. "A review of non-DLVO interactions in environmental colloidal systems," Reviews in Environmental Science and Biotechnology 1 (1), 17–38
20. ^ N. Ise and I. S. Sogami, Structure Formation in Solution: Ionic Polymers and Colloidal Particles, (Springer, New York, 2005). | 3,592 | 14,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "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} | 2.671875 | 3 | CC-MAIN-2019-13 | latest | en | 0.857635 |
http://encyclopedia.kids.net.au/page/bl/Blackbody | 1,718,297,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00306.warc.gz | 9,259,721 | 4,865 | ## Encyclopedia > Blackbody
Article Content
# Blackbody
In physics a blackbody is an object that absorbs all light that falls onto it (and thus reflects none). Despite the name, blackbodies do radiate light, but they just don't reflect any. The spectrum (amount of light emitted at each wavelength) of a blackbody is very characteristic, and depends entirely on its temperature.
• The light emitted by a blackbody is called blackbody radiation.
• The differences between an object's spectrum and that of an idealized blackbody are sufficient to allow one to determine the chemical composition of the object.
A blackbody is an ideal emitter which radiates energy at the maximum possible rate per unit area at each wavelength for any given temperature. A blackbody also absorbs all the radiant energy incident on it: i.e. no energy is reflected or transmitted.
The term "black body" was introduced by Gustav Kirchhoff in 1862. The spectrum of a blackbody was first derived by Max Planck, who had to assume that electromagnetic radiation could propagate only on discrete packets, or quanta.
The intensity of radiation from a blackbody at temperature T is given by the Planck's Law of Radiation:
$I(\nu) = \frac{2h\nu^{3}}{c^2}\frac{1}{\exp\left(\frac{h\nu}{kT}\right)-1}$
where I(ν)δν is the amount of energy per unit surface per unit time per unit solid angle emitted in the frequency range between ν and ν+δν; h is Planck's constant, c is the speed of light and k is Boltzmann's constant.
The wavelength at which the radiation is strongest is given by Wien's law, and the overall power emitted per unit area is given by the Stefan-Boltzmann law.
In the laboratory, the closest thing to a blackbody is a small hole to a cavity with a non-smooth, black surface. In astronomy, such objects as stars and planets are frequently regarded as blackbodies, though this may be a bad approximation. An almost perfect blackbody spectrum is exhibited by the cosmic microwave background radiation.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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Explorer ... at the South Pole, first to navigate the Northwest Passage in a single ship Roy Chapman Andrews, (1884-1960), US explorer B Vasco Nunez de Balboa, (c. 1475-1519), ... | 534 | 2,373 | {"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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.93013 |
https://www.tutorialspoint.com/Graham-Scan-Algorithm | 1,657,092,716,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00660.warc.gz | 1,112,457,853 | 9,962 | # Graham Scan Algorithm
Data StructureAlgorithmsMisc Algorithms
The convex hull is the minimum closed area which can cover all given data points.
Graham’s Scan algorithm will find the corner points of the convex hull. In this algorithm, at first, the lowest point is chosen. That point is the starting point of the convex hull. Remaining n-1 vertices are sorted based on the anti-clockwise direction from the start point. If two or more points are forming the same angle, then remove all points of the same angle except the farthest point from start.
From the remaining points, push them into the stack. And remove items from stack one by one, when orientation is not anti-clockwise for stack top point, second top point and newly selected point points[i], after checking, insert points[i] into the stack.
## Input and Output
Input:
Set of points: {(-7,8), (-4,6), (2,6), (6,4), (8,6), (7,-2), (4,-6), (8,-7),(0,0), (3,-2),(6,-10),(0,-6),(-9,-5),(-8,-2),(-8,0),(-10,3),(-2,2),(-10,4)}
Output:
Boundary points of convex hull are:
(-9, -5) (-10, 3) (-10, 4) (-7, 8) (8, 6) (8, -7) (6, -10)
## Algorithm
findConvexHull(points, n)
Input − The set of points, number of points.
Output − The boundary points of the convex hull.
Begin
minY := points[0].y
min := 0
for i := 1 to n-1 do
y := points[i].y
if y < minY or minY = y and points[i].x < points[min].x, then
minY := points[i].y
min := i
done
swap points[0] and points[min]
p0 := points[0]
sort points from points[1] to end
arrSize := 1
for i := 1 to n, do
when i < n-1 and (p0, points[i], points[i+1]) are collinear, do
i := i + 1
done
points[arrSize] := points[i]
arrSize := arrSize + 1
done
if arrSize < 3, then
return cHullPoints
push points[0] into stack
push points[1] into stack
push points[2] into stack
for i := 3 to arrSize, do
while top of stack, item below the top and points[i] is not in
anticlockwise rotation, do
delete top element from stack
done
push points[i] into stack
done
while stack is not empty, do
item stack top element into cHullPoints
pop from stack
done
End
## Example
#include<iostream>
#include<stack>
#include<algorithm>
#include<vector>
using namespace std;
struct point { //define points for 2d plane
int x, y;
};
point p0; //used to another two points
point secondTop(stack<point>&stk) {
point tempPoint = stk.top(); stk.pop();
point res = stk.top(); //get the second top element
stk.push(tempPoint); //push previous top again
return res;
}
int squaredDist(point p1, point p2) {
return ((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}
int direction(point a, point b, point c) {
int val = (b.y-a.y)*(c.x-b.x)-(b.x-a.x)*(c.y-b.y);
if (val == 0)
return 0; //colinear
else if(val < 0)
return 2; //anti-clockwise direction
return 1; //clockwise direction
}
int comp(const void *point1, const void*point2) {
point *p1 = (point*)point1;
point *p2 = (point*)point2;
int dir = direction(p0, *p1, *p2);
if(dir == 0)
return (squaredDist(p0, *p2) >= squaredDist(p0, *p1))?-1 : 1;
return (dir==2)? -1 : 1;
}
vector<point>findConvexHull(point points[], int n) {
vector<point> convexHullPoints;
int minY = points[0].y, min = 0;
for(int i = 1; i<n; i++) {
int y = points[i].y;
//find bottom most or left most point
if((y < minY) || (minY == y) && points[i].x < points[min].x) {
minY = points[i].y;
min = i;
}
}
swap(points[0], points[min]); //swap min point to 0th location
p0 = points[0];
qsort(&points[1], n-1, sizeof(point), comp); //sort points from 1 place to end
int arrSize = 1; //used to locate items in modified array
for(int i = 1; i<n; i++) {
//when the angle of ith and (i+1)th elements are same, remove points
while(i < n-1 && direction(p0, points[i], points[i+1]) == 0)
i++;
points[arrSize] = points[i];
arrSize++;
}
if(arrSize < 3)
return convexHullPoints; //there must be at least 3 points, return empty list.
//create a stack and add first three points in the stack
stack<point> stk;
stk.push(points[0]); stk.push(points[1]); stk.push(points[2]);
for(int i = 3; i<arrSize; i++) { //for remaining vertices
while(direction(secondTop(stk), stk.top(), points[i]) != 2)
stk.pop(); //when top, second top and ith point are not making left turn, remove point
stk.push(points[i]);
}
while(!stk.empty()) {
stk.pop();
}
}
int main() {
point points[] = {{-7,8},{-4,6},{2,6},{6,4},{8,6},{7,-2},{4,-6},{8,-7},{0,0},
{3,-2},{6,-10},{0,-6},{-9,-5},{-8,-2},{-8,0},{-10,3},{-2,2},{-10,4}};
int n = 18;
vector<point> result;
result = findConvexHull(points, n);
cout << "Boundary points of convex hull are: "<<endl;
vector<point>::iterator it;
for(it = result.begin(); it!=result.end(); it++)
cout << "(" << it->x << ", " <<it->y <<") ";
}
## Output
Boundary points of convex hull are:
(-9, -5) (-10, 3) (-10, 4) (-7, 8) (8, 6) (8, -7) (6, -10)
Updated on 17-Jun-2020 09:26:16 | 1,589 | 4,825 | {"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} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.810762 |
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1. You read the following information about the economy:Real GDP up three percent from a year ago.Unemployment rate of 6.1 percent.Consumer Price Index up six percent from a year ago.Index of Leading Indicators up for the last six months.Prime interest rate of ten percent, up from seven percent a year ago. a. Explain what each of these economic indicators measures...
• #### 1. Suppose there are only 3 different goods produced and consumed in a closed economy. The...
1. Suppose there are only 3 different goods produced and consumed in a closed economy. The following table shows the prices and quantities of each good consumed in 2002, 2003, 2004, and 2005. Price of a Quantity of Price Price of Quantity pair of a pair of of Quantity Year Apple of Apple shoes Breed of Breed 3001000 1500 shoes...
• #### Using the data in the table below, calculate the CPI and the inflation rate in each...
Using the data in the table below, calculate the CPI and the inflation rate in each year using 2005 as the base year. Instructions: Round your answers to one decimal place. Year Price of basket (\$) CPI Inflation rate 2005 20,000 — 2006 21,500 2007 22,800 2008 26,150 2009 28,825 2010 32,700
• #### Using the data in the table below, calculate the CPI and the inflation rate in each...
Using the data in the table below, calculate the CPI and the inflation rate in each year, using 2010 as a base year. Instructions: Round your answers to one decimal place. Year Cost of basket () 2010 2011 2012 2013 2014 2015 CPI Inflation rate 20,000 21,600 22,800 26,150 28,845 32,500
• #### Using the data in the table below, calculate the CPI and the inflation rate in each...
Using the data in the table below, calculate the CPI and the inflation rate in each year, using 2010 as a base year. Instructions: Round your answers to one decimal place. CPI Inflation rate Year 2010 2011 2012 2013 cost of basket (\$) 18,000 21,500 22,500 26,150 28,850 32,500 2014 2015
• #### Imagine you are asked to help in the calculation of the cost of living and inflation in a particular country. (show the formulas) The basket of goods and services to estimate the CPl in this count...
Imagine you are asked to help in the calculation of the cost of living and inflation in a particular country. (show the formulas) The basket of goods and services to estimate the CPl in this country is composed by 7kg of pasta and 3lt of wine BASKET of GOODS Price of pasta (S/kg) Price of wine (S/it) 2018 2017 2016...
• #### 24. Use the following data to compute the real GDP for 2017 using 2010 as the...
24. Use the following data to compute the real GDP for 2017 using 2010 as the base year 2010 2017 Item Price Quantity Price Quantity Catfish Pizza Taco S25 20 \$12 15 20 a. \$351. b. \$245 c. \$265 d. \$197. e. \$220 25. When there is a sharp increase in oil prices, the overall inflation rate is most likely...
• #### Based on the table below, calculate nominal GDP, real GDP, the GDP deflator, and the inflation...
Based on the table below, calculate nominal GDP, real GDP, the GDP deflator, and the inflation rate in each year and fill in the missing parts of the table. Use 2014 as the base year. Instructions: Round nominal and real GDP values to two decimal places. Round GDP deflator and inflation rate values to the nearest whole number. Price of...
• #### Use the table below on the Canadian Consumer Price Index (CPI) to answer the following questions....
Use the table below on the Canadian Consumer Price Index (CPI) to answer the following questions. The base year is 2002. (1 mark each) YEAR 2014 2015 2016 2017 2018 CPI 125.2 126.6 128.4 130.4 133.4 136.0 2019 1. How much is the CPI in 2002? 2. Which year had the highest inflation rate? What was the inflation rate in...
Free Homework App | 1,102 | 4,351 | {"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-2020-45 | latest | en | 0.895568 |
https://darkskiesfilm.com/what-is-an-example-of-a-means-ends-analysis/ | 1,680,132,018,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00085.warc.gz | 237,753,795 | 11,722 | # What is an example of A means-ends analysis?
## What is an example of A means-ends analysis?
In means-ends analysis, the problem solver begins by envisioning the end, or ultimate goal, and then determines the best strategy for attaining the goal in his current situation. If, for example, one wished to drive from New York to Boston in the minimum time possible, then,…
## What is mean end analysis in artificial intelligence?
Means end analysis is a technique used to solve problems in AI programs. This technique combines forward and backward strategies to solve complex problems. With these mixed strategies, complex problems can be tackled first, followed by smaller ones.
What is the means end analysis technique?
Means-ends analysis is a problem solving strategy that arose from the work on problem solving of Newell and Simon (1972). In means-ends analysis, one solves a problem by considering the obstacles that stand between the initial problem state and the goal state.
Who created means end analysis?
Means-End Analysis is essentially an early form of Gap Analysis, which we commonly use today . It was created by Allen Newell and Herbert Simon in the late 1950s, and published in their 1972 book, “Human Problem Solving.”
### What is the meaning of means ends?
Definition of a means to an end : something done only to produce a desired result Computers have always been a means to an end for [George] Lucas: telling stories.—
### What is a means ends test?
Means-end scrutiny is an analytical process involving examination. of the purposes (ends) which conduct is designed to serve and the meth- ods (means) chosen to further those purposes. When government action.
What is the problem space of means-end analysis?
Explanation: The problem space of means-end analysis has an initial state and one or more goal states.
What is the problem space of mean end analysis?
What is the problem space of means-end analysis? Explanation: The problem space of means-end analysis has an initial state and one or more goal states. Explanation: An algorithm A is admissible if It is guaranteed to return an optimal solution when one exists.
#### Why is Means-End Theory important?
The means-end theory sustains that the way consumers relate to products can be represented by a hierarchical model of three interconnected levels: product attributes, consequences of use and personal values (Copetti, 2005; Dibley & Baker, 2001; Grunert et al., 1995; Kaminski & Prado, 2005; Leão & Mello, 2001, 2002.
#### Is a means rather than an end?
The idiom “A means to an end” differentiates between an end goal and the means or methods and actions used to reach that goal. If I wanted to get a job, a means to that end might be writing my resume.
When to use means-end analysis?
Means–ends analysis (MEA) is a problem solving technique used commonly in artificial intelligence (AI) for limiting search in AI programs. It is also a technique used at least since the 1950s as a creativity tool, most frequently mentioned in engineering books on design methods.
What means-end relationship?
‘Means-end relationships denotes the actual causal relationship between x and y. Sentences of the type (ME2) occur in two versions. On the one hand, (ME2) is used of performed means behaviors and of given end. states/behaviors and of the relationship between these. (
## What is best first search algorithm in AI?
The best first search uses the concept of a priority queue and heuristic search. It is a search algorithm that works on a specific rule. The aim is to reach the goal from the initial state via the shortest path.
What is the means-end model?
The means-end chain model developed by Gutman (1982) sustains the supposition that values are dominant factors in the pattern of consumer purchase; that these consider products or services based on the function of satisfying values; that all actions have consequences (desired or not); and, finally, that there is a …
What is a means to end?
### What is an example of means-end behavior?
Examples of means-end tasks that infants learn to perform in the first years of life include removing a cover to retrieve a hidden object (Diamond, 1985; Piaget, 1953; Shinskey & Munakata, 2003), pulling a cloth to retrieve a distant object supported on the cloth (Munakata, McClelland, Johnson, & Siegler, 1997; … | 943 | 4,364 | {"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-2023-14 | latest | en | 0.950601 |
http://touchnerds.com/standard-error/standard-error-interpretation.html | 1,516,737,810,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892238.78/warc/CC-MAIN-20180123191341-20180123211341-00723.warc.gz | 324,749,657 | 5,240 | Home > Standard Error > Standard Error Interpretation
# Standard Error Interpretation
## Contents
If the standard error of the mean is 0.011, then the population mean number of bedsores will fall approximately between 0.04 and -0.0016. Means ±1 standard error of 100 random samples (N=20) from a population with a parametric mean of 5 (horizontal line). What the standard error gives in particular is an indication of the likely accuracy of the sample mean as compared with the population mean. Taken together with such measures as effect size, p-value and sample size, the effect size can be a very useful tool to the researcher who seeks to understand the reliability and Check This Out
Application of biological variation – a review Što treba znati kada izračunavamo koeficijent korelacije? Biochemia Medica 2008;18(1):7-13. The SPSS ANOVA command does not automatically provide a report of the Eta-square statistic, but the researcher can obtain the Eta-square as an optional test on the ANOVA menu. Usually, a larger standard deviation will result in a larger standard error of the mean and a less precise estimate.
## Standard Error Interpretation
And n equals 10, it's not going to be a perfect normal distribution, but it's going to be close. The standard error, or standard error of the mean, of multiple samples is the standard deviation of the sample means, and thus gives a measure of their spread. Normally when they talk about sample size, they're talking about n.
But then, the trade off is, you need to carry more information.23.6k Views · View UpvotesRelated QuestionsMore Answers BelowA graph is bimodal and is skewed to the right. And we saw that just by experimenting. Note the similarity of the formula for σest to the formula for σ.  It turns out that σest is the standard deviation of the errors of prediction (each Y - Standard Error Of The Mean Definition And let's see if it's 1.87.
Now let's look at this. Standard Error Formula We keep doing that. When the S.E.est is large, one would expect to see many of the observed values far away from the regression line as in Figures 1 and 2. Figure 1. Let's see if it conforms to our formula.
For examples, see the central tendency web page. Difference Between Standard Error And Standard Deviation Now, to show that this is the variance of our sampling distribution of our sample mean, we'll write it right here. Standard error. I'll show you that on the simulation app probably later in this video.
## Standard Error Formula
Allison PD. When n was equal to 16-- just doing the experiment, doing a bunch of trials and averaging and doing all the thing-- we got the standard deviation of the sampling distribution Standard Error Interpretation Statistical Methods in Education and Psychology. 3rd ed. Standard Error Vs Standard Deviation So let's see if this works out for these two things.
In fact, the level of probability selected for the study (typically P < 0.05) is an estimate of the probability of the mean falling within that interval. his comment is here Accessed September 10, 2007. 4. What significance does it have?If the standard deviation values of the two sets of data are 2.8 and 2.0, do the data disperse close or far from the mean?How do you Why are you given the average age and not the age of each person separately? Standard Error Regression
1. So it's going to be a very low standard deviation.
2. Standard error functions more as a way to determine the accuracy of the sample or the accuracy of multiple samples by analyzing deviation within the means.
3. Thus 68% of all sample means will be within one standard error of the population mean (and 95% within two standard errors).
4. Coefficient of determination The great value of the coefficient of determination is that through use of the Pearson R statistic and the standard error of the estimate, the researcher can
5. But anyway, hopefully this makes everything clear.
6. mean, or more simply as SEM.
7. In this way, the standard error of a statistic is related to the significance level of the finding.
8. Accessed: October 3, 2007 Related Articles The role of statistical reviewer in biomedical scientific journal Risk reduction statistics Selecting and interpreting diagnostic tests Clinical evaluation of medical tests: still a long
They don't want you to stare blankly at 10 different values. And if we did it with an even larger sample size-- let me do that in a different color. Recall that the regression line is the line that minimizes the sum of squared deviations of prediction (also called the sum of squares error). this contact form However, one is left with the question of how accurate are predictions based on the regression?
For some statistics, however, the associated effect size statistic is not available. What Is A Good Standard Error The standard error of the mean estimates the variability between samples whereas the standard deviation measures the variability within a single sample. You can see that in Graph A, the points are closer to the line than they are in Graph B.
## Assume the data in Table 1 are the data from a population of five X, Y pairs.
Statistics and probability Sampling distributionsSample meansCentral limit theoremSampling distribution of the sample meanSampling distribution of the sample mean 2Standard error of the meanSampling distribution example problemConfidence interval 1Difference of sample means It represents the standard deviation of the mean within a dataset. I...How do I find the standard deviation of a combined data set?Can I find the standard deviation over a day from monthly data?What does standard deviation higher than mean of a Standard Error Of Proportion In that case, the statistic provides no information about the location of the population parameter.
How to calculate the standard error Spreadsheet The descriptive statistics spreadsheet calculates the standard error of the mean for up to 1000 observations, using the function =STDEV(Ys)/SQRT(COUNT(Ys)). As long as you report one of them, plus the sample size (N), anyone who needs to can calculate the other one. In a town of 10 households, one has an income of \$1,000,000 USD and the other 9 make \$30,000 USD. navigate here So if I know the standard deviation-- so this is my standard deviation of just my original probability density function.
H. 1979. Large S.E. In the second case, all of them are exactly on mean and in the third case, the values are moderately separated away. Now, this is going to be a true distribution.
Means ±1 standard error of 100 random samples (n=3) from a population with a parametric mean of 5 (horizontal line). There's no point in reporting both standard error of the mean and standard deviation. | 1,450 | 6,741 | {"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} | 3.40625 | 3 | CC-MAIN-2018-05 | longest | en | 0.884881 |
https://puzzling.stackexchange.com/questions/48070/an-untimely-rebus-for-your-puzzlement?noredirect=1 | 1,722,975,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00288.warc.gz | 376,938,612 | 43,738 | # An Untimely Rebus For Your Puzzlement
Now with eye-popping graphics! (Hey, it's my first rebus.)
Text version:
$$\begin{matrix} light-speed & & & & here\\ & & ? & & \\ \end{matrix}$$
• Is the ? part of the rebus? Commented Jan 16, 2017 at 22:19
• @boboquack Yes, the question mark is a part of the rebus. Sorry for the slow response. Posted just before going to bed. Commented Jan 17, 2017 at 5:09
• untimely, hmm Commented Jan 17, 2017 at 10:40
Could it be a
$corpse$
$c$ (speed of light) + $or$ (deciding which ?) + $pse$ (here = pse site)
The title: An untimely rebus may refer to
- late which has a meaning as a deceased body, or theatrical slang for ill-timed stage behaviour
- the same answer 'corpse' to one of the OP's questions posted the same day - very suspicious!
• Great! You might say the second puzzle rose from the ashes of the first ;) Commented Jan 18, 2017 at 9:26
Could it be
CLEFT?
Explanation:
Light speed is usually denoted by $c$, and the fact that "Here" is on the right indicates that that's our frame of reference, i.e. c is on the left, giving "cleft".
• Nice, but the question mark is at the heart of the solution, I'm afraid. Commented Jan 17, 2017 at 19:26
• @HughMeyers Now I get what you meant by "at the heart of the solution". :) Commented Jan 18, 2017 at 15:12
• @Ankoganit ... as in "slightly to the left of center". ;) Commented Jan 18, 2017 at 15:16
• @HughMeyers You might say it filled the CLEFT in the middle of the solution ;-) Commented Jan 18, 2017 at 15:17
Is it -
Future .(as time-travel is there)--thinking about future--
As-
"Here" is front of "speed of light" , so you have crossed speed of light thus you are in traveling to future i.e time travel
Rayovac, a famous battery brand
Because
What is untimely light speed? Absent time its just distance, an infinite ray. A ray to here over a stick figure with an empty question? The ennui! It's a ray over a vacuous individual or, to stretch it, 'rayovac'.
Perhaps
Nothing to see here
From
speed of light $c$ = see
nothing between it and here
• Another good guess. The answer, however, is a single word. Commented Jan 17, 2017 at 17:27 | 646 | 2,159 | {"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": 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} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.915395 |
https://www.jiskha.com/display.cgi?id=1356434863 | 1,516,498,204,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00602.warc.gz | 909,320,426 | 4,134 | # college algebra
posted by .
Solve the following inequality.
x^3+9x^2-108 less or equal to 0.
Then write your solution in interval notation. I am sooo confused and do not have any idea how to do this. Can someone please show me step by step please???
Thanks in advance. Someone who is confused.
• college algebra -
This polynomial graphing might help sometimes:
http://mathportal.org/calculators/polynomials-solvers/polynomial-graphing-calculator.php
zeros at x = -6 (double zero) and x = +3
so we really have
(x+6)(x+6)(x-3)
It is negative for large negative x
it comes up and bounces off the x axis at x=-6, dropping down negative again
it does not actually come up and cross the x axis until x = 3
from then on it is positive,
therefore negative for
x</= +3 except for touching zero at x = -6
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solve the following inequality. write your solution in interval notation. x^3+9x^2-108 less than or equal to 0 please show all work | 597 | 2,225 | {"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} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.888475 |
https://socratic.org/questions/how-do-you-verify-the-identity-2sec-2x-2sec-2xsin-2x-sin-2x-cos-2x-1 | 1,582,838,627,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00332.warc.gz | 544,151,716 | 6,164 | # How do you verify the identity 2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x=1?
Jan 19, 2017
Let's do a little bit of factoring.
$2 {\sec}^{2} x \left(1 - {\sin}^{2} x\right) - {\sin}^{2} x - {\cos}^{2} x = 1$
Use the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$:
$2 {\sec}^{2} x \left({\cos}^{2} x\right) - {\sin}^{2} x - {\cos}^{2} x = 1$
Secant and cosine are inverses; their product is $1$.
$2 - {\sin}^{2} x - {\cos}^{2} x = 1$
You will want to convert all to sine or all to cosine, using the pythagorean identity given above.
$2 - \left(1 - {\cos}^{2} x\right) - {\cos}^{2} x = 1$
$2 - 1 + {\cos}^{2} x - {\cos}^{2} x = 1$
$1 + {\cos}^{2} x - {\cos}^{2} x = 1$
$1 = 1$
$L H S = R H S$
The identity is proved.
Hopefully this helps! | 326 | 751 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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} | 4.34375 | 4 | CC-MAIN-2020-10 | latest | en | 0.568837 |
https://www.true-telecom.com/what-is-nutation/ | 1,722,694,926,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00804.warc.gz | 823,002,167 | 13,672 | # What is nutation?
## What is nutation?
Nutation (from Latin nūtātiō ‘nodding, swaying’) is a rocking, swaying, or nodding motion in the axis of rotation of a largely axially symmetric object, such as a gyroscope, planet, or bullet in flight, or as an intended behaviour of a mechanism.
What is nutation in ballistics?
Nutation refers to small circular movement at the bullet tip. Yaw and precession decrease as the distance of the bullet from the barrel increases. Example of yaw.
What is nutation and precession?
Precession is the slow, toplike wobbling of the spinning Earth, with a period of about 25,772 years. Nutation (Latin nutare, “to nod”) superimposes a small oscillation, with a period of 18.6 years and an amplitude of 9.2 seconds of arc, upon this great slow movement.
### What is sacral nutation?
The movement of the sacrum between the ilia involves a nodding motion, known as nutation, which creates an anterior motion of the sacral promontory. Counternutation is the return to the neutral start position from a nutated position as well as a posterior motion of the sacral promontory.
Is nutation anterior pelvic tilt?
Nutation is defined as a relative anterior tilt of the sacral base (upper flat surface of the sacrum that articulates with L5) in relation to the ilium (Figure 1). It is described as a relative movement because it can occur with the sacrum rotating anteriorly, the ilium rotating posteriorly, (or both).
Why does nutation exist?
The nutation (etymologically a ‘nodding’) of a fast-spinning gyroscope reveals itself as (small) vibration and shivering of the precessing axis. Nutation is caused by a possible small deviation of the vector of own angular momentum from the axis of symmetry.
## Why is nutation and precession important?
An example of precession and nutation is the variation over time of the orientation of the axis of rotation of the Earth. This is important because the most commonly used frame of reference for measurement of the positions of astronomical objects is the Earth’s equator — the so-called equatorial coordinate system.
Is Nutation a extension?
Normal Sacral Flexion/Extension Starting from the neutral position, lumbar extension (backward bending) results in sacral flexion (nutation), while lumbar flexion (foreward bending) results in sacral extension (counternutation).
Is the sacrum a joint?
Sacroiliac joints They’re made up of the sacrum — the bony structure above your tailbone and below your lower vertebrae — and the top part (ilium) of your pelvis. There are sacroiliac joints in both the right and left sides of your lower back. Strong ligaments hold these joints in place.
### Is nutation a extension?
Begin typing your search term above and press enter to search. Press ESC to cancel. | 639 | 2,783 | {"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-2024-33 | latest | en | 0.909295 |
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# Some B1 Putnam Problems
## 18.S34, Fall 2009
(1995) For a partition of {1, 2, 3, 4, 5, 6, 7, 8, 9}, let (x) be the number of
elements in the part containing x. Prove that for any two partitions
and 0 , there are two distinct numbers x and y in {1, 2, 3, 4, 5, 6, 7, 8, 9}
such that (x) = (y) and 0 (x) = 0 (y). [A partition of a set S is a
collection of disjoint subsets (parts) whose union is S.]
(1996) Define a selfish set to be a set which has its own cardinality (number
of elements) as an element. Find, with proof, the number of subsets of
{1, 2, . . . , n} which are minimal selfish sets, that is, selfish sets none of
whose proper subsets is selfish.
(1997) Let {x} denote the distance between the real number x and the nearest
integer. For each positive integer n, evaluate
Fn =
6n1
X
min({
m=1
m
m
}, { }).
6n
3n
## (Here min(a, b) denotes the minimum of a and b.)
(1998) Find the minimum value of
(x + 1/x)6 (x6 + 1/x6 ) 2
(x + 1/x)3 + (x3 + 1/x3 )
for x > 0.
(1999) Right triangle ABC has right angle at C and BAC = ; the point D
is chosen on AB so that |AC| = |AD| = 1; the point E is chosen on
BC so that CDE = . The perpendicular to BC at E meets AB at
F . Evaluate lim0 |EF |.
(2000) Let aj , bj , cj be integers for 1 j N . Assume for each j, at least
one of aj , bj , cj is odd. Show that there exist integers r, s, t such that
raj + sbj + tcj is odd for at least 4N/7 values of j, 1 j N .
## (2001) Let n be an even positive integer. Write the numbers 1, 2, . . . , n2 in the
squares of an n n grid so that the k-th row, from left to right, is
(k 1)n + 1, (k 1)n + 2, . . . , (k 1)n + n.
Color the squares of the grid so that half of the squares in each row and
in each column are red and the other half are black (a checkerboard
coloring is one possibility). Prove that for each coloring, the sum of
the numbers on the red squares is equal to the sum of the numbers on
the black squares.
(2002) Shanille OKeal shoots free throws on a basketball court. She hits the
first and misses the second, and thereafter the probability that she hits
the next shot is equal to the proportion of shots she has hit so far.
What is the probability she hits exactly 50 of her first 100 shots?
(2003) Do there exist polynomials a(x), b(x), c(y), d(y) such that
1 + xy + x2 y 2 = a(x)c(y) + b(x)d(y)
holds identically?
(2004) Let P (x) = cn xn + cn1 xn1 + + c0 be a polynomial with integer
coefficients. Suppose that r is a rational number such that P (r) = 0.
Show that the n numbers
cn r, cn r 2 + cn1 r, cn r 3 + cn1 r 2 + cn2 r,
. . . , cn r n + cn1 r n1 + + c1 r
are integers.
(2005) Find a nonzero polynomial P (x, y) such that P (bac, b2ac) = 0 for all
real numbers a. (Note: bc is the greatest integer less than or equal to
.)
(2006) Show that the curve x3 +3xy +y 3 = 1 contains only one set of three distinct points, A, B, and C, which are vertices of an equilateral triangle,
and find its area.
(2007) Let f be a polynomial with positive integer coefficients. Prove that if n
is a positive integer, then f (n) divides f (f (n) + 1) if and only if n = 1.
[Editors note: one must assume f is nonconstant.]
2
(2008) What is the maximum number of rational points that can lie on a circle
in R2 whose center is not a rational point? (A rational point is a point
both of whose coordinates are rational numbers.) | 1,099 | 3,366 | {"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-2019-43 | latest | en | 0.862292 |
https://slideplayer.com/slide/6963277/ | 1,631,984,289,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00581.warc.gz | 571,289,981 | 27,088 | # Chapter 22 Heat Engines, Entropy and the Second Law of Thermodynamics.
## Presentation on theme: "Chapter 22 Heat Engines, Entropy and the Second Law of Thermodynamics."— Presentation transcript:
Chapter 22 Heat Engines, Entropy and the Second Law of Thermodynamics
First Law of Thermodynamics – Review The first law is a statement of Conservation of Energy The first law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both
First Law – Missing Pieces Only certain types of energy-conversion and energy-transfer processes actually take place in nature The first law makes no distinction between processes that occur spontaneously and those that do not An example is that it is impossible to design a device that takes in energy and converts it all to energy
The Second Law of Thermodynamics Establishes which processes do and which do not occur Some processes can occur in either direction according to the first law They are observed to occur only in one direction This directionality is governed by the second law
Irreversible Processes An irreversible process is one that occurs naturally in one direction only No irreversible process has been observed to run backwards An important engineering implication is the limited efficiency of heat engines
Heat Engine A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process
Heat Engine, cont. The working substance absorbs energy by heat from a high temperature energy reservoir (Q h ) Work is done by the engine (W eng ) Energy is expelled as heat to a lower temperature reservoir (Q c ) Use the active figure to change the efficiency of the engine and observe energy transfers
Heat Engine, cont. Since it is a cyclical process, ΔE int = 0 Its initial and final internal energies are the same Therefore, Q net = W eng The work done by the engine equals the net energy absorbed by the engine
Thermal Efficiency of a Heat Engine Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature We can think of the efficiency as the ratio of what you gain to what you give
More About Efficiency In practice, all heat engines expel only a fraction of the input energy by mechanical work Therefore, their efficiency is always less than 100% To have e = 100%, Q C must be 0
Second Law: Kelvin-Planck Form It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work W eng can never be equal to Q c Means that Qc cannot equal 0 Some Qc must be expelled to the environment Means that e cannot equal 100%
Perfect Heat Engine No energy is expelled to the cold reservoir It takes in some amount of energy and does an equal amount of work e = 100% It is impossible to construct such an engine
Heat Pumps and Refrigerators Heat engines can run in reverse This is not a natural direction of energy transfer Must put some energy into a device to do this Devices that do this are called heat pumps or refrigerators Examples A refrigerator is a common type of heat pump An air conditioner is another example of a heat pump
Heat Pump Process Energy is extracted from the cold reservoir, Q C Energy is transferred to the hot reservoir, Q h Work must be done on the engine, W Use the active figure to change the COP of the heat pump and observe the transfers of energy
Second Law – Clausius Form It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work Or – energy does not transfer spontaneously by heat from a cold object to a hot object
Perfect Heat Pump Takes energy from the cold reservoir Expels an equal amount of energy to the hot reservoir No work is done This is an impossible heat pump
Coefficient of Performance The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) Similar to thermal efficiency for a heat engine
COP, Cooling Mode In cooling mode, you “gain” energy removed from a cold temperature reservoir A good refrigerator should have a high COP Typical values are 5 or 6
COP, Heating Mode In heating mode, the COP is the ratio of the heat transferred in to the work required Q h is typically higher than W Values of COP are generally greater than 1 It is possible for them to be less than 1 The use of heat pumps that extract energy from the air are most satisfactory in moderate climates
Reversible and Irreversible Processes A reversible process is one in which every point along some path is an equilibrium state And one for which the system can be returned to its initial state along the same path An irreversible process does not meet these requirements All natural processes are known to be irreversible Reversible processes are an idealization, but some real processes are good approximations
Reversible and Irreversible Processes, cont A real process that is a good approximation of a reversible one will occur very slowly The system is always very nearly in an equilibrium state A general characteristic of a reversible process is that there are no dissipative effects that convert mechanical energy to internal energy present No friction or turbulence, for example
Reversible and Irreversible Processes, Summary The reversible process is an idealization All real processes on Earth are irreversible
Sadi Carnot 1796 – 1832 French engineer First to show quantitative relationship between work and heat Published Reflections on the Motive Power of Heat Reviewed industrial, political and economic importance of the steam engine
Carnot Engine A theoretical engine developed by Sadi Carnot A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible This sets an upper limit on the efficiencies of all other engines
Carnot’s Theorem No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.
Carnot Cycle Overview of the processes in a Carnot cycle
Carnot Cycle, A to B A → B is an isothermal expansion The gas is placed in contact with the high temperature reservoir, T h The gas absorbs heat |Q h | The gas does work W AB in raising the piston
Carnot Cycle, B to C B → C is an adiabatic expansion The base of the cylinder is replaced by a thermally nonconducting wall No heat enters or leaves the system The temperature falls from T h to T c The gas does work W BC
Carnot Cycle, C to D The gas is placed in contact with the cold temperature reservoir C → D is an isothermal compression The gas expels energy Q c Work W CD is done on the gas
Carnot Cycle, D to A D → A is an adiabatic compression The gas is again placed against a thermally nonconducting wall So no heat is exchanged with the surroundings The temperature of the gas increases from T c to T h The work done on the gas is W DA
Carnot Cycle, PV Diagram The work done by the engine is shown by the area enclosed by the curve, W eng The net work is equal to |Q h | – |Q c | E int = 0 for the entire cycle Use the active figures to observe the piston and the PV diagram
Efficiency of a Carnot Engine Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs Temperatures must be in Kelvins All Carnot engines operating between the same two temperatures will have the same efficiency
Notes About Carnot Efficiency Efficiency is 0 if T h = T c Efficiency is 100% only if T c = 0 K Such reservoirs are not available Efficiency is always less than 100% The efficiency increases as T c is lowered and as T h is raised In most practical cases, T c is near room temperature, 300 K So generally T h is raised to increase efficiency
Carnot Cycle in Reverse Theoretically, a Carnot-cycle heat engine can run in reverse This would constitute the most effective heat pump available This would determine the maximum possible COPs for a given combination of hot and cold reservoirs
Carnot Heat Pump COPs In heating mode: In cooling mode: In practice, the value of the COP is limited to below 10
Gasoline Engine In a gasoline engine, six processes occur during each cycle For a given cycle, the piston moves up and down twice This represents a four-stroke cycle The processes in the cycle can be approximated by the Otto cycle
Otto Cycle The PV diagram of an Otto cycle is shown at right The Otto cycle approximates the processes occurring in an internal combustion engine Use the active figures to observe the movement of the piston and the location on the PV diagram
The Conventional Gasoline Engine
Gasoline Engine – Intake Stroke During the intake stroke, the piston moves downward A gaseous mixture of air and fuel is drawn into the cylinder Energy enters the system as potential energy in the fuel O → A in the Otto cycle
Gasoline Engine – Compression Stroke The piston moves upward The air-fuel mixture is compressed adiabatically The temperature increases The work done on the gas is positive and equal to the negative area under the curve A → B in the Otto cycle
Gasoline Engine – Spark Combustion occurs when the spark plug fires This is not one of the strokes of the engine It occurs very quickly while the piston is at its highest position Conversion from potential energy of the fuel to internal energy B → C in the Otto cycle
Gasoline Engine – Power Stroke In the power stroke, the gas expands adiabatically This causes a temperature drop Work is done by the gas The work is equal to the area under the curve C → D in the Otto cycle
Gasoline Engine – Valve Opens This is process D → A in the Otto cycle An exhaust valve opens as the piston reaches its bottom position The pressure drops suddenly The volume is approximately constant So no work is done Energy begins to be expelled from the interior of the cylinder
Gasoline Engine – Exhaust Stroke In the exhaust stroke, the piston moves upward while the exhaust valve remains open Residual gases are expelled to the atmosphere The volume decreases A → O in the Otto cycle
Otto Cycle Efficiency If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is is the ratio of the molar specific heats V 1 / V 2 is called the compression ratio
Otto Cycle Efficiency, cont Typical values: Compression ratio of 8 = 1.4 e = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture
Diesel Engines Operate on a cycle similar to the Otto cycle without a spark plug The compression ratio is much greater and so the cylinder temperature at the end of the compression stroke is much higher Fuel is injected and the temperature is high enough for the mixture to ignite without the spark plug Diesel engines are more efficient than gasoline engines
Entropy Entropy, S, is a state variable related to the second law of thermodynamics The importance of entropy grew with the development of statistical mechanics A main result is isolated systems tend toward disorder and entropy is a natural measure of this disorder
Microstates vs. Macrostates A microstate is a particular configuration of the individual constituents of the system A macrostate is a description of the conditions from a macroscopic point of view It makes use of macroscopic variables such as pressure, density, and temperature for gases
Microstates vs. Macrostates, cont For a given macrostate, a number of microstates are possible It is assumed that all microstates are equally probable When all possible macrostates are examined, it is found that macrostates associated with disorder have far more microstates than those associated with order
Microstates vs. Macrostates, Probabilities The probability of a system moving in time from an ordered macrostate to a disordered macrostate is far greater than the probability of the reverse There are more microstates in a disordered macrostate If we consider a system and its surroundings to include the Universe, the Universe is always moving toward a macrostate corresponding to greater disorder
Entropy and the Second Law Entropy is a measure of disorder The entropy of the Universe increases in all real processes This is another statement of the second law of thermodynamics
Entropy and Heat The original formulation of entropy dealt with the transfer of energy by heat in a reversible process Let dQ r be the amount of energy transferred by heat when a system follows a reversible path The change in entropy, dS is
Entropy and Heat, cont The change in entropy depends only on the endpoints and is independent of the actual path followed The entropy change for an irreversible process can be determined by calculating the change in entropy for a reversible process that connects the same initial and final points
More About Change in Entropy dQ r is measured along a reversible path, even if the system may have followed an irreversible path The meaningful quantity is the change in entropy and not the entropy itself For a finite process,
Change in Entropy, cont The change in entropy of a system going from one state to another has the same value for all paths connecting the two states The finite change in entropy depends only on the properties of the initial and final equilibrium states Therefore we are free to choose a particular reversible path over which to evaluate the entropy the actual path as long as the initial and final states are the same
S for a Reversible Cycle S = 0 for any reversible cycle In general, This integral symbol indicates the integral is over a closed path
Entropy Changes in Irreversible Processes To calculate the change in entropy in a real system, remember that entropy depends only on the state of the system Do not use Q, the actual energy transfer in the process Distinguish this from Q r, the amount of energy that would have been transferred by heat along a reversible path Q r is the correct value to use for S
In general, the total entropy and therefore the total disorder always increases in an irreversible process The total entropy of an isolated system undergoes a change that cannot decrease This is another statement of the second law of thermodynamics Entropy Changes in Irreversible Processes, cont
Entropy Changes in Irreversible Processes, final If the process is irreversible, then the total entropy of an isolated system always increases In a reversible process, the total entropy of an isolated system remains constant The change in entropy of the Universe must be greater than zero for an irreversible process and equal to zero for a reversible process
Heat Death of the Universe Ultimately, the entropy of the Universe should reach a maximum value At this value, the Universe will be in a state of uniform temperature and density All physical, chemical, and biological processes will cease The state of perfect disorder implies that no energy is available for doing work This state is called the heat death of the Universe
S in Thermal Conduction The cold reservoir absorbs Q and its entropy changes by Q/T c At the same time, the hot reservoir loses Q and its entropy changes by -Q/T h Since T h > T c, the increase in entropy in the cold reservoir is greater than the decrease in entropy in the hot reservoir Therefore, S U > 0 For the system and the Universe
S in a Free Expansion Consider an adiabatic free expansion Q = 0 but cannot be used since that is for an irreversible process
S in Free Expansion, cont For an isothermal process, this becomes Since V f > V i, S is positive This indicates that both the entropy and the disorder of the gas increase as a result of the irreversible adiabatic expansion
Entropy on a Microscopic Scale We can treat entropy from a microscopic viewpoint through statistical analysis of molecular motions A connection between entropy and the number of microstates (W) for a given macrostate is S = k B ln W The more microstates that correspond to a given macrostate, the greater the entropy of that macrostate This shows that entropy is a measure of disorder
Entropy, Molecule Example One molecule in a two-sided container has a 1-in-2 chance of being on the left side Two molecules have a 1-in-4 chance of being on the left side at the same time Three molecules have a 1-in-8 chance of being on the left side at the same time
Entropy, Molecule Example Extended Consider 100 molecules in the container The probability of separating 50 fast molecules on one side and 50 slow molecules on the other side is (½) 100 If we have one mole of gas, this is found to be extremely improbable
Entropy, Marble Example Suppose you have a bag with 50 red marbles and 50 green marbles You draw a marble, record its color, return it to the bag, and draw another Continue until four marbles have been drawn What are possible macrostates and what are their probabilities?
Entropy, Marble Example, Results The most ordered are the least likely The most disorder is the most likely
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# What is 3 over 4 divided by 8 in a fraction?
Wiki User
2013-07-11 11:26:36
3/4 ÷ 8 = 3/32
Wiki User
2013-07-11 11:26:36
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Helper I
## Measure difference between rows in scoreboard
I'm trying to make a dynamic Scoreboard for a sales department and everything works fine except for one thing: calculating the differences for each line.
I know that i could do it with Power Query by using two index columns and then just merge them, but that would make it static. I want to be able to compare different groups of people and it should automaticly update the rank and differences in the current list.
The list looks something like this at the moment. In the example I have choosen 6 members to compare but I might need to switch them out and the values should then be calculated based on the new table.
Details
Rank = RANKX(ALLSELECTED('Sales people');[Sum];[Sum];DESC;Dense)
Sum = This is just a measure that adds a couple of sum-functions from various tables
The Diff Sum to Rank 1 is first a measure that calculates the top sum and then it subtracts the sales persons sum from that:
Sum for Rank 1 = CALCULATE(MAXX('Sales people';[Sum]);ALLSELECTED('Salespeople'))
Diff Sum to Rank 1 = [Sum]-[Sum for Rank 1]
So far so good! What I want now is a measure that calculates the difference between each sales person and the one above in the score list.
This is how I would have done it with Excel if I had a static list.
Any ideas?
1 ACCEPTED SOLUTION
Super User
Hi @Ville
https://www.sqlbi.com/articles/displaying-nth-element-in-dax/
```Diff to next person =
VAR CurrentRank = [Rank]
VAR CurrentSum = [Sum]
VAR NextPerson =
FILTER ( ALLSELECTED ( 'Sales people' ); [Rank] = CurrentRank - 1 )
VAR NextPerson_Sum =
MINX ( NextPerson; [Sum] )
RETURN
IF ( NOT ISBLANK ( NextPerson_Sum ); CurrentSum - NextPerson_Sum )```
This formula basically filters Sales People to the current rank minus 1, and evaluates Sum for that sales persion (using MINX in case of tied ranks for the next person).
Regards,
Owen
Owen Auger
My Blog
2 REPLIES 2
Super User
Hi @Ville
https://www.sqlbi.com/articles/displaying-nth-element-in-dax/
```Diff to next person =
VAR CurrentRank = [Rank]
VAR CurrentSum = [Sum]
VAR NextPerson =
FILTER ( ALLSELECTED ( 'Sales people' ); [Rank] = CurrentRank - 1 )
VAR NextPerson_Sum =
MINX ( NextPerson; [Sum] )
RETURN
IF ( NOT ISBLANK ( NextPerson_Sum ); CurrentSum - NextPerson_Sum )```
This formula basically filters Sales People to the current rank minus 1, and evaluates Sum for that sales persion (using MINX in case of tied ranks for the next person).
Regards,
Owen
Owen Auger
My Blog
Helper I
It works like a charm! Thanks!
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Grade:
## 2 Answers
Ashwin Sinha
520 Points
11 years ago
Dear Neeta Gupta,
second-order reaction depends on the concentrations of one second-order reactant, or two first-order reactants.
For a second order reaction, its reaction rate is given by:
$\ -\frac{d[A]}{dt} = 2k[A]^2$ or $\ -\frac{d[A]}{dt} = k[A][B]$ or $\ -\frac{d[A]}{dt} = 2k[B]^2$
In several popular kinetics books, the definition of the rate law for second-order reactions is $-\frac{d[A]}{dt} = k[A]^2$. Conflating the 2 inside the constant for the first, derivative, form will only make it required in the second, integrated form (presented below). The option of keeping the 2 out of the constant in the derivative form is considered more correct, as it is almost always used in peer-reviewed literature, tables of rate constants, and simulation software.[8]
The integrated second-order rate laws are respectively
$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$
or
$\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}$
[A]0 and [B]0 must be different to obtain that integrated equation.
The half-life equation for a second-order reaction dependent on one second-order reactant is $\ t_ \frac{1}{2} = \frac{1}{k[A]_0}$. For a second-order reaction half-lives progressively double.
Another way to present the above rate laws is to take the log of both sides: $\ln{}r = \ln{}k + 2\ln\left[A\right]$
Examples of a Second-order reaction
• $2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)$
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E.Nandhini ketha
115 Points
6 years ago
Chemical kinetics, also known as reaction kinetics, Chemical kinetics is the study of rates of chemical processes.
## ASK QUESTION
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Pack 1 • Paper 2 • Question 22
05:17
### Video Transcript
A circle has equation 𝑥 squared plus 𝑦 squared is equal to 45. Prove algebraically that the straight line two 𝑥 minus 𝑦 is equal to 15 is a tangent to the circle.
Now, I’ll come back to what it means for a line to be a tangent to a circle in a minute. But first, I’m just going to rearrange the equation of the straight line slightly. I want to make 𝑦 the subject. So I’ll begin by adding 𝑦 to each side, giving two 𝑥 is equal to 𝑦 plus 15. Next, I’ll subtract 15 from each side, giving two 𝑥 minus 15 is equal to 𝑦. I’ve done that rearranging just so that I can picture this line a little bit more accurately in the next stage. I now know that this is a straight line with a positive gradient as its gradient is two.
Let’s think about what it means for a straight line to be a tangent to a circle. Whenever a straight line and a circle exist in the same coordinate plane, there’re three possibilities for the relative position of the circle and the line. The first possibility is that the line cuts through the circle. This means that the line meets the circle in two places. There are two points of intersection. The second possibility is that the line just touches the circumference of the circle, but doesn’t pass inside, meaning that there’s one point of intersection between the line and the circle. The third possibility is that the line completely misses the circle and there are no points of intersection between the two.
The second possibility would mean that the line is a tangent to the circle, which is what we’ve been asked to prove in this question. We, therefore, need to prove that there is only one point of intersection between the line and the circle. To do so, we need to solve the equation of the line and the equation of the circle simultaneously.
We’ve already rearranged the equation of the line to give 𝑦 is equal to two 𝑥 minus 15. So we’re going to substitute this into the equation of the circle. 𝑥 squared plus 𝑦 squared is equal to 45. Therefore, becomes 𝑥 squared plus two 𝑥 minus 15 all squared is equal to 45. This is an equation in 𝑥 only, which we can solve.
We need to expand the bracket and we must be careful here because a really common mistake to make when squaring a bracket is to just square the two terms individually. Two 𝑥 minus 15 all squared is not equal to two 𝑥 squared minus 15 squared. Instead, let’s write the repeated bracket out twice and then we can use the FOIL method to expand.
Multiplying the first term in each bracket together, two 𝑥 multiplied by two 𝑥 is equal to four 𝑥 squared. Multiplying the outer terms together, we have two 𝑥 multiplied by negative 15 which is negative 30𝑥. Multiplying the inner terms together, we have negative 15 multiplied by two 𝑥 which is also negative 30𝑥. Finally, multiplying the last term in the two brackets together, we have negative 15 multiplied by negative 15 which is 225.
The equation therefore becomes 𝑥 squared plus four 𝑥 squared minus 60𝑥 plus 225 is equal to 45. And I’ve grouped the two lots of negative 30𝑥 together to make negative 60𝑥. Now, 𝑥 squared plus four 𝑥 squared is equal to five 𝑥 squared. And at the same time, I’ve subtracted 45 from both sides of the equation. So I now have five 𝑥 squared minus 60𝑥 plus 180 is equal to zero. I have a quadratic equation in 𝑥 which I’d like to solve. But I can make this simpler.
I noticed that all of the coefficients are multiples of five. So I can simplify by dividing the whole equation through by five. This gives a simple equation 𝑥 squared minus 12𝑥 plus 36 is equal to zero. I want to solve this equation for 𝑥. And to do so, I need to factorize. You maybe able to spot straightaway how this quadratic factorizes. But if not, let’s recall how to factorize a quadratic equation.
As the coefficient of 𝑥 squared is just one, the beginning terms in each bracket are each just 𝑥. To find the two numbers that go in the bracket, I need to find two numbers that sum to the coefficient of 𝑥 — so that’s negative 12 — and multiply it to the constant term, which is plus 36. The two numbers that do that are negative six and negative six. So the two brackets are 𝑥 minus six and 𝑥 minus six. Now, here’s the key point: these two brackets are identical, which means if I were to solve this equation, I would get 𝑥 minus six is equal to zero, meaning 𝑥 is equal to six. But there’s only one value for 𝑥.
If we think back to the three situations we described, this means that there’s only one point of intersection between the line and the circle. And therefore, the line is a tangent to the circle. By solving the equation of the line and the circle simultaneously, we’ve proved and algebraically that the straight line is a tangent to the circle. | 1,188 | 4,815 | {"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} | 4.53125 | 5 | CC-MAIN-2020-10 | latest | en | 0.948413 |
https://www.numberempire.com/612?number=612 | 1,563,450,378,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00099.warc.gz | 794,631,990 | 5,118 | Home | Menu | Get Involved | Contact webmaster
# Number 612
six hundred twelve
### Properties of the number 612
Factorization 2 * 2 * 3 * 3 * 17 Divisors 1, 2, 3, 4, 6, 9, 12, 17, 18, 34, 36, 51, 68, 102, 153, 204, 306, 612 Count of divisors 18 Sum of divisors 1638 Previous integer 611 Next integer 613 Is prime? NO Previous prime 607 Next prime 613 612th prime 4513 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 1001100100 Octal 1144 Duodecimal 430 Hexadecimal 264 Square 374544 Square root 24.738633753706 Natural logarithm 6.4167322825123 Decimal logarithm 2.7867514221456 Sine 0.57333247711966 Cosine -0.81932281225402 Tangent -0.69976383977688
Number 612 is pronounced six hundred twelve. Number 612 is a composite number. Factors of 612 are 2 * 2 * 3 * 3 * 17. Number 612 has 18 divisors: 1, 2, 3, 4, 6, 9, 12, 17, 18, 34, 36, 51, 68, 102, 153, 204, 306, 612. Sum of the divisors is 1638. Number 612 is not a Fibonacci number. It is not a Bell number. Number 612 is not a Catalan number. Number 612 is not a regular number (Hamming number). It is a not factorial of any number. Number 612 is an abundant number and therefore is not a perfect number. Binary numeral for number 612 is 1001100100. Octal numeral is 1144. Duodecimal value is 430. Hexadecimal representation is 264. Square of the number 612 is 374544. Square root of the number 612 is 24.738633753706. Natural logarithm of 612 is 6.4167322825123 Decimal logarithm of the number 612 is 2.7867514221456 Sine of 612 is 0.57333247711966. Cosine of the number 612 is -0.81932281225402. Tangent of the number 612 is -0.69976383977688 | 588 | 1,730 | {"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-2019-30 | latest | en | 0.67513 |
http://stackoverflow.com/questions/15370029/need-assistance-understanding-a-shift-calculation-and-trimming-some-fat-off-my-c | 1,387,635,863,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345775611/warc/CC-MAIN-20131218054935-00074-ip-10-33-133-15.ec2.internal.warc.gz | 187,863,947 | 15,978 | # Need assistance understanding a shift calculation and trimming some fat off my code
I was going over this code to read a file on my desktop that decrypts a cesar cipher code and I am stuck trying to figure out how the shift is calculated in this program.
From what I can tell, Max e's is the shifted letter with the highest frequency. Since e's are the most common letter in English, the program is trying to set the highest frequency char in the cipher, to an 'e' in English. Which is good, as far as it goes, but there are many phrases where e's are NOT the most frequent letter, and then it will fall on it's face.
So, how can I tell the program to guess at the most frequent cipher letter to be an e in plain text BUT in case it isn't, then proceed to try an e shifted to the second most frequent letter in the text, and so on, until I find it?
A friend helped with that part but has poor English so it is difficult for him to explain it to me. Can someone please elaborate? Assistance is greatly appreciated! Let me know what you think:
``````#include <iostream>
#include <string>
#include <cctype> // isalpha, islower, isupper, functions
#include <fstream>
using namespace std;
string caesarShift(string text, int shift);
int main()
{
int maxEs = 0; // # of e's in maxString
int currentEs = 0; // # of e'sin currentString
string maxString; // decrypted caesar shift with most e's
string currentString; //decrypted caesar shift
string cipher; // Stores cipher text
char ch; // Stores currentcharacter for reading
ifstream fin("/Users/jasonrodriguez/Desktop/encrypted.txt"); //opens "encrypted.txt" file
while( fin.get(ch) ) // readseach char into the cipher till EOF
{
cipher += ch;
}
fin.close(); // be safe andclose file
for(int i=0; i < 26; i++)
{
currentEs =0; // Reset counter
currentString =caesarShift(cipher, i); // get shifted text
for(unsigned int x=0; x <currentString.size(); x++) // check each character of stringarray
{
if(currentString[x] == 'e' || currentString[x] == 'E') // check fore's
{
currentEs++; // increment Ecounter
}
}
if(currentEs > maxEs) //if currentEs is greater than maxEs, replace max with current
{
maxEs =currentEs;
maxString= currentString;
}
}
cout << maxString << endl;
return 0;
}
/**
string caesarShift(string text, int shift)
Decrypts Caesar Shift using text and shift
*/
string caesarShift(string text, int shift)
{
shift = shift % 26; // Morethan 26 is redundant and unneeded
char ch = 0; // holds current character
char chs = 0; // holds shiftedcharacter
for(unsigned int i=0; i < text.size();i++)
{
ch = text[i];
if( isalpha(ch) )
{
chs = ch -shift; // reverse shifting
if( (islower(ch) && chs < 'a' ) // If is lowercase andshifted value is lower than 'a'
||
( isupper(ch) && chs < 'A' ) ) // Ifis uppercase and shifted value is lower than 'A'
{
chs += 26; // Add 26(number ofletters) to get back to the correct place in alphabet
}
text[i] =chs; // Set character to shifted character
}
}
return text;
}
``````
Questions:
1. From what I can tell, Max e's is the shifted letter with the highest frequency. Since e's are the most common letter in English, the program is trying to set the highest frequency char in the cipher, to an 'e' in English. Which is good, as far as it goes, but there are many phrases where e's are NOT the most frequent letter, and then it will fall on it's face. So, how can I tell the program to guess at the most frequent cipher letter to be an e in plain text BUT in case it isn't, then proceed to try an e shifted to the second most frequent letter in the text, and so on, until I find it?
2. I think if I shift a character by a shift amount the character may or may not be out of bounds. 'a' + 3 is 'd' ok, 'x' + 3 is '{' not ok. So if the character is over 'z' take away 26, if under 'a' add 26. can be a usable function. However, can you please explain to me how the shift is calculated in the program and is applied it to the file? It has me totally stumped :(
-
How do you keep track of the letter frequencies and the corresponding shift? – Code-Guru Mar 12 at 22:43
That is a big part of my question yes – Jason Mar 12 at 22:48
I suggest that you start by looking at std::map. You can use it to store the frequencies. – Code-Guru Mar 12 at 22:52
I think if I shift a character by a shift amount the character may or may not be out of bounds. 'a' + 3 is 'd' ok, 'x' + 3 is '{' not ok. So if the character is over 'z' take away 26, if under 'a' add 26. can be a usable function. However, can you please explain to me how the shift is calculated in the program and is applied it to the file? It has me totally stumped :( – Jason Mar 12 at 22:54
Calculating a shift:
First, you need to mapy the letters 'A' through 'Z' (assuming only uppercase letters) to the integers 0 through 25. C++ let's you do this easily with subtraction:
``````n = c - 'A';
``````
Now you can perform the shift with modulus arithmetic:
``````n = (n + shift) % 26;
``````
Finally map back to a letter:
``````p = n + 'A';
``````
(Note that you will need apprporiate declarations for the variables used in these examples. I suggest that you use more meaningful names than the single-letter variables names I use.)
-
Thank you, I will work with this – Jason Mar 12 at 23:55 | 1,420 | 5,310 | {"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-2013-48 | latest | en | 0.841828 |
http://en.wikipedia.org/wiki/Limit_point_compact | 1,405,014,232,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776420978.98/warc/CC-MAIN-20140707234020-00031-ip-10-180-212-248.ec2.internal.warc.gz | 43,882,520 | 11,208 | # Limit point compact
In mathematics, a topological space X is said to be limit point compact[1] or weakly countably compact if every infinite subset of X has a limit point in X. This property generalizes a property of compact spaces. In a metric space, limit point compactness, compactness, and sequential compactness are all equivalent. For general topological spaces, however, these three notions of compactness are not equivalent.
## Properties and Examples
• Even though a continuous function from a compact space X, to an ordered set Y in the order topology, must be bounded, the same thing does not hold if X is limit point compact. An example is given by the space $X\times\mathbb{Z}$ (where X = {1, 2} carries the indiscrete topology and $\mathbb{Z}$ is the set of all integers carrying the discrete topology) and the function $f=\pi_{\mathbb{Z}}$ given by projection onto the second coordinate. Clearly, ƒ is continuous and $X\times \mathbb{Z}$ is limit point compact (in fact, every nonempty subset of $X\times \mathbb{Z}$ has a limit point) but ƒ is not bounded, and in fact $f(X\times \mathbb{Z})=\mathbb{Z}$ is not even limit point compact.
• Every countably compact space (and hence every compact space) is weakly countably compact, but the converse is not true.
• The set of all real numbers, R, is not limit point compact; the integers are an infinite set but do not have a limit point in R.
• If (X, T) and (X, T*) are topological spaces with T* finer than T and (X, T*) is limit point compact, then so is (X, T).
• A finite space is vacuously limit point compact. | 400 | 1,585 | {"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": 6, "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} | 3.4375 | 3 | CC-MAIN-2014-23 | latest | en | 0.884186 |
http://www.texasoft.com/tutorial-time-series.html | 1,550,540,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489282.7/warc/CC-MAIN-20190219000551-20190219022551-00257.warc.gz | 460,708,528 | 5,544 | Home
What's New?
Order WINKS
Order WINKS 7
These WINKS statistics tutorials explain the use and interpretation of standard statistical analysis techniques for Medical, Pharmaceutical, Clinical Trials, Marketing or Scientific Research. The examples include how-to instructions for WINKS SDA Version 6.0 Software. Download evaluation copy of WINKS.
This tutorial illustrates how to model parameters for an ARMA Time Series and to create a forecast using WINKS SDA (Version 6.09 and later)
Estimate Parameters
Step 1: Open the time series data file. For this example, open SUN.SDA.
Step 2: Select Analyze -> Time Series -> Model Data. Select COUNT as the variable to analyze.
Step 3: In the dialog box section titled “Model the Data” enter a 3 in the “P” text box to specify that you want to model the data as an ARMA(3,0), and click Calculate.
The following information is reported:
Series Length = 176
Estimated parameters for P = 3 and Q = 0
Burg/Tiao/Tsay AR Estimates
---------------------------
Estimated Burg/Tiao/Tsay AR Parameters:
1) 1.266145 2) -0.498998
3) -0.112711
White Noise Variance = 232.8768
AIC = 965.2897
AR Factors for Burg-Tsay Estimates
ROOT ABS RECIP FREQUENCY OPERATOR COEFS
.9860 ± .6436i 0.8493 0.0920 1 - 1.422B + 0.721B^2
-6.3993 0.1563 0.5000 1 + 0.156B
Step 4: Click “Data View” to see the results – which includes spectrum and residuals.
Forecast
Once you have the estimates of a model, you can use this information to create a forecast. SDA will allow you to use your model to forecast future values of the series. An optional 95% confidence bound may be plotted for your estimated forecast.
Step 5: Select Analyze -> Time Series -> Forecast. Choose COUNT as the variable to forecast. The following dialog box appears:
At the bottom of the dialog box, select how you want to forecast. In this case select “Beyond Series N= 20.” Click Calculate. A forecast summary is reported for the model indicated.
Forecast data is based on the following information
Forecast 20 steps beyond series.
Series Length = 176
Differenced = 0
Estimated parameters for p = 3 and q = 0
AR Parameters:
1) 1.266145 2) -0.498998
3) -0.112711
Estimated WNV = 232.876837617018
Forecast Summary...
COUNT LOWER UPPER FORECAST SERIES PSI
1 80.9 80.9 80.9 80.9 1.2661
2 83.4 83.4 83.4 83.4 1.1041
3 47.7 47.7 47.7 47.7 0.6535
4 47.8 47.8 47.8 47.8 0.1337
5 30.7 30.7 30.7 30.7 -0.2812
Step 6: Click on “Data View.” The WINKS Data sheet contains columns for the forecast and upper and lower confidence levels.
Step 7: To plot the results, select Graph/Charts -> Line/Time Series Plots -> Overlay. Specify Lower, Upper and Forecast (in that order) to plot. The following plot appears indicating the original data, a forecast, and upper and lower limits for the forecast.
This procedure is available in the WINKS Professional edition. | 868 | 3,169 | {"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} | 3.203125 | 3 | CC-MAIN-2019-09 | latest | en | 0.712446 |
https://justaaa.com/statistics-and-probability/232957-ten-salespeople-were-surveyed-and-the-average | 1,713,632,810,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00505.warc.gz | 318,986,318 | 10,465 | Question
# Ten salespeople were surveyed and the average number of client contacts per month, x, and the...
Ten salespeople were surveyed and the average number of client contacts per month, x, and the sales volume, y (in thousands), were recorded for each: X 40 42 44 40 41 46 50 48 50 55 Y 50 55 42 45 30 80 90 95 110 130
a) Find the correlation coefficient r.
b) Find the regression equation
c) Construct the scatter plot and graph the regression equation.
SOLUTION:-
(A)
X Values
∑ = 456
Mean = 45.6
∑(X - Mx)2 = SSx = 232.4
Y Values
∑ = 727
Mean = 72.7
∑(Y - My)2 = SSy = 9886.1
X and Y Combined
N = 10
∑(X - Mx)(Y - My) = 1426.8
R Calculation
r = ∑((X - My)(Y - Mx)) / √((SSx)(SSy))
r = 1426.8 / √((232.4)(9886.1)) = 0.9413
r = 0.9413
This is a strong positive correlation, which means that high X variable scores go with high Y variable scores
(B)
Sum of X = 456
Sum of Y = 727
Mean X = 45.6
Mean Y = 72.7
Sum of squares (SSX) = 232.4
Sum of products (SP) = 1426.8
Regression Equation = ŷ = bX + a
b = SP/SSX = 1426.8/232.4 = 6.13941
a = MY - bMX = 72.7 - (6.14*45.6) = -207.25731
ŷ = 6.13941X - 207.2573
(C)
THIS IS A 100% VALID ANSWER
THANKS | 484 | 1,219 | {"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-2024-18 | latest | en | 0.758544 |
http://www.learnpython.org/es/Decorators | 1,513,005,361,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00120.warc.gz | 416,902,563 | 7,584 | Get started learning Python with DataCamp's free Intro to Python tutorial. Learn Data Science by completing interactive coding challenges and watching videos by expert instructors. Start Now!
# Decorators
Los decorqadores te permiten hacen modificaciones simples a objetos invocables como functions, methods, or classes. Nos ocuparemos de las funciones para este tutorial. La sintaxis
``````@decorator
def functions(arg):
return "Return"
``````
Es equivalente a:
``````def function(arg):
return "Return"
function=decorator(function) #this passes the function to the decorator, and reassigns it to the functions
``````
Como habrás visto, un decorador es simplemente otra función que toma varias funciones y devuelve una. Por ejemplo podrías hacer esto:
``````def repeater(old_function):
def new_function(*args, **kwds): #See learnpython.org/page/Multiple%20Function%20Arguments for how *args and **kwds works
old_function(*args, **kwds) #we run the old function
old_function(*args, **kwds) #we do it twice
return new_function #we have to return the new_function, or it wouldn't reassign it to the value
``````
Esto haría que una función se repita dos veces
``````>>> @repeater
def Multiply(num1, num2):
print(num1*num2)
>>> Multiply(2, 3)
6
6
``````
También puedes hacer que cambie la salida:
``````def Double_Out(old_function):
def new_function(*args, **kwds):
return 2*old_function(*args, **kwds) #modify the return value
return new_function
``````
``````def Double_In(old_function):
def new_function(arg): #only works if the old function has one argument
return old_function(arg*2) #modify the argument passed
return new_function
``````
y hacer comprobaciones
``````def Check(old_function):
def new_function(arg):
if arg<0: raise ValueError, "Negative Argument" #This causes an error, which is better than it doing the wrong thing
old_function(arg)
return new_function
``````
Digamos que quieres multiplicar la salida por una cantidad variable. Podrías hacerlo
``````def Multiply(multiplier):
def Multiply_Generator(old_function):
def new_function(*args, **kwds):
return multiplier*old_function(*args, **kwds)
return new_function
return Multiply_Generator #it returns the new generator
``````
Ahora, podrías hacer:
``````@Multiply(3) #Multiply no es un generador, pero Multiply(3) sí lo es
def Num(num):
return num
``````
Puedes hacer lo que quieras con la función antigua, ¡incluso ignorarla completamente! Los decoradores avanzados también pueden manipular el doc string y el número de argumentos Para conocer algunos decoradores elegantes, visita http://wiki.python.org/moin/PythonDecoratorLibrary. Exercise
Haz un manejador de decoradores que devuelva un decorador que decore funciones con un argumento. El manjeador debería coger un argumento, un tipo, y luego devolver un decorador que haga que la función verifique si la entrada es del tipo correcto. Si es errónea, debería imprimir "Bad Type" (En realidad, debería lanzar un error - y manejarlo - pero el manejo de errores no es parte de este tutorial). Mira el códugo del tutorial y la salida esperada si estás confundido respecto a esto (sé que es confuso). Usar isinstance(object, type_of_object) o type(object) tal vez te ayude.
This site generously supported by DataCamp. DataCamp offers online interactive Python Tutorials for Data Science. Join over a million other learners and get started learning Python for data science today! | 857 | 3,419 | {"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} | 2.640625 | 3 | CC-MAIN-2017-51 | latest | en | 0.287597 |
http://mathhelpforum.com/algebra/220531-quick-question-about-expressions.html | 1,527,325,472,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867374.98/warc/CC-MAIN-20180526073410-20180526093410-00343.warc.gz | 182,596,444 | 10,047 | 1. ## Quick question about expressions
The answer is A but idk how its A... Please explain. Thanks!!
2. ## Re: Quick question about expressions
Originally Posted by asilvester635
The answer is A but idk how its A... Please explain. Thanks!!
$\displaystyle 20-\tfrac{4}{5}x\ge 16$ is equivalent to $\displaystyle 100-4x\ge 80.$
3. ## Re: Quick question about expressions
Equivalently, $\displaystyle 20- \frac{4}{5}\ge 16$ is equivalent to $\displaystyle 20\ge \frac{4}{5}x+ 16$ (add $\displaystyle \frac{4}{5}x$ to both sides) which equivalent to $\displaystyle 20- 16= 4\ge \frac{4}{5}x$ (subtract 16 from both sides).
Now multiply both sides by the positive number $\displaystyle \frac{5}{4}$:
$\displaystyle \frac{5}{4}(4)= 5\ge x$.
4. ## Re: Quick question about expressions
Thank you so much!!! | 252 | 807 | {"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} | 3.890625 | 4 | CC-MAIN-2018-22 | latest | en | 0.805379 |
http://www.physicsforums.com/showthread.php?s=22fb1a497d26c13839ff04a37e34e6ca&p=4266256 | 1,371,635,549,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708664942/warc/CC-MAIN-20130516125104-00050-ip-10-60-113-184.ec2.internal.warc.gz | 625,967,327 | 7,238 | ## Dynamics - Normal and Tangential Motion
1. The problem statement, all variables and given/known data
The tires on a car are capable of exerting a maximum frictional force of 1753 lb. If the car is traveling at 75 ft.s and the curvature of the road is ρ=560 ft, what is the maximum acceleration that the car can have without sliding?
2. Relevant equations
ƩFn = man
3. The attempt at a solution
Ff = 1753 lb
v = 75 ft/s
ρ=560 ft
wcar = 3150 lb
an = $\frac{v^2}{ρ}$ = $\frac{75^2}{560}$ = 10.04 ft/s2
I believe that the acceleration would be the magnitude of the tangential and normal acceleration.
ƩFn = man = $\frac{3150}{32.2}$*10.04 = 982.2 lb
1753 = √Ft2 + 982.22
Solving for Ft = 1452 lb;
Now solving for at → 1452 = $\frac{3150}{32.2}$*at
at = 14.85 ft/s2
a = √at2 + an2 = √14.852 + 10.042 = 17.90 ft/s2
I'd appreciate it if someone could verify my work.
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I get the same answer as you. I think the question is asking for the maximum tangential acceleration which is 10.04 ft/sec^2.
Quote by LawrenceC I get the same answer as you. I think the question is asking for the maximum tangential acceleration which is 10.04 ft/sec^2.
Tangential was actually 14.85 ft/s^2.
And it makes sense too.
Thanks.
## Dynamics - Normal and Tangential Motion
Quote by aaronfue Tangential was actually 14.85 ft/s^2. And it makes sense too. Thanks.
I typed the wrong number.......should have typed 14.85 ft/s^2.
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https://www.learnatnoon.com/s/a-dice-is-thrown-once-what-is-the-probability/57019/ | 1,721,098,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00157.warc.gz | 740,719,000 | 22,409 | A dice is thrown once. What is the probability of getting a number: (i) greater than 2? (ii) less than or equal to 2?
A dice is thrown once. What is the probability of getting a number: (i) greater than 2? (ii) less than or equal to 2?
Solution:
The number of possible outcomes when dice is thrown = {1, 2, 3, 4, 5, 6}
So, n(S) = 6
(i) Event of getting a number greater than 2 = E = {3, 4, 5, 6}
So, n(E) = 4
Thus, probability of getting a number greater than 2 = n(E)/ n(S) = 4/6 = 2/3
(ii) Event of getting a number less than or equal to 2 = E = {1, 2}
So, n(E) = 2
Thus, probability of getting a number less than or equal to 2 = n(E)/ n(S) = 2/6 = 1/3 | 235 | 663 | {"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} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.937891 |
http://mrwhatis.com/44mm-conversion.html | 1,369,429,970,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705069221/warc/CC-MAIN-20130516115109-00026-ip-10-60-113-184.ec2.internal.warc.gz | 176,702,021 | 10,980 | Search What is?:
Latest Questions
# What is 44MM CONVERSION?
Answer: roughly 1.732 inches For reference: 1 inch = 25.4mm For quick conversions, try Google. It knows you are asking for a calulation if you start a search with the word ...
MM to Inch Conversion Table: MM. Inches
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There are 25.4 millimetres in one inch. Therefore, rounded to two decimal places, 44 millimetes is equal to 44 / 25.4 = 1.73 inches.
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Convert MM to Inches Automatically To convert MM to Inches, enter the number of MM to be converted into the MM box below. MM: Millimeters
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We know that 1 inches is 1.54 cm ; and you have 44mm so to make it easy to convert we will convert to from mm to cm and we will get 4.4 cm then we will do this : 1 inche ----- 1.54 cm
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Inches to millimeters (mm) conversion table shows the most common values for the quick reference. Alternatively, you may use the converter below to convert any other values if wondering how many millimeters in an inch.
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Best Answer: 44mm=4.4 cm, so it's shorter then 5 cm 50m=50,000mm, so it's more then 5,000mm you convert units by moving the decimal point. to convert m to cm, multiply by 100 ...
depends on power goal sir, and SC44, 44mm wastegate, 50mm bov doesn't equal any of that yet. you need to think about your manifold, intercooler,injectors to start out your turbo setup is only as efficient as your manifold IMO.
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To convert from metric to SAE, multiply the value in millimeters by 0.039, and to do the opposite conversion multiply the SAE value by 25.4. As an example, if you have a bolt that is 3/8-inch in SAE units and you have metric wrenches then you will want to convert it to millimeters: ...
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Tial wastegate spring chart & BAR to PSI conversion table. So, I figured this would be a couple nice charts to post up to aid others in making an educated decision on which wastegate spring combination to go with. I hope it helps.
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Cane Creek's new XX 44mm Traditional lower headset assemblies will let riders use tapered forks in frames equipped with a straight 44mm-bore 1-1/8in ZeroStack head tube. When launched in June, this should help breathe new life into countless older frames out there and allow their owners to take
Convert fractional and/or decimal inches to metric mm or vice versa
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Conversion chart for ring sizes; Inside diameter Inside circumference Sizes (in) (mm) (in) (mm) United States and Canada United Kingdom, Ireland, Australia and New Zealand India, China, Japan, South America Italy, Spain, Netherlands, Switzerland; 0.458: 11.63: 1.44: 36.5: 0:
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... a peltier device or thermoelectric device go here http://www.allelectronics.com/make-a-store/item/PJT-7/40MM-X-44MM-THERMOELECTRIC-COOLER//1.html. Reply. 23. Jun 20, ... Thermoelectric devices can convert heat into electricity directly.
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https://www.playtaptales.com/10-to-the-power-of-2256/ | 1,555,895,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532948.2/warc/CC-MAIN-20190421235818-20190422021818-00519.warc.gz | 774,980,358 | 3,886 | ## 10 to the power of 2,256
What is 10 to the power of 2256 (10^2256)? The answer is a Unquinquagintaseptingentillion. This is based on the Conway-Wechsler system for naming numbers.
## 10 to the power of 2,256 written out
10^2256 is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 | 1,588 | 3,263 | {"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-2019-18 | latest | en | 0.604544 |
sophwats.github.io | 1,721,403,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00429.warc.gz | 465,811,530 | 3,700 | # A gentle introduction to Alternating Least Squares
A successful recommendation engine is one which provides accurate and personalised predictions to users in a timely manner. Such systems are of increasing commercial importance. Collaborative Filtering algorithms allow us to give predictions by capturing the commonality between users.
In this post I’ll give you a gentle introduction to one such collaborative filtering algorithm - Alternating Least Squares, or ALS for short. Through a gentle example you will get a feel for what alternating least squares is doing. Disclaimer: We are going to do a simplified ALS to give you a feeling for what is going on, rather than pushing you into a deep, deep pool of mathematics and greek letters.
Consider the following matrix which holds opinions of two users on three products:
We wish to be able to predict how user 1 will rate product two. This is where Alternating Least Squares comes in.
The goal of Alternating Least Squares is to find two matrices, U and P, such that their product is approximately equal to the original matrix of users and products. Once such matrices have been found, we are able to predict what user i will think of product j by multiplying row i of U with row j of P.
## Computing U and P through ALS.
The method is called “alternating” since the following two steps are repeated:
1. fix matrix U and find the optimal matrix P
2. fix matrix P and find the optimal matrix U
The ‘least squares’ bit of ALS comes into play when we find the ‘optimal’ matrices, because ‘optimal’ in this case is taken to mean “minimises least squares error” (plus some regularisation term which prevents overfitting).
Let’s run through the first couple of steps by hand. For simplicity, I’m ignoring the term which prevents overfitting. I’m also making the choice that the number of features U and P is one. This corresponds to the number of columns in U, and rows in P. (In practise for a non-trivial example you will likely want a larger number of features. We will look at tuning for the optimal number of features in a future post.)
So our aim is to find matrices U and P such that:
We start by fixing matrix U. I could have picked any values for U, but I chose 1’s to make the maths easier.
If we multiply this out, we get 5 equations which involve the components of P:
There is a lone equation which governs the second component of P, thus we set that equal to 3. We need to select the remaining two components such that the mean squared error is minimised. Thus we compute:
and
From this, we get our first estimate of P:
Now we keep P fixed and optimise for matrix U. This similarly gives us the following equations for U:
We can then minimise mean squared error to solve for the components of U, as we did for P in equations (6) and (7).
Repeating this process we will, at some point, converge upon the optimal matrices U and P. In this example, after 20 iterations we find that U and P are given by:
Thus we are now able to predict what user 1 thinks of product 2. By multiplying the first component of U by the second component of P we reach an estimate of 1.897.
# Disclaimers:
1. I ignored the regularisation term. In practise, you always want to prevent overfitting when you are fitting a model to data.
2. This is a very simple example. Usually, for non trivial ratings matrices the minimisation is tricky.
3. This decomposition of the original matrix into matrices U and P is not exact - when you multiply U and P together you will not recover the exact ratings matrix. This needs to be taken into account when designing applications. (Don’t worry - we’ll talk about that another day.) | 799 | 3,677 | {"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} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.943548 |
https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/How-to-count-by-two-categories-with-DAX/m-p/3853759 | 1,716,871,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00645.warc.gz | 158,068,594 | 115,716 | cancel
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How to count by two categories with DAX ?
Hey everyone, I have my table one, who look like this :
Location X Y Location_1 A B Location_1 B C Location_2 A C Location_4 C A ... ... ...
I've created another table manually to which I'd like to add the count of each note by Location in another column, but I can't manage to code it in DAX to do this, can someone help me?
Expected output :
Location Rank Count_X Count_Y Location_1 A 56 10 Location_2 A 38 16 Location_3 A 23 63 Location_4 A 12 65 Location_1 B 75 26 Location_2 B 83 37 Location_3 B 94 74 Location_4 B 35 73 Location_1 C 64 15 Location_2 C 194 64 Location_3 C 285 25 Location_4 C 186 241
(All numbers are here for example)
Thanks all for any help !
2 ACCEPTED SOLUTIONS
Super User
Unpivot your "Table one" to bring it into usable format.
Then you can use implicit measures to achieve your result, no need for DAX.
Community Support
Hi @Ouhla
As @lbendlin mentioned, you need to do unpivot columns for column "X" and column "Y" and then do the calculations.
Here's some dummy data
“Table”
Select two columns in the power query and click unpivot columns.
Create measures.
``````Count_Y =
CALCULATE(
COUNTROWS('Table'),
FILTER(
ALL('Table'),
'Table'[Location] = MAX('Table'[Location])
&&
'Table'[Attribute] = "Y"
&&
'Table'[Rank] = MAX('Table'[Rank])
)
)``````
``````Count_X =
CALCULATE(
COUNTROWS('Table'),
FILTER(
ALL('Table'),
'Table'[Location] = MAX('Table'[Location])
&&
'Table'[Attribute] = "X"
&&
'Table'[Rank] = MAX('Table'[Rank])
)
)
``````
Here is the result.
Regards,
Nono Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 REPLIES 2
Community Support
Hi @Ouhla
As @lbendlin mentioned, you need to do unpivot columns for column "X" and column "Y" and then do the calculations.
Here's some dummy data
“Table”
Select two columns in the power query and click unpivot columns.
Create measures.
``````Count_Y =
CALCULATE(
COUNTROWS('Table'),
FILTER(
ALL('Table'),
'Table'[Location] = MAX('Table'[Location])
&&
'Table'[Attribute] = "Y"
&&
'Table'[Rank] = MAX('Table'[Rank])
)
)``````
``````Count_X =
CALCULATE(
COUNTROWS('Table'),
FILTER(
ALL('Table'),
'Table'[Location] = MAX('Table'[Location])
&&
'Table'[Attribute] = "X"
&&
'Table'[Rank] = MAX('Table'[Rank])
)
)
``````
Here is the result.
Regards,
Nono Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
Unpivot your "Table one" to bring it into usable format.
Then you can use implicit measures to achieve your result, no need for DAX.
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Top Kudoed Authors | 876 | 3,180 | {"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-2024-22 | latest | en | 0.778225 |
http://math.stackexchange.com/questions/169091/finding-product-of-scattered-variables | 1,448,587,623,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447881.87/warc/CC-MAIN-20151124205407-00250-ip-10-71-132-137.ec2.internal.warc.gz | 149,986,154 | 18,103 | # Finding Product of Scattered Variables
Hi I came across the following question where I need to find $$mk$$ from $$(x-2) (x+k) = x^2 + mx - 10$$ The answer is 15. Any suggestions on how I could do that ?
-
Use the relation between the roots of a polynomial and its coefficients. – Host-website-on-iPage Jul 10 '12 at 15:53
The sum of the roots $(2, -k)$ equals $-m$. The product of the roots $-2k=-10$. Therefore $k=5$ and $m=3$.
-
Thanks for the great solution . I realize that 2,-k are the two roots. However I did not know that sum of the roots (2,−k) equals −m. and the product of the roots -10 can be obtained from the equation. Could you show me how you got that or link to any references please.. – Rajeshwar Jul 10 '12 at 16:12
If you expand the left hand side, you get it. $(x-a)(x-b)=x^2+px+q$ $x^2-(a+b)x+ab=x^2+px+q$ It's true for polynomials over any field. – Host-website-on-iPage Jul 10 '12 at 16:13
Aneesh gave the full explanation! – PAD Jul 10 '12 at 16:19
You can find the product $(x-2)(x+k)$, getting $x^2+(k-2)x-2k$. This is supposed to be the same polynomial as $x^2+mx-10$.
So the constant terms must match, and the coefficients of $x$ must match. That gives us $-2k=-10$ amd $k-2=m$. From $-2k=10$, we conclude that $k=5$. Then from $k-2=m$ we conclude that $m=3$. It follows that $mk=15$.
-
Expression$$(x-2) (x+k) = x^2 + mx - 10$$ can be rewriten as $$x^2+(k-2)x-2k = x^2 + mx - 10$$ equating the coefficients next to same power of $x$ we get that $k-2=m$ and $-2k=-10$ or$k=5$ and $m=5-2=3$ that means $$mk=5\times 3=15$$
-
Your second equation (k-2) is missing an x – Rajeshwar Jul 10 '12 at 16:22
there is $x^0=1$ – Milingona Ana Jul 10 '12 at 16:25 | 598 | 1,690 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2015-48 | longest | en | 0.852103 |
cahgaming.com | 1,558,902,289,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259452.84/warc/CC-MAIN-20190526185417-20190526211417-00487.warc.gz | 414,358,225 | 15,355 | Berita Game Terupdate
A comprehension of discrete mathematics is necessary for students of computer science wishing to enhance www.paramountessays.com/ their programming competence. Discrete mathematics deals with discrete objects that are separated from one another. If you believe you are experiencing adult dyslexia then you should think about taking a dyslexia online tests.
## Understanding What Is a Ray in Math
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## The History of What Is a Ray in Math Refuted
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The consequence is fairly surprising. The very first letter has to be the endpoint. It is the same distance from each of the three sides of the triangle.
PHYSICIST An individual who studies physics. You could realize that you’ll just chance to be in the suitable place at the proper moment. UNIVERSE The huge expanse of space which comprises all the issue and energy in existence.
## The Battle Over What Is a Ray in Math and How to Win It
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At every game cycle, we loop by means of these strips, figure out the direction depending on the player’s rotation and cast a ray to get the nearest wall to render. It is very important to remember that the equation of a line in three dimensions isn’t unique. Light and LightIntensity classes are utilised to specify the shading of the object depending on the formula above.
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Nonetheless, it’s quite fantastic to understand when you have to receive your hands dirty because the physics engine isn’t performing properly. There are many websites that provide totally free games. Some search engines calculate the quantity of links to your website from different sites and utilize it to identify your ranking. | 965 | 4,996 | {"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} | 2.671875 | 3 | CC-MAIN-2019-22 | longest | en | 0.947455 |
https://xn--80akibgedmg0bnqw.xn--p1ai/life/fourier-serileri.php | 1,603,429,696,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00718.warc.gz | 1,005,169,138 | 8,302 | # FOURIER SERILERI PDF
Tygolrajas There were four terms. Here it extends the period therefore p is equal to two. Because not only you fourier serileri find both a both a and fourier serileri. So we see srrileri in many instances, integration between -5 and 5 x squared.
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Cosines Description Learn how to make waves of all different shapes by adding up sines or cosines. Make waves in space and time and measure their wavelengths and periods. See how changing the amplitudes of different harmonics changes the waves. Compare different mathematical expressions for your waves. Sample Learning Goals Explain qualitatively how sines and cosines add up to produce arbitrary periodic functions.
Recognize that each Fourier component corresponds to a sinusoidal wave with a different wavelength or period. Mentally map simple functions between Fourier space and real space. Describe sounds in terms of sinusoidal waves. Describe the difference between waves in space and waves in time. Become comfortable with various mathematical notations for writing Fourier transforms, and relate the mathematics to an intuitive picture of wave forms.
Determine which aspect of a graph of a wave is described by each of the symbols lambda, T, k, omega, and n. Translate an equation from summation notation to extended notation. Recognize that the width of a wave packet in position space is inversely related to the width of a wave packet in Fourier space. Explain how the Heisenberg Uncertainty principle results from the properties of waves. Recognize that the spacing between Fourier components is inversely related to the spacing between wave packets, and that a continuous distribution of fourier components leads to a single wave packet.
Version 3.
CONSTITUCION POLITICA DE COLOMBIA 1991 ACTUALIZADA PDF
## Fourier Dönüşümü
.
CONDILOMAS EN GESTANTES PDF
. | 470 | 2,127 | {"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} | 3.453125 | 3 | CC-MAIN-2020-45 | latest | en | 0.858759 |
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Learning Curve B
# Learning Curve B - What if the learning rate is actually 85...
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ESTIMATING PRICES USING LEARNING RATES A buyer has placed an order with a supplier for 100 pieces at a per-unit price of \$281 and has collected the following cost data: Material \$100 Direct labor \$50 (5 hours at \$10 per hour) Overhead \$75 (150% of direct labor) Total costs \$225 Profit (25%) \$56 Total per unit \$281 The buyer now wants to place an order for an additional 700 pieces. Assignment Questions 1. What do you estimate the buyer should pay per unit for the next 700 pieces assuming the supplier demonstrates a 75% learning curve?
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Unformatted text preview: What if the learning rate is actually 85%? What do you estimate is the per unit cost of the next 700 pieces? 2. Under what conditions can we use learning curves to estimate prices? In other words, when does the learning curve apply? 3. Why can we use rough estimates when applying learning curves? 4. Why do manual processes experience greater learning curves than automated processes? 5. Are there factors besides learning that can help reduce costs as volume increases?...
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04-21-2022, 03:01 PM
Post: #1
Dave Britten Senior Member Posts: 2,280 Joined: Dec 2013
(EL-5030) Prime Factors
I got one of these odd little programmable Sharp calculators recently, and worked out how to do the typical mod-30 factor finder with its somewhat-limited "AER" (Algebraic Expression Reserve) programming language.
Notable limitations of AER as seen on this model:
- No arbitrary GOTO, only a looping structure resembling nested WHILE loops.
- IF/THEN/ELSE structure, but no loops allowed inside the THEN/ELSE clauses.
- Can do subroutines, but the subroutines can't contain any loops or conditionals.
- Only comparison operators are >, >=, and <> (not equal). To do other comparisons, swap the order of the arguments, or swap your THEN/ELSE bodies.
It does have arrays with indirect addressing, and that's enough to make this workable.
To run this, use the TITLE key to call up the factor program, press COMP, input a number to factor, and press COMP. The program will stop and display the first factor; press COMP to continue finding additional factors. The last factor will be displayed as a negative number.
Some translation notes, because AER uses a lot of symbols that aren't necessarily present in all fonts:
#1 : Reverse-video numbers (subroutines, e.g. 2ndF+Alpha+1)
=> : STO
\-> : Loop start (2ndF 0)
<-\ : Loop end (2ndF .)
<> : Not equal (2ndF 6)
=Y=>[, =N=>[ : THEN and ELSE clauses (2ndF 1, 2ndF 2)
/ : Division
Note that the last step should be negative X (unary minus, not subtraction).
http://www.arithmomuseum.com/szamologep....25&lang=en
I couldn't find a manual for the EL-5030 anywhere, but the EL-9000 manual is sufficient. The EL-5030 is largely equivalent to the AER I mode of the EL-9000.
Program code has been heavily broken up and indented for clarity, but this should be entered into the calculator without any spacing.
Code:
Main Body M:f(X)= #1 \-> \-> C+A[L]=>C \-> X<>INT (X/C)*C =N=>[ C, X/C=>X <-\ ] L-1=>L L>0 =Y=>[ <-\ ] X>C*C =Y=>[ 8=>L <-\ ] DIM A[1] -X Subroutine 1 #1:DIM A[12] 2=>A[12] 1=>A[11] 2=>A[10] 2=>A[9] 4=>A[8] 2=>A[7] 4=>A[6] 2=>A[5] 4=>A[4] 6=>A[3] 2=>A[2] 6=>A[1] 12=>L 0=>C
04-22-2022, 02:02 AM
Post: #2
Jlouis Senior Member Posts: 764 Joined: Nov 2014
RE: (EL-5030) Prime Factors
Very interesting.
Thanks
04-22-2022, 04:17 PM
Post: #3
Didier Lachieze Senior Member Posts: 1,620 Joined: Dec 2013
RE: (EL-5030) Prime Factors
Nice program! I've entered it on my EL-5030 but it returns 0 for any input.
I double checked the program in my EL-5030 vs. your listing and I don't see any difference.
However I'm wondering if the following sequence is correct:
Code:
=N=>[ C, X/C=>X <-\ ]
What is the meaning of C, ?
04-22-2022, 04:27 PM
Post: #4
Dave Britten Senior Member Posts: 2,280 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-22-2022 04:17 PM)Didier Lachieze Wrote: Nice program! I've entered it on my EL-5030 but it returns 0 for any input.
I double checked the program in my EL-5030 vs. your listing and I don't see any difference.
However I'm wondering if the following sequence is correct:
Code:
=N=>[ C, X/C=>X <-\ ]
What is the meaning of C, ?
I just double-checked the listing to make sure I didn't fat-finger anything when typing it in, and the listing looks okay.
The comma acts as "stop and display the result of the previous expression", so "C," is basically "PRINT C". In this program, C is the trial divisor, and if X<>INT (X/C)*C is false, that means C is a factor of X, so the program displays C, stores X/C into X, then repeats the trial division loop with the same value of C.
04-22-2022, 06:48 PM (This post was last modified: 04-22-2022 08:21 PM by C.Ret.)
Post: #5
C.Ret Member Posts: 241 Joined: Dec 2013
RE: (EL-5030) Prime Factors
If we represent the Dave's nice AER program based on 3-loops and 3-conditionals by the following "inline-graphic":
Where the purple part is an outlaw AER syntax deliberately used for clarity by avoiding the element by element long list of A[..] affectations .
Code:
┌────────────────────────────────────────────────────────────────────────────────────────◄┐ │ ┌────────────────────────────────────────────────────────────────◄┐ │ │ │ ┌─────────────────────────────◄┐ │ │ M:f(X)= #1 └►└►C+A[L]=>C└►X≠INT(X/C)*C =N=>[ C, X/C=>X ┘] ˽ L-1=>L ˽ L>0 =Y=>[ ┘] ˽ X>C*C =Y=>[ 8=>L ┘] ˽ DIM A[1] ˽ -X ■ #1:DIM A[12] ˽ { 6 2 6 4 2 4 2 4 2 2 1 2 }=>A[] ˽ 12=>L ˽ 0=>C
Then, we may consider this close related but shorter and faster AER code using only 2-loops and 2-conditionals :
Code:
┌─────────────────────────────────────────────────────────────────────────────────────────────────◄┐ │ ┌───────────────────────────────────────────────────────────◄┐ │ M:f(X)= #1 └►F+10*FRAC(A[I])=>F ˽ INT(A[I])=>I└►Q=X/F ˽ F>Q =Y=>[ DIM A[1] ˽ -X ■ ] Q≠INT(Q) =N=>[ F, Q=>X ┘] ┘ #1:DIM A[12] ˽ { 2.2 3.1 4.2 5.2 6.4 7.2 8.4 9.2 10.4 11.6 12.2 5.6 }=>A[] ˽ I=1 ˽ F=0
Please fill free to test and correct my code, I (still) have no SHARP EAR hardware...
... but expecting one in a near future
04-22-2022, 07:02 PM
Post: #6
Dave Britten Senior Member Posts: 2,280 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-22-2022 06:48 PM)C.Ret Wrote: If we represent the Dave's nice AER program based on 3-loops and 3-conditionals by the following "inline-graphic":
Thanks, your graphical representation of the program looks good. Aside from the deliberate liberty taken on the array initialization, of course - would be nice if the calculator had a less-verbose method like that! In the Casio version that I based this on, I do 2->O~Z to initialize the whole array with 2 in one shot, then individually store the 6 elements that aren't 2, so it's a little bit shorter.
(04-22-2022 06:48 PM)C.Ret Wrote: Please fill free to test and correct my code, I (still) have no SHARP EAR hardware...
... but expecting one in a near future
Clever trick with stuffing the next array index into the registers! I'll have to try keying it in later to compare the speed and size.
04-22-2022, 08:07 PM (This post was last modified: 04-22-2022 08:12 PM by C.Ret.)
Post: #7
C.Ret Member Posts: 241 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-22-2022 07:02 PM)Dave Britten Wrote: In the Casio version that I based this on, I do 2->O~Z to initialize the whole array with 2 in one shot, then individually store the 6 elements that aren't 2, so it's a little bit shorter.
On the SHARP, you may try the following to short your #1 Subroutine: I found same AER listings using multiple embedded store instructions, but I don't know if it's possible on the EL 5030 ?
Code:
#1: 12=>L ˽ DIM A[L] ˽ 6=>A[3]=>A[1] ˽ 4=>A[8]=>A[6]=>A[4] ˽ 2=>A[L]=>A[10]=>A[9]=>A[7]=>A[5]=>A[2] ˽ 1=>A[11] ˽ 0=>C
04-22-2022, 09:37 PM
Post: #8
Didier Lachieze Senior Member Posts: 1,620 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-22-2022 04:17 PM)Didier Lachieze Wrote: Nice program! I've entered it on my EL-5030 but it returns 0 for any input.
Ok, I’ve found my mistake, I had entered a ◣ at the end of the subroutine, so the program just stopped there. After removing it it works correctly, I get the different factors until it returns -1. For 15 I get 3, 5 and -1.
(04-22-2022 08:07 PM)C.Ret Wrote: On the SHARP, you may try the following to short your #1 Subroutine: I found same AER listings using multiple embedded store instructions, but I don't know if it's possible on the EL 5030 ?
Code:
#1: 12=>L ˽ DIM A[L] ˽ 6=>A[3]=>A[1] ˽ 4=>A[8]=>A[6]=>A[4] ˽ 2=>A[L]=>A[10]=>A[9]=>A[7]=>A[5]=>A[2] ˽ 1=>A[11] ˽ 0=>C
Yes it works on the EL-5030 (without the spaces).
(04-22-2022 06:48 PM)C.Ret Wrote: Then, we may consider this close related but shorter and faster AER code using only 2-loops and 2-conditionals :
Code:
┌─────────────────────────────────────────────────────────────────────────────────────────────────◄┐ │ ┌───────────────────────────────────────────────────────────◄┐ │ M:f(X)= #1 └►F+10*FRAC(A[I])=>F ˽ INT(A[I])=>I└►Q=X/F ˽ F>Q =Y=>[ DIM A[1] ˽ -X ■ ] Q≠INT(Q) =N=>[ F, Q=>X ┘] ┘ #1:DIM A[12] ˽ { 2.2 3.1 4.2 5.2 6.4 7.2 8.4 9.2 10.4 11.6 12.2 5.6 }=>A[] ˽ I=1 ˽ F=0
Please fill free to test and correct my code, I (still) have no SHARP EAR hardware...
... but expecting one in a near future
This one works also with a few changes:
- Replacing Q=X/F by X/F=>Q , I=1 by 1=>I and F=0 by 0=>F
- Initializing individually each element of A[]
For 15 I get 3 and -5
04-23-2022, 04:27 AM (This post was last modified: 04-23-2022 09:45 AM by C.Ret.)
Post: #9
C.Ret Member Posts: 241 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-22-2022 09:37 PM)Didier Lachieze Wrote: This one works also with a few changes:
- Replacing Q=X/F by X/F=>Q , I=1 by 1=>I and F=0 by 0=>F
- Initializing individually each element of A[]
For 15 I get 3 and -5
Thanks for pointing out all those silly mistakes!
As I don't have an AER device, I try my code on BASIC SHARP and I accidentally mixed up codes in the notepad before copying and pasting.
Much better AER code here:
Code:
M:f(X)= #1 └►F+10*FRAC (A[I])=>F˽INT (A[I])=>I└►X/F=>Q˽F>Q =Y=>[DIM A[1]˽-X■]Q≠INT (Q)=N=>[F,Q=>X◄┐]◄┐ #1:DIM A[12]˽2.2=>A[1]˽3.1=>A[2]˽4.2=>A[3]˽5.2=>A[4]˽6.4=>A[5]˽7.2=>A[6]˽8.4=>A[7]˽9.2=>A[8]˽10.4=>A[9]˽11.6=>A[10]˽12.2=>A[11]˽5.6=>A[12]˽1=>I˽0=>F
This code is longer than it appears on the voluntarily reduced form!
I'm not sure it's shorter or faster anymore since the F>Q 'end-test' is done at each iteration; in the initial code the test X>C*C only takes place once after the 8th iteration. How much time does it save or lose?
I imagine the speed improvements will greatly depend on specific values for X.
Can the sequence INT (A[I])=>I be replaced by a simpler A[I]=>I ?
Did the EL-5030 accept the reduced syntax INT A[I] or FRAC A[I] instead of INT (A[I]) or FRAC (A[I]) like the PC-1211 does?
04-23-2022, 08:31 AM
Post: #10
Didier Lachieze Senior Member Posts: 1,620 Joined: Dec 2013
RE: (EL-5030) Prime Factors
(04-23-2022 04:27 AM)C.Ret Wrote: Can the sequence INT (A[I])=>I be replaced by a simpler A[I]=>I ?
Did the EL-5030 accept the reduced syntax INT A[I] or FRAC A[I] instead of INT (A[I]) or FRAC (A[I]) like the PC-1211 does?
Yes to both. So the updated code below works fine on the EL-5030:
Code:
M:f(X)= #1└►F+10*FRAC A[I]=>F˽A[I]=>I└►X/F=>Q˽F>Q =Y=>[DIM A[1]˽-X◣]Q≠INT Q=N=>[F,Q=>X◄┐]◄┐ #1:DIM A[12]˽2.2=>A[1]˽3.1=>A[2]˽4.2=>A[3]˽5.2=>A[4]˽6.4=>A[5]˽7.2=>A[6]˽8.4=>A[7]˽9.2=>A[8]˽10.4=>A[9]˽11.6=>A[10]˽12.2=>A[11]˽5.6=>A[12]˽1=>I˽0=>F
04-23-2022, 05:07 PM (This post was last modified: 04-23-2022 05:43 PM by C.Ret.)
Post: #11
C.Ret Member Posts: 241 Joined: Dec 2013
RE: (EL-5030) Prime Factors
We may spare one element in matrix/vector A[..]:
Code:
M:f(X)=#1└►└►X/F꞊>Q˽F>Q ꞊Y꞊>[-X◣]Q≠INT Q ꞊N꞊>[F,Q꞊>X◄┐]F+10*FRAC A[I]꞊>F˽A[I]꞊>I◄┐ 1:DIM A[11]˽2.1꞊>A[1]˽3.2꞊>A[2]˽4.2꞊>A[3]˽5.4꞊>A[4]˽6.2꞊>A[5]˽7.4꞊>A[6]˽8.2꞊>A[7]˽9.4꞊>A[8]˽10.6꞊>A[9]˽11.2꞊>A[10]˽4.6꞊>A[11]˽1꞊>I˽2꞊>F
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From Gene Weingarten’s recent column, “Rhymes Against Humanity,” in the January 28 edition of the Washington Post Magazine:
An infinite number of mathematicians
Walked into a bar on one recent night,
And, under the strangest of barroom conditions,
What followed quite nearly became a big fight.
“I’ll have a pint,” said the first to the ’tender.
“I’ll have a half,” said the next fellow down.
“I’ll have a quarter,” said the third (no big spender).
“Give me an eighth,” said the next, like a clown.
The bartender fumed and grew suddenly pale
Then, calmly, he turned and he went to the spout
Drew up two pints, set them down at the rail.
Said, “Enough of this nonsense — you all work it out.”
This is an MJ4MF original, though like Gene’s, it’s based on a stale, old joke:
With my head in an oven
And my feet on some ice,
I’d say that, on average,
I feel rather nice!
What other classic math jokes can be easily converted to poems? Or have already been?
### Stupid Stats
C’mon, now… really?
Uterine size in non-pregnant women varies in relation to age and gravidity [number of pregnancies]. The mean length-to-width ratio conformed to the golden ratio at the age of 21, coinciding with peak fertility.
Claiming that a uterine golden ratio coincides with peak fertility is highly suspect. The good folks at Ava Women claim that, “Most women reach their peak fertility rates between the ages of 23 and 31.” Information at Later Baby states, “Female fertility and egg quality peak around the age of 27.” And WebMD says, “A woman’s peak fertility is in her early 20s.” So, there seems to be some debate about when peak fertility actually occurs. Consequently, this strikes me as retro-fitting, and it seems that Dr. Verguts and his colleagues may have played loose with the age of peak fertility in order to make a connection to the golden ratio.
In their defense, though, it’s not the first time that folks have gone uptown trying to find a connection to the golden ratio. A claim by The Golden Number states, “[The DNA molecule] measures 34 angstroms long by 21 angstroms wide for each full cycle of its double helix spiral,” and 34/21 ≈ 1.6190476, which is approximately equal to φ, 1.6180339.
Though this guy — an honest-to-goodness biologist — seems to disagree:
I’ve also heard folks say that people are perceived as more beautiful if certain bodily proportions are in the golden ratio. The most extreme example of this that I’ve found involves the teeth:
…the most “beautiful” smiles are those in which central incisors are 1.618 wider than the lateral incisors, which are 1.618 wider than canines, and so on.
In a study of 4,572 extracted adult teeth, Dr. Julian Woelfel found the average width of the central incisor to be 8.6 mm. If the teeth in a beautiful smile follow the geometric progression described above, well, that would imply that the first molar would be just 8.6 × 0.6185 ≈ 0.8 mm wide, which isn’t reasonable and, moreover, is not even remotely close to the average width that Dr. Woelfel found for the first molar: approximately 10.4 mm.
But all of these claims involving the golden ratio are not even close to being the stupidest statistics I’ve heard in my life. Mary Anne Tebedo made a remark on the floor of the Colorado State Senate in 1995 that may hold that distinction:
Statistics show that teen pregnancy drops off significantly after age 25.
Of course, it’s hard to call that a statistic, since it’s completely nonsensical. Maybe it’s only the stupidest statement I’ve ever heard.
Then there’s this one, from the New York Times on August 8, 2016, which couldn’t be more useless:
No presidential candidate has secured a major party nomination after an FBI investigation into her use of a private email server.
Well, duh. Email didn’t even exist before the 1970’s. Moreover, besides Hillary Clinton, has any presidential candidate ever had their use of a private server investigated by the FBI? This is like saying, “No one has ever been named People‘s Sexiest Man Alive after writing a math joke book.” (Not yet, anyway.)
Randall Munroe made fun of these types of “no politician has ever…” claims in 2012 with his cartoon Election Precedents:
And it’s true:
But perhaps my all-time favorite is one that Frank Deford — may he rest in peace — included in his piece “The Stupidest Statistics in the Modern Era” on NPR’s Morning Edition:
He’s [Brandon Phillips] the first National League player to account for as many as 30 steals and 25 double plays in one season.
About this stat, Deford commented, “Steals and double plays together? This is like saying, ‘He’s the first archaeologist to find 23 dinosaur bones and 12 Spanish doubloons on the same hunt.'” (I sure am going to miss him.)
The preponderance of dumb stats shouldn’t come as a surprise, though. A recent study found that people deemed real news headlines to be accurate 83% of the time and fake news headlines to be accurate 75% of the time. So, if we can’t tell truth from fiction, how can we possibly distinguish useful statistics from inane?
If you’d like to test your ability to detect fake news, check out Factitious from American University.
### WODB, Quora Style
The following puzzle was recently posted on Quora:
Which of the following numbers don’t belong: 64, 16, 36, 32, 8, 4?
What I liked about this puzzle was the answer posted by Danny Mittal, a sophomore at the Thomas Jefferson High School for Science and Technology. Danny wrote:
64 doesn’t belong, as it’s the only one that can’t be represented by fewer than 7 binary bits.
36 doesn’t belong, as it’s the only one that isn’t a power of 2.
32 doesn’t belong, as it’s the only one whose number of factors has more than one prime factor.
16 doesn’t belong, as it’s the only one that can be written in the form xy, where x is an integer and y is a number in the list.
8 doesn’t belong, as it’s the only one that doesn’t share a digit with any other number in the list.
4 doesn’t belong, as it’s the only one that’s a factor of all other numbers in the list.
I suspect that Danny has visited Which One Doesn’t Belong or has read Christopher Danielson’s Which One Doesn’t Belong. Or maybe he’s just a math teacher groupie and trolls MTBoS.
But then Jim Simpson pointed out the use of “don’t” in the problem statement, which I had assumed was a grammatical error. Jim interpreted this to mean that there must be two or more numbers that don’t belong for the same reason, and with that interpretation, Jim suggested the answer was 32 and 8, since all of the others are square numbers.
Don’t get me wrong — I don’t think this is a great question. But I love that it was interpreted in many different ways. It could lead to a good classroom conversation, and it makes me consider all sorts of things, not the least of which is standardized assessments. How many times have students gotten the wrong answer for the right reason, because they interpreted an item on a state exam or the SAT differently than the author intended? And how many times have we bored students with antiseptic questions, only because we knew they’d be free from such alternate interpretations? Both scenarios make me sad.
### Four, or F**k You?
If you asked a student, “How many sides does a quadrilaterals have?” and you received the following response…
…well, you might be upset.
But perhaps the student learned to count in binary on her fingers, where the right thumb is the register for 1, the right index finger is the register for 2, the right middle finger is the register for 4, and so on. Then the response above would be appropriate, despite appearances.
If you then asked, “Into how many regions will a circle be divided if 6 points are placed randomly on a circle, and each point is connected to every other point?” the student might appear to wave at you — or, she may just be telling you (correctly) that 31 regions would be created by holding up all 5 fingers. (In binary counting, all five fingers add up to 1 + 2 + 4 + 8 + 16 = 31.)
My sons learned to count in binary when, at age 5, they asserted that the highest you can count on your fingers is 10. “Actually,” I told them, “You can count as high as 1,023 on your fingers. If you want, I can show you how.”
Of course, they wanted to learn, and I was happy to teach them. There are at least four good reasons for teaching students to count in other bases, and “Dr. Peterson” at the Math Forum had this to say:
I taught my son to multiply in binary before he really learned it in decimal, because it’s easier; you have only the algorithm (method) with no multiplication tables to learn.
Knowing how bases work helps to develop number sense while clarifying the concept of place value. And not understanding place value leads to things like this…
My former boss shared this video with me on Facebook recently, and he asked,
Does this work with other numbers?
I had a fun time playing with that question, so let me now give you a chance to think about it. Can you find another pair of numbers that produce analogous incorrect results when multiplying and dividing? And if you’re feeling really ambitious, can you generalize to determine what types of number pairs will always give these kinds of incorrect results?
The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.
## MJ4MF (offline version)
Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature. | 2,296 | 9,802 | {"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-2022-33 | latest | en | 0.960354 |
https://discuss.leetcode.com/topic/30166/easy-python-o-n-o-1-solution | 1,516,411,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00762.warc.gz | 653,864,685 | 11,187 | # Easy Python O(n) - O(1) solution
• ``````class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
xor = 0
a = 0
b = 0
for num in nums:
xor ^= num
for num in nums:
a ^= num
else:
b ^= num
return [a, b]``````
• class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
nums_map, result = collections.defaultdict(list), []
for num in nums:
nums_map[num].append(1)
`````` for key in nums_map:
if len(nums_map[key]) == 1:
result.append(key)
return result``````
• Your idea is quite delicate but somehow not easy to understand.
Here is how I understand your thought (let's take nums = [1, 2, 1, 3, 2, 5] as example):
1. The binary mask has only one digit equals to one (in this case, mask = 2 = 0010)
2. xor = 3^5 = 6 = 0110.
3. for all numbers in nums other than 3, 5, each pair will counteract themselves no matter this pair belongs to 'num&mask ==0' or 'num&mask !=0', what you wanna do is put 3 and 5 into these two different statements, respectively.
4. thus, for xor = 0110, each '1' digit comes from 3 (0011) or 5 (0101) according to the property of operator '^'.
5. based on the while loop, xor&mask != 0 means there is only one common digit of both xor and mask that equals to 1 (xor = 0110, mask = 0010) and this '1' digit either comes from 3 or 5.
6. Thus, 3&mask and 5&mask must have different results (0011&0010 = 0010, 0101&0010 = 0000), which means 'a ^=num' won't contain both the cases of (num=3, num=5).
7. based on the steps above, the 3 and 5 will be sure to be assigned to a and b, respectively.
• Great idea!
You first xor all numbers and then find a bit that is different in those two distinct numbers. This bit is used to divide all numbers into two group. Finally xor them individually, the two numbers left are the two distinct numbers. | 559 | 1,858 | {"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} | 3.140625 | 3 | CC-MAIN-2018-05 | latest | en | 0.820044 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Author Message
DJ Amdle
Registered: 04.01.2005
From: San Diego CA USA
Posted: Saturday 30th of Dec 07:58 I am going to college now. As math has always been my weakness, I purchased the course material in advance. I am plan studying a handful of topics before the classes begin . Any kind of tools would be much appreciated that could help me to start studying fundamental math poems myself.
AllejHat
Registered: 16.07.2003
From: Odense, Denmark
Posted: Monday 01st of Jan 07:02 I do understand what your problem is, but if you could elaborate a bit on the areas in which you are facing difficulty , then I might be in a better position to help you out . Anyhow I have a suggestion for you, try Algebrator. It can solve a wide variety of questions, and it can do so within minutes. And that’s not it , it also gives a detailed step-by-step description of how it arrived at a particular answer. That way you don’t just find a solution to your problem but also get to understand how to go about solving it. I found this software to be particularly useful for solving questions on fundamental math poems. But that’s just my experience, I’m sure it’ll be good for other topics as well.
daujk_vv7
Registered: 06.07.2001
From: I dunno, I've lost it.
Posted: Tuesday 02nd of Jan 07:57 I agree. Algebrator not only gets your homework done faster, it actually improves your understanding of the subject by providing very useful information on how to solve similar questions. It is a very popular online help tool among students so you should try it out.
Homuck
Registered: 05.07.2001
From: Toronto, Ontario
Posted: Wednesday 03rd of Jan 08:43 Algebrator is the program that I have used through several algebra classes - Basic Math, Pre Algebra and Basic Math. It is a really a great piece of algebra software. I remember of going through difficulties with least common denominator, multiplying fractions and linear algebra. I would simply type in a problem from the workbook , click on Solve – and step by step solution to my math homework. I highly recommend the program. | 729 | 2,944 | {"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-2018-05 | latest | en | 0.908415 |
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A241635 Number of compositions of n with exactly ten descents. 3
%I #8 Nov 28 2023 10:48:16
%S 89,1529,15072,109700,651755,3333143,15159290,62673818,239267852,
%T 853417204,2869778022,9163554647,27947044811,81799126374,230699706092,
%U 629085587197,1663426352840,4275866388599,10708451601656,26178837135330,62580214195713,146503017803381
%N Number of compositions of n with exactly ten descents.
%H Joerg Arndt and Alois P. Heinz, <a href="/A241635/b241635.txt">Table of n, a(n) for n = 30..1000</a>
%p b:= proc(n, i) option remember;
%p `if`(j<i, x, 1), j=1..n), x, 11), polynom))
%p end:
%p a:= n-> coeff(b(n, 0), x, 10):
%p seq(a(n), n=30..55);
%t b[n_, i_] := b[n, i] = If[n == 0, 1, Sum[b[n - j, j]*
%t If[j < i, x, 1], {j, 1, n}] // Expand];
%t a[n_] := Coefficient[b[n, 0], x, 10];
%t Table[a[n], {n, 30, 55}] (* _Jean-François Alcover_, Nov 28 2023, from Maple code *)
%Y Column k=10 of A238343 and of A238344.
%K nonn,changed
%O 30,1
%A _Joerg Arndt_ and _Alois P. Heinz_, Apr 26 2014
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Last modified December 3 18:58 EST 2023. Contains 367540 sequences. (Running on oeis4.) | 622 | 1,783 | {"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} | 3.328125 | 3 | CC-MAIN-2023-50 | latest | en | 0.66635 |
https://forums.wolfram.com/mathgroup/archive/2004/Dec/msg00142.html | 1,721,723,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00750.warc.gz | 228,932,945 | 7,906 | Re: Complex Analysis using Mathematica
• To: mathgroup at smc.vnet.net
• Subject: [mg52656] Re: Complex Analysis using Mathematica
• From: David Bailey <dave at Remove_Thisdbailey.co.uk>
• Date: Tue, 7 Dec 2004 04:09:43 -0500 (EST)
• References: <cos1eq\$dnm\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Pratik Desai wrote:
> Here we go again,
>
> I have to define a complex function
> So I go through this procedure to define that the variables are really "real"
>
> TagSet[p, Im[p], 0];
> TagSet[a, Im[a], 0];
> TagSet[b, Im[b], 0];
> TagSet[p, Re[p], p];
> TagSet[a, Re[a], a];
> TagSet[b, Re[b], b];
> lamda = a + I*b
> z = ComplexExpand[lamda*p]
> x=Re[z]
> y=Im[z]
> TagSet[u, Im[u[x, y]], 0];
> TagSet[v, Im[v[x, y]], 0];
> TagSet[x, Re[x], x];
> TagSet[y, Re[y], y];
> TagSet[u, Re[u[x, y]], u[x, y]];
> TagSet[v, Re[v[x, y]], v[x, y]];
>
>
> Then I define my actual function
>
> u1 = TrigToExp[Sinh[z]] (*By this time I have realized that
> Mathematica or for that matter most of the CAS work better with
> exponentials when it comes to complex analysis*)
>
> u[x, y] = Re[u1]
> v[x, y] = Im[u1]
>
> The problem I face is that the software is not able to identify x and y
> as I have defined above. May be I am making a trivial mistake. Please
>
>
>
>
> Pratik Desai
>
>
Pratik,
I think the most useful way to procede would be if you gave an explicit,
simple example of the kind of input you would like to give Mathematica
and the kind of output you would like it to produce. I am fairly certain
that if you do this you will get a whole bunch of replies telling you
how to achieve it and they will look a lot neater than what you have at
present!
Regards,
David Bailey
```
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# MTH301 Assignment No 03 Solution & Discusion Due Date:28-01-2013
MTH301 (Fall 2012)
Total marks: 20
Lecture # 23 to 33
Due Date: January 28, 2013
DON’T MISS THESE Important instructions:
• Two questions are graded. But you have to send the solution of all the 3 questions.
• All students are directed to use the font and style of text as is used in this document.
• This is an individual assignment, not group assignment, so keep in mind that you are supposed to submit your own, self made & different assignment even if you discuss the questions with your class fellows. All similar assignments (even with some meaningless modifications) will be awarded zero marks and no excuse will be accepted. This is your responsibility to keep your assignment safe from others.
• Solve the assignment on MS word document and upload your word (.doc) files only
Question # 1:
Find the volume of the solid in the first octant bounded by the coordinate planes, the plane, and the parabolic cylinder
Question 2:
If is the position of a particle in space at time .Find the angle between the velocity and acceleration vectors at time
Question 3:
For the complete revolution( 0≤ t≤ 2π ) of the helix curve: x = Cost, y = Sint and z = t, evaluate .
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MTH301_Assignment#03_Question#01_Solution
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Yes, 16
sir where is Q3
Tariq Bro.. x ki 2 wali limit kesi i...? Please explain
x ki nhi y ki aii hai coz humare pas 4-y^2 hai y mein hum 2 put karein toh 2^2 = 4 so, 4-4 = 0
agr 2 se zyada leinge limit toh answer -ive aega jo nhi aya chaheye so limit 2 aegi...
thanks dear me solved this question... But jis tarha ap keh rahi ho ye aise nahi hai
phr kaise hain
dear z ki value ap k paas hai let c
z=0 to z=4-y^2
then
4-y^2=0
y^2=4
n y=2 so y=0 to y=2 hope u understand
got it thnx
It's ok
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3 | 959 | 3,454 | {"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} | 3.09375 | 3 | CC-MAIN-2020-34 | latest | en | 0.894566 |
https://www.univerkov.com/a-car-leaves-the-city-at-a-speed-of-18m-s-after-20-minutes-a-second-car-leaves-after/ | 1,660,440,078,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00513.warc.gz | 922,878,421 | 6,334 | # A car leaves the city at a speed of 18m / s, after 20 minutes a second car leaves after
A car leaves the city at a speed of 18m / s, after 20 minutes a second car leaves after it, at what speed did the second car move if it caught up with the first one an hour after the start of its movement?
V1 = 18 m / s.
t1 = 20 min = 1200 s.
t2 = 1 h = 3600 s.
V2 -?
With uniform rectilinear movement, the path S traversed by the body is determined by the formula: S = V * t, where V is the speed of movement, t is the time of movement.
Before the meeting, the first car covered the path S1 = V1 * t, where V1 is the speed of the first car, t is the travel time of the first car before the meeting.
t = t1 + t2.
S1 = V1 * (t1 + t2).
Before the meeting, the second car covered the path S2 = V2 * t2, where V2 is the speed of the second car, t2 is the travel time of the second car before the meeting.
S1 = S2.
V1 * (t1 + t2) = V2 * t2.
The speed of the second car V2 will be determined by the formula: V2 = V1 * (t1 + t2) / t2.
V2 = 18 m / s * (1200 s + 3600 s) / 3600 s = 24 m / s.
Answer: the speed of the second car is V2 = 24 m / s.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 422 | 1,452 | {"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} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.940997 |
https://www.coursehero.com/file/6656266/HW3/ | 1,498,455,917,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00412.warc.gz | 840,578,607 | 23,052 | # HW3 - EEL 3105 Fall 2011 Homework 3 For grading and credit...
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Unformatted text preview: EEL 3105 Fall 2011 Homework 3 For grading and credit: 1. Suppose G ( s) 2s 3 s 2s 4s 2 2s 1 4 3 Use MATLAB to find the roots of the denominator. Then compute the partial fraction expansion for G(s). 2. Suppose G ( s) as b ( s 1)( s 2)( s 3) Roots of the denominator are called poles of G(s) while the roots of the numerator are called zeros of G(s). [A pole of G(s) corresponds to a value of s where G(s) becomes infinity; a zero corresponds to a value of s for which G(s) equals 0.] Suppose the residue corresponding to the pole at ‐1 is 1 and to the pole at ‐2 is 3. Find values of a and b. What is the residue corresponding to the pole at ‐3? What is the zero of G(s)? 3. Consider two phasors x(t ) 2e j (120 t /4) y (t ) 3e j (120 t /3) Suppose we add the two signals x(t) and y(t), let z(t)=x(t)+y(t). Find a phasor representation of z(t). What is its frequency? What are its magnitude and phase? Provide a graphical depiction of this addition operation in the complex plane. [For your own understanding, connect this to vector addition in the 2 dimensional plane.] 4. Let z1 and z2 be two unknown complex numbers. Suppose we are told that z1 e j z2 z1 z2 r 2 Assume r > 0. Determine z1 and z2 in terms of r and . ...
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## This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.
Ask a homework question - tutors are online | 475 | 1,660 | {"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} | 3.765625 | 4 | CC-MAIN-2017-26 | longest | en | 0.910053 |
https://twiki-edlab.cs.umass.edu/bin/view/H401/BuchiSTheorem?rev=2 | 1,575,658,947,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00400.warc.gz | 590,860,769 | 6,950 | Tags:
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Buchi's Theorem
A language is definable in second-order monadic logic if and only if it is regular.
Proof
Let L be a regular language with associated DFA D = (Q, q1, δ, F). We construct a formula Φ describing an accepting run of D on a string (Φ is true iff there is an accepting run of D on the string).
Let Q = {q1, q2,..., qn).
Φ can be described as: ∃C1∃C2...∃Cn: ∀x: ( should be in some state) ∧ ( can be in only 1 state) ∧ (zero(x) → start at q1) ∧ ( follow the right transitions) ∧ (last(x) → end at a final state)
• should be in some state: [C1(x) ∨ C2(x) ∨ ... ∨ Cn(x)]
• can be in only 1 state: for every pair of states qi, qj, add the following: [Ci(x) → ¬Cj(x)]. Connect these together with ∧.
• start at q1: C1(x)
• follow the right transitions: for every transition δ(qi, a) = qj, add the following: [(Ci(x) ∧ Ia(x)) → Cj(x+1)]. Connect these together with ∧.
• end at a final state: for every transition δ(qi, a) = qj such that qj∈F, add the following: [Ia(x) ∧ Ci(x)]. Connect these together with ∨.
zero(x) - x is the beginning of the string, (x+1), and last(x) - x is the end of the string, can be easily described in first-order logic. | 373 | 1,182 | {"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} | 3.453125 | 3 | CC-MAIN-2019-51 | latest | en | 0.839056 |
http://rostovmonetniy.ru/relativistic-electrodynamics-pdf.html | 1,563,330,205,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00303.warc.gz | 135,419,300 | 5,358 | # Relativistic Electrodynamics Pdf
## Classical Electrodynamics
The results show that the experiment is a pure expression of the transverse Doppler effect. Subsequently, a large number of scattering and radiation processes involving electrons, positrons, and photons are introduced and their theoretical treatment is presented in great detail.
Acceleration and Forces in Relativity. In particular, as is seen from Eqs. Thus, we will follow the approach developed in the previous section.
The quantity Ws is a relativistic invariant, and so Eq. It is no accident that the equations of electrodynamics can be written in the beautifully elegant form of Eq. We now discuss the application of the special theory of relativity to electrodynamics. Our approach rests on the metric-free integral formulation of the conservation laws of electrodynamics in the tradition of F. To solve this problem one needs to pass the complex plane, relational algebra in database pdf according to which we represent Eq.
Hence, the considered setup of coherent bremsstrahlung in a crystal may serve as a powerful mechanism for prebunched electron beam superradiation, at moderate relativistic energies of electron beams. Continuum Solid Fluid Acoustics. To show the role of initial conditions in Fig. The present text aims at a balance between basic theory and practical applications, and includes introductions to specific quantum mechanical effects.
According to the latter in Eqs. As the modulated particle beam radiates coherently this mechanism can be of interest in astrophysics where the radiating matter may be in a strongly nonstationary state.
We assume that initially ions are in the ground state, so that in Eq. The style pedagogical with clear and concise illustration followed by practise problems at the end of each chapter. Instinct should give the right answer. The contour integration where A in Eq. We can use this fact to simplify our calculations.
American Journal of Science. Considering U x as a perturbation, from the classical equations of motion one can obtain the perturbed velocity of the electron, which is responsible for the coherent bremsstrahlung. The emphasis in this text is on classical electromagnetic theory and electrodynamics, that is, dynamical solutions to the Lorentz-force and Maxwell's equations. At the discontinuity of the dielectric permittivity in general, properties of the medium only the derivatives of the physical quantities can have large values.
We thus give consideration to the induced channeling process in general aspects of coherent interaction of relativistic electrons and positrons with a plane electromagnetic wave in a crystal. The use of retarded potentials to describe electromagnetic fields from source-charges is an expression of relativistic electromagnetism.
The book emphasizes problem solving, contains abundant problem sets, and is conveniently organized to meet the needs of both student and instructor. Nikishov, Problemi Teoreticheskoi. Hence, we will keep the second-order derivatives in Eq. Our program is to do in the four dimensions of space-time all of the things we did with vectors for three dimensions. To illustrate the particle acceleration in the capture regime we will represent the results of numerical solution of Eqs.
It is especially important in this process, because at the abrupt temporal variation of the dielectric permittivity the hard quanta in the spectrum of the initial radiation arise. For the numerical calculations we have chosen the initial electron momentum p to be colinear with the laser propagation direction.
## Relativistic electromagnetism
Classical mechanics Continuum Solid Fluid Acoustics. The electromagnetic wave rays through the absorber increases, which is subsequently measured by the stationary counter beyond the absorber. But after the unitarian transformation for the transformed function the variables are separated. We would like to ask you for a moment of your time to fill in a short questionnaire, at the end of your visit. In this case the set of Eqs.
Theorems and Concepts of Vector Calculus E. What we call the scalar and vector potentials are really different aspects of the same physical thing. In addition, note that Eqs. The expressions of particle wave functions show that the degeneration of particle states over the spin projection that takes place in vacuum Volkov states vanishes in a dielectriclike medium.
Carefully constructed problems complement the material of the text, and introduce new topics. However, a moving observer looking at the same set of charges does perceive a current, and thus a magnetic field. It is a thorough introductory text providing all necessary mathematical tools together with many examples and worked problems. Consequently, substituting Eq. Yerevan, Armenia June Hamlet K.
However, there is more to the simplicity of the laws of electromagnetism written in the form of Eq. The main impending factor in the coherent bremsstrahlung process, which we have not taken into account in Eq. Can we from this equation deduce something that could not be deduced from the wave equations for the potentials in terms of the charges and currents? In this case it makes sense to take into account the transitions to the states with other energies in higher order.
The point is that it must be possible to write the equations of physics so that both sides transform the same way under rotations. Here is how you calculate the unworldliness. This book covers information relating to physics and classical mathematics that is necessary to understand electromagnetic fields in materials and at surfaces and interfaces.
This completely revised and corrected new edition provides several new examples and exercises to enable deeper insight in formalism and application of Quantum electrodynamics. The book should be of great value to all physicists, from first-year graduate students to senior researchers, and to all those interested in electrodynamics, field theory, and mathematical physics. Then, at these energies a few discrete energy levels in the transverse potential well of the channeled electron are formed that are not equidistant. At these densities the electron component of the superdense plasma is highly degenerate the dispersion of the transverse electromagnetic waves is determined by electrons. Then when you are moving at a uniform velocity in a spaceship, all of the laws of nature transform together in such a way that no new phenomenon will show up.
Classical Electrodynamics. Then any Lorentz transformation plus a rotation will leave the quantity unchanged. Quantitative analysis of Eqs.
## Theory of Light Emission and Application to Free Electron Lasers
Many applications are given in the text. We hope that our book can be seen in the classical tradition of the book by E. My apologies go to all authors whose works are not covered in this book. It means more, just as a theory of vector analysis means more.
Various problems are presented, together with their worked-out solutions. With this broad coverage of topics, the series is of use to all research scientists and engineers who need up-to-date reference books.
We have the result that if charge is conserved in one coordinate system, it is conserved in all coordinate systems moving with uniform velocity. Students in astro physics, astronomy and cosmology who want to understand general and special relativity theory. To compensate it we will consider the case when the electron beam current density is initially modulated. Whereas the experimental results on atomic systems frequently had preceded the theoretical ones, the experimental investigations on free electrons began gathering power only recently. | 1,475 | 7,749 | {"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-2019-30 | latest | en | 0.926777 |
https://metanumbers.com/132518 | 1,631,970,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00435.warc.gz | 451,957,951 | 7,371 | # 132518 (number)
132,518 (one hundred thirty-two thousand five hundred eighteen) is an even six-digits composite number following 132517 and preceding 132519. In scientific notation, it is written as 1.32518 × 105. The sum of its digits is 20. It has a total of 3 prime factors and 8 positive divisors. There are 65,704 positive integers (up to 132518) that are relatively prime to 132518.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 20
• Digital Root 2
## Name
Short name 132 thousand 518 one hundred thirty-two thousand five hundred eighteen
## Notation
Scientific notation 1.32518 × 105 132.518 × 103
## Prime Factorization of 132518
Prime Factorization 2 × 173 × 383
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 132518 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 132,518 is 2 × 173 × 383. Since it has a total of 3 prime factors, 132,518 is a composite number.
## Divisors of 132518
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 200448 Sum of all the positive divisors of n s(n) 67930 Sum of the proper positive divisors of n A(n) 25056 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 364.03 Returns the nth root of the product of n divisors H(n) 5.28887 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 132,518 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 132,518) is 200,448, the average is 25,056.
## Other Arithmetic Functions (n = 132518)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 65704 Total number of positive integers not greater than n that are coprime to n λ(n) 32852 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12341 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 65,704 positive integers (less than 132,518) that are coprime with 132,518. And there are approximately 12,341 prime numbers less than or equal to 132,518.
## Divisibility of 132518
m n mod m 2 3 4 5 6 7 8 9 0 2 2 3 2 1 6 2
The number 132,518 is divisible by 2.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (132518)
Base System Value
2 Binary 100000010110100110
3 Ternary 20201210002
4 Quaternary 200112212
5 Quinary 13220033
6 Senary 2501302
8 Octal 402646
10 Decimal 132518
12 Duodecimal 64832
20 Vigesimal gb5i
36 Base36 2u92
## Basic calculations (n = 132518)
### Multiplication
n×y
n×2 265036 397554 530072 662590
### Division
n÷y
n÷2 66259 44172.7 33129.5 26503.6
### Exponentiation
ny
n2 17561020324 2327151291295832 308389434819941064976 40867151123468950048489568
### Nth Root
y√n
2√n 364.03 50.9829 19.0796 10.5793
## 132518 as geometric shapes
### Circle
Diameter 265036 832635 5.51696e+10
### Sphere
Volume 9.74795e+15 2.20678e+11 832635
### Square
Length = n
Perimeter 530072 1.7561e+10 187409
### Cube
Length = n
Surface area 1.05366e+11 2.32715e+15 229528
### Equilateral Triangle
Length = n
Perimeter 397554 7.60414e+09 114764
### Triangular Pyramid
Length = n
Surface area 3.04166e+10 2.74257e+14 108200
## Cryptographic Hash Functions
md5 ace492111e44aabe42e9f52bbb4a325a 524db653169be86a031b97e15527d44c34e31212 e3483ca718f003592d78e87796cc0aa6520f2311ba1262c738347ce20cb21a4e 8d8c5f4c8d73ea2635029c19180584dae4dbac2d8ed2251ff2463ea68f4292edc03fcc60a5c46b7a3fa45a765377aeba6c57b9b27802c07454d79270f60debef 2f391ecd0d691b0b872c0ce96d57a6ebf23e7706 | 1,420 | 4,130 | {"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} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.809886 |
https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/gfanInterface/html/_gfan__To__Latex.html | 1,726,287,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00064.warc.gz | 345,115,054 | 3,083 | next | previous | forward | backward | up | index | toc
# gfanToLatex -- convert a list of polynomials to LaTeX
## Synopsis
• Usage:
gfanToLatex(L)
gfanToLatex(M)
• Inputs:
• L, a list, of marked polynomial lists or a marked polynomial list (for the polynomialset option)
• Optional inputs:
• h
• polynomialset
• polynomialsetlist
• Outputs:
• S, , LaTeX markup for L
## Description
This method converts marked polynomial lists and lists of marked polynomial lists to latex. If the given input is a list, the option polynomialsetlist is assumed. Similarly, if the input is a marked polynomial list then polynomialset is assumed.
i1 : QQ[x,y,z]; i2 : L = gfan{x^2 + y*z, z^2 + y*z} 2 2 2 2 2 4 2 2 2 o2 = {{(y*z) + z , (x ) - z }, {(x ) + y*z, (z ) + y*z}, {(x ) - x y , (x z) ------------------------------------------------------------------------ 2 2 2 2 2 2 2 2 2 2 4 + x y, (z ) - x , (y*z) + x }, {(x z) + x y, (z ) - x , (x y ) - x , ------------------------------------------------------------------------ 2 2 2 2 2 2 (y*z) + x }, {(x y) + x z, (y*z) + x , (z ) - x }} o2 : List i3 : gfanToLatex L o3 = $\{\{yz+z^{2}, x^{2}-z^{2}\} , \{x^{2}+yz, z^{2}+yz\} , \{x^{4}-x^{2}y^{2}, x^{2}z+x^{2}y, z^{2}-x^{2}, yz+x^{2}\} , \{x^{2}z+x^{2}y, z^{2}-x^{2}, x^{2}y^{2}-x^{4}, yz+x^{2}\} , \{x^{2}y+x^{2}z, yz+x^{2}, z^{2}-x^{2}\} \}$ i4 : gfanToLatex first L o4 = $\{yz+z^{2}, x^{2}-z^{2}\}$ i5 : gfanToLatex({{x,z}, {x+y, x+z}}, "polynomialset" => true) o5 = $\{x+y, z+x\}$
gfan Documentation
This program converts ASCII math to TeX math. The data-type is specified by the options.
Options:
-h:
Add a header to the output. Using this option the output will be LaTeXable right away.
--polynomialset_:
The data to be converted is a list of polynomials.
--polynomialsetlist_:
The data to be converted is a list of lists of polynomials.
## Ways to use gfanToLatex :
• gfanToLatex(List)
• gfanToLatex(MarkedPolynomialList)
## For the programmer
The object gfanToLatex is . | 712 | 1,984 | {"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} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.472612 |
https://ru.scribd.com/document/159543922/16100lectre12-cg | 1,591,331,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00520.warc.gz | 514,794,783 | 90,321 | Вы находитесь на странице: 1из 9
# Quick Visit to Bernoulli Land
Although we have seen the Bernoulli equation and seen it derived before, this
next note shows its derivation for an uncompressible & inviscid flow. The
derivation follows that of Kuethe &Chow most closely (I like it better than
Anderson).
1
Start from inviscid, incompressible momentum equation
1 u
u u p
t
+ =
There is a vector calculus identity:
( )
2
,
1
2
vorticity
u u u u u
| |
=
|
\ .
2 1 1
2
u
u p u
t
| |
+ + =
|
\ .
From here, we can make the final re-arrangement:
2 1
2
u
p u u
t
| |
+ =
|
\ .
Two common applications:
0 0
Irrotational
u
t
= =
2
2
1
0
2
1
.
2
p u
p u const for entire flow
| |
+ =
|
\ .
+ =
1
Kuethe and Chow, 5
th
Ed. Sec 3.3-3.5
Quick Visit to Bernoulli Land
16.100 2002 2
0 0
Rotational
u
t
2 1
2
p u u
| |
+ =
|
\ .
This is a vector equation. If we dot product this into the streamwise
direction:
u
u
s
streamwise direction
( )
( )
2
0,
2
2
1
2
1
0
2
1
.
2
u u
s p u s u
d
p u
ds
p u const along streamline
=
| |
+ =
|
\ .
| |
+ =
|
\ .
+ =
Vortex Panel Methods
2
Step#1: Replace airfoil surface with panels
Step #2: Distribute singularities on each panel with unknown strengths
In our case we will use vortices distributed such that their strength varies linearly
from node to node:
Recall a point vortex at the origin is:
1
tan
2 2
y
x
| |
= =
|
\ .
2
Kuethe and Chow, 5
th
Ed. Sec. 5.10
1 2
3 4
m
m+1
i-1
i
i+1
Original airfoil m-panels (m+1 nodes)
Quick Visit to Bernoulli Land
16.100 2002 3
A point vortex at y x , is:
1
tan
2
y y
x x
| |
=
|
\ .
Next, consider an arbitrary panel:
At any
j
s , we will place a vortex with strength
( )
j
s ds :
( )
1
( )
, tan
2
j j
j
s ds y y
d x y
x x
| |
=
|
|
\ .
where
( )
1
1
( )
j
j j j j
j
j
j j j j
j
s
x x x x
S
s
y y y y
S
+
+
+
+
Thus, the potential at any ( , ) x y due to the entire panel j is:
( )
( )
1
0
, tan
2
j
S
j j
j
j
s y y
x y ds
x x
| |
|
|
\ .
We will assume linear varying on each panel:
( ) ( )
1
j
j j j j
j
s
s
S
+
= +
s
j
S
j
(s
j
)
ds
j
Vortex of strength
(s
j
)ds
j+1
j
(x
j
,y
j
)
(x
j+1
,y
j+1
)
expanded
view
Midpoints will
be (x
j
,y
j
)
Quick Visit to Bernoulli Land
16.100 2002 4
With this type of panel, we have m+1 unknowns =
1 , , 1 ... 3 , 2 , 1 + m m m
, so we
need m+1 equations.
Step#3: Enforce Flow Tangency at Panel Midpoints
The next step is to enforce some approximation of the boundary conditions at the
airfoil surface. To do this, we will enforce flow tangency at the midpoint of each
panel.
Panel method lingo: control point is a location where o n u =
is enforced.
To do this, we need to find the potential and the velocity at each control point.
The potential has the following form:
#
panels
freestream individual panel
potential potential
| | | |
= +
| |
\ . \ .
Suppose freestream has angle :
( ) ( )
( )
1
1
0
## , cos sin tan
2
j
S
m
j j
j j
s y y
x y V x y ds
x x
=
| |
= +
|
|
\ .
The required boundary condition is
( )
,
0 1
i i
i
x y for all i m
n
= =
So, lets carry this out a little further:
( ) ( )
1
,
0
,
component of freestream normal
to surface of panel i
normal velocity due to panel j
at control poi
( )
cos sin tan
2
j
i i
S
j j
i i i
j i j
x y
s y y
x y V i j n ds
n n x x
(
| |
= +
( | `
|
(
\ .
)
0
nt of panel i
0
m
j=
=
And recall ( ) ( )
j
j
j j j j
S
s
s + =
+1
.
We can re-write these integrals in a compact notation:
( )
1
1
1 2 1
0
,
Influence of panel j due to
node j on control point of panel i
tan
2
j
ij ij
i i
n
ij
S
j j
n j n j
i j
x y
C
s y y
ds C C
n x x
+
=
(
| |
= +
( | `
|
(
\ .
)
i.e.
1
ij
n j
C =normal velocity from panel j due to node j on control point of panel i .
i i
y x ,
Quick Visit to Bernoulli Land
16.100 2002 5
2
ij
n
C =
Influence of panel j due to node 1 j + on control point at panel i
Total normal velocity at control point of panel i due to panel j
1 2 1
ij ij
n j n j
C C
+
= +
So, lets look at the control point normal velocity
So, for panel i , flow tangency looks like:
( ) ( )
1 2 1
1
cos sin , for all 1
ij ij
i
m
n j n j i
j
V n
C C V i j n i m
+
=
+ = + =
We can write this as a set of m equations for m+1 unknowns.
Question: What can we do for one more equation?
Step#4: Apply Kutta condition
We need to relate Kutta condition to the unknown vortex strengths
j
. To do this,
consider a portion of a vortex panel.
ds
Quick Visit to Bernoulli Land
16.100 2002 6
Put a contour about differential element ds
Recall: | |
( ) ( )
2 2 1 1
1 2 1 2
u ds
ds V dn U ds V dn U ds
V V dn U U ds
=
= +
=
Now let & 0 dn ds :
( )
1 2
2 1
,
, or
, in general
top bottom
dn O ds U U ds
U U
U U
=
=
=
So, since the Kutta condition requires
top bottom
U U = at TE:
. .
0, Kutta condition
t e
=
For the vortex panel method, this means:
1 1
0
m
+
+ =
Step#5: Set-up System of Equations & Solve
ds V
2
V
1
U
2
U
1
ds
dn
Quick Visit to Bernoulli Land
16.100 2002 7
1
2
8
11 12 19 1
21 22 2
31 32 33 3
41
51
61
71
8 81
9
0@ 1
0@ 2
.
. .
. .
.
0@
1 0 0 0 1 0
n
n
n
I
V
I I I
u n i
V
I I u n i
I I I
I
I
I
I
V u n i m I
Kutta
( = = (
( (
= =
( (
( (
( (
( (
( (
=
( (
( (
( (
( (
= =
( (
( (
i
i
i
n
V
(
(
(
(
(
(
(
(
(
(
(
(
(
(
Where =
ij
I total influence of node j at control point i
For example:
36 37
2 1 37 n n
C C I + =
The problem thus reduces to = Ax b , or, using our notation
n
I = V ,
which we solve to find the vector of s!
1
2
3 4
6
5
8
7
9 C
n1
37
C
n2
36
Nodes
Control points
Quick Visit to Bernoulli Land
16.100 2002 8
Step #6: Post-processing
The final step is to post-process the results to find the pressures and the lift
acting on the airfoil.
airfoil
L V V ds
= =
So, for our method, this reduces to:
( )
1
1
1
2
1
1
2
1
2
m
i i i
i
m
i i i
l
i
L V S
L S
C
V V c
V c
+
=
+
=
= +
| |
= = +
|
\ .
Vortex Panel Method Summary
In practice, the vortex panel method used for airfoil flows is a little different than
the strategy used in the windy city problem. Heres a summary:
Step #1: Replace airfoil surface with panels
Note: the trailing edge is double-numbered 1 points, panels m m + .
Step #2: Distribute vortex singularities with linear strength variables on each
panel
( )
1
( )
j
j j j j
j
s
s
S
+
= +
1 2
3 4
m
m+1
ds
S
j
j
j+1
(j)
(j+1)
(s
j
)
s
j
ds
Quick Visit to Bernoulli Land
16.100 2002 9
This means we have m+1 unknowns:
1 ,
.. ..........
, 3 , 2 , 1 + m m
Step #3: Enforce flow tangency at panel midpoints
0 u n =
at midpoint of every panel
equations m
Step#4: Apply Kutta condition
Kutta condition becomes:
. . 1 1
0 0
1 equations & 1 unknowns
t e m
m m
+
= + =
+ + | 2,560 | 6,591 | {"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-2020-24 | latest | en | 0.732134 |
https://sdrta.net/2-to-the-3rd-power/ | 1,638,516,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00067.warc.gz | 562,105,076 | 4,937 | The exponent that a number says how countless times to usage the number in a multiplication.
You are watching: -2 to the 3rd power
In 82 the "2" claims to use 8 twice in a multiplication,so 82 = 8 × 8 = 64
In words: 82 could be dubbed "8 come the strength 2" or "8 to the 2nd power", or merely "8 squared"
Exponents are also called powers or Indices.
Some an ext examples:
### Example: 53 = 5 × 5 × 5 = 125
In words: 53 could be referred to as "5 come the 3rd power", "5 come the power 3" or merely "5 cubed"
### Example: 24 = 2 × 2 × 2 × 2 = 16
In words: 24 might be referred to as "2 come the 4th power" or "2 come the strength 4" or merely "2 to the 4th"
So in general:
an speak you to multiply a by itself,so there are n that those a"s:
## Another method of creating It
Sometimes civilization use the ^ prize (above the 6 on your keyboard), together it is basic to type.
## Negative Exponents
Negative? What could be opposing of multiplying? Dividing!
So we division by the number every time, i m sorry is the very same as multiplying by 1number
## Negative? flip the Positive!
That last example showed one easier way to handle an adverse exponents: calculate the positive exponent (an)
More Examples:
an unfavorable Exponent reciprocal ofPositive Exponent prize
4-2 = 1 / 42 = 1/16 = 0.0625
10-3 = 1 / 103 = 1/1,000 = 0.001
(-2)-3 = 1 / (-2)3 = 1/(-8) = -0.125
## What if the Exponent is 1, or 0?
1 If the exponent is 1, then you just have actually the number itself (example 91 = 9) 0 If the exponent is 0, climate you get 1 (example 90 = 1) But what around 00 ? It can be either 1 or 0, and also so human being say it is "indeterminate".
## It All provides Sense
If friend look at the table, girlfriend will watch that positive, zero ornegative exponents space really part of the exact same (fairly simple) pattern:
Example: strength of 5
.. Etc..
See more: How Many Innings Are In High School Baseball Game? Well, It Depends
52 5 × 5 25
51 5 5
50 1 1
5-1 15 0.2
5-2 15 × 15 0.04
.. Etc.. | 624 | 2,025 | {"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} | 4.375 | 4 | CC-MAIN-2021-49 | latest | en | 0.891537 |
https://mathoverflow.net/questions/255974/automorphism-of-restriction-of-scalars | 1,642,603,668,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00264.warc.gz | 452,511,351 | 30,758 | # Automorphism of restriction of scalars
Let $E/F$ be a finite field extension. Let $G_E$ be a reductive linear algebraic group defined over $E$ and let $G=\mathrm{R}_{E/F}G_E$ be the Weil restriction of scalars. Then $G$ is a linear algebraic group defined over $F$ and $\mathrm{R}_{E/F}$ is a functor from the category of linear algebraic groups defined over $E$ to the category of linear algebraic groups defined over $F$.
Consider $\mathrm{Aut}_E(G_E)$, the group of automorphisms of $G_E$ as an algebraic group over $E$. The group $\mathrm{Aut}_F(G)$ is defined similarly. The functor provides a homomorphism $$\mathrm{R}_{E/F}:\mathrm{Aut}_E(G_E)\to\mathrm{Aut}_F(G).$$ My question is, what is the quotient $$\Gamma=\mathrm{Aut}_F(G)/\mathrm{R}_{E/F}(\mathrm{Aut}_E(G_E))?$$
If $G_E=\mathrm{GL}(n)$, then I would like to know in particular that $\Gamma$ is the Galois group $\mathrm{Gal}(E/F)$. To see concretely how the Galois group acts, we can construct the restriction of scalars by choosing an $F$-basis for $E$ that provides a map from $G_E(E)$ to $\mathrm{GL}_N(F)$ where $N=n[E:F]$. The image of this map defines an algebraic group over $F$ which represents the restriction of scalars and the Galois group action on $G_E(E)$ can be transferred to an action on the restriction of scalars.
• In general the image of your homomorphism between Aut groups is not normal, as you seem to assume. For instance, choose $F$ the reals, $E$ the complex numbers, $G_E=(GL_1)^2$ the (complex) 2-torus. Indeed define, in coordinates, $u(z,w)=(|w|\frac{z}{|z|},|z|\frac{w}{|w|})$ and $v(z,w)=(z^{-1},w)$. Then both $u,v$ are group automorphisms and $v$ is complex (i.e. belongs to $Aut_E(G_E)$ in your notation, while $u$ is in $Aut_F(G)$). Note that both are involutions. Then $uvu^{-1}(z,w)=(\bar{z},1/\bar{w})$. In particular $uvu^{-1}$ is not complex (that is, does not belong to $Aut_E(G_E)$).
– YCor
Nov 30 '16 at 4:14
• For $GL(2)$ the automorphism $A\mapsto \frac{\overline{\det A}}{|\det A|}A$ is neither holomorphic nor antiholomorphic (it acts on $SL_2$ as the identity and on scalar matrices by conjugation). This seems specific to $n=2$ though.
– YCor
Nov 30 '16 at 7:08
• @YCor: Thanks for this. I wonder if there is an assumption on $G$ that makes the subgroup normal, something that eliminates the possibility for $G_E$ to be a direct product. Nov 30 '16 at 15:13
• For arbitrary $n\ge 2$ (and real/complex) in $GL(n)$ we also have the involution $A\mapsto (1/|\det A|^2)A$, it is the identity on $SL(n)$ and maps a scalar matrix $tI_n$ to $(t/|t|^2)I_n$.
– YCor
Nov 30 '16 at 23:38
You surely meant to assume the finite extension of fields $E/F$ is separable (otherwise ${\rm{R}}_{E/F}(H)$ is never reductive for a smooth connected affine $E$-group $H \ne 1$). Also, this is one of those cases where more generality clarifies the situation. The reason I say this is that by limiting yourself only to Weil restriction from a field over $F$ rather than more generally from a finite etale $F$-algebra (i.e., a product of several fields over $F$, all finite and separable over $F$), you prevent yourself from accessing the technique of scalar extension on $F$ to simplify computations.
So let's pose the situation more broadly as follows. Let $F$ be a field, $E$ and $E'$ nonzero finite etale $F$-algebras, and $G$ and $G'$ smooth affine groups over $E$ and $E'$ respectively such that their fibers over the factor fields of $E$ and $E'$ are connected reductive and non-trivial. We want to describe $F$-isomorphisms $f: {\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ in terms of more concrete data, perhaps under some hypotheses on the groups (and in particular to address the case when $E = E'$ is a field and $G = G' \ne 1$).
Given an $F$-algebra isomorphism $\alpha:E \simeq E'$ and a group isomorphism $\varphi:G \simeq G'$ over $\alpha$ there is an evident associated $F$-group isomorphism ${\rm{R}}_{\alpha/F}(\varphi): {\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$. So we can ask the (easy) question of whether the pair $(\alpha, \varphi)$ is uniquely determined by the $F$-isomorphism ${\rm{R}}_{\alpha/F}(\varphi)$, and the more serious question of whether every $f$ arises from such a pair. The first key point, and the reason for recasting the setup in terms of the generality of (nonzero) finite etale $F$-algebras rather than only fields (of finite degree and separable) over $F$ is that if $F'/F$ is any extension field (such as a separable closure, or simply a big enough finite Galois extension, or whatever) then $${\rm{R}}_{E/F}(X) \otimes_F F' \simeq {\rm{R}}_{(E \otimes_F F')/F'}(X \otimes_E (E \otimes_F F'))$$ for any affine $E$-scheme $X$ of finite type. Note that if $E$ were a field then $E \otimes_F F'$ is typically not a field, but that since $E$ is finite etale over $F$ at least $E \otimes_F F'$ is finite etale over $F'$. More specifically, a nonzero finite etale $F$-algebra $E$ always has the form $E = \prod E_i$ for fields $E_i$ that are finite separable over $F$, any scheme $X$ over ${\rm{Spec}}(E) = \coprod {\rm{Spec}}(E_i)$ always has the form $X = \coprod X_i$ where $X_i$ is the $E_i$-fiber of $X$, and for $X$ affine and finite type over $E$ we have $${\rm{R}}_{E/F}(X) \simeq \prod {\rm{R}}_{E_i/F}(X_i);$$ this makes contact back with the more familiar world of Weil restriction through field extensions, but the language of finite etale $F$-algebras is much more efficient for bookkeeping purposes with the notation when we want to bring in ground field extensions (as we will do below!).
Let's now come back to the question of whether an $F$-isomorphism $f: {\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ having the form ${\rm{R}}_{\alpha/F}(\varphi)$ uniquely determines the $F$-algebra isomorphism $\alpha$ and the group isomorphism $\varphi$ over $\alpha$. Since we're only asking about uniqueness of such a pair, not existence, it suffices to check after making a ground field extension! And we were careful to set up our entire framework in a manner that is not destroyed by such an operation, so for the purpose of proving uniqueness we can apply ground field extension from $F$ to $F_s$ so that $F$ is separably closed and hence all finite etale $F$-algebras are split!
More to the point, now (for the purpose of proving uniqueness) we can assume as $F$-algebras that $E = F^I$ and $E' = F^J$ for non-empty finite sets $I$ and $J$, so $\alpha$ is nothing other than a bijection of sets $\tau:I \simeq J$ (with $\alpha((x_i)) = (x_{\tau^{-1}(j)})$). Likewise, now $G = \coprod G_i$ and $G' = \coprod G'_j$ with $G_i$ and $G'_j$ all non-trivial, and $\varphi$ is nothing other than a collection of $F$-isomorphisms $\varphi_i:G_i \simeq G'_{\tau(i)}$ for $i \in I$. Since ${\rm{R}}_{E/F}(G) = \prod G_i$ and ${\rm{R}}_{E'/F}(G') = \prod G'_j$, the uniqueness question comes down to the concrete problem of whether the product isomorphism of $F$-groups $$\prod_{i \in I} G_i \simeq \prod_{j \in J} G'_j$$ defined by $(g_i) \mapsto (\varphi_{\tau^{-1}(j)}(g_{\tau^{-1}(j)}))$ uniquely determines both $\tau$ and the collection of $\varphi_i$'s. Since all $G_i$ and $G'_j$ are nontrivial, the uniqueness of both is immediate by chasing the image of $G_i$ under the product isomorphism.
That settles the uniqueness aspect, so now we can turn to the much more interesting existence aspect. Obviously this cannot be affirmative in general (just imagine that $G$ is itself a Weil restriction, and consider that a composition of Weil restrictions is a Weil restriction for the composite extension). So the good result is as follows:
Theorem. Assume that the connected reductive fibers of $G$ and $G'$ over the respective factor fields of $E$ and $E'$ are semisimple, absolutely simple, and simply connected. Then any $F$-group isomorphism ${\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ has the form ${\rm{R}}_{\alpha/F}(\varphi)$ for a $($uniquely determined$)$ pair $(\alpha, \varphi)$ as above. The same holds with "adjoint type" in place of "simply connected".
The "absolutely simple" hypothesis rules out the obstruction caused by internal Weil restrictions as noted above, the "semisimple" condition is needed prior to making any meaningful/useful notion of "absolutely simple", and the "simply connected" (or alternatively "adjoint type") hypothesis ensures that the absolute root datum (not just root system!) is a direct product of those for its irreducible components.
For a proof of the above Theorem (including a reference to its historical antecedent in the older language of Weil restriction only through field extensions, at the cost of slightly heavier notation), see Proposition A.5.14 in the book Pseudo-reductive Groups.
Now let's finally come back to the original question concerning automorphisms of $H := {\rm{R}}_{E/F}({\rm{GL}}_n)$ for a finite separable extension of fields $E/F$ and $n \ge 1$. For $n = 1$ it is really hopeless to say anything in general (since passing to the Cartier dual Galois lattice ${\rm{Ind}}_{\Gamma_F}^{\Gamma_E}(\mathbf{Z})$ shows that (if $E \ne F$) there are a lot of automorphisms, as we can see by first considering the situation after scalar extension to $\mathbf{Q}$). So let's assume $n \ge 2$. Then the derived group $\mathscr{D}(H)$ coincides with ${\rm{R}}_{E/F}({\rm{SL}}_n)$, and we have $$H \simeq (\mathscr{D}(H) \times Z_H)/\mu = ({\rm{R}}_{E/F}({\rm{SL}}_n) \times {\rm{R}}_{E/F}({\rm{GL}}_1))/{\rm{R}}_{E/F}(\mu_n)$$ where $Z_H$ is the (scheme-theoretic) center of $H$ and $\mu = \mathscr{D}(H) \cap Z_H = Z_{\mathscr{D}(H)} \simeq {\rm{R}}_{E/F}(\mu_n)$.
Thus, any $F$-automorphism $f$ of $H$ is exactly the data of an $F$-automorphism $f_1$ of $\mathscr{D}(H) = {\rm{R}}_{E/F}({\rm{SL}}_n)$ and an $F$-automorphism $f_2$ of $Z_H = {\rm{R}}_{E/F}({\rm{GL}}_1)$ that coincide on their overlap ${\rm{R}}_{E/F}(\mu_n)$. The preceding stuff shows that $f_1$ is uniquely of the form ${\rm{R}}_{\alpha/F}(\varphi)$ for an $F$-algebra automorphism $\alpha$ of $E$ and a group automorphism $\varphi$ of ${\rm{SL}}_n$ over $\alpha$. Our knowledge of the automorphism group of ${\rm{SL}}_n$ over fields shows that the effect of $\varphi$ on $\mu_n$ is (up to the effect of $\alpha$) either trivial or inversion, and both options extend to automorphisms of ${\rm{GL}}_1$ over $\alpha$! Thus, whatever $f_1$ we might want to consider, a compatible $f_2$ can always be found.
Consequently, we see that the obstruction to an affirmative answer to your question is exactly the group of $F$-automorphisms of the induced torus ${\rm{R}}_{E/F}({\rm{GL}}_1)$ restricting to the identity on its $n$-torsion subgroup ${\rm{R}}_{E/F}(\mu_n)$. Since $n > 1$, whenever $E \ne F$ we can make infinitely many such extra automorphisms. I conjecture that you posed the question for ${\rm{GL}}_n$ under the mistaken belief that it would make the problem easier (whereas now you see that the central torus actually makes the situation harder), and that the positive answer for ${\rm{SL}}_n$ (and more generally as in the Theorem above) is the sort of thing with which you would be satisfied. So I'll leave it to you to mull over the extra junk automorphisms arising from the central torus if you really do need that information (which feels doubtful).
• Thank you very much for this answer. I am still working through parts of it, but I plan to accept it. Dec 2 '16 at 13:30
• Regarding your comment that the torus $H=\mathrm{R}_{E/F}\mathrm{GL}_1$ has many automorphisms. I am primarily interested in the cyclic case, in which case it seems to me that $H$ has $2n$ automorphisms for $n=[E:F]$. Am I missing something here? Dec 7 '16 at 23:10
• Ah, for $E/F$ Galois the relevant arithmetic group has only finitely many elements: the Cartier dual lattice is $\mathbf{Z}[{\rm{Gal}}(E/F)]$ as a left Galois module in such cases, and its automorphism group is the unit group of that ring (acting through right multiplications). By the classic theorem of Higman, the unit group of the group ring $\mathbf{Z}[G]$ for a finite group $G$ is $\{\pm g\}_{g \in G}$. I expect that for $E/F$ not Galois, the arithmetic group of $\Gamma_E$-linear automorphisms of ${\rm{Ind}}_{\Gamma_E}^{\Gamma_F}(\mathbf{Z})$ is typically infinite. Dec 8 '16 at 1:08
• Thanks, that's exactly the kind of reference I needed. Thank you so much for your help. Dec 8 '16 at 1:54
• I meant to say "$\Gamma_F$-linear" rather than $\Gamma_E$-linear near the end of my preceding comment. Dec 8 '16 at 8:21 | 3,841 | 12,480 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.781538 |
https://philoid.com/question/133109-light-falls-from-glass-15-to-air-find-the-angle-of-incidence-for-which-the-angle-of-deviation-is-90- | 1,723,779,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00478.warc.gz | 350,145,216 | 8,234 | ##### Light falls from glass (μ = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.
Refractive index of glass is
Angle of incidence=i
Light ray is passing from glass to air i.e. from denser to rarer medium.
We know that;
Critical Angle=
=
Maximum angle of deviation in refraction is=
In reflection,
Deviation =
Or,
So, the angle of incidence=.
1 | 99 | 389 | {"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-2024-33 | latest | en | 0.884745 |
https://www.physicsforums.com/threads/am-i-understanding-the-concept-of-suction-correctly.647985/ | 1,537,911,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.84/warc/CC-MAIN-20180925202648-20180925223048-00476.warc.gz | 816,209,435 | 20,769 | # Am I understanding the concept of suction correctly?
1. Oct 29, 2012
### spacediver
My understanding of suction is that it is a phenomenon describing the flow of fluid from a (relatively) high pressure volume into a (relatively) low pressure volume.
This makes sense to me, as I grasp that air pressure is a measurement that describes a large number of molecules, each with its own momentum, and that the higher the mean momentum, the higher the air pressure. Accordingly, when you introduce a volume that has a relatively low air pressure relative to its surrounding volume, and there is no barrier between these two volumes, the molecules in the higher pressure system will tend to distribute themselves to fill in the "gap". To me, this is statistically similar to the phenomenon of increasing the homogeneity of a cake batter by stirring it enough.
What I'm having difficulty understanding is how a vacuum cleaner is able to generate so much suction.
My understanding is that the maximum suction is a function of the difference in air pressure between the two "volumes". Even if a vacuum cleaner creates a perfect vacuum internally, surely the amount of suction force is limited by the air pressure of the environment "outside" the vacuum cleaner.
It also would seem that the area of the channel connecting the vacuum to the outside world plays an important role, similarly to how pressure is a function of the area over which a force is distributed. Perhaps the length of the channel also plays a role, but this is more speculative on my part.
So here are some questions to test my understanding:
1: Suppose you have a constant "external" air pressure, and you're using a vacuum pipe with a fixed diameter and length. Will the maximum suction physically possible be experienced when a perfect vacuum is created inside the cleaner?
2: Are there any other sort of mechanisms that can provide for even greater amounts of suction than that referenced in question 1?
3: What is a useful way of formally quantifying "suction"?
thanks!
2. Oct 29, 2012
### Staff: Mentor
That is correct. 14 psi = 1 bar = air pressure relative to vacuum at sea level is a lot of pressure. If you could pump all of the air out of a building it would collapse.
Again, you are right. See the answer to #3 below
1) Yes, that's about as good as it gets.
2) No. (It's possible to do a bit better under very special circumstances when the suction is applied for only a very short time and the air is moving fast enough as it is sucked in. High-performance automobile engines can use this effect to draw a bit more air into the cylinder during the few milliseconds that the intake valves are open. This effect is completely unhelpful for a vacuum cleaner or anything else that is generating steady flow).
3) The standard measure is volume of air moved per unit time at a given pressure drop, something like "500 cubic feet per minute at 3 psi" for example. This is also why the size, shape, and length of the intake matters. It's easy to maintain a very high pressure drop if you only let a small volume of air through, but then your vacuum cleaner isn't going to be picking much up.
3. Oct 29, 2012
### spacediver
thanks nugatory, those answers really consolidated my understanding!
4. Oct 29, 2012
### spacediver
very interesting, and that certainly gives a vivid impression of the forces involved.
Here's a somewhat related question then:
If the amount of force involved in a differential of 1 bar of air pressure is enough to implode a building, then this means that a building in equilibrium is essentially experiencing that level of force from both outside and inside (and the same can be said for our bodies).
But surely that means that any given membrane in a pressure equalized system of 1 bar should be experiencing catastrophic compression. I'm having trouble squaring the compressive resilience of ordinary matter at sea level with the idea of a building caving in!
5. Oct 30, 2012
At one atmosphere of pressure water will compress in volume by only 46.4 parts in one million, steel a factor of 70 less. Anything that is not a gas is surprisingly rigid as long as the pressure from all sides is the same. Why do you think there are animals on the bottom of the ocean where the pressure is 4000 metric tons per square meter.
6. Oct 30, 2012
### spacediver
I still don't understand how a force that could implode a building, under a pressure differential of 1 atmosphere would not cause catastrophic compression when both sides are in equilibrium.
Put it this way:
Imagine you had a giant vise and placed a building in between its grips. Assume the air pressure inside and outside of the building is the same. You then tighten the vise until the building implodes. If I am understanding Nugatory's post correctly, this amount of pressure that the vise is applying is equivalent to 1 bar of pressure.
Now if a building is in equilibrium (forget the vise for now), this means that the building is experiencing that huge amount of pressure, but on both sides of each surface.
How come if you take a piece of foam and just place it anywhere on earth, it doesn't immediately compress to a tiny thickness? If i were to place that piece of foam in a vice and apply the same amount of pressure as I needed to implode the building, the foam would surely be severely compressed.
I know something is wrong with my thinking, but I can't quite figure it out.
7. Oct 31, 2012
### JustinRyan
If you put the foam in a bag and sucked all the air out of the bag, it would compress to a much smaller size.
8. Oct 31, 2012
### A.T.
Which way should it deform, if both sides are in equilibrium?
Because the pressure in the foam is the same as outside the foam.
9. Oct 31, 2012
### torquil
A building, which is almost hollow, is actually incredibly frail compared to a solid material like one of the bricks that make up the wall of the building, when acted upon with a given pressure on the outsides, e.g. using your hypothetical vice.
10. Oct 31, 2012
### CWatters
So a small building with walls say 20ft x 8ft (and a vacuum inside) would have a force of about
14 x 20 x 8 x 144 = 322,560 lbs
acting on each wall.
11. Oct 31, 2012
### CWatters
That's because it's not just "both sides" that are in equilibrium. For example in a cavity wall there is also air in the wall at the same pressure. Where there isn't air (eg in the middle of a brick) the material the brick is made of brick is also under pressure and "pushes back". In other words there is no pressure differential to cause a collapse.
That doesn't mean buildings don't deform under the pressure.. look at a double glazed sealed window unit. It was made with the gas inside at a certain temperature and pressure. When the weather changes the pressure inside the unit changes relative to the pressure outside/inside. You can sometimes see the glass has gone concave, compressing the unit due to the pressure differential.
However something like a brick doesn't have to be compressed much before the pressure differential between the inside of the brick inside and outside the brick is equal. So the deformation is incredibly small. The brick was also made under 1 atmosphere to start with.
12. Oct 31, 2012
### sophiecentaur
That would need to be a pretty special building; hermetically sealed. At the rate atmospheric pressure changes, the pressure inside and out would equalise even by the air passing in or out of the keyhole.
The excess pressure of a supersonic shock wave blew out the windows ("mysteriously") in an airfield building during early test flights, so I heard on TV - so it must be true haha. But that was a pressure change in a very short time. [Edit - also see the effects of a brief drop in AP inside a tornado column]
Closed cell foam will change dramatically volume when placed in a decompression (hyperbaric) chamber. I remember a great kids' Science programme with balloons, cream and marshmallow giving very impressive results. If the mass of gas inside a cell is not allowed to change, then the volume will change with pressure.
13. Oct 31, 2012
### spacediver
If I understand Cwatters correctly, he was referring to the pressure inside the window unit itself relative to outside the window unit (double glazed window remember).
14. Oct 31, 2012
### spacediver
I guess that makes sense - i had failed to consider that all that force gets transferred to a finite number of joints in the building, which is where the structural vulnerabilities lie.
15. Oct 31, 2012
### spacediver
By equilibrium, I meant forces acting on the inner wall match forces acting on the outer wall. But I suppose there is another way of thinking about equilibrium: when the forces acting within the bricks against the air match the forces without the brick against the brick.
16. Oct 31, 2012
### AJ Bentley
I gotta say, despite the number of people who seem to be happy with it, suction as a concept has no place in physics.
You can talk about negative pressure if you like - but suction is a lay term more properly applied to house-wifery than physics.
The force that moves fluid particles around comes from mutual repulsion and collisions. There isn't a 'suction force'. It's just woolly thinking.
17. Oct 31, 2012
### spacediver
Interesting point AJ.
I think we'd both agree that it can be useful (both practically and theoretically) to employ a construct that efficiently describes the phenomenon. Perhaps suction isn't the most ontologically sound one, but surely you're not suggesting we revert to Newtonian mechanics every time we wish to describe/predict the tendency for matter to flow from a high to low pressure system. Given Nugatory's description of the way that "suction" is measured:
I'm not sure that the term "negative pressure" will adequately serve this purpose.
Last edited: Oct 31, 2012
18. Oct 31, 2012
### sophiecentaur
Oh yes, I see - very much hermitacally sealed. I have been thinking how much you could expect windows to bow inwards or outwards. You could expect a pressure range of, say 20% between extremes of high and low. That would represent 20kPa of pressure variation at sea level (+/- 10kPa around an average value). If the gap in the glass were 4mm then you could imagine each sheet could move by 0.2mm (Boyle's Law and assuming the sheets were frictionless pistons) If they were very flexible and fixed around the outside, you would get about twice this movement at the centre (0.4mm). The radius of curvature of such a mirror would be, I estimate, about 600m - giving a focal length of 300m. As it is, glass is fairly rigid so the flexing would be quite a bit less (half / quarter??) than that. I wonder just how much this amount of curvature would actually be visible. It's hard to tell. Perhaps distortion of images of nearby buildings would,in fact, be visible. We're quite sensitive to that sort of thing.
I'd expect a similar effect due to a +/-10% variation in temperature (60oC in 300K) - also significantly reduced by the rigidity of the glass.
I think my back of fag packet sums are reasonable.
19. Oct 31, 2012
### Staff: Mentor
It works better than you'd expect at first, because a pressure drop is the difference between two pressures, so doesn't care about the absolute number assigned to either pressure. I can think of a 4 psi drop as 14 psi outside, 9 psi inside, or as 0 psi outside, -4 psi inside, and I get pretty much the same flow behavior.
20. Oct 31, 2012
### AJ Bentley
The reason I have a 'down' on the idea of suction is because of a much admired Chemistry master at my school (many, many decades ago).
He would hand someone a test tube full of water with a bung and tube in the top and instruct them to suck the water out.
It's impossible of course - "There's no such thing as a 'suck'" he would say.
He was quite right - the idea is both unnecessary and fallacious. | 2,673 | 11,947 | {"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-2018-39 | latest | en | 0.95129 |
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CEE125-SPR08-HW5
# CEE125-SPR08-HW5 - Problem 5-3(5 points The strong motion...
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CEE 125 R. Nigbor Earthquake Engineering Spring 2008 Assignment No. 5 DUE WEDNESDAY MAY 28, 2008 Problem 5-1 (3 points) Use the data from Homework 4, Problem 1 to create two models of the soil column at the Arleta strong motion station. One model should be a multi-layer model based on the soil layers from the geological log. The other should be a two-layer model, with one soil layer over an elastic rock layer (“engineering rock”). For each model, get average Vp and Vs and estimate the density for each layer. Display your results graphically per Text Figure 7.9. These are interpreted models, so there are many “correct” answers. Use your engineering judgment here. Problem 5-2 (2 points ) For your simple two-layer model in Problem 4-1, calculate and plot the amplification (transfer function) versus frequency for the Arleta site. Use the appropriate text equation for soil over elastic rock. Repeat for 3%, 5%, and 10% damping, plotting the three curves on the same plot.
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Unformatted text preview: Problem 5-3 (5 points) The strong motion station at Arleta is inside a fire station. Assume you are designing a replacement 2-story, steel-moment-frame fire station building for that site. The building will be approximately the same size as the existing building (use Google Earth to estimate the area of the existing building if needed). Assume 5% damping. Assume that there are no horizontal irregularities. Develop the CBC (IBC2006) design spectrum for the building . List your assumptions & calculations needed to develop the design spectrum. Plot your design spectrum along with the earthquake motions measured during the 1994 Northridge Earthquake. Discuss the differences, especially at the estimated period of the new building. Assuming a repeat of the 1994 Northridge ground motions, how do you think your new building would perform?...
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{[ snackBarMessage ]} | 502 | 2,169 | {"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} | 3.296875 | 3 | CC-MAIN-2018-17 | latest | en | 0.87021 |
https://testbook.com/question-answer/during-a-high-cycle-fatigue-test-a-metallic-speci--5d8a2ccaf60d5d65747177c1 | 1,632,211,103,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00346.warc.gz | 592,426,807 | 30,233 | # During a high cycle fatigue test, a metallic specimen is subjected to cyclic loading with a mean stress of +140 MPa, and a minimum stress of −70 MPa. The R-ratio (minimum stress to maximum stress) for this cyclic loading is_______ (round off to one decimal place)
This question was previously asked in
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## Answer (Detailed Solution Below) -0.2
Free
CT 1: Ratio and Proportion
1963
10 Questions 16 Marks 30 Mins
## Detailed Solution
Concept:
$$R - ratio = \frac{{Minimum\;stress\left( {{\sigma _{min}}} \right)}}{{Maximum\;stress\;\left( {{\sigma _{max}}} \right)}}$$ …1)
Take care of signs of stresses.
Calculation:
Given data
$${\sigma _{mean}} = + 140MPa,\;{\sigma _{min}} = - 70MPa$$
$${\sigma _{mean}} = \frac{{{\sigma _{max}} + {\sigma _{min}}}}{2}$$ …2)
$$140 = \frac{{{\sigma _{max}} - 70}}{2} \Rightarrow {\sigma _{max}} = 70 + 280 = + 350MPa$$
$$R - ratio = - \frac{{70}}{{350}} = - 0.2$$
Key Point:
Take due account of the sign of stresses in the calculation. | 333 | 1,060 | {"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} | 3.703125 | 4 | CC-MAIN-2021-39 | latest | en | 0.645412 |
https://girlsaregeeks.wordpress.com/2010/10/25/mathematical-monday-10-25-10/ | 1,529,893,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00317.warc.gz | 624,773,900 | 18,341 | ## Mathematical Monday 10-25-10
This past Saturday was Mole day! For those who aren’t super nerdy science! dorks, a mole is 6.02 x 10^23 molecules, and Saturday was 10-23 (get it!). If you are in science! a mole is one of those numbers you memorize rather quickly. Actually, there are a whole bunch of those numbers depending on your field of science!, and here are a few more:
Acceleration due to gravity (at sea level) 9.8 m/s^2
Pi 3.14159 (I know many geeks know further digits, but this is a good minimum)
Gas constant 8.314 J/(mol-K)
The speed of light 3.00 x 10^8 m/s
Does having memorized the quadratic equation when I was about 10 count here too?
Well. I’m a biologist, so I’ve forgotten a lot of my good chemistry and physics (ick!) numbers, but I’m sure there are some of you out there who either remember the horror of memorizing these or actually still remember a few more!
1. Ooh, these are good ones. The other value I remember from chemistry is the specific heat of water, 4.186 J/(g degC), also useful for conversion because 4.186 J = 1 cal.
As far as oceanography goes, only two numbers come to mind at the moment:
1030 kg/m3 – approximate density of seawater
86400 seconds in a day – easy enough to calculate (60*60*24), but I find it helpful to know off the top of my head because it’s used in finding the Coriolis parameter, f = 4*pi*sin(latitude)/86400s.
• Oh, I like seconds in a day. I may need to learn that one just because.
Rosalind
2. ElZapato
Oct 26, 2010 @ 22:06:09
The speed of light is a teeny bit faster than 3.00 m/s.
more like 3.0 x 10^8 m/s :*)
• Whoops! Seems to have been left out. I’ll go fix that now!
Rosalind | 470 | 1,665 | {"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} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.931696 |
http://freedissertation.com/dissertation-examples/tests-of-significance-uses-and-limitations/ | 1,652,809,438,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00789.warc.gz | 26,598,149 | 27,820 | HELLO, GUEST
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## Tests of Significance: Uses and Limitations
### Abstract
Statistical tools are undoubtedly important in decision making. The use of these tools in everyday problems has led to a number of discoveries, conclusions and enhancement of knowledge. This ranges from direct calculations using general statistical formulas to formulas integrated in Statistical software to fasten the process of decision making.
Statistical tools for testing hypothesis, significance tests are strong but only if used correctly and in good understanding of their concepts and limitations. Some researchers have indulged into wrong usage of this tests leading to wrong conclusions.
This paper looks at the different significance tests (both parametric and non-parametric tests) their uses, when to be used and their limitations. It also evaluates the use of Statistical Significance tests in Information Retrieval and then proceeds to check the different significant tests used by researchers in the papers submitted to Special Interest Group on Information Retrieval (SIGR) in the period 2006, 2007 and 2008. For the combined period 2006-2008, including the years 2006 and 2008, of the papers submitted had statistical tests used and of these tests were used wrongly.
Key Words: Significance Test, Information Retrieval, Parametric Tests, Non-parametric Tests, Hypothesis Testing
### 1.0 Introduction
Statistical methods play a very important role in all aspects of research, ranging from data collection, recording, analysis, to making conclusions and inferences. The credibility of the research results and conclusions will depend on each and every step mentioned above; any fault made in these steps can render a research carried out for several years, spending millions of shillings to be worthless.
This does not mean carrying any test and mincing figures shows that statistics has been used in the given research; the researcher should be able support why he or she used that specific test or method.
Misuse of significance test is not new in the world of science. According to Campbell (1974), there are different types of statistical misuse:
### Discarding unfavorable portion of data
This occurs when the researcher selects only a portion of data which produces the results that he/she requires perfectly while discarding the other portion. After a well done research, the researcher might get values that are not consistent to what he/she was expecting. This researcher might decide to ignore this section of data during the analysis so as to get the “expected results”. This is a wrong take since the inconsistent data could give very new thoughts in that particular field that is if these irregularities are checked and explained why they occurred, more ideas abut that area can be explored..
### Overgeneralization
Sometimes the conclusions from a research can only work on that particular research problem but the researcher might blindly generalize the results obtained to other kinds of research similar or dissimilar. Overgeneralization is a common mistake in current research activities. A researcher after successfully completing a research on a particular field, he/she might be tempted to make generalizations reached in this research to other fields of study without regarding the different orientations of these different populations and assumptions in them.
### Non representative sample
This arises when the researcher selects a sample which produces results geared towards his/her liking. Sample selected for a particular study should be one that truly represents the entire population. The procedure of selecting the sample units to be used in the study should be done in an unbiased manner.
### Consciously manipulating data
Occurs when a researcher consciously changes the collected data in order to reach a particular conclusion. This is mainly noticed when the researcher knows exactly what the customers aim are, so the researcher changes part of the data so that the aim of that research is covered strongly. For example if a researcher is carrying out a regression analysis and does a scatter plot, if he/she sees that there are many out liers,the researcher might decide to change some values so that the scatter plot appears as a straight line or something very close to that. This act leads to results which are appealing to the customer and the eyes of other user but in real sense does not give a clear indicator of what is really happening in the population at large.
### 1.0.5 False correlation
This is observed when the researcher claims that one factor causes the other while in real sense both two factors are caused by another hidden factor which was not identified during the study. Correlation researches are common in social sciences and sometimes they are not adequately approached, this leads to wanting results. In correlation studies say to check if variable X causes variable Y, in real sense there are four possible things. The first one is that X causes Y,secondly Y causes X, third is X and Y are both caused by another unidentified variable say Z and lastly the correlation between X and Y occurred purely by sheer luck.
All these possibilities should be checked while doing these kinds of study to avoid rushing into wrong conclusions. False causality can be eliminated in studies by using two groups for the same experiment that is the “control group (the one receiving a placebo)” and the “treatment group (the one receiving the treatment)” .
Even though this method is efficient, implementing it raises very many challenges. There are ethical issues like when one patient is given a placebo (effect less drug) without his/her conscious and the other group given the right drug. One question comes to mind; is it ethical to do this to the first group? Carrying out the experiment in parallel for two different groups can also prove to be very expensive.
The questions used in survey can really affect the outcome of the survey. The structure of questions in a questionnaires and the method of formulating and asking the questions can influence the manner in which the respondent answers the questions. Long wordy questions in a questionnaire can be too boring to a respondent and he/she might just fill the questionnaire in a hurry so that he/she finishes it but does not really care about the answers that he/she has provided. The framing of questions can also yield leading questions. Some questions will just lead the respondent on what to answer for example “The government is not offering security to its citizens, do you agree to this? (Yes or No)”
Use of statistical significance has been with us for more than 300 years (Huberty, 1993).Despite being used for a long time, this field of decision making is cornered by criticism from all directions, which has led to many researchers writing materials digging into the problems of statistical significance testing. Harlow et. al (1997), discussed the controversy in significance testing in depth. Carver (1993) expressed dislike of significance tests and clearly advocated researchers to stop using them.
In his book, How to Lie with Statistics, Huff (1954) outlined errors both intentional and unintentional and misinterpretations made in statistical analyses in depth. Some journals e.g. American Psychological Association (APA) recommended minimum use of statistical significance test by researchers submitting papers for publications (APA, 1996), though not revoking the use of the tests.
With the relentless criticism, other researchers have not given up on using statistical significance testing but have clearly encourage users of the tests to have good knowledge in them before making conclusions using them. Mohr (1990) discussed the use of these tests and supported their use but warning researchers to know the limitations of each tests and correct application of the tests so as to make a correct inferences and conclusions. In his paper, Burr (1960) supported the use of statistical significance test but requested researchers to make allowances for existence of statistical errors in the data.
Amidst these controversies, statistical significance testing has been applied to many areas of research and remarkable achievements have been recorded. One such area is the information retrieval (IR). Significant tests have been used to compare different algorithms in information retrieval.
### 1.1.0 Information retrieval
Information retrieval is defined as the science of searching databases, World Wide Web and other documents looking for information on a particular subject. In order to get information, the user is required to enter keywords which are to be used for searching, a combination of objects containing the keywords are usually returned from which the user looking for information can single out and pick one which gives him or her the much required information.
The user usually progressively refines the search by narrowing down and using specific words. Information retrieval has developed as a highly dynamic and empirical discipline, requiring careful and thorough evaluation to show the superior performance of different new techniques on representative document collections.
There are many algorithms for Information Retrieval .It is usually important to measure the performance of different information retrieval systems so as to know which one gives the required information faster. In order to measure information retrieval effectiveness, three test items are required;
• (i) A collection of documents on which the different retrieval methods will be run on and compared.
• (ii) A test collection of information needs which are expressible in terms of queries
• (iii)A collection of “relevance judgment” that will distinguish on whether the results returned are relevant to the person doing the search or they are irrelevant.
### A question might arise on which collection of objects to be used in testing different systems. There are several standard test collections used universally, these include;
(i) Text Retrieval Conference (TREC). – This a standard collection comprising 6 CDs containing 1.89 million documents (mainly, but not exclusively, newswire articles) and relevance judgments for 450 information needs, which are called topics and specified in detailed text passages. Individual test collections are defined over different subsets of this data.
(ii)GOV2-This was developed by The U.S. National Institute of Standards and Technology (NIST).It is a 25 paged collection of web pages.
(iii) NII Test Collections for IR Systems (NTCIR)-This is also a large test collection focusing mainly on East Asian language and cross-language information retrieval, where queries are made in one language over a document collection containing documents in one or more other languages.
(iii) Cross Language Evaluation Forum (CLEF). This Test collection is mainly focused on European languages and cross-language information retrieval.
(iv) 20 Newsgroups. This text collection was collected by Ken Lang. It consists of 1000 articles from each of 20 Usenet newsgroups (the newsgroup name being regarded as the category). After the removal of duplicate articles, as it is usually used, it contains 18941 articles.
(v) The Cranfield collection. This is the oldest test collection in allowing precise quantitative measures of information retrieval effectiveness, but is nowadays too small for anything but the most elementary pilot experiments. It was collected in the United Kingdom starting in the late 1950s and it contains 1398 abstracts of aerodynamics journal articles, a set of 225 queries, and exhaustive relevance judgments of all (query, document) pairs.
### There exist several methods of measuring the performance of retrieval systems namely; Precision, Recall, Fall-Out, E-measure and F-measure just to mention a few since researchers are coming up with other new methods.
A brief description of each method will shade some light.
### 1.1.1 Recall
Recall in information retrieval is defined as the number of relevant documents returned from a search divided by the total number of documents that can be retrieved from a database. Recall can also be looked at as evaluating how well the method that is being used to retrieve information gets the required information.
### Recall(1.1)
As an example, if a database contains 500 documents, out of which 100 contain relevant information required by a researcher, the complement ,number of documents not required = 400.
If the researcher uses a system to search for the documents in this database and it return 100 documents of which all of them are relevant to the researcher, then the recall is given by:
Recall
Supposed that out of 120 returned documents, 30 are irrelevant, then the recall would be given by
Recall
### 1.1.2 Precision
Precision is defined as the number of relevant documents retrieved from the system over the total number of documents retrieved in that search. It valuates how well the method being used to retrieve information filters the unwanted information.
### Precision(1.2)
As an example, if a database contains 500 documents, out of which 100 contain relevant information required by a researcher, the complement ,number of documents not required = 400.
If the researcher uses a system to search for the documents in this database and it returns 100 documents of which all of them are relevant to the researcher, then the precision is given by:
### Precision
Supposed that out of 120 returned documents, 30 are irrelevant, then the precision would be given by
### Precision
Both precision and recall are based on one term; Relevance Oxford dictionary defines relevance as “connected to the issue being discussed”.
Yolanda Jones (2004) identified three types of relevance, namely;
Subject relevance which is the connection between the subject submitted via a query and subject covered by returned texts. Situational relevance: connection between the situation being considered and texts returned by database system. Motivational relevance: connection between the motivations of a researcher and texts returned by database system.
There are two measures of relevance;
• Novelty Ratio: This refers to the proportion of items returned from a search and acknowledged by the user as being relevant, of which they were previously unaware of.
• Coverage Ratio: This refers to the proportion of items returned from a search out of the total relevant documents that the user was aware of before he/she started the search.
Precision and recall affect each other i.e. increase in recall value decreases precision value.
If one increases a system’s ability to retrieve more documents, this implies increasing recall, this will have a drawback since the system will also be retrieving more irrelevant documents hence reducing the precision of that system. This means that a trade-off is required in these two measures so as to ensure better search results.
Precision and recall measures make use of the following assumptions
They make the assumption that either a system returns a document or doesn’t.
They make the assumption that either the document is relevant or not relevant, nothing in between.
New methods are being introduced by researchers which rank the degree of relevance of the documents.
### 1.1. 3 Receiver Operating Characteristics (ROC) Curve
This is the plot of the true positive rate or sensitivity against the false positive rate or (1 − specificity).Sensitivity is just another term for recall. The false positive rate is given by. An ROC curve always goes from the bottom left to the top right of the graph. For a good system, the graph climbs steeply on the left side. For unranked result sets, specificity, given bywas not seen as a very useful idea. Because the set of true negatives is always so large, its value would be almost 1 for all information needs (and, correspondingly, the value of the false positive rate would be almost 0).
### 1.1.4 F-measure and E-measure
This is defined as the weighted harmonic mean of the recall and precision. Numerically, it is defined as
(1.3)
Whereis the weight.
Ifis assumed to be 1, then
(1.4)
The E-measure is given by(1.5)
E –measure has a maximum value of 1.0, 1.0 being the best.
### 1.1.5 Fall-Out
This is defined as the proportion of irrelevant documents that are returned in a search out of all the possible irrelevant documents.
Fall out(1.6)
It can also be defined as the probability of a system retrieving an irrelevant document.
These are just a few methods of measuring performance of search systems. Then after looking after one system, there arise a problem of comparing two systems or algorithms, that is, is this system better than the other one?
To answer this question, scientist in Information retrieval use statistical significance tests to do the comparisons in order to establish if the difference in systems performance are not by chance. These tests are used to confirm beyond doubt that one system is better than another.
### Statement of the problem
Statistical inference tools like statistical significance tests are important in decision making. Their use has been on the rise in different areas of research. With their rise, novel users make use of these tools but in questionable manners. There are many researchers who do not understand the basic concepts in statistics leading to misuse of the tools. Any conclusions reached from a research might be termed bogus if the statistical tests used in it are shoddy.
More light needs to be shade in this area of research to ensure correct use of these tests. Researchers in Information Retrieval also use these tests to compare systems and algorithms, are the conclusions from these tests truly correct? Are there any other ways of comparison which minimize the use of statistical tests?
### Objectives of the study
The objectives of this study are:
Investigate use and misuse of statistical significance tests in scientific papers submitted by researchers to SIGIR.
Shade light on different statistical significance tests their use, assumptions and limitations.
Identify the most important statistical concepts that can provide solutions to the problems of statistical significance in scientific papers submitted by researchers to SIGIR.
Investigate the reality of the problems of statistical significance in scientific papers submitted by researchers to SIGIR.
Investigate the use of statistical significant tests used by researchers in Information Retrieval
Discover the availability of statistical concepts and methods that can provide solutions to the problems of statistical significance in scientific papers submitted by researchers to SIGIR
### Chapter Two
This section of this paper has been divided into three major parts, the sample selection and sample size choosing which will discusses methods of selecting a sample and the size of the sample to be used in a given research, the second part deals with statistical analysis methods and procedures, mainly in significance testing and the third part discusses other statistical methods that can be used in place of statistical significance test.
### 2.0.1 Sample selection
Sampling plays a major role in research, according to Cochran (1977), sampling is the process of selecting a portion of the population and using the information derived from this portion to make inferences about the entire population.
(i)Reduced cost
For example it is very expensive to carry out a census than just collecting information from a small portion of the population. This is because only a small number of measures will be made so only a few people will be hired to do the job compared to complete census which will require a large labor force.
(ii)Greater speed during the process(less time)
Since only a few people will be used or rather only a few items will be measured, the time for doing the measurement will be reduced and also summarization of the data will be quick as opposed to when measures are taken for the whole population.
(iii)Greater accuracy
Since only a few people will be considered in the process, the researchers will be very thorough as compared to the entire population which will see the researchers get tired in the middle of the process leading to lousy collection of data and shoddy analysis.
The choice of the sampling units in a given research may affect the credibility of the whole research. The researcher must make sure that the sample being used is not biased, that is it represents the whole population.
There are several methods of selecting samples to be used in a study. A researcher should always make sure that the sample drawn is large enough to be a representative of the population as a whole and at the same time manageable. In this section the two major types of sampling, random and non-random, will be examined.
### 2.0.1.1 Random sampling
In random sampling, all the items or individuals in the population have equal chances of being selected into the sample. This procedure ensures that no bias is introduced during the selection of sample units since a n items selection will be only by chance and will not depend on the person assigned with the duty of coming up with the sample. There exist five major random sampling techniques, namely; simple random sampling, multi-stage sampling, stratified sampling, cluster sampling and systematic sampling. The following section discusses each of these.
### 2.0.1.1.1 Simple random sampling
In simple random sampling, each item in the population has the same and equal chance of being included in the sample. Usually each sampling unit is assigned a unique number and then numbers are generated using a random number generator and a sampling unit is included in the sample if its corresponding number is generated from the random number generator.
One advantage attributed to simple random sampling is its simplicity and ease in application when dealing with small populations. Every entity in the population has to be enlisted and given a unique number then their respective random numbers be read. This makes this method of sampling very tedious and cumbersome especially where large populations are involved.
### 2.0.1.1.2 Stratified sampling
In stratified random sampling, the entire population is first divided into N disjoint subpopulations .Each sampling unit belongs to one and only one sub population. These sub populations are called strata, they might be of different sizes and they are homogenous within the strata and each stratum completely differs with the other strata. It is from these strata that samples are drawn for a particular study. Examples of strata that are commonly used include States, provinces, Age and Sex, religion, academic ability or marital status etc.
Stratification is most useful when the stratifying variables are simple to work with, easy to observe and closely related to the topic of the survey (Sheskin, 1997).
Stratification can be used to select more of one group than another. This may be done if it is felt that the responses obtained vary in one group than another. So, if the researcher knows that every entity in each group has much the same value, he/she will only need a small sample to get information for that group; whereas in another group, the values may differ widely and a bigger sample is needed.
If you want to combine group level information to get an answer for the whole population, you have to take account of what proportion you selected from each group. This method is mainly used when information is required for only a particular subdivision of the population, administrative convenience is an issue and the sampling problems differ greatly in different portions of the population of study.
### 2.0.1.1.3 Systematic sampling
Systematic sampling is quite different from the other methods of sampling, supposed the population contains N units and a sample of n units is required, a random number is generated using the random number generator, call it k, then a unit(represented as a number) is drown from the sample then the researcher picks every kth unit thereafter. Consider the example that k is 20 and the first unit that is drawn is 5, the subsequent units will be 25,45,65,85 and so on.
The implication of this method is that the selection of the whole sample will be determined by only the first item since the rest will be obtained sequentially. This type is called an every kth systematic sample. This technique can also be used when questioning people in a sample survey. A researcher might select every 15th person who enters a particular store, after selecting a person at random as a starting point; or interview the shopkeepers of every 3rd shop in a street, after selecting a starting shop at random.
It may be that a researcher wants to select a fixed size sample. In this case, it is first necessary to know the whole population size from which the sample is being selected. The appropriate sampling interval, I, is then calculated by dividing population size, N, by required sample size, n. This method is advantageous since it is easy and it is more precise than simple random sampling.
Also it is simpler in systematic sampling to select one random number and then every kth member on the list, than to select as many random numbers as sample size. It also gives a good spread right across the population. A disadvantage is that the researcher may be forced to have a starting list if he/she wishes to know the sample size and calculate the sampling interval.
### 2.0.1.1.4 Cluster sampling
The Austarlian Bureau of Statistics insinuates that cluster sampling divides the population into groups, or clusters. A number of clusters are selected randomly to represent the population, and then all units within selected clusters are included in the sample. No units from non-selected clusters are included in the sample. They are represented by those from selected clusters. This differs from stratified sampling, where some units are selected from each group.
The clusters are heterogeneous within each cluster (that is the sampling units inside a cluster vary from each other completely) and each cluster looks alike with the other clusters. Cluster sampling has several advantages which include reduced costs, simplified field work and administration is more convenient. Instead of having a sample scattered over the entire coverage region, the sample is more concentrated in relatively few collection points (clusters).
Cluster sampling provides results that are less accurate compared to stratified random sampling.
### 2.0.1.1.5 Multi-stage sampling
Multi-stage sampling is like cluster sampling, but involves selecting a sample within each chosen cluster, rather than including all units in the cluster. The Australian Bureau of Statistics postulates that multi-stage sampling involves selecting a sample in at least two stages. In the first stage, large groups or clusters are selected. These clusters are designed to contain more population units than are required for the final sample.
In the second stage, population units are chosen from selected clusters to derive a final sample. If more than two stages are used, the process of choosing population units within clusters continues until the final sample is achieved. If two stages are used then it will be called a two stage sampling, if three stages are used it will be called a three stage sampling and so on.
### 2.1 Statistical Analysis
In this section, different statistical tests are discussed in details in their general form, then move to discussed how each of them(the ones used in IR) are applied to information retrieval. Only some of these tests are used to compare systems or/and algorithms.
In this paper we look at three sections of statistical analysis, namely:
(i) Summarizing data using a single value.
(ii) Summarizing variability.
(iii) Summarizing data using an interval (no specific value)
In the first case, we have the mean, mode, median etc and in the second case, we look at variability in the data and in the third case we look at the confidence intervals, parametric and nonparametric tests of hypothesis testing
### 2.1.1 Summarizing data using a single value
In this case, the data being analyzed is represented by a single value, example for this scenario are discussed below:
### 2.1.1.1 Mean
There are three different kinds of mean:
(i)Arithmetic mean
(ii)Geometric Mean
(iii)Harmonic mean
(i) Arithmetic mean
This is computed by summing all the observations then dividing by the number of observations that you have collected.
Letbe n observations of a random variable X. The arithmetic mean is defined as
Arithmetic mean
When to use the arithmetic mean
The arithmetic mean is used when:
When the collected data is a numeric observation.
When the data has only one mode (uni-modal)
When the data is not skewed i.e. not concentrated to extreme values.
When the data does not have many outliers (very extreme values)
The arithmetic mean is not used when:
You have categorical data
When the data is extremely skewed.
(ii) Geometric mean
This is defined as the product of the observations, everything raised to power of, usually n.
Letbe n observations of a random variable X. The geometric mean is defined as
Geometric mean
The Geometric mean is used when:
The observations are numeric.
The item that we are interested in is the product of the observations.
(iii) Harmonic mean
This is defined as the number of observations divide be the sum of reciprocals of the observations.
Letbe n observations of a random variable X. The harmonic mean is defined as
Harmonic mean
The Harmonic mean is used when:
The average can be justified for the reciprocal of the observations.
### 2.1.1.2 Median
This is defined as the middle value of the observations. The observations are first arranged in ascending or descending order then the middle value is taken as the median.
The median is used when:
When the observations are skewed.
The observations have a single mode.
The observations are numerical.
The median is not used when:
We are interested in the total value.
### 2.1.1.3 Mode
This is defined as the largest value in the given dataset or the value that has the highest frequency of occurrence.
The mode is used when:
The dataset is categorical.
The dataset is both numeric and multimodal.
### 2.1.2 Summarizing variability
Variability in a data can be summarized using the following measures:
### 2.1.2.1 Sample variance
Letbe n observations of a random variable X, then the Sample variance, is given by
The standard deviation is used when:
The data is normally distributed.
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## Section40.2Microscopic View of Magnetic Materials
The magnetism of a material is due to the magnetic properties of the constituent particles, electrons, protons and neutrons of their atoms. Electrons, protons and neutrons each have their magnetic dipole moments, and their vector sum gives the net magnetic dipole moment of the atom.
However, due to the low mass of electrons compared to protons and neutrons the electronic contribution to the net magnetic dipole moment is dominant unless electronic magnetic dipole moment happens to be zero. Therefore, in the following, we will ignore the contributions of protons and neutrons.
The magnetic dipole moment of an electron in an atom comes from two sources:
1. Orbital magnetic dipole moment due to the motion of elctron about the nucleus. We will denote the corresponding magnetic dipole moment by $\vec \mu_l\text{.}$
2. Quantum spin. Electron is intrinsically magnetic with a magnetic dipole moment due to an intrinsic angular momentum, called spin. We will denote the corresponding magnetic dipole moment by $\vec \mu_s\text{.}$
The net angular momentum of an atom will be a vector sum of the two angular momenta of every electron in the atom.
$$\vec \mu_\text{atom} \approx \sum_\text{electrons} \left( \vec \mu_l + \vec \mu_s\right) \tag{40.2.1}$$
### Subsection40.2.1Orbital Angular Momentum and Magnetic Dipole Moment
Here, I will use fictitious current loop model of a magnetic dipole moment to show that when an electron moves in an orbit, it will develop a magnetic dipole moment due to its motion. The current loop model states that magnetic moment can be written as a product of current and an area of a fictitious closed circuit.
$$\mu = \left(I A\right)_\text{model}.\label{eq-mu-equals-IA}\tag{40.2.2}$$
Suppose the circuit is formed by an electron moving uniformly with speed $v$ in a circle of radius R. Therefore, $I$ and $A$ will be
\begin{equation*} I = \frac{|e|v}{2\pi R},\ \ A = \pi R^2. \end{equation*}
This will correspond to a magnetic dipole moment
\begin{equation*} \mu = \fract{|e|}{2}\, v R. \end{equation*}
In this formula, we can replace $vR$ by angular momentum $L$ divided by mass of electron, $m_e\text{.}$
\begin{equation*} \mu = \fract{|e|}{2m}\, L. \end{equation*}
We denote this dipole moment by $\mu_l\text{,}$ to indicate that it is from the angular momentum due to the orbital motion.
$$\mu_l = \dfrac{|e|}{2m_e} L.\tag{40.2.3}$$
In general, for any charged particle, we expect that there will be a magnetic moment due to its orbital angular momentum $L\text{,}$ with the proportionality depending on its charge $q$ and mass $m\text{.}$
$$\mu_l = \dfrac{|q|}{2m} L.\tag{40.2.4}$$
The quantity $|q|/2m$ is called the gyromagnetic ratio of the particle, usually denoted by $g_q\text{,}$ where the subscript is an identifier for the particle.
\begin{equation*} g_q = \dfrac{|q|}{2m}, \end{equation*}
For an electron the value of gyromagnetic ratio is
$$g_e = \dfrac{|e|}{2m_e} \approx 8.78 \times 10^{10}\text{ C/kg}.\tag{40.2.5}$$
For a proton, it is about $1/2000$ times smaller due to the mass of proton being about $2000$ times larger.
Since, conventional current is in the opposite direction to the motion of electron, the direction of magnetic moment vector of an electron will be in the opposite direction to its angular momentum. Thus, we will write the vector relation as
\begin{equation*} \vec \mu_l = - g_e\, \vec L, \end{equation*}
It turns out that in SI units, the value of $\mu$ due to electron's motion in atoms are very tiny numbers. More commonly, we express it in a unit called Bohr magneton, denoted by $\mu_B\text{.}$
$$\mu_B = \dfrac{|e|\hbar}{ 2 m_e} = g_e\hbar,\tag{40.2.6}$$
where $\hbar$ is Planck constant $h$ divided by $2\pi\text{.}$
\begin{equation*} \hbar = \dfrac{h}{2\pi} = 1.0545718 \times 10^{-34} \text{ m}^2\text{kg / s}. \end{equation*}
In terms of $\mu_B$ the magnetic dipole moment of electron that has angular momentum $\vec L$ will be
$$\vec \mu_l = - \dfrac{\vec L}{\hbar}\, \mu_B.\tag{40.2.7}$$
In atoms, angular momentum $L$ is usually a small multiple of $\hbar\text{,}$ and hence, magnetic dipole moments are a small multiple of $\mu_B\text{.}$
Classically, the magnitude of angular momentum $L$ can be any positive real number with units, of course. But, quantum mechanically, $L$ can only have certain values given by the following formula.
$$L = \sqrt{l(l+1)}\, \hbar,\ \ l = 0, 1, 2, 3, \cdots.\label{eq-orbital-angular-momentum}\tag{40.2.8}$$
That is, only $L=0\text{,}$ $L = \sqrt{2}\, \hbar\text{,}$ $L = \sqrt{6}\, \hbar\text{,}$ $L = \sqrt{12}\, \hbar\text{,}$ etc are allowed. Therefore, magnetic dipole moment will also be quantized in the same way and will have only certain values.
$$\mu_l = \sqrt{l(l+1)}\, \mu_B,\ \ l = 0, 1, 2, 3, \cdots.\label{eq-orbital-angular-momentum-magnetic-moment}\tag{40.2.9}$$
Furthermore, when we look at one of its Cartesian components, it can be integral values of $\hbar\text{.}$
$$L_z = l_z \hbar,\ \ l_z = 0, \pm 1, \pm 2, \pm 3, \cdots.\label{eq-orbital-angular-momentum-components}\tag{40.2.10}$$
This says that magnetic moment along a Cartesian axis will be integral multiples of a Bohr magneton.
$$\mu_{lz} = l_z \mu_B,\ \ l_z = 0, \pm 1, \pm 2, \pm 3, \cdots.\label{eq-orbital-angular-momentum-components-magnetic-moment}\tag{40.2.11}$$
### Subsection40.2.2Spin And Magnetic Dipole Moment
Every electron has an intrinsic angular momentum called spin regardless of the state of motion of the electron. The name “spin” for this property of an electron is a misnomer and suggests that the electron is in rotational motion about an axis through its center. If that were the case, angular momentum of such a spinning motion will be zero since moment of inertia about axis through a point particle would be zero. Hence, spin of an electron is best regarded as an intrinsic property and not to be taken literally.
The spin property accounts for the fact that electrons appear to have additional angular momentum than one can account for based only on the motion of the electron.
Unlike the magnitude of the orbital angular momentum, which can take many values, depending upon integer $l$ in Eq. (40.2.9), a measurement of the magnitude of the spin angular momentum of an electron, to be denoted by $\vec S\text{,}$ results in only one value for the magnitude. For an electron, the magnitude $S=\frac{\sqrt{3}}{2}\hbar\text{.}$ In general, for any particle, we can write the magnitude $S$ as
$$S = \sqrt{ s\left( s + 1 \right)}\, \hbar,\tag{40.2.12}$$
where $s$ is called the spin of the particle. For electron
\begin{equation*} s = \dfrac{1}{2}. \end{equation*}
Proton and neutrons also have $s=\frac{1}{2}\text{.}$ As a matter of fact, $s$ can only be positive integer multiples of $\frac{1}{2}\text{.}$ Particles with odd multiples are called Fermions and 0 and even multiples are called Bosons.
A measurement of the spin angular momentum of an electron along any direction, say the $z$-axis, results in only the following two values.
\begin{equation*} S_z = s_z\, \hbar,\ \ s_z = \pm\dfrac{1}{2}, \end{equation*}
similar to the way $L_z$ works out, except now the value is $1/2$ instead of integers.
Just as the orbital angular momentum creates a magnetic dipole moment for the electron, spin angular momentum creates additional magnetic dipole moment for the electron. However, the gyromagnetic ratio of the spin angular momentum is not the same as the gyromagnetic ratio of the orbital angular momentum. Actually, for electrons, each unit of spin angular momentum is actually about twice as effective in creating magnetic moment as orbital angular momentum.
The factor by which you will need to multiply gyromagnetic ratio of orbital angular momentum to get the g-factor of spin angular momentum is called the $g$ factor. It is often denoted by $g_S\text{.}$
$$g_{e,\text{spin}} = g_S\, g_e = g_S\, \dfrac{\mu_B}{\hbar}.\tag{40.2.13}$$
$g_S$ is one of most precisely known number in physics with reported value at NIST in May/2020 to be
\begin{equation*} g_S = 2.00231930436256 \left(1 \pm 1.7\times10^{−13} \right). \end{equation*}
We will just use $g_S=2.0$ for electron. I gave you the more precise value for you to marvel at how many decimals have actually been measured! The $g$ factor of protons, neutrons, and other particles are different.
Therefore, magnetic moment contribution of the spin angular moment will be
\begin{equation*} \vec \mu_S = g_{e,\text{spin}}\ \vec S = g_S\, \vec S\, \dfrac{\mu_B}{\hbar} , \end{equation*}
with Cartesian components given by
$${\mu_S}_z = \pm\dfrac{1}{2}\, g_S\, \mu_B = \pm\mu_B,\tag{40.2.14}$$
and total spin magnetic moment to be
$$\mu_S = \dfrac{\sqrt{3}} {2}\, g_S\, \mu_B = \sqrt{3}\,\mu_B .\tag{40.2.15}$$
This much magnetic moment will be produced by orbital angular momentum between $l=2$ and $l=3\text{.}$
#### Subsubsection40.2.2.1Pairing of Spins
The two values of the spin projections along a Cartesian axis are called the spin up and the spin down states respectively. The rules of quantum mechanics often forces the pairing of up and down spins so that atoms that have an even number of electrons often end up with zero net spin angular momentum in any direction. In these cases, the magnetism of a material comes from the orbital angular momentum.
Therefore, the effect of the spin magnetic moment will be present when spins of electrons cannot be paired up. This will definitely happen if a particular atom has either an odd number of electrons or where spins of different electrons are not paired up. From our formulas above, we see that the spin magnetic moment of a single elctron along any axis is either $+1\ \mu_B$ or $-1\ \mu_B\text{.}$
If we place atoms with magnetic dipole moment due to unpaired spins or non-zero orbital angular momentum in an external magnetic field, the alignment of atomic dipole moments give rise to macroscopically observable magnetic phenomenon.
For instance, if the magnetic dipole moments of $N$ atoms, each with a dipole moment of, say 1 $\mu_B\text{,}$ are line up in one direction, then the net magnetic dipole moment of the sample will be $N\mu_B\text{.}$ How large can these number be? Suppose $N$ is of the order of the Avogadro number, then the net magnetic dipole moment will be
\begin{equation*} \mu \sim 6\times 10^{23}\times 9\times 10^{-24} \text{A.m}^2 = 5\ \text{A.m}^2 . \end{equation*}
This is equivalent to the magnetic dipole moment of a whopping $500\text{ A}$ current in a $10 \text{ cm} \times 10 \text{ cm}$ loop.
Determine the magnetic moment of (a) a sodium atom which has an unpaired electron in $l_z = 0$ and the spin up state, and (b) an electron in the carbon atom that is in spin up state and has $l_z = 1\text{.}$ Use the $z$-axis as the axis for the angular momentum vector.
Hint
Add up the magnetic momenta of orbital and spin parts.
Answer
(a) $\mu_B\text{,}$ (b) $2\mu_B\text{.}$
Solution
(a) Since $l_z = 0\text{,}$ the orbital angular momentum along the $z$-axis is zero. That means magnetic moment will be all from the spin of the electron.
\begin{equation*} \mu_z = {\mu_S}_z = + \mu_B. \end{equation*}
(b) Now, both orbital and spin will contribute. Since they are in the same direcion, they will add together.
\begin{equation*} \mu_z = \mu_{\text{orbital},z} + \mu_{\text{spin},z} = 1\times \mu_B + \mu_B = 2\mu_B. \end{equation*}
An excited helium atom has two electrons in different states, one with $l_z = 0\text{,}$ and spin down, and the other with $l_z = + 2 \text{,}$ and spin up state. What is the magnetic moment of the atom? Use the $z$-axis as the axis of projections of the angular momentum vectors.
Hint
Find magnetic moment of each and then add them vectorally
Answer
$2\mu_B\text{.}$
Solution
Adding the $z$ magnetic moments of the two electrons we get
\begin{align*} \mu_z \amp = \mu_{1z} + \mu_{2z}\\ \amp = \left[ \mu_{\text{orbital},z} + \mu_{\text{spin},z} \right]_1 + (1\rightarrow 2)\\ \amp = \left[(0-1)\mu_B \right]_1 + \left[(2+1)\mu_B \right]_2\\ \amp = -\mu_B + 3 \mu_B = 2\mu_B. \end{align*} | 3,584 | 12,163 | {"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": 15, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-14 | latest | en | 0.815628 |
https://aprove.informatik.rwth-aachen.de/eval/FullComplexity/newtpdb/Runtime_Complexity_Full_Rewriting/AProVE_07/kabasci02.xml.AProVE%20Lower.html | 1,627,667,516,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00568.warc.gz | 125,412,925 | 7,242 | ### (0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))
Rewrite Strategy: FULL
### (1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
plus(s(x), y) →+ s(plus(x, y))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / s(x)].
The result substitution is [ ].
### (3) RenamingProof (EQUIVALENT transformation)
Renamed function symbols to avoid clashes with predefined symbol.
### (4) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
S is empty.
Rewrite Strategy: FULL
Infered types.
### (6) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
### (7) OrderProof (LOWER BOUND(ID) transformation)
Heuristically decided to analyse the following defined symbols:
plus, times, div, quot, zero, eq, divides, pr, if
They will be analysed ascendingly in the following order:
plus < times
plus < zero
times < div
times < divides
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
eq < zero
zero = divides
zero = pr
zero = if
eq < divides
divides = pr
divides = if
pr = if
### (8) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
plus, times, div, quot, zero, eq, divides, pr, if
They will be analysed ascendingly in the following order:
plus < times
plus < zero
times < div
times < divides
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
eq < zero
zero = divides
zero = pr
zero = if
eq < divides
divides = pr
divides = if
pr = if
### (9) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
Induction Base:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(0)) →RΩ(1)
gen_0':s:true:false2_0(a)
Induction Step:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(+(n4_0, 1))) →RΩ(1)
s(plus(gen_0':s:true:false2_0(a), p(s(gen_0':s:true:false2_0(n4_0))))) →RΩ(1)
s(plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0))) →IH
s(gen_0':s:true:false2_0(+(a, c5_0)))
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (11) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
times, div, quot, zero, eq, divides, pr, if
They will be analysed ascendingly in the following order:
times < div
times < divides
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
eq < zero
zero = divides
zero = pr
zero = if
eq < divides
divides = pr
divides = if
pr = if
### (12) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
Induction Base:
times(gen_0':s:true:false2_0(0), gen_0':s:true:false2_0(b)) →RΩ(1)
0'
Induction Step:
times(gen_0':s:true:false2_0(+(n983_0, 1)), gen_0':s:true:false2_0(b)) →RΩ(1)
plus(gen_0':s:true:false2_0(b), times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b))) →IH
plus(gen_0':s:true:false2_0(b), gen_0':s:true:false2_0(*(c984_0, b))) →LΩ(1 + b·n9830)
gen_0':s:true:false2_0(+(*(n983_0, b), b))
We have rt ∈ Ω(n3) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n3).
### (14) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
eq, div, quot, zero, divides, pr, if
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
eq < zero
zero = divides
zero = pr
zero = if
eq < divides
divides = pr
divides = if
pr = if
### (15) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
Induction Base:
eq(gen_0':s:true:false2_0(0), gen_0':s:true:false2_0(0)) →RΩ(1)
true
Induction Step:
eq(gen_0':s:true:false2_0(+(n2234_0, 1)), gen_0':s:true:false2_0(+(n2234_0, 1))) →RΩ(1)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) →IH
true
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (17) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
quot, div, zero, divides, pr, if
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (18) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Induction Base:
quot(gen_0':s:true:false2_0(0), gen_0':s:true:false2_0(0), gen_0':s:true:false2_0(1)) →RΩ(1)
s(div(gen_0':s:true:false2_0(0), s(gen_0':s:true:false2_0(0)))) →RΩ(1)
s(0')
Induction Step:
quot(gen_0':s:true:false2_0(+(n2877_0, 1)), gen_0':s:true:false2_0(+(n2877_0, 1)), gen_0':s:true:false2_0(1)) →RΩ(1)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) →IH
gen_0':s:true:false2_0(1)
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (20) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
div, zero, divides, pr, if
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (21) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol div.
### (22) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
zero, divides, pr, if
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (23) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol zero.
### (24) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
if, divides, pr
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (25) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol if.
### (26) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
pr, divides
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (27) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol pr.
### (28) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
The following defined symbols remain to be analysed:
divides
They will be analysed ascendingly in the following order:
div = quot
div = zero
div = divides
div = pr
div = if
quot = zero
quot = divides
quot = pr
quot = if
zero = divides
zero = pr
zero = if
divides = pr
divides = if
pr = if
### (29) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol divides.
### (30) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
No more defined symbols left to analyse.
### (31) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n3) was proven with the following lemma:
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
### (33) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
quot(gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(n2877_0), gen_0':s:true:false2_0(1)) → gen_0':s:true:false2_0(1), rt ∈ Ω(1 + n28770)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
No more defined symbols left to analyse.
### (34) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n3) was proven with the following lemma:
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
### (36) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
eq(gen_0':s:true:false2_0(n2234_0), gen_0':s:true:false2_0(n2234_0)) → true, rt ∈ Ω(1 + n22340)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
No more defined symbols left to analyse.
### (37) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n3) was proven with the following lemma:
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
### (39) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
No more defined symbols left to analyse.
### (40) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n3) was proven with the following lemma:
times(gen_0':s:true:false2_0(n983_0), gen_0':s:true:false2_0(b)) → gen_0':s:true:false2_0(*(n983_0, b)), rt ∈ Ω(1 + b·n98302 + n9830)
### (42) Obligation:
TRS:
Rules:
p(0') → 0'
p(s(x)) → x
plus(x, 0') → x
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0', y) → 0'
times(s(0'), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0', y) → 0'
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0'
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0', s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0', 0') → true
eq(s(x), 0') → false
eq(0', s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0')) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0')), plus(zero(0'), 0'), s(plus(0', zero(0'))))
Types:
p :: 0':s:true:false → 0':s:true:false
0' :: 0':s:true:false
s :: 0':s:true:false → 0':s:true:false
plus :: 0':s:true:false → 0':s:true:false → 0':s:true:false
times :: 0':s:true:false → 0':s:true:false → 0':s:true:false
div :: 0':s:true:false → 0':s:true:false → 0':s:true:false
quot :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
zero :: 0':s:true:false → 0':s:true:false
eq :: 0':s:true:false → 0':s:true:false → 0':s:true:false
true :: 0':s:true:false
false :: 0':s:true:false
divides :: 0':s:true:false → 0':s:true:false → 0':s:true:false
prime :: 0':s:true:false → 0':s:true:false
pr :: 0':s:true:false → 0':s:true:false → 0':s:true:false
if :: 0':s:true:false → 0':s:true:false → 0':s:true:false → 0':s:true:false
hole_0':s:true:false1_0 :: 0':s:true:false
gen_0':s:true:false2_0 :: Nat → 0':s:true:false
Lemmas:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40)
Generator Equations:
gen_0':s:true:false2_0(0) ⇔ 0'
gen_0':s:true:false2_0(+(x, 1)) ⇔ s(gen_0':s:true:false2_0(x))
No more defined symbols left to analyse.
### (43) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
plus(gen_0':s:true:false2_0(a), gen_0':s:true:false2_0(n4_0)) → gen_0':s:true:false2_0(+(n4_0, a)), rt ∈ Ω(1 + n40) | 18,651 | 43,398 | {"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} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.492296 |
http://ciphersbyritter.com/NEWS2/95060501.HTM | 1,493,416,262,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123097.48/warc/CC-MAIN-20170423031203-00615-ip-10-145-167-34.ec2.internal.warc.gz | 81,687,466 | 2,362 | ```Newsgroups: comp.security.misc,sci.crypt,alt.security
Path: news.io.com!news.tamu.edu!news.utdallas.edu!news.starnet.net!wupost!
+ ullisys.pond.sub.org!felix
From: felix@ullisys.pond.sub.org (Felix Schroeter)
Subject: Re: Random Number Generators
Message-ID: <1995Jun5.165619.175@ullisys.pond.sub.org>
Organization: K1, Germany
Date: Mon, 5 Jun 1995 16:56:19 GMT
References: <1995May28.131614.
+ 4713@ullisys.pond.sub.org> <3qfbkc\$pnp@sol.ctr.columbia.edu>
Followup-To: comp.security.misc,sci.crypt,alt.security
Lines: 81
Xref: news.io.com comp.security.misc:17563 sci.crypt:38298 alt.security:25411
Hello!
Seth Robertson (seth@soscorp.com) wrote:
: [...]
: >secret information: seed (64 bit), key (56 bit)
: >Generation of a random number:
: >seed = des_encrypt (seed, key);
: >return seed%100000000; /* this returns a number from 0 to 99 999 999 */
: Your last line is a *serious* flaw--since you used mod, you no longer
: have uniform random numbers.
In a strict sense, you are right. The numbers 0 to (2^64-1)%100000000
have probability (2^64 div 100000000+1)/(2^64), the numbers
from (2^64)%100000000 to 99999999 "only" probability
(2^64 div 100000000)/(2^64).
Let's start bc...
(< marks input, > marks output)
< scale=0
< a=2^64
< a
> 18446744073709551616
< b=a/100000000
< c=a%100000000
< scale=100
< (b+1)/a
> .00000001000000000004903216721530156974040437489748001098632812500000\
> 00000000000000000000000000000000
< b/a
> .00000000999999999999482205859102634804003173485398292541503906250000\
> 00000000000000000000000000000000
< c
> 9551616
So, the numbers from 0 to 9551615 have probability 0.00000001000000000005
(approx), those from 9551616 to 99999999 "only" 0.0000000099999999999948.
The difference of these probabilities is therefore 0.0000000000000000000552.
This is a relative error of 0.00000000000552, compared to the probability
of an ideal RNG (i.e. 1/100000000). So I think, your statement is of theortical
importance only.
: By way of example, let us say that des_encrypt produces numbers
: between 0 and 2^27 (it does not matter what exact range it produces as
: long as it is not a multiple of 100 million).
It doesn't matter for theory's sake, but for practical sake, because
the relative error of the probabilities is the smaller, the bigger the
range of "raw" numbers produced (e.g. by des_encrypt) is...
: [...]
: There are two solutions to this. The first is if the number is 100
: million or more, you loop back until you get a number less than 100
: million.
Or loop only if the number is greater or equal
((2^64 div 100000000)/100000000). Then the range produced after leaving
the loop is just a multiple of those 100000000 big. And you have a rather
low chance of having to loop anyway :-)
: The other is to use division to normalize the numbers.
: Something like: ((double)seed) / ((double)(1<<27)) * 100000000.0
: should work if I didn't make a stupid mistake. I prefer the former
: method since on the average it is faster and you will never get into
: trouble because of precision limitation on various machines.
Hmmm. Your former method is rather slow, because the probability that
des_encrypt yields a number <100000000 is 100000000/(2^64), which is
(bc)
< scale=100
< 100000000/(2^64)
> .00000000000542101086242752217003726400434970855712890625000000000000\
> 00000000000000000000000000000000
: [... other (P)RNG suggestions ...]
Regards, Felix.
``` | 1,074 | 3,442 | {"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-2017-17 | longest | en | 0.580949 |
http://math.stackexchange.com/questions/416088/help-with-a-trig-substitution-integral | 1,469,653,626,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00156-ip-10-185-27-174.ec2.internal.warc.gz | 156,112,436 | 20,729 | # Help with a trig-substitution integral
I'm in the chapter of trigonometric substitution for integrating different functions. I'm having a bit of trouble even starting this homework question: $$\int \frac{(x^2+3x+4)\,dx} {\sqrt{x^2-4x}}$$
-
A start: We have $x^2-4x=(x-2)^2-4$. If you want to use a trigonometric substitution, the natural one is $x-2=2\sec t$.
We unfortunately end up needing to integrate powers of $\sec t$, a messy business. Nicer is the hyperbolic function substitution $x-2=2\cosh t$.
-
Also note that $x^{2}+3x+4 = (x-2)^{2} + 7x = 4\sec^{2}(t) + 7 \cdot (2\sec(t)+2)$
-
Is this a comment to someone else's answer? It does not appear to be an answer. – robjohn Jun 13 '13 at 17:47
@robjohn, thank you, I will edit. – Brady Trainor Jun 14 '13 at 14:09
Make the substitution suggested by completing the square, $\sqrt{x^2-4x}=\sqrt{(x-2)^2-4}$, that is, $y=x-2$. Then, by the easily constructed triangle, we have
\begin{align} \frac{y}{2} & = \sec\theta\\ \frac{\sqrt{y^2-4}}{2} & = \tan\theta\\ x^2+3x+4 & =(y+2)^2+3(y+2)+4\\ & = y^2+7y+14\\ \frac{dy}{2} & = \sec\theta\tan\theta\ d\theta. \end{align} The integral becomes
\begin{align} \int \frac{(x^2+3x+4)} {\sqrt{x^2-4x}}dx & = \int\frac{y^2+7y+14}{\sqrt{y^2-4}}dy\\ & = \int\frac{4\sec^2\theta+14\sec\theta+14}{2\tan\theta}2\sec\theta\tan\theta\ d\theta\\ & = \int4\sec^3\theta+14\sec^2\theta+14\sec\theta\ d\theta. \end{align}
Andre Nicolas suggests in another response that the powers of secant are unsavory, and that a hyperbolic trig sub would be more apt. (Claude Leibovici explores this.)
(I vaguely recall methods for the $\sec^3\theta$ integrand in Stewart's Calculus. The $\sec\theta$ is in most trig integral lists, and perhaps the $\sec^2\theta$ is amenable to some fortuitous trig identity.)
Here are some resources on those:
http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_secant
http://en.wikipedia.org/wiki/Integral_of_the_secant_function
http://en.wikipedia.org/wiki/Integral_of_secant_cubed
-
Unfortunately $\sqrt{\vphantom{()^2}x^2-4x}\ne\sqrt{(x-2)^2+4}$ – robjohn Jun 13 '13 at 17:45
@robjohn, thank you, I will edit. – Brady Trainor Jun 14 '13 at 14:09
In order to make a proper substitution in integral calculus, the function that you are substituting must have a unique inverse function. However, there is such a case where the the derivative is present and you can make what I refer to as a "virtual substitution". This is not exactly the case here, we have to do other algebraic manipulations. Trigonometric functions such as sine, cosine and their variants have infinitely many inverse functions; inverse trigonometric functions ( i.e. arcsine, arccosine, etc...) have a unique inverse function, thus are fine. For example, if I made the substitution $y = \sin x$ ( where $-1≤y≤1$) , then $x = (-1)^n \cdot \arcsin y + n\pi$ ($n \in \mathbb Z$): this does not work, without bound. If anyone disagrees with my statement, please prove that the substitution is proper. Also, in my opinion, turning a rational/algebraic function into a transcendental function is ridiculous. There are very elementary ways to approach this integral; a good book to read on many of these methods is Calculus Made Easy by Silvanus P. Thompson.
-
May be I was too fast even if the answer is almost correct.
First make $x = 2 (\cosh(y) + 1)$. The integrand then becomes
$(\pm) 2 (8 + 7 \cosh(y) + \cosh(2 y))$
So, the integral is $(\pm) (16 y + 14 \sinh(y) + \sinh(2 y))$
If you replace $y$, you arrive to my formula (the $\pm$ was missing)
-
$%1/2 Sqrt[(-4 + x) x] (12 + x + (64 Log[Sqrt[-4 + x] + Sqrt[x]])/(Sqrt[-4 + x] Sqrt[x]))$ $$\frac12\sqrt{(-4+x)x}\left(12+x+\frac{64\log(\sqrt{-4+x}+\sqrt{x})}{\sqrt{-4+x}\sqrt{x}}\right)$$
-
The question was about how to do the problem, not what the answer is. – robjohn Jun 13 '13 at 17:43 | 1,278 | 3,892 | {"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": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.670969 |
https://edurev.in/course/quiz/attempt/1643_Test-Kirchhoff%E2%80%99S-Rules/5f5a3eff-82b7-4209-a413-fe8d6698641c | 1,627,549,847,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00633.warc.gz | 241,447,619 | 41,338 | Courses
# Test: Kirchhoff’S Rules
## 5 Questions MCQ Test Physics Class 12 | Test: Kirchhoff’S Rules
Description
This mock test of Test: Kirchhoff’S Rules for JEE helps you for every JEE entrance exam. This contains 5 Multiple Choice Questions for JEE Test: Kirchhoff’S Rules (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Kirchhoff’S Rules quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Kirchhoff’S Rules exercise for a better result in the exam. You can find other Test: Kirchhoff’S Rules extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1
### At any junction, the sum of the currents entering the junction is equal to the sum of _______
Solution:
Kirchhoff's Current Law or KCL, states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, I(exiting) + I(entering) = 0.
QUESTION: 2
### The ______ of changes in potential around any closed loop involving resistors and cells in a loop is zero.
Solution:
In accordance with Kirchhoff’s second law i.e. Kirchhoff’s voltage law (KVL), the algebraic sum of all the potential differences in a closed electric circuit or closed loop that contains one or more cells and resistors is always equal to zero.
This law is popularly called the law of conservation of voltage.
QUESTION: 3
### The Wheatstone bridge Principle is deduced using
Solution:
Kirchoff's Law states that the algebraic sum of currents at a junction of an electric circuit is zero. ... Wheatstone bridge is formed by connecting a battery B or an electric source, a plug key K and a variable resistor X between the junctions A and D and a galvanometer G between the junctions C and D.
QUESTION: 4
Given N resistors each of resistance R are first combined to get minimum possible resistance and then combined to get maximum possible resistance. The ratio of the minimum to maximum resistance is
Solution:
Minimum possible resistance can be found in a parallel arrangement and maximum possible resistance can be found in a series arrangement.
Thus, 1/Rmin=1/R+1/R+....+ n times
=n/R
⇒Rmin=R/n
and Rmax=R+R+.......+ n times
⇒Rmax=nR
∴Rmin/Rmax=(R/n)/nR=1/n2
QUESTION: 5
Two cells of 1.25 V and 0.75V are connected in parallel. The effective voltage will be
Solution:
When two cells are conneted in parallel , then
Effective Voltage,V = V1 − V2 = 1.25 − 0.75 = 0.5 V | 662 | 2,709 | {"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-2021-31 | latest | en | 0.915229 |
https://blog.solidsignal.com/tutorials/a-designer-power-bank/ | 1,627,516,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00676.warc.gz | 153,021,624 | 24,785 | # A designer power bank?
It seems like the power bank is the must-have accessory of 2018. I’ve had one for about five years but it seems like this year, everyone has one. Not terribly surprising when you think about how much time we spend on our phones.
## What does a power bank do?
Truth is, a power bank isn’t some magical thing. As I showed you not long ago, most power banks are garden variety rechargeable batteries with just enough circuitry to connect them to your phone. There’s nothing magical there. However, we’re starting to see them coming in all sorts of colors, sizes and shapes as if they were slender monoliths of pure awesomeness.
In fact, it was a few weeks ago that I read an article that none other than Elon Musk himself, purveyor of magical automobiles that go like crazy and never use a drop of gas, was offering a power bank for sale. This power bank not only charged your phone —wirelessly at that— but contained within it one of the very same batteries that make Tesla cars work. Wow, I thought. Then, I thought, \$65 for a power bank. That doesn’t exactly light me up, pun intended. It’s back on his site now, and still overpriced.
So what makes his power bank worth paying for? I’d say, nothing.
## Here’s the only thing you need to look for in a power bank.
Power banks are measured in mAh, or if you’re pronouncing it out loud, “milliamp-hours.” That’s the only thing you need to look for. Let’s unpack that a little bit.
Amps (really short for amperes but no one says amperes) are a measure of current. Amps measure the power flowing from one point to another.
Milliamps are thousandths of an amp. 1000 milliamps (mA) equals one amp. We use milliamps as a measurement for small batteries because small batteries carry small amounts of power.
Hours, that’s self explanatory hopefully.
One milliamp-hour is a flow rate of 1/1000 amp for a period of one hour. It doesn’t mean that it takes a full hour to do it, though. It could also mean a flow rate of 4/1000 amp for 15 minutes, or 60/1000 amp for 1 minute. All of these things are the same. A lot of people use water to describe how electricity flows. In that analogy, a gallon of water is a gallon of water, whether it takes an hour or a day to flow out.
## How does this translate out?
It takes about 3,000mAh to fill a typical cell phone battery. Of course some phones have bigger batteries and some smaller, but it’s a basic measurement. So if you want to completely recharge your phone you will need a power bank that can supply 3,000mAh. Most of them sold today supply more. The sharp looking one at the top of this article supplies over 10,000mAh, meaning that it can charge your phone three times over before it’s depleted. Or, it can charge the phone of you and two of your friends. Either is true.
But basically mAh rating is the only thing you really care about in a power bank. So Mr. Musk’s fancy Tesla power bank, at \$45, doesn’t supply any more power than this one we sell at Solid Signal for a lot less. It might look cooler, but that’s the end of the story.
## But wait, there’s a bit more.
Yes, some power banks will do other tricks. Some have flashlights in them. Some will charge wirelessly, some have suction cups to hold them to your phone. There are power banks that come with pre-attached Lightning or USB-C connectors, for those folks who will never, ever change from Apple to Android or vice versa. Some are pink. Some just look cool. If you want to get a power bank like this, that’s awesome and I honestly encourage it. But you’re not deciding to buy one for its capabilities. You’re buying one to be fashionable. That’s fine, of course. Just know that you can get a perfectly great power bank at an awesome price at Solid Signal. | 885 | 3,758 | {"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} | 2.625 | 3 | CC-MAIN-2021-31 | longest | en | 0.961372 |
http://www.instructables.com/id/Simple-USB-Socket-Charger/?comments=all | 1,369,116,266,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00013-ip-10-60-113-184.ec2.internal.warc.gz | 534,439,466 | 26,663 | Simple USB Socket Charger!
9 Steps
Most of us have iPods or cameras. This is instructable will show you How to make a wall adapter for them. Normally it would cost \$30 at the Apple store but I had all the components in my spare parts bin. For me it was free! IF YOU SOMEHOW MESS UP WHILE BUILDING THIS AND
SOMETHING BAD HAPPENS TO YOUR USB DEVICE, I AM NOT TO BLAME...
Just had to say that. Oh, and watch the video too!
Remove these ads by Signing Up
## Step 1: Video!
electro18 says: Oct 28, 2012. 6:59 AM
can we attach a 6v 120mA solar panel to it and make it a "solar usb phone charger" ?
nodoubtman says: Oct 27, 2011. 5:53 AM
i have a wall adapter 5 V 500 mA that doesnt work... all battery charger instructables didn't work for me! it's very frustrating!!
vitalzero says: Apr 26, 2010. 8:25 AM
Here is my setup, it's just a draft.
the resistors for data pins are 100k ohm, for the led is 10k ohm, the connector at the bottom left is for a nokia phone charger which delivers 5 v @ 360mA.
If I connect it to a usb hub, all the lights light up like if it were connected to the pc, but if I connect it to the ipod nothing works :(
fruitkid101 (author) says: Apr 26, 2010. 9:56 AM
Make sure that it is hooked up like this
vitalzero says: Apr 26, 2010. 10:34 AM
yep, it is.
I'm still thinking it has something to do with the phone charger giving 360mA instead of 500mA :(
nodoubtman says: Oct 25, 2011. 8:30 AM
did you see any battery with 5 V input?
thank yoU! :)
legless says: Jul 31, 2011. 2:26 PM
for resistors in series the value would be
R1 + R2 + R3 +... = R(total)
for resistors in parallel the value would be
1/R1 + 1/R2 + 1/R3 + .... = 1/R(total)
or
(1 / (1/R1 + 1/R2 + 1/R3 + ....) = R(total)
aekara says: Jun 6, 2011. 2:05 AM
can i use two 50k ohm resistor in place of the one?
legless says: Jul 31, 2011. 2:20 PM
or 51kohms depending on the resistor type (tolerance)
legless says: Jul 31, 2011. 2:15 PM
Do you have 50kohm resistors? The more likely "preferred" value would be 47kohms.
fruitkid101 (author) says: Jul 10, 2011. 3:56 PM
If you hook them up in seires, not parallel, I think it would work...maybe
ckrill1 says: Jul 13, 2011. 3:02 PM
Does this charge a 4th generation ipod touch? cause i've tried others that don't work with my ipod
mujtababhat says: Jul 4, 2010. 11:34 PM
can we we connect it to a notebook cooling pad
fruitkid101 (author) says: Jul 10, 2011. 3:57 PM
You can connect it to anything that gets power via usb
ckrill1 says: Jul 10, 2011. 10:11 AM
instead of a wall plug could i use a 9 volt battery hooked up to a 5 volt regulater?
fruitkid101 (author) says: Jul 10, 2011. 3:55 PM
Yes indeed, there are many instructables for that out here
ckrill1 says: Jul 10, 2011. 10:16 AM
also, do you really need the led?
fruitkid101 (author) says: Jul 10, 2011. 3:55 PM
Nope, not at all. It just helps sometimes.
EToft says: Jan 8, 2011. 10:25 PM
The schematic in step 3 has an error in it. Refer to these pictures, and the corrected schematic here...
.Unknown. says: Nov 7, 2010. 2:59 AM
I thought pin 4 was meant to be ground...
EToft says: Jan 8, 2011. 11:45 AM
I agree...pin 4 is ground.
I believe that this is actually born out in the pictures in the ensuing steps. You can see that pin 4 is at the "bottom" in the picture of the USB connector. The photos appear to be the rotated reverse of the schematic...with the ground/negative side of the circuit at the bottom rather than at the top.
Kirbsome! says: Aug 8, 2010. 6:30 AM
Oh...
My...
GOD!
This is AWESOME!!!
Rayadillo says: Jun 22, 2010. 10:18 PM
But my cellphone charges just fine :(
Rayadillo says: Jun 22, 2010. 10:14 PM
My ipod just turns its light on :/ whats going on? and its an ipod nano video
ratgod says: May 2, 2010. 2:14 PM
I tried something similar by connecting power to the power pins on the USB socket and certain devices would not charge, I assume the 2x 100K tell the USB data lines that there is a connection and allow the USB to use the power?
On your schematic, the LED is drawn in reverse polarity compared to the power input, is this a reverse polarity indicator or is this a mistake?
good instructable.
fruitkid101 (author) says: May 5, 2010. 3:01 PM
Yes, the 100ks are for the data lines. I seem to have missed the LED on the diagram, I'll try and fix that soon. Cheers!
fruitkid101 (author) says: Apr 26, 2010. 9:55 AM
Make sure that it is all hooked up like this
vitalzero says: May 4, 2010. 6:13 AM
Yep, it is
vitalzero says: Apr 23, 2010. 8:23 AM
Nice tut, I did it all the same, but i soldered the wires.
The thing is that didn't work and my wife ended up laughing at me :(
Just a note and dont know if that caused the failure: I used as the power source a nokia 5v 360mA charger and the female usb connector I had had 4 pins, but there was another kind of pin with a hole on it (in the pic, the left one), tried to reconnect again using other pins and the only thing I got, was an error msg from the ipod sayin that the device was unknown (and a beeping sound of course)
ratgod says: May 2, 2010. 4:27 PM
From what I can see,(if its the metal tab on the left side of the left connector in the pic that your questioning) then that is for attaching the shielding braid from the cable.
I personally would suggest it to be left unconnected in this project as you don't need to shield any data transfer from interference, unless you need it attached for some mechanical stability, but make sure its not connected to any of the wiring.
fruitkid101 (author) says: Apr 23, 2010. 9:20 AM
If it's not too much to ask, could you post some pictures of what your whole setup? thanks
vitalzero says: Apr 24, 2010. 7:24 AM
hehe, sorry, I disassembled it, but I can tell that Interchanged cables for pins 1 and 4 (+ -) and also for data pins, connected the resistors in different ways and none worked. I'm thinking it may have something to do with the power source givin 360mA instead 500mA
ppatches24 says: Apr 26, 2010. 5:13 PM
Hey this is really cool and i am going to make one of these i have wanted a nice simple good instructable for an ipod touch for a long time just one question i have made one of these and i pluged in my ipod then my battery explode so i had to buy a new one will this kill my ipod if i do it right.
fruitkid101 (author) says: Apr 26, 2010. 9:28 PM
Thanks. You know, some people wouldn't believe you when you say you fried your battery. I do however, being a victim of fried ipod batteries myself. | 1,991 | 6,443 | {"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-2013-20 | latest | en | 0.907561 |
https://www.physicsforums.com/threads/coulombs-law-and-net-electric-field.372318/ | 1,545,206,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00283.warc.gz | 972,086,089 | 12,810 | # Homework Help: Coulomb's Law and net electric field
1. Jan 24, 2010
### yb1013
1. The problem statement, all variables and given/known data
Two point charges q1 = -5.60 nC, and q2 = -14.0 nC are separated by 25.0 cm (see figure below).
http://www.webassign.net/yf12/21-p-031-alt.gif
(a) Find the net electric field these charges produce at point A
magnitude 10360 N/C
direction is to the right
(b) Find the net electric field these charges produce at point B.
magnitude 6068.6 N/C
direction is to the right
(c) What would be the magnitude and direction of the electric force this combination of charges would produce on a proton at A?
magnitude ______ N
direction is to the right
3. The attempt at a solution
Okay well you can see that I figured out all of the answers except for the last magnitude, it seems to be really easy but I think Im just doing something little wrong, not sure.
Thank you!
2. Jan 24, 2010
### Redbelly98
Staff Emeritus
There is a strong hint in the units of the electric field you calculated: Newtons per Coulomb.
3. Jan 24, 2010
### yb1013
I'm still not quite sure where you're going with that. I tried a couple things but I ended with the wrong answer..
4. Jan 24, 2010
### Redbelly98
Staff Emeritus
What did you try? | 345 | 1,264 | {"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-2018-51 | latest | en | 0.905187 |
https://kevintshoemaker.github.io/NRES-746/LAB3demo.html | 1,708,912,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00193.warc.gz | 368,060,432 | 9,151 | Maximum likelihood and optimization
These next few weeks are focused on fitting models, specifically estimating model parameters and confidence intervals, using likelihood-based techniques (maximum likelihood and Bayesian model fitting). Estimating model parameters means finding the values of a set of parameters that best ‘fit’ the data. Likelihood is a metric that represents the probability of drawing your particular data set given a fully specified model (e.g., a particular data-generating model with a particular set of parameter values). This lab is designed to take two lab sessions to complete.
As with all labs in this course, your answers will either take the form of R functions (submitted as an R script) or short written responses (submitted together in a Word document). The R functions (and only the functions- not your testing code) should be stored in an R script file (‘.R’ extension). You don’t need to follow any naming convention for your R script as long as you submit via WebCampus.
Please submit the R script and the Word document via WebCampus by midnight on the due date (one week after the final lab session allocated for this topic – here, Oct. 25, 2021). You can work in groups but please submit the materials individually.
First, take a little time to review the likelihood lecture!
Example: reed frog predation data
First, load the reed frog predation data from the Bolker book- it can be found here. Save this file to your working directory.
This dataset represents predation data for Hyperolius spinigularis (Vonesh and Bolker 2005). You can read more about this data set in the Bolker book.
###### Read in the reed frog data set
########
# alternatively, load the data using the 'emdbook' package:
library(emdbook)
rfp <- ReedfrogPred
head(rfp)
## density pred size surv propsurv
## 1 10 no big 9 0.9
## 2 10 no big 10 1.0
## 3 10 no big 7 0.7
## 4 10 no big 10 1.0
## 5 10 no small 9 0.9
## 6 10 no small 9 0.9
Because predation on tadpoles is size and density-dependent, we will subset these data to a single size class (‘small’) and density (10) for all treatments including a predator (this simplifies the problem!). Subset your data now:
##### Take a subset of the data
rfp_sub <- subset(rfp, (rfp$pred=='pred')&(rfp$size=="small")&(rfp$density==10)) rfp_sub ## density pred size surv propsurv ## 13 10 pred small 7 0.7 ## 14 10 pred small 5 0.5 ## 15 10 pred small 9 0.9 ## 16 10 pred small 9 0.9 For each individual, the per-trial probability of being eaten by a predator is a binomial process (i.e., they can survive or die during the interval). Recall that the likelihood that k out of N individuals are eaten as a function of the per capita predation probability p is: $$Prob(k|p,N) = \binom{N}{k}p^{k}(1-p)^{N-k}$$ Since the observations are independent, the joint likelihood of the whole data set is the product of the likelihood of each individual observation. So, if we have n observations, each with the same total number of tadpoles N, and the number of tadpoles killed in the ith observation is ki, then the likelihood is: $$L = \prod_{i=1}^{n}\binom{N}{k_{i}}p^{k_{i}}(1-p)^{N-k_{i}}$$ Here we assume the data are binomially distributed – the binomial distribution is the natural choice for data that are represented as k ‘successes’ out of N ’trials. We conventionally work in terms of the log-likelihood (LL), which is: $$LL = \sum_{i=1}^{n}\left [log\binom{N}{k}+k_{i}log(p)+(N-k_{i})log(1-p) \right ]$$ In R this would be killed <- rfp_sub$density-rfp_sub$surv N=rfp_sub$density
p=0.5
sum(dbinom(killed, size=N, prob=p, log=TRUE)) # expression of data likelihood (log scale)
There is only one parameter in this calculation, p, because we know how many individuals we started with (N = 10 for each trial) and how many survived in each trial (k = 7, 5, 9, and 9). So we want to solve for the most likely value of p given our observations of N and surv. In essence we do this by picking a possible value of p (which can only range from 0 to 1), calculating the log-likelihood using the equation above, picking another value of p, completing the equation, etc. until we exhaust all possible values of p and identify the one having the highest likelihood value. Of course R has useful built in functions to help us optimize the likelihood function!
The “dbinom()” function calculates the binomial likelihood for a specified data set, specifically a vector of the number of successes (or events) k, probability p, and number of trials N. Specify your vector of successes (here a success means being eaten by a predator!):
num_killed <- rfp_sub$density-rfp_sub$surv # specify vector of "successes" (being eaten!)
num_killed
## [1] 3 5 1 1
Given our observed k (number killed), and N = 10 for each trial, what is the likelihood that p = 0.5 for each of our trials?
dbinom(num_killed,size=10,prob=0.5) # evaluate data likelihood with p=0.5
## [1] 0.117187500 0.246093750 0.009765625 0.009765625
[1] 0.117187500 0.246093750 0.009765625 0.009765625
We can see that given our data, fixed sample size, and model (with p = 0.5), our observed outcomes are very unlikely.
What is the likelihood of observing all 4 of our outcomes, i.e, the joint probability of our data?
prod(dbinom(num_killed,size=10,prob=0.5)) # joint data likelihood
## [1] 2.750312e-06
The joint likelihood values will be less than 1, and gets smaller and smaller each time we add more data (can you see why?). This is why we prefer to work with log-likelihoods (which yield larger numbers having better mathematical properties). And taking the log of a value <1 yields a negative number, which is why we often see that our log likelihood values are negative.
For now, we can build on this above process to estimate the likelihood function over the entire possible parameter space (probability of being eaten- which can range from 0 to 1).
First we make a sequence of 100 possible parameter values from 0.01 to 1.
p <- seq(0.01, 1, length=100) # prepare for visualizing the likelihood across parameter space
Then we make an empty storage vector for the likelihoods we’ll calculate
Lik <- numeric(length=100)
Now for the for loop! For every value of p (a sequence of 100 values) we will calculate the binomial probability and store it in the ‘Lik’ vector.
#########
# plot out the likelihood
for(i in 1:100){
Lik[i] <- prod(dbinom(num_killed,size=10,prob=p[i]))
}
plot(Lik~p,lty="solid",type="l", xlab="Predation Probability", ylab="Likelihood")
But we want to maximize the log-likelihood:
########
# plot out the log-likelihood
p <- seq(0.01, 1, by=0.01)
LogLik <- numeric(length=100)
for(i in 1:100){
LogLik[i] <- sum(dbinom(num_killed, size=10,
prob=p[i],log=TRUE))
}
plot(LogLik~p,lty="solid",type="l", xlab="Predation Probability", ylab="Log Likelihood")
We can ask R to tell us at which value of p the Log-Likelihood is maximized:
p[which(LogLik==max(LogLik))] # MLE for probability of predation
## [1] 0.25
And we can add an “abline()” to indicate the maximum Log-Likelihood estimate:
plot(LogLik~p,lty="solid",type="l", xlab="Predation Probability", ylab="Log Likelihood")
abline(v=0.25,lwd=3)
Alternatively, we can use the optim() or mle2() functions to find the maximum likelihood estimate. Although we seek the most likely, or maximum likelihood estimate, in practice we generally minimize the negative log-likelihood. To do so, first write a function to calculate the binomial negative log-likelihood function and estimate parameter p.
###########
# Write a likelihood function
# p: probability of predation per trial (param to estimate)
# k: number killed per trial (data)
# N: number of tadpoles per trial (data)
binomNLL1 <- function(p, k, N) {
-sum(dbinom(k, size=N, prob=p, log=TRUE))
}
As we did in class, you can use the ‘optim()’ function to minimize your negative log-likelihood function (‘binomNLL1()’) given a vector of starting parameters and your data. The starting parameters need not be accurate, but do need to be reasonable for the function to work, that’s why we spent time in class eyeballing curves (also read the Bolker book for a discussion of the ‘method of moments’, which can help you get reasonable starting values!). Given that there is only one estimable parameter, p, in the binomial function, you need only provide a starting estimate for it. Calculate the negative log-likelihood:
#####
# use "optim()" to find the MLE
opt1 <- optim(fn=binomNLL1, par = c(p=0.5), N = 10, k = num_killed, method = "BFGS") # use "optim()" to estimate the parameter value that maximizes the likelihood function
## Warning in dbinom(k, size = N, prob = p, log = TRUE): NaNs produced
## Warning in dbinom(k, size = N, prob = p, log = TRUE): NaNs produced
## Warning in dbinom(k, size = N, prob = p, log = TRUE): NaNs produced
## Warning in dbinom(k, size = N, prob = p, log = TRUE): NaNs produced
## Warning in dbinom(k, size = N, prob = p, log = TRUE): NaNs produced
You may get several warning messages, can you think why? opt1 returns a list that stores information about your optimization process.
opt1 # check out the results of "optim()"
## $par ## p ## 0.2500002 ## ##$value
## [1] 7.571315
##
## $counts ## function gradient ## 17 7 ## ##$convergence
## [1] 0
##
## $message ## NULL The important bits are whether or not the process achieved convergence and the parameter estimate that was converged upon. opt1$convergence
## [1] 0
Here a value of 0 means convergence has been achieved, a value of 1 means the process failed to converge. There is more info about convergence and alternative optimization options in Chapter 7 of the Bolker book.
Your best fit estimate of p is:
opt1$par # MLE ## p ## 0.2500002 This numerically computed answer is (almost exactly) equal to the theoretical answer of 0.25. The value of the function you optimized, binomNLL1, is: opt1$value # max. likelihood (actually minimum negative-log-likelihood)
## [1] 7.571315
which is the negative log-likelihood for the model. And, as we already know, the absolute likelihood of this particular outcome (5, 3, 1 and 1 out of 10 tadpoles eaten in four replicates) is quite low, even for this simple four-observation scenario:
exp(-opt1$value) # convert to likelihood ## [1] 0.0005150149 Plot your observed outcomes against your predictions under the maximum likelihood model: hist(num_killed,xlim=c(0,10),freq=F) curve(dbinom(x,prob=opt1$par,size=10),add=T,from=0,to=10,n=11)
Note that “freq=F” scales the y-axis of a histogram to “density”, which allows us to overlay probability density functions.
–End of demo– | 2,867 | 10,725 | {"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": 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} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.810789 |
https://hiptv.tv/know-what-gurus-are-saying-about-what-is-a-math-factor/ | 1,702,243,672,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00512.warc.gz | 332,324,109 | 10,445 | # Finding the Best What Is a Math Factor
## All About What Is a Math Factor
In addition, everybody can begin a podcast. The math can be a little confusing if you attempt to do it manually. Few men and women would, as an example, argue that numbers ought to be defined differently.
All the lovestruck robots are rushing to find the best gifts. You may also click every one of my favourite episodes right here in the post! They will learn to calculate the root of each given number in this worksheet.
A number’s prime factorization is just the list of prime numbers you would multiply together to have a particular item. The effect of the prosperous synthetic division is going to be a quadratic component. There’s a selection of numbers. An additional way to define a factor is to thesis multiply a few numbers with each other to create a bigger number. The bilateral factor might have an effect on your rating and thus don’t dismiss it. There are several ways to come across the very best typical factor of numbers.
I only wanted to allow you to know I am glad I purchased your product. For large numbers, like 108, an easy division procedure is required to get the factors. For bigger numbers this might not be such an excellent method. For numbers this may not be a superb method. Finally, make certain to confirm the complete price tag of the program.
However, it still supplies a wide selection of courses. The LCM or least common multiple are available for a couple of numbers. That’s slowly beginning to change. The number 9 is called a perfect square as it’s a whole number’s square. One isn’t a composite number. There’s an endless number of prime numbers.
## The Secret to What Is a Math Factor
My concepts are very clear and I really enjoy the step-by-step strategy. essay topics It’s very likely to use a similar process to obtain the nth root of a number. You are looking for the root of 256.
Let us have a look at what these properties are and learn to identify them properly. However on scratching the face of the subject I discovered an enormous wealth of material. Let’s run through a good example, building on the preceding profile.
## What Is a Math Factor Explained
The result obtained is known as the quotient. You have to think about the moment the sum that’s negative is insidethe exponent and after it’s not. Because the expression does not include any symbolic parameters aside from the variable x, the outcome is the exact same as in complex factorization mode.
The term divisor sounds complicated, but it truly isn’t. The end result of dividing two rational numbers is another rational number once the divisor isn’t 0. Integer factorization for big integers is apparently a problem.
Not all of these are still running, and a few formats will interest you more than others, but all of them are wonderfully mathematical. Typically, if you’re on a high-speed connection, you will understand our content in under a second. academic writing services My concepts are extremely clear and I really like the step-by-step strategy.
To discover more about rep-tiles search on the net, there is lots to find, but nevertheless, it can be somewhat hard as for some reason the biologists dominate the google success. The solution is yes, and in fact there’s an incredible tiling. That is impossible in this instance, you can receive a glimpse of this by taking a look at the second portion of the picture where every tile is coloured by orientation.
## What Is a Math Factor – Dead or Alive?
The last stage is to attempt to classify all potential examples. Group the pairs individually in case the variables have some perfect square. There’s no difference between the 2 graphs.
It measures what the excellent bulk of the data do toward the middle of a set. Let us begin with an observation and a very simple example. That means you might wish to use the Combined Ratings Table to identify your general rating if you’ve got bilateral disabilities.
## What Everybody Dislikes About What Is a Math Factor and Why
This site is free for the users on account of the earnings made by the advertisements running on the site. It almost makes a cut-out section. Be sure to See My Score at the end of the quiz to understand how you step.
## What Is a Math Factor – Overview
Fleebur desperately wishes to create a small puppy! There are a lot of things in the area of math that seem perfectly intriguing but perhaps not particularly essential in the grand scheme of things. Your life will be a good deal easier when you can merely don’t forget the multiplication tables.
I hope 2017 is going to be a great year for everyone!! By way of example, 132 is an origin that’s huge, and it could be challenging to learn what things to do. Check back next week to learn!
A computer might actually work out this issue easily. You wish to try to find the problem. Let’s try a few easy factoring troubles and discover how.
You may choose the group of numbers and exceptional formats for the multiplication issues to use. Regardless, oftentimes, it’s possible to turn the polynomial into a simpler form by means of a process called factoring. The disabilities don’t need to mirror each other.
Research indicates that it might start sooner. It gives students a wide selection of courses to select from and works with different departments to provide students courses it can’t provide alone. It gives students a wide variety of courses to choose from and works with other departments to offer students courses it cannot provide on its own.
Cost isn’t everything in education, but nevertheless, it shouldn’t be ignored when you’re enrolling in a science or math degree. Students may also use repeated addition to get the item. It may also be pursued with honors. | 1,178 | 5,771 | {"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} | 4 | 4 | CC-MAIN-2023-50 | longest | en | 0.936442 |
https://www.grc.nasa.gov/WWW/K-12/p_test/math_9/math_p_q18_a.html | 1,618,158,897,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00515.warc.gz | 929,851,638 | 2,039 | # Sorry
Let's just look at each of the conversions and see where you may have gone wrong.
• 96 cups = 6 gallons
• You know that each gallon equals 4 quarts therefore 6 gallons would be 24 quarts. Then each quart is 2 pints therefore 24 quarts is equal to 48 pints. And each pint is made up of 2 cups, therefore 48 pints would be 96 cups.
• Or you could have done the problem the other way. There are 2 cups in one pint. There are 2 pints in one quart so that means that there are 4 cups in one quart. There are 4 quarts in a gallon, therefore there are 16 cups in a gallon. Therefore there are 96 cups in 6 gallons.
• 9 days = 216 hours
• You know that each day has 24 hours. Therefore 9 days times 24 hours would be 216 hours.
• Since there are 24 hours in a day you can say that 216 hours divided by 24 hours is 9 days
• 84 ounces = 7 pounds
• There are 16 ounces in each pound, therefore 7 pounds times 16 ounces is 112 ounces. Therefore the conversation of 84 ounces = 7 pounds is incorrect.
• Since there are 16 ounces in each pound you could say that 84 ounces divided by 16 equals 5.25 pounds. Therefore the conversation of 84 ounces = 7 pounds is incorrect.
• 2 miles = 10,560 feet
• There are 5,280 feet to a mile, therefore 2 miles times 5280 feet is equal to 10,560 feet.
• Since there are 5,280 feet in a mile you could say 10,560 feet divided by 5,280 feet is 2 miles.
Now let us try another one
Which statement is NOT correct?
Back to Proficiency Page | 410 | 1,470 | {"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-2021-17 | latest | en | 0.947636 |
https://homeworkslavecenter.com/in/question/question/89000.html | 1,624,507,857,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00036.warc.gz | 280,004,520 | 5,605 | Calculating Fees on a Loan Commitment You have approached your local bank for a start up loan commitment for 1 120 000 needed to open a car repair
Calculating Fees on a Loan Commitment You have approached your local bank for a start up loan commitment for
Calculating Fees on a Loan Commitment You have approached your local bank for a
Loan Commitment You have approached your local bank for a start up loan commitment for needed to open a car repair
Calculating Fees on a Loan Commitment You have approached your local bank
for a start up loan commitment for needed to open a car repair
Calculating Fees on a Loan Commitment You have approached
Calculating Fees on a
Calculating Fees on a Loan Commitment You have approached your local bank for a start-up loan commitment for \$1,120,000 needed to open a car repair...
Category: Words: Amount: \$20 Writer: 0
Paper instructions
Calculating Fees on a Loan Commitment You have approached your local bank for a start-up loan commitment for \$1,120,000 needed to open a car repair store. You have requested that the term of the loan be one-year. Your bank has offered you the following terms: size of loan commitment = \$1,120,000, term = 1 year, up-front fee = 20 basis points, back-end fee = 50 basis points, and rate on the loan = 10%. If you immediately take down \$870,000 and no more during the year, what is the total interest and fees you have paid on this loan commitment? | 327 | 1,432 | {"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-2021-25 | longest | en | 0.952324 |
https://algebrahomework.org/algebrahomework/adding-functions/how-to-solve-algeba-equations.html | 1,627,436,755,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00284.warc.gz | 109,873,202 | 12,874 | Algebra Tutorials!
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
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Equ. #4:
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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Author Message
Bliacx
Registered: 14.02.2005
From: UK
Posted: Friday 29th of Dec 12:16 Guys and Gals ! Ok, we’re doing how to solve algeba equations and I was not present in my last algebra class so I have no notes and my teacher discusses lessons way bad that’s why I didn’t get to understand it very well when I went to our math class a while ago. To make matters worse, our class will have our quiz on our next class so I can’t afford not to understand how to solve algeba equations. Can somebody please assist me try to understand how to answer couple of questions about how to solve algeba equations so that I can prepare for the quiz. I’m hoping that somebody would assist me as soon as possible .
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Saturday 30th of Dec 14:37 How about giving a little more details of what exactly is your problem with how to solve algeba equations? This would assist in finding out ways to hunt for a solution . Finding a coach these days fast enough and that too at a cost that you can afford can be a frustrating task. On the other hand, these days there are programs that are to be had to help you with your math problems. All you have to do is to pick the most suitable one. With just a click the right answer pops up. Not only this, it helps you to arriving at the answer. This way you also get to find out how to get at the right answer.
molbheus2matlih
Registered: 10.04.2002
From: France
Posted: Sunday 31st of Dec 09:21 I remember I faced similar difficulties with graphing inequalities, binomials and rational equations. This Algebrator is rightly a great piece of algebra software program. This would simply give step by step solution to any algebra problem that I copied from homework copy on clicking on Solve. I have been able to use the program through several Pre Algebra, Basic Math and College Algebra. I seriously recommend the program.
Majnatto
Registered: 17.10.2003
From: Ontario
Posted: Sunday 31st of Dec 16:02 I would recommend using Algebrator. It not only assists you with your math problems, but also displays all the necessary steps in detail so that you can enhance the understanding of the subject.
Somt Pxeldel
Registered: 20.02.2005 | 819 | 3,249 | {"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} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.904411 |
https://studyrankersonline.com/64256/four-equal-circles-each-radius-touch-each-other-shown-figure | 1,558,532,303,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256812.65/warc/CC-MAIN-20190522123236-20190522145236-00260.warc.gz | 653,390,886 | 15,112 | # Four equal circles, each of radius 5 cm, touch each other as shown in the figure.
1 view
Four equal circles, each of radius 5 cm, touch each other as shown in the figure. Find the area included between them. (Take π = 3.14)
answered Mar 13 by (-2,347 points)
Required area = Area of square ABCD – Area of 4 quadrants
= (10 cm × 10 cm) – 4 × 1/4πr2
= 100 – 4 × ¼ × 3.14 × 5 × 5
= 100 – 3.14 × 25
= 21.5 cm2 | 149 | 415 | {"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} | 3.75 | 4 | CC-MAIN-2019-22 | latest | en | 0.915622 |
https://www.doubtnut.com/qna/350234351 | 1,726,530,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00523.warc.gz | 693,917,319 | 35,197 | # The efficiency of a Carnot engine operating between temperatures of 100∘C and −23∘C will be
A
373+250373
B
373250373
C
100+23100
D
10023100
Video Solution
Text Solution
Verified by Experts
## The correct Answer is:B
|
Answer
Step by step video, text & image solution for The efficiency of a Carnot engine operating between temperatures of 100^(@)C and -23^(@)C will be by Physics experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## The efficiency of a Carnot heat engine
Ais independent of the temperature of the source and the sink
Bis independent of the working substance
Ccan be 10%
Din not affected by the thermal capacity of the source of the sink
• Question 2 - Select One
## The efficiency of a Carnot engine operating between temperatures of 100∘C and −23∘C will be
A373+250373
B373250373
C100+23100
D10023100
• Question 3 - Select One
## The efficiency of carnot engine depends on
Atemperature of sink only
Btemperature of source and sink
Cvolume of cylinder of engine
Dtemperature of source
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