url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://myhomeworkwriters.com/static-ge-equilibrium-and-welfare-gain-exercise-online-assignment-help/ | 1,638,121,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00121.warc.gz | 487,388,903 | 12,015 | # Static GE Equilibrium and Welfare Gain Exercise | Online Assignment Help
For exercises 1-3, use the benchmark model in class where the government taxes household
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labor income and lump sum transfers it back to them.
Exercise 1: What if we were Italian? Calibrate α so as to match the average work week for Italia
for the 1970-74 period as reported in Table 2 in Prescott (2004).
Exercise 2: (Laffer Curve) Use Excel to show how the equilibrium varies with the tax rate given
the value for from Exercise 1 above. Vary in steps of 0.01 from 0.00 to 1.00. Plot
c, h, and tax revenue variables, hw versus .
Exercise 3: Determine the welfare gain associated with lowering the tax rate in Europe from 0.60
to the U.S. tax rate of 0.40.
(Recall, the welfare gain in lifetime consumption equivalents of changing the tax rate from the
current U.S. rate of 0.40 to the European rate of 0.60 is the value of such that
Therefore, (1+) is the factor that consumption must be scaled if there is a change in the tax rate
from 40% to 60% to leave people indifferent to the change. The welfare loss in consumption
equivalents is c . In effect, find the values of . and ),60.0( ),60.0( ),40.0( hcc
Exercise 4.
Consider the model economy with an alternative government policy. The commodities are labor
h and consumption c. The price of c is normalized to 1. The price of labor is the wage rate w.
A household’s problem is to maximize utility
)100ln(ln sd hc
subject to its budget constraint sd whc , where is a lump sum transfer from the
government. Here the superscripts s and d indicate supply and demand
A firm’s problem is to maximize profits dd whhA )1( . Notice that the government
imposes a tax on the firm’s output or sales.
The government’s budget constraint is dAh .
The consumption and labor markets clear which means that ccc ds and hhh ds .
Find the equilibrium for this economy; that is equilibrium ( , , , )c h w as a function of utility
function parameter and tax policy parameter . Note set the TFP parameter A=1
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## Essay Writing Services
At My Homework Writers, we have highly qualified academic gurus who will offer great assistance towards completing your essays. Our homework writing service providers are well-versed with all the aspects of developing high-quality and relevant essays. | 1,106 | 4,880 | {"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-2021-49 | latest | en | 0.913833 |
https://youvegotthismath.com/gingerbread-man-counting/ | 1,723,524,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00304.warc.gz | 842,918,387 | 71,467 | # 5 Worksheets on Gingerbread Man Counting Game
These worksheets on the Gingerbread Man counting game will help to visualize and understand counting and number recognition. 1st-5th grade students will learn basic counting methods and can improve their basic math skills with our free printable Gingerbread Man counting game worksheets.
## 5 Exciting Practicing Gingerbread Man Counting Game Worksheets for Your Little Champ
• The numbers are written in a row.
• Choose the correct number.
• For example, if a student wants to solve the first problem, she will count the gingerbread man and choose the correct answer which is 5.
## Match the Dots for the Gingerbread Man Counting Game
This worksheet provides a matching dots game. You can follow the steps.
• Draw a poster board with the number of gingerbread men you need for matching.
• Add the number of dots you are currently working on to each.
• For instance, if you are just matching the numbers 6 -10, only trace 5 gingerbread men and 6–10 circles within each.
• Then, write down that number on the gingerbread men’s pages.
• Match the number dots on the poster board with the appropriate gingerbread man sheets.
## Gingerbread Man Running Cut and Paste Game
• Count the numbers.
• Provide random numbers.
• Cut the numbers and stick them beside the right pictures.
• For example, if they count four gingerbread men, they will cut the number 4 and paste it beside the picture.
## Gingerbread Man Treasure Hunt Game
This worksheet contains a treasure hunt game.
• Solve the problem so you can reach the gingerbread man safely.
• The game is divided into two parts.
• The students will be given a set of numbers which will be the clue for their treasure hunt game.
• First, they will rescue the gingerbread man from the bad people by finding the missing numbers.
• After, completing the first level the students need to solve the second part to save the gingerbread man from the fox by finding the missing numbers again.
## Count and Match with Gingerbread Man
In this activity, students count the numbers and match the gingerbread men with the correct number. | 449 | 2,128 | {"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} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.894136 |
https://dokumen.tips/documents/cs221-algorithms-and-data-structures-lecture-2-tail-recursion-induction.html | 1,620,416,446,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00426.warc.gz | 227,922,538 | 21,036 | # CS221: Algorithms and Data Structures Lecture #2 (Tail) Recursion, Induction, Loop Invariants, and Call Stacks Steve Wolfman 2014W1 1
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CSE 326: Data Structures Lecture #2 measuring hOw fast with a Big-Oh
CS221: Algorithms and Data StructuresLecture #2(Tail) Recursion, Induction, Loop Invariants, and Call StacksSteve Wolfman2014W111TODO (future terms): fix year/term!
TODO (future terms): prep fibonacci and shampoo code to illustrate call stack.Todays OutlineThinking RecursivelyRecursion ExamplesAnalyzing Recursion: Induction and RecurrencesAnalyzing Iteration: Loop InvariantsMythbusters:Recursions not as efficient as iteration??Recursion and the Call StackIteration and Explicit StacksTail Recursion (but our KW text is wrong about this!)22Random String Permutations(reigniouS mPRrtmnsdtan aot)Problem: Permute a string so that every reordering of the string is equally likely. You may use a function randrange(n), which selects a number [0,n) uniformly at random.3Random String PermutationsUnderstanding the ProblemA string is: an empty string or a letter plus the rest of the string.
We want every letter to have an equal chance to end up first. We want all permutations of the rest of the string to be equally likely to go after.
And.. theres only one empty string.
(Tests: tricky, but result should always have same letters as orginal.)4Random String PermutationsAlgorithmPERMUTE(s): if s is empty, just return s else: use randRange to choose a random first letter permute the rest of the string (minus that random letter) return a string that starts with the random letter and continues with the permuted rest of the string5Random String PermutationsConverting Algorithm to CodePERMUTE(s):
if s is empty, just return s
else:
choose random letter
permute the rest
return random letter + rest6C++11-style code (ignoring the fact that algorithm already has a permutation function in it!):
#include // for seeding w/time#include #include
int choose_random_index(std::string const & str) { static unsigned seed = std::chrono::system_clock::now().time_since_epoch().count(); static std::default_random_engine generator(seed); std::uniform_int_distribution distribution(0,str.size()-1); return distribution(generator);}
std::string permute(std::string str) { // if s is empty if (str.size() == 0) { // just return s return str; } else { // choose random letter int index = choose_random_index(str); char c = str[index];
// permute the rest std::string rest = str.erase(index, 1); std::string rest_permuted = permute(rest);
// return random letter + rest return c + rest_permuted; }}
int main(int argc, char *argv[]) { if (argc != 2) { std::cerr newIndex; j--) array[j] = array[j-1]; array[newIndex] = val; }
Inductive Step: bSearch gives the appropriate index at which to put array[i]. So, the new element ends up in sorted order, and the rest of array[0..i] stays in sorted order.(A bit hand-wavy what should we have done?)30(surprisingly: linear search may be a better choice here; ask after class!)Proving Insertion Sort Works // Invariant: before each test i < length (including the last // one), the elements in array[0..i-1] are in sorted order. for (int i = 1; i < length; i++) { // i is about to go up by 1 but array[i] may be out of order! int val = array[i]; int newIndex = bSearch(array, val, 0, i); for (int j = i; j > newIndex; j--) array[j] = array[j-1]; array[newIndex] = val; }
Loop termination: The loop ends when i == length (which it must be eventually since length is non-negative and i increases). At which point, array[0..i-1] is sorted which is array[0..length-1] or the whole array31Practice: Prove the Inner Loop Correctfor (int i = 1; i < length; i++){ // i is about to go up by 1 but array[i] may be out of order! int val = array[i]; int newIndex = bSearch(array, val, 0, i); // Whats the invariant? Maybe: just before j > newIndex, // array[0..j-1] + array[j+1..i] = the old array[0..i-1] for (int j = i; j > newIndex; j--) array[j] = array[j-1]; array[newIndex] = val;}
We just waved our hands at the inner loop. Prove its correct! (This may feel unrealistically easy!)Do note that j is going down, not up.32Todays OutlineThinking RecursivelyRecursion ExamplesAnalyzing Recursion: Induction and RecurrencesAnalyzing Iteration: Loop InvariantsMythbusters:Recursions not as efficient as iteration??Recursion and the Call StackIteration and Explicit StacksTail Recursion (but our KW text is wrong about this!)3333Mythbusters: Recursion vs. IterationWhich one can do more? Recursion or iteration?34Mythbusters: Simulating a Loop with Recursionint i = 0while (i < n) doFoo(i) i++
recDoFoo(0, n)
Where recDoFoo is:
void recDoFoo(int i, int n){ if (i < n) { doFoo(i) recDoFoo(i + 1, n) }}
35Anything we can do with iteration, we can do with recursion.Mythbusters: Simulating Recursion with a Stack(Going Quick.. Already Discussed)How does fib actually work? Each function call generates a stack frame (also known as activation record or, just between us, function pancake) holding local variables and the program point to return to, which is pushed on a stack (the call stack) that tracks the current chain of function calls.
int fib(int n) { if (n
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Here, we will be discussing about App that can do any math problem. A variable equation solver is a mathematical tool that is used to find the value of an unknown variable in an equation. This type of equation is usually represented by two lines that intersect at a point, with the unknown variable being represented by the letter x. To use a variable equation solver, simply input the values of the known variables into the tool and it will output the value of the unknown variable. This process can be repeated for multiple equations, allowing you to solve for multiple unknowns simultaneously. Variable equation solvers can be found online or in mathematical textbooks. However, it is important to note that these tools only provide approximate values for the unknowns; they cannot give exact solutions. Therefore, if you need an accurate answer, it is best to use a different method such as algebraic methods. Nevertheless, variable equation solvers can be a valuable tool for solving simple equations quickly and easily.
Solving trinomials is a process that can be broken down into a few simple steps. First, identify the coefficients of the terms. Next, use the quadratic formula to find the roots of the equation. Finally, plug the roots back into the original equation to verify your results. While this process may seem daunting at first, with a little practice it will become second nature. With so many trinomials to solve, there's no time to waste - get started today!
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Problem solving or problem solving Quadratic congruence solver App to scan math problems Midpoint formula solver Equation solver app | 734 | 3,667 | {"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-2022-49 | latest | en | 0.951692 |
https://studyadda.com/sample-papers/kvpy-stream-sx-model-paper-21_q108/1897/462991 | 1,643,293,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00595.warc.gz | 572,250,674 | 21,094 | • # question_answer If ${{K}_{sp}}$Ag $A{{g}_{2}}C{{O}_{3}}$ is $8\times {{10}^{-12}},$the molar solubilty of $A{{g}_{2}}C{{O}_{3}}$ in $AgN{{O}_{3}}$ is: A) $8\times {{10}^{-12}}M$ B) $8\times {{10}^{-11}}M$ C) $8\times {{10}^{-10}}M$ D) $8\times {{10}^{-13}}M$
$8\times {{10}^{-12}}={{(2S'+0.1)}^{2}}S'$ Or, $S'=8\times {{10}^{-10}}M.$ | 170 | 345 | {"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} | 3.203125 | 3 | CC-MAIN-2022-05 | latest | en | 0.384263 |
http://research.stlouisfed.org/fred2/graph/?g=kwO | 1,419,487,772,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447546043.1/warc/CC-MAIN-20141224185906-00040-ip-10-231-17-201.ec2.internal.warc.gz | 156,940,596 | 18,473 | # FRED Graph
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Release:
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) New Private Housing Units Authorized by Building Permits, Thousands of Units, Seasonally Adjusted Annual Rate (PERMIT)
Starting with the 2005-02-16 release, the series reflects an increase in the universe of permit-issuing places from 19,000 to 20,000 places.
New Private Housing Units Authorized by Building Permits
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Color:
(a) New Privately-Owned Housing Units Completed: Total, Thousands of Units, Seasonally Adjusted Annual Rate (COMPUTSA)
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` US. Bureau of the Census, New Private Housing Units Authorized by Building Permits [PERMIT], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/PERMIT/, December 25, 2014. ```
``` US. Bureau of the Census, New Privately-Owned Housing Units Completed: Total [COMPUTSA], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/COMPUTSA/, December 25, 2014. ```
Retrieving data.
Graph updated.
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Name: Email: | 727 | 2,883 | {"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-2014-52 | latest | en | 0.827306 |
https://www.physicsforums.com/threads/getting-ffts-to-behave.723903/ | 1,558,363,790,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00549.warc.gz | 904,020,968 | 15,959 | # Getting FFTs to behave
#### birdhen
Hi there,
I am having trouble with some Fast Fourier Transforms.
I have been using both matlab and FFTW in c.
The problem is that I want to calculate the potential of a system from the overdensity. The equation in Fourier space is simple
phi(k)= -del(k)/(k^2) ,
where del is the overdensity, phi is the potential and k is the wavenumber.
To test this I use a simple del(x) = x, and feed this in to the fft. I then run through the k values and divide through by k squared. When k =0, I am not sure what to do as obviously you can't divide by zero.
I have tried a variety of things such as setting phi(k) = 0 when k=0, or just leaving phi(k) = del(k)..both of which I am sure are wrong.
When I reverse FFT this phi(k), my result is always a sine wave with a wavelength the size of the x range, where as I am expecting (1/6)*x^3.
I am having trouble finding the answer online. Maybe I am searching for the wrong thing. Can anyone help?
Best wishes,
Related Programming and Computer Science News on Phys.org
#### DrClaude
Mentor
When k =0, I am not sure what to do as obviously you can't divide by zero.
I have tried a variety of things such as setting phi(k) = 0 when k=0, or just leaving phi(k) = del(k)..both of which I am sure are wrong.
The best choice here is to set up a grid of points such that $k=0$ is not included. This may require a shift that will add a complex phase to the result of the FFT, which can then be removed.
"Getting FFTs to behave"
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# The profit from the sale of a certain appliance increases
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The profit from the sale of a certain appliance increases [#permalink]
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The profit from the sale of a certain appliance increases, though not proportionally, with the number of units sold. did the profit exceed $4 million on sales of 380000 units? (1) The profit exceeded$2 million on sales of 200,000 units.
(2) The profit exceeded $5 million on sales of 350,000 units. OG12, 23 DS Are we supposed to assume that the remaining 150,000 in A and the remaining 30,000 in B will generate different profits?-that is, they are sold separately as different chunks? Can't we just say, 2 Million / 200,000 = price per unit? Therefore, answer is D. OA is [Reveal] Spoiler: B Some people calculate it profit for A this way [Reveal] Spoiler: So, for 380,000 units, the profit exceeded (380,000 * 2)/200,000 =$3.8 million, but it is not clear if the profit exceeded $4 million or not; NOT sufficient. Where did they get the 2 from? OPEN DISCUSSION OF THIS QUESTION IS HERE: the-profit-from-the-sale-of-a-certain-appliance-increases-137874.html [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 39045 Followers: 7753 Kudos [?]: 106510 [4] , given: 11627 Re: Profit increase [#permalink] ### Show Tags 20 Feb 2012, 16:07 4 This post received KUDOS Expert's post 3 This post was BOOKMARKED The profit from the sale of a certain appliance increases, though not proportionally, with the number of units sold. did the profit exceed$4 million on sales of 380000 units?
Two important notes from the stem:
A. The profit from the sale increases with the number of units sold, so the profit on sales of 2 units is more than the profit on sales of 1 unit;
B. That increase is not proportional, so even though the profit on sales of 2 units is more than the profit on sales of 1 unit, it's not twice as much (it can be 1.1 times more, 5 times more, ..., 1,000,000 times more).
(1) The profit exceeded $2 million on sales of 200,000 units --> since the increase is not proportional we can not say whether the profit exceeded$4 million on sales of 380,000 units. Not sufficient.
(2) The profit exceeded $5 million on sales of 350,000 units --> since the profit on sales of 350,000 units, which is less than 380,000 units, is ALREADY greater than$4 million, then the profit on sales of 380,000 units will definitely be greater than \$4 million. Sufficient.
Hope it's clear.
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Re: The profit from the sale of a certain appliance increases [#permalink]
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OPEN DISCUSSION OF THIS QUESTION IS HERE: the-profit-from-the-sale-of-a-certain-appliance-increases-137874.html
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Re: The profit from the sale of a certain appliance increases [#permalink] 07 Sep 2013, 10:34
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# multivariable_04_Transcendental_Functions_2up - 4...
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Unformatted text preview: 4 Transcendental Functions So far we have used only algebraic functions as examples when finding derivatives, that is, functions that can be built up by the usual algebraic operations of addition, subtraction, multiplication, division, and raising to constant powers. Both in theory and practice there are other functions, called transcendental, that are very useful. Most important among these are the trigonometric functions, the inverse trigonometric functions, exponential functions, and logarithms. 4Ö igÓÒÓÑ eØÖ iF ÙÒØ iÓÒ × When you first encountered the trigonometric functions it was probably in the context of “triangle trigonometry,” defining, for example, the sine of an angle as the “side opposite over the hypotenuse.” While this will still be useful in an informal way, we need to use a more expansive definition of the trigonometric functions. First an important note: while degree measure of angles is sometimes convenient because it is so familiar, it turns out to be ill-suited to mathematical calculation, so (almost) everything we do will be in terms of radian measure of angles. 59 60 Chapter 4 Transcendental Functions To define the radian measurement system, we consider the unit circle in the xy-plane: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................. ............................................. x (cos x, sin x ) y A B (1 , 0) An angle, x , at the center of the circle is associated with an arc of the circle which is said to subtend the angle. In the figure, this arc is the portion of the circle from point (1 , 0) to point A . The length of this arc is the radian measure of the angle x ; the fact that the radian measure is an actual geometric length is largely responsible for the usefulness of radian measure. The circumference of the unit circle is 2 πr = 2 π (1) = 2 π , so the radian measure of the full circular angle (that is, of the 360 degree angle) is 2 π . While an angle with a particular measure can appear anywhere around the circle, we need a fixed, conventional location so that we can use the coordinate system to define properties of the angle. The standard convention is to place the starting radius for the angle on the positive x-axis, and to measure positive angles counterclockwise around the circle. In the figure, x is the standard location of the angle π/ 6, that is, the length of the arc from (1 , 0) to A is π/ 6. The angle y in the picture is − π/ 6, because the distance from (1 , 0) to B along the circle is also π/ 6, but in a clockwise direction.6, but in a clockwise direction....
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## This note was uploaded on 12/01/2011 for the course MATH 305 taught by Professor Guichard during the Fall '11 term at Whitman.
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# Section _6--Bullseye - Section #6 Capital Structure Case...
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Section #6 “Capital Structure Case Analysis: The Bullseye Corporation” October 14th, 2010 Copyright 2010 by Rich Curtis
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The Bullseye Corporation
A. Valuing The Common Stock Of The Unlevered Firm (See “Bullseye” Case, pp. 1-2) 1. Assume that the Bullseye Corporation currently has no debt but has 1,000,000 shares of common stock outstanding. The corporate tax rate is 35%. Each year, all the corporation’s after-tax earnings are paid to the common stockholders as a cash dividend. The cash dividends are paid annually with the next cash dividend being paid one year from today. Also assume that EBIT = \$15,384,615.38 each year, Net Income = (1-t c )(\$15,384,615.38) = (1-.35)(\$15,384,615.38) = \$10,000,000 each year, and k e (0) = the after-tax cost of equity of the unlevered firm = .20
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2. Then the total market value of the unlevered firm equals the market value of the common stock since there is no debt outstanding. 3. Questions : What is the total market value of the common stock of the the Bullseye Corporation and what is the price per share?
Answers : Total Equity Value = Dividend k e (0) = \$10,000,000 .20 = \$50,000,000 Net Income P S = \$50,000,000 1,000,000 shares = \$50/share
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B. Substitute \$40,000,000 Of Debt For The Common Stock Equity (See “Bullseye” Case, pp. 3-6) 4. Now assume that the Bullseye Corporation issues \$40,000,000 of perpetual debt and pays the proceeds to the common stockholders as a common stock cash dividend. The cash dividend will therefore be \$40 per share. The assets of the firm will be unchanged since the cash raised is immediately paid out.
5. The market value of the levered firm, V L , is given by the following expression: V L = V U + t c B = \$50,000,000 + (.35)(\$40,000,000) = \$64,000,000. The overall value of the firm has increased because the debt shields some EBIT from taxes. More of EBIT goes to investors in the firm’s securities--less goes to the Internal Revenue Service.
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6. Since the market value of the levered firm, V L , equals the market value of the bonds outstanding plus the market value of the common stock outstanding, one can solve for the market value of the outstanding common stock. Questions : What is the total market value of the outstanding common stock of “The Bullseye Corporation” after the \$40 million bond issuance and the payment of the \$40 million cash dividend? What is the new common stock price per share?
Answers : S L = V L B L = \$64,000,000 \$40,000,000 = \$24,000,000 P S = \$24,000,000 1,000,000 shares = \$24/share
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Let’s look at the annual tax savings and also the dividends received by the stockholders to see if our results are consistent with a cash flow analysis. (I hope they are or my intuition is messed up!!)
EBIT = \$15,384,615.38 Interest @ 8% = \$3,200,000.00 (8% is Assumed) Taxable Income = \$12,184,615.38 Tax @ 35% = \$4,264,615.38 Net Income = \$7,920,000.00 = Annual Dividend Note : The annual tax saving is: \$5,384,615.38 - \$4,264,615.38 = \$1,120,000 Discounting the annual tax savings “perpetuity”
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## This note was uploaded on 09/24/2011 for the course HADM 2225 taught by Professor Wellman, j during the Spring '08 term at Cornell.
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Section _6--Bullseye - Section #6 Capital Structure Case...
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"string" Theory
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:(x
On Truth:
1 "String" is a string in six letters. (Truth by Exemplification.)
2 "x" is a string in six letters. (function having an instance: Truth by Abstraction)
3 "x" is a string (another such)
4 And again : xZ
Ending in a reminder :x (What waz all that fuzz about?)
The Truths of a Language forms a Model of Reality.
The tools (of logic) are Abstraction and Exemplification.
Obviously there are abstraction levels: x = "x"
1 this time this is this level! (my personal level)
2 SigurdV obviously is a nutcase (Said to myself: You! Intruded on My Territory!)
3 thoughts expressed in clusters (belonging to the powerset)
4 our minds are not constructed for much more then stepping three steps up on abstraction ladders.
We have the ordinary language level for expressing and comparing sentences, use of the signs """ , "(" , ")" , and ":" divides the string into subject level(beginning part = left) and object level (ending part = right). Right?
PS Im sorry if you thought the topic was string theory its ""string"Theory" :)
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• 2 weeks later...
x = x
x = "string"
"string" = "x"
x = "x",
thus, no x, no string...hence why all the fuss about x.
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I dont recall whether you are matematician logician, both or neither.
Looking at foundations makes me think on beginnings.
The further past you go the less distinktion there is between sciences themselves and philosophy.
x = x
x = "string"
"string" = "x"
x = "x",
thus, no x, no string...hence why all the fuss about x.
Eloquent! :)
I definitely get the feeling you can detect flaws in arguments.
1 x = x (The Law of Identity.)
2 x = "x" (Something of Abstraction?)
All strings are now divided into two sets , strings inside and outside quotation marks.(Minding the possibility of strings being both or neither.)
What obviously can be done with any divider of strings.(Living strings ending in a variable called "now " in time)
(Should be continued "I think"... what next? What are we doing and why? Who and how are we? And where are we? And where is there? (Yes! I am talking to You!))
Truth and negation:
1 x = "x is true" (Truth)
2 -x = "x is not true" (Negation)
Having made some progress, maybe I should consider strings not made up by letters... like strings of time and/or space? Or should I wait until everything necessary is done with the study of strings consisting of letters and/or numbers?
PS Do me a favour, please, if you can find the time... check the arguments in "The Final Solution of the Liar", and tell me what you think.
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Truth and negation:
1 x = "x is true" (Truth)
2 -x = "x is not true" (Negation)
However,
1. x = "man is beautiful"
2. x = ""man is beautiful" is true" (Affirmation-Truth)
3. -x = ""man is beautiful" is not true" (Denial-Truth)
thus,
x = -x = Truth
...maybe I should consider strings not made up by letters... like strings of time and/or space?
Let "---" = time' date=' and "| |" = space, and "." = moment, and "< >" = infinity, then
spacetime = "< .|---|.|---|.|---|. >" , thus
Time is that which is intermediate between moments and both are contained within the innermost boundary of space, yielding the strings "past" "present" "future".
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However,
1. x = "man is beautiful"
2. x = ""man is beautiful" is true" (Affirmation-Truth)
3. -x = ""man is beautiful" is not true" (Denial-Truth)
thus,
x = -x = Truth
Let "---" = time, and "| |" = space, and "." = moment, and "< >" = infinity, then
spacetime = "< .|---|.|---|.|---|. >" , thus
Time is that which is intermediate between moments and both are contained within the innermost boundary of space, yielding the strings "past" "present" "future".
(i) Whether your derivation of "x = -x" was correct or not,
Whatever "x" is, it is not its negation. (Negative definition of Negation.)
Form all x, subtract all y such that "y is true" is true, then what remains are nontruths all not identical with x.
So the foundation is:
1 Law of Identity
2 Definition of Negation
(ii) "..." is more traditional as infinity...
(iii) "Time is that" ... is there but one time? Time is necessarily singular?
Theres "parallell times" and there might be an eternity of time inside a moment. Thanx for the notation :)
(iv) A Being in Time "me" notes that living time strings end in one or more variables...What must preceed them? The past, free of variables?
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"string" = "x"
Huh?!
Unless you’re using the symbol “=” to mean “is not equal to”, this statement isn’t true in any sense that I know of involving strings!
I think this thread would benefit from introducing some conventional terms and concepts of string arithmetic and arithmetic (or algebra) in general, especially some from computer programming disciplines.
A string is an ordered set of elements. It’s common to call the elements of a string characters, but to avoid confusion with standard definitions such as ASCII, better I think to use a more general term from the literature, letter, and refer to the collection of all letters in a particular string system (sometimes, as in GEB, called a typographical axiomatic system) as its alphabet. The term symbol is more commonly used as a synonym for letter, but I prefer letter, to avoid confusion with the user of the term symbol as a synonym for variable common in computer science.
It’s important to note that an alphabet need not be limited to a finite number of letters - Some important proofs involving strings, Gödel's incompleteness theorems, begin with the description of a simple means of assuring an arbitrarily large alphabet known as Gödel numbering.
A distinct visual mark used to represent a letter is called a glyph. It’s important, in the same way it’s important not to confuse the map with the territory, not to confuse letters with their glyphs.
A string that has a truth value or is a command in some representative system is called an expression. An expression that asserts its own truth is known as an assertion, definition, or in special cases, an assignment. All of the lines that Sigurd and Rade have been interspersing between the natural language in their posts in this thread are expressions.
It’s convenient to divide strings into substrings, calling the substrings terms and operators. For example, in the expressions
x = “string”
and
“string” = “x”
, ‘x’ and ‘”string”’ and ‘”x”’ are terms, while ‘=’ is an operator
As with numbers in more common arithmetic, there are two kinds of terms: variables (also called symbols) and literals. For example, in
1 + 2 = 3
y = 0
x = “string”
and
name = “Alice”
,’y’, ‘x’ and ‘name’ are variables, while ‘1’, ‘2’, ‘3’, ‘0’ ‘string’ and ‘Alice’ are literals.
It’s common to indicate string literals by enclosing them with double and/or single quotes, though I know an obscure computer languages (Converse) in which they’re used to indicate variables.
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Hi Craig! Welcome to "String"Theory :)
I dont think you are wrong, but i dont think you are right either: You are on track but our direction is perhaps inversed. There is nothing wrong in introducing "Top to bottom concepts" but to us beginners/researchers "Bottom to top concepts" assimilates easier. Anyway, in the end theres not much else to do than reversing, if possible, so I agree with the proviso that one takes the steps carefully, one thing/concept/string at the time.
Also since "Truth" is my basic subject matter i want to start on a level where "Logic" and "Mathematics" might be "easily" modelled.
So lastly: What is needed to unify "String"Theory, "String Theory" and String Theory...
A slightly cheating answer is perhaps "Mind." and its constructing/analysing capability...
This climbing of Mount Semantics is tiresome to old legs, so thanx for your helping hand!
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http://www.charmpeach.com/stochastic-processes/solutions-to-stochastic-processes-ch-8/1004/ | 1,669,743,121,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00398.warc.gz | 61,912,025 | 29,856 | # Solutions to Stochastic Processes Ch.8
Solutions to Stochastic Processes Sheldon M. Ross Second Edition(pdf)
Since there is no official solution manual for this book, I
handcrafted the solutions by myself. Some solutions were referred from web, most copyright of which are implicit, can’t be listed clearly. Many thanks to those authors! Hope these solutions be helpful, but No Correctness or Accuracy Guaranteed. Comments are welcomed. Excerpts and links may be used, provided that full and clear credit is given.
In Problem 8.1, 8.2 and 8.3, let $$\{X(t), t \geq 0\}$$ denote a Brownian motion process.
8.1 Let $$Y(t) = tX(1/t)$$.
(a) What is distribution of $$Y(t)$$?
(b) Compute $$Cov(Y(s), Y(t)$$
(c) Argue that $$\{Y(t), t \geq 0\}$$ is also Brownian motion
(d) Let $$T = inf\{t>0: X(t)=0\}$$. Using (c) present an argument that $$P\{T = 0\} = 1$$.
(a) $$X(1/t) \sim N(0, 1/t)$$ then $$Y(t) = tX(1/t) \sim N(0, t)$$
(b) \begin{align} Cov(Y(s), Y(t)) &= Cov(sX(1/s), tX(1/t)) \\ &= st\cdot Cov(X(1/s), X(1/t)) \\ &= st\cdot min(1/s, 1/t) = min(s, t) \end{align}
(c) Since $$\{X(t)\}$$ is a Gaussian process so is $$\{Y(t)\}$$. Further from parts (a) and (b) above $$\{Y(t)\}$$ is a Brownian motion.
(d) Since $$Y(t)$$ is Brownian motion then $$T_1 \equiv sup\{t: Y(t) = 0\} = \infty$$ with probability 1. Note $$\{T = 0\} = \{T_1 = \infty\}$$. Thus, $$P\{T = 0\} = 1$$
8.2 Let $$W(t) = X(a^2t)/a$$ for $$a > 0$$. Verify that $$W(t)$$ is also Brownian motion.
$$W(0) = X(0)/a = 0$$. Non-overlapping increments of $$W(t)$$ map to non-overlapping increments of $$X(t)$$. Thus increments of $$W(t)$$ are independent. Further, for $$s < t$$,
$$W(t) – W(s) = \frac{X(a^2t) – X(a^2s)}{a} \sim N(0, t-s)$$
Thus $$W(t)$$ has stationary increments with required distribution. Therefore, $$W(t)$$ is a Brownian motion.
8.5 A stochastic process $$\{X(t), t \geq 0\}$$ is said to be stationary if $$X(t_1), \dots, X(t_n)$$ has the same joint distribution as $$X(t_1+a), \dots, X(t_n +a)$$ for all $$n, a, t_1, \dots, t_n$$.
(a) Prove that a necessary and sufficient condition for a Gaussian process to be stationary is that $$Cov(X(s), X(t))$$ depends only on $$t-s, s \leq t$$, and $$E[X(t)] = c$$.
(b) Let $$\{X(t), t \geq 0\}$$ be Brownian motion and define
$$V(t) = e^{-\alpha t/2}X(\alpha e^{\alpha t})$$
Show that $$\{V(t), t \geq 0\}$$ is a stationary Gaussian process. It is called Ornstein-Uhlenbeck process.
(a) If the Gaussian process is stationary then for $$t > s, (X(t), X(s))$$ and $$(X(t-s), X(0))$$ have same distribution. Thus, $$E[X(s)] = E[X(0)]$$ for all $$s$$ and $$Cov(X(t), X(s)) = Cov(X(t-s), X(0))$$, for all $$t < s$$. Now, assume $$E[X(t)] = c$$ and $$Cov(X(t), X(s)) = h(t-s)$$. For any $$T = (t_1, \dots, t_k)$$ define vector $$X_T \equiv (X(t_1), \dots, X(t_k))$$. Let $$\tilde{T} = (t_1-a, \dots, t_k -a)$$. If $$\{X(t)\}$$ is a Gaussian process then both $$X_T$$ and $$X_{\tilde{T}}$$ are multivariate normal and it suffices to show that they have the same mean and covariance. This follows directly from the fact that they have the same element-wise mean $$c$$ and the equal pair-wise covariances, $$Cov(X(t_i-a), X(t_j -a)) = h(t_i-t_j) = Cov(X(t_i), X(t_j))$$
(b) Since all finite dimensional distributions of $$\{V(t)\}$$ are normal, it is a Gaussian process. Thus from part (a) is suffices to show the following:
(i) $$E[V(t)] = e^{-at/2}E[X(\alpha e^{\alpha t})] = 0$$. Thus $$E[V(t)]$$ is constant.
(ii) For $$s \leq t$$, \begin{align} Cov(V(s), V(t)) &= e^{-\alpha(t+s)/2}Cov(X(\alpha e^{\alpha s}), X(\alpha e^{\alpha t}))\\ &= e^{-\alpha(t+s)/2}\alpha e^{\alpha s} = \alpha e^{-\alpha(t-s)/2} \end{align}
which depends only on $$t-s$$.
8.8 Suppose $$X(1) = B$$. Characterize, in the manner of Proposition 8.1.1, $$\{X(t), 0 \leq t \leq 1\}$$ given that $$X(1) = B$$.
Condition on $$X(1) = B, X(t) \sim N(Bt, t(1-t))$$, then $$Z(t) \sim N(0, t(1-t))$$ which is a Brownian motion.
8.9 Let $$M(t) = max_{0 \leq s \leq t} X(s)$$ and show that
$$P\{M(t) > a| M(t) = X(t)\} = e^{-a^2/2t}, \quad a > 0$$
From Section 8.3.1, we get
$$P\{M(t) > y, X(t) < x\} = \int_{2y-x}^{\infty}\frac{1}{\sqrt{2\pi t}}e^{-u^2/2t}du$$
By using Jacobian formula, we can derive the density function of $$M(t)$$ and $$W(t) = M(t) – X(t)$$, which we denote by $$f_{MW}$$. Thus
$$f_W(w) = 2\int_0^{\infty}f_{MW}(m, w)dm \\ P\{M(t) > a | W(t) = 0\} = 1 – \int_0^a \frac{f_{MW}(m, 0)}{f_W(0)}dm$$
The last equation can be computed, which equal $$e^{-a^2/2t}$$
8.10 Compute the density function of $$T_x$$, the time until Brownian motion hits $$x$$.
\begin{align} f_{T_x}(t) &= F_{T_x}^{\prime}(t) = (\frac{2}{\sqrt{2\pi}}\int_{|x|/\sqrt{t}}^{\infty}e^{-y^2/2}dy)^{\prime} \\ &= -\frac{2}{\sqrt{2\pi}} \cdot e^{-x^2/2t} \cdot \frac{x}{2}t^{-3/2}\\ &= -\frac{x}{\sqrt{2\pi}}e^{-x^2/2t}t^{-3/2} \end{align}
8.11 Let $$T_1$$ denote the largest zero of $$X(t)$$ that is less than $$t$$ and let $$T_2$$ be the smallest zero greater than $$t$$. Show that
(a) $$P\{T_2 < s\} = (2/\pi)\arccos\sqrt{t/s}, s> t$$.
(b) $$P\{T_1 < s, T_2 > y\} = (2/\pi)\arcsin\sqrt{s/y}, s < t< y$$.
(a) \begin{align} P\{T_2 < s\} &= 1 – P\{\text{no zeros in } (t, s)\} \\ &= 1 – \frac{2}{\pi}\arcsin\sqrt{t/s} \\ &= (2/\pi)\arccos\sqrt{t/s} \end{align}
(b) $$P\{T_1 < s, T_2 > y\} = P\{\text{no zeros in } (s, y)\} = \frac{2}{\pi}\arcsin\sqrt{s/y}$$
8.12 Verify the formulas given in (8.3.4) for the mean and variance of $$|X(t)|$$.
$$f_Z(y) = (\frac{2}{\sqrt{2\pi t}}\int_{-\infty}^y e^{-x^2/2t}dx – 1)^{\prime} = \frac{2}{\sqrt{2\pi t}}e^{-y^2/2t}\\ E[Z(t)] = \int_{0}^{\infty}yf_Z(y)dy = -\frac{2t}{\sqrt{2\pi t}}e^{-y^2/2t}|_0^{\infty} = \sqrt{2t/\pi} \\ Var(Z(t)) = E[Z^2(t)] – E^2[Z(t)] = E[X^2(t)] – E^2[Z(t)] = (1 – \frac{2}{\pi})t$$
8.13 For Brownian motion with drift coefficient $$\mu$$, show that for $$x>0$$
$$P\{max_{0 \leq s \leq h} |X(s)| > x\} = o(h).$$
8.18 Let $$\{X(t), t \geq 0\}$$ be a Brownian motion with drift coefficient $$\mu, \mu < 0$$, which is not allowed to become negative. Find the limiting distribution of $$X(t)$$.
8.19 Consider Brownian motion with reflecting barriers of $$-B$$ and $$A, A >0, B > 0$$. Let $$p_t(x)$$ denote the density function of $$X_t$$.
(a) Compute a differential equation satisfied by $$p_t(x)$$.
(b) Obtain $$p(x) = \lim_{t \to \infty} p_t(x)$$.
8.20 Prove that, with probability 1, for Brownian motion with drift $$\mu$$.
$$\frac{X(t)}{t} \to \mu, \quad \text{ as } t \to \infty$$
8.21 Verify that if $$\{B(t), t \geq 0\}$$ is standard Brownian motion then $$\{Y(t), t \geq 0\}$$ is a martingale with mean 1, when $$Y(t) = exp\{cB(t) – c^2t/2\}$$
\begin{align} E[Y(t)] &= \int_{-\infty}^{\infty} exp\{cx – c^2t/2\}\frac{1}{\sqrt{2\pi t}}exp\{-x^2/2t\}dx\\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}}exp\{-(x-ct)^2/2t\}dx = 1 \end{align} \begin{align} E[Y(t)|Y(u), 0 \leq u \leq s] &= Y(s) + E[Y(t) – Y(s)|Y(u), 0 \leq u \leq s ]\\ &= Y(s) \cdot E[exp\{c(B(t) – B(s)) – c^2(t-s)/2\}] \\ &= Y(s) \cdot E[Y(t-s)] = Y(s) \end{align}
8.22 In Problem 8.16, find $$Var(T_a)$$ by using a martingale argument.
8.23 Show that
$$p(x,t;y) \equiv \frac{1}{\sqrt{2\pi t}}e^{-(x – y – \mu t)^2/2t}$$
satisfies the backward and forward diffusion. Equations (8.5.1) and (8.5.2)
Just do it : )
8.24 Verify Equation (8.7.2)
Let $$f(s, y) = [\phi(se^{-\alpha y}) – 1]dy$$, then
\begin{align} E[X(t)] &=\frac{d}{ds}E[exp\{sX(t)\}]|_{s=0} \\ &= exp\{\lambda\int_0^t f(0, y)dy\} \lambda \int_0^t \frac{d}{ds}f(s, y)|_{s=0} dy \\ &= \lambda E[X](1 – e^{-\alpha t})/\alpha\\ Var(X(t)) &= E[X^2(t)] – E^2[X(t)] \\ &= \frac{d^2}{ds^2}E[exp\{sX(t)\}]|_{s=0} – E^2[X(t)] \\ &= \lambda E[X^2](1 – e^{-2\alpha t})/2\alpha \end{align}
8.25 Verify that $$\{X(t) = N(t + L) – N(t), t \geq 0\}$$ is stationary when $$\{N(t)\}$$ is a Poisson process.
For any $$t, X(t) = N(t + L) – N(t) = N(L)$$, thus
$$E[X(t)] = E[N(L)] = \lambda L\\ Cov(X(t), X(t+s)) = Var(N(L)) = \lambda L$$
8.26 Let $$U$$ be uniformly distributed over $$(-\pi, \pi)$$, and let $$X_n = cos(nU)$$. By using trigonometric identity
$$\cos x \cos y = \frac{1}{2} [\cos(x+y) + \cos(x-y)]$$
verify that $$\{X_n, n \geq 1\}$$ is a second-order stationary process.
\begin{align} E[X_n] &= \frac{1}{n\pi}\int_{-n\pi}^{n\pi} \cos xdx = 0\\ Cov(X_{n+L}, X_n) &= E[X_{n+L}X_n] – E[X_{n+L}]E[X_n] \\ &= \frac{1}{2}E[X_{2n+L} + X_L] = 0 \end{align}
8.27 Show that
$$\sum_{i=1}^n \frac{R(i)}{n} \to 0 \quad \text{implies} \quad {\sum\sum}_{i < j < n}\frac{R(j-i)}{n^2} \to 0$$
thus completing the proof of Proposition 8.8.1
8.28 Prove the Cauchy-Schwarz inequality:
$$(E[XY])^2 \leq E[X^2]E[Y^2]$$
(Hint: Start with the inequality $$2|xy| \leq x^2 + y^2$$ and then substitute $$X/\sqrt{E[X^2]}$$ for $$x$$ and $$Y/\sqrt{E[Y^2]}$$ for $$y$$)
Since $$2xy \leq x^2 + y^2$$, then
\begin{align} 2\frac{X}{\sqrt{E[X^2]}}\frac{Y}{\sqrt{E[Y^2]}} &\leq \frac{X^2}{E[X^2]} + \frac{Y^2}{E[Y^2]} \\ E[2\frac{X}{\sqrt{E[X^2]}}\frac{Y}{\sqrt{E[Y^2]}}] &\leq E[\frac{X^2}{E[X^2]} + \frac{Y^2}{E[Y^2]}]\\ 2\frac{E[XY]}{\sqrt{E[X^2]E[Y^2]}} &\leq 2\\ (E[XY])^2 &\leq E[X^2]E[Y^2] \end{align}
8.29 For a second-order stationary process with mean $$\mu$$ for which $$\sum_{i=0}^{n-1}R(i)/n \to 0$$, show that for any $$\varepsilon > 0$$
$$\sum_{i=0}^{n-1}P\{|\bar{X_n} – \mu| > \varepsilon \} \to 0 \quad \text{as } n \to \infty$$ | 3,974 | 9,314 | {"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": 2, "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} | 3.8125 | 4 | CC-MAIN-2022-49 | longest | en | 0.816312 |
https://www.hindawi.com/journals/mpe/2014/521712/ | 1,591,272,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00102.warc.gz | 728,635,452 | 128,218 | Research Article | Open Access
Volume 2014 |Article ID 521712 | 10 pages | https://doi.org/10.1155/2014/521712
# The ()-Expansion Method and Its Applications to Find the Exact Solutions of Nonlinear PDEs for Nanobiosciences
Accepted10 Feb 2014
Published25 Mar 2014
#### Abstract
The two-variable ()-expansion method is employed to construct exact traveling wave solutions with parameters of nanobiosciences partial differential equation. When the parameters are replaced by special values, the solitary wave solutions and the periodic wave solutions of this equation have been obtained from the traveling waves. This method can be thought of as the generalization of well-known original -expansion method proposed by M. Wang et al. It is shown that the two-variable ()-expansion method provides a more powerful mathematical tool for solving many other nonlinear PDEs in mathematical physics. Comparison between our results and the well-known results is given.
#### 1. Introduction
In recent years, investigations of exact solutions to nonlinear partial differential equations (PDEs) play an important role in the study of nonlinear physical phenomena. Many powerful methods for finding these exact solutions have been presented, such as the inverse scattering method [1], the Hirota bilinear transform method [2], the truncated Painleve expansion method [36], the Backlund transform method [7, 8], the exp-function method [913], the tanh-function method [1417], the Jacobi elliptic function expansion method [1820], the -expansion method [2130], the modified -expansion method [31], the -expansion method [3235], the modified simple equation method [36], the multiple exp-function algorithm method [37], the transformed rational function method [38], the local fractional variation iteration method [39], and the local fractional series expansion method [40]. Further exact solutions to some real-life physical problems were already given in the recent articles [4144]. The key idea of the one-variable -expansion method is that the exact solutions of nonlinear PDEs can be expressed by a polynomial in one variable in which satisfies the second-order linear ODE , where and are constants and . The key idea of the two variable -expansion method is that the exact traveling wave solutions of nonlinear PDEs can be expressed by a polynomial in two variables and in which satisfies the second-order linear ODE , where and are constants. The degree of this polynomial can be determined by considering the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in the given nonlinear PDEs. The coefficients of this polynomial can be obtained by solving a set of algebraic equations resulting from the process of using this method. Recently, Li et al. [32] have applied the -expansion method and determined the exact solutions of Zakharov equations, while Zayed et al. [3335] have used this method to find the exact solutions of the combined KdV-mKdV equation, the Kadomtsev-Petviashvili equation, and the potential YTSF equation, respectively.
The objective of this paper is to apply the two-variable -expansion method to find the exact traveling wave solutions of the following nonlinear partial differential equation of nanobiosciences [45, 46]: where is the resistance of the ER with length , is total maximal capacitance of the ER, is conductance of pertaining NPs, and is the characteristic impedance of our system. Parameters and describe nonlinearity of ER capacitor and conductance of NPs in RE, respectively.
The rest of this paper is organized as follows: In Section 2, we give the description of the two-variable -expansion method. In Section 3, we apply this method to solve (1). In Section 4, some conclusions are given.
#### 2. Description of the Two-Variable -Expansion Method
Before we describe the main steps of this method, we need the following remarks (see [3235]).
Remark 1. If we consider the second-order linear ODE and set , , then we get where and are constants.
Remark 2. If , then the general solution of (2) has the form where and are arbitrary constants. Consequently, we have where .
Remark 3. If , then the general solution of (2) has the form and hence where .
Remark 4. If , then the general solution of (2) has the form and hence Suppose we have the following nonlinear evolution equation: where is a polynomial in and its partial derivatives. In the following, we give the main steps of the -expansion method [3235].
Step 1. The traveling wave transformation where is a nonzero constant, reduces (10) to an ODE in the form where is a polynomial of and its total derivatives with respect to .
Step 2. Assume that the solution of (12) can be expressed by a polynomial in the two variables and as follows: where and are constants to be determined later.
Step 3. Determine the positive integer in (13) by using the homogeneous balance between the highest-order derivatives and the nonlinear terms in (12).
Step 4. Substitute (13) into (12) along with (3) and (5); the left-hand side of (12) can be converted into a polynomial in and , in which the degree of is not longer than one. Equating each coefficient of this polynomial to zero yields a system of algebraic equations which can be solved by using the Maple or Mathematica to get the values of , , , , , , and , where .
Step 5. Similar to Step 4, substituting (13) into (12) along with (3) and (7) for , (or (3) and (9) for ), we obtain the exact solutions of (12) expressed by trigonometric functions (or by rational functions), respectively.
#### 3. An Application
In this section, we will apply the method described in Section 2 to find the exact traveling wave solutions of (1) which describe models of microtubules (MTs) as nonlinear transmission lines. The physical details of derivation of (1) describing its ionic currents are elaborated in [45].
In order to solve (1) by the two-variable -expansion method, we see that the traveling wave transformation (11) with and the characteristic time of charging ER capacitor is . Thus (1) takes the form of ODE where , , .
By balancing with in (14), we get . Consequently, we get where , , , , and are constants to be determined later. There are three cases to be discussed as follows.
Case 1 (hyperbolic function solutions ). If , substituting (15) into (14) and using (3) and (5), the left-hand side of (14) becomes a polynomial in and . Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in , , , , , , , and as follows:
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider
From (4), (15), and (17), we deduce the traveling wave solution of (1) as follows: where is a nonzero constant, .
In particular, by setting , , and in (18), we have the solitary wave solution while if , , and , then we have the solitary wave solution
Result 2. Consider
In this result, we deduce the traveling wave solution of (1) as follows: where is an arbitrary constant and .
In particular, by setting and in (22), we have the solitary wave solution (Figure 1) while if and , then we have the solitary wave solution
Note that our results (23) and (24) are in agreement with the results () obtained in page 1251 of [47] with interchanges and with the results (28) and (29) in page 1544 of [48] with the interchanges .
We close this case with the remark that our results (18)–(20) and (22) are new and are not reported elsewhere.
Case 2 (trigonometric function solution ). If , substituting (15) into (14) and using (3) and (7), the left-hand side of (14) becomes a polynomial in and . Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in , , , , , , , and as follows:
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider
From (6), (15), and (26), we deduce the traveling wave solution of (1) as follows: where is a nonzero constant, .
In particular, by setting , , and in (27), we have the periodic solution while if , , and , then we have the periodic solution
Result 2. Consider
In this result, we deduce the traveling wave solution of (1) as follows: where is an arbitrary constant and .
In particular, by setting and in (31), we have the periodic solution (Figure 2) while if and , then we have the periodic solution
Note that our results (32) and (33) are in agreement with the results and obtained in page 1251 of [47] with interchanges and with the results (33) and (34) in page 1544 of [48] with the interchanges .
We close this case with the remark that our results (27)–(29) and (31) are new and are not reported elsewhere.
Case 3 (rational function solutions ). If , substituting (15) into (14) and using (3) and (9), the left-hand side of (14) becomes a polynomial in and . Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in , , , , , and as follows:
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider
From (8), (15), and (35), we deduce the traveling wave solution of (1) as follows:
Result 2. Consider
In this result, we deduce the traveling wave solution of (1) as follows: where in (36) and (38) is given by . Our result (38) is new and is not reported elsewhere.
Remark 5. In all the above results, we have noted the term which is equivalent to the condition (21) of [45]. The condition (39) enables that the ohmic loss could be balanced by fresh ions injected from the nanopores in microtubules since these act ionic pumps in suitable voltage regime. Also, the authors [46] have obtained the condition (39) when they have discussed (1) using the extended tanh-function method. The same remark has been happened in [47, 48] using the improved Riccati equation mapping method and the improved -expansion method, respectively.
Remark 6. Since , then (14) becomes Integrating (40) once with respect to with zero constant of integration, we get
Multiplying (41) by and integrating once with respect to with zero constant of integration, we have and hence we have Consequently, we have where is a constant of integration.
From (44) we have the following solution (Figure 3): which is in agreement with the solution (17) obtained in [48] when , using the improved -expansion method. After some simple calculation, the solution (45) is equivalent to the solution (34) obtained in [45].
Remark 7. Integrating (40) twice with respect to , we get where and are nonzero constants of integration.Setting we obtain the Weierstrass equation [49, 50] where , , while is the Weierstrass elliptic -function.
If , (48) reduces to the equation , which admits the elementary solution where is a constant of integration. From (47) and (49), we get which is equivalent to the solution (36) for , .
From the well-known solutions of (48), see, for example, [50], we rewrite the solution (47) in terms of the Jacobi elliptic functions as follows: where , is the modulus of the Jacobi elliptic functions and , , are three roots of the cubic equation . When , that is, , we get and . Finally, we note that the solutions (51) are new and are not reported elsewhere.
#### 4. Conclusions
The two-variable -expansion method is used in this paper to obtain some new solutions of a selected nonlinear partial differential equation of nanoionic currents along MTs (1). As the two constants and take special values, we obtain the solitary wave solutions. When and in (2) and (13), the two-variable -expansion method reduces to the -expansion method. So the two-variable -expansion method is an extension of the -expansion method. The proposed method in this paper is more effective and more general than the -expansion method because it gives exact solutions in more general forms. In summary, the advantage of the two-variable -expansion method over the -expansion method is that the solutions using the first method recover the solutions using the second one. On comparing our results obtained in this paper with the well-known results obtained in [45, 47, 48] we deduce that some of these results are the same while the others are new which are not reported elsewhere. Finally, all solutions obtained in this paper have been checked with the Maple by putting them back into the original equation (14).
#### Glossary
ER: elementary ring MT: microtubule NP: nanopore ODE: ordinary differential equation PDE: partial differential equation.
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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47. E. M. E. Zayed, Y. A. Amer, and R. M. A. Shohib, “The improved Riccati equation mapping method for constructing many families of exact solutions for a nonlinear partial differential equation of nanobiosciences,” International Journal of Physical Sciences, vol. 8, no. 22, pp. 1246–1255, 2013. View at: Publisher Site | | 6,352 | 24,087 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 23, "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-2020-24 | longest | en | 0.852668 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-9th-edition/chapter-3-stoichiometry-exercises-page-128/39 | 1,591,419,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00415.warc.gz | 731,806,041 | 11,424 | ## Chemistry 9th Edition
Mass of $^{185}Re$= 185.0 amu
Let m be the molar mass of $^{185}Re$: $186.207 = 0.6206*186.956+(1-0.6206)m$ $m=\frac{186.207-0.6206*186.956}{(1-0.6206}=185.0$ amu | 93 | 188 | {"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} | 3.25 | 3 | CC-MAIN-2020-24 | latest | en | 0.324871 |
https://www.answers.com/Q/What_is_the_formula_use_to_calculate_the_weight_of_an_object | 1,713,071,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00410.warc.gz | 621,875,396 | 48,230 | 0
# What is the formula use to calculate the weight of an object?
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A spring device can only measure an object's weight. In order to find its mass, you then have to either compare its weight with the weight of a known mass, or else use the value of gravitational acceleration to calculate the mass from the weight. | 470 | 2,112 | {"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} | 3.953125 | 4 | CC-MAIN-2024-18 | latest | en | 0.916078 |
http://mathhelpforum.com/number-theory/177476-congruence-proof-using-wilsons-theorem-print.html | 1,529,301,890,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00619.warc.gz | 194,980,262 | 3,228 | # Congruence Proof using Wilson's Theorem
• Apr 10th 2011, 02:15 PM
scherz0
Congruence Proof using Wilson's Theorem
Dear all,
I'm having trouble completing the proof for the following result, that requires Wilson's Theorem. I've shown this, along with my work, below.
Thank you!
scherz0
--------
For any odd prime $\displaystyle p$, show that:
$\displaystyle 1^2 \cdot 3^2 \cdot ... \cdot (p - 2)^2 \equiv 2^2 \cdot 4^2 \cdot ... \cdot (p - 1)^2 \equiv (-1)^{\frac{p+1}{2}}$
Work: By Wilson's Theorem, I know that: $\displaystyle p$ prime $\displaystyle \Longleftrightarrow (p - 1)! \equiv -1\pmod {p} \Longleftrightarrow (p - 2)! \equiv 1\pmod {p}$.
Also, for any $\displaystyle k$, I know that $\displaystyle p - k \equiv -k\pmod {p}$. This means that:
$\displaystyle p - 1 \equiv -1\pmod {p}, p - 2 \equiv -2\pmod {p}, p - 3 \equiv -3\pmod {p} \implies (p - 1)(p - 2)(p - 3)... \equiv (-1)(-2)(-3)... \pmod{p}$
But I can't see how to apply the results above to prove the congruence relation?
• Apr 10th 2011, 07:14 PM
tonio
Quote:
Originally Posted by scherz0
Dear all,
I'm having trouble completing the proof for the following result, that requires Wilson's Theorem. I've shown this, along with my work, below.
Thank you!
scherz0
--------
For any odd prime $\displaystyle p$, show that:
$\displaystyle 1^2 \cdot 3^2 \cdot ... \cdot (p - 2)^2 \equiv 2^2 \cdot 4^2 \cdot ... \cdot (p - 1)^2 \equiv (-1)^{\frac{p+1}{2}}$
Work: By Wilson's Theorem, I know that: $\displaystyle p$ prime $\displaystyle \Longleftrightarrow (p - 1)! \equiv -1\pmod {p} \Longleftrightarrow (p - 2)! \equiv 1\pmod {p}$.
Also, for any $\displaystyle k$, I know that $\displaystyle p - k \equiv -k\pmod {p}$. This means that:
$\displaystyle p - 1 \equiv -1\pmod {p}, p - 2 \equiv -2\pmod {p}, p - 3 \equiv -3\pmod {p} \implies (p - 1)(p - 2)(p - 3)... \equiv (-1)(-2)(-3)... \pmod{p}$
But I can't see how to apply the results above to prove the congruence relation?
RHS, working all the time with arithmetic modulo p:
$\displaystyle \displaystyle{2^2\cdot 4^2\cdot\ldots\cdot (p-1)^2=2\cdot 4\cdot\ldots\cdot (p-1)(p-1)(p-3)\cdot\ldots\cdot [-(p-2]=$
$\displaystyle \displaystyle{=(-1)^{\frac{p-1}{2}}(p-1)!=(-1)^{\frac{p+1}{2}}$
Now you do the LHS...
Tonio
• Apr 11th 2011, 08:08 AM
scherz0
Quote:
Originally Posted by tonio
RHS, working all the time with arithmetic modulo p:
$\displaystyle \displaystyle{2^2\cdot 4^2\cdot\ldots\cdot (p-1)^2=2\cdot 4\cdot\ldots\cdot (p-1)(p-1)(p-3)\cdot\ldots\cdot [-(p-2]$
$\displaystyle \displaystyle{=(-1)^{\frac{p-1}{2}}(p-1)!=(-1)^{\frac{p+1}{2}}$
Now you do the LHS...
Tonio
However, could you explain more on:
$\displaystyle \displaystyle{2^2\cdot 4^2\cdot\ldots\cdot (p-1)^2=2\cdot 4\cdot\ldots\cdot (p-1)(p-1)(p-3)\cdot\ldots\cdot [-(p-2]?$
From what I see, I think that we are separating the powers so that:
$\displaystyle 2^2 \cdot 4^2 \cdot ... \cdot (p - 1)^2 \equiv 2\cdot 4 ... \cdot (p - 1) \cdot 2 \cdot 4 \cdot ... \cdot (p - 1) \pmod {p}$.
But why are there:
1) a $\displaystyle (p - 3)$ after the two $\displaystyle (p - 1)$s and
2) $\displaystyle -(p - 2)$, since your post was about the RHS only? | 1,192 | 3,166 | {"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} | 3.9375 | 4 | CC-MAIN-2018-26 | latest | en | 0.669861 |
https://rdrr.io/cran/GPoM/man/gloMoId.html | 1,638,836,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00185.warc.gz | 549,407,988 | 13,234 | # gloMoId: Global Model Identification In GPoM: Generalized Polynomial Modelling
## Description
Algorithm for global modelling in polynomial and canonical formulation of Ordinary Differential Equations. Univariate Global modeling aims to obtain multidimensional models from single time series (Gouesbet & Letellier 1994, Mangiarotti et al. 2012). An example of such application can be found in Mangiarotti et al. (2014) For a multivariate application, see `GPoMo` (Mangiarotti 2015, Mangiarotti et al. 2016).
Example:
For a model dimension nVar=3, the global model will read:
dX1/dt = X2
dX2/dt = X3
dX3/dt = P(X1,X2,X3).
## Usage
``` 1 2 3 4 5 6 7 8 9 10 11``` ```gloMoId( series, tin = NULL, dt = NULL, nVar = NULL, dMax = 1, weight = NULL, show = 1, filterReg = NULL, winL = 9 ) ```
## Arguments
`series` The original data set: either a single vector corresponding to the original variable; Or a matrix containing the original variable in the first column and its successive derivatives in the next columns. In the latter case, for the construction of n-dimensional model, `series` should have nVar+1 columns since one more derivative will be necessary to identify the model parameters. Variable `nVar` will be set equal to n. In the former case, that is when only a single vector is provided, the derivatives will be automatically recomputed. Therefore, the dimension nVar expected for the model has to be provided. `tin` Input date vector which length should correspond to the input time series. `dt` Sampling time of the input time series. `nVar` Number of variables considered in the polynomial formulation. `dMax` Maximum degree of the polynomial formulation. `weight` A vector providing the binary weighting function of the input data series (0 or 1). By default, all the values are set to 1. `show` Provide (2) or not (0-1) visual output during the running process. `filterReg` A vector that specifies the template for the equation structure (for one single equation). The convention defined by `poLabs` is used. Value is 1 if the regressor is available, 0 if it is not. `winL` Total number of points used for computing the derivatives of the input time series. This parameter will be used as an input in function `drvSucc` to compute the derivatives.
## Value
`A` list of five elements :
`\$init` The original time series and the successive derivatives used for the modeling.
`\$filterReg` The structure of the output model. Value is 1 if the regressor is available, 0 if it is not. The terms order is given by function `poLabs`.
`\$K` Values of the identified coefficients corresponding to the regressors defined in `filterReg`.
`\$resTot` The variance of the residual signal of the model.
`\$resSsMod` The variance of the residual signal of the closer submodels.
`\$finalWeight` Weighting series after boundary values were removed.
## Author(s)
Sylvain Mangiarotti, Laurent Drapeau, Mireille Huc
## References
[1] Gouesbet G., Letellier C., Global vector-field reconstruction by using a multivariate polynomial L2 approximation on nets, Physical Review E, 49 (6), 4955-4972, 1994.
[2] Mangiarotti S., Coudret R., Drapeau L., & Jarlan L., Polynomial search and global modeling : Two algorithms for modeling chaos, Physical Review E, 86, 046205, 2012.
[3] Mangiarotti S., Drapeau L. & Letellier C., Two chaotic models for cereal crops observed from satellite in northern Morocco. Chaos, 24(2), 023130, 2014.
[4] Mangiarotti S., Low dimensional chaotic models for the plague epidemic in Bombay (1896-1911), Chaos, Solitons & Fractals, 81(A), 184-196, 2015.
[5] Mangiarotti S., Peyre M. & Huc M., A chaotic model for the epidemic of Ebola Virus Disease in West Africa (2013-2016). Chaos, 26, 113112, 2016.
`gPoMo`, `autoGPoMoSearch`, `autoGPoMoTest`, `poLabs`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59``` ```############# # Example 1 # ############# # load data data("Ross76") tin <- Ross76[,1] data <- Ross76[,2:3] # Polynomial identification reg <- gloMoId(data[0:500,2], dt=1/100, nVar=2, dMax=2, show=0) ############# # Example 2 # ############# # load data data(NDVI) # Definition of the Model structure terms <- c(1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1) poLabs(3,3,terms==1) reg <- gloMoId(NDVI [,1:1], dt=1/125, nVar=3, dMax=3, show=0, filterReg=terms==1) ## Not run: ############# # Example 3 # ############# # load data data("Ross76") # time vector tin <- Ross76[1:500,1] # single time series series <- Ross76[1:500,3] # some noise is added series[1:100] <- series[1:100] + 0.01 * runif(1:100, min = -1, max = 1) series[301:320] <- series[301:320] + 0.05 * runif(1:20, min = -1, max = 1) # weighting function W <- tin * 0 + 1 W[1:100] <- 0 # the first hundred values will not be considered W[301:320] <- 0 # twenty other values will not be considered either reg <- gloMoId(series, dt=1/100, weight = W, nVar=3, dMax=2, show=1) visuEq(reg\$K, 3, 2, approx = 4) # first weight which value not equal to zero: i1 = which(reg\$finalWeight == 1)[1] v0 <- reg\$init[i1,1:3] reconstr <- numicano(nVar=3, dMax=2, Istep=5000, onestep=1/250, PolyTerms=reg\$K, v0=v0, method="ode45") plot(reconstr\$reconstr[,2], reconstr\$reconstr[,3], type='l', lwd = 3, main='phase portrait', xlab='time t', ylab = 'x(t)', col='orange') # original data: lines(reg\$init[,1], reg\$init[,2], type='l', main='phase portrait', xlab='x', ylab = 'dx/dt', col='black') # initial condition lines(v0[1], v0[2], type = 'p', col = 'red') ## End(Not run) ``` | 1,765 | 5,611 | {"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-2021-49 | longest | en | 0.808189 |
https://dsp.stackexchange.com/questions/89790/formant-bandwidths | 1,708,949,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00142.warc.gz | 223,511,636 | 38,295 | # Formant bandwidths
I have some intuition about bandwidths from these videos (1 and 2) explaining how the frequency response corresponds to the Z-surface.
The only thing I could find on the internet about calculating the bandwidths is this matlab tutorial:
The bandwidths of the formants are represented by the distance of the prediction polynomial zeros from the unit circle.
[frqs,indices] = sort(angz.*(Fs/(2*pi)));
bw = -1/2*(Fs/(2*pi))*log(abs(rts(indices)));
but bw is not the distance to the unit circle but the log of the norm.
This is assuming the roots/poles are inside the unit circle or bw will be negative. But I'm unclear why the roots/poles shouldn't be outside the unit circle?
I can see that the relationship of the bandwidth is not linear with the root/pole position (relative to the unit circle), but why is it logarithmic?
Perhaps, if the root/pole is outside of the unit circle we should reflect log(x) around 1 i.e. log(2-x):
bw = -1/2*(Fs/(2*pi))*log(2-abs(rts(indices))); | 245 | 1,006 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.897248 |
http://www.ck12.org/probability/Permutation-Problems/rwa/Permutations/r2/ | 1,485,278,132,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00556-ip-10-171-10-70.ec2.internal.warc.gz | 384,472,832 | 30,499 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Permutation Problems
## Using the nPr function found in the Math menu under PRB on the TI calculator
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Practice Permutation Problems
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Permutations
Teacher Contributed
## Locker Combinations and Permutations
### Topic
Locker Combinations and Permutations
• Permutation
• Combination
### Student Exploration
#### How many different locker combos can a combination lock have?
It is Fall and school is back in session, Denise needs to buy a lock for her school locker. She wants her belongings to be safe, but she doesn’t want to carry around a key and would rather have a lock with a combination. Before going shopping she decides to research locks online and wants the deciding factor to be the number of locker combination codes that the lock has, that way a thief will have a nearly impossible task of breaking her lock’s combination code.
Research different locks online and find three different models with different designs and set-ups for the locker combos. Then determine which one you would recommend to Denise as the safest for protecting her belongings, using your math work as evidence!
### Extension Investigation
One of the locks Denise considers is a circular lock with 20 numbers. To open it she turns the dial to the first number, then she turns the dial to the second number, and then she turns it to the third number. She may not repeat numbers in her locker combo.
a. Why does this situation represent a Permutation?
b. Write an expression for the number of different combination codes that are possible with this lock?
c. If Denise could use four numbers in her locker combo, how would you approach this task?
d. If Denise could repeat numbers in her locker combo, how would you approach this task?
e. If Denise could choose from 40 different number, rather than 20, how would you approach this task?
f. What would need to change in the problem in order for it to represent a Combination?
g. Why doesn’t it make sense for locks to be “Combination Locks”?
### Connections to other CK-12 Subject Areas
• Combination Problems
• Permutations and Combinations Compared
• Permutations with Repetition
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes | 539 | 2,498 | {"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-2017-04 | longest | en | 0.895365 |
https://hsm.stackexchange.com/questions/6721/what-are-philolaos-even-odd-numbers?noredirect=1 | 1,718,676,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00333.warc.gz | 254,385,732 | 42,753 | # What are Philolaos' “even-odd” numbers?
Number, indeed, has two proper kinds (ιδια ειδη), odd and even, and a third mixed together from both, the even-odd(αρτιοπέριττον). Of each of the two kinds there are many shapes, of which each thing itself gives signs.
(Philolaus F5 = Stobaeus, Eclogae 1.21.7c; tr. Huffman 1993)
The quote is from Horky's Plato and Pythagoreanism, 2013, (p.141), it is commented twice further in text. On p185 the author writes " the so-called even-odd (αρτιοπέριττον), which appears to refer to the “one.” Aristotle, too, knew this fragment, and he is explicit in associating the “one” with the “even-odd” class and stating that it is derived from the “even” and the “odd.” On p190, disappointingly, he says "however It is beyond the scope of this study to examine in more detail the significance of the concept of “mixture” of Forms or classes".
The “one” is not the number preceding two which Ancient greeks apparently did not accept as a number. Philolaos quote might suggest that it is set apart as it has no shape but I am tempted to read it as suggesting that the mixed numbers have multiple shapes, while the numbers from the proper kinds have each just one.
So what are the 3 kinds of numbers? The odd numbers are perhaps the ones that we also call odd, (starting with 3), while the even numbers are seen as, so to say, 'strictly even', the powers of 2, that is 2,4,8,16 and the rest, that is 6,10,12 etc. as even-odd. References and comments (or should this be moved to Philosophy SE)?
• According to Liddell & Scott, perseus.tufts.edu/hopper/…, an even-odd is two times an odd number. Commented Nov 19, 2017 at 0:35
• Thanks but LS sends us back to F5 and Aristotle's "one" (Arist.Fr.199); to Plutarch (?= Plu.2.1139f ) and what Ph.1.3?. Commented Nov 19, 2017 at 9:25
• First, let me say that while I am familiar with a couple of relevant texts, I don't have a comprehensive knowledge of the area, and thus my opinions should not be thoroughly relied upon. Perhaps you're suggesting LS on even-odd has been superseded (plausible). I thought you were asking just about the definitions of three terms. I'm not sure of the context, the Pythagoreans or broadly "Ancient greeks." The Ancient greeks were a heterogeneous bunch, and perhaps the Pythagoreans were, too. For instance, Euclid gives a different classification of numbers (even-times odd etc.)..... Commented Nov 19, 2017 at 15:13
• ...The LS reference to an Aristotle fragment, which can be tracked down, seems at odds with the definition in LS. The fragment raises more questions than it answers, which is the way with fragments, I suppose. It seems to me that Aristotle might be criticizing the Pythagorean concept of even-odd, as opposed to stating or explaining it. Commented Nov 19, 2017 at 15:13
• I'd guess he means the classification scheme given by Euclid, book 7, definitions 8-10. Euclid classifies numbers according to this scheme in book 9, propositions 32-34. Commented Nov 19, 2017 at 19:00
Aristotle comments on the Pythagorean theory in Met, Book I (A), 986a14-986a22:
these thinkers also consider that number is the principle both as matter for things and as forming their modifications and states, and hold that the elements of number are the even and the odd, and of these the former is unlimited, and the latter limited; and the $$1$$ proceeds from both of these (for it is both even and odd), and number from the $$1$$; and the whole heaven, as has been said, is numbers.
But see also Thomas Heath, A History of Greek Mathematics. Volume I (1921), page 71:
The explanation of this strange view might apparently be that the unit, being the principle of all number, even as well as odd, cannott itself be odd and must therefore be called even-odd.
There is, however, another explanation, attributed by Theon of Smyrne to Aristotle, to the effect that the unit when added to an even number makes an odd number, but when added to an odd number makes an even number: which could not be the case if it did not partake of both species.
Philolaus' Fr.5 is discussed at lenght in: Carl Huffman, Philolaus of Croton: Pythagorean and Presocratic, Cambridge UP (1993), page 178-on. There are various interpretations, including the possibility of an interpolation.
See page 190:
In summary, I believe that the even-odd is a derived class of numbers whose first member is, as the ancient tradition indicates, the one, but which also includes numbers that consist of even and odd numbers combined in ratios (e.g. $$2 : 1 , 4 : 3$$, and $$3 : 2$$ ) . This class of numbers corresponds to the third class of things in Fr.2, which consists of members that are harmonized from both limiting and unlimited constituents. The even-odd numbers are the numbers by which these harmonized things are known. This connection of course remains conjectural, but I believe that it is a plausible way to make sense of both Fr.2 and Fr.5 of Philolaus and Aristotle's testimony that there was a connection between the even - odd dichotomy and the unlimited - limited dichotomy, although my suggestion does not identify the two as Aristotle does.
• Using today terms,one could say that for the ancient Greeks, before the discovery of irrationals, numbers were integers>1, even or odd, and rationals. So, including 1 among the even-odd makes sense but not the explanation given by Theon and others, as adding any odd number changes parity. Commented Nov 21, 2017 at 16:48
• @sand1 - Strictly speaking, for ancient Greeks numbers where only naturals. Then they have ratios between magnitudes, and magnitudes can be "measured" by numbers. You can see the post: irrationality-of-the-square-root-of-2. Commented Nov 21, 2017 at 16:50
• Notice also that for Euclid 'irrational' (ἄλογος – a better translation is 'ratioless') means something different from the modern 'irrational' – Euclid's term for his equivalent of the modern concept is 'incommensurable'. (See the definitions of book 10 of the Elements.) Commented Nov 21, 2017 at 21:15
• Also, in Euclid a number doesn't measure a magnitude – only objects that have a ratio to each other can measure each other (strictly, if the measuring one is equal to or less than the measured one). Commented Nov 22, 2017 at 14:06 | 1,587 | 6,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.95533 |
https://www.manchester.ac.uk/study/undergraduate/courses/2024/00075/ba-modern-language-and-business-and-management-french/course-details/ECON20192 | 1,716,364,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00474.warc.gz | 774,444,886 | 13,791 | # BA Modern Language and Business & Management (French) / Course details
Year of entry: 2024
## Course unit details:Introduction to Mathematical Economics
Unit code ECON20192 10 Level 2 Semester 2 Yes
### Overview
This course introduces the mathematical tools essential to study advanced topics in economic theory and mathematical economics. First part of the semester is dedicated to studying real analysis concepts relevant to economists. Emphasis is placed on how mathematical results are proved, how to use these results, and studying situations where the results cannot be applied. Rest of the semester is dedicated to applying these tools to study the theorems underlying unconstrained and constrained optimisation. Students will be provided with detailed material through lectures, tutorials, course notes, textbook references, and other resources. Regular feedback is provided through discussion boards and feedbacks from problem sets.
### Pre/co-requisites
Unit title Unit code Requirement type Description
ECON20192 Pre-requisites: ECON10071 or ECON20071
ECON10071 OR ECON20071
### Aims
The aim of this course is to help students develop the mathematical techniques to analyse advanced topics in economic theory. Topics covered in this course provide a foundation for many results in microeconomics and mathematical economics.
### Learning outcomes
By the end of the course, you will:
1. Understand the concepts of proof and counter examples.
2. Develop the toolbox of real analysis for economics.
3. In-depth knowledge of constrained optimisation theory and comparative statics
### Syllabus
• Logic and proof
• Metric spaces, Open & closed sets, Sequence & convergence
• Continuity, Compactness, Differentiability
• Concavity, Taylor’s theorem, Unconstrained Optimisation
• Constrained Optimisation and Envelope theorem.
### Teaching and learning methods
Synchronous activities (such as Lectures or Review and Q&A sessions, and tutorials), and guided self-study.
### Employability skills
Analytical skills
Problem solving
Other
Logical reasoning
### Assessment methods
70% Exam
15% Coursework (Weekly Online Quizzes)
15% Coursework (Problem Set)
### Feedback methods
• Solution and Feedback to Problem Sets and Quizzes
• Tutorial feedback and solutions
• Piazza Discussion Board
• Office Hours
• Weekly Open Study Sessions
1. Simon, C. and Blume, L. (2010) Mathematics for Economists, International Student Edition, Norton, NY.
2. Derek G. Ball (2014) An Introduction to Real Analysis, Pergamon.
### Teaching staff
Staff member Role
Shomak Chakrabarti Unit coordinator | 544 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-22 | latest | en | 0.844091 |
https://oaklandfhc.org/a-blackjack-strategy-chart-can-help-you-win/ | 1,718,792,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00294.warc.gz | 372,988,832 | 13,758 | Blackjack is one of the casino’s table games that uses math to give players a chance to beat the house. While the game can seem complicated and difficult to master, it’s actually quite simple. By using a blackjack strategy chart to guide you, you can make the best possible decisions in every situation. A blackjack strategy chart will help you know when to hit, stand, split, double down, and surrender according to the rules of the game. You should also learn about the different types of blackjack games and how they affect your chances of winning.
In the game of blackjack, you and the dealer each receive two cards. The goal is to get a hand value of 21 or higher, which will beat the dealer’s hand. If you do not have a 21-valued hand, you must continue to hit until you are satisfied with your total or go bust (hit too many times and exceed 21). In addition to hitting, there is another option known as standing, which means that you do not take any more cards in your hand.
During the game of blackjack, you can split your hand whenever you are dealt two cards of the same denomination. When you split your hand, you place a second bet equal to the original bet amount and play each card as a separate hand. The dealer will then deal you a single card to each split hand. Splitting your hand can improve your odds of getting a blackjack. However, it is important to remember that splitting aces and eights can increase the risk of going bust.
When playing blackjack, you can choose to split your hand when the dealer’s upcard is weak such as a five or six. Splitting your hand may also be a good idea when you are dealt two pairs such as two nines, two sevens, or two threes. However, you should avoid splitting hands such as two tens or two fours because they have a low chance of being improved by an additional card.
The blackjack table is usually semicircular and can accommodate a number of players, from five to seven. Generally, each player is assigned a seat and the dealer stands behind the chip rack and the blackjack table. Typically, you can sit anywhere at the table that has an empty seat (unless chips or a coat are holding it for someone else).
There are several different blackjack tables available at casinos, with some offering a 6 to 5 payout for Blackjacks. This change increases the house edge, but it is still better than the standard 3 to 2 payout. You should always read the rules of each blackjack table before sitting down to play.
Once all the players have finished their turns, the dealer will reveal his hole card. Then the game will proceed as normal. The dealer will try to get a Blackjack or push until his hand totals 21. The dealer will then check his own hand for a Blackjack before paying out any winnings. | 585 | 2,758 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | longest | en | 0.968601 |
https://brainly.in/question/158250 | 1,485,266,910,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284411.66/warc/CC-MAIN-20170116095124-00365-ip-10-171-10-70.ec2.internal.warc.gz | 784,771,047 | 10,830 | # Calculate the mole fraction of ethylene glycol in a solution containing 20% of ethylene glycol prepared in water. plsssssssssssssss answer it fasssssssssssstttttt,tmr is my exam
1
by Raksha1
Sir can u explain how to get mass and molar mass please sir explain it to me.
Sir can u explain how to get mass and molar mass please sir explain it to me
2015-08-03T19:27:40+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Chemical formula of Ehylene glycol
C2 H6 O2 = OH - CH2 - CH2 - OH
molar mass = 2 * 12 + 6* 1+ 16*2 = 62 and density = 1.113 gm/cm³
chemical formula of water : H2 O
molecular mass: 16+2 *1 = 18 Density = 1 gm/cm³
case 1) The solution of glycol is prepared by mixing water and glycol by mass.
So in 100 gms of the solution, we have 20 gms of glycol and 80 gm of water.
number of moles of glycol = mass/molar mass = 20/62 = 10/31
number of moles of water = 80/18 = 40/9
total number of moles in 100 grams of solution: N = 10/31 + 40/9 = 1330/279
Mole fraction of glycol = (10/31) /(1330/279) = 2790/(1330*31) =0.06767
mole fraction of water in the solution = 1 - 0.06767 =
This can be calculated directly as : [20/62] / [20/62 + 80/18]
2) if the liquids are mixed by volume then:
In 100 ml of solution we have 20 ml of glycol and 80 ml of water.
20 ml of glycol has a mass of 20 * 1.113 gm. and 80 ml of water has a mass of 80 gm.
mole fraction of glycol in this case :
[20*1.113/62] / [ 20*1.113/62 + 80/18] = 0.359 / 4.803 = 0.0747
mole fraction of water in the solution will be 1 - 0.0747 =
click on the blue thanks button above please
i clicked sir
sir answer my another question sir..
it is posted in ur messagebox
how much time it Sir for you to type this answer | 654 | 1,992 | {"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-2017-04 | latest | en | 0.861092 |
https://cracku.in/1-number-of-players-who-scored-less-than-or-equal-to-x-cmat-4-may-2023-slot-1 | 1,719,347,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00031.warc.gz | 156,385,253 | 25,960 | Instructions
The following table shows the percentage of Cricket players and scored runs by them in three different tournaments P, Q and R. Total number of players is 300 and all the 300 players played all the matches in each tournament. Based on the data in the table; answer-the questions 1-5.
Tournament-wise Percentage of Players scoring runs
Question 1
# Number of players who scored less than or equal to 40 runs in tournament Q is _____ % more than the number of players who scored more than 60 runs in tournament P and Q together.
Solution
It is given,
Number of players who scored less than or equal to 40 runs in tournament Q = 300 - 90 = 210
$$210$$ =Â $$\ \frac{\ 100+x}{100}\left(75+75\right)$$
$$\ 100+x=210\left(\frac{100}{150}\right)$$
x = 40%
• All Quant Formulas and shortcuts PDF
• 40+ previous papers with detau solutions PDF | 224 | 855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.934835 |
http://astronomy.stackexchange.com/questions/1249/what-is-the-exact-time-earth-takes-to-revolve-around-its-axis | 1,469,694,864,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.15/warc/CC-MAIN-20160723071028-00254-ip-10-185-27-174.ec2.internal.warc.gz | 15,810,083 | 18,065 | # What is the Exact time earth takes to revolve around its axis [closed]
I read that our day is not exactly 24 hours long. If it is not then why weather seasons come at the same time (Summer, Winter, Autumn Spring) ? how it is equalized ?
Update: Sorry I could not make it clear. My meaning was if 3.8 seconds are shorter of everyday the how is the calculation made that summer starts in May and winter start december everytime why it does not move further since. How time is neutralized ? I hope you guys understood this time.
-
## closed as unclear what you're asking by TildalWave, Eduardo Serra, Donald.McLean♦Dec 27 '13 at 20:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
Which weather changes did you have in mind? Did you mean the difference in time we measure and length of day from one solar noon to another? "Day" as a unit of time is by definition 24 hours long. Could you please clarify what you're asking about? There are many factors that are at play here, which ones in particular are you inquiring about? – TildalWave Dec 27 '13 at 9:23
Our day is 23 hours and 56 minutes long, and slowing by an infinitesimal (but measurable) amount each year due to tidal losses.
Our day has a connection with the weather, in that the sun drives all our weather systems, so heating over each part of the globe happens every day, but aside from that, your question doesn't make much sense.
Weather changes may come at the same time where you live (on a 24 hour cycle) but here in Scotland, we still have low reliability on even a 3 day weather forecast, because the weather systems that impact the UK are so complex as heating from the sun drives various air flows.
After your update, I still cannot understand what you mean. There is slippage in accuracy, but this is counteracted by leap seconds and leap years. It has nothing to do with weather. The shortest and longest days happen when they happen and are measurable. They help us define the year.
-
23h 56m is the sidereal day. The solar day is very close to 24 hours. – Keith Thompson Dec 27 '13 at 21:55
I meant the seasons (instead of weather) like summer, winter, autumn and spring which comes at the same time(at least in my country) sorry my bad... Yes that's what i wanted to know. Its mean that 24 hours(29th feb) are added to our year after every 4 years. means if everyday is 3:55.91 minutes(235.9084 seconds) shorter then every year is 86106.566 seconds shorter which is approximately equals to 24 hours then how after four years >.< confused – Asadullah Ali Dec 31 '13 at 1:51
No - that isn't how it works. Humans have chosen seconds, minutes, hours etc., but they don't quite match up to reality. The Earth will happily keep doing it's thing, but every now and then we have to realign our measures. The leap year adjustments do this. – Rory Alsop Dec 31 '13 at 11:45
The realignment part is what sucks. If the world is revolving around the sun at a steady rate, what is that rate in milliseconds? Milliseconds are the most basic unit used in computer programming; seasons be damned. Myself, I am interested in working out the math as it relates to the leap year. I'm surprised this question is closed. – Joshua Dannemann Mar 12 at 23:02
Joshua - the question is closed as it doesn't really make sense as it's written. Leap years/leap seconds only exist as a human construct. They aren't real. – Rory Alsop Mar 13 at 8:00
The length of a solar day is very close to 24 hours. A solar day is the time it takes the Sun to return to the same position in the sky. The exact length varies slightly over the course of the year, because the Earth's orbit around the Sun is not perfectly circular; the tilt of the Earth's axis with respect to its orbit also has an effect. See the linked article for details.
A sidereal day is slightly shorter. It's the time it takes for a distant star to return to the same position in the sky, and it differs from the solar day because the Earth revolves around the Sun as it rotates on its own axis; there is one more sidereal day than solar day each year. A sidereal day is approximately 23 hours, 56 minutes, 4.0916 seconds.
Weather can actually have a minuscule effect on the Earth's rotation, as the mass of the atmosphere is distributed differently. This effect is barely measurable. I don't think that's the kind of "weather changes" you were asking about; can you clarify? Obviously it gets colder at night, and that's controlled by the length of the solar day.
- | 1,125 | 4,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-30 | latest | en | 0.968457 |
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I have a constructor that assigns a single digit to each element of an array. For example if the user input (128) it would assign 1 to element 0, 2 to element 1, etc. I have a max size of 500 elements for this array. So there would be 497 0's after the 128.. How would I only output 128 and not the other 497 0's? Here's my code so far..
``123456789`` `````` bigint::bigint(int digit){ int numdigits = log10((double)digit) + 1; for (int i=numdigits-1; i >= 0; i--) { value[i] = digit%10; digit /= 10; }``````
``12345`` ``````void bigint::output(){ for(int i=0; i <= MAXINT; i++) std::cout << value [i]; }``````
I think you go at it the wrong way. Say you want to display 100. How do you know to display two 0, and not the rest? I would suggest that you store the information in reverse order. If you assign 8 to element 0, 2 to element 1, and 1 to element 2, you would not get this problem.
For the other part, if you use the above mentioned convention, you can do something like
``12345678910`` ``````#include using namespace std; int main() { int a[5],i=0; a[0]=1;a[1]=2;a[2]=3;a[3]=0;a[4]=0; while(a[4-i]==0) i++; for(;i<=4;i++) cout<
I still don't see how this would solve the problem. Okay so if the user did want to display 104 for example. And the max size they can input is 500 digits. Even if I assign 4 to element 0, 0 to element 1, and 1 to element 2, how would the program know to stop outputting the other 497 0's?
Look at the above code: on line 7 I skip all zeroes, starting from the end, until I get a non-zero integer, then output elements a[2],a[1],a[0], so I stored number 321. If I have number 104, I store a[0]=4,a[1]=0,a[2]=1, a[3]=...=a[499]=0. Now display the array backwards, skipping all zeroes at the end. If you feel uncomfortable storing the number backwards, store a[0]=...=a[496]=0, a[497]=1,a[498]=0,a[499]=4. Then increase i from 0, until a[i]==0. Then print a[i] from that i until 499
@ats15 Thank you for your help I figured it out now!
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- MCSE (http://www.velocityreviews.com/forums/f25-mcse.html)
- - Subnetting Problem (http://www.velocityreviews.com/forums/t45841-subnetting-problem.html)
Myrt Webb 07-14-2003 12:49 AM
Subnetting Problem
I need some help with the following subnet problem:
"You are the new administrator of a 2000 node network.
There is only one router on the entire network, which
provides all the computers with Internet access. The
company's ISP has assigned the following 8 network
10.24.32.0/24
10.24.33.0/24
10.24.34.0/24
10.30.35.0/24
10.30.36.0/24
10.30.37.0/24
10.30.38.0/24
10.30.39.9/24
What subnet mask could you use to minimize the complexity
of the routing tables while maintaining the existing
Internet connectivity?
a. 255.255.252.0
b. 255.255.255.252
c. 255.255.255.248
d. 255.255.248.0 "
The answer is 'd.' Quite frankly I do not understand what
the problem is getting at nor how 'd' gives the desired
results.
Any explainations would be appreciated.
moonlighting 07-14-2003 05:47 AM
Subnetting Problem
"/24" will tell you that the SNM is 255.255.254.0, but
not 255.255.248.0 (this is "/21").
>-----Original Message-----
>I need some help with the following subnet problem:
>
>"You are the new administrator of a 2000 node network.
>There is only one router on the entire network, which
>provides all the computers with Internet access. The
>company's ISP has assigned the following 8 network
>
>10.24.32.0/24
>10.24.33.0/24
>10.24.34.0/24
>10.30.35.0/24
>10.30.36.0/24
>10.30.37.0/24
>10.30.38.0/24
>10.30.39.9/24
>
>What subnet mask could you use to minimize the
complexity
>of the routing tables while maintaining the existing
>Internet connectivity?
>
>a. 255.255.252.0
>b. 255.255.255.252
>c. 255.255.255.248
>d. 255.255.248.0 "
>
>The answer is 'd.' Quite frankly I do not understand
what
>the problem is getting at nor how 'd' gives the desired
>results.
>
>Any explainations would be appreciated.
>
>
>.
>
Tom Helms 07-14-2003 05:57 AM
Re: Subnetting Problem
D. 255.255.248.0
BTW, /24 subnet mask is 255.255.255.0
"moonlighting" <moon@lighting.com> wrote in message
news:052801c349cb\$61ef0930\$a301280a@phx.gbl...
> "/24" will tell you that the SNM is 255.255.254.0, but
> not 255.255.248.0 (this is "/21").
>
>
>
>
> >-----Original Message-----
> >I need some help with the following subnet problem:
> >
> >"You are the new administrator of a 2000 node network.
> >There is only one router on the entire network, which
> >provides all the computers with Internet access. The
> >company's ISP has assigned the following 8 network
> >
> >10.24.32.0/24
> >10.24.33.0/24
> >10.24.34.0/24
> >10.30.35.0/24
> >10.30.36.0/24
> >10.30.37.0/24
> >10.30.38.0/24
> >10.30.39.9/24
> >
> >What subnet mask could you use to minimize the
> complexity
> >of the routing tables while maintaining the existing
> >Internet connectivity?
> >
> >a. 255.255.252.0
> >b. 255.255.255.252
> >c. 255.255.255.248
> >d. 255.255.248.0 "
> >
> >The answer is 'd.' Quite frankly I do not understand
> what
> >the problem is getting at nor how 'd' gives the desired
> >results.
> >
> >Any explainations would be appreciated.
> >
> >
> >.
> >
Myrt Webb 07-14-2003 03:39 PM
Re: Subnetting Problem
I thought the same thing. That /24 indicates a
255.255.255.? mask. So I figured that either b or c was
the answer. But according to the MS 70-216 trng kit pg 808
it is d.
I understand the basics of subnetting and figured out
correctly the other questions in this section but I do not
understand the logic of this question.
>-----Original Message-----
>You're right, Tom. /24 is 255.255.255.0. Any explanation
>
>Thanks!
>
>
>>-----Original Message-----
>>D. 255.255.248.0
>>BTW, /24 subnet mask is 255.255.255.0
>>
>>"moonlighting" <moon@lighting.com> wrote in message
>>news:052801c349cb\$61ef0930\$a301280a@phx.gbl...
>>> "/24" will tell you that the SNM is 255.255.254.0, but
>>> not 255.255.248.0 (this is "/21").
>>>
>>>
>>>
>>>
>>> >-----Original Message-----
>>> >I need some help with the following subnet problem:
>>> >
>>> >"You are the new administrator of a 2000 node network.
>>> >There is only one router on the entire network, which
>>> >provides all the computers with Internet access. The
>>> >company's ISP has assigned the following 8 network
>>> >
>>> >10.24.32.0/24
>>> >10.24.33.0/24
>>> >10.24.34.0/24
>>> >10.30.35.0/24
>>> >10.30.36.0/24
>>> >10.30.37.0/24
>>> >10.30.38.0/24
>>> >10.30.39.9/24
>>> >
>>> >What subnet mask could you use to minimize the
>>> complexity
>>> >of the routing tables while maintaining the existing
>>> >Internet connectivity?
>>> >
>>> >a. 255.255.252.0
>>> >b. 255.255.255.252
>>> >c. 255.255.255.248
>>> >d. 255.255.248.0 "
>>> >
>>> >The answer is 'd.' Quite frankly I do not understand
>>> what
>>> >the problem is getting at nor how 'd' gives the
>desired
>>> >results.
>>> >
>>> >Any explainations would be appreciated.
>>> >
>>> >
>>> >.
>>> >
>>
>>
>>.
>>
>.
>
Zenner 07-14-2003 03:53 PM
Re: Subnetting Problem
You "borrow" from the Host address to increase the client address range. You
really need to take the advice to visit the referenced site
(learntosubnet.com), lots of good stuff there!
"moonlighting" <moon@lighting.com> wrote in message
news:052801c349cb\$61ef0930\$a301280a@phx.gbl...
> "/24" will tell you that the SNM is 255.255.254.0, but
> not 255.255.248.0 (this is "/21").
>
>
>
>
> >-----Original Message-----
> >I need some help with the following subnet problem:
> >
> >"You are the new administrator of a 2000 node network.
> >There is only one router on the entire network, which
> >provides all the computers with Internet access. The
> >company's ISP has assigned the following 8 network
> >
> >10.24.32.0/24
> >10.24.33.0/24
> >10.24.34.0/24
> >10.30.35.0/24
> >10.30.36.0/24
> >10.30.37.0/24
> >10.30.38.0/24
> >10.30.39.9/24
> >
> >What subnet mask could you use to minimize the
> complexity
> >of the routing tables while maintaining the existing
> >Internet connectivity?
> >
> >a. 255.255.252.0
> >b. 255.255.255.252
> >c. 255.255.255.248
> >d. 255.255.248.0 "
> >
> >The answer is 'd.' Quite frankly I do not understand
> what
> >the problem is getting at nor how 'd' gives the desired
> >results.
> >
> >Any explainations would be appreciated.
> >
> >
> >.
> >
Jeffrey L. Woods 07-14-2003 04:27 PM
Re: Subnetting Problem
In article <08eb01c34a1a\$ff2229e0\$a601280a@phx.gbl>, moon@lighting.com
says...
> You're right, Tom. /24 is 255.255.255.0. Any explanation
> why the answer is 255.255.248.0?
Either the question has been mis-typed here or mis-remembered.
255.255.248 won't help cover all eight of those addresses, which jump
from 10.24.*.* to 10.30.*.*. You'd need at least a /13 (255.248.0.0)
The question is also bogus because 10.*.*.* is private NAT space, and
would not be allocated by an ISP -- but even overlooking that, the
suggestion that you could simply apply one large mask to the whole
network and call it a day is specious at best. If you were to do so,
then people on this network would be unable to reach people using the
same ISP from elsewhere, i.e. 10.27.5.0/24 would be unreachable.
> >> >-----Original Message-----
> >> >I need some help with the following subnet problem:
> >> >
> >> >"You are the new administrator of a 2000 node network.
> >> >There is only one router on the entire network, which
> >> >provides all the computers with Internet access. The
> >> >company's ISP has assigned the following 8 network
> >> >
> >> >10.24.32.0/24
> >> >10.24.33.0/24
> >> >10.24.34.0/24
> >> >10.30.35.0/24
> >> >10.30.36.0/24
> >> >10.30.37.0/24
> >> >10.30.38.0/24
> >> >10.30.39.9/24
> >> >
> >> >What subnet mask could you use to minimize the
> >> complexity
> >> >of the routing tables while maintaining the existing
> >> >Internet connectivity?
> >> >
> >> >a. 255.255.252.0
> >> >b. 255.255.255.252
> >> >c. 255.255.255.248
> >> >d. 255.255.248.0 "
> >> >
> >> >The answer is 'd.' Quite frankly I do not understand
> >> what
> >> >the problem is getting at nor how 'd' gives the
> desired
Tom Helms 07-14-2003 05:43 PM
Re: Subnetting Problem
Unless my calculations are incorrect, subnet mask 255.255.248.0 will allow
2048 or 2046 host addresses using a class A address to start.
"Jeffrey L. Woods" <jeff@telix.com> wrote in message
news:MPG.197ca3fdf29bd97d9898d8@news.usenetserver. com...
> In article <08eb01c34a1a\$ff2229e0\$a601280a@phx.gbl>, moon@lighting.com
> says...
>
> > You're right, Tom. /24 is 255.255.255.0. Any explanation
> > why the answer is 255.255.248.0?
>
> Either the question has been mis-typed here or mis-remembered.
> 255.255.248 won't help cover all eight of those addresses, which jump
> from 10.24.*.* to 10.30.*.*. You'd need at least a /13 (255.248.0.0)
> mask to cover all those.
>
> The question is also bogus because 10.*.*.* is private NAT space, and
> would not be allocated by an ISP -- but even overlooking that, the
> suggestion that you could simply apply one large mask to the whole
> network and call it a day is specious at best. If you were to do so,
> then people on this network would be unable to reach people using the
> same ISP from elsewhere, i.e. 10.27.5.0/24 would be unreachable.
>
>
> > >> >-----Original Message-----
> > >> >I need some help with the following subnet problem:
> > >> >
> > >> >"You are the new administrator of a 2000 node network.
> > >> >There is only one router on the entire network, which
> > >> >provides all the computers with Internet access. The
> > >> >company's ISP has assigned the following 8 network
> > >> >addresses to them:
> > >> >
> > >> >10.24.32.0/24
> > >> >10.24.33.0/24
> > >> >10.24.34.0/24
> > >> >10.30.35.0/24
> > >> >10.30.36.0/24
> > >> >10.30.37.0/24
> > >> >10.30.38.0/24
> > >> >10.30.39.9/24
> > >> >
> > >> >What subnet mask could you use to minimize the
> > >> complexity
> > >> >of the routing tables while maintaining the existing
> > >> >Internet connectivity?
> > >> >
> > >> >a. 255.255.252.0
> > >> >b. 255.255.255.252
> > >> >c. 255.255.255.248
> > >> >d. 255.255.248.0 "
> > >> >
> > >> >The answer is 'd.' Quite frankly I do not understand
> > >> what
> > >> >the problem is getting at nor how 'd' gives the
> > desired
>
Marko 07-15-2003 09:37 AM
Re: Subnetting Problem
Jeffrey Woods has provided the best answer so far.
Your ISP may not be a total idiot for "assigning" this IP
space in the first place. It would only work on your
10.24or30.y.z network and possibly to reach the ISP's
servers / routers and that is likely it. However, if the
ISP provides mail and proxy servers (possibly other
services?) then this IP assignment is OK.
10.24.x.y to 10.30.a.b would require a mask of 255.248.0.0
to cover all those IP ranges mentioned.
The 10.24.x.y ranges could be covered by "route add
10.24.32.0 mask 255.255.253.0 (gateway IP here)"
....Assuming 10.30.39.9/24 is a typo and should be
10.30.39.0/24 (cause the first is definitely wrong and
does not exist)...
HOWEVER, if you concede that it is highly probable that
all the class C subnets (those finishing with /24) start
with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
255.255.248.0 will cover all IPs in the range 10.x.32.0 to
10.x.39.255. Making "d" the correct choice....
It it says d is right, then I would suggest the above
paragraph is the solution.
Anyone else care to have a go?
....care to kudos?
....care to flame?
Marko Cosic
>-----Original Message-----
>In article <08eb01c34a1a\$ff2229e0\$a601280a@phx.gbl>,
moon@lighting.com
>says...
>
>> You're right, Tom. /24 is 255.255.255.0. Any
explanation
>> why the answer is 255.255.248.0?
>
>Either the question has been mis-typed here or mis-
remembered.
>255.255.248 won't help cover all eight of those
>from 10.24.*.* to 10.30.*.*. You'd need at least a /13
(255.248.0.0)
>
>The question is also bogus because 10.*.*.* is private
NAT space, and
>would not be allocated by an ISP -- but even overlooking
that, the
>suggestion that you could simply apply one large mask to
the whole
>network and call it a day is specious at best. If you
were to do so,
>then people on this network would be unable to reach
people using the
>same ISP from elsewhere, i.e. 10.27.5.0/24 would be
unreachable.
>
>
>> >> >-----Original Message-----
>> >> >I need some help with the following subnet problem:
>> >> >
>> >> >"You are the new administrator of a 2000 node
network.
>> >> >There is only one router on the entire network,
which
>> >> >provides all the computers with Internet access. The
>> >> >company's ISP has assigned the following 8 network
>> >> >
>> >> >10.24.32.0/24
>> >> >10.24.33.0/24
>> >> >10.24.34.0/24
>> >> >10.30.35.0/24
>> >> >10.30.36.0/24
>> >> >10.30.37.0/24
>> >> >10.30.38.0/24
>> >> >10.30.39.9/24
>> >> >
>> >> >What subnet mask could you use to minimize the
>> >> complexity
>> >> >of the routing tables while maintaining the existing
>> >> >Internet connectivity?
>> >> >
>> >> >a. 255.255.252.0
>> >> >b. 255.255.255.252
>> >> >c. 255.255.255.248
>> >> >d. 255.255.248.0 "
>> >> >
>> >> >The answer is 'd.' Quite frankly I do not understand
>> >> what
>> >> >the problem is getting at nor how 'd' gives the
>> desired
>
>.
>
anees.network 12-12-2010 09:25 AM
I have something to say here.
Let me try an answer for this
This questions involves something called "ROUTE Summarization".
10.24.32.0/24
10.24.33.0/24
10.24.34.0/24
10.30.35.0/24
10.30.36.0/24
10.30.37.0/24
10.30.38.0/24
10.30.39.9/24
these are the IP Addresses given. The questions says "What subnet mask could you use to minimize the complexity of the routing tables while maintaining the existing Internet connectivity?"
so instead of just sub-netting you have to take care of minimizing the routing table it generates.
So we have to make this pool of IPs into two groups. 10.24 and 10.30 group.
10.24 has 3 IPs in the pool
10.30 has 5 ips in the pool so we have to consider grouping 5 IP pool to minimize the routing table.
10.30. 35-39 is the range - group of 8 ip address will accommodate this range
let me do this in binary
To get a 8 ip range we have to start from the 5th binary digit 11111000=248 but we need this 248 in the 3rd octet mask to accomplish the range we needed because first two octets are same ( 10.30.)
so the mask should be 255.255.248.0.
eventually,
10.30.35.0/24
10.30.36.0/24
10.30.37.0/24 ---->255.255.248.0 is the mask you need
10.30.38.0/24
10.30.39.9/24
For 10.24 series you can use the same mask because it lies within the range but route summarization is not needed as there are only 3 IPs in this range.
What do you guys think?
Kirsteins 03-10-2012 07:38 PM
I really recommend using this too to do route summarize: http://www.subnet-calculator.org/supernets.php
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# prime number problem
Robert Fuhrman
Greenhorn
Posts: 3
Hi, hope you can help me out here. I need to generate all the prime number factors for the number 999999 (3, 7, 11, 13, 37). Here is my algorithm:
step n k comment
------------------------------------------------
0 999999 2 not divisible by 2, k = k + 1
1 999999 3 n is divisible by 3, n = n / 3, print 3
2 333333 3 n is divisible by 3, n = n / 3, print 3
3 111111 3 n is divisible by 3, n = n / 3, print 3
4 37037 3 not divisible by 3, k = k + 1
5 37037 4 not divisible by 4, k = k + 1
6 37037 5 not divisible by 5, k = k + 1
7 37037 6 not divisible by 6, k = k + 1
8 37037 7 n is divisible by 7, n = n / 7, print 7
9 5291 7 not divisible by 7, k = k + 1
10 5291 8 not divisible by 8, k = k + 1
11 5291 9 not divisible by 9, k = k + 1
12 5291 10 not divisible by 10, k = k + 1
13 5291 11 n is divisible by 11, n = n / 11, print 11
14 481 11 not divisible by 11, k = k + 1
15 481 12 not divisible by 12, k = k + 1
16 481 13 n is divisible by 13, n = n / 13, print 13
17 37 13 STOP since 13^2 > 37 print 37
I had no problem generating all the factors for 999999 - but I can't seem to get my isPrime method to work, included below is a copy of my code. It won't compile because apparently I have erros on lines 18 and 33, 18 is the beginning of the isPrime method, and 33 apparently I need a ; but I can't figure out why I could possibly need one there. Hope you all can help me out. Thanks in advance
{
public static void main (String[] args)
{
int n=999999;
System.out.println(n+ "'s factors are ");
for (int i = 1; i <= n / 2; i++)
{
if (n % i == 0 && isPrime(i))
{
System.out.print(i+ " ");
}
}
public isPrime()
{
int n=number;
int j;
for (j=2; j<i; j++)
if (j%i != 0)
System.out.print(" ");
j++;
else
j++;
}
}
}
Garrett Rowe
Ranch Hand
Posts: 1296
Hey Robert, there are a few things going on with this code. I've tried to highlight the errors I see, but there may be more.
Garrett
[ March 06, 2006: Message edited by: Garrett Rowe ]
Robert Fuhrman
Greenhorn
Posts: 3
Hi Garrett, thanks for your input. Am I correct in what I am attempting to do though? In the main method I am trying to get all the factors of 999999 which I think is correct...my isPrime method is probably totally screwed up lol - what I am trying to do is take the factor of 999999 that I found in the main method and test it to see if its prime - I thought I could accomplish this by that for statement i have in there. I used i in the statement because i is the factor of n , then using that loop to test for its factors. Once I do test for it's factors and find them, do I need to make another method for output?Is this the right way to approach this problem or am I way off base?
Garrett Rowe
Ranch Hand
Posts: 1296
The way your code is set up will work. For reusability though, I would consider making three seperate methods.
Robert Fuhrman
Greenhorn
Posts: 3
Garrett, I can't seem to get the factor from my main method to test in the isPrime method...any ideas?
Garrett Rowe
Ranch Hand
Posts: 1296
What does your main() method and isPrime() method look like. What problems are you having? | 1,014 | 3,215 | {"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} | 3.609375 | 4 | CC-MAIN-2016-44 | longest | en | 0.888098 |
https://us.metamath.org/qleuni/i2i1.html | 1,721,248,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00395.warc.gz | 522,114,641 | 3,197 | Quantum Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > QLE Home > Th. List > i2i1 GIF version
Theorem i2i1 267
Description: Correspondence between Sasaki and Dishkant conditionals. (Contributed by NM, 7-Feb-1999.)
Assertion
Ref Expression
i2i1 (a2 b) = (b1 a )
Proof of Theorem i2i1
StepHypRef Expression
1 ax-a1 30 . . 3 a = a
21ud2lem0b 259 . 2 (a2 b ) = (a 2 b )
3 ax-a1 30 . . 3 b = b
43ud2lem0a 258 . 2 (a2 b) = (a2 b )
5 i1i2 266 . 2 (b1 a ) = (a 2 b )
62, 4, 53tr1 63 1 (a2 b) = (b1 a )
Colors of variables: term Syntax hints: = wb 1 ⊥ wn 4 →1 wi1 12 →2 wi2 13 This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 This theorem depends on definitions: df-a 40 df-i1 44 df-i2 45 This theorem is referenced by: nom40 325 nom41 326 nom42 327 nom43 328 nom44 329 nom45 330 nom50 331 nom51 332 nom52 333 nom53 334 nom54 335 nom55 336 oal2 999
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https://byjus.com/question-answer/the-ratio-of-number-of-sides-of-two-regular-polygons-are-as-5-4-and/ | 1,680,312,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00553.warc.gz | 185,707,145 | 23,596 | Question
# The ratio of number of sides of two regular polygons are as 5:4 and the difference of there exterior angles is 9o. Find the number of sides of both the polygon.
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Solution
## Let n be the Greatest Common Divisor (GCD) of the numbers under the question.Then one polygon has 5n sides, while the other has 4n sides.It is well known fact that the sum of exterior angles of each (convex) polygon is 360o.So, the exterior angle of the regular 5n-sided polygon is 360o5n.Similarly, the exterior angle of the regular 4n-sided polygon is 360o4n.The difference between the corresponding exterior angles is 9o.⇒ 360o4n−360o5n=9o⇒ 14n−15n=9o360o⇒ 5n−4n20n2=140⇒ n20n2=140⇒ 1n=12∴ n=2⇒ Number of sides =4n=2×4=8 and 5n=2×5=10.
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# Chapter Four Fluid Dynamic
## Chapter Four Fluid Dynamic
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##### Presentation Transcript
1. Chapter FourFluid Dynamic What are the types of flow Conservation equation (mass, momentum, and energy) Types of friction Boundary Layer Flow in non-circular pipe Multiple pipe system Unsteady state flow References: Streeter,V. ”Fluid Mechanic”,3rd edition, Mc-Graw Hill, 1962. Frank M. White “Fluid Mechanics” 5th edition McGraw Hill. Coulson, J.M. and J.F. Richardson, “Chemical Engineering”, Vol.I “ Fluid Flow, Heat Transfer, and Mass Transfer” 5th edition, (1998).
2. . Fluid mechanics is the study of fluids and the forces on them • - The fluid motion is generated by pressure difference between two points and is constrained by the pipe walls. The direction of the flow is always from a point of high pressure to a point of low pressure. • - If the fluid does not completely fill the pipe, such as in a concrete sewer, the existence of any gas phase generates an almost constant pressure along the flow path. • - If the sewer is open to atmosphere, the flow is known as open-channel flow and is out of the scope of this chapter or in the whole course.
3. Types of Flow • Flow in pipes can be divided into two different regimes, i.e. laminar and turbulence. • The experiment to differentiate between both regimes was introduced in 1883 by Osborne Reynolds (1842–1912), an English physicist who is famous in fluid experiments in early days. • The velocity, together with fluid properties, namely density and dynamic viscosity , as well as pipe diameter D, forms the dimensionless Reynolds number, that is • From Reynolds’ experiment, he suggested that Re < 2100 for laminar flows and Re > 4000 for turbulent flows. The range of Re between 2100 and 4000 represents transitional flows
4. Types of Flow
5. Example • Consider a water flow in a pipe having a diameter of D = 20 mm which isintended to fill a 0.35 liter container. Calculate the minimum time required if the flow is laminar, and the maximum time required if the flow is turbulent. • Use density = 998 kg/m3 and dynamic viscosity = 1.1210–3 kg/ms.
6. Governing Equations • – Mass cannot be created or destroyed → Continuity Equation • – F=ma (Newton’s 2ndlaw) → Momentum Equation • – Energy cannot be created or destroyed → Energy Equation
7. 1- Continuity Equation ( Overall Mass Balance) • Its also called (conservation of mass) For incompressible fluid (the density is constant with velocity) then
8. 2 Momontume Equation and Bernoulli Equation • Its also called equation of motion • consider a small element of the flowing fluid as shown below, • Let: • dA: cross-sectional area of the fluid element, • dL: Length of the fluid element’ • dW: Weight of the fluid element’ • u: Velocity of the fluid element’ • P: Pressure of the fluid element dL
9. Assuming that:p;;l • the fluid is steady, • non-viscous (the frictional losses are zero), • incompressible (the density of fluid is constant) • .
10. The forces on the cylindrical fluid element are, • 1- Pressure force acting on the direction of flow (PdA) • 2- Pressure force acting on the opposite direction of flow [(P+dP)dA] • 3- A component of gravity force acting on the opposite direction of flow (dW sin θ) • Hence, the total force = gravity force + pressure force
11. dP/ ρ + udu + dz g = 0 ---- Euler’s equation of motion • Bernoulli’s equation could be obtain by integration the Euler’s equation • ∫dP/ ρ + ∫udu + ∫dz g = constant • ⇒ P/ ρ + u2/2 + z g = constant • ⇒ ΔP/ ρ + Δu2/2 + Δz g = 0 --------- Bernoulli’s equation
12. In general • In the fluid flow the following forces are present: - • 1- Fg ---------force due to gravity • 2- FP ---------force due to pressure • 3- FV ---------force due to viscosity • 4- Ft ---------force due to turbulence • 5- Fc ---------force due to compressibility • 6- Fσ ---------force due to surface tension
13. 3- Energy Equation and Bernoulli Equation • The total energy (E) per unit mass of fluid is given by the equation: - • E1 + ∆q + ∆w1 = E2 + ∆w2 • where • ∆q represents the heat added to the fluid • ∆w1 represents the work added to the fluid like a pump • ∆w2 represents the work done by the fluid like the work to overcome the viscose or friction force • E is energy consisting of:
14. Internal Energy (U) • This is the energy associated with the physical state of fluid, i.e. the energy of atoms and molecules resulting from their motion and configuration. Internal energy is a function of temperature. It can be written as (U) energy per unit mass of fluid. • Potential Energy (PE) • This is the energy that a fluid has because of its position in the earth’s field of gravity. The work required to raise a unit mass of fluid to a height (z) above a datum line is (zg), where (g) is gravitational acceleration. This work is equal to the potential energy per unit mass of fluid above the datum line. • Kinetic Energy (KE) • This is the energy associated with the physical state of fluid motion. The kinetic energy of unit mass of the fluid is (u2/2), where (u) is the linear velocity of the fluid relative to some fixed body. • Pressure Energy (Prss.E) • This is the energy or work required to introduce the fluid into the system without a change in volume. If (P) is the pressure and (V) is the volume of a mass (m) of fluid, then (PV/m ≡ Pυ) is the pressure energy per unit mass of fluid. The ratio (m/V) is the fluid density (ρ).
15. In the case of: • No heat added to the fluid • The fluid is ideal • There is no pump • The temperature is constant along the flow • Then • ⇒ ΔP/ ρ + Δu2/2 + Δz g = 0 --------- Bernoulli’s equation
16. Modification of Bernoulli’s Equation • 1- Correction of the kinetic energy term • α = 0.5 for laminar flow • - α = 1.0 for turbulent flow • 2- - Modification for real fluid • Thus the modified Bernoulli’s equation becomes, • P1/ ρ + u12/2 + z1 g = P2/ ρ + u22/2 + z2 g + F ---------(J/kg ≡ m2/s2)
17. 3- Pump work in Bernoulli’s equation • Frictions occurring within the pump are: - • Friction by fluid • Mechanical friction • Since the shaft work must be discounted by these frictional force (losses) to give net mechanical energy as actually delivered to the fluid by pump (Wp). • Thus, Wp= η Ws where η, is the efficiency of the pump.
18. P1/ ρ + u12/2 + z1 g + η Ws = P2/ ρ + u22/2 + z2 g + F ---------(J/kg ≡ m2/s2) • By dividing each term of this equation by (g), each term will have a length units, and the equation will be: - • P1/ ρg + u12/2g + z1 + η Ws /g = P2/ ρg + u22/2g + z2 + hf ---------(m) • where hF = F/g ≡ head losses due to friction.
19. 4 Friction in Pipes
20. Relation between Skin Friction and Wall Shear Stress – dPfs = 4(τ dL/d) = 4 (τ /ρ ux2) (dL/d) ρ ux2 where, (τ /ρ ux2) = Φ=Jf =f/2 =f′/2 Φ(or Jf): Basic friction Factor f: Fanning (or Darcy) friction Factor f′: Moody friction Factor –ΔPfs= 4f (L/d) (ρu2/2) ---------------------(Pa) The energy lost per unit mass Fs is then given by: Fs = (–ΔPfs/ρ) = 4f (L/d) (u2/2) -----------------(J/kg) or (m2/s2) The head loss due to skin friction (hFs) is given by: hFs= Fs/g = (–ΔPfs/ρg) = 4f (L/d) (u2/2g) ---------------(m)
21. Evaluation of Friction Factor in Straight Pipes • Velocity distribution • in laminar flow ⇒ ux = [(-ΔPfs R2)/(4L μ)][1– (r/R)2] velocity distribution (profile) in laminar flow ⇒ umax = [(–ΔPfs d2)/(16 L μ)] ----------centerline velocity in laminar flow ∴ ux / umax = [1–(r/R)2] ---------velocity distribution (profile)in laminar flow
22. ux / umax = [1–(r/R)]1/7Prandtl one-seventh law equation. (velocity distribution profile)in turbulent flow • Velocity distribution • in turbulent flow
23. Average (mean) linear velocity • in laminar flow • in Turbulent flow u = umax/2 = [(–ΔPfs R2)/(8L μ)] = [(–ΔPfs d2)/(32 L μ)] ∴ –ΔPfs= (32 L μ u) / d2 Hagen–Poiseuille equation ∴ u = 49/60 umax ≈ 0.82 umax ------------average velocity in turbulent flow
24. Friction factor • in laminar flow • in Turbulent flow • for 2,500 < Re <100,000 • Or • and, for 2,500 < Re <10,000,000 • These equations are for smooth pipes in turbulent flow ∴ f = 16 / Re Fanning or Darcy friction factor in laminar flow. Or F = 64/Re
25. For rough pipes, the ratio of (e/d) acts an important role in evaluating the friction factor in turbulent flow as shown in the following equation
26. Graphical evaluation of friction factor
27. Form Friction • - Sudden Expansion (Enlargement) Losses • Sudden Contraction Losses
28. Form Friction • Losses in Fittings and Valves
29. Butterfly valve Typical commercial valve geometries: (a) gate valve; (b) globe valve; (c) angle valve; (d) swing-check valve; (e) disk type gate valve. Total Friction Losses
30. Example 4 • Determine the velocity of efflux from the nozzle in the wall of the reservoir of Figure below. Then find the discharge through the nozzle. Neglect losses. • Example 5 • Oil, with 900 kg/m and 0.00001 m/s, flows at 0.2m3 /s through 500 m of 200-mmdiameter cast-iron pipe. Determine (a) the head loss and (b) the pressure drop if the pipe slopes down at 10° in the flow direction. • Example 6 • A pump draws 69.1 gal/min of liquid solution having a density of 114.8 lb/ft3 from an open storage feed tank of large cross-sectional area through a 3.068″I.D. suction pipe. The pump discharges its flow through a 2.067″I.D. line to an open over head tank. The end of the discharge line is 50′ above the level of the liquid in the feed tank. The friction losses in the piping system are F = 10 ft lbf/lb. what pressure must the pump develop and what is the horsepower of the pump if its efficiency is η=0.65.
31. Example 7 • The siphon of Fig. 3.14 is filled with water and discharging at 2.80 cfs. Find the losses from point 1 to point 3 in terms of velocity head u2/2g. find the pressure at point 2 if two-third of the Losses occur between points I and2 • Example 8 • A conical tube of 4 m length is fixed at an inclined angle of 30° with the horizontal-line and its small diameter upwards. The velocity at smaller end is (u1 = 5 m/s), while (u2 = 2 m/s) at other end. The head losses in the tub is [0.35 (u1-u2)2/2g]. Determine the pressure head at lower end if the flow takes place in down direction and the pressure head at smaller end is 2 m of liquid. • Example 9 • A pump developing a pressure of 800 kPa is used to pump water through a 150 mm pipe, 300 m long to a reservoir 60 m higher. With the valves fully open, the flow rate obtained is 0.05 m3/s. As a result of corrosion and scalling the effective absolute roughness of the pipe surface increases by a factor of 10 by what percentage is the flow rate reduced. μ= 1 mPa.s
32. Example 10 • A liquid of specific weight g 58 lb/ft3 flows by gravity through a 1-ft • tank and a 1-ft capillary tube at a rate of 0.15 ft3/h, as shown in • Fig. Sections 1 and 2 are at atmospheric pressure. • Neglecting entrance effects, compute the viscosity of the liquid. • Example 11 • 630 cm3/s water at 320 K is pumped in a 40 mm I.D. pipe through a length of 150 m in horizontal direction and up through a vertical height of 10 m. In the pipe there is a control valve which may be taken as equivalent to 200 pipe diameters and also other fittings equivalent to 60 pipe diameters. Also other pipe fittings equivalent to 60 pipe diameters. Also in the line there is a heat exchanger across which there is a loss in head of 1.5 m H2o. If the main pipe has a roughness of 0.0002 m, what power must supplied to the pump if η = 60%, μ= 0.65 mPa.s. Example 12 A pump driven by an electric motor is now added to the system. The motor delivers 10.5 hp. The flow rate and inlet pressure remain constant and the pump efficiency is 71.4 %, determine the new exit pressure.
33. Example 13 An elevated storage tank contains water at 82.2°C as shown in Figure below. It is desired to have a discharge rate at point 2 of 0.223 ft3/s. What must be the height H in ft of the surface of the water in the tank relative to discharge point? The pipe is schedule 40, e = 1.5 x10-4 ft. Take that ρ = 60.52 lb/ft3, μ= 2.33 x10-4 lb/ft.s. Example 14 • Water, 1.94 slugs/ft and 0.000011 ft/s, is pumped between two reservoirs at 0.2 ft/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig.. The roughness ratio is /d 0.001. Compute the pump horsepower required. Example 15 The pump in Fig. E3.20 delivers water (62.4 lbf/ft3) at 3 ft3/s to a machine at section 2, which is 20 ft higher than the reservoir surface. The losses between 1 and 2 are given by hf =_ Ku2 /(2g), where K _ 7.5 is a dimensionless loss coefficient. Take α= 1.07. Find the horsepower required for the pump if it is 80 percent efficient.
34. The Boundary Layer Boundary layer for flow on flat plate
35. The Boundary Layer Developing velocity profiles and pressure changes in the entrance of a duct flow
36. The Boundary Layer • For fully developed velocity profile to be formed in laminar flow, the approximate entry length (Le) of pipe having diameter d, is: - • Le/d = 0.0575 Re -------------------laminar • In turbulent flow the boundary layers grow faster, and Le is relatively shorter, according to the approximation for smooth walls • ------------------turbulent
37. The Boundary Layer • Example • A 0.5in-diameter water pipe is 60 ft long and delivers water at 5 gal/min at 20°C. What fraction of this pipe is taken up by the entrance region?
38. Flow in non-circular pipe • If the duct is noncircular, the analysis of fully developed flow follows that of the circular pipe but is more complicated algebraically. For laminar flow, one can solve the exact equations of continuity and momentum. For turbulent flow, the logarithm-law velocity profile can be used, or (better and simpler) the hydraulic diameter is an excellent approximation.
39. Multiple pipe system
40. Example: Given is a three-pipe series system. The total pressure drop is pA-pB=150,000 Pa, and the elevation drop is zA-zB=5m . The pipe data are • The fluid is water, . Calculate the flow rate Q in m3/h through the system. • Example: Assume that the same three pipes in above example are now in parallel with the same total head loss of 20.3 m. Compute the total flow rate Q, neglecting minor losses. | 3,969 | 14,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-25 | latest | en | 0.905469 |
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Unformatted text preview: xv Test Form A September 13, 2004 Principles of Microeconomics Name Qfl’imflzfl Ari. dM’ s Emmi, GT in bi (i) ii I i .i i L. i l 2 3 4 5 b urs per month) A w v Quantity (millions of ho Mzww v 1) In the figure above, if the minimum wage is \$4 per hour, then y—L \_r A) the quantity of labor supplied is 3 million hours and the quantity of labor demanded is 3 million hours. B) unemployment is 1 million hours. C) the quantity of labor demanded is 4 million hours and the quantity of labor supplied is 2 million hours. D) the quantity of labor supplied is 4 million hours and the quantity of labor demanded is 2 million hours. ” ‘ 3 : »~t’~.~\$“ ,zzmfiéfiiwmmm 'Awlfémafilfiwfly S + tax on seller 1:. O (a) 0 Price {doliors per CD) N O 10 L i .1 J... 1 _|.____...«._L 0 10 20 30 40 50 60 7O 80 Quantity (minions of CD5 per year) 24; t»: :‘z‘ firm A,» r71 2) In the above figure, the price paid by the buyer before the tax is per compact disc, and the 2) ,3 price paid by the buyer after the tax is per compact disc. A) \$30; \$30 B) \$20; \$30 C) \$20; \$20 D) \$30; \$20 3) In the above figure, what is the amount (size) of the tax per compact disc? 3) i A) \$40 B) \$30 C) \$20 D) \$10 WWWWIR Ln) 0 Price (dollars per pair of gloves) 25 ‘ \ ( H 20 X L- ,1 l5 \ , / D 3 MB /’v\ J. \ \ / \/ 3 4 5 (thousands of gloves per day) ~m»;»mW~/m 4) 1n the above figure, if the production of gloves was restricted to 2,000 a day, then the deadweight loss would equal A) \$0, because 2,000 gloves per day is an efficient quantity of gloves to produce. B) \$2,000. C) \$10,000. D) \$5,000. 5) In the above figure, when the efficient quantity of gloves is produced, die producer surplus equals A) \$22,500. B) \$15,000. C) \$45,000. D) \$3,000. 6) If the (marginal) gain that consumers receive from the fifth slice of pizza is greater than the marginal cost of the fifth slice of pizza then the output level is A) efficient and less pizza should be produced. B) inefficient and more pizza should be produced. C) efficient and more pizza should be produced. D) inefficient and less pizza should be produced. 7) A tax is imposed on the sellers of gasoline. This tax shifts A) the supply of gasoline curve downward. B) both. the supply of gasoline and demand for gasoline curves leftward. C) the demand for gasoline curve upward. D) the supply of gasoline curve upward. 8) Nick can purchase each milkshake for \$2. For the first milkshake purchased Nick is willing to pay \$4, for the second milkshake \$3, for the third milkshake \$2 and for the fourth milkshake \$1. What is the value of Nick's consumer surplus? A) \$10 is) (“>07 D) 9 I . 8) 9) A price floor set below the equilibrium price 9) b A) restricts the quantity supplied but not the quantity demanded. B) has no effect. C) restricts the quantity demanded but not the quantity supplied. D) restricts both the quantity demanded and the quantity supplied. 10) Suppose a country produces only bikes and clothing. The country achieves an efficient allocation of 10) /\/ resources when A) it produces equal amount of bikes and clothes. R\ if nan't rnAnr-n an u] 1 u \. n y mn‘ro In Lull u FLU H‘— ulvu, u in 1) C) the prices charged for the goods are as low as possible. D) there is no deadweight loss. ‘ ,‘T‘KT‘rD‘, XX“? "" 'Féhfifl<W“”""~"?\$3% 0:: C) C: 0 O 0 Rent (dollars per month) 3' o 500 1100 11) In the figure above, originally the apartment rental market is in equilibrium with a rental price of 11) A \$600 per month. Then the government imposes a rent ceiling of \$500 per month. If the law is enforced, the maximum for which an apartment will rent on the black market is strict! v J A \ r-Inn , A ,_ .1, n\ , .I,, mr—rnn , .1 A) IUU PEI HIUHUL DI IIIUIC “16111 31/ UU per (“Until \$ C) \$600 per month. D) less than \$600 per month. S=MC Price {dollars per purse} 1:. o T 2 a". r J e l f 1 O l 00 200 300 1100 500 60 Quantity (thousands of purses per month) <3 EL mm» , away“; 12) In the above figure, 300,000 purses per month is 12) A) the efficient amount to produce because consumer surplus is maximized. B) a is not maximized. . i efficient C) the efficient amount to produce because the sum of consumer surplus and producer surplus is maximized. D) an inefficient amount to produce because consumer surplus is not maximized. 13) In the market for CD3, the producer surplus will decrease if . 13) _.. A) the mar inal cost of a CD decreases B) twe market price of a CD ~ncreases O D C) the demand for CD5 decreases D) the supply of CD5 increases 'E 3 r -.-. o. n l/ 2 A . 1,, /, 2 t4 _ *5 7 *0 x w t. 0.) .2 a: t 6" ; _ a / i U 3 \{l ‘ t/ ,a : \\1 i / 4/ § L, l 1 i; O l O 20 30 40 50 60 Quantity (units) 14) In the above figure, suppose that the government sets a quota at 10 units of output and the price rises to \$4. In comparison to a competitive market the consumer surplus would fall by A) \$20. B) \$15. C) \$0. D) \$10. 15) Surnlus nrodi. ’ r r A) minus its marginal cost of production B) subtracted f‘or‘t the value of the good C) times the quantity sold D) plus 9 consumer surplus 3% Pryce (dollars per ton of wheat) 5 O Q j _,i i 3 4 5 6 Quantity (tons ol wheat per year) r " :msut' 4? ' W ,, 5w 16) In the above figure, the lowest price for which the firm will sell its second ton of wheat is A) \$25. B) \$50. C) \$100. D) \$75. m 14) J1 15> A I i ‘3 16) 17) The deadweight loss from producing an inefficient amount is 17) l J A) a gain to the consumer and the producer, but a loss to the rest of society. B) a loss to the producer but a gain to the consumer. C) a loss to the consumer but a gain to the producer. D) a loss to the consumer and to the producer. 5" Ln 0 I Price (dollars per bcgle} Ix) o O r; M B r i ; L. 0 lo 20 3 0 Quantity lbagles per hour) :‘ w - >2 re» s '1‘ J»; 4&5 u f‘) 18) The figure illustrates the market for bagels. If the number of bagels is cut from 20 to 10 an hour, the 18) D deadweight loss is . A) —\$5.00 an hour B) \$5.00 an hour C) \$0 an hour D) \$0.50 a bagel 19) As we move up along the supply curve for hot dogs, 19) f—«E A) the marginal cost of hot dogs increases. B) the value of hot dogs decreases. C) the maximum price that people are willing to pay for hot dogs increases. \ L r -rdmud 1,, H. MAI"; _ r I- A. .1. AA LA -1“ _ D} tilt: LUllbUlllfil bulplub U1 llUl U053 lIlL'l'Cd 'Cb- § 41“ MC / / .0 (.n O Marginal cost and marginal benefit // / MB 1 l J 2 4 6 8 Quontiiy [thousands of hot dogs per day) “‘3‘an /r”‘r.?\$“%‘i/;\ (dollars worth of goods and sewices) S o O “8‘ i 20) In the above figure, what is the efficient quantity of hotdogs to produce? A) 6 thousand per day B) 2 thousand per day C) 4 thousand per day D) The efficient quantity cannot be determined without knowing the income distribution of this economy. ()0 ...
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http://www.ehow.co.uk/how_5791925_use-percent-discount-excel.html | 1,480,758,601,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00418-ip-10-31-129-80.ec2.internal.warc.gz | 453,250,921 | 16,517 | # How to Use Percent Discount in Excel
Written by mark kennan
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Microsoft Excel is a spreadsheet program that can be very useful for rapidly entering and calculating percentage discounts. You can quickly and easily set up a spreadsheet where you can enter an item's original price and the per cent discount, and Microsoft Excel will automatically calculate the final price.
Skill level:
Moderately Easy
### Things you need
• Computer
• Microsoft Excel
## Instructions
1. 1
Enter "Original Price" in cell A1, "Percent Discount" in cell A2 and "Discounted Price" in cell A3.
2. 2
Enter the original price of the item in cell B1. For example, if the item was £32, you would enter "50."
3. 3
Enter the percentage discount as a decimal in cell B2. For example, if the discount was 35 per cent, you would enter 0.35.
4. 4
Enter "=B1*(1-B2)" in cell B3 to calculate the final price of the item.
#### Tips and warnings
• Make sure you enter the percentage as a decimal or your final price will be off.
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HomeTechnologyWhat is the Square Root of 8?
# What is the Square Root of 8?
When searching for the square root of 8, various methods are available. To quickly arrive at an answer, calculators may be an efficient method; alternatively, you could also try long division.
Parents and teachers can engage kids with math games to help them comprehend the concept of numbers, ultimately improving their ability to solve problems and learn.
Contents
## Rational or irrational?
A square root is any number that yields its original number when multiplied by itself. It may be represented as either fractions or radicals. For instance, 8’s square root is 2.828; this actual number can be found using either calculators or manual methods; students must know how to locate square roots when needed.
Irrational numbers are defined as numbers that cannot be expressed as fractions. They have non-terminating decimals with non-repeating digits, making them impossible to divide by the fraction they would create when written out fractionally. Although such irrational numbers are typically challenging to deal with in digital formats, they still exist and outnumber rational ones!
There are various approaches to identifying whether a number is rational or irrational, with the most accessible being to calculate its most common form on either a calculator or by hand, depending on your problem. Once this step has been completed, simplifying it may make the number more rational through long division and prime factorization methods, among others.
The square root of 8 is rational because it forms an ideal square. However, when expressed as a fraction, it does not adhere to logic as its non-terminating and non-repeating digits make the number non-rational and thus authentic.
Osmo’s Multiplication Table for Kids can help children develop strong math skills while forming great habits.
Real numbers can be written as ratios between integers or fractions with whole numbers as their denominators, including rational and irrational real numbers like 2 and 8. Specifically, square root eight can be considered unreasonable because its decimals never stop repeating themselves – something two does not do!
## Long division method
The long division method is an established way of dividing numbers. This approach involves first subdividing a dividend into smaller portions to form its divisor before multiplying it by itself and adding any remainder as fractions (r/m). This approach is constructive when working with more significant numbers; students learn the division process, an essential skill needed for algebra and advanced mathematics courses.
One of the critical aspects of long division is understanding and managing remainders, which may prove challenging for young children without much experience with division. If students cannot comprehend how to approach remainders, they could feel confused and frustrated, leading to poor confidence and performance if left alone to handle these problems. Therefore we must teach our young students how to tackle them using simple steps.
Students should first ensure the division problem is correctly set up. The dividend should be placed beneath a division symbol on the right, while the divisor should be on the left. They should write out the first term of the quotient on top and divide the second term of division by divisor for the third term of the quotient. They must ensure each step’s remainders are less than the divisor; otherwise, it indicates an incorrect multiplication operation or that a more considerable quotient is necessary.
Students can practice this method with Wiingy’s free online long division worksheets, which help develop their skills while increasing their understanding of decimals and place value. Once students have practiced with several numbers, they can move on to more complex problems.
Teaching students about rational numbers can also be invaluable. Rational numbers refer to those that can be divided into multiples of integers; the square root is an example of such a reasonable number because it can be divided into a perfect square.
One practical approach for teaching kids about the square roots of numbers is through approximation. This simple method involves guessing their square root by dividing non-perfect numbers by perfect squares that are either less or greater than them and taking an average. This gives them a rough estimate and helps prevent getting stuck on problems.
## Prime factorization method
Are You Searching For the Square Root Of A Number? The division and factorization methods can quickly help you estimate its square root. Both techniques can soon accurately estimate any number’s square root.
Utilizing the factorization method, begin by dividing any given number by a prime number that divides it ultimately and multiplying that quotient by each prime factor until all have been identified; once complete, add these back together for your answer – making this an extremely efficient method when working with large numbers.
This method can also calculate the HCF and LCM of two numbers. An HCF of a number is its most significant common factor in both numbers, while its LCM is its smallest common multiple.
A factor tree method of calculating square roots of numbers is a more accessible and more straightforward approach suitable for beginners. You need only identify the divisors of a number and write them as branches on your factor tree before dividing each number by its divisors, noting the remainder each time. You can repeat this process until you get to your desired result.
If you want to simplify a number, one way is to divide it by one of its perfect square roots identified. This will generate a number closer to its original form before taking an average between the result and root to produce your final number – making it simpler and easier to read/understand while helping solve complex problems.
## Repeated subtraction method
Trying to determine what the square root of 8 is can be made simple using these steps. First, divide by repeated subtraction before moving on to division with simple removal; you should soon have your answer! This method also teaches children the fundamentals of division in an accessible and straightforward manner and provides them with the practice for more extended division methods like long division.
A square root of a number is defined as the value that results from multiplying its original number by itself, represented by either a radical symbol () or radix. The number beneath the radical sign, the radicand, can be whole numbers or decimals depending on context, such as an answer to an equation or a remainder for calculations.
To calculate the square root of any number, start by identifying its dividend and divisor. Next, divide by subtracting the divisor from the tip until the result is less than the divisor, and finally, remove the divisor from the remainder until the result equals the square root of the original number.
The repeated subtraction method is a practical starting point when calculating a number’s square root. You can then go on to the other forms, which are more accessible and suitable for beginners as it’s more straightforward and user-friendly.
The repeated subtraction method is an invaluable way to solve complex equations and numbers, much like its chemical equivalent, but using square roots of numbers instead. Like chemistry uses fractions to divide more significant numbers into smaller ones, this technique can be applied across all kinds of numbers, from large with many digits to multivariable problems, and even find decimals’ roots! Furthermore, repeated subtraction provides an effective technique for manually dividing complex numbers using calculators.
RELATED ARTICLES | 1,439 | 7,838 | {"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} | 4.71875 | 5 | CC-MAIN-2023-40 | latest | en | 0.954394 |
https://metanumbers.com/23648 | 1,601,310,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00050.warc.gz | 468,987,505 | 7,495 | ## 23648
23,648 (twenty-three thousand six hundred forty-eight) is an even five-digits composite number following 23647 and preceding 23649. In scientific notation, it is written as 2.3648 × 104. The sum of its digits is 23. It has a total of 6 prime factors and 12 positive divisors. There are 11,808 positive integers (up to 23648) that are relatively prime to 23648.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 23
• Digital Root 5
## Name
Short name 23 thousand 648 twenty-three thousand six hundred forty-eight
## Notation
Scientific notation 2.3648 × 104 23.648 × 103
## Prime Factorization of 23648
Prime Factorization 25 × 739
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 1478 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 23,648 is 25 × 739. Since it has a total of 6 prime factors, 23,648 is a composite number.
## Divisors of 23648
1, 2, 4, 8, 16, 32, 739, 1478, 2956, 5912, 11824, 23648
12 divisors
Even divisors 10 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 46620 Sum of all the positive divisors of n s(n) 22972 Sum of the proper positive divisors of n A(n) 3885 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 153.779 Returns the nth root of the product of n divisors H(n) 6.087 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 23,648 can be divided by 12 positive divisors (out of which 10 are even, and 2 are odd). The sum of these divisors (counting 23,648) is 46,620, the average is 3,885.
## Other Arithmetic Functions (n = 23648)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 11808 Total number of positive integers not greater than n that are coprime to n λ(n) 5904 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2629 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 11,808 positive integers (less than 23,648) that are coprime with 23,648. And there are approximately 2,629 prime numbers less than or equal to 23,648.
## Divisibility of 23648
m n mod m 2 3 4 5 6 7 8 9 0 2 0 3 2 2 0 5
The number 23,648 is divisible by 2, 4 and 8.
## Classification of 23648
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Frugal
## Base conversion (23648)
Base System Value
2 Binary 101110001100000
3 Ternary 1012102212
4 Quaternary 11301200
5 Quinary 1224043
6 Senary 301252
8 Octal 56140
10 Decimal 23648
12 Duodecimal 11828
20 Vigesimal 2j28
36 Base36 i8w
## Basic calculations (n = 23648)
### Multiplication
n×i
n×2 47296 70944 94592 118240
### Division
ni
n⁄2 11824 7882.67 5912 4729.6
### Exponentiation
ni
n2 559227904 13224621473792 312735848612233216 7395577347982091091968
### Nth Root
i√n
2√n 153.779 28.7033 12.4008 7.49478
## 23648 as geometric shapes
### Circle
Diameter 47296 148585 1.75687e+09
### Sphere
Volume 5.53952e+13 7.02747e+09 148585
### Square
Length = n
Perimeter 94592 5.59228e+08 33443.3
### Cube
Length = n
Surface area 3.35537e+09 1.32246e+13 40959.5
### Equilateral Triangle
Length = n
Perimeter 70944 2.42153e+08 20479.8
### Triangular Pyramid
Length = n
Surface area 9.68611e+08 1.55854e+12 19308.5
## Cryptographic Hash Functions
md5 05a4be7f376456730bdcb93e2abc9e39 db9f4032171dc152a1309422c92247121d8123dc 20cba91025290d0e423eb4294b03363e0a73a0be3ef28385268af8717b02dfed 4a6c0bb362c82a292a0748552200f54cd1870c53999f8e55eb947b729bb8801f43eaf4ecf9f252c1f185bdcf2ebf101aaef190f1566d7f9ba0bb23012b53bcfb 347d688bab464215444ef9dcc1171596f9f6470d | 1,466 | 4,170 | {"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-2020-40 | longest | en | 0.81559 |
https://kmhalpern.com/2019/01/22/semidirect-products-split-exact-sequences-and-all-that/ | 1,686,101,641,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00407.warc.gz | 402,308,502 | 36,490 | # Semidirect Products, Split Exact Sequences, and all that
One of the things I’ve butted heads with in studying Lie Groups is the semidirect product and its relationship to split exact sequences. It quickly became apparent that this was a pretty sizeable hole in my basic knowledge, so I decided to clarify this stuff once and for all.
— Normal Subgroups and Quotient Groups —
First, a brief refresher on Normal subgroups and Quotient groups. We are given a group ${G}$ and subgroup ${H\subseteq G}$.
• Left cosets are written ${gH}$ and right cosets are written ${Hg}$. Each is a set of elements in ${G}$. Not all left cosets are distinct, but any two are either equal or disjoint. Ditto for right cosets.
• The left (right) cosets form a partition of ${G}$, but they do not in general form a group. We can try to imbue them with a suitable product, but there are obstructions to the group axioms. For example ${g^{-1}H}$ is not a useful inverse since ${(gh)^{-1}= h^{-1}g^{-1}}$, so neither left cosets nor right cosets multiply as desired. More generally ${(gg')H}$ does not consist of a product of an element of ${gH}$ and an element of ${g'H}$.
• We define the Quotient Set ${G/H}$ to be the set of left cosets. As mentioned, it is not a group in general. There is an equivalent definition for right cosets, written ${H\setminus{}G}$, but it doesn’t appear often. In most cases we care about the two are the same.
• It is easy to see that the condition which removes the obstruction is that ${gH=Hg}$ for all ${g}$. Equivalently, ${gHg^{-1}=H}$ for all ${g}$. If this holds, the cosets form a group. Often the stated condition is that the sets of left and right cosets are the same. But ${g\in gH,Hg}$ so this is the same exact condition.
• ${H}$ is a Normal Subgroup if it obeys the conditions which make the cosets into a group.
• Usually a Normal Subgroup is denoted ${N}$, and we write ${N\triangleleft G}$ (or ${N\trianglelefteq G}$).
• For a Normal subgroup ${N}$, the Quotient Set ${Q=G/N}$ has (by definition) the natural structure of a group. It is called the Quotient Group.
• We have two natural maps associated with a Normal Subgroup:
• ${N\xrightarrow{i} G}$ is an inclusion (i.e. injective), defined by ${h\rightarrow h}$ (where the righthand ${h}$ is viewed in ${G}$). This is a homomorphism defined for any subgroup, not just normal ones
• ${G\xrightarrow{q} Q}$ is the quotient map (surjective), defined by ${g\rightarrow gN}$ (with the righthand viewed as a coset, i.e. an element of ${G/N}$). This map is defined for any subgroup, with ${Q}$ the Quotient Set. For Normal Subgroups, it is a group homomorphism.
• We know there is a copy of ${N}$ in ${G}$. Though ${Q}$ is derived from ${G}$ and ${N}$, and possesses no new info, there may or may not be a copy of it in ${G}$. Two natural questions are when that is the case, and how ${G}$, ${N}$, and ${Q}$ are related in general.
Let’s also recall the First Isomorphism Theorem for groups. Given any two groups ${G}$ and ${H}$ and a homomorphism ${\phi:G\rightarrow H}$, the following hold:
• ${\ker \phi}$ is a Normal Subgroup of ${G}$
• ${\mathop{\text{im}} \phi}$ is a subgroup of ${H}$
• ${\mathop{\text{im}} \phi}$ is isomorphic to the Quotient Group ${G/\ker\phi}$.
Again, we have to ask: since ${\ker\phi}$ is a Normal Subgroup of ${G}$, and ${\mathop{\text{im}}\phi}$ is isomorphic to the Quotient Group ${G/\ker\phi}$ which “sort of” may have an image in ${G}$, is it meaningful to write something like (playing fast and loose with notation) ${G\stackrel{?}{=} \ker\phi \oplus \mathop{\text{im}} \phi}$ (being very loose with notation)? The answer is no, it’s more complicated.
— Exact Sequences —
Next, a very brief review of exact sequences. We’ll use ${1}$ for the trivial group. The usual convention is to use ${1}$ for general groups and ${0}$ for Abelian groups. An exact sequence is a sequence of homomorphisms between groups ${\cdots \rightarrow G_n \xrightarrow{f_n} G_{n-1}\xrightarrow{f_{n-1}} \cdots}$ where ${\mathop{\text{im}} f_n= \ker f_{n-1}}$ for every pair. Here are some basic properties:
• ${1\rightarrow A \xrightarrow{f} B\cdots}$ means that ${f}$ is injective.
• ${\cdots A\xrightarrow{f} B\rightarrow 1}$ means that ${f}$ is surjective.
• ${1\rightarrow A\rightarrow B\rightarrow 1}$ means ${A=B}$.
• Short Exact Sequence (SES): This is defined as an exact sequence of the form: ${1\rightarrow A\xrightarrow{f} B\xrightarrow{g} C\rightarrow 1}$.
• For an SES, ${f}$ is injective, ${g}$ is surjective, and ${C=B/\mathop{\text{im}} f}$
• SES’s arise all the time when dealing with groups, and the critical question is whether they “split”.
We’re now ready to define Split SES’s.
• Right Split SES: There exists a homomorphism ${h:C\rightarrow B}$ such that ${g\circ h=Id_C}$. Basically, we can move to ${B}$ and back from ${C}$ without losing info — which means ${C}$ is in some sense a subgroup of ${B}$.
• Left Split SES: There exists a homomorphism ${h:B\rightarrow A}$ such that ${h\circ g=Id_A}$. Basically, we can move to ${B}$ and back from ${A}$ without losing info — which means ${A}$ is in some sense a subgroup of ${B}$.
• These two conditions are not in general equal, or even equivalently restrictive. The Left Split condition is far more constraining than the Right Split one in general. The direction of the homomorphisms in the SES introduce an asymmetry. [My note: it seems likely that the two are dual in some sense.]
— External vs Internal View —
We’re going to described 3 types of group operations: the direct product, semi-direct product, and group extension. Each has a particular relationship to Normality and SES’s. There are two equivalent ways to approach this, depending whether we prefer to define a binary operation between two distinct groups or to consider the relationship amongst subgroups of a given group.
• External view: We define a binary operation on two distinct, unrelated groups. Two groups go in, and another group comes out.
• Internal view: We define a relationship between a group and various groups derived from it (ex. Normal or Quotient).
• These approaches are equivalent. The Internal view describes the relationship amongst the two groups involved in the External view and their issue. Conversely, the derived groups in the Internal view may be recombined via the External view operation.
We must be a little careful with notation and terminology. When we use the symbol ${HK}$, it can mean one of two things.
• Case 1: ${H}$ and ${K}$ are distinct groups. ${HK}$ is just the set of all pairs of elements ${(h,k)}$. I.e. it is the direct product set (but not group).
• Case 2: ${H}$ and ${K}$ are subgroups of a common group ${G}$ (or have some natural implicit isomorphisms to such subgroups). In this case, ${HK}$ is the set of all elements in ${G}$ obtained as a product of an element of ${H}$ and an element of ${K}$ under the group multiplication.
• Note that we may prefer cases where two subgroups cover ${G}$, but there are plenty of other possibilities. For example, consider ${Z_{30}}$ (the integers mod 30). This has several obvious subgroups (${Z_2}$, ${Z_3}$, ${Z_5}$, ${Z_6}$, ${Z_{10}}$, ${Z_{15}}$). ${Z_2}$ and ${Z_3}$ only intersect on ${0}$ (the additive identity). However, the two do not cover (or even generate) the group! Similarly, ${Z_2}$ and ${Z_{10}}$ do not cover the group (or even generate it) but intersect on a nontrivial subset!
• Going the other way, we’ll say that ${G=HK}$ if ${H}$ and ${K}$ are subgroups and every element ${g}$ can be written as ${hk}$ for some ${h\in H}$ and ${k\in K}$. Note that ${H}$ and ${K}$ need not be disjoint (or even cover ${G}$ set-wise).
Another potentially confusing point should be touched on. When we speak of “disjoint” subgroups ${H}$ and ${K}$ we mean that ${H\cap K=\{e\}}$, NOT that it is the null set. I.e., ${H\cap K= 1}$, the trivial group.
— Semidirect Product —
The semidirect product may seem a bit arbitrary at first but, as we will see, it is a natural part of a progression which begins with the Direct Product. Here are the two ways of defining it.
• External view (aka Outer Semidirect Product): Given two groups ${H}$ and ${K}$ and a map ${\phi:K\rightarrow Aut(H)}$, we define a new group ${H\rtimes K}$. We’ll denote by ${\phi_k(h)}$ the effect of the automorphism ${\phi(k)}$ on ${h}$ (and thus an element of ${H}$). Set-wise, ${H\rtimes K}$ is just ${H\times K}$ (i.e. all pairs ${(h,k)}$). The identity is ${(e,e)}$. Multiplication on ${H\rtimes K}$ is defined as ${(h,k)(h',k')= (h\phi_k(h'),kk')}$. The inverse is ${(h,k)^{-1}= (\phi_{k^{-1}}(h^{-1}),k^{-1})}$.
• Internal view (aka Inner Semidirect Product): Given a group ${G}$ and two disjoint subgroups ${N}$ and ${K}$, such that ${G=NK}$ and ${N}$ is a Normal Subgroup, ${G}$ is called the Semidirect product ${N\rtimes K}$. The normality of ${H}$ constrains ${K}$ to be isomorphic to the Quotient Group ${G/N}$.
• There are (potentially) many Semidirect products of two given groups, obtained via different choices of ${\phi}$. The notation is deceptive because it hides our choice of ${\phi}$. Given any ${H,K,\phi}$ there exists a Semidirect product ${H\rtimes K}$. The various Semidirect products may be isomorphic to one another, but in general need not be. I.e., a given ${H}$ and ${K}$ may have multiple distinct semidirect products. This actually happens. Wikipedia mentions that there are 4 non-isomorphic semidirect products of ${C_8}$ and ${C_2}$ (the former being the Normal Subgroup in each case). One is a Direct Product, and the other 3 are not.
• It also is possible for a given group ${G}$ to arise from several distinct Semidirect products (of different pairs of groups). Again from Wikipedia, there is a group of order 24 which can be written as 4 distinct semiproducts of groups.
• Yet another oddity is that a seemingly nontrivial ${H\rtimes K}$ can be isomorphic to ${H\oplus K}$.
• If ${\phi= Id}$ (i.e. every ${k}$ maps to the identify map on ${H}$), then ${G=H\oplus K}$.
• To go from the External view to the Internal one, we note that, by construction, ${H}$ is a Normal Subgroup of ${H\rtimes K}$ and ${K}$ is the Quotient Group ${G/H}$. To be precise, the Normal Subgroup is ${(N,e)}$, which is isomorphic to ${N}$, and the Quotient Group ${G/(N,e)}$ is isomorphic to ${K}$.
• To go from the Internal view to the External one, we choose ${\phi_k(h)= khk^{-1}}$ as our function. I.e., ${\phi}$ is just conjugation by the relevant element.
• It may seem like there is an imbalance here. For a specific choice of Normal Subgroup ${N}$, the External view offers complete freedom of ${\phi}$, while the Internal view has a fixed ${\phi}$. Surely the latter is a special case of the former. The fallacy in this is that we must consider the pair ${(G,N)}$. We very well could have non-isomorphic ${G,G'}$ with Normal Subgroups ${N,N'}$ where ${N\approx N'}$. I.e. they are the same Normal Subgroup, but with different parent groups. We then would have different ${\phi}$‘s via our Internal view procedure. The correspondence is between ${(H,K,\phi)}$ and ${(G,N,K)}$ choices. Put differently, the freedom in ${\phi}$ loosely corresponds to a freedom in ${G}$.
• Note that, given ${G}$ and a Normal Subgroup ${N}$ — with the automatic Quotient Group ${G/N}$ — we do NOT necessarily have a Semidirect product relationship. The condition of the Semidirect product is stricter than this. As we will see it requires not just isomorphism, but a specific isomorphism, between ${H}$ and ${G/N}$. Equivalently, it requires a Right-Split SES (as we will discuss).
• The multiplication defined in the External view may seem very strange and unintuitive. In essence, here is what’s happening. For a direct product, ${H}$ and ${K}$ are independent of one another. Each half of the pair acts only on its own elements. For a semidirect product, the non-normal half ${K}$ can twist the normal half ${H}$. Each element of ${K}$ can alter ${H}$ in some prescribed fashion, embodied in ${\phi(k)}$. So ${K}$ is unaffected by ${H}$ but ${H}$ can be twisted by ${K}$.
• It is interesting to compare the basic idea to that of a Fiber bundle. There, the fiber can twist (via a group of homeomorphisms) as we move around the base space. Here, the normal subgroup can twist as we move around the non-normal part. Each generalizes a direct product and measures our need to depart from it.
• The semidirect product of two groups is Abelian iff it’s just a direct product of abelian groups.
— Group Extensions —
As with Semidirect products, there are 2 ways to view these. To make matters confusing, the notation speaks to an Internal view, while the term “extension” speaks to an External view.
• External view: Given groups ${A}$ and ${C}$, we say that ${B}$ is an extension of ${C}$ by ${A}$ if there is a SES ${1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1}$.
• Internal view: Given a group ${G}$ and Normal Subgroup ${N\triangleleft G}$, we say that ${G}$ is an extension of ${Q}$ by ${N}$, where ${Q=G/N}$ is the Quotient Group.
• Note that the two are equivalent. If ${B}$ is an extension of ${A}$ by ${C}$, then ${A}$ is Normal in ${B}$ and ${C}$ is isomorphic to the Quotient Group ${B/A}$.
• Put simply, the most general form of the Group, Normal Subgroup, induced Quotient Group trio is the Group Extension.
— Direct Products, Semidirect Products, and Group Extensions —
In the External view, we’ve mentioned three means of getting a group ${B}$ from two groups ${A}$ and ${C}$:
• Direct Product: ${B=A\oplus C}$. This is unique.
• Semidirect Product: ${B=A\rtimes C}$. There may multiple of these, corresponding to different ${\phi}$‘s.
• Group Extension: A group ${B}$ for which there are 2 homomorphisms forming a SES ${1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1}$. There may be many of these, corresponding to different choices of the two homomorphisms.
Equivalently, we have several ways of describing the relationship between two subgroups ${H,K\subseteq G}$ which are disjoint (i.e. ${H\cap K=\{e\}}$).
• Direct Product: ${G=H\oplus K}$ requires that both be Normal Subgroups.
• Semidirect Product: ${G=H\rtimes K}$ requires that ${H}$ be normal (in which case, ${Q=G/H}$, and ${\phi}$ is determined by it). For a given ${H}$ there may be multiple, corresponding to different ${G}$‘s.
• Group Extension: Both ${H}$ and ${K}$ sit in ${G}$ to some extent. ${H}$ must be Normal.
Note that not every possible relationship amongst groups is captured by these. For example, we could have two non-normal subgroups or two homomorphisms which don’t form an SES, or no relationship at all.
An excellent hierarchy of conditions was provided by Arturo Magidin in answer to someone’s question on Stackoverflow. I roughly replicate it here. Unlike him, I’ll be sloppy and not distinguish between subgroups and groups isomorphic to subgroups.
• Direct Product (${G=H\oplus K}$): ${H,K}$ both Normal Subgroups. ${H,K}$ disjoint. ${G=HK}$
• Semidirect Products (${G=H\rtimes K}$): ${H}$ Normal Subgroup, ${K}$ Subgroup. ${H,K}$ disjoint. ${G=HK}$. I.e., we lose Normality of ${K}$.
• Group Extension (${G}$ is extension of ${H}$ by ${K}$): ${H}$ Normal Subgroup, ${G/H\approx K}$. I.e. ${K}$ remains the Quotient Group (as before), but the Quotient Group may no longer be a subgroup of ${G}$ at all!
Now is a good time to mention the relationship between the various SES Splitting conditions:
• For all groups: Left Split is equivalent to ${B=A\oplus C}$, and they imply Right Split. (LS=DP) => RS always.
• For abelian groups, the converse holds and Right split implies Left Split and Direct Sum. I.e. the conditions are equivalent. LS=DP=RS for Abelian.
• For nonabelian groups: Right Split implies ${B=A\rtimes C}$ (with ${\phi}$ depending on the SES map). We’ll discuss this shortly.
Back to the hierarchy, now from a SES standpoint:
• Most general case: There is no SES at all. Given groups ${A,B,C}$, there may be no homomorphisms between them. If there are homomorphisms, there may be none which form an SES. Consider a general pair of homomorphisms ${f:A\rightarrow B}$ and ${g:B\rightarrow C}$, with no assumptions. We may turn to the first isomorphism theorem for help, but that does us no good. The first isomorphism theorem says that ${\ker f \triangleleft B}$ and ${\mathop{\text{im}} f\approx A/\ker f}$, and ${\ker g \triangleleft C}$ and ${\mathop{\text{im}} g\approx B/\ker g}$. This places no constraints on ${A}$ or ${C}$.
• Group Extension: Any SES defines a group extension. They are the same thing.
• Semidirect Product: Any SES which right-splits corresponds to a Semidirect Product (with the right-split map determining ${\phi}$)
• Direct Product: Any SES which left-splits (and thus right-splits too) corresponds to a direct product.
So, when we see the standard SES: ${1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 1}$, this is a group extension. Only if it right splits can we write ${G= N\rtimes G/N}$, and only if it left splits can we write ${G= N\oplus G/N}$.
— Some Notes —
• Group Extensions are said to be equivalent if their ${B}$‘s are isomorphic and there exists an isomorphism between them which makes a diamond diagram commute. It is perfectly possible for the ${B}$‘s to be isomorphic but for two SES’s not to be equivalent extensions.
• Subtlety referred to above. A quotient group need not be isomorphic to a subgroup of ${G}$. It only is defined when ${N}$ is normal, and there automatically is a surjective homomorphism ${G\rightarrow Q}$. But we don’t have an injective homomorphism ${Q\rightarrow G}$, which is what would be need for it to be isomorphic to a subgroup of ${G}$. This is precisely what the right-split furnishes. In that case, it is indeed a subgroup of ${G}$. The semidirect product may be thought of as the statement that ${Q}$ is a subgroup of ${G}$.
• In the definition of right split and left split, the crucial aspect of the “inverse” maps is that they be homomorphisms. A simple injective (for right-split, or surjective for left-split) map is not enough!
• It is sometimes said that the concept of subgroup is dual to the concept of quotient group. This is intuitive in the following sense. A subgroup can be thought of as an injective homomorphism. By the SES for normal/quotient groups, we can think of a quotient group as a surjective homomorphism. Since injections and surjections are categorically dual, it makes sense to think of quotient groups and subgroups as similarly dual. Whether the more useful duality is subgroup quotient group or normal subgroup quotient group is unclear to me. | 4,997 | 18,635 | {"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": 283, "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} | 3.984375 | 4 | CC-MAIN-2023-23 | latest | en | 0.931107 |
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## #1951 2012-12-12 10:31:24
anonimnystefy
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From: Harlan's World
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### Re: What do you think?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1952 2012-12-12 10:34:06
bobbym
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Registered: 2009-04-12
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### Re: What do you think?
That is very good. It is a nice problem. Actually the answer is not that hard.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1953 2012-12-12 10:35:21
anonimnystefy
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### Re: What do you think?
What do you get for 3x3 unit squares?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1954 2012-12-12 10:38:40
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1955 2012-12-12 10:45:43
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
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### Re: What do you think?
That is not correct.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1956 2012-12-12 10:49:27
bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
What do you get?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1957 2012-12-12 10:51:47
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
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### Re: What do you think?
433.
I think you have misuderstood the problem. Rea it once again.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1958 2012-12-12 10:54:18
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
I was going to say the same thing. Post your method and maybe we can resolve this.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1959 2012-12-12 10:56:38
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
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### Re: What do you think?
I did it by counting all possibilities.
The thing is that, if a squares neighbour is a 0, the square itself must be either 0 or 1. Your counting method allows the squares to be greater than 1 even if neighbours are 0.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1960 2012-12-12 10:59:52
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
Give me a chance to show why I think my method is correct. I will post a diagram.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1961 2012-12-12 11:02:41
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
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### Re: What do you think?
Sure!
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1962 2012-12-12 11:07:38
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Here is the possibilities for each box.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1963 2012-12-12 11:13:56
anonimnystefy
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### Re: What do you think?
That is not correct. The top left-hand side box can contain a 2 only if both its neighbours are 1s. Other wise it can take values of 0 and 1 only.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1964 2012-12-12 11:23:24
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
Yes, I know. But that still means 3 choices for that box. The multiplication takes care of what is the next boxes.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1965 2012-12-12 11:26:00
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
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### Re: What do you think?
Your method allows the square to be filled like this:
2 0 2
0 0 0
2 0 2
which is not allowed by the problem!
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1966 2012-12-12 11:31:20
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
I do not think so, I worked from the inside out. I will try to enumerate all of them. You may be right.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1967 2012-12-12 11:38:08
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: What do you think?
My answer might not be correct, but I am sure that the one in #1954 isn't correct.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1968 2012-12-12 19:55:51
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
You are correct with the answer of 433 for a 3 x 3. Very good!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1969 2012-12-12 21:04:07
anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: What do you think?
Unfortunately, the problem is harder this way!
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1970 2012-12-12 21:09:44
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Well young fella, are you going to explain your method?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1971 2012-12-12 21:12:34
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: What do you think?
I did it by casework, the cases being tje possibilities for the squares around the center square.
Unfortunately, it cannot be applied to the 5x5...
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1972 2012-12-12 21:20:07
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Hi;
The approach I used is maybe possible with the 5 x 5 but certainly not for an 11 x 11.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #1973 2012-12-12 21:21:21
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: What do you think?
Which approach is that?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #1974 2012-12-12 21:25:54
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: What do you think?
Experimental of course!
How many times have even weak programmers solved problems when mathematicians just went, "Duh?"
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #1975 2012-12-12 21:29:08
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: What do you think?
How did you count all the possibilities?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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A063453 Multiplicative with a(p^e) = 1 - p^3. 8
1, -7, -26, -7, -124, 182, -342, -7, -26, 868, -1330, 182, -2196, 2394, 3224, -7, -4912, 182, -6858, 868, 8892, 9310, -12166, 182, -124, 15372, -26, 2394, -24388, -22568, -29790, -7, 34580, 34384, 42408, 182, -50652, 48006, 57096, 868, -68920, -62244, -79506, 9310, 3224, 85162, -103822, 182 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS More generally, Dirichlet g.f. for Sum_{d|n} mu(d)*d^k, the Dirichlet inverse of the Jordan function J_k, is zeta(s)/zeta(s-k). Apart from different signs also Sum_{d|n} core(d)^3*mu(n/d) where core(x) is the squarefree part of x. - Benoit Cloitre, May 31 2002 Dirichlet inverse of A059376. - R. J. Mathar, Jul 15 2010 REFERENCES T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1986. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 P. G. Brown, Some comments on inverse arithmetic functions, Math. Gaz. 89 (516) (2005) 403-408. FORMULA a(n) = Sum_{d|n} mu(d)*d^3. Dirichlet g.f.: zeta(s)/zeta(s-3). A023900(n) | a(n). - R. J. Mathar, Mar 30 2011 a(n)= product_{p|n}(1-p^3), n>=2, p prime, a(1)=1. a(n)= J_{-3}(n)*n^3, with the Jordan function J_k(n). See the Apostol reference, p. 48, exercise 17. - Wolfdieter Lang, Jun 16 2011. G.f.: Sum_{k>=1} mu(k)*k^3*x^k/(1 - x^k). - Ilya Gutkovskiy, Jan 15 2017 MAPLE Jinvk := proc(n, k) local a, f, p ; a := 1 ; for f in ifactors(n)[2] do p := op(1, f) ; a := a*(1-p^k) ; end do: a ; end proc: A063453 := proc(n) Jinvk(n, 3) ; end proc: # R. J. Mathar, Jul 04 2011 MATHEMATICA a[n_] := Total[MoebiusMu[#]*#^3& /@ Divisors[n]]; Table[a[n], {n, 1, 48}] (* Jean-François Alcover, Jul 26 2011 *) PROG (Haskell) a063453 = product . map ((1 -) . (^ 3)) . a027748_row -- Reinhard Zumkeller, Jan 19 2012 (PARI) a(n) = sumdiv(n, d, moebius(d) * d^3); \\ Indranil Ghosh, Mar 11 2017 CROSSREFS Cf. A023900, A046970. Cf. A027748. Sequence in context: A214593 A012490 A157702 * A284054 A284786 A262109 Adjacent sequences: A063450 A063451 A063452 * A063454 A063455 A063456 KEYWORD mult,sign AUTHOR Vladeta Jovovic, Jul 26 2001 STATUS approved
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http://www.ebay.co.uk/gds/How-to-measure-your-head-for-right-cap-and-hat-size-/10000000176186976/g.html | 1,503,368,220,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109803.8/warc/CC-MAIN-20170822011838-20170822031838-00648.warc.gz | 534,392,004 | 17,963 | # How to measure your head for right cap and hat size
How to measure your head for the correct size hat or cap
This guide is to help explain the methods used for hat size measuring and the conversion of the different types of measurements.
Many people purchase hats and caps without knowing what actual size measurement iis. It can be difficult as like with other apparel merchandises the size range will vary depending on the country of origin, the manufacturer, the brand and the type of hat or cap.
To know your hat or cap size there are three main methods:
1) The use of a measuring tape
2) By trying on other hats or caps
3) The use of a piece of string and a ruler
Method 1
Using a measuring tape is the most popular and useful method known.
By wrapping the tape measure around your head, placing it approximately an inch above your ears (the point that your hat usually sits) you need to hold the measuring tape securely but not too tight. Make sure you have wrapped it completely around your and the point where the tape meets is where you will find your hat measurement. This measurement can now be used as a reference when buying a cap or hat.
Method 2
This is the most easiest method by trying on a hat or cap from a friend or in a shop. With this you will know what sizes are too big, too small or just about right for yourself.
This method is good but not too reliable as different brands will have slightly different size measurements. Also the type of measurement used between brands can be confusing if you do not know the conversion.
Method 3
The use of a string and ruler is the most simplest method. It is similar to the measuring tape method and useful if you do not have a measuring tape to use.
This is done in the same way by wrapping the string around the head. When you have the size length on the string it can then be measured against a ruler to show what your correct head size is.
*The Best Measurement for Hats and Caps is in Centimetres (cm) which can be converted with ease*
Hat and Cap Size Conversions
Hat and cap sizes come in different measurement forms, five known ones are:
The Metric = measured in Centimetres (cm)
The Imperial = measured in Inches (in or ")
Easy fitted = Size range Small (S), Medium (M), Large (L) and Extra-Large (XL).
Stretch fitted = Size range Small/Medium (S/M), Medium/Large (M/L) and Large/Extra-Large (L/XL)
Fitted = Size measurements are more specific and are in the form of numbers plus a fraction. eg 6 3/4, 7, 71/2 and 8.
Below is a brief guide showing the different measurements and how these can be translated.
Measurement types
Fitted Caps | 6 3/4 | 6 7/8 | 7 | 7 1/8 | 7 1/4 | 7 3/8 | 7 1/2 | 7 5/8 | 7 3/4 | 7 7/8 | 8 |
Easy Fitted | Small (S) | Medium (M) | Large (L) | X-Large (XL) | Extra Extra-Large (XXL) |
Centimeters | 53.9 | 54.9 | 55.9 | 56.9 | 57.8 | 58.9 | 59.7 | 60.6 | 61.6 | 62.5 | 63.5 |
Inches | 21 1/4| 21 5/8| 22 | 22 3/8 | 22 3/4 | 23 1/8 | 23 1/2 | 23 7/8 | 24 1/4 | 24 5/8 | 25 |
Stretch Fitted | Small/Medium (S/M) | Medium/Large | Large/Extra -Large (L/XL) |
Thank you for taking time to read my guide.
Fitted Caps and other head wear are available form my eBay store KayCee's Urban Trading
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Explore more guides | 1,004 | 3,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-34 | latest | en | 0.930036 |
https://in.mathworks.com/help/control/ref/lti.zpkdata.html | 1,643,359,662,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00280.warc.gz | 375,214,301 | 17,815 | zpkdata
Access zero-pole-gain data
Syntax
```[z,p,k] = zpkdata(sys) [z,p,k,Ts] = zpkdata(sys) [z,p,k,Ts,covz,covp,covk] = zpkdata(sys) ```
Description
`[z,p,k] = zpkdata(sys) ` returns the zeros `z`, poles `p`, and gain(s) `k` of the zero-pole-gain model `sys`. The outputs `z` and `p` are cell arrays with the following characteristics:
• `z` and `p` have as many rows as outputs and as many columns as inputs.
• The `(i,j)` entries `z{i,j}` and `p{i,j}` are the (column) vectors of zeros and poles of the transfer function from input `j` to output `i`.
The output `k` is a matrix with as many rows as outputs and as many columns as inputs such that `k(i,j)` is the gain of the transfer function from input `j` to output `i`. If `sys` is a transfer function or state-space model, it is first converted to zero-pole-gain form using `zpk`.
For SISO zero-pole-gain models, the syntax
```[z,p,k] = zpkdata(sys,'v') ```
forces `zpkdata` to return the zeros and poles directly as column vectors rather than as cell arrays (see example below).
`[z,p,k,Ts] = zpkdata(sys) ` also returns the sample time `Ts`.
`[z,p,k,Ts,covz,covp,covk] = zpkdata(sys) ` also returns the covariances of the zeros, poles and gain of the identified model `sys`. `covz` is a cell array such that `covz{ky,ku}` contains the covariance information about the zeros in the vector `z{ky,ku}`. `covz{ky,ku}` is a 3-D array of dimension 2-by-2-by-Nz, where `Nz` is the length of `z{ky,ku}`, so that the `(1,1)` element is the variance of the real part, the `(2,2)` element is the variance of the imaginary part, and the `(1,2)` and `(2,1)` elements contain the covariance between the real and imaginary parts. `covp` has a similar relationship to `p.covk` is a matrix containing the variances of the elements of `k`.
You can access the remaining LTI properties of `sys` with `get` or by direct referencing, for example,
```sys.Ts sys.inputname ```
Examples
Example 1
Given a zero-pole-gain model with two outputs and one input
```H = zpk({[0];[-0.5]},{[0.3];[0.1+i 0.1-i]},[1;2],-1) Zero/pole/gain from input to output... z #1: ------- (z-0.3) 2 (z+0.5) #2: ------------------- (z^2 - 0.2z + 1.01) Sample time: unspecified ```
you can extract the zero/pole/gain data embedded in `H` with
```[z,p,k] = zpkdata(H) z = [ 0] [-0.5000] p = [ 0.3000] [2x1 double] k = 1 2 ```
To access the zeros and poles of the second output channel of `H`, get the content of the second cell in `z` and `p` by typing
```z{2,1} ans = -0.5000 p{2,1} ans = 0.1000+ 1.0000i 0.1000- 1.0000i ```
Example 2
Extract the ZPK matrices and their standard deviations for a 2-input, 1 output identified transfer function.
```load iddata7 ```
transfer function model
`sys1 = tfest(z7, 2, 1, 'InputDelay',[1 0]);`
an equivalent process model
```sys2 = procest(z7, {'P2UZ', 'P2UZ'}, 'InputDelay',[1 0]); 1, p1, k1, ~, dz1, dp1, dk1] = zpkdata(sys1); [z2, p2, k2, ~, dz2, dp2, dk2] = zpkdata(sys2);```
Use `iopzplot` to visualize the pole-zero locations and their covariances
```h = iopzplot(sys1, sys2); showConfidence(h)```
Introduced before R2006a | 1,035 | 3,111 | {"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-2022-05 | longest | en | 0.824394 |
https://www.jiskha.com/display.cgi?id=1327266614 | 1,498,667,148,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323711.85/warc/CC-MAIN-20170628153051-20170628173051-00021.warc.gz | 890,486,857 | 3,917 | # math
posted by .
how do you solve this.prove: cos2A plus cosA divided by sin2A minus sinA equal to cos2A plus 1 divided by sinA
• math -
I think there's a typo in your equation. For example , plug in A = pi/4. You then have
(0 + 1/√2)/(1 - 1/√2) = (0+1)/(1/√2)
1/√2 *√2 /(√2 - 1) = √2
1/(√2-1) = √2
√2+1 = √2
However, if you fix the right side to read (cosA+1)/sinA
you have the left side:
(cos^2 A - sin^2 A + cosA)/(2sinAcosA - sinA)
(cos^2 A - (1 - cos^2 A) + cosA)/(2sinAcosA - sinA)
(2cos^2 A + cosA - 1)/[sinA(2cosA - 1)
(2cosA-1)(cosA - 1)/[(2cosA - 1)(sinA)]
(cosA -1)/sinA
Now if you plug in any angle, the equality holds. | 276 | 646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-26 | longest | en | 0.67331 |
https://us.metamath.org/mpeuni/hvmapval.html | 1,721,511,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00453.warc.gz | 512,126,327 | 6,852 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > hvmapval Structured version Visualization version GIF version
Theorem hvmapval 39055
Description: Value of map from nonzero vectors to nonzero functionals in the closed kernel dual space. (Contributed by NM, 23-Mar-2015.)
Hypotheses
Ref Expression
hvmapval.h 𝐻 = (LHyp‘𝐾)
hvmapval.u 𝑈 = ((DVecH‘𝐾)‘𝑊)
hvmapval.o 𝑂 = ((ocH‘𝐾)‘𝑊)
hvmapval.v 𝑉 = (Base‘𝑈)
hvmapval.p + = (+g𝑈)
hvmapval.t · = ( ·𝑠𝑈)
hvmapval.z 0 = (0g𝑈)
hvmapval.s 𝑆 = (Scalar‘𝑈)
hvmapval.r 𝑅 = (Base‘𝑆)
hvmapval.m 𝑀 = ((HVMap‘𝐾)‘𝑊)
hvmapval.k (𝜑 → (𝐾𝐴𝑊𝐻))
hvmapval.x (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))
Assertion
Ref Expression
hvmapval (𝜑 → (𝑀𝑋) = (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))))
Distinct variable groups: 𝑡,𝑗,𝑣,𝐾 𝑡,𝑊 𝑡,𝑂 𝑅,𝑗 𝑗,𝑊,𝑣 𝑣,𝑉 𝑗,𝑋,𝑡,𝑣
Allowed substitution hints: 𝜑(𝑣,𝑡,𝑗) 𝐴(𝑣,𝑡,𝑗) + (𝑣,𝑡,𝑗) 𝑅(𝑣,𝑡) 𝑆(𝑣,𝑡,𝑗) · (𝑣,𝑡,𝑗) 𝑈(𝑣,𝑡,𝑗) 𝐻(𝑣,𝑡,𝑗) 𝑀(𝑣,𝑡,𝑗) 𝑂(𝑣,𝑗) 𝑉(𝑡,𝑗) 0 (𝑣,𝑡,𝑗)
Proof of Theorem hvmapval
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 hvmapval.h . . . 4 𝐻 = (LHyp‘𝐾)
2 hvmapval.u . . . 4 𝑈 = ((DVecH‘𝐾)‘𝑊)
3 hvmapval.o . . . 4 𝑂 = ((ocH‘𝐾)‘𝑊)
4 hvmapval.v . . . 4 𝑉 = (Base‘𝑈)
5 hvmapval.p . . . 4 + = (+g𝑈)
6 hvmapval.t . . . 4 · = ( ·𝑠𝑈)
7 hvmapval.z . . . 4 0 = (0g𝑈)
8 hvmapval.s . . . 4 𝑆 = (Scalar‘𝑈)
9 hvmapval.r . . . 4 𝑅 = (Base‘𝑆)
10 hvmapval.m . . . 4 𝑀 = ((HVMap‘𝐾)‘𝑊)
11 hvmapval.k . . . 4 (𝜑 → (𝐾𝐴𝑊𝐻))
121, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11hvmapfval 39054 . . 3 (𝜑𝑀 = (𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥))))))
1312fveq1d 6651 . 2 (𝜑 → (𝑀𝑋) = ((𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)))))‘𝑋))
14 hvmapval.x . . 3 (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))
154fvexi 6663 . . . 4 𝑉 ∈ V
1615mptex 6967 . . 3 (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))) ∈ V
17 sneq 4538 . . . . . . . 8 (𝑥 = 𝑋 → {𝑥} = {𝑋})
1817fveq2d 6653 . . . . . . 7 (𝑥 = 𝑋 → (𝑂‘{𝑥}) = (𝑂‘{𝑋}))
19 oveq2 7147 . . . . . . . . 9 (𝑥 = 𝑋 → (𝑗 · 𝑥) = (𝑗 · 𝑋))
2019oveq2d 7155 . . . . . . . 8 (𝑥 = 𝑋 → (𝑡 + (𝑗 · 𝑥)) = (𝑡 + (𝑗 · 𝑋)))
2120eqeq2d 2812 . . . . . . 7 (𝑥 = 𝑋 → (𝑣 = (𝑡 + (𝑗 · 𝑥)) ↔ 𝑣 = (𝑡 + (𝑗 · 𝑋))))
2218, 21rexeqbidv 3358 . . . . . 6 (𝑥 = 𝑋 → (∃𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)) ↔ ∃𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋))))
2322riotabidv 7099 . . . . 5 (𝑥 = 𝑋 → (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥))) = (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋))))
2423mpteq2dv 5129 . . . 4 (𝑥 = 𝑋 → (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)))) = (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))))
25 eqid 2801 . . . 4 (𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥))))) = (𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)))))
2624, 25fvmptg 6747 . . 3 ((𝑋 ∈ (𝑉 ∖ { 0 }) ∧ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))) ∈ V) → ((𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)))))‘𝑋) = (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))))
2714, 16, 26sylancl 589 . 2 (𝜑 → ((𝑥 ∈ (𝑉 ∖ { 0 }) ↦ (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑥})𝑣 = (𝑡 + (𝑗 · 𝑥)))))‘𝑋) = (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))))
2813, 27eqtrd 2836 1 (𝜑 → (𝑀𝑋) = (𝑣𝑉 ↦ (𝑗𝑅𝑡 ∈ (𝑂‘{𝑋})𝑣 = (𝑡 + (𝑗 · 𝑋)))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2112 ∃wrex 3110 Vcvv 3444 ∖ cdif 3881 {csn 4528 ↦ cmpt 5113 ‘cfv 6328 ℩crio 7096 (class class class)co 7139 Basecbs 16479 +gcplusg 16561 Scalarcsca 16564 ·𝑠 cvsca 16565 0gc0g 16709 LHypclh 37279 DVecHcdvh 38373 ocHcoch 38642 HVMapchvm 39051 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 ax-rep 5157 ax-sep 5170 ax-nul 5177 ax-pr 5298 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2601 df-eu 2632 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ne 2991 df-ral 3114 df-rex 3115 df-reu 3116 df-rab 3118 df-v 3446 df-sbc 3724 df-csb 3832 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-nul 4247 df-if 4429 df-sn 4529 df-pr 4531 df-op 4535 df-uni 4804 df-iun 4886 df-br 5034 df-opab 5096 df-mpt 5114 df-id 5428 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-f1 6333 df-fo 6334 df-f1o 6335 df-fv 6336 df-riota 7097 df-ov 7142 df-hvmap 39052 This theorem is referenced by: hvmapvalvalN 39056 hvmapidN 39057 hdmapevec2 39131
Copyright terms: Public domain W3C validator | 3,237 | 4,529 | {"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-2024-30 | latest | en | 0.299337 |
http://www.ask.com/web?qsrc=314&q=Luminosity+Formula&o=41647999&l=dir | 1,455,417,009,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701168076.20/warc/CC-MAIN-20160205193928-00073-ip-10-236-182-209.ec2.internal.warc.gz | 272,557,750 | 18,619 | Web Results
## Luminosity
en.wikipedia.org/wiki/Luminosity
. Point source S is radiating light equally in all directions. The amount passing through an area A varies with the distance ...
## Formulas - Luminosity - Astronomy Online
astronomyonline.org/Science/Luminosity.asp
Science - Formulas. ... Physics - Formulas - Luminosity ... if we know distance and brightness of a star, we can determine its Luminosity (or actual brightness):.
## Luminosity of Stars - Australia Telescope National Facility
www.atnf.csiro.au/outreach/education/senior/astrophysics/photometry_luminosity.html
How is this equation derived? It is simply an application of the luminosity ratio relationship (4.7).
## Luminosity - Hertzsprung-Russell Diagram - NAAP - UNL Astronomy
astro.unl.edu/naap/hr/hr_background2.html
Luminosity is the total energy that a star produces in one second. It depends on both the radius of the star and on its surface temperature. One can calculate ...
## Luminosity Formula - UNL Astronomy
astro.unl.edu/classaction/questions/stellarprops2/ca_stellarprops2_luminosity.html
A star with a radius like that of the sun and a surface temperature twice that of the sun, will have luminosity _____ times as great as the sun's luminosity.
## What is the Formula for Determining the luminosity of a star ...
Feb 10, 2010 ... Best Answer: The formula for the Luminosity L = Area * Energy Flux = 4pi R^2 * sigma T^4, where R is the radius and T the temperature.
## Luminosity and Apparent Brightness | Astronomy 801: Planets, Stars ...
www.e-education.psu.edu/astro801/content/l4_p4.html
The difference between luminosity and apparent brightness depends on distance . ... Thus, the equation for the apparent brightness of a light source is given by ...
## Luminosity of Stars: Radius & Surface Temperature
www.idialstars.com/april2013.htm
The luminosity of a star depends upon radius and surface temperature. ... In the equation below L = luminosity and R = radius. Let's say the radius of the star is 5 ...
## Finding the Radius of a Star - SkyServer
The luminosity of a star is given by the equation. L = 4pR<sup>2</sup>s T<sup>4</sup>,. Where L is the luminosity in Watts, R is the radius in meters, s is the Stefan-Boltzmann constant
Dec 26, 2012 ... Luminosity of a Star. ... Introductory Astronomy: Luminosity, Temperature, and Surface Area - Duration: 16:15. P.E. Robinson 3,999 views. 16:15.
Popular Q&A
Q: How would i use the luminosity formula with this?
A: Most guys ( gals) do the mistake of calculating in absolute quantities when only ratios suffice. Don't convert every quantity to absolute quantiies; do the comp... Read More » | 669 | 2,666 | {"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-2016-07 | longest | en | 0.818994 |
http://publish.illinois.edu/ymb/author/ymbillinois-edu/page/9/ | 1,537,381,383,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156270.42/warc/CC-MAIN-20180919180955-20180919200955-00082.warc.gz | 205,390,554 | 10,353 | Author Archive | yuliy
parabola construction
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Consider the optimal control problem
$\dot{x}=u; x(0)=x_o; \int_0^T u^2/2dt+M(x(T))\to\min.$
The task is to find the cost function $$S(x_o)$$ (the optimal achievable value).
It can be solved as follows. Assume for …
ece515, homework 5
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Due by midnight of december 4.
Use MATLAB or octave if needed.
1. Consider the optimal control problem
${x}”’=u; \int_0^\infty u^2+x^2 dt\to\min.$
• Write this system as a LQR;
• Solve the ARE;
november 21
1. Find sin transform of the function on $$[0,1]$$
given by
1. $$1$$;
2. $$\cos(\pi x)$$;
3. $$x$$.
2. Solve the heat equation
$u_t=\frac{1}{2} u_{xx},$
on $$[0,1]$$ with boundary conditions $$u(0,t)=u(1,t)=0$$ and initial values
1. u(x,0)=$$1$$;
2. u(x,0)=$$\cos(\pi x)$$;
3. u(x,0)=$$x$$.
3. Solve the heat equation
$math285, week of november 17 Read the textbook, chapters 9.5 and 9.7. Homework (due by Monday, 12.1): 1. Find the cosine transform of $$\sin(x)$$ on $$[0,\pi]$$. 2. Using separation of variables, solve the heat equation on the interval $$[0,\pi]$$: \[u_{t}=u_{xx}; \quad u_x(0,t)=u_x(\pi,t)=0; u(x,0)=\sin(x) november 14 1. Find the solutions of the wave equation at time $$t=1/6,1/3,1/2$$ \[ u_{tt}=u_{xx}, u(0,t)=u(1,t)=0,$
on $$[0,1]$$
with initial data $$u_t(x,0)=0$$ and $$u(x,0)$$ is as shown:
2. Same question, but the initial data are
$$u(x,0)=0$$ and $$u_t(x,0)$$ is as shown:
3. Sketch the
math 285, week of november 10
Read the textbook, chapters 9.4 and 9.6.
Videos to watch.
Homework (due by Monday, 11.17):
1. Find sine transform of the function (\T(x)\) on $$[0,1]$$ equal to $$x$$ for $$0\leq x\leq 1/2$$ and to $$1-x$$ for $$1/2\leq x \leq1$$.
2. Using
november 6
• Find Fourier series for $$2\pi$$-periodic function defined on $$-\pi,\pi$$ by
1. $\cos(2x)\sin^2(x);$
2. $\cos^3(x);$
3. $f(x)=\left\{\begin{array}{rcl} 1& \mathrm{if}& -\pi\leq x<0;\\ -1 & \mathrm{if}& 0\leq x \leq \pi.\\ \end{array}\right.$
• Find the Fourier series for the function which is
• notes for lectures 10-14
here
Many thanks to James!…
here
ece515, homework 4
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Due by midnight of november 8.
1. Consider the system $$\dot{x}=Ax+bu$$,
$A=\left( \begin{array}{ccc} 1&1&0\\ 1&1&1\\ 0&1&1\\ \end{array}\right); b=\left( \begin{array}{c} b_1\\ b_2\\ b_3\\ \end{array}\right).$
• Derive conditions on $$b$$ making the system controllable. | 1,120 | 2,817 | {"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} | 3.546875 | 4 | CC-MAIN-2018-39 | latest | en | 0.637893 |
https://www.coursehero.com/file/46439/HW7/ | 1,544,980,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00017.warc.gz | 829,477,152 | 110,081 | # HW7 - ma293hw7solns.nb 1 MA293 Homework 7 Solutions 11 yH4L...
• Notes
• 2
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MA293 Homework 7 Solutions 11. y H 4 L - 8 y H 3 L + 16 y H 2 L = 0 r 4 - 8 r 3 + 16 r 2 = 0 fl r 2 H r 2 - 8 r + 16 L = 0 fl r 2 H r - 4 L 2 = 0 So r = 0, 0, 4, 4 gives y H x L = c 1 + c 2 x + c 3 4 x + c 4 x 4 x 23. y '' - 6 y ' + 25 y = 0 ; y H 0 L = 3 ; y ' H 0 L = 1 r 2 - 6 r + 25 = 0 r 1,2 = 6 è!!!!!!!!!!!!!!!!!!!!!!! 36 - 25 * 4 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 = 3 4 i So our general solution is y H x L = ‰ 3 x H c 1 cos H 4 x L + c 2 sin H 4 x LL y H 0 L = c 1 = 3 y ' H x L = ‰ 3 x H 4 c 2 cos H 4 x L - 12 sin H 4 x LL + 3 3 x H 3 cos H 4 x L + c 2 sin H 4 x LL y ' H 0 L = 9 + 4 c 2 = 1 fl c 2 = - 2 y H x L = ‰ 3 x H 3 cos H 4 x L - 2 sin H 4 x LL 28. 2 y ''' - y '' - 5 y ' - 2 y = 0 2 r 3 - r 2 - 5 r - 2 = 0 One root is r = 2 H r - 2 L H 2 r 2 + 3 r + 1 L = 0 H r - 2 L H r + 1 L H r + 1 ê 2 L y H x L = c 1 2 x + c 2 x + c 3 x ê 2 34. 3 y ''' - 2 y '' + 12 y ' - 8 y = 0 ; y 1 = ‰ 2 ÅÅÅÅ 3 x 3 r 3 - 2 r 2 + 12 r - 8 = 0 One root is r = 2 ê 3 H r + 2 ê 3 L H 3 r 2 + 12 L = 0 H r + 2 ê 3 L H r + 2 I L H r - 2 I L = 0 y H x L = c 1 2 ÅÅÅÅ 3 x + c 2 cos H 2 x L + c 3 sin H 2 x L 49. y H 4 L = y ''' + y '' + y ' + 2 y ma293hw7solns.nb 1
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r 4 - r 3 - r 2 - r - 2 = 0 H r - 2 L H r + 1 L H r 2 + 1 L = 0 r = 2, - 1, i , - i y H x L = c 1 2 x + c 2 - x + c 3 cos H x L + c 4 sin H x L Now apply initial conditions: y H 0 L = 0 = c 1 + c 2 + c 3 y ' H 0 L = 0 = 2 c 1 - c 2 + c 4 y '' H 0 L = 0 = 4 c 1 + c 2 - c 3 y ''' H 0 L = 30 = 8 c 1 - c 2 - c 4 Solving this system of equations yields: c 1 = 2, c 2 = - 5, c 3 = 3, c 4 = - 9 Solution: y H x L = 2 2 x - 5 - x + 3 cos H x L - 9 sin H x L 54. x 3 y ''' + 6 x 2 y '' + 4 xy ' = 0 dy ÅÅÅÅÅÅÅ dx = dy ÅÅÅÅÅÅÅ dv 1 ÅÅÅÅ x d 2 y ÅÅÅÅÅÅÅÅÅÅ dx 2 = - dy ÅÅÅÅÅÅÅ
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Unformatted text preview: d v 2 d 3 y ÅÅÅÅÅÅÅÅÅÅ d x 3 = 1 ÅÅÅÅÅ x 3 I d 3 y ÅÅÅÅÅÅÅÅ d v 3-3 d 2 y ÅÅÅÅÅÅÅÅ d v 2 + 2 d y ÅÅÅÅÅÅ d v M Substituting this into the differential equation yields: y ''' H v L-3 y '' H v L + 2 y ' H v L + 6 y '' H v L-6 y ' H v L + 4 y ' H v L = y ''' H v L + 3 y '' H v L = 0. r 3 + 3 r 2 = r 2 H r + 3 L y H v L = c 1 + c 2 v + c 3 ‰-3 v So y H x L = c 1 + c 2 ln H x L + c 3 ‰-3 ln H v L = c 1 + c 2 ln H x L + c 3 x-3 ma293hw7solns.nb 2...
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Jill Tulane University ‘16, Course Hero Intern | 1,573 | 3,484 | {"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} | 4.15625 | 4 | CC-MAIN-2018-51 | latest | en | 0.439592 |
http://icpc.njust.edu.cn/Problem/CF/489B/ | 1,603,151,573,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00078.warc.gz | 54,573,473 | 9,917 | # BerSU Ball
Time Limit: 1 second
Memory Limit: 256 megabytes
## Description
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
## Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
## Output
Print a single number — the required maximum possible number of pairs.
## Sample Input
Input41 4 6 255 1 5 7 9Output3Input41 2 3 4410 11 12 13Output0Input51 1 1 1 131 2 3Output2
## Sample Output
None
None
None | 358 | 1,251 | {"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} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.847081 |
https://blenderartists.org/t/worldposition-with-linearvelocity/1213091 | 1,590,518,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00071.warc.gz | 276,648,729 | 6,433 | # WorldPosition with LinearVelocity
Hi guys. I have a question. How can I set world position for an object with setting linearVelocity as well??
This is example of my code:
from bge import logic, events
from mathutils import Vector
def floor():
cont = logic.getCurrentController()
own = cont.owner
scene = logic.getCurrentScene()
keyboard = logic.keyboard
``````first_floor = Vector([-0.35608, 0, -2.4525])
if keyboard.events[events.ZEROKEY] == logic.KX_INPUT_ACTIVE:
own.worldPosition = first_floor
own.setLinearVelocity((0.0,120.0,0.0),True)
``````
if name == ‘main’:
floor()
I get this world position after pressing “0” but get it imediatelly not with my velocity. Why? Can anybody help?
Is that object parented to something else?
Is it an ‘actor’?
Nope. It has only always actuator with python script.
Nope to which question? If you don’t have any physics enabled on it then it won’t move with velocity. You can still move a physics-less object, but by adjusting its position.
Hmmm think that nope to both. Could you tell me how i should add physic then? When i do getLinearVelocity i receive value when i change position constantly by myself
velocity get’s added every pulse so True pulse need to be on. Then your script waits till the button is pressed, once it’s pressed it will execute the code in single pulse, thus moving it a tiny bit or a lot (120 velocity is a lot).
so you can either write an other function, or get the velocity out of the function, or program a way so it can be run in true pulse.
also your code can be shortened to:
``````from bge import logic, events
def floor(cont):
own = cont.owner
scene = own.scene
keyboard = logic.keyboard
first_floor = [-0.35608, 0, -2.4525]
if keyboard.events[events.ZEROKEY] == logic.KX_INPUT_ACTIVE:
own.worldPosition = first_floor
own.setLinearVelocity([0.0,120.0,0.0],True)
``````
Thank you very much, will try to do this like that and will give you a feedback
Unfortunately it is not working as i wanted
i have “always” sensor with pulse==True and what i want to achieve is to by one keyboard click “4” get position from Vector e.g. [0, 0, 2] with seted velocity eg to 0.2. Is it possible?
i’m not sure what your goal is here.
so on hitting a key you want to grab a vector and set a speed? or move it to that vector by the speed? or how do i need to see this?
i guess you want to move it and set the 0.2 speed
``````from bge import logic, events
def floor(cont):
own = cont.owner
keyboard = logic.keyboard
#set the current position on startup into a property to be used
if not 'old_pos' in own:
own['old_pos'] = own.worldPosition
#default speed
x=0
y=0
z=0
#the vectors as worldPosition xyz
first_floor = [0.0, 1.0, 0.0]
#if the button is pressed
if keyboard.events[events.ZEROKEY] == logic.KX_INPUT_ACTIVE:
#set the position
own.worldPosition = first_floor
#set the new position as old
own['old_pos'] = own.worldPosition.copy()
#if current and old_pos is not the same
if own.worldPosition != own['old_pos']:
#set a speed on y axis
y = 0.2
# set/activate the speed/movement
own.localLinearVelocity = [x,y,z]
``````
#edit
code was not working so made it work
Thank you! This works for me
1 Like | 851 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-24 | latest | en | 0.884935 |
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# What is the Mode?
Learn about a third method for calculating average values, the mode, and find out exactly when and why it is useful.
By
Jason Marshall, PhD
Episode #25
## A Puzzle With the Mean, Median, and Mode
In recent articles, we’ve talked about two methods of calculating average values, the mean and the median, and we’ve talked about some everyday uses for them such as how to use median averaging to get better photos. But that’s not all there is to say about average values. Today, we’re talking about one additional average statistic: the mode.
## Mean, Median, and Mode
Mean, median, and mode. This trio of names rolls easily off the tongues of people who spend time dealing with statistical quantities. In my case, being an astrophysicist, I use these statistical values in one way or another almost every day. In fact, while writing this series of articles about the mean, median, and mode, I’ve been internally referring to it as “MMM”—which also happens to be the name of a computer program used by many astrophysicists to automatically calculate how bright the sky is (a task made much harder by all those interloping bright stars). All this is to say I couldn’t have talked about the mean and the median without saying a few words about the mode. Besides, it’s useful in a way that the mean and median can never be—which we’ll get to shortly.
## Why Another “Average?”
So, why do we need another average value? The short answer is that the mean and median alone can’t provide answers for all the statistical questions you may be confronted with. But that sort of “you need it because you need it” answer isn’t very satisfying to me. So let’s look at the reasoning in a little more detail. To really get to the bottom of answering “Why another average?” let’s start by taking a look at what the various types of averages we’ve talked about so far are actually good for.
## When Should You Use the Mean?
First, the mean. In the article on how to calculate mean values we learned that the mean is the most natural way to describe average values. To find the mean, we take all of whatever we’re averaging, and we spread it evenly amongst the various “containers” we’re averaging. A little more concretely, if we’re calculating the average weight of tomatoes in several bowls, the tomatoes are the “whatever,” and the bowls are the “containers.” Or, if we’re talking about the average number of points students have earned in a class, the student’s scores are the “whatever,” and the students themselves are the “containers.” The net result is a number that tells us the average number of whatever per container—tomatoes per bowl, points per student, and so on. It’s very intuitive.
## When Should You Use the Median?
Next, the median. In the article on how to calculate median values we learned that the median steps up to the plate to help out when the mean is overwhelmed by problematic data. In particular, if a couple of values in your set of data are outliers, meaning they’re either much larger or much smaller than most of the other values, the mean value will be thrown off. But the median will not be since it is resistant to these types of outlying values. It’s a very powerful statistic as we discovered in the last article on using median averaging to remove tourists from your photos.
## When Should You Use the Mode?
And now we’ve arrived at the crux of our question: Why do we need another average? Well, let me answer that question with a question. What is the average color of a cat? To answer, we obviously need to calculate an average value of some sort—should we use the mean or the median? Do either actually make any sense? No, not really. First, it’s not at all clear how to calculate the mean value. Take a brown tabby, a calico, and a black cat. We can’t even start to calculate the mean of these cat colors since it’s totally different than doing something like calculating the average number of coffee beans in jars—with cat colors, there are no numbers so there’s nothing to add up! And, if you think about it for a minute, you’ll see that you can’t calculate a median value either since there’s no way to put cat colors in order and find the one in the middle!
## What is the Mode and How is it Calculated?
Problems like this clearly show that we need a new way to calculate averages. And that new way is called the mode. Unlike the mean and median, the mode can be used to find the average of non-numerical data—like the color of cats. Here’s how it works. Imagine asking everybody in your city to write down the color of their cats. Then, take all these submissions and create a list of all the unique cat colors. Next, go through all the submissions and tally the number of cats of each particular color. The mode is the color that occurs most frequently. In other words, if there are 100 black cats, 50 calicos, and 150 brown tabbies, the mode would be 150, since brown tabby occurs most frequently. In this case, brown tabby is the “average” color of a cat—in the sense that it’s the most popular or common.
Of course, the mode works for things that are entirely numerical from the get-go too. Remember, the cat color problem didn’t start with numbers—it started with cat colors—although we did turn it into a numerical problem by tallying up the totals in each category. In general, the mode is the value that occurs most frequently in a list of numbers. The mode of the list of numbers (3, 5, 5, 7) is 5 since it occurs twice. If the list of numbers were instead (3, 5, 5, 7, 7, 9), you’ll notice that there isn’t a unique mode. This list has two modes—5 and 7—and is therefore called bi-modal.
## Brain-Teaser Problem on Averages
So, now you’re up-to-date on the big three of averages: the mean, median, and mode. Hopefully you have a good intuitive grasp of what these three quantities represent, and when they’re useful. And to make sure you have a good practical grasp of how to use them too, here’s a brain-teaser problem for you to try courtesy of the Ask Dr. Math website (which is a great resource with lots of answers to common math questions). Here’s the question: “Find a set of five data values with modes 0 and 2, median 2, and mean 2.” Give the problem a shot, and then check out this week’s Math Dude “Video Extra!” episode on YouTube for an explanation of the solution.
Next week we'll talk about another subject related to averages: range and standard deviation.
## Wrap Up
Okay, that’s all the math we have time for today. Now, the next time you’re at a party and somebody asks you to calculate the “average” name of partygoers, you can use your knowledge of the mode to help out. (Okay, this’ll probably never happen—but you get the idea.)
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http://sciencedocbox.com/Physics/73066853-1-the-heisenberg-model.html | 1,558,509,899,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00496.warc.gz | 183,755,443 | 36,498 | # 1 The Heisenberg model
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1 1 The Heisenberg model 1.1 Definition of the model The model we will focus on is called the Heisenberg model. It has the following Hamiltonian: H = 1 J ij S i S j. (1) 2 i,j i j Here i and j refer to sites on a lattice. The model can be defined on any lattice, but for concreteness (and to eep things simple) we will here limit ourselves to a hypercubic lattice in d dimensions (d = 1, 2, 3). 1,2 The S i are spin operators which live on the lattice sites. Spin components on the same lattice site obey the standard angular momentum commutation relations [Sj α, S β j ] = i ɛ αβγ S γ j (α, β, γ = x, y, z) (2) γ and spins on different sites commute with each other. The spin operators all have spin S, i.e. the operators Si 2 have eigenvalue S(S +1) where S is an integer or half-integer. The spin interaction in (1), which is of the form S i S j, is called an exchange interaction, and the coefficients J ij are called exchange constants. We will mae the simplifying assumption (which is often realistic) that the spin interactions are negligible between spins that are not nearest-neighbors, i.e. J ij is nonzero only if i and j are nearest-neighbor lattice sites, in which case we further assume J ij = J, where J is a constant. There are then two different cases to consider, J < 0 and J > 0. For J < 0 the interaction energy of two spins favors them to be parallel; this is the ferromagnetic case. For J > 0 antiparallel orientation is instead favored; this is the antiferromagnetic case. 1.2 The S = 1/2 Heisenberg antiferromagnet as an effective lowenergy description of the half-filled Hubbard model for U t It turns out that the magnetic properties of many insulating crystals can be quite well described by Heisenberg-type models of interacting spins. Let us consider an example based on assuming that the electrons can be described in terms of the so-called Hubbard model, with Hamiltonian H = t i,j,σ(c iσ c jσ + h.c.) + U n i n i. (3) i This is probably the most important (and famous) lattice model of interacting electrons. c iσ creates an electron on site i with spin σ, and n iσ = c iσ c iσ counts the number of electrons with spin σ on site i. In this model there is therefore one electronic orbital per site. The first term in (3) is the inetic energy describing electrons hopping between nearest-neighbor sites i and j, and the second term is the interaction energy describing the energy cost U > 0 1 In 2 dimensions a hypercubic lattice is a square lattice and in 3 dimensions it is a cubic lattice. 2 Periodic boundary conditions will be used in all spatial directions, so that the Hamiltonian is fully translationally invariant. 1
2 associated with having two electrons on the same site (these electrons must have opposite spin, as having two electrons on the same site with the same spin would violate the Pauli principle). ote that the interaction energy between electrons which are not on the same site is completely neglected in this model. The Hubbard model is the simplest model describing the fundamental competition between the inetic energy and the interaction energy of electrons on a lattice. Despite much research, there is still a great deal of controversy about many of its properties in two and three dimensions (the model has been solved exactly in one dimension, but even in that case the solution is extremely complicated) Physical picture ow consider the case when there is exactly /2 spin-up electrons and /2 spin-down electrons in the system, where is the number of lattice sites. Hence the total number of electrons is which equals the number of lattice sites. This is called the half-filled case because maximally the system could contain two electrons per site (one of each spin) and thus a total electron number of 2. Assume further that U t. This suggests that to find the low-energy states we should first minimize the interaction energy and then treat the inetic energy as a perturbation. The interaction energy is minimized by putting exactly one electron on each site; then no site is doubly occupied so the total interaction energy is zero. Moving an electron creates a doubly occupied site which is penalized by a large energy cost U. Thus as long as we re only interested in understanding the physics of the system for energies (and/or temperatures) much less than U, we can neglect configurations with double occupancies. Thus the interaction energy completely determines the charge distribution of the electrons. What about the spin distribution, i.e. how to put up- and down-spin electrons relative to each other? As far as the interaction energy is concerned, any spin distribution gives the same energy. However, if we now consider the inetic energy term ( t) as a perturbation, it is clear that (see Fig. 1(a)) if neighboring electrons have opposite spins, an electron can hop (virtually, in the sense of 2nd order perturbation theory) to a neighboring site and bac; this virtual delocalization reduces the inetic energy. 3 In contrast (see Fig. 1(b)), if neighboring spins are parallel such hopping is forbidden, as the intermediate state with two electrons on the same site with the same spin would violate the Pauli principle. Therefore in this situation an effective interaction is generated which favors neighboring electrons to have opposite spin, i.e. antiparallel orientation. It can be shown 4 that the resulting effective model, valid at temperatures and energies U, is the Heisenberg antiferromagnet for the S = 1/2 electron spins, with J = 4t 2 /U (this expression for J can be understood from 2nd order perturbation theory: there is one factor of t for each of the two hops (first to the neighboring site, then bac) and a factor U coming from the energy denominator due to the larger energy of the intermediate state). Therefore, the antiferromagnetic exhange interaction J comes about due to an interplay between electron hopping, the electron-electron interactions, and the Pauli principle; the effect of the antiferromagnetic exchange is to reduce the inetic energy of the electrons. 3 That the inetic energy is reduced by this process can be understood from the formula for the energy correction in 2nd order perturbation theory, which is always negative for the ground state. 4 See e.g. Auerbach, Interacting electrons and quantum magnetism, Secs. 2.3 and 3.2, or agaosa, Quantum field theory in strongly correlated electronic systems, Sec
3 Figure 1: Illustration of why an effective antiferromagnetic interaction is generated in the half-filled Hubbard model with U t. In both (a) and (b) the transition shown is between an initial state with an electron on each site and an intermediate state where the electron initially on site 2 has hopped to site 3. See text for further explanation. We will not here go into how other examples of Heisenberg models (e.g. with spin S > 1/2 on each site and/or with ferromagnetic interactions) can arise as low-energy descriptions of different systems. Thus in the remainder of these notes we will simply consider the Heisenberg model as an interesting effective model of interacting spins and explore its properties (such as the nature of its ground state and excitations) without dwelling more on the origin of the model itself. 1.3 Ferro- and antiferromagnetic order Let us discuss the behavior of the spins, as a function of temperature, in materials described by the Heisenberg model. At high temperatures there are strong thermal fluctuations so that the spins are disordered, meaning that the expectation value of each spin vanishes: Ŝi = 0. (Here the bracets represent both a thermal and quantum-mechanical expectation value.) However, below some critical temperature T c it may be that the spins order magnetically, meaning that the spins on average point in some definite direction in spin space, Ŝi = 0. Whether or not such magnetic order occurs depends on the dimensionality and type of lattice, and the range of the interactions (we will limit ourselves to hypercubic lattices and nearestneighbor interactions in our explicit investigation of the Heisenberg model). If magnetic order occurs with T c > 0, then, as the temperature is lowered from T c down to zero, S i will increase and reach some maximum value at zero temperature. The critical temperature T c is called the Curie temperature (T C ) in ferromagnets and the eel temperature (T ) in antiferromagnets. The spin ordering pattern in two magnetically ordered phases are illustrated for a square lattice in Fig. 2: In the ferromagnetic case (J < 0) all spins point in the same direction (Fig. 2(a)) while in the antiferromagnetic case (J > 0) neighboring spins point in opposite directions (Fig. 2(b)). 3
4 , Figure 2: Average spin directions in phases with (a) ferromagnetic and (b) antiferromagnetic order on the square lattice. (The ordering direction is arbitrary.) 2 The Holstein-Primaoff representation ote that the commutator of two spin operators is itself an operator (see (2)), rather than just a complex number (c-number). This maes it much more complicated 5 to wor directly with spin operators than with canonical bosonic (fermionic) creation and annihilation operators whose commutators (anticommutators) are just c-numbers. It would therefore be advantageous if one could represent the spin operators in terms of such canonical bosonic or fermionic operators and wor with these instead. Fortunately quite a few such representations are nown. In these lectures we will mae use of the so-called Holstein-Primaoff (HP) representation which expresses the spin operators on a site j in terms of canonical boson creation and annihilation operators a j and a j as follows: S + j = 2S ˆn j a j, (4) S j = a j 2S ˆnj, (5) S z j = S ˆn j. (6) Here we have introduced the raising and lowering operators S ± j = Sj x ± is y j. The operator ˆn j a j a j is the number operator for site j, i.e. it counts the number of bosons on this site. As the allowed eigenvalues of Sj z are S, S + 1,..., S you can see from the equation for Sj z that this boson number must satisfy the constraint ˆn j 2S. (7) Also note that the expressions for S + j and S j are Hermitian conjugates of each other, as they should be. It can be shown that the spin commutation relations (2) and the relation 5 At least as far as analytical (as opposed to numerical) approaches are concerned. 4
5 S 2 = S(S + 1) follow from the HP representation and the bosonic commutation relations [a j, a j ] = 1 etc. (You will be ased to verify this in a tutorial). As will hopefully become clear in the next couple of sections, the HP representation is very useful for studying magnetically ordered states and their excitations. The reason for this is related to the fact that the vacuum of the a j bosons (i.e. the state with no such bosons) is the state corresponding to the maximum eigenvalue S of Sj z. 3 Spin-wave theory of ferromagnets We will now begin our study of the Heisenberg model (1) using spin-wave theory. We will first consider the ferromagnetic case, with J < 0 (which can be written J = J ). It will be convenient to rewrite the Hamiltonian in terms of the raising and lowering operators Ŝ ± j = Ŝx j ± iŝy j. Also since we re only considering nearest neighbor interactions we can write j = i + δ where δ is a vector connecting nearest-neighbor sites. Actually to avoid counting each interaction twice we will tae δ to run over only half the nearest neighbor vectors. We will consider hypercubic lattices only. Thus in 1D, we tae δ = +ˆx, in 2D δ = +ˆx, +ŷ and in 3D δ = +ˆx, +ŷ, +ẑ. This gives H = J [ ] 1 2 (S+ i S i+δ + S i S+ i+δ ) + Sz i Si+δ z. (8) i,δ The state with all spins pointing along z, i.e. Sj z = S is a ground state of the Heisenberg ferromagnet. This should be intuitively quite reasonable, and it can be proven easily. First let us show that it is an eigenstate. When we apply (8) to this state, the part involving the ladder operators give exactly zero because the operators S + j ill the ground state since it already has maximum Sj z which therefore cannot be increased further. Acting with the Si z Sj z gives bac an energy J S 2 z/2 times the same state, where z is the number of nearest neighbors. This shows that this state is an eigenstate. To show that it is also the ground state, we note that the minimal energy possible is given by E 0 = J i,δ max S i S i+δ. (9) It can be shown 6 that max S i S i+δ = S 2 which gives E 0 = J S 2 z/2, the same as above. Hence the state in question is indeed a ground state. We will now use the HP representation to study the Heisenberg ferromagnet, especially its excitations. As noted earlier we will consider a hypercubic lattice in d spatial dimensions, 6 The material in this footnote was not discussed in the lectures. Let O be a Hermitian operator and let { Φ n } be its complete set of orthonormal eigenstates, with eigenvalues O n. We will consider the expectation value of O in an arbitrary state Ψ. We can expand Ψ in the set { Φ n }: Ψ = n c n Φ n. ormalization of Ψ gives that n c n 2 = 1. The expectation value of O can then be written O Ψ O Ψ = c mc n Φ m O Φ n = }{{} m,n O nδ mn c n 2 O n. This shows that the largest value of O is given by the largest eigenvalue of O, obtained by taing Ψ to be the eigenstate of O with the largest eigenvalue. 5 n
6 i.e. a standard lattice in one dimension, a square lattice in 2 dimensions, and a cubic lattice in 3 dimensions. The theory that will be developed is nown as spin-wave theory. A natural guess for the low-energy excitations would be that they just correspond to small collective oscillations of the spins around the ordering direction (which we choose to be the z direction). Thus these oscillations, which are called spin waves, mae Sj z less than the maximum value S. In terms of the HP representation this means that the boson number ˆn j is nonzero, see (6). If this boson number ˆn j is much smaller than S the reduction in Sj z is small (i.e. very wea oscillations) and one might expect that an expansion in a small parameter proportional to ˆn j /S would mae sense. One might expect this to wor better the larger S is, since one might guess that increasing S would mae this parameter smaller (we ll verify this explicitly later). On the other hand, if it should turn out that ˆn j /S is not small (this is something we ll have to chec at the end of our calculation), then our basic assumption, that spin-waves are just wea oscillations around an ordered state, is wrong or at least questionable, and we may have to conclude that the system is not magnetically ordered after all. For example, this will be the conclusion if ˆn j /S turns out to be divergent, which we ll see some examples of later. The above indicates that spin-wave theory is essentially a 1/S expansion. It is semiclassical in nature, which follows since the limit S corresponds to classical spins, which can be seen e.g. from the fact that the eigenvalues of Ŝ2 i are S(S + 1) = S 2 (1 + 1/S). If the spins were just classical vectors of length S, the square of their length should be just S 2. Instead we see that there is a correction factor (1 + 1/S) due to the quantum nature of the spins. As the correction factor goes to 1 in the limit S, this limit corresponds to classical spins. Let us now discuss the spin-wave theory for the Heisenberg ferromagnet in detail. As noted earlier, in both the quantum and classical ground state all the spins point along the same direction, which we will tae to be the z direction. We then rewrite the spin operators in the Heisenberg Hamiltonian in terms of boson operators using the HP representation. We write 2S ˆn j = 2S 1 ˆn j /(2S) and expand the last square root here in a series in the operator ˆn j /(2S). This gives H = J S 2 z/2 J S i,δ [a i+δ a i + a i a i+δ a i a i a i+δ a i+δ] + O(S 0 ). (10) Here is the total number of lattice sites and z is the number of nearest neighbors and ow we consider spins on two different sites i j. Since (S i + S j ) 2 = S 2 i + S2 j + 2S i S j, we can write S i S j = 1 2 [ ] (S i + S j ) 2 Si 2 Sj 2. The largest expectation value of S i S j is, from (6), given by the largest eigenvalue of the operator on the rhs of this equation. The three terms on the rhs commute with each other, so we can find simultaneous eigenstates for them. The eigenvalue of S 2 i and S2 j is S(S+1). The eigenvalues of (S i+s j ) 2 are S tot (S tot +1), where S tot = S S,..., S + S = 0, 1,..., 2S. Thus the largest expectation value of S i S j is obtained with S tot = 2S, which gives max S i S j = 1 2 [2S(2S + 1) S(S + 1) S(S + 1)] = S2. 6
7 given by z = 2d for a hypercubic lattice in d spatial dimensions. (ote that after doing the summations the two last terms inside the square bracets are in fact identical). We have only included terms of O(S 2 ) and O(S) in (10). To this order, the HP representation reduces to 7 S + j 2Sa j, S j 2Sa j (and Sz j = S a j a j). The term of O(S) is quadratic in the boson operators and can be straightforwardly diagonalized. As it is quadratic it is equivalent to noninteracting bosons. Terms in the Hamiltonian which are higher order in the 1/S expansion (not shown) contain four or more boson operators and thus represent interactions between bosons. However, these are suppressed at least by a factor 1/S compared to the O(S) noninteracting term and one can thus hope that their effects are small (at least at large S and when the boson number is small) so that they can either be neglected to a first approximation or be treated as wea perturbations on the noninteracting theory. Let us next diagonalize the quadratic, O(S) term. In this ferromagnetic case, this can be accomplished simply by introducing Fourier-transformed boson operators as follows: a = 1 e i r i a i. (11) This is just a variable transformation, so there are as many operators a as there are operators a i. The inverse transformation is i a i = 1 e i r i a. (12) Periodic boundary conditions imply that e.g. a i = a i+x ˆx where x is the number of sites in the x direction. This is satisfied if e ixx = 1, i.e. x taes the form x = 2πn x / x where n x is an integer. It is customary to choose n x to tae the x successive values x /2, x /2 + 1,..., x /2 1 (here we have assumed for simplicity that x is even so that x /2 is in fact an integer). Then x taes values in the interval [ π, π. Doing the same for all directions, the resulting values of lie within what is called the first Brillouin zone. An important aspect of the transformation (11) is that it is canonical, i.e. it preserves the commutation relations in the sense that the operators a obey the same ind of commutation relations as the original boson operators: [a, a ] = δ, etc. Inserting (12) in (10) and using that ) r i i ei( = δ, one gets H = E 0 + ω a a, (13) where and ω = 2 J S δ E 0 = J S 2 z/2 (14) (1 cos δ) S J z(1 γ ), (15) 7 ote that when the HP expressions for S ± j are expanded in a series in ˆn j /(2S), and this series is truncated, the result is just approximate expressions for S ± j. In particular they do not exactly obey the spin commutation relations. 7
8 where we have defined γ = 2 cos δ. (16) z δ Eq. (13) describes a Hamiltonian which is just a bunch of independent harmonic oscillators, each labeled by a wavevector. The quanta of the harmonic oscillators are called magnons; they are the quantized spin wave excitations (just lie phonons are the quantized lattice vibrations in a crystal) with energy ω. In the limit 0, we have ω J S 2. (17) As a magnon with wavevector costs an energy ω > 0, 8 the ground state has no magnons, i.e. ˆn = 0 at zero temperature (here ˆn = a a ). The ground state energy is therefore simply E 0 which is the interaction energy of all spins pointing in the same direction with maximal projection S along the z axis. As the temperature is increased, magnons will be thermally excited. Since they are just noninteracting bosons (when O(S 0 ) terms and higher are neglected in the Hamiltonian, as done so far), the mean number of magnons with momentum is given by the Bose-Einstein distribution function, ˆn = 1 e βω 1. (18) The magnetization M (1/) i S i is a natural measure of the strength of the putative magnetic order in the system (cf. the discussion in Sec. 1.3). If M M is positive (zero) we say that the system is (is not) ferromagnetically ordered. The larger M is, the stronger is the ferromagnetic order. We say that the magnetization is an order parameter for the ferromagnetic phase. By definition, an order parameter for a given type of order is a quantity that is nonzero in the phase(s) where that order is present and is zero in other phases. With the ordering direction being the z direction, we get M = 1 Si z = S 1 ˆn i = S 1 ˆn S M. (19) i i We d lie to loo at how M depends on temperature for low temperatures. We first introduce an artificial wavevector cutoff 0 which is the smallest wavevector in the sum; the real system is described by the limit 0 0. We also introduce another wavevector > 0 which is chosen such that ω B T J S; this means in particular that for < the quadratic form (17) is valid. This gives M = 1 1 e J S2 / BT (20) e ω / BT 1 0 < < The second term is independent of 0 and finite. For reasons that soon will become clear, we will therefore neglect it and focus on the first term. Converting the sum to an integral and expanding the exponential (using J S 2 2 B T ) we get M d d 1 BT 0 J S { BT 2 J S 1/0 +..., d = 1 (21) log , d = 2 8 Here (and also in the antiferromagnetic case to be considered later) we gloss over a minor subtlety associated with the = 0 wavevector which comes with an energy ω = 0. 8 >
9 Therefore we see that, at nonzero temperatures in one and two dimensions, M diverges as the cutoff 0 is sent to zero. Therefore our initial assumption that this correction is small, is found to be wrong for these cases (note that the quantity M/(2S) is the expectation value of the average over all sites of our original expansion parameter ˆn j /(2S), which we assumed to be small when we expanded the square roots in the HP expression). Thus we conclude that M = 0 (i.e. there is no ferromagnetic order) at finite (i.e. nonzero) temperatures for the Heisenberg model in one and two dimensions. For the case of three dimensions, it can be shown 9 that M T 3/2 as T 0. Thus spin-wave theory predicts that ferromagnetic order is stable (i.e. M > 0) at sufficiently low temperatures in three dimensions. 4 Spin-wave theory of antiferromagnets We next turn to the antiferromagnetic case (J > 0 in (1)). As the ground state for classical spins has the spins on neighbouring sites pointing in opposite directions, one might naively guess that the ground state in the quantum case is analogous, thus having maximal and opposite spin projections ±S on neighboring sites. Such a state can be written S j S l. (22) j A l B Here A and B denote the two sublattices such that the spins on A (B) sites have spin projection S ( S), i.e. the states ± S are eigenstates of S z for the given lattice site with eigenvalue ±S. It is however easy to see that can not be the ground state, and is in fact not even an eigenstate, for the Heisenberg model for finite values of S: Acting with H in (1) on this state, the quantum fluctuation terms involving the spin raising and lowering operators change the state so that H is not proportional to. ote that this did not happen in the ferromagnetic case because then the S + operators always illed the ferromagnetic ground 9 The material in this footnote was not covered in the lectures. In this case one can find the leading temperature dependence at low temperatures by writing the Bose-Einstein function in terms of a geometric sum as follows (x J S 2 / B T ): 1 e x 1 = 1 e x 1 e x = e x (e x ) n = e nx, and integrating over wavevectors up to infinity (i.e. 0 0 and ). This gives M 1 (2π) 3 2 2π 0 n=0 d 2 n=1 n=1 e n J S2 / B T. Changing integration variable to u = n J S 2 /( B T ) the integral to solve is du ue u = (1/2) πerf( u) e u u where erf is the so-called error function which satisfies erf(0) = 0 and erf( ) = 1. This gives finally where ζ(s) = n n s is the Riemann zeta function. ( ) 3/2 M 1 B T ζ(3/2), 8 π J S 9
10 state, leaving only the contribution from the S z S z part of the Hamiltonian. Consequently, quantum fluctuations play a much more important role in the antiferromagnetic case, as they change the ground state (and its energy) away from the classical result. Although the ground state is not given by, it may still be that the ground state has antiferromagnetic order, i.e. that the spins on sublattice A point predominantly in one direction (taen to be the z direction here) and the spins on sublattice B point predominantly in the opposite direction. (If so the state captures the structure of the true ground state at least in a qualitative sense.) To investigate this possibility we again develop a spin-wave theory based on expanding the square roots in the HP expansion. On the A sublattice where the spin projection in (22) is +S we use the standard expressions: S + Aj = 2S a j a j a j, (23) S Aj = a j 2S a j a j, (24) S z Aj = S a j a j. (25) However, on the B sublattice where the spin projection in (22) is S we must modify the HP expressions accordingly to reflect this: S + Bl = b l 2S b l b l, (26) S Bl = 2S b l b l b l, (27) S z Bl = S + b l b l. (28) Compared to the expressions on the A sublattice, these modified expressions correspond to the changes S + S, S z S z, which preserve the commutation relations, which shows that the HP expressions for sublattice B are indeed correct. ote that different boson operators a j and b l have been introduced for sublattices A and B respectively. The indices j and l run over the sites in A and B respectively. Inserting the HP expressions in the Hamiltonian, expanding the square roots and eeping terms to order S in the Hamiltonian, we get H = J [ ] 2S 2 (a jb j+δ + h.c.) + S(a j a j + b j+δ b j+δ) S 2 j A δ + J [ ] 2S 2 (b la l+δ + h.c.) + S(b l b l + a l+δ a l+δ) S 2. (29) l B δ ext we introduce Fourier-transformed operators a = b = 1 e i r j a j, (30) A j A 1 e i r l b l. (31) B l B 10
11 Figure 3: Crystal Brillouin zone and magnetic Brillouin zone in one dimension (a) and two dimensions (b). where A = B = /2 is the number of lattice sites in each sublattice. The inverse transformation is a j = b l = 1 e i r j a, (32) A 1 e i r l b. (33) B The commutation relations are standard bosonic, i.e. the only nonzero commutators are [a, a ] = [b, b ] = δ,. We choose the -vectors to lie in the Brillouin zone associated with each sublattice. (Since the two sublattices are identical, their Brillouin zone is also identical). This Brillouin zone is called the magnetic Brillouin zone to distinguish it from the Brillouin zone associated with the full lattice which is called the crystal Brillouin zone. In one dimension periodic boundary conditions gives a j+a 2ˆx = a j (the factor of 2 comes from the fact that the spacing between neigboring sites in the sublattice is 2, not 1) which gives x = 2πn x /(2 A ). Choosing the A values of n x which lie closest to 0 (i.e. n x = A /2,..., A /2 1) then gives x [π/2, π/2. Thus the length of the magnetic Brillouin zone is half the length of the crystal Brillouin zone [π, π. In two dimensions the two sublattices are oriented at a 45 degree angle with respect to the full lattice and have a lattice spacing which is 2 larger. It follows from this that the magnetic Brillouin zone is a square oriented at a 45 degree angle with respect to the crystal Brillouin zone and has half its area. These facts are illustrated in Fig. 3. In the following you should eep in mind that when we are discussing antiferromagnets, it is implicitly understood that all -sums are over the magnetic Brillouin zone. Inserting (32)-(33) in the Hamiltonian and using ) r j j ei( = A δ, etc., and also renaming the summation variable as where needed, we get H = JS 2 z/2 + JSz [γ (a b + a b ) + a a + b b ], (34) where γ was defined earlier in (16). While the last two terms in the O(S) part are in diagonal form, the first two terms are not; thus in contrast to the ferromagnetic case, Fourier transformation alone does not diagonalize the Hamiltonian in the antiferromagnetic case. To bring the Hamiltonian into diagonal form, we will perform another canonical transformation, 11
12 nown as a Bogoliubov transformation: α = u a v b, (35) β = u b v a. (36) Here u and v are two real functions of which are to be determined. From the requirement that [α, α ] = [β, β ] = δ, we get the condition u 2 v 2 = 1. (37) From the requirement [α, β ] = 0 we get another condition, u v = u v. This is satisfied if u = u and v = v, which will be assumed to hold in the following. The inverse transformation is given by a = u α + v β, (38) b = u β + v α. (39) By expressing the Hamiltonian in terms of the α and β bosons it can be rewritten as H = JS 2 z/2 + JSz {(2γ u v + u 2 + v 2 )(α α + β β ) + 2(γ u v + v 2 ) + [γ (u 2 + v) 2 + 2u v ](α β + α β )}. (40) ow we will choose u and v such that the term which is not on diagonal form, i.e. the term proportional to α β + α β, vanishes. Thus we require that γ (u 2 + v 2 ) + 2u v = 0. (41) Furthermore we note that the condition (37) is automatically satisfied if we set u = cosh θ, v = sinh θ. (42) Inserting this into (41) gives 10 an equation which determines θ : tanh 2θ = γ. (43) Since γ = γ it follows that θ = θ which is consistent with our earlier assumption that u and v were even functions of. Using (42) and (43), after some manipulations we get H = JS 2 z/2 JSz/2 + ω (α a + β β + 1) = E 0 + ω (α a + β β ), (44) where we have defined ω = JSz 1 γ 2 (45) 10 We use that cosh 2 x + sinh 2 x = cosh 2x and 2 cosh x sinh x = sinh 2x. 12
13 and E 0 = JS 2 z/2 JSz/2 + ω. (46) Eq. (44) expresses the Hamiltonian in its final, diagonal form. The operators α and β create magnon excitations with wavevector and energy ω. These magnons are bosons, and are noninteracting in the approximation used here (i.e. when only including terms of O(S 2 ) and O(S) in the Hamiltonian). ote that in this antiferromagnetic case, for each there are two types of magnons (α and β) which are degenerate in energy. On the other hand, the sum goes over the magnetic Brillouin zone which only has /2 vectors, so the total number of magnon modes is 2 /2 =, the same as for the ferromagnetic case. ote that as 0, ω 0 as in the ferromagnetic case, but unlie the ferromagnetic case, for which a quadratic dispersion ω 2 was found in this limit, in the antiferromagnetic case we have instead a linear dispersion, ω as 0. (47) In the ground state of H (call it G ) there are neither α nor β magnons as these cost an energy ω > 0. Thus G can be defined by the relations α G = 0, β G = 0, for all. (48) This gives H G = E 0 G, i.e. the ground state energy is E 0, given in Eq. (46). The first term JS 2 z/2, i.e. the term S 2, is just the ground state energy E class of a classical nearest-neighbor antiferromagnet of spins with length S. The other terms are S and represent quantum corrections to the classical ground state energy. ote that this quantum correction E is negative: E = E 0 E class = ω JSz/2 = JSz [ 1 γ 2 1]. (49) }{{} <0 Thus quantum fluctuations lower the energy of the system. (ote that E was 0 in the ferromagnetic case, i.e. there were no quantum fluctuations in the ferromagnetic ground state.) Let us next investigate the amount of magnetic order in the system. Thus we need to identify an order parameter for antiferromagnetic order. ote that the magnetization M = (1/) i S i can not be used since it is zero in the presence of antiferromagnetic order, because the two sublattices give equal-magnitude but opposite-sign contributions to M. Instead the natural order parameter is the so-called sublattice magnetization, defined by averaging S i only over the sites of one of the two sublattices. Without loss of generality, let s pic sublattice A, where the putative ordering is in the z direction. The magnitude of the sublattice magnetization is thus M A = 1 Sj z = S 1 a j A a j = S 1 a A a. (50) A j A j A Writing M A = S M A, the correction M A to the classical result S is therefore given by M A = 1 a a. (51) A 13
14 We need to rewrite this further in terms of the α and β type magnon operators, as it is in terms of them that the Hamiltonian taes its simple harmonic oscillator form. This gives M A = 2 [u 2 α α + v β 2 β + u v α β + h.c. ]. (52) ow we can calculate the expectation values. They are both thermal and quantum, i.e. O Z 1 m m O m e βem where Z = m e βem. Here the sum m is over the eigenstates { m } of H with energy eigenvalues E m and β = 1/( B T ) where T is the temperature and B is Boltzmann s constant. First consider α β = Z 1 m m α β m e βem (53) and its complex conjugate. The summand involves m α β m, which is the overlap of the states β m and α m. Both states are eigenstates of H, but as they clearly do not have identical sets of occupation numbers of the various α and β bosons, their overlap is zero. Hence α β and its c.c. are zero. Thus M A = 2 [u 2 α α + v β 2 β + v] 2 (54) = 2 [n cosh 2θ (cosh 2θ 1)] ( ) = n (55) 2 1 γ 2 Here we used that α α = β β = 1/(e βω 1) n and cosh 2θ = 1/ 1 γ 2 (the last result follows from cosh 2 x = 1/(1 tanh 2 x)). We will not analyze Eq. (55) in its full glory, but briefly consider a few important points. This expression has a temperature-dependent part coming from n (note that n = 0 at T = 0) and a temperature-independent part coming from the two terms containing the factor 1/2. Let us first consider the case of zero temperature. In one dimension the -sum becomes (note γ = cos in one dimension) 1 1 π/2 lim 1 γ d 1 cos2. (56) The most important contribution to this integral comes from the small- region where we can approximate cos 1 2 /2. Thus the leading term becomes 0 d/ = log 0 as 0 0. Thus M A diverges even at zero temperature. We must therefore conclude that for this one-dimensional case our assumption that the system was magnetically ordered is invalid and so is our truncated spin-wave expansion. ote that this conclusion holds for any S. In two dimensions at zero temperature M A is still nonzero so quantum fluctuations do reduce the magnetization, but the correction turns out to be small enough, M 0.2, so that even for the lowest spin, S = 1/2, spin-wave theory indicates that the system is ordered 14
15 at zero temperature for a square lattice. This is also in agreement with other methods. In three dimensions M is even smaller so the order is more robust then. ext we consider finite nonzero temperatures. We just summarize the results that are obtained by analyzing (55) (which, we stress, is valid for a hypercubic lattice). In one dimension there is of course no antiferromagnetic order since none existed even at zero temperature. In two dimensions it turns out that the order does not survive at finite temperatures, so the spin-wave approach again is invalid. In three dimensions the system is ordered at sufficiently low temperatures. 5 Summary of spin-wave results In the table below we summarize the main conclusions we have obtained from applying spin-wave theory to the Heisenberg ferromagnet and antiferromagnet on a hypercubic lattice in d spatial dimensions (d = 1, 2, 3). Ordered means spin-wave theory predicts magnetic ordering of the ferromagnetic/antiferromagnetic type, disordered means spin-wave theory predicts the absence of such order. Ferromagnet Antiferromagnet d = 1, T = 0 Ordered Disordered d = 1, T > 0 Disordered Disordered d = 2, T = 0 Ordered Ordered d = 2, T > 0 Disordered Disordered d = 3, T = 0 Ordered Ordered d = 3, T > 0 Ordered (at low T ) Ordered (at low T ) 6 Broen symmetry and Goldstone modes 6.1 Broen symmetry By definition, a symmetry transformation of a Hamiltonian is a transformation that leaves the Hamiltonian invariant. As an example, the Heisenberg Hamiltonian (1) considered in these notes is invariant under global spin rotations. Therefore we say that the Heisenberg Hamiltonian has a global spin rotation symmetry. A global spin rotation (w, φ) is a transformation in which all the spin operators S i are rotated by the angle φ around the axis w (more precisely, w is a unit vector that points in the direction of the rotation axis). The word global refers to the fact that all spin operators are rotated in the same way. The invariance of the Heisenberg Hamiltonian under global spin rotations can be understood intuitively by noting that if the spins had been just classical vectors, rotating them all in the same way preserves the angles between them and thus it also preserves the scalar products S i S j in the Heisenberg Hamiltonian. The same conclusion can be shown to hold for the quantum case The rotations can be effected using rotation operators, which we haven t covered in this course. See e.g. Ch. 3 in J. J. Saurai, Modern quantum mechanics, revised edition, Addison-Wesley,
16 When the ground state of the Heisenberg Hamiltonian is magnetically ordered (either ferromagnetically for J < 0 or antiferromagnetically for J > 0), so that S i = 0, this ground state is not invariant under a global spin rotation. For example, if we have a ferromagnetically ordered ground state with all the spins pointing in the z direction, rotating this state by an angle φ around a general axis w leads to a different ferromagnetically ordered ground state with all spins pointing in a different direction. As a concrete example, if we rotate all spins by 90 degrees around the negative x direction, after the rotation all the spins would point in the y direction. Clearly the nonzero order parameter M is also not invariant; it undergoes exactly the same rotation as the spins. The same conclusion also holds for the antiferromagnetic ground state and its order parameter. To describe this situation we say that the ground state spontaneously breas the global spin rotation symmetry of the Hamiltonian. More generally, spontaneous 12 symmetry breaing refers to the situation when the ground state of the Hamiltonian is less symmetric than the Hamiltonian, i.e. the ground state is not invariant under all transformations that leave the Hamiltonian invariant. For simplicity we focused on the ground state here. The notion of spontaneous symmetry breaing can however be generalized to systems at finite temperature: the equilibrium state of the Heisenberg model is said to exhibit spontaneous symmetry breaing if it has magnetic order, i.e. if its order parameter is nonzero. More generally, most nown ordered phases of matter can be described in terms of broen symmetries. 13, Goldstone modes Symmetries of a Hamiltonian can be classified as discrete or continuous. A symmetry is discrete if the associated symmetry transformations form a discrete set. In contrast, a symmetry is continuous if the associated symmetry transformations form a continuous set. The global spin-flip symmetry of an Ising model is an example of a discrete symmetry. The global spin-rotation symmetry of the Heisenberg model, on the other hand, is a continuous symmetry, since the rotations form a continuous set. We found that both in the ferromagnetic and antiferromagnetic Heisenberg model, the magnon excitations have an energy ω that goes to zero as 0. This implies that there is no gap in the excitation spectrum, i.e. there is no minimum nonzero energy cost to creating an excitation. We say that the magnon excitations are gapless. When the ground state of the Heisenberg model is magnetically ordered and thus breas the continuous symmetry of the Heisenberg model, these gapless excitations are a consequence of the so-called Goldstone theorem, 15 and are therefore also sometimes referred to as Goldstone modes or Goldstone bosons. The Goldstone theorem implies that when the ground state breas a continuous symmetry of the Hamiltonian, gapless bosonic excitations will exist in the energy spectrum. 12 To understand the meaning of the word spontaneous here, consider again the magnetically ordered ground state of the Heisenberg model. The ordering direction of the spins can be any direction; the actual ordering direction is chosen by the system spontaneously. 13 The fact that symmetries can be spontaneously broen is mathematically rather subtle. We haven t had time to go into that here; see e.g. the discussion in Ch. 2 in. Goldenfeld, Lectures on phase transitions and the renormalization group, Westview Press, Much research, both theoretical and experimental, is currently being done looing for novel phases of matter with more subtle types of order that cannot be described in terms of broen symmetries. 15 More precisely, a generalized version of this theorem due to ielsen and Chadha. 16
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### 7.4. Why we have two different types of materials: conductors and insulators?
Phys463.nb 55 7.3.5. Folding, Reduced Brillouin zone and extended Brillouin zone for free particles without lattices In the presence of a lattice, we can also unfold the extended Brillouin zone to get | 16,906 | 67,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | latest | en | 0.907153 |
https://www.numbersaplenty.com/66811140 | 1,716,064,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00573.warc.gz | 827,775,878 | 4,046 | Search a number
66811140 = 223251141823
BaseRepresentation
bin1111111011011…
…1010100000100
311122201100122200
43332313110010
5114100424030
610343554500
71440612444
oct376672404
9148640580
1066811140
1134793220
121a45ba30
1310ac3226
148c32124
hex3fb7504
66811140 has 144 divisors (see below), whose sum is σ = 226751616. Its totient is φ = 15782400.
The previous prime is 66811109. The next prime is 66811183. The reversal of 66811140 is 4111866.
It is an unprimeable number.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 80769 + ... + 81591.
It is an arithmetic number, because the mean of its divisors is an integer number (1574664).
Almost surely, 266811140 is an apocalyptic number.
66811140 is a gapful number since it is divisible by the number (60) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 66811140, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (113375808).
66811140 is an abundant number, since it is smaller than the sum of its proper divisors (159940476).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
66811140 is a wasteful number, since it uses less digits than its factorization.
66811140 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 890 (or 885 counting only the distinct ones).
The product of its (nonzero) digits is 1152, while the sum is 27.
The square root of 66811140 is about 8173.8081700025. The cubic root of 66811140 is about 405.7728270396.
The spelling of 66811140 in words is "sixty-six million, eight hundred eleven thousand, one hundred forty". | 518 | 1,807 | {"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-2024-22 | latest | en | 0.869782 |
https://mathoverflow.net/questions/402501/moduli-spaces-of-horizontal-curves | 1,723,722,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00765.warc.gz | 302,440,360 | 24,143 | # Moduli spaces of horizontal curves
Let $$f:X\rightarrow Y$$ be a morphism of projective varieties. We may assume that $$X$$ and $$Y$$ are smooth, and $$f$$ is flat of relative dimension one. Fix an ample divisor $$A$$ on $$X$$.
I would like to ask if there exists a compact moduli space $$\overline{M}_{g,d}(X)$$ such that all points of $$\overline{M}_{g,d}(X)$$ represent curves $$C\subset X$$ of arithmetic genus $$g$$ and degree $$d$$ (with respect to $$A$$) with the following additional property:
$$(\star)$$ none of the irreducible components of $$C$$ is contracted by $$f$$.
Thank you.
• As noted by@WillSawin, there is no fine module space for such curves. However, when $g$ equals zero, there is the space of quasi maps. This has some of the properties that you list. Commented Aug 25, 2021 at 15:11
It is not possible to have such a moduli space that contains all the smooth curves of genus $$g$$ and degree $$d$$ and over which the universal family of curves is proper.
Let $$X = \mathbb P^1 \times \mathbb P^1$$ with coordinates $$x,y$$, $$Y= \mathbb P^1$$, $$f$$ the projection onto the $$x$$ coordinate. Let $$C_t$$ be given by $$y=tx$$. In the limit as $$t \to \infty$$, this converges to the curve with the vertical component $$x=0$$ (in addition to the horizontal component $$y=\infty$$). So if the moduli space is compact and the universal family is proper, the limit as $$t \to \infty$$ of $$C_t$$ will contain that component.
You could try to do something non-proper but if you want the universal family to be flat you will still have problems, as then $$C_{\infty}$$ will have to be contained in the union of $$x=0$$ and $$y = \infty$$, so to have the same degree as $$C_t$$ according to an ample line bundle will have to contain both $$x=0$$ and $$y=\infty$$. | 500 | 1,789 | {"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": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-33 | latest | en | 0.906644 |
https://stats.stackexchange.com/questions/208666/estimating-normalization-constant-with-monte-carlo-integration | 1,725,719,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00089.warc.gz | 517,593,595 | 39,759 | # Estimating normalization constant with Monte Carlo integration
Be $f(x)$ a function. Suppose that $f(x)$ integrates to a finite value $k$:
$$\int_{-\infty}^{\infty}f(x)dx=k$$
The normalization constant of $f(x)$ is $1/k$. Monte Carlo integration can give an estimate $\hat{k}$ of $k$, so we can estimate the normalization constant simply by taking the reciprocal of this estimate ($1/\hat{k}$). According to the CLT, the Monte Carlo estimator has an asymptotic normal distribution, so a 95% confidence interval for $k$ is simply: $$[\hat{k}-1.96*SE,\hat{k}+1.96*SE]$$ The question is: how can I calculate a confidence interval for the normalization constant? Can I just calculate the reciprocal of the previous confidence interval?
Yes: the defining property of a 95% confidence interval $[L, U]$ for an unknown parameter $k$ is that $P(L < k < U) = 95\%$. Since $L < k < U \iff 1/U < 1/k < 1/L$, you are guaranteed that $P(1/U < 1/k < 1/L) = 95\%$, so $[1/U, 1/L]$ is a 95% CI for $1/k$. | 292 | 992 | {"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} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.741272 |
https://math.stackexchange.com/questions/2751322/a-set-contained-in-varepsilon-neighborhood-of-a-finite-dimensional-subspace-i | 1,566,479,787,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317130.77/warc/CC-MAIN-20190822130553-20190822152553-00288.warc.gz | 555,293,413 | 31,575 | # A set contained in $\varepsilon$-neighborhood of a finite dimensional subspace is totally bounded
I was thinking about the following: if $X$ is totally bounded, then we can always choose the space spanned by the centers of an $\varepsilon$-ball net so that $X$ is lies in a neighborhood of finite dimensional vector space. Is the converse true?
Let $V$ be a Banach space and $X \subseteq V$ a bounded subset. If for all $\varepsilon >0$, $$X \subseteq B_\varepsilon(W_{\varepsilon}):= \{ x \in V : ||x-y|| < \varepsilon, y \in W_{\varepsilon} \}$$ for some $W_{\varepsilon}$ finite dimensional subspace, then $X$ is totally bounded.
EDIT: I think I have a proof for the statment.
Suppose $X$ is not totally bounded. Exists $(x_i) \subseteq X$ such that $||x_i - x_j || \ge 2\varepsilon$ for all $i \not= j$.
1. Exists $y_i \in W$ such that $||x_i - y_i || < \varepsilon/2$ for each $i$.
2. $||y_i -y_j|| \ge ||x_i- x_j|| - \Big(||x_i-y_i||+||x_j-y_j||\Big) \ge \varepsilon$ for all $i \not= j$.
3. Note that $||y_i||$ is bounded as $||y_i|| \le ||x_i||+\varepsilon<2 \le M +\varepsilon/2$ .
4. But the closed ball $B_M(0) \subseteq W$ is closed, bounded, hence compact. Hence sequentially comapct. Contradiction.
Is this proof correct?
• Yes, because all bounded subsets of $W$ are totally bounded, since $W$ is finite-dimensional. – Giuseppe Negro Apr 24 '18 at 9:12
• Where can I get a reference for the proof of your statement? – CL. Apr 24 '18 at 9:18
• Sorry, but I am also failing to see how the statement might follow directly from your claim. – CL. Apr 24 '18 at 9:26
• I meant that, since $X$ is a bounded subset of $W$ which is a finite-dimensional space, it is totally bounded. This is the same as the proof you wrote, essentially. – Giuseppe Negro Apr 24 '18 at 9:28
• In the present form, your condition is just a complicated way to say that $X\subset W$. – MaoWao Apr 24 '18 at 10:10
Given $\epsilon > 0$, let $W$ be a finite-dimensional subspace such that $X$ is contained in the $(\epsilon/2)$-neighborhood of $W$. For each point of $X$, pick a point of $W$ at distance less than $\epsilon/2$ from it. Let $E$ be the set of all such points in $W$. Since $X$ is bounded, so is $E$. Being a bounded subset of a finite-dimensional linear space, $E$ is totally bounded. Thus, there exists a finite set $F$ such that $E$ is contained in the $(\epsilon/2)$-neighborhood of $F$. Consequently, $X$ is contained in the $\epsilon$-neighborhood of $F$, which shows it is totally bounded.
The above is taken from my blog where I mention a simple corollary: a subset of a Banach space is compact iff it is closed, bounded, and "flat" in the sense you described (contained in $\epsilon$-neighborhood of a finite-dimensional subspace, for every $\epsilon$).
• Great, I only dislike the adjective "flat": for example, with this notation we have that $\mathbb S^{d-1}$ is flat, because it is compact. That's just a matter of taste. – Giuseppe Negro Apr 25 '18 at 9:13 | 901 | 2,978 | {"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} | 3.390625 | 3 | CC-MAIN-2019-35 | latest | en | 0.866483 |
https://ask.learncbse.in/t/suppose-that-you-can-lift-no-more-than-650-n-around-150-lb-unaided/57660 | 1,701,189,217,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00430.warc.gz | 138,552,341 | 3,465 | # Suppose that you can lift no more than 650 N (around 150 lb) unaided
Suppose that you can lift no more than 650 N (around 150 lb) unaided.
a) How much can you lift using a 1.4-m-long wheelbarrow that weighs 80.0 N and whose center of gravity is 0.50 m from the center of the wheel? The center of gravity of the load carried in the wheelbarrow is also 0.50 m from the center of the wheel.
b) Where does the force come from to enable you to lift more than 650 N using the wheelbarrow? | 139 | 486 | {"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-2023-50 | latest | en | 0.952506 |
https://www.coursehero.com/file/77990567/FINAL-PROJEK-CALCULUS-GROUP8pdf/ | 1,653,397,187,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00188.warc.gz | 813,438,332 | 59,073 | # FINAL PROJEK CALCULUS (GROUP8).pdf - FACULTY OF CIVIL...
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FACULTY OF CIVIL ENGINEERING AND BUILTENVIRONMENTBFC 15003CIVIL CALCULUSPROJECT GROUP 8NAMEMATRIC NOMUHAMAD ZARIF EMAN BIN ZABRIAF200099MARWAN NASRALDEENAF200213SITI NOR AMALINI BINTI HASSANCF200249PALESLY BINIUSCF200021MAHAMAT OUMAR HAROUNAF200208SECTION: 8SEMESTER/SESSION: 1/2020LECTURER’S NAME: DR. ADEL ALI SAEED ABDUHALGHEETHIDATE SUBMIT: 7 JANUARY 2021MARKAHTUGASAN
3INTRODUCTIONIn this project, application of differentiation and integration in civil engineeringfield. The aim of this project is to find speed of traffic located at Site B, in front of BHPetrol,TamanUniversiti.DifferentiationandIntergrationaretwofundamentalconcepts in calculus, which studies the change. Differentiation, in mathematics,process of finding the derivative, or rate of change, of a function. In contrast to theabstract nature of the theory behind it, the practical technique of differentiation can becarried out by purely algebraic manipulations, using three basic derivatives, four rulesof operation, and a knowledge of how to manipulate functions. The three basicderivatives are for algebraic functions is which any real number, for trigonometricfunctions and for exponential functions.The other basic rule, called the chain rule,provides a way to differentiate a composite function. Also, derivative of a function isthe slope or the gradient of curve(graph) at any point that has given. Gradient of acurve at any given point is the gradient of the tangent drawn to that curve at the givenpoint. For non-linear curve, the gradient of the curve can be different points along theaxis. Differentiation process is useful in calculating the gradient of the curve at anypoint. After acquiring the understanding of differention in this chapter, the studentswill be exposed to some of applications on differentiation. Below is differentiationequation :-Furthermore,integration is the process of calculating either definite integral orindefinite integral. For a real function f(x) and a closed interval [a,b] on the real line,the definite integral is define as the area between the graph of the function, thehorizontal axis and the two vertical lines at the end points of an interval. Whenspecific reverse method to differentiation, and has wide applications, for instance, inthis case, areas beneath curves needs to be find. This section explains what's meant byintegration and provides several customary integration techniques.Below is integralequation: -
4Integration and Differentiation are two fundamental concepts in calculus, whichstudies the change. The knowledge of calculus functions is important because it iswidely used in science, engineering, computer science, economy, medicine and etc.
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# 03.25.08_Page_3 - Blank Section 1.2 Series 261”< so...
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Unformatted text preview: Blank: Section 1.2 Series 261” < so if {3”} is convergent (lirn Sn = S) nzl ngaw . , w 7 a Geometrlc Serles Zar” 1 = 1—, 'flrl <1 n:1 -r Zorn and 2 either both converge or diverge n:1 n=k Test for Divergence If lirn an at 0 then summation diverges Blank It means 20!” convergent 9 lim an = 0 But lim afl = 0 does not go to Earn convergent 41. 0.§=0.222=£+%=i3+m. geometric series with a=im=i 10 10 10 10 10 l 0_§=L=L=E l—r 1_i 9 20 Section 11.3 Integral Test “° 1 1 1 1 Exam 1e: Investi ate —=—+—+—+... p g ”2:1: :12 12 22 32 Let’s look at geometry 1 Consider curve y = —2 x Total area of rectangle (forget that) 612 +613 +614 +... < Laof(x)dx=foxi2dx=1 p:2 ...
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#1 2015-05-19 05:26:42
shyclaw
Member
From: Dagoba
Registered: 2015-03-23
Posts: 67
Guestbook
Tell me about yourself!
-Shyclaw
V.R.
"“A Jedi uses the Force for knowledge and defense, never for attack.”
-Yoda
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#2 2015-05-19 05:43:51
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
I am a Kaboobly Dooist.
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#3 2015-05-19 05:45:11
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
I am a Gemini, have dark hair, like hiking and spaghetti and I am not afraid to cry.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#4 2015-05-19 06:08:28
shyclaw
Member
From: Dagoba
Registered: 2015-03-23
Posts: 67
Re: Guestbook
I am a young girl, I have dark hair and dark eyes, I love music and pizza! I also love working with code and ALT codes.
ALT 1=☺
ALT 2=☻
ALT 3=♥
ALT 4=♦
ALT 5=♣
ALT 6=♠
ALT 7=•
ALT 8=◘
ALT 9=○
"“A Jedi uses the Force for knowledge and defense, never for attack.”
-Yoda
Offline
#5 2015-05-27 06:11:21
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
I am an Agnishom
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#6 2015-05-27 10:42:18
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
You are an Agnishom.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#7 2015-07-12 23:29:51
threep14rrr
Member
From: The country I made up.
Registered: 2015-07-12
Posts: 109
Re: Guestbook
Uhhh... I'm 5 ft 5 in (Lil Wayne's height), had visited Singapore before, and I like making metro maps.
Memorizing pi ain't no piece of PIe. You need to pray and have PIety, because it's an ePIc job.
Progress-120
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#8 2015-07-13 00:14:21
David
Member
From: Bumpkinland
Registered: 2014-04-23
Posts: 3,135
Re: Guestbook
Agnishom wrote:
I am an Agnishom
Name: Agnishom
Species: Immobile computer
Profession: Moderator @ Brilliant.org Self-proclaimed kaboobly dooist, NOT A bumpkin. //Edited 8:22 a.m. 7/13/15
Age: 14 //? Verification required... //?
Favorite OS : Linux Ubuntu
Kernel Version : Linux_Agnishom 1.0
Last edited by David (2015-07-13 00:22:38)
Meaningless, meaningless, Everything is meaningless! - Ecclesiastes 1:2
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#9 2015-07-13 00:18:02
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
That resume contains false data.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#10 2015-07-13 00:19:29
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
Yes, I am most likely not a bumpkin.
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#11 2015-07-13 00:25:24
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
You are not 14, you are not a kaboobly dooist. You use Ubuntu 14.04. Your species is human.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#12 2015-07-13 00:31:02
threep14rrr
Member
From: The country I made up.
Registered: 2015-07-12
Posts: 109
Re: Guestbook
Haha
But true
Memorizing pi ain't no piece of PIe. You need to pray and have PIety, because it's an ePIc job.
Progress-120
Offline
#13 2015-07-13 00:32:43
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
Recall what mommym said?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#14 2015-07-13 00:34:23
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
She did not say much but when she did, it made me laugh.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#15 2015-07-13 00:42:31
David
Member
From: Bumpkinland
Registered: 2014-04-23
Posts: 3,135
Re: Guestbook
bobbym wrote:
You are not 14, you are not a kaboobly dooist. You use Ubuntu 14.04. Your species is human.
//Edited 8:45 a.m. 7/13/15
Name: Agnishom
Species: Human
Profession: Moderator @ Brilliant.org
Age: ≈ 14
Favorite OS : Linux Ubuntu 14.04
Kernel Version : Linux_Agnishom 3.0
Meaningless, meaningless, Everything is meaningless! - Ecclesiastes 1:2
Offline
#16 2015-07-13 00:43:50
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
I think he uses a kernel version of around 3.14.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#17 2015-07-13 00:44:58
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
bobbym wrote:
She did not say much but when she did, it made me laugh.
When you say something, say something nice
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#18 2015-07-13 00:49:47
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
I did say lots of nice things.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#19 2015-07-13 00:52:28
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
You said I am not 14
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#20 2015-07-13 00:54:08
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
You are 16, are you not?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#21 2015-07-13 00:58:02
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
I have been 14 for around three years
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#22 2015-07-13 01:05:05
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
Why do you stay there?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#23 2015-07-13 01:06:42
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
Because 17 - 3 = 14
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
#24 2015-07-13 01:09:28
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Re: Guestbook
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
#25 2015-07-13 01:23:43
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
Re: Guestbook
If you cannot compute 17 - 3, you'd better check your program
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
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Powered by FluxBB | 3,097 | 9,803 | {"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-2017-43 | latest | en | 0.850364 |
https://researchhubs.com/post/ai/data-analysis-and-statistical-inference/accuracy-vs-precision.html | 1,709,382,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00422.warc.gz | 495,721,966 | 4,889 | ### Introduction
This section is going to talk about accuracy and precision of confidence intervals.
The accuracy is defined in terms of whether or not the confidence interval contains the true population parameter.
The precision refers to the width of a confidence interval.
### Confidence Level
First, let's define the confidence level.
• Suppose we took many samples and built a confidence interval from each sample using the equation:
`point estimate ± 1.96 × SE`
• Then about 95% of those intervals would contain the true population mean ( μ ).
• Commonly used confidence levels in practice are 90%, 95%, 98%, and 99%.
Remember that we saw earlier that changing the confidence level simply means adjusting the value of the critical value in the confidence interval formula.
For example, the vertical line represents the true population mean which we rarely known. And each horizontal line is an interval calculated based on different random sample. There are 25 total interval plotted, and 24 of them contain the true population mean, and one does not. Therefore, the confidence level for these intervals would be 24 over 25:
`24 / 25 = 0.96 = 96%`
This is not exactly 95%, but it is close enough. If we examine many more intervals, the percentage of those capturing the true population parameter will be closer to 95%.
#### Wider or narrow interval?
If we want to be very certain that we capture the true population parameter, shall we use a wider interval or a narrower interval? Looking at this figure, it seems like a wider interval would indeed be much better. You can think about the red interval that is plotted on this figure and imagine that it extends even further. It would be much likely for it to then capture the true population parameter which is shown here as the vertical dashed line.
Therefore, as the confidence level increase, so does the width of the confidence interval.
Another way of thinking about this is the width of the area that captures the middle 95% or 99% of the distribution.
• The middle 99% will inevitably span a larger area, and hence the 99% confidence interval is going to be wider. Therefore, as we increase the confidence level, the width of the interval increases as well.
• More accurate means a higher confidence level. So if we are saying that we want to increase accuracy, we also need to increase the confidence level, but this might come at a cost.
#### What is the drawback when using a wider interval
As the confidence level increase, the width of the confidence interval increase as well. Which then increase the accuracy . However, the precision goes down.
``` CL
↑
Width
↑
Accuracy
↑, but
Precision
↓
```
Example:
Suppose you are watching the weather forecase, and you are told that the next day, low is -20F and high is 110F.
• Is this accurate? Most likely, yes.
• Tomorrow's temperature is probably going to be somewhere between -20F and 100F, however is it informative? Or, in other wards, is it precise? Not really. It is nearly impossible to figure out what to wear tomorrow according to this information.
### Example
The General Social Survey (GSS) is a sociological survey used to collect data on demographic characteristics and attitudes of residents of the US. In 2010, the survey collected responses from 1,154 US residents. Based on the survey results, a 95% confidence interval for the average number of hours Americans have to relax or pursue activities that they enjoy after an average work day was found to be 3.53 to 3.83 hours. Determine if each of the following statements are true or false.
• (a) 95% of Americans spend 3.53 to 3.83 hours relaxing after a work day.
• (b) 95% of random samples of 1,154 Americans will yield confidence intervals that contain the true average number of hours Americans spend relaxing after a work day.
• (c) 95% of the time the true average number of hours Americans spend relaxing after a work day is between 3.53 and 3.83 hours.
• (d) We are 95% confident that Americans in this sample spend on average 3.53 to 3.83 hours relaxing after a work day.
• (a) is False , because the confidence interval is not about individuals in the population. But in stead, about the true population parameter.
• (b) is True , because it is the definition of the confidence level. The percentage of random samples that will yield confidence intervals that contain the true population parameter.
• (c) is False , because the population parameter is not this moving target that is sometimes within an interval and sometime outside of it.
• (d) is False , because the confidence interval is not about the sample mean but instead about the population mean.
• N/A | 1,014 | 4,696 | {"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-2024-10 | latest | en | 0.895816 |
https://ch.mathworks.com/matlabcentral/cody/problems/44946-solve-a-system-of-linear-equations/solutions/2809309 | 1,604,154,081,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00539.warc.gz | 229,666,987 | 16,966 | Cody
# Problem 44946. Solve a System of Linear Equations
Solution 2809309
Submitted on 8 Aug 2020 by Suganth R
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
theta = 2*pi; x_correct = [1; 1]; assert(sum(abs((solve_lin(theta)-x_correct)))< 1e-5)
2 Pass
theta = pi; x_correct = [-1; -1]; assert(sum(abs((solve_lin(theta)-x_correct)))< 1e-5)
3 Pass
theta = -0.5; x_correct = [1.357;0.3982]; assert(sum(abs((solve_lin(theta)-x_correct)))< 1e-4)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 208 | 698 | {"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-2020-45 | latest | en | 0.699649 |
https://blog.doublehelix.csiro.au/catching-a-colossal-prime/ | 1,720,966,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00791.warc.gz | 123,799,424 | 21,658 | # Blog
## Catching a colossal prime
By
,
Late last year, Jonathan Pace’s computer found something special. Jonathan is an electrical engineer who also manages computers for charities, so he has a lot of computers, but this one was nothing out of the ordinary. Except for one thing: the computer was running software from the Great Internet Mersenne Prime Search, an international volunteer group on the hunt for giant primes.
For six days, this computer analysed a very large number, doing calculation after calculation, trying to find a number that would divide into it without leaving a remainder. Eventually, it ran out of divisors to check. The very large number couldn’t be made by multiplying smaller whole number together. It was a prime number.
There are plenty of prime numbers out there, and one of the earliest facts ever discovered about prime numbers was that they never end. No matter how big a prime number you find, there are always bigger ones to discover.
Jonathan’s prime is special because it’s the largest prime ever found, with 23 249 425 digits. If you printed Jonathan’s prime in a book, it would have around 5200 pages. That’s more pages than all the Harry Potter books, plus The Hunger Games trilogy combined!
This newly discovered prime is so big it’s hard to imagine. It’s larger than the number of stars in the sky, and larger than the number of atoms in the ocean. If you packed the entire observable universe with atoms as tight as you possibly could, that number would only have around 180 digits – only about three lines on one page of a book!
Although Jonathan’s prime number is the largest we’ve ever discovered, we know there are even larger ones out there. And there’s plenty of incentive to look. The first person to discover a prime number with more than 100 million digits will win a \$150 000 prize!
If you’re after more maths news for kids, subscribe to Double Helix magazine!
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# Multiplication and Division games - Fall Themed
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Multiplication and Division No PREP games for 1 or 2 players contain 20 printable games to use during autumn or fall. Each game can be played in pairs or used as an interactive worksheet by a single player. A recording sheet is included. To play, players only require a colored pencil or pen and a paperclip.
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Lots of Leaves – multiply the numbers 2, 3, 4 or 5 with 2 or 3 then double
It's Turkey Time – multiply two numbers and add 20
Pumpkin Patch – divide from 60 then find out how many more are needed to make 40
Harvest Happenings – Find the missing dividend
Gobble Gobble – divide by 8 then add 20
Whoos Right? – add two numbers then multiply by 6
Falling Leaves – multiply two numbers 6 – 9
Squirreling Away Nuts– multiply 5, 6, 7, or 8 with 5, 6 or 7
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Something to Crow About – Multiply by 5
An Apple a Day – multiply 4 - 7
A Job Well Done – multiply 5 - 8 by 7 - 9 then find out how many more are needed to make 80
Spider's Lair – Divide 2 numbers
Haunted House – Find the missing dividend
It's in the Bag – Divide by 6 then find out how many more are needed to make 20
Trick or Treat - Multiply 3 numbers
Bubbling Away – Divide 2 numbers
Welcome to Spooksville– Multiply by 8 then find out how many more are needed to make 100
The Witching Hour - Multiply 2 numbers
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Multiplication Game Pack - 35 games
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http://www.braingle.com/palm/teaser.php?id=48750&op=0&comm=1 | 1,495,810,963,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608668.51/warc/CC-MAIN-20170526144316-20170526164316-00170.warc.gz | 536,779,554 | 3,286 | Lite
## Elemental Opposites
Category: Science Submitted By: logicgenius Fun: (2.38) Difficulty: (2.04)
You have a bowl of water. Placed in the center of it is a paraffin candle. The water level is below the top of the wax by about a few milliliters. How can you use any ONE common household object so that when the candle is lit, the wick will go below the water level with the fire still going? A lighter is provided.
Show Hint Show Answer
Comments on this teaser
Posted by princess2007 06/02/11 Wow. I didn't know that!! Gonna try it sometime! :D Posted by Junj13 10/16/11 Nice question. :D Posted by zembobo 11/18/11 I tried it and it didn't work :cry: but I'm sure I did something wrong. Posted by smartguyneil 01/04/13 although wax may float on water the fire will still go out in this situation once the candle has burned for a little bit Posted by LanceAlot 09/27/15 Although technically if the diameter of the candle is large enough then the flame will just burn down below the water and the edges of the candle will remain intact and prevent water from burning into the candle. Posted by RICHARDYAMA 09/17/16 Find a large bowl with 6" depth, an empty tomato can(4.75" tall), a 4" long candle and a 2.375" diameter flower arranging metal "frog". Press the bottom of the candle into the frog, then drop the candle/frog into the can. This allows the candle a solid base. Place the can into the center of the bowl. Weigh the can down, if necessary, with pebbles, marbles, fishing lead, etc. This ensures that the can does not float off the bottom of the bowl. Fill the bowl to the top of the can. If you overfill the bowl, it won't matter at all. A few extra ounces of water will just sit at the bottom of the can. Light the candle, and you'll have a flame and wick below the water's surface! Posted by RICHARDYAMA 09/17/16 Oops! I used more than ONE common household object, didn't I? Ok, let's use a VERY heavy clear glass vase, 5" tall. Melt the bottom of a 4" candle and press it into the bottom of the vase. Place the assembly into a large bowl that's 6" deep or more. Fill the bowl with 5" water. If the vase flounders, simply add water into the vase until it stabilizes. Light the candle. You'll have a flame and wick below the water surface that'll burn all the way down, like it should. This lit candle should make for an awesome conversation piece! Posted by RICHARDYAMA 09/17/16 Ok, one more try. Cut 2" off of a shower curtain rod/tube. Press the rod over a 4" long pillar candle at the wick's end. Melt the other end and press it into the bowl. Light the candle, then fill the bowl with water. Do NOT fill past the top of the metal tube, but just a fraction of an inch below the top of the band. The candle will burn for hours, but you'll have to keep pushing the tube down the candle shaft as it melts away. And of course, the water level must always be below the edge of the tube. It's not as aesthetically pleasing as the candled vase, but the candle will burn for hours!
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Privacy | Terms | 804 | 3,158 | {"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-2017-22 | latest | en | 0.933115 |
http://www.numbersaplenty.com/231112333202230 | 1,585,589,256,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00297.warc.gz | 271,503,044 | 3,443 | Search a number
231112333202230 = 2523111233320223
BaseRepresentation
bin110100100011001000001000…
…100001100001001100110110
31010022022010022212111011200001
4310203020020201201030312
5220243021024244432410
62135311312401341514
766452213031151264
oct6443101041411466
91108263285434601
10231112333202230
1167704210008807
1221b0715093789a
139bc5a5169239c
14410dc7a42d434
151babb6745633a
hexd23208861336
231112333202230 has 8 divisors (see below), whose sum is σ = 416002199764032. Its totient is φ = 92444933280888.
The previous prime is 231112333202203. The next prime is 231112333202239. The reversal of 231112333202230 is 32202333211132.
Adding to 231112333202230 its reverse (32202333211132), we get a palindrome (263314666413362).
It is a happy number.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (231112333202239) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 11555616660102 + ... + 11555616660121.
It is an arithmetic number, because the mean of its divisors is an integer number (52000274970504).
Almost surely, 2231112333202230 is an apocalyptic number.
231112333202230 is a deficient number, since it is larger than the sum of its proper divisors (184889866561802).
231112333202230 is a wasteful number, since it uses less digits than its factorization.
231112333202230 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 23111233320230.
The product of its (nonzero) digits is 7776, while the sum is 28.
The spelling of 231112333202230 in words is "two hundred thirty-one trillion, one hundred twelve billion, three hundred thirty-three million, two hundred two thousand, two hundred thirty". | 552 | 1,864 | {"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-2020-16 | latest | en | 0.781895 |
http://forum.allaboutcircuits.com/threads/realize-a-function-from-a-digital-circuit.112369/ | 1,484,613,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279379.41/warc/CC-MAIN-20170116095119-00174-ip-10-171-10-70.ec2.internal.warc.gz | 109,831,815 | 13,168 | Realize a function from a digital circuit
Discussion in 'Homework Help' started by Abdullah AlMarkhi, Jun 11, 2015.
1. Abdullah AlMarkhi Thread Starter New Member
Jun 11, 2015
1
0
Hello,
The following question asks to realize the following circuit as a (Sum of Products) and I really got stuck finding the solution.
The diagram given: http://i.imgur.com/qtx7wFB.jpg
My Attempt of solution (which is wrong): (Y'+X') (W+Z+P') (W'+Z'+P') (X')
Anyone that can possibly help?
Thanks
2. WBahn Moderator
Mar 31, 2012
18,079
4,917
You need to show your WORK, so that we can see WHERE you went wrong.
Either show a truth table (which would have 32 rows) that shows the value at each node or develop the Boolean expression step by step. | 205 | 737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-04 | latest | en | 0.918333 |
https://www.physicsforums.com/threads/problem-with-bose-einstein-condensation.899417/ | 1,713,906,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00425.warc.gz | 839,912,931 | 16,918 | # Problem with Bose-Einstein Condensation
• A
• ShayanJ
In summary, the formula ## N_e=V\frac{(2\pi m k T)^{\frac 3 2}}{h^3}g_{\frac 3 2}(z) ## gives the number of particles not in the ground state at a given temperature. The maximum of ## g_{\frac 3 2}(z) ## occurs at ## z=1 ## and ## N_e ## reaches its maximum at this point. For temperatures above the critical temperature ## T_c ##, the number of particles in the excited states can be calculated, but if more particles are added, Bose-Einstein condensation will occur again. However, this is not expected to happen for temperatures above ## T_c ##, indicating
ShayanJ
Gold Member
In section 7.1 of his statistical mechanics, Pathria derives the formula ## N_e=V\frac{(2\pi m k T)^{\frac 3 2}}{h^3}g_{\frac 3 2}(z) ## where ## \displaystyle g_{\frac 3 2}(z)=\sum_{l=1}^\infty \frac{z^l}{l^{\frac 3 2}} ## and ## z=e^{\frac \mu {kT}} ##. This formula gives the number of particles that are not in the ground state w.r.t. the temperature.
The maximum of ## g_{\frac 3 2}(z) ## happens at ## z=1 ## and is equal to ## \zeta(\frac 3 2) ##. So whenever ## z=1 ##, ## N_e ## reaches its maximum and any other particle has to go to the ground state and ## z=1 ## happens at any ## T<T_c ##.
My problem is with ## T>T_c ##. I can calculate ## N_e ## for any temperature which gives me the capacity of the excited states at the given temperature. Now I put ## N>N_e ## particles in the energy levels and so the excited states become full and the rest of the particles have to go to the ground state and I get Bose-Einstein condensation again, this time for ## T>T_c ## which can't be right because we're supposed to have condensation only for ## T<T_c ##.
What's wrong here?
Thanks
The critical temperature is dependent on the particle density, so if you keep the volume constant and add more particles, ##T_{c}## is going to change as well.
Yeah, good point. Thanks.
## 1. What is Bose-Einstein Condensation?
Bose-Einstein Condensation (BEC) is a phenomenon that occurs when a group of particles with integer spin, called bosons, are cooled to a very low temperature and begin to occupy the lowest energy state. This results in all of the particles behaving as one coherent entity, rather than individual particles.
## 2. How is Bose-Einstein Condensation different from other states of matter?
Bose-Einstein Condensation is different from other states of matter, such as solids, liquids, and gases, because it is a quantum phenomenon that only occurs at extremely low temperatures. In this state, particles behave as waves rather than distinct particles, and the entire group of particles behaves as a single entity.
## 3. What are some real-world applications of Bose-Einstein Condensation?
Bose-Einstein Condensation has been used in various fields, including atomic physics, quantum optics, and superconductivity. It has also been used in the development of highly sensitive sensors and atomic clocks, as well as in the creation of new materials with unique properties.
## 4. What are the challenges in achieving Bose-Einstein Condensation?
One of the main challenges in achieving Bose-Einstein Condensation is cooling the particles to a low enough temperature. This requires specialized equipment and techniques, such as laser cooling and evaporative cooling. Another challenge is controlling the interactions between the particles, as this can affect the formation and stability of the condensate.
## 5. How does Bose-Einstein Condensation contribute to our understanding of quantum mechanics?
Bose-Einstein Condensation is a manifestation of quantum mechanics at a macroscopic level, providing insights into the behavior of matter at the atomic and subatomic levels. It also allows for the observation of quantum phenomena, such as superposition and entanglement, on a larger scale. This contributes to our overall understanding and exploration of the fundamental principles of quantum mechanics.
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2K | 1,087 | 4,401 | {"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-2024-18 | latest | en | 0.847666 |
http://openstudy.com/updates/4f2a97c4e4b049df4e9e362f | 1,448,579,624,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447860.26/warc/CC-MAIN-20151124205407-00217-ip-10-71-132-137.ec2.internal.warc.gz | 177,356,889 | 11,671 | ## JamesJ 3 years ago Who's up for a challenge? Show that for all positive $$x, y, z$$ we have $\left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2}$
1. Ishaan94
I can reduce it down to $(x+y)^{1/2}(y+z)^{1/2} + (x+y)^{1/2}(x + z)^{1/2} + (y+z)^{1/2}(x+z)^{1/2} \le 2(x+y+z)$
2. Tala
x+y+z$\le$2$1/2$
3. cinar
x=2 y=2 z=2 just one example
4. JamesJ
Hint: Use the Cauchy-Schwarz Inequality
5. nikvist
$\frac{1}{2}\left(x+y+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{x+y}{x+y+z}}$$\frac{1}{2}\left(y+z+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{y+z}{x+y+z}}$$\frac{1}{2}\left(z+x+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{z+x}{x+y+z}}$$\sum\quad\Rightarrow\quad\frac{1}{2}\left(2(x+y+z)+\frac{3}{x+y+z}\right)\ge$$\ge\sqrt{\frac{x+y}{x+y+z}}+\sqrt{\frac{y+z}{x+y+z}}+\sqrt{\frac{z+x}{x+y+z}}$$u=x+y+z\quad,\quad f(u)=u+\frac{3}{2u}\quad,\quad f'(u)=1-\frac{3}{2u^2}=0$$u=\sqrt{\frac{3}{2}}\quad\Rightarrow\quad f_{\min}=\sqrt{6}$
6. JamesJ
Nicely done. I'll wait a bit and see what other proofs turn up, if any, and then I'll post my solution.
7. JamesJ
got this ffm?
8. FoolForMath
I think I would have used AM-GM as shown by nikvist.
9. JamesJ
I'm going to write this out. It's going to take a few minutes ...
10. JamesJ
$\left( \frac{x+y}{x+y+z} \right)^{1/2} + \left( \frac{y+z}{x+y+z} \right)^{1/2} + \left( \frac{z+x}{x+y+z} \right)^{1/2}$ $= 1.\left( \frac{x+y}{x+y+z} \right)^{1/2} + 1.\left( \frac{y+z}{x+y+z} \right)^{1/2} + 1.\left( \frac{z+x}{x+y+z} \right)^{1/2}$ Now by Cauchy-Schwatz this expression is $\leq (1^2 + 1^2 + 1^2)^{1/2} . \left( \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} \right)^{1/2}$ $= 3^{1/2} \left( \frac{2(x+y+z)}{x+y+z} \right)^{1/2}$ $= 6^{1/2}$ qed
11. FoolForMath
That's even more compact, well done James.
12. JamesJ
It's a nice problem.
13. FoolForMath
Indeed it is.
14. MathDude000
Jamesj?
15. MathDude000
Hello?
16. FoolForMath
To ping somone use @ feature, for example @JamesJ
17. Libniz | 906 | 2,037 | {"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 | 4 | CC-MAIN-2015-48 | longest | en | 0.590808 |
https://math.stackexchange.com/questions/354462/prove-that-l-ax-ax-xa-is-symmetric-if-and-only-if-a-is-symmetric | 1,576,340,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00226.warc.gz | 442,043,190 | 33,264 | # Prove that $L_A(X)=AX-XA$ is symmetric if and only if $A$ is symmetric.
Define for a fixed $A \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ the mapping:
$$L_A : \mathbb{M}^{2 \times 2}(\mathbb{R}) \to \mathbb{M}^{2 \times 2}(\mathbb{R}) : X \mapsto AX-XA.$$
Define on $\mathbb{M}^{2 \times 2}(\mathbb{R})$ the dotproduct $\langle \cdot , \cdot \rangle$ as follows: $\langle X, Y \rangle = [X]_{\xi}^t [Y]_{\xi}$.
Here is $[\cdot ]_\xi$ the coordinate map that belongs to the standard basis
$$\xi = \{ E_1 = {\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}}, E_2={\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}}, E_3={\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}}, E_4={\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}} \}$$ of $\mathbb{M}^{2 \times 2}(\mathbb{R})$.
Write $M_A$ for the matrix such that for all $X \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ it satisfies $[L_A (X)]_\xi=M_A [X]_\xi$.
Prove that $L_A$ is symmetric if and only if $A$ is symmetric.
In a previous exercise I had to determine all the matrices $M_{E_1}, M_{E_2}, M_{E_3}, M_{E_4}$. And this is the result:
$M_{E_1} = {\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}},~M_{E_2}={\begin{pmatrix} 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}}, ~M_{E_3}={\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}}, \\M_{E_4}={\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}}$
How am I to use this to prove the given problem? I already showed that $L_A$ is symmetric if and only if $M_A$ is symmetric, meaning $\langle L_A(X), Y \rangle = \langle X, L_A(Y) \rangle$ if and only if $M_A = M_A^t$.
I am at a loss. Thanks in advance.
• This makes little sense. Part one, why does your definition of the inner product has no letter $Y$ in the right hand side? Second, what do you mean by symmetric for your operator? And, mostly, where did you get this? – Will Jagy Apr 8 '13 at 2:25
• I noticed a few mistakes. What I mean by symmetric is that $\langle L_A (X), Y \rangle = \langle X, L_A (Y) \rangle$ for all $X \in \mathbb{M}^{2 \times 2} (\mathbb{R})$. This is one of my teacher's questions to refresh some of our linear algebra. – Jeroen Apr 8 '13 at 2:31
Let $A=\pmatrix{a&b\\ c&d}$. Then $M_A=aM_{E_1}+cM_{E_2}+bM_{E_3}+dM_{E_4}$. Here are some entries of $M_A$: $$M_A=\pmatrix{ 0 &b &-c &0\\ c &d-a &0 &-c\\ -b &0 &\ast &\ast\\ 0 &-b &\ast &\ast}.\tag{1}$$ And I will leave the rest to you. Now, the given inner product $\langle X,Y\rangle$ is the usual dot product of $[X]_{\xi}$ and $[Y]_{\xi}$ in $\mathbb{R}^4$. Therefore, $L_A$ is symmetric w.r.t. $\langle\cdot,\cdot\rangle$ if and only if $M_A$ is a symmetric matrix. More specifically, \begin{align*} L_A \text{ is symmetric } &\Leftrightarrow \langle L_A(X),Y\rangle=\langle X,L_A(Y)\rangle \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\ &\Leftrightarrow [X]_{\xi}^tM_A^t [Y]_{\xi}=[X]_{\xi}^t M_A[Y]_{\xi} \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\ &\Leftrightarrow u^tM_A^tv=u^t M_Av \text{ for all } u,v\in \mathbb{R}^4\\ &\Leftrightarrow M_A^t=M_A. \end{align*}
From $(1)$, we see that for $M_A$ to be symmetric, a necessary condition is $b=c$. If you could determine the unspecified entries of $M_A$ correctly (note that your $M_{E_3}$ is wrong; there's perhaps a typo), you will see that $b=c$ is also a sufficient condition. Hence the result.
• Thanks for doing the way the person who wrote the exercise had in mind, +1. – Julien Apr 8 '13 at 10:55
Note: I am sorry this is a little far from your notations. But lots of people would define this inner product as $\mbox{Tr}(X^*Y)$. This idea of using the trace to create bilinear objects is frequent and natural (see the Killing form, to mention one famous example). Last but not least, observe that the argument below applies to $M_n(\mathbb{R})$ and even $M_n(\mathbb{C})$ for any $n$. This is my main motivation for doing it the following way. If you are only interested in the $2\times 2$ case, then you should keep your notations and do like @user1551 did.
Observe that the inner product you are given on the matrices has a nice and tractable matrix trace expression:
$$(X,Y)=\mbox{Tr}(X^*Y)$$
where $X^*$ denotes the adjoint of $X$, i.e. the transpose in the real case.
Now, using the commutativity of the trace $\mbox{Tr}(MN)=\mbox{Tr}(NM)$, straightforward computations show that $$(L_A(X),Y)=\mbox{Tr}(A^*(YX^*-X^*Y))$$ while $$(X,L_A(Y))=\mbox{Tr}(A(YX^*-X^*Y)).$$
Easy direction: it follows at once that if $A=A^*$, then $(L_A(X),Y)=(X,L_A(Y))$ for every $X,Y$. Hence $L_A=L_A^*$, i.e. $L_A$ is symmetric.
Conversely: assume that $L_A=L_A^*$. Then the matrix $B=A-A^*$ satisfies $$\mbox{Tr}(B(YX^*-X^*Y))=\mbox{Tr}((BY-YB)X^*) =0$$ for every $X,Y$. First, this implies, with $X=BY-YB$, $$\|BY-YB\|^2=\mbox{Tr}((BY-YB)^*(BY-YB))=\mbox{Tr}((BY-YB)(BY-YB)^*) =0.$$ Where the norm is the one induced by your inner product. It follows that $BY-YB=0$ for every $Y$, i.e. $B$ commutes with every other matrix. It is well-know that the only such matrices are the scalar matrices (nice exercise, just compute the products with matrices of the canonical basis $E_{ij}$). So $B=A-A^*=\lambda I_2$. But taking the trace of the latter, we find $2\lambda=\mbox{Tr}(A)-\mbox{Tr}(A^*)=0$. Finally, $B=A-A^*=0$, so $A=A^*$ is symmetric.
"Another" way is the following:
Show that $\langle L_{E_2}(X),Y\rangle=\langle X,L_{E_3}(Y)\rangle$ and that $\langle L_{E_3}(X),Y\rangle=\langle X,L_{E_2}(Y)\rangle$ (actually these two are equivalent).
Therefore for $A=\begin{pmatrix} a_1&a_3\\a_2&a_4\end{pmatrix}$ we have (note that $M_{E_1},M_{E_4}$ are symmetric) $$\langle L_A(X),Y\rangle=\sum_{i=1}^4a_i\langle L_{E_i}(X),Y\rangle=\\ a_1\langle X,L_{E_1}(Y)\rangle+a_2\langle X,L_{E_3}(Y)\rangle+a_3\langle X,L_{E_2}(Y)\rangle+a_4\langle X,L_{E_4}(Y)\rangle=\\ \langle X,L_{A^t}(Y)\rangle.$$ It follows that $L_A$ is symmetric $\iff \langle X,L_{A}(Y)\rangle=\langle X,L_{A^t}(Y)\rangle, \ \forall X,Y\in M^{2\times2}(\mathbb R)\ldots$ | 2,359 | 6,107 | {"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} | 3.828125 | 4 | CC-MAIN-2019-51 | latest | en | 0.602076 |
http://forums.wolfram.com/mathgroup/archive/2010/Apr/msg00283.html | 1,721,055,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00027.warc.gz | 12,788,138 | 9,124 | Re: if using Mathematica to solve an algebraic problem
• To: mathgroup at smc.vnet.net
• Subject: [mg109070] Re: if using Mathematica to solve an algebraic problem
• From: David Bailey <dave at removedbailey.co.uk>
• Date: Mon, 12 Apr 2010 06:54:54 -0400 (EDT)
• References: <hpml5n\$9nu\$1@smc.vnet.net> <hppl95\$m08\$1@smc.vnet.net>
```Helen Read wrote:
> On 4/9/2010 3:32 AM, David Park wrote:
>> Sometimes I find it difficult to understand these discussions.
>>
>> For example, Richard's: "There is of course the possibility that something
>> really useful will be developed that will make it possible to teach all students
>> everything they need to know." What kind of something would that be, and in
>> what way would it make it possible? It seems like a rather vague but
>> expansive goal.
>>
>> Similarly, David suggests that Mathematica might be "terribly dangerous"
>> with the possibility of "becoming skilled in answering questions through
>> Mathematica, rather than actually learning the subject!" (Not always bad. Is
>> there anything wrong at becoming skilled at driving to various locations in
>> your city without actually learning how the internal combustion engine
>> works?)
>
> I don't get this either. I teach my students to use Mathematica as part
> of a mathematics course -- mainly calculus, but other classes as well.
> We use Mathematica differently and for different purposes depending on
> the subject matter. Using Mathematica does not subtract anything from
> their learning of the course content. Rather, it adds to it.
>
> As just one example, when we come to the topic of series in Calculus II,
> after a little chalk (well, whiteboard) and talk, I give the students
> some examples of series and have them make tables and plots of terms and
> partial sums in Mathematica, and try to guess whether the series
> converges (and if so, to what). I'll give them the following examples to
> work on.
>
> (a) A series that converges very quickly and obviously. (Usually I'll
> use a geometric series for this, and I'll remind them of this example
> example later on when we discuss geometric series.)
>
> (b) A series whose terms don't converge to 0 (and thus the series diverges).
>
> (c) The Harmonic series.
>
> They go through (a) and (b) pretty quickly. When they get to (c), they
> have a hard time deciding if the series converges or diverges. They will
> puzzle over it for quite a while. Some of them will make tables and
> plots of partial sums going out to thousands of terms in the series, and
> still they are unsure if the series converges or diverges.
>
> After letting them work while I walk around and answer questions, we
> stop to discuss the examples. The class will tell me that Example (a)
> has terms that converge to 0, and the partial sums converge to (whatever
> I have rigged it to). For (b), they will explain that the terms converge
> to let's say 1/2, and so the partial sums increase approximately
> linearly with a slope of 1/2, and the series diverges. From this, they
> make the observation that if a series has terms that do not converge to
> 0, there is no way the series could converge.
>
> Then we get to (c). The terms converge to 0, and it's tough to tell
> whether the partial sums converge or diverge, but most of them will lean
> toward thinking that they diverge. So then I lower the boom. Let's
> *prove* that it diverges. This leads us into a discussion of the
> Integral Test. We continue with this interplay between concrete examples
> in Mathematica and analytical work on the board as we progress through
> the chapter.
>
> I don't understand how anyone would think that this use of Mathematica
> is "dangerous" or "threatening" or somehow prevents my students from
> learning the subject. On the contrary, this use of Mathematica helps my
> students to gain some conceptual understanding of a topic that they
> otherwise find difficult and abstract.
>
> --
> University of Vermont
>
Peace Helen!
I am not in any way saying that how you teach is dangerous - I was more
time they wanted (student copy) and could use it to attack pencil and
paper problems that you had set them. In that situation, I think you
could end up with some students who became good at Mathematica, but
didn't learn enough maths.
One problem is that students may feel under time and grade pressure. If
they see a way to cheat with Mathematica, they may not like the idea,
but the competition might force it on them
David Bailey
http://www.dbaileyconsultancy.co.uk
```
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• Next by thread: Re: if using Mathematica to solve an algebraic problem | 1,177 | 4,803 | {"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-2024-30 | latest | en | 0.944565 |
https://ebrainanswer.com/mathematics/question13787952 | 1,627,330,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00222.warc.gz | 231,017,433 | 31,375 | , 08.11.2019 16:31, cupcake686
And show your work for both a. and b. And show your work for both a. and b.
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https://proofwiki.org/wiki/Triangular_Numbers_which_are_Sum_of_Two_Cubes | 1,582,053,837,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143805.13/warc/CC-MAIN-20200218180919-20200218210919-00423.warc.gz | 544,456,164 | 10,149 | Triangular Numbers which are Sum of Two Cubes
Theorem
$28$ is the smallest triangular number which is the sum of $2$ cubes:
$\displaystyle 28$ $=$ $\displaystyle 1 + 27$ $\displaystyle$ $=$ $\displaystyle 1^3 + 3^3$
Proof
Can be demonstrated by brute force. | 78 | 263 | {"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": 2, "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-2020-10 | latest | en | 0.812357 |
https://www.enotes.com/homework-help/motor-running-its-rated-power-provides-2-6-lb-ft-453212 | 1,621,384,966,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00528.warc.gz | 777,084,922 | 17,895 | # A motor running at its rated power provides 2.6 lb-ft torque at its rated speed of 2020 rpm. The motor draws 5.6-A from a 230-V supply at a phase angle of 37 degrees. Would the effciency of the motor be 82.7% or 1.6%
The relation between the torque `T` , the motor angular speed `omega` and its power `P` in circular motion is (similar to `P = F*v` in linear motion)
`P_m =T*omega`
we know that `1 lb*ft = 1.3558 N*m` and `omega=2*pi*F`
therefore
`P_m = (1.3558*2.6)*(2*pi*2200/60) =745.66 W`
This is the mechanical power that the motor generates.
The apparent electric power that the motor draws from the AC source is
`P_a =U*I =230*5.6 =1288 VA`
The real electrical power (the part of apparent power that can be transformed into mechanical power) is
`P = P_a*cos(theta) = 1288*cos(37) =1028.6 W`
The efficiency of the motor is just the rapport of the mechanical power produced to the real electric power.
`eta = P/P_a = 745.66/1028.6 =0.725 =72.5%`
The efficiency of the motor is 72.5%.
Approved by eNotes Editorial Team | 320 | 1,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-21 | latest | en | 0.905571 |
http://www.evi.com/q/137_cm_is_how_many_inches | 1,419,142,307,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770742.121/warc/CC-MAIN-20141217075250-00133-ip-10-231-17-201.ec2.internal.warc.gz | 486,226,965 | 6,170 | # 137 cm is how many inches
• 137 centimeters is 53.94 inches.
• tk10npubl tk10ncanl
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• 137cm into inches? | 1,510 | 4,895 | {"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-2014-52 | latest | en | 0.865433 |
https://www.jobilize.com/trigonometry/test/extensions-polar-coordinates-by-openstax?qcr=www.quizover.com | 1,561,373,151,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999482.38/warc/CC-MAIN-20190624104413-20190624130413-00307.warc.gz | 763,989,321 | 24,297 | # 10.3 Polar coordinates (Page 5/8)
Page 5 / 8
How are the polar axes different from the x - and y -axes of the Cartesian plane?
Explain how polar coordinates are graphed.
Determine $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point, then move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole to plot the point. If $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is negative, move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole in the opposite direction but along the same angle. The point is a distance of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ away from the origin at an angle of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ from the polar axis.
How are the points $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ related?
Explain why the points $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ are the same.
The point $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a positive angle but a negative radius and is plotted by moving to an angle of $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y -axis. The point $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a negative angle and a positive radius and is plotted by first moving to an angle of $\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y -axis.
## Algebraic
For the following exercises, convert the given polar coordinates to Cartesian coordinates with $\text{\hspace{0.17em}}r>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0\le \theta \le 2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located when determining $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point.
$\left(7,\frac{7\pi }{6}\right)$
$\left(5,\pi \right)$
$\left(-5,0\right)$
$\left(6,-\frac{\pi }{4}\right)$
$\left(-3,\frac{\pi }{6}\right)$
$\left(-\frac{3\sqrt{3}}{2},-\frac{3}{2}\right)$
$\left(4,\frac{7\pi }{4}\right)$
For the following exercises, convert the given Cartesian coordinates to polar coordinates with $\text{\hspace{0.17em}}r>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \theta <2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located.
$\left(4,2\right)$
$\left(-4,6\right)$
$\left(3,-5\right)$
$\left(\sqrt{34},5.253\right)$
$\left(-10,-13\right)$
$\left(8,8\right)$
$\left(8\sqrt{2},\frac{\pi }{4}\right)$
For the following exercises, convert the given Cartesian equation to a polar equation.
$x=3$
$y=4$
$r=4\mathrm{csc}\theta$
$y=4{x}^{2}$
$y=2{x}^{4}$
$r=\sqrt[3]{\frac{sin\theta }{2co{s}^{4}\theta }}$
${x}^{2}+{y}^{2}=4y$
${x}^{2}+{y}^{2}=3x$
$r=3\mathrm{cos}\theta$
${x}^{2}-{y}^{2}=x$
${x}^{2}-{y}^{2}=3y$
$r=\frac{3\mathrm{sin}\theta }{\mathrm{cos}\left(2\theta \right)}$
${x}^{2}+{y}^{2}=9$
${x}^{2}=9y$
$r=\frac{9\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta }$
${y}^{2}=9x$
$9xy=1$
$r=\sqrt{\frac{1}{9\mathrm{cos}\theta \mathrm{sin}\theta }}$
For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.
$r=3\mathrm{sin}\text{\hspace{0.17em}}\theta$
$r=4\mathrm{cos}\text{\hspace{0.17em}}\theta$
${x}^{2}+{y}^{2}=4x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{{\left(x-2\right)}^{2}}{4}+\frac{{y}^{2}}{4}=1;$ circle
$r=\frac{4}{\mathrm{sin}\text{\hspace{0.17em}}\theta +7\mathrm{cos}\text{\hspace{0.17em}}\theta }$
$r=\frac{6}{\mathrm{cos}\text{\hspace{0.17em}}\theta +3\mathrm{sin}\text{\hspace{0.17em}}\theta }$
$3y+x=6;\text{\hspace{0.17em}}$ line
$r=2\mathrm{sec}\text{\hspace{0.17em}}\theta$
$r=3\mathrm{csc}\text{\hspace{0.17em}}\theta$
$y=3;\text{\hspace{0.17em}}$ line
$r=\sqrt{r\mathrm{cos}\text{\hspace{0.17em}}\theta +2}$
${r}^{2}=4\mathrm{sec}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}\theta$
$xy=4;\text{\hspace{0.17em}}$ hyperbola
$r=4$
${r}^{2}=4$
${x}^{2}+{y}^{2}=4;\text{\hspace{0.17em}}$ circle
$r=\frac{1}{4\mathrm{cos}\text{\hspace{0.17em}}\theta -3\mathrm{sin}\text{\hspace{0.17em}}\theta }$
$r=\frac{3}{\mathrm{cos}\text{\hspace{0.17em}}\theta -5\mathrm{sin}\text{\hspace{0.17em}}\theta }$
$x-5y=3;\text{\hspace{0.17em}}$ line
## Graphical
For the following exercises, find the polar coordinates of the point.
$\left(3,\frac{3\pi }{4}\right)$
$\left(5,\pi \right)$
For the following exercises, plot the points.
$\left(-2,\frac{\pi }{3}\right)$
$\left(-1,-\frac{\pi }{2}\right)$
$\left(3.5,\frac{7\pi }{4}\right)$
$\left(-4,\frac{\pi }{3}\right)$
$\left(5,\frac{\pi }{2}\right)$
$\left(4,\frac{-5\pi }{4}\right)$
$\left(3,\frac{5\pi }{6}\right)$
$\left(-1.5,\frac{7\pi }{6}\right)$
$\left(-2,\frac{\pi }{4}\right)$
$\left(1,\frac{3\pi }{2}\right)$
For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis.
$5x-y=6$
$r=\frac{6}{5\mathrm{cos}\theta -\mathrm{sin}\theta }$
$2x+7y=-3$
${x}^{2}+{\left(y-1\right)}^{2}=1$
$r=2\mathrm{sin}\theta$
${\left(x+2\right)}^{2}+{\left(y+3\right)}^{2}=13$
$x=2$
$r=\frac{2}{\mathrm{cos}\theta }$
${x}^{2}+{y}^{2}=5y$
${x}^{2}+{y}^{2}=3x$
$r=3\mathrm{cos}\theta$
For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane.
$r=6$
$r=-4$
${x}^{2}+{y}^{2}=16$
$\theta =-\frac{2\pi }{3}$
$\theta =\frac{\pi }{4}$
$y=x$
$r=\mathrm{sec}\text{\hspace{0.17em}}\theta$
$r=-10\mathrm{sin}\text{\hspace{0.17em}}\theta$
${x}^{2}+{\left(y+5\right)}^{2}=25$
$r=3\mathrm{cos}\text{\hspace{0.17em}}\theta$
## Technology
Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(2,-\frac{\pi }{5}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.
$\left(1.618,-1.176\right)$
Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(-3,\frac{3\pi }{7}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.
Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-7,8\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest thousandth.
$\left(10.630,131.186°\right)$
Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(3,-4\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest hundredth.
Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ in radians. Round to the nearest hundredth.
$\text{\hspace{0.17em}}\left(2,3.14\right)or\left(2,\pi \right)\text{\hspace{0.17em}}$
## Extensions
Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a>0.$
Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a<0.$
A vertical line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units left of the y -axis.
Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a>0.$
Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a<0.$
A horizontal line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units below the x -axis.
What polar equations will give an oblique line?
For the following exercise, graph the polar inequality.
$r<4$
$0\le \theta \le \frac{\pi }{4}$
$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}2$
$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge -3$
$0\le \theta \le \frac{\pi }{3},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}<\text{\hspace{0.17em}}2$
$\frac{-\pi }{6}<\theta \le \frac{\pi }{3},-3
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions | 3,693 | 9,683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 122, "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} | 4.4375 | 4 | CC-MAIN-2019-26 | latest | en | 0.650955 |
http://homeguides.sfgate.com/cut-square-out-middle-sheet-wood-99350.html | 1,524,818,871,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524127095762.40/warc/CC-MAIN-20180427075937-20180427095937-00274.warc.gz | 144,219,632 | 15,285 | # How to Cut a Square Out of the Middle of a Sheet of Wood
Puzzles, toys and other projects sometimes require cutting specific shapes. Creating geometric figures in the middle of a wood sheet such as plywood or MDF have three requirements: orientating the shape, laying out out the shape and cutting out the shape while leaving the surrounding sheet intact. A square has four sides and it is easiest to make them parallel to the sides of the sheet of wood, but virtually any orientation is possible.
## Orientation and Layout
#### 1
Mark the location of the square's center on the wood sheet with a pencil. Place the framing square with the short leg flush with the sheet's edge and the long leg against the mark. Draw a pencil line through the mark to establish a reference line.
#### 2
Determine the orientation of the square. On a square with sides parallel to the sheet sides, a line drawn through two opposite corners runs at a 45-degree angle to the reference line. You can rotate the square up to 45 degrees in either direction.
#### 3
Place the protractor with the zero point on the center of the square and the straight edge parallel to the reference line. Mark the direction of two corners from the center of the square. For example: If the sides are parallel to the sheet sides, the marks are at 45 and 135 degrees. On a square rotated 45 degrees, the marks are at 90 and 180 degrees and the corners will point at the sides of the sheet.
#### 4
Find the distance from the center point to each corner. Divide the length of one side of the square by 2, and then by 0.707107, which is the cosine of 45 degrees. For example: If the side of the square is 3 inches, the distance to the corner is 3 divided by 2 divided by 0.70710. 3 / 2 = 1.5 1.5 / 0.707107 = 2.1213
#### 5
Place the square on the sheet of wood with one leg aligned with one corner angle mark and the center point. Draw a line through the corner angle mark, through the center point and across the reference line. The line extends the distance to the corner on each side of the reference line. Measure along the line just drawn in both directions from the center point. Make a mark at the distance to the corner on each side of the reference line to mark the two opposite corners. Repeat this step with the other corner angle mark for the other two corners.
#### 6
Align the square with two adjacent corner marks and draw a line through them. Add additional lines for the other three sides.
## Cut the Square
#### 1
Place the drill with the point of the bit at one inside corner of the square. The edges of the drill bit should touch the sides of the square, but not go past it. Drill a hole through the sheet.
#### 2
Install the cutting bit on the multipurpose tool or the blade on the jigsaw. Insert the bit or blade into the hole. Hold the tool firmly and turn it on. Follow the pencil lines that outline the square from corner to corner to cut out the shape.
#### 3
File the corners of the cutout to make the corners square if so desired. The round bit on the multipurpose tool leaves a small radius instead of a square cut at corners. Sand the edges of the cutout smooth to remove sharp or rough edges.
#### Things You Will Need
• Tape measure
• Framing square
• Protractor
• Calculator
• Drill with 1/4-inch bit
• Handheld multipurpose tool or jigsaw
• Cutting bit or jigsaw blade
• Wood file
• Sandpaper
• Sanding tool or block
#### Tip
• If you use a jigsaw to cut out the square, drill a hole in each corner to make turning the corners easy.
#### Warning
• Wear safety glasses and a dust mask while cutting with power tools. Follow all tool manufacturer safety instructions. | 844 | 3,682 | {"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-2018-17 | latest | en | 0.895934 |
https://www.coursesarchive.com/AnsweredQuestions/assume-that-the-demand-curve-d-p-given-below-is-the-market-demand-for-apples-q-d-p-320-12pq-d-p-320--ca126 | 1,695,338,173,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00442.warc.gz | 807,645,403 | 6,653 | #### (Solved): Assume That The Demand Curve D(p) Given Below Is The Market Demand For Apples: Q=D(p)=320?12pQ=D(p ...
Assume that the demand curve D(p) given below is the market demand for apples:
Q=D(p)=320?12pQ=D(p)=320-12p, p > 0
Let the market supply of apples be given by:
Q=S(p)=60+15pQ=S(p)=60+15p, p > 0
where p is the price (in dollars) and Q is the quantity. The functions D(p) and S(p) give the number of bushels demanded and supplied.
What is the consumer surplus at the equilibrium price and quantity?
Round the equilibrium price to the nearest cent, use that rounded price to compute the equilibrium quantity, and round the equilibrium quantity DOWN to its integer part.
Maintain full precision for the vertical intercept by carrying the full fraction into your consumer surplus calculation. | 203 | 813 | {"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} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.866602 |
https://forum.allaboutcircuits.com/threads/pcb-layout.175069/ | 1,610,746,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00148.warc.gz | 354,093,681 | 21,695 | # Pcb layout
#### Berzin
Joined Dec 17, 2020
19
Hi guys
Actually I am little new to the great world of pcb and I am working on a new project of my own so i have fews questions regarding it so please help me out if possible
I am making a machine which is been solely be working on pcb/ motherboard. So my first question is the pcb has atleast 6 to 7 main process ICs to function their individual process so how calculate the power requirements for the whole pcb as all processes are 12 volt each so the answer will be 12 volts or 72 volts
And secondly the main question that I am trying to trail my circuit design on the breadboard with 18volt and the circuit requires 12volts but the circuit is not working please any one explain me why is this happening and solutions for it
#### ericgibbs
Joined Jan 29, 2010
11,551
hi Berzin,
Welcome to AAC.
Do you have a circuit diagram of your circuit that you could post.?
E
#### jpanhalt
Joined Jan 18, 2008
10,913
Why do you thing the supply needs to tbe 72 volts when you have 6, 12-volt IC's?
Generally, one establishes a "12V Rail" and attaches the IC's in parallel to that rail, not in series.
#### Berzin
Joined Dec 17, 2020
19
hi Berzin,
Welcome to AAC.
Do you have a circuit diagram of your circuit that you could post.?
E
Thank you so much for your support
Please find the attachment sorry it's a rough and secondly I will separate the ground and positive on the pcb by using double layer pcb
All the ics are using 12 volts each please help me out
Mod: lightened your image.E
#### Attachments
• 1.6 MB Views: 15
Last edited by a moderator:
#### Berzin
Joined Dec 17, 2020
19
hi Berzin,
Welcome to AAC.
Do you have a circuit diagram of your circuit that you could post.?
E
Thank you so much for your support
Please find the attachment sorry it's a rough and secondly I will separate the ground and positive on the pcb by using double layer pcb
All the ics are using 12 volts each please help me out
#### Berzin
Joined Dec 17, 2020
19
hi Berzin,
Welcome to AAC.
Do you have a circuit diagram of your circuit that you could post.?
E
Thank you so much for your support
Please find the attachment sorry it's a rough and secondly I will separate the ground and positive on the pcb by using double layer pcb
All the ics are using 12 volts each please help me out
#### jpanhalt
Joined Jan 18, 2008
10,913
Your IC's appear to be in parallel. All you need is 12V, if that is what you want them to run at.
#### Berzin
Joined Dec 17, 2020
19
Your IC's appear to be in parallel. All you need is 12V, if that is what you want them to run at.
Thank you for your reply but then why it is not working on the 9 volt + 9 volt battery or it requires only 12 volts
#### jpanhalt
Joined Jan 18, 2008
10,913
All of the ICs/arrays should work at 9V. That means there is something wrong with your design or the way it is assembled.
Your design has several essentially identical "cells." Make one cell and see if it works. Get it to work, then duplicate as needed. | 794 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-04 | longest | en | 0.922901 |
https://oeis.org/A225204 | 1,721,130,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00837.warc.gz | 376,787,120 | 4,301 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A225204 Numerators of convergents to the square root of the golden ratio. 7
1, 4, 5, 14, 159, 491, 2614, 3105, 33664, 36769, 107202, 572779, 1825539, 9700474, 11526013, 32752500, 142536013, 175288513, 317824526, 810937565, 3561574786, 182451251651, 186012826437, 926502557399, 2039017941235, 5004538439869, 157179709577174 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Table of n, a(n) for n=0..26. I. J. Good, Complex Fibonacci and Lucas Numbers, Continued Fractions, and the Square Root of the Golden Ratio (Condensed Version), Journal of the Operational Research Society, 43 (1992), 837-842. I. J. Good, Complex Fibonacci and Lucas Numbers, Continued Fractions, and the Square Root of the Golden Ratio, The Fibonacci Quarterly 31.1 (1993):7-20. FORMULA a(n) = A331692(n)*a(n-1) + a(n-2) for n >= 2. - Jianing Song, Aug 18 2022 EXAMPLE 1, 4/3, 5/4, 14/11, 159/125, 491/386, 2614/2055, 3105/2441, ... = A225204/A225205 MATHEMATICA Numerator[Convergents[Sqrt[GoldenRatio], 20]] CROSSREFS Cf. A001622, A139339, A331692, A225205 (denominators). Sequence in context: A042535 A222489 A222501 * A363817 A308095 A321348 Adjacent sequences: A225201 A225202 A225203 * A225205 A225206 A225207 KEYWORD nonn,frac AUTHOR Eric W. Weisstein, May 01 2013 STATUS approved
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Last modified July 16 05:19 EDT 2024. Contains 374343 sequences. (Running on oeis4.) | 596 | 1,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.571642 |
http://www.reddit.com/user/TrippSkiggins/comments/ | 1,432,354,977,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927185.70/warc/CC-MAIN-20150521113207-00050-ip-10-180-206-219.ec2.internal.warc.gz | 680,445,139 | 17,763 | [–] 0 points1 point (0 children)
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Lubbock, TX: You can see for 30 miles in any direction. If you stand on a can of Coke, you can see for 60 miles in any direction.
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One way to convince yourself of the validity of this formula is through Taylor series. For real numbers x,
ex = \Sigma xn / n!
sin(x) = \Sigma (-1)n x2n+1 / (2n+1)!
cos(x) = \Sigma (-1)n x2n / (2n)!
Skipping the parts about making sure everything converges, you can just plug x=i\theta into the formula for ex , collect all the n odd and n even terms, and then Euler's formula pops right out.
Edit: Formatting is hard.
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Pass the petition around -- it needs 25,000 signatures before October 8.
Video of the vote.
Article explanation.
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I'm heading to the Goldendale Observatory in eastern Washington.
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It's pronounced "tomato".
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That's cute! Remids me of this little guy.
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Stockard Channing was 33 in Grease.
My Dad Said... by in funny
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The Full House drinking game...
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TV Funhouse had a good one for never forgetting taxonomic rank: Phylum, Class, Order, Family, Genus, Species.
Please come over for gay sex.
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Who wouldn't.
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By degree I'm guessing you mean the total number of edges in the graph.
You can do this by induction on n. For the base case n=1, there's not much to show. For the inductive step, consider what you can say about vertices in T which have the smallest degree (number of edges coming out of them).
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2 dislikes :(
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Do you have any opinions on the open source Sage project?
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91
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The last time this happened to me I had to actually do the math to figure it out.
"Let's see. I was born in 1984. That makes me...28."
I go home, tell my wife the funny story, and am politely reminded that I was in fact born in 1985.
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At first I thought you meant Fred Phelps.
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It's a regular heptagon!
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Gyros House at Division and Cooper.
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The third limit is taken at 3 explicitly, so you take the limit of the second equation, which is simply 2.
That is not correct. The third limit means you approach 3 not from just the left or just the right, but rather from both directions. To evaluate the third limit, you must see if the two limits before are equal. If they are, then this limit is the value of the third limit. If they are unequal, it means the function approaches different values from the left and the right of 3, meaning that the last limit does not exist.
The actual function value at 3 (i.e. 2 in this equation) does not matter when taking limits.
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This works nicely. As a nifty shortcut, you could also use this same way just to show that u is orthogonal to sv (and similarly to tw). Then part (a) implies part (b).
I'll never say what I did to get this autograph. by in pics
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Spoilers!
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My advice for basic integrals is to try to reduce everything to the stuff you already know how to do, say the power rule. Something like [; (x^2+3x)^3 ;] looks daunting, since there's no good chain rule for integrals -- try expanding it out first. A good way of dealing with denominators like sqrt(x) is to try splitting the numerator into a bunch of terms, all over the denominator.
Edit: Learning to format math...
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Elementary school for me, apparently. | 1,467 | 5,526 | {"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} | 3.375 | 3 | CC-MAIN-2015-22 | latest | en | 0.946136 |
https://mersenneforum.org/showthread.php?s=046ea76dd0683d3455221fb3886dd828&p=555610 | 1,603,926,215,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00567.warc.gz | 440,269,274 | 8,322 | mersenneforum.org a nice remark about mersene composite
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2020-09-01, 02:39 #1 baih Jun 2019 1000102 Posts a nice remark about mersene composite a nice remark Mersenne Number 2n-1 x2 = (2n-2)-1 mod 2n-1 there is no solution for x if 2n-1 composite
2020-09-01, 03:16 #2 carpetpool "Sam" Nov 2016 22·79 Posts Claim: If 2^n-1 is not prime, then 2^(n-2)-1 is a quadratic non-residue mod 2^n-1. The claim is false, however the contrapositive is true: If 2^n-1 is prime, then 2^(n-2)-1 is a quadratic residue mod 2^n-1. n = 2 is a special case implying that x^2 = 0 mod 3, which has solution x = 0 (trivial). In all other cases, n must be odd. For simplicity, suppose p = 2^n-1. 2^(n-2)-1 = (p - 3)/4 p = 1 mod 3, and since p is prime, quadratic reciprocity tells us that x^2 = -3 mod p is solvable. Hence, x^2 = (p - 3)/4 mod p, 4*x^2 = p - 3 mod p, or (2*x)^2 = -3 mod p. The map x --> 2*x is a bijection in Zp/Z. Done. Proven. Now what remains is to show when your claim is false (given n is composite): If every prime q | p is congruent to 1 mod 3, then the claim is false. Since q = 1 mod 3, x^2 = -3 mod q will have a solution, using the Chinese Remainder Theorem allows us to construct a solution x to x^2 = -3 mod p, so the conclusion follows. There are infinitely many composite numbers of the form 2^n-1 such that the claim is false. If n = 3^k, then 2^n-1 is a counterexample if k > 2, since every prime q | 2^(3^k) - 1, the order of 2 mod q by definition divides 3^k, and trivially cannot be 1, so q = 1 mod 3. I do not find your claim interesting at all, what I find interesting is finding the density of primes n such that each prime q | 2^n-1 is congruent to 1 mod 3: Any arbitrary integer N has about ln(ln(N)) prime factors, so 2^n-1 will have on average, ln(n) prime factors by this assumption. The probability that all of these factors are congruent to 1 mod 3 is about 1/2^(ln(n)). I suppose there is a better estimate?
2020-09-01, 11:32 #3
Dr Sardonicus
Feb 2017
Nowhere
3·1,193 Posts
Quote:
Originally Posted by baih a nice remark Mersenne Number 2n-1 x2 = (2n-2)-1 mod 2n-1 there is no solution for x if 2n-1 composite
The smallest prime exponent which furnishes a counterexample is n = 37. Thread closed.
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http://www.enotes.com/homework-help/find-derivative-y-11-x-sqrt-1-x-sin-3x-4-x-2-1-3-x-328983 | 1,477,153,269,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719027.25/warc/CC-MAIN-20161020183839-00559-ip-10-171-6-4.ec2.internal.warc.gz | 426,709,524 | 9,787 | Find the derivative of y =11^(x)sqrt(1-x)(sin(3x))^(4)/((x^2 + 1)^(3)*x^(1/5)).
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You need to differentiate the function with respect to x using chain rule, hence you should start to differentiate from outside to inside such that:
`y'= 11^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5))) ln 11* ^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))'`
You need to differentiate using quotient rule such that:
`(((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))' = ((sqrt(1-x)sin(3x))^4)'*(x^2 + 1)^3*x^(1/5) - sqrt(1-x)(sin(3x))^4*((x^2 + 1)^3*x^(1/5))')/((x^2 + 1)^(6)*x^(2/5))`
You need to differentiate each product from numerator using product rule such that:
`(((sqrt(1-x)(sin(3x))^4))'*((x^2 + 1)^(3)*x^(1/5))) = (-(sin 3x)^4/(2sqrt(1-x)) + 12sqrt(1-x)*(sin 3x)^3*(cos 3x))*((x^2 + 1)^(3)*x^(1/5)))`
`(((sqrt(1-x)(sin(3x))^4))*((x^2 + 1)^(3)*x^(1/5))') = (((sqrt(1-x)(sin(3x))^4))*((6x^(6/5)*(x^2+1)^2 + (1/5)(x^2 + 1)^(3)*x^(-4/5))`
Hence, differentiating with respect to x yields:
`y' = (((-(sin 3x)^4/(2sqrt(1-x)) + 12sqrt(1-x)*(sin 3x)^3*(cos 3x))*(x^2 + 1)^(3)*x^(1/5)) - (sqrt(1-x)sin(3x)^4)*(6x^(6/5)*(x^2+1)^2 + (1/5)(x^2 + 1)^3*x^(-4/5)))/((x^2 + 1)^(6)*x^(2/5)))` | 598 | 1,245 | {"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} | 4.09375 | 4 | CC-MAIN-2016-44 | latest | en | 0.63956 |
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Problem 11
# Find the periodic payments necessary to accumulate the given amount in an annuity account. (Assume end-of-period deposits and compounding at the same intervals as deposits.) [HINT: See Quick Example 2.] $$\ \ 20,000$$ in a fund paying $$5 \%$$ per year, with monthly payments for 5 years, if the fund contains $$\ 10,000$$ at the start
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## Step 4: Apply the FVOA formula and solve for P
We have all the necessary parameters, so we can now apply the FVOA formula to solve for the periodic payment P: $$FV' = P\frac{(1+r)^n-1}{r}$$ Substitute the values of FV', r, and n into the formula: $$\10,000 = P\frac{(1+0.004167)^{60}-1}{0.004167}$$ Now solve for P using algebra: $$P = \frac{\10,000 \times 0.004167}{(1+0.004167)^{60}-1}$$ $$P = \frac{\41.67}{(1.004167)^{60}-1}$$ $$P = \frac{\41.67}{1.28323-1}$$ $$P = \frac{\41.67}{0.28323}$$ $$P \approx \147.13$$
## Step 5: Conclusion
To accumulate \$20,000 in the annuity account with a starting balance of \$10,000, periodic payments of approximately \\$147.13 are necessary for the next 5 years, with monthly deposits and compounding interest at the same intervals.
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# The expected values, variances and standard Deviatiations for two random variables X and Y are given...
The expected values, variances and standard Deviatiations for two random variables X and Y are given in the following table
Variable expected value variance standard deviation X 20 9 3 Y 35 25 5
Find the expected value and standard deviation of the following combinations of the variable X and Y. Round to nearest whole number.
E(X+10) = ,
StDev(X+10) =
E(2X) = ,
StDev(2X) =
E(3X-2) = ,
StDev(3X-2) =
E(3X +4Y) = ,
StDev(3X+4Y) =
E(X-2Y) = ,
StDev(X-2Y) =
E(X + 10) = E(X) + 10 = 20 + 10 = 30
SD(X + 10) = SD(X) + SD(10) = 3 + 0 = 3
E(2X) = 2 * E(X) = 2 * 20 = 40
SD(2X) = 2 * SD(X) = 2 * 3 = 6
E(3X - 2) = 3 * E(X) - 2 = 3 * 20 - 2 = 58
SD(3X - 2) = 3 * SD(X) - 0 = 3 * 3 = 9
E(3X + 4Y) = 3 * E(X) + 4 * E(Y) = 3 * 20 + 4 * 35 = 200
SD(3X + 4Y) = sqrt(Var(3X + 4Y)) = sqrt(32 * Var(X) + 42 * Var(Y)) = sqrt(9 * 9 + 16 * 25) = 22
E(X - 2Y) = E(X) - 2 * E(Y) = 20 - 2 * 35 = -50
SD(X - 2Y) = sqrt(Var(X - 2Y)) = sqrt(Var(X) + (-2)2 * Var(Y)) = sqrt(9 + 4 * 25) = 10
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http://lacasadelespia.co.cr/y3kpb9/5zqfe.php?tag=rationalizing-a-denominator-using-conjugates-square-root-in-numerator-calculator-48b347 | 1,624,589,859,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00035.warc.gz | 28,393,672 | 9,687 | $$,$$ \frac{ 3 \color{red}{ - } \sqrt{5} }{ 3 \color{red}{ - } \sqrt{5}} 1. The reasoning and methodology are similar to the "difference of squares" conjugate process for square roots. \\ \\ 5 43 To rationalize a denominator with Two Terms – Multiply both numerator and denominator by a conjugate. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.. We know that multiplying by 1 … \frac{ 9}{ Formulate three word problems from day-to-day life that can be translated to quadratic equations. 1. \frac{ 3}{ \sqrt{7} +2 } The product of conjugates is always the square of … example of math word problems of "number relation " with the solutions. \frac{ 9}{ \frac{ \sqrt{7} -\sqrt{5} }{ \sqrt{5} + \sqrt{7} } {3(1)} more gifs . \frac{ \color{red}{\cancel{3}}(2 -1\sqrt{5} ) } { \color{red}{\cancel{3} } (3 ) } \frac{ (\sqrt{7} -\sqrt{5})( \sqrt{5} \color{red}{-} \sqrt{7})} \frac{ 27 - 9\sqrt{5} } To rationalize a denominator, multiply the fraction by a "clever" form of 1--that is, by a fraction whose numerator and denominator are both equal to the square root in the denominator. = Rationalizing the denominator with variables - Examples \frac{ 7 \sqrt{2} - 7\sqrt{5} } \frac{ 6 -3\sqrt{5} } { \\ \frac{ 14}{ 2 - \sqrt{7} } Multiply the numerator and denominator by the same number as the square root in the denominator. The most common used irrational numbers that are used are radical numbers, for example √3. 3 23+5 To rationalize an Nth Root Denominator – Multiply by the base raised to the power of the index minus the exponent. \\ \frac{ 3 \color{red}{+} 2\sqrt{6} }{ 3 \color{red}{+} 2\sqrt{6} } Real World Math Horror Stories from Real encounters. $$, Rationalize$$, Identify the conjugate of the denominator more gifs . $$.$$, \frac{ 5}{ \color{red}{ 3 - 2\sqrt{6}} } } square root fraction calculator online ; lowest common denominator calculator ; Algebra 1 Free Answers ; Radical Calculator ; GCF and monomials worksheets ; Symmetry worksheet first grade ; ti-86 finding slope ; online equation solver ; Square roots algebra ; mcdougal littell algebra 2 answers for chapter 5.2 worksheet ; college word equations Multiply the numerator and denominator by the conjugate. Both the top and bottom of the fraction must be multiplied by the same term, because what you are really doing is multiplying by 1. \\ "Rationalizing the denominator" is when we move a root (like a square root or cube root) from the bottom of a fraction to the top. \frac{ 7}{ \sqrt{2} + \sqrt{5}} $$, Identify the conjugate of the denominator { -2 (1)} When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. \frac{ 2\sqrt{3}(5\color{red}{-} 2\sqrt{ 7 } ) }{ (5 + 2\sqrt{7 })( 5 \color{red}{-} 2\sqrt{ 7 } ) } \\ =$$, Rationalize {7 \color{red}{ +2\sqrt{7}-2\sqrt{7}} -4 } Rationalizing the Denominator by Multiplying by a Conjugate Rationalizing the denominator of a radical expression is a method used to eliminate radicals from a denominator. In this case, reducing the number in the root sign by prime factorization first will reduce miscalculation. \frac{ 7}{ \sqrt{2} + \sqrt{5}} Remember that . \frac{\cancel{-2}( 6 -1 \sqrt{35}) } Check out the interactive simulations and rationalize using a calculator to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page. \frac{ 7 \sqrt{2} - 7\sqrt{5} } \cdot \frac{2 \color{red}{+} \sqrt{7} }{2 \color{red}{ + } \sqrt{7} } Simply type into the app below and edit the expression. \cdot \frac{ \sqrt{7} -\sqrt{5} }{ \color{red}{ \sqrt{5} + \sqrt{7} }} free online calculator for college algebra, free online calculater for clooege algebra, graphing linear equations free worksheets, how can i take the old algebra taks test online, "nonhomogeneous differential equation" Laplace, answer book to practice: skills workbook mathematic glencoe mathematics, adding and subtracting integers worksheets, compound interest and principal using ti83plus, adding subtracting multiplying and dividing fractions 6th grade, harcourt 3rd grade math virginia edition +teachers resources. $$,$$ $$, Identify the conjugate of the denominator \sqrt{2} \color{red}{-} \sqrt{5}$$, Rationalize 2) 5 5 3) − 6 2 4) 6 3 5) 2 3 6) 2 7 7) 7 6 8) 4 5 9) 4 6 10) 7 5 11) − 4 7 $$,$$ I could take a 3 out of the denominator of my radical fraction if I had two factors of 3 inside the radical. It can rationalize denominators with one or two radicals. Examples, formula and the Steps! How do you rewrite and simplify radicals using rational exponents? \\ \frac{ \sqrt{5} \color{red}{-} \sqrt{7}} 7 352 Evaluate the expression using a calculator. \frac{ 6 -3\sqrt{5} } { \frac{ \sqrt{7} -\sqrt{5} }{ \color{red}{ \sqrt{5} + \sqrt{7} }} When you encounter a fraction that contains a radical in the denominator, you can eliminate the radical by using a process called rationalizing the denominator. what number is 2% out of 2,550 +math help, McDougal Littell Chapter Review Games and Activities for chapter 6 algebra 2. gcsebitesize simultaneous equation power point, solve quadratic equations using absolute values, modern chemistry textbook holt rinehart and winston answers, dissolution of sodium chloride chemical equation, Rational Exponents, Radicals, Complex Numbers, free worksheets for trigonometry functions for 8th grade, Free online scientific calulator with radicals, quadratic formula step by step calculator, answers Algebra and trigonometry structure and method book 2, cartesian algebra solving for multiple variables, approximate area under a parabola by inscribed rectangles "negative", why does multiplying two negatives give a positive, quick way to find a number to the nth power, calculator for ordering fractions from least to greatest, vertical acceleration method of teaching maths, basic business statistics 9th edition past exam, how to do standard deviation on a ti 83 plus, converting decimal to fractions or mixed numbers, convert decimal to fraction form calculator, cubic equation solver online, from sequence to formula, glencoe teachers edition workbook answers, how to solve a second order differential equation, quadratic equation for the price of college, practice adding and subtracting equations, equations with two variables worksheet printouts, +free o'level past paper in english language, math problems,trivia for rd year high school, how do I convert a decimal measurement to a mixed number, statistics crib OR cheat sheet OR formulas, gallian chapter 10, #26 homework solutions, Linear Equations and Two Variables Worksheets, bittinger ellenbogen, elementary and intermediate algebra, graphs and models, syllabus, rules in adding subtracting multiplying and dividing scientific notation, free online property math games 6th grade, holt mathematics help with my homework lesson 5-2 ratios rates, unit rates u, high school math logarithms application practice exam, SUM OF NUMBERS IN DO WHILE LOOP JAVA SOURCE, glencoe mcgraw hill algebra 1 answers chapter 8, solving equation with addition and subtraction, turning fractions into decimals step by step, florida edition mcdougal littell middle school course 1, california math 6th grade fraction questions, how to find the LCM of polynomials easily, examples of distributive property with fractions, solving second order differential equation matlab, convert fraction to decimal using algebra, online Algebra and Trigonometry: Structure and Method Book 2, answer key for prentice hall physics book, trigonometry sample problems for grade 11, math work sheet for 5th grade greatest common factor, printable math pages for fith grade from houghton miffin math books, ti-89 solving equations with two variables, radius and diameter worksheets and elementary school, what is the difference between geometric and arithmetic graphs, printable worksheets one step equations seventh grade, Solving systems of equations with three variables calculator, partial differential equation software for TI89, formula,numerical answer,and unit problem solver, Multiplying and dividing decimals word problems worksheet, math homework help quadratic function not in standard form, first order nonlinear differential equation, square the first derivative. \frac{ 14}{ 2 - \sqrt{7} } $${ 4 }$$, Identify the conjugate of the denominator how do I get compound interest using a TI83 plus calculator? Explain the relevance and application of exponential functions in real-life situations. Example problems have radicals with variables and use conjugates to rationalize. \\ when are roots of variables absolute value? \cdot { 5 -7} Examples of Use. \frac{ 5}{ 3 - 2\sqrt{6} } Below is some background knowledge that you must remember in order to be able to understand the steps we are going to use. \frac{ 3 \color{red}{+} 2\sqrt{6} }{ 3 \color{red}{+} 2\sqrt{6} } As you know, rationalizing the denominator means to “rewrite the fraction so there are no radicals in the denominator”. = I will use "root" as a symbol for square root. 3. \\ \frac{ 7( \sqrt{2} \color{red}{-} \sqrt{5} )} \\ Thus, = . \frac{ \\ \frac{ \color{red}{3}(2 -1\sqrt{5}) } { \color{red}{3}(3) } Radical expressions in real life i get compound interest using a TI83 plus calculator part... Steps we are going to use it, replace square root of 2 imaginary. 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https://en.wikisource.org/wiki/Page:LorentzGravitation1916.djvu/32 | 1,550,521,808,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247488374.18/warc/CC-MAIN-20190218200135-20190218222135-00377.warc.gz | 562,427,033 | 12,276 | # Page:LorentzGravitation1916.djvu/32
derivatives and we shall determine the variation it undergoes by arbitrarily chosen variations ${\displaystyle \delta g_{ab}}$, these latter being continuous functions of the coordinates. We have evidently
${\displaystyle \delta Q=\sum (ab){\frac {\partial Q}{\partial g_{ab}}}\delta g_{ab}+\sum (abe){\frac {\partial Q}{\partial g_{ab,e}}}\delta g_{ab,e}+\sum (abef){\frac {\delta Q}{\partial g_{ab,ef}}}\delta g_{ab,ef}}$
By means of the equations
${\displaystyle \delta g_{ab,ef}={\frac {\partial }{\partial x_{f}}}\delta g_{ab,e}}$ and ${\displaystyle \delta g_{ab,e}={\frac {\partial }{\partial x_{e}}}\delta g_{ab}}$
this may be decomposed into two parts
${\displaystyle dQ=\delta _{1}Q+\delta _{2}Q}$ (42)
namely
${\displaystyle \delta _{1}Q=\sum (ab)\left\{{\frac {\partial Q}{\partial g_{ab}}}-\sum (e){\frac {\partial }{\partial x_{e}}}{\frac {\partial Q}{\partial g_{ab,e}}}+\sum (ef){\frac {\partial ^{2}}{\partial x_{e}\partial x_{f}}}{\frac {\partial Q}{\partial g_{ab,ef}}}\right\}\delta g_{ab}}$ (43)
${\displaystyle {\begin{array}{c}\delta _{2}Q=\sum (abe){\frac {\partial Q}{\partial x_{e}}}\left({\frac {\partial Q}{\partial g_{ab,e}}}\delta g_{ab}\right)+\sum (abef){\frac {\partial }{\partial x_{f}}}\left({\frac {\partial Q}{\partial g_{ab,ef}}}\delta g_{ab,e}\right)-\\\\-\sum (abef){\frac {\partial }{\partial x_{e}}}\left\{{\frac {\partial }{\partial x_{f}}}\left({\frac {\partial Q}{\partial g_{ab,ef}}}\right)\delta g_{ab}\right\}\end{array}}}$ (44)
The last equation shows that
${\displaystyle \int \delta _{2}QdS=0}$ (45)
if the variations ${\displaystyle \delta g_{ab}}$ and their first derivatives vanish at the boundary of the domain of integration.
§ 35. Equations of the same form may also be found if ${\displaystyle Q}$ is expressed in one of the two other ways mentioned in § 33. If e.g. we work with the quantities ${\displaystyle {\mathfrak {g}}^{ab}}$ we shall find
${\displaystyle (\delta Q)=\left(\delta _{1}Q\right)+\left(\delta _{2}Q\right)}$
where ${\displaystyle \left(\delta _{1}Q\right)}$ and ${\displaystyle \left(\delta _{2}Q\right)}$ are directly found from (43) and (44) by replacing ${\displaystyle g_{ab}}$, ${\displaystyle g_{ab,e}}$, ${\displaystyle g_{ab,ef}}$, ${\displaystyle \delta g_{ab}}$ and ${\displaystyle \delta g_{ab,e}}$ etc. by ${\displaystyle {\mathfrak {g}}^{ab}}$, ${\displaystyle {\mathfrak {g}}^{ab,e}}$ etc. If the variations chosen in the two cases correspond to each other we shall have of course
${\displaystyle (dQ)=\delta Q}$
Moreover we can show that the equalities
${\displaystyle \left(\delta _{1}Q\right)=\delta _{1}Q,\ \left(\delta _{2}Q\right)=\delta _{2}Q}$
exist separately.[1]
1. Suppose that at the boundary of the domain of integration ${\displaystyle \delta g_{ab}=0}$ and ${\displaystyle \delta g_{ab,e}=0}$. Then we have also ${\displaystyle \delta {\mathfrak {g}}^{ab}=0}$ and ${\displaystyle \delta {\mathfrak {g}}^{ab,e}=0}$, so that
${\displaystyle \int \left(\delta _{2}Q\right)dS=0,\ \int \delta _{2}QdS=0}$
and from | 1,017 | 3,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 28, "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-2019-09 | latest | en | 0.6798 |
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# ENMA 420/520 Statistical Processes Spring 2007 - PowerPoint PPT Presentation
ENMA 420/520 Statistical Processes Spring 2007. Michael F. Cochrane, Ph.D. Dept. of Engineering Management Old Dominion University. Class Eight Readings & Problems. Continuing assignment from last week! Reading assignment M & S Chapter 7 Sections 7.1 – 7.7; 7.9, 7.11 Recommended problems
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### ENMA 420/520Statistical ProcessesSpring 2007
Michael F. Cochrane, Ph.D.
Dept. of Engineering Management
Old Dominion University
• Continuing assignment from last week!
• M & S
• Chapter 7 Sections 7.1 – 7.7; 7.9, 7.11
• Recommended problems
• M & S Chapter 7
• 37, 40 (use Excel), 47, 61, 85, 98, 104
Estimating y2: Convenience of Normality Now Absent!
• Recall that s2 is a scaled X2 distribution
• Same approach for estimation
• Take sample of n observations
• Use s20 as basis for estimating 2y
• point estimate
• confidence interval
What are the cases
in which the sampling
distribution is
“conveniently” normal?
Now want to estimate 2y
reasoning
behind this?
Getting Confidence Interval for y2:Conceptually Same Approach
Recall from Section 6.11
Which variables are random variables?
Here is conceptual approach to be taken:
- sample n observations
- calculate s02 from sample
- substitute for X02 in terms of s02 and y2 in the following
p(X2(1-/2) X02 X2(/2) ) = 1 -
- the above range provides the (1-)100% CI for y2
Why?
Where do we get these?
The (1- )100% CI for y2: Working Through the Math
Notes: - the parent distribution y is assumed normal
- the CI is not necessarily symmetric about s2
a scaled X2
distribution
s^2 = y2
This area is 0.05
What are the critical values on the pdf?
Estimating CI for y2:Example Problem
• Problem summary
• Took n = 10 observations
• Found s0= 0.0098
• Want 95% CI for y2
Previous Example ProblemFinding the 95% CI
How do you interpret the above confidence interval?
Your sample variance was 0.00009604, do you see that the
the actual CI will depend on your sample.
This is THE pdf of s2,
a scaled X2 distribution.
For n=10, it exists and is
exact.
s^2 = y2
This is s^2 which
you do not know,
but you wish you did.
This area is 0.05,
how often will your sample s2
fall in this range?
Thinking About the Solutionto Example Problem
What keeps you from
determining it exactly?
• For means estimated differences between population means
• Why not estimate difference between population variances?
• Do you recall Section 6.11 in text?
is a “standard”
distribution
Which are the
random variables?
Ratio of VarianceTwo Populations
• F distribution has 2 associated degrees of freedom
• 1 = n1 - 1 ==> associated with numerator
• 2 = n2 - 1 ==> associated with denominator
• Have tabulated values of F (1, 2)
• Excel provides significantly more capability than tables
all variables
CI for the Ratio of VariancesFrom Two Populations
• Let’s discuss above CI and use of Table in text
• Problem 7.78 in M&S
Illustrating CI ofRatio of Population Variances
• Problem 7.79
• Comparing shear stress variances for two types of wood
• Southern Pine
• N = 100, y-bar = 1312, s = 422
• Ponderosa Pine
• N = 47, y-bar = 1352, s = 271
• Use interval estimation to
• Compare variation in shear stresses
• Draw inference from analysis
• How many measurements should we include in our sample??
• How wide do we want our CI to be?
• What confidence coefficient do we require?
Also a function of cost of sampling!
Choosing Sample Size
• Based on CI “half-width”, H
• We don’t know “s”, so we’ll have to approximate
• See example 7.17 on page 315
• If no estimate of “p” available, use p = q = 0.5
• If true p value differs substantially from 0.5, you’ll have a larger sample than needed
Recall our polling example… H is the “margin of error” | 1,288 | 4,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-47 | latest | en | 0.819181 |
http://www.justintools.com/unit-conversion/length.php?k1=fingerbreadth&k2=zeptometers | 1,571,214,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00505.warc.gz | 265,370,165 | 19,463 | 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 :)
Convert [Fingerbreadth] to [Zeptometers], (fb to zm)
LENGTH
= 2.2225E+19 Zeptometers
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: length
The base unit for length is meters (SI Unit)
[Zeptometers] symbol/abbrevation: (zm)
How to convert Fingerbreadth to Zeptometers (fb to zm)?
1 fb = 2.2225E+19 zm.
1 x 2.2225E+19 zm = 2.2225E+19 Zeptometers.
Always check the results; rounding errors may occur.
Definition:
A finger (sometimes fingerbreadth) is any of several units of measurement that are approximately the width of an adult human finger, including: The digit, also known a ..more definition+
In relation to the base unit of [length] => (meters), 1 Fingerbreadth (fb) is equal to 0.022225 meters, while 1 Zeptometers (zm) = 1.0E-21 meters.
1 Fingerbreadth to common length units
1 fb =0.022225 meters (m)
1 fb =2.2225E-5 kilometers (km)
1 fb =2.2225 centimeters (cm)
1 fb =0.072916666666667 feet (ft)
1 fb =0.875 inches (in)
1 fb =0.024305555555556 yards (yd)
1 fb =1.3809974747475E-5 miles (mi)
1 fb =2.3491174294472E-18 light years (ly)
1 fb =84.000010582679 pixels (PX)
1 fb =1.3890625E+33 planck length (pl)
1 fb =2.2225E+19 zm
2 fb =4.445E+19 zm
3 fb =6.6675E+19 zm
4 fb =8.89E+19 zm
5 fb =1.11125E+20 zm
6 fb =1.3335E+20 zm
7 fb =1.55575E+20 zm
8 fb =1.778E+20 zm
9 fb =2.00025E+20 zm
10 fb =2.2225E+20 zm
20 fb =4.445E+20 zm
30 fb =6.6675E+20 zm
40 fb =8.89E+20 zm
50 fb =1.11125E+21 zm
60 fb =1.3335E+21 zm
70 fb =1.55575E+21 zm
80 fb =1.778E+21 zm
90 fb =2.00025E+21 zm
100 fb =2.2225E+21 zm
200 fb =4.445E+21 zm
300 fb =6.6675E+21 zm
400 fb =8.89E+21 zm
500 fb =1.11125E+22 zm
600 fb =1.3335E+22 zm
700 fb =1.55575E+22 zm
800 fb =1.778E+22 zm
900 fb =2.00025E+22 zm
1000 fb =2.2225E+22 zm
2000 fb =4.445E+22 zm
4000 fb =8.89E+22 zm
5000 fb =1.11125E+23 zm
7500 fb =1.666875E+23 zm
10000 fb =2.2225E+23 zm
25000 fb =5.55625E+23 zm
50000 fb =1.11125E+24 zm
100000 fb =2.2225E+24 zm
1000000 fb =2.2225E+25 zm
1000000000 fb =2.2225E+28 zm | 925 | 2,371 | {"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-2019-43 | latest | en | 0.819344 |
https://electronics.stackexchange.com/questions/185117/solar-battery-charging-solar-iv-curve/185136#185136 | 1,632,368,533,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057416.67/warc/CC-MAIN-20210923013955-20210923043955-00547.warc.gz | 279,483,489 | 38,655 | # Solar battery charging, solar IV curve
I am thinking of making a solar battery charger for some NiMH batteries but am a bit confused as to how make sense of a solar panels IV curve in the application of battery charging.
1) For a direct connection of the battery to the solar cell, will the solar cell be forced to operate at the voltage of the battery?
2)For charging a battery, will higher currents reduce the charge time?
he maximum power point of the solar panel a point where power is maximum but this does not imply that the current is maximum at that point.
3) So if 1) and 2) are true then why is operation at the maximum power point desirable, if the current is not as high as it could be at the MPP? Wouldn't a battery with a lower voltage charge faster as the solar panel would be forced to operate at a point where the current was higher (left of the MPP)?
• (1) True until 1b)the battery is fully charged. Then the solar cell can destroy the battery. (2) True. (3) If the MPP is at a higher voltage than the battery, the MPPT charger steps it down, (a) avoiding 1b and (b) increasing charge current. Aug 11 '15 at 9:41
Connecting a battery to a PV cell without some kind of switcher to buck/boost or otherwise control the voltage of your panel will not make for the kind of results you're looking for.
If you just wire the battery and PV in parallel, the effect can be similar to putting an old battery in with a new battery; in short, the effect will be that current will flow from one source to the other and your battery might see the cell as a load, or vice versa.
As for point 2, more current would typically make for faster charging times, however, if we inspect the IV curve of a PV we see that as we draw more current, the voltage will begin to sag. Thus, we may run into the problem described above.
It is desirable to operate a PV at the MPP because solar can be quite an investment in certain cases. As such, we want to not necessarily draw the most current nor the most voltage, but the maximum amount of usable power with which to do work. Is MPP necessary for a battery charger? Debatable. I think the most important point is to make sure we're only loading the panel, and not loading the battery unintentionally. A SEPIC converter might be used in a battery charging scenario to get you where you need to be.
1) Yes. A pv panel connected directly to a battery will be "clamped" at the battery voltage.
2) Yes. More current at the same voltage means more "power".
3) No. At the maximum power point, there may be less current, but there is also more voltage and therefore more power. The trick is to convert the pv's higher-voltage power to the lower battery voltage using a switching dc-dc converter. Essentially, the extra volts are converted to extra amps. In other words, a dc-dc switch converter is an integral piece of an MPPT circuit. Without an efficient power converter, there is no gain. | 667 | 2,938 | {"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} | 3.234375 | 3 | CC-MAIN-2021-39 | latest | en | 0.949328 |
https://math.stackexchange.com/questions/3429655/identification-tangent-space-with-manifold | 1,656,169,936,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00147.warc.gz | 441,490,512 | 64,519 | # Identification tangent space with manifold
In the classical Euclidean setting $$\mathbb{R}^n$$ with cartesian coordinates $$x_i$$ one can identify a vector field with a map $$\mathbb{R^n}\to\mathbb{R}^n$$, because every tangent space at any point can be identified with $$\mathbb{R}^n$$ without ambiguity.
Suppose now that one has on $$\mathbb{R}^n$$ (or on an open subset of it, a ball for instance) a metric different from the classical one. My question is:
up to renormalise the basis $$\frac{\partial}{\partial x_i}$$ with respect to the new metric, can one identify $$\frac{\partial}{\partial x_i}$$ with $$x_i$$ so that a vector field can be identified once again with a map $$\mathbb{R}^n\to\mathbb{R}^n$$?
The identification will be something like $$b_i\frac{\partial}{\partial x^i}\mapsto b_ix^i$$.
• What does this have to do with a metric? This just depends on the structure of ${\mathbb R}^n$ as a differentiable manifold. As long as this structure is standard, the identification makes perfect sense. What is relevant here is that the tangent bundle is trivial. Nov 10, 2019 at 13:01
## 1 Answer
Suppose that $$M$$ is a differentiable $$n$$-manifold with trivial tangent bundle, $$TM\cong M\times R^n$$. For instance, $$M$$ can be any open subset of $$R^n$$.
A vector field on $$M$$ is just a smooth section of $$TM$$; in view of triviality of $$TM$$, it is just a smooth map $$M\to TM, x\mapsto (x, v(x)), x\in M, v(x)\in R^n$$. But it is the same thing as to say that you have a smooth map $$x\mapsto v(x)$$, i.e. a smooth map $$M\to R^n$$.
You do not need a metric for this, you also do not need explicit coordinates for this. | 481 | 1,653 | {"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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-27 | latest | en | 0.854117 |
https://numbermatics.com/n/225187539062500/ | 1,590,567,584,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392142.20/warc/CC-MAIN-20200527075559-20200527105559-00561.warc.gz | 485,765,730 | 9,422 | # 225187539062500
## 225,187,539,062,500 is an even composite number composed of three prime numbers multiplied together.
What does the number 225187539062500 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 297 divisors.
225187539062500 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of two hundred ninety-seven divisors.
## Prime factorization of 225187539062500:
### 22 × 510 × 78
(2 × 2 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7)
See below for interesting mathematical facts about the number 225187539062500 from the Numbermatics database.
### Names of 225187539062500
• Cardinal: 225187539062500 can be written as Two hundred twenty-five trillion, one hundred eighty-seven billion, five hundred thirty-nine million, sixty-two thousand, five hundred.
### Scientific notation
• Scientific notation: 2.251875390625 × 1014
### Factors of 225187539062500
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 20
• Sum of prime factors: 14
### Bases of 225187539062500
• Binary: 1100110011001110100011110100011000001110111001002
• Base-36: 27TLQ6L5US
### Squares and roots of 225187539062500
• 225187539062500 squared (2251875390625002) is 50709427749024963378906250000
• 225187539062500 cubed (2251875390625003) is 11419131242070580387175083160400390625000000
• 225187539062500 is a perfect square number. Its square root is 15006250
• The cube root of 225187539062500 is 60838.9137882303
### Scales and comparisons
How big is 225187539062500?
• 225,187,539,062,500 seconds is equal to 7,160,267 years, 9 weeks, 6 days, 16 hours, 1 minute, 40 seconds.
• To count from 1 to 225,187,539,062,500 would take you about seventeen million, nine hundred thousand, six hundred sixty-seven years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 225187539062500 cubic inches would be around 5069.9 feet tall.
### Recreational maths with 225187539062500
• 225187539062500 backwards is 005260935781522
• The number of decimal digits it has is: 15
• The sum of 225187539062500's digits is 55
• More coming soon! | 753 | 2,596 | {"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-2020-24 | latest | en | 0.812291 |
https://r02pro.github.io/normal-distribution.html | 1,695,320,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00214.warc.gz | 537,400,591 | 16,843 | ## 9.1 Normal Distribution
First, let’s review the definition of normal distribution, which is also called Gaussian distribution. If $$X\sim N(\mu, \sigma^2)$$, we say $$X$$ is a random variable following a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$.
In the following table, we list the four useful functions for normal distribution, and they will be introduced in the subsequent four parts, respectively.
Code Name Section
dnorm(x, mean, sd) probability density function 9.1.1
pnorm(q, mean, sd) cumulative distribution function 9.1.2
qnorm(p, mean, sd) quantile function 9.1.3
rnorm(n, mean, sd) random number generator 9.1.4
### 9.1.1 Probability Density Function (pdf)
To characterize the distribution of a continuous random variable, you can use the probability density function (pdf) . When $$X\sim N(\mu,\sigma^2)$$, its pdf is $f(x) = \frac{1}{\sqrt{2\pi \sigma}}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right].$
In R, you can use dnorm(x, mean, sd) to calculate the pdf of normal distribution.
• The argument x represent the location(s) at which to compute the pdf.
• The arguments mean and sd represent the mean and standard deviation of the normal distribution, respectively.
For example, dnorm(0, mean = 1, sd = 2) computes the pdf at location 0 of $$N(1, 4)$$, normal distribution with mean 1 and variance 4.
Note that the argument sd is the standard deviation, which is the square root of the variance.
In particular, dnorm() without specifying the mean and sd arguments will compute the pdf of $$N(0,1)$$, which is the standard normal distribution. Let’s see examples of computing the pdf at one location for three different normal distributions.
dnorm(0, mean = 1, sd = 2)
#> [1] 0.1760327
dnorm(1, mean = -1, sd = 0.5)
#> [1] 0.0002676605
dnorm(0) #standard normal
#> [1] 0.3989423
In addition to computing the pdf at one location for a single normal distribution, dnorm also accepts vectors with more than one elements in all three arguments. For example, you can use the following code to compute the three pdf values in the previous code block.
dnorm(c(0,1,0), mean = c(1, -1, 0), sd= c(2, 0.5, 1))
#> [1] 0.1760326634 0.0002676605 0.3989422804
If you want to compute the pdf at the same location 0 for distributions $$N(1,4)$$, $$N(-1, 0.25)$$, and $$N(0, 1)$$, you can use the following code.
dnorm(0, mean = c(1, -1, 0), sd= c(2, 0.5, 1))
#> [1] 0.1760327 0.1079819 0.3989423
If you want to compute the pdf at three different locations (-3, 2, and 5) for distribution $$N(3, 4)$$, you can use the following code.
dnorm(c(-3, 2, 5), mean = 3, sd = 2)
#> [1] 0.002215924 0.176032663 0.120985362
To get a better understanding on the shape of the normal pdf, let’s visualize the pdf of $$N(0,1)$$. You first need to create a equal-spaced vector x from -5 to 5 with increment 0.1. Then, you can compute the pdf value for each element of x using dnorm(). Finally, you can visualize the pdf using geom_line().
library(ggplot2)
x <- seq(from = -5, to = 5, by = 0.05)
norm_dat <- data.frame(x = x, pdf = dnorm(x))
ggplot(norm_dat) + geom_line(aes(x = x, y = pdf))
Next, you can take a step further to visualize three different normal distributions in the same plot, $$N(0,1)$$, $$N(1,4)$$, and $$N(-1, 0.25)$$. You can use the same vector x and compute the three pdfs on each element of x. geom_line() is still used with the variable dist mapped to the color aesthetic.
x <- seq(from = -5, to = 5, by = 0.05)
norm_dat_1 <- data.frame(dist = "N(0,1)", x = x, pdf = dnorm(x))
norm_dat_2 <- data.frame(dist = "N(1,4)", x = x, pdf = dnorm(x, mean = 1, sd = 2))
norm_dat_3 <- data.frame(dist = "N(-1, 0.25)", x = x, pdf = dnorm(x, mean = -1, sd = 0.5))
norm_dat <- rbind(norm_dat_1, norm_dat_2, norm_dat_3)
ggplot(norm_dat) + geom_line(aes(x = x, y = pdf, color = dist))
### 9.1.2 Cumulative Distribution Function (cdf)
In addition to pdf, you can compute the cumulative distribution function (cdf) of the normal distribution using the function pnorm(q, mean, sd). Generally speaking, the cdf of a random variable $$X$$ is defined as $F(x) = P(X\leq x).$ Similar to dnorm(), pnorm() also has two optional arguments, mean and sd, which represent the mean and standard deviation of the normal distribution, respectively. If you don’t specify these two arguments, pnorm() will compute the cdf of $$N(0,1)$$.
pnorm(0, mean = 1, sd = 2)
#> [1] 0.3085375
pnorm(0) # cdf at 0 of standard normal
#> [1] 0.5
You can also use pnorm() to visualize the cdf of the standard normal distribution.
q <- seq(from = -5, to = 5, by = 0.1)
norm_dat <- data.frame(q = q, cdf = pnorm(q))
ggplot(norm_dat) + geom_line(aes(x = q, y = cdf))
### 9.1.3 Quantile Function
The third useful function related to distributions is the quantile function. You can compute the quantile of the normal distribution using qnorm(p, mean, sd). The quantile function is the inverse function of the cdf. In particular, the $$p$$ quantile returns the value $$x$$ such that $F(x) = P(X\leq x) = p$
Let’s verify qnorm() is indeed the inverse function of pnorm() using the following example.
pnorm(qnorm(c(0.5,0.7)))
#> [1] 0.5 0.7
When $$p=0.5$$, qnorm() gives us the median of the normal distribution. Let’s see a few examples for computing the quantiles.
qnorm(0.5, mean = 1, sd = 2)
#> [1] 1
qnorm(0.5)
#> [1] 0
You can also visualize the shape of the quantile function.
p <- seq(from = 0.01, to = 0.99, by = 0.01)
norm_dat <- data.frame(p = p, quantile = qnorm(p))
ggplot(norm_dat) + geom_line(aes(x = p, y = quantile))
### 9.1.4 Random Number Generator
Lastly, to generate (pick up) random numbers from normal distributions, you can use the function rnorm(n, mean, sd) , with the argument n represents the number of random numbers to generate, the arguments mean and sd are the mean and standard deviation of the normal distribution you would like to generate from, respectively. Again, if you only supply the argument n, you will be generating random numbers from $$N(0,1)$$.
rnorm(3, mean = 0, sd = 1) #generate 3 random numbers from N(0, 1)
#> [1] -0.3393447 -0.1928847 0.2492699
rnorm(3) #generate another 3 random numbers from N(0,1)
#> [1] 0.1987472 -0.9778133 1.1789737
Since you are generating random numbers, the results may be different each time. In many applications, however, you may want to make the results reproducible. To do this, you can set random seed using the function set.seed() before generating the random numbers. Let’s see the following example.
set.seed(0)
rnorm(3)
#> [1] 1.2629543 -0.3262334 1.3297993
Now, let’s run it one more time.
set.seed(0)
rnorm(3)
#> [1] 1.2629543 -0.3262334 1.3297993
You can see that the exact 3 numbers are reproduced since you are using the same random seed 0. You can run these two lines of code on any machine and will get the exact same three random numbers.
Note that the code that involves randomness needs to be identical to reproduce the results. If you change the arguments in rnorm(), you will get totally different results. See the following example.
set.seed(0)
rnorm(1)
#> [1] 1.262954
rnorm(3)
#> [1] -0.3262334 1.3297993 1.2724293
By setting a different random seed, you will see different results as the following example.
set.seed(1)
rnorm(3)
#> [1] -0.6264538 0.1836433 -0.8356286
Lastly, let’s do a simple statistical exercise by checking the closeness of the sample mean and sample standard deviation to their population counterparts.
x <- rnorm(1e6, mean = 1, sd = 2)
mean(x) #sample mean
#> [1] 1.000093
var(x) #sample covariance
#> [1] 4.001485
sd(x) #sample standard deviation
#> [1] 2.000371 | 2,384 | 7,624 | {"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} | 4.3125 | 4 | CC-MAIN-2023-40 | latest | en | 0.786496 |
https://high-event.com/qa/why-is-acceleration-constant-in-free-fall.html | 1,611,458,789,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00237.warc.gz | 392,290,728 | 8,749 | # Why Is Acceleration Constant In Free Fall?
## What is constant acceleration equal to?
If we assume that the rate of change of velocity (acceleration) is a constant, then the constant acceleration is given by.
Acceleration=Change in velocityChange in time.
More precisely, the constant acceleration a is given by the formula.
a=v(t2)−v(t1)t2−t1, where v(ti) is the velocity at time ti..
## What is negative acceleration called?
If the speed of the vehicle decreases, this is an acceleration in the opposite direction and mathematically a negative, sometimes called deceleration, and passengers experience the reaction to deceleration as an inertial force pushing them forward.
## How do you calculate free fall acceleration?
How to use the free fall formula: an exampleDetermine the gravitational acceleration. … Decide whether the object has an initial velocity. … Choose how long the object is falling. … Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s .More items…
## Is Earth in free fall?
The Earth is in free-fall, but the pull of the Moon is not the same at the Earth’s surface as at its centre; the rise and fall of ocean tides occur because the oceans are not in perfect free-fall.
## Is acceleration constant in free fall?
An object that is moving only because of the action of gravity is said to be free falling and its motion is described by Newton’s second law of motion. … The acceleration is constant and equal to the gravitational acceleration g which is 9.8 meters per square second at sea level on the Earth.
## Why is acceleration constant?
Sometimes an accelerating object will change its velocity by the same amount each second. … This is referred to as a constant acceleration since the velocity is changing by a constant amount each second. An object with a constant acceleration should not be confused with an object with a constant velocity.
## Do heavier objects fall faster?
Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.
## Why is G negative free fall?
Any object affected only by gravity (a projectile or an object in free fall) has an acceleration of -9.81 m/s2, regardless of the direction. The acceleration is negative when going up because the speed is decreasing. … If the equation has g in it, like W = mg, direction is implied and the acceleration is positive.
## What is free fall acceleration?
A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. … The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.
## What are the 4 types of acceleration?
Any change in the velocity of an object results in an acceleration: increasing speed (what people usually mean when they say acceleration), decreasing speed (also called deceleration or retardation ), or changing direction (called centripetal acceleration ).
## What is the acceleration of free fall class 9?
Ans. The acceleration of free fall is; when the Body falls due to earth’s gravitational pull, its velocity changes and is said to be accelerated due to . the earth’s gravity and it falls freely called as free fall. This acceleration is calculated to be 9.8 m/s2.
## Why is free fall acceleration at Earth’s surface constant?
Explain why free-fall acceleration near Earth’s surface is constant. Near Earth’s surface, the distance between the object that is falling and the center of Earth does not change very much. … The extra mass of the heavy object exactly compensates fot the additional gravitational force.
## What is acceleration at maximum height?
Acceleration is the velocity increasing or decreasing(depend direction of acceleration ) in 1 second. … So at maximum height velocity becomes zero. then acceleration still acting on object and there is no balancing force at maximum height which can hold the object at max height.
## Does the initial velocity affect acceleration?
3. Initial velocity is independent of the slope of the graph; that is, the acceleration. An object thrown downward still accelerates after release at the same rate as an object that is dropped.
## What is the formula of free fall?
vf = g * t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. | 989 | 4,717 | {"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-2021-04 | longest | en | 0.911287 |
https://vrcacademy.com/tutorials/quartile-deviation-grouped-data/ | 1,685,962,068,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651815.80/warc/CC-MAIN-20230605085657-20230605115657-00513.warc.gz | 652,556,282 | 20,666 | ## Quartile Deviation for grouped data
Quartile deviation is half the difference between the thrid quartile $Q_3$ and the first quartile $Q_1$ of a frequency distribution. It is also known as Semi interquartile range.
Quartile Deviation is given by
$QD = \dfrac{Q_3-Q_1}{2}$
where,
• $Q_1$ is the first quartile
• $Q_3$ is the third quartile
## Quartile Deviation Formula for Grouped Data
The formula for $i^{th}$ quartile is
\begin{aligned} Q_i=l + \bigg(\frac{\frac{iN}{4} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned}
where
• $l :$ the lower limit of the $i^{th}$ quartile class
• $N=\sum f :$ total number of observations
• $f :$ frequency of the $i^{th}$ quartile class
• $F_< :$ cumulative frequency of the class previous to $i^{th}$ quartile class
• $h :$ the class width
The coefficient of quartile deviation is given by
### Coeff. of $QD=\dfrac{Q_3-Q_1}{Q_3+Q_1}$
where
• $Q_1$ is the first quartile of the data
• $Q_3$ is the third quartile of the data
## Quartile Deviation for Grouped Data Example 1
A class teacher has the following data about the number of absences of 35 students of a class. Solve five number summary for the following frequency distribution.
No.of days ($x$) 2 3 4 5 6
No. of Students ($f$) 1 15 10 5 4
### Solution
$x_i$ $f_i$ $cf$
2 1 1
3 15 16
4 10 26
5 5 31
6 4 35
Total 35
Inter-quartile range (IQR)
The inter-quartile range is given by $IQR= Q_3-Q_1$.
using the quartile deviation formula for grouped data,formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$ is calculated as
\begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(35)}{4}\bigg)^{th}\text{ value}\\ &=\big(8.75\big)^{th}\text{ value} \end{aligned}
The cumulative frequency just greater than or equal to $8.75$ is $16$. The corresponding value of $X$ is the $1^{st}$ quartile. That is, $Q_1 =3$ days.
Thus, $25$ % of the students had absences less than or equal to $3$ days.
Third Quartile $Q_3$ is calculated as
\begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(35)}{4}\bigg)^{th}\text{ value}\\ &=\big(26.25\big)^{th}\text{ value} \end{aligned}
The cumulative frequency just greater than or equal to $26.25$ is $31$. The corresponding value of $X$ is the $3^{rd}$ quartile. That is, $Q_3 =5$ days.
Thus, $75$ % of the students had absences less than or equal to $5$ days.
Quartile deviation for grouped data is calculated as
The quartile deviation ($QD$) is
\begin{aligned} QD & = \frac{Q_3 - Q_1}{2}\\ &= \frac{5 - 3}{2}\\ & = 1. \end{aligned}
Coefficient of quartile deviation is
\begin{aligned} \text{Coeff. of }QD & = \frac{Q_3 - Q_1}{Q_3+Q_1}\\ &= \frac{5 - 3}{5 + 3}\\ & = 0.25. \end{aligned}
## Quartile Deviation for Grouped Data Example 2
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. calculate Inter quartile range for the following frequency distribution.
Time spent on Internet ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 24 2
### Solution
Class Interval Class Boundries $f_i$ $cf$
10-12 9.5-12.5 3 3
13-15 12.5-15.5 12 15
16-18 15.5-18.5 15 30
19-21 18.5-21.5 24 54
22-24 21.5-24.5 2 56
Total 56
Quartiles
using quartile deviation formula for grouped data,formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$ is calculated as
\begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(14\big)^{th}\text{ value} \end{aligned}
The cumulative frequency just greater than or equal to $14$ is $15$. The corresponding class $12.5-15.5$ is the $1^{st}$ quartile class.
Thus
• $l = 12.5$, the lower limit of the $1^{st}$ quartile class
• $N=56$, total number of observations
• $f =12$, frequency of the $1^{st}$ quartile class
• $F_< = 3$, cumulative frequency of the class previous to $1^{st}$ quartile class
• $h =3$, the class width
The first quartile $Q_1$ can be computed as follows:
\begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{1*56}{4} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{14 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.9167\big)\times 3\\ &= 12.5 + 2.75\\ &= 15.25 \text{ minutes} \end{aligned} Thus, $25$ % of the students spent less than or equal to $15.25$ minutes on the internet.
Third Quartile $Q_3$
\begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(42\big)^{th}\text{ value} \end{aligned}
The cumulative frequency just greater than or equal to $42$ is $54$. The corresponding class $18.5-21.5$ is the $3^{rd}$ quartile class.
Thus
• $l = 18.5$, the lower limit of the $3^{rd}$ quartile class
• $N=56$, total number of observations
• $f =24$, frequency of the $3^{rd}$ quartile class
• $F_< = 30$, cumulative frequency of the class previous to $3^{rd}$ quartile class
• $h =3$, the class width
The third quartile $Q_3$ can be calculated as follows:
\begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{3*56}{4} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{42 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.5\big)\times 3\\ &= 18.5 + 1.5\\ &= 20 \text{ minutes} \end{aligned} Thus, $75$ % of the students spent less than or equal to $20$ minutes on the internet.
Quartile Deviation for grouped data is calculated as
The quartile deviation ($QD$) is
\begin{aligned} QD & = \frac{Q_3 - Q_1}{2}\\ &= \frac{20 - 15.25}{2}\\ & = 2.375\text{ minutes}. \end{aligned}
Coefficient of quartile deviation is
\begin{aligned} \text{Coeff. of }QD & = \frac{Q_3 - Q_1}{Q_3+Q_1}\\ &= \frac{20 - 15.25}{20 + 15.25}\\ & = 0.13475. \end{aligned}
Hope you like Quartile deviation for grouped data tutorial. Use quartlie deviation calculator to find the quartile deviation for grouped data (frequency distribution). | 2,239 | 6,150 | {"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} | 4.25 | 4 | CC-MAIN-2023-23 | longest | en | 0.790807 |
https://www.slingacademy.com/article/pandas-calculate-the-rolling-weighted-window-mean-of-a-dataframe/ | 1,716,098,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00301.warc.gz | 905,505,834 | 63,774 | # Pandas: Calculate the rolling weighted window mean of a DataFrame
Updated: February 23, 2024 By: Guest Contributor Post a comment
## Introduction
Working with time series data introduces specific statistical tools to efficiently analyze and transform the data, one of which is the rolling window operations. These operations are crucial when the data points are serially correlated. Particularly, in finance, economics, and weather forecasting, rolling window operations, such as weighted moving averages, are widely used for smoothing the data or generating trading signals.
This tutorial will guide you through the process of computing the rolling window weighted mean with the Pandas library in Python. By the end of this tutorial, you should be able to apply these techniques to your DataFrame and understand how to customize these for different analytical needs.
Prerequisite
Before diving into the calculations, ensure you have Pandas installed. If not, you can install it using pip:
``pip install pandas``
## Introduction to Pandas Rolling Window
Pandas provides robust methods for rolling window calculations, among them `.rolling()`, which sets the window and prepares the data for the operation. However, for weighted mean, we require an additional method: `.apply()`, with a lambda or predefined function to incorporate weights into our calculation.
## Basic Rolling Window Calculation
``````import pandas as pd
import numpy as np
# Sample Data
s = pd.Series([1, 2, 3, 4, 5])
# Simple rolling mean without weights
df_rolling = s.rolling(window=3).mean()
print(df_rolling)``````
The output,
``````0 NaN
1 NaN
2 2.0
3 3.0
4 4.0
dtype: float64``````
shows the simple moving average for a 3-day window. Since the first two data points do not have two preceding values, Pandas returns NaN values.
## Rolling Window with Weights
The concept of weighted means adds importance to certain values within your window, offering a more nuanced approach than the simple mean. In the case of a rolling weighted mean, weights are generally assigned so that more recent observations contribute more to the mean than older observations.
### Implementation Steps
1. Define your window size and weights.
2. Use the `.rolling()` with `.apply()` to implement the weighted mean calculation.
### Example
``````def weighted_mean(series, weights):
return np.average(series, weights=weights)
s = pd.Series([10, 20, 30, 40, 50])
window_size = 3
weights = np.array([0.5, 1, 1.5]) # More recent dates have higher weight
# Applying the weighted mean calculation
result = s.rolling(window=window_size).apply(lambda x: weighted_mean(x, weights), raw=True)
print(result)``````
The output,
``````0 NaN
1 NaN
2 28.0
3 38.0
4 48.0
dtype: float64``````
displays a weighted mean that increases over time, reflecting the increased weighting of more recent data. It’s crucial to ensure that the sum of your weights equals the window size or is normalized accordingly to reflect the distribution accurately.
## Advanced Scenarios
Rolling window calculations can be applied to DataFrame objects as well, enabling the examination of multiple time series or attributes simultaneously.
### Example with a DataFrame
``````df = pd.DataFrame({
'A': [1, 2, 3, 4, 5],
'B': [5, 4, 3, 2, 1]
})
# Assigning different weights to each column
weights_A = np.array([0.5, 1, 1.5])
weights_B = np.array([1, 1, 1]) # Equal weight for simplicity
# Function to apply different weights per column
def custom_weighted_mean(df, weights_dict):
results = {}
for column in df.columns:
weighted_means = df[column].rolling(window=3).apply(lambda x: np.average(x, weights=weights_dict[column]), raw=True)
results[column] = weighted_means
return pd.DataFrame(results)
weights_dict = {'A': weights_A, 'B': weights_B}
result_df = custom_weighted_mean(df, weights_dict)
print(result_df)``````
Such a tailored application allows for complex analysis across multiple columns with different weighting schemes, demonstrating the versatility of Pandas for rolling window operations.
## Handling Missing Data
Rolling window operations with Pandas handle missing data by default as NaN values. However, you might want to adjust this behavior depending on your analysis goals. Using parameters such as `min_periods` within `.rolling()` can control how the method deals with NaN values.
## Conclusion
Calculating the rolling weighted window mean with Pandas is an effective method to analyze time series data, offering insights into the data’s trends and patterns by assigning different importance to various points in the series. By mastering these techniques, you can unlock powerful data analysis capabilities for your projects, making your analyses more nuanced and impactful.
Search tutorials, examples, and resources | 1,079 | 4,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-22 | longest | en | 0.849626 |
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6 D. Optional: There are several helpful web tutorials, example models, and YouTube channel which could be useful to students through the course E. Optional: You may also want to read a good book about how companies are using what we cover in this course to gain a competitive edge in the business world: Competing on Analytics - The New Science of Winning (Hardcover) by Thomas H. Davenport and Jeanne G. Harris, Harvard Business School Press. Class Schedule* The Schedule provides a map of the course. The schedule is organized by Week Number and Week Start Date so you can organize and plan your studies accordingly. The critical item to remember is that project assignment and exam due dates cannot be delayed. Week Topic Description number and date 1 8/19 Chapter 1 Course overview Introduction to Analytics and Excel Decision Models Problem Solving and Decision Making Spreadsheet Add-ins for Business Analytics Software requirement: Solver Platform 6
7 2 8/25 & 8/27 3 9/1 & 9/3 Chapter 4 Chapter 5 Chapter 6 Review Online material: instructional videos in Descriptive Measures: D-1, D-2, D-3, and D-4 In-class applications: Descriptive Statistical Measures (Chap. 4): Populations and Samples Measures of Location, Dispersion and Shape Measures of Association Excel functions Review online instructional videos in probability distribution: Dist-1 In-class review: probability distribution and applications (Chap. 5) Random variable and probability distribution Continuous and discrete distributions Random Sampling Due: Assignment 1 8/25: Course expectations (upload to Canvas drop box) Assignment 2 handout Last day to drop 8/25 (100% refund) Review online material: Sampling and estimation: confidence-r1 In class applications: Sampling and estimation (Chap 6) Estimating population parameters Sampling error (margin of error) Confidence interval for decision making 4 9/8 & 9/10 Chapter 7 Review online material: Hypothesis Testing HT-1s, HT-2s Optional: Download (in Canvas Files) PowerPoint slides (PP-1s, PP2s) In class applications: Statistical inference (Chap 7) Hypothesis Testing One-Sample Hypothesis Tests Two-Sample Hypothesis Tests ANOVA 5 9/15 & 9/17 Chapter 7 In class applications: Statistical inference Hypothesis Testing One-Sample Hypothesis Tests Two-Sample Hypothesis Tests ANOVA 6 9/22 & 9/24 Exam week EXAMS 1&2 (theory & practice) Due: Assignment 2-9/24 7
8 7 9/29 & 10/1 8 10/6 & 10/8 9 10/13 & 10/15 Chapter 8 Chapter 8 Chapter 8 Chapter 9 Regression Analysis Simple Linear Regression Evaluating Good Regression Models Assignment 3 handout Regression Analysis Simple and Multiple Linear Regression Building Good Regression Models Regression with Categorical Independent Variables Regression Analysis (Chap 8) Simple and Multiple Linear Regression Building Good Regression Models Regression with Categorical Independent Variables Forecasting (Chap 9) Forecasting Techniques Qualitative and Judgmental Forecasting Forecasting Models for Time Series with a Linear Trend Forecasting Time Series with Seasonality Measures of Forecasting Accuracy 10 10/20 & 10/22 Chapter 9 Forecasting (Chap 9) Forecasting Techniques Qualitative and Judgmental Forecasting Forecasting Models for Time Series with a Linear Trend Forecasting Time Series with Seasonality Measures of Forecasting Accuracy 11 10/27 & 10/ /3 & 11/ /10 & 11/12 Exam week Chapter 10 Chapter 10 Exam 3&4: 10/27 & 10/29 (theory and practice) Due: Assignment 3-10/29 Introduction to data mining The Scope of Data Mining Data Exploration and Reduction Classification Techniques Association Rule Mining Assignment 4 handout Last day to drop w/o penalty (11/3) Introduction to data mining The Scope of Data Mining Data Exploration and Reduction Classification Techniques Association Rule Mining 8
10 * This is a planned course structure and may change if necessary to meet learning goals 10
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IE 6.0 Download and Set-up To use the Web Commerce e-commerce service, you need to: Gain access to the Internet Install Microsoft Internet Explorer 6.0 Configure Temporary Internet files in Internet Explorer. | 8,217 | 31,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-39 | latest | en | 0.832064 |
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# 11-8 Exponential Growth and Decay
## 11-8 Exponential Growth and Decay
Warm Up Simplify each expression. 2. 25(1 + 0.02)3 1. (4 + 0.05)2 16.4025 26.5302 3. 1.0075 4. The first term of a geometric sequence is 3 and the common ratio is 2. What is the 5th term of the sequence? 48 5. The function f(x) = 2(4)x models an insect population after x days. What is the 128 insects population after 3 days?
## 11-8 Exponential Growth and Decay
California Standards
Preview of Algebra ll 12.0 Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay.
## 11-8 Exponential Growth and Decay
Vocabulary
exponential growth compound interest exponential decay half-life
## 11-8 Exponential Growth and Decay
Exponential growth occurs when a quantity increases by the same rate r in each period t. When this happens, the value of the quantity at any given time can be calculated as a function of the rate and the original amount.
## 11-8 Exponential Growth and Decay
The original value of a painting is \$9000 and the value increases by 7% each year. Write an exponential growth function to model this situation. Then find the paintings value in 15 years. Step 1 Write the exponential growth function for this situation. y = a(1 + r)t = 9000(1 + 0.07)t = 9000(1.07)t Write the formula. Substitute 9000 for a and 0.07 for r. Simplify.
## 11-8 Exponential Growth and Decay
The original value of a painting is \$9,000 and the value increases by 7% each year. Write an exponential growth function to model this situation. Then find the paintings value in 15 years. Step 2 Find the value in 15 years.
y = 9000(1.07)t = 9000(1.07)15
24,831.28
## 11-8 Exponential Growth and Decay
Check It Out! Example 1 An investment is increasing in value at a rate of 8% per year, and its value in 2000 was \$1200. Write an exponential growth function to model this situation. Then find the investments value in 2006.
## Step 1 Write the exponential growth function for this situation.
y = a(1 + r)t Write the formula.
= 1200(1 + 0.08)t
= 1200(1.08)t
## 11-8 Exponential Growth and Decay
Check It Out! Example 1 Continued A sculpture is increasing in value at a rate of 8% per year, and its value in 2000 was \$1200. Write an exponential growth function to model this situation. Then find the sculptures value in 2006. Step 2 Find the value in 6 years.
y = 1200(1.08)t
= 1200(1.08)6 1,904.25 Substitute 6 for t. Use a calculator and round to the nearest hundredth.
## 11-8 Exponential Growth and Decay
A common application of exponential growth is compound interest. Recall that simple interest is earned or paid only on the principal. Compound interest is interest earned or paid on both the principal and previously earned interest.
## Reading Math For compound interest
annually means once per year (n = 1).
## quarterly means 4 times per year (n = 4).
monthly means 12 times per year (n = 12).
## 11-8 Exponential Growth and Decay
Write a compound interest function to model the situation. Then find the balance after the given number of years. \$1200 invested at a rate of 2% compounded quarterly; 3 years Step 1 Write the compound interest function for this situation. Write the formula.
## Substitute 1200 for P, 0.02 for r, and 4 for n.
= 1200(1.005)4t Simplify.
## 11-8 Exponential Growth and Decay
Additional Example 2A Continued Write a compound interest function to model the situation. Then find the balance after the given number of years. \$1200 invested at a rate of 2% compounded quarterly; 3 years Step 2 Find the balance after 3 years. A = 1200(1.005)4(3) = 1200(1.005)12 Substitute 3 for t.
Use a calculator and round 1274.01 to the nearest hundredth. The balance after 3 years is \$1274.01.
## 11-8 Exponential Growth and Decay
Write a compound interest function to model the situation. Then find the balance after the given number of years. \$15,000 invested at a rate of 4.8% compounded monthly; 2 years Step 1 Write the compound interest function for this situation. Write the formula.
## Substitute 15,000 for P, 0.048 for r, and 12 for n.
= 15,000(1.004)12t Simplify.
## 11-8 Exponential Growth and Decay
Additional Example 2B Continued Write a compound interest function to model the situation. Then find the balance after the given number of years. \$15,000 invested at a rate of 4.8% compounded monthly; 2 years
Step 2 Find the balance after 2 years. A = 15,000(1.004)12(2) Substitute 2 for t. = 15,000(1.004)24 16,508.22
## 11-8 Exponential Growth and Decay
Check It Out! Example 2a
Write a compound interest function to model the situation. Then find the balance after the given number of years. \$1200 invested at a rate of 3.5% compounded quarterly; 4 years Step 1 Write the compound interest function for this situation. Write the formula.
= 1200(1.00875)4t
## 11-8 Exponential Growth and Decay
Check It Out! Example 2a Continued
Write a compound interest function to model each situation. Then find the balance after the given number of years. \$1200 invested at a rate of 3.5% compounded quarterly; 4 years Step 2 Find the balance after 4 years. A = 1200(1.00875)4(4) Substitute 4 for t. = 1200(1.00875)16
1379.49
## 11-8 Exponential Growth and Decay
Check It Out! Example 2b
Write a compound interest function to model the situation. Then find the balance after the given number of years. \$4000 invested at a rate of 3% compounded monthly; 8 years Step 1 Write the compound interest function for this situation. Write the formula.
## Substitute 4000 for P, 0.03 for r, and 12 for n.
= 4000(1.0025)12t Simplify.
## 11-8 Exponential Growth and Decay
Check It Out! Example 2b Continued
Write a compound interest function to model each situation. Then find the balance after the given number of years. \$4000 invested at a rate of 3% compounded monthly; 8 years Step 2 Find the balance after 8 years. A = 4000(1.0025)12(8) Substitute 8 for t. = 4000(1.0025)96
5083.47
## 11-8 Exponential Growth and Decay
Exponential decay occurs when a quantity decreases by the same rate r in each time period t. Just like exponential growth, the value of the quantity at any given time can be calculated by using the rate and the original amount.
## 11-8 Exponential Growth and Decay
Notice an important difference between exponential growth functions and exponential decay functions. For exponential growth, the value inside the parentheses will be greater than 1 because r is added to 1. For exponential decay, the value inside the parentheses will be less than 1 because r is subtracted from 1.
## 11-8 Exponential Growth and Decay
Additional Example 3: Exponential Decay The population of a town is decreasing at a rate of 3% per year. In 2000 there were 1700 people. Write an exponential decay function to model this situation. Then find the population in 2012.
## Step 1 Write the exponential decay function for this situation.
y = a(1 r)t Write the formula. Substitute 1700 for a and 0.03 for r. Simplify.
= 1700(1 0.03)t
= 1700(0.97)t
## 11-8 Exponential Growth and Decay
The population of a town is decreasing at a rate of 3% per year. In 2000 there were 1700 people. Write an exponential decay function to model this situation. Then find the population in 2012. Step 2 Find the population in 2012. Substitute 12 for t. Use a calculator and round 1180 to the nearest whole number. The population in 2012 will be approximately 1180 people. y = 1700(0.97)12
## 11-8 Exponential Growth and Decay
Check It Out! Example 3 The fish population in a local stream is decreasing at a rate of 3% per year. The original population was 48,000. Write an exponential decay function to model this situation. Then find the population after 7 years.
## Step 1 Write the exponential decay function for this situation.
y = a(1 r)t Write the formula. Substitute 48,000 for a and 0.03 for r. Simplify.
= 48,000(1 0.03)t
= 48,000(0.97)t
## 11-8 Exponential Growth and Decay
Check It Out! Example 3 Continued The fish population in a local stream is decreasing at a rate of 3% per year. The original population was 48,000. Write an exponential decay function to model this situation. Then find the population after 7 years. Step 2 Find the population in 7 years.
y = 48,000(0.97)7 Substitute 7 for t. Use a calculator and round to the nearest whole 38,783 number. The population after 7 years will be approximately 38,783 people.
## 11-8 Exponential Growth and Decay
A common application of exponential decay is half-life. The half-life of a substance is the time it takes for one-half of the substance to decay into another substance.
## 11-8 Exponential Growth and Decay
Additional Example 4A: Science Application Astatine-218 has a half-life of 2 seconds.
Find the amount left from a 500 gram sample of astatine-218 after 10 seconds. Step 1 Find t, the number of half-lives in the given time period. Divide the time period by the half-life. The value of t is 5. Write the formula. Step 2 A = P(0.5)t = 500(0.5)5 Substitute 500 for P and 5 for t. Use a calculator. = 15.625 There are 15.625 grams of Astatine-218 remaining after 10 seconds.
## 11-8 Exponential Growth and Decay
Additional Example 4B: Science Application Astatine-218 has a half-life of 2 seconds. Find the amount left from a 500-gram sample of Astatine-218 after 1 minute. Step 1 Find t, the number of half-lives in the given time period. 1(60) = 60 Find the number of seconds in 1 minute.
## 11-8 Exponential Growth and Decay
Additional Example 4B Continued Astatine-218 has a half-life of 2 seconds.
Find the amount left from a 500-gram sample of Astatine-218 after 1 minute.
Step 2 A = P(0.5)t Write the formula.
= 500(0.5)30
## Substitute 500 for P and 30 for t.
Use a calculator.
= 0.00000047 g
## 11-8 Exponential Growth and Decay
Check It Out! Example 4a Cesium-137 has a half-life of 30 years. Find the amount of Cesium-137 left from a 100 milligram sample after 180 years. Step 1 Find t, the number of half-lives in the given time period. Divide the time period by the half-life. The value of t is 6.
## 11-8 Exponential Growth and Decay
Check It Out! Example 4a Continued Cesium-137 has a half-life of 30 years. Find the amount of Cesium-137 left from a 100 milligram sample after 180 years. Step 2 A = P(0.5)t Write the formula. Substitute 100 for P 6 = 100(0.5) and 6 for t. = 1.5625 mg Use a calculator.
## 11-8 Exponential Growth and Decay
Check It Out! Example 4b Bismuth-210 has a half-life of 5 days. Find the amount of Bismuth-210 left from a 100-gram sample after 5 weeks. (Hint: Change 5 weeks to days.)
## Step 1 Find t, the number of half-lives in the given time period.
5 weeks = 35 days Find the number of days in 5 weeks. Divide the time period by the half-life. The value of t is 5.
## 11-8 Exponential Growth and Decay
Check It Out! Example 4b Continued Bismuth-210 has a half-life of 5 days. Find the amount of bismuth-210 left from a 100-gram sample after 5 weeks. (Hint: Change 5 weeks to days.)
Step 2 A = P(0.5)t
## Write the formula.
= 100(0.5)7 Substitute 100 for P and 7 for t. = 0.78125 g Use a calculator. There are 0.78125 grams of Bismuth-210 remaining after 5 weeks.
## 11-8 Exponential Growth and Decay
Lesson Quiz: Part I 1. The number of employees at a certain company is 1440 and is increasing at a rate of 1.5% per year. Write an exponential growth function to model this situation. Then find the number of employees in the company after 9 years. y = 1440(1.015)t; 1646 Write a compound interest function to model each situation. Then find the balance after the given number of years. 2. \$12,000 invested at a rate of 6% compounded quarterly; 15 years A = 12,000(1.015)4t, \$29,318.64
## 11-8 Exponential Growth and Decay
Lesson Quiz: Part II 3. \$500 invested at a rate of 2.5% compounded annually; 10 years A = 500(1.025)t; \$640.04 4. The deer population of a game preserve is decreasing by 2% per year. The original population was 1850. Write an exponential decay function to model the situation. Then find the population after 4 years. y = 1850(0.98)t; 1706
5. Iodine-131 has a half-life of about 8 days. Find the amount left from a 30-gram sample of Iodine-131 after 40 days. 0.9375 g | 3,434 | 12,307 | {"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} | 4.78125 | 5 | CC-MAIN-2020-05 | latest | en | 0.838931 |
http://www.jiskha.com/display.cgi?id=1337645948 | 1,495,825,199,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608676.72/warc/CC-MAIN-20170526184113-20170526204113-00420.warc.gz | 680,019,794 | 3,772 | # algebra
posted by on .
two motorcyclists started at the same point and travel in opposite directions. one travels 6 mph faster than the other.In 4 hours they are 432miles apart. how fast is each traveling?
speed of slower mph
speed of faster mph
• algebra - ,
let speed of slower be x mph
then speed of the faster is x+6 mph
distance covered by slower = 4x
distance covered by faster = 4(x+6)
4x + 4(x+6) = 432
8x + 24=432
8x = 408
x = 51
the slower went 51 mph, the faster went 57 mph | 148 | 495 | {"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-2017-22 | latest | en | 0.944383 |
https://qanda.ai/en/solutions/rTdUonaqyj | 1,627,723,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00474.warc.gz | 473,239,544 | 14,347 | Symbol
Problem
$8.$ GIVEN: $x^{2}=x+12$ PROVE: $x=4orx=-3$ Statements Reasons | 33 | 77 | {"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} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.394 |
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# Grammar School Bucks Entrance Exams, Maths and Verbal Reasoning - Week 5
## Welcome to week 5 of my interactive 11+ blog for Bucks children.
#### This week I am looking at multiples, factors, another Verbal Reasoning technique and some essential vocabulary to add to your growing knowledge. It includes exclusive example materials, based on real 11+ style questions.
A multiple is the answer you get when two numbers are multiplied.
Lowest common multiple (LCM): and what is it for?
Eg LCM of 10 and 25
The multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100. The multiples of 25 are: 25, 50, 75, 100
Multiples they have 'in common' are 50 and 100 50 100 So the LCM is 50
Here's a clip.
This is useful for finding the lowest common denominator of two fractions. (more of this in later weeks.)
1. Find the LCM of 8 and 7.
List the multiples of 8. __________________ List the multiples of 7. __________________
List the common multiples. __________________ What is the LCM? __________________
2. Find the LCM of 15 and 20.
List the multiples of 15. __________________ List the multiples of 20. __________________
List the common multiples. __________________ What is the LCM? ___________________
3. Find the LCM of 21 and 35.
List the multiples of 21. __________________ List the multiples of 35. __________________
List the common multiples. __________________ What is the LCM? ___________________
4. Find the LCM of 6 and 18.
List the multiples of 6. __________________ List the multiples of 8. __________________
List the common multiples. __________________ What is the LCM? ___________________
5. Find the LCM of 10 and 11.
List the multiples of 10. __________________ List the multiples of 11. __________________
List the common multiples. __________________ What is the LCM? __________________
6. Find the LCM of 2 and 12.
List the multiples of 2. __________________ List the multiples of 12. __________________
List the common multiples. __________________ What is the LCM? ___________________
7. Find the LCM of 3 and 9.
List the multiples of 3. __________________ List the multiples of 9. __________________
List the common multiples. __________________ What is the LCM? ___________________
8. Find the LCM of 7 and 5.
List the multiples of 7. __________________ List the multiples of 5. __________________
List the common multiples. __________________ What is the LCM? ___________________
A factor is a number that divides into a number.
It is useful to be able to quickly work out the highest common factor of numbers when we are simplifying.
For Greatest Common Factor, (GCF)
Find the GCF of 12 and 20
List the factors of 12: 12, 1, 6, 2, 3, 4________ List the factors of 20: 10, 2, 5, 4.
Find the common factors 2, 4 GCF is 4
1. Find the GCF of 8 and 12.
List the factors of 8. __________________ List the factors of 12. __________________
List the common factors. __________________ What is the GCF? __________________
2. Find the GCF of 15 and 20.
List the factors of 15. __________________ List the factors of 20. __________________
List the common factors. __________________ What is the GCF? ___________________
3. Find the GCF of 21 and 35.
List the factors of 21. __________________ List the factors of 35. __________________
List the common factors. __________________ What is the GCF? ___________________
4. Find the GCF of 6 and 18.
List the factors of 6. __________________ List the factors of 18. __________________
List the common factors. __________________ What is the GCF? ___________________
5 Find the GCF of 6 and 3.
List the factors of 6. __________________ List the factors of 3. __________________
List the common factors. __________________ What is the GCF? __________________
6. Find the GCF of 14 and 21.
List the factors of 14. __________________ List the factors of 21. __________________
List the common factors. __________________ What is the GCF? ___________________
7. Find the GCF of 64 and 18.
List the factors of 64. __________________ List the factors of 18. __________________
List the common factors. __________________ What is the GCF? ___________________
8. Find the GCF of 9 and 45.
List the factors of 9. __________________ List the factors of 45. __________________
List the common factors. __________________ What is the GCF? ___________________
http://www.bbc.co.uk/bitesize/ks2/maths/number/factors_multiples/play/
Primes are numbers that don't appear in the times tables, except the one times tables. It is useful to know your primes because you can't cancel down or simplify prime numbers: when you reach a prime, you know it is time to stop. Here's a reminder of your primes. You need to learn these.
The link below covers Primes, LCM, HCF. Do all pages, and the test section.
If you still don't get it, look at the links below.
https://www.mathsisfun.com/greatest-common-factor.html
https://www.mathsisfun.com/least-common-multiple.html
Verbal Reasoning
Find the missing word.
Example: The LIEST I can be there is noon becomes the EARliest I can be there is noon.
The three letter word was EAR.
Here's another example:
Do your work CFULLY. What word 'fits' that sentence? It has to be three letters long and go somewhere in the word CFULLY.
The word ends in ly so is probably an adverb, a word that describes a verb.
C---FULLY
I'll give you a clue, it goes in the gap there. So you know the three-letter-word does not always go right at the beginning of the word in capitals.
CAREFULLY
The word is ARE. (It doesn't have to have the same pronunciation in the long word and in the short word.)
Here's some for practice:
1) Her HING is getting worse as she gets older.
2) She said her FAVRITE food was cheese.
3) The PERMER loved applause.
4) My HBRUSH is broken..
5) My shoes are KING me.
6) WHATR goes around, comes around.
7) I HD that you were doing well.
8) What OCATION would you like when you grow up?
9) The girl WPERED a wish.
10) The puppy WPERED for its mum.
11) That is an IMPROVET on your first effort.
This weeks 10(ish) words:
amble, stroll, saunter, mosey, ramble, meander, dawdle, wander, (all mean to walk at a slow pace)
march, trek, hike, stride, (all mean to walk a medium pace)
trudge, plod, slog, (all mean to walk with difficulty)
run, jog, race, sprint, dart, dash, hurry, canter, gallop, (to move quickly)
NEXT WEEK -
Verbal Reasoning: Find the Hidden Word
Maths: Square Numbers, Triangular Numbers, and Fractions. | 1,609 | 6,475 | {"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} | 4.34375 | 4 | CC-MAIN-2022-05 | latest | en | 0.777436 |
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