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http://www.mrexcel.com/forum/excel-questions/343229-calculate-length-service-multiple-dates.html | 1,480,724,599,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00487-ip-10-31-129-80.ec2.internal.warc.gz | 619,899,414 | 9,917 | calculate length of service with multiple dates
This is a discussion on calculate length of service with multiple dates within the Excel Questions forums, part of the Question Forums category; I need to be able to determine an employee's seniority date (based on actual time worked) between several dates. The ...
1. calculate length of service with multiple dates
I need to be able to determine an employee's seniority date (based on actual time worked) between several dates.
The example is that the original hire date is:
hire(rehire) dates term dates
3/31/2000 2/6/2004
1/5/2005 10/5/2007
9/22/2008
The length of service the amount of time served between the hire(re) dates and the term dates and also between the last hire and today.
The seniority date would have to be moved from 3/31/00 to another date that incorporates the length of his absences (something around 22 months and 17 days later).
Thank you!
2. Re: calculate length of service with multiple dates
Welcome to the board.
If you're OK with the seniority being in days, try:
=SUMPRODUCT(B1:B3-A1:A3)
where A1:A3 and B1:B2 are as you posted and B3 has =TODAY()
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• | 317 | 1,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-50 | latest | en | 0.939492 |
https://garmentsmerchandising.com/standard-time-calculation-in-industrial-engineering/ | 1,726,693,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00821.warc.gz | 245,470,999 | 22,798 | # Standard Time Calculation in Industrial Engineering
Standard Time in Industrial Engineering:
Standard time of an operator is the time which a qualified operator should take to accomplish it if the she works at standard performance (at the rating of 100) and provided she takes extra time allowed to her as allowance.
If time is counted in second it is called standard time (or Standard Second) but if time is calculated in minutes, Standard time is referred to as Standard Minute or Standard Minute Value (SMV). In case of jackets or garments which take standard time to the tune of more than 2,000 seconds standard time may be calculated as Standard Minute or SMV. Standard time of an operation is also a constant value though different operators may take different observed time due to their different rating or efficiencies and if the allowance rate is fixed throughout the enterprise. Allowance is calculated on basic time.
Operator Observed Time Operator Rating Basic Time Allowance @ 15% Standard Time Julie 16” sec 75% 16 × 75% = 12” 1.8” 13.8” Jesmin 12” sec 100% 12 × 100% = 12” 1.8” 13.8” Kariman 10” sec 120% 12 × 120% = 12” 1.8” 13.8”
This table shows that two operators were observed to have taken different time to complete the same job due to their different efficiencies. But the basic and standard time for the operation are constant.
Standard time is calculated by adding allowance time to the basic time.
Example:
An operation has been calculated to have an average observed time of 20 seconds. The operator’s rating is 90%. Allowance rate to apply is 15%. Calculate standard time. | 366 | 1,611 | {"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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.943791 |
http://hackage.haskell.org/package/mtl-2.2.0.1/docs/src/Control-Monad-Cont.html | 1,540,299,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00165.warc.gz | 153,017,531 | 3,886 | {- |
Copyright : (c) The University of Glasgow 2001,
(c) Jeff Newbern 2003-2007,
(c) Andriy Palamarchuk 2007
License : BSD-style (see the file LICENSE)
Stability : experimental
Portability : portable
[Computation type:] Computations which can be interrupted and resumed.
[Binding strategy:] Binding a function to a monadic value creates
a new continuation which uses the function as the continuation of the monadic
computation.
[Useful for:] Complex control structures, error handling,
and creating co-routines.
[Zero and plus:] None.
[Example type:] @'Cont' r a@
The Continuation monad represents computations in continuation-passing style
(CPS).
In continuation-passing style function result is not returned,
but instead is passed to another function,
received as a parameter (continuation).
Computations are built up from sequences
of nested continuations, terminated by a final continuation (often @id@)
which produces the final result.
Since continuations are functions which represent the future of a computation,
manipulation of the continuation functions can achieve complex manipulations
of the future of the computation,
such as interrupting a computation in the middle, aborting a portion
of a computation, restarting a computation, and interleaving execution of
computations.
The Continuation monad adapts CPS to the structure of a monad.
Before using the Continuation monad, be sure that you have
a firm understanding of continuation-passing style
and that continuations represent the best solution to your particular
design problem.
Many algorithms which require continuations in other languages do not require
them in Haskell, due to Haskell's lazy semantics.
Abuse of the Continuation monad can produce code that is impossible
to understand and maintain.
-}
-- * MonadCont class
-- * The Cont monad
Cont,
cont,
runCont,
mapCont,
withCont,
-- * The ContT monad transformer
ContT(ContT),
runContT,
mapContT,
withContT,
-- * Example 1: Simple Continuation Usage
-- \$simpleContExample
-- * Example 2: Using @callCC@
-- \$callCCExample
-- * Example 3: Using @ContT@ Monad Transformer
-- \$ContTExample
) where
{- \$simpleContExample
Calculating length of a list continuation-style:
>calculateLength :: [a] -> Cont r Int
>calculateLength l = return (length l)
Here we use @calculateLength@ by making it to pass its result to @print@:
>main = do
> runCont (calculateLength "123") print
> -- result: 3
It is possible to chain 'Cont' blocks with @>>=@.
>double :: Int -> Cont r Int
>double n = return (n * 2)
>
>main = do
> runCont (calculateLength "123" >>= double) print
> -- result: 6
-}
{- \$callCCExample
This example gives a taste of how escape continuations work, shows a typical
pattern for their usage.
>-- Returns a string depending on the length of the name parameter.
>-- If the provided string is empty, returns an error.
>-- Otherwise, returns a welcome message.
>whatsYourName :: String -> String
>whatsYourName name =
> (`runCont` id) \$ do -- 1
> response <- callCC \$ \exit -> do -- 2
> validateName name exit -- 3
> return \$ "Welcome, " ++ name ++ "!" -- 4
> return response -- 5
>
>validateName name exit = do
> when (null name) (exit "You forgot to tell me your name!")
Here is what this example does:
(1) Runs an anonymous 'Cont' block and extracts value from it with
@(\`runCont\` id)@. Here @id@ is the continuation, passed to the @Cont@ block.
(1) Binds @response@ to the result of the following 'Control.Monad.Cont.Class.callCC' block,
binds @exit@ to the continuation.
(1) Validates @name@.
This approach illustrates advantage of using 'Control.Monad.Cont.Class.callCC' over @return@.
We pass the continuation to @validateName@,
and interrupt execution of the @Cont@ block from /inside/ of @validateName@.
(1) Returns the welcome message from the 'Control.Monad.Cont.Class.callCC' block.
This line is not executed if @validateName@ fails.
(1) Returns from the @Cont@ block.
-}
{-\$ContTExample
'ContT' can be used to add continuation handling to other monads.
Here is an example how to combine it with @IO@ monad:
>import System.IO
>
>main = do
> hSetBuffering stdout NoBuffering
> runContT (callCC askString) reportResult
>
>askString :: (String -> ContT () IO String) -> ContT () IO String
>askString next = do
> liftIO \$ putStrLn "Please enter a string"
> s <- liftIO \$ getLine
> next s
>
>reportResult :: String -> IO ()
>reportResult s = do
> putStrLn ("You entered: " ++ s)
Action @askString@ requests user to enter a string,
and passes it to the continuation.
@askString@ takes as a parameter a continuation taking a string parameter,
and returning @IO ()@.
Compare its signature to 'runContT' definition.
-} | 1,210 | 4,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-43 | latest | en | 0.816746 |
https://www.physicsforums.com/threads/basic-kinematic-equations.427796/ | 1,544,495,863,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823550.42/warc/CC-MAIN-20181211015030-20181211040530-00534.warc.gz | 1,016,203,888 | 12,473 | # Homework Help: Basic Kinematic Equations
1. Sep 9, 2010
### Siege
I already typed a long drawn out post, so I am going to simplfy it.
I am having trouble using algebra to rearrange equation.
For Example,
v=d/t
And so, to solve for both t and d, I do the following
vt=d/t x t
vt=d
vt/v=d/v
t=d/v
I have difficulty with most if not all the other equations. For example, the formula for acceleration.
a=Vf-Vi/t
Solve for t
a=Vf-Vi/t
at=Vf-Vi/t x t
at + Vi=Vf -Vi + Vi
at+Vi=Vf
Vf-Vi/a=t
This is my attempt at it. I multiplied by t, then eliminated the negative, and finally divided by a.
2. Sep 9, 2010
### rock.freak667
Yes those are correct.
For the second one, you could have done it simpler instead of expanding out:
a=(vf-vi)t
multiplying both sides by 't'
at=(vf-vi)
divide both sides by 'a'
t=(vf-vi)/a
3. Sep 10, 2010
### Siege
Thank you very much.
Are there any tricks to rearranging equations? or rules to follow? | 298 | 946 | {"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} | 3.375 | 3 | CC-MAIN-2018-51 | latest | en | 0.889826 |
http://westerncalendar.uwo.ca/Archive/2013/2013/pg883.html | 1,513,367,046,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00011.warc.gz | 304,206,298 | 6,631 | Actuarial Science (S)
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A course description will be available from the department at the time of registration.
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Extra Information: 3 lecture hours, 0.5 course. | 1,748 | 7,396 | {"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-2017-51 | latest | en | 0.806556 |
http://ticehurst.info/rank-of-matrices-39.html | 1,571,337,510,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00166.warc.gz | 194,394,003 | 5,680 | # RANK OF MATRICES PDF
The rank of a matrix A is the number of independent columns of A. A square matrix is full rank if all of its columns are independent. That is, a. The Rank of a Matrix. The fact that the vectors r 3 and r 4 can be written as linear combinations of the other two (r 1 and r 2, which are independent) means that the maximum number of independent rows is 2. Thus, the row rank—and therefore the rank—of this matrix is 2. (and checking that c 1 and c 3 are independent). However, naive approaches for minimizing functions over the set of low-rank matrices are either prohibitively time consuming (repeated singular value.
Author: Shannon Powlowski Country: Eritrea Language: English Genre: Education Published: 11 April 2016 Pages: 877 PDF File Size: 3.2 Mb ePub File Size: 16.39 Mb ISBN: 732-8-76488-184-8 Downloads: 25981 Price: Free Uploader: Shannon Powlowski
Gal Chechik Abstract When learning models that are represented in matrix forms, enforcing a low-rank constraint can dramatically improve the memory and rank of matrices time complexity, rank of matrices providing a natural regularization of the model.
However, its performance and applicability in real scenarios are limited by a lack of robustness to outlying or corrupted observations. Let the column rank of A be r and let c1, Every column of A can be expressed as a linear combination of the r columns in C.
R is the matrix whose i-th column is formed from the coefficients giving the i-th column of A as a linear combination of the r columns of C.
Though the new vector after addition is a different point [1. Now, imagine you took all possible combinations of the two vectors.
The set of all possible points you could reach by scaling and adding these two vectors including zero of each rank of matrices the entire plane that passes through the two rank of matrices and the origin.
When we do this, we are imagining the space that the vectors "span". In this case, it's a plane.
## Rank (linear algebra) - Wikipedia
Imagine the two vectors we started with were actually pointed in the same direction. If r is greater than c, then the maximum rank of the matrix is c.
The rank of a matrix would be zero only if the matrix had no elements. | 508 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-43 | latest | en | 0.911999 |
https://kr.mathworks.com/matlabcentral/cody/problems/62-elapsed-time/solutions/480491 | 1,576,204,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548537.21/warc/CC-MAIN-20191213020114-20191213044114-00114.warc.gz | 416,891,799 | 15,594 | Cody
# Problem 62. Elapsed Time
Solution 480491
Submitted on 30 Jul 2014 by Jon
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
%% d1 = '2010/12/14 12:00:00'; d2 = '2010/12/17 17:06:00'; elapsed = 77.1; assert(abs(elapsed_time(d1,d2) - elapsed) < 1e-6)
2 Pass
%% d1 = '2010/12/14 12:00:00'; d2 = '2010/12/14 12:00:01'; elapsed = 1/3600; assert(abs(elapsed_time(d1,d2) - elapsed) < 1e-6)
3 Pass
%% d1 = '1924/1/7 12:45:00'; d2 = '2010/12/14 17:06:00'; elapsed = 762076.35; assert(abs(elapsed_time(d1,d2) - elapsed) < 1e-6)
4 Pass
%% d1 = '2010/12/14 12:00:00' d2 = '2010/12/14 13:06:36' elapsed = 1.11; assert(abs(elapsed_time(d1,d2) - elapsed) < 1e-6)
d1 = 2010/12/14 12:00:00 d2 = 2010/12/14 13:06:36 | 361 | 834 | {"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-2019-51 | latest | en | 0.487466 |
http://www.freemathhelp.com/forum/threads/58088-Where-should-a-pilot-start-descent | 1,480,970,102,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541839.36/warc/CC-MAIN-20161202170901-00109-ip-10-31-129-80.ec2.internal.warc.gz | 476,161,678 | 10,624 | # Thread: Where should a pilot start descent
1. ## Where should a pilot start descent
Here is a diagram of the problem.
http://www.stewartcalculus.com/data/...p_0205_stu.pdf
Where Should a Pilot Start Descent?
An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:
i. The cruising altitude is h when descent starts at a horizontal
distance l from touchdown at the origin.
ii. The pilot must maintain a constant horizontal speed v
throughout descent.
iii. The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity).
1. Find a cubic polynomial P(x) = ax^3 + bx^2 + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and touchdown.
2. Use conditions (ii) and (iii) to show that .
(6hv^2)/l^2?k
i completed the first part, and i got p(x) = -2hx^3/l^3 + 3hx^2/l^2, but im a bit confused about the second part. I know how to get the 6h/l^2 but i dont know how to place the v^2 (or how it even got there).
to get the 6h/l^2, i simply took the third derivative of the polynomial p''(x) = 6ax + 2b, and i subbed in my a and b values from question one to get
p''(x) = 6(-2h/l^3)x + 2(3h/l^2)
i then subbed in x=0, since that is when the maximum accelreation will occur, and this value cannot be greater than a value k,
p''(0) = 6h/l^2
i dont know how i would include the v^2,
thanks,
2. ## Re: Where should a pilot start descent
Originally Posted by evox
Here is a diagram of the problem.
http://www.stewartcalculus.com/data/...p_0205_stu.pdf
Where Should a Pilot Start Descent?
An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:
i. The cruising altitude is h when descent starts at a horizontal
distance l from touchdown at the origin.
ii. The pilot must maintain a constant horizontal speed v
throughout descent.
iii. The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity).
1. Find a cubic polynomial P(x) = ax^3 + bx^2 + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and touchdown.
2. Use conditions (ii) and (iii) to show that .
(6hv^2)/l^2?k
i completed the first part, and i got p(x) = -2hx^3/l^3 + 3hx^2/l^2, but im a bit confused about the second part. I know how to get the 6h/l^2 but i dont know how to place the v^2 (or how it even got there).
to get the 6h/l^2, i simply took the third derivative of the polynomial p''(x) = 6ax + 2b, and i subbed in my a and b values from question one to get
p''(x) = 6(-2h/l^3)x + 2(3h/l^2)
i then subbed in x=0, since that is when the maximum accelreation will occur, and this value cannot be greater than a value k,
p''(0) = 6h/l^2
i dont know how i would include the v^2,
thanks,
This is a classic problem
http://www.freemathhelp.com/forum/viewt ... =3&t=26449
However, do not be tempted to copy the answers - your instructor might be looking at these reolies too.
3. ## Re: Where should a pilot start descent
Does my answer to question 1 look correct? it seems it is correct because if i sub in x=l, i get h and thats satisfies condition i
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• | 960 | 3,441 | {"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} | 3.84375 | 4 | CC-MAIN-2016-50 | latest | en | 0.884996 |
https://forums.developer.nvidia.com/t/how-to-get-the-boundary-position-of-the-top-box/263180 | 1,717,087,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00515.warc.gz | 216,213,432 | 6,531 | # How to get the boundary position of the top box？
How to get the boundary position of the top box like the picture bleow (blue mark) ？Thanks
2 Likes
I’m not exactly sure what you mean by “boundary position”
If those prims are labelled then you can use the `bounding_box_3d_fast` annotator to retrieve each prim’s transform.
The `transform[3, :3]` corresponds to its xyz position in world space. You can also look at the min/max for whichever axis you are interested in if you wish to get the “edge” from this view
Here is the `bounding_box_3d` annotator’s output of one of my box:
(0, `# Semantic identifier which can be transformed into a readable label using the idToLabels mapping`
-0.50000006, `# Minimum bound in x axis in local reference frame (in world units)`
-0.5000001, `# Minimum bound in y axis in local reference frame (in world units)`
-0.5000001, `# Maximum bound in x axis in local reference frame (in world units)`
0.50000006, `# Maximum bound in y axis in local reference frame (in world units)`
0.5, `# Minimum bound in z axis in local reference frame (in world units)`
0.5, `# Maximum bound in z axis in local reference frame (in world units)`
[[ 9.9999748e-02, -2.2479685e-04, 4.1298188e-07, 0.0000000e+00],
[ 2.2479685e-04, 9.9999748e-02, -3.2997058e-09, 0.0000000e+00],
[-4.1297341e-07, 4.2280677e-09, 1.0000000e-01, 0.0000000e+00],
[ 6.6031649e-04, -1.2652274e-03, 2.0099999e-01, 1.0000000e+00]], `# Local to world transformation matrix (transforms the bounds from local frame to world frame)`
0.9475 `# **EXPERIMENTAL** Occlusion percentage, where 0.0 is fully visible and 1.0 is fully occluded. See additional notes below.`
)
I am curious about how to get the minimum and maximum bounding box pixel coordinate in x,y,z axis from this annotation information, cause I am not really understand why the value of minimum and maximum bound in x and z axis in local reference frame (in world units) is nearly same.
Additional, Is that possible to transfer the `transform[3, :3]` (its xyz position in world space) to pixel coordinate?
If you can provide me the code snippets for visulizing this `bounding_box_3d` annotator, I will be truly appriciate, Many Thanks !!!
how can I get d 2D image annotation from bounding_box_3d.npy | 660 | 2,258 | {"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} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.773311 |
https://www.slideserve.com/curran-farrell/david-roseman-august-8-2012 | 1,511,034,059,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00295.warc.gz | 861,171,175 | 13,354 | 1 / 7
# David Roseman August 8, 2012 - PowerPoint PPT Presentation
FAQ: Why doesn’t the sun look like it is setting in the center of the channel on the day that the PYC celebrates Sunset Down the Channel? Answer: It’s all a matter of your perspective. David Roseman August 8, 2012.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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FAQ: Why doesn’t the sun look like it is setting in the center of the channel on the day that the PYC celebrates Sunset Down the Channel?
David Roseman
August 8, 2012
The sun is so far away that, for all practical purposes, its rays are parallel
The bearing of the channel is about 293.6 degrees true
The Sunset Down the Channel date is selected as the date when the sun sets at about 293.6
Assume three people observing the sunset on August 6, 2012. One is standing on the east end of the north pier, the second is standing on the east end of the south pier, and the third is standing at the SW corner of the PYC property.
With some help from Google Earth… One is standing on the east end of the north pier, the second is standing on the east end of the south pier, and the third is standing at the SW corner of the PYC property.
1. Look!! The sun is setting at the North pierhead!
2. No!! The sun is setting at the South pierhead!
3. Actually, the sun is setting down the center of the channel
The centerline view One is standing on the east end of the north pier, the second is standing on the east end of the south pier, and the third is standing at the SW corner of the PYC property.
August 6, 2012
The 15' Boston Whaler Summer Visit cruises inbound. Aboard are Kim, Jennifer, Katelyn and Lacey Jacobs, and (Granddad)David Roseman.Photo by Rilla and Joe Nameth, setup by David Roseman.
So… One is standing on the east end of the north pier, the second is standing on the east end of the south pier, and the third is standing at the SW corner of the PYC property.
Almost all of the PYC property is north of the centerline. If you are sitting in a blue chair on the PYC deck, or anywhere else on PYC property north of the centerline, the sun will appear to be setting north of the channel.
Perhaps we should calculate the angles for each deck chair, and celebrate for many days.
For more information, visit One is standing on the east end of the north pier, the second is standing on the east end of the south pier, and the third is standing at the SW corner of the PYC property.
http://pentwateryachtclub.com/sunset/sunset_calculations_2012.htm
Errors or comments to [email protected] | 777 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-47 | latest | en | 0.906828 |
https://www.physicsforums.com/threads/wind-pressure.799947/ | 1,679,480,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00278.warc.gz | 1,063,407,868 | 17,036 | # Wind pressure
Can.i.say?
I am a skydiving coach and I am trying to figure out how many pounds of force wind moving approximately 120 mph has on something with a surface area of 25 square inches. I know turbulence and compression play into it as we'll but I just want a general formula to figure it out. The object is flat. I have spent over 500 hours in a vertical wind tunnel teaching people how to manipulate the wind to fly their bodies. I want to know how slow I can fall
Mentor
120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
Can.i.say?
120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
What do you mean cross sectional area?
Can.i.say?
120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
I know I have been clocked at 210 mph fastest and 90 mph approx. slowest. I am not sure of average area of someone skydiving on with their stomach to the wind. But I was more thinking about the lift I get from my hands. Approx. 25 square inches
Can.i.say?
Is there something more simple? Area, wind speed=lbs?
Homework Helper
Gold Member
I'm not a skydiver, but you fall slower at terminal velocity when you expose as much of your body (including hands) to the 'wind'. Belly down is slower than feet first, because you expose greater surface area to the air, which increases the drag. At terminal velocity, the skydivers weight is equal to the air drag force. Without getting into the actual Physics too much, the drag force at terminal velocity V on a cylindrical surface is 0.00256V^2( A) , where V is in mph and A is the area of the body exposed to the wind, in square feet. So if a person with gear weighs 200 pounds and exposes full body to the wind (say 10 square feet to use a number) the air drag force at terminal velocity is 200 pounds, solve for V and you get about 90 mph. If you fall feet first if that is possible, then your exposed surface area is about say I don't know about 2 square feet, so V works out to about 200 mph . These are real rough numbers, depends on shape factor whether surface area is flat or cylindrical or rolled up a ball, and other factors.
zoki85
And which one of these profiles is his body in a horizontal position?
He didn't ask about his body, but something that is flat and has a surface area of 25 square inches.
zoki85
He didn't ask about his body, but something that is flat and has a surface area of 25 square inches.
He wants to know how slow he can fall. In the wiki page you linked to, there's actually listed drag coefficient range for a "man in upright position"
(= 1-1.3). If it is measured, numerically calculated or somehow aproximated I don't know, but maybe could be of use. | 680 | 2,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.951284 |
https://number.academy/10377 | 1,656,406,050,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00572.warc.gz | 483,023,888 | 12,370 | # Number 10377
Number 10,377 spell 🔊, write in words: ten thousand, three hundred and seventy-seven . Ordinal number 10377th is said 🔊 and write: ten thousand, three hundred and seventy-seventh. The meaning of number 10377 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 10377. What is 10377 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 10377.
## What is 10,377 in other units
The decimal (Arabic) number 10377 converted to a Roman number is (X)CCCLXXVII. Roman and decimal number conversions.
#### Weight conversion
10377 kilograms (kg) = 22877.1 pounds (lbs)
10377 pounds (lbs) = 4707.0 kilograms (kg)
#### Length conversion
10377 kilometers (km) equals to 6448 miles (mi).
10377 miles (mi) equals to 16701 kilometers (km).
10377 meters (m) equals to 34045 feet (ft).
10377 feet (ft) equals 3163 meters (m).
10377 centimeters (cm) equals to 4085.4 inches (in).
10377 inches (in) equals to 26357.6 centimeters (cm).
#### Temperature conversion
10377° Fahrenheit (°F) equals to 5747.2° Celsius (°C)
10377° Celsius (°C) equals to 18710.6° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
10377 seconds equals to 2 hours, 52 minutes, 57 seconds
10377 minutes equals to 1 week, 4 hours, 57 minutes
### Codes and images of the number 10377
Number 10377 morse code: .---- ----- ...-- --... --...
Sign language for number 10377:
Number 10377 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 10377
### Multiplications
#### Multiplication table of 10377
10377 multiplied by two equals 20754 (10377 x 2 = 20754).
10377 multiplied by three equals 31131 (10377 x 3 = 31131).
10377 multiplied by four equals 41508 (10377 x 4 = 41508).
10377 multiplied by five equals 51885 (10377 x 5 = 51885).
10377 multiplied by six equals 62262 (10377 x 6 = 62262).
10377 multiplied by seven equals 72639 (10377 x 7 = 72639).
10377 multiplied by eight equals 83016 (10377 x 8 = 83016).
10377 multiplied by nine equals 93393 (10377 x 9 = 93393).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 10377
Half of 10377 is 5188,5 (10377 / 2 = 5188,5 = 5188 1/2).
One third of 10377 is 3459 (10377 / 3 = 3459).
One quarter of 10377 is 2594,25 (10377 / 4 = 2594,25 = 2594 1/4).
One fifth of 10377 is 2075,4 (10377 / 5 = 2075,4 = 2075 2/5).
One sixth of 10377 is 1729,5 (10377 / 6 = 1729,5 = 1729 1/2).
One seventh of 10377 is 1482,4286 (10377 / 7 = 1482,4286 = 1482 3/7).
One eighth of 10377 is 1297,125 (10377 / 8 = 1297,125 = 1297 1/8).
One ninth of 10377 is 1153 (10377 / 9 = 1153).
show fractions by 6, 7, 8, 9 ...
### Calculator
10377
#### Is Prime?
The number 10377 is not a prime number. The closest prime numbers are 10369, 10391.
#### Factorization and factors (dividers)
The prime factors of 10377 are 3 * 3 * 1153
The factors of 10377 are 1 , 3 , 9 , 1153 , 3459 , 10377
Total factors 6.
Sum of factors 15002 (4625).
#### Powers
The second power of 103772 is 107.682.129.
The third power of 103773 is 1.117.417.452.633.
#### Roots
The square root √10377 is 101,867561.
The cube root of 310377 is 21,811755.
#### Logarithms
The natural logarithm of No. ln 10377 = loge 10377 = 9,247347.
The logarithm to base 10 of No. log10 10377 = 4,016072.
The Napierian logarithm of No. log1/e 10377 = -9,247347.
### Trigonometric functions
The cosine of 10377 is -0,949404.
The sine of 10377 is -0,314059.
The tangent of 10377 is 0,330796.
### Properties of the number 10377
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 10377 in Computer Science
Code typeCode value
10377 Number of bytes10.1KB
Unix timeUnix time 10377 is equal to Thursday Jan. 1, 1970, 2:52:57 a.m. GMT
IPv4, IPv6Number 10377 internet address in dotted format v4 0.0.40.137, v6 ::2889
10377 Decimal = 10100010001001 Binary
10377 Decimal = 112020100 Ternary
10377 Decimal = 24211 Octal
10377 Decimal = 2889 Hexadecimal (0x2889 hex)
10377 BASE64MTAzNzc=
10377 MD5294e09f267683c7ddc6cc5134a7e68a8
10377 SHA1a0f8707f7196bc498f3b3424c2e22e5dab28f162
10377 SHA2241b483095e9b42b44ae1ef9c0d8138d2e0913ca94df0b244780433c9f
10377 SHA25619ff64bd9bf35b03a7a8e67a10478a319a16ecc8083e8edba8052811f8db3777
10377 SHA384b404c6397d8c5a3725d4f777779fe622dfc7067c92a50155d426f77e3ef67f594ff91bb45731149e3dc6b9143b361bc0
More SHA codes related to the number 10377 ...
If you know something interesting about the 10377 number that you did not find on this page, do not hesitate to write us here.
## Numerology 10377
### Character frequency in number 10377
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 1 3 1 7 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 10377, the numbers 1+0+3+7+7 = 1+8 = 9 are added and the meaning of the number 9 is sought.
## Interesting facts about the number 10377
### Asteroids
• (10377) Kilimanjaro is asteroid number 10377. It was discovered by E. W. Elst from La Silla Observatory on 7/14/1996.
### Distances between cities
• There is a 6,448 miles (10,377 km) direct distance between Addis Ababa (Ethiopia) and Yono (Japan).
• There is a 6,448 miles (10,377 km) direct distance between Amman (Jordan) and Duque de Caxias (Brazil).
• There is a 10,377 miles (16,699 km) direct distance between Cochabamba (Bolivia) and Varanasi (India).
• There is a 6,448 miles (10,377 km) direct distance between Dallas (USA) and Yono (Japan).
• More distances between cities ...
• There is a 6,448 miles (10,377 km) direct distance between Depok (Indonesia) and Łódź (Poland).
• There is a 6,448 miles (10,377 km) direct distance between Douala (Cameroon) and Nanchong (China).
• There is a 6,448 miles (10,377 km) direct distance between Duque de Caxias (Brazil) and Konya (Turkey).
• There is a 6,448 miles (10,377 km) direct distance between Kaduna (Nigeria) and Xi’an (China).
• There is a 10,377 miles (16,699 km) direct distance between Kathmandu (Nepal) and Rosario (Argentina).
• There is a 6,448 miles (10,377 km) direct distance between Konya (Turkey) and Rio de Janeiro (Brazil).
• There is a 6,448 miles (10,377 km) direct distance between Luanda (Angola) and Phnom Penh (Cambodia).
• There is a 6,448 miles (10,377 km) direct distance between Maiduguri (Nigeria) and Yunfu (China).
• There is a 10,377 miles (16,700 km) direct distance between Puebla (Mexico) and Singapore (Singapore).
## Number 10,377 in other languages
How to say or write the number ten thousand, three hundred and seventy-seven in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 10.377) diez mil trescientos setenta y siete German: 🔊 (Anzahl 10.377) zehntausenddreihundertsiebenundsiebzig French: 🔊 (nombre 10 377) dix mille trois cent soixante-dix-sept Portuguese: 🔊 (número 10 377) dez mil, trezentos e setenta e sete Chinese: 🔊 (数 10 377) 一万零三百七十七 Arabian: 🔊 (عدد 10,377) عشرة آلاف و ثلاثمائةسبعة و سبعون Czech: 🔊 (číslo 10 377) deset tisíc třista sedmdesát sedm Korean: 🔊 (번호 10,377) 만 삼백칠십칠 Danish: 🔊 (nummer 10 377) titusinde og trehundrede og syvoghalvfjerds Dutch: 🔊 (nummer 10 377) tienduizenddriehonderdzevenenzeventig Japanese: 🔊 (数 10,377) 一万三百七十七 Indonesian: 🔊 (jumlah 10.377) sepuluh ribu tiga ratus tujuh puluh tujuh Italian: 🔊 (numero 10 377) diecimilatrecentosettantasette Norwegian: 🔊 (nummer 10 377) ti tusen, tre hundre og sytti-syv Polish: 🔊 (liczba 10 377) dziesięć tysięcy trzysta siedemdziesiąt siedem Russian: 🔊 (номер 10 377) десять тысяч триста семьдесят семь Turkish: 🔊 (numara 10,377) onbinüçyüzyetmişyedi Thai: 🔊 (จำนวน 10 377) หนึ่งหมื่นสามร้อยเจ็ดสิบเจ็ด Ukrainian: 🔊 (номер 10 377) десять тисяч триста сiмдесят сiм Vietnamese: 🔊 (con số 10.377) mười nghìn ba trăm bảy mươi bảy Other languages ...
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If you know something interesting about the number 10377 or any natural number (positive integer) please write us here or on facebook. | 2,844 | 8,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.640785 |
https://ro.scribd.com/document/343119134/18-01-ja-p1p2-a | 1,582,706,386,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00337.warc.gz | 521,185,024 | 89,406 | Sunteți pe pagina 1din 15
# ALLEN
TM
PAPER CODE 0 1 C T 2 1 4 0 7 8
CAREER INSTITUTE
Path to Success KOTA (RAJASTHAN)
CLASSROOM CONTACT PROGRAMME
SCORE-I TEST # 08
ENTHUSIAST COURSE PATTERN : JEE (Advanced)
TARGET : JEE (Main + Advanced) 2015 Date : 18 - 01 - 2015
PAPER-1
PART-1 : PHYSICS SOLUTION
SECTION-I 2q A + q B 4
1. Ans. (A,C) =
qA + q B 3
r
6q A + 3q B = 4q A + 4q B 2qA = qB
B
C qA 1
Sol. q =2 (A) True
A B
D Final : q'A = qA ....(iii)
T vB = 0
PV = nRT kq 'A kq 'B
+ =0
1. A B, T = constant, PV = constant. 2R 2R
Now r increases q 'B
V decreases q ' = -1 (B) False
A
\ P increases. kq A kq 'B kq A kq A
\ out of A and B, A is true. VA = + = -
R 2R R 2R
B C: rT = constant
P = constant kq A
VA = ....(iv)
\ out of C and D ; C is true. 2R
\ Ans. A, C Put qB = 2qA in ....(1)
2. Ans. (A,D) kq A k(2q A )
+ = 2V
B R 2R
B
2R q'B kq A V
A
A
+
2R 2
R R
V
\ VA = ; VB = 0
q'A
Sol. qB 2R
qA
2
V
Initial Final
\ VA - VB =
2
kq A kq B
Initial : + = 2V ....(i) \ (C) is False
R 2R (D) is True
kq A kq B 3V 3. Ans. (B, C)
+ = ....(ii) 4. Ans. (A,C)
2R 2R 2 R1 R2
(i) (ii); 1 3 1 1 air air
Sol. = - 1 +
f 2 R 1 R 2
m=1 m=1
qB
qA +
2 =4 1 1 1 1
= +
qA qB 3
+ f 2 R1 R 2 m=
3
2 2 2
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7. Ans. (B,C)
3 Sol. Path difference
1 2 1 1
= - 1 + Dx = {(S2P t2) + m2t2} {(S1P t1) + m1t1}
f ' 4 / 3 R1 R 2 R1 R2 = S2P S1P + (m2 1)t2 (m1 1)t1
water water For nth order maxima
1 9 1 1 m=4/3 m=4/3
= - 1 + dy
f ' 8 R1 R 2 nl = {(m2 - 1)t 2 - (m1 - 1)t1} +
D
For zero or+der maxima
1 1 1 1 1 3 D
= + = m= y0 = {(m 2 - 1) t 2 - (m1 - 1) t1}
f ' 8 R 1 R 2 4f 2 d
(a) When both sheets have same average
f' = 4f
Initial : t1 + t 2
thickness and R.I. m1 and m2
1 1 1 2
{ }
d
= +
f v d D t +t
O y1 = (m 2 - m1 ) 1 2
1 1 1 fd d 2
= - v= -3 -3
v f d d-f 5 10 110 t +t
= (1.6 - 1.4) 1 2
Image is inverted, real and magnified. 1 2
\ t1 + t2 = 5 105 ....(i)
\ v>0 d>f ....(1)
m + m2
f (b) When both sheets have same R.I. = 1
and >1 2
d-f and thickness t1 and t2
f>df
2f > d ....(2)
m1 + m 2
2
y2 =
d {
D m 2 + m1
2 }
- 1 (t1 - t 2 )
1 1 1 8 10-3 110 -3
Final = + = t1 - t 2
4f v ' d 1(1.5 - 1)
t1 t2 = 1.6 105 ....(ii)
From (i) and (ii)
O
2t1 = 6.6 106
d t1 = 3.3 105 m and t2 = 1.7 105 m
8. Ans. (D)
9. Ans. (B,D)
1 1 1 Sol. Consider squirrel at distance x from midpoint C
= -
v ' 4f d of platform.
4fd v'
v' = <0 & >1
d - 4f u'
O
\ image is virtual & magnified. R R
37 h
5. Ans. (A,C)
C x
F
Sol. v = F.B.D. of squirrel :
r
f = ma ....(1)
500 N
v= = 50 ms -1
0.2
f
Pmean = 2p A F = 197 W
2 2 3/ 2
(friction)
l2 r mg
## 6. Ans. (D) N = mg ....(2)
HS-2/8 Kota/01CT214078
Enthusiast Course/Score-I/18-01-2015/Paper-1
F.B.D. of platform + Wheel cage Radius of x2 + y2 4x 21 = 0 is 5m.
Platform + wheel cage remain at equilibrium Radius of x2 + y2 = 25 is 5m.
only if
2pm 2p 1
tpivot = 0 , which is possible according to T= = = 3.14sec.
Bq 1 2
following F.B.D. only.
fh = Nx \ A, B, D
mah = mg x (from (1) & (2)) SECTION-IV
1. Ans. 4
2. Ans. 1
N' Sol. Let ball be dropped from origin at t = 0.
O O x
z
h w
N
x f Path
w'
y
gx
a= towards C (restoring acceleration)
g = gj
h B = Bk;
\ Squirrel performs SHM about mean
At = 0, v = 0 , force due to B is in xy plane
g always, force due to gravity is along Y.
position C, angular frequency w = and
h \ Ball always moves in XY plane
amplitude
\ v = v x i + v y j (at any time)
3R
A= (half the platform's length)
5 F = mg j + q(v x i + v y j) (Bx i + B y j)
In SHM, velocity at extreme should be zero.
Squirrel is at rest at one end of the platform
F = qBv yi + (mg - qBv x )j
at t = 0 .
\ Components of acceleration :
3R g
Maximum velocity = Aw = qBVy
5 h ax = ....(1)
m
3R g
vmax = qBVx
5 4R / 5 ay = g - ....(2)
m
9Rg 5
Vmax = d
25 4 of (2) :
dt
9Rg &&
v y = -qBv x
Vmax =
20
Maximum acceleration = w2A -qB qBv y
&&
vy = from (1)
m m
gA g 3R / 5 3g
a max = = = 2
h 4R / 5 4 qB
10. Ans. (A,B,D) &&
vy = - vy
m
Sol. v ^ B particle is performing circular motion in qBt
xy plane v y = A sin + f ....(3)
m
mv 110 AqB qBt
r= = = 5m ay = cos + f ....(4)
Bq 2 m m
Kota/01CT214078 HS-3/8
Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-1
At t = 0, vy = 0 and ay = g 4. Ans. 2
AqB di1
\ From (3); sin f = 0, & g = cos f Sol. E 2 = M = 2ktM
m dt
E 2ktM
mg i2 = 2 =
\ f = 0 and A = R R
Bq T
2Mk kMT 2
mg qBt
2
dq = i dt =
R 0
tdt =
R
\ vy = sin ....(5)
Bq m l 1 1
\ = 1 l = = 2 .
mg qBt 1- l 2 l
vy = sin 5. Ans. 3
Bq m
Sol. Colm : mV0 = (M + m)v
y
mg t qBt mV0
v= ....(1)
dy =
0
Bq 0
sin
m
dt M+m
v
2
m qBt at rest
y= g 1 - cos ....(6)
Bq m V0
M
qBt m
y is max. when vy = 0 i.e. sin =0
m a = g sin q ....(2)
2
m 2v 2mV0
\ y max = g(1 - (-1)) Time elapsed = =
Bq g sin q (M + m)g sin q
m
2
2 5 5 2
\ y max = 2 = = n=3
g 15 5 3
Bq
\ 3k 1 = 2 6. Ans. 0
\ k=1 Sol. B depends upon r.
3. Ans. 2 \ Flux linked with ring :
Y
2 1
1 Ring
Sol. E = - (13.6eV)Z - 2
(Z + 2) Z
2 B
X
Z2 - (Z + 2)3 4(Z + 1) 13.6 O
= -13.6 = eV
(Z + 2) (Z + 2) 2
2
...(i)
[Now energy of electron is R R
E0 t
f = B 2prdr = 2pr dr
h2 h 0 0 r
k= ; we have l =
2l 2m 2mk f = 2pE 0 tR
or, k = 6eV df
= 2pE 0 R = (induced electric field) 2pR
4(Z + 1) 13.6 dt
So, = 6 + 4.2 = 10.2eV \ Induced electric field = E0 (tangential)
(Z + 2) 2
Z +1 3
or, = (Z - 2)(3Z + 2) = 0
(Z + 2) 16
2
v
qvB
N
So, the value of Z = 2 (neglecting the negative/
fractional value) Ft=qE0
HS-4/8 Kota/01CT214078
Enthusiast Course/Score-I/18-01-2015/Paper-1
Circular motion of bead : 8. Ans. 9
Sol. Given body
mdV qE t
Ft = qE 0 = v = 0 m m
dt m P P
Centripetal force equation :
2R 5R
mv 2 2
N + qVB =
R
qE t E 0 t m q 2 E02 t 2 r
N+q 0 = R C1
R/2
C
m R R m2 r 2
N=0 M1 M2
7. Ans. 2
By superposition :
Sol. Dimension of wall = l;
P0 = atmospheric pressure GM1M GM 2 M
F= +
(2R) 2 5R 2
2
h F2
F1 GM M1 4M 2
F3 F= +
R 2 4 25
4 3
r pR 3
Pressure at bottom of meniscus = P0 GM 3 4 4 R
F= 2 + r p
Pressure at top of meniscus = P0 rgh R 4 25 3 2
\ Average pressure exerted on wall GM 4 3 1 4
F= pR r +
r gh 2
R 3 4 25 8
= P0 -
2 4 1 1
Force corresponding to this value F = pRr GM +
3 4 50
rgh 54
F1 = P0 - hl 4
2 F = pRrGM
3 200
Consider horizontal forces on volume of water
enclosed in dashed lines. 18 9
F = pRr GM = pRr GM
F3 + F1 = F2 50 25
where F1 = force by wall of vessel K
F = pRr GM
F2 = force by atmosphere 25
F3 = force by surface tension by rest of \ K=9
the liquid 9. Ans. 9
Sol. Acceleration of train = a
rgh
\ Sl + P0 - lh = P0 hl Motion of bob w.r.t. train :
2
rgh 2l Initially V bob / train = 0
Sl = From A B, pseudo force F = ma
2
acts on bob.
2S Let bob turns through 180 and has speed v' at
h=
rg highest position
W/E theorem A B :
KS K
h= \ 2= K=2 1
(K - 1)r g K -1 FR = mv 2 + mgh
2
Kota/01CT214078 HS-5/8
Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-1
C 10. Ans. 3
v' Sol. m1
released
m2
v B 180 R
F q
F=ma A
(pseudo) at rest
m2
R
v= 2gR (By C.O.M.E.) at rest
mv 2
= 2(ma - mg) ....(1)
R m1 m1
w
B C : Now train has stopped, a = 0, Conservation of angular momentum about axis:
## Conservation of energy m2R 2
m1 2gR R = w
3
1 1
mgR + mv '2 = mv2 3m1 2g / R
2 2 w=
m2
mv '2 mv 2
= - 2mg Coefficient of restitution :
R R
Rw
T + mg = 2(ma mg) 2mg from (1) = e =1
v
T = 2ma 5mg 0 Rw = v
(to complete vertical circular motion) R 3m1 2g
= 2gR
5g m2 R
a or a 25 ms2
2 m1 1 m 2
m = 3 m =3
\ 3K 2 = 25; K = 9 2 1
## PART2 : CHEMISTRY SOLUTION
SECTION - I 3. Ans. 8
1. Ans. (C, D) 4. Ans. 8
2. Ans. (A) i.e. a = 12, b = 4.
3. Ans. (C) 5. Ans. 6
4. Ans. (A,B,C) 6. Ans. 8
5. Ans. (A,C) 7. Ans. 3
6. Ans. (B)
7. Ans. (A,B,C) CHO
8. Ans. (A,C,D)
9. Ans. (A,B,D)
10. Ans. (A,B,C,D) 8. Ans. 6
SECTION - IV 9. Ans. 3
1. Ans. 10 [OMR Ans. 1] 10. Ans. 2
2. Ans. 2
HS-6/8 Kota/01CT214078
Enthusiast Course/Score-I/18-01-2015/Paper-1
PART-3 : MATHEMATICS SOLUTION
SECTION-I 8. Ans. (A, C)
B(z 2)
1. Ans. (A, B, D)
P(z)
f(x) must be surjective
D(z1) A(z1)
h(x) must be injective
C(z3)
2. Ans. (A, B)
From graph option (A, B) is correct PD||BC
3. Ans. (C, D) z + z1
\ arg = 0 or p
f -1:R + R + is given by y = xe
x z2 - z3
9. Ans. (B)
e ee
-1 For 2 selected elements there is only one
Now f (x)dx + f(x)dx = e.e - e
e
## 1 e options and for rest there will be 3 options =
n n2
ee +1 e C2 3
f(x)dx = e.ee +1 - e - x e x dx = ee +1(e - 1) + (ee - e)
e 1 10. Ans. (A, C)
4. Ans. (A, D) 5
5
C 2 3C2
n (s) = 5 , n(E) = 1 + C 2 + 2
= 26
The DE of orthogonal trajectory is
dx x-y 26 13
- = Required probability = =
dy x + y - 1 120 60
dy x + y - 1 SECTION-IV
=
dx -x + y 1. Ans. 1
(x dy + y dx) + y dy (x1) dx = 0 det (A) = 1, det (B) = 1
1 1
det (A ) = 1, det (B ) = 1
y2 x 2
-xy + - +x = c 2. Ans. 9
2 2
Angle between the given line and
y 2 - 2xy - x 2 + 2x + l = 0 which is a
1
Rectangular Hyperbola x axis = cos -1 so one of the variable line
3
5. Ans. (C, D) will be perpendicular to the x-axis. Required
The circle with AB as diameter intersect circle shortest distance = distance between (5, 0, 0)
with AA1 as diameter at M and N. Then line
and (5, 8, 8) = 128
MN will pass through the orthocentre of the
triangle. 3. Ans. 1
Hence line PQ will pass through the 1 2 3
orthocentre. 2 3 1 a b c = - 18 [a b c] = 1
6. Ans. (A, B) 3 1 2
## SP = SM and PT = MN (where PT is 4. Ans. 7
perpendicular from P to the axis) 180
sin 1 sin 2 ....... sin 179 =
NP = semi latus rectum = length of sub 2179
normal 90
sin 2 sin 4 ....... sin 178 =
7. Ans. (C, D) 289
1
The exponent of 2 in prime factorisation of sin 1 sin 3 ....... sin 179 =
289
15 , 1 14 , 2 13 ........ 7 8 are equal. l = 89
Kota/01CT214078 HS-7/8
Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-1
5. Ans. 5 9. Ans. 9
2
3 cos A sin C = 2 - sin A A must be acute AT + 10 T + 1 = 0 has roots m1, m2
Now max value of 2 1
and T + 2HT 9 = 0 has roots m1, -
3 cos A + sin A = 2 sin C = 1 m2
m
C = 90 1
Now m = 9
2
Answer l = 90 m1 + m 2 1
6. Ans. 7 and m m = - 10 m1 = - 1, m 2 = - 9
1 2
1000 = 1 + (n1) d A=9
3
(n1) d = 999 = 3 37 10. Ans. 1
Hence number of factors = 8 we get 8 DABC is a right angled triangle with area = 6
ordered pair of (n, d) but n > 3 unit and semi perimeter = 6 units
there are 7 AP's. circum radius of DDEF
7. Ans. 3 = inradius of DABC = 1 unit
p
cos 2A = 0 A = a : b : c =1 : 2 : 1
4
8. Ans. 0
F '(x) = f(x) - f(10 - x)
F"(x) = '(x) + '(10 x)
= 10 + cosx + cos(10 x) > 0
\ '(x) increasing function.
+
5
in (0,10) (x) has no maxima.
HS-8/8 Kota/01CT214078
ALLEN
TM
PAPER CODE 0 1 C T 2 1 4 0 7 9
CAREER INSTITUTE
Path to Success KOTA (RAJASTHAN)
CLASSROOM CONTACT PROGRAMME
SCORE-I TEST # 08
ENTHUSIAST COURSE PATTERN : JEE (Advanced)
TARGET : JEE (Main + Advanced) 2015 Date : 18 - 01 - 2015
PAPER-2
PART-1 : PHYSICS SOLUTION
SECTION-I
v u 2 + 300
1. Ans. (B) =
2 2
Sol. Acceleration of ball w.r.t. plane
If ball just rises by 15m then by conservation of
a bp = a bg - a pg = 2g (downward)
energy:
2
1 v
2gsin300 m = mg(15)
10 sin 300 2gcos300 2 2
2g
2
300 1 u 2 + 300
10cos300
m = mg(15) u=30m/s
2 2
300
1 5. Ans. (B)
s = ut + at 2
2
Sol. Conserving moment of system along rod.
1
0 = (10 sin 30 0
) t - ( 2g cos 30 0 ) t 2
2 mv cos45 = 2mv1
1 v1= v/22
t=
3 By work energy theorem:
2. Ans. (A) 2
1 1 v
mgcosq
f - kx 2 = - ( 2m )
mg q 2 2 2 2
q
in
Sol. m
gs
1 m
q g x = v
2 k
q
mg
6. Ans. (A)
f
y
Angle between friction will be p2q
p p Sol. q x
If q < . If q > , there will be no friction.
2 2
g
a=
3 dy 2
3. Ans. (A) = x
dx 5
4. Ans. (B)
Sol. Ball hits ground with velocity dy 2 x g
a = g sinq g tanq = g = = w2x
dx 5
v = u 2 + 2g(15) = u 2 + 300
Ball bounces back with velocity w2 = 4 w = 2 rad/sec
Corporate Office : ALLEN CAREER INSTITUTE, SANKALP, CP-6, Indra Vihar, Kota (Rajasthan)-324005
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Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-2
7. Ans. (B) 11. Ans. (B)
a
P
v
Sol. q
5P0 Sol.
P0
V
V0 (180 - q ) q
a= = 90 -
2 2
Dw q
h= DP = 2 mv sin ;
QS 2
2 q
time between successive collision = r sin
h=
16 P0V0 v 2
5R 7R
( 4 P0V0 ) + ( 20 P0V0 )
2 2 q
DP 2 mv sin 2 mv 2
Favg = = =
h = 0.2 Dt 2
v
(
r sin q
2 )
r
8. Ans. (C)
Sol. Calculate flux through curve surface and circular 12. Ans. (B)
cross section both. Sol. tnet about centre = 0 angular momentum about
9. Ans. (A) centre of circle is conserved
13. Ans. (B)
R
Sol. Resistance of each part = . For 'n' such parts Sol. If light ray strike mirror perpendicularly, it will
2n
retrace its path and image will be formed on
connected in series, equivalent resistances, say
object itself
R R 10
4
R1 = n = 10 3
2n 2 u = (208) + m = 12 +
( 3 / 2)
Similarly, equivalent resistance say R2 for 80 188
= 12 + = cm
9 9
another set of n identical respectively in parallel
14. Ans. (A)
1 R R 8 8
would be -1
= Sol. Critical angle qC = sin 9 = tan -1
n 2n 2n 2 17
therefore radius of circle from which light ray
For getting maximum of R1 & R2, they should
can enter is more than 10 cm. Hence complete
be connected in series & hence,
base area of the cube will let pass light ray.
## R 1 15. Ans. (C)
Req = R1 + R2 = 1 + 2
2 n
16. Ans. (C)
10. Ans. (D)
HS-2/7 Kota/01CT214079
Enthusiast Course/Score-I/18-01-2015/Paper-2
17. Ans. (A)
kQ2
For (A) KE of system =
R
Sol. Buoyancy B = rVsub g & spring balance
7 25kQ2
Electronic balance reading = costant + B For (B) KE of A = mv12 =
5 67R
## 18. Ans. (B)
5 42kQ2
19. Ans. (A) For (C) KE of B = mv 22 =
6 67R
Sol. a2
a1 a1 a2 For (D) Rotational KE of system
R R
F F
f1 f2 2 1 838kQ2
= mv12 + mv 22 =
5 3 2345R
## Let v1 = Velocity of A and v2 = velocity of B
20. Ans. (D)
just before they strike by energy conservation
Sol. For A :
k(2q)(-q) 1 2
0= + (2m)v12 1 +
2R 2 5 m0 I p p 2 2m 0 I
B P1 = 4 sin + sin =
4p ( L / 2) 4 4 pL
1 2 For B :
+ (m)r 2 1 +
2 3
m
2 0 ( IL2 )
2kM 4p m 0 IL2
7 5 kQ2 B P2 = 3 = =
mv12 + mv 22 = r a3 2 pa 3
5 6 R
## By taking torque about contact points For C :
7 a 5 a m0 ( 2 )
FR = (2m)R 2 1 2
& FR = 3 (m)R R2 IL
5 R kM 4 p m IL2
B P3 = 3 = = 0 3
r a 3
4 pa
## 42a1 = 25a2 42v1 = 25v2
For D :
2
7 5 42 kQ2
mv12 + v12 =
B P4 =
kM
1 + 3cos2 q where cos q =
1
5 6 25 R 3
r 2
## 469 kQ2 125kQ2
mv12 = mv12 = m0 ( 2 )
125 R 469R IL 2
4p 1 m 5IL
BP = 1 + 3 = 0
2
4
(a 2 ) 3
16pa3
Therefore
Kota/01CT214079 HS-3/7
Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-2
## PART2 : CHEMISTRY SOLUTION
SECTION-I 12. Ans. (C)
1. Ans. (C) Sol : PhN2Cl show coupling reaction only with highly
2. Ans. (C) activated ring.
## Sol : W and X are CH3CH2OH and (CH3)2CHOH. 1
H + + e-
H 2
On dry distillation it produces 2
## (HCOO)2Ca and (CH3COO)2Ca. 1
H+ + e
H
2 2
6. Ans. (B)
1/ 2
7. Ans. (C) PH 2C H+ A
Q= +
8. Ans.(B) PH2 A H C
9. Ans.(C) 0.1
1/ 2
10-1
= -2
Sol. (A) DG = DH TDS < O as DS < O so DH has 0.4 10
to be negative
1
(B) micelles formation will take place above Tk = (10) = 5
2
and above CMC
E = 0.50 (0.7)
(C) this solution will be negatively charged.
E = 0.6 (0.7)
(D) Fe3+ ions will have greater flocculatibility
power so smaller flocculating value. E = 0.42
## 10. Ans.(A) 16. Ans. (B)
(A) = 76 2.37
0.849
(B) 0.76 + 0.8 = 1.56
Sol. 1.24 = 34.3 M M = 469.68
0.05 (C) 0.34 + 0.8
(D) 0.8 2.37
\ as Hg2Cl2 molecules. = 1.57
11. Ans. (B)
17. Ans. (B)
## NO2 NH2 18. Ans. (B)
(i) Sn/HCl 19. Ans. (D)
Sol :
(ii) OH 20. Ans.(C)
HS-4/7 Kota/01CT214079
Enthusiast Course/Score-I/18-01-2015/Paper-2
PART-3 : MATHEMATICS SOLUTION
SECTION-I 5. Ans. (C)
1. Ans. (D) p-1
r a = b 2
(r)
P a
- ( p- 2)
sin -1 (sin(2x + 1)) dx
4
r Let 2x + 1 = t
p
1
O
2 -p
sin -1 (sin t) dt
Taking cross product with a , +2
2
a r - (a.r )a = a b
2
p2
a b a.r 1
p
r = 2 + 2
a a
a
.....(i) =
2 -p [t] dt + [ p - t] dt
2 + 2 p
r moves along a line parallel to a
2
for r to be minimum r .a = 0 1 p
2 p-1 p
1
=
2 -p
0.dt + 1.dt + 1.dt + 0.dt
ab 1 p p-1
from (i) rmin = 2 2 +2 2
a
1 p p
- 1 + - 1
2 2 2
2. Ans. (B) =
x 2 + 3a , x 1 1 p
f(x) = = [ p - 2] = - 1
2x + a , x =1 2 2
f(1) = f(1+) = f(1) y 6. Ans. (B)
a
3
3a + 1 = 2 + a f(x) f(y) = (x3 y 3) g(x y) "x, y R ..(1)
3a = a + 1 a +1 f(x + h) - f(x)
f '(x) = lim
x h
h0
f ' (1) = 2 replace x x + h and y x
3. Ans. (A)
[(x + h)3 - x 3 ]g(x + h - x)
= lim
S13 / 2 83 / 2 h 0 h
Area = = =8 2
2a 2
(x + h - x)((x + h)2 + x 2 + x(x + h))g(h)
4. Ans. (C) = lim
h 0 h
(x 2 + x + 1) - (x 2 - x + 1) 2
f ' (x) = 3x g(0)
y=
x2 + x + 1 + x2 - x + 1 let g(0) = l
2x f ' (x) = 3lx2
f(x) =
x2 + x + 1 + x 2 - x + 1 f(x) = lx3 + b
It is an odd function. Now from (1) put y = 0 f(x) f(0) = x 3g(x)
lx3 + b b = x3 g(x)
lim f(x) 1 and f(0) = 0 & lim -1
x x - g(x) = l, l 0
range is (1, 1) \ f is one-one onto and g is many one into
Kota/01CT214079 HS-5/7
Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-2
y
7. Ans. (A)
n6 25 n 6 35 n 6 ex
lim + + + ..... +
n 1 + n12 212 + n12 312 + n12 x
2
2n
r 5 n6 x 5dx ex
= lim 12 12 =
n
r =1 r + n 0
1 + x12
0
1
required area = 2 e dx = 2 e x - = 2
x 0
-1
put x = t = tan (64)
6
6 -
## 8. Ans. (C) 11. Ans. (D)
A Q chance of P 4 winning (or losing) the
B tournament is independent of the location of P2,
2
c1 5 c2 x
q
q P \ Events A : P2 & P4 are in different groups
and B : P 4 wins the tournament are
C
D
independent events
x 1 6 3
= x=5 \ P(A / B) = P(A) = =
x+5 2 8 4
12. Ans. (A)
1 1 A : P1 is not post of the new group
= PA.PD sin 2q - PB.PCsin 2q B : P2 is the winner of the tournament
2 2
1 1 2 1 1
= 96 sin 2q - 24 sin 2 q P(A B) = P(A), P(B/A) = =
2 2 3 8 12
= 36 sinq cosq 13. Ans. (A)
1 2 6 144 6 z3 - 1 p
= 72 = arg 3 =
5 5 25 z +1 2
9. Ans. (D) let z3 = w
e- x w moves on a semicircle as shown below
f(x) =
x w
-xe - x - e - x e - x (x + 1)
f '(x) = = -
x2 x2 q
+ 1 0 1
1 \ let w = e , 0 q p
iq
q+ 2 p
i ( q- 2 p )
i
z =e 3 iq
z - e iq / 3 ,e 3
,e 3
1
14. Ans. (B)
locus of z is as shown below
e
p/3
p/3
10. Ans. (A)
x (y ex) (y + ex) = 0
x = 0, y = ex, y = ex
shaded arc denotes the locus of z
HS-6/7 Kota/01CT214079
Enthusiast Course/Score-I/18-01-2015/Paper-2
15. Ans. (C) 18. Ans. (D)
Let P(cosq, sinq, 0) be a point on the circle then f + f '' = 2 f '
its distance d from the f f ' = (f f ')'
let f f ' = u
p u = kex f ' (x) f(x) = ke x
3 - 2 sin q +
3 - cos q - sin q 4
plane = d = = exf(x) = kdx + c
3 3
exf(x) = ku + c f(x) = kxex + cex
-3p f(0) = 0 c = 0
for maxima, q = f(1) = e k = 1 f(x) = xex
4
p 19. Ans. (B)
and for minima, q =
4 (P) General term of (1 + x 2 + x3)40 is
1 -1 1 1 40 l 2 m 3 n
A- , ,0 and B , ,0 1 .(x ) (x )
2 2 2 2 lmn
x1 + y 1 - x 2 - y 2 = 2 2 let r = exponent of x = 2m + 3n where m, n W
If r is odd then put n = 1 r = 2m + 3
16. Ans. (D)
can take all odd values except 1
Plane P is x + y +z = 3
If r is even, put n = 0
get the image of P(cosq, sinq, 0) in x + y + z =3
which comes out as r = 2m can take all even values
cos q - 2 sin q sin q - 2 cos q
a1 = 0
2
P ' 2 + ,2 + , 2 - (sin q + cos q) (Q),(R)
3 3 3
cos q - 2 sin q Put x = i
i.e. x = 2 + (a0 a2 + a4 ) + i(a1 a3 + a5 .....) = 1
3
sin q - 2 cos q a0 a2 + a4 ..... = 1
y = 2+ and a1 a3 + a5 ..... 0
3
(S) Differentiate the given identity thice and
2
and z = 2 - (sin q + cos q) then put x = 0 to get a2 = 40
3
20. Ans. (C)
eliminating q, we get
Area of triangle is minimum when (4, 6) is the
2x + 2y z = 6
mid-point of AB
17. Ans. (A)
Let axsecq bycosecq = a2 b2 be the normal (0,a) A
4xsecq 2 3 ycosecq = 4 .....(i) (4,6)
1 2
Q it passes through ,
2 2 2 3
B
2 sec q - 2 cos ecq = 4 (b,0)
a = 12 and b = 8
secq cosecq = 2 2 1
A min = 8 12 = 48
p 5p 13p 2
q=- , , \ (P)(S) No. of lines with area < 48 will be
4 12 12
2 (in II & IV quadrants)
4 sec q 2 tan q (Q) No. of lines with area = 48 will be 3
Now m = = (from (i))
2 3 cos ecq 3 (in I, II & IV quadrants)
2 4 4 (R) No. of lines with area = 60 will be 4
m=- , - 2, +2
3 3 3 (two in I & one each in II & IV quadrants)
Kota/01CT214079 HS-7/7 | 9,703 | 19,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-10 | latest | en | 0.625561 |
http://mathoverflow.net/questions/153700/probability-of-random-geodesics-on-the-round-sphere-intersecting?answertab=active | 1,462,185,905,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461863352151.52/warc/CC-MAIN-20160428170912-00056-ip-10-239-7-51.ec2.internal.warc.gz | 188,203,022 | 14,298 | # Probability of random geodesics on the round sphere intersecting
4 end points (a,b,c,d say) are chosen uniformly randomly and connected a to b and c to d by two geodesics on the 2-dim round sphere. Here uniformly is in the obvious sense of the volume form associated with the round metric. What is the probability the two curves cross?
-
Perhaps the question is about the shortest geodesics, in which case the answer is not so immediate, though not very hard either. – Wlodek Kuperberg Jan 6 '14 at 15:41
## 2 Answers
The probability is 1/8. With each (generic) pair of geodesic segments you can associate another 15 pairs which you get by replacing each of the 4 end points by its antipodal point. You can easily check that of these 16 pairs, exactly 2 intersect.
-
If you extend to complete circular geodesics then the probability is 1. So it is just a matter of finding the probability that the point of intersection is in both segments chosen, which is an easier linear (not spherical) problem.
- | 242 | 1,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-18 | latest | en | 0.931352 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-1-section-1-6-properties-of-integral-exponents-exercise-set-page-79/13 | 1,695,946,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00496.warc.gz | 852,014,303 | 13,554 | ## Intermediate Algebra for College Students (7th Edition)
$21x^{5}y^{}z^{4}$
RECALL: The product rule of exponents states that: $a^m \cdot a^n=a^{m+n}$. Multiply the numerical coefficients, and use the rule above to find: $\\=21x^{4+1}y^{0+1}z^{1+3} \\=21x^{5}y^{1}z^{4} \\=21x^{5}y^{}z^{4}$ | 116 | 293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.626037 |
https://www.hackmath.net/en/calculator/fraction?input=3%2F6-4%2F18 | 1,709,076,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00398.warc.gz | 799,555,624 | 7,617 | # Fraction calculator
This calculator subtracts two fractions. First, convert all fractions to a common denominator when fractions have different denominators. Find Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, simply subtract the numerators and place the result over the common denominator. Then simplify the result to the lowest terms or a mixed number.
## The result:
### 3/6 - 4/18 = 5/18 ≅ 0.2777778
Spelled result in words is five eighteenths.
### How do we solve fractions step by step?
1. Subtract: 3/6 - 4/18 = 3 · 3/6 · 3 - 4/18 = 9/18 - 4/18 = 9 - 4/18 = 5/18
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 18) = 18. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 18 = 108. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three sixths minus four eighteenths is five eighteenths.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. | 783 | 3,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-10 | latest | en | 0.833915 |
http://www.authorstream.com/Presentation/juliapeters-2543271-micro-ch6-part-cost-theory/ | 1,576,292,941,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00019.warc.gz | 164,941,958 | 30,330 | # Micro Ch6 part 1 cost theory
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Category: Entertainment
IB Econ Micro
## Presentation Transcript
### Cost Theory :
Cost Theory IB Economics
### Slide2:
Learning Objectives
### Slide4:
Long run and Short run When we talk about production costs we have to distinguish between short run and long run When a firm is producing some of its factors of production will be fixed in the short run The firm will not be able to quickly increase the quantity of them that it has In the short run a firm that makes furniture will only have a limited amount of factory space In the long run it could plan to build another factory but in the short term it has to make do with what it has Often the fixed factor is some element of capital or land It could also be a a type of highly skilled labour such as a specialist machine worker If a firm wants to produce more in the short run it can only apply more units of its variable factors to the fixed factors while it plans ahead Short run – the period of time in which at least one factor of production is fixed. All production takes place in the short run Long run – the period of time in which all factors of production are variable, but the state of technology is fixed. All planning takes place in the long run
### Slide5:
Long run and short run The length of the short run for the firm will be determined by the time it takes to increase the quantity of the fixed factor The furniture factory’s short run may be as long as it takes to build another factory (perhaps a year) A small firm involved in gardening’s fixed factor may be the number of lawn mowers it has Its short run will be the time it takes to order and take delivery of a new lawn mower which may only be a week The firm is in the long-run when it is planning (all the factors of production are variable) As soon as it makes the change the firm is once again in the short run but with a different number of fixed factors Once again the only way they can increase output is by applying more units of variable factors Complete Student workpoint 6.1
### Slide6:
Total, average and marginal product A firm has four machines (fixed factors) and increases its output by using more operators to work the machines (variable factors) Make a note of the AP and MP definitions and then complete the table on the next slide Once you have completed the table you need to plot 2 graphs The total product curve (total product on the Y axis and quantity of variable factor on the X axis) The AP and MP curve (same axes but scale will be different) Plot the MP on the mid points just as you drew the table e.g. you would plot 10 between 0 and 1 Now look at the curves and think about what we can deduce Average product (AP) – the output produced, on average, by each unit of variable factor Marginal product (MP) – the extra output that is produced by using an extra unit of the variable factor
### Slide7:
Quantity of Labour (V) Total product (TP) Average Product (AP) Marginal Product (MP) 0 0 10 1 10 10 15 2 25 12.5 20 3 45 4 70 5 90 6 105 7 115 8 120
### Slide8:
The Tennis Ball Game Now plot your AP and MP curves What is happening? What is your product? What are you fixed factors? What are your variable factors?
### Slide9:
Total, average and marginal product Watch the mjm foodie video Producer Theory
### Slide10:
The law of diminishing returns The hypothesis of eventually diminishing marginal returns – As extra units of a variable factor are added to a given quantity of a fixed factor, the output from each additional unit of the variable factor will eventually diminish The hypothesis of eventually diminishing average returns – As extra units of a variable factor are added to a given quantity of a fixed factor, the output per unit of the variable factor will eventually diminish
### Slide11:
Economic costs / profit Watch the mjm foodie video Accounting vs Economics Profit
### Slide12:
Time for you to do some work!! Read through pages 77-82 / make notes Complete the student workpoint P82
### Slide13:
Short run costs Firm’s have many different costs when producing whatever good or service they provide We need to understand the different types and where they originate There are 3 main types Fixed Variable Marginal
### Slide14:
Short run fixed costs Fixed costs (in the short run) are costs of production that do not vary as output changes It makes no difference how much the factory makes the fixed costs will be the same e.g. payment for buildings, rent, salaries (not wages), depreciation, machinery etc Fixed costs have to be paid for regardless of whether the firm produces nothing or works 24 hours Fixed costs are never zero Drawn as a horizontal line If there are high fixed to variable costs there is an opportunity for economies of scale because the fixed costs are spread over the output to give average fixed costs and these will reduce as output increases
### Slide15:
Variable costs Variable costs are costs of production that vary with output the more that is produced the more the variable costs Examples - raw materials, power, labour When there is no output the variable costs are zero Semi variable costs are costs that have a fixed and variable element e.g. telephone – there will be a fixed rental cost and a variable usage cost We add the variable costs to the fixed costs to get the total costs (TC) Average variable costs (AVC) are the total variable costs divided by the amount of output Watch this foodie video
### Slide16:
Cost Activity Using the same scenario that we used to create the MP and AP curves we are going to work out the costs and plot the curves (on excel) We are going to assume that there are 4 machines that cost \$100 each per week (fixed costs) Each worker costs \$200 per week (variable costs) For this scenario there are no other costs
### Slide17:
Fill in the spreadsheet and then use excel to create the curves Use 2 separate diagrams and include the following on each TFC,TVC, TC and AFC, AVC, ATC, ATC, MC Output should be on the x axis and costs on the y axis
### Slide21:
What do you notice about the TC curve? The TC curve is just the TVC curve shifted up by the fixed cost amount
### Slide22:
What do you notice about the marginal cost curve? It cuts through the ATC and the AVC at their lowest points (always!!)
### Slide23:
Average total costs (ATC) / Marginal Costs ATC is the total cost divided by the number of units produced These will decline as the fixed costs are spread over more unit It will then increase as the growth in variable costs is greater than the fall in fixed costs Marginal costs are the cost of producing one extra unit of output The average fixed costs fall as output rises The fall is very rapid as the fixed cost is spread over more units This will reduce the cost of producing the extra unit The falling AFC pull the marginal cost curve down But..the firm will be taking on more labour At some point the falling fixed costs will be unable to compensate for the increased labour costs Marginal costs will start to increase Watch this foodie video
### Slide24:
Note that the MC intersects the AC curve at the lowest point of the AC curve Extra units of labour raise marginal product (increasing returns) Extra units of labour reduce marginal product (decreasing returns) Whenever the marginal cost is below the average total costs then the average must be falling Even if the MC is increasing if it is less than the AC it will mean that AC are falling. When the MC is higher than the AC the AC will start to rise
### Slide25:
Note that the MC intersects the AC curve at the lowest point of the AC curve Extra units of labour raise marginal product (increasing returns) Extra units of labour reduce marginal product (decreasing returns) If there is a room full of people that are all 2m tall the average height will be 2m. If someone else comes into the room (the marginal person) that is 1.5m tall the average will fall (because they are lower than the average). If the next marginal person is 1.8m the average will still fall (still lower than the average). It is only when the next marginal person who is 2.2m tall that the average starts to increase (because the marginal person is taller than the average).
### Slide26:
These are the curves you need to remember Watch this foodie video
### Slide27:
Watch Pajholden’s video on costs http://www.youtube.com/watch?v=jNL9PNfrKZI
### Slide28:
Long Run Costs This diagram shows what happens in theory in the long run (in the planning stage) The LRAC is an envelope curve (it envelops an infinite number of short run average cost curves) Any point on the SRAC curve that is tangential to the LRAC is the lowest possible cost of producing the output (in the short run)
### Slide29:
Long Run Costs Say a firm is operating at C3 on SRAC1 Demand increases and the firm wants to produce q2 They can employ more variable factors and move along the SRAC They are producing even more cheaply The firm knows that it can produce this output even more cheaply if it alters all of its factors of production They will plan to make the change and move onto SRAC2 They will now be producing Q2 at C2 (an even lower cost), again at the lowest point on the SRAC2 because it is tangential to the LRAC
### Slide30:
Long Run Costs The whole LRAC curve is made up of an infinite number of single points from SRAC curves These curves represent all of the possible combinations of fixed and variable factors that could be used to produce different levels of output The LRAC is the boundary between unit cost levels that are attainable and those that are unattainable If possible the firm would wish to produce different output levels at points on the LRAC curve in order to minimise their cost per unit of output This may not always be possible in the short run
### Slide31:
Long Run Costs When the long run average costs are falling as output increases we say that the firm is experiencing increasing returns to scale This means that a given percentage increase in all factors of production will lead to a greater percentage increase in output When the LRAC are constant as output increases we call it constant returns to scale This means that a given percentage increase in all factors of production will lead to the same percentage increase in output When the LRAC are increasing with increased output we call it decreasing returns to scale This means that a given percentage increase in all factors of production will lead to a lower percentage increase in output thus increasing long run average costs Don’t confuse this with the law of diminishing returns; that is always the short run. This is the long run
### Slide32:
Long Run Costs There are two factors that make long run average costs increase or decrease Economies of scale and Diseconomies of scale There are a number of different economies of scale Specialisation – as the firm grows managers can concentrate on areas that they specialise in e.g. finance, marketing etc and become more efficient Division of Labour – breaking the production process down into small jobs and sometimes replacing people with machinery Bulk buying – getting discount for buying larger supplies Financial economies – getting a better rate of interest on loans because the company is less of a risk Transport economies – have their own transport fleet Large machines – when a firm is small they may not be able to afford large machinery and may have to hire it but as they grow they can buy it Promotional economies – the costs of promotion do not tend to increase in the same proportion as output – the cost of promotion per unit of output falls Economies of scale – any decreases in long run average costs that come about when a firm alters all of its factors of production in order to increase its scale of output
### Slide33:
Long Run Costs There are 2 main diseconomies of scale Control and communication problems – as firms grow the management will find it harder to control and coordinate the activities of the firm This may lead to inefficiency and increases in unit costs of production Alienation and loss of identity – as firms grow workers and managers may begin to feel that they are only a very small part of the organisation They may be less motivated and therefore work less hard and become less productive All of these economies and diseconomies of scale relate to the unit cost decreases or increases that may be encountered by a single firm – they are internal economies and diseconomies of scale When the whole industry size increases and it effects the unit costs these are known as external economies and diseconomies of scale Growth of industry leads to local universities with courses related to the industry – graduates will leave ready trained so there is less cost to the firms Rapid growth of the industry may lead to more competition for raw materials, capital and qualified labour which forces up the pries of these factors Diseconomies of scale – any increases in long run average costs that come about when a firm alters all of its factors of production in order to increase its scale of output
### Slide34:
Final note on cost theory Short run cost curves are U shaped because of the hypothesis of diminishing returns Long run cost curves are U shaped because of the existence of economies and diseconomies of scale In reality, economists have not yet found evidence of a firm becoming so large that the diseconomies of scale start to outweigh the economies of scale in the long run Actual long run cost curves may be drawn like this
### Slide35:
Time for you to do some work!! Read through pages 83-88 / make notes Complete exam questions 2 and 5 on P100 and the example paper 3 question on P101 | 2,864 | 13,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-51 | latest | en | 0.950809 |
https://www.jiskha.com/display.cgi?id=1162818228 | 1,516,596,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00332.warc.gz | 889,347,751 | 4,134 | # physics
posted by .
A 1-kg block is lifted vertically 1m by a boy. The work done by the boy would be one jule or would no work be done. I'm confused, I thought lifting wasn't considered work.
The work done is actually about 9.81 joules. Think of gravity pulling on the block as an invisible string connecting the block to the Earth. If you lift the block you store energy in the gravitational field (you "strech the string" a bit more) which you can release by dropping the block.
You are correct, Count.
• physics -
4
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You take a 10 N block and lift it 1 meter vertically from the floor and then place it back down on the floor. How much work do you do during this motion?
7. ### Physics
15. A refrigerator weighing 1500 N is to be lifted onto a truck bed that is 1.0 m above the ground. When pushed up a slanting ramp 2.0 m in length a force parallel to the ramp of only 800 N is required to move it at constant velocity. …
8. ### Physics
Alright, I can't figure it out, here's the problem. 3. A boy is giving his little sister a ride in the wagon. He pulls with an applied force of 20N. The frictional force against him is 20N and the sled is traveling 2m/s. If the boy …
9. ### Englsih
1. He bade the boy (to) work on the farm. 2. He ordered the boy to work on the farm. 3. He told the boy to work on the farm. ---------------------- Q1: Does #1 mean #2 or #3?
10. ### فيزياء
A 8.4-kg block is lifted vertically 9.0 m at constant speed by a boy. The work done by the boy is about:
More Similar Questions | 649 | 2,380 | {"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-2018-05 | latest | en | 0.944186 |
https://math.stackexchange.com/questions/2096574/conditional-probability-with-balls | 1,571,418,237,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00317.warc.gz | 589,586,608 | 30,864 | Conditional Probability with Balls
Hi I am having a difficult time to understand how to solve this problem although I get the concept of conditional probability but this is a tricky one:
suppose we have 4 balls that can be placed into 4 boxes one after another with equally likely possibilities. the question is what is the probability that a box contains exactly 3 balls given that the first 2 balls have been placed into different boxes. The solution I tried is as below but I am not convince that is done correctly: I take these events: PA(2 balls in the box) = 2/4 = 1/2 PB(1 ball in the box) = 1/4 then, calculate the prob that these two events happened PA and PB = 1/2* 1/4=1/8
is this correct? and appreciate for some explanation as I don't quite understand how this works.
Thanks.
There are $2$ ways to do this, and $4^2 = 16$ ways to distribute the remaining two balls without restriction. So your probability is $2/16 = 1/8$. | 232 | 940 | {"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} | 3.921875 | 4 | CC-MAIN-2019-43 | latest | en | 0.973194 |
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A109054 Numbers n such that the continued fraction expansion of sqrt(n) is multiplicative. 1
0, 1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 22, 23, 24, 25, 32, 33, 34, 35, 36, 44, 47, 48, 49, 58, 59, 60, 62, 63, 64, 74, 75, 78, 79, 80, 81, 95, 96, 98, 99, 100, 114, 119, 120, 121, 135, 136, 138, 140, 141, 142, 143, 144, 160, 162, 164, 167, 168, 169, 185, 187, 189, 192 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Perfect squares are assumed to have a continued fraction expansion of all zeros after a(0) and so are trivially multiplicative For nonsquares, a(1) must be 1 and so n must satisfy k+1/2 < sqrt(n) <= k+1, for some integer k. LINKS EXAMPLE The continued fraction of sqrt(22) is (4; 1, 2, 4, 2, 1, 8, ...), which is multiplicative with a(2^e) = 2, a(3^e) = 4, a(p^e) = 1 otherwise. CROSSREFS Cf. A040001, A010121, A040005, etc. Sequence in context: A319736 A100452 A004201 * A129142 A075752 A046541 Adjacent sequences: A109051 A109052 A109053 * A109055 A109056 A109057 KEYWORD nonn,easy AUTHOR Mitch Harris, Jun 18 2005 STATUS approved
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Last modified May 27 05:48 EDT 2020. Contains 334649 sequences. (Running on oeis4.) | 576 | 1,589 | {"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-2020-24 | latest | en | 0.75326 |
https://physics.stackexchange.com/questions/505514/paulis-exclusion-principle-in-elementary-particles | 1,579,733,975,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00112.warc.gz | 607,855,188 | 31,307 | # Pauli's exclusion principle in elementary particles
Elementary particles such as Quarks obey Pauli's exclusion principle since they exist in three colors (RBG). Where as electrons which is also elementary that does not have any color quantum numbers obey Pauli's exclusion principle. Any one please explain why.
• The Pauli principle is about being in the same quantum state. If a state is determined by a set of quantum numbers such as position, momentum, spin, color,+++ then the forbidden thing is for two fermions to share all these quantum numbers. Color is just one such quantum number, there is nothing special about it wrt. Pauli. – Marius Ladegård Meyer Sep 30 '19 at 6:52
• The question seems to be a non sequitur. Please edit the question to explain why you think there is an issue. – Ben Crowell Sep 30 '19 at 13:20
Pauli's exclusion principle applies to any particle which has a half integer spin.
Pauli first proposed his principle to explain the behaviour of electrons in atoms, and it was later found to apply to other spin 1/2 particles.
Spin 1/2 particles form a category called 'Fermions' (named after Enrico Fermi). Particles with integer spin are called 'Bosons' (named after Satyendra Bose).
Fermions can behave like Bosons in certain combinations. For example, a 'Cooper pair' of electrons.
Pauli Exclusion Principle:
No two electrons in an atom can have identical quantum numbers. This is an example of a general principle which applies not only to electrons but also to other particles of half-integer spin (fermions). It does not apply to particles of integer spin (bosons).
You state
Elementary particles such as Quarks obey Pauli's exclusion principle since
they are fermions. That is the correct continuation of the sentence.
they exist in three colors (RBG).
That is correct, the colors are three quantum numbers that can distinguish quarks ( which are fermions) between them, so if they have a different color they are allowed to be in the same quantum state because the color quantum number makes them non identical.
Whereas electrons which is also elementary that does not have any color quantum numbers obey Pauli's exclusion principle. Any one please explain why.
Pauli's exclusion principle for electrons as applied to atomic physics is one of the cases of the exclusion principle in quantum mechanics, two fermions of exactly the same quantum numbers cannot occupy the same energy level, in a quantum mechanical system. A red quark is as different from a blue quark as an electron from a muon . | 555 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-05 | latest | en | 0.93243 |
https://math.answers.com/math-and-arithmetic/How_do_you_cube_the_square_root_of_imaginary_number_ex_square_root_of_2i_cubed | 1,721,346,130,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00305.warc.gz | 333,165,871 | 48,815 | 0
# How do you cube the square root of imaginary number ex square root of 2i cubed?
Updated: 9/24/2023
Wiki User
10y ago
From the question, it seems you already calculated the square root, or know how to get it. You can cube complex numbers just like you cube normal numbers: multiply them by themselves; the number must appear three times as a factor. For example, the cube of (2 + i) is (2+i) x (2+i) x (2+i). Another method - usually faster - to calculate any power is to express the complex number in polar form (absolute value and angle). For the specific case of a cube, the cube of such a number is the cube of the absolute value, at an angle that is three times the angle of the original number.
Wiki User
10y ago | 181 | 728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.91689 |
http://mathhelpforum.com/pre-calculus/72958-distance-time-word-problem.html | 1,524,424,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00451.warc.gz | 198,462,148 | 11,432 | # Thread: distance and time word problem
1. ## distance and time word problem
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 1 hour and half.Find the rule of the function that expresses this distance traveled as a function of time [ Caution: Don't mix up the units of time ; use either minutes or hours , not both]
could someone help me with this problem? I sketched the graph ,and the ranges I got are from 0 to .75 , .75 to 2, 2 to 2.5 and 2.5 to 4 . I know that the first function is 3t and the third one is 6.25.However, I don't know how to write the other two ones.
2. Originally Posted by vance
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 1 hour and half.Find the rule of the function that expresses this distance traveled as a function of time [ Caution: Don't mix up the units of time ; use either minutes or hours , not both]
could someone help me with this problem? I sketched the graph ,and the ranges I got are from 0 to .75 , .75 to 2, 2 to 2.5 and 2.5 to 4 . I know that the first rule is 3t and the third one is 6.25.However, I don't know how to write the other two ones.
Hello!
The function rule for this problem is
distance = rate x time
d = rt
To write the rule as the distance traveled as a function of the time it took to travel that distance you will need to evaluate each set of data using this function. Since you have 3 different rates of speed, and 4 time periods, you will have to calculate each one.
I changed the times to hours.
RATE TIME
0 $\displaystyle \rightarrow$ .5
3 $\displaystyle \rightarrow$ .75
5 $\displaystyle \rightarrow$ 1.25
3 $\displaystyle \rightarrow$ 1.5
So, distance as a function of time:
d= rt
1) $\displaystyle d(t) = 0t$ so $\displaystyle d(.5) = (0)(.5) = 0 miles$
2) $\displaystyle d(t) = 3t$ so $\displaystyle d(.75) = (3)(.75) = 2.25 miles$
3) $\displaystyle d(t) = 5(t)$ so $\displaystyle d(1.25) = (5)(1.25) = 6.25 miles$
4) $\displaystyle d(t) = 3t$ so $\displaystyle d(1.5) = (3)(1.5) = 4.5 miles$
Hope this is what you needed! Good luck!
3. Hello, vance!
This is a tricky one!
Luckily, I've done one of these before . . .
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at 5 mph,
then sits and rests for 30 minutes, and finally walks for 1 hour and half.
Find the rule of the function that expresses this distance traveled as a function of time.
This is a piece-wise function.
The distance function depends on what time period is involved.
From $\displaystyle t=0\text{ to }t = 0.75$, the distance function is: .$\displaystyle 3t$
. . By the end of the 45 minutes, he has gone 2.25 miles.
From $\displaystyle t = 0.75\text{ to } t = 2$, the distance function is: .$\displaystyle 2.25 + 5t$
. . By the end of the 75 minutes, he has gone another 6.25 miles, a total of 8.5 miles.
From $\displaystyle t = 2\text{ to }t = 2.5$, the distance function is: .$\displaystyle 8.5 + 0t \:=\:8.5$
. . By the end of the 30 minutes, he hasn't moved; his total is still 8.5 miles.
From $\displaystyle t = 2.5\text{ to }4$, the distance function is: .$\displaystyle 8.5+3t$
. . By the end of the 90 minutes, he has gone another 4.5 miles, a total of 13 miles.
The function would be written like this:
. . $\displaystyle d(t) \;=\;\begin{Bmatrix}3t & & 0 \leq t \leq 0.75 \\ 2.25 + 5t & & 0.75 \leq t < 2 \\ 8.5 & & 2 \leq t \leq 2.5 \\ 8.5+3t & & 2.5 < t < 4 \end{Bmatrix}$
The graph would look like this:
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4. Originally Posted by Soroban
Hello, vance!
This is a tricky one!
Luckily, I've done one of these before . . .
This is a piece-wise function.
The distance function depends on what time period is involved.
From $\displaystyle t=0\text{ to }t = 0.75$, the distance function is: .$\displaystyle 3t$
. . By the end of the 45 minutes, he has gone 2.25 miles.
From $\displaystyle t = 0.75\text{ to } t = 2$, the distance function is: .$\displaystyle 2.25 + 5t$
. . By the end of the 75 minutes, he has gone another 6.25 miles, a total of 8.5 miles.
From $\displaystyle t = 2\text{ to }t = 2.5$, the distance function is: .$\displaystyle 8.5 + 0t \:=\:8.5$
. . By the end of the 30 minutes, he hasn't moved; his total is still 8.5 miles.
From $\displaystyle t = 2.5\text{ to }4$, the distance function is: .$\displaystyle 8.5+3t$
. . By the end of the 90 minutes, he has gone another 4.5 miles, a total of 13 miles.
The function would be written like this:
. . $\displaystyle d(t) \;=\;\begin{Bmatrix}3t & & 0 \leq t \leq 0.75 \\ 2.25 + 5t & & 0.75 \leq t < 2 \\ 8.5 & & 2 \leq t \leq 2.5 \\ 8.5+3t & & 2.5 < t < 4 \end{Bmatrix}$
The graph would look like this:
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- - * - - + - - - - + - + - - - - - + - -
| ¾ 2 2½ 4
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Thanks soroban. This is what I was looking for!!! | 1,812 | 5,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-17 | latest | en | 0.92502 |
https://www.marottaonmoney.com/how-to-use-a-marotta-capital-gains-tax-hurdle-report/ | 1,674,785,444,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00790.warc.gz | 887,673,817 | 19,723 | How To Use a Marotta Capital Gains Tax Hurdle Report
While many investment managers require their clients to liquidate their holdings to cash, we assume management of new client’s existing positions. From there, we evaluate holding by holding whether it is more advantageous to sell or hold.
Many of our new clients come to us after having worked with a commission-based advisor or a “mutual fund guy.” Because of this, their existing holdings include several funds with higher than desired expense ratios, purchased a long time ago, and never trimmed. Now, this means that they have a holding with a high expense ratio (costs a lot to hold) which also has very large unrealized capital gains (costs a lot to sell).
While this problem is hard to eyeball, it is relatively easy to math out.
We use a set of complex formulas to help decide when highly appreciated investments should be sold and the proceeds put in a different investment philosophy.
As far as I know these formulas are unique to our firm, as we formulated the premises and worked out the algebra by hand. They are significant enough, I would hope that one day they are part of financial planning textbooks.
We use these formulas to determine how much additional future return you must earn in a new investment in order to overcome the cost of realizing gains at the current rate. In order to run the analysis, you must choose:
1. How long will you hold the investment before we should decide whether sell or hold wins?
2. At the end of the analysis, will your held security get a step-up in cost basis from death?
3. What should we assume is the annual rate of return of the held security?
4. If you sell, what is the tax rate your realized capital gains will experience? (Take into consideration state taxes, federal qualified taxes, and Net Investment Income taxation.)
An Example with Expense Ratio
For a long time, iShares was the only fund company that had reasonably priced country-specific exchange-traded funds (ETFs). As they advertise on their website , “With 66 funds across 42 countries, iShares offers the world’s broadest range of country ETFs.” The expense ratios on their country-specific ETFs sit right around 0.51%.
Recently, Franklin Templeton created 23 country-specific ETFs with expense ratios around 0.09%. In addition to being cheaper, the Franklin Templeton funds typically have more holdings, making them more diversified in their sector.
Take iShares MSCI Switzerland ETF (EWL) and Franklin FTSE Switzerland ETF (FLSW).
EWL has 47 holdings and an expense ratio of 0.51%. FLSW has 54 holdings and an expense ratio of 0.09%. They are near identical funds except that the Franklin fund is 42 basis points cheaper.
When compared in this manner, one would imagine that the Franklin fund would be the overwhelming winner. Surely, the Franklin funds would have more investors than iShares. Surely, iShares would have pressure to make their fund better or lower the cost. Alas, capital gains taxation has made the opposite true.
EWL was opened in 1996 and has \$1.57 Billion invested as of September 2021. FLSW was opened in 2018 and has \$38.38 Million invested as of September 2021.
Franklin funds isn’t competing with iShares for investors; they are competing with the capital gains tax.
This is how the capital gains tax is economically destructive. It traps capital in inefficient investments and forces taxpayers to make odd decisions to engage in smart tax planning.
Some investors should not realize the gains in their iShares fund even though they could have 0.42% higher returns each year by making the switch. Others can benefit from biting the tax bullet to make the swap.
Here is how we can use hurdle rates to decide.
Let’s imagine a forty-year-old investor who resides in Virginia. Imagine that this investor owns two trade lots of EWL. The trade lots have capital gains of 77.67% and 32.03% respectively. If the investor sells these trade lots, they are going to be subject to a 15% marginal federal capital gains rate and a 5.75% marginal Virginia state tax rate. If they continue to hold EWL, they suspect they might be able to get an annual return of 8% over the next 10 years.
Using our formula, this puts the hurdle return for the two trade lots at 0.56% and 0.30%. The hurdle return is the amount of additional return we would need to suspect that our new purchase would be able to earn over our sale in order to breakeven.
In this case, EWL and FLSW are tracking the same country as a benchmark. As a result, we would suspect EWL and FLSW to have approximately the same return with one exception. We would suspect FLSW to have an advantage of 0.42% annually thanks to its lower expense ratio.
When we compare this advantage to our hurdles, we find that the trade lot with a 77.67% unrealized gain should stay unrealized. It’s hurdle is 0.56% and we do not have reason to suspect that our new purchase will beat our current holding by that much. If we want to sell this holding, we would need to do so through other means such as pairing its gains with capital losses or donating it to charity.
However, for the trade lot with a 32.03% gain, the hurdle is 0.30%. As we suspect an additional annual return of 0.42%, we benefit from selling this trade lot, realizing the capital gains on our taxes, and purchasing the new investment.
The Same Example with Step-Up in Basis
Using the same example as the first one, this time let’s imagine that the investor is 85 years old. At 85 years old, it is easy to imagine holding a security until you receive a step-up in cost basis at death.
However, if in the scenario where you hold your security you never have to pay tax on your gains, then the hurdle rates become harder to meet.
For the 85-year-old investor, holding the securities for 10-years before getting a step-up in basis, makes the hurdle returns jump to 1.03% for the lot with a 77.67% gain and 0.56% for the lot with the 32.03% gain.
Now, both positions should be held or an alternate strategy employed.
You can see that the Hurdle Report doesn’t directly tell you what you should sell and reinvest. Instead, it allows an experienced advisor to evaluate holdings where a lower cost alternative is known. Funds in some investment categories can be purchased with very low expense ratios. Other categories have funds with higher expense ratios. These categories may or may not be justified with higher expected returns.
Other Scenarios
In this article, our example of realizing the capital gains was justified by a reduced expense ratio alone. However, there are other criteria that can be used to justify realizing gains. Other examples include the rebalancing bonus; expected extra return from factors such as Freedom Investing, specific sectors, or investment types.
Also, in these examples we are running the hurdle analysis alone. In the case of a real investor, this type of hurdle analysis is best completed as part of a larger tax planning strategy. How much gain to realize each year is an important part of effectively navigating IRMAA Medicare surcharges and creating a systematic Roth conversion plan.
Lastly, these hurdle rates are unnecessary when selecting which securities to sell in order to fuel your cash withdrawal needs. That decision should be made more based on the projected taxation owed for the specific dollar amount of unrealized gains you would be realizing.
In a tax planning strategy, you may want to intentionally realize a little capital gains each year or in years when you find yourself in the 0% federal capital gains tax bracket.
These complex formulas help decide when highly appreciated investments should be sold and the proceeds put in a different investment philosophy. If you are interested in becoming a client, we would be happy to meet with you. Our prospective client meetings are free and it is easy to get started as a client.
Photo by Kaizen Nguyễn on Unsplash
David John Marotta
President, CFP®, AIF®, AAMS®
David John Marotta is the Founder and President of Marotta Wealth Management. He played for the State Department chess team at age 11, graduated from Stanford, taught Computer and Information Science, and still loves math and strategy games. In addition to his financial writing, David is a co-author of The Haunting of Bob Cratchit. | 1,778 | 8,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-06 | longest | en | 0.96862 |
http://mathhelpforum.com/algebra/26184-reciprocal-complex-number.html | 1,521,347,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645513.14/warc/CC-MAIN-20180318032649-20180318052649-00736.warc.gz | 196,116,178 | 10,732 | # Thread: Reciprocal of a complex number
1. ## Reciprocal of a complex number
I am asked to solve this
$\displaystyle \frac{10}{3-4i}$
I looked over the book to see how this is done...
according to the book
$\displaystyle \frac{1}{3+4i}$ is to be written in stnrd form a+bi, so then one must multiply both the denominator and numerator by the opposite of the denominator in this case 3-4i.
$\displaystyle \frac{1}{3+4i}.\frac{3-4i}{3-4i}$=$\displaystyle \frac{3-4i}{9+16}$
Ok here is where I don't get it, what happened(s) to the $\displaystyle i$ and the signs in the denominator of $\displaystyle \frac{3-4i}{9+16}$
is the multiplication process of the denominator the same as using foil?! or am I just full of complex numbers!?
$\displaystyle \frac{3-4i}{9+16}$ = $\displaystyle \frac{3}{25}-\frac{4}{25}$$\displaystyle i Where did the 25 come from?! Thanks in advance to all!! 2. Originally Posted by Morzilla I am asked to solve this \displaystyle \frac{10}{3-4i} I looked over the book to see how this is done... according to the book \displaystyle \frac{1}{3+4i} is to be written in stnrd form a+bi, so then one must multiply both the denominator and numerator by the opposite of the denominator in this case 3-4i. \displaystyle \frac{1}{3+4i}.\frac{3-4i}{3-4i}=\displaystyle \frac{3-4i}{9+16} Ok here is where I don't get it, what happened(s) to the \displaystyle i and the signs in the denominator of \displaystyle \frac{3-4i}{9+16} is the multiplication process of the denominator the same as using foil?! or am I just full of complex numbers!? \displaystyle \frac{3-4i}{9+16} = \displaystyle \frac{3}{25}-\frac{4}{25}$$\displaystyle i$
Where did the 25 come from?!
Yes, you FOIL out the denominator as if "i" were a variable.
$\displaystyle (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 - 16i^2 = 9 + 16$
(Or you could just note that this is a $\displaystyle (a + b)(a - b) = a^2 - b^2$ form.)
-Dan
3. Originally Posted by topsquark
Yes, you FOIL out the denominator as if "i" were a variable.
$\displaystyle (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 - 16i^2 = 9 + 16$
(Or you could just note that this is a $\displaystyle (a + b)(a - b) = a^2 - b^2$ form.)
-Dan
and what about the 25 in the denominator, where did come from?!
4. Originally Posted by Morzilla
and what about the 25 in the denominator, where did come from?!
$\displaystyle (3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$
5. Originally Posted by colby2152
$\displaystyle (3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$
yes, a math tutor showed me this in the math lab......boy I felt like such a Duh!
Much Thanks!! | 889 | 2,660 | {"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} | 4.3125 | 4 | CC-MAIN-2018-13 | latest | en | 0.869239 |
https://www.udemy.com/tutorial/dax-power-pivot-time-intelligence-10-easy-steps/adding-a-fiscal-year-end-date-for-running-totals/ | 1,591,112,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00442.warc.gz | 926,791,217 | 47,260 | # Adding a FISCAL Year End Date for Running Totals
A free video tutorial from Gilly Dow
IT Training Consultant, founder gdteach.co.uk
4.6 instructor rating • 7 courses • 4,415 students
## Lecture description
If you require Fiscal year end dates this lecture shows how to utilise the optional filter parameters to get exactly what you need.
### Learn more from the full course
DAX Power Pivot Time Intelligence - 10 Easy Steps
Support the Needs of Business Intelligence Using Custom Calendars and Time Period Functions in DAX.
05:48:34 of on-demand video • Updated October 2019
• Support the needs of Business Intelligence using Time Period Functions in DAX
• Create running-totals with DATESYTD(), DATESQTD() & DATESMTD()
• Learn how to use PREVIOUSYEAR() and related DAX functions including PREVIOUSQUARTER(), PREVIOUSMONTH() & PREVIOUSDAY()
• Customise DAX ready Time Intelligence functions with DATEADD()
• Learn how to create base measures including FIRSTDATE(), LASTDATE(), ENDOFMONTH() & STARTOFYEAR()
• Learn how to use CLOSINGBALANCE() family of DAX functions
• Understand the benefits of FILTER() to super-charge measures in place of CALCULATE()'s "simple" filters
• Create a 4-4-5 Custom Calendar in Excel and relate to the existing Data Model
• Learn how to recreate DAX in-built Time Intelligence functions for custom calendars
• Use DAX Patterns to master navigating the custom calendar
• Learn how to over-ride the incoming filter context with the ALL() function
• Suppress totals and subtotals with HASONEVALUE()
• Create % growth calculations and generate reports to impress
English [Auto] A cable just pick up where we left off. Guys I'm currently on page 14 of these guys and we're going to have a look at the fiscal year. Today it is an optional argument that enables you to customize the can and if you Year-End is done on thirty first of December so we're going to insert into it table into so see 26. So let's zoom over to see 26. And then from the insert type just insert a pivot table and. OK. So it's come into month in the very drop down and then the week number underneath month and it's just total sales. So do a search for total sales. And there it is just there. We're going to add a Canada slice to a calendar year slice so I should say this happens because we've used this search for it to find omega here. Then we can only see the results of our search. So we just have to cancel out of it in order to get all of our tables back from a white year and add a slicer and we're just positioned not just on you it's roughly about I think we're going to add a new measure into our pivot table. So in the sales table and the measure name is total sales for school. And it's going to be the third first of March or in upper case. So again we need our trusty friend our calculate and again the total sales we enter comma and this is our Filcher expression using our IT TAKES here today. All good so far. And again it's from our calendar table it's the date field and this is where we put in the optional parameter. So it's a comma and then it's surrounded by quote marks and it's going to be in the thirty first of the third. And just quote marks. Now we don't need to here. I mean if you add the the function will accept it but it's just ignored and it makes an awful lot of sense if you think about it especially when our year span over several several years if we put a day in there then it would only be subject to that particular year and that's certainly what we don't want so we'll close up with a couple of closing parenthesis. Again it's can't see the zero decimal places. And again we can check up on me now just to make absolutely certain. OK. But it looks good. And again I'm still having this problem and I'll just put in the search for 30 first and it finds it so perfect. And then we'll just drop that undertows with Souths and it is our fiscal year. And now when I'm inserting a fiscal year and normally takes me a while just to make sure that everything is as it should be. First of all I've got to sort out this filter of business so I should say so I equate that to four. And then if we go to 2002 and apply some conditional formatting so I'm going to select some 20 A's and from our hometown we just click on conditional formatting. Select the data box or Archie can choose any one you like us going to choose the blue. And the reason I do this this way is that when the pivot table is sliced the conditional formatting will apply to all the cells. If I highlight a block of cells and apply conditional formatting to the block then it's subject to that range that block that I've highlighted. So if I choose to do it this way. Choose the third option. It says all cells showing told to us with this thirty first of March this week number so does it matter how tall the pivot table is we will also always apply to just that block of cells so choose that third option and conditional formatting makes it an awful lot easier. So is the thirty first of March and this is our greatest value which is the last week in March and the first week starts in April. And we can see our use getting progressively larger and larger so conditional formatting is perfect for this. Now we need to put some formatting on our pivot table. So let's do our text wrap and center both horizontally and vertically will increase the height of the depth of our. See if we can just squish that and bit more. No probably not looking. OK. And I'll show you outline for us here. So we've got a couple of levels in almost drop zone. Now if we go to design and invite say to go in outline form it gives us an extra column because we've got levels underneath of those. So if that suits you that's great. If you prefer not to have it that way then I just add a control set to undo that. And you just have to do this manually. So you could put month and week number if you prefer or just leave it month it's up to you. So outline form is great but it will give you that extra column if you've got levels in your drop zone you both drop so so looking great guys and let's just have another look at that measure. Now I can go into measure's manage Mengesha and get all measure that way or another easy way of doing it looks slightly different but you'll be able to see if I put in 30 first of March here I'm just searching for all measure. You can just right click and it measure OK. So it looks slightly different. There's no coloring on the tax functions or anything. OK so it's calculate total sales dates year to date calendar dates. And that's what we did previously we just stopped it. We didn't add in the optional parameter at the end but here in the fiscal year we just surround our dates with out the year the end in quotes and that's how you apply fiscal energy to your dates to date. Perfect. Now the last thing I want to do here you can see that I've got some subtotals going on and I really don't like subtotals being at the top. So again from the Design Menu subtotals you can choose not to show them at all or you can put them at the bottom and turn them off completely. That's looking great. Okay guys. So that brings us to the end of step one coming up in the practice exercises. You'll have an opportunity to look at the other two members of this family when we're generating money totals. So have a go at month to date in court today. And then in the next video be solutions and we'll go through it together there. I urge you to have a go on your own first apply all the knowledge that you've learned so far and are seeing even the solutions based on guys. | 1,706 | 7,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.791896 |
https://www.coursehero.com/file/p3tnh79/Important-It-is-important-to-realise-that-the-water-cycle-is-all-about-energy/ | 1,542,406,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743216.58/warc/CC-MAIN-20181116214920-20181117000920-00161.warc.gz | 841,181,629 | 73,537 | # Important it is important to realise that the water
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Important: It is important to realise that the water cycle is all about energy exchanges . The sun is the original energy source . Energy from the sun heats the water and causes evaporation. This energy is stored in water vapour as latent heat . When the water vapour condenses again, the latent heat is released, and helps to drive circulation in the atmosphere. The liquid water falls to earth, and will evaporate again at a later stage. The atmospheric circulation patterns that occur because of these exchanges of heat are very important in influencing climate patterns. Activity :: Experiment : The Water Cycle Materials: Tile or piece of plastic (e.g. lid of ice-cream container) to make a hill slope; glass fish tank with a lid; beaker with ice cubes; lamp; water Set up a model of the water cycle as follows: lamp ice cubes water slope glass tank 1. Lean the plastic against one side so that it creates a ’hill slope’ as shown in the diagram. 2. Pour water into the bottom of the tank until about a quarter of the hill slope is covered. 3. Close the fish tank lid. 4. Place the beaker with ice on the lid directly above the hill slope. 5. Turn the lamp on and position it so that it shines over the water. 6. Leave the model like this for 20-30 minutes and then observe what happens. Make sure that you don’t touch the lamp as it will be very hot! Observation questions: 1. Which parts of the water cycle can you see taking place in the model? 2. Which parts of the water cycle are not represented in the model? 358
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CHAPTER 18. THE WATER CYCLE - GRADE 10 18.4 3. Can you think of how those parts that are not shown could be represented? 4. What is the energy source in the model? What would the energy source be in reality? 5. What do you think the function of the ice is in the beaker? 18.4 The microscopic structure of water In many ways, water behaves very differently from other liquids. These properties are directly related to the microscopic structure of water, and more specifically to the shape of the molecule and its polar nature , and to the bonds that hold water molecules together. 18.4.1 The polar nature of water Every water molecule is made up of one oxygen atom that is bonded to two hydrogen atoms. When atoms bond, the nucleus of each atom has an attractive force on the electrons of the other atoms. This ’pull’ is stronger in some atoms than in others and is called the electronegativity of the atom. In a water molecule, the oxygen atom has a higher electronegativty than the hydrogen atoms and therefore attracts the electrons more strongly. The result is that the oxygen atom has a slightly negative charge and the two hydrogen atoms each have a slightly positive charge. The water molecule is said to be polar because the electrical charge is not evenly distributed in the molecule. One part of the molecule has a different charge to other parts. You will learn more about this in chapter 4.
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=RegularChains/ChainTools/IsAlgebraic | 1,498,702,266,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00413.warc.gz | 566,486,950 | 22,939 | RegularChains[ChainTools] - Maple Programming Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : ChainTools Subpackage : RegularChains/ChainTools/IsAlgebraic
RegularChains[ChainTools]
IsAlgebraic
test if a variable is algebraic with respect to a regular chain
Calling Sequence IsAlgebraic(v, rc, R)
Parameters
v - variable of of R rc - regular chain of R R - polynomial ring
Description
• The command IsAlgebraic(v, rc, R) returns true if v is algebraic with respect to rc, false otherwise.
• This command is part of the RegularChains[ChainTools] package, so it can be used in the form IsAlgebraic(..) only after executing the command with(RegularChains[ChainTools]). However, it can always be accessed through the long form of the command by using RegularChains[ChainTools][IsAlgebraic](..).
Examples
> $\mathrm{with}\left(\mathrm{RegularChains}\right):$
> $R≔\mathrm{PolynomialRing}\left(\left[x,y,z,t\right]\right)$
${R}{≔}{\mathrm{polynomial_ring}}$ (1)
> $\mathrm{with}\left(\mathrm{ChainTools}\right):$
> $T≔\mathrm{Empty}\left(R\right):$
> $T≔\mathrm{Chain}\left(\left[\left(z+1\right)\left(z+2\right),{y}^{2}+z,\left(x-z\right)\left(x-y\right)\right],T,R\right)$
${T}{≔}{\mathrm{regular_chain}}$ (2)
> $\mathrm{Equations}\left(T,R\right)$
$\left[{{x}}^{{2}}{+}\left({-}{y}{-}{z}\right){}{x}{+}{z}{}{y}{,}{{y}}^{{2}}{+}{z}{,}{{z}}^{{2}}{+}{3}{}{z}{+}{2}\right]$ (3)
> $\mathrm{IsAlgebraic}\left(x,T,R\right)$
${\mathrm{true}}$ (4)
> $\mathrm{IsAlgebraic}\left(t,T,R\right)$
${\mathrm{false}}$ (5) | 531 | 1,586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-26 | latest | en | 0.456018 |
https://www.hpmuseum.org/forum/archive/index.php?thread-11792.html | 1,638,485,744,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00022.warc.gz | 819,187,082 | 2,624 | HP Forums
Full Version: (11C) Loan Repayments
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Program to calculate for "Loan Repayments"
n = number of payments
i = Interest Rate in Percent
PMT = Payment
PV = Principal
Formula Used:
[attachment=6594]
Procedure:
Interest Rate [A] Principal [R/S] Payment [R/S] --> Answer
Example:
This Example is from HP-38C Owner's Handbook.
Bill Buckskin wished to invest in a $22,000 log carbin to keep the rain off. A local merchant has offered to loan Bill the$22,000 at 10.5% interest.
Making $200 monthly payments, how long will it take Bill to repay his mortgage? 10.5 [A] 22000 [R/S] 200 [R/S] --> 376.89 Answer: 376.89 // Months 12 ÷ Answer 31.41 // Years Program: Code: LBL A EEX 2 ÷ 12 ÷ STO 0 R/S x R/S ÷ 1 X<>Y - 1/x LN RCL 0 1 + LN ÷ RTN Gamo (11-13-2018 01:37 PM)Gamo Wrote: [ -> ]This Example is from HP-38C Owner's Handbook. Bill Buckskin wished to invest in a$22,000 log carbin to keep the rain off.
A local merchant has offered to loan Bill the $22,000 at 10.5% interest. Making$200 monthly payments, how long will it take Bill to repay his mortgage?
10.5 [A] 22000 [R/S] 200 [R/S] --> 376.89
You might want to rearrange the order, to get more information from same calculation.
22000 [Enter] 200 / --> 110.00, number of payments if no interest
10.5 [R/S] --> 376.89, effect of 10.5% interest. 277 payments (23 years) just for interest !
Thanks Albert Chan
Very good touch up.
Gamo
(11-13-2018 01:37 PM)Gamo Wrote: [ -> ]Formula Used:
image: "n = log 1 + i · [1 / (1 – PV · i / PMT)]"
No, that's not the formula. The "1+i" part must be a subscript after the log, indicating that the base 1+i logarithm of the bracketed term is calculated.
But the program uses a correct formula. I'd write it this way:
\begin{align} n~&=~-log_{~1+i}~(1-\frac{i \cdot PV}{PMT}) \\ \\ &=~ -\frac{ln(1-\frac{i \cdot PV}{PMT})}{ln(1+~i)} \end{align}
This is what the program correctly implements. Assuming that the user enters the monthly payment and the annual interest rate.
Dieter
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | 693 | 2,198 | {"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": 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} | 3.625 | 4 | CC-MAIN-2021-49 | latest | en | 0.852291 |
https://www.physicsforums.com/threads/nontrivial-solutions-for-ax-0.4286/ | 1,571,356,737,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00535.warc.gz | 1,030,083,868 | 21,889 | # Nontrivial Solutions for AX=0
#### StephenPrivitera
Theorem 2.3.3
A homogeneous system of m linear algebraic equations in m unknowns has no nontrivial solutions iff the reduced coefficient matrix has no rows consisting entirely of zeros.
So in other words, if no row of the matrix consists entirely of zeros, then the only solution is X=0.
This is what my book teaches me before determinants. When it introduces determinants, it asks as an exerice to rewrite the theorem using the determinant.
The book rewrites it,
...iff the determinant does not equal zero.
p->q
p->r
q->p
r->p
Thus
r->q and q->r
Which means that if there are no rows consisting entirely of zeros, then the determinant is not zero. This is clearly false, since a determinant can be zero if two rows are proportional. I feel like my book has lied to me again.
I at first trusted my book over my own belief, and when it told me to determine for what values of a (some coefficient) does this system has nontrivial solutions, I set up an upper-triangular determinant, multiplied the elements along the principal diagonal and set it equal to zero. This process yields solutions which are trivial. Admittedly, it would have been a lot easier to set the last element in the last row equal to zero, but I thought this would omit some possibilities.
For instance,
x+y=ax
-x+y=ay
Using the above procedure, I get a=1 as one of the possibilities.
Am I completely off with this accusation?
Last edited:
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#### arcnets
Originally posted by StephenPrivitera
Which means that if there are no rows consisting entirely of zeros, then the determinant is not zero.
The theorem states:
reduced coefficient matrix has no rows consisting entirely of zeros
=>
no nontrivial solutions
What you must prove, is:
determinant does not equal zero
=>
no nontrivial solutions
So, what's left to prove is:
determinant does not equal zero
=>
reduced coefficient matrix has no rows consisting entirely of zeros
This is logically the same as
reduced coefficient matrix has at least one row consisting entirely of zeros
=>
determinant equals zero.
This should be easy if you use Cramer's rule.
#### StephenPrivitera
So in other words, the fact that a system has a nontrivial solution does not imply that the determinant of the coefficient matrix is zero and the fact that the determinant is zero does not imply that the system has a nontrivial solution. But if the determinant is not zero then its only solution is trivial.
My problem was that when a=1, the system only has the solution x=y=0 and the determinant is also zero. I saw this as a contradiction to the theorem, but if I am understanding you correctly, my logic was wrong. And so this is not a contradiction to the theorem.
Looking back on the problem, the determinant isn't zero. It's one. That's very perturbing. When I was setting up the determinant I multiplied a row by 1-a, but if a=1, then that's equivalent to multiplying by zero.
BTW, I don't know Cramer's rule.
#### HallsofIvy
Homework Helper
"For instance,
x+y=ax
-x+y=ay
Using the above procedure, I get a=1 as one of the possibilities."
I assume this is the "a" you are referring to when you say:
"My problem was that when a=1, the system only has the solution x=y=0 and the determinant is also zero."
If a= 1, then x+ y= x and -x+ y= y. The matrix form of this is
[ 1 1][x] = [x] but this is NOT of the form Ax= 0.
[-1 1][y] [y]
You need to rewrite it as [1 1][x]= [1 0][x]
[-1 1][y]= [0 1][y]
which is the same as [0 1][x] = [0] (you would get the same thing
[-1 0][y] [0]
if you rewrite the equations as y= 0 and -x= 0).
Yes, that has the single solution x= y= 0 but the determinant is
1, not 0.
The general problem x+y= ax, -x+ y= ay can be rewritten as
(1-a)x+ y= 0, -x+ (1-a)y= 0 which, in matrix form, is
[1-a 1][x] = [0]
[-1 1-a][y] [0].
The determinant is (1-a)2+ 1. Setting that equal to 0,
(1-a)2+ 1= 0 is the same as (1-a)2= -1 and so has roots a= 1+ i and 1- i, not a= 1.
It IS true that a system of the form Ax= 0 has a non-trivial solution (in fact, has an infinite number of solutions) if and only if the determinant of the coefficient matrix is 0. And THAT is true if and only if the row reduced form of the matrix has at least one row of 0s.
The point is this: if you row-reduce a matrix getting it to either diagonal or triangular form, WITHOUT dividing or multiplying a row by a number (only using "add a multiple of one row to another"), then the determinant is the product of the numbers on the diagonal.
(If you swap two rows, the determinant is multiplied by -1. If you multiply a row by a number, the determinant is multiplied by that number. If you add any multiple of one row to another, the determinant is not changed.)
If you can row-reduce the matrix to diagonal or triangular form so that it has a 0 on the diagonal, then it's determinant is 0. Of course, that happens if one of the rows is all 0s.
#### StephenPrivitera
Ok,ok, just one more thing.
Is this true:
The determinant of the coefficient matrix is zero if and only if the reduced coefficient matrix has at least one row of zeros.
(q<-->r)
p= AX=0 has nontrivial solutions
q= the determinant of the coefficient matrix is zero
r= the reduced coefficient matrix has at least one rows of zeros
We know that,
r<-->p
r->q
q<-->p
Therefore, since q->p and p->r, q->r
So it seems right to me that q<-->r, but it's been years since I've studied logic, so who knows.
I just want to know because if q->r is true, then if there is a zero in the prin. diagonal in the reduced matrix, then there must also be a row of zeros. Correct?
QUOTE:
if you row-reduce a matrix getting it to either diagonal or triangular form, WITHOUT dividing or multiplying a row by a number (only using "add a multiple of one row to another"), then the determinant is the product of the numbers on the diagonal.
____
What's wrong with multiplying a row by a number as long as I also multiple by the reciprocal? Or do you mean when there are unknowns, I should avoid this practice so I don't accidentally multiply by zero? Kind of like you're not supposed to divide an equation by x?
#### arcnets
Originally posted by StephenPrivitera
The determinant of the coefficient matrix is zero if and only if the reduced coefficient matrix has at least one row of zeros.
I think this is true. HallsofIvy said so, too.
What's wrong with multiplying a row by a number...
As long as you're only interested in the determinant being zero or not, you may multiply (of course not by zero) and swap as you like. The determinant will never change from zero to non-zero or vice versa.
#### HallsofIvy
Homework Helper
Well, if I said it, it MUST be true!
What's wrong with multiplying a row by a number as long as I also multiple by the reciprocal?
Not sure I understand this. If you multiply a row by a number and the multiply the SAME row by the reciprocal- you haven't changed it.
Here's and example:
Find the determinant of [2 2]
[4 3] by row-reduction.
Some people (especially if they've just learned row-reduction to find inverse matrices) prefer to reduce to a 1 on the diagonal (that makes it easier to decide what to multiply by to get 0s on the other rows).
Here, you can divide the first row by 2, then subtract 4 time that (new) first row from the second row to get
[1 1]
[0 -1]
Now, we can divide the second row by -1 then subtract the new second row from the first row to get the identity matrix:
[1 0]
[0 1]
Obviously, that by itself doesn't tell us the determinant of the original matrix. However if we kept track of the numbers we divided by we can say that we, essentially, "factored out" a 2 and a -1 so that the determinant is 2(-1)= -2.
The way I would prefer to do this problem is to simply subtract twice the first row from the second row to get
[2 2]
[0 -1]- a triangular matrix. BECAUSE I have not multiplied or divided a row by a number (or swapped two rows), I know that the determinant of this matrix is the same as the determinant of the original matrix- and the determinant of a triangular matrix is just the product of the numbers on the main diagonal: -2.
#### StephenPrivitera
The way I would prefer to do this problem is to simply subtract twice the first row from the second row to get
...
Yes, you're right. This is a much better approach.
I had been multiplying rows by number so that I could add them together and make some elements disappear, but I might as well do that in the same step.
As for the multiplying by the reciprocal part, what I meant was this:
If you multiply a row of the determinant by a factor of n, then the value of the determinant has changed by a factor of n. So why can't I multiply a row by n and the determinant by 1/n, so that the value isn't changed? From what you said, I can, it's just more unnecessary work.
I'm sorry for not believing you... But I really wanted to make that other part really clear. For instance, it bothers me that this determinant is zero, but doesn't have a row of zeros.
1 2 3 4 5
0 1 2 3 4
0 0 0 1 2
0 0 0 1 2
0 0 0 0 1
My only guess is that this doesn't fit the definition of "reduced."
__________
(especially if they've just learned row-reduction to find inverse matrices)
That's me - Topic for next week!
Last edited:
#### HallsofIvy
Homework Helper
Here's an even easier example:
[ 1 1]
[ 1 1] has determinant 0 but does not have a zero row.
You are right. The difference is that it is not "row-reduced".
We can "row reduce" it to
[ 1 1]
[ 0 0] which does have a row of zeros.
The example you give:
[1 2 3 4 5]
[0 1 2 3 4]
[0 0 0 1 2]
[0 0 0 1 2]
[0 0 0 0 1] is not row-reduced. It's determinant is 0 because it has two rows exactly the same- and because it has two rows exactly the same, we can row reduce it to
[1 2 3 4 5]
[0 1 2 3 4]
[0 0 0 0 1]
[0 0 0 0 0]
[0 0 0 0 0]
A matrix has determinant 0 if and only if it can be row-reduced to a triangular form that has at least one row consisting entirely of 0s.
That follows from:
1) Dividing an entire row by a number divides the determinant by that number.
2) Swapping two rows multiplies the determinant by 0.
3) Adding a multiple of one row to another does not change the determinant.
together with: The determinant of a triangular matrix is the product of the numbers on the main diagonal.
#### HallsofIvy
Homework Helper
Blasted typo:
The fifth line from the bottom should be:
2) Swapping two rows of a matrix multiplies the determinant by -1
(not 0).
#### Hurkyl
Staff Emeritus
Gold Member
I think that reduced matrix should be
[1 2 3 4 5]
[0 1 2 3 4]
[0 0 0 1 2]
[0 0 0 0 1]
[0 0 0 0 0]
#### HallsofIvy
Homework Helper
Yes, I just noticed that. Thank you.
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• Solo and co-op problem solving | 2,947 | 11,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | longest | en | 0.920277 |
https://cumbremundialafro.org/homework-296 | 1,674,832,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00557.warc.gz | 214,432,871 | 4,506 | How Do I Calculate the Percentage of a Salary Increase
How do I calculate salary increases? - 1. Determine the annual salary before the increase. - 2. Calculate the salary increase based on the percentage or the
Explain math equation
The math equation is a way of representing a problem or situation using numbers and symbols.
Clear up math problems
If you're struggling to clear up math problems, try these tips.
Explain math problems
Math can be a difficult subject for many people, but it doesn't have to be! By taking the time to explain the problem and break it down into smaller pieces, anyone can learn to solve math problems.
## How to calculate a salary increase or decrease
Learn how to CALCULATE YOUR SALARY INCREASE with a simple formula. Also download an Excel template that will do the calculation for you.
## How to Calculate Your Worth
Divide the salary difference by your old salary. To convert the amount of the raise into a percentage, you must first calculate it as a decimal. To | 210 | 1,010 | {"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-2023-06 | latest | en | 0.900759 |
https://leetcode.com/articles/smallest-range/ | 1,576,454,140,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00501.warc.gz | 434,792,661 | 18,852 | ## Solution
#### Approach #1 Brute Force [Time Limit Exceeded]
The naive approach is to consider every pair of elements, and from amongst the given lists and consider the range formed by these elements. For every range currently considered, we can traverse over all the lists to find if atleast one element from these lists can be included in the current range. If so, we store the end-points of the current range and compare it with the previous minimum range found, if any, satisfying the required criteria, to find the smaller range from among them.
Once all the element pairs have been considered as the ranges, we can obtain the required minimum range.
Complexity Analysis
• Time complexity : . Considering every possible range(element pair) requires time. For each range considered, we need to traverse over all the elements of the given lists in the worst case requiring another time. Here, refers to the total number of elements in the given lists.
• Space complexity : . Constant extra space is used.
#### Approach #2 Better Brute Force [Time Limit Exceeded]
Algorithm
In the last approach, we consider every possible range and then traverse over every list to check if atleast one of the elements from these lists lies in the required range. Instead of doing this traversal for every range, we can make use of Binary Search to find the index of the element just larger than(or equal to) the lower limit of the range currently considered.
If all the elements in the current list are lesser than this lower limit, we'll get the index as for the list being currently checked. In this case, none of the elements of the current list lies in the current range.
On the other hand, if all the elements in this list are larger than this lower limit, we'll get the index of the first element(minimum) in the current list. If this element happens to be larger than the upper limit of the range currently considered, then also, none of the elements of the current list lies within the current range.
Whenever a range is found which satisfies the required criteria, we can compare it with the minimum range found so far to determine the required minimum range.
Complexity Analysis
• Time complexity : . The time required to consider every possible range is . For every range currently considered, a Binary Search requiring time is required. Here, refers to the total number of elements in the given lists and refers to the average length of each list.
• Space complexity : . Constant extra space is used.
#### Approach #3 Using Pointers [Time Limit Exceeded]
Algorithm
We'll discuss about the implementation used in the current approach along with the idea behind it.
This approach makes use of an array of pointers, , whose length is equal to the number of given lists. In this array, refers to the element which needs to be considered next in the list. The meaning of this will become more clearer when we look at the process.
We start by initializing all the elements of to 0. Thus, currently, we are considering the first(minimum) element among all the lists. Now, we find out the index of the list containing the maximum() and minimum() elements from amongst the elements currently pointed by . The range formed by these maximum and minimum elements surely
contains atleast one element from each list.
But, now our objective is to minimize this range. To do so, there are two options: Either decrease the maximum value or increase the minimum value.
Now, the maximum value can't be reduced any further, since it already corresponds to the minimum value in one of the lists. Reducing it any further will lead to the exclusion of all the elements of this list(containing the last maximum value) from the new range.
Thus, the only option left in our hand is to try to increase the minimum value. To do so, we now need to consider the next element in the list containing the last minimum value. Thus, we increment the entry at the corresponding index in , to indicate that the next element in this list now needs to be considered.
Thus, at every step, we find the maximum and minimum values being pointed currently, update the values appropriately, and also find out the range formed by these maximum and minimum values to find out the smallest range satisfying the given criteria.
While doing this process, if any of the lists gets completely exhausted, it means that the minimum value being increased for minimizing the range being considered can't be increased any further, without causing the exclusion of all the elements in atleast one of the lists. Thus, we can stop the search process whenever any list gets completely exhausted.
We can also stop the process, when all the elements of the given lists have been exhausted.
Summarizing the statements above, the process becomes:
1. Initialize array(pointers) with all 0's.
2. Find the indices of the lists containing the minimum() and the maximum() elements amongst the elements pointed by the array.
3. If the range formed by the maximum and minimum elements found above is larger than the previous maximum range, update the boundary values used for the maximum range.
4. Increment the pointer .
5. Repeat steps 2 to 4 till any of the lists gets exhausted.
The animation below illustrates the process for a visual understanding of the process.
!?!../Documents/632_Smallest_Range.json:1000,563!?!
Complexity Analysis
• Time complexity : . In the worst case, we need to traverse over array(of length ) for all the elements of the given lists. Here, refers to the total number of elements in all the lists. refers to the total number of lists.
• Space complexity : . array of size is used.
#### Approach #4 Using Priority Queue [Accepted]:
Algorithm
In the last approach, at each step, we update the pointer corresponding to the current minimum element and traverse over the whole array to determine the new maximum and minimum values. We can do some optimization here, by making use of a simple observation.
Whenever we update a single entry of to consider the new maximum and minimum values(if we already know the last maximum and minimum values), all the elements to be considered for finding the maximum and minimum values remain the same except the new element being pointed by a single updated entry in . This new entry is certainly larger than the last minimum value(since that was the reasoning behind the updation).
Thus, we can't be sure whether this is the new minimum element or not. But, since it is larger than the last value being considered, it could be a potential competitor for the new maximum value. Thus, we can directly compare it with the last maximum value to determine the current maximum value.
Now, we're left with finding the minimum value iteratively at every step. To avoid this iterative process, a better idea is to make use of a Min-Heap, which stores the values being pointed currently by the array. Thus, the minimum value always lies at the top of this heap, and we need not do the iterative search process.
At every step, we remove the minimum element from this heap and find out the range formed by the current maximum and minimum values, and compare it with the minimum range found so far to determine the required minimum range. We also update the increment the index in corresponding to the list containing this minimum entry and add this element to the heap as well.
The rest of the process remains the same as the last approach.
Complexity Analysis
• Time complexity : . Heapification of elements requires time. This step could be done for all the elements of the given lists in the worst case. Here, refers to the total number of elements in all the lists. refers to the total number of lists.
• Space complexity : . array of size is used. A Min-Heap with elements is also used.
Analysis written by: @vinod23 | 1,584 | 7,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-51 | latest | en | 0.878145 |
https://elteoremadecuales.com/courcelles-theorem/ | 1,670,663,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00725.warc.gz | 269,563,821 | 15,762 | # Courcelle's theorem
Courcelle's theorem In the study of graph algorithms, Courcelle's theorem is the statement that every graph property definable in the monadic second-order logic of graphs can be decided in linear time on graphs of bounded treewidth.[1][2][3] The result was first proved by Bruno Courcelle in 1990[4] and independently rediscovered by Borie, Parker & Tovey (1992).[5] It is considered the archetype of algorithmic meta-theorems.[6][7] Contents 1 Formulations 1.1 Vertex sets 1.2 Edge sets 1.3 Labeled graphs 1.4 Modular congruences 1.5 Decision versus optimization 1.6 Space complexity 2 Proof strategy and complexity 3 Bojańczyk-Pilipczuk's theorem 4 Satisfiability and Seese's theorem 5 Applications 6 References Formulations Vertex sets In one variation of monadic second-order graph logic known as MSO1, the graph is described by a set of vertices and a binary adjacency relation {displaystyle operatorname {adj} (.,.)} , and the restriction to monadic logic means that the graph property in question may be defined in terms of sets of vertices of the given graph, but not in terms of sets of edges, or sets of tuples of vertices.
As an example, the property of a graph being colorable with three colors (represented by three sets of vertices {displaystyle R} , {displaystyle G} , and {displaystyle B} ) may be defined by the monadic second-order formula {displaystyle {begin{aligned}exists R exists G exists B {Bigl (}&forall v (vin Rvee vin Gvee vin B)wedge {}\&forall u forall v {bigl (}(uin Rwedge vin R)rightarrow lnot operatorname {adj} (u,v){bigr )}wedge {}\&forall u forall v {bigl (}(uin Gwedge vin G)rightarrow lnot operatorname {adj} (u,v){bigr )}wedge {}\&forall u forall v {bigl (}(uin Bwedge vin B)rightarrow lnot operatorname {adj} (u,v){bigr )}{Bigr )}end{aligned}}} with the naming convention that uppercase variables denote sets of vertices and lowercase variables denote individual vertices (so that an explicit declaration of which is which can be omitted from the formula). The first part of this formula ensures that the three color classes cover all the vertices of the graph, and the rest ensures that they each form an independent set. (It would also be possible to add clauses to the formula to ensure that the three color classes are disjoint, but this makes no difference to the result.) Thus, by Courcelle's theorem, 3-colorability of graphs of bounded treewidth may be tested in linear time.
For this variation of graph logic, Courcelle's theorem can be extended from treewidth to clique-width: for every fixed MSO1 property {displaystyle Pi } , and every fixed bound {displaystyle b} on the clique-width of a graph, there is a linear-time algorithm for testing whether a graph of clique-width at most {displaystyle b} has property {displaystyle Pi } .[8] The original formulation of this result required the input graph to be given together with a construction proving that it has bounded clique-width, but later approximation algorithms for clique-width removed this requirement.[9] Edge sets Courcelle's theorem may also be used with a stronger variation of monadic second-order logic known as MSO2. In this formulation, a graph is represented by a set V of vertices, a set E of edges, and an incidence relation between vertices and edges. This variation allows quantification over sets of vertices or edges, but not over more complex relations on tuples of vertices or edges.
For instance, the property of having a Hamiltonian cycle may be expressed in MSO2 by describing the cycle as a set of edges that includes exactly two edges incident to each vertex, such that every nonempty proper subset of vertices has an edge in the putative cycle with exactly one endpoint in the subset. However, Hamiltonicity cannot be expressed in MSO1.[10] Labeled graphs It is possible to apply the same results to graphs in which the vertices or edges have labels from a fixed finite set, either by augmenting the graph logic to incorporate predicates describing the labels, or by representing the labels by unquantified vertex set or edge set variables.[11] Modular congruences Another direction for extending Courcelle's theorem concerns logical formulas that include predicates for counting the size of the test. In this context, it is not possible to perform arbitrary arithmetic operations on set sizes, nor even to test whether two sets have the same size. However, MSO1 and MSO2 can be extended to logics called CMSO1 and CMSO2, that include for every two constants q and r a predicate {displaystyle operatorname {card} _{q,r}(S)} which tests whether the cardinality of set S is congruent to r modulo q. Courcelle's theorem can be extended to these logics.[4] Decision versus optimization As stated above, Courcelle's theorem applies primarily to decision problems: does a graph have a property or not. However, the same methods also allow the solution to optimization problems in which the vertices or edges of a graph have integer weights, and one seeks the minimum or maximum weight vertex set that satisfies a given property, expressed in second-order logic. These optimization problems can be solved in linear time on graphs of bounded clique-width.[8][11] Space complexity Rather than bounding the time complexity of an algorithm that recognizes an MSO property on bounded-treewidth graphs, it is also possible to analyze the space complexity of such an algorithm; that is, the amount of memory needed above and beyond the size of the input itself (which is assumed to be represented in a read-only way so that its space requirements cannot be put to other purposes). In particular, it is possible to recognize the graphs of bounded treewidth, and any MSO property on these graphs, by a deterministic Turing machine that uses only logarithmic space.[12] Proof strategy and complexity The typical approach to proving Courcelle's theorem involves the construction of a finite bottom-up tree automaton that acts on the tree decompositions of the given graph.[6] In more detail, two graphs G1 and G2, each with a specified subset T of vertices called terminals, may be defined to be equivalent with respect to an MSO formula F if, for all other graphs H whose intersection with G1 and G2 consists only of vertices in T, the two graphs G1 ∪ H and G2 ∪ H behave the same with respect to F: either they both model F or they both do not model F. This is an equivalence relation, and it can be shown by induction on the length of F that (when the sizes of T and F are both bounded) it has finitely many equivalence classes.[13] A tree decomposition of a given graph G consists of a tree and, for each tree node, a subset of the vertices of G called a bag. It must satisfy two properties: for each vertex v of G, the bags containing v must be associated with a contiguous subtree of the tree, and for each edge uv of G, there must be a bag containing both u and v. The vertices in a bag can be thought of as the terminals of a subgraph of G, represented by the subtree of the tree decomposition descending from that bag. When G has bounded treewidth, it has a tree decomposition in which all bags have bounded size, and such a decomposition can be found in fixed-parameter tractable time.[14] Moreover, it is possible to choose this tree decomposition so that it forms a binary tree, with only two child subtrees per bag. Therefore, it is possible to perform a bottom-up computation on this tree decomposition, computing an identifier for the equivalence class of the subtree rooted at each bag by combining the edges represented within the bag with the two identifiers for the equivalence classes of its two children.[15] The size of the automaton constructed in this way is not an elementary function of the size of the input MSO formula. This non-elementary complexity is necessary, in the sense that (unless P = NP) it is not possible to test MSO properties on trees in a time that is fixed-parameter tractable with an elementary dependence on the parameter.[16] Bojańczyk-Pilipczuk's theorem The proofs of Courcelle's theorem show a stronger result: not only can every (counting) monadic second-order property be recognized in linear time for graphs of bounded treewidth, but also it can be recognized by a finite-state tree automaton. Courcelle conjectured a converse to this: if a property of graphs of bounded treewidth is recognized by a tree automaton, then it can be defined in counting monadic second-order logic. In 1998 Lapoire (1998), claimed a resolution of the conjecture.[17] However, the proof is widely regarded as unsatisfactory.[18][19] Until 2016, only a few special cases were resolved: in particular, the conjecture has been proved for graphs of treewidth at most three,[20] for k-connected graphs of treewidth k, for graphs of constant treewidth and chordality, and for k-outerplanar graphs. The general version of the conjecture was finally proved by Mikołaj Bojańczyk and Michał Pilipczuk.[21] Moreover, for Halin graphs (a special case of treewidth three graphs) counting is not needed: for these graphs, every property that can be recognized by a tree automaton can also be defined in monadic second-order logic. The same is true more generally for certain classes of graphs in which a tree decomposition can itself be described in MSOL. However, it cannot be true for all graphs of bounded treewidth, because in general counting adds extra power over monadic second-order logic without counting. For instance, the graphs with an even number of vertices can be recognized using counting, but not without.[19] Satisfiability and Seese's theorem The satisfiability problem for a formula of monadic second-order logic is the problem of determining whether there exists at least one graph (possibly within a restricted family of graphs) for which the formula is true. For arbitrary graph families, and arbitrary formulas, this problem is undecidable. However, satisfiability of MSO2 formulas is decidable for the graphs of bounded treewidth, and satisfiability of MSO1 formulas is decidable for graphs of bounded clique-width. The proof involves building a tree automaton for the formula and then testing whether the automaton has an accepting path.
Si quieres conocer otros artículos parecidos a Courcelle's theorem puedes visitar la categoría Graph algorithms.
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Utilizamos cookies propias y de terceros para mejorar la experiencia de usuario Más información | 2,296 | 10,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-49 | latest | en | 0.827929 |
https://mathoverflow.net/questions/214937/gagliardo-nirenberg-inequality-for-the-laplacian/214946 | 1,632,267,568,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00149.warc.gz | 445,179,394 | 29,511 | # Gagliardo Nirenberg inequality for the laplacian
It is a classical result due to Gagliardo and Nirenberg that there exists a constant C such that it holds $$||\nabla \psi|| _{L ^\infty (\mathbb{R}^2)} ^2 \le ||D ^2 \psi|| _{L ^\infty (\mathbb{R}^2)} ||\psi|| _{L ^\infty (\mathbb{R}^2)},$$ for any $\psi \in C ^2 (\mathbb{R} ^2)$.
My question would be if anyone knows if it is possible to obtain the same result or a similar one with the laplacian instead of the second derivative, namely if it holds that $$||\nabla \psi|| _{L ^\infty (\mathbb{R}^2)} ^2 \le ||\Delta \psi|| _{L ^\infty (\mathbb{R}^2)} ||\psi|| _{L ^\infty (\mathbb{R}^2)}.$$
Any help or reference will be greatly appreciated!
## 1 Answer
By the classical regularity theory for the Poisson equation, you have $$\Vert \nabla \psi \Vert_{L^\infty (B_1)} \le C \bigl(\Vert \Delta \psi \Vert_{L^\infty (B_2)} + \Vert \psi \Vert_{L^\infty (B_2)} \bigr).$$ See for example Gilbarg and Trudinger, Elliptic partial differential equations of second order, 1983, theorem 3.9, where it is proved by the maximum principle. (An alternative would be to use a Green representation formula of the solution in a ball.) By scaling this gives $$R\Vert \nabla \psi \Vert_{L^\infty (B_R)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (B_{2 R})} + \Vert \psi \Vert_{L^\infty (B_{2 R})} \bigr).$$ By translating the estimate, you obtain thus for every $R > 0$, $$R\Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)} + \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} \bigr).$$ By taking $R = \sqrt{\Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} / \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}}$, we conclude that $$\Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le 2C \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2} \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2}.$$ | 701 | 1,893 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-39 | latest | en | 0.625356 |
https://pty.vanderbilt.edu/2024/06/summer-savy-session-3-day-4the-one-to-beat-3rd-4th/ | 1,721,522,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00236.warc.gz | 411,995,726 | 9,846 | # Summer SAVY, Session 3 Day 4, The One to Beat (3rd – 4th)
Posted by on Thursday, June 27, 2024 in blog, SAVY.
Hello SAVY parents! Students were able to create a foldable that would allow them to remember the various types of graphs we have been using to record data. Today, we added new vocabulary terms to our word wall: slope, positive slope, and negative slope.
The emphasis today was on analyzing change. We were able to study how change in one variable is related to change in another. Tables and coordinate graphs were used to assist students in the data analysis process as they organized information to aid in the search for number patterns.
In the morning lesson, students investigated linear relationships by assuming a constant rate of speed for a man walking across the ocean. They completed a table and a scatter plot (a graph of unconnected dots representing the ordered pairs from the table) showing the relationship of time to distance. The students noticed that the rate of change was constant, which meant that all points would lie on the same line. Next, they compared Mr. Bricka’s record to the record of Mr. Strel by plotting the data on the same graph. In the new scenario, a man swims for the same amount of time as the man walking across the ocean (60 days) but travels half the distance. Students made a table for this scenario and then graphed the distance traveled against the time for both scenarios and compared the two graphs, paying careful attention to the steepness or slope of the lines and what the difference in the steepness indicated. In both cases, because of the assumed constant rate of travel, students were able to explain why the points for each graph all lie on the same line when they were plotted demonstrating that they understood the term, linear relationships.
Tomorrow’s challenge will involve a record that the students will investigate dubbed, “The Chain Gang.” In this lesson, we will investigate linear relationships that result from building a chain at a constant rate. We will graph the data by looking at the steepness of a line (slope) to determine what would happen if the rate changed. We will discuss differences in the starting point (y-intercept), as well as the differences in steepness (slope).
Thank you so much for another successful day, and I look forward to our final session of fun and learning tomorrow!
Questions to Consider:
• What kind of graph would you use to represent change over time?
• What are three ways to write a ratio?
• How can you determine if a rate of change is constant or variable?
• Can you explain the difference between a positive slope and a negative slope?
With children first,
–D. Fuller | 572 | 2,711 | {"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} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.952271 |
https://discourse.mc-stan.org/t/generate-quantities-using-bernoulli-rng-gives-real-value-not-int/13520 | 1,653,080,225,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00763.warc.gz | 265,971,749 | 7,835 | # Generate quantities using bernoulli_rng gives real value not int
Hi,
I’m confused about what bernoulli_rng function does. Based on Stan manual (p.306), it takes a real value (probability) and returns an integer (0 or 1).
int bernoulli_rng(real theta) Generate a Bernoulli variate with chance of success theta; may only be used
ingenerated quantities block
So I assume if I use it as they do here:
y_pred[i, t] = bernoulli_rng(pGamble);
I should feed it with a probability and get 0 or 1 (nothing or succes). However, I get real numbers \in (0,1). Why is that? What am I missing?
How can I predict the behaviour based on those generated quantities? Ie., how do I get the prediction of the model (which are discrete, in this case accept/reject)?
Thank you.
I just tested it on my computer and I get integer values as expected. Could you post a minimal example where you get the real numbers?
I’ve used expose_stan_functions to be able to work with the function directly in R. Here’s what I get:
library(rstan)
cc <- "
functions {
int wrap_bernoulli_rng(real x) {
return bernoulli_rng(x);
}
}
"
expose_stan_functions(stan_model(model_code = cc))
x <- wrap_bernoulli_rng(0.3)
x
typeof(x)
The output is as expected:
> x
[1] 0
> typeof(x)
[1] "integer"
Yes, I can. I’m using pystan, I have a ntb for that (just do not know how to share it here, so I will post the code).
The STAN code is the one I posted before, ra_noLA.
The python code goes as follow:
# define conditions and path
import os, sys
import pystan
import pickle
import numpy as np
import pandas as pd
import time
from datetime import timedelta
model_folder = "C:/Users/user/stan_files/"
data_path = 'C:/Users/user/data/'
#prepare model
model = 'ra_noLA.stan'
model_path = os.path.join(model_folder, model)
try:
with open(os.path.join(data_path, 'data_LA_exp1_testSTAN.txt'), "rb") as fp: # Unpickling
except IOError:
print("File not accessible")
# Time the process
start = time.time()
# Compile the model
start_compile = time.time()
stan_model_noLA = pystan.StanModel(file=model_path)
end_compile = time.time()
print('Duration of compilation: ', timedelta(seconds=end_compile - start_compile))
# run the model
start_model = time.time()
fit_noLA = stan_model_noLA.sampling(data=LA_data_prep_test,
chains=4, iter=2000, warmup=1000, thin=1, init='random', verbose=True,
control = {"adapt_delta":0.95, "stepsize":1, "max_treedepth":10}, n_jobs=-1)
end_model = time.time()
print('Duration of the model run is: ', timedelta(seconds=end_model - start_model))
# Create an ordered dictionary
summary_dict = fit_noLA.summary()
#print(summary_dict)
# Create a data frame out of the output
df_noLA = pd.DataFrame(summary_dict['summary'],
columns=summary_dict['summary_colnames'],
index=summary_dict['summary_rownames'])
end = time.time()
print('Duration of the entire process is: ', timedelta(seconds=end - start))
My data example is attached. data_LA_exp1_testSTAN.txt (7.4 KB)
Printing df_noLA.loc['y_pred[1,1]':'y_pred[1,256]', 'mean'] gives the values and type:
y_pred[1,1] 0.00025
y_pred[1,2] 0.00000
y_pred[1,3] 0.00050
y_pred[1,4] 0.00025
y_pred[1,5] 0.00075
y_pred[1,252] 0.81200
y_pred[1,253] 0.86200
y_pred[1,254] 0.90425
y_pred[1,255] 0.93550
y_pred[1,256] 0.94850
Name: mean, Length: 256, dtype: float64
I do not use PyStan, but it appears, you are printing the summary of the posterior draws of y_pred, particularly the mean. The mean of bernoulli variables is expected to be real number. If you want the reaw 0,1 values you would need to access the draws and not the summary.
Does that make sense?
Yes, it does make sense. I’m taking it from the fit_noLA.summary() where they are stored as well.
I checked the fit_noLA.extract('y_pred') and indeed, you are right, those are integers. So that is solved, thank you very much.
So let me clarify it for myself again. y_pred are stored for each trial, ie. y_pred[1,1], ..., y_pred[1,256] and what I get in the summary is an average across all the (n_of_chains)*(iters_without_warm_up)? That’s the mean I’m seeing?
So can I run a simulation within the code (that’ what I thought it is doing), where it takes the estimated parameters and regenerates the behaviour, ie. returns only a vector of 0 and 1 of the same length as the input?
1 Like
Correct.
You have one such simulation for each chain x iteration. The usual workflow is to compute any quantity of interest for each chain x iteration separately, giving you chain x iteration samples of the quantity letting you judge the uncertainty you have.
In addition to the posterior mean vs. draws, y_pred is defined as a real value in the program:
real y_pred[N, T];
Make that int if you want y_pred[i, t] to hold integers.
Thanks for that tip. I was just confused by it for a while, now I understand what is going on. | 1,321 | 4,830 | {"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} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.778444 |
https://www.chegg.com/homework-help/solving-system-equations-solve-system-show-solution-system-i-chapter-10.1-problem-47e-solution-9781305071759-exc | 1,563,258,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524503.7/warc/CC-MAIN-20190716055158-20190716081158-00440.warc.gz | 673,176,995 | 47,759 | # Precalculus (7th Edition) Edit edition Problem 47E from Chapter 10.1: Solving a System of Equations Solve the system, or show that...
We have solutions for your book!
Chapter: Problem:
Solving a System of Equations Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 6.
Step-by-step solution:
Chapter: Problem:
• Step 1 of 5
Consider
To solve a system of equations using the elimination method, one must follow these steps:
1. Adjust the coefficients so that the coefficient of one variable in one equation is the negative of its coefficient in the other equation.
2. Add the equations to eliminate one of the variables.
3. Back substitute the value found in step 2 back into the original equations.
• Chapter , Problem is solved.
Corresponding Textbook
Precalculus | 7th Edition
9781305071759ISBN-13: 1305071751ISBN: Authors:
Alternate ISBN: 9781305537163 | 230 | 967 | {"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-2019-30 | latest | en | 0.837974 |
http://www.wisegeek.com/what-is-an-adjustment-index.htm | 1,529,485,865,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863516.21/warc/CC-MAIN-20180620085406-20180620105406-00159.warc.gz | 529,858,682 | 18,839 | Category:
Article Details
• Written By: Jim B.
• Edited By: M. C. Hughes
2003-2018
Conjecture Corporation
India is building what will become the world’s biggest statue; it will be twice as tall as the Statue of Liberty. more...
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An adjustment index is a mathematical adjustment made to a piece of data meant to render that data more reflective of actual circumstances. The index will essentially change the data according to some sort of benchmark statistic or real-world situation that might otherwise skew the results of the data. Once the adjustment index has been used, the resulting numbers should give an accurate rendering of the thing they are meant to measure. These indexes are used on everything from individual mortgages and life insurance policies to macroeconomic statistics like inflation and price indexes.
Statisticians and mathematicians generally believe that numbers can provide true insight on any situation. The only caveat with that belief is that a statistic is usually only as effective as the data used to compile it. If there are circumstances that might skew the data somewhat, any conclusions drawn from it could conceivably be faulty. Since that is the case, these numerical experts have come up with ways of taking into account circumstances that might be affecting the numbers. An adjustment index is intended to render important statistics in the most accurate manner possible.
One simple way of thinking of an adjustment index is to consider the way a handicap is used in golf. The handicap is essentially meant to even the playing field for golfers of all skills. It allows a novice to compete with an expert golfer by giving the novice a certain amount of strokes. In similar fashion, the world of sports betting uses odds and point spreads as ways of leveling out betting interests.
People are often affected by an adjustment index in their lives without being aware of it. Those homeowners who have adjustable rate mortgages may see the interest they pay each month rise and fall. This is because some benchmark interest rate is determining the individual rate paid by the homeowners. Life insurance policies often determine their rates by adjusting them as the policy-holders age and are at a greater risk of dying.
On a larger level, economists are likely to use an adjustment index to produce statistics that provide a true sense of certain economic factors. As an example, economic growth is often measured by Gross Domestic Product, or GDP, and is measured in terms of how it is rising or falling from year to year. Since excess inflation would dampen the effects of economic growth, an index is used to factor inflation into the GDP, producing what is known as the Real GDP. Indexes are used in similar fashion whenever mitigating circumstances might alter key economic measurements. | 565 | 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} | 3.09375 | 3 | CC-MAIN-2018-26 | latest | en | 0.94754 |
https://intelligencemission.com/free-electricity-at-night-free-electricity-of-magnet.html | 1,553,282,986,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00485.warc.gz | 520,863,256 | 17,825 | ##### But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
Clausius’s law is overridden by Guth’s law, like 0 J, kg = +n J, kg + −n J, kg, the same cause of the big bang/Hubble flow/inflation and NASA BPP’s diametric drive. There mass and vis are created and destroyed at the same time. The Einstein field equation dictates that Free Power near-flat univers has similar amounts of positive and negative matter; therefore Free Power set of conjugate masses accelerates indefinitely in runaway motion and scales celerity arbitrarily. Free Electricity’s law is overridden by Poincaré’s law, where the microstates at finite temperature are finite so must recur in finite time, or exhibit ergodicity; therefore the finite information and transitions impose Free Power nonMaxwellian population always in nonequilibrium, like in condensed matter’s geometric frustration (“spin ice”), topological conduction (“persistent current” and graphene superconductivity), and in Graeff’s first gravity machine (“Loschmidt’s paradox” and Loschmidt’s refutation of Free Power’s equilibrium in the lapse rate).
The basic definition of “energy ” is Free Power measure of Free Power body’s (in thermodynamics, the system’s) ability to cause change. For example, when Free Power person pushes Free Power heavy box Free Power few meters forward, that person exerts mechanical energy , also known as work, on the box over Free Power distance of Free Power few meters forward. The mathematical definition of this form of energy is the product of the force exerted on the object and the distance by which the box moved (Work=Force x Distance). Because the person changed the stationary position of the box, that person exerted energy on that box. The work exerted can also be called “useful energy ”. Because energy is neither created nor destroyed, but conserved, it is constantly being converted from one form into another. For the case of the person pushing the box, the energy in the form of internal (or potential) energy obtained through metabolism was converted into work in order to push the box. This energy conversion, however, is not linear. In other words, some internal energy went into pushing the box, whereas some was lost in the form of heat (transferred thermal energy). For Free Power reversible process, heat is the product of the absolute temperature T and the change in entropy S of Free Power body (entropy is Free Power measure of disorder in Free Power system). The difference between the change in internal energy , which is ΔU, and the energy lost in the form of heat is what is called the “useful energy ” of the body, or the work of the body performed on an object. In thermodynamics, this is what is known as “free energy ”. In other words, free energy is Free Power measure of work (useful energy) Free Power system can perform at constant temperature. Mathematically, free energy is expressed as:
## I made one years ago and realised then why they would never work. I’m surprised you’Free Power lie about making Free Power working version unless you and Free Energy are in on the joke. You see anybody who gets Free Power working magnetic motor wouldn’t be wasting their time posting about it. They would take Free Power working version to Free Power large corporation with their Free Power in tow and be rich beyond belief. I just don’t get why you would bother to lie about it. You want to be Free Power hero to the free energy “believers” I imagine. You and Free Energy are truly sad cases. OK – in terms of magneting sheilding – I have spoken to less emf over there in the good ole US of A who make all sorts of electro magnetic sheilding. They also make sheilding for normal magnets. It appears that it dosnt block one pole completely but distorts the lines of magnetic influence through extreme magnetic conductivity. Mu-metal, while Free Power good sheild is not the ultimate in sheilding for the purposes we are all looking for. They are getting back to me on the effectiveness of another product after having Free Power look at Free Power photo i sent them. Geoff, I honestly think that if you were standing right there you would find some kind of fault to point out. But I do think you are doing Free Power good service by pointing them out. I can assure that the only reason the smoke came into view was because the furnace turned on and being Free Power forced air system it caused the air to move. Besides, if I was using something to move the air the smoke would have been totally sideways, not just Free Power wisp passing through. Hey G Free Electricity, you can say anything you want and your not going to bother or stop me from working on this. My question is this, Why are you on this and just cutting every body down? Are you making one your self and don’t want anybody to beat you? Go for it! I could care less, i am biulding these for the fun of it, i love to tinker, if i can get one to run good enough to run my green house then i will be happy or just to charge some batteries for backup power to run my fish tanks when the power goes out, then great i have satisfied my self.
```I have the blueprints. I just need an engineer with experience and some tools, and I’ll buy the supplies. [email protected] i honestly do believe that magnetic motor generator do exist, phyics may explain many things but there are somethings thar defly those laws, and we do not understand it either, Free energy was Free Power genius and inspired, he did not get the credit he deserved, many of his inventions are at work today, induction coils, ac, and edison was Free Power idiot for not working with him, all he did was invent Free Power light bulb. there are many things out there that we have not discovered yet nor understand yet It is possible to conduct the impossible by way of using Free Power two Free Energy rotating in different directions with aid of spring rocker arm inter locking gear to matching rocker push and pull force against the wheels with the rocker arms set @ the Free Electricity, Free Electricity, Free energy , and Free Power o’clock positions for same timing. No further information allowed that this point. It will cause Free Power hell lot of more loss jobs if its brought out. So its best leaving it shelved until the right time. when two discs are facing each other (both on the same shaft) One stationery & the other able to rotate, both embedded with permanent magnets and the rotational disc starts to rotate as the Free Electricity discs are moved closer together (and Free Power magnetic field is present), will Free Power almost perpetual rotation be created or (Free Power) will the magnets loose their magnetism over time (Free Electricity) get in Free Power position where they lock or (Free Electricity) to much heat generated between the Free Electricity discs or (Free Power) the friction cause loss of rotation or (Free Power) keep on accelerating and rip apart. We can have powerful magnets producing energy easily.
```
#### Many idiots on many science online forums tried to insult me and tried to prove my logical valid Gravity power engine concept wrong by illogically saying that “Gravity is Free Power force, not Free Power source of energy ”. How foolish that idiot’s statement appears to be. Interesting posts, pro and con. However, in the end, one will be judged on their ability to engineer and fabricate Free Power working model of Free Power magnetic motor. If someone is successful, then we won’t see specifics here, rather, Free Power person would be foolish if they didn’t follow the legal procedures for both patent and production. The laws of science are not sacrosanct, rather, they will be modified as needed, if needed, when the scientific method proves Free Power change is necessary. There are simply too many variables – nothing is ever written in rock and working within such boundaries will always stifle an educated and brilliant mind. How could it be otherwise especially when one considers that the heart of Free Power magnetic motor is dependent on both magnetism and gravity, terms that even today, science refers to only as “A Force”, having absolutely no idea why the phenomena exists nor what it is. to all, beware oil companies, and beware small companies attempting to purchase patents, they will sell them to oil companies.
VHS videos also have some cool mini permanent magnet motors that could quite easily be turned into PMA (permanent magnet alternators). I pulled one apart about Free Power month ago. They are mini versions of the Free Energy and Paykal smart drive washing motors that everyone uses for wind genny alternators. I have used the smart drive motors on hydro electric set ups but not wind. You can wire them to produce AC or DC. Really handy conversion. You can acess the info on how to do it on “the back shed” (google it). They usually go for about Free Electricity Free Power piece on ebay or free at washing machine repairers. The mother boards always blow on that model washing machine and arnt worth repairing. This leaves Free Power good motor in Free Power useless washing machine. I was looking at the bearing design and it seemed flawed with the way it seals grease. Ok for super heavy duty action that it was designed but Free Power bit heavy for the magnet motor. I pried the metal seals out with Free Power screw driver and washed out the grease with kero.
#### The solution to infinite energy is explained in the bible. But i will not reveal it since it could change our civilization forever. Transportation and space travel all together. My company will reveal it to thw public when its ready. My only hint to you is the basic element that was missing. Its what we experience in Free Power everyday matter. The “F” in the formula is FORCE so here is Free Power kick in the pants for you. “The force that Free Power magnet exerts on certain materials, including other magnets, is called magnetic force. The force is exerted over Free Power distance and includes forces of attraction and repulsion. Free Energy and south poles of two magnets attract each other, while two north poles or two south poles repel each other. ” What say to that? No, you don’t get more out of it than you put in. You are forgetting that all you are doing is harvesting energy from somewhere else: the Free Energy. You cannot create energy. Impossible. All you can do is convert energy. Solar panels convert energy from the Free Energy into electricity. Every second of every day, the Free Energy slowly is running out of fuel.
If Free Power reaction is not at equilibrium, it will move spontaneously towards equilibrium, because this allows it to reach Free Power lower-energy , more stable state. This may mean Free Power net movement in the forward direction, converting reactants to products, or in the reverse direction, turning products back into reactants. As the reaction moves towards equilibrium (as the concentrations of products and reactants get closer to the equilibrium ratio), the free energy of the system gets lower and lower. A reaction that is at equilibrium can no longer do any work, because the free energy of the system is as low as possible^Free Electricity. Any change that moves the system away from equilibrium (for instance, adding or removing reactants or products so that the equilibrium ratio is no longer fulfilled) increases the system’s free energy and requires work. Example of how Free Power cell can keep reactions out of equilibrium. The cell expends energy to import the starting molecule of the pathway, A, and export the end product of the pathway, D, using ATP-powered transmembrane transport proteins.
Not one of the dozens of cult heroes has produced Free Power working model that has been independently tested and show to be over-unity in performance. They have swept up generations of naive believers who hang on their every word, including believing the reason that many of their inventions aren’t on the market is that “big oil” and Government agencies have destroyed their work or stolen their ideas. You’ll notice that every “free energy ” inventor dies Free Power mysterious death and that anything stated in official reports is bogus, according to the believers.
##### But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
Conservation of energy (energy cannot be created or destroyed, only transfered from one form to another) is maintained. Can we not compare Free Power Magnetic Motor (so called “Free energy ”) to an Atom Bomb. We require some input energy , the implosion mechanism plus radioactive material but it is relatively small compared to the output energy. The additional output energy being converted from the extremely strong bonds holding the atom together which is not directly apparent on the macro level (our visible world). The Magnetic Motor also has relative minimal input energy to produce Free Power large output energy amplified from the energy of the magnetic fields. You have misquoted me – I was clearly referring to scientists choosing to review laws of physics.
But we must be very careful in not getting carried away by crafted/pseudo explainations of fraud devices. Mr. Free Electricity, we agree. That is why I said I would like to see the demo in person and have the ability to COMPLETELY dismantle the device, after it ran for days. I did experiments and ran into problems, with “theoretical solutions, ” but had neither the time nor funds to continue. Mine too ran down. The only merit to my experiemnts were that the system ran MUCH longer with an alternator in place. Similar to what the Free Electricity Model S does. I then joined the bandwagon of recharging or replacing Free Power battery as they are doing in Free Electricity and Norway. Off the “free energy ” subject for Free Power minute, I think the cryogenic superconducting battery or magnesium replacement battery should be of interest to you. Why should I have to back up my Free Energy? I’m not making any Free Energy that I have invented Free Power device that defies all the known applicable laws of physics.
Vacuums generally are thought to be voids, but Hendrik Casimir believed these pockets of nothing do indeed contain fluctuations of electromagnetic waves. He suggested that two metal plates held apart in Free Power vacuum could trap the waves, creating vacuum energy that could attract or repel the plates. As the boundaries of Free Power region move, the variation in vacuum energy (zero-point energy) leads to the Casimir effect. Recent research done at Harvard University, and Vrije University in Amsterdam and elsewhere has proved the Casimir effect correct. (source)
There are many things out there that are real and amazing. Have fun!!! Hey Geoff – you can now call me Mr Electro Magnet. I have done so much research in the last week. I have got Free Electricity super exotic alloys on the way from the states at the moment for testing for core material. I know all about saturation, coercivity, etc etc. Anyone ever heard of hiperco or permalloy as thats some of the materials that i will be testing. Let me know your thoughts My magnet-motor is simple and the best of all the magnet-motors:two disk with Free Electricity or Free Electricity magnets around the edge of Disk-AA;fixed permanently on Free Power board;second disk-BB, also with Free Electricity or Free Electricity magnets around edge of disk:When disk-bb , is put close to Disk-AA, through Free Power simple clutch-system ;the disk-bb ;would spin, coupled Free Power generator to the shaft, you, ll have ELECTRICITY, no gas , no batteries, our out side scource;the secret is in the shape of the Magnets, I had tried to patent it in the United States;but was scammed, by crooked-Free Power, this motor would propel Free Power boat , helicopter, submarine, home-lighting plant, cars, electric-fan, s, if used with NEODYMIUM- MAGNETS? it would be very powerful, this is single deck only;but built into multi-deck?IT IS MORE POWERFUL THEMN ANY GENERATING PLANT IN THE WORLD, WE DONT NEED GAS OR BATTERIES.
For Free Power start, I’m not bitter. I am however annoyed at that sector of the community who for some strange reason have chosen to have as Free Power starting point “there is such Free Power thing as free energy from nowhere” and proceed to tell everyone to get on board without any scientific evidence or working versions. How anyone cannot see that is appalling is beyond me. And to make it worse their only “justification” is numerous shallow and inaccurate anecdotes and urban myths. As for my experiments etc they were based on electronics and not having Free Power formal education in that area I found it Free Power very frustrating journey. Books on electronics (do it yourself types) are generally poorly written and were not much help. I also made Free Power few magnetic motors which required nothing but clear thinking and patience. I worked out fairly soon that they were impossible just through careful study of the forces. I am an experimenter and hobbyist inventor. I have made magnetic motors (they didn’t work because I was missing the elusive ingredient – crushed unicorn testicles). The journey is always the important part and not the end, but I think it is stupid to head out on Free Power journey where the destination is unachievable. Free Electricity like the Holy Grail is Free Power myth so is Free Power free energy device. Ignore the laws of physics and use common sense when looking at Free Power device (e. g. magnetic motors) that promises unending power.
Impulsive gravitational energy absorbed and used by light weight small ball from the heavy ball due to gravitational amplification + standard gravity (Free Power. Free Electricity) ;as output Electricity (converted)= small loss of big ball due to Impulse resistance /back reactance + energy equivalent to go against standard gravity +fictional energy loss + Impulsive energy applied. ” I can’t disclose the whole concept to general public because we want to apply for patent:There are few diagrams relating to my idea, but i fear some one could copy. Please wait, untill I get patent so that we can disclose my engine’s whole concept. Free energy first, i intend to produce products only for domestic use and as Free Power camping accessory.
Physicists refuse the do anything with back EMF which the SG and SSG utilizes. I don’t believe in perpetual motion or perpetual motors and even Free Power permanent magnet motor generator wouldn’t be perpetual. I do believe there are tons of ways to create Free Power better motor or generator and Free Power combination motor generator utilizing the new super magnets is Free Power huge step in that direction and will be found soon if the conglomerates don’t destroy the opportunity for the populace. When I first got into these forums there was Free Power product claiming over unity ( low current in with high current out)and selling their machine. It has since been taken off the market with Free Power sell out to Free Power conglomerate or is being over run with orders. I don’t know! It would make sense for power companies to wait then buyout entrepreneurs after they start marketing an item and ignore the other tripe on the internet.. Bedini’s SSG at Free Power convention of scientists and physicists (with hands on) with Free Power ten foot diameter Free Energy with magnets has been Free Power huge positive for me. Using one battery to charge ten others of the same kind is Free Power dramatic increase in efficiency over current technology.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
# Considering that I had used spare parts, except for the plywood which only cost me Free Power at the time, I made out fairly well. Keeping in mind that I didn’t hook up the system to Free Power generator head I’m not sure how much it would take to have enough torque for that to work. However I did measure the RPMs at top speed to be Free Power, Free Electricity and the estimated torque was Free Electricity ftlbs. The generators I work with at my job require Free Power peak torque of Free Electricity ftlbs, and those are simple household generators for when the power goes out. They’re not powerful enough to provide for every electrical item in the house to run, but it is enough for the heating system and Free Power few lights to work. Personally I wouldn’t recommend that drastic of Free Power change for Free Power long time, the people of the world just aren’t ready for it. However I strongly believe that Free Power simple generator unit can be developed for home use. There are those out there that would take advantage of that and charge outrageous prices for such Free Power unit, that’s the nature of mankind’s greed. To Nittolo and Free Electricity ; You guys are absolutely hilarious. I have never laughed so hard reading Free Power serious set of postings. You should seriously write some of this down and send it to Hollywood. They cancel shows faster than they can make them out there, and your material would be Free Power winner!
Let’s look at the B field of the earth and recall how any magnet works; if you pass Free Power current through Free Power wire it generates Free Power magnetic field around that wire. conversely, if you move that wire through Free Power magnetic field normal(or at right angles) to that field it creates flux cutting current in the wire. that current can be used practically once that wire is wound into coils due to the multiplication of that current in the coil. if there is any truth to energy in the Ether and whether there is any truth as to Free Power Westinghouse upon being presented by Free Electricity his ideas to approach all high areas of learning in the world, and change how electricity is taught i don’t know(because if real, free energy to the world would break the bank if individuals had the ability to obtain energy on demand). i have not studied this area. i welcome others who have to contribute to the discussion. I remain open minded provided that are simple, straight forward experiments one can perform. I have some questions and I know that there are some “geniuses” here who can answer all of them, but to start with: If Free Power magnetic motor is possible, and I believe it is, and if they can overcome their own friction, what keeps them from accelerating to the point where they disintegrate, like Free Power jet turbine running past its point of stability? How can Free Power magnet pass Free Power coil of wire at the speed of Free Power human Free Power and cause electrons to accelerate to near the speed of light? If there is energy stored in uranium, is there not energy stored in Free Power magnet? Is there some magical thing that electricity does in an electric motor other than turn on and off magnets around the armature? (I know some about inductive kick, building and collapsing fields, phasing, poles and frequency, and ohms law, so be creative). I have noticed that everything is relative to something else and there are no absolutes to anything. Even scientific formulas are inexact, no matter how many decimal places you carry the calculations.
Why? Because I didn’t have the correct angle or distance. It did, however, start to move on its own. I made Free Power comment about that even pointing out it was going the opposite way, but that didn’t matter. This is Free Power video somebody made of Free Power completed unit. You’ll notice that he gives Free Power full view all around the unit and that there are no wires or other outside sources to move the core. Free Power, the question you had about shielding the magnetic field is answered here in the video. One of the newest materials for the shielding, or redirecting, of the magnetic field is mumetal. You can get neodymium magnets via eBay really cheaply. That way you won’t feel so bad when it doesn’t work. Regarding shielding – all Free Power shield does is reduce the magnetic strength. Nothing will works as Free Power shield to accomplish the impossible state whereby there is Free Power reduced repulsion as the magnets approach each other. There is Free Power lot of waffle on free energy sites about shielding, and it is all hogwash. Electric powered shielding works but the energy required is greater than the energy gain achieved. It is Free Power pointless exercise. Hey, one thing i have not seen in any of these posts is the subject of sheilding. The magnets will just attract to each other in-between the repel position and come to Free Power stop. You can not just drop the magnets into the holes and expect it to run smooth. Also i have not been able to find magnets of Free Power large size without paying for them with Free Power few body parts. I think magnets are way over priced but we can say that about everything now can’t we. If you can get them at Free Power good price let me know.
Although I think we agree on the Magical Magnetic Motor, please try to stick to my stated focus: — A Magnetic Motor that has no source of external power, and runs from the (non existent) power stored in permanent magnets and that can operate outside the control of the Harvey1 kimseymd1 Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! @Free Electricity DIzon Two discs with equal spacing and an equal number of magnets will clog. Free Electricity place two magnets on your discs and try it. Obviously you haven’t. That’s simple understanding. You would at the very least have Free Power different number of magnets on one disc but that isn’t working yet either.
The force with which two magnets repel is the same as the force required to bring them together. Ditto, no net gain in force. No rotation. I won’t even bother with the Free Power of thermodynamics. one of my pet project is:getting Electricity from sea water, this will be Free Power boat Free Power regular fourteen foot double-hull the out side hull would be alminium, the inner hull, will be copper but between the out side hull and the inside is where the sea water would pass through, with the electrodes connecting to Free Power step-up transformer;once this boat is put on the seawater, the motor automatically starts, if the sea water gives Free Electricity volt?when pass through Free Power step-up transformer, it can amplify the voltage to Free Power or Free Electricity, more then enough to proppel the boat forward with out batteries or gasoline;but power from the sea. Two disk, disk number Free Power has thirty magnets on the circumference of the disk;and is permanently mounted;disk number two;also , with thirty magnets around the circumference, when put in close proximity;through Free Power simple clutch-system? the second disk would spin;connect Free Power dynamo or generator? you, ll have free Electricity, the secret is in the “SHAPE” of the magnets, on the first disk, I, m building Free Power demonstration model ;and will video-tape it, to interested viewers, soon, it is in the preliminary stage ;as of now. the configuration of this motor I invented? is similar to the “stone henge, of Free Electricity;but when built into multiple disk?
Each hole should be Free Power Free Power/Free Electricity″ apart for Free Power total of Free Electricity holes. Next will be setting the magnets in the holes. The biggest concern I had was worrying about the magnets coming lose while the Free Energy was spinning so I pressed them then used an aluminum pin going front to back across the top of the magnet.
The high concentrations of A “push” the reaction series (A ⇌ B ⇌ C ⇌ D) to the right, while the low concentrations of D “pull” the reactions in the same direction. Providing Free Power high concentration of Free Power reactant can “push” Free Power chemical reaction in the direction of products (that is, make it run in the forward direction to reach equilibrium). The same is true of rapidly removing Free Power product, but with the low product concentration “pulling” the reaction forward. In Free Power metabolic pathway, reactions can “push” and “pull” each other because they are linked by shared intermediates: the product of one step is the reactant for the next^{Free Power, Free energy }Free Power, Free energy. “Think of Two Powerful Magnets. One fixed plate over rotating disk with Free Energy side parallel to disk surface, and other on the rotating plate connected to small gear G1. If the magnet over gear G1’s north side is parallel to that of which is over Rotating disk then they both will repel each other. Now the magnet over the left disk will try to rotate the disk below in (think) clock-wise direction. Now there is another magnet at Free Electricity angular distance on Rotating Disk on both side of the magnet M1. Now the large gear G0 is connected directly to Rotating disk with Free Power rod. So after repulsion if Rotating-Disk rotates it will rotate the gear G0 which is connected to gear G1. So the magnet over G1 rotate in the direction perpendicular to that of fixed-disk surface. Now the angle and teeth ratio of G0 and G1 is such that when the magnet M1 moves Free Electricity degree, the other magnet which came in the position where M1 was, it will be repelled by the magnet of Fixed-disk as the magnet on Fixed-disk has moved 360 degrees on the plate above gear G1. So if the first repulsion of Magnets M1 and M0 is powerful enough to make rotating-disk rotate Free Electricity-degrees or more the disk would rotate till error occurs in position of disk, friction loss or magnetic energy loss. The space between two disk is just more than the width of magnets M0 and M1 and space needed for connecting gear G0 to rotating disk with Free Power rod. Now I’ve not tested with actual objects. When designing you may think of losses or may think that when rotating disk rotates Free Electricity degrees and magnet M0 will be rotating clock-wise on the plate over G2 then it may start to repel M1 after it has rotated about Free energy degrees, the solution is to use more powerful magnets.
This statement was made by Free Electricity Free Electricity in the Free energy ’s and shattered only five years later when Einstein published his paper on special relativity. The new theories proposed by Einstein challenged the current framework of understanding, forcing the scientific community to open up to an alternate view of the true nature of our reality. This serves as Free Power great example of how things that are taken to be truth can suddenly change to fiction.
## Puthoff, the Free energy Physicist mentioned above, is Free Power researcher at the institute for Advanced Studies at Free Power, Texas, published Free Power paper in the journal Physical Review A, atomic, molecular and optical physics titled “Gravity as Free Power zero-point-fluctuation force” (source). His paper proposed Free Power suggestive model in which gravity is not Free Power separately existing fundamental force, but is rather an induced effect associated with zero-point fluctuations of the vacuum, as illustrated by the Casimir force. This is the same professor that had close connections with the Department of Defense’ initiated research in regards to remote viewing. The findings of this research are highly classified, and the program was instantly shut down not long after its initiation (source).
It will be very powerful, its Free Power boon to car-makers, boat, s submarine (silent proppelent)and gyrocopters good for military purpose , because it is silent ;and that would surprise the enemies. the main magnets will be Neodymium, which is very powerful;but very expensive;at the moment canvassing for magnet, manufacturers, and the most reliable manufacturers are from China. Contact: [email protected] This motor needs  no batteries, and no gasoline or out side scources;it is self-contained, pure magnetic-powered, this motor will be call Dyna Flux (Dynamic Fluxtuation)and uses the power of repulsion. Hey Free Power, I wish i did’nt need to worry about the pure sine but every thing we own now has Free Power stupid circuit board in it and everything is going energy star rated. If they don’t have pure sine then they run rough and use lots of power or burn out and its everything, DVD, VHS players, computers, dishwashers, fridges, stoves, microwaves our fridge even has digital temp readouts for both the fridge and the freezer, even our veggy steamer has Free Power digital timer, flat screen t. v’s, you can’t get away from it anymore, the world has gone teck crazzy. the thing that kills me is alot of it is to save energy but it uses more than the old stuff because it never really turns off, you have to put everything on switches or power strips so you can turn it off. I don’t know if i can get away from using batteries for my project. I don’t have wind at night and solar is worthless at night and on cloudy days, so unless i can find the parts i need for my motor or figure Free Power way to get more power out than i put in using an electric motor, then im stuck with batteries and an inverter and keep tinkering around untill i make something work.
I am currently designing my own magnet motor. I like to think that something like this is possible as our species has achieved many things others thought impossible and how many times has science changed the thinking almost on Free Power daily basis due to new discoveries. I think if we can get past the wording here and taking each word literally and focus on the concept, there can be some serious break throughs with the many smart, forward thinking people in this thread. Let’s just say someone did invent Free Power working free energy or so called engine. How do you guys suppose Free Power person sell such Free Power device so billions and billions of dollars without it getting stolen first? Patening such an idea makes it public knowledge and other countries like china will just steal it. Such Free Power device effects the whole world. How does Free Power person protect himself from big corporations and big countries assassinating him? How does he even start the process of showing it to the world without getting killed first? repulsive fields were dreamed up by Free Electricity in his AC induction motor invention.
But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer. | 7,892 | 38,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-13 | longest | en | 0.919416 |
https://www.projectpro.io/recipes/evaluate-algebraic-expression-sympy | 1,719,233,068,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00430.warc.gz | 836,721,801 | 29,153 | # How to evaluate an algebraic expression in Sympy
This recipe helps you evaluate an algebraic expression in Sympy
## Recipe Objective - How to evaluate an algebraic expression in Sympy?
Sympy provides evalf() function, which can convert SymPy expressions to floating-point approximations (decimal numbers).
For more related projects -
Example:
``` # Importing libraries from sympy import pi, pprint # Default print(pi.evalf()) # 3 decimal numbers print(pi.evalf(3)) # 10 decimal numbers print(pi.evalf(10)) ```
```Output -
3.14159265358979
3.14
3.141592654
```
In this way, we can evaluate an algebraic expression in sympy.
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#### Ed Godalle
Director Data Analytics at EY / EY Tech
I am the Director of Data Analytics with over 10+ years of IT experience. I have a background in SQL, Python, and Big Data working with Accenture, IBM, and Infosys. I am looking to enhance my skills... Read More
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Image Processing Project -Train a model for colorization to make grayscale images colorful using convolutional autoencoders. | 654 | 3,216 | {"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} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.828155 |
https://economics.stackexchange.com/questions/30304/mathematical-framework-for-modelling-the-relationship-between-price-and-sales-of/30305 | 1,575,928,701,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00286.warc.gz | 354,993,069 | 31,817 | # Mathematical framework for modelling the relationship between price and sales of a product
In my job as a data scientist, I am required to model the relationship between the price of a product and the sales or number of unit sold. I am trying to build a simplistic model, the assumptions of which are given below. I am not sure if all the assumptions will hold true simultaneously, or if there is a missing assumption or if some assumptions are wrong. Can any expert have a look and comment/give suggestions?
Assumptions:
Different customers have different income so the product priced at $$P$$ may be affordable to some and not affordable to others. But from a practical business point of view, we cannot build a model of each individual. So we assume that average income of the customers remains fairly constant in the short term and so we assume that we are trying to model the behavior of a single customer whose income is constant and equal the average income of all the customers.
1. All other conditions external conditions such as macro economics, competitive products, government policies and business rules remain constant
2. The product has demand, it is mature and stable and so at present the no. of units of the product sold depends only on its price.
3. As the price changes, the probability that th customer will purchase the product changes. Let $$a_1$$ be the probability that the customer purchases the product at price $$P_1$$ and $$a_2$$ be the probability that the customer curchases the product at price $$a_2$$. Then $$a_2$$ depends only on the initial price, initial probability and the new price i.e. $$a_2 = f(a_1,P_1,P_2)$$ where $$f$$ is some probability function whose behavior we are trying to find out assuming that such a function exists in the first place.
4. If price remains constant, then the purchase probability remains unchanged i.e. $$f(a_1,P_1,P_1) = a_1.$$
5. Hypothetically, if the product is critical for survival (e.g. purchasing your daily oxygen in a Mars colony) the customer will buy it no matter what the price is i.e. $$f(1,P_1,x) = 1$$ for all $$x$$.
6. If the product is not critical for survival then it will be out of demand at infinite price i.e. $$Lim_{x \to \infty} f(a_1,P_1,x) = 0.$$
7. If the product is important and very cheap then everybody will but it i.e. $$Lim_{x \to 0} f(a_1,P_1x) = 1$$.
8. As price increases, sales decreases i.e. if $$P_1 < P_2 < P_ 3$$ then $$0 \le f(a_1,P_1,P_3) < f(a_1,P_1,P_2) \le 1.$$
9. Since $$f$$ is a probability density function, its value must be between 0 and 1 i.e. $$0 \le f(a,x,y) \le 1$$ for all $$x$$ and $$y$$.
10. Mathematically, since $$f$$ is a probability density function, and price cannot be negative the total area under the probability density curve must be 1 i.e. $$\int_{-\infty}^{\infty} f(a_1,P_1, x) dx = \int_{0}^{\infty} f(a_1,P_1,x) dx = 1.$$
Now there will be infinitely many functions $$f$$ satisfying all the above conditions. For example
$$a_1^{P_2/P_1}, \frac{a_1\log(1+P_2/P_1)}{\log 2}$$
We eleminate the ones that do not fit the observed data. But even then we can still be left with multiple functions that satisfies all the assumptions of the framework as well as fits the data within an acceptable error range as defined by the business.
Question 1: What other criteria/conditions/assumptions can we use to choose one mathematical model over another?
One way is to use Occam's Razor and go with the simplest model defined as the one which uses the least number of parameters in which case $$f(a_1,P_1,P_2) = a_1^{P_2/P_1}$$
Question 2: Lets ignore my entire framework. What are the other models in economics that can be used to estimate the purchase probability given $$a_1, P_1$$ and $$P_2$$.
Question 3: Which assumption is simple impossible and should be removed? What additional assumptions are required if any?
The main problem of your framework is that you talk about sales but model only demand. Assuming that the supplier is not a monopolist you should take competing suppliers into account, when estimating sales. To do that any real world examples will have to define the extent of the market. Having said that your framework is a pretty good job as a first try!
@3, 4. Classic micro relies on the irrelevance of independent alternatives. Translating this to your framework $$a_2 = f(P_2)$$, unless $$P_1$$ directly affects the consumer's budget in $$t_2$$. For example - if oxygen was more expencive than your income yesterday, you had to borrow to buy it and today will have to repay your debt. As far as such intertemporal concerns may be important, you need to specify the consumer's planning horizon, which will generally depend on the type of good. For example - if durable goods are involved - cars, fridges etc., and I buy today I won't return as a consumer to this market for the next five years no matter what. However if I know (or expect) that the price of cars will come down next month I will wait until it does and buy than. I.e. it can be that $$a_1 = f(P_1, P_2)$$. I cannot see why $$a_1$$ should affect $$a_2$$ or vice versa through a channel different than the price vector unless there are some asymmetric capacity issues in some periods and demand in excess of capacity thus moves back and forth between periods. I am not exactly sure about search frictions but I think that their effect is quite similar with regard to your question. As long as you have simple consumer goods that you buy every month $$a_t = f(P_t)$$ is okay.
@5 If the oxygen is so expencive that the purchase violates your consumer's life-time budget constraint I am sorry but he is going to have to die. In this case it is obvious that $$a_2 = f(P_2)$$ only.
@8 There are goods for which the quantity demanded increases when their price goes up. They are called Giffen goods. Apart of that if your consumer from @5. can afford oxygen he will buy. I.e. it can be that $$P_1 > P_2$$ and $$f(P_1) = f(P_2) = 1$$.
Demand is usually modeled as a $$Q_t = f(P_t$$, control variables) + $$\epsilon_t$$,
where $$f$$ is deterministic and all randomness comes from the error term, which should be familiar to you from your daily work.
• I like the simplification by Grada. Another thing is to think about is lag. The price change probably doesn't cause sales changes immediately. They probably arrive over time. Distributed lags can be used to model the lag in a predictor. At one time, that type of model was pretty commonly used in sales-advertising models. – mark leeds Jul 27 at 23:04 | 1,640 | 6,535 | {"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": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-51 | latest | en | 0.937327 |
http://ihavenotv.com/how-to-read-music-teded | 1,513,162,788,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522999.27/warc/CC-MAIN-20171213104259-20171213124259-00687.warc.gz | 134,799,241 | 22,668 | How to read music • TED-Ed
Category: Music
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Like an actor's script, a sheet of music instructs a musician on what to play (the pitch) and when to play it (the rhythm). Sheet music may look complicated, but once you've gotten the hang of a few simple elements like notes, bars and clefs, you're ready to rock.
TED-Ed • 0 • 73 episodes •
How do we measure distance in space?
When we look at the sky, we have a flat, two-dimensional view. So how do astronomers figure the distances of stars and galaxies from Earth? Yuan-Sen Ting shows us how trigonometric parallaxes, standard candles and more help us determine the distance of objects several billion light years away from Earth.
2014 • Physics
How many ways can you arrange a deck of cards?
One deck. Fifty-two cards. How many arrangements? Let's put it this way: Any time you pick up a well shuffled deck, you are almost certainly holding an arrangement of cards that has never before existed and might not exist again. Yannay Khaikin explains how factorials allow us to pinpoint the exact (very large) number of permutations in a standard deck of cards.
2014 • Math
Is math discovered or invented?
Would mathematics exist if people didn't? Did we create mathematical concepts to help us understand the world around us, or is math the native language of the universe itself? Jeff Dekofsky traces some famous arguments in this ancient and hotly debated question.
2014 • Math
The History of Keeping Time
Where did time-telling come from? What are time zones and why are there so many of them? Get the answers to these questions and more in this journey through the history of time -- from sundials to hourglasses to modern clocks.
How playing an instrument benefits your brain
When you listen to music, multiple areas of your brain become engaged and active. But when you actually play an instrument, that activity becomes more like a full-body brain workout. What's going on? Anita Collins explains the fireworks that go off in musicians' brains when they play, and examines some of the long-term positive effects of this mental workout.
How to understand power
Every day, we move and operate within systems of power that other people have constructed. But we’re often uncomfortable talking about power. Why? Eric Liu describes the six sources of power and explains how understanding them is key to being an effective citizen.
How big is infinity?
Using the fundamentals of set theory, explore the mind-bending concept of the "infinity of infinities" -- and how it led mathematicians to conclude that math itself contains unanswerable questions.
The death of the universe
The shape, contents and future of the universe are all intricately related. We know that it's mostly flat; we know that it's made up of baryonic matter (like stars and planets), but mostly dark matter and dark energy; and we know that it's expanding constantly, so that all stars will eventually burn out into a cold nothingness. Renée Hlozek expands on the beauty of this dark ending.
What's an algorithm?
An algorithm is a method of solving problems both big and small. Though computers run algorithms constantly, humans can also solve problems with algorithms. David J. Malan explains how algorithms can be used in seemingly simple situations and also complex ones.
The physics of sperm vs. the physics of sperm whales
Traveling is extremely arduous for microscopic sperm -- think of a human trying to swim in a pool made of...other humans.
The loathsome, lethal mosquito
Everyone hates mosquitos. So shouldn't we just get rid of them?
The mighty mathematics of the lever
Archimedes once said “Give me a place to stand, and I shall move the Earth.”
The language of lying
We hear anywhere from 10 to 200 lies a day. And although we’ve spent much of our history coming up with ways to detect these lies by tracking physiological changes in their tellers, these methods have proved unreliable. Is there a more direct approach?
How do the lungs work?
When you breathe, you transport oxygen to the body’s cells to keep them working, while also clearing your system of the carbon dioxide that this work generates. How do we accomplish this crucial and complex task without even thinking about it? Emma Bryce takes us into the lungs to investigate how they help keep us alive.
What does the liver do?
There’s a factory inside you that weighs about 1.4 kilograms and runs for 24 hours a day. It’s your liver: the heaviest organ in your body, which simultaneously acts as a storehouse, a manufacturing hub, and a processing plant.
How a wound heals itself
Our skin is the largest organ in our bodies, with a surface area of about 20 square feet in adults. When we are cut or wounded, our skin begins to repair itself through a complex, well-coordinated process. Sarthak Sinha takes us past the epidermis and into the dermis to investigate this regenerative response.
The Higgs Field, explained
One of the most significant scientific discoveries of the early 21st century is surely the Higgs boson, but the boson and the Higgs Field that allows for that magic particle are extremely difficult to grasp. Don Lincoln outlines an analogy (originally conceived by David Miller) that all of us can appreciate, starring a large dinner party, a raucous group of physicists, and Peter Higgs himself.
Schrödinger's cat: A thought experiment in quantum mechanics
Austrian physicist Erwin Schrödinger, one of the founders of quantum mechanics, posed this famous question: If you put a cat in a sealed box with a device that has a 50% chance of killing the cat in the next hour, what will be the state of the cat when that time is up?
Particles and waves: The central mystery of quantum mechanics
One of the most amazing facts in physics is that everything in the universe, from light to electrons to atoms, behaves like both a particle and a wave at the same time. But how did physicists arrive at this mind-boggling conclusion?
Einstein's brilliant mistake: Entangled states
When you think about Einstein and physics, E=mc^2 is probably the first thing that comes to mind. But one of his greatest contributions to the field actually came in the form of an odd philosophical footnote in a 1935 paper he co-wrote -- which ended up being wrong.
What is the Heisenberg Uncertainty Principle?
The Heisenberg Uncertainty Principle states that you can never simultaneously know the exact position and the exact speed of an object.
What can Schrödinger's cat teach us about quantum mechanics?
The classical physics that we encounter in our everyday, macroscopic world is very different from the quantum physics that governs systems on a much smaller scale (like atoms).
Music and Math: The genius of Beethoven
How is it that Beethoven, who is celebrated as one of the most significant composers of all time, wrote many of his most beloved songs while going deaf?
The 2,400-year search for the atom
How do we know what matter is made of? The quest for the atom has been a long one, beginning 2,400 years ago with the work of a Greek philosopher and later continued by a Quaker and a few Nobel Prize-winning scientists. Theresa Doud details the history of atomic theory.?
What is Fat?
As the narrative goes, fat is bad. Well, it's actually more nuanced than that. The type of fat you eat is more impactful on your health than the quantity. George Zaidan examines triglycerides, the varied molecules that make up fat, and how to identify which types of fat you are consuming.
Why we love repetition in music
How many times does the chorus repeat in your favorite song? How many times have you listened to that chorus? Repetition in music isn’t just a feature of Western pop songs, either; it’s a global phenomenon. Why? Elizabeth Hellmuth Margulis walks us through the basic principles of the ‘exposure effect,’ detailing how repetition invites us into music as active participants, rather than passive listeners.
A different way to visualize rhythm
In standard notation, rhythm is indicated on a musical bar line. But there are other ways to visualize rhythm that can be more intuitive. John Varney describes the ‘wheel method’ of tracing rhythm and uses it to take us on a musical journey around the world.
Dead stuff: The secret ingredient in our food chain
When you picture the lowest levels of the food chain, you might imagine herbivores happily munching on lush, living green plants. But this idyllic image leaves out a huge (and slightly less appetizing) source of nourishment: dead stuff. John C. Moore details the "brown food chain," explaining how such unlikely delicacies as pond scum and animal poop contribute enormous amounts of energy to our ecosystems.
How do you know you exist?
How do you know you’re real? Is existence all just a big dream? Has some mad scientist duped us into simply believing that we exist? James Zucker investigates all of these questions (and more) in this mind-boggling tribute to René Descartes’s "Meditations on First Philosophy."
Population pyramids: Powerful predictors of the future
Population statistics are like crystal balls -- when examined closely, they can help predict a country's future (and give important clues about the past). Kim Preshoff explains how using a visual tool called a population pyramid helps policymakers and social scientists make sense of the statistics, using three different countries' pyramids as examples.
From the top of the food chain down: Rewilding our world
Our planet was once populated by megafauna, big top-of-the-food-chain predators that played their part in balancing our ecosystems. When those megafauna disappear, the result is a "trophic cascade," where every part of the ecosystem reacts to the loss. How can we stay in balance? George Monbiot suggests rewilding: putting wolves, lions and other predators back on top -- with surprising results.
The fundamentals of space-time (Part 2)
Light always travels at a speed of 299,792,458 meters per second. But if you're in motion too, you're going to perceive it as traveling even faster -- which isn't possible! In this second installment of a three-part series on space-time, CERN scientists Andrew Pontzen and Tom Whyntie use a space-time diagram to analyze the sometimes confounding motion of light.
The evolution of the human eye
The human eye is an amazing mechanism, able to detect anywhere from a few photons to a few quadrillion, or switch focus from the screen in front of you to the distant horizon in a third of a second. How did these complex structures evolve? Joshua Harvey details the 500 million year story of the human eye.
The Benefits of a Good Night's Sleep
It’s 4am, and the big test is in 8 hours. You’ve been studying for days, but you still don’t feel ready. Should you drink another cup of coffee and spend the next few hours cramming? Or should you go to sleep? Shai Marcu defends the latter option, showing how sleep restructures your brain in a way that’s crucial for how our memory works.
8 traits of successful people
Ten years of research and 500 face-to-face-interviews led Richard St. John to a collection of eight common traits in successful leaders around the world.
How spontaneous Brain Activity Keeps You Alive
The wheels in your brain are constantly turning, even when you're asleep or not paying attention. In fact, most of your brain’s activities are ones you’d never be aware of … unless they suddenly stopped. Nathan S. Jacobs takes us inside the always active, surprisingly spontaneous brain.
GPS location apps on a smartphone can be very handy when mapping a travel route or finding nearby events. But how does your smartphone know where you are?
The Distracted Mind
Everyone knows we're not supposed to multitask while driving, but do you know why? It turns out your brain literally can't focus on too much at once.
Why do buildings fall in earthquakes?
Earthquakes have always been a terrifying phenomenon, and they’ve become more deadly as our cities have grown — with collapsing buildings posing one of the largest risks. But why do buildings collapse in an earthquake? And how can it be prevented? Vicki V. May explains the physics of why it is not the sturdiest buildings, but the smartest, that will remain standing.
The Five Major World Religions
It's perfectly human to grapple with questions, like 'Where do we come from?' and 'How do I live a life of meaning?' These existential questions are central to the five major world religions -- and that's not all that connects these faiths. John Bellaimey explains the intertwined histories and cultures of Hinduism, Judaism, Buddhism, Christianity and Islam.
After drinking a few glasses of water on a hot day, you might be struck with a sudden ... urge. Behind that feeling are two bean-shaped organs that work as fine-tuned internal sensors.
Should you trust your first impression?
You can't help it; sometimes, you just get a bad feeling about someone that's hard to shake. So, what's happening in your brain when you make that critical (and often lasting) first judgment? Peter Mende-Siedlecki shares the social psychology of first impressions -- and why they may indicate that, deep down, people are basically good.
The last banana: A thought experiment in probability
Imagine a game of dice: if the biggest number rolled is one, two, three, or four, player 1 wins. If the biggest number rolled is five or six, player 2 wins. Who has the best probability of winning the game? Leonardo Barichello explains how probability holds the answer to this seemingly counterintuitive puzzle.
What does the pancreas do?
Beneath your ribs, you’ll find, among other things, the pancreas -- an organ that works a lot like a personal health coach.
The operating system of life
Humans, octopi and pine trees alike are all made up of cells, tiny but sophisticated systems that keep life going. Cells are almost like tiny factories run by robots, with the nucleus, DNA, proteins, lipids, and vitamins and minerals all playing critical roles. George Zaidan and Charles Morton lay out the blueprint of a cell and explain how biochemistry binds all life together.
Like an actor's script, a sheet of music instructs a musician on what to play (the pitch) and when to play it (the rhythm). Sheet music may look complicated, but once you've gotten the hang of a few simple elements like notes, bars and clefs, you're ready to rock.
Why sitting is bad for you
Sitting down for brief periods can help us recover from stress or recuperate from exercise. But nowadays, our lifestyles make us sit much more than we move around. Are our bodies built for such a sedentary existence?
What is Alzheimer's disease?
Alzheimer's disease is the most common cause of dementia, affecting over 40 million people worldwide. And though it was discovered over a century ago, scientists are still grappling for a cure.
What Happened to Antimatter?
Particles come in pairs, which is why there should be an equal amount of matter and antimatter in the universe. Yet, scientists have not been able to detect any in the visible universe. Where is this missing antimatter?
The real story behind Archimedes’ Eureka!
When you think of Archimedes’ Eureka moment, you probably imagine a man in a bathtub, right? As it turns out, there's much more to the story. Armand D'Angour tells the story of Archimedes' biggest assignment -- an enormous floating palace commissioned by a king -- that helped him find Eureka.
Underwater astonishments
David Gallo shows jaw-dropping footage of amazing sea creatures, including a color-shifting cuttlefish, a perfectly camouflaged octopus, and a Times Square's worth of neon light displays from fish who live in the blackest depths of the ocean.
Why are manhole covers round?
Why are most manhole covers round? Sure it makes them easy to roll, and slide into place in any alignment. But there’s another, more compelling reason, involving a peculiar geometric property of circles and other shapes. Marc Chamberland explains curves of constant width and Barbier’s theorem.
Questions of good and evil, right and wrong are commonly thought unanswerable by science. But Sam Harris argues that science can -- and should -- be an authority on moral issues, shaping human values and setting out what constitutes a good life.
What is leukemia?
Stem cells found in the bone marrow are crucial for our health because they are needed to become new blood cells that sustain and protect our bodies. But when the transformation goes wrong, harmful mutations can cause the cells to start replicating without control -- a type of cancer known as leukemia. Danilo Allegra and Dania Puggioni explain how this happens and how certain treatments provide hope for those suffering from the disease.
How to Make a Baby (In a Lab)
Infertility affects 1 in 8 couples worldwide. But in the last 40 years, more than 5 million babies have been born using in vitro fertilization (IVF). How does it work? Nassim Assefi and Brian A. Levine detail the science behind making a baby in a lab.
Some people love the feeling of cracking their knuckles, while others cringe at the sound. But what causes that trademark pop? And is it dangerous? Eleanor Nelsen gives the facts behind joint popping.
How does the thyroid manage your metabolism?
Nestled in the tissues of your neck is a small, unassuming organ that wields enormous power over your body: the thyroid. Emma Bryce explains how the thyroid, like the operations manager in a company, is tasked with making sure that all the cells in your body are working properly.
What really happens to the plastic you throw away
We’ve all been told that we should recycle plastic bottles and containers. But what actually happens to the plastic if we just throw it away? Emma Bryce traces the life cycles of three different plastic bottles, shedding light on the dangers these disposables present to our world.
The famously difficult green-eyed logic puzzle
One hundred green-eyed logicians have been imprisoned on an island by a mad dictator. Their only hope for freedom lies in the answer to one famously difficult logic puzzle. Can you solve it? Alex Gendler walks us through this green-eyed riddle.
What is a Calorie?
We hear about calories all the time: How many calories are in this cookie? How many are burned by doing 100 jumping jacks, or long-distance running, or fidgeting? But what is a calorie, really? And how many of them do we actually need? Emma Bryce explains how a few different factors should go into determining the recommended amount for each person.
2015 • Health
The benefits of good posture
Has anyone ever told you, “Stand up straight!” or scolded you for slouching at a family dinner? Comments like that might be annoying—but they’re not wrong. Your posture is the foundation for every movement your body makes and can determine how well your body adapts to the stresses on it. Murat Dalkilinç gives the pros of good posture.
2015 • Health
Who am I?
Throughout the history of mankind, the subject of identity has sent poets to the blank page, philosophers to the agora and seekers to the oracles. These murky waters of abstract thinking are tricky to navigate, so it’s probably fitting that to demonstrate the complexity, the Greek historian Plutarch used the story of a ship.
2015 • Brain
How smart are dolphins?
Dolphins are one of the smartest animal species on Earth. In fact, their encephalization quotient (their brain size compared to the average for their body size) is second only to humans. But exactly how smart are they? Lori Marino details some incredible facts about dolphins.
2015 • Nature
In previous decades, most news with global reach came from several major newspapers and networks with the resources to gather information directly. The speed with which information spreads now, however, has created the ideal conditions for something called circular reporting. Noah Tavlin sheds light on this phenomenon.
2015 • People
Can you ever travel from one place to another? Ancient Greek philosopher Zeno of Elea gave a convincing argument that all motion is impossible - but where's the flaw in his logic? Colm Kelleher illustrates how to resolve Zeno's Dichotomy Paradox.
2013 • Math
The mathematical secrets of Pascal’s triangle
Pascal’s triangle, which at first may just look like a neatly arranged stack of numbers, is actually a mathematical treasure trove. But what about it has so intrigued mathematicians the world over?
2015 • Math
3 tips to boost your confidence
When faced with a big challenge where potential failure seems to lurk at every corner, you’ve probably heard the advice, “Be more confident!” But where does confidence come from, and how can you get more of it? Here are three easy tips to boost your confidence.
2015 • Lifehack
The moon illusion
Have you noticed how the full moon looks bigger on the horizon than high overhead? Actually, the two images are exactly the same size -- so why do we perceive them differently? Scientists aren't sure, but there are plenty of intriguing theories. Andrew Vanden Heuvel unravels the details of focus, distance and proportion that contribute to this mystifying optical illusion.
2013 • Brain
The Arctic vs. the Antarctic
How can you tell the two poles apart? Where are the penguins? What about the bears? The Arctic pole is located in the Northern Hemisphere within the deep Arctic Ocean, while the Antarctic pole is smack in the middle of the ice-covered Antarctica. Camille Seaman describes how enterprising people and organisms have found ways to reside around both poles despite the frigid temperatures.
2013 • Nature
Would you opt for a life with no pain?
Imagine if you could plug your brain into a machine that would bring you ultimate pleasure for the rest of your life. The only catch? You have to permanently leave reality behind. Hayley Levitt and Bethany Rickwald explore Robert Nozick’s thought experiment that he called the Experience Machine.
2015 • Brain
The Sun’s surprising movement across the sky
Suppose you placed a camera at a fixed position, took a picture of the sky at the same time every day for an entire year, and overlaid all of the photos on top of each other. What would the sun look like in that combined image? A stationary dot? A circular path? Neither. Oddly enough, it makes a ‘figure 8’ pattern, known as the Sun’s analemma.
2015 • Astronomy
What would happen if you didn’t sleep?
In the United States, it’s estimated that 30 percent of adults and 66 percent of adolescents are regularly sleep-deprived. This isn’t just a minor inconvenience: staying awake can cause serious bodily harm. Claudia Aguirre shows what happens to your body and brain when you skip sleep.
2015 • Health
Our hard-wired stress response is designed to gives us the quick burst of heightened alertness and energy needed to perform our best. But stress isn’t all good. When activated too long or too often, stress can damage virtually every part of our body. Sharon Horesh Bergquist gives us a look at what goes on inside our body when we are chronically stressed.
2015 • Health
You might also like
Concerto - A Beethoven Journey with Leif Ove Andsnes
Filmed over the course of four years, award-winning director Phil Grabsky follows one of the world's greatest pianists, Leif Ove Andsnes, as he attempts, in a series of sold-out worldwide performances, to interpret one of the greatest sets of works for piano ever written - Beethoven's five piano concertos.
2015 • Music
Jazz Legends in Their Own Words
A journey into the BBC archives unearthing glorious performances and candid interviews from the golden age of jazz. Featuring some of the greatest names in American music, including the godfather of New Orleans jazz Louis Armstrong, the King of Swing Count Basie, Duke Ellington, Oscar Peterson, Dizzy Gillespie and Ella Fitzgerald. Broadcasted as part of the Old Masters, Rising Stars: Jazz On BBC Four season, this film unlocks the BBC archives to explore the words and music of some of the greatest names in jazz. The BBC soon moved on from Lord Reith's proclamation, made in the 1930s, that jazz was "a filthy product of modernity", and invited some of the legends from the golden age of American jazz to perform and talk on British television. This film is a series of revealing portraits, from Louis Armstrong, jazz's first great soloist and global ambassador, to Duke Ellington, the ever-suave bandleader and composer who brought a new sophistication and ambition to the music. Count Basie is sheer swing, Dizzy Gillespie provoked a musical revolution with bebop, and Ella Fitzgerald is just incomparable. Through long-forgotten archive and specially shot interviews, Jazz Legends In Their Own Words tells the story of an art form that has been called "America's gift to the world".
2014 • Music
The Joy of Mozart
Tom Service plunges into the life and times of Mozart to try and rediscover the greatness and humanity of the living man in his moment.
The Power of the Pentatonic Scale
We don't know much about the human brain on music. Do people instinctively know the sound patterns of the pentatonic scale? Is there a base level of musical knowledge in all of us, just waiting to be tapped? Or is the pentatonic scale simply so common in Western music that it has become ingrained in all of our minds? Improvisational genius Bobby McFerrin uses audience participation to demonstrate the power of the pentatonic scale - or at least the audience's familiarity with it.
2011 • Music
Why we love repetition in music
How many times does the chorus repeat in your favorite song? How many times have you listened to that chorus? Repetition in music isn’t just a feature of Western pop songs, either; it’s a global phenomenon. Why? Elizabeth Hellmuth Margulis walks us through the basic principles of the ‘exposure effect,’ detailing how repetition invites us into music as active participants, rather than passive listeners.
A different way to visualize rhythm
In standard notation, rhythm is indicated on a musical bar line. But there are other ways to visualize rhythm that can be more intuitive. John Varney describes the ‘wheel method’ of tracing rhythm and uses it to take us on a musical journey around the world. | 5,589 | 26,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.936147 |
https://www.jiskha.com/questions/509161/the-earths-atmospheric-pressure-p-in-terms-of-height-above-sea-level-is-often-modeled | 1,618,470,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00396.warc.gz | 945,161,262 | 5,346 | # precalculus
The earth's atmospheric pressure, P, in terms of height above sea level is often modeled by an exponential decay function. The pressure at sea level is 1013 millibars and the pressure decreases by 14% for every kilometer above sea level.
(a) What is the atmospheric pressure at 50 km?
(b) Estimate the altitude h at which the atmospheric pressure equals 950 millibars.
1. 👍
2. 👎
3. 👁
1. .86=1e^k(1km)
so the drill is to find k
ln .86=k
k=-0.15082289
so at 50km
pressure= 1013e^(-.150*50)=.538millbars
check that.
1. 👍
2. 👎
👤
bobpursley
2. nope their both wrong.
1. 👍
2. 👎
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A163152 Primes of the form PartitionsP[p], p are prime numbers. 1
2, 3, 7, 101, 80630964769, 1394313503224447816939, 87674799670795146675673859587, 62607220478448273296879161314388228250413, 79074320470247928120049519839632230336234433216761019, 77355497906663686399579348109210219558359416885618588905259034616641337958059 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS MATHEMATICA f[n_]:=PartitionsP[n]; lst={}; Do[p=Prime[n]; If[PrimeQ[f[p]], AppendTo[lst, f[p]]], {n, 6!}]; lst Select[PartitionsP[Prime[Range[1000]]], PrimeQ] (* Harvey P. Dale, May 16 2020 *) PROG (PARI) forprime(p=2, 1e4, k=numbpart(p); if(isprime(k), print1(k", "))) CROSSREFS Cf. A098145, A163150, A163151, A049575. Sequence in context: A062529 A308917 A058443 * A088120 A230778 A111870 Adjacent sequences: A163149 A163150 A163151 * A163153 A163154 A163155 KEYWORD nonn AUTHOR Vladimir Joseph Stephan Orlovsky, Jul 21 2009 EXTENSIONS Program by Charles R Greathouse IV, Oct 12 2009 More terms from Harvey P. Dale, May 16 2020 STATUS approved
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Last modified April 16 12:45 EDT 2021. Contains 343037 sequences. (Running on oeis4.) | 506 | 1,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-17 | latest | en | 0.58461 |
https://gitlab.tue.nl/stem/tflab/-/commit/af04fa2c70f0ac86dfdba001e2cafe868d6c6665 | 1,643,076,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00392.warc.gz | 325,176,725 | 71,704 | Commit af04fa2c by Turnhout, M.C. van
### first ani results and analysis
parent 0f41d0d4
... ... @@ -422,4 +422,13 @@ In isotropic films, the radius of curvature $R$ as estimated from projected edge \normalsize \center ~~\def\svgwidth{0.9\linewidth}\includesvg{pics/lkaniexamcolorbar}\\ \caption{The three components of the local curvature vector along the deformed edges of the films in figure \ref{lgmu}. With the left edge on the left, the right edge at the right, and from top to bottom $\kappa_x$, $\kappa_y$, and $\kappa_z$. \label{fig:anikappaxyz}} \end{figure} \ No newline at end of file \end{figure} \input{anispace_msig_eq.tex} \begin{table} \center\small \caption{Multiple linear regression model fitted to $R_\varepsilon$ for $\sigma_x = 0.025$\,MPa for the 132 random models. With the coefficients, their 95\,\% confidence intervals (CI), the amount of variation in the model explained by the parameter (SS/SS$_\text{tot}$), and the $p$-values for the coefficients. The last two columns contain the predicted range in the simulations based on the parameter space, and that predicted range divided by the complete range of the variable found in the simulations.\label{isospace_dR_tab}} \input{anispace_msig_tab.tex} \end{table} \ No newline at end of file
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2, 1.000000e-03, 3.910929, 10.906506, 4.000763, 4.001625, 0.064671, 0.108419, 0.026309, 0.033722, NaN, NaN 3, 2.000000e-03, 3.668385, 5.599594, 4.001358, 4.003055, 0.129783, 0.211638, 0.052089, 0.061928, NaN, NaN 4, 3.000000e-03, 3.343532, 3.927612, 4.001818, 4.004314, 0.189468, 0.306880, 0.077310, 0.085714, NaN, NaN 5, 4.000000e-03, 3.004458, 3.143215, 4.002157, 4.005409, 0.242954, 0.393518, 0.101809, 0.105648, NaN, NaN 6, 5.000000e-03, 2.693982, 2.705479, 4.002396, 4.006366, 0.291239, 0.472383, 0.124817, 0.122482, NaN, NaN 7, 6.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 8, 7.000000e-03, 2.246932, 2.246932, 4.002659, 4.007953, 0.377405, 0.610484, 0.162348, 0.149306, NaN, NaN 9, 8.000000e-03, 2.092803, 2.092803, 4.002716, 4.008618, 0.416728, 0.671484, 0.177226, 0.160121, NaN, NaN 10, 9.000000e-03, 1.970938, 1.970938, 4.002745, 4.009216, 0.454047, 0.728178, 0.190454, 0.169622, NaN, NaN 11, 1.000000e-02, 1.903921, 1.903921, 4.002756, 4.009760, 0.490624, 0.781330, 0.200073, 0.177950, NaN, NaN 12, 1.100000e-02, 1.870092, 1.870092, 4.002756, 4.010250, 0.525475, 0.830916, 0.208897, 0.185546, NaN, NaN 13, 1.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 14, 1.300000e-02, 1.868425, 1.868425, 4.002742, 4.011114, 0.590437, 0.922560, 0.226309, 0.198664, NaN, NaN 15, 1.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 16, 1.500000e-02, 1.910050, 1.910050, 4.002709, 4.011849, 0.649400, 1.005941, 0.244449, 0.209964, NaN, NaN 17, 1.600000e-02, 1.932138, 1.932138, 4.002683, 4.012197, 0.679065, 1.046991, 0.251975, 0.216462, NaN, NaN 18, 1.700000e-02, 1.961181, 1.961181, 4.002649, 4.012474, 0.705296, 1.082641, 0.257040, 0.220038, NaN, NaN 19, 1.800000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 20, 1.900000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 21, 2.000000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 22, 2.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 23, 2.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.944142, 13.789845, 4.000316, 4.000819, 0.056641, 0.080919, 0.014412, 0.026392, NaN, NaN 3, 2.000000e-03, 3.786185, 7.005787, 4.000540, 4.001604, 0.110710, 0.160214, 0.029637, 0.051109, NaN, NaN 4, 3.000000e-03, 3.558036, 4.829121, 4.000689, 4.002321, 0.160565, 0.234956, 0.044834, 0.073366, NaN, NaN 5, 4.000000e-03, 3.296326, 3.786087, 4.000765, 4.002959, 0.205651, 0.304235, 0.060096, 0.093120, NaN, NaN 6, 5.000000e-03, 3.029858, 3.187656, 4.000773, 4.003519, 0.246297, 0.368189, 0.075478, 0.110587, NaN, NaN 7, 6.000000e-03, 2.777290, 2.807045, 4.000727, 4.004013, 0.283149, 0.427393, 0.091033, 0.126038, NaN, NaN 8, 7.000000e-03, 2.551742, 2.551746, 4.000636, 4.004449, 0.316694, 0.482299, 0.106837, 0.139729, NaN, NaN 9, 8.000000e-03, 2.369324, 2.369324, 4.000510, 4.004836, 0.347585, 0.533581, 0.122985, 0.151905, NaN, NaN 10, 9.000000e-03, 2.221432, 2.221432, 4.000360, 4.005180, 0.376181, 0.581619, 0.139607, 0.162744, NaN, NaN 11, 1.000000e-02, 2.098770, 2.098770, 4.000191, 4.005487, 0.403390, 0.626719, 0.155226, 0.172360, NaN, NaN 12, 1.100000e-02, 1.995546, 1.995546, 4.000014, 4.005762, 0.430820, 0.669201, 0.166910, 0.180845, NaN, NaN 13, 1.200000e-02, 1.908322, 1.908322, 3.999840, 4.006008, 0.457940, 0.709168, 0.177145, 0.188197, NaN, NaN 14, 1.300000e-02, 1.836102, 1.836102, 3.999685, 4.006228, 0.486268, 0.746833, 0.183205, 0.194353, NaN, NaN 15, 1.400000e-02, 1.787801, 1.787801, 3.999568, 4.006427, 0.516083, 0.782474, 0.184961, 0.199359, NaN, NaN 16, 1.500000e-02, 1.774235, 1.774235, 3.999490, 4.006603, 0.544649, 0.816110, 0.188984, 0.203883, NaN, NaN 17, 1.600000e-02, 1.772079, 1.772079, 3.999431, 4.006764, 0.571769, 0.848622, 0.194289, 0.208131, NaN, NaN 18, 1.700000e-02, 1.778666, 1.778666, 3.999374, 4.006905, 0.598546, 0.879883, 0.196088, 0.212247, NaN, NaN 19, 1.800000e-02, 1.789949, 1.789949, 3.999306, 4.007031, 0.623267, 0.910330, 0.200184, 0.216566, NaN, NaN 20, 1.900000e-02, 1.805229, 1.805229, 3.999228, 4.007151, 0.646589, 0.940438, 0.205791, 0.220841, NaN, NaN 21, 2.000000e-02, 1.824772, 1.824772, 3.999144, 4.007249, 0.668296, 0.968956, 0.212233, 0.224935, NaN, NaN 22, 2.100000e-02, 1.844634, 1.844634, 3.999047, 4.007341, 0.689252, 0.997340, 0.219283, 0.229173, NaN, NaN 23, 2.200000e-02, 1.865212, 1.865212, 3.998940, 4.007424, 0.709518, 1.025282, 0.226642, 0.233379, NaN, NaN 24, 2.300000e-02, 1.887046, 1.887046, 3.998834, 4.007489, 0.728381, 1.051536, 0.234354, 0.237180, NaN, NaN 25, 2.400000e-02, 1.907471, 1.907471, 3.998718, 4.007547, 0.746946, 1.077687, 0.241967, 0.241026, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.914303, 11.120605, 4.000596, 4.001064, 0.075266, 0.101490, 0.018330, 0.021633, NaN, NaN 3, 2.000000e-03, 3.683164, 5.731964, 4.001059, 4.002021, 0.145290, 0.197488, 0.036226, 0.042425, NaN, NaN 4, 3.000000e-03, 3.369944, 4.013538, 4.001379, 4.002842, 0.207035, 0.284870, 0.052985, 0.061800, NaN, NaN 5, 4.000000e-03, 3.033954, 3.195000, 4.001561, 4.003526, 0.260330, 0.363109, 0.068645, 0.079792, NaN, NaN 6, 5.000000e-03, 2.708392, 2.722359, 4.001640, 4.004103, 0.307175, 0.434020, 0.083432, 0.096478, NaN, NaN 7, 6.000000e-03, 2.443841, 2.443841, 4.001628, 4.004586, 0.348244, 0.498169, 0.097525, 0.111872, NaN, NaN 8, 7.000000e-03, 2.238729, 2.238729, 4.001550, 4.004996, 0.385068, 0.557034, 0.111081, 0.126074, NaN, NaN 9, 8.000000e-03, 2.073256, 2.073256, 4.001420, 4.005346, 0.418482, 0.611448, 0.124213, 0.139155, NaN, NaN 10, 9.000000e-03, 1.937101, 1.937101, 4.001249, 4.005645, 0.449231, 0.662202, 0.137000, 0.151221, NaN, NaN 11, 1.000000e-02, 1.827920, 1.827920, 4.001046, 4.005902, 0.477793, 0.709809, 0.149495, 0.162382, NaN, NaN 12, 1.100000e-02, 1.738702, 1.738702, 4.000815, 4.006121, 0.504448, 0.754582, 0.161729, 0.172723, NaN, NaN 13, 1.200000e-02, 1.664441, 1.664441, 4.000565, 4.006311, 0.529670, 0.797202, 0.173803, 0.182311, NaN, NaN 14, 1.300000e-02, 1.601864, 1.601864, 4.000294, 4.006469, 0.553397, 0.837472, 0.185667, 0.191250, NaN, NaN 15, 1.400000e-02, 1.547354, 1.547354, 4.000012, 4.006609, 0.576249, 0.876315, 0.197452, 0.199629, NaN, NaN 16, 1.500000e-02, 1.500292, 1.500292, 3.999715, 4.006722, 0.597862, 0.913204, 0.209118, 0.207460, NaN, NaN 17, 1.600000e-02, 1.504944, 1.504944, 3.999406, 4.006814, 0.618522, 0.948536, 0.220719, 0.214820, NaN, NaN 18, 1.700000e-02, 1.524582, 1.524582, 3.999092, 4.006893, 0.638571, 0.982844, 0.232330, 0.221766, NaN, NaN 19, 1.800000e-02, 1.548946, 1.548946, 3.998766, 4.006948, 0.657576, 1.015467, 0.243933, 0.228291, NaN, NaN 20, 1.900000e-02, 1.576254, 1.576254, 3.998441, 4.006999, 0.676370, 1.047673, 0.255599, 0.234490, NaN, NaN 21, 2.000000e-02, 1.605680, 1.605680, 3.998103, 4.007026, 0.694053, 1.078411, 0.267369, 0.239947, NaN, NaN 22, 2.100000e-02, 1.633078, 1.633078, 3.997767, 4.007039, 0.712002, 1.107684, 0.277318, 0.245778, NaN, NaN 23, 2.200000e-02, 1.660137, 1.660137, 3.997435, 4.007056, 0.730582, 1.137315, 0.286049, 0.251031, NaN, NaN 24, 2.300000e-02, 1.685575, 1.685575, 3.997098, 4.007052, 0.748809, 1.165514, 0.293790, 0.255921, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.865093, 8.846569, 4.001009, 4.001056, 0.114925, 0.118027, 0.014357, 0.011832, NaN, NaN 3, 2.000000e-03, 3.512287, 4.588509, 4.001839, 4.001939, 0.221003, 0.227659, 0.027585, 0.022454, NaN, NaN 4, 3.000000e-03, 3.051655, 3.227139, 4.002427, 4.002591, 0.313354, 0.324316, 0.038420, 0.030657, NaN, NaN 5, 4.000000e-03, 2.567975, 2.568245, 4.002798, 4.003040, 0.392782, 0.409047, 0.047278, 0.036755, NaN, NaN 6, 5.000000e-03, 2.201456, 2.201456, 4.002988, 4.003329, 0.460754, 0.483745, 0.054668, 0.041262, NaN, NaN 7, 6.000000e-03, 1.955026, 1.955026, 4.003037, 4.003504, 0.519462, 0.550995, 0.061104, 0.044664, NaN, NaN 8, 7.000000e-03, 1.778073, 1.778073, 4.002970, 4.003599, 0.570010, 0.612395, 0.066927, 0.047433, NaN, NaN 9, 8.000000e-03, 1.645225, 1.645225, 4.002802, 4.003634, 0.613012, 0.668942, 0.072362, 0.050120, NaN, NaN 10, 9.000000e-03, 1.541614, 1.541614, 4.002550, 4.003632, 0.649284, 0.721589, 0.077586, 0.053465, NaN, NaN 11, 1.000000e-02, 1.459977, 1.459977, 4.002222, 4.003603, 0.678972, 0.770449, 0.083000, 0.058393, NaN, NaN 12, 1.100000e-02, 1.391661, 1.391661, 4.001851, 4.003551, 0.704870, 0.816213, 0.087698, 0.064960, NaN, NaN 13, 1.200000e-02, 1.335007, 1.335007, 4.001434, 4.003481, 0.726390, 0.858643, 0.092754, 0.073642, NaN, NaN 14, 1.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 15, 1.400000e-02, 1.247117, 1.247117, 4.000533, 4.003285, 0.762650, 0.935226, 0.102892, 0.094751, NaN, NaN 16, 1.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 17, 1.600000e-02, 1.182199, 1.182199, 3.999576, 4.003024, 0.792912, 1.002593, 0.113984, 0.118390, NaN, NaN 18, 1.700000e-02, 1.151880, 1.151880, 3.999104, 4.002869, 0.809187, 1.035186, 0.118463, 0.128607, NaN, NaN 19, 1.800000e-02, 1.126687, 1.126687, 3.998618, 4.002702, 0.823788, 1.065603, 0.123712, 0.139350, NaN, NaN 20, 1.900000e-02, 1.103093, 1.103093, 3.998130, 4.002529, 0.838041, 1.095069, 0.129085, 0.149932, NaN, NaN 21, 2.000000e-02, 1.082803, 1.082803, 3.997640, 4.002344, 0.852111, 1.123526, 0.134434, 0.160003, NaN, NaN 22, 2.100000e-02, 1.062929, 1.062929, 3.997149, 4.002151, 0.865999, 1.151159, 0.139787, 0.169689, NaN, NaN 23, 2.200000e-02, 1.044826, 1.044826, 3.996655, 4.001947, 0.879706, 1.177943, 0.145119, 0.178929, NaN, NaN 24, 2.300000e-02, 1.027845, 1.027845, 3.996161, 4.001737, 0.893283, 1.204163, 0.150457, 0.187842, NaN, NaN 25, 2.400000e-02, 1.011797, 1.011797, 3.995665, 4.001519, 0.906701, 1.229713, 0.155765, 0.196384, NaN, NaN 26, 2.500000e-02, 0.997409, 0.997409, 3.995167, 4.001294, 0.919978, 1.254689, 0.161042, 0.204601, NaN, NaN
2, 1.000000e-03, 3.975163, 20.704460, 4.000240, 4.001033, 0.030672, 0.055631, 0.013744, 0.036227, NaN, NaN 3, 2.000000e-03, 3.901391, 10.361862, 4.000486, 4.002029, 0.071245, 0.112861, 0.027957, 0.063285, NaN, NaN 4, 3.000000e-03, 3.783976, 6.969261, 4.000810, 4.002963, 0.113252, 0.170836, 0.041679, 0.085245, NaN, NaN 5, 4.000000e-03, 3.633049, 5.315685, 4.001162, 4.003852, 0.154420, 0.228805, 0.054869, 0.103493, NaN, NaN 6, 5.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 7, 6.000000e-03, 3.288552, 3.764153, 4.001863, 4.005498, 0.231947, 0.341039, 0.080865, 0.131435, NaN, NaN 8, 7.000000e-03, 3.120057, 3.360271, 4.002197, 4.006252, 0.268030, 0.394208, 0.093752, 0.142338, NaN, NaN 9, 8.000000e-03, 2.965333, 3.077793, 4.002516, 4.006962, 0.302351, 0.445126, 0.106463, 0.151808, NaN, NaN 10, 9.000000e-03, 2.830092, 2.876759, 4.002817, 4.007627, 0.334770, 0.493483, 0.118847, 0.160096, NaN, NaN 11, 1.000000e-02, 2.714939, 2.730179, 4.003102, 4.008253, 0.365601, 0.539621, 0.130897, 0.167512, NaN, NaN 12, 1.100000e-02, 2.620403, 2.623177, 4.003371, 4.008843, 0.394869, 0.583512, 0.142546, 0.174216, NaN, NaN 13, 1.200000e-02, 2.545547, 2.545547, 4.003625, 4.009398, 0.422616, 0.625160, 0.153745, 0.180350, NaN, NaN 14, 1.300000e-02, 2.489713, 2.489713, 4.003866, 4.009922, 0.448750, 0.664676, 0.164447, 0.185735, NaN, NaN 15, 1.400000e-02, 2.444606, 2.444606, 4.004092, 4.010429, 0.474590, 0.703372, 0.175021, 0.191257, NaN, NaN 16, 1.500000e-02, 2.413808, 2.413808, 4.004308, 4.010911, 0.499137, 0.740335, 0.185158, 0.196223, NaN, NaN 17, 1.600000e-02, 2.393949, 2.393949, 4.004512, 4.011370, 0.522635, 0.775760, 0.194904, 0.200869, NaN, NaN 18, 1.700000e-02, 2.382673, 2.382673, 4.004707, 4.011809, 0.545200, 0.809758, 0.204289, 0.205274, NaN, NaN 19, 1.800000e-02, 2.377057, 2.377057, 4.004892, 4.012234, 0.567328, 0.843043, 0.213489, 0.209573, NaN, NaN 20, 1.900000e-02, 2.378234, 2.378234, 4.005069, 4.012630, 0.587925, 0.874092, 0.222096, 0.213481, NaN, NaN 21, 2.000000e-02, 2.381722, 2.381722, 4.005238, 4.013020, 0.608362, 0.904904, 0.230624, 0.217312, NaN, NaN 22, 2.100000e-02, 2.387542, 2.387542, 4.005401, 4.013393, 0.628117, 0.934740, 0.238874, 0.220934, NaN, NaN 23, 2.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.882954, 9.504128, 4.000923, 4.001636, 0.082547, 0.122868, 0.026461, 0.028289, NaN, NaN 3, 2.000000e-03, 3.578323, 4.947948, 4.001641, 4.003035, 0.160743, 0.236518, 0.051336, 0.052796, NaN, NaN 4, 3.000000e-03, 3.187829, 3.507900, 4.002163, 4.004208, 0.229542, 0.338440, 0.074536, 0.074357, NaN, NaN 5, 4.000000e-03, 2.793807, 2.828407, 4.002515, 4.005185, 0.289216, 0.429193, 0.095985, 0.093407, NaN, NaN 6, 5.000000e-03, 2.467429, 2.467429, 4.002730, 4.006003, 0.341290, 0.510407, 0.115882, 0.110395, NaN, NaN 7, 6.000000e-03, 2.224873, 2.224873, 4.002839, 4.006696, 0.387422, 0.583893, 0.134503, 0.125655, NaN, NaN 8, 7.000000e-03, 2.036441, 2.036441, 4.002864, 4.007284, 0.428657, 0.650762, 0.152064, 0.139398, NaN, NaN 9, 8.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 10, 9.000000e-03, 1.772687, 1.772687, 4.002735, 4.008228, 0.500678, 0.769552, 0.184751, 0.163301, NaN, NaN 11, 1.000000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 12, 1.100000e-02, 1.605993, 1.605993, 4.002443, 4.008947, 0.562800, 0.873598, 0.215162, 0.183352, NaN, NaN 13, 1.200000e-02, 1.577325, 1.577325, 4.002249, 4.009235, 0.590862, 0.920933, 0.229679, 0.192259, NaN, NaN 14, 1.300000e-02, 1.592194, 1.592194, 4.002038, 4.009494, 0.617772, 0.966385, 0.243910, 0.200596, NaN, NaN 15, 1.400000e-02, 1.617594, 1.617594, 4.001804, 4.009717, 0.643175, 1.009455, 0.257851, 0.208342, NaN, NaN 16, 1.500000e-02, 1.648465, 1.648465, 4.001553, 4.009910, 0.667402, 1.050613, 0.271566, 0.215590, NaN, NaN 17, 1.600000e-02, 1.681391, 1.681391, 4.001291, 4.010081, 0.690770, 1.090314, 0.285082, 0.222418, NaN, NaN 18, 1.700000e-02, 1.713632, 1.713632, 4.001016, 4.010227, 0.713418, 1.128549, 0.298023, 0.228823, NaN, NaN 19, 1.800000e-02, 1.742692, 1.742692, 4.000731, 4.010351, 0.736124, 1.165261, 0.308537, 0.234878, NaN, NaN 20, 1.900000e-02, 1.768126, 1.768126, 4.000440, 4.010460, 0.758800, 1.200978, 0.317720, 0.240615, NaN, NaN 21, 2.000000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 22, 2.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 23, 2.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.990740, 33.928299, 4.000207, 4.000832, 0.021286, 0.031308, 0.018739, 0.036749, NaN, NaN 3, 2.000000e-03, 3.962489, 16.839123, 4.000558, 4.001725, 0.048872, 0.065501, 0.036598, 0.064015, NaN, NaN 4, 3.000000e-03, 3.914439, 11.129475, 4.000991, 4.002580, 0.075976, 0.102483, 0.050020, 0.086528, NaN, NaN 5, 4.000000e-03, 3.846467, 8.286762, 4.001438, 4.003387, 0.103184, 0.141852, 0.060915, 0.105143, NaN, NaN 6, 5.000000e-03, 3.760466, 6.612396, 4.001880, 4.004153, 0.131087, 0.182884, 0.070363, 0.120619, NaN, NaN 7, 6.000000e-03, 3.660207, 5.529976, 4.002310, 4.004891, 0.159426, 0.224867, 0.079320, 0.133628, NaN, NaN 8, 7.000000e-03, 3.550701, 4.788113, 4.002726, 4.005605, 0.187862, 0.267160, 0.088164, 0.144707, NaN, NaN 9, 8.000000e-03, 3.437338, 4.258919, 4.003127, 4.006298, 0.216106, 0.309226, 0.096926, 0.154272, NaN, NaN 10, 9.000000e-03, 3.325262, 3.871155, 4.003515, 4.006969, 0.243551, 0.350613, 0.106335, 0.162631, NaN, NaN 11, 1.000000e-02, 3.218208, 3.580093, 4.003890, 4.007618, 0.270172, 0.391108, 0.116212, 0.170056, NaN, NaN 12, 1.100000e-02, 3.119606, 3.359413, 4.004253, 4.008244, 0.296187, 0.430434, 0.125463, 0.176686, NaN, NaN 13, 1.200000e-02, 3.031657, 3.190899, 4.004604, 4.008844, 0.321227, 0.468332, 0.134582, 0.182686, NaN, NaN 14, 1.300000e-02, 2.955485, 3.061867, 4.004946, 4.009418, 0.345093, 0.504670, 0.143781, 0.188144, NaN, NaN 15, 1.400000e-02, 2.888236, 2.958870, 4.005277, 4.009982, 0.368551, 0.540462, 0.153133, 0.193283, NaN, NaN 16, 1.500000e-02, 2.835084, 2.883530, 4.005603, 4.010514, 0.390351, 0.574118, 0.162188, 0.197779, NaN, NaN 17, 1.600000e-02, 2.785711, 2.817917, 4.005914, 4.011056, 0.413096, 0.608731, 0.171389, 0.202402, NaN, NaN 18, 1.700000e-02, 2.751426, 2.774525, 4.006222, 4.011553, 0.433518, 0.640115, 0.179834, 0.206393, NaN, NaN 19, 1.800000e-02, 2.723119, 2.740002, 4.006520, 4.012042, 0.453871, 0.671033, 0.188087, 0.210405, NaN, NaN 20, 1.900000e-02, 2.701451, 2.714252, 4.006813, 4.012525, 0.473842, 0.701514, 0.196169, 0.214113, NaN, NaN 21, 2.000000e-02, 2.686561, 2.696896, 4.007098, 4.012986, 0.492869, 0.730574, 0.203989, 0.217632, NaN, NaN 22, 2.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 23, 2.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.858676, 8.641338, 4.001021, 4.001311, 0.108563, 0.127543, 0.021257, 0.012408, NaN, NaN 3, 2.000000e-03, 3.499338, 4.526424, 4.001811, 4.002396, 0.205923, 0.243727, 0.040256, 0.024252, NaN, NaN 4, 3.000000e-03, 3.044893, 3.214741, 4.002342, 4.003233, 0.288295, 0.345474, 0.056673, 0.035659, NaN, NaN 5, 4.000000e-03, 2.593164, 2.594280, 4.002652, 4.003858, 0.356932, 0.433952, 0.070840, 0.047019, NaN, NaN 6, 5.000000e-03, 2.248868, 2.248868, 4.002792, 4.004322, 0.414521, 0.511661, 0.083311, 0.058525, NaN, NaN 7, 6.000000e-03, 2.007389, 2.007389, 4.002803, 4.004666, 0.463597, 0.580878, 0.094536, 0.070215, NaN, NaN 8, 7.000000e-03, 1.834800, 1.834800, 4.002719, 4.004919, 0.506154, 0.643390, 0.104846, 0.081999, NaN, NaN 9, 8.000000e-03, 1.704589, 1.704589, 4.002561, 4.005100, 0.543701, 0.700525, 0.114472, 0.093760, NaN, NaN 10, 9.000000e-03, 1.603162, 1.603162, 4.002346, 4.005225, 0.577357, 0.753235, 0.123572, 0.105396, NaN, NaN 11, 1.000000e-02, 1.521265, 1.521265, 4.002087, 4.005305, 0.607982, 0.802420, 0.132286, 0.116753, NaN, NaN 12, 1.100000e-02, 1.453470, 1.453470, 4.001792, 4.005344, 0.636120, 0.848397, 0.140657, 0.127794, NaN, NaN 13, 1.200000e-02, 1.396188, 1.396188, 4.001470, 4.005352, 0.662354, 0.891879, 0.148787, 0.138474, NaN, NaN 14, 1.300000e-02, 1.346677, 1.346677, 4.001125, 4.005334, 0.687078, 0.933269, 0.156726, 0.148791, NaN, NaN 15, 1.400000e-02, 1.304000, 1.304000, 4.000761, 4.005291, 0.710443, 0.972653, 0.164466, 0.158697, NaN, NaN 16, 1.500000e-02, 1.265918, 1.265918, 4.000381, 4.005227, 0.732712, 1.010355, 0.172051, 0.168213, NaN, NaN 17, 1.600000e-02, 1.232672, 1.232672, 3.999986, 4.005143, 0.753996, 1.046492, 0.179483, 0.177331, NaN, NaN 18, 1.700000e-02, 1.202492, 1.202492, 3.999579, 4.005041, 0.774471, 1.081341, 0.186796, 0.186062, NaN, NaN 19, 1.800000e-02, 1.175391, 1.175391, 3.999162, 4.004926, 0.794276, 1.114992, 0.193979, 0.194509, NaN, NaN 20, 1.900000e-02, 1.149969, 1.149969, 3.998736, 4.004795, 0.813412, 1.147528, 0.201052, 0.202600, NaN, NaN 21, 2.000000e-02, 1.161370, 1.161370, 3.998299, 4.004648, 0.831870, 1.178907, 0.208005, 0.210344, NaN, NaN 22, 2.100000e-02, 1.201099, 1.201099, 3.997857, 4.004492, 0.849948, 1.209588, 0.214881, 0.217852, NaN, NaN 23, 2.200000e-02, 1.237733, 1.237733, 3.997402, 4.004316, 0.867268, 1.238977, 0.221609, 0.225002, NaN, NaN 24, 2.300000e-02, 1.272642, 1.272642, 3.996951, 4.004145, 0.884728, 1.268500, 0.228363, 0.232086, NaN, NaN 25, 2.400000e-02, 1.303434, 1.303434, 3.996491, 4.003959, 0.901659, 1.297154, 0.235000, 0.238853, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.962829, 16.916275, 4.000020, 4.000441, 0.046118, 0.061285, 0.008456, 0.028807, NaN, NaN 3, 2.000000e-03, 3.853430, 8.483505, 4.000001, 4.000957, 0.089659, 0.124438, 0.017190, 0.056531, NaN, NaN 4, 3.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 5, 4.000000e-03, 3.485900, 4.464436, 3.999862, 4.001887, 0.168968, 0.243653, 0.037520, 0.103134, NaN, NaN 6, 5.000000e-03, 3.270176, 3.713608, 3.999743, 4.002293, 0.204000, 0.298467, 0.049420, 0.121677, NaN, NaN 7, 6.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 8, 7.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 9, 8.000000e-03, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 10, 9.000000e-03, 2.498433, 2.498433, 3.999019, 4.003526, 0.318407, 0.486649, 0.111510, 0.173309, NaN, NaN 11, 1.000000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 12, 1.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 13, 1.200000e-02, 2.276686, 2.276686, 3.999166, 4.004074, 0.418237, 0.583843, 0.132333, 0.178767, NaN, NaN 14, 1.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 15, 1.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 16, 1.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 17, 1.600000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 18, 1.700000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 19, 1.800000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 20, 1.900000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 21, 2.000000e-02, 1.950987, 1.950987, 3.998888, 4.005037, 0.599954, 0.827197, 0.192883, 0.217193, NaN, NaN 22, 2.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 23, 2.200000e-02, 1.948603, 1.948603, 3.998715, 4.005159, 0.637114, 0.879443, 0.205730, 0.225591, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.992382, 37.407494, 4.000024, 4.000511, 0.020597, 0.027037, 0.018685, 0.035111, NaN, NaN 3, 2.000000e-03, 3.971206, 19.226074, 4.000211, 4.001135, 0.046047, 0.055362, 0.037663, 0.061525, NaN, NaN 4, 3.000000e-03, 3.934102, 12.691116, 4.000484, 4.001762, 0.070906, 0.087437, 0.051827, 0.083518, NaN, NaN 5, 4.000000e-03, 3.878478, 9.325886, 4.000775, 4.002360, 0.096547, 0.123178, 0.062523, 0.101885, NaN, NaN 6, 5.000000e-03, 3.803329, 7.309532, 4.001065, 4.002928, 0.123367, 0.161837, 0.071779, 0.117290, NaN, NaN 7, 6.000000e-03, 3.710204, 5.999851, 4.001345, 4.003476, 0.151387, 0.202463, 0.080342, 0.130336, NaN, NaN 8, 7.000000e-03, 3.602898, 5.103779, 4.001612, 4.004011, 0.180045, 0.244172, 0.088831, 0.141521, NaN, NaN 9, 8.000000e-03, 3.486524, 4.467297, 4.001864, 4.004534, 0.209052, 0.286212, 0.096956, 0.151247, NaN, NaN 10, 9.000000e-03, 3.366863, 4.003239, 4.002104, 4.005043, 0.237417, 0.327907, 0.105870, 0.159817, NaN, NaN 11, 1.000000e-02, 3.249327, 3.658390, 4.002333, 4.005534, 0.265393, 0.368724, 0.114118, 0.167453, NaN, NaN 12, 1.100000e-02, 3.137623, 3.396893, 4.002551, 4.006006, 0.292341, 0.408458, 0.122857, 0.174337, NaN, NaN 13, 1.200000e-02, 3.037310, 3.201008, 4.002759, 4.006446, 0.317626, 0.446004, 0.131874, 0.180669, NaN, NaN 14, 1.300000e-02, 2.942164, 3.040791, 4.002960, 4.006887, 0.343135, 0.483988, 0.140773, 0.186374, NaN, NaN 15, 1.400000e-02, 2.860789, 2.919388, 4.003152, 4.007297, 0.367259, 0.519769, 0.149155, 0.191689, NaN, NaN 16, 1.500000e-02, 2.790549, 2.824211, 4.003336, 4.007689, 0.390447, 0.554223, 0.157471, 0.196670, NaN, NaN 17, 1.600000e-02, 2.725393, 2.742672, 4.003499, 4.008066, 0.413924, 0.588331, 0.165904, 0.202069, NaN, NaN 18, 1.700000e-02, 2.680308, 2.689743, 4.003680, 4.008426, 0.434461, 0.619989, 0.174204, 0.205820, NaN, NaN 19, 1.800000e-02, 2.637444, 2.641678, 4.003837, 4.008775, 0.455699, 0.651603, 0.182595, 0.210300, NaN, NaN 20, 1.900000e-02, 2.604417, 2.606106, 4.003997, 4.009115, 0.476078, 0.682403, 0.191028, 0.214145, NaN, NaN 21, 2.000000e-02, 2.584895, 2.585602, 4.004149, 4.009404, 0.493613, 0.709198, 0.198567, 0.217754, NaN, NaN 22, 2.100000e-02, 2.566409, 2.566624, 4.004293, 4.009706, 0.512109, 0.737156, 0.206435, 0.221428, NaN, NaN 23, 2.200000e-02, 2.552915, 2.552891, 4.004431, 4.009998, 0.530150, 0.764462, 0.214151, 0.224954, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.876975, 9.268188, 4.000913, 4.001419, 0.092287, 0.122966, 0.023526, 0.020772, NaN, NaN 3, 2.000000e-03, 3.559639, 4.838181, 4.001621, 4.002625, 0.176835, 0.236009, 0.045352, 0.039891, NaN, NaN 4, 3.000000e-03, 3.153905, 3.431892, 4.002112, 4.003604, 0.249593, 0.336173, 0.065092, 0.057731, NaN, NaN 5, 4.000000e-03, 2.741364, 2.762127, 4.002418, 4.004389, 0.311453, 0.424435, 0.082876, 0.074466, NaN, NaN 6, 5.000000e-03, 2.409396, 2.409396, 4.002577, 4.005020, 0.364452, 0.502768, 0.099053, 0.090186, NaN, NaN 7, 6.000000e-03, 2.160967, 2.160967, 4.002624, 4.005534, 0.410667, 0.573229, 0.113993, 0.104983, NaN, NaN 8, 7.000000e-03, 1.970646, 1.970646, 4.002584, 4.005951, 0.451527, 0.637183, 0.127945, 0.118835, NaN, NaN 9, 8.000000e-03, 1.828043, 1.828043, 4.002479, 4.006295, 0.488411, 0.696067, 0.141130, 0.131815, NaN, NaN 10, 9.000000e-03, 1.717286, 1.717286, 4.002322, 4.006577, 0.522102, 0.750686, 0.153690, 0.143963, NaN, NaN 11, 1.000000e-02, 1.628050, 1.628050, 4.002124, 4.006811, 0.553265, 0.801966, 0.165776, 0.155200, NaN, NaN 12, 1.100000e-02, 1.554835, 1.554835, 4.001892, 4.006997, 0.582304, 0.849813, 0.177349, 0.165981, NaN, NaN 13, 1.200000e-02, 1.493588, 1.493588, 4.001629, 4.007141, 0.609392, 0.894891, 0.188514, 0.175979, NaN, NaN 14, 1.300000e-02, 1.440728, 1.440728, 4.001347, 4.007261, 0.635275, 0.938122, 0.199434, 0.185395, NaN, NaN 15, 1.400000e-02, 1.394684, 1.394684, 4.001048, 4.007355, 0.660037, 0.979591, 0.210103, 0.194297, NaN, NaN 16, 1.500000e-02, 1.372772, 1.372772, 4.000725, 4.007416, 0.683367, 1.018780, 0.220425, 0.202686, NaN, NaN 17, 1.600000e-02, 1.404442, 1.404442, 4.000392, 4.007460, 0.705989, 1.056807, 0.230589, 0.210668, NaN, NaN 18, 1.700000e-02, 1.439334, 1.439334, 4.000042, 4.007479, 0.727550, 1.093128, 0.240504, 0.218221, NaN, NaN 19, 1.800000e-02, 1.474868, 1.474868, 3.999686, 4.007486, 0.748709, 1.128686, 0.250292, 0.225507, NaN, NaN 20, 1.900000e-02, 1.509282, 1.509282, 3.999315, 4.007472, 0.768973, 1.162819, 0.259885, 0.232411, NaN, NaN 21, 2.000000e-02, 1.540500, 1.540500, 3.998929, 4.007432, 0.788219, 1.195294, 0.269274, 0.238962, NaN, NaN 22, 2.100000e-02, 1.569112, 1.569112, 3.998546, 4.007395, 0.807643, 1.228017, 0.278672, 0.245323, NaN, NaN 23, 2.200000e-02, 1.592512, 1.592512, 3.998150, 4.007338, 0.826253, 1.259293, 0.287879, 0.251473, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.931733, 12.467766, 4.000350, 4.000556, 0.074023, 0.084361, 0.009686, 0.015106, NaN, NaN 3, 2.000000e-03, 3.758764, 6.588554, 4.000515, 4.001118, 0.132670, 0.161719, 0.019410, 0.037126, NaN, NaN 4, 3.000000e-03, 3.534939, 4.703236, 4.000524, 4.001598, 0.179039, 0.228644, 0.027523, 0.060967, NaN, NaN 5, 4.000000e-03, 3.293098, 3.776932, 4.000435, 4.002004, 0.217910, 0.288072, 0.035501, 0.083617, NaN, NaN 6, 5.000000e-03, 3.052115, 3.227997, 4.000280, 4.002350, 0.251818, 0.342021, 0.043984, 0.104088, NaN, NaN 7, 6.000000e-03, 2.820776, 2.864170, 4.000083, 4.002644, 0.282581, 0.392068, 0.052932, 0.122090, NaN, NaN 8, 7.000000e-03, 2.608590, 2.610493, 3.999860, 4.002914, 0.310471, 0.438737, 0.063372, 0.138518, NaN, NaN 9, 8.000000e-03, 2.423270, 2.423270, 3.999611, 4.003140, 0.336879, 0.483083, 0.073610, 0.152446, NaN, NaN 10, 9.000000e-03, 2.273938, 2.273938, 3.999345, 4.003341, 0.361534, 0.524994, 0.084618, 0.165108, NaN, NaN 11, 1.000000e-02, 2.148346, 2.148346, 3.999067, 4.003519, 0.384906, 0.564983, 0.096078, 0.176487, NaN, NaN 12, 1.100000e-02, 2.040881, 2.040881, 3.998777, 4.003675, 0.407024, 0.603056, 0.107946, 0.186774, NaN, NaN 13, 1.200000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 14, 1.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 15, 1.400000e-02, 1.796736, 1.796736, 3.997860, 4.004031, 0.467025, 0.707476, 0.146856, 0.212299, NaN, NaN 16, 1.500000e-02, 1.737632, 1.737632, 3.997548, 4.004117, 0.485018, 0.739215, 0.161735, 0.219217, NaN, NaN 17, 1.600000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 18, 1.700000e-02, 1.648319, 1.648319, 3.996959, 4.004251, 0.521115, 0.797931, 0.190567, 0.229940, NaN, NaN 19, 1.800000e-02, 1.661389, 1.661389, 3.996925, 4.004297, 0.552587, 0.819851, 0.189986, 0.227996, NaN, NaN 20, 1.900000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 21, 2.000000e-02, 1.754348, 1.754348, 3.997482, 4.004348, 0.613800, 0.859638, 0.192997, 0.222440, NaN, NaN 22, 2.100000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 23, 2.200000e-02, 1.779541, 1.779541, 3.997356, 4.004387, 0.653000, 0.908633, 0.206938, 0.229695, NaN, NaN 24, 2.300000e-02, 1.796295, 1.796295, 3.997266, 4.004390, 0.671557, 0.932180, 0.211524, 0.233175, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, 1.826042, 1.826042, 3.997003, 4.004397, 0.707464, 0.980320, 0.222403, 0.241451, NaN, NaN
2, 1.000000e-03, 3.974106, 20.276451, 4.000086, 4.000732, 0.031525, 0.055100, 0.014967, 0.033539, NaN, NaN 3, 2.000000e-03, 3.898347, 10.204411, 4.000137, 4.001474, 0.070029, 0.111473, 0.028146, 0.060093, NaN, NaN 4, 3.000000e-03, 3.779526, 6.897354, 4.000253, 4.002168, 0.111911, 0.167818, 0.040302, 0.081495, NaN, NaN 5, 4.000000e-03, 3.628505, 5.282116, 4.000419, 4.002831, 0.153687, 0.223523, 0.052098, 0.099164, NaN, NaN 6, 5.000000e-03, 3.456905, 4.338455, 4.000604, 4.003463, 0.193857, 0.278243, 0.064141, 0.114107, NaN, NaN 7, 6.000000e-03, 3.276197, 3.729916, 4.000790, 4.004063, 0.232253, 0.331635, 0.076294, 0.126983, NaN, NaN 8, 7.000000e-03, 3.101341, 3.322411, 4.000968, 4.004615, 0.268335, 0.382453, 0.088182, 0.137961, NaN, NaN 9, 8.000000e-03, 2.936056, 3.031158, 4.001134, 4.005134, 0.302673, 0.431370, 0.099910, 0.147685, NaN, NaN 10, 9.000000e-03, 2.787060, 2.819633, 4.001287, 4.005616, 0.335155, 0.477998, 0.111320, 0.156323, NaN, NaN 11, 1.000000e-02, 2.655599, 2.661800, 4.001426, 4.006071, 0.366113, 0.522946, 0.122466, 0.163990, NaN, NaN 12, 1.100000e-02, 2.546122, 2.546122, 4.001550, 4.006480, 0.395077, 0.564715, 0.133076, 0.171177, NaN, NaN 13, 1.200000e-02, 2.454157, 2.454157, 4.001662, 4.006868, 0.422830, 0.605120, 0.143429, 0.177696, NaN, NaN 14, 1.300000e-02, 2.383235, 2.383235, 4.001762, 4.007218, 0.448685, 0.642878, 0.153253, 0.183447, NaN, NaN 15, 1.400000e-02, 2.323068, 2.323068, 4.001850, 4.007571, 0.474330, 0.680875, 0.163066, 0.189020, NaN, NaN 16, 1.500000e-02, 2.277155, 2.277155, 4.001925, 4.007884, 0.498651, 0.716048, 0.172439, 0.194761, NaN, NaN 17, 1.600000e-02, 2.245813, 2.245813, 4.001991, 4.008176, 0.521587, 0.749722, 0.181411, 0.199748, NaN, NaN 18, 1.700000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 19, 1.800000e-02, 2.209070, 2.209070, 4.002099, 4.008722, 0.565597, 0.814497, 0.198746, 0.209175, NaN, NaN 20, 1.900000e-02, 2.202247, 2.202247, 4.002140, 4.008968, 0.586295, 0.845005, 0.206978, 0.213542, NaN, NaN 21, 2.000000e-02, 2.200701, 2.200701, 4.002175, 4.009200, 0.606356, 0.874637, 0.214990, 0.217710, NaN, NaN 22, 2.100000e-02, 2.203158, 2.203158, 4.002203, 4.009421, 0.625756, 0.903220, 0.222759, 0.221789, NaN, NaN 23, 2.200000e-02, 2.208580, 2.208580, 4.002225, 4.009628, 0.644610, 0.931089, 0.230337, 0.225667, NaN, NaN 24, 2.300000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 25, 2.400000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
2, 1.000000e-03, 3.955852, 15.518093, 4.000192, 4.000715, 0.048261, 0.071227, 0.013124, 0.028636, NaN, NaN 3, 2.000000e-03, 3.827637, 7.815361, 4.000308, 4.001426, 0.095248, 0.142510, 0.027759, 0.054977, NaN, NaN 4, 3.000000e-03, 3.635580, 5.334663, 4.000378, 4.002091, 0.139833, 0.211159, 0.042939, 0.078295, NaN, NaN 5, 4.000000e-03, 3.407786, 4.146290, 4.000402, 4.002700, 0.181126, 0.275966, 0.058855, 0.098533, NaN, NaN 6, 5.000000e-03, 3.169001, 3.465176, 4.000382, 4.003248, 0.219035, 0.336638, 0.075651, 0.115965, NaN, NaN 7, 6.000000e-03, 2.937280, 3.033066, 4.000325, 4.003738, 0.253794, 0.393300, 0.093490, 0.130946, NaN, NaN 8, 7.000000e-03, 2.725064, 2.742290, 4.000241, 4.004178, 0.286476, 0.446108, 0.110518, 0.143743, NaN, NaN 9, 8.000000e-03, 2.539220, 2.539220, 4.000143, 4.004572, 0.319070, 0.495354, 0.123384, 0.154538, NaN, NaN 10, 9.000000e-03, 2.386842, 2.386842, 4.000055, 4.004925, 0.352600, 0.541119, 0.131357, 0.163329, NaN, NaN 11, 1.000000e-02, 2.267980, 2.267980, 4.000008, 4.005242, 0.385646, 0.583531, 0.140099, 0.170154, NaN, NaN 12, 1.100000e-02, 2.171247, 2.171247, 4.000012, 4.005528, 0.419857, 0.623278, 0.143656, 0.175745, NaN, NaN 13, 1.200000e-02, 2.089354, 2.089354, 4.000031, 4.005788, 0.451411, 0.661291, 0.149602, 0.181093, NaN, NaN 14, 1.300000e-02, 2.028968, 2.028968, 4.000042, 4.006023, 0.479945, 0.697701, 0.158236, 0.186467, NaN, NaN 15, 1.400000e-02, 1.982175, 1.982175, 4.000038, 4.006243, 0.506562, 0.733133, 0.167677, 0.191749, NaN, NaN 16, 1.500000e-02, 1.949183, 1.949183, 4.000021, 4.006436, 0.532331, 0.766843, 0.174268, 0.196926, NaN, NaN 17, 1.600000e-02, 1.926045, 1.926045, 3.999991, 4.006623, 0.557184, 0.800088, 0.180842, 0.201996, NaN, NaN 18, 1.700000e-02, 1.914005, 1.914005, 3.999950, 4.006790, 0.580680, 0.831953, 0.187646, 0.206913, NaN, NaN 19, 1.800000e-02, 1.911314, 1.911314, 3.999899, 4.006934, 0.602844, 0.862268, 0.194579, 0.211637, NaN, NaN 20, 1.900000e-02, 1.913120, 1.913120, 3.999837, 4.007080, 0.624758, 0.892872, 0.201821, 0.216383, NaN, NaN 21, 2.000000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN 22, 2.100000e-02, 1.934992, 1.934992, 3.999699, 4.007312, 0.664879, 0.949332, 0.216230, 0.224959, NaN, NaN 23, 2.200000e-02, 1.949551, 1.949551, 3.999618, 4.007412, 0.684057, 0.976629, 0.223450, 0.229109, NaN, NaN 24, 2.300000e-02, 1.965224, 1.965224, 3.999532, 4.007503, 0.702623, 1.003172, 0.230626, 0.233149, NaN, NaN 25, 2.400000e-02, 1.981732, 1.981732, 3.999441, 4.007585, 0.720695, 1.029132, 0.237735, 0.237084, NaN, NaN 26, 2.500000e-02, -1.000000, -1.000000, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN
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Finish editing this message first! | 27,951 | 49,365 | {"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} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.551233 |
https://beyond-universe.fandom.com/wiki/Shavicuneifillion | 1,591,262,147,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439213.69/warc/CC-MAIN-20200604063532-20200604093532-00191.warc.gz | 259,931,093 | 83,802 | FANDOM
4,464 Pages
The shavicuneifillion (formerly fiftyillion) is equal to 103x103x103x103x103x10150+3.[1] The term was coined by UniversePoker777.
Expressions in other notations Edit
Notation Expression
Up-arrow notation $$10 \uparrow (3 \times (10 \uparrow (3 \times (10 \uparrow (3 \times (10 \uparrow (3 \times (10 \uparrow (3 \times (10 \uparrow 150)))))))+3)$$
Chained arrow notation $$10 \rightarrow (3 \times (10 \rightarrow (3 \times (10 \rightarrow (3 \times (10 \rightarrow (3 \times (10 \rightarrow (3 \times (10 \rightarrow 150))))))+3)$$
BEAF $$\{10,\{3\times \{10,3 \times \{10,3 \times \{10,3 \times \{10,3 \times \{10,150\}\}\}\}\}+3\}\}$$
Munafo's ASCII-lexicographic ordering p3_c3000_477121254720
1. [1] | 263 | 729 | {"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} | 2.640625 | 3 | CC-MAIN-2020-24 | latest | en | 0.608247 |
https://puzzling.stackexchange.com/questions/89507/paint-7-cells-of-a-7x7-grid | 1,709,648,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00848.warc.gz | 460,472,538 | 41,222 | # Paint 7 cells of a 7x7 grid
Can you paint 7 cells of a 7x7 grid such that the largest unpainted rectangle with grid-aligned sides has an area of 6 cells?
Good luck!
• There is a skew rectangle with area more than 6 in the intended solution. Jan 31 at 2:48
• A very elegant puzzle! Jan 31 at 15:52
Since the largest unpainted rectangle has an area of 6,
there must be no empty row or column. So each row and column has exactly one painted cell.
So,
consider the columns with the bottom four cells empty. (Let's call these "top-heavy columns".) No two of these can be adjacent, because that would make a 2x4 rectangle. Similarly, there can't be two adjacent "bottom-heavy" columns (with the top four cells empty).
The same applies to the rows, of course: no two "left-heavy" or "right-heavy" rows can be adjacent.
Time for some case bashing!
Let's say the central cell is shaded. Then the pattern of top- and bottom-heavy columns must give this...
...but now to block these four rectangles, we need to shade both of these two cells, and we can't do that.
So the central cell is unshaded.
Continuing from there,
let's arbitrarily say that the central column is bottom-heavy and the central row is right-heavy: that is, the first four cells of row 4 and column 4 are unshaded.
Then column 3 must be top-heavy, column 2 must be bottom-heavy, and column 1 must be top-heavy (and same for the rows 1-3).
The cell in row 3 column 3 must be shaded to avoid a 3x3.
Then, so must the first available cells in row and column 2...
And then, so must the first available cells in row and column 6, and the final cell is shaded in the bottom right corner.
• Doh you beat me to it. I was struggling with keeping my <pre> text hidden lol!
– JS1
Sep 25, 2019 at 5:54
• That's a brilliant answer and plenty of details! Sep 25, 2019 at 6:00
• Follow up question: How many possible arrangements are there that meet the criteria (excluding rotations and mirroring)? The two answers given are not the same, but both are valid. So we know there are at least 2. Are they the only ones? Sep 25, 2019 at 14:10
• @DarrelHoffman The answers are the same - JS1's is just mine rotated 90 degrees clockwise. And because the only assumption I made was the one that broke symmetry, the solution is unique up to rotation/reflection.
– Deusovi
Sep 25, 2019 at 14:39
• It is probably cheating, (and useless), but I did a quick computer search. This is the only solution, modulo a rotation. Sep 30, 2019 at 20:56
I got this:
. . . . . . x
. . x . . . .
. . . . x . .
. x . . . . .
. . . . . x .
. . . x . . .
x . . . . . .
Method:
1. Each row and column must have 1 painted square because otherwise one unpainted row/column would be a 1x7 rectangle.
2. I started with two opposite corners because that seemed optimal for taking care of all four edges.
3. I tried placing the 2nd row square at (2,2) but it didn't work out because it left too much open space to its right. I moved it to (2,3).
4. From there it was a bit of trial and error to place the next 4 squares, but there weren't too many possibilities left. | 834 | 3,100 | {"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} | 4.21875 | 4 | CC-MAIN-2024-10 | latest | en | 0.931797 |
https://www.physicsforums.com/threads/wave-equation-pde-problem.133778/ | 1,553,395,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00000.warc.gz | 863,859,839 | 14,107 | # Wave equation pde problem (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### sarahisme
Hi everyone,
I'm having a bit of trouble with this pde problem:
http://img243.imageshack.us/img243/9313/picture3ui3.png [Broken]
i get the answer to be u(x,t)=0 but i am guessing thats not right.
is the general solution to this problem: u(x,t) = f(x+ct) + g(x-ct) ??
thanks
sarah :)
Last edited by a moderator:
#### HallsofIvy
No, it's not. That would be the general solution to the homogeneous wave equation, uxx= c2utt, without a "source" term. Likewise, u(x,t)= 0 is the solution to the homogeneous equation with those conditions.
Remember that with a linear equation, you can construct a general solution to a non-homogeneous equation by adding the general solution to the related homogeneous equation to a single specific solution to the entire equation.
#### sarahisme
hmmm ok, i tried again and this is what i get...
for (a) http://img291.imageshack.us/img291/6120/picture4td7.png [Broken]
for (b) http://img301.imageshack.us/img301/5572/picture5ii4.png [Broken]
what do you think?
Last edited by a moderator:
#### sarahisme
lol, well i think its right anyway, thanks for the help HallsofIvy! your a legend! :D
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• Solo and co-op problem solving | 404 | 1,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-13 | latest | en | 0.880689 |
https://www.convertunits.com/from/gigahertz/to/revolution/second | 1,686,234,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00578.warc.gz | 771,559,596 | 12,707 | ## Convert gigahertz to revolution/second
gigahertz revolution/second
How many gigahertz in 1 revolution/second? The answer is 1.0E-9. We assume you are converting between gigahertz and revolution/second. You can view more details on each measurement unit: gigahertz or revolution/second The SI derived unit for frequency is the hertz. 1 hertz is equal to 1.0E-9 gigahertz, or 1 revolution/second. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between gigahertz and revolutions/second. Type in your own numbers in the form to convert the units!
## Quick conversion chart of gigahertz to revolution/second
1 gigahertz to revolution/second = 1000000000 revolution/second
2 gigahertz to revolution/second = 2000000000 revolution/second
3 gigahertz to revolution/second = 3000000000 revolution/second
4 gigahertz to revolution/second = 4000000000 revolution/second
5 gigahertz to revolution/second = 5000000000 revolution/second
6 gigahertz to revolution/second = 6000000000 revolution/second
7 gigahertz to revolution/second = 7000000000 revolution/second
8 gigahertz to revolution/second = 8000000000 revolution/second
9 gigahertz to revolution/second = 9000000000 revolution/second
10 gigahertz to revolution/second = 10000000000 revolution/second
## Want other units?
You can do the reverse unit conversion from revolution/second to gigahertz, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Gigahertz
The SI prefix "giga" represents a factor of 109, or in exponential notation, 1E9.
So 1 gigahertz = 109 .
The definition of a hertz is as follows:
The hertz (symbol Hz) is the SI unit of frequency. It is named in honour of the German physicist Heinrich Rudolf Hertz who made some important contributions to science in the field of electromagnetism.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 591 | 2,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.853946 |
https://nrich.maths.org/public/topic.php?code=-317&cl=3&cldcmpid=2160 | 1,597,062,947,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00413.warc.gz | 435,724,548 | 5,335 | # Resources tagged with: Sine rule & cosine rule
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Broad Topics > Pythagoras and Trigonometry > Sine rule & cosine rule
### Raising the Roof
##### Age 14 to 16 Challenge Level:
How far should the roof overhang to shade windows from the mid-day sun?
### Hexy-metry
##### Age 14 to 16 Challenge Level:
A hexagon, with sides alternately a and b units in length, is inscribed in a circle. How big is the radius of the circle?
### Get Cross
##### Age 14 to 16 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Xtra
##### Age 14 to 18 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
##### Age 14 to 16 Challenge Level:
Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.
##### Age 14 to 16 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Darts and Kites
##### Age 14 to 16 Challenge Level:
Explore the geometry of these dart and kite shapes!
### Calculating with Cosines
##### Age 14 to 18 Challenge Level:
If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle? | 435 | 1,733 | {"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} | 3.9375 | 4 | CC-MAIN-2020-34 | latest | en | 0.841394 |
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# ch5-6 - SECTION 5.12 Beams with Axial Loads 363 Problem...
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Problem 5.12-10 A flying buttress transmits a load P 5 25 kN, acting at an angle of 60° to the horizontal, to the top of a vertical buttress AB (see figure). The vertical buttress has height h 5 5.0 m and rectangular cross section of thickness t 5 1.5 m and width b 5 1.0 m (perpendicular to the plane of the figure). The stone used in the construction weighs g 5 26 kN/m 3 . What is the required weight W of the pedestal and statue above the vertical buttress (that is, above section A ) to avoid any tensile stresses in the vertical buttress? Solution 5.12-10 Flying buttress SECTION 5.12 Beams with Axial Loads 363 60 ° P A A B B W h h t 2 Flying buttress t t F REE - BODY DIAGRAM OF VERTICAL BUTTRESS P 5 25 kN h 5 5.0 m t 5 1.5 m b 5 width of buttress perpendicular to the figure b 5 1.0 m g 5 26 kN/m 3 W B 5 weight of vertical buttress 5 bth g 5 195 kN C ROSS SECTION A 5 bt 5 (1.0 m)(1.5 m) 5 1.5 m 2 A T THE BASE N 5 W 1 W B 1 P sin 60º 5 W 1 195 kN 1 (25 kN) sin 60º 5 W 1 216.651 kN M 5 ( P cos 60º) h 5 (25 kN)(cos 60º)(5.0 m) 5 62.5 kN ? m T ENSILE STRESS ( EQUAL TO ZERO ) or 2 W 2 216.651 kN 1 250 kN 5 0 W 5 33.3 kN 52 W 1 216.651 kN 1.5 m 2 1 62.5 kN ? m 0.375 m 3 5 0 s t N A 1 M S S 5 1 6 bt 2 5 1 6 (1.0 m)(1.5 m) 2 5 0.375 m 3 60 ° P W h W B t N V M Problem 5.12-11 A plain concrete wall (i.e., a wall with no steel reinforcement) rests on a secure foundation and serves as a small dam on a creek (see figure). The height of the wall is h 5 6.0 ft and the thickness of the wall is t 5 1.0 ft. (a) Determine the maximum tensile and compressive stresses s t and s c , respectively, at the base of the wall when the water level reaches the top ( d 5 h ). Assume plain concrete has weight density g c 5 145 lb/ft 3 . (b) Determine the maximum permissible depth d max of the water if there is to be no tension in the concrete. h d t
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Solution 5.12-11 Concrete wall 364 CHAPTER 5 Stresses in Beams h 5 height of wall t 5 thickness of wall b 5 width of wall (perpendicular to the figure) g c 5 weight density of concrete g w 5 weight density of water d 5 depth of water W 5 weight of wall W 5 bht g c F 5 resultant force for the water pressure M AXIMUM WATER PRESSURE 5 g w d A 5 bt S TRESSES AT THE BASE OF THE WALL ( d 5 DEPTH OF WATER ) Eq. (1) Eq. (2) s c 52 W A 2 M S h g c 2 d 3 g w t 2 s t W A 1 M S h g c 1 d 3 g w t 2 S 5 1 6 bt 2 M 5 F ¢ d 3 5 1 6 bd 3 g w F 5 1 2 ( d )( g w d b ) 5 1 2 bd 2 g w (a) S TRESSES AT THE BASE WHEN d 5 h h 5 6.0 ft 5 72 in. d 5 72 in. t 5 1.0 ft 5 12 in. Substitute numerical values into Eqs. (1) and (2): s t 6.042 psi 1 93.600 psi 5 87.6 psi s c 6.042 psi 2 93.600 psi 99.6 psi (b) M AXIMUM DEPTH FOR NO TENSION Set s t 5 0 in Eq. (1): d max 5 28.9 in. d 3 5 (72 in.)(12 in.) 2 ¢ 145 62.4 5 24,092 in. 3 d 3 5 ht 2 ¢ g c g w 2 h g c 1 d 3 g w t 2 5 0 g w 5 62.4 lb / ft 3 5 62.4 1728 lb / in. 3 g c 5 145 lb / ft 3 5 145 1728 lb / in. 3 h d t M W F d/ 3 V
Eccentric Axial Loads Problem 5.12-12 A circular post and a rectangular post are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depth of the rectangular post are the same.
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ch5-6 - SECTION 5.12 Beams with Axial Loads 363 Problem...
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Ask a homework question - tutors are online | 1,425 | 3,895 | {"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} | 3.578125 | 4 | CC-MAIN-2018-09 | latest | en | 0.708118 |
https://gamedev.stackexchange.com/questions/183051/cube-doesnt-render-as-expected-when-rotated-representing-3d-in-2d | 1,709,266,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00400.warc.gz | 276,558,831 | 35,743 | # Cube doesn't render as expected when rotated (representing 3D in 2D)
Cube. When I rotate the object around OZ moves to the left. Rotation around OX - the cube rises up. Rotation around the OY-the cube is compressed. The function I use for rotation:
void rotation(double angleX, double angleY, double angleZ, int x, int y, int z)
{
angleX = angleX * M_PI / 180;
angleY = angleY * M_PI / 180;
angleZ = angleZ * M_PI / 180;
double cX = cos(angleX);
double sX = sin(angleX);
double cY = cos(angleY);
double sY = sin(angleY);
double cZ = cos(angleZ);
double sZ = sin(angleZ);
double x0 = x;
double y0 = y * cX + z * sX;
double z0 = z * cX - y * sX;
double x1 = x0 * cY - z0 * sY;
double y1 = y0;
double z1 = z0 * cY + x0 * sY;
double x2 = x1 * cZ + y1 * sZ;
double y2 = y1 * cZ - x1 * sZ;
SDL_RenderDrawPoint(ren, x2, y2);
}
Used matrices. However, the result was the same.
I don't have SDL installed, so I couldn't run your code. However, from what I can see the problem is not the calculation of the rotations (not 100% certain if they do not contain any mistakes, but they look fine), but the object you are rotating.
The problematic line is this function call in your main function:
rotation(angleX, angleY, angleZ, v[i].x, v[i].y, 100);
What you want to do, is to rotate a cube, a 3-dimensional object, and draw it to your screen. But what you are actually doing is the equivalent of rotating a piece of paper (2-dimensional) with a drawn cube on it and printing this to your screen. Every point you pass to your function has the same z-value.
With the paper example in mind, let's go through your images and the effects you are seeing (take a real one and draw your cube on it, if you can't imagine it). Note that rotations always rotate around the coordinate origin (the point at x=0, y=0, z=0).
Rotation around the z-axis (second image)
This is like putting your finger onto the point where your coordinate origin is and starting to rotate the paper around this axis. Since your cube is not at the coordinate center:
int centerX = SCREEN_WIDTH / 3;
int centerY = SCREEN_HEIGHT / 2 + 100;
it is not only rotated, but also translated. The farther away your finger is from the cube, the larger the translations.
Rotation around the x-axis (third image)
This is basically like gluing the piece of paper to your fingertip (don't do that) and tilting your wrist joint so that your finger goes upwards. The paper moves upwards, and so does the cube. Additionally, the paper also gets tilted which has the effect, that your cube gets visually compressed in the y-direction since you can see less of the papers surface. In the extreme case, you would only see the upper side of the paper, degenerating your cube into a line.
Rotation around the y-axis (fourth image)
Same as the x-axis but tilt your wrist joint to the left instead upwards. So the cube moves to the left and gets visually compressed in the x-direction.
Solution
The solution should be obvious: Use 3d points by adding a z-component.
What you did with your cube is that you already manually applied a parallel projection to it before all other transformations. But the projection usually has to be the last step after all other transformations. I say usually since I can't rule out, that there are some fancy effects/techniques, that do it differently.
However, I think it might be helpful for you, to read at least some tutorials about transformations. Read for example those two:
They might be related to OpenGL, but the information you get from those two are universal. Make sure you read them in the order I linked them.
One remark regarding this part of your title: "(representing 3D in 2D)". Every "3d game" calculates a representation of your 3d scene in 2d since your screen is 2 dimensional. This is also true for VR. In VR you just do it twice, one representation for each eye, tricking your brain into believing you see in 3d.
Don't do using namespace std;. See this Stackoverflow question to understand why. | 1,000 | 4,012 | {"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} | 3.0625 | 3 | CC-MAIN-2024-10 | latest | en | 0.881016 |
https://chronicdb.com/solution/find-the-derivative-of-fx-1-5x2x-x2-in-two-ways-use-the-product-168 | 1,701,416,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00883.warc.gz | 197,192,302 | 10,449 | admin
# Find the derivative of f(x) = (1 + 5x2)(x − x2) in two ways. use the product rule. f '(x) = perform the multiplication first. f '(x) = do your answers agree? yes no
2 months ago
## Solution 1
Guest #10154
2 months ago
Answer:
The answers agree.
Step-by-step explanation:
The product rule is:
In this problem, we have that:
So
.
Multiplication first:
## Solution 2
Guest #10155
2 months ago
The product rule states that given a function:
f(x)=g(x)h(x)
f'(x)=g'(x)h(x)+h'(x)g(x)
thus the derivative of the expression will be given by:
f(x) = (1 + 5x2)(x − x2)
g(x)=(1+5x^2)
g'(x)=10x
h(x)=(x-x^2)
h'(x)=1-2x
thus:
f'(x)=10x(x-x^2)+(1-2x)(1+5x^2)
simplifying this we get:
f'(x)=-20x^3+15x^2-12x+1
## 📚 Related Questions
Question
Leon deposited $30000 in a bank which paid 2% interest per year. After 1 year, Leon withdrew all his money including the interest. how much money did Leon withdraw? Solution 1 The amount of money Leon had after 1 year will be given by: A=P(1+r)^n where: P=principle r=rate n=time But from the information given: P=$30,000, r=2%, n=1 year
thus
A=30000(1+2/100)^1
A=30000(1.02)^1
A=$30600 Thus amount after 1 year was$30600
Question
Find the equation for the plane through the points upper p 0 left parenthesis negative 2 comma negative 5 comma negative 4 right parenthesis, upper q 0 left parenthesis 5 comma 1 comma negative 4 right parenthesis, and upper r 0 left parenthesis negative 1 comma 2 comma 5 right parenthesis.
Solution 1
We can suppose the equation of the plane is in the form ...
ax +by +cz = 1
By using the given point coordinates for x, y, and z, we end up with three equations in 3 unknowns. These can be described by the augmented matrix ...
Putting this in reduced row-echelon form, we find
a = 54/35
b = -63/35
c = 43/35
The equation of the plane through is ...
54x -63y +43z = 35
Question
In a family with 8 children, excluding multiple births, what is the probability of having 7 boys and 1 girl, in any order? assume that a boy is as likely as a girl at each birth
Solution 1
There is a 7/8 chance there will be boys and a 1/8 chance there will be a girl
Question
For the particular problem raised in the introduction, assume that the total bill is $44. to answer the question "how should the bill be split?" we will create a linear equation. the unknown is how much money a single person (besides anika) must pay, so call that x. although four people (you plus three friends) went to dinner, only three are paying the unknown amount x for a total of 3x. since anika is paying$2, the total amount paid is 3x+2 dollars, which must equal the amount of the bill, $44. thus, the equation to find x is 3x+2=44. the steps for solving a linear equation are as follows: move all of the constants to the right side. move all of the variable terms (terms containing x) to the left side. divide both sides by the coefficient of the variable to isolate the variable. you will go through these steps one at a time to solve the equation and determine how much each person should pay. Solution 1 3x+2=44 3x+2(-2)=44-2 3x=42 3x/3=42/3 x=14 So each person would have to pay$14, except Anika
Question
A pair of two distinct dice are rolled six times. suppose none of the ordered pairs of values (1, 5), (2, 6), (3, 4), (5, 5), (5, 3), (6, 1), (6, 2) occur. what is the probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice?
Solution 1
The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice is 1/6.
### What is probability?
It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.
It is given that:
A pair of two distinct dice are rolled six times. suppose none of the ordered pairs of values (1, 5), (2, 6), (3, 4), (5, 5), (5, 3), (6, 1), (6, 2) occur.
Total number of outcomes = 6x6
Total number of outcomes = 36
Total favourable outcome = 6
The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice:
P = 6/36 = 1/6
Thus, the probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice is 1/6.
Learn more about the probability here:
brainly.com/question/11234923
#SPJ2
Solution 2
But them in order of add them all up i guess
Question
If the numerator of a fraction is increased by 3, the fraction becomes 3/4. If the denominator is decreased by 7, the fraction becomes 1. Determine the original fraction. Which of the following equations represents "If the numerator of a fraction is increased by 3, the fraction becomes 3/4"?
Solution 1
For this case, the original fraction is:
x / y
Where,
x = numerator
y = denominator:
If the numerator of a fraction is increased by 3, the fraction becomes 3/4:
(x + 3) / y = 3/4
If the denominator is decreased by 7, the fraction becomes 1:
x / (y-7) = 1
Solving the system of equations we have:
x = 9
y = 16
The original fraction is:
9/16
Answer:
the original fraction is:
9/16
"If the numerator of a fraction is increased by 3, the fraction becomes 3/4" is:
(x + 3) / y = 3/4
Solution 2
Let n = numerator
d = denominator
n+3/d = 3/4
n/(d-7) = 1
Which of the following equations represents "If the numerator of a fraction is increased by 3, the fraction becomes 3/4"?
The equation is (
n+3)/d = 3/4
Question
A single fair die is tossed. find the probability of rolling a number greater than 55.
Solution 1
0% no # greater than 6
Question
A jar contains 8 large red marbles, 5 small red marbles, 7 large blue marbles, and 4 small blue marbles. if a marble is chosen at random, which of the conditional probabilities is larger, p(red|large) or p(large|red)?
Solution 1
8 large red, 5 small red, 7 large blue, 4 small blue:
totally = 24 marbles
P(red)= 13/24
let A red and B large
P(A/B)= P(A∩B)/P(B)
P(B)= 15/24 =.625
P(A∩B)= 8/24 = .333
P(A/B) =.333/.625 =.533
P(B/A) =.625/.333= 1.875
so the probability of p(large/red) is larger.
Question
"a fair coin is tossed 5 times. what is the probability of exactly 4 heads?"
Solution 1
5 flips render40 % 4 will turn up heads
Question
Evaluate the surface integral. s xyz ds, s is the cone with parametric equations x = u cos(v), y = u sin(v), z = u, 0 ≤ u ≤ 3, 0 ≤ v ≤ π 2
Solution 1
Answer:
Step-by-step explanation:
Let
Differentiate partially with respect to u, v.
To find f, put in xyz.
So,
Integrate with respect to u.
Integrate with respect to v.
Solution 2
Let's capture the parameterization of the surface by the vector function
Then the surface element is given by
So the surface integral is equivalent to
2647929
842281
748681
586256
406852
368373
348603
324927
199835
130075
112100
106146
77164
23213
22589
19607
17108
13966
10987
3389
3423409
2092250
1518097 | 2,107 | 7,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | latest | en | 0.866009 |
https://www.namastekadapa.com/2016/10/26/aptitude-quiz-for-rrb-ssc-cgl-appsc-tspsc-other-competitive-exams/ | 1,726,752,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00290.warc.gz | 839,137,742 | 51,918 | Current Affairs
# Aptitude Quiz for RRB / SSC CGL / APPSC / TSPSC & Other Competitive Exams
1. 4/5 1.th of the voters in a constituency promised to vote for Rahul and the rest promised to vote for Varun. Of these voters, 10% of the voters who had promised to vote for Rahul, did not vote on the election day, while 20% of the voters who had promised to vote for Varun did not vote on the election day. What is the total number of votes polled if Rahul got 216 votes?
a) 200
b) 300
c) 264
d) 100
2.Areduction of 10% in the price of wheat enables a man to buy 50gram of wheat more for a rupee. How much wheat could originally be had for a rupee?
a) 400 grams
b) 500 grams
c) 450 grams
d) 350 grams
3.Afruit seller makes a profit of20% by selling mangoes at a certain price. If he charges Rs.1 more for each mango, he can make a profit of 40%. What is the selling price of a mango in the first case?
a) Rs. 5
b) Rs. 6
c) Rs. 5.50
d) Rs. 7
4.Aman sells an article at 5% above its cost price. If he had bought it at 5% less than what he paid for it and sold it for Rs. 2 less, he would have gained 10%. What is the cost price of the article?
a) Rs. 500
b) Rs. 360
c) Rs. 425
d) Rs. 400
5.Two dealers A and B selling the same model of fan mark them under the same selling price. A gives successive discounts of 5%and 25% and B gives successive discounts of 16% and 12%. From whom is it more profitable to purchase the fan?
a) From B
b) From A
c) Indifferent between the two
d) Cannot be determined
6.If there is a reduction in the number of workers in a factory in the ratio 15 : 11 and an incrementing their wages in the ratio 22 : 25,then the ratio by which the total wages of the workers should be decreased is:
a) 6 : 5
b) 5 : 6
c) 3 : 7
d) 3 : 5
7.What quarterly payment will discharge a debt of Rs. 2280 due in two years at 16% per annum simple interest?
a) Rs. 500
b) Rs. 450
c) Rs. 550
d) Rs. 250
8.Asum of Rs. 3310 is to be paid back in 3 equal annual installments. How much is each installment if the interest is compounded annually at 10% per annum?
a) Rs. 1321
b) Rs. 1343
c) Rs. 1325
d) Rs. 1331
9.Aand B together can complete a work in 12 days. A alone can complete in 20 days. If B does the work only half a day daily,then in how many days A and B together will complete the work?
a) 10 days
b) 20 days
c) 11 days
d) 15 days
10.Two bullets were fired at a place at an interval of 28 minutes 30seconds. A person approaching the firing point in his car hears the two sounds at an interval of27 minutes. The speed of sound is 330 m/sec. What is the speed of the car?
a) 44 km/hr
b) 66 km/hr
c) 64 km/hr
d) 54 km/hr
11.In a race of 600 metres, Sonu beats Monu by 60 metres and in a race of 500 metres, Monu beats Bablu by 25 metres. By how many metres will Sonu beat Bablu in a 400 metre race?
a) 48 m
b) 52 m
c) 56 m
d) 58 m
12.A boat goes downstream at x km/h and upstream at z km/h. The speed of the boat in stillwater is:
a) 0.5(x + y)
b) 0.5(x– y)
c) x+ y
d) x– y | 959 | 2,975 | {"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} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.948309 |
https://interestingengineering.com/culture/first-graders-homework-confusing-adults | 1,680,324,583,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00287.warc.gz | 358,689,212 | 24,435 | For parents, math can be confusing especially when they've been out of the school system for decades. But even we think this is a bit much for the typical first grader.
The logic puzzle, seen below, uses a key of letters to replace numbers in no discernible pattern.
So, we decided to give this a shot and "show your work" like the hashtag requested. We first replaced all letters with their representative numerical value:
We then decided to add the diagonals. Assuming the homework is truly intended for younger kids, any pattern more complex than that would be straight up cruel.
Adding up the diagonal with all three gave us a sum of 40. Thus (assuming our theory is correct), the alternative diagonal will also equal 40.
Basic math kicks in here. 40 - 26 = 14. A quick check on the key tells us that 14 = J. Thus, J is our final answer.
However, not everyone appreciated the brainteaser. One commenter said:
Well, like most math, there isn't a ton to "hone" other than, ya know, basic logic and pattern skills...
Most Popular
And then there's one commenter who doesn't agree with the majority's answer of 40:
We don't quite see that pattern. However, the answer key hasn't been released, so the answer could be something as equally odd as the question.
Source : The Holderness Family
SHOW COMMENT (1) | 296 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.959681 |
https://textbook.prob140.org/ch23/Least_Squares.html | 1,508,318,174,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00133.warc.gz | 851,948,997 | 10,019 | ## Least Squares
### Least Squares¶
We now turn to the conditional expectation $E(Y \mid X)$ viewed as an estimate or predictor of $Y$ given the value of $X$. As you saw in Data 8, the mean squared error of prediction can be used to compare predictors: those with small mean squared errors are better.
In this section we will identify least squares predictors, that is, predictors that minimize mean squared error among all predictors in a specified class.
### Minimizing the MSE¶
Suppose you are trying to estimate or predict the value of $Y$ based on the value of $X$. The predictor $E(Y \mid X) = b(X)$ seems to be a good one to use, based on the scatter plots we examined in the previous section.
It turns out that $b(X)$ is the best predictor of $Y$ based on $X$, according the least squares criterion.
Let $h(X)$ be any function of $X$, and consider using $h(X)$ to predict $Y$. Define the mean squared error of the predictor $h(X)$ to be
$$MSE(h) ~ = ~ E\Big{(}\big{(}Y - h(X)\big{)}^2\Big{)}$$
We will now show that $b(X)$ is the best predictor of $Y$ based on $X$, in the sense that it minimizes this mean squared error over all functions $h(X)$.
To do so, we will use a fact we proved in the previous section:
• If $g(X)$ is any function of $X$ then $E\big{(}(Y - b(X))g(X)\big{)} = 0$.
\begin{align*} MSE(h) ~ &= ~ E\Big{(}\big{(}Y - h(X)\big{)}^2\Big{)} \\ &= ~ E\Big{(}\big{(}Y - b(X)\big{)}^2\Big{)} + E\Big{(}\big{(}b(X) - h(X)\big{)}^2\Big{)} + 2E\Big{(}\big{(}Y - b(X)\big{)}\big{(}b(X) - h(X)\big{)}\Big{)} \\ &= ~ E\Big{(}\big{(}Y - b(X)\big{)}^2\Big{)} + E\Big{(}\big{(}b(X) - h(X)\big{)}^2\Big{)} \\ &\ge ~ E\Big{(}\big{(}Y - b(X)\big{)}^2\Big{)} \\ &= ~ MSE(b) \end{align*}
### Least Squares Predictor¶
The calculations in this section include much of the theory behind least squares prediction familiar to you from Data 8. The result above shows that the least squares predictor of $Y$ based on $X$ is the conditional expectation $b(X) = E(Y \mid X)$.
In terms of the scatter diagram of observed values of $X$ and $Y$, the result is saying that the best predictor of $Y$ given $X$, by the criterion of smallest mean squared error, is the average of the vertical strip at the given value of $X$.
Given $X$, the root mean squared error of this estimate is the SD of the strip, that is, the conditional SD of $Y$ given $X$:
$$SD(Y \mid X) ~ = ~ \sqrt{Var(Y \mid X)}$$
This is a random variable; its value is determined by the variation within the strip at the given value of $X$.
Overall across the entire scatter diagram, the root mean squared error of the estimate $E(Y \mid X)$ is
$$RMSE(b) ~ = ~ \sqrt{E\Big{(}\big{(}Y - b(X)\big{)}^2\Big{)}} ~ = ~ \sqrt{E\big{(} Var(Y \mid X) \big{)}}$$
Notice that the result makes no assumption about the joint distribution of $X$ and $Y$. The scatter diagram of the generated $(X, Y)$ points can have any arbitrary shape. So the result can be impractical, as there isn't always a recognizable functional form for $E(Y \mid X)$.
Sometimes we want to restrict our attention to a class of predictor functions of a specified type, and find the best one among those. The most important example of such a class is the set of all linear functions $aX + b$.
### Least Squares Linear Predictor¶
Let $h(X) = aX + b$ for constants $a$ and $b$, and let $MSE(a, b)$ denote $MSE(h)$.
$$MSE(a, b) ~ = ~ E\big{(} (Y - (aX + b))^2 \big{)} ~ = ~ E(Y^2) + a^2E(X^2) + b^2 -2aE(XY) - 2bE(Y) + 2abE(X)$$
To find the least squares linear predictor, we have to minimize this MSE over all $a$ and $b$. We will do this using calculus, in two steps:
• Fix $a$ and find the value $b_a^*$ that minimizes $MSE(a, b)$ for that fixed value of $a$.
• Then plug in the minimizing value $b_a^*$ in place of $b$ and minimize $MSE(a, b_a^*)$ with respect to $a$.
#### Step 1.¶
Fix $a$ and minimize $MSE(a, b)$ with respect to $b$.
$$\frac{d}{db} MSE(a, b) ~ = ~ 2b - 2E(Y) + 2aE(X)$$
Set this equal to 0 and solve to see that the minimizing value of $b$ for the fixed value of $a$ is
$$b_a^* ~ = ~ E(Y) - aE(X)$$
#### Step 2.¶
Now we have to minimize the following function with respect to $a$:
\begin{align*} E\big{(} (Y - (aX + b_a^*))^2 \big{)} ~ &= ~ E\big{(} (Y - (aX + E(Y) - aE(X)))^2 \big{)} \\ &= ~ E\Big{(} \big{(} (Y - E(Y)) - a(X - E(X))\big{)}^2 \Big{)} \\ &= ~ E\big{(} (Y - E(Y))^2 \big{)} - 2aE\big{(} (Y - E(Y))(X - E(X)) \big{)} + a^2E\big{(} (X - E(X))^2 \big{)} \\ &= ~ Var(Y) - 2aCov(X, Y) + a^2Var(X) \end{align*}
The derivative with respect to $a$ is $2Cov(X, Y) + 2aVar(X)$. Thus the minimizing value of $a$ is
$$a^* ~ = ~ \frac{Cov(X, Y)}{Var(X)}$$
### Slope and Intercept of the Regression Line¶
The least squares straight line is called the regression line.You now have a proof of its equation, familiar to you from Data 8. The slope and intercept are given by
\begin{align*} \text{slope of regression line} ~ &= ~ \frac{Cov(X,Y)}{Var(X)} ~ = ~ r(X, Y) \frac{SD(Y)}{SD(X)} \\ \\ \text{intercept of regression line} ~ &= ~ E(Y) - \text{slope} \cdot E(X) \end{align*}
To derive the second expression for the slope, recall that in exercises you defined the correlation between $X$ and $Y$ to be
$$r(X, Y) ~ = ~ \frac{Cov(X, Y)}{SD(X)SD(Y)}$$
#### Regression in Standard Units¶
If both $X$ and $Y$ are measured in standard units, then the slope of the regression line is the correlation $r(X, Y)$ and the intercept is 0.
In other words, given that $X = x$ standard units, the predicted value of $Y$ is $r(X, Y)x$ standard units. When $r(X, Y)$ is positive but not 1, this result is called the regression effect: the predicted value of $Y$ is closer to 0 than the given value of $X$.
It is important to note that the equation of the regression line holds regardless of the shape of the joint distribution of $X$ and $Y$. Also note that there is always a best straight line predictor among all straight lines, regardless of the relation between $X$ and $Y$. If the relation isn't roughly linear you won't want to use the best straight line for predictions, because the best straight line is best among a bad class of predictors. But it exists. | 1,946 | 6,135 | {"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} | 4.09375 | 4 | CC-MAIN-2017-43 | latest | en | 0.791619 |
https://www.ato.gov.au/Individuals/Ind/Paying-it-Forward---Primary-school-resource/Years-5-6/?page=9 | 1,696,075,779,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00529.warc.gz | 706,771,953 | 20,546 | • #### Topic 3 Collecting tax fairly (Year 5)
Activity 2
Achievement Standard
Mathematics
What students do
Students order decimals and unit fractions and locate them on number lines
A
correctly represents, orders and compares decimals and unit fractions, locates them on number lines and uses mathematical language to justify conclusions about relative fairness of tax contributions
B
correctly represents, orders and compares unit fractions, locates them on number lines and compares numbers to draw conclusions about the relative fairness of tax contributions
C
represents and orders decimals and unit fractions and locates them on a number line
D
with support, represents and orders decimals and unit fractions and locates them on a number line
E
with direction, represents and orders decimals and unit fractions and locates them on a number line
Activity 4
Achievement Standard
Mathematics
What students do
Students solve simple problems involving the four operations using a range of strategies
A
correctly solves problems involving the four operations using a range of efficient strategies and explains and justifies strategies used
B
correctly solves problems involving the four operations using a range of efficient strategies and explains strategies used
C
solves simple problems involving the four operations using a range of strategies
D
with support, solves simple problems involving the four operations using one or more strategy
E
solves simple problems using strategies as directed
Activity 5
Achievement Standard
HASS
What students do
Students reflect on their learning to independently propose action, describing the possible effects of their proposed action
Students suggest conclusions based on evidence
A
uses criteria and evidence to propose appropriate tax models for different items in a hypothetical situation and effectively justifies these choices in detail and with evidence
B
uses criteria and evidence to propose appropriate tax models for different items in a hypothetical situation and justifies these choices with evidence
C
uses evidence to propose tax models for different items in a hypothetical situation and justifies these choices
D
with support, uses evidence to propose tax models for different items in a hypothetical situation and partially justifies these choices
E
with direction, uses evidence to propose tax models for different items in a hypothetical situation and provides fragmented justifications for these choices | 458 | 2,511 | {"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-2023-40 | latest | en | 0.889726 |
http://www.chegg.com/homework-help/questions-and-answers/a-0140-rm-kg-cube-of-ice-frozen-water-is-floating-in-glycerine-the-gylcerine-is-in-a-tall--q3399398 | 1,369,530,395,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706474776/warc/CC-MAIN-20130516121434-00010-ip-10-60-113-184.ec2.internal.warc.gz | 382,583,115 | 8,064 | ## urgent Help !!!
A 0.140-{\rm kg} cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.30cm . The level of the glycerine is well below the top of the cylinder. If the ice completely melts, by what distance does the height of liquid in the cylinder change? Express your answer with the appropriate units.
## Answers (1)
• here , mass of ice (m) = 0.14 kg density of ice (p1) = 0.9167 *10^3 kg/m^3 volume of ice (V) = m/p1 = 1.52*10^-4 m^3 density of water ( p ) = 1000 kg/m^3 volume of water (V1) = m / p { since mass conservation is valid } = 1.4*10^-4 m^3 therefore, height displaced (h) = change in volume / pi*r^2 r: radius of cylinder = 0.033 m h = V - V1 / pi*r^2 on substituting h = 0.37 cm
Comments
• Anonymous commented
This is incorrect.
Get homework help
More than 200 experts are waiting to help you now... | 276 | 885 | {"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} | 3.78125 | 4 | CC-MAIN-2013-20 | latest | en | 0.867254 |
http://ncatlab.org/nlab/show/Isbell+duality | 1,397,972,804,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00083-ip-10-147-4-33.ec2.internal.warc.gz | 169,230,418 | 30,094 | nLab Isbell duality
Context
Higher algebra
higher algebra
universal algebra
duality
In QFT and String theory
higher geometry $\leftarrow$ Isbell duality $\rightarrow$ higher algebra
Contents
Idea
$(\mathcal{O} \dashv Spec) : CoPresheaves \stackrel{\overset{\mathcal{O}}{\leftarrow}}{\underset{Spec}{\to}} Presheaves$
relates (higher) presheaves with (higher) copresheaves on a given (higher) category $C$: this is called Isbell conjugation or Isbell duality (after John Isbell).
To the extent that this adjunction descends to presheaves that are (higher) sheaves and copresheaves that are (higher) algebras this duality relates higher geometry with higher algebra.
Objects preserved by the monad of this adjunction are called Isbell self-dual.
Definition
Let $\mathcal{V}$ be a good enriching category (a cosmos, i.e. a complete and cocomplete closed symmetric monoidal category).
Let $\mathcal{C}$ be a small $\mathcal{V}$-enriched category.
Write $[\mathcal{C}^{op}, \mathcal{V}]$ and $[\mathcal{C}, \mathcal{V}]$ for the enriched functor categories.
Proposition
There is a $V$-adjunction
$(\mathcal{O} \dashv Spec) : [C, \mathcal{V}]^{op} \stackrel{\overset{\mathcal{O}}{\leftarrow}}{\underset{Spec}{\to}} [C^{op}, \mathcal{V}]$
where
$\mathcal{O}(X) : c \mapsto [C^{op}, \mathcal{V}](X, \mathcal{V}(-,c)) \,,$
and
$Spec(A) : c \mapsto [C, \mathcal{V}]^{op}(\mathcal{V}(c,-),A) \,.$
Remark
This is also called Isbell duality. Objects which are preserved by $\mathcal{O} \circ Spec$ or $Spec \mathcal{O}$ are called Isbell self-dual .
The proof is mostly a tautology after the notation is unwinded. The mechanism of the proof may still be of interest and be relevant for generalizations and for less tautological variations of the setup. We therefore spell out several proofs.
Proof A
Use the end-expression for the hom-objects of the enriched functor categories to compute
\begin{aligned} [C,\mathcal{V}]^{op}(\mathcal{O}(X), A) & := \int_{c \in C} \mathcal{V}(A(c), \mathcal{O}(X)(c)) \\ & := \int_{c \in C} \mathcal{V}(A(c), [C^{op}, \mathcal{V}](X, \mathcal{V}(-,c))) \\ & := \int_{c \in C} \int_{d \in C} \mathcal{V}(A(c), \mathcal{V}(X(d), \mathcal{V}(d,c))) \\ & \simeq \int_{d \in C} \int_{c \in C} \mathcal{V}(X(d), \mathcal{V}(A(c), \mathcal{V}(d,c))) \\ & =: \int_{d \in C} \mathcal{V}(X(d), [C,\mathcal{V}]^{op}(\mathcal{V}(d,-),A)) \\ & =: \int_{d \in C} \mathcal{V}(X(d), Spec(A)(d)) \\ & =: [C^{op}, \mathcal{V}](X, Spec(A)) \end{aligned} \,.
Remark
Here apart from writing out or hiding the ends, the only step that is not a definition is precisely the middle one, where we used that $\mathcal{V}$ is a symmetric closed monoidal category.
The following proof does not use ends and needs instead slightly more preparation, but has then the advantage that its structure goes through also in great generality in higher category theory.
Proof B
Notice that
Lemma 1: $Spec(\mathcal{V}(c,-)) \simeq \mathcal{V}(-,c)$
because we have a natural isomorphism
\begin{aligned} Spec(\mathcal{V}(c,-))(d) & := [C,\mathcal{V}](\mathcal{V}(c,-), \mathcal{V}(d,-)) \\ & \simeq \mathcal{V}(d,c) \end{aligned}
by the Yoneda lemma.
From this we get
Lemma 2: $[C^{op}, \mathcal{V}](Spec \mathcal{V}(c,-), Spec A) \simeq [C,\mathcal{V}](A, \mathcal{V}(c,-))$
by the sequence of natural isomorphisms
\begin{aligned} [C^{op}, \mathcal{V}](Spec \mathcal{V}(c,-), Spec A) & \simeq [C^{op}, \mathcal{V}](\mathcal{V}(-,c), Spec A) \\ & \simeq (Spec A)(c) \\ & := [C, \mathcal{V}](A, \mathcal{V}(c,-)) \end{aligned} \,,
where the first is Lemma 1 and the second the Yoneda lemma.
Since (by what is sometimes called the co-Yoneda lemma) every object $X \in [C^{op}, \mathcal{V}]$ may be written as a colimit
$X \simeq {\lim_\to}_i \mathcal{V}(-,c_i)$
over representables $\mathcal{V}(-,c_i)$ we have
$X \simeq {\lim_\to}_i Spec(\mathcal{V}(c_i,-)) \,.$
In terms of the same diagram of representables it then follows that
Lemma 3:
$\mathcal{O}(X) \simeq {\lim_{\leftarrow}}_i \mathcal{V}(c_i,-)$
because using the above colimit representation and the Yoneda lemma we have natural isomorphisms
\begin{aligned} \mathcal{O}(X)(d) &= [C^{op}, \mathcal{V}](X, \mathcal{V}(-,c)) \\ & \simeq [C^{op}, \mathcal{V}]({\lim_\to}_i \mathcal{V}(-,c_i), \mathcal{V}(-,c)) \\ & \simeq {\lim_\leftarrow}_i [C^{op}, \mathcal{V}](\mathcal{V}(-,c_i), \mathcal{V}(-,c)) \\ & \simeq {\lim_\leftarrow}_i \mathcal{V}(c_i,c) \end{aligned} \,.
Using all this we can finally obtain the adjunction in question by the following sequence of natural isomorphisms
\begin{aligned} [C,\mathcal{V}]^{op}(\mathcal{O}(X), A) & \simeq [C,\mathcal{V}](A, {\lim_\leftarrow}_i \mathcal{V}(c_i,-), ) \\ & \simeq {\lim_{\leftarrow}}_i [C, \mathcal{V}](A, \mathcal{V}(c_i,-)) \\ & \simeq {\lim_{\leftarrow}}_i [C^{op}, \mathcal{V}](Spec \mathcal{V}(c_i,-), Spec A) \\ & \simeq [C^{op}, \mathcal{V}]({\lim_{\to}}_i Spec \mathcal{V}(c_i,-), Spec A) \\ & \simeq [C^{op}, \mathcal{V}](X, Spec A) \end{aligned} \,.
The pattern of this proof has the advantage that it goes through in great generality also on higher category theory without reference to a higher notion of enriched category theory.
Definition
An object $X$ or $A$ is Isbell-self-dual if
• $A \stackrel{}{\to} \mathcal{O} Spec(A)$ is an isomorphism in $[C,\mathcal{V}]$;
• $X \to Spec \mathcal{O} X$ is an isomorphism in $[C^{op}, \mathcal{V}]$, respectively.
Remark
Under certain circumstances, Isbell duality can be extended to large $\mathcal{V}$-enriched categories $C$. For example, if $C$ has a small generating ssubcategory $S$ and a small cogenerating subcategory $T$, then for each $F: C^{op} \to \mathcal{V}$ and $G: C \to \mathcal{V}$, one may construct $\mathcal{O}(F)$ and $Spec(G)$ objectwise as appropriate subobjects in $\mathcal{V}$:
$\mathcal{O}(F)(c) = [C^{op}, \mathcal{V}](F, C(-, c)) \hookrightarrow \int_{s: S} \mathcal{V}(F s, \hom(s, c))$
$Spec(G)(c) = [C, \mathcal{V}](G, C(c, -)) \hookrightarrow \int_{t: T} \mathcal{V}(G t, \hom(c, t))$
Properties
Respect for limits
Choose any class $L$ of limits in $C$ and write $[C,\mathcal{V}]_\times \subset [C,\mathcal{V}]$ for the full subcategory consisting of those functors preserving these limits.
Proposition
The $(\mathcal{O} \dashv Spec)$-adjunction does descend to this inclusion, in that we have an adjunction
$(\mathcal{O} \dashv Spec) : [C, \mathcal{V}]_{\times}^{op} \stackrel{\overset{\mathcal{O}}{\leftarrow}}{\underset{Spec}{\to}} [C^{op}, \mathcal{V}]$
Proof
Because the hom-functors preserves all limits:
\begin{aligned} \mathcal{O}(X)({\lim_{\leftarrow}}_j c_j) & := [C^{op}, \mathcal{V}](X,\mathcal{V}(-,{\lim_{\leftarrow}}_j c_j)) \\ & \simeq [C^{op}, \mathcal{V}](X,{\lim_{\leftarrow}}_j \mathcal{V}(-,c_j)) \\ & \simeq {\lim_{\leftarrow}}_j [C^{op}, \mathcal{V}](X,\mathcal{V}(-,c_j)) \\ & =: {\lim_{\leftarrow}}_j \mathcal{O}(X)(c_j) \end{aligned} \,.
Isbell self-dual objects
Proposition
All representables are Isbell self-dual.
Proof
By Proof B , lemma 1 we have a natural isomorphisms in $c \in C$
$Spec(\mathcal{V}(c,-)) \simeq \mathcal{V}(-,c) \,.$
Therefore we have also the natural isomorphism
\begin{aligned} \mathcal{O} Spec \mathcal{V}(c,-)(d) & \simeq \mathcal{O} \mathcal{V}(-,c) (d) \\ & := [C^{op}, \mathcal{V}](\mathcal{V}(-,c), \mathcal{V}(-,d)) \\ & \simeq \mathcal{V}(c,d) \end{aligned} \,,
where the second step is the Yoneda lemma. Similarly the other way round.
Isbell envelope
See Isbell envelope.
Examples and similar dualities
Isbell duality is a template for many other space/algebra-dualities in mathematics.
Function $T$-Algebras on presheaves
Let $\mathcal{V}$ be any cartesian closed category.
Let $C := T$ be the syntactic category of a $\mathcal{V}$-enriched Lawvere theory, that is a $\mathcal{V}$-category with finite products such that all objects are generated under products from a single object $1$.
Then write $T Alg := [C,\mathcal{V}]_\times$ for category of product-preserving functors: the category of $T$-algebras. This comes with the canonical forgetful functor
$U_T : T Alg \to \mathcal{V} : A \mapsto A(1)$
Write
$F_T : T^{op} \hookrightarrow T Alg$
for the Yoneda embedding.
Definition
Call
$\mathbb{A}_T := Spec(F_T(1)) \in [C^{op}, \mathcal{V}]$
the $T$-line object.
Observation
For all $X \in [C^{op}, \mathcal{V}]$ we have
$\mathcal{O}(X) \simeq [C^{op}, \mathcal{V}](X, Spec(F_T(-))) \,.$
In particular
$U_T(\mathcal{O}(X)) \simeq [C^{op}, \mathcal{V}](X,\mathbb{A}_T) \,.$
Proof
We have isomorphisms natural in $k \in T$
\begin{aligned} [C^{op}, \mathcal{V}](X, Spec(F_T(k))) & \simeq T Alg(F_T(k), \mathcal{O}(X)) \\ & \simeq \mathcal{O}(X)(k) \end{aligned}
by the above adjunction and then by the Yoneda lemma.
All this generalizes to the following case:
instead of setting $C := T$ let more generally
$T \subset C \subset T Alg^{op}$
be a small full subcategory of $T$-algebras, containing all the free $T$-algebras.
Then the original construction of $\mathcal{O} \dashv Spec$ no longer makes sense, but that in terms of the line object still does
Proposition
Set
$Spec A : B \mapsto T Alg(A,B)$
and
$\mathcal{O}(X) : k \mapsto [C^{op}, \mathcal{V}](X, Spec(F_T(k))) \,.$
Then we still have an adjunction
$(\mathcal{O} \dashv Spec) : T Alg^{op} \stackrel{\overset{\mathcal{O}}{\leftarrow}}{\underset{Spec}{\to}} [C^{op}, \mathcal{V}] \,.$
Proof
\begin{aligned} T Alg^{op}(\mathcal{O}(X), A) & := \int_{k \in T} \mathcal{V}( A(k), \mathcal{O}(X)(k) ) \\ & := \int_{k \in T} \mathcal{V}( A(k), [C^{op}, \mathcal{V}](X, Spec(F_T(k))) ) \\ & := \int_{k \in T} \int_{B \in C} \mathcal{V}(A(k), \mathcal{V}(X(B), T Alg(F_T(k), B) )) \\ & \simeq \int_{k \in T} \int_{B \in C} \mathcal{V}(A(k), \mathcal{V}(X(B), B(k) )) \\ & \simeq \int_{k \in T} \int_{B \in C} \mathcal{V}(X(B), \mathcal{V}(A(k), B(k) )) \\ & =: \int_{B \in C} \mathcal{V}(X(B), T Alg(A,B)) \\ & =: \int_{B \in C} \mathcal{V}(X(B), Spec(A)(B)) \\ & =: [C^{op}, Set](X,Spec(A)) \end{aligned} \,.
The first step that is not a definition is the Yoneda lemma. The step after that is the symmetric-closed-monoidal structure of $\mathcal{V}$.
Function $k$-algebras on derived $\infty$-stacks
The structure of our Proof B above goes through in higher category theory.
Formulated in terms of derived stacks over the (∞,1)-category of dg-algebras, this is essentially the argument appearing on page 23 of (Ben-ZviNadler).
Function $T$-algebras on $\infty$-stacks
for the moment see at function algebras on ∞-stacks.
Gelfand duality
Gelfand duality is the equivalence of categories between (nonunital) commutative C-star algebras and (locally) compact topological spaces. See there for more details.
Serre-Swan theorem
The Serre-Swan theorem says that suitable modules over an commutative C-star algebra are equivalently modules of sections of vector bundles over the Gelfand-dual topological space.
duality between algebra and geometry in physics:
References
• Michael Barr, John Kennison, R. Raphael, Isbell Duality, PDF
Isbell conjugation is reviewed on page 17 of
Isbell conjugacy for (∞,1)-presheaves over the (∞,1)-category of duals of dg-algebras is discussed around page 32 of
in
Isbell self-dual ∞-stacks over duals of commutative associative algebras are called affine stacks . They are characterized as those objects that are small in a sense and local with respect to the cohomology with coefficients in the canonical line object.
A generalization of this latter to $\infty$-stacks over duals of algebras over arbitrary abelian Lawvere theories is the content of
• Herman Stel, $\infty$-Stacks and their function algebras – with applications to $\infty$-Lie theory , master thesis (2010) (web) | 4,102 | 11,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 94, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2014-15 | longest | en | 0.768779 |
https://engineersedge.com/plastic/plastic-screw-design.htm | 1,702,276,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00677.warc.gz | 260,516,108 | 6,230 | ### Plastic Screw Design Equations and Calculator
Theoretical equations to calculate the strength of plastic screwed joints.
This calculator requires a Premium Membership
Where:
F = axial force in screw [lbs, N]
f1 = friction between screw head and part
fa = friction derived
fb = friction derived
p = thread pitch [in, mm]
Mh = Torque applied to screw head [in-lbs, N-mm]
MT = Torque applied to screw thread [in-lbs, N-mm]
Stresses in screw shaft formula:
Where:
ro = outside radius of screw core [in, mm]
ri= inside radius of hollow screw; (solid: ri = 0) [in, mm]
σy = tensile yield strength [psi, N/mm2]
A = Area of Screw [in2, mm2]
Ip = Polar moment of Inertia [in4, mm4] | 186 | 683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | latest | en | 0.885847 |
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# Ragini Ravichandren
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## Know Your Sales Tax Rate
Before purchasing anything an essential step in calculating sales tax is finding out what the tax rate in your area is. Different states have varying rates, as do most major cities. Other areas meanwhile have no sales tax at all, and the difference can be as much as 9.4% of the total value of most goods.
Furthermore different types of goods are taxed at a different rates than others. For example prepared foods are often taxed at a different rate than groceries. Similarly items considered to be luxury rather than staple goods are often charged at higher rates.
States without any sales tax are currently Alaska, Delaware, Montana, New Hampshire and Oregon.
Most state government websites have specific sales tax available, but the general rates can be found below or on the table at the bottom of the page:
## Methods of Calculating SalesTax
To calculate the added sales tax required, the first step is to add together the cost of any goods you want to purchase.
Secondly, multiply the total of the goods being purchased by the rate of tax.
For example a goods total of \$150 and a sales tax rate of 7%, would look something like this:
150 x 1.07 = 160.5 (\$160.50)
An alternative method that some people find easier is to divide the goods total by one hundred and then multiply the result by the sales tax rate, and then adding the answer to the original goods value. For example:
150 / 100 = 1.5
1.5 x 7 = 10.5
150 + 10.5 = 160.50 (\$160.50)
## Factoring in Discounts
Where sales tax can become slightly more difficult is when a discount is to be applied to the goods before the tax is added. The process of calculating the discount is the same as calculating tax however, and can be worked out as follows:
Using the same example from before of a \$150 purchase and sales tax of 7%, except this time with a 20% discount applied:
150 / 100 = 1.5
1.5 x 20 = 30
150 - 30 = 120 (\$120)
120 x 1.07 = 128.4 (\$128.40)
Credit: TaxFoundation.org | 477 | 2,017 | {"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} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.951868 |
https://www.101computing.net/bcd-to-7-segment-display/ | 1,726,263,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00642.warc.gz | 574,700,519 | 35,887 | # BCD to 7-Segment Display
#### What is BCD code?
In computing and electronic systems, binary-coded decimal (BCD) is used to encode decimal numbers (base-10 numbers) in a binary form where each decimal digit is represented by a nibble (4 bits).
For instance decimal number 5 is represented as 0101 in BCD as 5 = 4 + 1
8 4 2 1 0 1 0 1
The 10 digits in BCD:
Decimal Digit BCD Nibble 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001
Numbers larger than 9, having two or more digits in the decimal system, are expressed one digit at a time using on nibble per digit. For example, the decimal number 365 is encoded as:
0011 0110 0101
#### 7-Segment Display
A lot of electronic devices use 7-segment displays (Watch, alarm clock, calculator etc.).
Typically 7-segment displays consist of seven individual coloured LED’s (called the segments). Each segment can be turned on or off to create a unique pattern/combination. Each segment is identified using a letter between A to G as follows:
The 10 descimal digits can be displayed on a seven-segment display as follows:
#### BCD to 7-Segment Display: Truth Tables & Karnaugh Maps
We will use four inputs A,B,C and D to represent the four BCD digits as ABCD (A is the most significant digit, D is the least significant digit). When creating an electronic circuit we could use 4 switches to represent these 4 inputs.
We will need 7 outputs one for each segment. So let’s investigate each segment one at a time.
Segment ABCDEFG
Segment A should be on for the following values:
BCD: 0000 BCD: 0010 BCD: 0011 BCD: 0101 BCD: 0110 BCD: 0111 BCD: 1000 BCD: 1001
Segment A should be off for the following values:
BCD: 0001 BCD: 0100
Segment A should also be off for the following BCD values which are not used to represent decimal digits values from 0 to 9:
• 1010
• 1011
• 1100
• 1101
• 1110
• 1111
Note that these values could be used to represent hexadecimal values A to F (10 to 15). Here we will not display anything instead.
Hence the Truth Table for Segment A is as follows:
Inputs Output A B C D Segment A 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 0
This Truth table can be represented using a Karnaugh Map:
Karnaugh Map for Segment A
Follow the same process to define the Truth Table and Karnaugh Map of the B segment:
Follow the same process to define the Truth Table and Karnaugh Map of the C segment:
Follow the same process to define the Truth Table and Karnaugh Map of the D segment:
Follow the same process to define the Truth Table and Karnaugh Map of the E segment:
Follow the same process to define the Truth Table and Karnaugh Map of the F segment:
Follow the same process to define the Truth Table and Karnaugh Map of the G segment:
#### BCD to 7-Segment Display: Boolean Expressions
The Karnaugh maps will help us define the Boolean Expressions associated with each of the 7 segments.
Segment ABCDEFG
Karnaugh Map:
Karnaugh Map – Segment A
Boolean Expression:
Boolean Expression for Segment A
Use the Karnaugh Map for the B segment to define the Boolean Expression of the B segment.
Use the Karnaugh Map for the C segment to define the Boolean Expression of the C segment.
Use the Karnaugh Map for the D segment to define the Boolean Expression of the D segment.
Use the Karnaugh Map for the E segment to define the Boolean Expression of the E segment.
Use the Karnaugh Map for the F segment to define the Boolean Expression of the F segment.
Use the Karnaugh Map for the G segment to define the Boolean Expression of the G segment.
#### BCD to 7-Segment Display: Logic Gates Diagrams
We can now convert each Boolean expression into a Logic Gates circuit to link our 4 inputs (switches) to our 7-segment display using a range of logic gates.
Segment ABCDEFG
Boolean Expression:
Booelan Expression for Segment A
Logic Gates Diagram:
Segment A – Logic Gates Diagram
Use the Boolean Expression of the B segment to draw the logic gates diagram required to control the LED of the B segment.
Use the Boolean Expression of the C segment to draw the logic gates diagram required to control the LED of the C segment.
Use the Boolean Expression of the D segment to draw the logic gates diagram required to control the LED of the D segment.
Use the Boolean Expression of the E segment to draw the logic gates diagram required to control the LED of the E segment.
Use the Boolean Expression of the F segment to draw the logic gates diagram required to control the LED of the F segment.
Use the Boolean Expression of the G segment to draw the logic gates diagram required to control the LED of the G segment.
#### Testing
You can now recreate your logic gates circuit using logic.ly to test if it behaves as expected for all 16 BCD entries.
e.g. For Segment A:
Logic Gates Circuit for Segment A
#### BCD to 7-Segment Display Integrated Circuit
All these 7 logic gates diagrams can all be integrated into one single integrated circuit: The CD74HCT4511E is a CMOS logic high-speed BCD to 7-segment Latch/Decoder/Driver with four inputs and is used to use these 4 inputs (BCD nibble) to control the display of a 7-segment display.
#### Hex/BCD to 7-Segment Display Integrated Circuit
A very similar approach can be used to display hexadecimal digits as these are also based on a nibble per digit.
The extra values that we discarded previously (BCD: 1010, 1011, 1100, 1101, 1110, 1111) can be used to represent the extra 6 digits available in hexadecimal:
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Newton's Second Law of Motion Newton's Second Law of Motion Newton's Second Law of Motion Professor Mac Explains Newton's Second Law of Motion Julius Sumner Miller - Physics - Newton's Second Law pt. 1 Newton's Laws Of Motion (2) : Force, Mass And Acceleration Julius Sumner Miller - Physics - Newton's Second Law pt. 2 Physics: Newton's Second Law for Rotational Motion Physics: Using Newton's second law (2) Newton's Second Law of Motion
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https://inter-base.net/the-atomic-mass-of-an-atom-equals-the-sum-of-the-masses-of-the/ | 1,643,007,775,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00672.warc.gz | 374,060,221 | 5,408 | Learning ObjectiveCalculate the mean atomic fixed of an aspect given that is isotopes and their herbal abundance
Key PointsAn element can have differing number of neutron in that nucleus, but it always has the same variety of protons. The execution of an aspect with various neutrons have different masses and also are called isotopes.The average atomic mass because that an facet is calculation by summing the masses the the element’s isotopes, every multiplied through its organic abundance on Earth.When doing any mass calculations involving elements or compounds, always use mean atomic mass, which can be discovered on the regular table.
You are watching: The atomic mass of an atom equals the sum of the masses of the _____.
Termsnatural abundanceThe variety of a certain isotope naturally uncovered on the planet.average atom massThe fixed calculated by summing the masses of an element’s isotopes, every multiplied by its natural abundance top top Earth.mass numberthe total variety of protons and neutrons in an atomic nucleus.
The atomic number of an aspect defines the element’s identity and signifies the variety of protons in the nucleus of one atom. For example, the aspect hydrogen (the lightest element) will always have one proton in that is nucleus. The element helium will constantly have 2 protons in that is nucleus.
## Isotopes
Atoms that the same element can, however, have differing number of neutron in your nucleus. Because that example, secure helium atom exist that contain one of two people one or two neutrons, but both atoms have actually two protons. This different species of helium atoms have various masses (3 or 4 atom mass units), and they are dubbed isotopes. For any given isotope, the amount of the numbers of protons and neutrons in the cell core is called the mass number. This is since each proton and each neutron sweet one atom mass unit (amu). By including together the number of protons and also neutrons and multiplying through 1 amu, you deserve to calculate the mass of the atom. All elements exist as a collection of isotopes. The word ‘isotope’ comes from the Greek ‘isos’ (meaning ‘same’) and ‘topes’ (meaning ‘place’) since the facets can accounting the same place on the periodic table while being different in subatomic construction.
Lithium AtomStylized lithium-7 atom: 3 proton (red), 4 neutron (blue), and 3 electrons (black). (Lithium likewise has another, rarer isotope with just 2 neutrons. )
## Calculating average Atomic Mass
The typical atomic mass of an element is the amount of the masses the its isotopes, each multiplied by its natural abundance (the decimal connected with percent of atoms of that aspect that room of a provided isotope).
Average atomic mass = f1M1 + f2M2 + … + fnMn wherein f is the portion representing the herbal abundance the the isotope and M is the massive number (weight) of the isotope.
The median atomic mass of an aspect can be discovered on the periodic table, generally under the elemental symbol. Once data are available about the herbal abundance of various isotopes of an element, it is straightforward to calculate the average atomic mass.
For helium, over there is approximately one isotope the Helium-3 because that every million isotopes of Helium-4; therefore, the mean atomic mass is really close to 4 amu (4.002602 amu).Chlorine consists of two significant isotopes, one with 18 neutrons (75.77 percent of natural chlorine atoms) and also one with 20 neutron (24.23 percent of natural chlorine atoms). The atomic variety of chlorine is 17 (it has 17 protons in that is nucleus).
To calculate the mean mass, an initial convert the percentages right into fractions (divide castle by 100). Then, calculate the mass numbers. The chlorine isotope v 18 neutron has an abundance of 0.7577 and a mass variety of 35 amu. To calculation the mean atomic mass, multiply the portion by the fixed number because that each isotope, then include them together.
See more: Fallout: A Look Back At The Height Of The Cold War In America, Circa 1960
Average atom mass of chlorine = (0.7577 \cdot 35 amu) + (0.2423 \cdot 37 amu) = 35.48 amu
Another example is to calculate the atom mass the boron (B), which has actually two isotopes: B-10 with 19.9% organic abundance, and also B-11 through 80.1% abundance. Therefore,
Average atom mass of boron = (0.199\cdot
10 amu) + (0.801\cdot
11 amu) = 10.80 amu
Whenever we do mass calculations involving elements or compounds (combinations that elements), we always use typical atomic masses.
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https://www.mathhomeworkanswers.org/250869/can-the-following-system-be-solved-algebraically | 1,590,848,157,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409337.38/warc/CC-MAIN-20200530133926-20200530163926-00185.warc.gz | 791,990,387 | 21,426 | Can the following system be solved algebraically?
2ln(x) + y= 4
e^x - y = 0
where ln(x) is loge(x)
If so, how would you do it?
Yes. You need to work out x in terms of y or y in terms of x in one equation, then substitute in the other equation. Or simply add the equations together: 2ln(x)+eˣ=4.
Draw a graph of 2ln(x)+eˣ-4. The graph crosses the x axis at around x=1.26, so this an approximate solution. [But without a graph, by substituting x=1 in the equation this function is negative, and by substituting x=2 it’s positive, so the solution lies between x=1 and 2. Continuing in this way the solution can be found (the solution lies between x=1.2 and 1.3 is the next step).]
Newton’s method can then be used to get a more accurate solution. Then y can be found from y=eˣ:
x[n+1]=x[n]−(2ln(x[n])+e^(x[n])-4)/(2/x[n]+e^(x[n])) where x[0]=1.26 and square brackets mean subscript. After only two or three iterations very accurate results are obtained, using a calculator, of course. Other algebraic solutions are possible using iterative procedures, independent of Newton’s method, but they may take longer to produce the same degree of accuracy.
Applying this method, x=1.262415239263 and y=3.53394651461 approx.
Another method is to solve for y by putting x=ln(y) in the first equation:
2ln(ln(y))+y=4, so y=4-2ln(ln(y)). If we start with y=e, then when we work out the right-hand side we get y=4. We then substitute y=4 in the right-hand expression and, using a calculator, we get y=3.3467... Now we put this result into the right-hand side again and the next value of y=3.6220... And so on. Eventually we get the value found above. To find x we use x=ln(y). This method is probably the easiest but it takes longer than Newton’s method.
by Top Rated User (764k points) | 505 | 1,782 | {"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} | 4.5625 | 5 | CC-MAIN-2020-24 | latest | en | 0.854564 |
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# AD 7822 Confusing ~PD and ~CONVST
Good day everyone!
I'm about to use AD 7822 but as I read the datasheet, I got confused regarding the ~PD and ~CONVST. The AD 7822 powers up (getting ready) if ~PD is logic high OR ~CONVST is logic high (after #ns of ~EOC duration). The word OR here confuses me.
1) If I tied ~PD permanently high to Vcc, does that mean the AD 7822 will not power down at all even if after ~EOC, ~CONVST is low? Is the state of ~CONVST ignored regarding power down if ~PD is always high?
2) Does that mean that a 1µs power up time will always be needed before every conversion start even if ~PD is tied to Vcc?
3) Can I use a continuous crystal oscillator (50/50) for ~CONVST, or does ~CONVST just accepts pulses or "per request" conversion? If yes, does that mean frequency of ~CONVST and sampling frequency is equal?
I've wanted to use 7822 with a continuous crystal oscillator for clock (50% duty cycle) and I am trying to compute for the frequency to achieve a little more than 1.5MSPS (my signal is 750kHz). I need a continuous ADC and the 1µs power up time (if it powered down during the signal) will mean that AD 7822 is not for the job.
Thank you very much.
ps. I will use ad 7822 with arduino.
Parents
• Hi Miek3886,
Please refer to response to each question below.
1) If I tied ~PD permanently high to Vcc, does that mean the AD 7822 will not power down at all even if after ~EOC, ~CONVST is low? Is the state of ~CONVST ignored regarding power down if ~PD is always high?
The AD7822 can still enter power down with PD always high. The AD7822 can enter mode 2, automatic power down at the end of conversion. You can refer to figure 25 of the datasheet.
2) Does that mean that a 1µs power up time will always be needed before every conversion start even if ~PD is tied to
Vcc? power up time is not needed every conversion. Power up time is needed during when VDD is first connected and when it has been from a power down (like mode 2).
3) Can I use a continuous crystal oscillator (50/50) for ~CONVST, or does ~CONVST just accepts pulses or "per request" conversion? If yes, does that mean frequency of ~CONVST and sampling frequency is equal?
I am not sure on how would you do it. But if you are going to use crystal oscillator at 50% duty cycle that means high time and low time are equal. Checking on the figure 24 and figure 25, there would be different high time and low time of the CONVST. The low time is much longer as it will consider the conversion time plus the reading of the data time. CONVST can accept pulses, but proper timing must be followed for conversion and reading of data.
Regards,
Jonathan
• Hi Miek3886,
Please refer to response to each question below.
1) If I tied ~PD permanently high to Vcc, does that mean the AD 7822 will not power down at all even if after ~EOC, ~CONVST is low? Is the state of ~CONVST ignored regarding power down if ~PD is always high?
The AD7822 can still enter power down with PD always high. The AD7822 can enter mode 2, automatic power down at the end of conversion. You can refer to figure 25 of the datasheet.
2) Does that mean that a 1µs power up time will always be needed before every conversion start even if ~PD is tied to
Vcc? power up time is not needed every conversion. Power up time is needed during when VDD is first connected and when it has been from a power down (like mode 2).
3) Can I use a continuous crystal oscillator (50/50) for ~CONVST, or does ~CONVST just accepts pulses or "per request" conversion? If yes, does that mean frequency of ~CONVST and sampling frequency is equal?
I am not sure on how would you do it. But if you are going to use crystal oscillator at 50% duty cycle that means high time and low time are equal. Checking on the figure 24 and figure 25, there would be different high time and low time of the CONVST. The low time is much longer as it will consider the conversion time plus the reading of the data time. CONVST can accept pulses, but proper timing must be followed for conversion and reading of data.
Regards,
Jonathan
Children
No Data | 1,064 | 4,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.938825 |
https://discuss.codechef.com/t/fctrl2-wrong-answer/10122 | 1,600,440,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00615.warc.gz | 365,343,919 | 6,289 | **
Here is my block of code kindly tell me what i am doing wrong
**
``````#include <iostream>
using namespace std;
/*
*
*/
int main(int argc, char** argv) {
int testcase, n, i = 1;
long fctrl = 1;
scanf("%d", &testcase);
while(testcase--) {
cin >> n;
while(i <= n) {
fctrl *= i;
i++;
}
printf("%d\n", fctrl);
}
return 0;
}``````
What you are missing is the integer range overflow.
Go through this link for a detailed approach towards calculating factorials of large numbers .
import java.util.*;
class factorial
{
public static void main(String args[])
{
Scanner I=new Scanner(System.in);
int b=I.nextInt();
``````long ar[]=new long[b];
int i;
for(i=0;i<b;i++)
{
int c=I.nextInt();
long fact=1;
for(int j=c;j>1;j--)
{
fact=fact*j;
}
ar[i]=fact;
}
for(i=0;i<b;i++)
{
System.out.println(+ar[i]);
}
``````
}
}
You made at least three mistakes. First is variable overflow. Even using unsigned long long int you can store factorial only up to 22!. What is more you are trying to print long type using %d. %d is for integer, for long you should use %l. Also there is one more - I believe you should set fctrl to 1 for each testcase.
To solve problem do what @mediocoder told you.
#include
using namespace std;
int main() {
int t;
long long int fact=1;
std::cin >> t;
int n[t];
for(int i=0;i<t;i++)
{
std::cin >> n[i];
}
for(int k=0;k<t;k++)
{
int fact=1;
for(int j=1;j<=n[k];j++)
{
`````` fact=fact*j;
}
``````
std::cout << fact <<"\n";
}
``````return 0;
``````
}
What is the problem in this code? Kindly suggest me…
use BigInteger class since the answer can be very big. in c++ try to use a data type that is suitable for holding very large numbers. you can also study the karatsuba algorithm for implementing the same.
1 Like
using namespace std;
int main()
{
int times;
double num,totale;
cin>>times;
double a[times];
double numb[times];
totale = 1;
for(int i =0;i<times;i++)
{
cin>>num;
`````` for(int s=0;s<num;s++)
{
if(num-s == 0)
{
break;
}
totale *= (num-s);
}
a[i] = totale;
totale = 1;
}
for(int i =0;i<times;i++)
{
cout<<a[i]<<endl;
}
return 0;
``````
}
I dont know what is wrong in here!
Please let me know what is wrong with this
https://www.codechef.com/viewsolution/13594809
I have gone through the big computation tutorial and is obviously a good choice but still why wrong answer with this?
#include <stdio.h>
#include <math.h>
#include <string.h>
#define LL long long
#define ULL unsigned long long
#define max(a,b) a>b?a:b;
#define min(a,b) a<b?a:b;
#define fr(i,a) for(i = 0; i < a; i++)
#define fre(i,a) for(i = 1; i <= a; i++)
#define si(a) scanf("%d",&a)
#define sd(a) scanf("%lf",&a)
#define sl(a) scanf("%Ld",&a)
#define ss(a) scanf("%s",a)
LL gcd(LL a, LL b){
if (b == 0) return a;
else return gcd(b,a % b);
}
LL lcm(LL a,LL b){
return (a * b)/ gcd(a,b);
}
LL a[109];
double fact(double n) {
int x = n;
if(a[x]) return a[x];
else return n*fact(n-1);
}
int main() {
int t;
si(t);
while(t–) {
int n;
double ans;
si(n);
a[0] = a[1] = 1;
ans = fact(n);
printf("%.0lf\n", ans);
}
}
You cannot store more than 20! this way. After that it causes overflow.
Remember factorial of numbes like 90, 100 … are to large for a primitive datatypes to handle so there will be an overflow, try using BigInteger Class of Java which helps to handle huge numbers
Check my Solution here
Check this
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# Search results
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Friction Loss (hydraulic slope) - related to pressure change
In fluid flow, friction loss (or skin friction) is the loss of pressure or “head” that occurs in pipe or duct flow due to the effect of the fluid’s ... more
Friction Loss (hydraulic slope)
In fluid flow, friction loss (or skin friction) is the loss of pressure or “head” that occurs in pipe or duct flow due to the effect of the fluid’s ... more
Schmidt Number
Schmidt number (Sc) is a dimensionless number defined as the ratio of momentum diffusivity (viscosity) and mass diffusivity, and is used to characterize ... more
Head loss in terms of volumetric flow rate
Hydraulic head or piezometric head is a specific measurement of liquid pressure above a geodetic datum.
In any real moving fluid, energy is dissipated ... more
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In dimensional analysis, the Strouhal number is a dimensionless number describing oscillating flow mechanisms.
In metrrology, specifically axial-flow ... more
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The constant volume adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat ... more
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In fluid dynamics, the Darcy friction factor ormulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more
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In fluid dynamics, the Darcy friction factor formulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more
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Often during their use, heat exchangers collect a layer of fouling on the surface which, in addition to potentially contaminating a stream, reduces the ... more
Darcy friction factor - Serghides's solution (Variable A)
In fluid dynamics, the Darcy friction factor formulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more
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### Author Topic: Superlumina Paradox - Faster than light? (Read 4117 times)
#### Gasparri
• Jr. Member
• Posts: 21
##### Superlumina Paradox - Faster than light?
« on: 06/09/2009 00:50:31 »
Spaceship A and spaceship B are identical. They are on a mission
to far space. They separate until they are 300,000 Km apart at
which point they lock course and maintain formation. To preserve
secrecy they communicate with each other with lasers using standard
Morse code. They begin their mission and accelerate to 99.9 percent
the speed of light. As they communicate a laser beam takes one
second to reach the other ship. During that same time the ships
have moved ~300,000 Km. This defines a square in space. Actually
a c^2. During it's journey the laser beam traversed the hypotenuse
of the square. The hypotenuse is a bit longer than a side of the c^2
and apparently the laser beams are quite happy to traverse it in
one second flat.
Now, before you get too excited over what you see as a flaw just
remember one thing, the space ships are at rest with respect to
each other, there is no violation. Relativity is preserved yet...
There's that darn hypotenuse thing.
Any suggestions? My brain has gone hypercritical trying to figure this one out.
#### lyner
• Guest
##### Superlumina Paradox - Faster than light?
« Reply #1 on: 06/09/2009 12:11:01 »
The hypotenuse doesn't exist for the two ships - they will not 'observe' one. They are the same distance as ever, as far as they are concerned. They will observe c to be c using their local timing and distance measurements..
For a 'stationary' observer, the light will take an appropriate length of time to traverse the hypotenuse (which he will observe) and his observed time intervals between transmission and reception will also be modified by SR. He will also observe c to be c.
#### Gasparri
• Jr. Member
• Posts: 21
##### Superlumina Paradox - Faster than light?
« Reply #2 on: 06/09/2009 18:15:23 »
The hypotenuse doesn't exist for the two ships - they will not 'observe' one.
Yes, they can only 'know' it's there.
They are the same distance as ever, as far as they are concerned. They will observe c to be c using their local timing and distance measurements..
For a 'stationary' observer, the light will take an appropriate length of time to traverse the hypotenuse (which he will observe) and his observed time intervals between transmission and reception will also be modified by SR. He will also observe c to be c.
I was expecting a space curve like a negative lens which would
shorten the hypotenuse to the proper length.
#### lyner
• Guest
##### Superlumina Paradox - Faster than light?
« Reply #3 on: 07/09/2009 11:10:29 »
When you say "know it's there" - they can deduce what the 'stationary' observer will see if they know his speed relative to them. They know he will see a hypotenuse.
#### Gasparri
• Jr. Member
• Posts: 21
##### Superlumina Paradox - Faster than light?
« Reply #4 on: 07/09/2009 21:40:02 »
When you say "know it's there" - they can deduce what the 'stationary' observer will see if they know his speed relative to them. They know he will see a hypotenuse.
None the less in true space the hypotenuse is there independent
of any observer. The conundrum is what is is. The path is virtual
so the answer must be virtual. Perhaps it's not space that's warped
but velocity that is curved into the Higgs field.
#### Vern
• Neilep Level Member
• Posts: 2072
##### Superlumina Paradox - Faster than light?
« Reply #5 on: 11/09/2009 13:26:20 »
Perhaps it is neither space nor velocity that is distorted. Perhaps it is as Lorentz suggested. Perhaps matter is distorted when it is forced to move relative to a special fixed inertial frame in space. We use material things in all of our measurements. The material things we use for measure are distorted by movement. Maybe their distortion due to movement is due to H. Ziegler's concept as he explained the Lorentz phenomena to Einstein in 1909. Matter must distort when it moves because of its composition.
« Last Edit: 11/09/2009 13:28:33 by Vern »
#### Gasparri
• Jr. Member
• Posts: 21
##### Superlumina Paradox - Faster than light?
« Reply #6 on: 13/09/2009 19:33:11 »
Perhaps it is neither space nor velocity that is distorted...
Matter must distort when it moves because of its composition.
Only under stress of acceleration, either G or g.
When there is neither G or g present the 'mass' is in stasis.
The condition of radiation is rather timeless and seemingly
devoid of real mass. Probably unhooked strings whose only real
option is to fall into the Higgs omni transverse and carry a
a tidbit of energy along at the universal constant of c.
Perhaps the whole universe vibrates at one wavelength of
somewhere around three hundred thousand kilometers. Odd length
that. Nonetheless there the damn thing is and outside of
fiddling with transfer delays in gasses and masses only
the Boaz Einstein condensates so far can fold space.
#### yor_on
• Naked Science Forum GOD!
• Posts: 11731
• Thanked: 1 times
• (Ah, yes:) *a table is always good to hide under*
##### Superlumina Paradox - Faster than light?
« Reply #7 on: 09/03/2010 21:38:05 »
Very nice one. if we assume A and B being at rest with each other and also assume that they are at a uniformly accelerating pace at f.ex one Gravitation, then the equivalence principle say that they will observe it just the same as if they both were on earth.
If we then assume an observer C at rest versus the place A and B once left (Mars:) somehow observing that signals journey through space, he should see it take a much longer path as the vectors would be different from his frame of observation. Not only that, it would also according to him take a longer time than that second those two ships would measure. So who would be right?
Both would be right. That's the new paradigm of relativity, that distances and time are relative the observer. One way to look at it is to assume a plasticity to SpaceTime, not only to time but also to the 'distances' in it. And what changes it is gravity/mass, and acceleration that then becomes equivalent to mass (relative mass/momentum).
There is a hidden 'gold standard' though. you can fit the observations to each other if knowing both frames observations, that as the 'distant observer' will observe the acceleration even if A and B would be inside 'black boxes' themselves, so assuming that his communication can reach them there will be a possibility to fit their observations into a whole. Well, as I see it.
And another thing, that light traveling the longer path, according to our 'distant observer'. Wouldn't it have a red shift too, as observed by that 'distant observer'?
« Last Edit: 10/03/2010 01:24:48 by yor_on »
#### The Naked Scientists Forum
##### Superlumina Paradox - Faster than light?
« Reply #7 on: 09/03/2010 21:38:05 » | 1,723 | 6,915 | {"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-2016-44 | longest | en | 0.931245 |
https://gitlab.inria.fr/why3/why3/-/blame/dadb290880f971cfe61c1c35da32f475c2b032ee/plugins/python/test.py | 1,610,724,996,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00026.warc.gz | 404,927,767 | 13,512 | test.py 818 Bytes
Jean-Christophe Filliâtre committed Jan 29, 2017 1 `````` `````` Jean-Christophe Filliâtre committed Jan 29, 2017 2 3 4 ``````a = 0 b = 1 while b < 100: `````` Jean-Christophe Filliâtre committed Jan 30, 2017 5 6 7 `````` #@ invariant b >= a >= 0 #@ invariant b >= 1 #@ variant 200 - b - a `````` Jean-Christophe Filliâtre committed Jan 29, 2017 8 9 `````` print(a) b = a+b `````` Jean-Christophe Filliâtre committed Jan 30, 2017 10 `````` #@ assert b >= 1 `````` Jean-Christophe Filliâtre committed Jan 29, 2017 11 `````` a = b-a `````` Jean-Christophe Filliâtre committed Jan 30, 2017 12 13 `````` # lists `````` Jean-Christophe Filliâtre committed Jan 30, 2017 14 ``````l = range(0, 10) `````` Jean-Christophe Filliâtre committed Jan 30, 2017 15 ``````#@ assert forall i. 0 <= i < 10 -> l[i] >= 0 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 16 ``````l[2] = 42 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 17 ``````#@ assert l[2] == 42 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 18 19 ``````i = 0 while i < 10: `````` Jean-Christophe Filliâtre committed Jan 30, 2017 20 21 22 `````` #@ invariant 0 <= i #@ invariant forall j. 0 <= j < i -> l[j] == 0 #@ variant 10 - i `````` Jean-Christophe Filliâtre committed Jan 30, 2017 23 24 `````` l[i] = 0 i = i+1 `````` Jean-Christophe Filliâtre committed Jan 29, 2017 25 `````` `````` Jean-Christophe Filliâtre committed Jan 30, 2017 26 ``````sum = 0 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 27 ``````for x in l: `````` Jean-Christophe Filliâtre committed Jan 30, 2017 28 `````` #@ invariant sum >= 0 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 29 `````` print(x) `````` Jean-Christophe Filliâtre committed Jan 30, 2017 30 31 `````` #@ assert x >= 0 sum = sum+x `````` Jean-Christophe Filliâtre committed Jan 30, 2017 32 `````` `````` Jean-Christophe Filliâtre committed Jan 30, 2017 33 34 35 36 37 ``````# Python's division is neither Euclidean division, nor computer division #@ assert 4 // 3 == 1 and 4 % 3 == 1 #@ assert -4 // 3 == -2 and -4 % 3 == 2 #@ assert 4 // -3 == -2 and 4 % -3 == -2 #@ assert -4 // -3 == 1 and -4 % -3 == -1 `````` Jean-Christophe Filliâtre committed Jan 30, 2017 38 `````` `````` Jean-Christophe Filliâtre committed Jan 29, 2017 39 ``````# Local Variables: `````` Jean-Christophe Filliâtre committed Jan 30, 2017 40 ``````# compile-command: "make -C ../.. && why3 prove -P alt-ergo test.py" `````` Jean-Christophe Filliâtre committed Jan 29, 2017 41 ``````# End: `````` Jean-Christophe Filliâtre committed Jan 30, 2017 42 | 976 | 2,578 | {"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} | 2.765625 | 3 | CC-MAIN-2021-04 | latest | en | 0.669786 |
https://www.itl.nist.gov/div898/handbook/mpc/section3/mpc345.htm | 1,718,484,494,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00560.warc.gz | 743,574,463 | 5,610 | 2. Measurement Process Characterization
2.3. Calibration
2.3.4. Catalog of calibration designs
## Designs for angle blocks
Purpose The purpose of this section is to explain why calibration of angle blocks of the same size in groups is more efficient than calibration of angle blocks individually.
Calibration schematic for five angle blocks showing the reference as block 1 in the center of the diagram, the check standard as block 2 at the top; and the test blocks as blocks 3, 4, and 5. A schematic of a calibration scheme for one reference block, one check standard, and three test blocks is shown below. The reference block, R, is shown in the center of the diagram and the check standard, C, is shown at the top of the diagram.
Block sizes Angle blocks normally come in sets of
1, 3, 5, 20, and 30 seconds
1, 3, 5, 20, 30 minutes
1, 3, 5, 15, 30, 45 degrees
and blocks of the same nominal size from 4, 5 or 6 different sets can be calibrated simultaneously using one of the designs shown in this catalog.
Restraint The solution to the calibration design depends on the known value of a reference block, which is compared with the test blocks. The reference block is designated as block 1 for the purpose of this discussion.
Check standard It is suggested that block 2 be reserved for a check standard that is maintained in the laboratory for quality control purposes.
Calibration scheme A calibration scheme developed by Charles Reeve (Reeve) at the National Institute of Standards and Technology for calibrating customer angle blocks is explained on this page. The reader is encouraged to obtain a copy of the publication for details on the calibration setup and quality control checks for angle block calibrations.
Series of measurements for calibrating 4, 5, and 6 angle blocks simultaneously For all of the designs, the measurements are made in groups of seven starting with the measurements of blocks in the following order: 2-3-2-1-2-4-2. Schematically, the calibration design is completed by counter-clockwise rotation of the test blocks about the reference block, one-at-a-time, with 7 readings for each series reduced to 3 difference measurements. For n angle blocks (including the reference block), this amounts to n - 1 series of 7 readings. The series for 4, 5, and 6 angle blocks are shown below.
Measurements for 4 angle blocks
Series 1: 2-3-2-1-2-4-2
Series 2: 4-2-4-1-4-3-4
Series 3: 3-4-3-1-3-2-3
Measurements for 5 angle blocks (see diagram)
Series 1: 2-3-2-1-2-4-2
Series 2: 5-2-5-1-5-3-5
Series 3: 4-5-4-1-4-2-4
Series 4: 3-4-3-1-3-5-3
Measurements for 6 angle blocks
Series 1: 2-3-2-1-2-4-2
Series 2: 6-2-6-1-6-3-6
Series 3: 5-6-5-1-5-2-5
Series 4: 4-5-4-1-4-6-4
Series 5: 3-4-3-1-3-5-3
Equations for the measurements in the first series showing error sources The equations explaining the seven measurements for the first series in terms of the errors in the measurement system are: \begin{array} $$Z_{11} = B + X_1 + \,\,\,\,\,\,\,\,\,\,\,\,\,\, error_{11} \\ Z_{12} = B + X_2 + \,\,\, d + error_{12} \\ Z_{13} = B + X_3 + 2d + error_{13} \\ Z_{14} = B + X_4 + 3d + error_{14} \\ Z_{15} = B + X_5 + 4d + error_{15} \\ Z_{16} = B + X_6 + 5d + error_{16} \\ Z_{17} = B + X_7 + 6d + error_{17} \end{array} with B a bias associated with the instrument, d is a linear drift factor, X is the value of the angle block to be determined; and the error terms relate to random errors of measurement. Calibration procedure depends on difference measurements The check block, C, is measured before and after each test block, and the difference measurements (which are not the same as the difference measurements for calibrations of mass weights, gage blocks, etc.) are constructed to take advantage of this situation. Thus, the 7 readings are reduced to 3 difference measurements for the first series as follows: \begin{array} \( Y_{11} = (Z_{11} - 2Z_{12} + Z_{13}) / 2 \\ Y_{12} = (Z_{13} - 2Z_{14} + Z_{15}) / 2 \\ Y_{13} = (Z_{15} - 2Z_{16} + Z_{17}) / 2 \\ \end{array} For all series, there are 3(n - 1) difference measurements, with the first subscript in the equations above referring to the series number. The difference measurements are free of drift and instrument bias. Design matrix As an example, the design matrix for n = 4 angle blocks is shown below. 1 1 1 1 0 1 -1 0 -1 1 0 0 0 1 0 -1 0 -1 0 1 -1 0 0 1 0 0 -1 1 0 0 1 -1 -1 0 1 0 0 -1 1 0 The design matrix is shown with the solution matrix for identification purposes only because the least-squares solution is weighted (Reeve) to account for the fact that test blocks are measured twice as many times as the reference block. The weight matrix is not shown. Solutions to the calibration designs measurements Solutions to the angle block designs are shown on the following pages. The solution matrix and factors for the repeatability standard deviation are to be interpreted as explained in solutions to calibration designs . As an example, the solution for the design for n=4 angle blocks is as follows: The solution for the reference standard is shown under the first column of the solution matrix; for the check standard under the second column; for the first test block under the third column; and for the second test block under the fourth column. Notice that the estimate for the reference block is guaranteed to be R*, regardless of the measurement results, because of the restraint that is imposed on the design. Specifically, \widehat{Reference} = 0 + R^* \widehat{Check} = \frac{1}{24} \left( \begin{array}{l} \,\,\,\,\, 2.272 Y_{11} + 9.352 Y_{12} + 2.272 Y_{13} \\ -5.052 Y_{21} + 7.324 Y_{22} - 1.221 Y_{23}\\ -1.221 Y_{31} + 7.324 Y_{32} - 5.052 Y_{33} \end{array} \right) + R^* \widehat{Test}_1 = \frac{1}{24} \left( \begin{array}{l} -5.052 Y_{11} + 7.324 Y_{12} - 1.221 Y_{13} \\ -1.221 Y_{21} + 7.324 Y_{22} - 5.052 Y_{23}\\ \,\,\,\,\, 2.272 Y_{31} + 9.352 Y_{32} + 2.272 Y_{33} \end{array} \right) + R^* \widehat{Test}_2 = \frac{1}{24} \left( \begin{array}{l} -1.221 Y_{11} + 7.324 Y_{12} - 5.052 Y_{13} \\ \,\,\,\,\, 2.272 Y_{21} + 9.352 Y_{22} + 2.272 Y_{23}\\ -5.052 Y_{31} + 7.324 Y_{32} - 1.221 Y_{33} \end{array} \right) + R^* Solutions are correct only for the restraint as shown. Calibrations can be run for top and bottom faces of blocks The calibration series is run with the blocks all face "up" and is then repeated with the blocks all face "down", and the results averaged. The difference between the two series can be large compared to the repeatability standard deviation, in which case a between-series component of variability must be included in the calculation of the standard deviation of the reported average. Calculation of standard deviations when the blocks are measured in two orientations For n blocks, the differences between the values for the blocks measured in the top ( denoted by "t") and bottom (denoted by "b") positions are denoted by: \delta_i = \widehat{X}_i^{\, t} - \widehat{X}_i^{\, b} \hspace{0.25in} \mbox{for} \,\,\, i=I, \, \ldots , \, n\,. The standard deviation of the average (for each block) is calculated from these differences to be: {\large s}_{avg \,\, test} = \sqrt{\frac{1}{4(n-1)}\sum_{i=1}^n \delta_i^2} \,\, . Standard deviations when the blocks are measured in only one orientation If the blocks are measured in only one orientation, there is no way to estimate the between-series component of variability and the standard deviation for the value of each block is computed as {\large s}_{test} = K_1 {\large s}_1 where \( K_1$$ is shown under "Factors for computing repeatability standard deviations" for each design and $${\large s}_1$$ is the repeatability standard deviation as estimated from the design. Because this standard deviation may seriously underestimate the uncertainty, a better approach is to estimate the standard deviation from the data on the check standard over time. An expanded uncertainty is computed according to the ISO guidelines. | 2,322 | 7,940 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.901051 |
http://www.instructables.com/id/Electric-Longboard-Tips-and-Tricks/ | 1,527,110,816,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00008.warc.gz | 392,300,292 | 13,608 | # Electric Longboard Tips and Tricks
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## Introduction: Electric Longboard Tips and Tricks
I have built an Electric Longboard and I'm going to share my experiences with you along with some tips and tricks I found neat. There are alot of guides out there with people sharing their builds so what makes this one different? I will mostly focus och the mechanical build. What motor to use, what voltage, and what transmission in a general manner.
## Step 1: Battery and Voltage ⚡
Okey. When I decided on building myself an electric longboard i found this area to be overwhelming. There are alot of people on forums etc. just screaming out what to use and what to avoid but I wanted to dig deeper into the mathematics to fully understand how it all comes together.
### Power
First off we need to discuss power. Everybody wants alot of power for their board. Power is measured in Watts (P) and is calculated as the product of Voltage (U) times Current (I), P = U * I. High current results in high losses. Think of it like a river. Voltage is the width of the river and Current is the speed of the water. With a wide river, lots of water can flow even though the speed of the water is low. To have the same amount of water flowing (Power) on a narrow river, the speed of the water have to be high, splashing all over (losses). To summerize, we want to have a high voltage so that the amperage can be minimized. Losses are bad for both longitivity of the components (through heat) and range of the longboard (through low efficiency).
But beware! In Sweden where i live, voltages above 48 volt is classified as lethal. A 12S battery is about 50,4 volts fully charged.
You want to have a capacitor with a large value (typically 2000uF+) close to your ESC. This is for reducing the risk of voltage spikes on the esc when connecting it to the power supply, and also to help deliver current spikes demanded by the esc.
Here's some LiPo batteries. Get three of these and your good to go! These are thin which makes it easy to
LiPo
Heres some capacitors:
Capacitors
### Switches
Having high voltages in combination with a large capacitor close to the ESC requires another type of switches than what would normaly be used in cars for example. I started out with a 12 V car switch on my e-board and it worked... For one time. The spark that hit the connector inside the switch welded it together. The instant the switch is closed, a very large current will flow to fill the capacitor. Very large currents are good for welding but nothing else. So how do we solve this problem? One solution is to use a so called "sparkless-switch". These use a normal small switch to enable the gate of one or several mosfets. It works like a charm but they are rather expensive.
Another way to go is to use thesesmall connectors. When plugging them in, a resistor of 5 ohms or so is connected for a brief moment before the switch is fully closed. For a 12S battery, it would mean that about 10 amps is flowing through the resistor for a blink of an eye before the capacitor is charged and the switch is closed. Using any online calculator you can see that it takes about 10 milliseconds to fill the capacitor to 67% whith 50 volt supply and 5 ohm series resistance. They also work like a charm and I have been using mine for some time now.
I use my spark free xt90 connector both as an on/off switch and as a safety feature. The positive wire to the ESC goes through the xt90 connector in a loop, enabling the rider to easily disconnect the Power to the esc when riding if something goes wrong.
Spark-free XT90 connectors to use as on/off switch
XT90-S
### Capacity
One thing that seems to confuse people is the amount of energy stored in batteries. When talking car batteries, amp-hours are often used and a battery with 110 Ah has higher capacity than one with 60 Ah, right? Not always. It depends on the voltage of the battery. Car batteries are always 12 V and can therefore be compared using Ah:s. But now that we are talking Lithium-Cells (cause this is probably the kind of cells that you will be using), we have a number of different voltages that people are using, ranging from 6S to 12S. BTW, this means 6 cells in series and 12 cells in series. Since the nominal voltage of a regular lithium cell is 3.7 volt, we could instead of 6S and 12S say 22,2 volt and 44,4 volt. Its just easier to say voltages in terms of how many cells are connected in series.
The best way to measure the capacity of the battery is to use the unit watt-hours Wh. This is simply the product of the Ah value and the nominal voltage of the battery. The battery for my longboard is a 5,2 Ah 9S battery, it has 173 Wh. If I drained it with 173 W it would last for 1 hour.
### Electronics in a Longboard
Building a battery system for a longboard is a mess. There is not a lot of room for components and there are extensive vibrations when riding. Therefore, it is crucial to pay extra attention to isolating all cables and connections. A short can cause explosions in your lithium cells and/or ruin your components. Use heat shrink everywhere you can and be generous when applying hot glue on all open connections. When doing electrical maintenance in a building, the law requires two layers of isolation on all wires. This is a good way to go here as well!
I would really recommend you getting a kit like this. It has all dimensions you need and plenty of it!
Shrink tube
And a glue-gun!
Glue gun
And Another one:
Glue gun #2
### Summary!
1. Use as high voltage as you can to reduce the current, but beware of shocks.
2. Have a large capacitor (2000uF+) in parallel and close to your ESC.
3. Use a spark-free switch or connector.
4. Charge and discharge your cells properly to increase lifetime and safety.
5. Use double layers of isolation on all cables and use hot glue on all open connections.
## Step 2: Transmission Ratio, Motor KV and Speed
Depending on the goal of your build you will want different characteristics. Do you want it for the daily commute or do you want to outrun road bikers? The easiest way, I think, to explain the relation between voltage, motor kV, transmission ratio, speed and torque is to make an example.
### Example build
Let's say you have decided on using a 9S battery (like mine). The voltage is about 35 volt and let's say I want a top speed of 35 km/h. My longboard have wheels with 69 mm diameter. You could make all calculations by hand or you can use this app for example. Mix and match motor kV, transmission ratio and find the combo that will work for you. Most transmissions out there are between 1:2 to 1:3 so there is not much wiggle room there. Motors comes in alot of different kV ratings however, so it may be easier to first settle for a transmission and then chose the motor. In the example above, a 1:3 transmission is used and a 245 kV motor. The top speed is about 37 km/h in theory.
The Power rating of my motor is 2700 W and thats way more than I use it for. About 1300 W is my max setting on my ESC and that gives me just the perfect amount of torque. But if you weight more than me (63kg) you might want more power. However, if your using a one wheel drive like me, you'll find that you need a really sturdy transmission to handle that kind of torque. More of that under the step "My personal experiences". If you got the resources, go for a dual motor build. Better traction, less wear on each transmisson, less unwanted turning when braking/accelerating.
Remember that if you want a high top speed, you will have to input alot of current in the motor to get good torque at slow speeds. It is a trade-off.
EDIT:
I have now switched to using chain instead. So much more reliable. The first belt I used wore down in a few weeks. Just slipped during slightest acceleration. With the chain mounted, no more slipping. Some say that it is more noisy, but I don't hear any of it while riding. The chain need lubrication once in a while but its worth it. The transmission ratio is now close to 1:2 so I need to input more current to get the same acceleration as before, my esc is set at 50A for now. If I were to build a new board, I would definitely go for a chain right away! I bought this one: Chain
Benjamin Vedder has made an excellent guide on this subject. Here it is!
## Step 3: My Personal Experiences
Where I live there are almost no smooth roads. I quickly found that my 69mm 82A wheels are both to small and too hard for a pleasant ride. I will upgrade to at least 80mm and preferably 78a or less in the future. This affects the top speed and low speed torque of the board so I may have to change either the motor (or its kV rating by ching from a delta configuration of the windings to a star configuration) or the transmission ratio.
I also found that the Htd3m (10mm wide) belt is way too weak and handle way to little torque. Belt slips at every hard acceleration and braking, resulting in quick wear. As a quick and easy fix for this I will upgrade the motor pulley and the belt to fit a 15mm belt. 50% larger contact area may solve the belt slip problem. The motor pulley also has a few more teeth then the one I have installed now.
If I were to redo all purchases of components from the start, I would buy a motor with a lower kV rating and get a htd5m belt (they can handle more torque!) or a chain instead. The top speed of my current build is according to the calculator about 37 km/h. And that i found to be just perfect for me.
HTD 5m transmission kit
Kit
HTD 5m motor pulley:
Motor pulley
HTD 5m wheel pulley
Sprocket chain
Chain
EDIT:
I have ordered 90 mm 76a wheels and a transmission with chain. I will update when they are installed and report on my experiences with those.
EDIT:
The 90 mm 76A wheels that i bought are the best! For me anyway ;) Smooth riding even though there's rough asphalt or cobble stones. Combined with the chain I got myself a reliable commute board. The downside of big wheels and changing transmission ratio is that I need to input lots of current to get the amount of acceleration I want. In some future I may rewire the motor from a D-connection to a Y-connection, lowering the kv-rating by the square-root of 3 (~1.73). That should reduce the amount of current needed by the same factor.
Another great upgrade I've done is to replace the box for the battery pack with one made of glass and carbon fibre. The old was made of SAN-plastic and was too brittle. It was a lot of work (and money) making the new one, but it was definitely worth it!
## Recommendations
• ### Large Motors Class
8,186 Enrolled
• ### Casting Contest
We have a be nice policy. | 2,488 | 10,685 | {"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-2018-22 | latest | en | 0.959862 |
https://proofwiki.org/wiki/Product_Space_is_T2_iff_Factor_Spaces_are_T2 | 1,653,495,965,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00768.warc.gz | 526,861,364 | 10,986 | # Product Space is T2 iff Factor Spaces are T2
## Theorem
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_2$ (Hausdorff) space if and only if each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space.
## Proof
### Necessary Condition
This is shown in Factor Spaces of Hausdorff Product Space are Hausdorff.
$\Box$
### Sufficient Condition
This is shown in Product of Hausdorff Factor Spaces is Hausdorff:General Result.
$\blacksquare$ | 231 | 724 | {"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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.686906 |
http://uncyclopedia.wikia.com/wiki/Machine_gun_that_shoots_all_the_digits_of_pi?oldid=5574618 | 1,462,464,811,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127870.1/warc/CC-MAIN-20160428161527-00043-ip-10-239-7-51.ec2.internal.warc.gz | 300,517,136 | 14,242 | # Machine gun that shoots all the digits of pi
Whoops! Maybe you were looking for Pi?
Because of their so-called intelligence, the so-called experts at Wikipedia will never have a proper article about Machine gun that shoots all the digits of pi. We are sorry for their blatant retardedness.
The Machine gun that shoots out all the digits of pi is a type of gun made in Greece (Possibly used in the Hellenic Army) to kill people while it shoots out the digits of pi.
## Old Kingdom Egypt
This gun dates back to the Old Kingdom Egypt, when they used it in the war between Lybia and the Phillistines, With its radius of 3+1/7 Inches. This was a great gun.
## Ancient History
The Greeks remodeled the Egyptian gun and made it a Circle. This was used by the Babylonians, Egyptians and Greeks The oldest model known to date is 1900BC. They were given the circles of 25/8 and 256/81 Both of which are 1% the same with todays version. The Indian text "The weapon guide" shows the value of the weapon is 339/108 or 3.139. Archimedes (287-212BC) Was the first to estimate what the circles is, So he made his gun. By using the equivalent of 96 pieces of other guns, He proved that $3 10/71 < Pi < 3 1/7$ The average of this value was 3.1485
Ptolemy Gave the value of 3.1416 which he may have come from Apollonius of Perga.
Around AD265 The Wei Kingdom mathematician Liu Hu Provided a simple version to calculate the circle to any degree of accuracy. The He carried the calculation of a 3072-gon and put it into his gun. With the value of 3.1416. Later, he developed another way of making his Machine, It was quicker though, with the value of 3.14 with only a 96-gon.
Around AD480 The Chinese mathematician Zu Chong did Pi as 355/113 and showed that 3.1415926 < Pi 3.1415927. Using Liu's method applying to a 12288-gon. This style of gun would last for about 900 years.
## Later History and Machine gun Companies
In 1400AD Madhava of Sangamagrama First known in such series. This was known as the Madhava-Lebiniz Line Then Sangamagra found The Machine-Pi. He even set a record Pi=3.14159265359. Just after he created his new gun. The ML-1. This record was there until beaten by Jamasid Al-kashi Who gave the estimate of pi to 16 digests. The First major european movement in this machine was by German Mathematician Ludolph van Ceulen (1540-1610) Who did the digits to 34 Digits. (Also founder of the Ceulen Machine Gun Co.) And now for the other founders of This machines Companies were:
• Francois Viete 1593
• John Wallis 1655
• John Machin 1706
Formulas after 1706 where known as the Machin Formulas.
## Today
It is now known that the circle of this machine is now about 1 Million digits known. (Not Kidding.)
## How Popular is it, Anyway?
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# Standard Error Mean Difference
## Contents
The researchers report that candidate A is expected to receive 52% of the final vote, with a margin of error of 2%. Figure 1. The standard error (SE) is the standard deviation of the sampling distribution of a statistic,[1] most commonly of the mean. That is used to compute the confidence interval for the difference between the two means, shown just below. http://activews.com/standard-error/standard-error-difference.html
Hutchinson, Essentials of statistical methods in 41 pages ^ Gurland, J; Tripathi RC (1971). "A simple approximation for unbiased estimation of the standard deviation". The critical value is a factor used to compute the margin of error. The sample mean x ¯ {\displaystyle {\bar {x}}} = 37.25 is greater than the true population mean μ {\displaystyle \mu } = 33.88 years. In lieu of taking many samples one can estimate the standard error from a single sample.
## Standard Error Of Difference Calculator
Gurland and Tripathi (1971)[6] provide a correction and equation for this effect. The standard error of the mean (SEM) can be seen to depict the relationship between the dispersion of individual observations around the population mean (the standard deviation), and the dispersion of On a standardized test, the sample from school A has an average score of 1000 with a standard deviation of 100. The unbiased standard error plots as the ρ=0 diagonal line with log-log slope -½.
As a result, we need to use a distribution that takes into account that spread of possible σ's. The difference between the two sample means is 2.98-2.90 = .08. It is rare that the true population standard deviation is known. Standard Deviation Of Difference The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error.
Standard deviation. Standard error of the mean Further information: Variance §Sum of uncorrelated variables (Bienaymé formula) The standard error of the mean (SEM) is the standard deviation of the sample-mean's estimate of a Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners. The sampling distribution of the difference between means.
Select a confidence level. Standard Deviation Of The Difference Between Two Means For any random sample from a population, the sample mean will very rarely be equal to the population mean. The sampling distribution of the difference between sample means has a mean µ1 – µ2 and a standard deviation (standard error). Notice that s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} is only an estimate of the true standard error, σ x ¯ = σ n
## Standard Error Of Difference Between Two Means Calculator
Without doing any calculations, you probably know that the probability is pretty high since the difference in population means is 10. They may be used to calculate confidence intervals. Standard Error Of Difference Calculator In fact, data organizations often set reliability standards that their data must reach before publication. Standard Error Of Difference Definition There is a second procedure that is preferable when either n1 or n2 or both are small.
The problem states that test scores in each population are normally distributed, so the difference between test scores will also be normally distributed. check over here These formulas, which should only be used under special circumstances, are described below. With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%. Casio fx-9860GII Graphing Calculator, BlackList Price: $67.05Buy Used:$56.99Buy New: $67.05Approved for AP Statistics and CalculusBarron's AP StatisticsMartin Sternstein Ph.D.List Price:$18.99Buy Used: $0.01Buy New:$5.18Statistics & Probability with the TI-89Brendan Standard Error Of The Difference Between Means Definition
Well....first we need to account for the fact that 2.98 and 2.90 are not the true averages, but are computed from random samples. Because the 5,534 women are the entire population, 23.44 years is the population mean, μ {\displaystyle \mu } , and 4.72 years is the population standard deviation, σ {\displaystyle \sigma } The formula for the obtained t for a difference between means test (which is also Formula 9.6 on page 274 in the textbook) is: We also need to calculate the degrees his comment is here Elsewhere on this site, we show how to compute the margin of error when the sampling distribution is approximately normal.
What is the 90% confidence interval for the difference in test scores at the two schools, assuming that test scores came from normal distributions in both schools? (Hint: Since the sample Sample Mean Difference Formula The Variability of the Difference Between Sample Means To construct a confidence interval, we need to know the variability of the difference between sample means. The sampling distribution of the difference between means is approximately normally distributed.
## AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots
1. The distribution of the differences between means is the sampling distribution of the difference between means.
2. Keywords: SE of difference Need to learnPrism 7?
3. Find the margin of error.
4. In this analysis, the confidence level is defined for us in the problem.
5. From the variance sum law, we know that: which says that the variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution
The mean age was 33.88 years. In each of these scenarios, a sample of observations is drawn from a large population. This simplified version of the formula can be used for the following problem: The mean height of 15-year-old boys (in cm) is 175 and the variance is 64. Standard Error Of Difference Between Two Proportions Or decreasing standard error by a factor of ten requires a hundred times as many observations.
doi:10.2307/2340569. Using either a Z table or the normal calculator, the area can be determined to be 0.934. We calculate the mean of each of these samples and now have a sample (usually called a sampling distribution) of means. weblink The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years.
Standard Error of the Difference Between the Means of Two Samples The logic and computational details of this procedure are described in Chapter 9 of Concepts and Applications. | 1,438 | 6,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-05 | latest | en | 0.857764 |
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# Same Perimeter, Different Areas pgs. 26 & 27 (Common Core)
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Area and Perimeter: Same Perimeter, Different Areas
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# Negative Percent Error
Yes No Sorry, way to express a change in a variable. Our Story Advertise With Us Site Map Help Write for About Careers at Chemistry, and some other sciences,
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AP EAPCET 2022 - 4th July Morning Shift
+1
-0
A coil of inductance 0.1 H and resistance $$110 \Omega$$ is connected to a source of 110 V and 350 Hz . The phase difference between the voltage maximum and the current maximum is
A
$$\tan ^{-1}(1.5)$$
B
$$\tan ^{-1}(0.5)$$
C
$$\tan ^{-1}(1.73)$$
D
$$\tan ^{-1}(2)$$
2
AP EAPCET 2021 - 20th August Morning Shift
+1
-0
A 20 V AC is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is 12 V, the voltage across the coil is
A
16 V
B
10 V
C
8 V
D
6 V
3
AP EAPCET 2021 - 19th August Evening Shift
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A bulb of resistance $$280 \Omega$$ is supplied with a 200 V AC supply. What is the peak current?
A
Nearly $$1 \mathrm{~A}$$
B
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C
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D
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4
AP EAPCET 2021 - 19th August Morning Shift
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-0
A resonant frequency of a current is $$f$$. If the capacitance is made four times the initial value, then the resonant frequency will become
A
$$\frac{f}{2}$$
B
$$2f$$
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D
$$\frac{f}{4}$$
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https://stackoverflow.com/questions/14260126/how-python-levenshtein-ratio-is-computed | 1,597,133,649,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00029.warc.gz | 498,075,961 | 37,442 | # How python-Levenshtein.ratio is computed
According to the `python-Levenshtein.ratio` source:
https://github.com/miohtama/python-Levenshtein/blob/master/Levenshtein.c#L722
it's computed as `(lensum - ldist) / lensum`. This works for
``````# pip install python-Levenshtein
import Levenshtein
Levenshtein.distance('ab', 'a') # returns 1
Levenshtein.ratio('ab', 'a') # returns 0.666666
``````
However, it seems to break with
``````Levenshtein.distance('ab', 'ac') # returns 1
Levenshtein.ratio('ab', 'ac') # returns 0.5
``````
I feel I must be missing something very simple.. but why not `0.75`?
• I checked the library (at link you given) I am also confuse why he is using `sum`.Also `(1-1/3) = .666..` this correct according to code but also `(1-1/4) = 0.75` How its .5 ? not clear even in the documentation.... But The actual formula to calculate Levenshtein Distance is in my answer. – Grijesh Chauhan Jan 10 '13 at 17:13
Levenshtein distance for `'ab'` and `'ac'` as below:
so alignment is:
`````` a c
a b
``````
Alignment length = 2
number of mismatch = 1
`Levenshtein Distance` is `1` because only one substitutions is required to transfer `ac` into `ab` (or reverse)
Distance ratio = (Levenshtein Distance)/(Alignment length ) = 0.5
EDIT
you are writing
`(lensum - ldist) / lensum` = `(1 - ldist/lensum)` = 1 - 0.5 = 0.5.
But this is matching (not distance)
REFFRENCE, you may notice its written
`Matching %`
``````p = (1 - l/m) × 100
``````
Where `l` is the `levenshtein distance` and `m` is the `length of the longest of the two` words:
(notice: some author use longest of the two, I used alignment length)
``````(1 - 3/7) × 100 = 57.14...
(Word 1 Word 2 RATIO Mis-Match Match%
AB AB 0 0 (1 - 0/2 )*100 = 100%
CD AB 1 2 (1 - 2/2 )*100 = 0%
AB AC .5 1 (1 - 1/2 )*100 = 50%
``````
Why some authors divide by alignment length,other by max length of one of both?.., because Levenshtein don't consider gap. Distance = number of edits (insertion + deletion + replacement), While Needleman–Wunsch algorithm that is standard global alignment consider gap. This is (gap) difference between Needleman–Wunsch and Levenshtein, so much of paper use max distance between two sequences (BUT THIS IS MY OWN UNDERSTANDING, AND IAM NOT SURE 100%)
Here is IEEE TRANSACTIONS ON PAITERN ANALYSIS : Computation of Normalized Edit Distance and Applications In this paper Normalized Edit Distance as followed:
Given two strings X and Y over a finite alphabet, the normalized edit distance between X and Y, d( X , Y ) is defined as the minimum of W( P ) / L ( P )w, here P is an editing path between X and Y , W ( P ) is the sum of the weights of the elementary edit operations of P, and L(P) is the number of these operations (length of P).
• Thanks for the answer, it makes sense, but it doesn't address the aspect that really bothers me: that the two results (obtained with the same code) don't seem to be consistent with each other (i.e. they suggest two different ways of computing the ratio). How can this be? – cjauvin Jan 11 '13 at 14:38
• @cjauvin Did you read my comment to your question ...I have checked and I have the same impression that according to documentation it should be `.75` but two results in your example are contradicts. – Grijesh Chauhan Jan 11 '13 at 18:34
• Yes I've seen your comment, and that's why, although good and interesting, I cannot accept your answer as the solution, because what I'm really after is the reason for the contradiction in this particular piece of code. Maybe I should ask the PL maintainer. – cjauvin Jan 11 '13 at 19:19
• @cjauvin By the change I am working for you ..I meaning i am looking at that file(u liked)..If in case I find something I will answer you back...give me some time... – Grijesh Chauhan Jan 11 '13 at 19:31
• @cjauvin what is your email id? – Grijesh Chauhan Jan 11 '13 at 19:57
By looking more carefully at the C code, I found that this apparent contradiction is due to the fact that `ratio` treats the "replace" edit operation differently than the other operations (i.e. with a cost of 2), whereas `distance` treats them all the same with a cost of 1.
This can be seen in the calls to the internal `levenshtein_common` function made within `ratio_py` function:
``````static PyObject*
ratio_py(PyObject *self, PyObject *args)
{
size_t lensum;
long int ldist;
if ((ldist = levenshtein_common(args, "ratio", 1, &lensum)) < 0) //Call
return NULL;
if (lensum == 0)
return PyFloat_FromDouble(1.0);
return PyFloat_FromDouble((double)(lensum - ldist)/(lensum));
}
``````
and by `distance_py` function:
``````static PyObject*
distance_py(PyObject *self, PyObject *args)
{
size_t lensum;
long int ldist;
if ((ldist = levenshtein_common(args, "distance", 0, &lensum)) < 0)
return NULL;
return PyInt_FromLong((long)ldist);
}
``````
which ultimately results in different cost arguments being sent to another internal function, `lev_edit_distance`, which has the following doc snippet:
``````@xcost: If nonzero, the replace operation has weight 2, otherwise all
edit operations have equal weights of 1.
``````
Code of lev_edit_distance():
``````/**
* lev_edit_distance:
* @len1: The length of @string1.
* @string1: A sequence of bytes of length @len1, may contain NUL characters.
* @len2: The length of @string2.
* @string2: A sequence of bytes of length @len2, may contain NUL characters.
* @xcost: If nonzero, the replace operation has weight 2, otherwise all
* edit operations have equal weights of 1.
*
* Computes Levenshtein edit distance of two strings.
*
* Returns: The edit distance.
**/
_LEV_STATIC_PY size_t
lev_edit_distance(size_t len1, const lev_byte *string1,
size_t len2, const lev_byte *string2,
int xcost)
{
size_t i;
``````
So in my example,
`ratio('ab', 'ac')` implies a replacement operation (cost of 2), over the total length of the strings (4), hence `2/4 = 0.5`.
That explains the "how", I guess the only remaining aspect would be the "why", but for the moment I'm satisfied with this understanding.
• The why: a guess, but since the sum of lengths is used in calculation of the ratio: r = (1- d/L)*100 , with d= lev.distance and L=sum of lengths, you need replacement to be 2 to get a ratio of 0 with a complete replacement. e.g. "abcd" -> "dcba" has a ratio of 0 ( = (1-8/8)*100 ) – dstibbe May 28 '19 at 6:17
• Correction: "abcd" -> "xyzw" will result in a ratio of 0, "abcd" -> "dcba" will result in 25, because the distance is 6, not 8. – dstibbe May 28 '19 at 6:42
`(lensum - ldist) / lensum`
ldist is not the distance, is the sum of costs
Each number of the array that is not match comes from above, from left or diagonal
If the number comes from the left he is an Insertion, it comes from above it is a deletion, it comes from the diagonal it is a replacement
The insert and delete have cost 1, and the substitution has cost 2. The replacement cost is 2 because it is a delete and insert
ab ac cost is 2 because it is a replacement
``````>>> import Levenshtein as lev
>>> lev.distance("ab","ac")
1
>>> lev.ratio("ab","ac")
0.5
>>> (4.0-1.0)/4.0 #Erro, the distance is 1 but the cost is 2 to be a replacement
0.75
>>> lev.ratio("ab","a")
0.6666666666666666
>>> lev.distance("ab","a")
1
>>> (3.0-1.0)/3.0 #Coincidence, the distance equal to the cost of insertion that is 1
0.6666666666666666
>>> x="ab"
>>> y="ac"
>>> lev.editops(x,y)
[('replace', 1, 1)]
>>> ldist = sum([2 for item in lev.editops(x,y) if item[0] == 'replace'])+ sum([1 for item in lev.editops(x,y) if item[0] != 'replace'])
>>> ldist
2
>>> ln=len(x)+len(y)
>>> ln
4
>>> (4.0-2.0)/4.0
0.5
``````
Another example:
The cost is 9 (4 replace => 4*2=8 and 1 delete 1*1=1, 8+1=9)
``````str1=len("google") #6
str2=len("look-at") #7
str1 + str2 #13
``````
distance = 5 (According the vector (7, 6) = 5 of matrix)
ratio is (13-9)/13 = 0.3076923076923077
``````>>> c="look-at"
Although there's no absolute standard, normalized Levensthein distance is most commonly defined `ldist / max(len(a), len(b))`. That would yield .5 for both examples.
The `max` makes sense since it is the lowest upper bound on Levenshtein distance: to obtain `a` from `b` where `len(a) > len(b)`, you can always substitute the first `len(b)` elements of `b` with the corresponding ones from `a`, then insert the missing part `a[len(b):]`, for a total of `len(a)` edit operations.
This argument extends in the obvious way to the case where `len(a) <= len(b)`. To turn normalized distance into a similarity measure, subtract it from one: `1 - ldist / max(len(a), len(b))`. | 2,585 | 8,658 | {"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} | 3.34375 | 3 | CC-MAIN-2020-34 | latest | en | 0.850658 |
https://physics.stackexchange.com/questions/503096/is-magnetic-flux-always-zero | 1,581,965,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00104.warc.gz | 533,797,904 | 31,854 | # Is magnetic flux always zero?
So I have a doubt regarding the equations of Magnetic flux and Gauss' Law for magnetic fields.
According to Gauss' Law for magnetic fields, integral form of div(B) = $$0$$ can be written as:
$$\oint \mathbf{B}.d\mathbf{a} = 0$$
And magnetic flux through a surface is also defined as $$\iint \mathbf{B}.d\mathbf{S} = \Phi _{B}$$
If we combine the two, does it mean that flux is always zero through the surface? I believe $$\iint$$ and $$\oint$$ are the same thing (correct me if I am wrong) ? What is the significance of $$\Phi _{B}$$ then?
• You can think of the Gauss Law flux as a count of the number field lines leaving a closed surface. Magnetic field lines always form closed loops. So each line that comes in ( - flux ) must also go out. – R.W. Bird Sep 16 '19 at 18:40
I believe ∬ and ∮ are the same thing (correct me if I am wrong)?
$$\oint$$ is a closed contour integral symbol while $$\iint$$ is a double integral symbol.
Gauss's Law for magnetism in integral form is
$$\unicode{x222F}_S\,\mathbf{B}\cdot\mathrm{d}\mathbf{A}=0$$
which states that the magnetic flux through a closed (no boundary) surface is zero (there are no magnetic monopoles). The symbol $$\unicode{x222F}_S$$ indicates integration over the closed surface $$S$$.
The magnetic flux through a surface bounded by a closed countour is not necessarily zero. For example, the Maxwell-Faraday equation is (for a surface unchanging with time)
$$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{l} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{A}$$
where $$\Sigma$$ is the surface of integration bounded by the closed countour $$\partial\Sigma$$.
• Shouldn't the surface integral over $\Sigma$ in the last equation be an $\iint$ symbol, for consistency with the notation conventions established earlier in this post? – Emilio Pisanty Sep 16 '19 at 17:35 | 533 | 1,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-10 | latest | en | 0.883696 |
https://www.qrisp.de/general/tutorial/TSP.html | 1,725,809,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00124.warc.gz | 920,195,430 | 14,151 | # Solving the Traveling Salesman Problem (TSP)#
While the previous examples contained a complexity, that could in principle still be managed with a low level QuantumCircuit creation framework, our next problem contains much more conceptual complexity. We believe that this example clearly displays the neccessity of a higher-level language like Qrisp in order to tackle problems of similar or even higher complexity.
The traveling salesman problem is a common example of an NP-hard problem. The task is to find the shortest round trip through a set of cities, which are described by a weighted graph. As shown in this tutorial the traveling salesman problem can be reformulated in terms of a QUBO problem instance and then solved using either VQE or QAOA. A certain disadvantage of this approach is that the amount of binary variables grows as $$\mathcal{O}(n^2)$$ where $$n$$ is the amount of cities. Another approach has been demonstrated in this paper (which has also been implemented in a Qiskit tutorial). The idea is to use quantum phase estimation on successive diagonal operators to retrieve the route distance. The amount of qubits here grows as $$\mathcal{O}(n log(n))$$. Unfortunately the paper provides no details on how to perform the actual quantum search.
We will now introduce a solution that resembles the second method including the quantum search. This involves creating a superposition of all possible routes (excluding those that repeat cities), assessing the distance of the routes, and determining whether the distance is below a certain threshold. The use of Qrisp’s high-level language features simplifies many of these steps. Moreover, employing Qrisp not only streamlines the process, but also greatly improves resource efficiency: The presented code includes more than 10 distinct QuantumVariables distributed across 41 qubits which compiles into a QuantumCircuit that uses only 23 qubits.
The first step is to create a problem instance. We do this by specifying the amount of cities and an adjancecy matrix.
import numpy as np
from qrisp import *
city_amount = 4
distance_matrix = np.array([[0, 0.25, 0.125, 0.5],
[0.25, 0, 0.625, 0.375],
[0.125, 0.625, 0, 0.75],
[0.5, 0.375, 0.75, 0]])/city_amount
## Setting up a superposition of all routes#
Unlike Qiskit’s solution, we will not search the space of every combination of cities but only the permutations (ie. the combinations, which contain each city only once). To do this we need to set up a superposition, where only the permutations are present. We achieve this by bringing a set of QuantumVariables $$s_0, s_1, .. s_k$$, which we call perm_specifiers into uniform superposition. Each computational basis-state of these variables specifies a permutation. We then call a function eval_perm, which creates a QuantumArray containing that permutation.
$\begin{split}U_{\text{eval_perm}} \sum_{s_0 = 0}^n \sum_{s_1 = 0}^n .. \sum_{s_k = 0}^n \left( \bigotimes_{i = 0}^k \ket{s_i} \right) \ket{0}\\ = \sum_{s_0 = 0}^n \sum_{s_1 = 0}^n .. \sum_{s_k = 0}^n \left( \bigotimes_{i = 0}^k \ket{s_i} \right) \ket{\text{eval_perm}(s_0, s_1, .. s_n)}\end{split}$
The first task in creating the oracle is therefore to code the function eval_perm. The idea here is to use “reverse” selection sort. What does that mean? A sorting algorithm is basically a procedure that reverts every permutation to its ordered state. The reversal of this procedure therefore transforms the ordered state (ie. (1,2,3..)) to a permutation, based on a set of variables. For selection sort this set of variables is the index of the minimum element of each iteration. This index specifies, which element is swapped to the first position. Swapping a QuantumArray entry specified by a QuantumFloat to the first position can be achieved by inverting the demux function :
#Receives a QuantumArray qa and a QuantumFloat index and
#then swaps the entry specified by index to the first position of the QuantumArray
def swap_to_front(qa, index):
with invert():
demux(qa[0], index, qa, permit_mismatching_size = True)
We can now define eval_perm:
#Receives a list of QuantumFloats, which specify a permuation
#and then returns QuantumArray holding that permutation
def eval_perm(perm_specifiers):
#Specify the size of the QuantumFloats, which will represent the cities
city_specifier_size = int(np.ceil(np.log2(city_amount)))
#Create the QuantumArray, which will hold the permutations
qa = QuantumArray(QuantumFloat(city_specifier_size), city_amount)
#Initiate the QuantumArray with the identity permutation, ie. (0,1,2..)
qa[:] = np.arange(city_amount)
#Iteratively swap
for i in range(city_amount - 1):
swap_to_front(qa[i:], perm_specifiers[i])
return qa
In this function we first determine, how big the QuantumFloat needs to be in order to represent every city. If we had $$8 = 2^3$$ cities, we would need 3 qubits. We then create a QuantumArray with qtype QuantumFloat and size city_amount. This array will hold the resulting permutation, which is why we initiate the identity permutation. The final part of the function is to iterate through the array and swap the entry specified by perm_specifiers to the i-th entry.
Before we test this function, we create another helper function, which creates the perm_specifiers.
#Function that returns a list of QuantumFloats,
#specifying the permutations (these will be in uniform superposition)
def create_perm_specifiers(city_amount):
perm_specifiers = []
for i in range(city_amount - 1):
qf_size = int(np.ceil(np.log2(city_amount-i)))
temp_qf = QuantumFloat(qf_size)
perm_specifiers.append(temp_qf)
return perm_specifiers
Note that we dynamically redetermine the size of the QuantumFloat before each iteration. To understand why we can do this, consider the last iteration of eval_perm. Here we treat a QuantumArray with only 2 entries. Therefore the last specifier only needs the size of a single qubit.
We can now test our results so far:
>>> perm_specifiers = create_perm_specifiers(city_amount)
>>> for qv in perm_specifiers: h(qv)
>>> perm = eval_perm(perm_specifiers)
>>> multi_measurement(perm_specifiers + [perm])
{(0, 0, 0, OutcomeArray([0, 1, 2, 3])): 0.0312,
(0, 0, 1, OutcomeArray([0, 1, 3, 2])): 0.0312,
(0, 1, 0, OutcomeArray([0, 2, 1, 3])): 0.0312,
(0, 1, 1, OutcomeArray([0, 2, 3, 1])): 0.0312,
(0, 2, 0, OutcomeArray([0, 3, 2, 1])): 0.0312,
(0, 2, 1, OutcomeArray([0, 3, 1, 2])): 0.0312,
(0, 3, 0, OutcomeArray([0, 3, 1, 2])): 0.0312,
(0, 3, 1, OutcomeArray([0, 3, 2, 1])): 0.0312,
(1, 0, 0, OutcomeArray([1, 0, 3, 2])): 0.0312,
(1, 0, 1, OutcomeArray([1, 0, 2, 3])): 0.0312,
(1, 1, 0, OutcomeArray([1, 3, 0, 2])): 0.0312,
(1, 1, 1, OutcomeArray([1, 3, 2, 0])): 0.0312,
(1, 2, 0, OutcomeArray([1, 2, 3, 0])): 0.0312,
(1, 2, 1, OutcomeArray([1, 2, 0, 3])): 0.0312,
(1, 3, 0, OutcomeArray([1, 2, 0, 3])): 0.0312,
(1, 3, 1, OutcomeArray([1, 2, 3, 0])): 0.0312,
(2, 0, 0, OutcomeArray([2, 1, 0, 3])): 0.0312,
(2, 0, 1, OutcomeArray([2, 1, 3, 0])): 0.0312,
(2, 1, 0, OutcomeArray([2, 0, 1, 3])): 0.0312,
(2, 1, 1, OutcomeArray([2, 0, 3, 1])): 0.0312,
(2, 2, 0, OutcomeArray([2, 3, 0, 1])): 0.0312,
(2, 2, 1, OutcomeArray([2, 3, 1, 0])): 0.0312,
(2, 3, 0, OutcomeArray([2, 3, 1, 0])): 0.0312,
(2, 3, 1, OutcomeArray([2, 3, 0, 1])): 0.0312,
(3, 0, 0, OutcomeArray([3, 0, 1, 2])): 0.0312,
(3, 0, 1, OutcomeArray([3, 0, 2, 1])): 0.0312,
(3, 1, 0, OutcomeArray([3, 1, 0, 2])): 0.0312,
(3, 1, 1, OutcomeArray([3, 1, 2, 0])): 0.0312,
(3, 2, 0, OutcomeArray([3, 2, 1, 0])): 0.0312,
(3, 2, 1, OutcomeArray([3, 2, 0, 1])): 0.0312,
(3, 3, 0, OutcomeArray([3, 2, 0, 1])): 0.0312,
(3, 3, 1, OutcomeArray([3, 2, 1, 0])): 0.0312}
In this code snipped, we first create the perm_specifiers and set them to uniform superposition by applying an H gate onto every participating qubit. We then call the eval_perm function to receive a QuantumArray containing the corresponding permutations. Finally we measure all of these variables together using the multi_measurement function. This function allows to measure multiple QuantumVariables at once and returns the results as a dictionary of tuples.
The first 3 numbers within the keys of this dictionary are the permutation specifiers. We see that the 3rd number only varies between 0 and 1, which is due to the previously mentioned dynamic size determination. The values denoted with OutcomeArray denote the permutation. Note that these permutations indeed never contain the same city twice.
## Evaluating route distance#
Now that we have a superposition of states describing every possible route, we have to evaluate the distance of them.
On a classical computer, this function is rather simple
def cl_route_distance(itinerary):
res = 0
for i in range(city_amount):
res += distance_matrix[itinerary[i], itinerary[(i+1)%city_amount]]
return res
We present two approaches to perform this function on a quantum computer. One is based one quantum phase estimation and the other one on loading the distance matrix from a QuantumDictionary.
Quantum Phase Estimation based evaluation
The idea here is to make use of quantum phase estimation while abusing the additive properties of successive phase functions.
We realize these phase functions using the as_hamiltonian decorator. This decorator turns Python functions of ordinary variables into functions of QuantumVariables. The effect of these functions is to apply the result of the original function as a phase onto the quantum state.
@as_hamiltonian
def trip_distance(i, j, iter = 1):
return distance_matrix[i, j]*2*np.pi*iter
In mathematical terms, this function performs the following operation
$\mathrm{U}_{\text{trip_distance}} \ket{i} \ket{j} = \exp(i \cdot \text{trip_distance}(i, j, \text{iter})) \ket{i} \ket{j}$
Therefore, having several of these applied iteratively yields the summed distance as a phase:
$\begin{split}\mathrm{U}_{\text{td}}\mathrm{U}_{\text{td}}\mathrm{U}_{\text{td}}\mathrm{U}_{\text{td}} \ket{\sigma(0)} \ket{\sigma(1)} \ket{\sigma(2)} \ket{\sigma(3)}\\ = \exp(i \sum_{i = 0}^{3} \text{trip_distance}(\sigma(i), \sigma((i+1)\%4), \text{iter})) \ket{\sigma(0)} \ket{\sigma(1)} \ket{\sigma(2)} \ket{\sigma(3)}\end{split}$
Where $$\sigma$$ is the permutation.
We set up the function for performing the successive phase application
def phase_apply_summed_distance(itinerary, iter = 1):
n = len(itinerary)
for i in range(n):
trip_distance(itinerary[i], itinerary[(i+1)%n], iter = iter)
This function can now be used as an input for the quantum phase estimation algorithm.
@lifted
def qpe_calc_perm_travel_distance(itinerary, precision):
return QPE(itinerary,
phase_apply_summed_distance,
precision = precision,
iter_spec = True)
Note the keyword iter_spec which indicates that, instead of repeatedly evaluating phase_apply_summed_distance, the algorithm supplies the amount of iterations as the keyword iter. This reduces the required resources significantly. Of further interest is the lifted decorator. This decorator tells the compiler that this function is qfree and permeable on its inputs and thus can be automatically uncomputed, even though it contains subroutines (QFT!) which aren’t qfree. For more information about these concepts, visit our uncomputation documentation page.
To verify our results we evaluate our function on a randomly chosen itinerary and compare with the classical result:
>>> test_itinerary = QuantumArray(qtype = QuantumFloat(2))
>>> test_itinerary[:] = [2,3,0,1]
>>> cl_route_distance(test_itinerary.most_likely())
0.53125
>>> qpe_res = qpe_calc_perm_travel_distance(test_itinerary, 5)
>>> print(qpe_res)
{0.53125: 1.0}
Quantum Dictionary based evaluation
Another approach to evaluate the route distance is to load the data of the distance matrix into the superposition using a QuantumDictionary.
def qdict_calc_perm_travel_distance(itinerary, precision):
#A QuantumFloat with n qubits and exponent -n
#can represent values between 0 and 1
res = QuantumFloat(precision, -precision)
#Fill QuantumDictionary with values
qd = QuantumDictionary(return_type = res)
for i in range(city_amount):
for j in range(city_amount):
qd[i, j] = distance_matrix[i, j]
#Evaluate result
for i in range(city_amount):
trip_distance = qd[itinerary[i], itinerary[(i+1)%city_amount]]
res += trip_distance
trip_distance.uncompute(recompute = True)
return res
Here, trip_distance is a QuantumFloat which contains the distance between the i-th city and its successor. This QuantumFloat then gets added to the result and afterwards uncomputed. The uncomputation allows the qubit manager to reuse the qubits of trip_distance after each iteration.
Note
The keyword recompute = True is a bit more involved: Since this function will be embedded into an oracle, the result will be uncomputed at some later point. There are now two ways to deal with the uncomputation of the trip_distance QuantumVariables inside the uncomputation of the res QuantumVariable.
• Delay the uncomputation of trip_distance such that the information can be used to uncompute res.
• Recompute trip_distance and use the recomputed value for the uncomputation of res.
There is pros and cons for both strategies. To get a better understanding of this problem you can check our documentation on recomputation. By default the underlying algorithm of Qrisps automatic uncomputation will always go for option 1. The main drawback of this approach is that delaying the uncomputation blocks the involved qubits for the time of the delay. This implies that the qubits of the trip_distance QuantumFloat from each iteration will be blocked, which is not the behavior we want. Much rather we would like to reuse the qubits of iteration 0 for iteration 1 and so on. Because of this, we set recompute = True which will perform the uncomputation immediately and automatically recompute if needed at a later point. All of this might seem complicated but simply setting recompute = True is enough - the rest is handled by the compiler. Once you reach the end of the tutorial, we invite you to compare the qubit count when setting this keyword to False!
We verify our function on our established example.
>>> test_itinerary = QuantumArray(qtype = QuantumFloat(2))
>>> test_itinerary[:] = [2,3,0,1]
>>> qdict_res = qdict_calc_perm_travel_distance(test_itinerary, 5)
>>> print(qdict_res)
{0.53125: 1.0}
We can now compare the required resources for both approaches. We do this by calling the compile method of the QuantumSession, the result is registered in. This method dynamically (de)allocates qubits and returns a QuantumCircuit.
>>> qdict_compiled_qc = qdict_res.qs.compile()
>>> qdict_compiled_qc.cnot_count()
560
>>> qdict_compiled_qc.depth()
299
>>> qdict_compiled_qc.num_qubits()
18
>>> qpe_compiled_qc = qpe_res.qs.compile()
>>> qpe_compiled_qc.cnot_count()
626
>>> qpe_compiled_qc.depth()
742
>>> qpe_compiled_qc.num_qubits()
13
We see that the QuantumDictionary based function is about twice as fast, takes a comparable amount of CNOT gates and needs about 50% more qubits. The qubit overhead is due to the fact that during each iteration, the QuantumFloat trip_distance is allocated and eventually uncomputed. These extra qubits are not present in the QPE based approach.
## Creating the oracle#
The next step is to create the oracle function. For this, we put everything together we have create so far:
from qrisp import auto_uncompute, z
@auto_uncompute
def eval_distance_threshold(perm_specifiers, precision, threshold, method = "qpe"):
itinerary = eval_perm(perm_specifiers)
if method == "qdict":
distance = qdict_calc_perm_travel_distance(itinerary, precision)
elif method == "qpe":
distance = qpe_calc_perm_travel_distance(itinerary, precision)
else:
raise Exception(f"Don't know method {method}")
is_below_treshold = (distance <= threshold)
z(is_below_treshold)
The line distance <= threshold returns a QuantumBool that is in the state $$\ket{\text{True}}$$ if the condition is met. The last line applies a z-gate, to perform the necessary phase flip. Note the auto_uncompute decorator, which uncomputes all local QuantumVariables that have been created inside this function.
## Evaluating the oracle#
Finally, we can evaluate the oracle and find a solution to our pet problem. We do this by calling the prebuild Grover's algorithm function.
#Create permutation specifiers
perm_specifiers = create_perm_specifiers(city_amount)
from math import factorial
#Determine an estimate for the amount of winner states
#(Average number of computational basis states per permutation) * (4 cyclic shifts)*(2 directions)
winner_state_amount = (2**sum([qv.size for qv in perm_specifiers])/factorial(city_amount))*city_amount*2
from qrisp.grover import grovers_alg
#Evaluate Grovers algorithm
grovers_alg(perm_specifiers, #Permutation specifiers
eval_distance_threshold, #Oracle function
kwargs = {"threshold" : 0.4, "precision" : 5, "method" : "qpe"}, #Specify the keyword arguments for the Oracle
winner_state_amount = winner_state_amount) #Specify the estimated amount of winners
#Retrieve measurement
res = multi_measurement(perm_specifiers)
>>> res
{(0, 0, 1): 0.096,
(0, 1, 1): 0.096,
(1, 0, 1): 0.096,
(1, 1, 1): 0.096,
(2, 1, 0): 0.096,
(2, 2, 1): 0.096,
(2, 3, 0): 0.096,
(3, 1, 0): 0.096,
(3, 2, 1): 0.096,
(3, 3, 0): 0.096,
(0, 0, 0): 0.0032,
(0, 1, 0): 0.0032,
(0, 2, 0): 0.0032,
(0, 2, 1): 0.0032,
(0, 3, 0): 0.0032,
(0, 3, 1): 0.0032,
(1, 0, 0): 0.0032,
(1, 1, 0): 0.0032,
(1, 2, 0): 0.0032,
(1, 2, 1): 0.0032,
(1, 3, 0): 0.0032,
(1, 3, 1): 0.0032,
(2, 0, 0): 0.0032,
(2, 0, 1): 0.0032,
(2, 1, 1): 0.0032,
(2, 2, 0): 0.0032,
(2, 3, 1): 0.0032,
(3, 0, 0): 0.0032,
(3, 0, 1): 0.0032,
(3, 1, 1): 0.0032,
(3, 2, 0): 0.0032,
(3, 3, 1): 0.0032}
We see that we have 10 states that have been amplified. This is no surprise as different permutations can represent the same route (and therefore result in the same travel distance). Indeed our estimate for the amount of winner states was pretty good:
>>> winner_state_amount
10.666666666666666
To extract the final solution, we need to evaluate the permuation given by one of the solutions for the permutation_specifiers. A resource efficient possibility would be to recreate the algorithm classically, however we will just evaluate it once more on the simulator
winning_specifiers = create_perm_specifiers(city_amount)
winning_specifiers[0][:] = 0
winning_specifiers[1][:] = 0
winning_specifiers[2][:] = 1
winning_itinerary = eval_perm(winning_specifiers)
>>> winning_itinerary.most_likely()
OutcomeArray([0, 1, 3, 2])
## Benchmarking Performance#
Last but not least, we evaluate some performance indicators. For this, we again compile the QuantumSession to a QuantumCircuit.
>>> qpe_compiled_qc = perm_specifiers[0].qs.compile()
>>> qpe_compiled_qc.depth()
4679
>>> qpe_compiled_qc.cnot_count()
3484
>>> qdict_compiled_qc.num_qubits()
21
These are the values for the quantum phase estimation based approach. To evaluate the QuantumDictionary based algorithm, we set "method" : "qdict" in the grovers_alg call.
>>> qdict_compiled_qc = perm_specifiers[0].qs.compile()
>>> qdict_compiled_qc.depth()
1065
>>> qdict_compiled_qc.cnot_count()
1632
>>> qdict_compiled_qc.num_qubits()
23
If you are interested in ways to improve the performance even further, we recommend checking the Efficient Solution for the TSP. This solution slightly tweeks the one presented here but roughly halves the required resources. | 5,515 | 19,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.916688 |
https://www.auditexcel.co.za/video-tutorial/excel-offset-formula/ | 1,701,659,007,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00315.warc.gz | 754,818,114 | 17,611 | # Excel Offset Formula
How to use the Excel Offset Formula to make excel refer to specific cells or ranges of cells based on inputs within the spreadsheet.
## Excel Offset Formula
The Excel OFFSET formula allows you to change the cell you refer to by moving the reference either up, down, left or right as you require. This can be done automatically and is very useful for cash flow projections.
In this segment you will learn how to activate and use this function.
In this example we have Sales figures for a five month period. But what we know is that these Sales don’t get paid until either a month or two later, and what we’d like to build is when the cash is likely to be received, but to build in the option that we can change how long it takes to get the money. This can be achieved by using the OFFSET function. To activate the OFFSET function,
you click on the cell,
activate the Function Wizard
and find the OFFSET function,
and click Ok,
The first question is, what is its original reference point, and that’ll be just in the same time period so
we’ll click up here
and then you can choose if you want it to move along rows, or do you want to move along columns, in this case we want it to look along columns
And what we want it to do is to look at this column, and depending on the payment terms get the column coupled back. So if in Month 1 there were sales of 100, and there’s a one month payment period we expect that Month 2 we’ll receive the money.
· so in this case it’ll be a negative,
· and then we’ll refer to this column
· and make it absolute
You can also tell the OFFSET function to combine a number of rows, if you put a number in the height
or a number of the columns, if you put a number in the width. However for these purposes we are just going to keep it to a simple OFFSET.
if you click ok
and copy it across,
what you’ll see is at the moment it refers in the same month
but if we say that it takes one month to receive payment
if you just watch here
you’ll see then it immediately, automatically updates the fact that sales in this month will only be paid here
and we can put a 2 here
and that’ll automatically update,
so Month 2 sales of 200, are only going to be received in month 4. | 505 | 2,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-50 | longest | en | 0.932109 |
https://code.activestate.com:443/recipes/475188-minmax-peaks-with-numeric/ | 1,686,239,415,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00016.warc.gz | 195,531,389 | 8,263 | Welcome, guest | Sign In | My Account | Store | Cart
Given a large one-dimensional array, break it into blocks of contstant length and compute min and max for each block, the so-called "peaks data".
Python, 21 lines
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21``` ```from Numeric import * def peaks(data, step): n = len(data) - len(data)%step # ignore tail slices = [ data[i:n:step] for i in range(step) ] peak_max = reduce(maximum, slices) peak_min = reduce(minimum, slices) return transpose(array([peak_max, peak_min])) """example of use: >>> x = sin(arrayrange(0, 3.14, 1e-5)) >>> len(x) 314000 >>> peaks(x,10000) array([[ 0.09982347, 0. ], [ 0.19865953, 0.09983342], ... [ 0.23924933, 0.14112991], [ 0.14112001, 0.04159065]]) """ ```
In a typical waveform display widget one needs to display a sampled sound of length ~1e6 in a GUI window of 800-1000 pixels width. Thus there are around 1024 samples for pixel, and it seems reasonable to draw both min and max values (there are alternatives, such as RMS energy).
Assuming the data is stored in a Numeric.array (numarray.array would work as well) of length n=m*step, the peaks() function returns (m,2)-shaped array containing min and max for each block of length step in data.
Note that the arrays created by slices = ... assignment do not copy the contents of data, so there is little (or no) memory allocation overhead.
The main disadvantage of the recipe is that it requires (steps-1) passes over the array and the CPU cache usage is non-optimal. The obvious one-pass algorithm should perform better, but i can't figure out how to write the one-pass algorithm in Numeric.
Alexander Ross 17 years, 2 months ago
This is quite a bit faster. The following is quite a bit faster. And should be directly translatable to Numeric code.
``````import numpy as np
def peaks(data, step):
data = data.ravel()
length = len(data)
if length % step == 0:
data.shape = (length/step, step)
else:
data.resize((length/step, step))
max_data = np.maximum.reduce(data,1)
min_data = np.minimum.reduce(data,1)
return np.concatenate((max_data[:,np.NewAxis], min_data[:,np.NewAxis]), 1)
``````
Profiling this function on my machine gives 0.015 seconds per call:
``````>>> x = np.sin(np.arrayrange(0, 3.14, 1e-5))
>>> profiler.run("run_n_times('peaks(x,1000)', 1000)").print_stats()
ncalls tottime percall cumtime percall filename:lineno(function)
999 0.190 0.000 14.880 0.015 :1(peaks)
``````
Compare with 0.104 seconds per call:
``````>>> x = sin(arrayrange(0, 3.14, 1e-5))
>>> profiler.run("run_n_times('peaksN(x,1000)', 1000)").print_stats()
ncalls tottime percall cumtime percall filename:lineno(function)
999 2.160 0.002 104.190 0.104 :1(peaksN)
``````
I don't think it is possilbe to combine the two operations (min and max) with functions that numpy provides, but you could probably write an algorithm in C and use weave to inline the code.
Maxim Krikun (author) 17 years, 2 months ago
so what i missed was ufunc's own .reduce() method
thanks a lot for pointing out!
Created by Maxim Krikun on Mon, 27 Mar 2006 (PSF) | 917 | 3,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.784722 |
http://stackoverflow.com/questions/13745241/maximal-sequence-of-equal-elements-in-an-array?answertab=active | 1,427,958,956,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131317570.85/warc/CC-MAIN-20150323172157-00074-ip-10-168-14-71.ec2.internal.warc.gz | 259,883,671 | 20,216 | # maximal sequence of equal elements in an array
I have for homework the exercise: Write a program that finds the maximal sequence of equal elements in an array. Example: {2, 1, 1, 2, 3, 3, 2, 2, 2, 1} = {2, 2, 2}. I came up with this:
``````Console.WriteLine("Enter array lenght");
int arrLenght = int.Parse(Console.ReadLine());
int[] arr = new int[arrLenght];
Console.WriteLine("Enter array elements");
for (int i = 0; i < arr.Length; i++)
{
arr[i] = int.Parse(Console.ReadLine());
}
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == arr[i + 1] && arr[i] == arr[i + 2])
{
Console.WriteLine("Maximal sequence of numbers is: {0},{1},{2}",arr[i],arr[i+1],arr[i+2]);
break;
}
}
``````
This works only if the sequence is exactly 3 numbers long. I have to search the array and find the largest sequence but i don't know how to code this. I'm sorry if the question is silly but i am a newbie and i couldn't find solution anywhere else. Thanks
-
## 7 Answers
``````static void Main(string[] args)
{
Console.WriteLine("enter length of the array");
int n = int.Parse(Console.ReadLine());
int[] myarray = new int[n];
for (int i = 0; i < n ; i++)
{
Console.WriteLine("enter value of array" + " " + i);
myarray[i]=int.Parse(Console.ReadLine());
}
int length = 1;
int start = 0;
int bestlength=0;
int beststart=0;
for (int i = 1; i < n; i++)
{
if (myarray[i - 1] == myarray[i])
{
length++;
continue;
}
if (bestlength<length)
{
bestlength = length;
beststart = start;
}
length = 1;
start = i;
}
Console.WriteLine("the best sequence is");
for (int j = beststart; j < bestlength + beststart; j++)
{
Console.Write(myarray[j] + ",");
}
}
}
}
``````
-
If you are looking for elegance then use Linq
``````var seq = new int[] {2, 1, 1, 2, 3, 3, 2, 2, 2, 1};
int[] max = seq.Select((n, i) => new { Value = n, Index = i})
.OrderBy(s => s.Value)
.Select((o, i) => new { Value = o.Value, Diff = i - o.Index } )
.GroupBy(s => new { s.Value, s.Diff})
.OrderByDescending(g => g.Count())
.First()
.Select(f => f.Value)
.ToArray();
``````
That's why I ♥ Linq
-
LINQ is basicly SQL in C# right? It makes life a lot easier :) – D_Andreev Dec 7 '12 at 13:13
No. Linq is a language of itself. However you can use Linq to SQL to run SQL queries from Linq. msdn.microsoft.com/en-us/library/bb386976.aspx. Also make sure you accept an answer so that people know this is closed. – Adam Nathan Dec 9 '12 at 21:21
With Linq:
``````int count = seq.Count();
int[] maxSeq = seq
.Select((i, index) => new{
Item = i, index,
PrevEqual = index == 0 || seq.ElementAt(index - 1) == i,
NextEqual = index == count - 1 || seq.ElementAt(index + 1) == i,
})
.Where(x => x.PrevEqual || x.NextEqual)
.GroupBy(x => x.Item)
.OrderByDescending(g => g.Count())
.First().Select(x => x.Item).ToArray();
``````
Explanation.
• select an anonymous type with a `bool` property that indicates if it's the same value as the previous
• since we're only interested in those, restrict the query with `Where`
• `GroupBy` the elements with equal values
• then order by the count of each group(descending)
• select the values of the first group(the largest)
• create a new array from the values
Demo
-
This algorithm does not work when the max sequence is at the end of the sequence. Example: { 2, 1, 1, 1 } – Andrea Angella Dec 6 '12 at 14:58
@AndreaAngella: Good catch. Corrected. – Tim Schmelter Dec 6 '12 at 15:49
This is a problem that can be solved with a single iteration of the sequence. It is important to make sure that the algorithm works in all the cases, included when the maximal sequence is at the end of the sequence.
`````` private static IEnumerable<int> GetMaxSequence(IList<int> seq)
{
if (seq == null || seq.Count == 0)
{
return new List<int>();
}
int value = seq[0];
int currentSequenceStartIndex = 0;
int currentSequenceLength = 1;
int maxSequenceStartIndex = 0;
int maxSequenceLength = 0;
for (int i = 1; i < seq.Count; i++)
{
if (seq[i] == value)
{
currentSequenceLength++;
continue;
}
if (currentSequenceLength > maxSequenceLength)
{
maxSequenceLength = currentSequenceLength;
maxSequenceStartIndex = currentSequenceStartIndex;
}
currentSequenceStartIndex = i;
currentSequenceLength = 1;
value = seq[i];
}
if (currentSequenceLength > maxSequenceLength)
{
maxSequenceLength = currentSequenceLength;
maxSequenceStartIndex = currentSequenceStartIndex;
}
return seq.Skip(maxSequenceStartIndex).Take(maxSequenceLength);
}
``````
-
Since that is a homework maybe there is a need to build an algorythm
`````` int[] arr = new int[30];//your array
Random rand = new Random(100);
int maxCount = 0, curCount, value = 0;
for (int i = 0; i < arr.Length; i++)
arr[i] = rand.Next(15);//fill the aray with random values
arr = arr.OrderBy(a => a).ToArray();
for (int i = 0; i < arr.Length; i++)
{
curCount = 1;//found new value. and now count == 1
for (int j = i+1/*search from next array element*/;
j < arr.Length-1/*to the end*/;
j++)
{
if (arr[i] == arr[j])
curCount++;//counts the count
else
break;//met new value
}
if (curCount > maxCount)//we've found new sequence
{
maxCount = curCount;//new sequence length
value = arr[i];//sequence values
i += maxCount;//we don't need to watch the sequence again
}
}
``````
I have no VS in my hands right now to check this, so I hope that works =) Anyway there is an idea
-
Yes this works. Thanks a lot – D_Andreev Dec 6 '12 at 14:38
Great to hear it. You're welcome – heq Dec 6 '12 at 14:51
Without completely giving you the answer to your question I can suggest that you take a look at LINQ as there would definitely be some simple ways of figuring this out. You could start by looking at the `Group`, `Count`, `OrderBy` and `Select` extension methods which could get you the answer.
-
I haven't learned those yet but I will look into them. Thanks for the tip. – D_Andreev Dec 6 '12 at 14:39
Try with this method:
``````int MaximimalSequence<T>(IList<T> list, out T value)
{
T aux = default(T);
value = default(T);
int max = 0, hist = 0;
bool first = true;
foreach (var i in list)
{
if (!first && aux.Equals(i))
{
max++;
}
else
{
first = false;
max = 1;
}
if (hist < max)
{
hist = max;
value = i;
}
aux = i;
}
return hist;
}
``````
To call it:
``````int value;
var maximumSequence = MaximimalSequence<int>(new List<int> { 2, 1, 1, 2, 3, 3, 2, 2, 2, 1 }, out i);
``````
- | 1,858 | 6,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2015-14 | latest | en | 0.437702 |
https://space.stackexchange.com/questions/32286/with-current-or-near-future-projectable-technology-how-long-would-it-take-a-shi?noredirect=1 | 1,716,086,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00310.warc.gz | 483,554,778 | 40,982 | # With current or near-future projectable technology, how long would it take a ship to fly from Earth to a theoretical Earth-Sun L4 or L5 space station
Lets assume there's a space station already built, so this would be a flight off earth, brief acceleration, coasting most of the way, brief deceleration and connect with the station. I'm assuming one or two captured asteroids are already at L4 providing basic materials like Water, Oxygen, Nitrogen, CO2, Metals and Silicon for solar panels, so it doesn't have to be an ultra heavy launch from Earth, but moderately heavy, maybe 2 astronauts and some equipment.
And the theoretical station could orbit L4 or L5 so it would have closer and further from the Earth. The journey wouldn't need to go all the way to the L4 point, 60 degrees ahead but could meet up at say, 45 or 50 degrees.
Using this article as a baseline, it says 260 days is a standard estimate to fly to Mars though faster is possible but difficult. It doesn't say how long the journey is, but eyeballing it, it's a bit more than half way around earth's orbit, so probably a little over 500 million km.
Using the times listed in this article, NASA has already sent ships to land on Mars in as little as 200 days - Mars Odyssey and taking less time, the distance was probably shorter as well. I'm not an expert in such things, but I'm guessing the weight of the payload was probably on the light side to make it in 200 days, for carrying passengers and some equipment current Mars estimates are probably more like the 260 day estimate (as a guess).
A journey to about 50 degrees ahead or behind Earth's orbit would be considerably shorter. maybe as little 75 or 80 million km to dock with my theoretical space station. I get a ballpark estimate of about 50 days. My question is, is that about right or is there something about the L4 or L5 trip that I'm overlooking? Like, it might take longer because the ship would have to fly closer to the sun before it coasts to L4 due to it's higher velocity.
The STEREO missions are probably not relevant as they took much longer.
Any additional relevant information on current or under development ship/engine designs and how that might affect time of travel would be appreciated. Thanks.
• Thank you. I read October 26 and December 15, but didn't read the rest. oops. I'll edit the sentence as I think it's no longer relevant. Nov 26, 2018 at 3:10
• This is a cool question! btw STEREO took about three years to do +/- 60 degrees, and Kepler has taken almost ten years to fall behind by that amount, but it could certainly be done much faster.
– uhoh
Nov 26, 2018 at 3:16
You can't really compare the time for interplanetary voyages just by comparing distances, because everything is moving around the Sun, and the challenge is at least as much to end up moving at the same speed as your target as to be in the same place as it. Also the Sun's gravity has a huge effect on the trajectory.
A journey to L4 or L5 is an interesting one, in that you can get there using as little fuel as you like (once you have escaped from Earth's gravity) at the price of it taking a long time. Let's consider L4, 60 degrees ahead of the Earth. What you do is, strangely, use your engines to slow your motion around the Sun a little. This leaves you in an elliptical orbit whose aphelion (furthest point from the Sun) is at the distance of the Earth, while your perihelion is closer. You start close to the Earth at aphelion. Such an orbit takes a bit less than one year, so when you come to your next aphelion, the Earth hasn't quite completed one full orbit, so you are ahead of it. After enough orbits, you are 60 degree ahead of it, at L4 and you use your engines again to speed up your motion around the sun so that you remain at L4.
So the fuel you need is determined by how much you slow down at the start of this, and speed up at the end, which determines how many orbits of the Sun it takes to get to L4. The fastest version of this uses just one orbit, so you drop to a 10 month elliptical orbit, and 10 months later you are 60 degrees ahead of the Earth -- welcome to L4.
Getting back or going to L5 is the opposite, you accelerate forwards, shifting to an elliptical orbit with perihelion near Earth's orbit, which takes longer than a year. The fastest version takes 14 months.
Of course if you use still more fuel, you can get there faster. The online pork chop plotter is helpful here. For L4 or L5 bring up "Earth" on the menu and then switch to "Custom" and change the MJD field by 1/6 year (61 days). For instance you get:
showing that you can get there in a little over 160 days using about 13 km/s of total delta-V or about 300 days using about 3 km/s. (another variation of the chart shows that in 10000 days I 0.12 km./s is enough, so parcel post can be really cheap, but really slow!).
This finally leaves the question of how much delta-V is feasible with current or near future technology. Of course it depends how much you want to spend. Various source suggest a delta-V of around 9 km/s for a fully refueled SpaceX Starship. Of that 3.2 km/s is needed to escape Earth (in theory you could replace some of that by refueling in elliptical orbit, but that means a lot more launches). Giving about 5.8 km/s for the trip to L4/L5. From the pork chop plot that suggests about 240 or 250 days. Of course that leaves no more propellant for the return journey. | 1,287 | 5,431 | {"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-2024-22 | latest | en | 0.971581 |
https://phifel.com/category/languages/python/ | 1,679,503,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00638.warc.gz | 539,769,991 | 42,094 | # An almost-from-scratch Python example of a simple neural network
### Introduction
A rite of passage in understanding machine learning is writing your own network from scratch. This isn’t usually about making a better framework, it’s about figuring out what’s going on in all those frameworks. What follows is my contribution to this small but growing genre of programming literature.
The code is based on Andrew Trask’s Grokking Deep Learning (Github), which I’ll refer to as GDL below. No, I haven’t finished it, but this is the first major milestone, so I’m documenting it before I forget.
My background is that of a developer. I have been programming for a living since the 1980’s, mostly across the Object-Oriented landscape. I like classes and generalized solutions. It’s how I think, so this will be the framing that I use for this example.
There are two files: SimpleLayer.py, a class that handles the particulars of what a layer in a network needs to do, and simple_nn.py, a file that exercises that class by building a three-layer NN. We’ll walk through simple_nn.py first, which sets up and runs the network. Then we’ll walk through SimpleLayer, which handles training and backpropagation. At the bottom of the post are full code listings, which are also available on GitHub if you want to use this as a basis for experimentation.
### simple_nn.py
Let’s start at the beginning:
```import numpy as np
import matplotlib.pyplot as plt
import src.SimpleLayer as sl```
One of the things that I try to do in these sort of exercises is to keep the amount of libraries to a minimum. For this I use two very vanilla imports, NumPy for math, and Matplotlib for diagrams of the weights changing over time.
Next, some global variables:
```# variables ------------------------------------------
# The samples. Columns are the things we're sampling, rows are the samples
streetlights_array = np.array( [[ 1, 0, 1 ],
[ 0, 1, 1 ],
[ 0, 0, 1 ],
[ 1, 1, 1 ]])
num_streetlights = len(streetlights_array[0])
num_samples = len(streetlights_array)
# The data set we want to map to. Each entry in the array matches the corresponding streetlights_array row
walk_vs_stop_array = np.array([[1],
[1],
[0],
[0]])```
Here is all the data about the lights. Each row is a sample. Each element in the row is a light. There are four rows in this set. They are matched to a classification of each row in the walk/stop array. These values come from GDL, where the premise is that you have a set of samples from three lights (streetlight_array), and a set of samples of actions that happen (walk_vs_stop_array).
Figure 1: The lights and the behaviors from GDL
The goal is to train a network from the input streetlights that produces the right walk/stop output. Now in a real network, we’d worry about overfitting and other related issues, but we’re going to ignore that here.
The next two variables are not in GDL. These are layer_array, which will contain the instances of the SimpleLayer class, and error_plot_mat, which will be used by pyplot to draw a chart of the error converging to zero. Or failing to, as the case may be.
```# set up the dictionary that will store the numpy weight matrices
layer_array = []
error_plot_mat = [] # for drawing plots```
There is one last bit of setup before we start doing things. There are three methods that will be used later in the program:
```# Methods ---------------------------------------------
# activation function: sets all negative numbers to zero
# Otherwise returns x
def relu(x: np.array) -> np.array :
return (x > 0) * x
# This is the derivative of the above relu function, since the derivative of 1x is 1
def relu2deriv(output: np.array) -> np.array:
return 1.0*(output > 0) # returns 1 for input > 0
# return 0 otherwise
# create a layer
def create_layer(layer_name: str, neuron_count: int, target: sl.SimpleLayer = None) -> 'SimpleLayer':
layer = sl.SimpleLayer(layer_name, neuron_count, relu, relu2deriv, target)
layer_array.append(layer)
return layer```
Let’s go through these one at a time. First, I’d like to say that as someone who likes compiling, strong typing, and all those components that keep me from doing dumb things, I am a fan of Python’s accommodation of typing. It could be better, but it helps:
• def relu(x: np.array) -> np.array :
• This is an example of an activation function. It will be used in the layer to determine whether or not a value propagates through the neuron. In this case, all it does is clamp negative values to zero.
• def relu2deriv(output: np.array) -> np.array:
• This is the inverse of the above function, and returns a one (the slope of the line in relu()) if the value is greater than zero.
• def create_layer(layer_name: str, neuron_count: int, target: sl.SimpleLayer = None) -> ‘SimpleLayer’:
• This is what I use to create a layer. You pass in the name for your layer, how many neurons it has, and its ‘target’, or the layer below it. One of the things that I discovered when writing the SimpleLayer class is how intimately layers are connected. In this case, building any other layer than the last requires a target layer. This allows the weights that will manage the influence between the neurons in each layer to be set up properly. It could have just as easily been built from top to bottom, and pointed at the ‘source’ layer.
• The other thing that this method does is to store the newly created layer in the layer_array, which makes experimenting with adding and deleting layers trivial.
Ok, let’s set up the layers in our network! Again, this is a reimplementation of the network built in GDL (chapter 6)
```np.random.seed(0)
#set up the layers from last to first, so that there is a target layer
output = create_layer("output", 1)
output.set_neurons([1])
hidden = create_layer("hidden", 4, output)
hidden.set_neurons([1, 2, 3, 4])
input = create_layer("input", 3, hidden)
input.set_neurons([1, 2, 3])
for layer in reversed(layer_array):
print ("--------------")
print (layer.to_string())```
First, we seed the random generator with a value so that these results are repeatable. It turns out that this network can be made to converge very fast or not to converge simply by picking a different seed. We’ll discuss what this implies later.
So what we’ve created is a stack of layers, built from bottom to top that looks like this:
• Input layer (3 neurons): This is where the streetlights information will be loaded
• Hidden layer (4 neurons): This layer mediates the interactions between the input and output layers, making these interactions nonlinear. It’s what allows deep neural networks to learn nonlinear, discontinuous functions from examples (And remember, that’s all that neural networks do. Though to be fair, it may be all our brains do, too…
• Output layer (1 neuron): This is where the walk/stop values will be used to adjust the output that is generated starting with the original, random weights.
We then print out the contents of each layer. One quick note – I load the neurons up with sequential integers (e.g. [[1. 2. 3.]]). These values get overridden when the system is run, so it’s just a way to quickly verify the as-built neurons :
```--------------
layer input:
target = hidden
source = no source
neurons (row) = [[1. 2. 3.]]
weights (row) =
[[-0.1526904 0.29178823 -0.12482558 0.783546 ]
[ 0.92732552 -0.23311696 0.58345008 0.05778984]
[ 0.13608912 0.85119328 -0.85792788 -0.8257414 ]]
--------------
layer hidden:
target = output
source = input
neurons (row) = [[1. 2. 3. 4.]]
weights (row) =
[[0.09762701]
[0.43037873]
[0.20552675]
[0.08976637]]
--------------
layer output:
target = no target
source = hidden
neurons (row) = [[1.]]
weights (row) =
None
```
I think this output is pretty obvious, aside from the weights, so let’s look at them more closely. But first, a short digression.
Normally when I see diagrams and descriptions of connected layers of neurons, I usually see something like this:
Figure 2: The typical neural network diagram
As you can see, each neuron in the input layer is connected to each neuron in the hidden layer and so on through to the output layer. And that’s nice conceptually, but as a developer, I have no understanding of the mechanics of what’s happening. Here’s what it really looks like. For clarity, only the interactions involving input neuron 1 and hidden neuron 4 are shown, but the process is identical:
Figure 3: How the input and hidden layer are actually connected
In this case, we’re looking at the mapping between the input layer and the hidden layer. Each neuron in the input layer gets its own row of weights, let’s say [0.1, 0.2, 0.0, 0.5] for neuron one. If that neuron is set to “10”, then a value of 1.0 will go to hidden neuron 1, a value of 2 to hidden neuron 2, and a value of 5 to hidden neuron 4. This process is repeated for each neuron in the input layer, and the value is added to the associated hidden neuron.
That’s what we mean when we talk about fully connected layers. Everything is mediated through an adjacency matrix of weights. We’ll revisit this in more detail when we walk through SimpleLayer. So in the listing above the two figures, the randomly initialized weights are organized so that the each neuron in the layer has its own row. Each entry in that row is the scalar value that the row’s neuron value will be multiplied by as it is accumulated in the target’s neuron. Source neurons are the row component. Target neurons are the column component.
Next is the body of the program:
```alpha = 0.2
iter = 0
max_iter = 1000
epsilon = 0.001
error = 2 * epsilon
while error > epsilon:
error = 0
for sample_index in range(num_samples):
input.set_neurons(streetlights_array[sample_index])
for layer in reversed(layer_array):
layer.train()
delta = output.calc_delta(walk_vs_stop_array[sample_index])
sample_error = np.sum(delta ** 2)
error += sample_error
for layer in layer_array:
layer.learn(alpha)
# Gather data for the plots
error_plot_mat.append([sample_error])
# print("{}.{} Error = {:.5f}".format(iter, sample_index, sample_error))
error /= num_samples
if (iter % 10) == 0 :
print("{} Error = {:.5f}".format(iter, error))
iter += 1
# stop even if we don't converge
if iter > max_iter:
break```
Let’s talk about the local variables first. The first variable, alpha, is the learning rate that we pass in. It’s a scalar that limits the step size in the change in weights. The bigger the scalar, the more likely to overshoot the goal and go into oscillation around it. The smaller the goal, the longer the approach will take, but the greater the chance that it will stabilize. Like the seed we use to set the random number generator, the number of layers, and the number of neurons per layer, this is a hyperparameter. Some, like alpha, result in predictable behavior. Others, like seed, do not. There is a lot of this in deep learning, and you need to be careful about it. In particular, testing the resiliency of the solution by running it with a variety of nonlinear hyperparameters to see if the results are consistent is probably a good idea, though it sucks up compute resources.
The rest of the variables are used for loop control:
• iter is the current count of times through the loop
• max_iter is the maximum times we’ll run through the loop, even if we don’t converge
• epsilon is the error threshold. If the error drops below that, we’re done.
• error is the sum of the squares of all the output neurons (in this case, one). We initialize it to a value that gets us into the loop
```while error > epsilon:
error = 0
sample_error_array = []
for sample_index in range(num_samples):
input.set_neurons(streetlights_array[sample_index])```
This is the main loop. First, we’re going to loop until our error is small. Error is computed by sample, so we need to know what the average (or max – I use average here) error is for each iteration. We also want to save the individual errors by sample for later plotting.
Within each loop, we’re going to evaluate the input streetlights sample against the output walk/stop sample. The first step in this process is to set the input neurons. This is where the input.set_neurons([1, 2, 3]) that we did when we were creating the layers gets overridden. In the training, the output from this layer will overwrite the values in the next layer and so on.
```for layer in reversed(layer_array):
layer.train()```
This is the training step. We’ll go into more detail when we walk through SimpleLayer, but for now not that we set through all the layers from the top to the bottom, in reversed order from how they were created and loaded into layer_array.
```delta = output.calc_delta(walk_vs_stop_array[sample_index])
sample_error = np.sum(delta ** 2)
error += sample_error```
This is where we calculate the array of deltas that are the difference between the goal of the walk/stop array and the output neurons. the error is the sum of the squares of all those deltas. SoS is nice because it’s always positive.
```for layer in layer_array:
layer.learn(alpha)
# Gather data for the plots
sample_error_array.append(sample_error)
# print("{}.{} Error = {:.5f}".format(iter, sample_index, sample_error))```
Learning is done from bottom to top, using the deltas stored in the output layer. These are backpropagated through the layers, and the changes in the weights are scaled to 20% of the calculated values so we settle nicely.
We also gather the error data (for each streetlight-walk/stop sample) into a matrix that we can print out when we’re done. If we want to, we can print the error for each sample in the training. Some converge faster than others, but this is not the best way to see that.
```error /= num_samples
if (iter % 10) == 0 :
print("{} Error = {:.5f}".format(iter, error))
iter += 1
# stop even if we don't converge
if iter > max_iter:
break```
At the bottom of the loop, we calculate the average error over all the samples. We then see if we’ve been here too long and break if we are, regardless of whether we’ve converged or not. And lastly, this is how I like to print formatted strings in Python (essentially the same as “%.5f” in Java/C/etc).
Once the loop terminates, we need to see how well the network has learned. As I said earlier, in a real machine learning situation we would be careful about issues such as overfitting by, for example, training against one set of data and testing against another. But since this is a toy problem, so we are simply going to see how it did with the training data. I’ve added some explicit variables for clarity:
• prediction: The contents of the single neuron in the output layer
• observed: The value in the walk/stop array that we’re evaluating against
• accuracy: how close did we get?
```print("\n--------------evaluation")
for sample_index in range(len(streetlights_array)):
input.set_neurons(streetlights_array[sample_index])
for layer in reversed(layer_array):
layer.train()
prediction = float(output.neuron_row_array)
observed = float(walk_vs_stop_array[sample_index])
accuracy = 1.0 - abs(prediction - observed)
print("sample {} - input: {} = pred: {:.3f} vs. actual:{} ({:.2f}% accuracy)".
format(sample_index, input.neuron_row_array, prediction, observed, accuracy*100.0))```
Since the network is already set up with weights, all we need to do is to see how well our inputs match to our outputs. All this means is to take a set of inputs and run them forward to the model. There will be no learning via backpropagation.
So let’s see how we did!
```0 Error = 0.35238
10 Error = 0.29001
20 Error = 0.19074
30 Error = 0.12883
40 Error = 0.04666
50 Error = 0.00544
--------------evaluation
sample 0 - input: [[1. 0. 1.]] = pred: 0.978 vs. actual:1.0 (97.78% accuracy)
sample 1 - input: [[0. 1. 1.]] = pred: 1.000 vs. actual:1.0 (100.00% accuracy)
sample 2 - input: [[0. 0. 1.]] = pred: 0.037 vs. actual:0.0 (96.27% accuracy)
sample 3 - input: [[1. 1. 1.]] = pred: 0.000 vs. actual:0.0 (99.95% accuracy)```
As you can see, the values converge in less than 60 iterations, and the predictions are quite close. For the second and fourth stoplight pattern, the results are basically exact (100% and 99.95%). That’s not bad for a bunch of random numbers and two simple rules.
These are the kinds of outputs that you get with heavyweight packages like Keras. It’s helpful (We trained successfully! Horay!). And these types of outputs make sense when models are huge – or even bigger toy problems like MNIST (which we will explore in a future post).
But this is toy code for a toy problem so we can show more than that. Being able to visualize what’s going on is very helpful. That’s why the error for each step has been saved in error_plot_mat.
Plotting data like this in Python is one of the joys of using the language. Here’s what it takes:
```# plots ----------------------------------------------
fig_num = 1
f1 = plt.figure(fig_num)
plt.plot(error_plot_mat)
plt.title("error")
names = []
for i in range(num_samples):
names.append("sample_{}".format(i))
names.append("average")
plt.legend(names)
for layer in reversed(layer_array):
if layer.target != None:
fig_num += 1
layer.plot_weight_matrix(var_name='sample_{}'.format(fig_num),fig_num=fig_num)
for layer in reversed(layer_array):
fig_num += 1
layer.plot_neuron_matrix(fig_num)
plt.show()```
We are going to be creating a bunch of plots. One for the error, and then one for each set of neurons and their weights. We’ll get back to the layer plots when we’re walking through SimpleLayer, but here’s a plot of all the errors, by sample and average for the entire training session:
Figure 4: Error for each sample
Some things worth noting are this is not a linear process. There are times where the learning process is pretty slow, particularly at the beginning in this example. The second observation is that zero error happens much sooner for some samples than others. The first sample with zero error happens around step 150 (iteration 37 or so of the main loop). If the exit condition were based on looking at one sample instead of the average of all the sample errors, the system could exit early. I had this happen when I was using sample_error rather than error in the exit condition. It took a while to figure out why some seed values behaved so differently from others….
And that ends the tour of the main loop. Next, we’ll look at how a layers interact to train and learn.
### SimpleLayer
The previous section is roughly equivalent to a Keras, Torch, or other machine learning framework. You get an idea of the behavior of a system and how the construction affects the output, but the details of the implementation are hidden. In this section, we’re going to look at the creation of a layer in detail – the ways they are connected and the ways that they communicate. As with the walkthrough of the main loop, we’ll start with the construction of the layer, then the forward learning process, the training backpropagation process, and graph what’s going on.
#### Construction
As with simple_nn.py, SimpleLayer is written to have very few dependencies. I actually struggled with whether or not to write my own matrix math, but I think NumPy is pretty clear, and it would get distracting with all the additional code.
```import numpy as np
import matplotlib.pyplot as plt
import types
import typing```
There are some class-wide variables that we should describe:
```class SimpleLayer:
name = "unset"
neuron_row_array = None
neuron_col_array = None
weight_row_mat = None
weight_col_mat = None
plot_mat = [] # for drawing plots
num_neurons = 0
delta = 0 # the 'movement' scalar
target = None
source = None
activation_func = None
derivative_func = None```
In order of declaration, these are
• name: the string name of the layer. Used in printing and surprisingly useful in debugging
• neuron_row_array: the neurons in row form (i.e. [[n1, n2, n3, … , nN])
• neuron_col_array: the transpose of neuron_row_array (i.e. [[n1], [n2], [n3], … ,[nN]]. We need the data in both forms for interactions between layers
• weight_row_mat: the weights in row format, as above
• weight_col_mat: the weights in column format, as above
• weight_history_mat: where the weight data from each training pass is stored for plotting
• neuron_history_mat: where the neuron data from each training pass is stored for plotting
• num_neurons: the number of neurons in this layer
• delta: the scalar that changes the size of the “step” this layer takes as it tries to converge on the goal. Passed in as alpha in simple_nn.py
• target: the layer “below” this layer. May be NULL
• source: the layer “above” this layer. May be NULL
• activation_func: the function that controls the nonlinearity of the training process. Passed in as relu() from simple_nn
• derivative_func: the function used in backpropagation that is the derivative of the activation function. Passed in as relu2deriv() in simple_nn
Next is the initialization, which is done through the constructor:
```# set up the layer with the number of neurons, the next layer in the sequence, and the activation/backprop functions
def __init__(self, name, num_neurons: int, activation_ptr: types.FunctionType, deriv_ptr: types.FunctionType, target: 'SimpleLayer' = None):
self.reset()
self.activation_func = activation_ptr
self.derivative_func = deriv_ptr
self.name = name
self.num_neurons = num_neurons
self.neuron_row_array = np.zeros((1, num_neurons))
self.neuron_col_array = np.zeros((num_neurons, 1))
# We only have weights if there is another layer below us
for i in range(num_neurons):
self.neuron_history_mat.append([])
if(target != None):
self.target = target
target.source = self
self.weight_row_mat = 2 * np.random.random((num_neurons, target.num_neurons)) - 1
self.weight_col_mat = self.weight_row_mat.T```
This takes the values supplied in the create_layer() method in simple_nn.py and bulds the layer. Once the local variables are set, the matricies of neurons are created.
If there is a target, the two layers are connected. What this means is that the source layer creates a numpy matrix that has as many rows as the source neurons and as many columns as the target neurons (See figure 3). This matrix is the weights that are used to uniquely distribute the value of each neuron in the source layer to each neuron in the target layer. As with the neurons, this is stored in row and column form.
Once each layer is set up, we are ready to begin the training process.
#### Training
Training a neural network is the process of take a set of input values and sending them through the entire network to get an output. We can compare that output to the desired value, and then adjust. Using the mechanism of a deep neural network allows us to build a system that can map many input values to a desired output value. In this case, we’re looking at three values in an array, but using exactly the same structure, we can increase the number of values to be the pixels in an image and the output to be the label for that image:
Figure 5: The CFAR-10 Dataset
That takes more layers and some other tricks, but the basic technique is the same.
Ok, back to three values in an array that represent some streetlights. To get this into the input layer, we use the set_neurons() method:
```# Fill neurons with values
def set_neurons(self, val_list: typing.List):
# print("cur = {}, input = {}".format(self.neuron_array, val_list))
for i in range(0, len(val_list)):
self.neuron_row_array[0][i] = val_list[i]
self.neuron_col_array = self.neuron_row_array.T```
The numpy neuron arrays are actually two-dimensional arrays that are one element deep. This supports numpy array math like dot product and transpose. That’s why the awkward syntax where we take the val_list and set the neurons to those values. We then take the transpose immediately so that I don’t have to wonder if it’s been done already.
The next step is to ripple the values through the network layers:
```def train(self):
# if not the bottom layer, we can record values for plotting
if(self.target != None):
self.weight_history_mat.append(self.nparray_to_list(self.weight_row_mat))
# if we're not the top layer, propagate weights
if self.source != None:
src = self.source
# set our neuron values as the dot product of the source neurons, and the source weights
self.neuron_row_array = np.dot(src.neuron_row_array, src.weight_row_mat)
# No activation function to output layer
if(self.target != None):
# Adjust the values based on the activation function. This introduces nonlinearity.
# For example, the relu function clamps all negative values to zero
self.neuron_row_array = self.activation_func(self.neuron_row_array)
# Transpose the neuron array and save for learn()
self.neuron_col_array = self.neuron_row_array.T
# record values for plotting
for i in range(self.num_neurons):
self.neuron_history_mat[i].append(self.neuron_row_array[0][i])```
We start to see how intimately the layers are connected in this method. We look to the target and source layers to adjust our behaviors and set values.
Since this is the top layer, we have no source. That means that record our weights for later plotting and we’re done. The layer below us will set its neurons based this layer’s weights and neurons, as handled in this line:
`self.neuron_row_array = np.dot(src.neuron_row_array, src.weight_row_mat)`
This is just the first step. If we’re not the bottom layer, we have to see if the neuron values make it past the activation function that we set in simple_nn.py:
```# activation function: sets all negative numbers to zero
# Otherwise returns x
def relu(x: np.array) -> np.array :
return (x > 0) * x```
This is done with these lines:
```# No activation function to output layer
if(self.target != None):
# Adjust the values based on the activation function. This introduces nonlinearity.
# For example, the relu function clamps all negative values to zero
self.neuron_row_array = self.activation_func(self.neuron_row_array)```
By running these same methods on each successive layer object, the streetlight values are slowly, and nonlinearly (in multi-layer networks, which is critical) modified to produce a single output. Unfortunately, that output is guaranteed to be wrong, since it’s based on multiplying the input values by a bunch of random values that we set up each layer with.
Time to fix that.
#### Learning
Back in simple_nn.py, between the train() and the learn() loops is this line:
`delta = output.calc_delta(walk_vs_stop_array[sample_index])`
The delta saves out the error for the plotting. The function sets up the values for the learning step:
```def calc_delta(self, goal: np.array) -> float:
self.delta = goal - self.neuron_row_array
return self.delta```
self.delta is a numpy array that stores the difference between the goal(s) and the current value. In this case, there is only one value, but this also works with multiple values. That’s another trick that gets used in training networks. For example, in handling the CIFAR images, there is an output neuron for each category (e.g. horse, automobile, truck, ship, etc.). In out toy example and in the CIFAR case, the goal is a one or zero in the output neuron(s). The delta is the difference between the computed value and the goal. That delta is what we will now backpropagate through the layers, from back to front. And that’s the learning process.
In learning, the basic goal is to adjust the weights that set this layer’s neurons (in this implementation, the source layer). This is done by backpropagating the error delta from this layer to the source layer. Since we only want to adjust the weights that participated in the training, we need to take the derivative of the activation function in train(). Again, the weight matrix is simply the source neurons times this layer’s neurons. For example, if the source layer had three neurons and this layer had four, then the (source) weight matrix would be 3*4 = 12 weights. The whole method is shown below.
```def learn(self, alpha):
if self.source != None:
src = self.source
delta_scalar = np.dot(self.delta, src.weight_col_mat)
delta_threshold = self.derivative_func(src.neuron_row_array)
src.delta = delta_scalar * delta_threshold
mat = np.dot(src.neuron_col_array, self.delta)
src.weight_row_mat += alpha * mat
src.weight_col_mat = src.weight_row_mat.T```
There’s a lot going on here, so let’s go through it slowly:
```def learn(self, alpha):
# if there is a layer above us
if self.source != None:
src = self.source```
Since weights exist between neurons, we once more have the intimate relationship between this layer’s neurons and the layer above this layer. If there is no layer above us, there is literally nothing to do, which is why this test is first.
`delta_scalar = np.dot(self.delta, src.weight_col_mat)`
Next, we calculate the error delta scalar array, which is the amount the source layer needs to change (set initially in the output layer’s calc_delta(), then rippled up through the layers), multiplied across the weights used to set this layer’s neurons (in the source).
`delta_threshold = self.derivative_func(src.neuron_row_array)`
In the train() process, we distributed the values in a non-linear way – any neuron value below zero was not distributed. (the relu() function from simple_nn.py). That process needs to be mirrored in the backpropagation process. There is always a matched pair of methods that make the core of a neural network – the activation function, and the derivative function.
`src.delta = delta_scalar * delta_threshold`
This is where the actual change for the source layer is calculated. It’s the product of the delta_scalar and the delta_threshold that we’ve just calculated. This is where the decision process of the derivitive_func() is scaled to the desired amount (the alpha value that we pass in from simple_nn.py). This value will be used when the learn() method is called for the source layer. Like I said, layers are intimately connected.
We now take the self.delta that was calculated in our target layer’s learn() method, and use it to adjust the weights in the source layer that will be used to set our neuron’s values on the next train() pass.
```mat = np.dot(src.neuron_col_array, self.delta)
src.weight_row_mat += alpha * mat
src.weight_col_mat = src.weight_row_mat.T```
This matrix (mat) contains the adjustments for the source layer’s weights. We want to add a fraction of these (or we won’t converge) values, so we multiply by alpha. The last step is simply making the transpose of the weight matrix.
And that’s pretty much the guts of this implementation. The important things to remember are:
• Input and output layers are special cases. The neurons are explicitly set in the input layer and there is no activation or derivative function applied to the output neurons (no, I don’t know why yet. When I figure that out, I’ll explain why here)
• In training, the current layer’s neuron’s values are set by multiplying the source neurons by the source weights.
• In learning, the source layer’s weights are adjusted by the current layer’s deltas, but thresholded by the derivative of the source layer’s neurons
This is pretty complicated, and I’ve split out the steps so that it’s possible to step through the running code in the debugger and see what’s going on with the values. But that only gives a level of insight at a single step. how can we show the global behavior of a layer?
#### Graphing
We are going to graph both the changing value of the neurons and the evolving weights. The neurons are an easier problem so we’ll start there:
```def plot_neuron_matrix(self, fig_num: int):
title = "{} neuron history".format(self.name)
plt.figure(fig_num)
np_mat = np.array(self.neuron_history_mat)
plt.plot(np_mat.T, '-o', linestyle=' ', ms=2)
names = []
for i in range(self.num_neurons):
names.append("neuron {}".format(i))
plt.legend(names)
plt.title(title)```
This method simply takes the history matrix (where we had a column for each time sample), turns it into a numpy array for easier manipulation and plotting, and plots the transpose (where each neuron’s history is a row). Because the neuron’s values change for each sample, the history of how they converge towards the final values doesn’t show up well with lines, so I set the drawing arguments to points ‘-o’, no line ‘ ‘ , with a point size of 2 pixels (ms=2):
Figure 6: Neuron Histories by layer (click to embiggen)
In the input layer, we see that the neurons are either one or zero, just as we set them. These values are then multiplied by the (initially random) weights and further adjusted by the activation function. Those values ripple through the hidden layer, where they are initially random overthe (0, 1) interval where they are then used to adjust the output neurons (which does not involve an activation function). Over time, you can see the system settle into a state where all neurons are either one or zero, depending on the inputs. So how do the weights achieve this?
Even in this toy system, there are still a lot of weights to keep track of, and I’m still working on a way of visualizing the process. I’m visualizing the weights instead of the neurons, because the weights are the “factors in the equation” that manipulate the “x” values to get a “y”. On other words, I’m watching how the “m” and “b” converge on their values in “y = mx + b”, rather than looking at a particular “x” or “y”.
The method that does this is plot_weight_matrix(), which assembles a chart for each set of weights and is called at the end of simple_nn.py:
```def plot_weight_matrix(self, fig_num: int):
var_name = "weight"
title = "{} to {} {}".format(self.name, self.target.name, var_name)
plt.figure(fig_num)
np_mat = np.array(self.weight_history_mat)
i = 0
for row in np_mat.T:
cstr = "C{}".format(i % self.num_neurons)
plt.plot(row, linewidth = int(i / self.num_neurons)+1, color=cstr)
i += 1
names = []
num_weights = self.num_neurons * self.target.num_neurons
for i in range(num_weights):
src_n = i % self.num_neurons
targ_n = int(i/self.num_neurons)
names.append("{} s:t[{}:{}]".format(var_name, src_n, targ_n))
plt.legend(names)
plt.title(title)```
One of the reasons that I really like OO programming is that so much useful data is associated with the object. You don’t have to go looking for it, or scope things in peculiar ways. As a result, for example, generating the title is simply assembling some strings that I already have lying around.
`title = "{} to {} {}".format(self.name, self.target.name, var_name)`
The next important step is to get the data that we’ve been assembling in train() into a form that the plotting library likes. The data has been assembled in an list of lists, where each individual list is a snapshot of the weights at one step in the training process. I do it this way because of two reasons:
1. I don’t know how many steps this process is going to take, and python lists handle dynamic memory allocation nicely.
2. The weight matrix is a 2D NumPy array, and dealing with a series of matricies is something that PyPlot has no idea how to handle.
Here’s the line from train():
```if(self.target != None):
self.plot_mat.append(self.nparray_to_list(self.weight_row_mat))```
PyPlot doesn’t really like to handle lists of lists, but it does know how to handle one big NumPy array, so we convert the list of lists to a matrix where the rows are the weights, and the columns are the timesteps:
`np_mat = np.array(self.plot_mat)`
At this point we could simply plot everything:
`plt.plot(np_mat)`
That produces a pretty chart:
Figure 7: First pass at drawing a lot of weights
But it’s pretty confusing. There are a lot of lines. So I did two related things. I set line thickness to be a function of which target neuron and the color of the line to be a function of which source neuron. Using the same scheme, I built the a legend to indicate the source and target neurons that identify each weight, using the coordinates of the matrix – basically treating it as an adjacency matrix.
```i = 0
for row in np_mat.T:
cstr = "C{}".format(i % self.num_neurons)
plt.plot(row, linewidth = int(i / self.num_neurons)+1, color=cstr)
i += 1
names = []
num_weights = self.num_neurons * self.target.num_neurons
for i in range(num_weights):
src_n = i % self.num_neurons
targ_n = int(i/self.num_neurons)
names.append("{} s:t[{}:{}]".format(var_name, src_n, targ_n))
plt.legend(names)
plt.title(title)```
And that gives a chart that lets us examine what’s going on. All the blue lines are the weights that adjust the value coming from source neuron one, distributed over target neurons [0, 1, 2, 3]. All the thin lines are all the weights that set the value of target neuron one from source neurons [0, 1, 2]:
Figure 8: Weights between input and hidden layers
We now have a way to visualize the whole process inside the layers. Let’s see if we can learn anything by looking at how the neurons and weights coevolve over time.
### Some final thoughts
I think the fundamental lesson here is one of gradient descent (or hill climbing if you prefer) from a random initial state to a stable set of values that will set the variables in a function. Once those values are found, the function can do it’s job, which in this case is taking a set of observations – ([ 1, 0, 1 ], [ 0, 1, 1 ], [ 0, 0, 1 ], [ 1, 1, 1 ]) and transforming them to a different set of values – ([1], [1], [0], [0]).
Figure 9: Weights influencing neurons
This is at its core stochastic, a mechanism for harnessing randomness by using rules. The weights and neurons exist in a constrained, multidimensional space. Much of this is fixed before a single iteration – the number of neurons and how they are arranged. The types of connections (activation and derivative functions). The initial value of the weights. Even the manner of input and the “fitness test” that determines the error that is measured. Within these constraints, the weights move slowly under multiple influences until they settle into places that they are no longer forced to move. That’s it.
Variations in this system can be used for all kinds of things, ranging from image recognition to generating words, but the basic process is always the same.
I hope this helped you to read as much it helped me to write!
### Full code listings
For the most current versions, please use the GitHub repo, but these are up to date as of January 10, 2019
#### simple_nn.py
```'''
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and
associated documentation files (the "Software"), to deal in the Software without restriction, including
without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the
following conditions:
The above copyright notice and this permission notice shall be included in all copies or substantial
portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT
LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO
EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER
IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR
THE USE OR OTHER DEALINGS IN THE SOFTWARE.
'''
import numpy as np
import matplotlib.pyplot as plt
import src.SimpleLayer as sl
# Methods ---------------------------------------------
# activation function: sets all negative numbers to zero
# Otherwise returns x
def relu(x: np.array) -> np.array :
return (x > 0) * x
# This is the derivative of the above relu function, since the derivative of 1x is 1
def relu2deriv(output: np.array) -> np.array:
return 1.0*(output > 0) # returns 1 for input > 0
# return 0 otherwise
# create a layer
def create_layer(layer_name: str, neuron_count: int, target: sl.SimpleLayer = None) -> 'SimpleLayer':
layer = sl.SimpleLayer(layer_name, neuron_count, relu, relu2deriv, target)
layer_array.append(layer)
return layer
# variables ------------------------------------------
np.random.seed(0)
alpha = 0.2
# the samples. Columns are the things we're sampling, rows are the samples
streetlights_array = np.array( [[ 1, 0, 1 ],
[ 0, 1, 1 ],
[ 0, 0, 1 ],
[ 1, 1, 1 ]])
num_streetlights = len(streetlights_array[0])
num_samples = len(streetlights_array)
# The data set we want to map to. Each entry in the array matches the corresponding streetlights_array row
walk_vs_stop_array = np.array([[1],
[1],
[0],
[0]])
# set up the dictionary that will store the numpy weight matrices
layer_array = []
error_plot_mat = [] # for drawing plots
#set up the layers from last to first, so that there is a target layer
output = create_layer("output", 1)
output.set_neurons([1])
''' # If we want to have four layers (two hidden), use this and comment out the other hidden code below
hidden2 = create_layer("hidden2", 2, output)
hidden2.set_neurons([1, 2])
hidden = create_layer("hidden", 4, hidden2)
hidden.set_neurons([1, 2, 3, 4])
'''
# If we want to have three layers (one hidden), use this and comment out the other hidden code above
hidden = create_layer("hidden", 4, output)
hidden.set_neurons([1, 2, 3, 4])
input = create_layer("input", 3, hidden)
input.set_neurons([1, 2, 3])
for layer in reversed(layer_array):
print ("--------------")
print (layer.to_string())
iter = 0
max_iter = 1000
epsilon = 0.001
error = 2 * epsilon
while error > epsilon:
error = 0
sample_error_array = []
for sample_index in range(num_samples):
input.set_neurons(streetlights_array[sample_index])
for layer in reversed(layer_array):
layer.train()
delta = output.calc_delta(walk_vs_stop_array[sample_index])
sample_error = np.sum(delta ** 2)
error += sample_error
for layer in layer_array:
layer.learn(alpha)
# Gather data for the plots
sample_error_array.append(sample_error)
# print("{}.{} Error = {:.5f}".format(iter, sample_index, sample_error))
error /= num_samples
sample_error_array.append(error)
error_plot_mat.append(sample_error_array)
if (iter % 10) == 0 :
print("{} Error = {:.5f}".format(iter, error))
iter += 1
# stop even if we don't converge
if iter > max_iter:
break
print("\n--------------evaluation")
for sample_index in range(len(streetlights_array)):
input.set_neurons(streetlights_array[sample_index])
for layer in reversed(layer_array):
layer.train()
prediction = float(output.neuron_row_array)
observed = float(walk_vs_stop_array[sample_index])
accuracy = 1.0 - abs(prediction - observed)
print("sample {} - input: {} = pred: {:.3f} vs. actual:{} ({:.2f}% accuracy)".
format(sample_index, input.neuron_row_array, prediction, observed, accuracy*100.0))
# plots ----------------------------------------------
fig_num = 1
f1 = plt.figure(fig_num)
plt.plot(error_plot_mat)
plt.title("error")
names = []
for i in range(num_samples):
names.append("sample_{}".format(i))
names.append("average")
plt.legend(names)
for layer in reversed(layer_array):
if layer.target != None:
fig_num += 1
layer.plot_weight_matrix(fig_num)
for layer in reversed(layer_array):
fig_num += 1
layer.plot_neuron_matrix(fig_num)
plt.show()```
#### SimpleLayer.py
```'''
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and
associated documentation files (the "Software"), to deal in the Software without restriction, including
without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the
following conditions:
The above copyright notice and this permission notice shall be included in all copies or substantial
portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT
LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO
EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER
IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR
THE USE OR OTHER DEALINGS IN THE SOFTWARE.
'''
import numpy as np
import matplotlib.pyplot as plt
import types
import typing
# methods --------------------------------------------
class SimpleLayer:
name = "unset"
neuron_row_array = None
neuron_col_array = None
weight_row_mat = None
weight_col_mat = None
weight_history_mat = [] # for drawing plots
neuron_history_mat = []
num_neurons = 0
delta = 0 # the amount to move the source layer
target = None
source = None
activation_func = None
derivative_func = None
# set up the layer with the number of neurons, the next layer in the sequence, and the activation/backprop functions
def __init__(self, name, num_neurons: int, activation_ptr: types.FunctionType, deriv_ptr: types.FunctionType, target: 'SimpleLayer' = None):
self.reset()
self.activation_func = activation_ptr
self.derivative_func = deriv_ptr
self.name = name
self.num_neurons = num_neurons
self.neuron_row_array = np.zeros((1, num_neurons))
self.neuron_col_array = np.zeros((num_neurons, 1))
# We only have weights if there is another layer below us
for i in range(num_neurons):
self.neuron_history_mat.append([])
if(target != None):
self.target = target
target.source = self
self.weight_row_mat = 2 * np.random.random((num_neurons, target.num_neurons)) - 1
self.weight_col_mat = self.weight_row_mat.T
def reset(self):
self.name = "unset"
self.target = None
self.neuron_row_array = None
self.neuron_col_array = None
self.weight_row_mat = None
self.weight_col_mat = None
self.weight_history_mat = [] # for drawing plots
self. neuron_history_mat = []
self.num_neurons = 0
self.delta = 0 # the amount to move the source layer
self.target = None
self.source = None
# Fill neurons with values
def set_neurons(self, val_list: typing.List):
# print("cur = {}, input = {}".format(self.neuron_array, val_list))
for i in range(0, len(val_list)):
self.neuron_row_array[0][i] = val_list[i]
self.neuron_col_array = self.neuron_row_array.T
def get_plot_mat(self) -> typing.List:
return self.weight_history_mat
# In training, the basic goal is to set a value for the layer's neurons, based on the weights in the source layer mediated by an activation function.
# This matrix is simply the source neurons times this layer's neurons. For example, if the source layer had three neurons and this layer had four, then
# the (source) weight matrix would be 3*4 = 12 weights.
def train(self):
# if not the bottom layer, we can record values for plotting
if(self.target != None):
self.weight_history_mat.append(self.nparray_to_list(self.weight_row_mat))
# if we're not the top layer, propagate weights
if self.source != None:
src = self.source
# set our neuron values as the dot product of the source neurons, and the source weights
self.neuron_row_array = np.dot(src.neuron_row_array, src.weight_row_mat)
# No activation function to output layer
if(self.target != None):
# Adjust the values based on the activation function. This introduces nonlinearity.
# For example, the relu function clamps all negative values to zero
self.neuron_row_array = self.activation_func(self.neuron_row_array)
# Transpose the neuron array and save for learn()
self.neuron_col_array = self.neuron_row_array.T
# record values for plotting
for i in range(self.num_neurons):
self.neuron_history_mat[i].append(self.neuron_row_array[0][i])
# In learning, the basic goal is to adjust the weights that set this layer's neurons (in this implementation, the source layer). This is done
# by backpropagating the error delta from this layer to the source layer. Since we only want to adjust the weights that participated in the
# training, we need to take the derivative of the activation function in train(). Again, the weight matrix is simply the source neurons times
# this layer's neurons. For example, if the source layer had three neurons and this layer had four, then the (source) weight matrix would be 3*4 = 12 weights.
def learn(self, alpha):
# if there is a layer above us
if self.source != None:
src = self.source
# calculate the error delta scalar array, which is the amount this layer needs to change,
# multiplied across the weights used to set this layer (in the source)
delta_scalar = np.dot(self.delta, src.weight_col_mat)
# determine the backpropagation distribution. In the case of Relu, it's just one or zero
delta_threshold = self.derivative_func(src.neuron_row_array)
# set the amount the source layer needs to change, based on this layer's delta distributed over the source
# neurons
src.delta = delta_scalar * delta_threshold
# create the weight adjustment matrix by taking the dot product of the source layer's neurons (as columns) and the
# scaled, thresholded row of deltas based on this layer's error delta and the source's weight layer
mat = np.dot(src.neuron_col_array, self.delta)
# add some percentage of the weight adjustment matrix to the source weight matrix
src.weight_row_mat += alpha * mat
src.weight_col_mat = src.weight_row_mat.T
# given one or more goals (that match the number of neurons in this layer), determine the delta that, when added to the
# neurons, would reach that goal
def calc_delta(self, goal: np.array) -> float:
self.delta = goal - self.neuron_row_array
return self.delta
# helper function to turn a NumPy array to a Python list
def nparray_to_list(self, vals: np.array) -> typing.List[float]:
data = []
for x in np.nditer(vals):
data.append(float(x))
return data
def to_string(self):
target_name = "no target"
source_name = "no source"
if self.target != None:
target_name = self.target.name
if self.source != None:
source_name = self.source.name
return "layer {}: \ntarget = {}\nsource = {}\nneurons (row) = {}\nweights (row) = \n{}".format(self.name, target_name, source_name, self.neuron_row_array, self.weight_row_mat)
# create a line chart of the plot matrix that we've been building
def plot_weight_matrix(self, fig_num: int):
var_name = "weight"
title = "{} to {} {}".format(self.name, self.target.name, var_name)
plt.figure(fig_num)
np_mat = np.array(self.weight_history_mat)
i = 0
for row in np_mat.T:
cstr = "C{}".format(i % self.num_neurons)
plt.plot(row, linewidth = int(i / self.num_neurons)+1, color=cstr)
i += 1
names = []
num_weights = self.num_neurons * self.target.num_neurons
for i in range(num_weights):
src_n = i % self.num_neurons
targ_n = int(i/self.num_neurons)
names.append("{} s:t[{}:{}]".format(var_name, src_n, targ_n))
plt.legend(names)
plt.title(title)
def plot_neuron_matrix(self, fig_num: int):
title = "{} neuron history".format(self.name)
plt.figure(fig_num)
np_mat = np.array(self.neuron_history_mat)
plt.plot(np_mat.T, '-o', linestyle=' ', ms=2)
names = []
for i in range(self.num_neurons):
names.append("neuron {}".format(i))
plt.legend(names)
plt.title(title)``` | 12,114 | 51,062 | {"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-2023-14 | latest | en | 0.887552 |
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# 106-20103-MT1 - PROBLEM(1(25 points A rod of length L...
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Phys 106 First Midterm Examination Page 1 Saturday, 16-July, 2011 PROBLEM (1) (25 points) A rod of length L carries uniformly distributed positive total charge Q and it lies along the x axis, as shown in the figure. A charged particle with positive charge Q is located at the point x = 0, y = r . Express your answers in terms of L , Q , r , k e and unit vectors . (a) (10 pts) Find the electric field ! E 1 due to the rod at a point P that is located along the axis of the rod and a distance r from one end. Q L r P r x y Divide the rod into small segments of charge dq . P x x dx dq dE E 1 = dE = k e " " dq x 2 = k e # dx x 2 where # = Q L r r + L " E 1 = k e Q L dx x 2 r r + L " = k e Q L # 1 x \$ % & ' ( ) r r + L E 1 = k e Q L 1 r " 1 r + L # \$ % & ' ( = k e Q r ( r + L ) ! E 1 = k e Q r ( r + L ) ( " " i ) (b) (5 pts) Find the electric field ! E 2 at the point P due to the charged particle. E 2 = k e Q r 2 ! E 2 = k e Q r 2 ( " " j ) (c) (5 pts) Find the resultant field ! E at the point P . ! E = ! E 1 + ! E 2 = k e Q r ( r + L ) ( " " i ) + k e Q r 2 ( " " j ) ! E = k e Q 1 r ( r + L ) ( " " i ) + 1 r 2 ( " " j ) # \$ % & ' ( (d) (5 pts) An electron with mass m and charge – e is located at P and released from rest. Ignore any gravitational effects and find the electron's acceleration a just after it is released. Express your answer in terms of the given quantities and constants, including m and e , as needed.
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Ask a homework question - tutors are online | 652 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-13 | latest | en | 0.876208 |
https://de.mathworks.com/matlabcentral/cody/problems/42892-transpose-of-matrix/solutions/1670102 | 1,603,357,954,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00247.warc.gz | 292,932,177 | 16,974 | Cody
# Problem 42892. Transpose of matrix
Solution 1670102
Submitted on 9 Nov 2018 by Sabine Ramp
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
x = 1; y_correct = 1; assert(isequal(tran_matrix(x),y_correct))
2 Pass
x = [1 2]; y_correct = [1 2]'; assert(isequal(tran_matrix(x),y_correct))
3 Pass
x = [1;2]; y_correct = [1 ;2]'; assert(isequal(tran_matrix(x),y_correct))
4 Pass
x = [1; 2+3i]; y_correct = [1 2+3i]; assert(isequal(tran_matrix(x),y_correct))
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Start Hunting! | 220 | 716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.727724 |
https://everything.explained.today/Hoare_logic/ | 1,642,998,370,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00132.warc.gz | 279,281,855 | 10,900 | # Hoare logic explained
Hoare logic (also known as Floyd - Hoare logic or Hoare rules) is a formal system with a set of logical rules for reasoning rigorously about the correctness of computer programs. It was proposed in 1969 by the British computer scientist and logician Tony Hoare, and subsequently refined by Hoare and other researchers.[1] The original ideas were seeded by the work of Robert W. Floyd, who had published a similar system[2] for flowcharts.
## Hoare triple
The central feature of Hoare logic is the Hoare triple. A triple describes how the execution of a piece of code changes the state of the computation. A Hoare triple is of the form
\{P\}C\{Q\}
where
P
and
Q
are assertions and
C
is a command.[3]
P
is named the precondition and
Q
the postcondition: when the precondition is met, executing the command establishes the postcondition. Assertions are formulae in predicate logic.
Hoare logic provides axioms and inference rules for all the constructs of a simple imperative programming language. In addition to the rules for the simple language in Hoare's original paper, rules for other language constructs have been developed since then by Hoare and many other researchers. There are rules for concurrency, procedures, jumps, and pointers.
## Partial and total correctness
Using standard Hoare logic, only partial correctness can be proven, while termination needs to be proved separately. Thus the intuitive reading of a Hoare triple is: Whenever
P
holds of the state before the execution of
C
, then
Q
will hold afterwards, or
C
does not terminate. In the latter case, there is no "after", so
Q
can be any statement at all. Indeed, one can choose
Q
to be false to express that
C
does not terminate.
Total correctness can also be proven with an extended version of the While rule.
In his 1969 paper, Hoare used a narrower notion of termination which also entailed absence of any run-time errors:"Failure to terminate may be due to an infinite loop; or it may be due to violation of an implementation-defined limit, for example, the range of numeric operands, the size of storage, or an operating system time limit."[4]
## Rules
### Empty statement axiom schema
The empty statement rule asserts that the statement does not change the state of the program, thus whatever holds true before also holds true afterwards.[5]
\dfrac{}{\{P\}tt{skip}\{P\}}
### Assignment axiom schema
The assignment axiom states that, after the assignment, any predicate that was previously true for the right-hand side of the assignment now holds for the variable. Formally, let be an assertion in which the variable is free. Then:
\dfrac{}{\{P[E/x]\}x:=E\{P\}}
where
P[E/x]
denotes the assertion in which each free occurrence of has been replaced by the expression .
The assignment axiom scheme means that the truth of
P[E/x]
is equivalent to the after-assignment truth of . Thus were
P[E/x]
true prior to the assignment, by the assignment axiom, then would be true subsequent to which. Conversely, were
P[E/x]
false (i.e.
\negP[E/x]
true) prior to the assignment statement, must then be false afterwards.
Examples of valid triples include:
\{x+1=43\}y:=x+1\{y=43\}
\{x+1\leqN\}x:=x+1\{x\leqN\}
All preconditions that are not modified by the expression can be carried over to the postcondition. In the first example, assigning
y:=x+1
does not change the fact that
x+1=43
, so both statements may appear in the postcondition. Formally, this result is obtained by applying the axiom schema with being (
y=43
and
x+1=43
), which yields
P[(x+1)/y]
being (
x+1=43
and
x+1=43
), which can in turn be simplified to the given precondition
x+1=43
.
The assignment axiom scheme is equivalent to saying that to find the precondition, first take the post-condition and replace all occurrences of the left-hand side of the assignment with the right-hand side of the assignment. Be careful not to try to do this backwards by following this incorrect way of thinking:
\{P\}x:=E\{P[E/x]\}
;this rule leads to nonsensical examples like:
\{x=5\}x:=3\{3=5\}
Another incorrect rule looking tempting at first glance is
\{P\}x:=E\{P\wedgex=E\}
; it leads to nonsensical examples like:
\{x=5\}x:=x+1\{x=5\wedgex=x+1\}
While a given postcondition uniquely determines the precondition
P[E/x]
, the converse is not true. For example:
\{0\leqyy\wedgeyy\leq9\}x:=yy\{0\leqx\wedgex\leq9\}
,
\{0\leqyy\wedgeyy\leq9\}x:=yy\{0\leqx\wedgeyy\leq9\}
,
\{0\leqyy\wedgeyy\leq9\}x:=yy\{0\leqyy\wedgex\leq9\}
, and
\{0\leqyy\wedgeyy\leq9\}x:=yy\{0\leqyy\wedgeyy\leq9\}
are valid instances of the assignment axiom scheme.
The assignment axiom proposed by Hoare does not apply when more than one name may refer to the same stored value. For example,
\{y=3\}x:=2\{y=3\}
is wrong if and refer to the same variable (aliasing), although it is a proper instance of the assignment axiom scheme (with both
\{P\}
and
\{P[2/x]\}
being
\{y=3\}
).
### Rule of composition
Hoare's rule of composition applies to sequentially executed programs and, where executes prior to and is written
S;T
(is called the midcondition):[6]
\dfrac{\{P\}S\{Q\},\{Q\}T\{R\}}{\{P\}S;T\{R\}}
For example, consider the following two instances of the assignment axiom:
\{x+1=43\}y:=x+1\{y=43\}
and
\{y=43\}z:=y\{z=43\}
By the sequencing rule, one concludes:
\{x+1=43\}y:=x+1;z:=y\{z=43\}
Another example is shown in the right box.
### Conditional rule
\dfrac{\{B\wedgeP\}S\{Q\},\{\negB\wedgeP\}T\{Q\}}{\{P\}tt{if} B tt{then} S tt{else} T tt{endif}\{Q\}}
The conditional rule states that a postcondition common to and part is also a postcondition of the whole statement.In the and the part, the unnegated and negated condition can be added to the precondition, respectively.The condition,, must not have side effects.An example is given in the next section.
This rule was not contained in Hoare's original publication.[1] However, since a statement
tt{if} B tt{then} S tt{else} T tt{endif}
has the same effect as a one-time loop construct
tt{bool} b:=tt{true};tt{while} B\wedgebtt{do} S;b:=tt{false} tt{done};b:=tt{true};tt{while} \negB\wedgebtt{do} T;b:=tt{false} tt{done}
the conditional rule can be derived from the other Hoare rules.In a similar way, rules for other derived program constructs, like loop, loop,,, can be reduced by program transformation to the rules from Hoare's original paper.
### Consequence rule
\dfrac{P1P2 ,\{P2\}S\{Q2\},Q2Q1}{\{P1\}S\{Q1\}}
This rule allows to strengthen the precondition
P2
and/or to weaken the postcondition
Q2
.It is used e.g. to achieve literally identical postconditions for the and the part.
For example, a proof of
\{0\leqx\leq15\}tt{if} x<15 tt{then} x:=x+1 tt{else} x:=0 tt{endif}\{0\leqx\leq15\}
needs to apply the conditional rule, which in turn requires to prove
\{0\leqx\leq15\wedgex<15\}x:=x+1\{0\leqx\leq15\}
, or simplified
\{0\leqx<15\}x:=x+1\{0\leqx\leq15\}
for the part, and
\{0\leqx\leq15\wedgex\geq15\}x:=0\{0\leqx\leq15\}
, or simplified
\{x=15\}x:=0\{0\leqx\leq15\}
for the part.
However, the assignment rule for the part requires to choose as
0\leqx\leq15
; rule application hence yields
\{0\leqx+1\leq15\}x:=x+1\{0\leqx\leq15\}
, which is logically equivalent to
\{-1\leqx<15\}x:=x+1\{0\leqx\leq15\}
.The consequence rule is needed to strengthen the precondition
\{-1\leqx<15\}
obtained from the assignment rule to
\{0\leqx<15\}
required for the conditional rule.
Similarly, for the part, the assignment rule yields
\{0\leq0\leq15\}x:=0\{0\leqx\leq15\}
, or equivalently
\{tt{true}\}x:=0\{0\leqx\leq15\}
,hence the consequence rule has to be applied with
P1
and
P2
being
\{x=15\}
and
\{tt{true}\}
, respectively, to strengthen again the precondition. Informally, the effect of the consequence rule is to "forget" that
\{x=15\}
is known at the entry of the part, since the assignment rule used for the part doesn't need that information.
### While rule
\dfrac{\{P\wedgeB\}S\{P\}}{\{P\}tt{while} B tt{do} S tt{done}\{\negB\wedgeP\}}
Here is the loop invariant, which is to be preserved by the loop body .After the loop is finished, this invariant still holds, and moreover
\negB
must have caused the loop to end.As in the conditional rule, must not have side effects.
For example, a proof of
\{x\leq10\}tt{while} x<10 tt{do} x:=x+1 tt{done}\{\negx<10\wedgex\leq10\}
by the while rule requires to prove
\{x\leq10\wedgex<10\}x:=x+1\{x\leq10\}
, or simplified
\{x<10\}x:=x+1\{x\leq10\}
,which is easily obtained by the assignment rule.Finally, the postcondition
\{\negx<10\wedgex\leq10\}
can be simplified to
\{x=10\}
.
For another example, the while rule can be used to formally verify the following strange program to compute the exact square root of an arbitrary number —even if is an integer variable and is not a square number:
\{tt{true}\}tt{while} x ⋅ xatt{do} tt{skip} tt{done}\{xx=a\wedgett{true}\}
After applying the while rule with being, it remains to prove
\{tt{true}\wedgexxa\}tt{skip}\{tt{true}\}
,which follows from the skip rule and the consequence rule.
In fact, the strange program is partially correct: if it happened to terminate, it is certain that must have contained (by chance) the value of 's square root.In all other cases, it will not terminate; therefore it is not totally correct.
### While rule for total correctness
If the above ordinary while rule is replaced by the following one, the Hoare calculus can also be used to prove total correctness, i.e. termination[7] as well as partial correctness. Commonly, square brackets are used here instead of curly braces to indicate the different notion of program correctness.
\dfrac{<isawell-foundedorderingonthesetD,[P\wedgeB\wedget\inD\wedget=z]S[P\wedget\inD\wedget<z]}{[P\wedget\inD]tt{while} B tt{do} S tt{done}[\negB\wedgeP\wedget\inD]}
In this rule, in addition to maintaining the loop invariant, one also proves termination by way of an expression, called the loop variant, whose value strictly decreases with respect to a well-founded relation on some domain set during each iteration. Since is well-founded, a strictly decreasing chain of members of can have only finite length, so cannot keep decreasing forever. (For example, the usual order is well-founded on positive integers
N
, but neither on the integers
Z
nor on positive real numbers
R+
; all these sets are meant in the mathematical, not in the computing sense, they are all infinite in particular.)
Given the loop invariant, the condition must imply that is not a minimal element of, for otherwise the body could not decrease any further, i.e. the premise of the rule would be false. (This is one of various notations for total correctness.)[8]
Resuming the first example of the previous section, for a total-correctness proof of
[x\leq10]tt{while} x<10 tt{do} x:=x+1 tt{done}[\negx<10\wedgex\leq10]
the while rule for total correctness can be applied with e.g. being the non-negative integers with the usual order, and the expression being
10-x
, which then in turn requires to prove
[x\leq10\wedgex<10\wedge10-x\geq0\wedge10-x=z]x:=x+1[x\leq10\wedge10-x\geq0\wedge10-x<z]
Informally speaking, we have to prove that the distance
10-x
decreases in every loop cycle, while it always remains non-negative; this process can go on only for a finite number of cycles.
The previous proof goal can be simplified to
[x<10\wedge10-x=z]x:=x+1[x\leq10\wedge10-x<z]
,which can be proven as follows:
[x+1\leq10\wedge10-x-1<z]x:=x+1[x\leq10\wedge10-x<z]
is obtained by the assignment rule, and
[x+1\leq10\wedge10-x-1<z]
can be strengthened to
[x<10\wedge10-x=z]
by the consequence rule.
For the second example of the previous section, of course no expression can be found that is decreased by the empty loop body, hence termination cannot be proved.
• Robert D. Tennent. Specifying Software (a textbook that includes an introduction to Hoare logic, written in 2002)
• KeY-Hoare is a semi-automatic verification system built on top of the KeY theorem prover. It features a Hoare calculus for a simple while language.
• j-Algo-modul Hoare calculus - A visualisation of the Hoare calculus in the algorithm visualisation program j-Algo
## Notes and References
1. Hoare . C. A. R. . C.A.R. Hoare . An axiomatic basis for computer programming . 10.1145/363235.363259 . . 12 . 10 . 576 - 580 . October 1969 .
2. [Robert Floyd|R. W. Floyd]
3. Hoare originally wrote "
P\{C\}Q
" rather than "
\{P\}C\{Q\}
".
4. p.579 upper left
5. This article uses a natural deduction style notation for rules. For example,
\dfrac{\alpha,\beta}{\phi}
informally means "If both and hold, then also holds"; and are called antecedents of the rule, is called its succedent. A rule without antecedents is called an axiom, and written as
\dfrac{}{\phi}
.
6. Book: Michael. Huth. Mark. Ryan. CUP. Logic in Computer Science. 978-0521543101. second . 276. 2004-08-26.
7. "Termination" here is meant in the broader sense that computation will eventually be finished; it does not imply that no limit violation (e.g. zero divide) can stop the program prematurely.
8. Hoare's 1969 paper didn't provide a total correctness rule; cf. his discussion on p.579 (top left). For example Reynolds' textbook (Book: Theory of Programming Languages. John C. Reynolds. Cambridge University Press. 2009.), Sect.3.4, p.64 gives the following version of a total correctness rule:
\dfrac{P\wedgeB0\leqt,[P\wedgeB\wedget=z]S[P\wedget<z]}{[P]tt{while} B tt{do} S tt{done}[P\wedge\negB]}
when is an integer variable that doesn't occur free in,,, or, and is an integer expression (Reynolds' variables renamed to fit with this article's settings). | 4,092 | 13,758 | {"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} | 3.34375 | 3 | CC-MAIN-2022-05 | longest | en | 0.946555 |
https://oeis.org/A337494 | 1,725,870,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00792.warc.gz | 410,217,663 | 4,160 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A337494 Smallest m such that prime(3*n)# can be written as a product of n sphenic numbers each <= m. 0
30, 182, 627, 1705, 3741, 7285, 13039, 21889, 33611, 51389, 74497, 104081, 140491, 188641, 246089, 312547, 394831, 491713, 604283, 736189, 886937, 1058581, 1249331, 1474531 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) >= ceiling((prime(3*n)#)^(1/n)). - Chai Wah Wu, Sep 24 2020 LINKS Table of n, a(n) for n=1..24. EXAMPLE a(4) = 1705. p(3*4)#, which is the product of the first 12 primes, can be written as s1 * s2 * s3 * s4 with s1 = 5 * 11 * 31 = 1705, s2 = 2 * 23 * 37 = 1702, s3 = 3 * 19 * 29 = 1653, s4 = 7 * 13 * 17 = 1547. No such factorization is possible in sphenic numbers that are all < 1705. CROSSREFS Cf. A002110, A007304. Sequence in context: A156318 A042758 A071311 * A265037 A249001 A249466 Adjacent sequences: A337491 A337492 A337493 * A337495 A337496 A337497 KEYWORD nonn,more AUTHOR Bert Dobbelaere, Aug 29 2020 STATUS approved
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Last modified September 9 03:44 EDT 2024. Contains 375759 sequences. (Running on oeis4.) | 522 | 1,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.683419 |
https://iwillsolvethat.com/tag/math-tutor/ | 1,680,162,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00200.warc.gz | 380,062,328 | 27,459 | ## Rewriting Expressions with Radicals to Expressions with Rational Exponents and Simplifying – HSN.RN.A.2
Let’s say you had the following problem √(x3)and you wanted to change that to an expression with a rational exponent. Since there is no number written in… Read more “Rewriting Expressions with Radicals to Expressions with Rational Exponents and Simplifying – HSN.RN.A.2”
## Fractional Exponents – N.RN.1
In Integrated Mathematics II students will begin the year by learning about fractional exponents. In this section they will learn that with fractional exponents the: 1. Numerator,… Read more “Fractional Exponents – N.RN.1”
## CCSS.MATH.CONTENT.7.RP.A.3 – Tax and Tip
A problem for Common Core standard CCSS.MATH.CONTENT.7.RP.A.3 McPuckles Restaurant menu is listed below. They charge a 6% sales tax and an 18% percent gratuity based on the pre-tax… Read more “CCSS.MATH.CONTENT.7.RP.A.3 – Tax and Tip” | 245 | 917 | {"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} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.8512 |
https://mail.python.org/pipermail/python-list/2001-March/107632.html | 1,501,264,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00252.warc.gz | 665,122,659 | 2,128 | # This math scares me
Aahz Maruch aahz at panix.com
Thu Mar 15 16:40:48 CET 2001
```In article <HR4s6.877\$lz1.73512 at ruti.visi.com>,
Grant Edwards <grante at visi.com> wrote:
>In article <mailman.984664520.26463.python-list at python.org>, Fabio Olive Leite wrote:
>>
>>The usual 1/3 example is a problem for base 10, but not so for
>>base 12, for example, since 12 contains 3 2 2. Thus:
>>
>>1_{12} / 3_{12} == 0.4_{12}
>>
>>And then base 30 would let us count exactly fractions of base
>>2, 3, 5, 6, 10, 12, 15, 16 and all the others you can think of
>>that have only 2, 3 and 5 as factors of the base. But since no
>>one (usualy) has 30 fingers, that base is not terribly popular.
>>:)
>
>One step further is base-60, which is still used for many
>things. Time and angular measurement for example is partially
>a base-60 system. I believe that one or more of the ancient
>Aztecs/Incans/Mayans used a base-60 number system.
Paging Ivan Van Laningham.....
--
--- Aahz <*> (Copyright 2001 by aahz at pobox.com)
Androgynous poly kinky vanilla queer het Pythonista http://www.rahul.net/aahz/
Hugs and backrubs -- I break Rule 6
"The overexamined life sure is boring." --Loyal Mini Onion
``` | 395 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-30 | longest | en | 0.854641 |
https://www.bartleby.com/questions-and-answers/q4.-you-conduct-a-poll-comparing-two-candidates.-in-a-poll-of-400-people-190-prefer-candidate-a-and-/ba95596a-c6cc-446a-8944-35bd51bdfcd4 | 1,627,524,663,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00052.warc.gz | 685,932,492 | 34,465 | # Q4. You conduct a poll comparing two candidates. In a poll of 400 people, 190 prefer candidate A, and 210 prefer B. Q4e. If you want to shrink one stdev error to 1%, how many people do you need to survey?
Question
Q4. You conduct a poll comparing two candidates. In a poll of 400 people, 190 prefer candidate A, and 210 prefer B.
Q4e. If you want to shrink one stdev error to 1%, how many people do you need to survey? | 118 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.907872 |
https://numberworld.info/3600 | 1,726,124,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00251.warc.gz | 385,108,663 | 4,047 | # Number 3600
### Properties of number 3600
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 3 * 3 * 5 * 5
Divisors:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36, 40, 45, 48, 50, 60, 72, 75, 80, 90, 100, 120, 144, 150, 180, 200, 225, 240, 300, 360, 400, 450, 600, 720, 900, 1200, 1800, 3600
Count of divisors:
Sum of divisors:
12493
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
e10
Base 32:
3gg
sin(3600)
-0.2620839590181
cos(3600)
0.96504507585159
tan(3600)
-0.27157690928253
ln(3600)
8.1886891244442
lg(3600)
3.5563025007673
sqrt(3600)
60
Square(3600)
### Number Look Up
Look Up
3600 which is pronounced (three thousand six hundred) is a very impressive figure. The cross sum of 3600 is 9. If you factorisate the number 3600 you will get these result 2 * 2 * 2 * 2 * 3 * 3 * 5 * 5. The number 3600 has 45 divisors ( 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36, 40, 45, 48, 50, 60, 72, 75, 80, 90, 100, 120, 144, 150, 180, 200, 225, 240, 300, 360, 400, 450, 600, 720, 900, 1200, 1800, 3600 ) whith a sum of 12493. 3600 is not a prime number. The figure 3600 is not a fibonacci number. 3600 is not a Bell Number. 3600 is not a Catalan Number. The convertion of 3600 to base 2 (Binary) is 111000010000. The convertion of 3600 to base 3 (Ternary) is 11221100. The convertion of 3600 to base 4 (Quaternary) is 320100. The convertion of 3600 to base 5 (Quintal) is 103400. The convertion of 3600 to base 8 (Octal) is 7020. The convertion of 3600 to base 16 (Hexadecimal) is e10. The convertion of 3600 to base 32 is 3gg. The sine of the number 3600 is -0.2620839590181. The cosine of the figure 3600 is 0.96504507585159. The tangent of the number 3600 is -0.27157690928253. The square root of 3600 is 60.
If you square 3600 you will get the following result 12960000. The natural logarithm of 3600 is 8.1886891244442 and the decimal logarithm is 3.5563025007673. that 3600 is very amazing figure! | 881 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.741031 |
https://www.mersenneforum.org/showthread.php?s=8957ff843a0e7807b6bb390548195c29&t=13860 | 1,603,751,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892062.70/warc/CC-MAIN-20201026204531-20201026234531-00635.warc.gz | 794,406,356 | 8,685 | mersenneforum.org Twin Prime Question
Register FAQ Search Today's Posts Mark Forums Read
2010-09-07, 23:51 #1 Unregistered 22·17·73 Posts Twin Prime Question I'm interested in twins of the form k*2^n+/-1. One question: What is the probablity that there is a twin of that form with k<1M and n>1M?
2010-09-08, 02:38 #2
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts
Quote:
Originally Posted by Unregistered One question: What is the probablity that there is a twin of that form with k<1M and n>1M?
Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)
It's just quite hard to discover with current technology.
2010-09-08, 02:59 #3
CRGreathouse
Aug 2006
32·659 Posts
Quote:
Originally Posted by Mini-Geek Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)
I don't think that's known. Heuristically, we expect something like
$\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$
examples, right?
Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k.
Edit:
$\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$
takes the factors of the k-values into account.
Last fiddled with by CRGreathouse on 2010-09-08 at 03:04 Reason: make approximation clearer
2010-09-08, 05:04 #4
Unregistered
13×467 Posts
Quote:
Originally Posted by CRGreathouse I don't think that's known. Heuristically, we expect something like $\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$ examples, right? Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k. Edit: $\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$ takes the factors of the k-values into account.
Where do the 21503 (second figure) and 4000 (first figure) numbers come from?
2010-09-08, 13:00 #5
CRGreathouse
Aug 2006
32·659 Posts
Quote:
Originally Posted by Unregistered Where do the 21503 (second figure) and 4000 (first figure) numbers come from?
4000 is about 1000 candidates * 2^2, where the 2 is because only odd numbers are used (which twice as likely to be prime). Oh! You want k < 1000000, not k < 1000; in that case it's 4000000.
The other one is a calculated number based on the factorizations of k < 1000. Let me try for k < 1000000. OK, I get 14756135.8...
Code:
sum(k=2,10^6-1,ff(k+k)^2,0.)/log(2)^2
So the expectations are
Code:
4e6/log(2)^2*(zeta(2)-sum(n=1,10^6,n^-2,0.))
8.32 (naive)
14.75 (including factorizations)
So with this larger range, the heuristic probability is high: with the Poisson model, it's something like 1 - e-14.75 = 99.99996% likely that such a pair of primes exist.
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Happy International Sudoku Day! Page-A-Day
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Sudoku Solving Algorithms – Wikipedia
Sudoku is a Japanese phrase for cross-section, because the layouts are cut throughout squares to ensure that they can be arranged. There are different approaches to reduce the squares out of the board. The styles are essential, as the patterns will dictate the styles for your other squares which can be cut in the board.
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A Finnish Mathematician Claimed That This Is The Most
Whenever you possess the patterns down, you’ll find that there are several different designs of puzzle. Some puzzles have styles that look only once, and after that other individuals have styles that appear over. Whenever you learn to handle these distinct designs, you’ll find which they will probably be easier to solve.
With the World wide web, it is also feasible to locate Sudoku Puzzles Printable. Sudoku Printable contains solutions to Sudoku puzzles in the form of very distinct and easy to go through graphics. All of the feasible patterns are demonstrated, so that you can see what is possible.
At the end of the day, the very best suggestions that anybody can give for studying Sudoku is always to get heading. Discover the basic principles, develop your abilities, and don’t quit until finally you get the puzzle that you’re happy with. One factor to recollect, the net has the biggest and most up to date Sudoku puzzles. | 565 | 2,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-05 | latest | en | 0.941601 |
https://www.teachoo.com/2830/1802/Ex-13.1--6---Evaluate--lim-x--0--(x---1)5--1-x---Class-11/category/Chapter-13-Class-11th-Limits-and-Derivatives/ | 1,547,906,748,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583668324.55/warc/CC-MAIN-20190119135934-20190119161934-00529.warc.gz | 945,938,226 | 11,921 | 1. Class 11
2. Important Question for exams Class 11
3. Chapter 13 Class 11 Limits and Derivatives
Transcript
Ex 13.1, 6 Evaluate the Given limit: lim x 0 x +1 5 1 x lim x 0 x + 1 5 1 x = 0 + 1 5 1 0 = 15 1 0 = 1 1 0 = 0 0 Since it is of from 0 0 Hence, we simplify lim x 0 x +1 5 1 x Putting y = x + 1 x = y 1 As x 0 y 0 + 1 y 1 Our equation becomes lim x 0 x +1 5 1 x = lim y 1 5 1 y 1 =
Chapter 13 Class 11 Limits and Derivatives
Class 11
Important Question for exams Class 11 | 204 | 483 | {"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} | 3.84375 | 4 | CC-MAIN-2019-04 | latest | en | 0.839477 |
https://homework.cpm.org/category/ACC/textbook/ccaa8/chapter/5%20Unit%206/lesson/CCA:%205.1.2/problem/5-25 | 1,726,675,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00489.warc.gz | 266,070,386 | 15,661 | Home > CCAA8 > Chapter 5 Unit 6 > Lesson CCA: 5.1.2 > Problem5-25
5-25.
Determine the domain and range of each of the following graphs.
• To determine the range, locate the vertex of the parabola.
• Domain: all numbers
Range: $y\le1$
• See part (a).
• See part (a).
• Domain: all numbers
Range: $y\ge−1$ | 103 | 310 | {"found_math": true, "script_math_tex": 2, "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-2024-38 | latest | en | 0.53203 |
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### Re: Accuracy of Random Pi
by Fingo (Monk)
on Feb 16, 2001 at 07:47 UTC ( #58781=note: print w/replies, xml ) Need Help??
in reply to Accuracy of Random Pi
```#!/usr/bin/perl
# This is a quick program to calculate pi using the Monte Carlo method
+.
# I recomend inputting a value for \$cycles greater that 1000.
# I am working on a detailed explanation of how and why this works.
# I will add it as soon as I'm done.
use strict;
open(PI, ">>pi.dat") || die "pi.dat";
my (\$i, \$j, \$yespi, \$pi) = 1;
my \$cycles = 1;
srand;
while (\$j <= 100000) {
\$cycles = \$j;
while (\$i <= \$cycles) {
my (\$x, \$y, \$cdnt) = 1;
\$x = rand;
\$y = rand;
\$cdnt = \$x**2 + \$y**2;
if (\$cdnt <= 1) {
++\$yespi;
}
\$i=\$i + 10;
\$pi = (\$yespi / (\$cycles / 10)) * 4; # since I add 10 every time.
print PI "\$cycles \$pi\n";
}
\$j = \$j + 10;
}
close(PI) || die "pi.dat";
Here is the modified script I was talking about. If you look at pi.dat it starts getting inacurate at one point. I will try using one of the ways to genereate seeds that are slightly more random than plain srand. Note: you need to make a file pi.dat for the script to run.
Replies are listed 'Best First'.
Srand versus rand
by gryng (Hermit) on Feb 16, 2001 at 11:21 UTC
I'm not sure if setting srand would really help the situation. In order for the Monte Carlo method to be most effective the random numbers produced must be the most evenly spread out. In fact, you can be more accurate without random numbers and instead going through all four corners, then their mid-points, and their mid-points' mid-points, e.g. :
```. . . . .
-> . . . -> etc.
. . . . .
However, if you would like to use random numbers you need one of two types of random numbers. You either need true even-bias random numbers (pseudo-random can work but often don't :) ). Or you can use a particular kind of random numbers that are designed not to repeat and are garunteed to be spread evenly and more and more closely together (very similar to the systematic approach above).
Two examples are Hamilton and Sobol sequences. If you use them, you get a 1/N convergence, instead of a 1/N^2 (which you get for uniform numbers since there is nothing that keeps them from clumping up). I was going to give below "Numerical Recipe's in C"'s version of Antonov-Saleev's variant on Sobol's sequence. However it's simply too long. Also, I have to wake up for work tomorrow, and the first version I typed in (converting it to Perl from C) didn't work just right. Oh well.
Good luck, Hamilton sequences are much easier to do.
Ciao,
Gryn
Nope, letting perl set srand is good enough. And you are right about Monte Carlo needing a purer random base than a pseudo-random generator. Most Monte Carlo's use scads of randoms per cycle and you loop the psuedo random in 2**31 calls on most systems and 2**15 on some.
Look to CPAN and you will find some of what you need. First off, if you are going to write your own "random" sequence generator you my find PDL handy. If you want someone else to do the work on random numbers try Math::Random or Math::TrulyRandom but I would recommend you find an OS specific random source like Linux's /dev/random and /dev/urandom
```#!/usr/bin/perl -w
use strict;
#use Linux; ## I wish. =) (yeah yeah, \$^O, etc etc)
open UR, "</dev/urandom" or die "Oh man, your system sucks, \$!";
my \$pages= shift()+0 || 1;
die "Give me a number greater than 0 or nothing, bub.\n" unless \$pages
+>0;
while (\$pages-->0) {
my \$buf;
read UR, \$buf, 512 or die "Ouch that shouldn't happen, \$!";
for (0..31) {
print vec(\$buf,\$_*4,32), "\t",
vec(\$buf,\$_*4+1,32), "\t",
vec(\$buf,\$_*4+2,32), "\t",
vec(\$buf,\$_*4+3,32), "\n";
}
}
--
\$you = new YOU;
honk() if \$you->love(perl)
It shoulden't matter if there are repeats or not. As long as they do not apper in the same order, it woulden't really matter since I am looking to see if A^2 + B^2 is below 1. Actualy if I make 2 pairs of numbers one pair that A^2 + B^2 is less than 1, and another where it is not. Than I can generate the order in which they appear by some random factor (take a random number and each digit is used to see which of the 2 pairs I choose depending on if it's even or not). UPDATE: I now realize what I said was wrong randomly generating true or false will not generate Pi. Random numbers are sometimes very hard to visualize
A trick I have mentioned before and may again.
If you need a large supply of random looking data, one of the best approaches is to grab a large file produced out of dynamic data (/dev/mem is often a good bet, there are plenty of choices though), compress it, and then encrypt it with a good algorithm. Samples from the resulting file are for all intents and purposes, random.
In fact, this is a good source of random data. However, because it is a good source, it could easily be bad for Monte-Carlo searching. As I've been suggesting in other posts, Monte-Carlo searching benefits from uniform-distributed numbers.
But truely random numbers are not statistically garunteed to be uniformly distributed (no, the law of averages does not work that way :) ), and so they cause Monte-Carlo searching to converge more slowly (but do not keep it from converging).
This is why psuedo-random numbers can, theoretically, be better than truely-random numbers, because often they are crafted to be uniformly distributed -- statistically. However, it often occurs in practice that pseudo-random numbers are not perfectly statistically uniformly distributed (what a mouthful), and so can easily lead to a mis-convergance.
Enter Quasi-random numbers. These are uniformly-distributed and have a bias towards non-repetition. This means that you still get a garunteed convergence and you get it faster (since there would be no clumps in your set).
Ok, time to go.
Ciao,
Gryn
I did a search on CPAN for a quasi-random number generation module, but sadly could not find one. I will try the systematic dot approch that you mentioned, but this has interested me very much. Can you please tell me how I could go about writing a quasi-random number generator? Building a module will be good practice anyway, even if I can go without it ;)
Welp I can't remember off the top of my head how to do anything but Hamilton sequences. (And I'm fuzzy on one point that I'll mention below).
Hamilton sequences aren't very random looking for quasi-random numbers, but they are better than nothing :) . You choose a base b and a number n. You then write n in base b, reverse it's digits and add a decimal point in front. Convert the number back into whatever base you want to view it in, and you are done.
I gave an example with b = 2 and n going from 0 to 7 before (here). And here is where I'm fuzzy. I believe the best way to use Hamilton sequences is to not keep b constant and vary n.
Rather you should vary both n and b. I beleive you increment n sequencially, but increment b to the nth prime number. This seems right, but it could also be that you make b the next larger prime than n.
If you notice, if you keep b constant you fill in your area in a regular grid-like patern, with b dictating the initial fineness (but any b will give you an infinite sequence). This is why I'm fairly certain you should probably use one of the two techniques above to vary b, I just can't remember which one at the moment :) .
Here is a snippet (which hopefully works) that should generate the nth number for a base b in a Hamilton sequence. There will probably be faster, more robust, and/or shorter approaches (perl golf is encouraged :) ) :
```sub hamilton {
my (\$n, \$b) = @_;
my (\$a, \$x) = (0, \$b);
while (\$n) {
\$a += (\$n % \$b)/\$x;
\$x = \$x * \$b;
\$n = int (\$n / \$b);
}
return \$a;
}
Back to work. Good luck.
Ciao,
Gryn
p.s. Note that it would be more accurate to calculate \$a by reversing this loop (that is (\$n % \$b)/\$x gets smaller as the loop progresses, and it would be better to add the smaller values of this term first, then proceed towards larger ones). This is a floating-point rounding issue only.
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